url
stringlengths 90
342
| html
stringlengths 602
98.8k
| text_length
int64 602
98.8k
| __index_level_0__
int64 0
5.02k
|
|---|---|---|---|
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/07%3A_Obtaining_and_Preparing_Samples_for_Analysis/7.06%3A_Classifying_Separation_Techniques |
We can separate an analyte and an interferent if there is a significant difference in at least one of their chemical or physical properties. Table 7.6.1
provides a partial list of separation techniques, organized by the chemical or physical property affecting the separation. Size is the simplest physical property we can exploit in a separation. To accomplish the separation we use a porous medium through which only the analyte or the interferent can pass. Examples of size-based separations include filtration, dialysis, and size-exclusion. In a we separate a particulate interferent from soluble analytes using a filter with a pore size that will retain the interferent. The solution that passes through the filter is called the , and the material retained by the filter is the . Gravity filtration and suction filtration using filter paper are techniques with which you should already be familiar. A membrane filter is the method of choice for particulates that are too small to be retained by filter paper. Figure 7.6.1
provides information about three types of membrane filters. For applications of gravity filtration and suction filtration in gravimetric methods of analysis, see . is another example of a separation technique in which size is used to separate the analyte and the interferent. A dialysis membrane usually is made using cellulose and fashioned into tubing, bags, or cassettes. Figure 7.6.2
shows an example of a commercially available dialysis cassette. The sample is injected into the dialysis membrane, which is sealed tightly by a gasket, and the unit is placed in a container filled with a solution with a composition different from the sample. If there is a difference in a species’ concentration on the membrane’s two sides, the resulting concentration gradient provides a driving force for its diffusion across the membrane. While small species freely pass through the membrane, larger species are unable to pass. Dialysis frequently is used to purify proteins, hormones, and enzymes. During kidney dialysis, metabolic waste products, such as urea, uric acid, and creatinine, are removed from blood by passing it over a dialysis membrane. is a third example of a separation technique that uses size as a means to effect a separation. In this technique a column is packed with small, approximately 10-μm, porous polymer beads of cross-linked dextrin or polyacrylamide. The pore size of the particles is controlled by the degree of cross-linking, with more cross-linking producing smaller pore sizes. The sample is placed into a stream of solvent that is pumped through the column at a fixed flow rate. Those species too large to enter the pores pass through the column at the same rate as the solvent. Species that enter into the pores take longer to pass through the column, with smaller species requiring more time to pass through the column. Size-exclusion chromatography is widely used in the analysis of polymers, and in biochemistry, where it is used for the separation of proteins. A more detailed treatment of size-exclusion chromatography, which also is called gel permeation chromatography, is in . If the analyte and the interferent have different masses or densities, then a separation using may be possible. The sample is placed in a centrifuge tube and spun at a high angular velocity, measured in revolutions per minute (rpm). The sample’s constituents experience a centrifugal force that pulls them toward the bottom of the centrifuge tube. Those species that experience the greatest centrifugal force have the fastest sedimentation rate and are the first to reach the bottom of the centrifuge tube. If two species have the same density, their separation is based on a difference in mass, with the heavier species having the greater sedimentation rate. If the species are of equal mass, then the species with the larger density has the greatest sedimentation rate. Centrifugation is an important separation technique in biochemistry. Table 7.6.2
, for example, lists conditions for separating selected cellular components. We can separate lysosomes from other cellular components by several differential centrifugations, in which we divide the sample into a solid residue and a supernatant solution. After destroying the cells, the solution is centrifuged for 20 minutes at \(15000 \times g\) (a centrifugal force that is 15 000 times the earth’s gravitational force), leaving a solid residue of cell membranes and mitochondria. The supernatant, which contains the lysosomes, is isolated by decanting it from the residue and then centrifuged for 30 minutes at \(30000 \times g\), leaving a solid residue of lysosomes. Figure 7.6.3
shows a typical centrifuge capable of producing the centrifugal forces needed for biochemical separations. Adapted from Zubay, G. , 2nd ed. Macmillan: New York, 1988, p.120. An alternative approach to differential centrifugation is a density gradient centrifugation. To prepare a sucrose density gradient, for example, a solution with a smaller concentration of sucrose—and, thus, of lower density—is gently layered upon a solution with a higher concentration of sucrose. Repeating this process several times, fills the centrifuge tube with a multi-layer density gradient. The sample is placed on top of the density gradient and centrifuged using a force greater than \(150000 \times g\). During centrifugation, each of the sample’s components moves through the gradient until it reaches a position where its density matches the surrounding sucrose solution. Each component is isolated as a separate band positioned where its density is equal to that of the local density within the gradient. Figure 7.6.4
provides an example of a typical sucrose density centrifugation for separating plant thylakoid membranes. One widely used technique for preventing an interference is to bind the interferent in a strong, soluble complex that prevents it from interfering in the analyte’s determination. This process is known as . As shown in Table 7.6.3
, a wide variety of ions and molecules are useful , and, as a result, selectivity is usually not a problem. Technically, masking is not a separation technique because we do not physically separate the analyte and the interferent. We do, however, chemically isolate the interferent from the analyte, resulting in a pseudo-separation. Ag, Au, Cd, Co, Cu, Fe, Hg, Mn, Ni, Pd, Pt, Zn Ag, Cd, Co, Cu, Fe, Ni, Pd, Pt, Zn Ag, Co, Ni, Cu, Zn Au, Ce, Co, Cu, Fe, Hg, Mn, Pb, Pd, Pt, Sb, Sn, Zn Al, Fe, Mg, Mn Cu, Fe, Sn Meites, L. , McGraw-Hill: New York, 1963. Using Table 7.6.3
, suggest a masking agent for the analysis of aluminum in the presence of iron. A suitable masking agent must form a complex with the interferent, but not with the analyte. Oxalate, for example, is not a suitable masking agent because it binds both Al and Fe. Thioglycolic acid, on the other hand, is a selective masking agent for Fe in the presence of Al. Other acceptable masking agents are cyanide (CN ) thiocyanate (SCN ), and thiosulfate (\(\text{S}_2\text{O}_3^{2-}\)). Using Table 7.6, suggest a masking agent for the analysis of Fe in the presence of Al. The fluoride ion, F , is a suitable masking agent as it binds with Al to form the stable \(\text{AlF}_6^{3-}\) complex, leaving iron in solution. As shown in Example 7.6.2
, we can judge a masking agent’s effectiveness by considering the relevant equilibrium constants. Show that CN is an appropriate masking agent for Ni in a method where nickel’s complexation with EDTA is an interference. The relevant reactions and formation constants are \[\mathrm{Ni}^{2+}(a q)+\mathrm{Y}^{4-}(a q)\rightleftharpoons \mathrm{NiY}^{2-}(a q) \quad K_{1}=4.2 \times 10^{18} \nonumber\] \[\mathrm{Ni}^{2+}(a q)+4 \mathrm{CN}^{-}(a q)\rightleftharpoons \mathrm{Ni}(\mathrm{CN})_{4}^{2-}(a q) \quad \beta_{4}=1.7 \times 10^{30} \nonumber\] where Y is an abbreviation for EDTA. Cyanide is an appropriate masking agent because the formation constant for \(\text{Ni(CN)}_4^{2-}\) is greater than that for the Ni–EDTA complex. In fact, the equilibrium constant for the reaction in which EDTA displaces the masking agent \[\mathrm{Ni}(\mathrm{CN})_{4}^{2-}(a q)+\mathrm{Y}^{4-}(a q) \rightleftharpoons \mathrm{NiY}^{2-}(a q)+4 \mathrm{CN}^{-}(a q) \nonumber\] \[K=\frac{K_{1}}{\beta_{4}}=\frac{4.2 \times 10^{18}}{1.7 \times 10^{30}}=2.5 \times 10^{-12} \nonumber\] is sufficiently small that \(\text{Ni(CN)}_4^{2-}\) is relatively inert in the presence of EDTA. Use the formation constants in to show that 1,10-phenanthroline is a suitable masking agent for Fe in the presence of Fe . Use a ladder diagram to define any limitations on using 1,10-phenanthroline as a masking agent. See for a review of ladder diagrams. The relevant reactions and equilibrium constants are \[\begin{array}{ll}{\mathrm{Fe}^{2+}(a q)+3 \mathrm{phen}(a q)} & {\rightleftharpoons\mathrm{Fe}(\mathrm{phen})_{3}^{2+}(a q) \quad \beta_{3}=5 \times 10^{20}} \\ {\mathrm{Fe}^{3+}(a q)+3 \mathrm{phen}(a q)} & {\rightleftharpoons \mathrm{Fe}(\mathrm{phen})_{3}^{3+}(a q) \quad \beta_{3}=6 \times 10^{13}}\end{array} \nonumber\] where phen is an abbreviation for 1,10-phenanthroline. Because \(\beta_3\) is larger for the complex with Fe than it is for the complex with Fe ,1,10-phenanthroline will bind Fe before it binds Fe . A ladder diagram for this system (as shown below) suggests that an equilibrium p(phen) between 5.6 and 5.9 will fully complex Fe without any significant formation of the \(\text{Fe(phen)}_3^{3+}\) complex. Adding a stoichiometrically equivalent amount of 1,10-phenanthroline to a solution of Fe is sufficient to mask Fe in the presence of Fe . A large excess of 1,10-phenanthroline, however, decreases p(phen) and allows for the formation of both metal–ligand complexes. Because an analyte and its interferent are usually in the same phase, we can achieve a separation if one of them undergoes a change in its physical state or its chemical state. When the analyte and the interferent are miscible liquids, separation by is possible if their boiling points are significantly different. Figure 7.6.5
shows the progress of a distillation as a plot of temperature versus the composition of mixture’s vapor-phase and liquid-phase. The initial liquid mixture (point A), contains more interferent than analyte. When this solution is brought to its boiling point, the vapor phase in equilibrium with the liquid phase is enriched in analyte (point B). The horizontal line that connects points A and B represents this vaporization equilibrium. Condensing the vapor phase at point B, by lowering the temperature, creates a new liquid phase with a composition identical to that in the vapor phase (point C). The vertical line that connects points B and C represents this condensation equilibrium. The liquid phase at point C has a lower boiling point than the original mixture, and is in equilibrium with the vapor phase at point D. This process of repeated vaporization and condensation gradually separates the analyte and the interferent. Two experimental set-ups for distillations are shown in Figure 7.6.6
. The simple distillation apparatus shown in Figure 7.6.6
a is useful only for separating a volatile analyte (or interferent) from a non-volatile interferent (or analyte), or for separating an analyte and an interferent whose boiling points differ by more than 150 C. A more efficient separation is achieved using the fractional distillation apparatus in Figure 7.6.6
b. Packing the fractionating column with a high surface area material, such as a steel sponge or glass beads, provides more opportunity for the repeated process of vaporization and condensation necessary to effect a complete separation. When the sample is a solid, may provide a useful separation of the analyte and the interferent. The sample is heated at a temperature and pressure below the analyte’s triple point, allowing it to vaporize without passing through a liquid state. Condensing the vapor recovers the purified analyte (Figure 7.6.7
). A useful analytical example of sublimation is the isolation of amino acids from fossil mollusk shells and deep-sea sediments [Glavin, D. P.; Bada, J. L. , , 3119–3122]. is another method for purifying a solid. A solvent is chosen in which the analyte’s solubility is significant when the solvent is hot and minimal when the solvent is cold. The interferents must be less soluble in the hot solvent than the analyte or present in much smaller amounts. After heating a portion of the solvent in an Erlenmeyer flask, small amounts of sample are added until undissolved sample is visible. Additional hot solvent is added until the sample redissolves, or until only insoluble impurities remain. This process of adding sample and solvent is repeated until the entire sample is added to the Erlenmeyer flask. Any insoluble impurities are removed by filtering the hot solution. The solution is allowed to cool slowly, which promotes the growth of large, pure crystals, and then cooled in an ice bath to minimize solubility losses. The purified sample is isolated by filtration and rinsed to remove any soluble impurities. Finally, the sample is dried to remove any remaining traces of the solvent. Further purification, if necessary, is accomplished by additional recrystallizations. Distillation, sublimation, and recrystallization use a change in physical state to effect a separation. Chemical reactivity also is a useful tool for separating analytes and interferents. For example, we can separate SiO from a sample by reacting it with HF to form SiF . Because SiF is volatile, it is easy to remove by evaporation. If we wish to collect the reaction’s volatile product, then a distillation is possible. For example, we can isolate the \(\text{NH}_4^+\) in a sample by making the solution basic and converting it to NH . The ammonia is then removed by distillation. Table 7.6.4
provides additional examples of this approach for isolating inorganic ions. Another reaction for separating analytes and interferents is precipitation. Two important examples of using a precipitation reaction in a separation are the pH-dependent solubility of metal oxides and hydroxides, and the pH-dependent solubility of metal sulfides. Separations based on the pH-dependent solubility of oxides and hydroxides usually use a strong acid, a strong base, or an NH /NH Cl buffer to adjust the pH. Most metal oxides and hydroxides are soluble in hot concentrated HNO , although a few oxides, such as WO , SiO , and SnO remain insoluble even under these harsh conditions. To determine the amount of Cu in brass, for example, we can avoid an interference from Sn by dissolving the sample with a strong acid and filtering to remove the solid residue of SnO . Most metals form a hydroxide precipitate in the presence of concentrated NaOH. Those metals that form amphoteric hydroxides, however, do not precipitate because they react to form higher-order hydroxo-complexes. For example, Zn and Al do not precipitate in concentrated NaOH because they form the soluble complexes \(\text{Zn(OH)}_3^-\) and \(\text{Al(OH)}_4^-\). The solubility of Al in concentrated NaOH allows us to isolate aluminum from impure samples of bauxite, an ore of Al O . After crushing the ore, we place it in a solution of concentrated NaOH, dissolving the Al O and forming \(\text{Al(OH)}_4^-\). Other oxides in the ore, such as Fe O and SiO , remain insoluble. After filtering, we recover the aluminum as a precipitate of Al(OH) by neutralizing some of the OH with acid. The pH of an NH /NH Cl buffer (p = 9.26) is sufficient to precipitate most metals as the hydroxide. The alkaline earths and alkaline metals, however, do not precipitate at this pH. In addition, metal ions that form soluble complexes with NH , such as Cu , Zn , Ni , and Co also do not precipitate under these conditions. The use of S as a precipitating reagent is one of the earliest examples of a separation technique. In Fresenius’s 1881 text , sulfide frequently is used to separate metal ions from the remainder of the sample’s matrix [Fresenius. C. R. , John Wiley and Sons: New York, 1881]. Sulfide is a useful reagent for separating metal ions for two reasons: (1) most metal ions, except for the alkaline earths and alkaline metals, form insoluble sulfides; and (2) these metal sulfides show a substantial variation in solubility. Because the concentration of S is pH-dependent, we can control which metal ions precipitate by adjusting the pH. For example, in Fresenius’s gravimetric procedure for the determination of Ni in ore samples (see for a schematic diagram of this procedure), sulfide is used three times to separate Co and Ni from Cu and, to a lesser extent, from Pb . The most important group of separation techniques uses a selective partitioning of the analyte or interferent between two immiscible phases. If we bring a phase that contains the solute, , into contact with a second phase, the solute will partition itself between the two phases, as shown by the following equilibrium reaction. \[S_{\text { phase } 1} \rightleftharpoons S_{\text { phase } 2} \label{7.1}\] The equilibrium constant for reaction \ref{7.1} \[K_{\mathrm{D}}=\frac{\left[S_{\mathrm{phase} \ 2}\right]}{\left[S_{\mathrm{phase} \ 1}\right]} \nonumber\] is called the distribution constant or the . If is sufficiently large, then the solute moves from phase 1 to phase 2. The solute will remain in phase 1 if the partition coefficient is sufficiently small. When we bring a phase that contains two solutes into contact with a second phase, a separation of the solutes is possible if is favorable for only one of the solutes. The physical states of the phases are identified when we describe the separation process, with the phase that contains the sample listed first. For example, if the sample is in a liquid phase and the second phase is a solid, then the separation involves liquid–solid partitioning. We call the process of moving a species from one phase to another phase an . Simple extractions are particularly useful for separations where only one component has a favorable partition coefficient. Several important separation techniques are based on a simple extraction, including liquid–liquid, liquid–solid, solid–liquid, and gas–solid extractions. A liquid–liquid extraction usually is accomplished using a separatory funnel (Figure 7.6.8
). After placing the two liquids in the separatory funnel, we shake the funnel to increase the surface area between the phases. When the extraction is complete, we allow the liquids to separate. The stopcock at the bottom of the separatory funnel allows us to remove the two phases. We also can carry out a liquid–liquid extraction without a separatory funnel by adding the extracting solvent to the sample’s container. Pesticides in water, for example, are preserved in the field by extracting them into a small volume of hexane. A liquid–liquid microextraction, in which the extracting phase is a 1-µL drop suspended from a microsyringe (Figure 7.6.9
), also has been described [Jeannot, M. A.; Cantwell, F. F. , , 235–239]. Because of its importance, a more thorough discussion of liquid–liquid extractions is in . In a solid phase extraction of a liquid sample, we pass the sample through a cartridge that contains a solid adsorbent, several examples of which are shown in Figure 7.6.10
. The choice of adsorbent is determined by the species we wish to separate. Table 7.6.5
provides several representative examples of solid adsorbents and their applications. As an example, let’s examine a procedure for isolating the sedatives secobarbital and phenobarbital from serum samples using a C-18 solid adsorbent [Alltech Associates , Bulletin 83]. Before adding the sample, the solid phase cartridge is rinsed with 6 mL each of methanol and water. Next, a 500-μL sample of serum is pulled through the cartridge, with the sedatives and matrix interferents retained following a liquid–solid extraction (Figure 7.6.11
a). Washing the cartridge with distilled water removes any interferents (Figure 7.6.11
b). Finally, we elute the sedatives using 500 μL of acetone (Figure 7.6.11
c). In comparison to a liquid–liquid extraction, a solid phase extraction has the advantage of being easier, faster, and requires less solvent. An extraction is possible even if the analyte has an unfavorable partition coefficient, provided that the sample’s other components have significantly smaller partition coefficients. Because the analyte’s partition coefficient is unfavorable, a single extraction will not recover all the analyte. Instead we continuously pass the extracting phase through the sample until we achieve a quantitative extraction. A continuous extraction of a solid sample is carried out using a (Figure 7.6.12
). The extracting solvent is placed in the lower reservoir and heated to its boiling point. Solvent in the vapor phase moves upward through the tube on the far right side of the apparatus, reaching the condenser where it condenses back to the liquid state. The solvent then passes through the sample, which is held in a porous cellulose filter thimble, collecting in the upper reservoir. When the solvent in the upper reservoir reaches the return tube’s upper bend, the solvent and extracted analyte are siphoned back to the lower reservoir. Over time the analyte’s concentration in the lower reservoir increases. Microwave-assisted extractions have replaced Soxhlet extractions in some applications [Renoe, B. W. August , 34–40]. The process is the same as that described earlier for a microwave digestion. After placing the sample and the solvent in a sealed digestion vessel, a microwave oven is used to heat the mixture. Using a sealed digestion vessel allows the extraction to take place at a higher temperature and pressure, reducing the amount of time needed for a quantitative extraction. In a Soxhlet extraction the temperature is limited by the solvent’s boiling point at atmospheric pressure. When acetone is the solvent, for example, a Soxhlet extraction is limited to 56 C, but a microwave extraction can reach 150 C. Two other continuous extractions deserve mention. Volatile organic compounds (VOCs) can be quantitatively removed from a liquid sample by a liquid–gas extraction. As shown in Figure 7.6.13
, an inert purging gas, such as He, is passed through the sample. The purge gas removes the VOCs, which are swept to a primary trap where they collect on a solid absorbent. When the extraction is complete, the VOCs are removed from the primary trap by rapidly heating the tube while flushing with He. This technique is known as a . Because the analyte’s recovery may not be reproducible, an internal standard is required for quantitative work. Continuous extractions also can be accomplished using supercritical fluids [McNally, M. E. , , 308A–315A]. If we heat a substance above its critical temperature and pressure it forms a whose properties are between those of a gas and a liquid. A supercritical fluid is a better solvent than a gas, which makes it a better reagent for extractions. In addition, a supercritical fluid’s viscosity is significantly less than that of a liquid, which makes it easier to push it through a particulate sample. One example of a supercritical fluid extraction is the determination of total petroleum hydrocarbons (TPHs) in soils, sediments, and sludges using supercritical CO [“TPH Extraction by SFE,” ISCO, Inc. Lincoln, NE, Revised Nov. 1992]. An approximately 3-g sample is placed in a 10-mL stainless steel cartridge and supercritical CO at a pressure of 340 atm and a temperature of 80 C is passed through the cartridge for 30 minutes at flow rate of 1–2 mL/min. To collect the TPHs, the effluent from the cartridge is passed through 3 mL of tetrachloroethylene at room temperature. At this temperature the CO reverts to the gas phase and is released to the atmosphere. In an extraction, the sample originally is in one phase and we extract the analyte or the interferent into a second phase. We also can separate the analyte and interferents by continuously passing one sample-free phase, called the mobile phase, over a second sample-free phase that remains fixed or stationary. The sample is injected into the mobile phase and the sample’s components partition themselves between the mobile phase and the stationary phase. Those components with larger partition coefficients are more likely to move into the stationary phase and take longer time to pass through the system. This is the basis of all chromatographic separations. Chromatography provides both a separation of analytes and interferents, and a means for performing a qualitative or quantitative analysis for the analyte. For this reason a more thorough treatment of chromatography is found in . | 24,942 | 2,428 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Amides/Reactivity_of_Amides/General_Mechanism_of_Amide_Reactions?sectionId=3 |
Carboxylic acid derivatives are a group of functional groups whose chemistry is closely related. The main difference is the presence of an electronegative substituent that can act as a leaving group during nucleophile substitution reactions. Although there are many types of carboxylic acid derivatives known we will be focusing on just four: Acid halides, Acid anhydrides, Esters, and Amides. 1) Nucleophilic attack on the carbonyl 2) Leaving group is removed Although aldehydes and ketones also contain a carbonyl their chemistry is distinctly different because they do not contain a suitable leaving group. Once the tetrahedral intermediate is formed aldehydes and ketones cannot reform the carbonyl. Because of this aldehydes and ketones typically undergo nucleophilic additions and not substitutions. The relative reactivity of carboxylic acid derivatives toward nucleophile substitutions is related to the electronegative leaving group’s ability to activate the carbonyl. The more electronegative leaving groups withdrawn electron density from the carbonyl, thereby, increasing its electrophilicity. ) | 1,127 | 2,430 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Concepts_in_Biophysical_Chemistry_(Tokmakoff)/06%3A_Dynamics_and_Kinetics/20%3A_Protein_Folding/20.01%3A_HelixCoil_Transition |
Our study of folding mechanism and the statistical mechanical relationship between structure and stability have been guided by models. Of these, simple reductionist models guided the conceptual development from the statistical mechanics side, since full atom simulations were initially intractable. We will focus on the simple models. Hierarchy of various models that reduce protein structure to a set of interacting beads. Gō models and Gō-like models refer to a class of coarse-grained models in which formation of structure is driven by a minimalist interaction potential that drives the system to its native structure. The folded state must be known _________________________________________ | 708 | 2,431 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Amides/Reactivity_of_Amides/General_Mechanism_of_Amide_Reactions?sectionId=2 |
Carboxylic acid derivatives are a group of functional groups whose chemistry is closely related. The main difference is the presence of an electronegative substituent that can act as a leaving group during nucleophile substitution reactions. Although there are many types of carboxylic acid derivatives known we will be focusing on just four: Acid halides, Acid anhydrides, Esters, and Amides. 1) Nucleophilic attack on the carbonyl 2) Leaving group is removed Although aldehydes and ketones also contain a carbonyl their chemistry is distinctly different because they do not contain a suitable leaving group. Once the tetrahedral intermediate is formed aldehydes and ketones cannot reform the carbonyl. Because of this aldehydes and ketones typically undergo nucleophilic additions and not substitutions. The relative reactivity of carboxylic acid derivatives toward nucleophile substitutions is related to the electronegative leaving group’s ability to activate the carbonyl. The more electronegative leaving groups withdrawn electron density from the carbonyl, thereby, increasing its electrophilicity. ) | 1,127 | 2,435 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/16%3A_Aqueous_AcidBase_Equilibriums/16.06%3A_Buffers |
Buffers are solutions that maintain a relatively constant pH when an acid or a base is added. They therefore protect, or “buffer,” other molecules in solution from the effects of the added acid or base. Buffers contain either a weak acid (HA) and its conjugate base (A ) or a weak base (B) and its conjugate acid (BH ), and they are critically important for the proper functioning of biological systems. In fact, every biological fluid is buffered to maintain its physiological pH. To understand how buffers work, let’s look first at how the ionization equilibrium of a weak acid is affected by adding either the conjugate base of the acid or a strong acid (a source of H ). Le Chatelier’s principle can be used to predict the effect on the equilibrium position of the solution. A typical buffer used in biochemistry laboratories contains acetic acid and a salt such as sodium acetate. Recall that the dissociation reaction of acetic acid is as follows:
\[CH_3CO_2H_{(aq)} \leftrightharpoons CH_3CO^−_{(aq)}+H^+_{(aq)} \tag{16.6.1}\] and the equilibrium constant expression is as follows:
\[K_a=\dfrac{[H^+,CH_3CO_2]}{[CH_3CO_2H]} \tag{16.6.2}\] Sodium acetate (CH CO Na) is a strong electrolyte that ionizes completely in aqueous solution to produce Na and CH CO ions. If sodium acetate is added to a solution of acetic acid, Le Chatelier’s principle predicts that the equilibrium in .1 will shift to the left, consuming some of the added CH CO and some of the H ions originally present in solution: Because Na is a spectator ion, it has no effect on the position of the equilibrium and can be ignored. The addition of sodium acetate produces a new equilibrium composition, in which [H ] is less than the initial value. Because [H ] has decreased, the pH will be higher. Thus adding a salt of the conjugate base to a solution of a weak acid increases the pH. This makes sense because sodium acetate is a base, and adding base to a solution of a weak acid should increase the pH. If we instead add a strong acid such as HCl to the system, [H ] increases. Once again the equilibrium is temporarily disturbed, but the excess H ions react with the conjugate base (CH CO ), whether from the parent acid or sodium acetate, to drive the equilibrium to the left. The net result is a new equilibrium composition that has a lower [CH CO ] than before. In both cases, . Adding a strong electrolyte that contains one ion in common with a reaction system that is at equilibrium, in this case CH CO , will therefore shift the equilibrium in the direction that reduces the concentration of the common ion. The shift in equilibrium is called the common ion effect . Adding a common ion to a system at equilibrium affects the equilibrium composition but not the ionization constant. In , we calculated that a 0.150 M solution of formic acid at 25°C (p = 3.75) has a pH of 2.28 and is 3.5% ionized. solution concentration and pH, p , and percent ionization of acid; final concentration of conjugate base or strong acid added pH and percent ionization of formic acid Write a balanced equilibrium equation for the ionization equilibrium of formic acid. Tabulate the initial concentrations, the changes, and the final concentrations. Substitute the expressions for the final concentrations into the expression for . Calculate [H ] and the pH of the solution. Construct a table of concentrations for the dissociation of formic acid. To determine the percent ionization, determine the anion concentration, divide it by the initial concentration of formic acid, and multiply the result by 100. The initial concentrations, the changes in concentration that occur as equilibrium is reached, and the final concentrations can be tabulated. We substitute the expressions for the final concentrations into the equilibrium constant expression and make our usual simplifying assumptions, so Rearranging and solving for , The value of is small compared with 0.150 or 0.100 M, so our assumption about the extent of ionization is justified. Moreover, = (1.8 × 10 )(0.150) = 2.7 × 10 , which is greater than 1.0 × 10 , so again, our assumption is justified. The final pH is −log(2.7 × 10 ) = 3.57, compared with the initial value of 2.29. Thus adding a salt containing the conjugate base of the acid has increased the pH of the solution, as we expect based on Le Chatelier’s principle; the stress on the system has been relieved by the consumption of H ions, driving the equilibrium to the left. To calculate the percentage of formic acid that is ionized under these conditions, we have to determine the final [HCO ]. We substitute final concentrations into the equilibrium constant expression and make the usual simplifying assumptions, so Rearranging and solving for , Once again, our simplifying assumptions are justified. The percent ionization of formic acid is as follows: Adding the strong acid to the solution, as shown in the table, decreased the percent ionization of formic acid by a factor of approximately 38 (3.45%/0.0900%). Again, this is consistent with Le Chatelier’s principle: adding H ions drives the dissociation equlibrium to the left. Exercise As you learned in Example 8, a 0.225 M solution of ethylamine (CH CH NH , p = 3.19) has a pH of 12.08 and a percent ionization of 5.4% at 20°C. Calculate the following: Now let’s suppose we have a buffer solution that contains equimolar concentrations of a weak base (B) and its conjugate acid (BH ). The general equation for the ionization of a weak base is as follows:
\[B_{(aq)}+H_2O_{(l)} \leftrightharpoons BH^+_{(aq)}+OH^−_{(aq)} \tag{16.6.3} \] If the equilibrium constant for the reaction as written in is small, for example = 10 , then the equilibrium constant for the reverse reaction is very large: = 1/ = 10 . Adding a strong base such as OH to the solution therefore causes the equilibrium in to shift to the left, consuming the added OH . As a result, the OH ion concentration in solution remains relatively constant, and the pH of the solution changes very little. Le Chatelier’s principle predicts the same outcome: when the system is stressed by an increase in the OH ion concentration, the reaction will proceed to the left to counteract the stress. If the p of the base is 5.0, the p of its conjugate acid is p = p − p = 14.0 – 5.0 = 9.0. Thus the equilibrium constant for ionization of the conjugate acid is even smaller than that for ionization of the base. The ionization reaction for the conjugate acid of a weak base is written as follows:
\[BH^+_{(aq)}+H_2O_{(l)} \leftrightharpoons B_{(aq)}+H_3O^+_{(aq)} \tag{16.6.4}\] Again, the equilibrium constant for the reverse of this reaction is very large: = 1/ = 10 . If a strong acid is added, it is neutralized by reaction with the base as the reaction in shifts to the left. As a result, the H ion concentration does not increase very much, and the pH changes only slightly. In effect, a buffer solution behaves somewhat like a sponge that can absorb H and OH ions, thereby preventing large changes in pH when appreciable amounts of strong acid or base are added to a solution. Buffers are characterized by the over which they can maintain a more or less constant pH and by their buffer capacity , the amount of strong acid or base that can be absorbed before the pH changes significantly. Although the useful pH range of a buffer depends strongly on the chemical properties of the weak acid and weak base used to prepare the buffer (i.e., on ), its buffer capacity depends solely on the concentrations of the species in the buffered solution. The more concentrated the buffer solution, the greater its buffer capacity. As illustrated in , when NaOH is added to solutions that contain different concentrations of an acetic acid/sodium acetate buffer, the observed change in the pH of the buffer is inversely proportional to the concentration of the buffer. If the buffer capacity is 10 times larger, then the buffer solution can absorb 10 times more strong acid or base before undergoing a significant change in pH. The pH of a buffer can be calculated from the concentrations of the weak acid and the weak base used to prepare it, the concentration of the conjugate base and conjugate acid, and the p or p of the weak acid or weak base. The procedure is analogous to that used in Example 14 to calculate the pH of a solution containing known concentrations of formic acid and formate. An alternative method frequently used to calculate the pH of a buffer solution is based on a rearrangement of the equilibrium equation for the dissociation of a weak acid. The simplified ionization reaction is for which the equilibrium constant expression is as follows:
This equation can be rearranged as follows: \[[H^+]=K_a\dfrac{[HA]}{[A^−]} \tag{16.6.6}\] Taking the logarithm of both sides and multiplying both sides by −1, \[−\log[H^+]=−\log K_a−\log\left(\dfrac{[HA]}{[A^−]}\right)=−\log{K_a}+\log\left(\dfrac{[A^−]}{[HA]}\right) \tag{16.6.7}\] Replacing the negative logarithms in Equation 16.60, \[pH=pK_a+\log \left( \dfrac{[A^−]}{[HA]} \right) \tag{16.6.8}\] or, more generally,
\[pH=pK_a+\log\left(\dfrac{[base]}{[acid]}\right) \tag{16.6.9}\] and are both forms of the Henderson-Hasselbalch equation , named after the two early-20th-century chemists who first noticed that this rearranged version of the equilibrium constant expression provides an easy way to calculate the pH of a buffer solution. In general, the validity of the Henderson-Hasselbalch equation may be limited to solutions whose concentrations are at least 100 times greater than their values. There are three special cases where the Henderson-Hasselbalch equation is easily interpreted without the need for calculations: Each time we increase the [base]/[acid] ratio by 10, the pH of the solution increases by 1 pH unit. Conversely, if the [base]/[acid] ratio is 0.1, then pH = p − 1. Each additional factor-of-10 decrease in the [base]/[acid] ratio causes the pH to decrease by 1 pH unit. If [base] = [acid] for a buffer, then pH = p . Changing this ratio by a factor of 10 either way changes the pH by ±1 unit. What is the pH of a solution that contains concentration of acid, conjugate base, and p ; concentration of base, conjugate acid, and p pH Substitute values into either form of the Henderson-Hasselbalch equation ( or ) to calculate the pH. This result makes sense because the [A ]/[HA] ratio is between 1 and 10, so the pH of the buffer must be between the p (3.75) and p + 1, or 4.75. This result is identical to the result in part (a), which emphasizes the point that the pH of a buffer depends on the of the concentrations of the conjugate base and the acid, on the magnitude of the concentrations. Because the [A ]/[HA] ratio is the same as in part (a), the pH of the buffer must also be the same (3.95). Once again, this result makes sense: the [B]/[BH ] ratio is about 1/2, which is between 1 and 0.1, so the final pH must be between the p (5.23) and p − 1, or 4.23. Exercise What is the pH of a solution that contains The p of benzoic acid is 4.20, and the p of trimethylamine is also 4.20. The Henderson-Hasselbalch equation can also be used to calculate the pH of a buffer solution after adding a given amount of strong acid or strong base, as demonstrated in Example 16. The buffer solution in Example 15 contained 0.135 M HCO H and 0.215 M HCO Na and had a pH of 3.95. composition and pH of buffer; concentration and volume of added acid or base final pH Calculate the amounts of formic acid and formate present in the buffer solution using the procedure from Example 14. Then calculate the amount of acid or base added. Construct a table showing the amounts of all species after the neutralization reaction. Use the final volume of the solution to calculate the concentrations of all species. Finally, substitute the appropriate values into the Henderson-Hasselbalch equation ( ) to obtain the pH. The added HCl (a strong acid) or NaOH (a strong base) will react completely with formate (a weak base) or formic acid (a weak acid), respectively, to give formic acid or formate and water. We must therefore calculate the amounts of formic acid and formate present after the neutralization reaction. \[ 100\; \cancel{mL}\left ( \dfrac{0.215 \; mmol\; HCO_2^{-}}{\cancel{mL}} \right )=21.5 \; mmol\; HCO_2^{-} \notag \] The millimoles of H in 5.00 mL of 1.00 M HCl is as follows: \[ 5.00\; \cancel{mL}\left ( \dfrac{1.00 \; mmol\; H^{+}}{\cancel{mL}} \right )=5.00 \; mmol\; H^{+} \notag \] Next, we construct a table of initial amounts, changes in amounts, and final amounts: \[HCO^{2−}_{(aq)} + H^+_{(aq)} \rightarrow HCO_2H_{(aq)} \notag \] The final amount of H in solution is given as “∼0 mmol.” For the purposes of the stoichiometry calculation, this is essentially true, but remember that the point of the problem is to calculate the final [H ] and thus the pH. We now have all the information we need to calculate the pH. We can use either the lengthy procedure of Example 14 or the Henderson–Hasselbach equation. Because we have performed equilibrium calculations in this chapter, we’ll take the latter approach. The Henderson-Hasselbalch equation requires the concentrations of HCO and HCO H, which can be calculated using the number of millimoles ( ) of each and the total volume ( ). Substituting these values into the Henderson-Hasselbalch equation, \[pH=pK_a+\log \left( \dfrac{[HCO_2^−]}{[HCO_2H]} \right)=pK_a+\log\left(\dfrac{n_{HCO_2^−}/V_f}{n_{HCO_2H}/V_f}\right)=pK_a+\log \left(\dfrac{n_{HCO_2^−}}{n_{HCO_2H}}\right) \notag \] Because the total volume appears in both the numerator and denominator, it cancels. We therefore need to use only the ratio of the number of millimoles of the conjugate base to the number of millimoles of the weak acid. So \[pH=pK_a+\log\left(\dfrac{n_{HCO_2^−}}{n_{HCO_2H}}\right)=3.75+\log\left(\dfrac{16.5\; mmol}{18.5\; mmol}\right)=3.75 −0.050=3.70 \notag \] Once again, this result makes sense on two levels. First, the addition of HCl has decreased the pH from 3.95, as expected. Second, the ratio of HCO to HCO H is slightly less than 1, so the pH should be between the p and p − 1. With this information, we can construct a table of initial amounts, changes in amounts, and final amounts. The final amount of OH in solution is not actually zero; this is only approximately true based on the stoichiometric calculation. We can calculate the final pH by inserting the numbers of millimoles of both HCO and HCO H into the simplified Henderson-Hasselbalch expression used in part (a) because the volume cancels: Once again, this result makes chemical sense: the pH has increased, as would be expected after adding a strong base, and the final pH is between the p and p + 1, as expected for a solution with a HCO /HCO H ratio between 1 and 10. Exercise The buffer solution from Example 15 contained 0.119 M pyridine and 0.234 M pyridine hydrochloride and had a pH of 4.94. Only the (in moles or millimoles) of the acidic and basic components of the buffer are needed to use the Henderson-Hasselbalch equation, their concentrations. The most effective buffers contain equal concentrations of an acid and its conjugate base. The results obtained in Example 16 and its corresponding exercise demonstrate how little the pH of a well-chosen buffer solution changes despite the addition of a significant quantity of strong acid or strong base. Suppose we had added the same amount of HCl or NaOH solution to 100 mL of an buffered solution at pH 3.95 (corresponding to 1.1 × 10 M HCl). In this case, adding 5.00 mL of 1.00 M HCl would lower the final pH to 1.32 instead of 3.70, whereas adding 5.00 mL of 1.00 M NaOH would raise the final pH to 12.68 rather than 4.24. (Try verifying these values by doing the calculations yourself.) Thus the presence of a buffer significantly increases the ability of a solution to maintain an almost constant pH. A buffer that contains approximately equal amounts of a weak acid and its conjugate base in solution is equally effective at neutralizing either added base or added acid. This is shown in for an acetic acid/sodium acetate buffer. Adding a given amount of strong acid shifts the system along the horizontal axis to the left, whereas adding the same amount of strong base shifts the system the same distance to the right. In either case, the change in the ratio of CH CO to CH CO H from 1:1 reduces the buffer capacity of the solution. There is a strong correlation between the effectiveness of a buffer solution and the titration curves discussed in . Consider the schematic titration curve of a weak acid with a strong base shown in . As indicated by the labels, the region around p corresponds to the midpoint of the titration, when approximately half the weak acid has been neutralized. This portion of the titration curve corresponds to a buffer: it exhibits the smallest change in pH per increment of added strong base, as shown by the nearly horizontal nature of the curve in this region. The nearly flat portion of the curve extends only from approximately a pH value of 1 unit less than the p to approximately a pH value of 1 unit greater than the p , which is why buffer solutions usually have a pH that is within ±1 pH units of the p of the acid component of the buffer. In the region of the titration curve at the lower left, before the midpoint, the acid–base properties of the solution are dominated by the equilibrium for dissociation of the weak acid, corresponding to . In the region of the titration curve at the upper right, after the midpoint, the acid–base properties of the solution are dominated by the equilibrium for reaction of the conjugate base of the weak acid with water, corresponding to . However, we can calculate either or from the other because they are related by . Metabolic processes produce large amounts of acids and bases, yet organisms are able to maintain an almost constant internal pH because their fluids contain buffers. This is not to say that the pH is uniform throughout all cells and tissues of a mammal. The internal pH of a red blood cell is about 7.2, but the pH of most other kinds of cells is lower, around 7.0. Even within a single cell, different compartments can have very different pH values. For example, one intracellular compartment in white blood cells has a pH of around 5.0. Because no single buffer system can effectively maintain a constant pH value over the entire physiological range of approximately pH 5.0 to 7.4, biochemical systems use a set of buffers with overlapping ranges. The most important of these is the CO /HCO system, which dominates the buffering action of blood plasma. The acid–base equilibrium in the CO /HCO buffer system is usually written as follows:
\[H_2CO_{3(aq)} \leftrightharpoons H^+_{(aq)}+HCO^−_{3(aq)} \tag{16.6.10}\] with = 4.5 × 10 and p = 6.35 at 25°C. In fact, is a grossly oversimplified version of the CO /HCO system because a solution of CO in water contains only rather small amounts of H CO . Thus does not allow us to understand how blood is actually buffered, particularly at a physiological temperature of 37°C. As shown in , CO is in equilibrium with H CO , but the equilibrium lies far to the left, with an H CO /CO ratio less than 0.01 under most conditions:
\[CO_{2(aq)}+H_2O_{(l)} \leftrightharpoons H_2CO_{3(aq)} \tag{16.6.11}\] with ′ = 4.0 × 10 at 37°C. The true p of carbonic acid at 37°C is therefore 3.70, not 6.35, corresponding to a of 2.0 × 10 , which makes it a much stronger acid than suggests. Adding and and canceling H CO from both sides give the following overall equation for the reaction of CO with water to give a proton and the bicarbonate ion:
\[CO_{2(aq)}+H_2O_{(l)} \leftrightharpoons H_2CO_{3(aq)}\tag{16.6.12a}\] with \(K'=4.0 \times 10^{−3} (37°C)\) \[ H_2CO_{3(aq)} \leftrightharpoons H^+_{(aq)} + HCO^−_{3(aq)}\tag{16.6.12b}\] with \(K_a=2.0 \times 10^{−4} (37°C) \) \[CO_{2(aq)}+H_2O_{(l)} \leftrightharpoons H^+_{(aq)}+HCO^−_{3(aq)} \tag{16.6.12c}\] with \(K=8.0 \times 10^{−7} (37°C)\) The value for the reaction in is the product of the ionization constant for carbonic acid ( ) and the equilibrium constant ( ) for the reaction of CO (aq) with water to give carbonic acid. The equilibrium equation for the reaction of CO with water to give bicarbonate and a proton is therefore
\[K=\dfrac{[H^+,HCO_3^−]}{[CO_2]}=8.0 \times 10^{−7} \tag{16.6.13}\] The presence of a gas in the equilibrium constant expression for a buffer is unusual. According to Henry’s law, CO ) where is the Henry’s law constant for CO , which is 3.0 × 10 M/mmHg at 37°C. (For more information about Henry’s law, see .) Substituting this expression for [CO ] in ,
\[K=\dfrac{[H^+,HCO_3^−]}{(3.0 \times 10^{−5}\; M/mmHg)(P_{CO_2})} \tag{16.6.14}\] is in mmHg. Taking the negative logarithm of both sides and rearranging,
\[pH=6.10+\log \left( \dfrac{ [HCO_3^−]}{(3.0 \times 10^{−5} M/mm \;Hg)\; (P_{CO_2}) } \right) \tag{16.6.15}\] Thus the pH of the solution depends on both the CO pressure over the solution and [HCO ]. plots the relationship between pH and [HCO ] under physiological conditions for several different values of P(CO ) with normal pH and [HCO ] values indicated by the dashed lines. According to , adding a strong acid to the CO /HCO system causes [HCO ] to decrease as HCO is converted to CO . Excess CO is released in the lungs and exhaled into the atmosphere, however, so there is essentially no change in CO ). Because the change in [HCO ]/P(CO ) is small, predicts that the change in pH will also be rather small. Conversely, if a strong base is added, the OH reacts with CO to form [HCO ], but CO is replenished by the body, again limiting the change in both [HCO ]/P(CO ) and pH. The CO /HCO buffer system is an example of an system, in which the total concentration of the components of the buffer change to keep the pH at a nearly constant value. If a passenger steps out of an airplane in Denver, Colorado, for example, the lower P(CO ) at higher elevations (typically 31 mmHg at an elevation of 2000 m versus 40 mmHg at sea level) causes a shift to a new pH and [HCO ]. The increase in pH and decrease in [HCO ] in response to the decrease in P(CO ) are responsible for the general malaise that many people experience at high altitudes. If their blood pH does not adjust rapidly, the condition can develop into the life-threatening phenomenon known as altitude sickness. are solutions that resist a change in pH after adding an acid or a base. Buffers contain a weak acid (HA) and its conjugate weak base (A ). Adding a strong electrolyte that contains one ion in common with a reaction system that is at equilibrium shifts the equilibrium in such a way as to reduce the concentration of the common ion. The shift in equilibrium is called the . Buffers are characterized by their pH range and buffer capacity. The useful pH range of a buffer depends strongly on the chemical properties of the conjugate weak acid–base pair used to prepare the buffer (the or ), whereas its depends solely on the concentrations of the species in the solution. The pH of a buffer can be calculated using the , which is valid for solutions whose concentrations are at least 100 times greater than their values. Because no single buffer system can effectively maintain a constant pH value over the physiological range of approximately 5 to 7.4, biochemical systems use a set of buffers with overlapping ranges. The most important of these is the CO /HCO system, which dominates the buffering action of blood plasma. : \[pH=pK_a +\log \left(\dfrac{[A^−]}{[HA]} \right) \notag \] .9: \[pH=pK_a +\log \left(\dfrac{[base]}{[acid]} \right) \notag \] Explain why buffers are crucial for the proper functioning of biological systems. What is the role of a buffer in chemistry and biology? Is it correct to say that buffers prevent a change in [H O ]? Explain your reasoning. Explain why the most effective buffers are those that contain approximately equal amounts of the weak acid and its conjugate base. Which region of the titration curve of a weak acid or a weak base corresponds to the region of the smallest change in pH per amount of added strong acid or strong base? If you were given a solution of sodium acetate, describe two ways you could convert the solution to a buffer. Why are buffers usually used only within approximately one pH unit of the p or p of the parent weak acid or base? The titration curve for a monoprotic acid can be divided into four regions: the starting point, the region around the midpoint of the titration, the equivalence point, and the region after the equivalence point. For which region would you use each approach to describe the behavior of the solution? Which of the following will produce a buffer solution? Explain your reasoning in each case. Which of the following will produce a buffer solution? Explain your reasoning in each case. Use the definition of for a weak base to derive the following expression, which is analogous to the Henderson-Hasselbalch equation but for a weak base (B) rather than a weak acid (HA): \[pOH=pK_b−\log\left(\dfrac{[base]}{[acid]}\right)\] Why do biological systems use overlapping buffer systems to maintain a constant pH? The CO /HCO buffer system of blood has an effective p of approximately 6.1, yet the normal pH of blood is 7.4. Why is CO /HCO an effective buffer when the p is more than 1 unit below the pH of blood? What happens to the pH of blood when the CO pressure increases? when the O pressure increases? Carbon dioxide produced during respiration is converted to carbonic acid (H CO ). The p of carbonic acid is 6.35, and its p is 10.33. Write the equations corresponding to each p value and predict the equilibrium position for each reaction. Benzenesulfonic acid (p = 0.70) is synthesized by treating benzene with concentrated sulfuric acid. Calculate the following: Phenol has a p of 9.99. Calculate the following: Salicylic acid is used in the synthesis of acetylsalicylic acid, or aspirin. One gram dissolves in 460 mL of water to create a saturated solution with a pH of 2.40. An intermediate used in the synthesis of perfumes is valeric acid, also called pentanoic acid. The p of pentanoic acid is 4.84 at 25°C. | 26,458 | 2,437 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/22%3A_Helmholtz_and_Gibbs_Energies/22.03%3A_The_Maxwell_Relations |
Modeling the dependence of the Gibbs and Helmholtz functions behave with varying temperature, pressure, and volume is fundamentally useful. But in order to do that, a little bit more development is necessary. To see the power and utility of these functions, it is useful to combine the First and Second Laws into a single mathematical statement. In order to do that, one notes that since \[dS = \dfrac{dq}{T} \nonumber \] for a reversible change, it follows that \[dq= TdS \nonumber \] And since \[dw = TdS - pdV \nonumber \] for a reversible expansion in which only p-V works is done, it also follows that (since \(dU=dq+dw\)): \[dU = TdS - pdV \nonumber \] This is an extraordinarily powerful result. This differential for \(dU\) can be used to simplify the differentials for \(H\), \(A\), and \(G\). But even more useful are the constraints it places on the variables T, S, p, and V due to the mathematics of exact differentials! The above result suggests that the natural variables of internal energy are \(S\) and \(V\) (or the function can be considered as \(U(S, V)\)). So the total differential (\(dU\)) can be expressed: \[dU = \left( \dfrac{\partial U}{\partial S} \right)_V dS + \left( \dfrac{\partial U}{\partial V} \right)_S dV \nonumber \] Also, by inspection (comparing the two expressions for \(dU\)) it is apparent that: \[\left( \dfrac{\partial U}{\partial S} \right)_V = T \label{eq5A} \] and \[\left( \dfrac{\partial U}{\partial V} \right)_S = -p \label{eq5B} \] But the value doesn’t stop there! Since \(dU\) is an exact differential, the Euler relation must hold that \[ \left[ \dfrac{\partial}{\partial V} \left( \dfrac{\partial U}{\partial S} \right)_V \right]_S= \left[ \dfrac{\partial}{\partial S} \left( \dfrac{\partial U}{\partial V} \right)_S \right]_V \nonumber \] By substituting Equations \ref{eq5A} and \ref{eq5B}, we see that \[ \left[ \dfrac{\partial}{\partial V} \left( T \right)_V \right]_S= \left[ \dfrac{\partial}{\partial S} \left( -p \right)_S \right]_V \nonumber \] or \[ \left( \dfrac{\partial T}{\partial V} \right)_S = - \left( \dfrac{\partial p}{\partial S} \right)_V \nonumber \] This is an example of a . These are very powerful relationship that allows one to substitute partial derivatives when one is more convenient (perhaps it can be expressed entirely in terms of \(\alpha\) and/or \(\kappa_T\) for example.) A similar result can be derived based on the definition of \(H\). \[ H \equiv U +pV \nonumber \] Differentiating (and using the chain rule on \(d(pV)\)) yields \[ dH = dU +pdV + Vdp \nonumber \] Making the substitution using the combined first and second laws (\(dU = TdS – pdV\)) for a reversible change involving on expansion (p-V) work \[ dH = TdS – \cancel{pdV} + \cancel{pdV} + Vdp \nonumber \] This expression can be simplified by canceling the \(pdV\) terms. \[ dH = TdS + Vdp \label{eq2A} \] And much as in the case of internal energy, this suggests that the natural variables of \(H\) are \(S\) and \(p\). Or \[dH = \left( \dfrac{\partial H}{\partial S} \right)_p dS + \left( \dfrac{\partial H}{\partial p} \right)_S dV \label{eq2B} \] Comparing Equations \ref{eq2A} and \ref{eq2B} show that \[\left( \dfrac{\partial H}{\partial S} \right)_p= T \label{eq6A} \] and \[\left( \dfrac{\partial H}{\partial p} \right)_S = V \label{eq6B} \] It is worth noting at this point that both (Equation \ref{eq5A}) \[\left( \dfrac{\partial U}{\partial S} \right)_V \nonumber \] and (Equation \ref{eq6A}) \[\left( \dfrac{\partial H}{\partial S} \right)_p \nonumber \] are equation to \(T\). So they are equation to each other \[\left( \dfrac{\partial U}{\partial S} \right)_V = \left( \dfrac{\partial H}{\partial S} \right)_p \nonumber \] Euler \[ \left[ \dfrac{\partial}{\partial p} \left( \dfrac{\partial H}{\partial S} \right)_p \right]_S= \left[ \dfrac{\partial}{\partial S} \left( \dfrac{\partial H}{\partial p} \right)_S \right]_p \nonumber \] so \[ \left( \dfrac{\partial T}{\partial p} \right)_S = \left( \dfrac{\partial V}{\partial S} \right)_p \nonumber \] This is the Maxwell relation on \(H\). Maxwell relations can also be developed based on \(A\) and \(G\). The results of those derivations are summarized in Table 6.2.1.. The Maxwell relations are extraordinarily useful in deriving the dependence of thermodynamic variables on the state variables of p, T, and V. Show that \[ \left( \dfrac{\partial V}{\partial T} \right)_p = T\dfrac{\alpha}{\kappa_T} - p \nonumber \] Start with the combined first and second laws: \[dU = TdS - pdV \nonumber \] Divide both sides by \(dV\) and constraint to constant \(T\): \[\left.\dfrac{dU}{dV}\right|_{T} = \left.\dfrac{TdS}{dV}\right|_{T} - p \left.\dfrac{dV}{dV} \right|_{T} \nonumber \] Noting that \[\left.\dfrac{dU}{dV}\right|_{T} =\left( \dfrac{\partial U}{\partial V} \right)_T \nonumber \] \[ \left.\dfrac{TdS}{dV}\right|_{T} = \left( \dfrac{\partial S}{\partial V} \right)_T \nonumber \] \[\left.\dfrac{dV}{dV} \right|_{T} = 1 \nonumber \] The result is \[ \left( \dfrac{\partial U}{\partial V} \right)_T = T \left( \dfrac{\partial S}{\partial V} \right)_T -p \nonumber \] Now, employ the Maxwell relation on \(A\) (Table 6.2.1) \[ \left( \dfrac{\partial p}{\partial T} \right)_V = \left( \dfrac{\partial S}{\partial V} \right)_T \nonumber \] to get \[ \left( \dfrac{\partial U}{\partial V} \right)_T = T \left( \dfrac{\partial p}{\partial T} \right)_V -p \nonumber \] and since \[\left( \dfrac{\partial p}{\partial T} \right)_V = \dfrac{\alpha}{\kappa_T} \nonumber \] It is apparent that \[ \left( \dfrac{\partial V}{\partial T} \right)_p = T\dfrac{\alpha}{\kappa_T} - p \nonumber \] : How cool is that? This result was given without proof in Chapter 4, but can now be proven analytically using the Maxwell Relations! | 5,764 | 2,438 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/23%3A_Phase_Equilibria/23.02%3A_Gibbs_Energies_and_Phase_Diagrams |
Although the \(G\) curve is continuous, its first order derivatives (\(-S\)) is discontinuous at the phase changes. This is why this transition it called a . We could say that: More subtle transitions where \(G\) is continuous, \(H\) and \(S\) are also continuous but have a kink and the discontinuity is only found in the second order derivatives (such as \(C_P\)) also exist. They are called . In such a case: This classification goes back to Ehrenfest. Obviously it based on the question: what order derivative is the first to go discontinuous? Of course we could extend this principle and define but there are reasons to be doubtful that such things exist. Another problem is that it is assumed that the order must be integer: 1,2, etc. Is it possible to have a transition of intermediate non-integer order, say 1.3? Although derivatives of fractional order are beyond the scope of the chemistry curriculum the mathematics does exist (Liouville). The Gibbs free energy is a particularly important function in the study of phases and phase transitions. The behavior of \(G(N, P,T)\), particularly as a function of \(P\) and \(T\), can signify a phase transition and can tell us some of the thermodynamic properties of different phases. Consider, first, the behavior of \(G\) vs. \(T\) between the solid and liquid phases of benzene: We immediately notice several things. First, although the free energy is continuous across the phase transition, its first derivative, \(\partial G/\partial T\) is not: The slope of \(G(T)\) in the solid region is different from the slope in the liquid region. When the first derivative of the free energy with respect to one of its dependent thermodynamic variables is discontinuous across a phase transition, this is an example of what is called a . The solid-liquid-gas phase transition of most substances is first order. When the free energy exhibits continuous first derivatives but discontinuous second derivatives, the phase transition is called . Examples of this type of phase transition are the order-disorder transition in paramagnetic materials. Now, recall that \[S = -\dfrac{\partial G}{\partial T} \label{13.1}\] \[\dfrac{\partial G^\text{(solid)}}{\partial T} = -S^\text{(solid)}, \: \: \: \: \: \: \: \dfrac{\partial G^\text{(liquid)}}{\partial T} = -S^\text{(liquid)} \label{13.2}\] \[\dfrac{\partial G^\text{(liquid)}}{\partial T} < \dfrac{\partial G^\text{(solid)}}{\partial T} \label{13.3}\] \[\bar{V} = \dfrac{\partial \bar{G}}{\partial P} \label{13.4}\] \[\bar{V}^\text{(gas)} \gg \bar{V}^\text{(liquid)} > \bar{V}^\text{(solid)} \label{13.5}\] \[\bar{V}^\text{(gas)} \gg \bar{V}^\text{(solid)} > \bar{V}^\text{(liquid)} \label{13.6}\] As discussed before there are many other forms of work possible, such as electrical work, magnetic work or elastic work. These they are commonly incorporated in the formalism of thermodynamics by adding other terms, e.g: \[dG = -SdT + VdP + ℰde + MdH + FdL + γdA \nonumber \] The terms always appear in a pair of what is known as conjugate variables. That is even clearer if we write out the state function rather than its differential form: \[G = U + PV -TS + ℰe + MH + FL + γA + ... \nonumber \] The PV term can also be generalized -and needs to be so- for a viscous fluid to a stress-strain conjugate pair. It then involves a stress tensor. We will soon encounter another conjugate pair: μdn that deals with changes in composition (n) and the thermodynamic potential μ. | 3,491 | 2,439 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/27%3A_More_about_Spectroscopy/27.02%3A_Line-Width_Differences_in_NMR |
If you look at the NMR spectra of many different kinds of organic compounds, you will notice that some resonances are sharp and others are broad. In a few spectra, all of the peaks may be broad as the result of poor spectrometer performance, but this is not true for the spectra of Figures 9-29 and 24-2 where, within a given spectrum, some resonances will be seen to be sharp and others broad. We can understand these differences by consideration of the of the magnetic states between which the NMR transitions occur.\(^1\) The lifetimes of the states can be related to the width of the lines by the . You may have heard of the uncertainty principle, but if you have not studied chemical physics you may have little idea of its possible importance to organic chemistry. The usual statement of the principle is that there are limits to how precisely we can specify the momentum and the position of a particle at the same time. An alternative statement has more relevance to spectroscopy and chemistry, namely, that . The the lifetime, the with which we can define the energy.\(^2\) Let us consider an example. Suppose a magnetic nucleus in a ground state with a long lifetime and rather precisely defined energy goes to an excited state with a short lifetime, \(\Delta t\).\(^3\) The uncertainty principle tells us that the energy of the excited state cannot be defined precisely. It will have an inherent uncertainty in its energy so that an \(\nu\), having uncertainty in frequency \(\Delta \nu\), will take the nucleus from the ground state to the excited state. The imprecision of the energy \(\Delta \Delta E\), or the imprecision \(\Delta \nu\) in the transition frequency, \(\nu\), depends on \(\Delta t\), and is given approximately by the relationship \[\Delta \Delta E \sim \frac{h}{2 \pi} \times \frac{1}{\Delta t} \sim h \Delta \nu \tag{27-1}\] in which \(h\) is Planck's constant. What this means is that the absorption line corresponding to the transition will have an uncertainty in that is inversely proportional to \(\Delta t\) (see Figure 27-1). It is most convenient to think of line widths in frequency units because most of our spectra are plotted this way. If the scale is wavelength or energy, it can be converted to frequency by the procedures given previously ( ). Division of Equation 27-1 by \(h\) leads to the relationship \(\Delta \nu \sim 1/ \left(2 \pi \times \Delta t \right)\). In NMR spectroscopy, we may wish to consider spin-spin splittings or chemical shifts involving lines no farther than \(1 \: \text{Hz}\) apart. However, two lines \(1 \: \text{Hz}\) apart will not be clearly distinguishable unless \(\Delta \nu\) of each is less than about \(1 \: \text{Hz}\), which corresponds to a \(\Delta t\), the lifetime of the excited state, of \(1/ \left( 2 \pi \right) \cong 0.16 \: \text{sec}\). If \(\Delta \nu\) is \(\geq 2 \: \text{Hz}\), lines that are \(1 \: \text{Hz}\) apart will be so poorly resolved as to appear as one line (cf. Figure 27-2). A \(\Delta \nu\) of \(2 \: \text{Hz}\) corresponds to a \(\Delta t\) of \(1/ \left( 2 \times 2 \pi \right) \cong 0.08 \: \text{sec}\). Clearly, line separations observed in NMR spectroscopy and, in fact, in all forms of spectroscopy, depend on the lifetimes of the states between which transitions take place. The lifetime of \(0.16 \: \text{sec}\) required for \(\Delta \nu\) to be \(1 \: \text{Hz}\) is a time for a molecule! During \(0.16 \: \text{sec}\), a molecule such as ethanol in the liquid phase may undergo \(10^{11}\) collisions with other molecules, \(10^{10}\) rotations about the \(\ce{C-C}\) bonds, and \(10^{12}\) vibrations of each of the various bonds, and may even undergo a number of chemical changes. The properties of magnetic states that have lifetimes of this order clearly must be an over all of these happenings. It is possible to shorten the lifetime of an excited nuclear magnetic state (or increase its ) in a number of ways. For a liquid, the simplest way is to dissolve in it paramagnetic metal ions, such as \(\ce{Cu}\)(II), \(\ce{Fe}\)(III), \(\ce{Mn}\)(II), and the like, or other substances (\(\ce{O_2}\), \(\ce{NO}\), and so on) that have . Another way is to reduce the rate of motion of magnetic nuclei in different molecules with respect to one another, which is easily done by . Without going into details of the mechanisms by which substances with unpaired electrons or increased viscosity shorten the lifetime of excited nuclear magnetic states, it is important to know that dramatic line broadening thereby can be produced. Thus the proton resonance line of water is enormously broadened by adding paramagnetic \(\ce{Mn}\)(II) ions or by freezing (water molecules in ice move much more slowly relative to one another than in liquid water). and (1977) | 4,820 | 2,440 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/08%3A_Properties_of_Organic_Compounds/8.16%3A_Aldehydes_and_Ketones/8.16.01%3A_Biology-_Ketosis |
Prevalent biomolecules processed by the human body every day are carbohydrates, fatty acids, and amino acids, and the ingestion of ethanol in alcoholic beverages is, in moderation, somewhat harmless. These seemingly perfectly edible compounds undergo chemical changes in the body that produce molecules that are not so biologically friendly especially in large amounts. The removal of an amine group from an amino acid helps with the decomposition of fatty acids into acetone, acetoacetic acid, and beta-hydroxybutyric acid . Acetone (and the other two molecules to a lesser extent) are members of the family. Ethanol is easily oxidized to acetaldehyde, one of the most important . Ketones and aldehydes are large families of organic compounds that have the functional group, called a . Rotate the jmol structures to see the similarities and differences between each model. Just as ethanol is oxidized to acetaldehyde, methanol is oxidized to formaldehyde. The loss of the hydrogen atom connected to methanol's oxygen atom forces it to double bond with the adjacent carbon atom, creating the carbonyl group. A ketone such as acetone shares this characteristic, except that the carbonyl group is attached to two carbons. Examine what type of alcohol methanol is and how it oxidizes to formaldehyde. Using this information, what type of alcohol do you think would oxidize to form a ketone? Methanol is a primary alcohol, with its hydroxyl group at the end of a carbon chain. Because the carbonyl group of a ketone is bonded to a carbon atom connected to two other carbon atoms, we should deduce that secondary alcohols oxidize to produce ketones. Using what you know about hydrogen bonding, qualitatively compare the melting and boiling points of aldehydes and ketones to their alcohol derivatives. Strong hydrogen bonding occurs between alcohols because of their hydroxyl (OH) groups. When these alcohols are oxidized, the hydrogen atom bonded to the oxygen atom is lost in order to form the double bond and carbonyl group. Therefore, aldehydes and ketones have weaker lower intermolecular forces and thus lower melting and boiling points compared to similarly sized alcohols. The carbonyl group is rather polar, however, since the difference between the electronegativities of carbon (2.5) and oxygen (3.5) is rather large, and there are usually no other dipoles in an aldehyde or ketone molecule to cancel the effect of C==O. Therefore the boiling points of aldehydes and ketones are intermediate between those of alkanes or ethers on the one hand and alcohols on the other. Acetaldehyde, CH CH CHO, boils at 20.8°C midway between propane (–42°C) and ethanol (78.5°C). The boiling points of propanal and acetone are compared with other organic compounds in the table of the boiling points of comparable organic compounds which shows the same trend. The endings and signify dehyde and ket , respectively. Compounds with lower molecular weights have common names, such as formaldehyde, but standard IUPAC nomenclature can also be applied. Aldehydes are named just as the alcohols from which they are oxidized, just replacing the "ol" with "al". For example, the primary alcohol butanol would oxidize to butanal. The names of ketones depend on where the carbonyl group is located in the chain. When counting the carbon atoms in the longest carbon backbone, the number of the carbon included in the carbonyl group is inserted between the base name and the suffix "one". An organic molecule with five carbon atoms with the carbonyl group consisting of an oxygen atom double bonded to the third carbon would be termed pentan-3-one. Alternatively, the digit can be placed in front, giving 3-pentanone. The general formula for an aldehyde is , while for a ketone it is . Note that every ketone is isomeric with at least one aldehyde. Acetone, for example, has the same molecular formula (C H O) as propanal. Name the organic molecules shown below. a) b) a) Propanal b) 2-Hexanone or Hexan-2-one Chemical tests may be performed to determine the presence of a carbonyl group. Immunochemically, it can be found by adding 2,4-dinitrophenylhydrazine to test tubes containing 2-propanol, an alcohol, 2-propanone (acetone), a ketone; and propionic acid, a carboxylic acid. 2,4-dinitrophenylhydrazine only reacts with the carbonyl group of 2-propanone, forming an orange precipitate. The reaction that occurs is: Another test can distinguish between aldehydes and ketones, using Tollens' reagent, which is an aqueous solution of silver nitrate, sodium hydroxide, and a little ammonia. If an aldehyde, in this case, glucose, is added to the solution, the Ag is reduced by the aldehyde, and the aldehyde is oxidized into a carboxylic acid. This produces silver metal, which coats the flask and creates the mirror. Ketones are not as easily oxidized, so only aldehydes produce the silver mirror. The equation for the reaction in the video is: The simplest aldehydes and ketones are the most commercially important ones. Formaldehyde, the simplest aldehyde, is one of the most widely used biological preservatives; you have probably seen a jar of it containing a random animal organ. It kills most bacteria and fungi, but its true value lies in its size. Many preservatives are chosen because they react with the covalent bonds in chemical tissue, hardening it. Because formaldehyde is smaller compared to other common preservatives, it can penetrate the outer layer of organic tissue and react away more easily. The speed of this reaction is especially important when the tissue is in an environment in which formaldehyde itself is produced by a chemical reaction that is not especially product-favored. For example, when formaldehyde is in an equilibrium reaction with methanediol or amines, the equilibrium constant is not favored towards formaldehyde. However, as the aldehyde reacts away with organic matter, the equilibrium shifts, creating more formaldehyde to partially replenish the used product . This is consistent with Le Chatelier's Principle. Acetone, the simplest ketone, is very miscible with water and is common in the body. As mentioned above, large ketones are produced naturally in the liver through the decomposition of fatty acids. These molecules will lose most of their carbonyl groups in a process called decarboxylation, where the carbonyl breaks away to make carbon dioxide. Acetone is the end product and excreted through breath and urine. When the metabolism is unable to quickly flush out ketones, the body enters a state of ketosis. This often happens when not enough carbohydrates are available in the bloodstream. Does this deprivation of carbs sound familar? When ketosis is induced, the body relies more of stores of fat for energy. Forced exclusion of carbohydrates from one's diet in order to lose - yes, this is the Atkins diet, chemically dependant on ketosis. While this process is normal, carbohydrate exclusion can lead to prolonged states of ketosis that induces dehydration and stresses the liver. In extreme diabetic or alcoholic cases, the uncontrolled production of ketones, especially acetoacetic acid and beta-hydroxybutyric acid, leads to ketoacidosis. In this state, the pH of a person's bloodstream drops below 7.2, and in some cases may be fatal . Ketoacidosis can be identified by a person's breath; the high levels of acetone exhaled can be detected as having a slightly fruity scent, though some say the smell is more like nail polish remover. | 7,511 | 2,441 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Book3A_Bioinorganic_Chemistry_(Bertini_et_al.)/05%3A_Dioxygen_Reactions/5.05%3A_Dioxygen_Toxicity |
Before we consider the enzymatically controlled reactions of dioxygen in living systems, it is instructive to consider the uncontrolled and deleterious reactions that must also occur in aerobic organisms. Life originally appeared on Earth at a time when the atmosphere contained only low concentrations of dioxygen, and was reducing rather than oxidizing, as it is today. With the appearance of photosynthetic organisms approximately 2.5 billion years ago, however, the conversion to an aerobic, oxidizing atmosphere exposed the existing anaerobic organisms to a gradually increasing level of oxidative stress. Modern-day anaerobic bacteria, the descendants of the original primitive anaerobic organisms, evolved in ways that enabled them to avoid contact with normal atmospheric concentrations of dioxygen. Modern-day aerobic organisms, by contrast, evolved by developing aerobic metabolism to harness the oxidizing power of dioxygen and thus to obtain usable metabolic energy. This remarkably successful adaptation enabled life to survive and flourish as the atmosphere became aerobic, and also allowed larger, multicellular organisms to evolve. An important aspect of dioxygen chemistry that enabled the development of aerobic metabolism is the relatively slow rate of dioxygen reactions in the absence of catalysts. Thus, enzymes could be used to direct and control the oxidation of substrates either for energy generation or for biosynthesis. Nevertheless, the balance achieved between constructive and destructive oxidation is a delicate one, maintained in aerobic organisms by several means, e.g.: compartmentalization of oxidative reactions in mitochondria, peroxisomes, and chloroplasts; scavenging or detoxification of toxic byproducts of dioxygen reactions; repair of some types of oxidatively damaged species; and degradation and replacement of other species. The classification "anaerobic" actually includes organisms with varying degrees of tolerance for dioxygen: strict anaerobes, for which even small concentrations of O are toxic; moderate anaerobes, which can tolerate low levels of dioxygen; and microaerophiles, which require low concentrations of O for growth, but cannot tolerate normal atmospheric concentrations, i.e., 21 percent O , 1 atm pressure. Anaerobic organisms thrive in places protected from the atmosphere, for example, in rotting organic material, decaying teeth, the colon, and gangrenous wounds. Dioxygen appears to be toxic to anaerobic organisms largely because it depletes the reducing equivalents in the cell that are needed for normal biosynthetic reactions. Aerobic organisms can, of course, live in environments in which they are exposed to normal atmospheric concentrations of O . Nevertheless, there is much evidence that O is toxic to these organisms as well. For example, plants grown in varying concentrations of O have been observed to grow faster in lower than normal concentrations of O . grown under 5 atm of O ceased to grow unless the growth medium was supplemented with branched-chain amino acids or precursors. High concentrations of O damaged the enzyme dihydroxy acid dehydratase, an important component in the biosynthetic pathway for those amino acids. In mammals, elevated levels of O are clearly toxic, leading first to coughing and soreness of the throat, and then to convulsions when the level of 5 atm of 100 percent O is reached. Eventually, elevated concentrations of O lead to pulmonary edema and irreversible lung damage, with obvious damage to other tissues as well. The effects of high concentrations of O on humans is of some medical interest, since dioxygen is used therapeutically for patients experiencing difficulty breathing, or for those suffering from infection by anaerobic organisms. The major biochemical targets of O toxicity appear to be lipids, DNA, and proteins. The chemical reactions accounting for the damage to each type of target are probably different, not only because of the different reactivities of these three classes of molecules, but also because of the different environment for each one inside the cell. Lipids, for example, are essential components of membranes and are extremely hydrophobic. The oxidative damage that is observed is due to free-radical autoxidation (see Reactions 5.16 to 5.21), and the products observed are lipid hydroperoxides (see Reaction 5.23). The introduction of the hydroperoxide group into the interior of the lipid bilayer apparently causes that structure to be disrupted, as the configuration of the lipid rearranges in order to bring that polar group out of the hydrophobic membrane interior and up to the membrane-water interface. DNA, by contrast, is in the interior of the cell, and its exposed portions are surrounded by an aqueous medium. It is particularly vulnerable to oxidative attack at the base or at the sugar, and multiple products are formed when samples are exposed to oxidants . Since oxidation of DNA may lead to mutations, this type of damage is potentially very serious. Proteins also suffer oxidative damage, with amino-acid side chains, particularly the sulfur-containing residues cysteine and methionine, appearing to be the most vulnerable sites. The biological defense systems protecting against oxidative damage and its consequences are summarized below. Some examples of small-molecule antioxidants are \(\alpha\)-tocopherol (vitamin E; 5.24), which is found dissolved in cell membranes and protects them against lipid peroxidation, and ascorbate (vitamin C; 5.25) and glutathione (5.26), which are found in the cytosol of many cells. Several others are known as well. \(\tag{5.24}\) \(\tag{5.25}\) \(\tag{5.26}\) The enzymatic antioxidants are (a) catalase and the various peroxidases, whose presence lowers the concentration of hydrogen peroxide, thereby preventing it from entering into potentially damaging reactions with various cell components (see Section VI and Reactions 5.82 and 5.83), and (b) the superoxide dismutases, whose presence provides protection against dioxygen toxicity that is believed to be mediated by the superoxide anion, O (see Section VII and Reaction 5.95). Some of the enzymatic and nonenzymatic antioxidants in the cell are illustrated in Figure 5.1. Redox-active metal ions are present in the cell in their free, uncomplexed state only in extremely low concentrations. They are instead sequestered by metal-ion storage and transport proteins, such as ferritin and transferrin for iron (see Chapter 1) and ceruloplasmin for copper. This arrangement prevents such metal ions from catalyzing deleterious oxidative reactions, but makes them available for incorporation into metalloenzymes as they are needed. In vitro experiments have shown quite clearly that redox-active metal ions such as Fe or Cu are extremely good catalysts for oxidation of sulfhydryl groups by O (Reaction 5.27). \[4RSH + O_{2} \xrightarrow{M^{n+}} 2RSSR + 2H_{2}O \tag{5.27}\] In addition, in the reducing environment of the cell, redox-active metal ions catalyze a very efficient one-electron reduction of hydrogen peroxide to produce hydroxyl radical, one of the most potent and reactive oxidants known (Reactions 5.28 to 5.30). \[M^{n+} + Red^{-} \rightarrow M^{(n-1)+} + Red \tag{5.28}\] \[M^{(n-1)+} + H_{2}O_{2} \rightarrow M^{n+} + OH^{-} + HO \cdotp \tag{5.29}\] \[Red^{-} + H_{2}O_{2} \rightarrow Red + OH^{-} + HO \cdotp \tag{5.30}\] \[(Red^{-} = reducing\; agent)\] Binding those metal ions in a metalloprotein usually prevents them from entering into these types of reactions. For example, transferrin, the iron-transport enzyme in serum, is normally only 30 percent saturated with iron. Under conditions of increasing iron overload, the empty iron-binding sites on transferrin are observed to fill, and symptoms of iron poisoning are not observed until after transferrin has been totally saturated with iron. Ceruloplasmin and metallothionein may playa similar role in preventing copper toxicity. It is very likely that both iron and copper toxicity are largely due to catalysis of oxidation reactions by those metal ions. Repair of oxidative damage must go on constantly, even under normal conditions of aerobic metabolism. For lipids, repair of peroxidized fatty-acid chains is catalyzed by phospholipase A , which recognizes the structural changes at the lipid-water interface caused by the fatty-acid hydroperoxide, and catalyzes removal of the fatty acid at that site. The repair is then completed by enzymatic reacylation. Although some oxidatively damaged proteins are repaired, more commonly such proteins are recognized, degraded at accelerated rates, and then replaced. For DNA, several multi-enzyme systems exist whose function is to repair oxidatively damaged DNA. For example, one such system catalyzes recognition and removal of damaged bases, removal of the damaged part of the strand, synthesis of new DNA to fill in the gaps, and religation to restore the DNA to its original, undamaged state. Mutant organisms that lack these repair enzymes are found to be hypersensitive to O , H O , or other oxidants. One particularly interesting aspect of oxidant stress is that most aerobic organisms can survive in the presence of normally lethal levels of oxidants if they have first been exposed to lower, nontoxic levels of oxidants. This phenomenon has been observed in animals, plants, yeast, and bacteria, and suggests that low levels of oxidants cause antioxidant systems to be induced . In certain bacteria, the mechanism of this induction is at least partially understood. A DNA-binding regulatory protein named OxyR that exists in two redox states has been identified in these systems. Increased oxidant stress presumably increases concentration of the oxidized form, which then acts to turn on the transcription of the genes for some of the antioxidant enzymes. A related phenomenon may occur when bacteria and yeast switch from anaerobic to aerobic metabolism. When dioxygen is absent, these microorganisms live by fermentation, and do not waste energy by synthesizing the enzymes and other proteins needed for aerobic metabolism. However, when they are exposed to dioxygen, the synthesis of the respiratory apparatus is turned on. The details of this induction are not known completely, but some steps at least depend on the presence of heme, the prosthetic group of hemoglobin and other heme proteins, whose synthesis requires the presence of dioxygen. What has been left out of the preceding discussion is the identification of the species responsible for oxidative damage, i.e., the agents that directly attack the various vulnerable targets in the cell. They were left out because the details of the chemistry responsible for dioxygen toxicity are largely unknown. In 1954, Rebeca Gerschman formulated the "free-radical theory of oxygen toxicity" after noting that tissues subjected to ionizing radiation resemble those exposed to elevated levels of dioxygen. Fourteen years later, Irwin Fridovich proposed that the free radical responsible for dioxygen toxicity was superoxide, O , based on his identification of the first of the superoxide dismutase enzymes. Today it is still not known if superoxide is the principal agent of dioxygen toxicity, and, if so, what the chemistry responsible for that toxicity is. There is no question that superoxide is formed during the normal course of aerobic metabolism, although it is difficult to obtain estimates of the amount under varying conditions, because, even in the absence of a catalyst, superoxide disproportionates quite rapidly to dioxygen and hydrogen peroxide (Reaction 5.4) and therefore never accumulates to any great extent in the cell under normal conditions of pH. One major problem in this area is that a satisfactory chemical explanation for the purported toxicity of superoxide has never been found, despite much indirect evidence from experiments that the presence of superoxide can lead to undesirable oxidation of various cell components and that such oxidation can be inhibited by superoxide dismutase. The mechanism most commonly proposed is production of hydroxyl radicals via Reactions (5.28) to (5.30) with Red = O , which is referred to as the "Metal-Catalyzed Haber-Weiss Reaction". The role of superoxide in this mechanism is to reduce oxidized metal ions, such as Cu or Fe , present in the cell in trace amounts, to a lower oxidation state. Hydroxyl radical is an extremely powerful and indiscriminate oxidant. It can abstract hydrogen atoms from organic substrates, and oxidize most reducing agents very rapidly. It is also a very effective initiator of free-radical autoxidation reactions (see Section II.C above). Therefore, reactions that produce hydroxyl radical in a living cell will probably be very deleterious. The problem with this explanation for superoxide toxicity is that the only role played by superoxide here is that of a reducing agent of trace metal ions. The interior of a cell is a highly reducing environment, however, and other reducing agents naturally present in the cell such as, for example, ascorbate anion can also act as Red in Reaction (5.28), and the resulting oxidation reactions due to hydroxyl radical are therefore no longer inhibitable by SOD. Other possible explanations for superoxide toxicity exist, of course, but none has ever been demonstrated experimentally. Superoxide might bind to a specific enzyme and inhibit it, much as cytochrome oxidase is inhibited by cyanide or hemoglobin by carbon monoxide. Certain enzymes may be extraordinarily sensitive to direct oxidation by superoxide, as has been suggested for the enzyme aconitase, an iron-sulfur enzyme that contains an exposed iron atom. Another possibility is that the protonated and therefore neutral form of superoxide, HO , dissolves in membranes and acts as an initiator of lipid peroxidation. It has also been suggested that superoxide may react with nitric oxide, NO, in the cell producing peroxynitrite, a very potent oxidant. One particularly appealing mechanism for superoxide toxicity that has gained favor in recent years is the "Site-Specific Haber-Weiss Mechanism." The idea here is that traces of redox-active metal ions such as copper and iron are bound to macromolecules under normal conditions in the cell. Most reducing agents in the cell are too bulky to come into close proximity to these sequestered metal ions. Superoxide, however, in addition to being an excellent reducing agent, is very small, and could penetrate to these metal ions and reduce them. The reduced metal ions could then react with hydrogen peroxide, generating hydroxyl radical, which would immediately attack at a site near the location of the bound metal ion. This mechanism is very similar to that of the metal complexes that cause DNA cleavage; by reacting with hydrogen peroxide while bound to DNA, they generate powerful oxidants that react with DNA with high efficiency because of their proximity to it (see Chapter 8). Although we are unsure what specific chemical reactions superoxide might undergo inside of the cell, there nevertheless does exist strong evidence that the superoxide dismutases play an important role in protection against dioxygen-induced damage. Mutant strains of bacteria and yeast that lack superoxide dismutases are killed by elevated concentrations of dioxygen that have no effect on the wild-type cells. This extreme sensitivity to dioxygen is alleviated when the gene coding for a superoxide dismutase is reinserted into the cell, even if the new SOD is of another type and from a different organism. In summary, we know a great deal about the sites that are vulnerable to oxidative damage in biological systems, about the agents that protect against such damage, and about the mechanisms that repair such damage. Metal ions are involved in all this chemistry, both as catalysts of deleterious oxidative reactions and as cofactors in the enzymes that protect against and repair such damage. What we still do not know at this time, however, is how dioxygen initiates the sequence of chemical reactions that produce the agents that attack the vulnerable biological targets | 16,226 | 2,442 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Photoacoustic_Spectroscopy |
\[1 = \alpha +T+R \label{1}\] e is \(\alpha\) t Light hits a sample of aluminum. The data collected shows that the sample had a transmittance 0.12 and reflectance value of 0.8196. Assuming light is conserved, what is the absorbance of the material? Using Equation \ref{1}, light must be conserved and so \[1-0.12-0.8196 = α \nonumber\] \[α = 0.0604 \nonumber\] Does vibrational or electronic excitation of electrons produce more heat? Why is that? Vibrational excitation of electrons produce better heat as all the energy from the absorbed light transfers into heat. This is because the vibrational lifetimes of the atoms is long enough so that the transfer of energy to heat is not interrupted by other processes. Why does PAS require pulses of light instead of a continuous steady source of light to hit the sample? To obtain a signal, it must come in the form of a wave. Having a constant source of light will prevent waves from forming and the spectrum will only show up as a straight line. The point of the pulsing light is to have the material absorb the energy, convert it to heat, send out the acoustic waves, and let it rest before allowing it to absorb another amount of energy. | 1,206 | 2,443 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkyl_Halides/Reactivity_of_Alkyl_Halides/Alkyl_Halide_Reactions/Reactions_of_Dihalides |
If two halogen atoms are present in a given compound, reactions with reducing metals may take different paths depending on how close the carbon-halogen bonds are to each other. If they are separated by four or more carbons, as in the first example below, a bis-organometallic compound may be formed. However, if the halogens are bonded to adjacent ( ) carbons, an elimination takes place with formation of a double bond. Since vicinal-dihalides are usually made by adding a halogen to a double bond, this reaction is mainly useful for relating structures to each other. The last example, in which two halogens are bonded to the same carbon, referred to as (twinned), gives an unusual reagent which may either react as a carbon nucleophile or, by elimination, as a . Such reagents are often termed . The solution structure of the Simmons-Smith reagent is less well understood than that of the Grignard reagent, but the formula given here is as useful as any that have been proposed. Other alpha-halogenated organometallic reagents, such as ClCH Li, BrCH Li, Cl CHLi and Cl CLi, have been prepared, but they are substantially less stable and must be maintained at very low temperature (ca. -100 º C) to avoid loss of LiX. The stability and usefulness of the Simmons-Smith reagent may be attributed in part to the higher covalency of the carbon-zinc bond together with solvation and internal coordination of the zinc. Hydrolysis (reaction with water) gives methyl iodide, confirming the basicity of the carbon; and reaction with alkenes gives cyclopropane derivatives, demonstrating the carbene-like nature of the reagent. The latter transformation is illustrated by the equation below. Elimination reactions of the stereoisomeric 1,2-dibromo-1,2-diphenylethanes provide a nice summary of the principles discussed above. The following illustration shows first the meso-diastereomer and below it one enantiomer of the racemic-diastereomer. In each case two conformers are drawn within parentheses, and the anti-relationship of selected vicinal groups in each is colored green. The reaction proceeding to the left is a dehydrohalogenation induced by treatment with KOH in alcohol. Since this is a elimination, each diastereomer gives a different stereoisomeric product. The reaction to the right is a dehalogenation (the reverse of halogen addition to an alkene), caused by treatment with iodide anion. Zinc dust effects the same reaction, but with a lower degree of stereospecificity. The mechanism of the iodide anion reaction is shown by red arrows in the top example. A similar mechanism explains the comparable elimination of the racemic isomer. In both reactions an anti-transition state is observed. The two stereoisomers of 1-bromo-1,2-diphenylethene (shown on the left of the diagram) undergo a second dehydrobromination reaction on more vigorous treatment with base, as shown in the following equation. This elimination generates the same alkyne (carbon-carbon triple bond) from each of the bromo-alkenes. Interestingly, the (Z)-isomer (lower structure) eliminates more rapidly than the (E)-isomer (upper structure), again showing a preference for anti-orientation of eliminating groups. C H CH=C C H + KOH → C H C≡CC H + K + H O The last reaction shown above suggests that alkynes might be prepared from alkenes by a two stage procedure, consisting first of chlorine or bromine addition to the double bond, and secondly a base induced double dehydrohalogenation. For example, reaction of 1-butene with bromine would give 1,2-dibromobutane, and on treatment with base this vicinal dibromide would be expected to yield 1-bromo-1-butene followed by a second elimination to 1-butyne. CH CH CH=CH + → CH CH CH –CH + base → CH CH CH=CH + base → CH CH C≡CH In practice this strategy works, but it requires care in the selection of the base and solvent. If KOH in alcohol is used, the first elimination is much faster than the second, so the bromoalkene may be isolated if desired. Under more extreme conditions the second elimination takes place, but isomerization of the triple bond also occurs, with the more stable isomer (2-butyne) being formed along with 1-butyne, even becoming the chief product. To facilitate the second elimination and avoid isomerization the very strong base sodium amide, NaNH , may be used. Since ammonia is a much weaker acid than water (by a factor of 10 ), its conjugate base is proportionally stronger than hydroxide anion (the conjugate base of water), and the elimination of HBr from the bromoalkene may be conducted at relatively low temperature. Also, the acidity of the sp-hybridized C-H bond of the terminal alkyne traps the initially formed 1-butyne in the form of its sodium salt. CH CH C≡CH + NaNH → CH CH C≡C: Na + NH An additional complication of this procedure is that the 1-bromo-1-butene product of the first elimination (see previous equations) is accompanied by its 2-bromo-1-butene isomer, CH CH C =CH , and elimination of HBr from this bromoalkene not only gives 1-butyne (base attack at C-1) but also 1,2-butadiene, CH CH=C=CH , by base attack at C-3. Dienes of this kind, in which the central carbon is sp-hybridized, are called and are said to have . They are usually less stable than their alkyne isomers. | 5,290 | 2,447 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/07%3A_Chemical_Bonding_and_Molecular_Geometry/7.E%3A_Chemical_Bonding_and_Molecular_Geometry_(Exercises) |
Does a cation gain protons to form a positive charge or does it lose electrons? The protons in the nucleus do not change during normal chemical reactions. Only the outer electrons move. Positive charges form when electrons are lost. Iron(III) sulfate [Fe (SO ) ] is composed of Fe and \(\ce{SO4^2-}\) ions. Explain why a sample of iron(III) sulfate is uncharged. Which of the following atoms would be expected to form negative ions in binary ionic compounds and which would be expected to form positive ions: P, I, Mg, Cl, In, Cs, O, Pb, Co? P, I, Cl, and O would form anions because they are nonmetals. Mg, In, Cs, Pb, and Co would form cations because they are metals. Which of the following atoms would be expected to form negative ions in binary ionic compounds and which would be expected to form positive ions: Br, Ca, Na, N, F, Al, Sn, S, Cd? Predict the charge on the monatomic ions formed from the following atoms in binary ionic compounds: P ; Mg ; Al ; O ; Cl ; Cs Predict the charge on the monatomic ions formed from the following atoms in binary ionic compounds: Write the electron configuration for each of the following ions: [Ar]4 3 4 ; [Kr]4 5 5 1 [Kr]4 ; [He]2 2 ; [Ar]3 ; 1 (h) [He]2 2 (i) [Kr]4 5 (j) [Ar]3 (k) [Ar]3 , (l) [Ar]3 4 Write the electron configuration for the monatomic ions formed from the following elements (which form the greatest concentration of monatomic ions in seawater): Write out the full electron configuration for each of the following atoms and for the monatomic ion found in binary ionic compounds containing the element: 1 2 2 3 3 ; Al : 1 2 2 ; 1 2 2 3 3 3 4 4 ; 1 2 2 3 3 3 4 4 ; 1 2 2 3 3 3 4 4 5 ; Sr : 1 2 2 3 3 3 4 4 ; 1 2 ; Li : 1 ; 1 2 2 3 3 3 4 4 ; 1 2 2 3 3 3 4 4 ; 1 2 2 3 3 ; 1 2 2 3 3 From the labels of several commercial products, prepare a list of six ionic compounds in the products. For each compound, write the formula. (You may need to look up some formulas in a suitable reference.) Why is it incorrect to speak of a molecule of solid NaCl? NaCl consists of discrete ions arranged in a crystal lattice, not covalently bonded molecules. What information can you use to predict whether a bond between two atoms is covalent or ionic? Predict which of the following compounds are ionic and which are covalent, based on the location of their constituent atoms in the periodic table: ionic: (b), (d), (e), (g), and (i); covalent: (a), (c), (f), (h), (j), and (k) Explain the difference between a nonpolar covalent bond, a polar covalent bond, and an ionic bond. From its position in the periodic table, determine which atom in each pair is more electronegative: Cl; O; O; S; N; P; N From its position in the periodic table, determine which atom in each pair is more electronegative: From their positions in the periodic table, arrange the atoms in each of the following series in order of increasing electronegativity: H, C, N, O, F; H, I, Br, Cl, F; H, P, S, O, F; Na, Al, H, P, O; Ba, H, As, N, O From their positions in the periodic table, arrange the atoms in each of the following series in order of increasing electronegativity: Which atoms can bond to sulfur so as to produce a positive partial charge on the sulfur atom? N, O, F, and Cl Which is the most polar bond? Identify the more polar bond in each of the following pairs of bonds: HF; CO; OH; PCl; NH; PO; CN Which of the following molecules or ions contain polar bonds? Write the Lewis symbols for each of the following ions: eight electrons: eight electrons: no electrons Be ; eight electrons: no electrons Ga ; no electrons Li ; eight electrons: Many monatomic ions are found in seawater, including the ions formed from the following list of elements. Write the Lewis symbols for the monatomic ions formed from the following elements: Write the Lewis symbols of the ions in each of the following ionic compounds and the Lewis symbols of the atom from which they are formed: (a) ; (b) ; (c) ; (d) ; (e) ; (f) In the Lewis structures listed here, M and X represent various elements in the third period of the periodic table. Write the formula of each compound using the chemical symbols of each element: (a) (b) (c) (d) Write the Lewis structure for the diatomic molecule P , an unstable form of phosphorus found in high-temperature phosphorus vapor. Write Lewis structures for the following: Write Lewis structures for the following: (a) In this case, the Lewis structure is inadequate to depict the fact that experimental studies have shown two unpaired electrons in each oxygen molecule (b) ; (c) ; (d) ; (e) ; (f) ; (g) ; (h) ; (i) ; (j) ; (k) Write Lewis structures for the following: Write Lewis structures for the following: SeF : ; XeF : ; \(\ce{SeCl3+}\): ; Cl BBCl : Write Lewis structures for: Correct the following statement: “The bonds in solid PbCl are ionic; the bond in a HCl molecule is covalent. Thus, all of the valence electrons in PbCl are located on the Cl ions, and all of the valence electrons in a HCl molecule are shared between the H and Cl atoms.” Two valence electrons per Pb atom are transferred to Cl atoms; the resulting Pb ion has a 6 valence shell configuration. Two of the valence electrons in the HCl molecule are shared, and the other six are located on the Cl atom as lone pairs of electrons. Write Lewis structures for the following molecules or ions: Methanol, H COH, is used as the fuel in some race cars. Ethanol, C H OH, is used extensively as motor fuel in Brazil. Both methanol and ethanol produce CO and H O when they burn. Write the chemical equations for these combustion reactions using Lewis structures instead of chemical formulas. Many planets in our solar system contain organic chemicals including methane (CH ) and traces of ethylene (C H ), ethane (C H ), propyne (H CCCH), and diacetylene (HCCCCH). Write the Lewis structures for each of these molecules. Carbon tetrachloride was formerly used in fire extinguishers for electrical fires. It is no longer used for this purpose because of the formation of the toxic gas phosgene, Cl CO. Write the Lewis structures for carbon tetrachloride and phosgene. Identify the atoms that correspond to each of the following electron configurations. Then, write the Lewis symbol for the common ion formed from each atom: The arrangement of atoms in several biologically important molecules is given here. Complete the Lewis structures of these molecules by adding multiple bonds and lone pairs. Do not add any more atoms. the amino acid serine: urea: pyruvic acid: uracil: carbonic acid: (a) ; (b) ; (c) ; (d) ; (e) A compound with a molar mass of about 28 g/mol contains 85.7% carbon and 14.3% hydrogen by mass. Write the Lewis structure for a molecule of the compound. A compound with a molar mass of about 42 g/mol contains 85.7% carbon and 14.3% hydrogen by mass. Write the Lewis structure for a molecule of the compound. Two arrangements of atoms are possible for a compound with a molar mass of about 45 g/mol that contains 52.2% C, 13.1% H, and 34.7% O by mass. Write the Lewis structures for the two molecules. How are single, double, and triple bonds similar? How do they differ? Each bond includes a sharing of electrons between atoms. Two electrons are shared in a single bond; four electrons are shared in a double bond; and six electrons are shared in a triple bond. Write resonance forms that describe the distribution of electrons in each of these molecules or ions. the formate ion: Write resonance forms that describe the distribution of electrons in each of these molecules or ions. the allyl ion: (a) ; (b) ; (c) ; (d) ; (e) Write the resonance forms of ozone, O , the component of the upper atmosphere that protects the Earth from ultraviolet radiation. Sodium nitrite, which has been used to preserve bacon and other meats, is an ionic compound. Write the resonance forms of the nitrite ion, \(\ce{NO2-}\). In terms of the bonds present, explain why acetic acid, CH CO H, contains two distinct types of carbon-oxygen bonds, whereas the acetate ion, formed by loss of a hydrogen ion from acetic acid, only contains one type of carbon-oxygen bond. The skeleton structures of these species are shown: Write the Lewis structures for the following, and include resonance structures where appropriate. Indicate which has the strongest carbon-oxygen bond. (a) (b) CO has the strongest carbon-oxygen bond because there is a triple bond joining C and O. CO has double bonds. Toothpastes containing sodium hydrogen carbonate (sodium bicarbonate) and hydrogen peroxide are widely used. Write Lewis structures for the hydrogen carbonate ion and hydrogen peroxide molecule, with resonance forms where appropriate. Determine the formal charge of each element in the following: H: 0, Cl: 0; C: 0, F: 0; P: 0, Cl 0; P: 0, F: 0 Determine the formal charge of each element in the following: Calculate the formal charge of chlorine in the molecules Cl , BeCl , and ClF . Cl in Cl : 0; Cl in BeCl : 0; Cl in ClF : 0 Calculate the formal charge of each element in the following compounds and ions: Draw all possible resonance structures for each of these compounds. Determine the formal charge on each atom in each of the resonance structures: ; (b) ; (c) ; (d) Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in nitrosyl chloride: ClNO or ClON? Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in hypochlorous acid: HOCl or OClH? HOCl Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in sulfur dioxide: OSO or SOO? Draw the structure of hydroxylamine, H NO, and assign formal charges; look up the structure. Is the actual structure consistent with the formal charges? The structure that gives zero formal charges is consistent with the actual structure: Iodine forms a series of fluorides (listed here). Write Lewis structures for each of the four compounds and determine the formal charge of the iodine atom in each molecule: Write the Lewis structure and chemical formula of the compound with a molar mass of about 70 g/mol that contains 19.7% nitrogen and 80.3% fluorine by mass, and determine the formal charge of the atoms in this compound. NF ; Which of the following structures would we expect for nitrous acid? Determine the formal charges: Sulfuric acid is the industrial chemical produced in greatest quantity worldwide. About 90 billion pounds are produced each year in the United States alone. Write the Lewis structure for sulfuric acid, H SO , which has two oxygen atoms and two OH groups bonded to the sulfur. Which bond in each of the following pairs of bonds is the strongest? Using the bond energies in , determine the approximate enthalpy change for each of the following reactions: Using the bond energies in , determine the approximate enthalpy change for each of the following reactions: When a molecule can form two different structures, the structure with the stronger bonds is usually the more stable form. Use bond energies to predict the correct structure of the hydroxylamine molecule: The greater bond energy is in the figure on the left. It is the more stable form. How does the bond energy of HCldiffer from the standard enthalpy of formation of HCl( )? Using the standard enthalpy of formation data in , show how the standard enthalpy of formation of HCl( ) can be used to determine the bond energy. \(\ce{HCl}(g)⟶\dfrac{1}{2}\ce{H2}(g)+\dfrac{1}{2}\ce{Cl2}(g)\hspace{20px}ΔH^\circ_1=−ΔH^\circ_{\ce f[\ce{HCl}(g)]}\\
\dfrac{1}{2}\ce{H2}(g)⟶\ce{H}(g)\hspace{105px}ΔH^\circ_2=ΔH^\circ_{\ce f[\ce H(g)]}\\
\underline{\dfrac{1}{2}\ce{Cl2}(g)⟶\ce{Cl}(g)\hspace{99px}ΔH^\circ_3=ΔH^\circ_{\ce f[\ce{Cl}(g)]}}\\
\ce{HCl}(g)⟶\ce{H}(g)+\ce{Cl}(g)\hspace{58px}ΔH^\circ_{298}=ΔH^\circ_1+ΔH^\circ_2+ΔH^\circ_3\) \(\begin{align}
D_\ce{HCl}=ΔH^\circ_{298}&=ΔH^\circ_{\ce f[\ce{HCl}(g)]}+ΔH^\circ_{\ce f[\ce H(g)]}+ΔH^\circ_{\ce f[\ce{Cl}(g)]}\\
&=\mathrm{−(−92.307\:kJ)+217.97\:kJ+121.3\:kJ}\\
&=\mathrm{431.6\:kJ}
\end{align}\) Using the standard enthalpy of formation data in , calculate the bond energy of the carbon-sulfur double bond in CS . Using the standard enthalpy of formation data in , determine which bond is stronger: the S–F bond in SF ( ) or in SF ( )? The S–F bond in SF is stronger. Using the standard enthalpy of formation data in , determine which bond is stronger: the P–Cl bond in PCl ( ) or in PCl ( )? Complete the following Lewis structure by adding bonds (not atoms), and then indicate the longest bond: The C–C single bonds are longest. Use the bond energy to calculate an approximate value of Δ for the following reaction. Which is the more stable form of FNO ? Use principles of atomic structure to answer each of the following: The first ionization energy of Mg is 738 kJ/mol and that of Al is 578 kJ/mol. Account for this difference. When two electrons are removed from the valence shell, the Ca radius loses the outermost energy level and reverts to the lower = 3 level, which is much smaller in radius. The +2 charge on calcium pulls the oxygen much closer compared with K, thereby increasing the lattice energy relative to a less charged ion. (c) Removal of the 4 electron in Ca requires more energy than removal of the 4 electron in K because of the stronger attraction of the nucleus and the extra energy required to break the pairing of the electrons. The second ionization energy for K requires that an electron be removed from a lower energy level, where the attraction is much stronger from the nucleus for the electron. In addition, energy is required to unpair two electrons in a full orbital. For Ca, the second ionization potential requires removing only a lone electron in the exposed outer energy level. In Al, the removed electron is relatively unprotected and unpaired in a orbital. The higher energy for Mg mainly reflects the unpairing of the 2 electron. The lattice energy of LiF is 1023 kJ/mol, and the Li–F distance is 200.8 pm. NaF crystallizes in the same structure as LiF but with a Na–F distance of 231 pm. Which of the following values most closely approximates the lattice energy of NaF: 510, 890, 1023, 1175, or 4090 kJ/mol? Explain your choice. For which of the following substances is the least energy required to convert one mole of the solid into separate ions? (d) The reaction of a metal, M, with a halogen, X , proceeds by an exothermic reaction as indicated by this equation: \(\ce{M}(s)+\ce{X2}(g)⟶\ce{MX2}(s)\). For each of the following, indicate which option will make the reaction more exothermic. Explain your answers. The lattice energy of LiF is 1023 kJ/mol, and the Li–F distance is 201 pm. MgO crystallizes in the same structure as LiF but with a Mg–O distance of 205 pm. Which of the following values most closely approximates the lattice energy of MgO: 256 kJ/mol, 512 kJ/mol, 1023 kJ/mol, 2046 kJ/mol, or 4008 kJ/mol? Explain your choice. 4008 kJ/mol; both ions in MgO have twice the charge of the ions in LiF; the bond length is very similar and both have the same structure; a quadrupling of the energy is expected based on the equation for lattice energy Which compound in each of the following pairs has the larger lattice energy? Note: Mg and Li have similar radii; O and F have similar radii. Explain your choices. Which compound in each of the following pairs has the larger lattice energy? Note: Ba and K have similar radii; S and Cl have similar radii. Explain your choices. Na O; Na has a smaller radius than K ; BaS; Ba has a larger charge than K; (c) BaS; Ba and S have larger charges; BaS; S has a larger charge Which of the following compounds requires the most energy to convert one mole of the solid into separate ions? Which of the following compounds requires the most energy to convert one mole of the solid into separate ions? (e) The lattice energy of KF is 794 kJ/mol, and the interionic distance is 269 pm. The Na–F distance in NaF, which has the same structure as KF, is 231 pm. Which of the following values is the closest approximation of the lattice energy of NaF: 682 kJ/mol, 794 kJ/mol, 924 kJ/mol, 1588 kJ/mol, or 3175 kJ/mol? Explain your answer. Explain why the HOH molecule is bent, whereas the HBeH molecule is linear. The placement of the two sets of unpaired electrons in water forces the bonds to assume a tetrahedral arrangement, and the resulting HOH molecule is bent. The HBeH molecule (in which Be has only two electrons to bond with the two electrons from the hydrogens) must have the electron pairs as far from one another as possible and is therefore linear. What feature of a Lewis structure can be used to tell if a molecule’s (or ion’s) electron-pair geometry and molecular structure will be identical? Explain the difference between electron-pair geometry and molecular structure. Space must be provided for each pair of electrons whether they are in a bond or are present as lone pairs. Electron-pair geometry considers the placement of all electrons. Molecular structure considers only the bonding-pair geometry. Why is the H–N–H angle in NH smaller than the H–C–H bond angle in CH ? Why is the H–N–H angle in \(\ce{NH4+}\) identical to the H–C–H bond angle in CH ? Explain how a molecule that contains polar bonds can be nonpolar. As long as the polar bonds are compensated (for example. two identical atoms are found directly across the central atom from one another), the molecule can be nonpolar. As a general rule, MX molecules (where M represents a central atom and X represents terminal atoms; n = 2 – 5) are polar if there is one or more lone pairs of electrons on M. NH (M = N, X = H, n = 3) is an example. There are two molecular structures with lone pairs that are exceptions to this rule. What are they? Predict the electron pair geometry and the molecular structure of each of the following molecules or ions: Identify the electron pair geometry and the molecular structure of each of the following molecules or ions: What are the electron-pair geometry and the molecular structure of each of the following molecules or ions? electron-pair geometry: octahedral, molecular structure: square pyramidal; electron-pair geometry: tetrahedral, molecular structure: bent; (c) electron-pair geometry: octahedral, molecular structure: square planar; electron-pair geometry: tetrahedral, molecular structure: trigonal pyramidal; electron-pair geometry: trigonal bypyramidal, molecular structure: seesaw; electron-pair geometry: tetrahedral, molecular structure: bent (109°) Predict the electron pair geometry and the molecular structure of each of the following ions: Identify the electron pair geometry and the molecular structure of each of the following molecules: electron-pair geometry: trigonal planar, molecular structure: bent (120°); electron-pair geometry: linear, molecular structure: linear; (c) electron-pair geometry: trigonal planar, molecular structure: trigonal planar; electron-pair geometry: tetrahedral, molecular structure: trigonal pyramidal; electron-pair geometry: tetrahedral, molecular structure: tetrahedral; electron-pair geometry: trigonal bipyramidal, molecular structure: seesaw; (g) electron-pair geometry: tetrahedral, molecular structure: trigonal pyramidal Predict the electron pair geometry and the molecular structure of each of the following: Which of the following molecules and ions contain polar bonds? Which of these molecules and ions have dipole moments? All of these molecules and ions contain polar bonds. Only ClF , \(\ce{ClO2-}\), PCl , SeF , and \(\ce{PH2-}\) have dipole moments. Which of the molecules and ions in contain polar bonds? Which of these molecules and ions have dipole moments? Which of the following molecules have dipole moments? SeS , CCl F , PCl , and ClNO all have dipole moments. Identify the molecules with a dipole moment: The molecule XF has a dipole moment. Is X boron or phosphorus? P The molecule XCl has a dipole moment. Is X beryllium or sulfur? Is the Cl BBCl molecule polar or nonpolar? nonpolar There are three possible structures for PCl F with phosphorus as the central atom. Draw them and discuss how measurements of dipole moments could help distinguish among them. Describe the molecular structure around the indicated atom or atoms: tetrahedral; trigonal pyramidal; (c) bent (109°); trigonal planar; bent (109°); bent (109°); (g) CH CCH tetrahedral, CH CCH linear; (h) tetrahedral; (i) H CCCH linear; H CCCH trigonal planar Draw the Lewis structures and predict the shape of each compound or ion: A molecule with the formula AB , in which A and B represent different atoms, could have one of three different shapes. Sketch and name the three different shapes that this molecule might have. Give an example of a molecule or ion for each shape. A molecule with the formula AB , in which A and B represent different atoms, could have one of three different shapes. Sketch and name the three different shapes that this molecule might have. Give an example of a molecule or ion that has each shape. Draw the Lewis electron dot structures for these molecules, including resonance structures where appropriate: predict the molecular shapes for \(\ce{CS3^2-}\) and CS and explain how you arrived at your predictions (a) ; (b) ; (c) ; \(\ce{CS3^2-}\) includes three regions of electron density (all are bonds with no lone pairs); the shape is trigonal planar; CS has only two regions of electron density (all bonds with no lone pairs); the shape is linear What is the molecular structure of the stable form of FNO ? (N is the central atom.) A compound with a molar mass of about 42 g/mol contains 85.7% carbon and 14.3% hydrogen. What is its molecular structure? The Lewis structure is made from three units, but the atoms must be rearranged: Use the to perform the following exercises for a two-atom molecule: Use the to perform the following exercises for a real molecule. You may need to rotate the molecules in three dimensions to see certain dipoles. The molecular dipole points away from the hydrogen atoms. Use the to build a molecule. Starting with the central atom, click on the double bond to add one double bond. Then add one single bond and one lone pair. Rotate the molecule to observe the complete geometry. Name the electron group geometry and molecular structure and predict the bond angle. Then click the check boxes at the bottom and right of the simulator to check your answers. Use the to explore real molecules. On the Real Molecules tab, select H O. Switch between the “real” and “model” modes. Explain the difference observed. The structures are very similar. In the model mode, each electron group occupies the same amount of space, so the bond angle is shown as 109.5°. In the “real” mode, the lone pairs are larger, causing the hydrogens to be compressed. This leads to the smaller angle of 104.5°. Use the to explore real molecules. On the Real Molecules tab, select “model” mode and S O. What is the model bond angle? Explain whether the “real” bond angle should be larger or smaller than the ideal model angle. | 23,156 | 2,448 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/18%3A_Solubility_and_Complex-Ion_Equilibria/18.8%3A_Equilibria_Involving_Complex_Ions |
Previously, you learned that metal ions in aqueous solution are hydrated—that is, surrounded by a shell of usually four or six water molecules. A hydrated ion is one kind of a (or, simply, complex), a species formed between a central metal ion and one or more surrounding , molecules or ions that contain at least one lone pair of electrons, such as the [Al(H O) ] ion. A complex ion forms from a metal ion and a ligand because of a Lewis acid–base interaction. The positively charged metal ion acts as a Lewis acid, and the ligand, with one or more lone pairs of electrons, acts as a Lewis base. Small, highly charged metal ions, such as Cu or Ru , have the greatest tendency to act as Lewis acids, and consequently, they have the greatest tendency to form complex ions. As an example of the formation of complex ions, consider the addition of ammonia to an aqueous solution of the hydrated Cu ion {[Cu(H O) ] }. Because it is a stronger base than H O, ammonia replaces the water molecules in the hydrated ion to form the [Cu(NH ) (H O) ] ion. Formation of the [Cu(NH ) (H O) ] complex is accompanied by a dramatic color change, as shown in . The solution changes from the light blue of [Cu(H O) ] to the blue-violet characteristic of the [Cu(NH ) (H O) ] ion. The replacement of water molecules from [Cu(H O) ] by ammonia occurs in sequential steps. Omitting the water molecules bound to Cu for simplicity, we can write the equilibrium reactions as follows: The sum of the stepwise reactions is the overall equation for the formation of the complex ion: The hydrated Cu ion contains six H O ligands, but the complex ion that is produced contains only four \(NH_3\) ligands, not six. \[Cu^{2+}_{(aq)} + 4NH_{3(aq)} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)} \label{Eq2}\] The equilibrium constant for the formation of the complex ion from the hydrated ion is called the . The equilibrium constant expression for has the same general form as any other equilibrium constant expression. In this case, the expression is as follows: The formation constant ( ) has the same general form as any other equilibrium constant expression. Water, a pure liquid, does not appear explicitly in the equilibrium constant expression, and the hydrated Cu (aq) ion is represented as Cu for simplicity. As for any equilibrium, the larger the value of the equilibrium constant (in this case, ), the more stable the product. With = 2.1 × 10 , the [Cu(NH ) (H O) ] complex ion is very stable. The formation constants for some common complex ions are listed in . If 12.5 g of Cu(NO ) •6H O is added to 500 mL of 1.00 M aqueous ammonia, what is the equilibrium concentration of Cu (aq)? mass of Cu salt and volume and concentration of ammonia solution equilibrium concentration of Cu (aq) Adding an ionic compound that contains Cu to an aqueous ammonia solution will result in the formation of [Cu(NH ) ] (aq), as shown in . We assume that the volume change caused by adding solid copper(II) nitrate to aqueous ammonia is negligible. The initial concentration of Cu from the amount of added copper nitrate prior to any reaction is as follows: \[12.5\mathrm{\;\cancel{g}\;Cu(NO_3)_2}\cdot\mathrm{6H_2O}\left(\dfrac{\textrm{1 mol}}{\textrm{295.65} \cancel{g}} \right )\left(\dfrac{1}{\textrm{500}\; \cancel{mL}} \right )\left(\dfrac{\textrm{1000}\; \cancel{mL}}{\textrm{1 L}} \right )=\textrm{0.0846 M}\] Because the stoichiometry of the reaction is four NH to one Cu , the amount of NH required to react completely with the Cu is 4(0.0846) = 0.338 M. The concentration of ammonia after complete reaction is 1.00 M − 0.338 M = 0.66 M. These results are summarized in the first two lines of the following table. Because the equilibrium constant for the reaction is large (2.1 × 10 ), the equilibrium will lie far to the right. Thus we will assume that the formation of [Cu(NH ) ] in the first step is complete and allow some of it to dissociate into Cu and NH until equilibrium has been reached. If we define as the amount of Cu produced by the dissociation reaction, then the stoichiometry of the reaction tells us that the change in the concentration of [Cu(NH ) ] is − , and the change in the concentration of ammonia is +4 , as indicated in the table. The final concentrations of all species (in the bottom row of the table) are the sums of the concentrations after complete reaction and the changes in concentrations. Substituting the final concentrations into the expression for the formation constant ( ) and assuming that << 0.0846, which allows us to remove from the sum and difference, The value of indicates that our assumption was justified. The equilibrium concentration of Cu (aq) in a 1.00 M ammonia solution is therefore 2.1 × 10 M. The ferrocyanide ion {[Fe(CN) ] } is very stable, with a of 1 × 10 . Calculate the concentration of cyanide ion in equilibrium with a 0.65 M solution of K [Fe(CN) ]. 2 × 10 M What happens to the solubility of a sparingly soluble salt if a ligand that forms a stable complex ion is added to the solution? One such example occurs in conventional black-and-white photography. Recall that black-and-white photographic film contains light-sensitive microcrystals of AgBr, or mixtures of AgBr and other silver halides. AgBr is a sparingly soluble salt, with a of 5.35 × 10 at 25°C. When the shutter of the camera opens, the light from the object being photographed strikes some of the crystals on the film and initiates a photochemical reaction that converts AgBr to black Ag metal. Well-formed, stable negative images appear in tones of gray, corresponding to the number of grains of AgBr converted, with the areas exposed to the most light being darkest. To fix the image and prevent more AgBr crystals from being converted to Ag metal during processing of the film, the unreacted AgBr on the film is removed using a complexation reaction to dissolve the sparingly soluble salt. The reaction for the dissolution of silver bromide is as follows: \[AgBr_{(s)} \rightleftharpoons Ag^+_{(aq)} + Br^{−}_{(aq)} \label{Eq4a}\] with \[K_{sp} = 5.35 \times 10^{−13} \text{ at 25°C} \label{Eq4b}\] The equilibrium lies far to the left, and the equilibrium concentrations of Ag and Br ions are very low (7.31 × 10 M). As a result, removing unreacted AgBr from even a single roll of film using pure water would require tens of thousands of liters of water and a great deal of time. Le Chatelier’s principle tells us, however, that we can drive the reaction to the right by removing one of the products, which will cause more AgBr to dissolve. Bromide ion is difficult to remove chemically, but silver ion forms a variety of stable two-coordinate complexes with neutral ligands, such as ammonia, or with anionic ligands, such as cyanide or thiosulfate (S O ). In photographic processing, excess AgBr is dissolved using a concentrated solution of sodium thiosulfate. The reaction of Ag with thiosulfate is as follows: \[Ag^+_{(aq)} + 2S_2O^{2−}_{3(aq)} \rightleftharpoons [Ag(S_2O_3)_2]^{3−}_{(aq)} \label{Eq5a}\] with \[K_f = 2.9 \times 10^{13} \label{Eq5b}\] The magnitude of the equilibrium constant indicates that almost all Ag ions in solution will be immediately complexed by thiosulfate to form [Ag(S O ) ] . We can see the effect of thiosulfate on the solubility of AgBr by writing the appropriate reactions and adding them together: Comparing with shows that the formation of the complex ion increases the solubility of AgBr by approximately 3 × 10 . The dramatic increase in solubility combined with the low cost and the low toxicity explains why sodium thiosulfate is almost universally used for developing black-and-white film. If desired, the silver can be recovered from the thiosulfate solution using any of several methods and recycled. If a complex ion has a large , the formation of a complex ion can dramatically increase the solubility of sparingly soluble salts. Due to the common ion effect, we might expect a salt such as AgCl to be much less soluble in a concentrated solution of KCl than in water. Such an assumption would be incorrect, however, because it ignores the fact that silver ion tends to form a two-coordinate complex with chloride ions (AgCl ). Calculate the solubility of AgCl in each situation: At 25°C, = 1.77 × 10 for AgCl and = 1.1 × 10 for AgCl . of AgCl, of AgCl , and KCl concentration solubility of AgCl in water and in KCl solution with and without the formation of complex ions Thus the solubility of AgCl in pure water at 25°C is 1.33 × 10 M. If the common ion effect were the only important factor, we would predict that AgCl is approximately five orders of magnitude less soluble in a 1.0 M KCl solution than in water. If we let equal the solubility of AgCl in the KCl solution, then at equilibrium [AgCl ] = and [Cl ] = 1.0 − . Substituting these quantities into the equilibrium constant expression for the net reaction and assuming that << 1.0, That is, AgCl dissolves in 1.0 M KCl to produce a 1.9 × 10 M solution of the AgCl complex ion. Thus we predict that AgCl has approximately the same solubility in a 1.0 M KCl solution as it does in pure water, which is 10 times greater than that predicted based on the common ion effect. (In fact, the measured solubility of AgCl in 1.0 M KCl is almost a factor of 10 greater than that in pure water, largely due to the formation of other chloride-containing complexes.) Calculate the solubility of mercury(II) iodide (HgI ) in each situation: = 2.9 × 10 for HgI and = 6.8 × 10 for [HgI ] . Solubility of Complex Ions: Complexing agents, molecules or ions that increase the solubility of metal salts by forming soluble metal complexes, are common components of laundry detergents. Long-chain carboxylic acids, the major components of soaps, form insoluble salts with Ca and Mg , which are present in high concentrations in “hard” water. The precipitation of these salts produces a bathtub ring and gives a gray tinge to clothing. Adding a complexing agent such as pyrophosphate (O POPO , or P O ) or triphosphate (P O ) to detergents prevents the magnesium and calcium salts from precipitating because the equilibrium constant for complex-ion formation is large: \[Ca^{2+}_{(aq)} + O_3POPO^{4−}_{4(aq)} \rightleftharpoons [Ca(O_3POPO_3)]^{2−}_{(aq)} \label{Eq7a}\] with However, phosphates can cause environmental damage by promoting eutrophication, the growth of excessive amounts of algae in a body of water, which can eventually lead to large decreases in levels of dissolved oxygen that kill fish and other aquatic organisms. Consequently, many states in the United States have banned the use of phosphate-containing detergents, and France has banned their use beginning in 2007. “Phosphate-free” detergents contain different kinds of complexing agents, such as derivatives of acetic acid or other carboxylic acids. The development of phosphate substitutes is an area of intense research. Commercial water softeners also use a complexing agent to treat hard water by passing the water over ion-exchange resins, which are complex sodium salts. When water flows over the resin, sodium ion is dissolved, and insoluble salts precipitate onto the resin surface. Water treated in this way has a saltier taste due to the presence of Na , but it contains fewer dissolved minerals. Another application of complexing agents is found in medicine. Unlike x-rays, magnetic resonance imaging (MRI) can give relatively good images of soft tissues such as internal organs. MRI is based on the magnetic properties of the H nucleus of hydrogen atoms in water, which is a major component of soft tissues. Because the properties of water do not depend very much on whether it is inside a cell or in the blood, it is hard to get detailed images of these tissues that have good contrast. To solve this problem, scientists have developed a class of metal complexes known as “MRI contrast agents.” Injecting an MRI contrast agent into a patient selectively affects the magnetic properties of water in cells of normal tissues, in tumors, or in blood vessels and allows doctors to “see” each of these separately ( ). One of the most important metal ions for this application is Gd , which with seven unpaired electrons is highly paramagnetic. Because Gd (aq) is quite toxic, it must be administered as a very stable complex that does not dissociate in the body and can be excreted intact by the kidneys. The complexing agents used for gadolinium are ligands such as DTPA (diethylene triamine pentaacetic acid), whose fully protonated form is shown here. Figure \(\Page {2}\): An MRI Image of the Heart, Arteries, and Veins. When a patient is injected with a paramagnetic metal cation in the form of a stable complex known as an MRI contrast agent, the magnetic properties of water in cells are altered. Because the different environments in different types of cells respond differently, a physician can obtain detailed images of soft tissues. A complex ion is a species formed between a central metal ion and one or more surrounding ligands, molecules or ions that contain at least one lone pair of electrons. Small, highly charged metal ions have the greatest tendency to act as Lewis acids and form complex ions. The equilibrium constant for the formation of the complex ion is the formation constant ( ). The formation of a complex ion by adding a complexing agent increases the solubility of a compound. | 13,500 | 2,449 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/04%3A_Putting_the_First_Law_to_Work/4.05%3A_The_Joule-Thomson_Effect |
In 1852, working with William Thomson (who would later become Lord Kelvin), Joule conducted an experiment in which they pumped gas at a steady rate through a lead pipe that was cinched to create a construction. On the upstream side of the constriction, the gas was at a higher pressure than on the downstream side of the constriction. Also, the temperature of the gas was carefully monitored on either side of the construction. The cooling that they observed as the gas expanded from a high pressure region to a lower pressure region was extremely important and lead to a common design of modern refrigerators. Not all gases undergo a cooling effect upon expansion. Some gases, such as hydrogen and helium, will experience a warming effect upon expansion under conditions near room temperature and pressure. The direction of temperature change can be determined by measuring the Joule-Thomson coefficient, \(\mu_{JT}\). This coefficient has the definition \[ \mu_{JT} \equiv \left( \dfrac{\partial T}{\partial p} \right)_H \nonumber \] Schematically, the Joule-Thomson coefficient can be measured by measuring the temperature drop or increase a gas undergoes for a given pressure drop (Figure \(\Page {1}\)). The apparatus is insulated so that no heat can be transferred in or out, making the expansion isenthalpic. The typical behavior of the Joule-Thomson coefficient can be summarized in Figure \(\Page {2}\). At the combinations of \(T\) and \(p\) for which \(\mu_{JT} > 0\) (inside the shaded region), the sample will cool upon expansion. At those \(p\) and \(T\) conditions outside of the shaded region, where \(\mu_{JT} < 0\), the gas will undergo a temperature increase upon expansion. And along the boundary, a gas will undergo neither a temperature increase not decrease upon expansion. For a given pressure, there are typically two temperatures at which \(\mu_{JT}\) changes sign. These are the upper and lower inversion temperatures. Using the tools of mathematics, it is possible to express the Joule-Thomson coefficient in terms of measurable properties. Consider enthalpy as a function of pressure and temperature: \(H(p, T)\). This suggests that the total differential \(dH\) can be expressed \[dH= \left( \dfrac{\partial H}{\partial p} \right)_T dp+ \left( \dfrac{\partial H}{\partial T} \right)_p dT \label{totalH} \] It will be shown later (again, once we develop the ) that \[\left( \dfrac{\partial H}{\partial p} \right)_T dp = -T \left( \dfrac{\partial V}{\partial T} \right)_p + V \nonumber \] A simple substitution shows \[\left( \dfrac{\partial H}{\partial p} \right)_T dp = - TV \alpha + V = V(1-T\alpha) \nonumber \] So \[ dH = V(1-T\alpha) dP + C_p dT \nonumber \] For an ideal gas, \(\alpha = 1/T\), so \[ dH = \cancelto{0}{V\left(1-T\dfrac{1}{T}\right) dP} + C_p dT \nonumber \] which causes the first term to vanish. So for constant enthalpy expansion (\(dH = 0\)), there can be no change in temperature (\(dT = 0\)). This will mean that gases will only show non-zero values for \(\mu_{JT}\) only because they deviate from ideal behavior! Derive an expression for \(\mu_{JT}\) in terms of \(\alpha\), \(C_p\), \(V\), and \(T\). Using the total differential for \(H(p, T)\) (Equation \ref{totalH}): \[dH= \left( \dfrac{\partial H}{\partial p} \right)_T dp+ \left( \dfrac{\partial H}{\partial T} \right)_p dT \nonumber \] Dividing by \(dp\) and constraining to constant \(H\): \[\left.\dfrac{dH}{dp} \right\rvert_{H}= \left( \dfrac{\partial H}{\partial p} \right)_T \left.\dfrac{dp}{dp} \right\rvert_{H} + \left( \dfrac{\partial H}{\partial T} \right)_p \left.\dfrac{dT}{dp} \right\rvert_{H} \nonumber \] Noting that \[\left.\dfrac{dH}{dp} \right\rvert_{H} = 0 \nonumber \] \[\left.\dfrac{dp}{dp} \right\rvert_{H} = 1 \nonumber \] and \[\left.\dfrac{dT}{dp} \right\rvert_{H} = \left(\dfrac{\partial T}{\partial p} \right)_{H} \nonumber \] so \[ 0 = \left( \dfrac{\partial H}{\partial p} \right)_T + \left( \dfrac{\partial H}{\partial T} \right)_p \left(\dfrac{\partial T}{\partial p} \right)_{H} \nonumber \] We can then use the following substitutions: \[\left( \dfrac{\partial H}{\partial p} \right)_T = V(1-T \alpha) \nonumber \] \[\left( \dfrac{\partial H}{\partial T} \right)_p = C_p \nonumber \] \[\left(\dfrac{\partial T}{\partial p} \right)_{H} = \mu_{JT} \nonumber \] To get \[ 0 = V(1-T \alpha) + C_p \mu_{JT} \nonumber \] And solving for \(\mu_{JT}\) gives \[\mu_{JT} = \dfrac{V}{C_p}(T \alpha -1) \nonumber \] | 4,466 | 2,450 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Elimination_Reactions/Elimination_Reactions/D._The_Dehydration_of_Propan-2-ol |
This page looks at the mechanism for the acid catalysed dehydration of propan-2-ol. The dehydration of propan-2-ol is taken as an simple example of the way that secondary and tertiary alcohols dehydrate. Primary alcohols like ethanol use a different mechanism, and ethanol is discussed separately on another page. You will find a link to this from the elimination mechanisms menu. You will also find a link there to a page on the dehydration of more complicated alcohols where more than one product may be formed. Propan-2-ol can be dehydrated to give propene by heating it with an excess of concentrated sulphuric acid at about 170°C. Concentrated phosphoric(V) acid, H PO , can be used instead. The acids aren't written into the equation because they serve as catalysts. If you like, you could write, for example, "conc H SO " over the top of the arrow. Notice that the -OH group is lost, together with a hydrogen from a next-door carbon - it doesn't matter which one. If you chose the other one, you would get CH CH=CH . That's the same molecule flipped over. We are going to discuss the mechanism using sulphuric acid. Afterwards, we'll describe how you can use a simplified version which will work for any acid, including phosphoric(V) acid. In the first stage, one of the lone pairs of electrons on the oxygen picks up a hydrogen ion from the sulphuric acid. The alcohol is said to be protonated. The negative ion produced is the hydrogensulphate ion, HSO . Notice that the oxygen atom in the alcohol has gained a positive charge. That charge has to be there for two reasons: In the second stage of the reaction the protonated propan-2-ol loses a water molecule to leave a carbocation (previously known as a carbonium ion) - an ion with a positive charge on a carbon atom. The carbon atom is positive because it has lost the electron that it originally contributed to the carbon-oxygen bond. Both of the electrons in that bond have moved onto the oxygen atom, neutralising the oxygen's charge. Finally, a hydrogensulphate ion (from the sulphuric acid) pulls off a hydrogen ion from the carbocation, and a double bond forms. People normally quote a simplified version of this mechanism. Instead of showing the full structure of the sulphuric acid, you write it as if it were simply a hydrogen ion, H . That leaves the full mechanism: An advantage of this (apart from the fact that it doesn't require you to draw the structure of sulphuric acid) is that it can be used for any acid catalyst without changing it at all. For example, if you use this version, you wouldn't need to worry about the structure of phosphoric(V) acid. | 2,644 | 2,451 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/15%3A_Aromaticity_(Reactions_of_Benzene)/15.08%3A_The_Nomenclature_of_Monosubstituted_Benzenes |
Benzene, C H , is an organic aromatic compound with many interesting properties. Unlike aliphatic (straight chain carbons) or other cyclic organic compounds, the structure of benzene (3 conjugated π bonds) allows benzene and its derived products to be useful in fields such as health, laboratory, and other applications such as rubber synthesis. Benzene derived products are well known to be pleasantly fragrant. For this reason, organic compounds containing benzene rings were classified as being "aromatic" (sweet smelling) amongst scientists in the early 19th century when a relation was established between benzene derived compounds and sweet/spicy fragrances. There is a misconception amongst the scientific community, however, that all aromatics are sweet smelling and that all sweet smelling compounds would have a benzene ring in its structure. This is false, since non-aromatic compounds, such as camphor, extracted from the camphor laurel tree, release a strong, minty aroma, yet it lacks the benzene ring in its structure (Figure 1). On the other hand, benzene itself gives off a rather strong and unpleasant smell that would otherwise invalidate the definition of an aromatic (sweet-smelling) compound. Despite this inconsistency, however, the term aromatic continues to be used today in order to designate molecules with benzene-like rings in their structures. For a modern, chemical definition of aromaticity, refer to sections and Hückel's Rule.
Many aromatic compounds are however, sweet/pleasant smelling. Eugenol, for example, is extracted from essential oils of cloves and it releases a spicy, clove-like aroma used in perfumes. In addition, it is also used in dentistry as an analgesic. Due to the similarity between benzene and , the two is often confused with each other in beginning organic chemistry students. If you were to count the number of carbons and hydrogens in , you will notice that its molecular formula is . Since the carbons in the ring is with hydrogens (carbon is bound to 2 hydrogens and 2 adjacent carbons), no double bonds are formed in the cyclic ring. In contrast, is only with one hydrogen per carbon, leading to its molecular formula of . In order to stabilize this structure, 3 conjugated π (double) bonds are formed in the benzene ring in order for carbon to have four adjacent bonds. In other words, cyclohexane is not the same as benzene! These two compounds have different molecular formulas and their chemical and physical properties are not the same. The technique can be used by chemists to convert from benzene to cyclohexane by saturating the benzene ring with missing hydrogens. A special catalyst is required to hydrogenate benzene rings due to its unusual stability and configuration. Normal catalytic hydrogenation techniques will not hydrogenate benzene and yield any meaningful products. Benzene can be drawn a number of different ways. This is because benzene's conjugated pi electrons freely resonate within the cyclic ring, thus resulting in its two resonance forms. As the electrons in the benzene ring can resonate within the ring at a fairly high rate, a simplified notation is often used to designate the two different resonance forms. This notation is shown above, with the initial three pi bonds (#1, #2) replaced with an inner ring circle (#3). Alternatively, the circle within the benzene ring can also be dashed to show the same resonance forms (#4). The phenyl group can be formed by taking benzene, and removing a hydrogen from it. The resulting molecular formula for the fragment is C H . NOTE: Although the molecular formula of the phenyl group is C H , the phenyl group would always have something attached to where the hydrogen was removed. Thus, the formula is often written as Ph-R, where Ph refers to the Phenyl group, and R refers to the R group attached to where the hydrogen was removed.
Different R groups on the phenyl group allows different benzene derivatives to be formed. Phenol, Ph-OH, or C H OH, for example, is formed when an alcohol (-OH) group displaces a hydrogen atom on the benzene ring. Benzene, for this very same reason, can be formed from the phenyl group by reattaching the hydrogen back its place of removal. Thus benzene, similar to phenol, can be abbreviated Ph-H, or C H . As you can see above, these are only some of the many possibilities of the benzene derived products that have special uses in human health and other industrial fields. Unlike aliphatic organics, nomenclature of benzene-derived compounds can be confusing because a single aromatic compound can have multiple possible names (such as common and systematic names) be associated with its structure. In these sections, we will analyze some of the ways these compounds can be named. Some common substituents, like NO , Br, and Cl, can be named this way when it is attached to a phenyl group. Long chain carbons attached can also be named this way. The general format for this kind of naming is: . For example, chlorine (Cl) attached to a phenyl group would be named . Since there is only one substituent on the benzene ring, we do not have to indicate its position on the benzene ring (as it can freely rotate around and you would end up getting the same compound.)
Example of simple benzene naming with chlorine and NO as substituents.
Instead of using numbers to indicate substituents on a benzene ring, can be used in place of positional markers when there are substituents on the benzene ring (disubstituted benzenes). They are defined as the following: Using the same example above in Figure 9a (1,3-dichlorobenzene), we can use the ortho-, meta-, para- nomenclature to transform the chemical name into m-dichlorobenzene, as shown in the Figure below.
Here are some other examples of ortho-, meta-, para- nomenclature used in context: However, the substituents used in ortho-, meta-, para- nomenclature do not have to be the same. For example, we can use chlorine and a nitro group as substituents in the benzene ring. In conclusion, these can be pieced together into a summary diagram, as shown below: In addition to simple benzene naming and OMP nomenclature, benzene derived compounds are also sometimes used as . The concept of a base is similar to the nomenclature of aliphatic and cyclic compounds, where the parent for the organic compound is used as a base (a name for its chemical name. For example, the following compounds have the base names and , respectively. See Nomenclature of Organic Compounds for a review on naming organic compounds. Benzene, similar to these compounds shown above, also has base names from its derived compounds. , as introduced previously in this article, for example, serves as a base when other substituents are attached to it. This is best illustrated in the diagram below.
Alternatively, we can use the numbering system to indicate this compound. When the numbering system is used, the carbon where the substituent is attached on the base will be given the first priority and named as carbon #1 (C ). The normal priority rules then apply in the nomenclature process (give the rest of the substituents the lowest numbering as you could).
Below is a list of commonly seen benzene-derived compounds. Some of these mono-substituted compounds (labeled in red and green), such as phenol or toluene, can be used in place of benzene for the chemical's base name.
Common benzene derived compounds with various substituents. According to the indexing preferences of the , are some of the common names that are retained in the IUPAC (systematic) nomenclature. Other names such as toluene, styrene, naphthalene, or phenanthrene can also be seen in the IUPAC system in the same way. While the use of other common names are usually acceptable in IUPAC, their use are discouraged in the nomenclature of compounds. Nomenclature for compounds which has such discouraged names will be named by the simple benzene naming system. An example of this would include (Note that toluene by itself is retained by the IUPAC nomenclature, but its derivatives, which contains additional substituents on the benzene ring, might be excluded from the convention). For this reason, the 2,4,6-trinitrotoluene, or TNT, as shown in Figure 17, would not be advisable under the IUPAC ( ) nomenclature. In order to correctly name TNT under the IUPAC system, the simple benzene naming system should be used:
Since the IUPAC nomenclature primarily rely on the simple benzene naming system for the nomenclature of different benzene derived compounds, the OMP (ortho-, meta-, para-) system is not accepted in the IUPAC nomenclature. For this reason, the OMP system will yield common names that can be converted to systematic names by using the same method as above. For example, o-Xylene from the OMP system can be named 1,2-dimethylbenzene by using simple benzene naming (IUPAC standard). As mentioned previously, the phenyl group (Ph-R, C H -R) can be formed by removing a hydrogen from benzene and attaching a substituent to where the hydrogen was removed. To this phenomenon, we can name compounds formed this way by applying this rule: . For example, a chlorine attached in this manner would be named , and a bromine attached in this manner would be named . (See below diagram)
While compounds like these are usually named by simple benzene type naming (chlorobenzene and bromobenzene), the phenyl group naming is usually applied to benzene rings where a substituent with six or more carbons is attached, such as in the diagram below.
Although the diagram above might be a little daunting to understand at first, it is not as difficult as it seems after careful analysis of the structure is made. By looking for the longest chain in the compound, it should be clear that the longest chain is eight (8) carbons long (octane, as shown in green) and that a benzene ring is attached to the second position of this longest chain (labeled in red). As this rule suggests that the benzene ring will act as a function group (a substituent) whenever a substituent of more than six (6) carbons is attached to it, the name "benzene" is changed to and is used the same way as any other substituents, such as Putting it all together, the name can be derived as: (phenyl is attached at the second position of the longest carbon chain, octane). The benzyl group (abbv. Bn), similar to the phenyl group, is formed by manipulating the benzene ring. In the case of the benzyl group, it is formed by taking the phenyl group and adding a CH group to where the hydrogen was removed. Its molecular fragment can be written as C H CH -R, PhCH -R, or Bn-R. Nomenclature of benzyl group based compounds are very similar to the phenyl group compounds. For example, a chlorine attached to a benzyl group would simply be called benzyl chloride, whereas an OH group attached to a benzyl group would simply be called benzyl alcohol.
Additionally, other substituents can attach on the benzene ring in the presence of the benzyl group. An example of this can be seen in the Figure below:
Similar to the base name nomenclature system, the carbon in which the base substituent is attached on the benzene ring is given the first priority and the rest of the substituents are given the lowest number order possible. Under this consideration, the above compound can be named:
As benzene derived compounds can be extremely complex, only compounds covered in this article and other commonly named compounds can be named using this flowchart. To demonstrate how this flowchart can be used to name TNT in its common and systematic (IUPAC) name, a replica of the flowchart with the appropriate flow paths are shown below:
The compound above contains a benzene ring and thus is aromatic. Benzene unusual stability is caused by how many conjugated pi bonds in its cyclic ring? ____ Menthol, a topical analgesic used in many ointments for the relief of pain, releases a peppermint aroma upon exposure to the air. Based on this conclusion, can you imply that a benzene ring is present in its chemical structure? Why or why not? At normal conditions, benzene has ___ resonance structures. Which of the following name(s) is/are correct for the following compound?
a) nitrohydride benzene
b) phenylamine
c) phenylamide
d) aniline
e) nitrogenhydrogen benzene
f) All of the above is correct Convert 1,4-dimethylbenzene into its common name. TNT's common name is: ______________________________ Name the following compound using OMP nomenclature:
Draw the structure of 2,4-dinitrotoluene. Name the following compound:
Which of the following is the correct name for the following compound?
a) 3,4-difluorobenzyl bromide
b) 1,2-difluorobenzyl bromide
c) 4,5-difluorobenzyl bromide
d) 1,2-difluoroethyl bromide
e) 5,6-difluoroethyl bromide
f) 4,5-difluoroethyl bromide Benzyl chloride can be abbreviated Bz-Cl. Benzoic Acid has what R group attached to its phenyl functional group? A single aromatic compound can have multiple names indicating its structure. List the corresponding positions for the OMP system (o-, m-, p-). A scientist has conducted an experiment on an unknown compound. He was able to determine that the unknown compound contains a cyclic ring in its structure as well as an alcohol (-OH) group attached to the ring. What is the unknown compound?
a) Cyclohexanol
b) Cyclicheptanol
c) Phenol
d) Methanol
e) Bleach
f) Cannot determine from the above information Which of the following statements is for the compound, phenol?
a) Phenol is a benzene derived compound.
b) Phenol can be made by attaching an -OH group to a phenyl group.
c) Phenol is highly toxic to the body even in small doses.
d) Phenol can be used as a catalyst in the hydrogenation of benzene into cyclohexane.
e) Phenol is used as an antiseptic in minute doses.
f) Phenol is amongst one of the three common names retained in the IUPAC nomenclature. False, this compound does not contain a benzene ring in its structure. 3 No, a substance that is fragrant does not imply a benzene ring is in its structure. See camphor example (Figure 1) No reaction, benzene requires a special catalyst to be hydrogenated due to its unusual stability given by its three conjugated pi bonds. 2 b, d p-Xylene 2,4,6-trinitrotoluene p-chloronitrobenzene 4-phenylheptane a False, the correct abbreviation for the benzyl group is Bn, not Bz. The correct abbreviation for Benzyl chloride is Bn-Cl. COOH True. TNT, for example, has the common name 2,4,6-trinitrotoluene and its systematic name is 2-methyl-1,3,5-trinitrobenzene. Ortho - 1,2 ; Meta - 1,3 ; Para - 1,4 The correct answer is f). We cannot determine what structure this is since the question does not tell us what kind of cyclic ring the -OH group is attached on. Just as cyclohexane can be cyclic, benzene and cycloheptane can also be cyclic. d | 14,933 | 2,452 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/15%3A_Thermodynamics-_Atoms_Molecules_and_Energy/15.03%3A_Heat_Capacity_and_Microscopic_Changes |
Let us turn our attention from the macroscopic to the microscopic level. According to the first law of thermodynamics, the heat energy absorbed as we raise the temperature of a substance cannot be destroyed. But where does it go? In the case of a monatomic gas, like neon, this question is easy to answer. All the energy absorbed is converted into the kinetic energy of the neon molecules (atoms). , we found that the kinetic energy of the molecules in a sample of gas is given by the expression \[E_{k}=\tfrac{\text{3}}{\text{2}}nRT \nonumber \] Thus if the temperature of a sample of neon gas is raised from to , the kinetic energy of the molecules increases from / to / , a total change of Inserting the value of in appropriate units, we obtain This is the same quantity that is obtained by substituting the experimental value of for neon (calculated in ) into . In other words the quantity of heat found experimentally exactly matches the increase in kinetic energy of the molecules required by the kinetic theory of gases. Table \(\Page {1}\) lists the values not only for neon but for some other gases as well. We immediately notice that only the noble gases and other mon-atomic gases such as Hg and Na have molar heat capacities equal to / , or 12.47 J K mol . All other gases have higher molar heat capacities than this. Moreover, as the table shows, the more complex the molecule, the higher the molar heat capacity of the gas. There is a simple reason for this behavior. A molecule which has two or more atoms is not only capable of moving from one place to another ( ), it can also about itself, and it can change its shape by . When we heat a mole of Cl molecules, for example, we not only need to supply them with enough energy to make them move around faster (increase their translational kinetic energy), we must also supply an additional quantity of energy to make them rotate and vibrate more strongly than before. For a mole of more complex molecules like -butane even more energy is required since the molecule is capable of changing its shape in all sorts of ways. In the butane molecule there are three C—C bonds around which segments of the molecule can rotate freely. All the bonds can bend or stretch, and the whole molecule can rotate as well. Such a molecule is constantly flexing and writhing at room temperature. As we raise the temperature, this kind of movement occurs more rapidly and extra energy must be absorbed in order to make this possible. When we heat and , the situation is somewhat different than for . The rapid increase of vapor pressure with temperature makes it virtually impossible to heat a solid or liquid in a closed container, and so heat capacities are always measured at constant pressure rather than at constant volume. Some values for selected simple liquids and solids at the melting point are shown in Table \(\Page {2}\). In general the of solids and liquids are higher than those of gases. This is because of the intermolecular forces operating in solids and liquids. When we heat solids and liquids, we need to supply them with potential energy as well as kinetic energy. Among the solids, the heat capacities of the metals are easiest to explain since the solid consists of individual atoms. Each atom can only vibrate in three dimensions. According to a theory first suggested by Einstein, this vibrational energy has the value 3 , while the heat capacity is given by 3 = 24.9 J K mol . As can be seen from the table, most monatomic solids have values slightly larger than this. This is because solids expand slightly on heating. The atoms get farther apart and thus increase in potential as well as vibrational energy. Solids which contain molecules rather than atoms have much higher heat capacities than 3 . In addition to the vibration of the whole molecule about its site in the crystal lattice, the individual atoms can also vibrate with respect to each other. Occasionally molecules can rotate in the crystal, but usually rotation is only possible when the solid melts. As can be seen from the values for molecular liquids in Table \(\Page {2}\), this sudden ability to rotate causes a sharp increase in the heat capacity. For monatomic substances, where there is no motion corresponding to the rotation of atoms around each other, the heat capacity of the liquid is only very slightly higher than that of the solid. | 4,431 | 2,453 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/22%3A_Chemistry_of_The_Main-Group_Elements_II/22.5%3A_Group_15%3A_The_Nitrogen_Family |
Like the group 14 elements, the lightest member of , nitrogen, is found in nature as the free element, and the heaviest elements have been known for centuries because they are easily isolated from their ores. Antimony (Sb) was probably the first of the pnicogens to be obtained in elemental form and recognized as an element. Its atomic symbol comes from its Roman name: stibium. It is found in stibnite (Sb S ), a black mineral that has been used as a cosmetic (an early form of mascara) since biblical times, and it is easily reduced to the metal in a charcoal fire (Figure \(\Page {1}\)). The Egyptians used antimony to coat copper objects as early as the third millennium BC, and antimony is still used in alloys to improve the tonal quality of bells. In the form of its yellow sulfide ore, orpiment (As S ), arsenic (As) has been known to physicians and professional assassins since ancient Greece, although elemental arsenic was not isolated until centuries later. The history of bismuth (Bi), in contrast, is more difficult to follow because early alchemists often confused it with other metals, such as lead, tin, antimony, and even silver (due to its slightly pinkish-white luster). Its name comes from the old German wismut, meaning “white metal.” Bismuth was finally isolated in the 15th century, and it was used to make movable type for printing shortly after the invention of the Gutenberg printing process in 1440. Bismuth is used in printing because it is one of the few substances known whose solid state is less dense than the liquid. Consequently, its alloys expand as they cool, filling a mold completely and producing crisp, clear letters for typesetting. Phosphorus was discovered in 1669 by the German alchemist Hennig Brandt, who was looking for the “philosophers’ stone,” a mythical substance capable of converting base metals to silver or gold. Believing that human urine was the source of the key ingredient, Brandt obtained several dozen buckets of urine, which he allowed to putrefy. The urine was distilled to dryness at high temperature and then condensed; the last fumes were collected under water, giving a waxy white solid that had unusual properties. For example, it glowed in the dark and burst into flames when removed from the water. (Unfortunately for Brandt, however, it did not turn lead into gold.) The element was given its current name (from the Greek phos, meaning “light,” and phoros, meaning “bringing”) in the 17th century. For more than a century, the only way to obtain phosphorus was the distillation of urine, but in 1769 it was discovered that phosphorus could be obtained more easily from bones. During the 19th century, the demand for phosphorus for matches was so great that battlefields and paupers’ graveyards were systematically scavenged for bones. Early matches were pieces of wood coated with elemental phosphorus that were stored in an evacuated glass tube and ignited when the tube was broken (which could cause unfortunate accidents if the matches were kept in a pocket!). Unfortunately, elemental phosphorus is volatile and highly toxic. It is absorbed by the teeth and destroys bone in the jaw, leading to a painful and fatal condition called “phossy jaw,” which for many years was accepted as an occupational hazard of working in the match industry. Although nitrogen is the most abundant element in the atmosphere, it was the last of the pnicogens to be obtained in pure form. In 1772, Daniel Rutherford, working with Joseph Black (who discovered CO ), noticed that a gas remained when CO was removed from a combustion reaction. Antoine Lavoisier called the gas azote, meaning “no life,” because it did not support life. When it was discovered that the same element was also present in nitric acid and nitrate salts such as KNO (nitre), it was named nitrogen. About 90% of the nitrogen produced today is used to provide an inert atmosphere for processes or reactions that are oxygen sensitive, such as the production of steel, petroleum refining, and the packaging of foods and pharmaceuticals. Because the atmosphere contains several trillion tons of elemental nitrogen with a purity of about 80%, it is a huge source of nitrogen gas. Distillation of liquefied air yields nitrogen gas that is more than 99.99% pure, but small amounts of very pure nitrogen gas can be obtained from the thermal decomposition of sodium azide: \[\mathrm{2NaN_3(s)\xrightarrow{\Delta}2Na(l)+3N_2(g)} \label{Eq1}\] In contrast, Earth’s crust is relatively poor in nitrogen. The only important nitrogen ores are large deposits of KNO and NaNO in the deserts of Chile and Russia, which were apparently formed when ancient alkaline lakes evaporated. Consequently, virtually all nitrogen compounds produced on an industrial scale use atmospheric nitrogen as the starting material. Phosphorus, which constitutes only about 0.1% of Earth’s crust, is much more abundant in ores than nitrogen. Like aluminum and silicon, phosphorus is always found in combination with oxygen, and large inputs of energy are required to isolate it. The other three pnicogens are much less abundant: arsenic is found in Earth’s crust at a concentration of about 2 ppm, antimony is an order of magnitude less abundant, and bismuth is almost as rare as gold. All three elements have a high affinity for the chalcogens and are usually found as the sulfide ores (M S ), often in combination with sulfides of other heavy elements, such as copper, silver, and lead. Hence a major source of antimony and bismuth is flue dust obtained by smelting the sulfide ores of the more abundant metals. In group 15, as elsewhere in the p block, we see large differences between the lightest element (N) and its congeners in size, ionization energy, electron affinity, and electronegativity (Table \(\Page {1}\)). The chemical behavior of the elements can be summarized rather simply: nitrogen and phosphorus behave chemically like nonmetals, arsenic and antimony behave like semimetals, and bismuth behaves like a metal. With their ns np valence electron configurations, all form compounds by losing either the three np valence electrons to form the +3 oxidation state or the three np and the two ns valence electrons to give the +5 oxidation state, whose stability decreases smoothly from phosphorus to bismuth. In addition, the relatively large magnitude of the electron affinity of the lighter pnicogens enables them to form compounds in the −3 oxidation state (such as NH and PH ), in which three electrons are formally added to the neutral atom to give a filled np subshell. Nitrogen has the unusual ability to form compounds in nine different oxidation states, including −3, +3, and +5. Because neutral covalent compounds of the trivalent pnicogens contain a lone pair of electrons on the central atom, they tend to behave as Lewis bases. In group 15, the stability of the +5 oxidation state decreases from P to Bi. Because neutral covalent compounds of the trivalent group 15 elements have a lone pair of electrons on the central atom, they tend to be Lewis bases. Like carbon, nitrogen has four valence orbitals (one 2s and three 2p), so it can participate in at most four electron-pair bonds by using sp hybrid orbitals. Unlike carbon, however, nitrogen does not form long chains because of repulsive interactions between lone pairs of electrons on adjacent atoms. These interactions become important at the shorter internuclear distances encountered with the smaller, second-period elements of groups 15, 16, and 17. Stable compounds with N–N bonds are limited to chains of no more than three N atoms, such as the azide ion (N ). Nitrogen is the only pnicogen that normally forms multiple bonds with itself and other second-period elements, using π overlap of adjacent np orbitals. Thus the stable form of elemental nitrogen is N , whose N≡N bond is so strong (D = 942 kJ/mol) compared with the N–N and N=N bonds (D = 167 kJ/mol; D = 418 kJ/mol) that all compounds containing N–N and N=N bonds are thermodynamically unstable with respect to the formation of N . In fact, the formation of the N≡N bond is so thermodynamically favored that virtually all compounds containing N–N bonds are potentially explosive. Again in contrast to carbon, nitrogen undergoes only two important chemical reactions at room temperature: it reacts with metallic lithium to form lithium nitride, and it is reduced to ammonia by certain microorganisms. At higher temperatures, however, N reacts with more electropositive elements, such as those in group 13, to give binary nitrides, which range from covalent to ionic in character. Like the corresponding compounds of carbon, binary compounds of nitrogen with oxygen, hydrogen, or other nonmetals are usually covalent molecular substances. Few binary molecular compounds of nitrogen are formed by direct reaction of the elements. At elevated temperatures, N reacts with H to form ammonia, with O to form a mixture of NO and NO , and with carbon to form cyanogen (N≡C–C≡N); elemental nitrogen does not react with the halogens or the other chalcogens. Nonetheless, all the binary nitrogen halides (NX ) are known. Except for NF , all are toxic, thermodynamically unstable, and potentially explosive, and all are prepared by reacting the halogen with NH rather than N . Both nitrogen monoxide (NO) and nitrogen dioxide (NO ) are thermodynamically unstable, with positive free energies of formation. Unlike NO, NO reacts readily with excess water, forming a 1:1 mixture of nitrous acid (HNO ) and nitric acid (HNO ): \[\ce{2NO2(g) + H2O(l) \rightarrow HNO2(aq) + HNO3(aq)} \label{Eq2}\] Nitrogen also forms N O (dinitrogen monoxide, or nitrous oxide), a linear molecule that is isoelectronic with CO and can be represented as N=N =O. Like the other two oxides of nitrogen, nitrous oxide is thermodynamically unstable. The structures of the three common oxides of nitrogen are as follows: Few binary molecular compounds of nitrogen are formed by the direct reaction of the elements. At elevated temperatures, nitrogen reacts with highly electropositive metals to form ionic nitrides, such as Li N and Ca N . These compounds consist of ionic lattices formed by M and N ions. Just as boron forms interstitial borides and carbon forms interstitial carbides, with less electropositive metals nitrogen forms a range of interstitial nitrides, in which nitrogen occupies holes in a close-packed metallic structure. Like the interstitial carbides and borides, these substances are typically very hard, high-melting materials that have metallic luster and conductivity. Nitrogen also reacts with semimetals at very high temperatures to produce covalent nitrides, such as Si N and BN, which are solids with extended covalent network structures similar to those of graphite or diamond. Consequently, they are usually high melting and chemically inert materials. Ammonia (NH ) is one of the few thermodynamically stable binary compounds of nitrogen with a nonmetal. It is not flammable in air, but it burns in an O atmosphere: \[\ce{4NH3(g) + 3O2(g) \rightarrow 2N2(g) + 6H2O(g)} \label{Eq3}\] About 10% of the ammonia produced annually is used to make fibers and plastics that contain amide bonds, such as nylons and polyurethanes, while 5% is used in explosives, such as ammonium nitrate, TNT (trinitrotoluene), and nitroglycerine. Large amounts of anhydrous liquid ammonia are used as fertilizer. Nitrogen forms two other important binary compounds with hydrogen. Hydrazoic acid (HN ), also called hydrogen azide, is a colorless, highly toxic, and explosive substance. Hydrazine (N H ) is also potentially explosive; it is used as a rocket propellant and to inhibit corrosion in boilers. B, C, and N all react with transition metals to form interstitial compounds that are hard, high-melting materials. For each reaction, explain why the given products form when the reactants are heated. balanced chemical equations why the given products form Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the observed reaction products form. Predict the product(s) of each reaction and write a balanced chemical equation for each reaction. Like the heavier elements of group 14, the heavier pnicogens form catenated compounds that contain only single bonds, whose stability decreases rapidly as we go down the group. For example, phosphorus exists as multiple allotropes, the most common of which is white phosphorus, which consists of P tetrahedra and behaves like a typical nonmetal. As is typical of a molecular solid, white phosphorus is volatile, has a low melting point (44.1°C), and is soluble in nonpolar solvents. It is highly strained, with bond angles of only 60°, which partially explains why it is so reactive and so easily converted to more stable allotropes. Heating white phosphorus for several days converts it to red phosphorus, a polymer that is air stable, virtually insoluble, denser than white phosphorus, and higher melting, properties that make it much safer to handle. A third allotrope of phosphorus, black phosphorus, is prepared by heating the other allotropes under high pressure; it is even less reactive, denser, and higher melting than red phosphorus. As expected from their structures, white phosphorus is an electrical insulator, and red and black phosphorus are semiconductors. The three heaviest pnicogens—arsenic, antimony, and bismuth—all have a metallic luster, but they are brittle (not ductile) and relatively poor electrical conductors. As in group 14, the heavier group 15 elements form catenated compounds that contain only single bonds, whose stability decreases as we go down the group. The reactivity of the heavier pnicogens decreases as we go down the column. Phosphorus is by far the most reactive of the pnicogens, forming binary compounds with every element in the periodic table except antimony, bismuth, and the noble gases. Phosphorus reacts rapidly with O , whereas arsenic burns in pure O if ignited, and antimony and bismuth react with O only when heated. None of the pnicogens reacts with nonoxidizing acids such as aqueous HCl, but all dissolve in oxidizing acids such as HNO . Only bismuth behaves like a metal, dissolving in HNO to give the hydrated Bi cation. The reactivity of the heavier group 15 elements decreases as we go down the column. The heavier pnicogens can use energetically accessible 3d, 4d, or 5d orbitals to form dsp or d sp hybrid orbitals for bonding. Consequently, these elements often have coordination numbers of 5 or higher. Phosphorus and arsenic form halides (e.g., AsCl ) that are generally covalent molecular species and behave like typical nonmetal halides, reacting with water to form the corresponding oxoacids (in this case, H AsO ). All the pentahalides are potent Lewis acids that can expand their coordination to accommodate the lone pair of a Lewis base: \[\ce{AsF5(soln) + F^{−}(soln) \rightarrow AsF^{−}6(soln)} \label{Eq4}\] In contrast, bismuth halides have extended lattice structures and dissolve in water to produce hydrated ions, consistent with the stronger metallic character of bismuth. Except for BiF , which is essentially an ionic compound, the trihalides are volatile covalent molecules with a lone pair of electrons on the central atom. Like the pentahalides, the trihalides react rapidly with water. In the cases of phosphorus and arsenic, the products are the corresponding acids, \(\ce{H3PO3}\) and \(\ce{H3AsO3}\), where E is P or As: \[\ce{EX3(l) + 3H2O(l) \rightarrow H3EO3(aq) + 3HX(aq)} \label{Eq5}\] Phosphorus halides are also used to produce insecticides, flame retardants, and plasticizers. Phosphorus has the greatest ability to form π bonds with elements such as O, N, and C. With energetically accessible d orbitals, phosphorus and, to a lesser extent, arsenic are able to form π bonds with second-period atoms such as N and O. This effect is even more important for phosphorus than for silicon, resulting in very strong P–O bonds and even stronger P=O bonds. The first four elements in group 15 also react with oxygen to produce the corresponding oxide in the +3 oxidation state. Of these oxides, P O and As O have cage structures formed by inserting an oxygen atom into each edge of the P or As tetrahedron (part (a) in Figure \(\Page {2}\)), and they behave like typical nonmetal oxides. For example, P O reacts with water to form phosphorous acid (H PO ). Consistent with its position between the nonmetal and metallic oxides, Sb O is amphoteric, dissolving in either acid or base. In contrast, Bi O behaves like a basic metallic oxide, dissolving in acid to give solutions that contain the hydrated Bi ion. The two least metallic elements of the heavier pnicogens, phosphorus and arsenic, form very stable oxides with the formula E O in the +5 oxidation state (part (b) in Figure \(\Page {2}\). In contrast, Bi O is so unstable that there is no absolute proof it exists. The heavier pnicogens form sulfides that range from molecular species with three-dimensional cage structures, such as P S (part (c) in Figure \(\Page {2}\)), to layered or ribbon structures, such as Sb S and Bi S , which are semiconductors. Reacting the heavier pnicogens with metals produces substances whose properties vary with the metal content. Metal-rich phosphides (such as M P) are hard, high-melting, electrically conductive solids with a metallic luster, whereas phosphorus-rich phosphides (such as MP ) are lower melting and less thermally stable because they contain catenated P units. Many organic or organometallic compounds of the heavier pnicogens containing one to five alkyl or aryl groups are also known. Because of the decreasing strength of the pnicogen–carbon bond, their thermal stability decreases from phosphorus to bismuth. The thermal stability of organic or organometallic compounds of group 15 decreases down the group due to the decreasing strength of the pnicogen–carbon bond. For each reaction, explain why the given products form. balanced chemical equations why the given products form Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the reaction products form. Predict the products of each reaction and write a balanced chemical equation for each reaction. The reactivity of the heavier group 15 elements decreases down the group, as does the stability of their catenated compounds. In group 15, nitrogen and phosphorus behave chemically like nonmetals, arsenic and antimony behave like semimetals, and bismuth behaves like a metal. Nitrogen forms compounds in nine different oxidation states. The stability of the +5 oxidation state decreases from phosphorus to bismuth because of the inert-pair effect. Due to their higher electronegativity, the lighter pnicogens form compounds in the −3 oxidation state. Because of the presence of a lone pair of electrons on the pnicogen, neutral covalent compounds of the trivalent pnicogens are Lewis bases. Nitrogen does not form stable catenated compounds because of repulsions between lone pairs of electrons on adjacent atoms, but it does form multiple bonds with other second-period atoms. Nitrogen reacts with electropositive elements to produce solids that range from covalent to ionic in character. Reaction with electropositive metals produces ionic nitrides, reaction with less electropositive metals produces interstitial nitrides, and reaction with semimetals produces covalent nitrides. The reactivity of the pnicogens decreases with increasing atomic number. Compounds of the heavier pnicogens often have coordination numbers of 5 or higher and use dsp or d sp hybrid orbitals for bonding. Because phosphorus and arsenic have energetically accessible d orbitals, these elements form π bonds with second-period atoms such as O and N. Phosphorus reacts with metals to produce phosphides. Metal-rich phosphides are hard, high-melting, electrically conductive solids with metallic luster, whereas phosphorus-rich phosphides, which contain catenated phosphorus units, are lower melting and less thermally stable. | 20,219 | 2,454 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria/15.07%3A_Some_Applications_of_Entropy_and_Free_Energy |
Thermodynamics may appear at first to be a rather esoteric subject, but when you think about it, almost every chemical (and biological) process is governed by changes in entropy and free energy. Examples such as those given below should help you connect these concepts with the real world. Vapor pressure lowering, boiling point elevation, freezing point depression and osmosis are well-known phenomena that occur when a non-volatile solute such as sugar or a salt is dissolved in a volatile solvent such as water. , and because of this commonality they are referred to as (Lat. , connected to.) The key role of the solvent concentration is obscured by the expressions used to calculate the magnitude of these effects, in which only the solute concentration appears. The details of how to carry out these calculations and the many important applications of colligative properties are covered in the unit on solutions. Our purpose here is to offer a more complete explanation of why these phenomena occur. The conventional explanation is that dilution of a liquid by a non-volatile solute reduces the vapor pressure or "escaping tendency" of the liquid in that phase, leading to a net transport of material into that phase. Equilibrium can then be re-established by adjusting the temperature of applying a hydrostatic pressure to the solution. A more fundamental approach is to recall that dilution of a liquid creates uncountable numbers of new microstates, increasing the density of quantum states in the solution compared to that in the pure liquid. To the extent that these new states are thermally accessible, they will become populated at the expense of some of the microstates of the other phase. Equilibrium between two phases can be restored by adjusting the temperature or pressure so that equal numbers of microstates are occupied in each phase. These effects are readily understood in terms of the schematic diagrams shown below. The red shading indicates the temperatures required to make equal numbers of microstates thermally accessible in the two phases. Dilution of the solvent adds new energy states to the liquid, but does not affect the vapor phase. This raises the temperature required to make equal numbers of microstates accessible in the two phases. Dilution of the solvent adds new energy states to the liquid, but does not affect the solid phase. This reduces the temperature required to make equal numbers of states accessible in the two phases. As mentioned previously, the more conventional explanation of bp elevation and fp depression is given in terms of vapor pressures. The latter are measures of the free energies of molecules in a phase, and the relationships are best understood by showing plots of – Δ ° for each phase in terms of the temperature:
.
The solute "dilutes" the solvent, reducing its free energy (purple line) and shifting the transition temperatures in opposite directions. When a liquid is subjected to hydrostatic pressure— for example, by an inert, non-dissolving gas that occupies the vapor space above the surface, the vapor pressure of the liquid is raised. The pressure acts to compress the liquid very slightly, effectively narrowing the potential energy well in which the individual molecules reside and thus increasing their tendency to escape from the liquid phase. (Because liquids are not very compressible, the effect is quite small; a 100-atm applied pressure will raise the vapor pressure of water at 25° C by only about 2 torr.) In terms of the entropy, we can say that the applied pressure reduces the dimensions of the "box" within which the principal translational motions of the molecules are confined within the liquid, thus reducing the density of the quantized energy states in the liquid phase. . Applying hydrostatic pressure to a liquid increases the spacing of its microstates, so that the number of energetically accessible states in the gas, although unchanged, is relatively greater— thus increasing the tendency of molecules to escape into the vapor phase. In terms of free energy, the higher pressure raises the free energy of the liquid, but does not affect that of the gas phase. Molecules in a fluid are always subject to random thermal motions. If a given kind of molecule is present in greater concentration in one region of a fluid, the effect of these random motions will be to produce a net migration into the region of lower concentration; this migration is known as . (commonly known simply as ) occurs when molecules of a solvent diffuse through a membrane that is permeable only to those molecules, and into a solution in which the solvent is "diluted" by the presence of solute molecules. In the simplest example, there is pure solvent on the left side of the membrane; on the other side the same solvent contains a solute whose molecules are too large to pass through the membrane. Because the solvent concentration on the right side will always be smaller than that on the left, osmotic flow will continue indefinitely if the right side can accommodate the increased volume. If, however, the liquid on the right side of the membrane is placed under hydrostatic pressure, the driving force for osmotic flow will diminish. If the pressure is raised sufficiently high, osmotic flow will come to a halt; the system is then said to be in osmotic equilibrium. The pressure required to bring about osmotic equilibrium (and thus to stop osmotic flow) is known as the , commonly denoted by Π (Greek capital ). Osmotic equilibrium, like any kind of equilibrium, occurs when the free energies (that is, the escaping tendencies) of the diffusible molecules are the same on the two sides of the membrane. The free energy on the right side, initially lower due to the lower solvent concentration, is raised by the hydrostatic pressure, making it identical with that of the pure liquid on the left. From the standpoint of microscopic energy states, the effect of the applied pressure is to slightly increase the spacing of solvent energy states on right side of the membrane so as to equalize the number of accessible states on the two sides, as shown here in a very schematic way. Since ancient times, the recovery of metals from their ores has been one of the most important applications of chemistry to civilization and culture. The oldest, and still the most common smelting process for oxide ores involves heating them in the presence of carbon. Originally, charcoal was used, but industrial-scale smelting uses coke, a crude form of carbon prepared by pyrolysis (heating) of coal. The basic reactions are: \[MO + C \rightleftharpoons M + CO \label{2.1}\] \[MO + ½ O_2 \rightleftharpoons M + ½ CO_2 \label{2.2}\] \[MO + CO \rightleftharpoons M + CO_2 \label{2.3}\] Each of these can be regarded as a pair of in which the metal M and the carbon are effectively competing for the oxygen atom. Using reaction \(\ref{2.1}\) as an example, it can be broken down into the following two parts: \[MO \rightleftharpoons M + ½ O_2 \;\;\; ΔG^o > 0 \label{2.4}\] \[C + ½ O_2 \rightleftharpoons CO \;\;\; ΔG^o < 0 \label{2.5}\] At ordinary environmental temperatures, reaction \(\ref{2.4}\) is always spontaneous in the reverse direction (that is why ores form in the first place!), so Δ ° of Reaction \(\ref{2.4}\) will be positive. Δ ° for reaction \(\ref{2.5}\) is always negative, but at low temperatures it will not be sufficiently negative to drive \(\ref{2.4}\). The process depends on the different ways in which the free energies of reactions like \(\ref{2.4}\) and \(\ref{2.4}\) vary with the temperature. This temperature dependence is almost entirely dominated by the Δ ° term in the Gibbs function, and thus by the entropy change. The latter depends mainly on Δ , the change in the number of moles of gas in the reaction. Removal of oxygen from the ore is always accompanied by a large increase in the system volume so Δ for this step is always positive and the reaction becomes more spontaneous at higher temperatures. The temperature dependences of the reactions that take up oxygen vary, however (Table 23.6.X). A plot of the temperature dependences of the free energies of these reactions, superimposed on similar plots for the oxygen removal reactions \(\ref{2.4}\) is called an . For a given oxide MO to be smeltable, the temperature must be high enough that reaction \(\ref{2.4}\) falls below that of at least one of the oxygen-consuming reactions. The slopes of the lines on this diagram are determined by the sign of the entropy change. Examination of the Ellingham diagram shown above illustrates why the metals known to the ancients were mainly those such as copper and lead, which can be obtained by smelting at the relatively low temperatures that were obtainable by the methods available at the time in which a charcoal fire supplied both the heat and the carbon. Thus the bronze age preceded the iron age; the latter had to await the development of technology capable of producing higher temperatures, such as the blast furnace. Smelting of aluminum oxide by carbon requires temperatures too high to be practical; commercial production of aluminum is accomplished by electrolysis of the molten ore. Many of the reactions that take place in living organisms require a source of free energy to drive them. The immediate source of this energy in heterotrophic organisms, which include animals, fungi, and most bacteria, is the sugar . Oxidation of glucose to carbon dioxide and water is accompanied by a large negative free energy change \[\ce{C6H12O6 + 6 O2 \rightarrow 6 CO2 + 6 H2O} \quad ΔG^o = – 2,880 \,kJ\, mol^{–1} \label{6-1}\] Of course it would not do to simply “burn” the glucose in the normal way; the energy change would be wasted as heat, and rather too quickly for the well-being of the organism! Effective utilization of this free energy requires a means of capturing it from the glucose and then releasing it in small amounts when and where it is needed. This is accomplished by breaking down the glucose in a series of a dozen or more steps in which the energy liberated in each stage is captured by an “energy carrier” molecule, of which the most important is , known as . At each step in the breakdown of glucose, an ADP molecule reacts with inorganic phosphate (denoted by P ) and changes into : \[ADP + P_i \rightarrow ATP\] \[ΔG^° = +30 \;kJ\; mol^{–1} \label{(6-2}\] The 30 kJ mol of free energy stored in each ATP molecule is released when the molecule travels to a site where it is needed and loses one of its phosphate groups, yielding inorganic phosphate and ADP, which eventually finds its way back the site of glucose metabolism for recycling back into ATP. The complete breakdown of one molecule of glucose is coupled with the production of 38 molecules of ATP according to the overall reaction \[\ce{C6H12O6 + 6 O2 + 38 Pi + 38 ADP 38 ATP + 6CO2 + 44 H2O} \label{6-3}\] For each mole of glucose metabolized, 38 × (30 kJ) = 1140 kJ of free energy is captured as ATP, representing an energy efficiency of 1140/2880 = 0.4. That is, Where does the glucose come from? Animals obtain their glucose from their food, especially cellulose and starches that, like glucose, have the empirical formula {CH O}. Animals obtain this food by eating plants or other animals. Ultimately, all food comes from plants, most of which are able to make their own glucose from CO and H O through the process of photosynthesis. This is just the reverse of Eq 40 in which the free energy is supplied by the quanta of light absorbed by chlorophyll and other photosynthetic pigments. This describes , which evolved after the development of photosynthetic life on Earth began to raise the concentration of atmospheric oxygen. Oxygen is a poison to most life processes at the cellular level, and it is believed that aerobic respiration developed as a means to protect organisms from this peril. Those that did not adapt have literally “gone underground” and constitute the more primitive anaerobic bacteria. The function of oxygen in respiration is to serve as an acceptor of the electrons that glucose loses when it undergoes oxidation. Other electron acceptors can fulfill the same function when oxygen is not available, but none yields nearly as much free energy. For example, if oxygen cannot be supplied to mammalian muscle cells as rapidly as it is needed, they switch over to an anaerobic process yielding lactic acid instead of CO : \[\ce{C6H12O6 + 2 ADP → 2 CH3CH(OH)COOH} \quad ΔG^o = –218\, kJ\, mol^{–1} \label{6-4}\] In this process, only (2 × 30 kJ) = 60 kJ of free energy is captured, so the efficiency is only 28% on the basis of this reaction, and it is even lower in relation to glucose. In “aerobic” exercising, one tries to maintain sufficient lung capacity and cardiac output to supply oxygen to muscle cells at a rate that promotes the aerobic pathway. Respiration and many other metabolic processes involve electron-transfer reactions. According to the widely useful , an acid is a proton donor and a base is a proton acceptor. In 1953, Gurney showed how this idea could be made even more useful by placing acid-base conjugate pairs on a . In this view, acids are and bases are . Protons fall spontaneously from acids to fill sinks in which the proton free energy levels are lower. The pH is a measure of the average proton free energy in the solution; when this quantity is the same as the proton free energy level of a conjugate pair, the two species are present in equal concentrations (this corresponds, of course to the equality of pH and pK in the conventional theory.) The proton-free energy concept is commonly employed in aquatic environmental chemistry in which multiple acid-base systems must be dealt with on a semi-quantitative bases. It is, however, admirably adapted to any presentation of acid-base chemistry, even at the first-year college level, and it seems a shame that it never seems to have made its way into the ordinary curriculum. Oxidation-reduction reactions proceed in a direction that allows the electron to “fall” (in free energy) from a “source” to a “sink”. Later on when you study electrochemistry you will see how this free energy can manifest itself as an electrical voltage and be extracted from the system as electrical work. You may already have seen an electromotive force table that shows the relative tendencies of different reducing agents to donate electrons. The same information can be presented in another way that relates electron-donating (reducing) power to the "fall" in free energy that electrons undergo when they are transferred to an oxidizing agent. Any available sink on the right side will tend to drain electrons from a source above it. For example, immersion of metallic zinc in a solution of CuSO allow electrons to "fall" from the high free energy they possess in Zn down to the lower free energy level in the newly-introduced Cu ions. This will result in the reduction of Cu to metallic copper and the oxidation of the zinc (red arrows.) Similarly, addition of chlorine to water will introduce a new electron sink (Cl ) that lies below the free energy of the accessible electrons in H O, draining them away and producing O and Cl (blue arrows.) Note especially the positions of the H / H and H O/O ,H couples on this chart, as they define the range of °s for substances that will not decompose water. The zero for ° is arbitrarily set at the electron activity at which the H /H couple is at equilibrium; this corresponds to ° = 0 volts on the ordinary electromotive scale. \[\{CH_2O\} → CO_2 + H_2O \label{6-5}\] In the above chart, the zero for ° is arbitrarily set at the electron free energy required to maintain the H /H couple is at equilibrium; this corresponds to E° = 0 volts on the ordinary electromotive scale. Organisms that live in environments where oxygen is lacking, such as marshes, muddy soils, and the intestinal tracts of animals, must utilize other electron acceptors to extract free energy from carbohydrate. A wide variety of inorganic ions such as sulfate and nitrate, as well as other carbon compounds can serve as electron acceptors, yielding the gaseous products like H S, NH and CH which are commonly noticed in such locations. From the location of these acceptors on the scale, it is apparent that the amount of energy they can extract from a given quantity of carbohydrate is much less than for O . One reason that aerobic organisms have dominated the earth is believed to be the much greater energy-efficiency of oxygen as an electron acceptor. What did organisms use for food before there was a widespread supply of carbohydrate in the world? Any of the electron sources near the upper left of the table can in theory serve this function, although at reduced energy efficiency. As a matter of fact, there are still a number of these autotrophic bacteria around whose “food” is CH , CH OH, FeCO , and even H ! Rubber is composed of random-length chains of polymerized molecules. The poly(isoprene) chains are held together partly by weak intermolecular forces, but are joined at irregular intervals by covalent disulfide bonds so as to form a network. The intermolecular forces between the chain fragments tend to curl them up, but application of a tensile force can cause them to elongate. The disulfide cross-links prevent the chains from slipping apart from one another, thus maintaining the physical integrity of the material. Without this cross-linking, the polymer chains would behave more like a pile of spaghetti. The ability of rubber bands and other elastic substances to undergo a change in physical dimensions in response to a change in the applied stretching force is subject to the same laws of thermodynamics as any other physical process. You can investigate this for yourself. Hold a rubber band (the thicker the better) against your upper lip, and notice how the temperature changes when the band is stretched, and then again when it is allowed to contract. When an ordinary material is placed under tension, the strain energy is taken up by bond distortions and is stored as electrostatic potential energy which rises very rapidly so as to greatly inhibit further elongation. In rubber-like polymers, this does not happen; the strain energy is instead stored as thermal (kinetic) energy. Free polymer chains naturally tend to curl up in complex and ever-changing ways as thermal energy allows random bond rotations to take place. In a rubber-like material in its relaxed state, the portions of the polymer chains between cross-links are continually jumping between different randomly-coiled configurations. When the rubber is stretched, the polymer segments straighten out as the applied force overcomes the weak dispersion force interactions that caused the strands to curl. Each chain segment is pulled into an almost-straight conformation, thus greatly reducing the quantity of thermal energy it can store. The excess thermal energy spreads into the material and is lost in the form of heat. When the rubber relaxes, the polymer strands curl up again and soak up thermal energy. The spontaneous contraction of rubber is largely an entropy-driven process. The number of energetically-equivalent ways of distributing thermal energy amongst the nearly-linear polymer chains of the stretched state of rubber is insignificant compared to those available when the chains are curled up in random ways, so the un-stretched form of rubber is statistically the most likely one by overwhelming odds. As noted in part (b) of the above problem example, the gain in entropy when the rubber contracts drives Δ more negative at higher temperatures. This means that a rubber band, held at constant tension in stretched state, will contract when it is heated. This fact can be put to use in an interesting way. Replace the spokes of a bicycle wheel with rubber bands, and shine a heat lamp on one side of the wheel. The contraction of the heated bands will shift the wheel off-center, causing it to rotate. This rotation will continue indefinitely as long as the heat source is present. The device has become a heat engine whose working "fluid" is rubber! This recalls the classic perpetual motion machine design in which a wheel is caused to rotate by continually-shifting unbalanced weights (the one depicted here uses hinged vials of mercury). That, as we saw, would violate the Second Law by producing work in a cyclic process without degrading heat to a lower temperature. The rubber-band heat engine avoids this pitfall by absorbing heat from an external source on one side of the wheel, and releasing it at a lower temperature on the unheated side. | 20,746 | 2,455 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/01%3A_Introduction_to_Analytical_Chemistry/1.01%3A_What_is_Analytical_Chemistry |
This quote is attributed to C. N. Reilly (1925-1981) on receipt of the 1965 Fisher Award in Analytical Chemistry. Reilly, who was a professor of chemistry at the University of North Carolina at Chapel Hill, was one of the most influential analytical chemists of the last half of the twentieth century. For another view of what constitutes analytical chemistry, see the article “Quo Vadis, Analytical Chemistry?”, the full reference for which is Valcárcel, M. , , 13-21. Let’s begin with a deceptively simple question: What is analytical chemistry? Like all areas of chemistry, analytical chemistry is so broad in scope and so much in flux that it is difficult to find a simple definition more revealing than that quoted above. In this chapter we will try to expand upon this simple definition by saying a little about what analytical chemistry is, as well as a little about what analytical chemistry is not. Analytical chemistry often is described as the area of chemistry responsible for characterizing the composition of matter, both qualitatively (Is there lead in this paint chip?) and quantitatively (How much lead is in this paint chip?). As we shall see, this description is misleading. Most chemists routinely make qualitative and quantitative measurements. For this reason, some scientists suggest that analytical chemistry is not a separate branch of chemistry, but simply the application of chemical knowledge [Ravey, M. , , , 11]. In fact, you probably have performed many such quantitative and qualitative analyses in other chemistry courses. You might, for example, have determined the concentration of acetic acid in vinegar using an acid–base titration, or used a qual scheme to identify which of several metal ions are in an aqueous sample. Defining analytical chemistry as the application of chemical knowledge ignores the unique perspective that an analytical chemist bring to the study of chemistry. The craft of analytical chemistry is found not in performing a routine analysis on a routine sample—a task we appropriately call chemical analysis—but in improving established analytical methods, in extending these analytical methods to new types of samples, and in developing new analytical methods to measure chemical phenomena [de Haseth, J. , , , 11]. Here is one example of the distinction between analytical chemistry and chemical analysis. A mining engineers evaluates an ore by comparing the cost of removing the ore from the earth with the value of its contents, which they estimate by analyzing a sample of the ore. The challenge of developing and validating a quantitative analytical method is the analytical chemist’s responsibility; the routine, daily application of the analytical method is the job of the chemical analyst. Steps 1–3 and 5 are the province of analytical chemistry; step 4 is the realm of chemical analysis. The seven stages of an analytical method listed here are modified from Fassel, V. A. , , 511–518 and Hieftje, G. M. , , 577–583. Another difference between analytical chemistry and chemical analysis is that an analytical chemist works to improve and to extend established analytical methods. For example, several factors complicate the quantitative analysis of nickel in ores, including nickel’s unequal distribution within the ore, the ore’s complex matrix of silicates and oxides, and the presence of other metals that may interfere with the analysis. Figure 1.1.1
outlines one standard analytical method in use during the late nineteenth century [Fresenius. C. R. ; John Wiley and Sons: New York, 1881]. The need for many reactions, digestions, and filtrations makes this analytical method both time-consuming and difficult to perform accurately. The discovery, in 1905, that dimethylglyoxime (dmg) selectively precipitates Ni and Pd led to an improved analytical method for the quantitative analysis of nickel [Kolthoff, I. M.; Sandell, E. B. , 3rd Ed., The Macmillan Company: New York, 1952]. The resulting analysis, which is outlined in Figure 1.1.2
, requires fewer manipulations and less time. By the 1970s, flame atomic absorption spectrometry replaced gravimetry as the standard method for analyzing nickel in ores, resulting in an even more rapid analysis [Van Loon, J. C. , Academic Press: New York, 1980]. Today, the standard analytical method utilizes an inductively coupled plasma optical emission spectrometer. Perhaps a more appropriate description of analytical chemistry is “the science of inventing and applying the concepts, principles, and...strategies for measuring the characteristics of chemical systems” [Murray, R. W. , , 271A]. Analytical chemists often work at the extreme edges of analysis, extending and improving the ability of all chemists to make meaningful measurements on smaller samples, on more complex samples, on shorter time scales, and on species present at lower concentrations. Throughout its history, analytical chemistry has provided many of the tools and methods necessary for research in other traditional areas of chemistry, as well as fostering multidisciplinary research in, to name a few, medicinal chemistry, clinical chemistry, toxicology, forensic chemistry, materials science, geochemistry, and environmental chemistry. To an analytical chemist, the process of making a useful measurement is critical; if the measurement is not of central importance to the work, then it is not analytical chemistry. You will come across numerous examples of analytical methods in this textbook, most of which are routine examples of chemical analysis. It is important to remember, however, that nonroutine problems prompted analytical chemists to develop these methods. An editorial in entitled “Some Words about Categories of Manuscripts” highlights nicely what makes a research endeavor relevant to modern analytical chemistry. The full citation is Murray, R. W. , , 4775; for a more recent editorial, see “The Scope of ” by Sweedler, J. V. et. al. , , 6425. | 5,997 | 2,457 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/Carboxyl_Substitution/CX2._General_Reactivity_Patterns |
Just as a simple carbonyl is an electrophile, so is a carboxyloid. Carboxyloids react with many of the same nucleophiles that react with aldehydes and ketones. The overall result of the reaction is very different. However, the mechanism of the reaction is really quite similar. Nucleophiles have the overall effect of adding across the carbonyl group of an aldehyde or ketone. Addition of a proton produces an alcohol. The C=O bond is parlty broken to a C-O bond. In some cases, the C=O bond is completely broken and after several steps the carbonyl oxygen is replaced by some other heteroatom. On the other hand, nucleophiles add to carboxyloids and end up replacing the heteroatom next to the carbonyl. The carbonyl itself remains intact after the reaction is complete. The general pattern in carboxyloid chemistry is for nucleophiles to substitute for the heteroatomic group next to the carbonyl. In these reactions, the group next to the carbonyl is sometimes referred to as a "leaving group". That term simply means that this group has been replaced by the end of the reaction. Note that there is no reason to believe that the initial elementary reaction between a carboxyloid and a nucleophie is any different than that of a simple carbonyl with a nucleophile. An examination of the frontier orbitals in a carboxyloid suggests the pi antibonding level would be the site of population by a nucleophilic lone pair. The carbonyl pi bond would break as the nucleophile approaches. However, an important feature of carbonyl chemistry is that two heteroatoms on one tetrahedral carbon cannot last. One always pushes the other off. If the former carbonyl oxygen pushes off the nucleophile, the system returns to the starting materials. However, if the former carbonyl oxygen displaces the heteroatomic group next to it, there is an overall change in bonding and a different product is formed. The net result is replacement of the group next to the carbonyl. The overall result of reaction with a carboxyloid is to displace the heteroatom group next to the carbonyl. This group is typically liberated as an anion. Suggest an order of reactivity for the carboxyloids: rank them from most reactive to least reactive. Provide a reason for your trend. , | 2,260 | 2,458 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/01%3A_Introduction_to_Analytical_Chemistry/1.01%3A_What_is_Analytical_Chemistry |
This quote is attributed to C. N. Reilly (1925-1981) on receipt of the 1965 Fisher Award in Analytical Chemistry. Reilly, who was a professor of chemistry at the University of North Carolina at Chapel Hill, was one of the most influential analytical chemists of the last half of the twentieth century. For another view of what constitutes analytical chemistry, see the article “Quo Vadis, Analytical Chemistry?”, the full reference for which is Valcárcel, M. , , 13-21. Let’s begin with a deceptively simple question: What is analytical chemistry? Like all areas of chemistry, analytical chemistry is so broad in scope and so much in flux that it is difficult to find a simple definition more revealing than that quoted above. In this chapter we will try to expand upon this simple definition by saying a little about what analytical chemistry is, as well as a little about what analytical chemistry is not. Analytical chemistry often is described as the area of chemistry responsible for characterizing the composition of matter, both qualitatively (Is there lead in this paint chip?) and quantitatively (How much lead is in this paint chip?). As we shall see, this description is misleading. Most chemists routinely make qualitative and quantitative measurements. For this reason, some scientists suggest that analytical chemistry is not a separate branch of chemistry, but simply the application of chemical knowledge [Ravey, M. , , , 11]. In fact, you probably have performed many such quantitative and qualitative analyses in other chemistry courses. You might, for example, have determined the concentration of acetic acid in vinegar using an acid–base titration, or used a qual scheme to identify which of several metal ions are in an aqueous sample. Defining analytical chemistry as the application of chemical knowledge ignores the unique perspective that an analytical chemist bring to the study of chemistry. The craft of analytical chemistry is found not in performing a routine analysis on a routine sample—a task we appropriately call chemical analysis—but in improving established analytical methods, in extending these analytical methods to new types of samples, and in developing new analytical methods to measure chemical phenomena [de Haseth, J. , , , 11]. Here is one example of the distinction between analytical chemistry and chemical analysis. A mining engineers evaluates an ore by comparing the cost of removing the ore from the earth with the value of its contents, which they estimate by analyzing a sample of the ore. The challenge of developing and validating a quantitative analytical method is the analytical chemist’s responsibility; the routine, daily application of the analytical method is the job of the chemical analyst. Steps 1–3 and 5 are the province of analytical chemistry; step 4 is the realm of chemical analysis. The seven stages of an analytical method listed here are modified from Fassel, V. A. , , 511–518 and Hieftje, G. M. , , 577–583. Another difference between analytical chemistry and chemical analysis is that an analytical chemist works to improve and to extend established analytical methods. For example, several factors complicate the quantitative analysis of nickel in ores, including nickel’s unequal distribution within the ore, the ore’s complex matrix of silicates and oxides, and the presence of other metals that may interfere with the analysis. Figure 1.1.1
outlines one standard analytical method in use during the late nineteenth century [Fresenius. C. R. ; John Wiley and Sons: New York, 1881]. The need for many reactions, digestions, and filtrations makes this analytical method both time-consuming and difficult to perform accurately. The discovery, in 1905, that dimethylglyoxime (dmg) selectively precipitates Ni and Pd led to an improved analytical method for the quantitative analysis of nickel [Kolthoff, I. M.; Sandell, E. B. , 3rd Ed., The Macmillan Company: New York, 1952]. The resulting analysis, which is outlined in Figure 1.1.2
, requires fewer manipulations and less time. By the 1970s, flame atomic absorption spectrometry replaced gravimetry as the standard method for analyzing nickel in ores, resulting in an even more rapid analysis [Van Loon, J. C. , Academic Press: New York, 1980]. Today, the standard analytical method utilizes an inductively coupled plasma optical emission spectrometer. Perhaps a more appropriate description of analytical chemistry is “the science of inventing and applying the concepts, principles, and...strategies for measuring the characteristics of chemical systems” [Murray, R. W. , , 271A]. Analytical chemists often work at the extreme edges of analysis, extending and improving the ability of all chemists to make meaningful measurements on smaller samples, on more complex samples, on shorter time scales, and on species present at lower concentrations. Throughout its history, analytical chemistry has provided many of the tools and methods necessary for research in other traditional areas of chemistry, as well as fostering multidisciplinary research in, to name a few, medicinal chemistry, clinical chemistry, toxicology, forensic chemistry, materials science, geochemistry, and environmental chemistry. To an analytical chemist, the process of making a useful measurement is critical; if the measurement is not of central importance to the work, then it is not analytical chemistry. You will come across numerous examples of analytical methods in this textbook, most of which are routine examples of chemical analysis. It is important to remember, however, that nonroutine problems prompted analytical chemists to develop these methods. An editorial in entitled “Some Words about Categories of Manuscripts” highlights nicely what makes a research endeavor relevant to modern analytical chemistry. The full citation is Murray, R. W. , , 4775; for a more recent editorial, see “The Scope of ” by Sweedler, J. V. et. al. , , 6425. | 5,997 | 2,459 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/09%3A_Food_to_energy_metabolic_pathways/9.05%3A_Oxidative_phosphorylation |
In the first and the second stage of food catabolism, the energy is transferred from food to high-energy molecules, such as energetic electron donors \(\ce{NADH}\) and \(\ce{FADH}\). The electrons in \(\ce{NADH}\) and \(\ce{FADH}\) are transported through a series of enzymes located in the inner membrane of mitochondria and ultimately reduce reducing \(\ce{O2}\) into \(\ce{H2O}\). In each step, the electrons are transferred from a substance with a higher reduction potential to one with a lower reduction potential. Energy is released in each step proportional to the difference in the reduction potential. The energy is converted to electrochemical potential energy by pumping protons (\(\ce{H^{+}}\)) from the matrix to intermembrane spaces through sets of enzymes called complexes I, III, & IV. The electrochemical energy is harnessed by \(\ce{ATP}\) synthase which allows protons to flow back from the intermembrane space to the matrix and uses the energy released to synthesize higher energy \(\ce{ATP}\) by phosphorylation lower energy \(\ce{ADP}\) with inorganic phosphate (\(\ce{P_i}\)). These interconnected processes of electron transport and \(\ce{ATP}\) synthesis are collectively called . It happens in the mitochondria of cells as illustrated in Figure \(\Page {1}\) and is explained in the following sections. In addition to \(\ce{NAD^{+}}\) and \(\ce{FAD}\) that transport electrons and protons in metabolic pathways, the following molecules are involved in the transport of electrons and protons in the electron transport chain. The first electron acceptor in the electron transport chain is which changes from its oxidized form represented as \(\ce{FMN}\) to its reduced form represented as \(\ce{FMNH2}\) as illustrated in figure on the right. \(\ce{FMNH2}\) pass on the electrons to iron-sulfur clusters, which include \(\ce{[2Fe-2S]}\) and \(\ce{[4Fe-4S]}\) clusters. The redox couple in iron-sulfur clusters is \(\ce{Fe^{3+}/Fe^{2+}}\), i.e., the following redox reaction: \(\ce{Fe^{3+} + e^{-} <=>[Reduction,Oxidation] Fe^{2+}}\). Finally, the electrons are transferred to coenzyme-Q which in oxidized form is ubiquinone represented as \(\ce{Q}\) and in its reduced form is ubiquinol represented as \(\ce{QH2}\), as illustrated in the figure above on left. Coenzyme-Q is hydrophobic, i.e., lipid-soluble, and moves freely within the inner membrane to pass on electrons from complex I and II to complex III. Coenzyme-Q not only can accept two electrons and two protons, but it can also accept one electron tp become semubiquinone which is a resonance stabilized radical anion. Cytochrome c is another electron carrier protein with a heme group attached. Heme group is illustrated in the figure on the right. Iron in the heme group of cytochrome c is the redox couple (\(\ce{Fe^{3+}/Fe^{2+}}\)) as in the case of iron-sulfur clusters. \(\ce{Fe^{3+}}\)) changes to \(\ce{Fe^{2+}}\)) when an electron is added to it and the reverse happens when an electron is removed. Cytochrome c is water soluble and travels between complex III and complex IV along the surface of the inner membrane facing intermembrane space. This is stage 3 of food catabolism. In this stage, electrons are removed from \(\ce{NADH}\) and \(\ce{FADH}\) and passed on to reduce \(\ce{O2}\) into \(\ce{H2O}\) through a series of enzyme complexes, i.e., complexes I through IV and mobile electron carriers like ubiquinone (\(\ce{Q}\)) and cytochrome c (\(\ce{Cyt c}\)) by a process called . At each step, electrons pass from a substance of higher reduction potential to one with a lower reduction potential and energy is released proportional to the difference in the reduction potentials. The complex I, II, and III extend from the matrix to the intermembrane space through the inner membrane, as illustrated in Figure \(\Page {1}\). These three complexes utilize the energy released as the electrons go through the series of redox reactions to pump protons (\(\ce{H^{+}}\)) from the matrix to the intermembrane space. Complex II extends to the matrix from the inner membrane. Complex II transports electrons but do not pump protons. Coenzyme-Q (\(\ce{Q}\)) is a hydrophobic electron carrier molecule that moves inside the lipid bilayer of the inner membrane. Cytochrome c (\(\ce{Cyt c}\)) is a hydrophilic electron carrier molecule that moves along the surface of the inner membrane facing the intermembrane space. The enzyme complexes and the mobile electron carriers involved in the electron transport chain and \(\ce{ATP}\) synthesis are described in the following sub-sections where the chemical equations and figures are taken from the public domain source via . Complex I, also known as NADH-coenzyme Q oxidoreductase or NADH dehydrogenase, is the first enzyme complex in the electron transport chain. Complex I is illustrated in the figure on the right. The electron transport begins when an NADH binds with and donates two electrons to complex I. The electrons are received by flavin mononucleotide (\(\ce{FMN}\)) attached to the complex which is reduced to \(\ce{FMNH2}\) form. Then the electrons are transferred through a series of iron-sulfur clusters to coenzyme-Q (\(\ce{Q}\)) that, in turn, reduces to its reduced form \(\ce{QH2}\). As the electrons move through complex I, four protons (\(\ce{4H^{+}}\)) are pumped from the matrix to intermembrane space. It generated polarity across the inner membrane with the positive side (P-side) facing intermembrane space and the negative side (N-side) facing the matrix. The exact mechanics of proton pumping is not yet clearly understood, but the research indicates that it happens through conformational changes in complex I such that the protein bind protons on the N-side and releases them on the P-side of the membrane. The overall redox reaction in complex I is the following. \(\ce{NAD^{+}}\) librated in this reaction becomes available to oxidize more substrates in the metabolic pathways. Complex II, also known as succinate-Q oxidoreductase or succinate dehydrogenase, is a second entry point to the electron transport chain. It is part of both the citric acid cycle and the electron transport chain. It contains protein subunits bonded with flavin adenine dinucleotide \(\ce{FAD}\), iron-sulfur clusters, and a heme group as illustrated in the figure on the right. Succinate is oxidized to fumarate in the citric acid cycle at the expense of oxidation of \(\ce{FAD}\) to \(\ce{FADH2}\), which, in turn, passes on the electrons and protons through iron-sulfur clusters to coenzyme-Q by the following overall reaction. Since oxidation \(\ce{FADH2}\) to \(\ce{FAD}\) at complex II releases less energy than oxidation of \(\ce{NADH}\) to \(\ce{NAD^{+}}\) at complex I, complex II do not pump protons from matrix to intermembrane space. Coenzyme-Q passes on the electrons from complex II to complex III as it does from complex I to complex III. Complex III, also know as cytochrome C reductase or cytochrome bc complex contains protein subunits, an iron-sulfur cluster, and cytochromes. A cytochrome is a protein with at least one heme group that transfers one electron by accepting and releasing an electron based on the following reaction: \(\ce{Fe^{3+} + e^{-} <=> Fe^{2+}}\) Complex III passes on electrons from coenzyme-Q (\(\ce{QH2}\)) to cytochrome c (\(\ce{Cyt~ c}\)). Since \(\ce{QH2}\) delivers two electrons (\(\ce{e^{-}}\)) while (\(\ce{Cyt~ c}\) can accept only one, the process happens in two steps, as illustrated in the figure on the right. In the first step, \(\ce{QH2}\) transfers one \(\ce{e^{-}}\) to \(\ce{Cyt~ c}\) and one to another coenzyme-Q in quinone (\(\ce{Q}\)) form and releases (\(\ce{2H^{+}}\). \(\ce{Q}\) converts to a semiubiquinone (\(\ce{Q^{.-}}\)). In the second step, another \(\ce{QH2}\) transfers one electron to a second \(\ce{Cyt~ c}\) and one to \(\ce{Q^{.-}}\) converting it to \(\ce{Q}\) which is released from the complex and repeats the above two steps. In the overall reaction, two electrons are transferred from a \(\ce{Q}\) to two \(\ce{Cyt~ c}\) and \(\ce{4H^{+}}\) are pumped out into intermembrane space by complex III as shown in the following reaction. \(\ce{Cyt~ c}\) carries the electron to complex IV. Complex IV, also known as cytochrome c oxidase, is the final complex in the electron transport chain. It contains several subunits, two heme groups, and several metal ion cofactors that include three atoms of copper, one magnesium, and one zinc. Complex IV reduces oxygen into water at the expense of oxidation of \(\ce{Cyt~c}\) and pumps \(\ce{4H^{+}}\) from matrix to intermembrane space as illustrated in the figure on the right and shown in the following overall reaction. \(\ce{ATP}\)-synthase, also known as complex V in the final complex in the oxidative phosphorylation process which is stage 4 of food catabolism. It is composed of two parts in the shape of a mushroom as illustrated in the figure on the left. The first part, called F is the hydrophobic part embedded in the inner membrane (blue and purple region in the model image on the right (Copyright; Alex.X, CC BY- , via Wikimedia Commons). The F acts as a pore for the flow of protons (\(\ce{H^{+}}\)) across the membrane. The second part is F which is hydrophilic and spheroidal and protrudes into the matrix (red area in the image on the right). \(\ce{ATP}\) synthesis takes place in the F part of \(\ce{ATP}\)-synthase. It operates on the principle of chemiosmosis which is described below. Water movement from a higher to lower water concentration direction across a semipermeable membrane is called When protons (\(\ce{H^{+}}\)) are pumped from the matrix to intermembrane space by complex I, III, & IV, a proton concentration gradient is developed across the membrane. Since protons are positive charge ions (\(\ce{H^{+}}\)), there is also an electrical gradient. Collectively, it is called an electrochemical gradient. Like the flow of water during osmosis, ions can move across a semipermeable membrane down their electrochemical gradient which is called . An electrochemical gradient developed due to protons (\(\ce{H^{+}}\)) is a form of potential energy that is harnessed by \(\ce{ATP}\)-synthase during chemiosmosis to synthesize higher energy \(\ce{ATP}\) by condensing lower energy \(\ce{ADP}\) and phosphate \(\ce{P_{i}}\) in a process called . It takes three to four \(\ce{H^{+}}\) to synthesize one \(\ce{ATP}\) by the following overall reaction. This is an equilibrium reaction, i.e., when the electrochemical gradient is low, \(\ce{ATP}\)-synthase consumes \(\ce{ATP}\) and returns protons from the matrix to the intermembrane space. \(\ce{ATP}\)-synthase is composed of several protein sub-units. It is in the shape of a mushroom with two major parts: a stem-like part which is a hydrophobic portion embedded in the inner membrane is called and a mushroom head-like part that protrudes into the matrix is called as illustrated in Figure \(\Page {2}\) a. Note: the subscript in F in the letter O not the number zero. F part consists of six c subunits (F c) arranged in a ring shape with proton channels between them that make the rotor part of this molecular machinery, as illustrated in Figure \(\Page {2}\) a & b. The b subunit (F b) connects to F portion and prevents it from rotating. The a subunit (F a) connects F b to F c ring. F c portion is embedded in the membrane and couples the potential energy released as the protons move from higher to lower concentration across the membrane to the rotation of F c ring, i.e., convert the potential energy in the form of the electrochemical gradient to kinetic energy in this molecular motor. Other subunits are not described here. F portion has a set of three \(\alpha\) and thee \(\beta\) subunits arranged alternately like carpels of an orange (F (\(\alpha\)\(\beta\)) ) with F \(\gamma\) in the middle. The F \(\gamma\) is alike an axil connected with the rotar F c, as shown in Figure \(\Page {2}\) a. The F (\(\alpha\)\(\beta\)) is prevented from rotating by F b subunit, but F (\(\gamma\) rotates along with the rotor F c ring. The rotation of F (\(\gamma\) within F (\(\alpha\)\(\beta\)) causes conformation changes in the \(\beta\) subunits as illustrated in the . The conformational changes in the \(\beta\) subunits are linked to the mechanism of \(\ce{ATP}\) synthesis. Other subunits of F portion are not described here. The steps of the rotation of Steps in the rotation mechanism of Asw-hamburg at German ., CC BY- , via Wikimedia Commons) one (labelled#1 in step a) which, being near a cation group (\(\ce{peptide-NH3^{+}}\) of arg residue of The simulation of the movement described above is illustrated in Figure \(\Page {2}\) c. The F (\(\alpha\)\(\beta\)) \(\alpha\)\(\beta\) dimers in the shape of carpels in an orange. It is the \(\beta\) subunits that catalyze the F (\(\gamma\) subunit inside the F (\(\alpha\)\(\beta\)) F (\(\beta\) subunits that are linked with the mechanism of F (\(\beta\) switched between three states with each 360 rotation of F \(\gamma\) subunit inside the F (\(\alpha\)\(\beta\)) Figure \(\Page {2}\) d. First is the , shown in brown, where F (\(\beta\) closes up around the substrate F (\(\beta\) tightens around the substrate forcing them to condense into an F (\(\beta\) reverts to an open-state that releases the Since one \(\ce{H^{+}}\) cause turn, a complete ( ) turn of F (\(\alpha\)\(\beta\)) \(\ce{12H^{+}}\) and produces three F (\(\beta\) subunits in it, that gives the following overall reaction. The following in a summary form. \(\ce{NADH}\) is reduced in the electron transport chain at the expense of reduction of \(\ce{O2}\) by the following overall reaction. \(\ce{1/2O2 + NADH + H^{+} -> H2O + NAD^{+}}\) The potential difference between these redox pairs is 1.14 volts, equivalent to -218 kJ/mol. Reduction of one \(\ce{NADH}\) can produce three \(\ce{ATP}\). Production of \(\ce{ATP}\) costs 30.5 kJ/mole, which is equivalent to 30.5 kJ/mole x 3 = 91.5 kJ/mole of \(\ce{NADH}\). So the percentage of energy conserved as \(\ce{ATP}\), i.e., the is: \(\frac{91.5}{218}\times{100} = 42\text{%}\). The remaining 58% of energy ends up heating the body. Some compounds, known as , uncouple electron transport from \(\ce{ATP}\) synthase, i.e., the electron transport pumps \(\ce{H^{+}}\) as usual but \(\ce{H^{+}}\) return to the matrix without producing \(\ce{ATP's}\). So, all of the energy released during the electron transport process becomes heat energy. Examples of uncouplers are 2,4-dinitrophenol and dicumarol that combine with \(\ce{H^{+}}\) and, being hydrophobic, carry them through the inner membrane. Some compounds, like oligomycin A prevent \(\ce{ATP}\) synthesis by blocking \(\ce{H^{+}}\)-channels in \(\ce{ATP}\) synthase. Salicylic acid (aspirin), if taken in extreme excess, also blocks \(\ce{H^{+}}\)-channels in \(\ce{ATP}\) synthase. When the electron transport chain operates without \(\ce{ATP}\) synthesis, all of the energy of electron transport is used as heat. Certain animals adapted to the cold environment have developed uncoupling systems to generate more heat for heating their body. They contain a large amount of which is a tissue having a large number of mitochondria and are brown due to iron in the cytochrome in the mitochondria. The electron transport works in the brown fat but the mitochondria have certain proteins embedded in the inner membrane that allow \(\ce{H^{+}}\) to return to the matrix without \(\ce{ATP}\) production. The body of new born babies loses more heat per unit mass because of the larger surface area to mass ratio. Newborn babies have brown fat deposits on arteries that heat the blood circulating in their bodies. Adults usually do not have brown fat except for those who live and work in a cold climate. The electron transport chain uses four electrons and four protons to reduce oxygen by the following overall reaction, \(\ce{O2 + 4e^{-} + 4H^{+} -> 2H2O}\). However, some electrons, particularly during the reduction of coenzyme-\(\ce{Q}\) in the complex-III leak and cause the following reaction. Species like superoxide (\(\ce{O2^{\bullet{-}}}\)), peroxide (\(\ce{O2^{2-}}\)) and their product hydroxyl radical (\(\ce{HO^{\bullet}}\) are called which are harmful. They damage proteins, cause mutations by reacting with DNA, and are considered responsible for the aging process, as illustrated in the figure on the right (taken from ). The body has mechanisms to suppress producing of reactive species and also to destroy them if they are formed. Production of reactive oxygen species increases with an increase in the inter-member potential. The reactive oxygen species, which are oxidants, activate uncoupling proteins that reduce the membrane potential. Some substances in the body, e.g., vitamin C, vitamin E, and antioxidant enzymes react with and destroy the reactive oxygen species. Overall reaction aerobic catabolism of a glucose molecule converts a \(\ce{6C}\) glucose molecule to six of \(\ce{2C}\) carbon dioxide along with a sent of \(\ce{ATP's}\), \(\ce{GTP's}\), and reduced coenzyme \(\ce{NADH's}\), and \(\ce{FADH2{'s}}\), as shown in the figure on the right. Glucose first goes through glycolysis that produces two pyruvate and overall two \(\ce{ATP}\) and two \(\ce{NADH}\). Two pyruvates go through oxidative decarboxylation producing two acetyl-\(\ce{CoA}\) and two \(\ce{NADH}\). Each of the acetyl-\(\ce{CoA}\) goes through one turn of the citric acid cycle. So, the two turns of the citric acid cycles convert two acetyl-\(\ce{CoA}\) into four \(\ce{CO2}\). These steps are summarized in the figure shown below. Figure \(\Page {3}\) illustrates the summary of catabolism of glucose and production of \(\ce{ATP's}\), \(\ce{GTP's}\), and reduced coenzyme \(\ce{NADH's}\), and \(\ce{FADH2{'s}}\) at different stages. Insulin is a hormone that regulates blood glucose levels. The glucose level increase after a meal and the insulin promotes its absorption into the liver, fat, and muscle cells where it is converted into glycogen or fats. High levels of insulin also inhibit the production of glucose by the liver. Low insulin levels have the opposite effects which is happen when a person is suffering from diabetes. Anaerobic catabolism, i.e., the two forms of fermentation shown in Figure \(\Page {3}\) yields a net two \(\ce{ATP's}\) from a glucose molecule. Aerobic catabolism yields about thirty \(\ce{ATP's}\). Two \(\ce{ATP's}\) are produced before the citric acid cycle and two \(\ce{GTP's}\), the \(\ce{NADH's}\), and two \(\ce{FADH2{'s}}\) are produced during citric acid cycle. \(\ce{GTP's}\) later convert to \(\ce{ATP's}\). the reduced coenzymes, i.e., \(\ce{NADH's}\) and \(\ce{FADH2{'s}}\) enter the electron transport chain and produce heat and \(\ce{ATP's}\) as explained in the next section. Two \(\ce{ATP's}\) are produced before the citric acid cycle along with four \(\ce{NADH}\). The citric acid cycle yields \(\ce{GTP's}\), six \(\ce{NADH's}\), and two \(\ce{FADH2{'s}}\). \(\ce{GTP's}\) convert to \(\ce{ATP's}\), i.e., one \(\ce{GTP}\) is equivalent to one \(\ce{ATP}\). \(\ce{NADH's}\) and \(\ce{FADH2{'s}}\) enter the electron transport chain and produce heat and \(\ce{ATP's}\). Earlier books state that one \(\ce{NADH's}\) is equivalent to three \(\ce{ATP's}\), but currently accepted average values are the following. \(\ce{GTP}~ \approx~ 1\ce{ATP}\) \(\ce{NADH} ~\approx~ 2.5\ce{ATP}\) \(\ce{FADH2} ~\approx~ 1.5\ce{ATP}\) Based on these conversions, complete aerobic respiration of glucose produces approximately \(\ce{32ATP's}\), as detailed in Table 1. | 19,716 | 2,461 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/08%3A_Solids_Liquids_and_Gases/8.02%3A_Solids_and_Liquids |
Solids and liquids are collectively called because their particles are in virtual contact. The two states share little else, however. In the solid state, the individual particles of a substance are in fixed positions with respect to each other because there is not enough thermal energy to overcome the intermolecular interactions between the particles. As a result, solids have a definite shape and volume. Most solids are hard, but some (like waxes) are relatively soft. Many solids composed of ions can also be quite brittle. Solids usually have their constituent particles arranged in a regular, three-dimensional array of alternating positive and negative ions called a crystal. The effect of this regular arrangement of particles is sometimes visible macroscopically, as shown in Figure \(\Page {1}\). Some solids, especially those composed of large molecules, cannot easily organize their particles in such regular crystals and exist as amorphous (literally, “without form”) solids. Glass is one example of an amorphous solid. If the particles of a substance have enough energy to partially overcome intermolecular interactions, then the particles can move about each other while remaining in contact. This describes the liquid state. In a liquid, the particles are still in close contact, so liquids have a definite volume. However, because the particles can move about each other rather freely, a liquid has no definite shape and takes a shape dictated by its container. If the particles of a substance have enough energy to completely overcome intermolecular interactions, then the particles can separate from each other and move about randomly in space. Like liquids, gases have no definite shape, but unlike solids and liquids, gases have no definite volume either. The change from solid to liquid usually does not significantly change the volume of a substance. However, the change from a liquid to a gas significantly increases the volume of a substance, by a factor of 1,000 or more. Figure \(\Page {3}\) shows the differences among solids, liquids, and gases at the molecular level, while Table \(\Page {1}\) lists the different characteristics of these states. What state or states of matter does each statement, describe? What state or states of matter does each statement describe? a. solid b. gas c. solid Earth is the only known body in our solar system that has liquid water existing freely on its surface. That is a good thing because life on Earth would not be possible without the presence of liquid water. Water has several properties that make it a unique substance among substances. It is an excellent solvent; it dissolves many other substances and allows those substances to react when in solution. In fact, water is sometimes called the because of this ability. Water has unusually high melting and boiling points (0°C and 100°C, respectively) for such a small molecule. The boiling points for similar-sized molecules, such as methane (BP = −162°C) and ammonia (BP = −33°C), are more than 100° lower. Though a liquid at normal temperatures, water molecules experience a relatively strong intermolecular interaction that allows them to maintain the liquid phase at higher temperatures than expected. Unlike most substances, the solid form of water is less dense than its liquid form, which allows ice to float on water. The most energetically favorable configuration of H O molecules is one in which each molecule is hydrogen-bonded to four neighboring molecules. Owing to the thermal motions, this ideal is never achieved in the liquid, but when water freezes to ice, the molecules settle into exactly this kind of an arrangement in the ice crystal. This arrangement requires that the molecules be somewhat farther apart than would otherwise be the case; as a consequence, ice, in which hydrogen bonding is at its maximum, has a more open structure, and thus a lower density than water. Here are three-dimensional views of a typical local structure of water (left) and ice (right.) Notice the greater openness of the ice structure which is necessary to ensure the strongest degree of hydrogen bonding in a uniform, extended crystal lattice. The structure of liquid water is very similar, but in the liquid, the hydrogen bonds are continually broken and formed because of rapid molecular motion. Because ice is less dense than liquid water, rivers, lakes, and oceans freeze from the top down. In fact, the ice forms a protective surface layer that insulates the rest of the water, allowing fish and other organisms to survive in the lower levels of a frozen lake or sea. If ice were denser than the liquid, the ice formed at the surface in cold weather would sink as fast as it formed. Bodies of water would freeze from the bottom up, which would be lethal for most aquatic creatures. The expansion of water when freezing also explains why automobile or boat engines must be protected by “antifreeze” and why unprotected pipes in houses break if they are allowed to freeze. Water also requires an unusually large amount of energy to change temperature. While 100 J of energy will change the temperature of 1 g of Fe by 230°C, this same amount of energy will change the temperature of 1 g of H O by only 100°C. Thus, water changes its temperature slowly as heat is added or removed. This has a major impact on weather, as storm systems like hurricanes can be impacted by the amount of heat that ocean water can store. Water’s influence on the world around us is affected by these properties. Isn’t it fascinating that such a small molecule can have such a big impact? | 5,612 | 2,463 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/08%3A_Solids_Liquids_and_Gases/8.03%3A_Gases_and_Pressure |
The gas phase is unique among the three states of matter in that there are some simple models we can use to predict the physical behavior of all gases—independent of their identities. We cannot do this for the solid and liquid states. In fact, the development of this understanding of the behavior of gases represents the historical dividing point between alchemy and modern chemistry. Initial advances in the understanding of gas behavior were made in the mid 1600s by Robert Boyle, an English scientist who founded the Royal Society (one of the world’s oldest scientific organizations). How is it that we can model all gases independent of their chemical identity? The answer is in a group of statements called the kinetic theory of gases: Did you notice that none of these statements relates to the identity of the gas? This means that all gases should behave similarly. A gas that follows these statements perfectly is called an . Most gases show slight deviations from these statements and are called . However, the existence of real gases does not diminish the importance of the kinetic theory of gases. One of the statements of the kinetic theory mentions collisions. As gas particles are constantly moving, they are also constantly colliding with each other and with the walls of their container. There are forces involved as gas particles bounce off the container walls (Figure \(\Page {1}\)). The force generated by gas particles divided by the area of the container walls yields pressure. Pressure is a property we can measure for a gas, but we typically do not consider pressure for solids or liquids. The basic unit of pressure is the newton per square meter (N/m ). This combined unit is redefined as a pascal (Pa). One pascal is not a very large amount of pressure. A more useful unit of pressure is the bar, which is 100,000 Pa (1 bar = 100,000 Pa). Other common units of pressure are the atmosphere (atm), which was originally defined as the average pressure of Earth’s atmosphere at sea level; and mmHg (millimeters of mercury), which is the pressure generated by a column of mercury 1 mm high. The unit millimeters of mercury is also called a torr, named after the Italian scientist Evangelista Torricelli, who invented the barometer in the mid-1600s. A more precise definition of atmosphere, in terms of torr, is that there are exactly 760 torr in 1 atm. A bar equals 1.01325 atm. Given all the relationships between these pressure units, the ability to convert from one pressure unit to another is a useful skill. Write a conversion factor to determine how many atmospheres are in 1,547 mmHg. Because 1 mmHg equals 1 torr, the given pressure is also equal to 1,547 torr. Because there are 760 torr in 1 atm, we can use this conversion factor to do the mathematical conversion: \(\mathrm{1,547\: torr\times \dfrac{1\: atm}{760\: torr}=2.04\: atm}\) Note how the torr units cancel algebraically. Write a conversion factor to determine how many millimeters of mercury are in 9.65 atm. \(\mathrm{9.65\: atm\times \dfrac{760\: mm Hg}{1\: atm}=7,334 \: mm Hg}\). The kinetic theory also states that there is no interaction between individual gas particles. Although we know that there are, in fact, intermolecular interactions in real gases, the kinetic theory assumes that gas particles are so far apart that the individual particles don’t “feel” each other. Thus, we can treat gas particles as tiny bits of matter whose identity isn’t important to certain physical properties. | 3,507 | 2,464 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Solubilty/Common_Ion_Effect |
The common-ion effect is used to describe the effect on an equilibrium when one or more species in the reaction is shared with another reaction. This results in a shifitng of the equilibrium properties. The solubility products 's are equilibrium constants in hetergeneous equilibria (i.e., between two different phases). If several salts are present in a system, they all ionize in the solution. If the salts contain a common cation or anion, these salts contribute to the concentration of the common ion. Contributions from all salts must be included in the calculation of concentration of the common ion. For example, a solution containing sodium chloride and potassium chloride will have the following relationship: \[\mathrm{[Na^+] + [K^+] = [Cl^-]} \label{1}\nonumber \] Consideration of or or both leads to the same conclusion. When \(\ce{NaCl}\) and \(\ce{KCl}\) are dissolved in the same solution, the \(\mathrm{ {\color{Green} Cl^-}}\) ions are to both salts. In a system containing \(\ce{NaCl}\) and \(\ce{KCl}\), the \(\mathrm{ {\color{Green} Cl^-}}\) ions are common ions. \[\begin{align*} \ce{NaCl &\rightleftharpoons Na^{+}} + \color{Green} \ce{Cl^{-}}\\[4pt]
\ce{KCl &\rightleftharpoons K^{+}} + \color{Green} \ce{Cl^{-}} \\[4pt]
\ce{CaCl_2 &\rightleftharpoons Ca^{2+}} + \color{Green} \ce{2 Cl^{-}}\\[4pt]
\ce{AlCl_3 &\rightleftharpoons Al^{3+}} + \color{Green} \ce{3 Cl^{-}}\\[4pt]
\ce{AgCl & \rightleftharpoons Ag^{+}} + \color{Green} \ce{Cl^{-}} \end{align*}\] For example, when \(\ce{AgCl}\) is dissolved into a solution already containing \(\ce{NaCl}\) (actually \(\ce{Na+}\) and \(\ce{Cl-}\) ions), the \(\ce{Cl-}\) ions come from the ionization of both \(\ce{AgCl}\) and \(\ce{NaCl}\). Thus, \(\ce{[Cl- ]}\) differs from \(\ce{[Ag+]}\). The following examples show how the concentration of the common ion is calculated. What are \(\ce{[Na+]}\), \(\ce{[Cl- ]}\), \(\ce{[Ca^2+]}\), and \(\ce{[H+]}\) in a solution containing 0.10 M each of \(\ce{NaCl}\), \(\ce{CaCl2}\), and \(\ce{HCl}\)? Due to the conservation of ions, we have \[\ce{[Na^{+}] = [Ca^{2+}] = [H^{+}] = $0.10$\, \ce M}. \nonumber\] but \[\begin{align*} \ce{[Cl^{-}]} &= 0.10 \, \ce{(due\: to\: NaCl)}\\[4pt]
&+ 0.20\, \ce{(due\: to\: CaCl_2)} \\[4pt]
&+ 0.10\, \ce{(due\: to\: HCl)} \\[4pt]
&= 0.40\, \ce{M} \end{align*}\] John poured 10.0 mL of 0.10 M \(\ce{NaCl}\), 10.0 mL of 0.10 M \(\ce{KOH}\), and 5.0 mL of 0.20 M \(\ce{HCl}\) solutions together and then he made the total volume to be 100.0 mL. What is \(\ce{[Cl- ]}\) in the final solution? \(\mathrm{[Cl^-] = \dfrac{0.1\: M\times 10\: mL+0.2\: M\times 5.0\: mL}{100.0\: mL} = 0.020\: M}\) Principle states that if an equilibrium becomes unbalanced, the reaction will shift to restore the balance. If a common ion is added to a weak acid or weak base equilibrium, then the equilibrium will shift towards the reactants, in this case the weak acid or base. Consider the lead(II) ion concentration in this solution of \(\ce{PbCl2}\). The balanced reaction is \[\ce{ PbCl2 (s) <=> Pb^{2+}(aq) + 2Cl^{-}(aq)} \label{Ex1.1} \] Defining \(s\) as the concentration of dissolved lead(II) chloride, then: \[[Pb^{2+}] = s \nonumber \] \[[Cl^- ] = 2s\nonumber \] These values can be substituted into the solubility product expression, which can be solved for \(s\): \[\begin{align*} K_{sp} &= [Pb^{2+}] [Cl^{-}]^2 \\[4pt] &= s \times (2s)^2 \\[4pt] 1.7 \times 10^{-5} &= 4s^3 \\[4pt] s^3 &= \dfrac{1.7 \times 10^{-5}}{4} \\[4pt] &= 4.25 \times 10^{-6} \\[4pt] s &= \sqrt[3]{4.25 \times 10^{-6}} \\[4pt] &= 1.62 \times 10^{-2}\ mol\ dm^{-3} \end{align*}\] The concentration of lead(II) ions in the solution is 1.62 x 10 M. Consider what happens if sodium chloride is added to this saturated solution. Sodium chloride shares an ion with lead(II) chloride. The chloride ion is to both of them; this is the origin of the term "common ion effect". Look at the original equilibrium expression in Equation \ref{Ex1.1}. What happens to that equilibrium if extra chloride ions are added? According to a the position of equilibrium will shift to counter the change, in this case, by removing the chloride ions by making extra solid lead(II) chloride. Of course, the concentration of lead(II) ions in the solution is so small that only a tiny proportion of the extra chloride ions can be converted into solid lead(II) chloride. The lead(II) chloride becomes even , and the concentration of lead(II) ions in the solution . This type of response occurs with any sparingly soluble substance: it is less soluble in a solution which contains any ion which it has in common. This is the common ion effect. If an attempt is made to dissolve some lead(II) chloride in some 0.100 M sodium chloride solution instead of in water, what is the equilibrium concentration of the lead(II) ions this time? As before, define s to be the concentration of the lead(II) ions. \[\ce{[Pb^{2+}]} = s \label{2}\nonumber \] The calculations are different from before. This time the concentration of the chloride ions is governed by the concentration of the sodium chloride solution. The number of ions coming from the lead(II) chloride is going to be tiny compared with the 0.100 M coming from the sodium chloride solution. In calculations like this, it can be assumed that the concentration of the common ion is entirely due to the other solution. This simplifies the calculation. So we assume: \[\ce{[Cl^{-} ]} = 0.100\; M \label{3}\nonumber \] The rest of the mathematics looks like this: \[ \begin{align*} K_{sp}& = [Pb^{2+},Cl^-]^2 \\[4pt] & = s \times (0.100)^2 \\[4pt] 1.7 \times 10^{-5} & = s \times 0.00100 \end{align*}\] therefore: \[ \begin{align*} s & = \dfrac{1.7 \times 10^{-5}}{0.0100} \\[4pt] & = 1.7 \times 10^{-3} \, \text{M} \end{align*}\] Finally, compare that value with the simple saturated solution: Original solution: \[\ce{[Pb^{2+}]} = 0.0162 \, M \label{5}\nonumber \] Solution in 0.100 M \(\ce{NaCl}\) solution: \[\ce{[Pb^{2+}]} = 0.0017 \, M \label{6}\nonumber \] The concentration of the lead(II) ions has decreased by a factor of about 10. If more concentrated solutions of sodium chloride are used, the solubility decreases further. Adding a common ion to a system at equilibrium affects the equilibrium composition, but the ionization constant. Adding a common ion prevents the weak acid or weak base from ionizing as much as it would without the added common ion. The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium. The common ion effect of \(\ce{H3O^{+}}\) on the ionization of acetic acid The common ion effect suppresses the ionization of a weak base by adding more of an ion that is a product of this equilibrium. Consider the common ion effect of \(\ce{OH^{-}}\) on the ionization of ammonia Adding the common ion of hydroxide shifts the reaction towards the left to decrease the stress (in accordance with Principle), forming more reactants. This decreases the reaction quotient, because the reaction is being pushed towards the left to reach equilibrium. The equilibrium constant, \(K_b=1.8 \times 10^{-5}\), does not change. The reaction is put out of balance, or equilibrium. \[Q_a = \dfrac{[\ce{NH_4^{+}},\ce{OH^{-}}]}{[\ce{NH_3}]} \nonumber \] At first, when more hydroxide is added, the quotient is greater than the equilibrium constant. The reaction then shifts right, causing the denominator to increase, decreasing the reaction quotient and pulling towards equilibrium and causing \(Q\) to decrease towards \(K\). Adding a common ion decreases solubility, as the reaction shifts toward the left to relieve the stress of the excess product. Adding a common ion to a dissociation reaction causes the equilibrium to shift left, toward the reactants, causing precipitation. Consider the reaction: \[\ce{ PbCl_2(s) <=> Pb^{2+}(aq) + 2Cl^{-}(aq)} \nonumber \] What happens to the solubility of \(\ce{PbCl2(s)}\) when 0.1 M \(\ce{NaCl}\) is added? \[K_{sp}=1.7 \times 10^{-5} \nonumber \] \[Q_{sp}= 1.8 \times 10^{-5} \nonumber \] Identify the common ion: \(\ce{Cl^{-}}\) Notice: \(Q_{sp} > K_{sp}\) The addition of \(\ce{NaCl}\) has caused the reaction to shift out of equilibrium because there are more dissociated ions. Typically, solving for the molarities requires the assumption that the solubility of \(\ce{PbCl2(s)}\) is equivalent to the concentration of \(\ce{Pb^{2+}}\) produced because they are in a 1:1 ratio. Because \(K_{sp}\) for the reaction is \(1.7 \times 10^{-5}\), the overall reaction would be \[(s)(2s)^2= 1.7 \times 10^{-5}. \nonumber\] Solving the equation for \(s\) gives \(s= 1.62 \times 10^{-2}\, \text{M}\). The coefficient on \(\ce{Cl^{-}}\) is 2, so it is assumed that twice as much \(\ce{Cl^{-}}\) is produced as \(\ce{Pb^{2+}}\), hence the '2s.' The solubility equilibrium constant can be used to solve for the molarities of the ions at equilibrium. The molarity of Cl added would be 0.1 M because \(\ce{Na^{+}}\) and \(\ce{Cl^{-}}\) are in a 1:1 ratio in the ionic salt, \(\ce{NaCl}\). Therefore, the overall molarity of \(\ce{Cl^{-}}\) would be \(2s + 0.1\), with \(2s\) referring to the contribution of the chloride ion from the dissociation of lead chloride. \[\begin{align*} Q_{sp} &= [\ce{Pb^{2+}},\ce{Cl^{-}}]^2 \\[4pt] &= 1.8 \times 10^{-5} \\[4pt] &= (s)(2s + 0.1)^2 \\[4pt] s &= [Pb^{2+}] \\[4pt] &= 1.8 \times 10^{-3} M \\[4pt] 2s &= [\ce{Cl^{-}}] \\[4pt] &\approx 0.1 M \end{align*} \] Notice that the molarity of \(\ce{Pb^{2+}}\) is lower when \(\ce{NaCl}\) is added. The equilibrium constant remains the same because of the increased concentration of the chloride ion. To simplify the reaction, it can be assumed that \([\ce{Cl^{-}}]\) is approximately 0.1 M since the formation of the chloride ion from the dissociation of lead chloride is so small. The reaction quotient for \(\ce{PbCl2(s)}\) is greater than the equilibrium constant because of the added \(\ce{Cl^{-}}\). This therefore shift the reaction left towards equilibrium, causing precipitation and lowering the current solubility of the reaction. Overall, the solubility of the reaction decreases with the added sodium chloride. | 10,194 | 2,466 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Reactivity_of_Aldehydes_and_Ketones/Conjugate_Addition_Reactions |
One of the largest and most diverse classes of reactions is composed of nucleophilic additions to a carbonyl group. Conjugation of a double bond to a carbonyl group transmits the electrophilic character of the carbonyl carbon to the beta-carbon of the double bond. These conjugated carbonyl are called enones or α, β unsaturated carbonyls. A resonance description of this transmission is shown below. From this formula it should be clear that nucleophiles may attack either at the carbonyl carbon, as for any aldehyde, ketone or carboxylic acid derivative, or at the beta-carbon. These two modes of reaction are referred to as 1,2-addition and 1,4-addition respectively. A 1,4-addition is also called a conjugate addition. Here the nucleophile adds to the carbon which is in the one position. The hydrogen adds to the oxygen which is in the two position. In 1,4 addition the Nucleophile is added to the carbon β to the carbonyl while the hydrogen is added to the carbon α to the carbonyl. 1) Nucleophilic attack on the carbon β to the carbonyl 2) Proton Transfer Here we can see why this addition is called 1,4. The nucleophile bonds to the carbon in the one position and the hydrogen adds to the oxygen in the four position. 3) Tautomerization Going from reactant to products simplified Whether 1,2 or 1,4-addition occurs depends on multiple variables but mostly it is determined by the nature of the nucleophile. During the addition of a nucleophile there is a competition between 1,2 and 1,4 addition products. If the nucleophile is a strong base, such as , both the 1,2 and 1,4 reactions are irreversible and therefor are under kinetic control. Since 1,2-additions to the carbonyl group are fast, we would expect to find a predominance of 1,2-products from these reactions. If the nucleophile is a weak base, such as alcohols or amines, then the 1,2 addition is usually reversible. This means the competition between 1,2 and 1,4 addition is under thermodynamic control. In this case 1,4-addition dominates because the stable carbonyl group is retained. Water Alcohols Thiols 1 Amines 2 Amines HBr Cyanides Another important reaction exhibited by organometallic reagents is metal exchange. Organolithium reagents react with cuprous iodide to give a lithium dimethylcopper reagent, which is referred to as a Gilman reagent. Gilman reagents are a source of carbanion like nucleophiles similar to Grignard and Organo lithium reagents. However, the reactivity of organocuprate reagents is slightly different and this difference will be exploited in different situations. In the case of α, β unsaturated carbonyls organocuprate reagents allow for an 1,4 addition of an alkyl group. As we will see later Grignard and Organolithium reagents add alkyl groups 1,2 to α, β unsaturated carbonyls Organocuprate reagents are made from the reaction of organolithium reagents and \(CuI\) \[ 2 RLi + CuI \rightarrow R_2CuLi + LiI\] This acts as a source of R: \[2 CH_3Li + CuI \rightarrow (CH_3)_2CuLi + LiI\] Example Metal Hydrides Grignard Reagents Organolithium Reagents ) ), | 3,091 | 2,467 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Elimination_Reactions/Elimination_Reactions/F._The_Dehydration_of_Butan-2-ol |
This page builds on your understanding of the acid catalysed dehydration of alcohols. You have to be wary with more complicated alcohols in case there is the possibility of more than one alkene being formed. Butan-2-ol is a good example of this, with no less than three different alkenes being formed when it is dehydrated. To make the diagrams less cluttered, we'll use the simplified version of the mechanism showing gain and loss of H . Remember that the mechanism takes place in three stages: So, in the case of the dehydration of propan-2-ol: There is nothing new at all in these stages. In the first stage, the alcohol is protonated by picking up a hydrogen ion from the sulphuric acid. In the second stage, the positive ion then sheds a water molecule and produces a carbocation. The complication arises in the next step. When the carbocation loses a hydrogen ion, where is it going to come from? So that a double bond can form, it will have to come from one of the carbons next door to the one with the positive charge. But-1-ene is formed. This time the product is but-2-ene, CH CH=CHCH . In fact the situation is even more complicated than it looks, because but-2-ene exhibits geometric isomerism. You get a mixture of two isomers formed - cis-but-2-ene and trans-but-2-ene. Cis-but-2-ene is also known as (Z)-but-2-ene; trans-but-2-ene is also known as (E)-but-2-ene. For an explanation of the two ways of naming these two compounds, follow the link in the box below. Dehydration of butan-2-ol leads to a mixture containing: | 1,549 | 2,468 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Complex_Ion_Chemistry/Geometry_of_Complex_Ions |
This page describes the shapes of some common complex metal ions. These shapes are for complex ions formed using monodentate ligands - ligands which only form one bond to the central metal ion. You will probably be familiar with working out the shapes of simple compounds using the electron pair repulsion theory. Unfortunately that does not work for most complex metal ions involving transition metals. These are complex ions in which the central metal ion is forming six bonds. In the simple cases we are talking about, that means that it will be attached to six ligands. These ions have an octahedral shape. Four of the ligands are in one plane, with the fifth one above the plane, and the sixth one below the plane. The diagram shows four fairly random examples of octahedral ions. It does not matter the nature of the ligands. If you have six of them, this is the shape they will take up. Easy! These are far less common, and they can take up one of two different shapes. There are two very similar ions which crop up commonly at this level: [CuCl ] and [CoCl ] . The copper(II) and cobalt(II) ions have four chloride ions bonded to them rather than six, because the chloride ions are too big to fit any more around the central metal ion. Occasionally a 4-coordinated complex turns out to be square planar. There is no easy way of predicting that this is going to happen. The only one you might possibly come across at this level is cisplatin which is used as an anti-cancer drug. is a neutral complex, Pt(NH ) Cl . It is neutral because the 2+ charge of the original platinum(II) ion is exactly canceled by the two negative charges supplied by the chloride ions. The platinum, the two chlorines, and the two nitrogens are all in the same plane. We will have more to say about cisplatin immediately below. Jim Clark ( ) | 1,838 | 2,470 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Proteins/Amino_Acids/Reactions_of_Amino_Acids/Acid-Base_Reactions_of_Amino_Acids |
This page looks at what happens to amino acids as you change the pH by adding either acids or alkalis to their solutions. For simplicity, the page only looks at amino acids which contain a single -NH group and a single -COOH group. An amino acid has both a basic amine group and an acidic carboxylic acid group. There is an internal transfer of a hydrogen ion from the -COOH group to the -NH group to leave an ion with both a negative charge and a positive charge. This is called a zwitterion. This is the form that amino acids exist in even in the solid state. If you dissolve the amino acid in water, a simple solution also contains this ion. A zwitterion is a compound with no overall electrical charge, but which contains separate parts which are positively and negatively charged. If you increase the pH of a solution of an amino acid by adding hydroxide ions, the hydrogen ion is removed from the -NH group. You could show that the amino acid now existed as a negative ion using . In its simplest form, electrophoresis can just consist of a piece of moistened filter paper on a microscope slide with a crocodile clip at each end attached to a battery. A drop of amino acid solution is placed in the center of the paper. Although the amino acid solution is colourless, its position after a time can be found by spraying it with a solution of ninhydrin. If the paper is allowed to dry and then heated gently, the amino acid shows up as a coloured spot. The amino acid would be found to travel towards the anode (the positive electrode). If you decrease the pH by adding an acid to a solution of an amino acid, the -COO part of the zwitterion picks up a hydrogen ion. This time, during electrophoresis, the amino acid would move towards the cathode (the negative electrode). Suppose you start with the ion we've just produced under acidic conditions and slowly add alkali to it. That ion contains two acidic hydrogens - the one in the -COOH group and the one in the -NH group. The more acidic of these is the one in the -COOH group, and so that is removed first - and you get back to the zwitterion. So when you have added just the right amount of alkali, the amino acid no longer has a net positive or negative charge. That means that it wouldn't move towards either the cathode or anode during electrophoresis. The pH at which this lack of movement during electrophoresis happens is known as the isoelectric point of the amino acid. This pH varies from amino acid to amino acid. If you go on adding hydroxide ions, you will get the reaction we've already seen, in which a hydrogen ion is removed from the -NH group. You can, of course, reverse the whole process by adding an acid to the ion we've just finished up with. That ion contains two basic groups - the -NH group and the -COO group. The -NH group is the stronger base, and so picks up hydrogen ions first. That leads you back to the zwitterion again. . . . and, of course, you can keep going by then adding a hydrogen ion to the -COO group. When an amino acid dissolves in water, the situation is a little bit more complicated than we tend to pretend at this level. The zwitterion interacts with water molecules - acting as both an acid and a base. As an acid: The -NH group is a weak acid and donates a hydrogen ion to a water molecule. Because it is only a weak acid, the position of equilibrium will lie to the left. As a base: The -COO group is a weak base and takes a hydrogen ion from a water molecule. Again, the equilibrium lies to the left. When you dissolve an amino acid in water, both of these reactions are happening. However, the positions of the two equilibria aren't identical - they vary depending on the influence of the "R" group. In practice, for the simple amino acids we have been talking about, the position of the first equilibrium lies a bit further to the right than the second one. That means that there will be rather more of the negative ion from the amino acid in the solution than the positive one. In those circumstances, if you carried out electrophoresis on the unmodified solution, there would be a slight drift of amino acid towards the positive electrode (the anode). To stop that, you need to cut down the amount of the negative ion so that the concentrations of the two ions are identical. You can do that by adding a very small amount of acid to the solution, moving the position of the first equilibrium further to the left. Typically, the pH has to be lowered to about 6 to achieve this. For glycine, for example, the isoelectric point is pH 6.07; for alanine, 6.11; and for serine, 5.68. Jim Clark ( ) | 4,632 | 2,471 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/25%3A_Amino_Acids_Peptides_and_Proteins/25.14%3A_Chemical_Evolution |
A problem of great interest to those curious about the evolution of life concerns the origins of biological molecules. When and how were the molecules of life, such as proteins, nucleic acids, and polysaccharides, first synthesized? In the course of geological history, there must have been a prebiotic period when organic compounds were formed and converted to complex molecules similar to those we encounter in living systems. The composition of the earth's atmosphere in prebiotic times was almost certainly very different from what it is today. Probably it was a reducing atmosphere consisting primarily of methane, ammonia, water, and because there was little or no free oxygen, there was no stratosphere ozone layer and little, if any, screening from the sun's ultraviolet radiation. Starting with \(\ce{CH_4}\), \(\ce{NH_3}\), and \(\ce{H_2O}\), it is plausible that photochemical processes would result in formation of hydrogen cyanide, \(\ce{HCN}\), and methanal, \(\ce{CH_2O}\), by reactions such as the following: \[\ce{CH_4} + \ce{NH_3} \overset{h \nu}{\longrightarrow} \ce{HCN} + 3 \ce{H_2}\] \[\ce{CH_4} + \ce{H_2O} \overset{h \nu}{\longrightarrow} \ce{CH_2=O} + 2 \ce{H_2}\] Hydrogen cyanide and methanal are especially reasonable starting materials for the prebiotic synthesis of amino acids, purine and pyrimidine bases, ribose, and other sugars. Formation of glycine, for example, could have occurred by a Strecker synthesis ( ), whereby ammonia adds to methanal in the presence of \(\ce{HCN}\). Subsequent hydrolysis of the intermediate aminoethanenitrile would produce glycine: The plausibility of these reactions is strongly supported by the classic experiments of S. Miller (1953), who showed that a mixture of methane or ethane, ammonia, and water, on prolonged ultraviolet irradiation or exposure to an electric discharge, produced a wide range of compounds including racemic \(\alpha\)- and \(\beta\)-amino acids. It is not difficult to visualize how sugars such as ribose may be formed. Methanal is known to be converted by bases through a series of aldol-type additions to a mixture of sugarlike molecules called "formose". Formation of racemic ribose along with its stereoisomers could occur as follows: It is more difficult to conceive how the nucleic acid bases such as adenosine may be achieved in prebiotic syntheses. However, adenine, \(\ce{C_5H_5N_5}\), corresponds in composition to a pentamer of hydrogen cyanide and could result by way of a trimer of \(\ce{HCN}\), \(20\), and the adduct of ammonia with \(\ce{HCN}\), \(21\): Without worrying about the mechanistic details of how adenine could be formed from \(20\) and \(21\), you can see that the adenine ring system is equivalent to \(20\) plus two \(21\) minus two \(\ce{NH_3}\): How did these small prebiotic organic molecules grow into large polymeric substances such as peptides, RNA, and so on? It is important to recognize that by whatever reactions polymerization occurred, they had to be reactions that would occur in an essentially aqueous environment. This presents difficulties because condensation of amino acids to form peptides, or of nucleotides to form RNA or DNA, is not thermodynamically favorable in aqueous solution. It is quite possible that primitive polypeptides were formed by the polymerization of aminoethanenitriles produced by the addition of ammonia and \(\ce{HCN}\) to methanal or other aldehydes. The resulting imino polymers certainly would hydrolyze to polypeptides: There are many other questions regarding the origin of the molecules of life for which we have only partial answers, or no answers at all. It is difficult to imagine just how these molecules, once formed, somehow evolved further into the extraordinarily complex systems afforded by even the simplest bacterium able to utilize energy from the sun to support and reproduce itself. Nonetheless, synthetic peptides do coil and aggregate rather like natural proteins, and some also have shown catalytic activities characteristic of natural enzymes. One would hope that some kind of life would be found elsewhere in the solar system, the analysis of which would help us to better understand how life began on earth. and (1977) | 4,229 | 2,473 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumentation_and_Analysis/Diffraction_Scattering_Techniques/X-ray_Diffraction |
The construction of a simple powder diffractometer was first described by Hull in 1917 which was shortly after the discovery of X-rays by Wilhelm Conrad Röntgen in 1895 . Diffractometer measures the angles at which X-rays get reflected and thus get the structural information they contains. Nowadays resolution of this technique get significant improvement and it is widely used as a tool to analyze the phase information and solve crystal structures of solid-state materials. Since the wavelength of X-rays is similar to the distance between crystal layers, incident X-rays will be diffracted, interacting with certain crystalline layers and diffraction patterns containing important structural information about the crystal can be obtained. The diffraction pattern is considered the fingerprint of the crystal because each crystal structures produce unique diffraction patterns and every phase in a mixture produces its diffraction pattern independently. We can use grinded bulk sample into fine powders, which are typical under 10 µm, as samples in powder X-ray Diffraction (XRD). Unlike single crystal X-ray diffraction (X-ray Crystallography) technique, the sample will distribute evenly at every possible orientation and powder XRD collects one-dimensional information, which is a diagram of diffracted beam intensity vs. Bragg angle θ, rather than three-dimensional information. In this section, let us take a look at the theoretical basis of powder X-ray diffraction technique. (e.g. lattice structures and how X-rays interacts with crystal structures) “Crystals are built up of regular arrangements of atoms in three dimensions; these arrangements can be represented by a repeat unit or motif called the unit cell.” In crystallography, all the crystal unit cells can be classied into 230 space groups. Some basic knowledge about crystallography is necessary for a well understanding of powder XRD technique. In crystallography, the basic possible classifications are: 6 crystal families, 7 crystal systems, 5 centering position, 14 Bravais lattices and 32 crystal classes. Based on the angles and the length of the axes sides, unit cell can be divided into 6 crystal families, which are cubic, tetragonal, hexagonal, orthorhombic, monoclinic and triclinic. As the hexagonal family can have two different appearances, we can divide it into two systems which are trigonal lattice and hexagonal lattice. That is how the 7 crystal systems generate. If forget the shape of lattice and just consider the atoms' positions, we can divide the lattices into primitive lattices and non-primitive ones. A primitive lattice (also defined as simple) is the lattice with the smallest possible atomic coordination number , e.g.wheneight atoms lie in the eight corners. And all the other lattices are called non-primitive lattice. Based on the three-dimension position of the atoms in the unit cell, we can divided the non-primitive lattice into three types: face centered (F), side centered (C), body centered (I) and based centered(R). Bravais lattice is a “combination of lattice type and crystal systems” .And you can find a chart of examples of all the 14 Bravais lattice in outside link. 32 crystal classes refer to 32 crystallographic point group classfied by the possible symmetric operations, which are rotation, reflection and inversion. You may wonder why only 32 possible point group. The answer is crystallographic restriction, which means crystal system can only have 5 kinds of rotation axises: 1-fold, 2-fold, 3-fold, 4-fold and 6-fold. To be simplify, only the permissible rotation axises allow unit cells grow uniformly without any openings among them. 230 space groups are combinations of 14 Bravais lattice and 32 crystal classes. Those space groups are generated from translations of related Bravais lattice and glide plane and/or scew axis of relative crystal classes. They are represented by Hermann-Mauguin. For example, the NO.62 space group, Pnma is derived from D2h crystal class. P indicated it is primitive structure and n, m, a stand for a diagonal glide plane, a mirror plane and a axial glide planes. The space group belongs to orthorhombic crystal family. Miller indices and reciprocal lattice are essential to understanding the geometry of lattice planes and X-ray diffraction technique, because they are widely used to index the planes and orientations in crystallography and allow data handling in a simple and mathematical method. To assign the Miller indices (h,k,l) to a certain set of parallel planes which are defined as a plane family , first we need to find the first plane next to the plane, passing through the origin. Then we can find the three intersection of this plane on the unit cell vectors, a, b, c. “The Miller indices would be the reciprocals of the fractional intersections.” Why we want the reciprocal of fraction instead of the fraction directly? To get the answer, we need get to understand what reciprocal lattice is. “Geometrically, the planes can be specified by two quantities: (1) their orientation in the crystal and (2) their d-spacing.” (d-spacing is the interplanar distance) This allows us to use a vector d , which is normal to the planes and whose length is inverse to d-spacing, i.e. d =K/d to stand for a certain family of planes. d is called reciprocal lattice vector and similarly in three dimension system, reciprocal lattice vector d stand for (h,k,l). And the end point of reciprocal lattice vector form a grid or lattice- reciprocal lattice unit cell . Reciprocal lattice cell vector a , b , c is reciprocal form of direct unit cell vector a, b, c. Then it is easy to find out that d =ha +kb +lc . By take the reciprocal number of the intercepts of Miller indices, those two notation systems are very consistent and straightforward in indexing the crystal lattice. Bragg’ s law is the theoretical basis of X-ray diffractometer. Let us consider the crystalline as built up in planes. As shown in the diagram, X-ray beam shines into the planes and is reflected by different planes. The beam reflected by the lower plane will travel an extra distance （shown in Figure 2.2.1 in red）than that reflected by the upper one,which is 2dsinθ. If that distance equals nλ(n is an integer), we will get constructive interference, which corresponds to the bright contrast in diffraction pattern. So the Bragg equation as shown below defines the position of the existence of constructive diffraction at different orders. D-spacing, which is the inter plane distance d in Bragg’s equation, is decided by the lattice parameter a, b, c, as shown below. So after finding out d-spacing from detected Bragg’s angle, we can figure out the lattice parameter which contains vital structural information. Also we can reconstruct the unknown structure by figuring out all the possible d-spacing. Powder XRD can define the phase contained in a mixture on the basis of separating and recognizing characteristic diffraction pattern. The sample of powder X-ray diffraction will distribute evenly at every possible orientation, so after diffracted, the diffraction pattern appears as circles with same center point instead of dots in single crystal diffraction patterns. The circles in the diffraction patterns with smaller radius correspond to smaller h, k, l. In certain types of unit cells, not all the lattice planes will have their diffraction observed, which is usually called systematic absence, because the diffracted beam may happen to be out of phase by 180°and the overall intensity would be zero. Structure factor F can decide the systematic absences and intensity.Systematic absences arise when F=0, so no diffraction will be observed. For example: For a fcc crystal, F =f{1+e +e +e }. When h, k, l are all odd or all even, F=4f. For the other situation, F=0 and thus diffraction intensity will also be zero. Structure factor is important in the structure determination step because it helps understand the Miller indices and intensities of diffraction peaks. The other common rules for reflection to be observed are listed as follows: Lattice type Rule for reflection to be observed Primitive, P None Body centered, I hkl: h+k+l=2n Face centered, F hkl: h, k, l either all odd or all even Side centered, C hkl: h+k=2n Rhombohedral, R hkl: -h+k+l=3n or (h-k+l=3n) Powder X-ray diffractometer consists of three components: X-ray source, sample holder and detector. Possible X-ray sources are X-ray tube, Synchrotron radiation and cyclotron radiation. X-ray tube equipped with filter is commonly used in laboratory diffractometer. Synchrotron radiation is a brighter source and as a result can increase the resolution. The cathode part of X-ray tube generated electrons under electric current. Electrons travel from cathode to anode through a high acceleration voltage, typically 30~150kV. In this process, most of the energy is released as heat and X-ray only account for approximately 1% of total energy. The X-ray tube needs lasing cooling water to protect it from over heat while working. After X rays hit the anode(red part in the schematic), the anode generates characteristic X-rays, which comes from the process of excited electrons falling down to lower electron shell and correspond to the energy difference between electron shells. In Bruker D8 diffractometer, the anode is made of Cu, so the X-ray souce is Cu-Ka1 and Cu-Ka2. K means the electrons falls to K shell from higher shells. α means the excited electrons lies in L shell, one shell higher than K shell. If the excited electron comes from M shell which is two electron shell higher, then what we have is defined as K . The difference between Cu-K and Cu-K is that they come from different subshell, Cu-Ka1 corresponds to 2p2/3 to 1s shell while Cu-K corresponds to 2p1/2 to 1s shell. In a reflection geometry instrumentation, X-ray tube usually contains a side window made of Be to allow the generated X-rays to emit at the demanding angle. The reason that Be is used as a X-ray window, is that the fluorescence yield (ration between characteristic X ray and Auger emission) of Be is close to zero, so it can make sure the X-ray source is monochromatic and does not contain introduced chracteristic X-rays from other metals. There are many holder options to holder all kinds of samples and meet people’s requirement. Usually, evenly grinded sample powder is dissolved in organic solvent such as acetone or pressed into a plain on a glass slide to make sure the sampel is flat. The sample holder has a press ring to fix the slide. At low angles, the signal to noise ratio can be relatively larger and we can use a zero background sample holder to avoid that. It is usually made of single crystal silicon. Photographic film serves as detector in earlier methods, Debye-Scherrer camera and Guinier camera methods . The film is placed around the sample as a circle and records the diffracted X-ray beams. The positions of diffraction lines correspond to Bragg angle. And Photographic film can record both the reflected and transmitted X-ray beams. Nowadays, people tend to only select the reflected beam and use radiation counter as detector. Compared to film, scintillationcounter can measure diffraction intensities and Bragg angles more accurately. And it is very convenient use computer to analyze data.To prevent the X-ray beam from going though the sample, we need more powder sample. X-ray diffractometer is most widely used in the phase analysis because compare to other characterization method, XRD gives a fast and reliable measurement(measurement time is determined by the step size, angle range and the number of second per step) and easy sample preperation(well grinded powder). What people do after get their raw data is opening the data in a XRD data handling software (JADE, WinXPOW, ect.) and compare the raw data with the standard pattern in ICDD database. In many cases, the database may not contain the pattern of a specific compound you are working on, then you can easily generate a calculated pattern based on a crystal informtion file or the space group and lattice parameters. The diffraction pattern of a well-prepared sample should be very reliable (all the peaks should match the peaks in the reference pattern) and contains much information. When the sample has impurities, every kinds of substance will generate their own pattern independently and allows us to analysize seperately and help people control and optimize the reactions. Other factor that can influence the pattern may be the X-ray souce, sample crystallinity (smaller crystallinity can broaden the peaks), ect. XRD technique is also capable for quantitative analysis of mixtures. The XRD data will not give the quantitative imformation because the intensity is not directly related to weight percantage. However, we can make a serie of controlled samples with different weight percentage of the impurity. Then XRD is performed on each of these sample and we can get a linear calibration curve which is the intensity ratio vs. weight percentage. The weight percentage in the sample can be determined based on the curve. If all the atomic and crystalline information are known, we can also conduct qualitative analysis with Rietveld method. A square least approximation method will be applied to modify all the parameters so that the difference between experimental point and the fitted pattern can be decreased to a least amount. During this process, the scale factor can also be determined. Rietveld method is widely used in samples containing more than one impurities. Powder X-ray diffraction can not only be used to analyze phase information,but can also be used to determine the structure of unknown substances. However, since in powder XRD we can only get one-dimensional information rather than three-dimensional information, resolution in powder XRD is much lower than that from single crystal method and data refinement process is more sophisticated.“If representative single crystal method are available, then single crystal diffraction is the preferred method.” As shown in Figure 4.1, one Bragg Ring corresponds to a certain Miller plane. With the detected Bragg angle and equation 2.1, we can figure out the lattice parameters. The common method to manage data is direct method, Patterson method and Fourier method . We can also use the square least approximation method, i.e. Rietveld method to refine the data, to refine the data and it has increased the resolution a lot. Common single crystal methods are Laue method, four-circle diffractometer and rotating-crystal method. | 14,641 | 2,474 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Proteins/Amino_Acids/Reactions_of_Amino_Acids/Acid-base_Chemistry_of_Amino_Acids |
Amino acids by themselves have amino (pKa ~9.0-10.5) and carboxyl groups (pKa ~2.0-2.4) that can be titrated. At neutral pH the amino group is protonated, and the carboxyl group is deprotonated. The side chains of acid and basic amino acids, and some polar amino acids can also be titrated: Physiological pH is near neutral. It would appear that only histidine is of physiological relevance. However, pKa values can be shifted significantly by neighboring charged groups in complex molecular structures. Some common carboxyl-group reactivities: A common side chain reaction involving cysteine: NOTE: This can covalently link two polypeptide chains in a "disulfide bond" crosslink
| 699 | 2,475 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/05%3A_The_Electronic_Structure_of_Atoms/5.17%3A_Electron_Configurations_and_the_Periodic_Table |
The commonly used long form of the periodic table is designed to emphasize . Since it is the outermost which are primarily involved in chemical interactions between atoms, the electron added to an atom in the building-up process is of far more interest to a chemist than the first. This last electron is called the because it distinguishes an atom from the one immediately preceding it in the periodic table. The type of subshell ( )into which the distinguishing electron is placed is very closely related to the chemical behavior of an element and gives rise to the classification shown by the color-coding on the periodic table seen here. The are those in which the distinguishing electron enter an or subshell. Most of the elements whose chemistry and valence we have discussed so far fall into this category. Many of the chemical properties of the representative elements can be explained on the basis of . That is, the valences of the representative elements may be predicted on the basis of the number of valence electrons they have, or from the number of electrons that would have to be added in order to attain the same electron configuration as an atom of a noble gas. For representative elements the number of valence electrons is the same as the periodic group number, and the number needed to match the next noble-gas configuration is 8 minus the group number. This agrees with the valence rules derived from the , and results in formulas for chlorides of the first dozen elements that show the periodic variation of valence. The first three horizontal rows or periods in the modern periodic table consist entirely of representative elements. In the first period the distinguishing electrons for H and He are in the 1 subshell. Across the second period Li and Be have distinguishing electrons in the 2 subshell, and electrons are being added to the 2 subshell in the atoms from B to Ne. In the third period the 3 subshell is filling for Na and Mg, and therefore Al, Si, P, S, Cl, and Ar. As a general rule, in the case of the representative elements, the distinguishing electron will be in an or subshell. The value of , the principal quantum number for the distinguishing electron, can be quickly determined by counting down from the top of the periodic table. For example, iodine is a representative element in the period. Therefore the distinguishing electron must occupy either the 5 or 5 subshell. Since I is on the right side of the table, 5 is the correct choice. When the principal quantum number is three or more, -type subshells are also possible. The or are those elements whose distinguishing electron is found in a orbital. The first examples of transition metals (Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn) are found in the period even though the distinguishing electron in each case is a 3 electron and belongs to the shell. This hiatus results, as we have already seen, because the 4 is lower in energy than the 3 . The 4 orbital thus starts to fill up, beginning the fourth period before any of the 3 orbitals can become occupied. Figure \(\Page {1}\) compares the probability distributions of a 4 and a 3 electron in a V atom. Although the 4 electron cloud lies farther from the nucleus on average than does the 3 cloud, a small portion of the 4 electron density is found very close to the nucleus where it is hardly shielded from the total nuclear charge of +23. It is the very strong attractive force of this small fraction of the total 4 electron density that lowers the energy of the 4 electron below that of the 3 . The fact that the 4 electron cloud is more extensive than the 3 has an important influence on the chemistry of the transition elements. When an atom such as V (Figure \(\Page {1}\) ) interacts with another atom, it is the 4 electrons extending farthest from the nucleus which first contact the other atom. Thus the 4 electrons are often more significant than the 3 in determining valence and the formulas of compounds. The 3 electrons are “buried” under the surfaces of the atoms of the transition metals. Adding one more 3 electron has considerably less effect on their chemical properties than adding one more 3 or 3 electron did in the case of the representative elements. Hence there is a slow but steady in properties from one transition element to another. Notice, for example, that except for Sc, all of the transition metals form chlorides, MCl2, where the metal has a valence of 2; examples are TiCl2, VCl2, CrCl2, and so on. This can be seen in the table found at the top of this page. The valence of 2 corresponds with the two 4s valence electrons. Each of the transition metals also exhibits other valences where one or more of the 3 electrons are also involved. For example, in some compounds V (vanadium) has a valence of 2 (VO, VCl ) in others it has a valence of 3 (V O , VCl ), in still others it has a valence of 4 (VO , VCl ), and in at least one case (V O ) it has a valence of 5. The chemistry of the transition metals is more complicated and a wider variety of formulas for transition-metal compounds is possible because of this variable valence. In some cases electrons in the subshells act as valence electrons, while in other cases they do not. Although the 3 electron clouds do not extend farther from the nucleus than 3 and 3 (and hence do not constitute another shell as the 4 electrons do), they are thoroughly shielded from the nuclear charge and thus often act as valence electrons. This Jekyll and Hyde behavior of 3 electrons makes life more complicated (and often far more interesting) for chemists who study the transition elements. The third major category of elements arises when the distinguishing electron occupies an subshell. The first example occurs in the case of the (elements having atomic numbers between 57 and 71).The lanthanoids have the general electron configuration where is a number between 0 and 14. Thus in the building-up process for the lanthanoids, electrons are being added to a subshell (4 ) whose principal quantum number is two less than that of the outermost orbital (6 ). Addition of another electron to an inner shell buried as deeply as the 4 has little or no effect on the chemical properties of these elements. All are quite similar to lanthanum (La) and might fit into exactly the same space in the periodic table as La. The lanthanoid elements are so similar to one another that special techniques are required to separate them. As a result, even approximately pure samples of most of them were not prepared until the 1870s. Following the element actinium (Ac) is a series of atoms in which the 5 subshell is filling. The are somewhat less similar to Ac than the lanthanoids are to La because some exceptions to the usual order of filling orbitals occur in the case of Th, Pa, and U (Table \(\Page {1}\) ). Because the lanthanoids and most of the actinoids behave chemically as if they should fit in group IIIB of the periodic table (where Lu and Lr are found), both groups are separated from the rest of the table and placed together in a block below. Taken together, the lanthanoids and actinoids are called because the subshells being filled lie so deep within the remaining electronic structure of their atoms. Figure \(\Page {2}\) summarizes the type of subshell in which the distinguishing electron is to be found for atoms of elements in various regions of the periodic table. This summary information makes it relatively simple to use the periodic table to obtain electron configurations, as the following example shows. Obtain the electron configuration for (a) Nb; (b) Pr. Nb, element number 41, is found in the fifth period and in a region of the periodic table where a subshell is filling (the second transition series). Moving backward (toward lower atomic numbers) through the periodic table, the nearest noble gas is Kr, and so we use the Kr kernel: Nb [Kr] _____ One more point needs to be emphasized about the relationship between electron configuration and the periodic table. . For example, consider the alkaline-earth elements (group IIA). Using our rules for deriving electron configurations (Example 1) we have Thus the similarities of chemical behavior and valence noted earlier for these elements correlate with the similarities of their outermost electron clouds. Such similarities account for the success of Mendeleev’s predictions of the properties of undiscovered elements. | 8,514 | 2,476 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/03%3A_First_Law_of_Thermodynamics/3.01%3A_Prelude_to_Thermodynamics |
Albert Einstein, a noted physicist, said of thermodynamics (Einstein, 1979) Thermodynamics is the study of how energy flows into and out of systems and how it flows through the universe. People have been studying thermodynamics for a very long time and have developed the field a great deal, including the incorporation of high-level mathematics into the process. Many of the relationships may look cumbersome or complicated, but they are always describing the same basic thing: the flow of energy through the universe. Energy, of course, can be used to do many useful things, such as allow us to drive our cars, use electronic devices, heat our homes, and cook our food. Chemistry is important as well since many of the processes in which we generate energy depend on chemical reactions (such as the combustion of hydrocarbons to generate heat or electron transfer reactions to generate electron flow.) The previous chapter investigated gases which are convenient systems to use to frame many discussions of thermodynamics since they can be modeled using specific equations of state such as the ideal gas law or the van der Waals law. These relationships depend on an important class of variables known as . are those variables which depend only upon the current conditions affecting a system. Pressure, temperature and molar volume are examples of state variables. A number of variables required to describe the flow of energy in a system do depend on the pathway a system follows to come into its current state. To illustrate the difference, consider climbing a mountain. You may choose to walk straight up the side of the mountain, or you may choose to circle the mountain several times in order to get to the top. These two pathways will differ in terms of how far you actually walk (a path-dependent variable) to attain the same change in altitude (an example of a state variable.) | 1,901 | 2,479 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Stereoisomers/Chirality_and_Symmetry/Enantiomorphism/Resolution_of_Racemates |
As noted earlier, chiral compounds synthesized from achiral starting materials and reagents are generally racemic (i.e. a 50:50 mixture of enantiomers). Separation of racemates into their component enantiomers is a process called . Since enantiomers have identical physical properties, such as solubility and melting point, resolution is extremely difficult. Diastereomers, on the other hand, have different physical properties, and this fact is used to achieve resolution of racemates. Reaction of a racemate with an enantiomerically pure chiral reagent gives a mixture of diastereomers, which can be separated. Reversing the first reaction then leads to the separated enantiomers plus the recovered reagent. Many kinds of chemical and physical reactions, including salt formation, may be used to achieve the diastereomeric intermediates needed for separation. The following diagram illustrates this general principle by showing how a nut having a right-handed thread (R) could serve as a "reagent" to discriminate and separate a mixture of right- and left-handed bolts of identical size and weight. Only the two right-handed partners can interact to give a fully-threaded intermediate, so separation is fairly simple. The resolving moiety, i.e. the nut, is then removed, leaving the bolts separated into their right and left-handed forms. Chemical reactions of enantiomers are normally not so dramatically different, but a practical distinction is nevertheless possible. To learn more about chemical procedures for achieving resolution . | 1,553 | 2,480 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria/15.08%3A_Quantum_states_Microstates_and_Energy_spreading_in_Reactions |
Entropy is a measure of the degree of spreading and sharing of thermal energy within a system. This “spreading and sharing” can be spreading of the thermal energy into a larger volume of or its sharing amongst previously inaccessible microstates of the system. The following table shows how this concept applies to a number of common processes. Entropy is an quantity; that is, it is proportional to the quantity of matter in a system; thus 100 g of metallic copper has twice the entropy of 50 g at the same temperature. This makes sense because the larger piece of copper contains twice as many quantized energy levels able to contain the thermal energy. is the portion of a molecule's energy that is proportional to its , and thus relates to motion at the molecular scale. What kinds of molecular motions are possible? For monatomic molecules, there is only one: actual movement from one location to another, which we call . Since there are three directions in space, all molecules possess three modes of . For polyatomic molecules, two additional kinds of motions are possible. One of these is ; a linear molecule such as CO in which the atoms are all laid out along the x-axis can rotate along the y- and z-axes, while molecules having less symmetry can rotate about all three axes. Thus linear molecules possess two modes of rotational motion, while non-linear ones have three rotational modes. Finally, molecules consisting of two or more atoms can undergo internal . For freely moving molecules in a gas, the number of vibrational modes or patterns depends on both the number of atoms and the shape of the molecule, and it increases rapidly as the molecule becomes more complicated. The relative populations of the quantized translational, rotational and vibrational energy states of a typical diatomic molecule are depicted by the thickness of the lines in this schematic (not-to-scale!) diagram. The colored shading indicates the total thermal energy available at an arbitrary temperature. The numbers at the top show order-of-magnitude spacings between adjacent levels. Notice that the spacing between the quantized translational levels is so minute that they can be considered nearly continuous. This means that at all temperatures, the thermal energy of a collection of molecules resides almost exclusively in translational microstates. At ordinary temperatures (around 25° C), most of the molecules are in their zero-level vibrational and rotational states (corresponding to the bottom-most bars in the diagram.) The prevalence of vibrational states is so overwhelming that we can effectively equate the thermal energy of molecules with translational motions alone. The number of ways in which thermal energy can be distributed amongst the allowed states within a collection of molecules is easily calculated from simple statistics. A very important point to bear in mind is that the number of discrete microstates that can be populated by an arbitrary quantity of energy depends on the spacing of the states. As a very simple example, suppose that we have two molecules (depicted by the orange dots) in a system total available thermal energy is indicated by the yellow shading. In the system with the more closely-spaced energy levels, there are three possible microstates, while in the one with the more widely-spaced levels, only two possibilities are available. The spacing of molecular energy states becomes closer as the mass and number of bonds in the molecule increases, so we can generally say that the more complex the molecule, the greater the density of its energy states. At the atomic and molecular level, all energy is ; each particle possesses discrete states of kinetic energy and is able to accept thermal energy only in packets whose values correspond to the energies of one or more of these states. Polyatomic molecules can store energy in rotational and vibrational motions, and all molecules (even monatomic ones) will possess translational kinetic energy (thermal energy) at all temperatures above absolute zero. The energy difference between adjacent translational states is so minute that translational kinetic energy can be regarded as (non-quantized) for most practical purposes. The number of ways in which thermal energy can be distributed amongst the allowed states within a collection of molecules is easily calculated from simple statistics, but we will confine ourselves to an example here. Suppose that we have a system consisting of three molecules and three quanta of energy to share among them. We can give all the kinetic energy to any one molecule, leaving the others with none, we can give two units to one molecule and one unit to another, or we can share out the energy equally and give one unit to each molecule. All told, there are ten possible ways of distributing three units of energy among three identical molecules as shown here: Each of these ten possibilities represents a distinct microstate that will describe the system at any instant in time. Those microstates that possess identical distributions of energy among the accessible quantum levels (and differ only in which particular molecules occupy the levels) are known as . Because all microstates are equally probable, the probability of any one configuration is proportional to the number of microstates that can produce it. Thus in the system shown above, the configuration labeled will be observed 60% of the time, while will occur only 10% of the time. As the number of molecules and the number of quanta increases, the number of accessible microstates grows explosively; if 1000 quanta of energy are shared by 1000 molecules, the number of available microstates will be around 10 — a number that greatly exceeds the number of atoms in the observable universe! The number of possible configurations (as defined above) also increases, but in such a way as to greatly reduce the probability of all but the most probable configurations. Thus for a sample of a gas large enough to be observable under normal conditions, only a single configuration (energy distribution amongst the quantum states) need be considered; even the second-most-probable configuration can be neglected. : any collection of molecules large enough in numbers to have chemical significance will have its therrmal energy distributed over an unimaginably large number of microstates. The number of microstates increases exponentially as more energy states ("configurations" as defined above) become accessible owing to Energy is ; if you lift a book off the table, and let it fall, the total amount of energy in the world remains unchanged. All you have done is transferred it from the form in which it was stored within the glucose in your body to your muscles, and then to the book (that is, you did work on the book by moving it up against the earth’s gravitational field). After the book has fallen, this same quantity of energy exists as thermal energy (heat) in the book and table top. What changed, however, is the availability of this energy. Once the energy has spread into the huge number of thermal microstates in the warmed objects, the probability of its spontaneously (that is, by chance) becoming un-dispersed is essentially zero. Thus although the energy is still “there”, it is forever beyond utilization or recovery. The profundity of this conclusion was recognized around 1900, when it was first described at the “heat death” of the world. This refers to the fact that every spontaneous process (essentially every change that occurs) is accompanied by the “dilution” of energy. The obvious implication is that all of the molecular-level kinetic energy will be spread out completely, and nothing more will ever happen. Everybody knows that a gas, if left to itself, will tend to expand and fill the volume within which it is confined completely and uniformly. What “drives” this expansion? At the simplest level it is clear that with more space available, random motions of the individual molecules will inevitably disperse them throughout the space. But as we mentioned above, the allowed energy states that molecules can occupy are spaced more closely in a larger volume than in a smaller one. The larger the volume available to the gas, the greater the number of microstates its thermal energy can occupy. Since all such states within the thermally accessible range of energies are equally probable, the expansion of the gas can be viewed as a consequence of the tendency of thermal energy to be spread and shared as widely as possible. Once this has happened, the probability that this sharing of energy will reverse itself (that is, that the gas will spontaneously contract) is so minute as to be unthinkable. Imagine a gas initially confined to one half of a box (Figure \(\Page {4}\)). The barrier is then removed so that it can expand into the full volume of the container. We know that the entropy of the gas will increase as the thermal energy of its molecules spreads into the enlarged space. In terms of the spreading of thermal energy, Figure \(\Page {5}\) may be helpful. The tendency of a gas to expand is due to the more closely-spaced thermal energy states in the larger volume . Mixing and dilution really amount to the same thing, especially for idea gases. Replace the pair of containers shown above with one containing two kinds of molecules in the separate sections (Figure \(\Page {6}\)). When we remove the barrier, the "red" and "blue" molecules will each expand into the space of the other. (Recall Dalton's Law that "each gas is a vacuum to the other gas".) However, notice that although each gas underwent an expansion, the overall process amounts to what we call "mixing". What is true for gaseous molecules can, in principle, apply also to solute molecules dissolved in a solvent. But bear in mind that whereas the enthalpy associated with the expansion of a perfect gas is by definition zero, Δ 's of mixing of two liquids or of dissolving a solute in a solvent have finite values which may limit the miscibility of liquids or the solubility of a solute. It's unfortunate the the simplified diagrams we are using to illustrate the greater numbers of energetically accessible microstates in an expanded gas or a mixture of gases fail to convey the immensity of this increase. Only by working through the statistical mathematics of these processes (beyond the scope of first-year Chemistry!) can one gain an appreciation of the magnitude of the probabilities of these spontaneous processes. It turns out that when just of a second gas is inroduced into the container of another gas, an unimaginably huge number of new configurations becom available. This happens because the added molecule (indicated by the blue arrow in the diagram) can in principle replace any one of the old (red) ones, each case giving rise to a new microstate. ust as gases spontaneously change their volumes from “smaller-to-larger”, the flow of heat from a warmer body to a cooler one always operates in the direction “warmer-to-cooler” because this allows thermal energy to populate a larger number of energy microstates as new ones are made available by bringing the cooler body into contact with the warmer one; in effect, the thermal energy becomes more “diluted”. In this simplified schematic diagram, the "cold" and "hot" bodies differ in the numbers of translational microstates that are occupied, as indicated by the shading. When they are brought into thermal contact, a hugely greater number of microstates are created, as is indicated by their closer spacing in the rightmost section of the diagram, which represents the combined bodies in thermal equilibrium. The thermal energy in the initial two bodies fills these new microstates to a level (and thus, temperature) that is somewhere between those of the two original bodies. Note that this explanation applies equally well to the case of two solids brought into thermal contact, or two the mixing of two fluids having different temperatures. As you might expect, the increase in the amount of energy spreading and sharing is proportional to the amount of heat transferred , but there is one other factor involved, and that is the at which the transfer occurs. When a quantity of heat passes into a system at temperature , the degree of dilution of the thermal energy is given by \[\dfrac{q}{T}\] To understand why we have to divide by the temperature, consider the effect of very large and very small values of \(T\) in the denominator. If the body receiving the heat is initially at a very low temperature, relatively few thermal energy states are initially occupied, so the amount of energy spreading into vacant states can be very great. Conversely, if the temperature is initially large, more thermal energy is already spread around within it, and absorption of the additional energy will have a relatively small effect on the degree of thermal disorder within the body. When a chemical reaction takes place, two kinds of changes relating to thermal energy are involved: Figure \(\Page {8}\) a The ability of energy to spread into the product molecules is constrained by the availability of sufficient thermal energy to produce these molecules. This is where the temperature comes in. At absolute zero the situation is very simple; no thermal energy is available to bring about dissociation, so the only component present will be dihydrogen. The result is exactly what the Le Chatelier Principle predicts: the equilibrium state for an endothermic reaction is shifted to the right at higher temperatures. The following table generalizes these relations for the four sign-combinations of ΔH and ΔS. (Note that use of the standard ΔH° and ΔS° values in the example reactions is not strictly correct here, and can yield misleading results when used generally.) This , like most such reactions, is . The positive entropy change is due mainly to the greater mass of CO molecules compared to those of O .
The decrease in moles of gas in the drives the entropy change negative, making the reaction . Thus higher , which speeds up the reaction, also reduces its extent.
are typically endothermic with positive entropy change, and are therefore .
This reaction is , meaning that . But because the reverse reaction is kinetically inhibited, NO can exist indefinitely at ordinary temperatures even though it is thermodynamically unstable. Everybody knows that the is the stable form of a substance at low temperatures, while the gaseous state prevails at high temperatures. Why should this be? The diagram in Figure \(\Page {9}\) shows that involve exchange of energy with the surroundings (whose energy content relative to the system is indicated (with much exaggeration!) by the height of the yellow vertical bars in Figure \(\Page {13}\). When solid and liquid are in equilibrium (middle section of diagram below), there is sufficient thermal energy (indicated by pink shading) to populate the energy states of both phases. If heat is allowed to flow into the surroundings, it is withdrawn selectively from the more abundantly populated levels of the liquid phase, causing the quantity of this phase to decrease in favor of the solid. The temperature remains constant as the heat of fusion is returned to the system in exact compensation for the heat lost to the surroundings. Finally, after the last trace of liquid has disappeared, the only states remaining are those of the solid. Any further withdrawal of heat results in a temperature drop as the states of the solid become depopulated. Vapor pressure lowering, boiling point elevation, freezing point depression and osmosis are well-known phenomena that occur when a non-volatile solute such as sugar or a salt is dissolved in a volatile solvent such as water. All these effects result from “dilution” of the solvent by the added solute, and because of this commonality they are referred to as (Lat. , connected to.) The key role of the solvent concentration is obscured by the greatly-simplified expressions used to calculate the magnitude of these effects, in which only the solute concentration appears. The details of how to carry out these calculations and the many important applications of colligative properties are covered elsewhere. Our purpose here is to offer a more complete explanation of these phenomena occur. Basically, these all result from the effect of on its entropy, and thus in the increase in the density of energy states of the system in the solution compared to that in the pure liquid. Equilibrium between two phases (liquid-gas for boiling and solid-liquid for freezing) occurs when the energy states in each phase can be populated at . The temperatures at which this occurs are depicted by the shading. Dilution of the solvent adds new energy states to the liquid, but does not affect the vapor phase. This raises the temperature required to make equal numbers of microstates accessible in the two phases. Dilution of the solvent adds new energy states to the liquid, but does not affect the solid phase. This reduces the temperature required to make equal numbers of states accessible in the two phases. When a liquid is subjected to hydrostatic pressure— for example, by an inert, non-dissolving gas that occupies the vapor space above the surface, the vapor pressure of the liquid is raised (Figure \(\Page {13}\)). The pressure acts to compress the liquid very slightly, effectively narrowing the potential energy well in which the individual molecules reside and thus increasing their tendency to escape from the liquid phase. (Because liquids are not very compressible, the effect is quite small; a 100-atm applied pressure will raise the vapor pressure of water at 25°C by only about 2 torr.) In terms of the entropy, we can say that the applied pressure reduces the dimensions of the "box" within which the principal translational motions of the molecules are confined within the liquid, thus reducing the density of energy states in the liquid phase. Applying hydrostatic pressure to a liquid increases the spacing of its microstates, so that the number of energetically accessible states in the gas, although unchanged, is relatively greater— thus increasing the tendency of molecules to escape into the vapor phase. In terms of free energy, the higher pressure raises the free energy of the liquid, but does not affect that of the gas phase. This phenomenon can explain . Osmotic pressure, students must be reminded, is not what drives osmosis, but is rather the hydrostatic pressure that must be applied to the more concentrated solution (more dilute solvent) in order to osmotic flow of solvent into the solution. The effect of this pressure \(\Pi\) is to slightly increase the spacing of solvent energy states on the high-pressure (dilute-solvent) side of the membrane to match that of the pure solvent, restoring osmotic equilibrium. | 18,993 | 2,481 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/27%3A_More_about_Spectroscopy/27.01%3A_Prelude_to_More_Spectroscopy |
In Chapter 9, we gave an exposition of the most generally useful and practical spectroscopic methods currently employed in modern organic laboratories. However, in our discussions of nmr spectra, we passed rather quickly over the basis of understanding why some lines are broad and others sharp, why rate effects can cause chemical shifts to be averaged, and how to correlate spin-spin splitting with the energies of nmr transitions. These topics will be discussed in this chapter along with a brief explanation of the remarkable effects on nmr spectra associated with some kinds of chemical reactions, namely, (CIDNP). In addition to the spectroscopic methods covered in Chapter 9, there are a number of other spectroscopic techniques that are less generally used, but can provide, and have provided, critical information with regard to specialized problems. Because some of these are relatively new and may become more widely used in the next few years, it is important that you be aware of them and their potentialities. However, because, they may be peripheral to your present course of study, we have reserved consideration of them to this chapter. and (1977) | 1,185 | 2,482 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/21%3A_Spectra_and_Structure_of_Atoms_and_Molecules/21.05%3A_The_Spectra_of_Molecules-_Infrared |
When we turn from the to those of molecules, we find that the region of most interest to chemists is no longer the visible and ultraviolet but rather the infrared. As its name implies, the infrared extends beyond the red end of the visible spectrum, from the limit of visibility at roughly 0.8 μm (800 nm) up to about 100 μm where the microwave region begins. Another difference from the spectra discussed in is that infrared spectra are all absorption spectra rather than emission spectra. Infrared light is passed through the sample, and the intensity of light emerging is measured electronically. The energies of infrared photons are very much less than those of visible and ultraviolet photons. A photon of wavelength 10 μm has an energy of only 0.02 aJ (about 12 kJ m )—not even enough to break a hydrogen bond, let alone a normal . It does have enough energy to make a molecule more strongly, however, and since vibrational energy is quantized, this can only happen at certain discrete frequencies and not at others. Figure 1 shows the infrared spectra of two triatomic molecules, H O and CO , and also that of a more complex molecule, C H OH (ethanol). Each of the peaks in these spectra corresponds to a strong absorption of infrared radiation on the macroscopic level and a sudden increase in the amplitude with which the molecule vibrates on the microscopic level. Since a polyatomic molecule can vibrate in a variety of ways, there are several peaks for each molecule. The more complex the molecule, the larger the number of peaks. Note also that not all the vibrations correspond to the stretching and unstretching of bonds. A vibration in a polyatomic molecule is defined as any periodic motion which changes the shape or size of the molecule. In this sense bending and twisting motions also count as vibrations. A useful feature of the vibrations which occur in is that many bonds and some small groups of atoms vibrate in much the same way no matter what molecule they are in. In Figure 1, for example, stretching of the O—H bond gives a peak between 2 and 3 μm for both H O and C H OH. Because of this it is possible to identify many of the functional groups in merely by looking at its infrared spectrum. Figure \(\Page {2}\) shows characteristic wavelengths by which some common functional groups can be identified. On the other hand, each molecule is a unique combination of chemical bonds and functional groups. Quite minor differences in molecular structure can result in noticeable differences in the infrared spectrum. Thus these spectra can be used in the same way the police use fingerprints. When an unknown compound is prepared, one of the first things that is usually measured is its infrared spectrum. If this spectrum should happen to match that of a previously prepared compound, the unknown compound can be readily identified. If not, it may still be possible to identify some of the functional groups that are present. In order for a molecular vibration to interact with , of the molecule must change as the vibration takes place. The larger this change in dipole moment, the more strongly the substance absorbs the incident radiation. Thus very polar bonds like O—H and C==O usually produce very prominent peaks in an infrared spectrum. Conversely some vibrations do not feature in the infrared at all. In particular, diatomic molecules like N and O , in which both atoms are identical, have zero dipole moment at any stage in a vibration. They produce no absorption in the infrared. Since N and O , are the chief constituents of the air, it is just as well that they do not absorb infrared radiation. The atmosphere would become intolerably hot if they did! As it is, only the minor constituents of the atmosphere, CO and H O, absorb in the infrared. Nevertheless this absorption still plays an important role in maintaining the surface of the earth at its current temperature. The earth absorbs energy from the sun by day, and radiates this energy away at night. The inflow and outflow must balance on average, otherwise the earth would heat up or cool down. Most of the sun’s radiation is in the visible region of the spectrum, but the radiation which escapes from the much cooler earth is mainly in the infrared, centered around 10 to 12 μm. As you can see from Figure \(\Page {1}\), water absorbs infrared radiation between 2 to 3 μm and 6 to 7.5 μm. Water also absorbs strongly above 18 μm. Thus much of the outgoing infrared radiation is absorbed by water vapor in the earth’s atmosphere and prevented from escaping. You may have noticed that after a really humid summer day the temperature does not fall very fast at night. Excess water vapor in the atmosphere prevents radiation from escaping the earth’s surface. On the other hand, in a desert area the low humidity allows rapid heat loss. Although rocks may become hot enough to fry an egg by day, temperatures often drop to freezing overnight. While local concentrations of water vapor may vary from time to time, the total quantity in the earth’s atmosphere is buffered by the vast areas of ocean and remains nearly constant. Thus the average absorption of outgoing radiation by water seldom changes. The quantity of CO in the atmosphere is not so well regulated, however, and it appears that human activities are causing it to increase. (Using the data provided in , you can calculate that about 9.4 Pg (9.4 × 10 g) of CO results from the combustion of fossil fuels in the world each year.) Even in a relatively non- industrial area such as Hawaii, there has been a steady increase in CO concentration for many years. Referring to Figure \(\Page {1}\) again, we can clearly see that infrared absorption by CO occurs in just those parts of the spectrum that were not blocked by H O absorption. Thus increasing the concentration of CO should decrease earth’s radiation to outer space and might increase the average surface temperature. On a global scale this is called the —the CO and H O act like the glass in a greenhouse, allowing visible light to pass in but blocking the loss of infrared. Climatologists have predicted that during the hundred years human beings have been using fossil fuels, the greenhouse effect should have raised surface temperatures by 0.5 to 1.0 K. Until 1950 that prediction appeared to have been borne out, but measured temperatures have since fallen back to about the 1900 level. Attempts to explain this drop on the basis of additional particulate matter in the atmosphere have met with varying degrees of success and failure. All that can be said for certain is that we know far less about the atmosphere and world climate than we would like. | 6,711 | 2,483 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/10%3A_Electrolyte_Solutions/10.04%3A_The_Debye-Huckel_Theory |
\( \newcommand{\tx}[1]{\text{#1}} % text in math mode\)
\( \newcommand{\subs}[1]{_{\text{#1}}} % subscript text\)
\( \newcommand{\sups}[1]{^{\text{#1}}} % superscript text\)
\( \newcommand{\st}{^\circ} % standard state symbol\)
\( \newcommand{\id}{^{\text{id}}} % ideal\)
\( \newcommand{\rf}{^{\text{ref}}} % reference state\)
\( \newcommand{\units}[1]{\mbox{$\thinspace$#1}}\)
\( \newcommand{\K}{\units{K}} % kelvins\)
\( \newcommand{\degC}{^\circ\text{C}} % degrees Celsius\)
\( \newcommand{\br}{\units{bar}} % bar (\bar is already defined)\)
\( \newcommand{\Pa}{\units{Pa}}\)
\( \newcommand{\mol}{\units{mol}} % mole\)
\( \newcommand{\V}{\units{V}} % volts\)
\( \newcommand{\timesten}[1]{\mbox{$\,\times\,10^{#1}$}}\)
\( \newcommand{\per}{^{-1}} % minus one power\)
\( \newcommand{\m}{_{\text{m}}} % subscript m for molar quantity\)
\( \newcommand{\CVm}{C_{V,\text{m}}} % molar heat capacity at const.V\)
\( \newcommand{\Cpm}{C_{p,\text{m}}} % molar heat capacity at const.p\)
\( \newcommand{\kT}{\kappa_T} % isothermal compressibility\)
\( \newcommand{\A}{_{\text{A}}} % subscript A for solvent or state A\)
\( \newcommand{\B}{_{\text{B}}} % subscript B for solute or state B\)
\( \newcommand{\bd}{_{\text{b}}} % subscript b for boundary or boiling point\)
\( \newcommand{\C}{_{\text{C}}} % subscript C\)
\( \newcommand{\f}{_{\text{f}}} % subscript f for freezing point\)
\( \newcommand{\mA}{_{\text{m},\text{A}}} % subscript m,A (m=molar)\)
\( \newcommand{\mB}{_{\text{m},\text{B}}} % subscript m,B (m=molar)\)
\( \newcommand{\mi}{_{\text{m},i}} % subscript m,i (m=molar)\)
\( \newcommand{\fA}{_{\text{f},\text{A}}} % subscript f,A (for fr. pt.)\)
\( \newcommand{\fB}{_{\text{f},\text{B}}} % subscript f,B (for fr. pt.)\)
\( \newcommand{\xbB}{_{x,\text{B}}} % x basis, B\)
\( \newcommand{\xbC}{_{x,\text{C}}} % x basis, C\)
\( \newcommand{\cbB}{_{c,\text{B}}} % c basis, B\)
\( \newcommand{\mbB}{_{m,\text{B}}} % m basis, B\)
\( \newcommand{\kHi}{k_{\text{H},i}} % Henry's law constant, x basis, i\)
\( \newcommand{\kHB}{k_{\text{H,B}}} % Henry's law constant, x basis, B\)
\( \newcommand{\arrow}{\,\rightarrow\,} % right arrow with extra spaces\)
\( \newcommand{\arrows}{\,\rightleftharpoons\,} % double arrows with extra spaces\)
\( \newcommand{\ra}{\rightarrow} % right arrow (can be used in text mode)\)
\( \newcommand{\eq}{\subs{eq}} % equilibrium state\)
\( \newcommand{\onehalf}{\textstyle\frac{1}{2}\D} % small 1/2 for display equation\)
\( \newcommand{\sys}{\subs{sys}} % system property\)
\( \newcommand{\sur}{\sups{sur}} % surroundings\)
\( \renewcommand{\in}{\sups{int}} % internal\)
\( \newcommand{\lab}{\subs{lab}} % lab frame\)
\( \newcommand{\cm}{\subs{cm}} % center of mass\)
\( \newcommand{\rev}{\subs{rev}} % reversible\)
\( \newcommand{\irr}{\subs{irr}} % irreversible\)
\( \newcommand{\fric}{\subs{fric}} % friction\)
\( \newcommand{\diss}{\subs{diss}} % dissipation\)
\( \newcommand{\el}{\subs{el}} % electrical\)
\( \newcommand{\cell}{\subs{cell}} % cell\)
\( \newcommand{\As}{A\subs{s}} % surface area\)
\( \newcommand{\E}{^\mathsf{E}} % excess quantity (superscript)\)
\( \newcommand{\allni}{\{n_i \}} % set of all n_i\)
\( \newcommand{\sol}{\hspace{-.1em}\tx{(sol)}}\)
\( \newcommand{\solmB}{\tx{(sol,$\,$$m\B$)}}\)
\( \newcommand{\dil}{\tx{(dil)}}\)
\( \newcommand{\sln}{\tx{(sln)}}\)
\( \newcommand{\mix}{\tx{(mix)}}\)
\( \newcommand{\rxn}{\tx{(rxn)}}\)
\( \newcommand{\expt}{\tx{(expt)}}\)
\( \newcommand{\solid}{\tx{(s)}}\)
\( \newcommand{\liquid}{\tx{(l)}}\)
\( \newcommand{\gas}{\tx{(g)}}\)
\( \newcommand{\pha}{\alpha} % phase alpha\)
\( \newcommand{\phb}{\beta} % phase beta\)
\( \newcommand{\phg}{\gamma} % phase gamma\)
\( \newcommand{\aph}{^{\alpha}} % alpha phase superscript\)
\( \newcommand{\bph}{^{\beta}} % beta phase superscript\)
\( \newcommand{\gph}{^{\gamma}} % gamma phase superscript\)
\( \newcommand{\aphp}{^{\alpha'}} % alpha prime phase superscript\)
\( \newcommand{\bphp}{^{\beta'}} % beta prime phase superscript\)
\( \newcommand{\gphp}{^{\gamma'}} % gamma prime phase superscript\)
\( \newcommand{\apht}{\small\aph} % alpha phase tiny superscript\)
\( \newcommand{\bpht}{\small\bph} % beta phase tiny superscript\)
\( \newcommand{\gpht}{\small\gph} % gamma phase tiny superscript\) \( \newcommand{\upOmega}{\Omega}\) \( \newcommand{\dif}{\mathop{}\!\mathrm{d}} % roman d in math mode, preceded by space\)
\( \newcommand{\Dif}{\mathop{}\!\mathrm{D}} % roman D in math mode, preceded by space\)
\( \newcommand{\df}{\dif\hspace{0.05em} f} % df\) \(\newcommand{\dBar}{\mathop{}\!\mathrm{d}\hspace-.3em\raise1.05ex{\Rule{.8ex}{.125ex}{0ex}}} % inexact differential \)
\( \newcommand{\dq}{\dBar q} % heat differential\)
\( \newcommand{\dw}{\dBar w} % work differential\)
\( \newcommand{\dQ}{\dBar Q} % infinitesimal charge\)
\( \newcommand{\dx}{\dif\hspace{0.05em} x} % dx\)
\( \newcommand{\dt}{\dif\hspace{0.05em} t} % dt\)
\( \newcommand{\difp}{\dif\hspace{0.05em} p} % dp\)
\( \newcommand{\Del}{\Delta}\)
\( \newcommand{\Delsub}[1]{\Delta_{\text{#1}}}\)
\( \newcommand{\pd}[3]{(\partial #1 / \partial #2 )_{#3}} % \pd{}{}{} - partial derivative, one line\)
\( \newcommand{\Pd}[3]{\left( \dfrac {\partial #1} {\partial #2}\right)_{#3}} % Pd{}{}{} - Partial derivative, built-up\)
\( \newcommand{\bpd}[3]{[ \partial #1 / \partial #2 ]_{#3}}\)
\( \newcommand{\bPd}[3]{\left[ \dfrac {\partial #1} {\partial #2}\right]_{#3}}\)
\( \newcommand{\dotprod}{\small\bullet}\)
\( \newcommand{\fug}{f} % fugacity\)
\( \newcommand{\g}{\gamma} % solute activity coefficient, or gamma in general\)
\( \newcommand{\G}{\varGamma} % activity coefficient of a reference state (pressure factor)\)
\( \newcommand{\ecp}{\widetilde{\mu}} % electrochemical or total potential\)
\( \newcommand{\Eeq}{E\subs{cell, eq}} % equilibrium cell potential\)
\( \newcommand{\Ej}{E\subs{j}} % liquid junction potential\)
\( \newcommand{\mue}{\mu\subs{e}} % electron chemical potential\)
\( \newcommand{\defn}{\,\stackrel{\mathrm{def}}{=}\,} % "equal by definition" symbol\) \( \newcommand{\D}{\displaystyle} % for a line in built-up\)
\( \newcommand{\s}{\smash[b]} % use in equations with conditions of validity\)
\( \newcommand{\cond}[1]{\\[-2.5pt]{}\tag*{#1}}\)
\( \newcommand{\nextcond}[1]{\\[-5pt]{}\tag*{#1}}\)
\( \newcommand{\R}{8.3145\units{J$\,$K$\per\,$mol$\per$}} % gas constant value\)
\( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \)
\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) The theory of Peter Debye and Erich Hückel (1923) provides theoretical expressions for single-ion activity coefficients and mean ionic activity coefficients in electrolyte solutions. The expressions in one form or another are very useful for extrapolation of quantities that include mean ionic activity coefficients to low solute molality or infinite dilution. The only interactions the theory considers are the electrostatic interactions between ions. These interactions are much stronger than those between uncharged molecules, and they die off more slowly with distance. If the positions of ions in an electrolyte solution were completely random, the net effect of electrostatic ion–ion interactions would be zero, because each cation–cation or anion–anion repulsion would be balanced by a cation–anion attraction. The positions are not random, however: each cation has a surplus of anions in its immediate environment, and each anion has a surplus of neighboring cations. Each ion therefore has a net attractive interaction with the surrounding ion atmosphere. The result for a cation species at low electrolyte molality is a decrease of \(\mu_+\) compared to the cation at same molality in the absence of ion–ion interactions, meaning that the single-ion activity coefficient \(\g_+\) becomes less than \(1\) as the electrolyte molality is increased beyond the ideal-dilute range. Similarly, \(\g_-\) also becomes less than \(1\). According to the Debye–Hückel theory, the single-ion activity coefficient \(\g_i\) of ion \(i\) in a solution of one or more electrolytes is given by \begin{equation} \ln\g_i = -\frac{A\subs{DH}z_i^2\sqrt{I_m}}{1 + B\subs{DH}a\sqrt{I_m}} \tag{10.4.1} \end{equation} where Lewis and Randall ( , 1112–1154, 1921) introduced the term , defined by Eq. 10.4.2, two years before the Debye–Hückel theory was published. They found empirically that in dilute solutions, the mean ionic activity coefficient of a given strong electrolyte is the same in all solutions having the same ionic strength. From Eqs. 10.3.8 and 10.4.1 and the electroneutrality condition \(\nu_+ z_+ {=} \nu_- z_-\), we obtain the following expression for the logarithm of the mean ionic activity coefficient of an electrolyte solute: \begin{equation} \ln\g_{\pm} = -\frac{A\subs{DH} \left| z_{+}z_{-} \right| \sqrt{I_m}} {1 + B\subs{DH}a\sqrt{I_m}} \tag{10.4.7} \end{equation} In this equation, \(z_+\) and \(z_-\) are the charge numbers of the cation and anion of the solute. Since the right side of Eq. 10.4.7 is negative at finite solute molalities, and zero at infinite dilution, the theory predicts that \(\g_{\pm}\) is less than \(1\) at finite solute molalities and approaches \(1\) at infinite dilution. Figure 10.4 shows \(\ln\g_{\pm}\) as a function of \(\sqrt{I_m}\) for aqueous HCl and CaCl\(_2\). The experimental curves have the limiting slopes predicted by the Debye–Hückel limiting law (Eq. 10.4.8), but at a low ionic strength the curves begin to deviate significantly from the linear relations predicted by that law. The full Debye–Hückel equation (Eq. 10.4.7) fits the experimental curves over a wider range of ionic strength. | 10,516 | 2,485 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/10%3A_Electrolyte_Solutions/10.03%3A_Electrolytes_in_General |
\( \newcommand{\tx}[1]{\text{#1}} % text in math mode\)
\( \newcommand{\subs}[1]{_{\text{#1}}} % subscript text\)
\( \newcommand{\sups}[1]{^{\text{#1}}} % superscript text\)
\( \newcommand{\st}{^\circ} % standard state symbol\)
\( \newcommand{\id}{^{\text{id}}} % ideal\)
\( \newcommand{\rf}{^{\text{ref}}} % reference state\)
\( \newcommand{\units}[1]{\mbox{$\thinspace$#1}}\)
\( \newcommand{\K}{\units{K}} % kelvins\)
\( \newcommand{\degC}{^\circ\text{C}} % degrees Celsius\)
\( \newcommand{\br}{\units{bar}} % bar (\bar is already defined)\)
\( \newcommand{\Pa}{\units{Pa}}\)
\( \newcommand{\mol}{\units{mol}} % mole\)
\( \newcommand{\V}{\units{V}} % volts\)
\( \newcommand{\timesten}[1]{\mbox{$\,\times\,10^{#1}$}}\)
\( \newcommand{\per}{^{-1}} % minus one power\)
\( \newcommand{\m}{_{\text{m}}} % subscript m for molar quantity\)
\( \newcommand{\CVm}{C_{V,\text{m}}} % molar heat capacity at const.V\)
\( \newcommand{\Cpm}{C_{p,\text{m}}} % molar heat capacity at const.p\)
\( \newcommand{\kT}{\kappa_T} % isothermal compressibility\)
\( \newcommand{\A}{_{\text{A}}} % subscript A for solvent or state A\)
\( \newcommand{\B}{_{\text{B}}} % subscript B for solute or state B\)
\( \newcommand{\bd}{_{\text{b}}} % subscript b for boundary or boiling point\)
\( \newcommand{\C}{_{\text{C}}} % subscript C\)
\( \newcommand{\f}{_{\text{f}}} % subscript f for freezing point\)
\( \newcommand{\mA}{_{\text{m},\text{A}}} % subscript m,A (m=molar)\)
\( \newcommand{\mB}{_{\text{m},\text{B}}} % subscript m,B (m=molar)\)
\( \newcommand{\mi}{_{\text{m},i}} % subscript m,i (m=molar)\)
\( \newcommand{\fA}{_{\text{f},\text{A}}} % subscript f,A (for fr. pt.)\)
\( \newcommand{\fB}{_{\text{f},\text{B}}} % subscript f,B (for fr. pt.)\)
\( \newcommand{\xbB}{_{x,\text{B}}} % x basis, B\)
\( \newcommand{\xbC}{_{x,\text{C}}} % x basis, C\)
\( \newcommand{\cbB}{_{c,\text{B}}} % c basis, B\)
\( \newcommand{\mbB}{_{m,\text{B}}} % m basis, B\)
\( \newcommand{\kHi}{k_{\text{H},i}} % Henry's law constant, x basis, i\)
\( \newcommand{\kHB}{k_{\text{H,B}}} % Henry's law constant, x basis, B\)
\( \newcommand{\arrow}{\,\rightarrow\,} % right arrow with extra spaces\)
\( \newcommand{\arrows}{\,\rightleftharpoons\,} % double arrows with extra spaces\)
\( \newcommand{\ra}{\rightarrow} % right arrow (can be used in text mode)\)
\( \newcommand{\eq}{\subs{eq}} % equilibrium state\)
\( \newcommand{\onehalf}{\textstyle\frac{1}{2}\D} % small 1/2 for display equation\)
\( \newcommand{\sys}{\subs{sys}} % system property\)
\( \newcommand{\sur}{\sups{sur}} % surroundings\)
\( \renewcommand{\in}{\sups{int}} % internal\)
\( \newcommand{\lab}{\subs{lab}} % lab frame\)
\( \newcommand{\cm}{\subs{cm}} % center of mass\)
\( \newcommand{\rev}{\subs{rev}} % reversible\)
\( \newcommand{\irr}{\subs{irr}} % irreversible\)
\( \newcommand{\fric}{\subs{fric}} % friction\)
\( \newcommand{\diss}{\subs{diss}} % dissipation\)
\( \newcommand{\el}{\subs{el}} % electrical\)
\( \newcommand{\cell}{\subs{cell}} % cell\)
\( \newcommand{\As}{A\subs{s}} % surface area\)
\( \newcommand{\E}{^\mathsf{E}} % excess quantity (superscript)\)
\( \newcommand{\allni}{\{n_i \}} % set of all n_i\)
\( \newcommand{\sol}{\hspace{-.1em}\tx{(sol)}}\)
\( \newcommand{\solmB}{\tx{(sol,$\,$$m\B$)}}\)
\( \newcommand{\dil}{\tx{(dil)}}\)
\( \newcommand{\sln}{\tx{(sln)}}\)
\( \newcommand{\mix}{\tx{(mix)}}\)
\( \newcommand{\rxn}{\tx{(rxn)}}\)
\( \newcommand{\expt}{\tx{(expt)}}\)
\( \newcommand{\solid}{\tx{(s)}}\)
\( \newcommand{\liquid}{\tx{(l)}}\)
\( \newcommand{\gas}{\tx{(g)}}\)
\( \newcommand{\pha}{\alpha} % phase alpha\)
\( \newcommand{\phb}{\beta} % phase beta\)
\( \newcommand{\phg}{\gamma} % phase gamma\)
\( \newcommand{\aph}{^{\alpha}} % alpha phase superscript\)
\( \newcommand{\bph}{^{\beta}} % beta phase superscript\)
\( \newcommand{\gph}{^{\gamma}} % gamma phase superscript\)
\( \newcommand{\aphp}{^{\alpha'}} % alpha prime phase superscript\)
\( \newcommand{\bphp}{^{\beta'}} % beta prime phase superscript\)
\( \newcommand{\gphp}{^{\gamma'}} % gamma prime phase superscript\)
\( \newcommand{\apht}{\small\aph} % alpha phase tiny superscript\)
\( \newcommand{\bpht}{\small\bph} % beta phase tiny superscript\)
\( \newcommand{\gpht}{\small\gph} % gamma phase tiny superscript\) \( \newcommand{\upOmega}{\Omega}\) \( \newcommand{\dif}{\mathop{}\!\mathrm{d}} % roman d in math mode, preceded by space\)
\( \newcommand{\Dif}{\mathop{}\!\mathrm{D}} % roman D in math mode, preceded by space\)
\( \newcommand{\df}{\dif\hspace{0.05em} f} % df\) \(\newcommand{\dBar}{\mathop{}\!\mathrm{d}\hspace-.3em\raise1.05ex{\Rule{.8ex}{.125ex}{0ex}}} % inexact differential \)
\( \newcommand{\dq}{\dBar q} % heat differential\)
\( \newcommand{\dw}{\dBar w} % work differential\)
\( \newcommand{\dQ}{\dBar Q} % infinitesimal charge\)
\( \newcommand{\dx}{\dif\hspace{0.05em} x} % dx\)
\( \newcommand{\dt}{\dif\hspace{0.05em} t} % dt\)
\( \newcommand{\difp}{\dif\hspace{0.05em} p} % dp\)
\( \newcommand{\Del}{\Delta}\)
\( \newcommand{\Delsub}[1]{\Delta_{\text{#1}}}\)
\( \newcommand{\pd}[3]{(\partial #1 / \partial #2 )_{#3}} % \pd{}{}{} - partial derivative, one line\)
\( \newcommand{\Pd}[3]{\left( \dfrac {\partial #1} {\partial #2}\right)_{#3}} % Pd{}{}{} - Partial derivative, built-up\)
\( \newcommand{\bpd}[3]{[ \partial #1 / \partial #2 ]_{#3}}\)
\( \newcommand{\bPd}[3]{\left[ \dfrac {\partial #1} {\partial #2}\right]_{#3}}\)
\( \newcommand{\dotprod}{\small\bullet}\)
\( \newcommand{\fug}{f} % fugacity\)
\( \newcommand{\g}{\gamma} % solute activity coefficient, or gamma in general\)
\( \newcommand{\G}{\varGamma} % activity coefficient of a reference state (pressure factor)\)
\( \newcommand{\ecp}{\widetilde{\mu}} % electrochemical or total potential\)
\( \newcommand{\Eeq}{E\subs{cell, eq}} % equilibrium cell potential\)
\( \newcommand{\Ej}{E\subs{j}} % liquid junction potential\)
\( \newcommand{\mue}{\mu\subs{e}} % electron chemical potential\)
\( \newcommand{\defn}{\,\stackrel{\mathrm{def}}{=}\,} % "equal by definition" symbol\) \( \newcommand{\D}{\displaystyle} % for a line in built-up\)
\( \newcommand{\s}{\smash[b]} % use in equations with conditions of validity\)
\( \newcommand{\cond}[1]{\\[-2.5pt]{}\tag*{#1}}\)
\( \newcommand{\nextcond}[1]{\\[-5pt]{}\tag*{#1}}\)
\( \newcommand{\R}{8.3145\units{J$\,$K$\per\,$mol$\per$}} % gas constant value\)
\( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \)
\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) The formula unit of a symmetrical electrolyte solute has more than two ions. General formulas for the solute as a whole are more complicated than those for the symmetrical case treated in the preceding section, but are derived by the same reasoning. Again we assume the solute dissociates completely into its constituent ions. We define the following symbols: \(\nu_+ =\) the number of cations per solute formula unit \(\nu_- =\) the number of anions per solute formula unit \(\nu =\) the sum \(\nu_+ + \nu_-\) For example, if the solute formula is Al\(_2\)(SO\(_4\))\(_3\), the values are \(\nu_+ {=} 2\), \(\nu_- {=} 3\), and \(\nu {=} 5\). In a solution of a single electrolyte solute that is not necessarily symmetrical, the ion molalities are related to the overall solute molality by \begin{equation} m_+ = \nu_+m\B \qquad m_- = \nu_-m\B \tag{10.3.1} \end{equation} From the additivity rule for the Gibbs energy, we have \begin{equation} \begin{split} G & =n\A\mu\A + n\B\mu\B\cr & = n\A\mu\A + \nu_+n\B\mu_{+}+\nu_-n\B\mu_{-} \end{split} \tag{10.3.2} \end{equation} giving the relation \begin{equation} \mu\B=\nu_+\mu_+ + \nu_-\mu_- \tag{10.3.3} \end{equation} in place of Eq. 10.2.4. The cations and anions are in the same phase of electric potential \(\phi\). We use Eqs. 10.1.4 and 10.1.5 to obtain \begin{equation} \nu_+\mu_+(\phi) + \nu_-\mu_-(\phi) = \nu_+\mu_+(0) + \nu_-\mu_-(0) + (\nu_+z_+ + \nu_-z_-)F\phi \tag{10.3.4} \end{equation} Electrical neutrality requires that \((\nu_+z_+ + \nu_-z_-)\) be zero, giving \begin{equation} \mu\B = \nu_+\mu_+(0) + \nu_-\mu_-(0) \tag{10.3.5} \end{equation} By combining Eq. 10.3.5 with Eqs. 10.1.10, 10.3.1, and 10.3.3, we obtain \begin{equation} \mu\B = \mu\B\rf + RT \ln \left[ \left(\nu_+^{\nu_+}\nu_-^{\nu_-}\right) \left(\g_+^{\nu_+}\right)\left(\g_-^{\nu_-}\right) \left( \frac{m\B}{m\st} \right)^{\nu} \right] \tag{10.3.6} \end{equation} where \(\mu\B\rf=\nu_+\mu_+\rf+\nu_-\mu_-\rf\) is the chemical potential of the solute in the hypothetical reference state at \(\phi{=}0\) in which B is at the standard molality and behaves as at infinite dilution. Equation 10.3.6 is the generalization of Eq. 10.2.6. It shows that although \(\mu_+\) and \(\mu_-\) depend on \(\phi\), \(\mu\B\) does not. The mean ionic activity coefficient \(\g_{\pm}\) is defined in general by \begin{equation} \g_{\pm}^{\nu} = \left( \g_{+}^{\nu_{+}} \right) \left( \g_{-}^{\nu_{-}} \right) \tag{10.3.7} \end{equation} or \begin{equation} \g_{\pm} = \left( \g_{+}^{\nu_{+}}\g_{-}^{\nu_{-}} \right)^{1/\nu} \tag{10.3.8} \end{equation} Thus \(\g_{\pm}\) is a geometric average of \(\g_{+}\) and \(\g_{-}\) weighted by the numbers of the cations and anions in the solute formula unit. With a substitution from Eq. 10.3.7, Eq. 10.3.6 becomes \begin{equation} \mu\B = \mu\B\rf + RT \ln \left[ \left(\nu_+^{\nu_+}\nu_-^{\nu_-}\right) \g_{\pm}^{\nu} \left( \frac{m\B}{m\st} \right)^{\nu} \right] \tag{10.3.9} \end{equation} Since \(\mu\B-\mu\B\rf\) is a measurable quantity, so also is \(\g_{\pm}\). The solute activity, defined by \(\mu\B=\mu\mbB\st+RT\ln a\mbB\), is \begin{equation} a\mbB = \left( \nu_{+}^{\nu_{+}} \nu_{-}^{\nu_{-}} \right) \G\mbB \g_{\pm}^{\nu} \left( \frac{m\B}{m\st} \right)^{\nu} \tag{10.3.10} \end{equation} where \(\G\mbB\) is the pressure factor that we can evaluate with Eq. 10.2.12. Equation 10.3.10 is the generalization of Eq. 10.2.10. From Eqs. 10.1.12, 10.1.13, and 10.2.11 and the relations \(\mu\B\rf=\nu_+\mu_+\rf+\nu_-\mu_-\rf\) and \(\mu\B\st=\nu_+\mu_+\st+\nu_-\mu_-\st\), we obtain the relation \begin{equation} \G\mbB = \G_+^{\nu_+}\G_-^{\nu_-} \tag{10.3.11} \end{equation} Equation 10.3.3 relates the chemical potential of electrolyte B in a binary solution to the single-ion chemical potentials of its constituent ions: \begin{equation} \mu\B=\nu_+\mu_+ + \nu_-\mu_- \tag{10.3.12} \end{equation} This relation is valid for each individual solute substance in a multisolute solution, even when two or more of the electrolyte solutes have an ion species in common. As an illustration of this principle, consider a solution prepared by dissolving amounts \(n\B\) of BaI\(_2\) and \(n\C\) of CsI in an amount \(n\A\) of H\(_2\)O. Assume the dissolved salts are completely dissociated into ions, with the I\(^-\) ion common to both. The additivity rule for the Gibbs energy of this solution can be written in the form \begin{equation} G = n\A\mu\A + n\B\mu\B + n\C\mu\C \tag{10.3.13} \end{equation} and also, using single-ion quantities, in the form \begin{equation} G = n\A\mu\A + n\B\mu(\tx{Ba\(^{2+}\)}) + 2n\B\mu(\tx{I\(^-\)}) + n\C\mu(\tx{Cs\(^+\)}) + n\C\mu(\tx{I\(^-\)}) \tag{10.3.14} \end{equation} Comparing Eqs. 10.3.13 and 10.3.14, we find the following relations must exist between the chemical potentials of the solute substances and the ion species: \begin{equation} \mu\B = \mu(\tx{Ba\(^{2+}\)})+2\mu(\tx{I\(^-\)}) \qquad \mu\C = \mu(\tx{Cs\(^+\)})+\mu(\tx{I\(^-\)}) \tag{10.3.15} \end{equation} These relations agree with Eq. 10.3.12. Note that \(\mu(\tx{I\(^-\)})\), the chemical potential of the ion common to both salts, appears in both relations. The solute activity \(a\mbB\) is defined by the relation \(\mu\B = \mu\B\st + RT\ln a\mbB\) (Eq. 10.2.9). Using this relation together with Eqs. 10.1.7 and 10.1.14, we find that the solute activity is related to ion molalities by \begin{equation} a\mbB = \G\mbB \g_{\pm}^\nu \left( \frac{m_+}{m\st} \right)^{\nu_+} \left( \frac{m_-}{m\st} \right)^{\nu_-} \tag{10.3.16} \end{equation} where the pressure factor \(\G\mbB\) is defined in Eq. 10.2.11. The ion molalities in this expression refer to the constituent ions of solute B, which in a multisolute solution are not necessarily present in the same stoichiometric ratio as in the solute substance. For instance, suppose we apply Eq. 10.3.16 to the solution of BaI\(_2\) and CsI used above as an illustration of a multisolute solution, letting \(a\mbB\) be the activity of solute substance BaI\(_2\). The quantities \(m_+\) and \(m_-\) in the equation are then the molalities of the Ba\(^{2+}\) and I\(^-\) ions, and \(\g_{\pm}\) is the mean ionic activity coefficient of the dissolved BaI\(_2\). Note that in this solution the Ba\(^{2+}\) and I\(^-\) ions are not present in the 1:2 ratio found in BaI\(_2\), because I\(^-\) is a constituent of both solutes. In the preceding sections of this chapter, the electrolyte solute or solutes have been assumed to be completely dissociated into their constituent ions at all molalities. Some solutions, however, contain —closely associated ions of opposite charge. Furthermore, in solutions of some electrolytes (often called “weak” electrolytes), an equilibrium is established between ions and electrically-neutral molecules. In these kinds of solutions, the relations between solute molality and ion molalities given by Eq. 10.3.1 are no longer valid. When dissociation is not complete, the expression for \(\mu\B\) given by Eq. 10.3.9 can still be used. However, the quantity \(\g_{\pm}\) appearing in the expression no longer has the physical significance of being the geometric average of the activity coefficients of the actual dissociated ions, and is called the of the electrolyte. | 14,618 | 2,486 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/01%3A_General_Techniques/1.04%3A_Heating_and_Cooling_Methods/1.4G%3A_Heating_Mantles |
Heating mantles are a relatively safe way to heat flammable organic liquids in a round bottomed flask (Figure 1.52). The mantles are cup-shaped and designed for different sizes of round bottomed flask (Figure 1.51a). If a mantle does not fit a round bottomed flask perfectly, sand can be added to ensure good thermal contact (Figure 1.51c). The mantles should be connected directly to the outlet, but first to a " " (blue piece of equipment in Figure 1.51b) which then connects to the outlet and delivers variable voltage to the mantle. A Variac set to "100" would be equivalent to plugging the mantle directly into the wall \(\left( 100\% \right)\), while a setting of "50" means the delivered voltage is halved \(\left( 50\% \right)\). By controlling the delivered voltage, Variacs are used to regulate the temperature of a heating mantle. There is variation between devices, and settings must be experimented with to determine appropriate heating rates. Heating mantles take some time to warm up (so may be pre-heated during setup of an apparatus), and also take some time to cool down. The mantle will remain warm even after turning off the Variac, and therefore flasks have to be removed from the mantle in order to cool (Figure 1.52c). the main hazard with heating mantles is that flammable organic liquids spilled on the surface of a hot mantle do have the possibility of ignition. | 1,403 | 2,487 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/02%3A_Chromatography/2.03%3A_Thin_Layer_Chromatography_(TLC)/2.3A%3A_Overview_of_TLC |
Two commonly used stationary phases in chromatography are silica (\(\ce{SiO2} \cdot x \ce{H2O}\)) and alumina (\(\ce{Al2O3} \cdot x \ce{H2O}\)). Both of these materials are fine white powders that, unlike paper, do not stand up by themselves, so they are deposited in a thin layer on some sort of backing (either glass, aluminum, or plastic). Using these thin layers of stationary phase for separations is called " " (TLC), and is procedurally performed much the same way as paper chromatography (Figure 2.5). It is very common for organic compounds to appear colorless on the white adsorbent background, which poses the challenge of the separation that occurs in a TLC. TLC plates often have to be "visualized," meaning something has to be done to the plate in order to temporarily see the compounds. Common methods of visualization are to use UV light (Figures 2.6 a+b) or a chemical stain (Figure 2.6c). (Butte Community College). is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International . Complete text is available . | 1,080 | 2,488 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/07%3A_Periodic_Properties_of_the_Elements/7.03%3A_Sizes_of_Atoms_and_Ions |
Although some people fall into the trap of visualizing atoms and ions as small, hard spheres similar to miniature table-tennis balls or marbles, the quantum mechanical model tells us that their shapes and boundaries are much less definite than those images suggest. As a result, atoms and ions cannot be said to have exact sizes; however, some atoms are larger or smaller than others, and this influences their chemistry. In this section, we discuss how atomic and ion “sizes” are defined and obtained. Recall that the probability of finding an electron in the various available orbitals falls off slowly as the distance from the nucleus increases. This point is illustrated in Figure \(\Page {1}\) which shows a plot of total electron density for all occupied orbitals for three noble gases as a function of their distance from the nucleus. Electron density diminishes gradually with increasing distance, which makes it impossible to draw a sharp line marking the boundary of an atom. Figure \(\Page {1}\) also shows that there are distinct peaks in the total electron density at particular distances and that these peaks occur at different distances from the nucleus for each element. Each peak in a given plot corresponds to the electron density in a given principal shell. Because helium has only one filled shell ( = 1), it shows only a single peak. In contrast, neon, with filled = 1 and 2 principal shells, has two peaks. Argon, with filled = 1, 2, and 3 principal shells, has three peaks. The peak for the filled = 1 shell occurs at successively shorter distances for neon ( = 10) and argon ( = 18) because, with a greater number of protons, their nuclei are more positively charged than that of helium. Because the 1 shell is closest to the nucleus, its electrons are very poorly shielded by electrons in filled shells with larger values of . Consequently, the two electrons in the = 1 shell experience nearly the full nuclear charge, resulting in a strong electrostatic interaction between the electrons and the nucleus. The energy of the = 1 shell also decreases tremendously (the filled 1 orbital becomes more stable) as the nuclear charge increases. For similar reasons, the filled = 2 shell in argon is located closer to the nucleus and has a lower energy than the = 2 shell in neon. Figure \(\Page {1}\) illustrates the difficulty of measuring the dimensions of an individual atom. Because distances between the nuclei in pairs of covalently bonded atoms can be measured quite precisely, however, chemists use these distances as a basis for describing the approximate sizes of atoms. For example, the internuclear distance in the diatomic Cl molecule is known to be 198 pm. We assign half of this distance to each chlorine atom, giving chlorine a (\(r_{cov}\)), of 99 pm or 0.99 Å (Figure \(\Page {2a}\)). In a similar approach, we can use the lengths of carbon–carbon single bonds in organic compounds, which are remarkably uniform at 154 pm, to assign a value of 77 pm as the covalent atomic radius for carbon. If these values do indeed reflect the actual sizes of the atoms, then we should be able to predict the lengths of covalent bonds formed between different elements by adding them. For example, we would predict a carbon–chlorine distance of 77 pm + 99 pm = 176 pm for a C–Cl bond, which is very close to the average value observed in many organochlorine compounds. Covalent atomic radii can be determined for most of the nonmetals, but how do chemists obtain atomic radii for elements that do not form covalent bonds? For these elements, a variety of other methods have been developed. With a metal, for example, the (\(r_{met}\)) is defined as half the distance between the nuclei of two adjacent metal atoms in the solid (Figure \(\Page {2b}\)). For elements such as the noble gases, most of which form no stable compounds, we can use what is called the (\(r_{vdW}\)), which is half the internuclear distance between two nonbonded atoms in the solid (Figure \(\Page {2c}\)). This is somewhat difficult for helium which does not form a solid at any temperature. An atom such as chlorine has both a covalent radius (the distance between the two atoms in a \(\ce{Cl2}\) molecule) and a van der Waals radius (the distance between two Cl atoms in different molecules in, for example, \(\ce{Cl2(s)}\) at low temperatures). These radii are generally not the same (Figure \(\Page {2d}\)). Because it is impossible to measure the sizes of both metallic and nonmetallic elements using any one method, chemists have developed a self-consistent way of calculating atomic radii using the quantum mechanical functions. Although the radii values obtained by such calculations are not identical to any of the experimentally measured sets of values, they do provide a way to compare the intrinsic sizes of all the elements and clearly show that atomic size varies in a periodic fashion (Figure \(\Page {3}\)). In the periodic table, atomic radii decrease from left to right across a row and increase from top to bottom down a column. Because of these two trends, the largest atoms are found in the lower left corner of the periodic table, and the smallest are found in the upper right corner (Figure \(\Page {4}\)). Trends in atomic size result from differences in the (\(Z_{eff}\)) experienced by electrons in the outermost orbitals of the elements. For all elements except H, the effective nuclear charge is always than the actual nuclear charge because of shielding effects. The greater the effective nuclear charge, the more strongly the outermost electrons are attracted to the nucleus and the smaller the atomic radius. Atomic radii decrease from left to right across a row and increase from top to bottom down a column. The atoms in the second row of the periodic table (Li through Ne) illustrate the effect of electron shielding. All have a filled 1 inner shell, but as we go from left to right across the row, the nuclear charge increases from +3 to +10. Although electrons are being added to the 2 and 2 orbitals, . Thus the single 2 electron in lithium experiences an effective nuclear charge of approximately +1 because the electrons in the filled 1 shell effectively neutralize two of the three positive charges in the nucleus. (More detailed calculations give a value of = +1.26 for Li.) In contrast, the two 2 electrons in beryllium do not shield each other very well, although the filled 1 shell effectively neutralizes two of the four positive charges in the nucleus. This means that the effective nuclear charge experienced by the 2 electrons in beryllium is between +1 and +2 (the calculated value is +1.66). Consequently, beryllium is significantly smaller than lithium. Similarly, as we proceed across the row, the increasing nuclear charge is not effectively neutralized by the electrons being added to the 2 and 2 orbitals. The result is a steady increase in the effective nuclear charge and a steady decrease in atomic size (Figure \(\Page {5}\)). The increase in atomic size going down a column is also due to electron shielding, but the situation is more complex because the principal quantum number is not constant. As we saw in Chapter 2, the size of the orbitals increases as increases, . In group 1, for example, the size of the atoms increases substantially going down the column. It may at first seem reasonable to attribute this effect to the successive addition of electrons to orbitals with increasing values of . However, it is important to remember that the radius of an orbital depends dramatically on the nuclear charge. As we go down the column of the group 1 elements, the principal quantum number increases from 2 to 6, but the nuclear charge increases from +3 to +55! As a consequence the radii of the in Cesium are much smaller than those in lithium and the electrons in those orbitals experience a much larger force of attraction to the nucleus. That force depends on the effective nuclear charge experienced by the the inner electrons. If the outermost electrons in cesium experienced the full nuclear charge of +55, a cesium atom would be very small indeed. In fact, the effective nuclear charge felt by the outermost electrons in cesium is much less than expected (6 rather than 55). This means that cesium, with a 6 valence electron configuration, is much larger than lithium, with a 2 valence electron configuration. The effective nuclear charge changes relatively little for electrons in the outermost, or valence shell, from lithium to cesium because . Even though cesium has a nuclear charge of +55, it has 54 electrons in its filled 1 2 2 3 3 4 3 4 5 4 5 shells, abbreviated as [Xe]5 4 5 , which effectively neutralize most of the 55 positive charges in the nucleus. The same dynamic is responsible for the steady increase in size observed as we go down the other columns of the periodic table. Irregularities can usually be explained by variations in effective nuclear charge. Electrons in the same principal shell are not very effective at shielding one another from the nuclear charge, whereas electrons in filled inner shells are highly effective at shielding electrons in outer shells from the nuclear charge. On the basis of their positions in the periodic table, arrange these elements in order of increasing atomic radius: aluminum, carbon, and silicon. three elements arrange in order of increasing atomic radius These elements are not all in the same column or row, so we must use pairwise comparisons. Carbon and silicon are both in group 14 with carbon lying above, so carbon is smaller than silicon (C < Si). Aluminum and silicon are both in the third row with aluminum lying to the left, so silicon is smaller than aluminum (Si < Al) because its effective nuclear charge is greater. Combining the two inequalities gives the overall order: C < Si < Al. On the basis of their positions in the periodic table, arrange these elements in order of increasing size: oxygen, phosphorus, potassium, and sulfur. O < S < P < K Atomic Radius: An ion is formed when either one or more electrons are removed from a neutral atom to form a positive ion (cation) or when additional electrons attach themselves to neutral atoms to form a negative one (anion). The designations cation or anion come from the early experiments with electricity which found that positively charged particles were attracted to the negative pole of a battery, the cathode, while negatively charged ones were attracted to the positive pole, the anode. Ionic compounds consist of regular repeating arrays of alternating positively charged cations and negatively charges anions. Although it is not possible to measure an ionic radius directly for the same reason it is not possible to directly measure an atom’s radius, it possible to measure the distance between the nuclei of a cation and an adjacent anion in an ionic compound to determine the ionic radius (the radius of a cation or anion) of one or both. As illustrated in Figure \(\Page {6}\), the internuclear distance corresponds to the of the radii of the cation and anion. A variety of methods have been developed to divide the experimentally measured distance proportionally between the smaller cation and larger anion. These methods produce sets of ionic radii that are internally consistent from one ionic compound to another, although each method gives slightly different values. For example, the radius of the Na ion is essentially the same in NaCl and Na S, as long as the same method is used to measure it. Thus despite minor differences due to methodology, certain trends can be observed. A comparison of ionic radii with atomic radii (Figure \(\Page {7}\)) shows that . When one or more electrons is removed from a neutral atom, two things happen: (1) repulsions between electrons in the same principal shell decrease because fewer electrons are present, and (2) the effective nuclear charge felt by the remaining electrons increases because there are fewer electrons to shield one another from the nucleus. Consequently, the size of the region of space occupied by electrons decreases and the ion shrinks (compare Li at 167 pm with Li at 76 pm). If different numbers of electrons can be removed to produce ions with different charges, the ion with the greatest positive charge is the smallest (compare Fe at 78 pm with Fe at 64.5 pm). Conversely, adding one or more electrons to a neutral atom causes electron–electron repulsions to increase and the effective nuclear charge to decrease, so the size of the probability region increases and the ion expands (compare F at 42 pm with F at 133 pm). Cations are smaller than the neutral atom and anions are larger. Because most elements form either a cation or an anion but not both, there are few opportunities to compare the sizes of a cation and an anion derived from the same neutral atom. A few compounds of sodium, however, contain the Na ion, allowing comparison of its size with that of the far more familiar Na ion, which is found in compounds. The radius of sodium in each of its three known oxidation states is given in Table \(\Page {1}\). All three species have a nuclear charge of +11, but they contain 10 (Na ), 11 (Na ), and 12 (Na ) electrons. The Na ion is significantly smaller than the neutral Na atom because the 3 electron has been removed to give a closed shell with = 2. The Na ion is larger than the parent Na atom because the additional electron produces a 3 valence electron configuration, while the nuclear charge remains the same. Ionic radii follow the same vertical trend as atomic radii; that is, for ions with the same charge, the ionic radius increases going down a column. The reason is the same as for atomic radii: shielding by filled inner shells produces little change in the effective nuclear charge felt by the outermost electrons. Again, principal shells with larger values of lie at successively greater distances from the nucleus. Because elements in different columns tend to form ions with different charges, it is not possible to compare ions of the same charge across a row of the periodic table. Instead, elements that are next to each other tend to form ions with the same number of electrons but with different overall charges because of their different atomic numbers. Such a set of species is known as an . For example, the isoelectronic series of species with the neon closed-shell configuration (1 2 2 ) is shown in Table \(\Page {3}\). The sizes of the ions in this series decrease smoothly from N to Al . All six of the ions contain 10 electrons in the 1 , 2 , and 2 orbitals, but the nuclear charge varies from +7 (N) to +13 (Al). As the positive charge of the nucleus increases while the number of electrons remains the same, there is a greater electrostatic attraction between the electrons and the nucleus, which causes a decrease in radius. Consequently, the ion with the greatest nuclear charge (Al ) is the smallest, and the ion with the smallest nuclear charge (N ) is the largest. The neon atom in this isoelectronic series is not listed in Table \(\Page {3}\), because neon forms no covalent or ionic compounds and hence its radius is difficult to measure. Based on their positions in the periodic table, arrange these ions in order of increasing radius: Cl , K , S , and Se . four ions order by increasing radius We see that S and Cl are at the right of the third row, while K and Se are at the far left and right ends of the fourth row, respectively. K , Cl , and S form an isoelectronic series with the [Ar] closed-shell electron configuration; that is, all three ions contain 18 electrons but have different nuclear charges. Because K has the greatest nuclear charge ( = 19), its radius is smallest, and S with = 16 has the largest radius. Because selenium is directly below sulfur, we expect the Se ion to be even larger than S . The order must therefore be K < Cl < S < Se . Based on their positions in the periodic table, arrange these ions in order of increasing size: Br , Ca , Rb , and Sr . Ca < Sr < Rb < Br Ionic radii share the same vertical trend as atomic radii, but the horizontal trends differ due to differences in ionic charges. A variety of methods have been established to measure the size of a single atom or ion. The is half the internuclear distance in a molecule with two identical atoms bonded to each other, whereas the is defined as half the distance between the nuclei of two adjacent atoms in a metallic element. The of an element is half the internuclear distance between two nonbonded atoms in a solid. Atomic radii decrease from left to right across a row because of the increase in effective nuclear charge due to poor electron screening by other electrons in the same principal shell. Moreover, atomic radii increase from top to bottom down a column because the effective nuclear charge remains relatively constant as the principal quantum number increases. The of cations and anions are always smaller or larger, respectively, than the parent atom due to changes in electron–electron repulsions, and the trends in ionic radius parallel those in atomic size. A comparison of the dimensions of atoms or ions that have the same number of electrons but different nuclear charges, called an , shows a clear correlation between increasing nuclear charge and decreasing size. ( ) | 17,466 | 2,489 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkyl_Halides/Properties_of_Alkyl_Halides/Introduction_to_Alkyl_Halides/Leaving_groups |
In our general discussion of nucleophilic substitution reactions, we have until now been designating the leaving group simply as “X". As you may imagine, however, the nature of the leaving group is an important consideration: if the C-X bond does not break, the new bond between the nucleophile and electrophilic carbon cannot form, regardless of whether the substitution is S 1 or S 2. In this module, we are focusing on substitution reactions in which the leaving group is a halogen ion, although many reactions are known, both in the laboratory and in biochemical processes, in which the leaving group is something other than a halogen. In order to understand the nature of the leaving group, it is important to first discuss factors that help determine whether a species will be a strong base or weak base. If you remember from general chemistry, a Lewis base is defined as a species that donates a pair of electrons to form a covalent bond. The factors that will determine whether a species wants to share its electrons or not include electronegativity, size, and resonance. : In general, if we move from the left of the periodic table to the right of the periodic table as shown in the diagram below, increases. As electronegativity increases, basicity will decrease, meaning a species will be less likely to act as base; that is, the species will be less likely to share its electrons. In general, if we move from the top of the periodic table to the bottom of the periodic table as shown in the diagram below, the size of an atom will increase. As size increases, basicity will decrease, meaning a species will be less likely to act as a base; that is, the species will be less likely to share its electrons. The third factor to consider in determining whether or not a species will be a strong or weak base is resonance. As you may remember from general chemistry, the formation of a resonance stabilized structure results in a species that is less willing to share its electrons. Since strong bases, by definition, want to share their electrons, resonance stabilized structures are weak bases. Now that we understand how electronegativity, size, and resonance affect basicity, we can combine these concepts with the fact that weak bases make the best leaving groups. Think about why this might be true. In order for a leaving group to leave, it must be able to accept electrons. A strong bases wants to donate electrons; therefore, the leaving group must be a weak base. We will now revisit electronegativity, size, and resonance, moving our focus to the leaving group, as well providing actual examples. As mentioned previously, if we move from left to right on the periodic table, electronegativity increases. With an increase in electronegativity, basisity decreases, and the ability of the leaving group to leave increases. This is because an increase in electronegativity results in a species that wants to hold onto its electrons rather than donate them. The following diagram illustrates this concept, showing CH to be the worst leaving group and F to be the best leaving group. This particular example should only be used to facilitate your understanding of this concept. In real reaction mechanisms, these groups are not good leaving groups at all. For example, fluoride is such a poor leaving group that S 2 reactions of fluoroalkanes are rarely observed. Here we revisit the effect size has on basicity. If we move down the periodic table, size increases. With an increase in size, basicity decreases, and the ability of the leaving group to leave increases. The relationship among the following halogens, unlike the previous example, is true to what we will see in upcoming reaction mechanisms. | 3,738 | 2,490 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/10%3A_Gases/10.01%3A_Gaseous_Elements_and_Compounds |
The three common phases (or states) of matter are gases, liquids, and solids. Gases have the lowest density of the three, are highly compressible, and completely fill any container in which they are placed. Gases behave this way because their intermolecular forces are relatively weak, so their molecules are constantly moving independently of the other molecules present. Solids, in contrast, are relatively dense, rigid, and incompressible because their intermolecular forces are so strong that the molecules are essentially locked in place. Liquids are relatively dense and incompressible, like solids, but they flow readily to adapt to the shape of their containers, like gases. We can therefore conclude that the sum of the intermolecular forces in liquids are between those of gases and solids. compares the three states of matter and illustrates the differences at the molecular level. (a) Solid O has a fixed volume and shape, and the molecules are packed tightly together. (b) Liquid O conforms to the shape of its container but has a fixed volume; it contains relatively densely packed molecules. (c) Gaseous O fills its container completely—regardless of the container’s size or shape—and consists of widely separated molecules. The state of a given substance depends strongly on conditions. For example, H O is commonly found in all three states: solid ice, liquid water, and water vapor (its gaseous form). Under most conditions, we encounter water as the liquid that is essential for life; we drink it, cook with it, and bathe in it. When the temperature is cold enough to transform the liquid to ice, we can ski or skate on it, pack it into a snowball or snow cone, and even build dwellings with it. Water vapor is a component of the air we breathe, and it is produced whenever we heat water for cooking food or making coffee or tea. Water vapor at temperatures greater than 100°C is called steam. Steam is used to drive large machinery, including turbines that generate electricity. The properties of the three states of water are summarized in .1 The geometric structure and the physical and chemical properties of atoms, ions, and molecules usually do depend on their physical state; the individual water molecules in ice, liquid water, and steam, for example, are all identical. In contrast, the macroscopic properties of a substance depend strongly on its physical state, which is determined by intermolecular forces and conditions such as temperature and pressure. shows the locations in the periodic table of those elements that are commonly found in the gaseous, liquid, and solid states. Except for hydrogen, the elements that occur naturally as gases are on the right side of the periodic table. Of these, all the noble gases (group 18) are monatomic gases, whereas the other gaseous elements are diatomic molecules (H , N , O , F , and Cl ). Oxygen can also form a second allotrope, the highly reactive triatomic molecule ozone (O ), which is also a gas. In contrast, bromine (as Br ) and mercury (Hg) are liquids under normal conditions (25°C and 1.0 atm, commonly referred to as “room temperature and pressure”). Gallium (Ga), which melts at only 29.76°C, can be converted to a liquid simply by holding a container of it in your hand or keeping it in a non-air-conditioned room on a hot summer day. The rest of the elements are all solids under normal conditions. Many of the elements and compounds we have encountered so far are typically found as gases; some of the more common ones are listed in . Gaseous substances include many binary hydrides, such as the hydrogen halides (HX); hydrides of the chalcogens; hydrides of the group 15 elements N, P, and As; hydrides of the group 14 elements C, Si, and Ge; and diborane (B H ). In addition, many of the simple covalent oxides of the nonmetals are gases, such as CO, CO , NO, NO , SO , SO , and ClO . Many low-molecular-mass organic compounds are gases as well, including all the hydrocarbons with four or fewer carbon atoms and simple molecules such as dimethyl ether [(CH ) O], methyl chloride (CH Cl), formaldehyde (CH O), and acetaldehyde (CH CHO). Finally, refrigerants, such as the chlorofluorocarbons (CFCs) and the hydrochlorofluorocarbons (HCFCs) are gases which can be easily liquified by compression and in turn the liquids can be turned into gases by decreasing the pressure on the liquids. The phase change from liquid to gas in tubes inside the refrigerator cools, while the compression in coils at the bottom or back of the refrigerator warms the room. Ammonia and SO are other compressible gases that have been used as refrigerants but cannot be used in houses because of their poisonous nature. because of its efficiency and low cost. All of the gaseous substances mentioned previously (other than the monatomic noble gases) contain covalent or polar covalent bonds and are nonpolar or polar molecules. In contrast, the strong electrostatic attractions in ionic compounds, such as NaBr (boiling point = 1390°C) or LiF (boiling point = 1673°C), prevent them from existing as gases at room temperature and pressure. In addition, the lightest members of any given family of compounds are most likely gases, and the boiling points of polar compounds are generally greater than those of nonpolar compounds of similar molecular mass. Therefore, in a given series of compounds, the lightest and least polar members are the ones most likely to be gases. With relatively few exceptions, however, compounds with more than about five atoms from period 2 or below are too heavy to exist as gases under normal conditions. Gaseous substances often contain covalent or polar covalent bonds, exist as nonpolar or slightly polar molecules, have relatively low molecular masses, and contain five or fewer atoms from periods 1 or 2. While gases have a wide array of uses, a particularly grim use of a gaseous substance is believed to have been employed by the Persians on the Roman city of Dura in eastern Syria in the third century AD. The Persians dug a tunnel underneath the city wall to enter and conquer the city. Archeological evidence suggests that when the Romans responded with counter-tunnels to stop the siege, the Persians ignited bitumen and sulfur crystals to produce a dense, poisonous gas. It is likely that bellows or chimneys distributed the toxic fumes. The remains of about 20 Roman soldiers were discovered at the base of the city wall at the entrance to a tunnel that was less than 2 m high and 11 m long. Because it is highly unlikely that the Persians could have slaughtered so many Romans at the entrance to such a confined space, archeologists speculate that the ancient Persians used chemical warfare to successfully conquer the city. Which compounds would you predict to be gases at room temperature and pressure? compounds physical state Decide whether each compound is ionic or covalent. An ionic compound is most likely a solid at room temperature and pressure, whereas a covalent compound may be a solid, a liquid, or a gas. Among the covalent compounds, those that are relatively nonpolar and have low molecular masses are most likely gases at room temperature and pressure. Lithium carbonate is Li CO , containing Li and CO ions, and vanadium(III) oxide is V O , containing V and O ions. Both are primarily ionic compounds that are expected to be solids. The remaining three compounds are all covalent. Benzoic acid has more than four carbon atoms and is polar, so it is not likely to be a gas. Both cyclohexene and cyclobutane are essentially nonpolar molecules, but cyclobutane (C H ) has a significantly lower molecular mass than cyclohexene (C H ), which again has more than four carbon atoms. We therefore predict that cyclobutane is most likely a gas at room temperature and pressure, while cyclohexene is a liquid. In fact, with a boiling point of only 12°C, compared to 83°C for cyclohexene, cyclobutane is indeed a gas at room temperature and pressure. Exercise Which compounds would you predict to be gases at room temperature and pressure? c; d Bulk matter can exist in three states: gas, liquid, and solid. Gases have the lowest density of the three, are highly compressible, and fill their containers completely. Elements that exist as gases at room temperature and pressure are clustered on the right side of the periodic table; they occur as either monatomic gases (the noble gases) or diatomic molecules (some halogens, N , O ). Many inorganic and organic compounds with four or fewer nonhydrogen atoms are also gases at room temperature and pressure. All gaseous substances are characterized by weak interactions between the constituent molecules or atoms. Explain the differences between the microscopic and the macroscopic properties of matter. Is the boiling point of a compound a microscopic or macroscopic property? molecular mass? Why? Determine whether the melting point, the dipole moment, and electrical conductivity are macroscopic or microscopic properties of matter and explain your reasoning. How do the microscopic properties of matter influence the macroscopic properties? Can you relate molecular mass to boiling point? Why or why not? Can polarity be related to boiling point? For a substance that has gas, liquid, and solid phases, arrange these phases in order of increasing Explain what is wrong with this statement: “The state of matter largely determines the molecular properties of a substance.” Describe the most important factors that determine the state of a given compound. What external conditions influence whether a substance exists in any one of the three states of matter? Which elements of the periodic table exist as gases at room temperature and pressure? Of these, which are diatomic molecules and which are monatomic? Which elements are liquids at room temperature and pressure? Which portion of the periodic table contains elements whose binary hydrides are most likely gases at room temperature? Is the following observation correct? “Almost all nonmetal binary hydrides are gases at room temperature, but metal hydrides are all solids.” Explain your reasoning. Is the following observation correct? “All the hydrides of the chalcogens are gases at room temperature and pressure except the binary hydride of oxygen, which is a liquid.” Explain your reasoning. Would you expect 1-chloropropane to be a gas? iodopropane? Why? Explain why ionic compounds are not gases under normal conditions. The molecular properties of a substance control its state of matter under a given set of conditions, the other way around. The presence of strong intermolecular forces favors a condensed state of matter (liquid or solid), while very weak intermolecular interaction favor the gaseous state. In addition, the shape of the molecules dictates whether a condensed phase is a liquid or a solid. Elements that exist as gases are mainly found in the upper right corner and on the right side of the periodic table. The following elements exist as gases: H, He, N, O, F, Ne, Cl, Ar, Kr, Xe, and Rn. Thus, half of the halogens, all of the noble gases, and the lightest chalcogens and picnogens are gases. Of these, all except the noble gases exist as diatomic molecules. Only two elements exist as liquids at a normal room temperature of 20°C–25°C: mercury and bromine. The upper right portion of the periodic table also includes most of the elements whose binary hydrides are gases. In addition, the binary hydrides of the elements of Groups 14–16 are gases. | 11,545 | 2,491 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/21%3A_Spectra_and_Structure_of_Atoms_and_Molecules/21.03%3A_Atomic_Spectra_and_the_Bohr_Theory |
If the emission spectrum of an element is passed through a prism and allowed to strike a photographic film, the intensity of emission as a function of wavelength can be measured. The equipment for making such measurements is called a and is shown schematically in Figure \(\Page {1}\). Typical emission spectra obtained in this way are shown in Figure \(\Page {2}\) where the intensity is plotted against wavelength. Note from this figure how the light emitted is confined to a few very specific wavelengths. At all other wavelengths there is no emission at all. Spectra of this kind are usually referred to as . The wavelengths of the lines in a line spectrum are unique to each element and are often used, especially in metallurgy, both to identify an element and to measure the amount present. These wavelengths can often be measured to an accuracy as great as one part in a billion (1 in 10 ). Because the spectrum of an element is readily reproducible and can be measured so accurately, it is often used to determine lengths. For example, the meter was once defined as the distance between two marks on a platinum bar kept at Sevrés in France, but now it is taken as 1 650 763.73 wavelengths of a particular line in the spectrum of Kr. For example, you can see that lines in the hydrogen spectrum occur very close together in the region just above 365 nm and then are spaced farther and farther apart as λ increases. In 1885 a Swiss high-school mathematics teacher, J. J. Balmer (1825 to 1898), discovered a formula which accounted for this regularity: Other similar series of lines are found in the ultraviolet (Lyman series) and infrared (Paschen, Brackett, Pfund series) regions of the spectrum. In each case the wavelengths of the lines can be predicted by an equation similar in form to Balmer’s. These equations were combined by Rydberg into the general form | 1,887 | 2,492 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/02%3A_Gas_Laws/2.13%3A_Virial_Equations |
It is often useful to fit accurate pressure-volume-temperature data to polynomial equations. The experimental data can be used to compute a quantity called the , \(Z\), which is defined as the pressure–volume product for the real gas divided by the pressure–volume product for an ideal gas at the same temperature. We have \[{\left(PV\right)}_{ideal\ gas}=nRT\] Letting and represent the pressure and volume of the real gas, and introducing the molar volume, \(\overline{V}={V}/{n}\), we have \[Z=\frac{\left(PV\right)_{real\ gas}}{\left(PV\right)_{ideal\ gas}}=\frac{PV}{nRT}=\frac{P\overline{V}}{RT}\] Since \(Z=1\) if the real gas behaves exactly like an ideal gas, experimental values of will tend toward unity under conditions in which the density of the real gas becomes low and its behavior approaches that of an ideal gas. At a given temperature, we can conveniently ensure that this condition is met by fitting the values to a polynomial in or a polynomial in \({\overline{V}}^{-1}\). The coefficients are functions of temperature. If the data are fit to a polynomial in the pressure, the equation is \[Z=1+B^*\left(T\right)P+C^*\left(T\right)P^2+D^*\left(T\right)P^3+\dots\] For a polynomial in \({\overline{V}}^{-1}\), the equation is \[Z=1+\frac{B\left(T\right)}{\overline{V}}+\frac{C\left(T\right)}{\overline{V}^2}+\frac{D\left(T\right)}{\overline{V}^3}+\dots\] These empirical equations are called . As indicated, the parameters are functions of temperature. The values of \(B^*\left(T\right)\), \(C^*\left(T\right)\), \(D^*\left(T\right)\), , and \(B\left(T\right)\), \(C\left(T\right)\), \(D\left(T\right)\), , must be at every temperature. (Note also that \(B^*\left(T\right)\neq B\left(T\right)\), \(C^*\left(T\right)\neq C\left(T\right)\), \(D^*\left(T\right)\neq D\left(T\right)\), . However, it is true that \(B^*={B}/{RT}\).) Values for these parameters are tabulated in various compilations of physical data. In these tabulations, \(B\left(T\right)\) and \(C\left(T\right)\) are called the and third virial coefficient, respectively. | 2,086 | 2,493 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/15%3A_Principles_of_Chemical_Equilibrium/15.2%3A_The_Equilibrium_Constant_Expression |
Because an equilibrium state is achieved when the forward reaction rate equals the reverse reaction rate, under a given set of conditions there must be a relationship between the composition of the system at equilibrium and the kinetics of a reaction (represented by rate constants). We can show this relationship using the system described in Equation \ref{15.2.1}, the decomposition of \(N_2O_4\) to \(NO_2\). Both the forward and reverse reactions for this system consist of a single elementary reaction, so the reaction rates are as follows: \[\text{forward rate} = k_f[N_2O_4] \label{15.2.1}\] and \[\text{reverse rate} = k_r[NO_2]^2 \label{15.2.2}\] At equilibrium, the forward rate equals the reverse rate (definition of equilibrium): \[ k_f[N_2O_4] = k_r[NO_2]^2 \label{15.2.3}\] so \[\dfrac{k_f}{k_r}=\dfrac{[NO_2]^2}{[N_2O_4]} \label{15.2.4}\] The ratio of the rate constants gives us a new constant, the equilibrium constant (\(K\)), which is defined as follows: \[K=\dfrac{k_f}{k_r} \label{15.2.5}\] Hence there is a fundamental relationship between chemical kinetics and chemical equilibrium: under a given set of conditions, the composition of the equilibrium mixture is determined by the magnitudes of the rate constants for the forward and the reverse reactions. The equilibrium constant is equal to the rate constant for the forward reaction divided by the rate constant for the reverse reaction. Table \(\Page {1}\) lists the initial and equilibrium concentrations from five different experiments using the reaction system described by Equation \ref{15.1}. At equilibrium the magnitude of the quantity \([NO_2]^2/[N_2O_4]\) is essentially the same for all five experiments. In fact, no matter what the initial concentrations of \(NO_2\) and \(N_2O_4\) are, at equilibrium the quantity \([NO_2]^2/[N_2O_4]\) will always be \(6.53 \pm 0.03 \times 10^{−3}\) at 25°C, which corresponds to the ratio of the rate constants for the forward and reverse reactions. That is, at a given temperature, the equilibrium constant for a reaction always has the same value, even though the specific concentrations of the reactants and products vary depending on their initial concentrations. Introduction to Dynamic Equilibrium: In 1864, the Norwegian chemists Cato Guldberg (1836–1902) and Peter Waage (1833–1900) carefully measured the compositions of many reaction systems at equilibrium. They discovered that for any reversible reaction of the general form \[aA+bB \rightleftharpoons cC+dD \label{15.2.6}\] where A and B are reactants, C and D are products, and a, b, c, and d are the stoichiometric coefficients in the balanced chemical equation for the reaction, the ratio of the product of the equilibrium concentrations of the products (raised to their coefficients in the balanced chemical equation) to the product of the equilibrium concentrations of the reactants (raised to their coefficients in the balanced chemical equation) is always a constant under a given set of conditions. This relationship is known as the and can be stated as follows: \[K=\dfrac{[C]^c[D]^d}{[A]^a[B]^b} \label{15.2.7}\] where \(K\) is the equilibrium constant for the reaction. Equation \ref{15.2.6} is called the equilibrium equation, and the right side of Equation \ref{15.2.7} is called the equilibrium constant expression. The relationship shown in Equation \ref{15.2.7} is true for any pair of opposing reactions regardless of the mechanism of the reaction or the number of steps in the mechanism. The equilibrium constant can vary over a wide range of values. The values of \(K\) shown in Table \(\Page {2}\), for example, vary by 60 orders of magnitude. Because products are in the numerator of the equilibrium constant expression and reactants are in the denominator, values of K greater than \(10^3\) indicate a strong tendency for reactants to form products. In this case, chemists say that equilibrium lies to the right as written, favoring the formation of products. An example is the reaction between \(H_2\) and \(Cl_2\) to produce \(HCl\), which has an equilibrium constant of \(1.6 \times 10^{33}\) at 300 K. Because \(H_2\) is a good reductant and \(Cl_2\) is a good oxidant, the reaction proceeds essentially to completion. In contrast, values of \(K\) less than \(10^{-3}\) indicate that the ratio of products to reactants at equilibrium is very small. That is, reactants do not tend to form products readily, and the equilibrium lies to the left as written, favoring the formation of reactants. You will also notice in Table \(\Page {2}\) that equilibrium constants have no units, even though Equation \ref{15.2.7} suggests that the units of concentration might not always cancel because the exponents may vary. As shown in Equation \ref{15.2.8}, the units of concentration cancel, which makes \(K\) unitless as well: \[ \dfrac{[A]_{measured}}{[A]_{standard\; state}}=\dfrac{\cancel{M}}{\cancel{M}} = \dfrac{\cancel{\frac{mol}{L}}}{\cancel{\frac{mol}{L}}} \label{15.2.8}\] Many reactions have equilibrium constants between 1000 and 0.001 (\(10^3 \ge K \ge 10^{−3}\)), neither very large nor very small. At equilibrium, these systems tend to contain significant amounts of both products and reactants, indicating that there is not a strong tendency to form either products from reactants or reactants from products. An example of this type of system is the reaction of gaseous hydrogen and deuterium, a component of high-stability fiber-optic light sources used in ocean studies, to form HD: \[H_{2(g)}+D_{2(g)} \rightleftharpoons 2HD_{(g)} \label{15.2.9}\] The equilibrium constant expression for this reaction is \[K= \dfrac{[HD]^2}{[H_2,D_2]}\] with \(K\) varying between 1.9 and 4 over a wide temperature range (100–1000 K). Thus an equilibrium mixture of \(H_2\), \(D_2\), and \(HD\) contains significant concentrations of both product and reactants. Figure \(\Page {3}\) summarizes the relationship between the magnitude of K and the relative concentrations of reactants and products at equilibrium for a general reaction, written as \[\text{reactants} \rightleftharpoons \text{products}.\] Because there is a direct relationship between the kinetics of a reaction and the equilibrium concentrations of products and reactants (Equations \ref{15.2.8} and \ref{15.2.7}), when \(k_f \gg k_r\), \(K\) is a number, and the concentration of products at equilibrium predominate. This corresponds to an essentially irreversible reaction. Conversely, when \(k_f \ll k_r\), \(K\) is a very number, and the reaction produces almost no products as written. Systems for which \(k_f ≈ k_r\) have significant concentrations of both reactants and products at equilibrium. A large value of the equilibrium constant \(K\) means that products predominate at equilibrium; a small value means that reactants predominate at equilibrium. Write the equilibrium constant expression for each reaction. : balanced chemical equations : equilibrium constant expressions : Refer to Equation \ref{15.2.7}. Place the arithmetic product of the concentrations of the products (raised to their stoichiometric coefficients) in the numerator and the product of the concentrations of the reactants (raised to their stoichiometric coefficients) in the denominator. : The only product is ammonia, which has a coefficient of 2. For the reactants, \(N_2\) has a coefficient of 1 and H2 has a coefficient of 3. The equilibrium constant expression is as follows: \[\dfrac{[NH_3]^2}{[N_2,H_2]^3}\] The only product is carbon dioxide, which has a coefficient of 1. The reactants are \(CO\), with a coefficient of 1, and \(O_2\), with a coefficient of \(\frac{1}{2}\). Thus the equilibrium constant expression is as follows: \[\dfrac{[CO_2]}{[CO,O_2]^{1/2}}\] This reaction is the reverse of the reaction in part b, with all coefficients multiplied by 2 to remove the fractional coefficient for \(O_2\). The equilibrium constant expression is therefore the inverse of the expression in part b, with all exponents multiplied by 2: \[\dfrac{[CO]^2[O_2]}{[CO_2]^2}\] Write the equilibrium constant expression for each reaction. Answer: Predict which systems at equilibrium will (a) contain essentially only products, (b) contain essentially only reactants, and (c) contain appreciable amounts of both products and reactants. : systems and values of \(K\) : composition of systems at equilibrium : Use the value of the equilibrium constant to determine whether the equilibrium mixture will contain essentially only products, essentially only reactants, or significant amounts of both. : Hydrogen and nitrogen react to form ammonia according to the following balanced chemical equation: \[3H_{2(g)}+N_{2(g)} \rightleftharpoons 2NH_{3(g)}\] Values of the equilibrium constant at various temperatures were reported as At which temperature would you expect to find the highest proportion of \(H_2\) and \(N_2\) in the equilibrium mixture? Assuming that the reaction rates are fast enough so that equilibrium is reached quickly, at what temperature would you design a commercial reactor to operate to maximize the yield of ammonia? : Because equilibrium can be approached from either direction in a chemical reaction, the equilibrium constant expression and thus the magnitude of the equilibrium constant depend on the form in which the chemical reaction is written. For example, if we write the reaction described in Equation 15.2.6 in reverse, we obtain the following: \[cC+dD \rightleftharpoons aA+bB \label{15.2.10}\] The corresponding equilibrium constant \(K′\) is as follows: \[K'=\dfrac{[A]^a[B]^b}{[C]^c[D]^d} \label{15.2.11}\] This expression is the inverse of the expression for the original equilibrium constant, so \(K′ = 1/K\). That is, when we write a reaction in the reverse direction, the equilibrium constant expression is inverted. For instance, the equilibrium constant for the reaction \(N_2O_4\) \rightleftharpoons 2NO_2\) is as follows: \[K=\dfrac{[NO_2]^2}{[N_2O_4]} \label{15.2.12}\] but for the opposite reaction, \(2 NO_2 \rightleftharpoons N_2O_4\), the equilibrium constant K′ is given by the inverse expression: \[K'=\dfrac{[N_2O_4]}{[NO_2]^2} \label{15.2.13}\] Consider another example, the formation of water: \(2H_{2(g)}+O_{2(g)} \rightleftharpoons 2H_2O_{(g)}\). Because \(H_2\) is a good reductant and \(O_2\) is a good oxidant, this reaction has a very large equilibrium constant (\(K = 2.4 \times 10^{47}\) at 500 K). Consequently, the equilibrium constant for the reverse reaction, the decomposition of water to form \(O_2\) and \(H_2\), is very small: \(K′ = 1/K = 1/(2.4 \times 10^{47}) = 4.2 \times 10^{−48}\). As suggested by the very small equilibrium constant, and fortunately for life as we know it, a substantial amount of energy is indeed needed to dissociate water into \(H_2\) and \(O_2\). The equilibrium constant for a reaction written in reverse is the of the equilibrium constant for the reaction as written originally. Writing an equation in different but chemically equivalent forms also causes both the equilibrium constant expression and the magnitude of the equilibrium constant to be different. For example, we could write the equation for the reaction \[2NO_2 \rightleftharpoons N_2O_4\] as \[NO_2 \rightleftharpoons \frac{1}{2}N_2O_4\] with the equilibrium constant K″ is as follows: \[ K′′=\dfrac{[N_2O_4]^{1/2}}{[NO_2]} \label{15.2.14}\] The values for K′ (Equation \ref{15.2.13}) and K″ are related as follows: \[ K′′=(K')^{1/2}=\sqrt{K'} \label{15.2.15}\] In general, if all the coefficients in a balanced chemical equation were subsequently multiplied by \(n\), then the new equilibrium constant is the original equilibrium constant raised to the \(n^{th}\) power. At 745 K, K is 0.118 for the following reaction: \[\ce{N2(g) + 3H2(g) <=> 2NH3(g)} \nonumber\] What is the equilibrium constant for each related reaction at 745 K? : balanced equilibrium equation, K at a given temperature, and equations of related reactions : values of \(K\) for related reactions : Write the equilibrium constant expression for the given reaction and for each related reaction. From these expressions, calculate \(K\) for each reaction. : The equilibrium constant expression for the given reaction of \(N_{2(g)}\) with \(H_{2(g)}\) to produce \(NH_{3(g)}\) at 745 K is as follows: \[K=\dfrac{[NH_3]^2}{[N_2,H_2]^3}=0.118\] This reaction is the reverse of the one given, so its equilibrium constant expression is as follows: \[K'=\dfrac{1}{K}=\dfrac{[N_2,H_2]^3}{[NH_3]^2}=\dfrac{1}{0.118}=8.47\] In this reaction, the stoichiometric coefficients of the given reaction are divided by 2, so the equilibrium constant is calculated as follows: \[K′′=\dfrac{[NH_3]}{[N_2]^{1/2}[H_2]^{3/2}}=K^{1/2}=\sqrt{K}=\sqrt{0.118} = 0.344\] At 527°C, the equilibrium constant for the reaction \[2SO_{2(g)}+O_{2(g)} \rightleftharpoons 2SO_{3(g)}\] is \(7.9 \times 10^4\). Calculate the equilibrium constant for the following reaction at the same temperature: \[SO_{3(g)} \rightleftharpoons SO_{2(g)}+\frac{1}{2}O_{2(g)}\] : \(3.6 \times 10^{−3}\) Determining the Equilibrium Expression: The ratio of the rate constants for the forward and reverse reactions at equilibrium is the equilibrium constant (K), a unitless quantity. The composition of the equilibrium mixture is therefore determined by the magnitudes of the forward and reverse rate constants at equilibrium. Under a given set of conditions, a reaction will always have the same \(K\). For a system at equilibrium, the law of mass action relates \(K\) to the ratio of the equilibrium concentrations of the products to the concentrations of the reactants raised to their respective powers to match the coefficients in the equilibrium equation. The ratio is called the equilibrium constant expression. When a reaction is written in the reverse direction, \(K\) and the equilibrium constant expression are inverted. For gases, the equilibrium constant expression can be written as the ratio of the partial pressures of the products to the partial pressures of the reactants, each raised to a power matching its coefficient in the chemical equation. An equilibrium constant calculated from partial pressures (\(K_p\)) is related to \(K\) by the ideal gas constant (\(R\)), the temperature (\(T\)), and the change in the number of moles of gas during the reaction. An equilibrium system that contains products and reactants in a single phase is a homogeneous equilibrium; a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibrium. When a reaction can be expressed as the sum of two or more reactions, its equilibrium constant is equal to the product of the equilibrium constants for the individual reactions. | 14,724 | 2,495 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Book3A_Bioinorganic_Chemistry_(Bertini_et_al.)/04%3A_Biological_and_Synthetic_Dioxygen_Carriers/4.20%3A_Thermodynamic_Factors |
The equilibrium constant K was defined in Equation (4.3) in terms of the activity a of component i. The a may be expressed as a function of concentration as \[a_{i} = \gamma_{i}[i], \tag{4.7}\] where for species i, \(\gamma\) is its activity coefficient and [i] is its concentration (strictly molality, but usually as molarity in mol L ). At infinite dilution \(\gamma\) = 1. Provided that the charge and size of species M and MO are similar and that O forms an ideal solution, then the activities of Equation (4.3) may be approximated by concentrations to give the expression \[K_{c} = \frac{[MO_{2}]}{[M,O_{2}]} \ldotp \tag{4.8}\] However, Equation (4.8) does not permit a direct comparison of the oxygen-binding behavior of one species in some solvent with that of a second in some other solvent. First, for a given partial pressure of dioxygen, the concentration of O in the solution varies considerably with temperature and from one solvent to another. Second, reliable measurements of oxygen solubilities are not always available, and it is only relatively recently that oxygen electrodes have been developed to measure directly oxygen concentrations (strictly, activities). However, oxygen-binding measurements are normally made with a solution of M in equilibrium with gaseous dioxygen. At equilibrium the molar Gibbs' free energies (chemical potentials) of the dissolved and gaseous dioxygen are identical—if they are not, gaseous O would dissolve, or dissolved O would be released. Thus the solvent-dependent quantity [O ] in Equation (4.8) may be replaced by the solvent-independent quantity P(O ), the partial pressure of dioxygen. Under almost all experimental conditions the quantity P(O ) is a very good approximation to the gas-phase activity (fugacity) of dioxygen; hence we obtain for the equilibrium constant* \[K_{p} = \frac{[MO_{2}]}{[M]P(O_{2})} \ldotp \tag{4.9}\] * There has been considerable discussion as to whether K (4.8) or K (4.9) should be used to compare dioxygen binding under different solvent conditions We believe that the latter is more appropriate, since for a system al equilibrium, the chemical potential of gaseous O must be identical with that of dissolved O On the other hand, the concentration of O varies from one solvent to another. It is very convenient to express the affinity as the partial pressure of dioxygen required for half-saturation of the species M, P (O ). Under such conditions, [M] = [MO ], one obtains \[P_{1/2}(O_{2}) = 1/K_{p}, \tag{4.10}\] where P (O ) is usually given in Torr or mm Hg.* As will be detailed shortly, values for P (O ) are typically in the range 0.5 to 40 Torr. The dioxygen affinity is composed of enthalpic (\(\Delta\)H) and entropic (\(\Delta\)S) components, with \[\Delta G^{o} = -RTlnK = \Delta H^{o} - T \Delta S^{o} \ldotp \tag{4.11}\] Within a family of oxygen carriers the values of \(\Delta\)Sº and \(\Delta\)H° are usually similar. Large deviations (such as a change of sign) are therefore indicative of a change in the nature of the oxygen-binding process. * Many authors use the symbol P (corresponding to 50% saturation) for P . If the oxygen-binding sites M are mutually independent and noninteracting, as in moderately dilute solutions of monomeric molecules, then the concentration of species MO as a function of the partial pressure of O is generally well fit by a Langmuir isotherm. Here a plot of the fractional saturation of dioxygen binding sites, \(\theta\), where \[\theta = \frac{[MO_{2}]}{[M] + [MO_{2}]} = \frac{K_{p}P(O_{2})}{1+K_{p}P(O_{2})} \tag{4.12}\] versus P(O ) gives the hyberbolic curve labeled "non-cooperative" in Figure 4.4A. Alternatively, a plot of log (\(\theta\)/(1 - \(\theta\))) versus log (P(O )), the so-called "Hill plot," gives a straight line with a slope of unity and an intercept of -log P (O ) (Figure 4.4B). A differential form is shown as the dotted line in Figure 4.4C. Such binding, where the dioxygen sites are independent of each other, is termed . Many dioxygen-binding proteins are not independent monomers, with only one dioxygen-binding site, but oligomeric species with the protein comprising two or more similar subunits. The subunits may be held together by van der Waals' forces or by stronger interactions, such as hydrogen bonds or salt bridges, or even by covalent bonds. For example, most mammalian hemoglobins are tetramers, consisting of two pairs [\(\alpha \beta\)] of myoglobin-like subunits denoted as \(\alpha\) and \(\beta\). Either none, one, two, three, or all four sites may be occupied by dioxygen. This situation is illustrated schematically in Figure 4.5, which also shows the statistical weighting of each level of saturation, treating the \(\alpha\) and \(\beta\) subunits as identical. Thus the binding or release of dioxygen at one site affect the affinity and kinetics of ligand binding and release at a neighboring site. As a result, the saturation curve becomes sigmoidal in shape, as illustrated in Figure 4.4A. The dioxygen binding is . When cooperativity is positive, the affinity of a vacant site is increased by occupancy of an adjacent one. This behavior, where the binding of one molecule influences the binding of successive molecules , is referred to as a interaction. A interaction occurs when the interaction with the protein of a second molecule, for instance, an organic polyphosphate for human hemoglobins, influences the binding of the first molecule (e.g., dioxygen). Such molecules are often termed . A commonly observed heterotropic allosteric interaction is the Bohr effect, named after the biologist Christian Bohr, father of physicist Niels Bohr. This effect, which relates the change in partial pressure of O to a change in pH at constant saturation of binding sites (\(\theta\)), is related thermodynamically to the Haldane effect, which relates the number of protons released (#H ) with a change in \(\theta\) at constant pH (Equation 4.13). A very large Bohr effect, where O affinity decreases sharply with pH, is often called the Root effect. It is physiologically important for fish such as trout, probably in maintaining buoyancy, but its molecular basis in trout hemoglobin IV remains to be discovered. \[\bigg[\frac{\partial (\#H^{+})}{\partial \theta}\bigg]_{pH} = \bigg[\frac{\partial (log P(O_{2}))}{\partial pH}\bigg]_{\theta} \tag{4.13}\] The degree of cooperativity can be characterized in a number of ways. By means of a Hill plot of log (\(\theta\)/(1 - \(\theta\))) versus log (P(O )), the limiting slopes (which should be unity) at high O pressure and low O pressure may be extrapolated as shown in Figure 4.4B to log (\(\theta\)/(1 - \(\theta\))) = 0, where \(\theta\) = 0.5. Two limiting values for P (O ) are obtained, one characterizing the regime of high partial pressure of dioxygen, where the O affinity is high (for the case illustrated of positive cooperativity). The other P (O ) value characterizes the regime of low partial pressure of dioxygen, where affinity is relatively low. This difference in affinities can be converted into a difference between the free-energy change upon O binding in the low-affinity state (K ) and the high-affinity state (K ) [the designations T and R will be described in subsection d]: \[\delta \Delta G^{o} = -RTln \left(\dfrac{K_{p}^{T}}{K_{p}^{R}}\right) \ldotp \tag{4.14}\] A second way to characterize cooperativity involves fitting the oxygen-binding data at intermediate saturation (0.2 < \(\theta\) < 0.8)—that is, about the inflection point in a Hill plot—to the Hill equation \[\frac{\theta}{1-\theta} = K_{p} P^{n}(O_{2})\] or \[log \left(\dfrac{\theta}{1-\theta}\right) = -log(P_{1/2}(O_{2})) + nlog(P(O_{2})) \ldotp \tag{4.15}\] The Hill coefficient (n) is an empirical coefficient that has a value of unity for non-cooperative binding, where Equation (4.15) reduces to the Langmuir isotherm, Equation (4.12). Any number greater than unity indicates positive cooperativity. If O binding is an all-or-nothing affair, where dioxygen binding sites are either all occupied or all vacant, n equals the number of subunits in the molecule. The fit is only approximate, since the Hill plot is only approximately linear about the inflection point, as may be seen in Figure 4.4B. A more precise value of n may be obtained by plotting the slope in the Hill plot (n') as a function \[n' = \frac{d\bigg[log\left(\dfrac{\theta}{1-\theta}\right)\bigg]}{d[log(P(O_{2}))]} \tag{4.16}\] of log (P(O )) (Figure 4.4C). The maximum value of n' is taken as the Hill coefficient n. Note that the maximum in this first-derivative plot of the binding curve will occur at P (O ) only if the Hill plot is symmetric about its inflection point. For tetrameric hemoglobins, a maximum Hill coefficient of around 3.0 is seen, and for hemocyanins n may be as high as 9. These values, like P (O ) values, are sensitive to the nature and concentrations of allosteric effectors. In general, oxygen-carrier proteins, being oligomeric, coordinate dioxygen cooperatively, whereas oxygen-storage proteins, being monomeric, do not. Oligomerization and cooperative binding confer enormous physiological benefits to an organism. The first benefit derives directly from oligomerization. Oxygen carriers either form small oligomers that are encapsulated into cells or erythrocytes (such hemoglobins are referred to as intracellular hemoglobins) or associate into large oligomers of 100 or more subunits. Such encapsulation and association reduce by orders of magnitude the number of independent particles in the blood, with consequent reductions in the osmotic pressure of the solution and in strain on vascular membranes. The second benefit derives from cooperative binding of ligands and the abilities of heterotropic allosteric effectors to optimize exquisitely the oxygen-binding behavior in response to the external and internal environment. The situation is illustrated in general terms in Figure 4.6. Most organisms that require O live in an environment where the activity of O corresponds to about 21 percent of an atmosphere, that is, to about 160 Torr, although usually the effective availability, because of incomplete exchange of gases in the lungs, for example, is around 100 Torr. The concentration of O in vertebrate tissues at rest is equivalent to a partial pressure of about 35-40 Torr dioxygen; lower values obtain at times of exertion. Now consider a noncooperative oxygen binder with an affinity expressed as P (O ) of 60 Torr (Figure 4.6, curve a). Then, at 100 Torr the fractional saturation \(\theta\) is 0.625. In other words, in a realm of high O availability, only 62.5 percent of the oxygen-binding capacity is used, which is not particularly efficient if the organism wished to climb Mt. Everest, where the partial pressure of O is less than half that at sea level. In the tissues, where P(O ) = 40 Torr, the fractional saturation is about 40 percent. Thus, only about one third of the coordinated dioxygen is released to the tissues, and total effciency is only 22.5 percent. Consider now a noncooperative oxygen carrier with a much higher affinity, P (O ) = 1.0 Torr (Figure 4.6, curve b). If we assume the same ambient pressure of O in the tissues, the fractional saturation is 97.6 percent. Note that at 100 Torr of O the carrier is 99.0 percent saturated. In other words, only about 1.4 percent of the available oxygen is delivered. With an oligomeric protein that binds dioxygen cooperatively, the problem of inefficient and inflexible oxygen delivery disappears. For example, the tetrameric protein hemoglobin has a mean affinity for O of P (O ) ≈ 26 Torr at 37 °C and pH 7.4. If hemoglobin bound O noncooperatively, then the hyberbolic binding curve (c) in Figure 4.6 would represent the O binding. Instead, the observed binding follows curve (d). Since the partial pressure of dioxygen in the lungs and arterial blood of vertebrates is around 100 Torr, but in the tissues and venous blood it is around 40 Torr, then at these pressures a typical myoglobin (P (O ) = 1 Torr) remains effectively saturated. On the other hand, about 25 percent of the available dioxygen can be delivered, even in the absence of myoglobin. With venous blood remaining 75 percent oxygenated, hemoglobin has substantial capacity to deliver more O at times of exertion or stress when P(O ) in the tissues falls below 40 Torr. The net result is that whole blood, which contains about 15 g of hemoglobin per 100 mL, can carry the equivalent of 20 mL of O (at 760 Torr) per 100 mL, whereas blood plasma (no hemoglobin) has a carrying capacity of only 0.3 mL of O per 100 mL. Oxygen binding is modulated by allosteric effectors that through interaction with the protein change the affinity and degree of cooperativity. For hemoglobin A (adult human hemoglobin), naturally occurring allosteric effectors include the proton, carbon dioxide, and 2,3-diphosphoglycerate (2,3-DPG). Increasing concentrations of these species progressively lower the affinity of free hemoglobin A, thereby enhancing the release of coordinated O (Figure 4.6, curve e). For example, 2,3-DPG is part of a subtle mechanism by which dioxygen is transferred from mother to fetus across the placenta. The subunits comprising fetal hemoglobin and adult hemoglobin are slightly different. In the absence of allosteric effectors (referred to as stripped hemoglobin), the oxygen-binding curves are identical. However, 2,3-DPG binds less strongly to fetal hemoglobin than to adult hemoglobin. Thus fetal hemoglobin has a slightly higher affinity for dioxygen, thereby enabling dioxygen to be transferred. The proton and carbon dioxide are part of a short-term feedback mechanism. When O consumption outpaces O delivery, glucose is incompletely oxidized to lactic acid (instead of CO ). The lactic acid produced lowers the pH, and O release from oxyhemoglobin is stimulated (Figure 4.6, curve e). The CO produced in respiration forms carbamates with the amino terminals, preferentially of deoxy hemoglobin. \[R-NH_{2} + CO_{2} \rightleftharpoons R-NH-COO^{-} + H^{+}\] Thus hemoglobin not only delivers O but also facilitates removal of CO to the lungs or gills, where CO is exhaled. The binding of Oz to hemoglobin can be described as four successive equilibria: \[Hb + O_{2} \xrightleftharpoons {K^{(1)}} Hb(O_{2}) \qquad P_{1/2}^{(1)}(O_{2}) = 123[46] Torr\] \[Hb(O_{2})_{1} + O_{2} \xrightleftharpoons {K^{(2)}} Hb(O_{2}) _{2} \qquad P_{1/2}^{(2)}(O_{2}) = 30[16] Torr \tag{4.17}\] \[Hb(O_{2})_{2} + O_{2} \xrightleftharpoons {K^{(3)}} Hb(O_{2}) _{3} \qquad P_{1/2}^{(3)}(O_{2}) = 33[3.3] Torr\] \[Hb(O_{2})_{3} + O_{2} \xrightleftharpoons {K^{(4)}} Hb(O_{2}) _{4} \qquad P_{1/2}^{(4)}(O_{2}) = 0.26[0.29] Torr\] This simple scheme proposed by Adair assumes that each of the four binding sites is identical. The P (O ) values given come from fitting the binding curve to this scheme. When 2,3-DPG is removed, the affinity of hemoglobin for the first three molecules of Oz is substantially increased, and the degree of cooperativity is lowered (values in square parentheses). For progressively stronger binding, the following inequalities, reflecting the proper statistical weighting illustrated in Figure 4.5, should hold: \[\frac{1}{4}K^{(1)} > \frac{4}{6}K^{(2)} > \frac{6}{4}K^{(3)} > \frac{4}{1}K^{(4)} \tag{4.18}\] The \(\frac{6}{4}\) ratio, for example, reflects the six equivalent forms of the doubly and the four equivalent forms of the triply ligated species. In other words, relative to a noncooperative system, at low O availability dioxygen is facilitated; at high O availability dioxygen is facilitated. The scheme is readily extended to higher orders of oligomerization. A simple model for analyzing cooperative ligand binding was proposed by Monod, Wyman, and Changeux in 1965, and is usually referred to as the MWC two-state concerted model. Molecules are assumed to be in equilibrium between two conformations or quaternary structures, one that has a low ligand affinity and a second that has a high ligand affinity. The low-affinity conformation is often designated the T or tense state, and the high-affinity conformation the R or relaxed state. The equilibrium between the two conformations is characterized by the allosteric constant \[L_{0} = \frac{[R_{0}]}{[T_{0}]} \tag{4.19}\] where the subscript denotes the unliganded Rand T states. The free-energy change upon binding a ligand to the R state, irrespective of saturation, is assumed to be a constant, and the associated equilibrium constant is designated K ; a third constant, K , characterizes binding to the T state. Figure 4.7 illustrates this model, and introduces the terminology conventionally used. To a reasonable approximation, the cooperative binding of dioxygen can be summarized by these three parameters, L , K , and K . The Adair constants may be expressed in terms of these parameters: \[K^{(1)} = \frac{(1+L_{0}C)K_{T}}{1+L_{0}}, \qquad K^{(2)} = \frac{(1+L_{0}C^{2})K_{T}}{1+L_{0}}, \tag{4.20}\] \[K^{(3)} = \frac{(1+L_{0}C^{3})K_{T}}{1+L_{0}C^{2}}, \qquad K^{(4)} = \frac{(1+L_{0}C^{4})K_{T}}{1+L_{0}C^{3}},\] where C = K /K . The fractional saturation is given as \[\theta = \frac{\alpha(1+\alpha)^{3} + L_{0}\alpha C(1+\alpha C)^{3}}{(1+\alpha)^{4} + L_{0}\alpha C(1+\alpha C)^{4}}, \tag{4.21}\] where a = K [X], and [X] is the concentration of the free ligand (e.g., O ) in the same units (M or Torr) in which K is expressed. Figure 4.7B illustrates how the allosteric parameters, C = K /K and L = [R ]/[T ], are extracted from a plot of saturation (as log [\(\theta\)/(1 - \(\theta\))]) versus partial pressure of dioxygen (as log [P(O )]). Notice how the two-state model (Figure 4.7B) matches very closely the form of the binding curve for hemoglobin (Figure 4.4B). Equations (4.20) and (4.21) may be generalized to an oligomer with n subunits. In the case of hemoglobin, Perutz and coworkers, through the determination of the crystal structures of a variety of hemoglobin derivatives, have given subsequently a sound structural basis to the MWC model of two basic quaternary states (see below). A more exact treatment of ligand-binding data would allow for different affinities for different binding sites (called subunit heterogeneity) and different intrinsic affinities for ligand binding to the R-state conformation compared with the T-state conformation, for each level of ligand saturation—that is, for tertiary structure change within subunits upon ligation. This more exact treatment requires 25 separate equilibrium constants. Statistical thermodynamical approaches exist. These explicitly incorporate the different types of subunit interactions that structural studies have revealed, and give improved fits to oxygen-binding data and to the Bohr effect. The key element of two basic quaternary states is preserved, at least for dioxygen binding. For some modified hemoglobins, for example [\(\alpha\)-Fe(II) [\(\beta\)-Mn(III)] , where in the \(\beta\) subunits the heme iron is replaced by Mn(III), there is now strong evidence for three quaternary states, with the singly and several of the doubly ligated species having an energy state intermediate between the T (unliganded) and R (fully, triply, and the other doubly liganded) states. | 19,339 | 2,496 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/12%3A_Intermolecular_Forces%3A_Liquids_And_Solids/12.2%3A_Some_Properties_of_Liquids |
Although you have been introduced to some of the interactions that hold molecules together in a liquid, we have not yet discussed the consequences of those interactions for the bulk properties of liquids. We now turn our attention to three unique properties of liquids that intimately depend on the nature of intermolecular interactions: If liquids tend to adopt the shapes of their containers, then, do small amounts of water on a freshly waxed car form raised droplets instead of a thin, continuous film? The answer lies in a property called , which depends on intermolecular forces. Surface tension is the energy required to increase the surface area of a liquid by a unit amount and varies greatly from liquid to liquid based on the nature of the intermolecular forces, e.g., water with hydrogen bonds has a surface tension of 7.29 x 10 J/m (at 20°C), while mercury with metallic (electrostatic) bonds has as surface tension that is 15-times lower: 4.6 x 10 J/m (at 20°C). Figure \(\Page {1}\) presents a microscopic view of a liquid droplet. A typical molecule in the of the droplet is surrounded by other molecules that exert attractive forces from all directions. Consequently, there is no force on the molecule that would cause it to move in a particular direction. In contrast, a molecule on the experiences a net attraction toward the drop because there are no molecules on the outside to balance the forces exerted by adjacent molecules in the interior. Because a sphere has the smallest possible surface area for a given volume, intermolecular attractive interactions between water molecules cause the droplet to adopt a spherical shape. This maximizes the number of attractive interactions and minimizes the number of water molecules at the surface. Hence raindrops are almost spherical, and drops of water on a waxed (nonpolar) surface, which does not interact strongly with water, form round beads (see the chapter opener photo). A dirty car is covered with a mixture of substances, some of which are polar. Attractive interactions between the polar substances and water cause the water to spread out into a thin film instead of forming beads. The same phenomenon holds molecules together at the surface of a bulk sample of water, almost as if they formed a skin. When filling a glass with water, the glass can be overfilled so that the level of the liquid actually extends the rim. Similarly, a sewing needle or a paper clip can be placed on the surface of a glass of water where it “floats,” even though steel is much denser than water. Many insects take advantage of this property to walk on the surface of puddles or ponds without sinking. This is even better describe in the zero gravity condictions of space as Figure \(\Page {2}\) indicates (and more so in the video link). Such phenomena are manifestations of surface tension, which is defined as the energy required to increase the surface area of a liquid by a specific amount. Surface tension is therefore measured as energy per unit area, such as joules per square meter (J/m ) or dyne per centimeter (dyn/cm), where 1 dyn = 1 × 10 N. The values of the surface tension of some representative liquids are listed in Table \(\Page {1}\). Note the correlation between the surface tension of a liquid and the strength of the intermolecular forces: the stronger the intermolecular forces, the higher the surface tension. For example, water, with its strong intermolecular hydrogen bonding, has one of the highest surface tension values of any liquid, whereas low-boiling-point organic molecules, which have relatively weak intermolecular forces, have much lower surface tensions. Mercury is an apparent anomaly, but its very high surface tension is due to the presence of strong metallic bonding. Adding soaps and detergents that disrupt the intermolecular attractions between adjacent water molecules can reduce the surface tension of water. Because they affect the surface properties of a liquid, soaps and detergents are called surface-active agents, or surfactants. In the 1960s, US Navy researchers developed a method of fighting fires aboard aircraft carriers using “foams,” which are aqueous solutions of fluorinated surfactants. The surfactants reduce the surface tension of water below that of fuel, so the fluorinated solution is able to spread across the burning surface and extinguish the fire. Such foams are now used universally to fight large-scale fires of organic liquids. Surface Tension, Viscosity, & Melting Point: Intermolecular forces also cause a phenomenon called capillary action, which is the tendency of a polar liquid to rise against gravity into a small-diameter tube (a ), as shown in Figure \(\Page {3}\). When a glass capillary is put into a dish of water, water is drawn up into the tube. The height to which the water rises depends on the diameter of the tube and the temperature of the water but on the angle at which the tube enters the water. The smaller the diameter, the higher the liquid rises. When a glass capillary is placed in liquid water, water rises up into the capillary. The smaller the diameter of the capillary, the higher the water rises. The height of the water does depend on the angle at which the capillary is tilted. The same phenomenon holds molecules together at the surface of a bulk sample of water, almost as if they formed a skin. When filling a glass with water, the glass can be overfilled so that the level of the liquid actually extends Capillary action is the net result of two opposing sets of forces: cohesive forces, which are the intermolecular forces that hold a liquid together, and adhesive forces, which are the attractive forces between a liquid and the substance that composes the capillary. Water has both strong adhesion to glass, which contains polar SiOH groups, and strong intermolecular cohesion. When a glass capillary is put into water, the surface tension due to cohesive forces constricts the surface area of water within the tube, while adhesion between the water and the glass creates an upward force that maximizes the amount of glass surface in contact with the water. If the adhesive forces are stronger than the cohesive forces, as is the case for water, then the liquid in the capillary rises to the level where the downward force of gravity exactly balances this upward force. If, however, the cohesive forces are stronger than the adhesive forces, as is the case for mercury and glass, the liquid pulls itself down into the capillary below the surface of the bulk liquid to minimize contact with the glass (part (a) in Figure \(\Page {4}\)). The upper surface of a liquid in a tube is called the meniscus, and the shape of the meniscus depends on the relative strengths of the cohesive and adhesive forces. In liquids such as water, the meniscus is concave; in liquids such as mercury, however, which have very strong cohesive forces and weak adhesion to glass, the meniscus is convex (part (b) in Figure \(\Page {4}\)). Polar substances are drawn up a glass capillary and generally have a concave meniscus. Fluids and nutrients are transported up the stems of plants or the trunks of trees by capillary action. Plants contain tiny rigid tubes composed of cellulose, to which water has strong adhesion. Because of the strong adhesive forces, nutrients can be transported from the roots to the tops of trees that are more than 50 m tall. Cotton towels are also made of cellulose; they absorb water because the tiny tubes act like capillaries and “wick” the water away from your skin. The moisture is absorbed by the entire fabric, not just the layer in contact with your body. Viscosity (η) is the resistance of a liquid to flow. Some liquids, such as gasoline, ethanol, and water, flow very readily and hence have a . Others, such as motor oil, molasses, and maple syrup, flow very slowly and have a . The two most common methods for evaluating the viscosity of a liquid are (1) to measure the time it takes for a quantity of liquid to flow through a narrow vertical tube and (2) to measure the time it takes steel balls to fall through a given volume of the liquid. The higher the viscosity, the slower the liquid flows through the tube and the steel balls fall. Viscosity is expressed in units of the poise (mPa•s); the higher the number, the higher the viscosity. The viscosities of some representative liquids are listed in Table \(\Page {1}\) and show a correlation between viscosity and intermolecular forces. Because a liquid can flow only if the molecules can move past one another with minimal resistance, strong intermolecular attractive forces make it more difficult for molecules to move with respect to one another. The addition of a second hydroxyl group to ethanol, for example, which produces ethylene glycol (HOCH CH OH), increases the viscosity 15-fold. This effect is due to the increased number of hydrogen bonds that can form between hydroxyl groups in adjacent molecules, resulting in dramatically stronger intermolecular attractive forces. There is also a correlation between viscosity and molecular shape. Liquids consisting of long, flexible molecules tend to have higher viscosities than those composed of more spherical or shorter-chain molecules. The longer the molecules, the easier it is for them to become “tangled” with one another, making it more difficult for them to move past one another. London dispersion forces also increase with chain length. Due to a combination of these two effects, long-chain hydrocarbons (such as motor oils) are highly viscous. Viscosity increases as intermolecular interactions or molecular size increases. Motor oils and other lubricants demonstrate the practical importance of controlling viscosity. The oil in an automobile engine must effectively lubricate under a wide range of conditions, from subzero starting temperatures to the 200°C that oil can reach in an engine in the heat of the Mojave Desert in August. Viscosity decreases rapidly with increasing temperatures because the kinetic energy of the molecules increases, and higher kinetic energy enables the molecules to overcome the attractive forces that prevent the liquid from flowing. As a result, an oil that is thin enough to be a good lubricant in a cold engine will become too “thin” (have too low a viscosity) to be effective at high temperatures. The viscosity of motor oils is described by an SAE (Society of Automotive Engineers) rating ranging from SAE 5 to SAE 50 for engine oils: the lower the number, the lower the viscosity. So-called can cause major problems. If they are viscous enough to work at high operating temperatures (SAE 50, for example), then at low temperatures, they can be so viscous that a car is difficult to start or an engine is not properly lubricated. Consequently, most modern oils are , with designations such as SAE 20W/50 (a grade used in high-performance sports cars), in which case the oil has the viscosity of an SAE 20 oil at subzero temperatures (hence the W for winter) and the viscosity of an SAE 50 oil at high temperatures. These properties are achieved by a careful blend of additives that modulate the intermolecular interactions in the oil, thereby controlling the temperature dependence of the viscosity. Many of the commercially available oil additives “for improved engine performance” are highly viscous materials that increase the viscosity and effective SAE rating of the oil, but overusing these additives can cause the same problems experienced with highly viscous single-grade oils. Based on the nature and strength of the intermolecular cohesive forces and the probable nature of the liquid–glass adhesive forces, predict what will happen when a glass capillary is put into a beaker of SAE 20 motor oil. Will the oil be pulled up into the tube by capillary action or pushed down below the surface of the liquid in the beaker? What will be the shape of the meniscus (convex or concave)? (Hint: the surface of glass is lined with Si–OH groups.) substance and composition of the glass surface behavior of oil and the shape of meniscus Motor oil is a nonpolar liquid consisting largely of hydrocarbon chains. The cohesive forces responsible for its high boiling point are almost solely London dispersion forces between the hydrocarbon chains. Such a liquid cannot form strong interactions with the polar Si–OH groups of glass, so the surface of the oil inside the capillary will be lower than the level of the liquid in the beaker. The oil will have a convex meniscus similar to that of mercury. Predict what will happen when a glass capillary is put into a beaker of ethylene glycol. Will the ethylene glycol be pulled up into the tube by capillary action or pushed down below the surface of the liquid in the beaker? What will be the shape of the meniscus (convex or concave)? Capillary action will pull the ethylene glycol up into the capillary. The meniscus will be concave. Nearly all of us have heated a pan of water with the lid in place and shortly thereafter heard the sounds of the lid rattling and hot water spilling onto the stovetop. When a liquid is heated, its molecules obtain sufficient kinetic energy to overcome the forces holding them in the liquid and they escape into the gaseous phase. By doing so, they generate a population of molecules in the vapor phase above the liquid that produces a pressure—the vapor pressure of the liquid. In the situation we described, enough pressure was generated to move the lid, which allowed the vapor to escape. If the vapor is contained in a sealed vessel, however, such as an unvented flask, and the vapor pressure becomes too high, the flask will explode (as many students have unfortunately discovered). In this section, we describe vapor pressure in more detail and explain how to quantitatively determine the vapor pressure of a liquid. Vapor Pressure & Boiling Point: Because the molecules of a liquid are in constant motion, we can plot the fraction of molecules with a given kinetic energy ( ) against their kinetic energy to obtain the kinetic energy distribution of the molecules in the liquid (Figure \(\Page {5}\)), just as we did for a gas (Figure 10.19). As for gases, increasing the temperature increases both the average kinetic energy of the particles in a liquid and the range of kinetic energy of the individual molecules. If we assume that a minimum amount of energy ( ) is needed to overcome the intermolecular attractive forces that hold a liquid together, then some fraction of molecules in the liquid always has a kinetic energy greater than . The fraction of molecules with a kinetic energy greater than this minimum value increases with increasing temperature. Any molecule with a kinetic energy greater than has enough energy to overcome the forces holding it in the liquid and escape into the vapor phase. Before it can do so, however, a molecule must also be at the surface of the liquid, where it is physically possible for it to leave the liquid surface; that is, only molecules at the surface can undergo evaporation (or vaporization), where molecules gain sufficient energy to enter a gaseous state above a liquid’s surface, thereby creating a vapor pressure. To understand the causes of vapor pressure, consider the apparatus shown in Figure \(\Page {6}\). When a liquid is introduced into an evacuated chamber (part (a) in Figure \(\Page {6}\)), the initial pressure above the liquid is approximately zero because there are as yet no molecules in the vapor phase. Some molecules at the surface, however, will have sufficient kinetic energy to escape from the liquid and form a vapor, thus increasing the pressure inside the container. As long as the temperature of the liquid is held constant, the fraction of molecules with > will not change, and the rate at which molecules escape from the liquid into the vapor phase will depend only on the surface area of the liquid phase. As soon as some vapor has formed, a fraction of the molecules in the vapor phase will collide with the surface of the liquid and reenter the liquid phase in a process known as condensation (part (b) in Figure \(\Page {6}\)). As the number of molecules in the vapor phase increases, the number of collisions between vapor-phase molecules and the surface will also increase. Eventually, a will be reached in which exactly as many molecules per unit time leave the surface of the liquid (vaporize) as collide with it (condense). At this point, the pressure over the liquid stops increasing and remains constant at a particular value that is characteristic of the liquid at a given temperature. The rates of evaporation and condensation over time for a system such as this are shown graphically in Figure \(\Page {7}\). Two opposing processes (such as evaporation and condensation) that occur at the same rate and thus produce no change in a system, constitute a dynamic equilibrium. In the case of a liquid enclosed in a chamber, the molecules continuously evaporate and condense, but the amounts of liquid and vapor do not change with time. The pressure exerted by a vapor in dynamic equilibrium with a liquid is the equilibrium vapor pressure of the liquid. If a liquid is in an container, however, most of the molecules that escape into the vapor phase will collide with the surface of the liquid and return to the liquid phase. Instead, they will diffuse through the gas phase away from the container, and an equilibrium will never be established. Under these conditions, the liquid will continue to evaporate until it has “disappeared.” The speed with which this occurs depends on the vapor pressure of the liquid and the temperature. Volatile liquids have relatively high vapor pressures and tend to evaporate readily; nonvolatile liquids have low vapor pressures and evaporate more slowly. Although the dividing line between volatile and nonvolatile liquids is not clear-cut, as a general guideline, we can say that substances with vapor pressures greater than that of water (Table 11.4) are relatively volatile, whereas those with vapor pressures less than that of water are relatively nonvolatile. Thus diethyl ether (ethyl ether), acetone, and gasoline are volatile, but mercury, ethylene glycol, and motor oil are nonvolatile. The equilibrium vapor pressure of a substance at a particular temperature is a characteristic of the material, like its molecular mass, melting point, and boiling point (Table 11.4). It does depend on the amount of liquid as long as at least a tiny amount of liquid is present in equilibrium with the vapor. The equilibrium vapor pressure does, however, depend very strongly on the temperature and the intermolecular forces present, as shown for several substances in Figure \(\Page {8}\). Molecules that can hydrogen bond, such as ethylene glycol, have a much lower equilibrium vapor pressure than those that cannot, such as octane. The nonlinear increase in vapor pressure with increasing temperature is steeper than the increase in pressure expected for an ideal gas over the corresponding temperature range. The temperature dependence is so strong because the vapor pressure depends on the fraction of molecules that have a kinetic energy greater than that needed to escape from the liquid, and this fraction increases exponentially with temperature. As a result, sealed containers of volatile liquids are potential bombs if subjected to large increases in temperature. The gas tanks on automobiles are vented, for example, so that a car won’t explode when parked in the sun. Similarly, the small cans (1–5 gallons) used to transport gasoline are required by law to have a pop-off pressure release. Volatile substances have low boiling points and relatively weak intermolecular interactions; nonvolatile substances have high boiling points and relatively strong intermolecular interactions. The exponential rise in vapor pressure with increasing temperature in Figure \(\Page {8}\) allows us to use natural logarithms to express the nonlinear relationship as a linear one. \[ \ln\left ( P \right)=\dfrac{-\Delta H_{vap}}{R}\left ( \dfrac{1}{T} \right) + C \label{Eq1} \] where A plot of \ln versus the inverse of the absolute temperature (1/ ) is a straight line with a slope of −Δ / . Equation \(\ref{Eq1}\), called the Clausius–Clapeyron equation, can be used to calculate the Δ of a liquid from its measured vapor pressure at two or more temperatures. The simplest way to determine Δ is to measure the vapor pressure of a liquid at two temperatures and insert the values of and for these points into Equation \(\ref{Eq1}\), which is derived from the Clausius–Clapeyron equation: \[ \ln\left ( \dfrac{P_{1}}{P_{2}} \right)=\dfrac{-\Delta H_{vap}}{R}\left ( \dfrac{1}{T_{2}}-\dfrac{1}{T_{1}} \right) \label{11.5.2} \] Conversely, if we know Δ and the vapor pressure at any temperature , we can use Equation \(\ref{Eq1}\) to calculate the vapor pressure at any other temperature , as shown in Example \(\Page {2}\). The experimentally measured vapor pressures of liquid Hg at four temperatures are listed in the following table: From these data, calculate the enthalpy of vaporization (Δ ) of mercury and predict the vapor pressure of the liquid at 160°C. (Safety note: mercury is highly toxic; when it is spilled, its vapor pressure generates hazardous levels of mercury vapor.) vapor pressures at four temperatures Δ of mercury and vapor pressure at 160°C The table gives the measured vapor pressures of liquid Hg for four temperatures. Although one way to proceed would be to plot the data using Equation \(\ref{Eq1}\) and find the value of Δ from the slope of the line, an alternative approach is to use Equation \(\ref{Eq1}\) to obtain Δ directly from two pairs of values listed in the table, assuming no errors in our measurement. We therefore select two sets of values from the table and convert the temperatures from degrees Celsius to kelvins because the equation requires absolute temperatures. Substituting the values measured at 80.0°C ( ) and 120.0°C ( ) into Equation \(\ref{Eq1}\) gives \[ \ln\left ( \dfrac{0.7457 \; \cancel{Torr}}{0.0888 \; \cancel{Torr}} \right)=\dfrac{-\Delta H_{vap}}{8.314 \; J/mol\cdot K}\left ( \dfrac{1}{\left ( 120+273 \right)K}-\dfrac{1}{\left ( 80.0+273 \right)K} \right) \] \[ \ln\left ( 8.398 \right)=\dfrac{-\Delta H_{vap}}{8.314 \; J/mol\cdot \cancel{K}}\left ( -2.88\times 10^{-4} \; \cancel{K^{-1}} \right) \] \[ 2.13=-\Delta H_{vap} \left ( -3.46 \times 10^{-4} \right) J^{-1}\cdot mol \] \[ \Delta H_{vap} =61,400 \; J/mol = 61.4 \; kJ/mol \] We can now use this value of Δ to calculate the vapor pressure of the liquid ( ) at 160.0°C ( ): \[ \ln\left ( \dfrac{P_{2} }{0.0888 \; torr} \right)=\dfrac{-61,400 \; \cancel{J/mol}}{8.314 \; \cancel{J/mol} \; K^{-1}}\left ( \dfrac{1}{\left ( 160+273 \right)K}-\dfrac{1}{\left ( 80.0+273 \right) K} \right) \] Using the relationship = , we have \[ \ln\left ( \dfrac{P_{2} }{0.0888 \; Torr} \right)=3.86 \] \[ \dfrac{P_{2} }{0.0888 \; Torr} =e^{3.86} = 47.5 \] \[ P_{2} = 4.21 Torr \] At 160°C, liquid Hg has a vapor pressure of 4.21 torr, substantially greater than the pressure at 80.0°C, as we would expect. The vapor pressure of liquid nickel at 1606°C is 0.100 torr, whereas at 1805°C, its vapor pressure is 1.000 torr. At what temperature does the liquid have a vapor pressure of 2.500 torr? 1896°C As the temperature of a liquid increases, the vapor pressure of the liquid increases until it equals the external pressure, or the atmospheric pressure in the case of an open container. Bubbles of vapor begin to form throughout the liquid, and the liquid begins to boil. The temperature at which a liquid boils at exactly 1 atm pressure is the normal boiling point of the liquid. For water, the normal boiling point is exactly 100°C. The normal boiling points of the other liquids in Figure \(\Page {8}\) are represented by the points at which the vapor pressure curves cross the line corresponding to a pressure of 1 atm. Although we usually cite the normal boiling point of a liquid, the boiling point depends on the pressure. At a pressure greater than 1 atm, water boils at a temperature greater than 100°C because the increased pressure forces vapor molecules above the surface to condense. Hence the molecules must have greater kinetic energy to escape from the surface. Conversely, at pressures less than 1 atm, water boils below 100°C. Typical variations in atmospheric pressure at sea level are relatively small, causing only minor changes in the boiling point of water. For example, the highest recorded atmospheric pressure at sea level is 813 mmHg, recorded during a Siberian winter; the lowest sea-level pressure ever measured was 658 mmHg in a Pacific typhoon. At these pressures, the boiling point of water changes minimally, to 102°C and 96°C, respectively. At high altitudes, on the other hand, the dependence of the boiling point of water on pressure becomes significant. Table \(\Page {2}\) lists the boiling points of water at several locations with different altitudes. At an elevation of only 5000 ft, for example, the boiling point of water is already lower than the lowest ever recorded at sea level. The lower boiling point of water has major consequences for cooking everything from soft-boiled eggs (a “three-minute egg” may well take four or more minutes in the Rockies and even longer in the Himalayas) to cakes (cake mixes are often sold with separate high-altitude instructions). Conversely, pressure cookers, which have a seal that allows the pressure inside them to exceed 1 atm, are used to cook food more rapidly by raising the boiling point of water and thus the temperature at which the food is being cooked. As pressure , the boiling point of a liquid and vice versa. Use Figure \(\Page {8}\) to estimate the following. data in Figure \(\Page {8}\), pressure, and boiling point corresponding boiling point and pressure Ethylene glycol is an organic compound primarily used as a raw material in the manufacture of polyester fibers and fabric industry, and polyethylene terephthalate resins (PET) used in bottling. Use the data in Figure \(\Page {8}\) to estimate the following. is the energy required to increase the surface area of a liquid by a given amount. The stronger the intermolecular interactions, the greater the surface tension. are molecules, such as soaps and detergents, that reduce the surface tension of polar liquids like water. is the phenomenon in which liquids rise up into a narrow tube called a capillary. It results when , the intermolecular forces in the liquid, are weaker than , the attraction between a liquid and the surface of the capillary. The shape of the , the upper surface of a liquid in a tube, also reflects the balance between adhesive and cohesive forces. The of a liquid is its resistance to flow. Liquids that have strong intermolecular forces tend to have high viscosities. Because the molecules of a liquid are in constant motion and possess a wide range of kinetic energies, at any moment some fraction of them has enough energy to escape from the surface of the liquid to enter the gas or vapor phase. This process, called or , generates a above the liquid. Molecules in the gas phase can collide with the liquid surface and reenter the liquid via . Eventually, a is reached in which the number of molecules evaporating and condensing per unit time is the same, and the system is in a state of . Under these conditions, a liquid exhibits a characteristic that depends only on the temperature. We can express the nonlinear relationship between vapor pressure and temperature as a linear relationship using the . This equation can be used to calculate the enthalpy of vaporization of a liquid from its measured vapor pressure at two or more temperatures. are liquids with high vapor pressures, which tend to evaporate readily from an open container; have low vapor pressures. When the vapor pressure equals the external pressure, bubbles of vapor form within the liquid, and it boils. The temperature at which a substance boils at a pressure of 1 atm is its . ( ) | 28,246 | 2,497 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Carboxylic_Acids/Reactivity_of_Carboxylic_Acids/Making_Acyl_Chlorides_(Acid_Chlorides) |
This page looks at ways of swapping the -OH group in the -COOH group of a carboxylic acid for a chlorine atom. This produces useful compounds called acyl chlorides (acid chlorides). It covers the use of phosphorus(V) chloride and phosphorus(III) chloride as well as sulfur dichloride oxide (thionyl chloride). We are going to be looking at converting a carboxylic acid, RCOOH, into an acyl chloride, RCOCl. Acyl chlorides are also known as acid chlorides. By far the most commonly used example of the conversion of a carboxylic acid into an acyl chloride is ethanoic acid to ethanoyl chloride. Acyl chlorides are very reactive, and can be used to make a wide range of other things. That's why they are important. Phosphorus(V) chloride is a solid which reacts with carboxylic acids in the cold to give steamy acidic fumes of hydrogen chloride. It leaves a liquid mixture of the acyl chloride and a phosphorus compound, phosphorus trichloride oxide (phosphorus oxychloride) - POCl . \[CH_3COOH + PCl_5 \rightarrow CH_3COCl + POCl_3 + HCl\] The acyl chloride can be separated by . Phosphorus(III) chloride is a liquid at room temperature. Its reaction with a carboxylic acid is less dramatic than that of phosphorus(V) chloride because there is no hydrogen chloride produced. You end up with a mixture of the acyl chloride and phosphoric(III) acid (old names: phosphorous acid or orthophosphorous acid), H PO . For example: \[CH_3COOH + PCl_3 \rightarrow 3CH_3COCl + H_3PO3 \] Again, the ethanoyl chloride can be separated by fractional distillation. Sulfur dichloride oxide (thionyl chloride) is a liquid at room temperature and has the formula SOCl . Traditionally, the formula is written as shown, despite the fact that the modern name writes the chlorine before the oxygen (alphabetical order). The sulfur dichloride oxide reacts with carboxylic acids to produce an acyl chloride, and sulfur dioxide and hydrogen chloride gases are given off. For example: \[CH_3COOH + SOCl_2 \rightarrow CH_3COCl + SO_2 + HCl\] The separation is simplified to an extent because the by-products are both gases. You would obviously still have to fractionally distil the mixture to separate the acyl chloride from any excess acid or sulfur dichloride oxide. Jim Clark ( ) | 2,267 | 2,498 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/13%3A_Properties_of_Solutions/13.02%3A_Saturated_Solutions_and_Solubility |
When a solute dissolves, its individual atoms, molecules, or ions interact with the solvent, become solvated, and are able to diffuse independently throughout the solution (Figure \(\Page {1a}\)). This is not, however, a unidirectional process. If the molecule or ion happens to collide with the surface of a particle of the undissolved solute, it may adhere to the particle in a process called crystallization. Dissolution and crystallization continue as long as excess solid is present, resulting in a dynamic equilibrium analogous to the equilibrium that maintains the vapor pressure of a liquid. We can represent these opposing processes as follows: \[ solute + solvent \underset{crystallization}{\stackrel{dissolution}{\longrightleftharpoons}} solution \nonumber \] Although the terms precipitation and crystallization are both used to describe the separation of solid solute from a solution, crystallization refers to the formation of a solid with a well-defined crystalline structure, whereas precipitation refers to the formation of any solid phase, often one with very small particles. The maximum amount of a solute that can dissolve in a solvent at a specified temperature and pressure is its solubility. Solubility is often expressed as the mass of solute per volume (g/L) or mass of solute per mass of solvent (g/g), or as the moles of solute per volume (mol/L). Even for very soluble substances, however, there is usually a limit to how much solute can dissolve in a given quantity of solvent. In general, the solubility of a substance depends on not only the energetic factors we have discussed but also the temperature and, for gases, the pressure. At 20°C, for example, 177 g of NaI, 91.2 g of NaBr, 35.9 g of NaCl, and only 4.1 g of NaF dissolve in 100 g of water. At 70°C, however, the solubilities increase to 295 g of NaI, 119 g of NaBr, 37.5 g of NaCl, and 4.8 g of NaF. As you learned in Chapter 12, the lattice energies of the sodium halides increase from NaI to NaF. The fact that the solubilities decrease as the lattice energy increases suggests that the \(ΔH_2\) term in Figure 13.1 dominates for this series of compounds. A solution with the maximum possible amount of solute is saturated. If a solution contains less than the maximum amount of solute, it is unsaturated. When a solution is saturated and excess solute is present, the rate of dissolution is exactly equal to the rate of crystallization (Figure \(\Page {1b}\)). Using the value just stated, a saturated aqueous solution of NaCl, for example, contains 35.9 g of NaCl per 100 mL of water at 20°C. We can prepare a homogeneous saturated solution by adding excess solute (in this case, greater than 35.9 g of NaCl) to the solvent (water), stirring until the maximum possible amount of solute has dissolved, and then removing undissolved solute by filtration. The solubility of most solids increases with increasing temperature. Because the solubility of most solids increases with increasing temperature, a saturated solution that was prepared at a higher temperature usually contains more dissolved solute than it would contain at a lower temperature. When the solution is cooled, it can therefore become supersaturated (Figure \(\Page {1c}\)). Like a supercooled or superheated liquid, a supersaturated solution is unstable. Consequently, adding a small particle of the solute, a seed crystal, will usually cause the excess solute to rapidly precipitate or crystallize, sometimes with spectacular results. The rate of crystallization in Equation \(\ref{13.2.1}\) is greater than the rate of dissolution, so crystals or a precipitate form (Figure \(\Page {1d}\)). In contrast, adding a seed crystal to a saturated solution reestablishes the dynamic equilibrium, and the net quantity of dissolved solute no longer changes. Because crystallization is the reverse of dissolution, a substance that requires an input of heat to form a solution (\(ΔH_{soln} > 0\)) releases that heat when it crystallizes from solution (\(ΔH_{crys} < 0\)). The amount of heat released is proportional to the amount of solute that exceeds its solubility. Two substances that have a positive enthalpy of solution are sodium thiosulfate (\(Na_2S_2O_3\)) and sodium acetate (\(CH_3CO_2Na\)), both of which are used in commercial hot packs, small bags of supersaturated solutions used to warm hands (see Figure 13.1.3). The interactions that determine the solubility of a substance in a liquid depend largely on the chemical nature of the solute (such as whether it is ionic or molecular) rather than on its physical state (solid, liquid, or gas). We will first describe the general case of forming a solution of a molecular species in a liquid solvent and then describe the formation of a solution of an ionic compound. The London dispersion forces, dipole–dipole interactions, and hydrogen bonds that hold molecules to other molecules are generally weak. Even so, energy is required to disrupt these interactions. As described in Section 13.1, unless some of that energy is recovered in the formation of new, favorable solute–solvent interactions, the increase in entropy on solution formation is not enough for a solution to form. For solutions of gases in liquids, we can safely ignore the energy required to separate the solute molecules (\(ΔH_2 = 0\)) because the molecules are already separated. Thus we need to consider only the energy required to separate the solvent molecules (\(ΔH_1\)) and the energy released by new solute–solvent interactions (\(ΔH_3\)). Nonpolar gases such as \(N_2\), \(O_2\), and \(Ar\) have no dipole moment and cannot engage in dipole–dipole interactions or hydrogen bonding. Consequently, the only way they can interact with a solvent is by means of London dispersion forces, which may be weaker than the solvent–solvent interactions in a polar solvent. It is not surprising, then, that nonpolar gases are most soluble in nonpolar solvents. In this case, \(ΔH_1\) and \(ΔH_3\) are both small and of similar magnitude. In contrast, for a solution of a nonpolar gas in a polar solvent, \(ΔH_1\) is far greater than \(ΔH_3\). As a result, nonpolar gases are less soluble in polar solvents than in nonpolar solvents. For example, the concentration of \(N_2\) in a saturated solution of \(N_2\) in water, a polar solvent, is only \(7.07 \times 10^{-4}\; M\) compared with \(4.5 \times 10^{-3}\; M\) for a saturated solution of \(N_2\) in benzene, a nonpolar solvent. The solubilities of nonpolar gases in water generally increase as the molecular mass of the gas increases, as shown in Table \(\Page {1}\). This is precisely the trend expected: as the gas molecules become larger, the strength of the solvent–solute interactions due to London dispersion forces increases, approaching the strength of the solvent–solvent interactions. Virtually all common organic liquids, whether polar or not, are miscible. The strengths of the intermolecular attractions are comparable; thus the enthalpy of solution is expected to be small (\(ΔH_{soln} \approx 0\)), and the increase in entropy drives the formation of a solution. If the predominant intermolecular interactions in two liquids are very different from one another, however, they may be immiscible. For example, organic liquids such as benzene, hexane, \(CCl_4\), and \(CS_2\) (S=C=S) are nonpolar and have no ability to act as hydrogen bond donors or acceptors with hydrogen-bonding solvents such as \(H_2O\), \(HF\), and \(NH_3\); hence they are immiscible in these solvents. When shaken with water, they form separate phases or layers separated by an interface (Figure \(\Page {2}\)), the region between the two layers. Just because two liquids are immiscible, however, does not mean that they are completely insoluble in each other. For example, 188 mg of benzene dissolves in 100 mL of water at 23.5°C. Adding more benzene results in the separation of an upper layer consisting of benzene with a small amount of dissolved water (the solubility of water in benzene is only 178 mg/100 mL of benzene). The solubilities of simple alcohols in water are given in Table \(\Page {2}\). Only the three lightest alcohols (methanol, ethanol, and n-propanol) are completely miscible with water. As the molecular mass of the alcohol increases, so does the proportion of hydrocarbon in the molecule. Correspondingly, the importance of hydrogen bonding and dipole–dipole interactions in the pure alcohol decreases, while the importance of London dispersion forces increases, which leads to progressively fewer favorable electrostatic interactions with water. Organic liquids such as acetone, ethanol, and tetrahydrofuran are sufficiently polar to be completely miscible with water yet sufficiently nonpolar to be completely miscible with all organic solvents. The same principles govern the solubilities of molecular solids in liquids. For example, elemental sulfur is a solid consisting of cyclic \(S_8\) molecules that have no dipole moment. Because the \(S_8\) rings in solid sulfur are held to other rings by London dispersion forces, elemental sulfur is insoluble in water. It is, however, soluble in nonpolar solvents that have comparable London dispersion forces, such as \(CS_2\) (23 g/100 mL). In contrast, glucose contains five –OH groups that can form hydrogen bonds. Consequently, glucose is very soluble in water (91 g/120 mL of water) but essentially insoluble in nonpolar solvents such as benzene. The structure of one isomer of glucose is shown here. Low-molecular-mass hydrocarbons with highly electronegative and polarizable halogen atoms, such as chloroform (\(CHCl_3\)) and methylene chloride (\(CH_2Cl_2\)), have both significant dipole moments and relatively strong London dispersion forces. These hydrocarbons are therefore powerful solvents for a wide range of polar and nonpolar compounds. Naphthalene, which is nonpolar, and phenol (\(C_6H_5OH\)), which is polar, are very soluble in chloroform. In contrast, the solubility of ionic compounds is largely determined not by the polarity of the solvent but rather by its dielectric constant, a measure of its ability to separate ions in solution, as you will soon see. Identify the most important solute–solvent interactions in each solution. : components of solutions : predominant solute–solvent interactions : Identify all possible intermolecular interactions for both the solute and the solvent: London dispersion forces, dipole–dipole interactions, or hydrogen bonding. Determine which is likely to be the most important factor in solution formation. Identify the most important interactions in each solution: hydrogen bonding London interactions London dispersion forces A solute can be classified as hydrophilic (literally, “water loving”), meaning that it has an electrostatic attraction to water, or hydrophobic (“water fearing”), meaning that it repels water. A hydrophilic substance is polar and often contains O–H or N–H groups that can form hydrogen bonds to water. For example, glucose with its five O–H groups is hydrophilic. In contrast, a hydrophobic substance may be polar but usually contains C–H bonds that do not interact favorably with water, as is the case with naphthalene and n-octane. Hydrophilic substances tend to be very soluble in water and other strongly polar solvents, whereas hydrophobic substances are essentially insoluble in water and soluble in nonpolar solvents such as benzene and cyclohexane. The difference between hydrophilic and hydrophobic substances has substantial consequences in biological systems. For example, vitamins can be classified as either fat soluble or water soluble. Fat-soluble vitamins, such as vitamin A, are mostly nonpolar, hydrophobic molecules. As a result, they tend to be absorbed into fatty tissues and stored there. In contrast, water-soluble vitamins, such as vitamin C, are polar, hydrophilic molecules that circulate in the blood and intracellular fluids, which are primarily aqueous. Water-soluble vitamins are therefore excreted much more rapidly from the body and must be replenished in our daily diet. A comparison of the chemical structures of vitamin A and vitamin C quickly reveals why one is hydrophobic and the other hydrophilic. Because water-soluble vitamins are rapidly excreted, the risk of consuming them in excess is relatively small. Eating a dozen oranges a day is likely to make you tired of oranges long before you suffer any ill effects due to their high vitamin C content. In contrast, fat-soluble vitamins constitute a significant health hazard when consumed in large amounts. For example, the livers of polar bears and other large animals that live in cold climates contain large amounts of vitamin A, which have occasionally proven fatal to humans who have eaten them. The following substances are essential components of the human diet: Using what you know of hydrophilic and hydrophobic solutes, classify each as water soluble or fat soluble and predict which are likely to be required in the diet on a daily basis. : chemical structures classification as water soluble or fat soluble; dietary requirement : Based on the structure of each compound, decide whether it is hydrophilic or hydrophobic. If it is hydrophilic, it is likely to be required on a daily basis. : These compounds are consumed by humans: caffeine, acetaminophen, and vitamin D. Identify each as primarily hydrophilic (water soluble) or hydrophobic (fat soluble), and predict whether each is likely to be excreted from the body rapidly or slowly. Caffeine and acetaminophen are water soluble and rapidly excreted, whereas vitamin D is fat soluble and slowly excreted Solutions are not limited to gases and liquids; solid solutions also exist. For example, amalgams, which are usually solids, are solutions of metals in liquid mercury. Because most metals are soluble in mercury, amalgams are used in gold mining, dentistry, and many other applications. A major difficulty when mining gold is separating very small particles of pure gold from tons of crushed rock. One way to accomplish this is to agitate a suspension of the crushed rock with liquid mercury, which dissolves the gold (as well as any metallic silver that might be present). The very dense liquid gold–mercury amalgam is then isolated and the mercury distilled away. An alloy is a solid or liquid solution that consists of one or more elements in a metallic matrix. A solid alloy has a single homogeneous phase in which the crystal structure of the solvent remains unchanged by the presence of the solute. Thus the microstructure of the alloy is uniform throughout the sample. Examples are substitutional and interstitial alloys such as brass or solder. Liquid alloys include sodium/potassium and gold/mercury. In contrast, a partial alloy solution has two or more phases that can be homogeneous in the distribution of the components, but the microstructures of the two phases are not the same. As a liquid solution of lead and tin is cooled, for example, different crystalline phases form at different cooling temperatures. Alloys usually have properties that differ from those of the component elements. Network solids such as diamond, graphite, and \(\ce{SiO_2}\) are insoluble in all solvents with which they do not react chemically. The covalent bonds that hold the network or lattice together are simply too strong to be broken under normal conditions. They are certainly much stronger than any conceivable combination of intermolecular interactions that might occur in solution. Most metals are insoluble in virtually all solvents for the same reason: the delocalized metallic bonding is much stronger than any favorable metal atom–solvent interactions. Many metals react with solutions such as aqueous acids or bases to produce a solution. However, as we saw in Section 13.1, in these instances the metal undergoes a chemical transformation that cannot be reversed by simply removing the solvent. Solids with very strong intermolecular bonding tend to be insoluble. , you were introduced to guidelines for predicting the solubility of ionic compounds in water. Ionic substances are generally most soluble in polar solvents; the higher the lattice energy, the more polar the solvent must be to overcome the lattice energy and dissolve the substance. Because of its high polarity, water is the most common solvent for ionic compounds. Many ionic compounds are soluble in other polar solvents, however, such as liquid ammonia, liquid hydrogen fluoride, and methanol. Because all these solvents consist of molecules that have relatively large dipole moments, they can interact favorably with the dissolved ions. The ion–dipole interactions between \(\ce{Li^{+}}\) ions and acetone molecules in a solution of LiCl in acetone are shown in Figure \(\Page {3}\). The energetically favorable \(\ce{Li^{+}}\)–acetone interactions make \(ΔH_3\) sufficiently negative to overcome the positive \(ΔH_1\) and \(ΔH_2\). Because the dipole moment of acetone (2.88 D), and thus its polarity, is actually larger than that of water (1.85 D), one might even expect that LiCl would be more soluble in acetone than in water. In fact, the opposite is true: 83 g of LiCl dissolve in 100 mL of water at 20°C, but only about 4.1 g of \(\ce{LiCl}\) dissolve in 100 mL of acetone. This apparent contradiction arises from the fact that the dipole moment is a property of a single molecule in the gas phase. A more useful measure of the ability of a solvent to dissolve ionic compounds is its (ε), which is the ability of a bulk substance to decrease the electrostatic forces between two charged particles. By definition, the dielectric constant of a vacuum is 1. In essence, a solvent with a high dielectric constant causes the charged particles to behave as if they have been moved farther apart. At 25°C, the dielectric constant of water is 80.1, one of the highest known, and that of acetone is only 21.0. Hence water is better able to decrease the electrostatic attraction between \(\ce{Li^{+}}\) and \(\ce{Cl^{-}}\) ions, so \(\ce{LiCl}\) is more soluble in water than in acetone. This behavior is in contrast to that of molecular substances, for which polarity is the dominant factor governing solubility. A solvent’s dielectric constant is the most useful measure of its ability to dissolve ionic compounds. A solvent’s polarity is the dominant factor in dissolving molecular substances. It is also possible to dissolve ionic compounds in organic solvents using crown ethers, cyclic compounds with the general formula \((OCH_2CH_2)_n\). Crown ethers are named using both the total number of atoms in the ring and the number of oxygen atoms. Thus 18-crown-6 is an 18-membered ring with six oxygen atoms (Figure \(\Page {1a}\)). The cavity in the center of the crown ether molecule is lined with oxygen atoms and is large enough to be occupied by a cation, such as \(K^+\). The cation is stabilized by interacting with lone pairs of electrons on the surrounding oxygen atoms. Thus crown ethers solvate cations inside a hydrophilic cavity, whereas the outer shell, consisting of C–H bonds, is hydrophobic. Crown ethers are useful for dissolving ionic substances such as \(KMnO_4\) in organic solvents such as isopropanol \([(CH_3)_2CHOH]\) (Figure \(\Page {5}\)). The availability of crown ethers with cavities of different sizes allows specific cations to be solvated with a high degree of selectivity. Cryptands (from the Greek kryptós, meaning “hidden”) are compounds that can completely surround a cation with lone pairs of electrons on oxygen and nitrogen atoms (Figure \(\Page {4b}\)). The number in the name of the cryptand is the number of oxygen atoms in each strand of the molecule. Like crown ethers, cryptands can be used to prepare solutions of ionic compounds in solvents that are otherwise too nonpolar to dissolve them. The solubility of a substance is the maximum amount of a solute that can dissolve in a given quantity of solvent; it depends on the chemical nature of both the solute and the solvent and on the temperature and pressure. When a solution contains the maximum amount of solute that can dissolve under a given set of conditions, it is a saturated solution. Otherwise, it is unsaturated. Supersaturated solutions, which contain more dissolved solute than allowed under particular conditions, are not stable; the addition of a seed crystal, a small particle of solute, will usually cause the excess solute to crystallize. A system in which crystallization and dissolution occur at the same rate is in dynamic equilibrium. The solubility of a substance in a liquid is determined by intermolecular interactions, which also determine whether two liquids are miscible. Solutes can be classified as hydrophilic (water loving) or hydrophobic (water fearing). Vitamins with hydrophilic structures are water soluble, whereas those with hydrophobic structures are fat soluble. Many metals dissolve in liquid mercury to form amalgams. Covalent network solids and most metals are insoluble in nearly all solvents. The solubility of ionic compounds is largely determined by the dielectric constant (ε) of the solvent, a measure of its ability to decrease the electrostatic forces between charged particles. Solutions of many ionic compounds in organic solvents can be dissolved using crown ethers, cyclic polyethers large enough to accommodate a metal ion in the center, or cryptands, compounds that completely surround a cation. | 21,494 | 2,499 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/12%3A_Chromatographic_and_Electrophoretic_Methods/12.04%3A_Gas_Chromatography |
In (GC) we inject the sample, which may be a gas or a liquid, into an gaseous mobile phase (often called the carrier gas). The mobile phase carries the sample through a packed or a capillary column that separates the sample’s components based on their ability to partition between the mobile phase and the stationary phase. Figure 12.4.1
shows an example of a typical gas chromatograph, which consists of several key components: a supply of compressed gas for the mobile phase; a heated injector, which rapidly volatilizes the components in a liquid sample; a column, which is placed within an oven whose temperature we can control during the separation; and a detector to monitor the eluent as it comes off the column. Let’s consider each of these components. The most common mobile phases for gas chromatography are He, Ar, and N , which have the advantage of being chemically inert toward both the sample and the stationary phase. The choice of carrier gas often is determined by the needs of instrument’s detector. For a packed column the mobile phase velocity usually is 25–150 mL/min. The typical flow rate for a capillary column is 1–25 mL/min. There are two broad classes of chromatographic columns: packed columns and capillary columns. In general, a packed column can handle larger samples and a capillary column can separate more complex mixtures. are constructed from glass, stainless steel, copper, or aluminum, and typically are 2–6 m in length with internal diameters of 2–4 mm. The column is filled with a particulate solid support, with particle diameters ranging from 37–44 μm to 250–354 μm. Figure 12.4.2
shows a typical example of a packed column. The most widely used particulate support is diatomaceous earth, which is composed of the silica skeletons of diatoms. These particles are very porous, with surface areas ranging from 0.5–7.5 m /g, which provides ample contact between the mobile phase and the stationary phase. When hydrolyzed, the surface of a diatomaceous earth contains silanol groups (–SiOH), that serve as active sites for absorbing solute molecules in gas-solid chromatography (GSC). In (GLC), we coat the packing material with a liquid mobile phase. To prevent uncoated packing material from adsorbing solutes, which degrades the quality of the separation, surface silanols are deactivated by reacting them with dimethyldichlorosilane and rinsing with an alcohol—typically methanol—before coating the particles with stationary phase. , for example, has approximately 1800 plates/m, or a total of approximately 3600 theoretical plates. If we assume a / ≈ 50, then it has a peak capacity (equation 12.2.16) of \[n_{c}=1+\frac{\sqrt{3600}}{4} \ln (50) \approx 60 \nonumber\] A capillary, or is constructed from fused silica and is coated with a protective polymer coating. Columns range from 15–100 m in length with an internal diameter of approximately 150–300 μm. Figure 12.4.3
shows an example of a typical capillary column. Capillary columns are of three principal types. In a (WCOT) a thin layer of stationary phase, typically 0.25 nm thick, is coated on the capillary’s inner wall. In a (PLOT), a porous solid support—alumina, silica gel, and molecular sieves are typical examples—is attached to the capillary’s inner wall. A (SCOT) is a PLOT column that includes a liquid stationary phase. Figure 12.4.4
shows the differences between these types of capillary columns. A capillary column provides a significant improvement in separation efficiency because it has more theoretical plates per meter and is longer than a packed column. For example, the capillary column in Figure 12.4.3
has almost 4300 plates/m, or a total of 129 000 theoretical plates. If we assume a / ≈ 50, then it has a peak capacity of approximately 350. On the other hand, a packed column can handle a larger sample. Because of its smaller diameter, a capillary column requires a smaller sample, typically less than 10 μL. Elution order in gas–liquid chromatography depends on two factors: the boiling point of the solutes, and the interaction between the solutes and the stationary phase. If a mixture’s components have significantly different boiling points, then the choice of stationary phase is less critical. If two solutes have similar boiling points, then a separation is possible only if the stationary phase selectively interacts with one of the solutes. As a general rule, nonpolar solutes are separated more easily when using a nonpolar stationary phase, and polar solutes are easier to separate when using a polar stationary phase. There are several important criteria for choosing a stationary phase: it must not react with the solutes, it must be thermally stable, it must have a low volatility, and it must have a polarity that is appropriate for the sample’s components. Table 12.4.1
summarizes the properties of several popular stationary phases. low-boiling aliphatics hydrocarbons amides, fatty acid methyl esters, terpenoids polydimethyl siloxane alkaloids, amino acid derivatives, drugs, pesticides, phenols, steroids phenylmethyl polysiloxane (50% phenyl, 50% methyl) alkaloids, drugs, pesticides, polyaromatic hydrocarbons, polychlorinated biphenyls trifluoropropylmethyl polysiloxane (50% trifluoropropyl, 50% methyl) alkaloids, amino acid derivatives, drugs, halogenated compounds, ketones cyanopropylphenylmethyl polysiloxane (50%cyanopropyl, 50% phenylmethyl) nitriles, pesticides, steroids polyethylene glycol aldehydes, esters, ethers, phenols Many stationary phases have the general structure shown in Figure 12.4.5
a. A stationary phase of polydimethyl siloxane, in which all the –R groups are methyl groups, –CH , is nonpolar and often makes a good first choice for a new separation. The order of elution when using polydimethyl siloxane usually follows the boiling points of the solutes, with lower boiling solutes eluting first. Replacing some of the methyl groups with other substituents increases the stationary phase’s polarity and provides greater selectivity. For example, replacing 50% of the –CH groups with phenyl groups, –C H , produces a slightly polar stationary phase. Increasing polarity is provided by substituting trifluoropropyl, –C H CF, and cyanopropyl, –C H CN, functional groups, or by using a stationary phase of polyethylene glycol (Figure 12.4.5
b). An important problem with all liquid stationary phases is their tendency to elute, or from the column when it is heated. The temperature limits in Table 12.4.1
minimize this loss of stationary phase. Capillary columns with bonded or cross-linked stationary phases provide superior stability. A bonded stationary phase is attached chemically to the capillary’s silica surface. Cross-linking, which is done after the stationary phase is in the capillary column, links together separate polymer chains to provide greater stability. Another important consideration is the thickness of the stationary phase. From equation we know that separation efficiency improves with thinner films of stationary phase. The most common thickness is 0.25 μm, although a thicker films is useful for highly volatile solutes, such as gases, because it has a greater capacity for retaining such solutes. Thinner films are used when separating low volatility solutes, such as steroids. A few stationary phases take advantage of chemical selectivity. The most notable are stationary phases that contain chiral functional groups, which are used to separate enantiomers [Hinshaw, J. V. , , 644–648]. Three factors determine how we introduce a sample to the gas chromatograph. First, all of the sample’s constituents must be volatile. Second, the analytes must be present at an appropriate concentration. Finally, the physical process of injecting the sample must not degrade the separation. Each of these needs is considered in this section. Not every sample can be injected directly into a gas chromatograph. To move through the column, the sample’s constituents must be sufficiently volatile. A solute of low volatility, for example, may be retained by the column and continue to elute during the analysis of subsequent samples. A nonvolatile solute will condense at the top of the column, degrading the column’s performance. We can separate a sample’s volatile analytes from its nonvolatile components using any of the extraction techniques described in . A liquid–liquid extraction of analytes from an aqueous matrix into methylene chloride or another organic solvent is a common choice. Solid-phase extractions also are used to remove a sample’s nonvolatile components. An attractive approach to isolating analytes is a (SPME). In one approach, which is illustrated in Figure 12.4.6
, a fused-silica fiber is placed inside a syringe needle. The fiber, which is coated with a thin film of an adsorbent material, such as polydimethyl siloxane, is lowered into the sample by depressing a plunger and is exposed to the sample for a predetermined time. After withdrawing the fiber into the needle, it is transferred to the gas chromatograph for analysis. Two additional methods for isolating volatile analytes are a purge-and-trap and headspace sampling. In a , we bubble an inert gas, such as He or N , through the sample, releasing—or purging—the volatile compounds. These compounds are carried by the purge gas through a trap that contains an absorbent material, such as Tenax, where they are retained. Heating the trap and back-flushing with carrier gas transfers the volatile compounds to the gas chromatograph. In we place the sample in a closed vial with an overlying air space. After allowing time for the volatile analytes to equilibrate between the sample and the overlying air, we use a syringe to extract a portion of the vapor phase and inject it into the gas chromatograph. Alternatively, we can sample the headspace with an SPME. Thermal desorption is a useful method for releasing volatile analytes from solids. We place a portion of the solid in a glass-lined, stainless steel tube. After purging with carrier gas to remove any O that might be present, we heat the sample. Volatile analytes are swept from the tube by an inert gas and carried to the GC. Because volatilization is not a rapid process, the volatile analytes often are concentrated at the top of the column by cooling the column inlet below room temperature, a process known as . Once volatilization is complete, the column inlet is heated rapidly, releasing the analytes to travel through the column. The reason for removing O is to prevent the sample from undergoing an oxidation reaction when it is heated. To analyze a nonvolatile analyte we must convert it to a volatile form. For example, amino acids are not sufficiently volatile to analyze directly by gas chromatography. Reacting an amino acid, such as valine, with 1-butanol and acetyl chloride produces an esterified amino acid. Subsequent treatment with trifluoroacetic acid gives the amino acid’s volatile N-trifluoroacetyl- -butyl ester derivative. In an analyte’s concentration is too small to give an adequate signal, then we must concentrate the analyte before we inject the sample into the gas chromatograph. A side benefit of many extraction methods is that they often concentrate the analytes. Volatile organic materials isolated from an aqueous sample by a purge-and-trap, for example, are concentrated by as much as \(1000 \times\). If an analyte is too concentrated, it is easy to overload the column, resulting in peak fronting (see ) and a poor separation. In addition, the analyte’s concentration may exceed the detector’s linear response. Injecting less sample or diluting the sample with a volatile solvent, such as methylene chloride, are two possible solutions to this problem. In we examined several explanations for why a solute’s band increases in width as it passes through the column, a process we called band broadening. We also introduce an additional source of band broadening if we fail to inject the sample into the minimum possible volume of mobile phase. There are two principal sources of this precolumn band broadening: injecting the sample into a moving stream of mobile phase and injecting a liquid sample instead of a gaseous sample. The design of a gas chromatograph’s injector helps minimize these problems. An example of a simple injection port for a packed column is shown in Figure 12.4.7
. The top of the column fits within a heated injector block, with carrier gas entering from the bottom. The sample is injected through a rubber septum using a microliter syringe, such as the one shown in in Figure 12.4.8
. Injecting the sample directly into the column minimizes band broadening because it mixes the sample with the smallest possible amount of carrier gas. The injector block is heated to a temperature at least 50 C above the boiling point of the least volatile solute, which ensures a rapid vaporization of the sample’s components. Because a capillary column’s volume is significantly smaller than that for a packed column, it requires a different style of injector to avoid overloading the column with sample. Figure 12.4.9
shows a schematic diagram of a typical split/splitless injector for use with a capillary column. In a we inject the sample through a rubber septum using a microliter syringe. Instead of injecting the sample directly into the column, it is injected into a glass liner where it mixes with the carrier gas. At the split point, a small fraction of the carrier gas and sample enters the capillary column with the remainder exiting through the split vent. By controlling the flow rate of the carrier gas as it enters the injector, and its flow rate through the septum purge and the split vent, we can control the fraction of sample that enters the capillary column, typically 0.1–10%. For example, if the carrier gas flow rate is 50 mL/min, and the flow rates for the septum purge and the split vent are 2 mL/min and 47 mL/min, respectively, then the flow rate through the column is 1 mL/min (= 50 – 2 – 47). The ratio of sample entering the column is 1/50, or 2%. In a , which is useful for trace analysis, we close the split vent and allow all the carrier gas that passes through the glass liner to enter the column—this allows virtually all the sample to enters the column. Because the flow rate through the injector is low, significant precolumn band broadening is a problem. Holding the column’s temperature approximately 20–25 C below the solvent’s boiling point allows the solvent to condense at the entry to the capillary column, forming a barrier that traps the solutes. After allowing the solutes to concentrate, the column’s temperature is increased and the separation begins. For samples that decompose easily, an may be necessary. In this method the sample is injected directly into the column without heating. The column temperature is then increased, volatilizing the sample with as low a temperature as is practical. Control of the column’s temperature is critical to attaining a good separation when using gas chromatography. For this reason the column is placed inside a thermostated oven (see ). In an separation we maintain the column at a constant temperature. To increase the interaction between the solutes and the stationary phase, the temperature usually is set slightly below that of the lowest-boiling solute. One difficulty with an isothermal separation is that a temperature that favors the separation of a low-boiling solute may lead to an unacceptably long retention time for a higher-boiling solute. provides a solution to this problem. At the beginning of the analysis we set the column’s initial temperature below that for the lowest-boiling solute. As the separation progresses, we slowly increase the temperature at either a uniform rate or in a series of steps. The final part of a gas chromatograph is the detector. The ideal detector has several desirable features: a low detection limit, a linear response over a wide range of solute concentrations (which makes quantitative work easier), sensitivity for all solutes or selectivity for a specific class of solutes, and an insensitivity to a change in flow rate or temperature. One of the earliest gas chromatography detectors takes advantage of the mobile phase’s thermal conductivity. As the mobile phase exits the column it passes over a tungsten-rhenium wire filament (see Figure 12.4.10
. The filament’s electrical resistance depends on its temperature, which, in turn, depends on the thermal conductivity of the mobile phase. Because of its high thermal conductivity, helium is the mobile phase of choice when using a (TCD). Thermal conductivity, as the name suggests, is a measure of how easily a substance conducts heat. A gas with a high thermal conductivity moves heat away from the filament—and, thus, cools the filament—more quickly than does a gas with a low thermal conductivity. When a solute elutes from the column, the thermal conductivity of the mobile phase in the TCD cell decreases and the temperature of the wire filament, and thus it resistance, increases. A reference cell, through which only the mobile phase passes, corrects for any time-dependent variations in flow rate, pressure, or electrical power, all of which affect the filament’s resistance. Because all solutes affect the mobile phase’s thermal conductivity, the thermal conductivity detector is a universal detector. Another advantage is the TCD’s linear response over a concentration range spanning 10 –10 orders of magnitude. The detector also is non-destructive, which allows us to recover analytes using a postdetector cold trap. One significant disadvantage of the TCD detector is its poor detection limit for most analytes. The combustion of an organic compound in an H /air flame results in a flame that contains electrons and organic cations, presumably CHO . Applying a potential of approximately 300 volts across the flame creates a small current of roughly 10 to 10 amps. When amplified, this current provides a useful analytical signal. This is the basis of the popular a schematic diagram of which is shown in Figure 12.4.11
. Most carbon atoms—except those in carbonyl and carboxylic groups—generate a signal, which makes the FID an almost universal detector for organic compounds. Most inorganic compounds and many gases, such as H O and CO , are not detected, which makes the FID detector a useful detector for the analysis of organic analytes in atmospheric and aqueous environmental samples. Advantages of the FID include a detection limit that is approximately two to three orders of magnitude smaller than that for a thermal conductivity detector, and a linear response over 10 –10 orders of magnitude in the amount of analyte injected. The sample, of course, is destroyed when using a flame ionization detector. The is an example of a selective detector. As shown in Figure 12.4.12
, the detector consists of a \(\beta\)-emitter, such as Ni. The emitted electrons ionize the mobile phase, usually N , generating a standing current between a pair of electrodes. When a solute with a high affinity for capturing electrons elutes from the column, the current decreases, which serves as the signal. The ECD is highly selective toward solutes with electronegative functional groups, such as halogens and nitro groups, and is relatively insensitive to amines, alcohols, and hydrocarbons. Although its detection limit is excellent, its linear range extends over only about two orders of magnitude. A \(\beta\)-particle is an electron. A is an instrument that ionizes a gaseous molecule using sufficient energy that the resulting ion breaks apart into smaller ions. Because these ions have different mass-to-charge ratios, it is possible to separate them using a magnetic field or an electrical field. The resulting mass spectrum contains both quantitative and qualitative information about the analyte. Figure 12.4.13
shows a mass spectrum for toluene. Figure 12.4.14
shows a block diagram of a typical gas chromatography-mass spectrometer (GC–MS) instrument. The effluent from the column enters the mass spectrometer’s ion source in a manner that eliminates the majority of the carrier gas. In the ionization chamber the remaining molecules—a mixture of carrier gas, solvent, and solutes—undergo ionization and fragmentation. The mass spectrometer’s mass analyzer separates the ions by their mass-to-charge ratio and a detector counts the ions and displays the mass spectrum. There are several options for monitoring a chromatogram when using a mass spectrometer as the detector. The most common method is to continuously scan the entire mass spectrum and report the total signal for all ions that reach the detector during each scan. This total ion scan provides universal detection for all analytes. We can achieve some degree of selectivity by monitoring one or more specific mass-to-charge ratios, a process called selective-ion monitoring. A mass spectrometer provides excellent detection limits, typically 25 fg to 100 pg, with a linear range of 10 orders of magnitude. Because we continuously record the mass spectrum of the column’s eluent, we can go back and examine the mass spectrum for any time increment. This is a distinct advantage for GC–MS because we can use the mass spectrum to help identify a mixture’s components. For more details on mass spectrometry see Introduction to Mass Spectrometry by Michael Samide and Olujide Akinbo, a resource that is part of the Analytical Sciences Digital Library. Two additional detectors are similar in design to a flame ionization detector. In the flame photometric detector, optical emission from phosphorous and sulfur provides a detector selective for compounds that contain these elements. The thermionic detector responds to compounds that contain nitrogen or phosphorous. A Fourier transform infrared spectrophotometer (FT–IR) also can serve as a detector. In GC–FT–IR, effluent from the column flows through an optical cell constructed from a 10–40 cm Pyrex tube with an internal diameter of 1–3 mm. The cell’s interior surface is coated with a reflecting layer of gold. Multiple reflections of the source radiation as it is transmit- ted through the cell increase the optical path length through the sample. As is the case with GC–MS, an FT–IR detector continuously records the column eluent’s spectrum, which allows us to examine the IR spectrum for any time increment. See for a discussion of FT-IR spectroscopy and instrumentation. Gas chromatography is widely used for the analysis of a diverse array of samples in environmental, clinical, pharmaceutical, biochemical, forensic, food science and petrochemical laboratories. Table 12.4.2
provides some representative examples of applications. green house gases (CO , CH , NO ) in air pesticides in water, wastewater, and soil vehicle emissions trihalomethanes in drinking water drugs blood alcohols analysis of arson accelerants detection of explosives volatile organics in spices and fragrances trace organics in whiskey monomers in latex paint purity of solvents refinery gas composition of gasoline In a GC analysis the area under the peak is proportional to the amount of analyte injected onto the column. A peak’s area is determined by integration, which usually is handled by the instrument’s computer or by an electronic integrating recorder. If two peak are resolved fully, the determination of their respective areas is straightforward. Before electronic integrating recorders and computers, two methods were used to find the area under a curve. One method used a manual planimeter; as you use the planimeter to trace an object’s perimeter, it records the area. A second approach for finding a peak’s area is the cut-and-weigh method. The chromatogram is recorded on a piece of paper and each peak of interest is cut out and weighed. Assuming the paper is uniform in thickness and density of fibers, the ratio of weights for two peaks is the same as the ratio of areas. Of course, this approach destroys your chromatogram. Overlapping peaks, however, require a choice between one of several options for dividing up the area shared by the two peaks (Figure 12.4.15
). Which method we use depends on the relative size of the two peaks and their resolution. In some cases, the use of peak heights provides more accurate results [(a) Bicking, M. K. L. Chromatography Online, April 2006; (b) Bicking, M. K. L. Chromatography Online, June 2006]. For quantitative work we need to establish a calibration curve that relates the detector’s response to the analyte’s concentration. If the injection volume is identical for every standard and sample, then an external standardization provides both accurate and precise results. Unfortunately,even under the best conditions the relative precision for replicate injections may differ by 5%; often it is substantially worse. For quantitative work that requires high accuracy and precision, the use of internal standards is recommended. To review the method of internal standards, see . Marriott and Carpenter report the following data for five replicate injections of a mixture that contains 1% v/v methyl isobutyl ketone and 1% v/v -xylene in dichloromethane [Marriott, P. J.; Carpenter, P. D. , , 96–99]. Assume that -xylene (peak 2) is the analyte, and that methyl isobutyl ketone (peak 1) is the internal standard. Determine the 95% confidence interval for a single-point standardization with and without using the internal standard. For a single-point external standardization we ignore the internal standard and determine the relationship between the peak area for -xylene, , and the concentration, , of -xylene. \[A_{2}=k C_{2} \nonumber\] Substituting the known concentration for -xylene (1% v/v) and the appropriate peak areas, gives the following values for the constant . \[78112 \quad 135404 \quad 132332 \quad 112889 \quad 91287 \nonumber\] The average value for is 110 000 with a standard deviation of 25 100 (a relative standard deviation of 22.8%). The 95% confidence interval is \[\mu=\overline{X} \pm \frac{t s}{\sqrt{n}}=111000 \pm \frac{(2.78)(25100)}{\sqrt{5}}=111000 \pm 31200 \nonumber\] For an internal standardization, the relationship between the analyte’s peak area, , the internal standard’s peak area, , and their respective concentrations, and , is \[\frac{A_{2}}{A_{1}}=k \frac{C_{2}}{C_{1}} \nonumber\] Substituting in the known concentrations and the appropriate peak areas gives the following values for the constant . \[1.5917 \quad 1.5776 \quad 1.5728 \quad 1.5749 \quad 1.5724 \nonumber\] The average value for is 1.5779 with a standard deviation of 0.0080 (a relative standard deviation of 0.507%). The 95% confidence interval is \[\mu=\overline{X} \pm \frac{t s}{\sqrt{n}}=1.5779 \pm \frac{(2.78)(0.0080)}{\sqrt{5}}=1.5779 \pm 0.0099 \nonumber\] Although there is a substantial variation in the individual peak areas for this set of replicate injections, the internal standard compensates for these variations, providing a more accurate and precise calibration. Figure 12.4.16
shows chromatograms for five standards and for one sample. Each standard and sample contains the same concentration of an internal standard, which is 2.50 mg/mL. For the five standards, the concentrations of analyte are 0.20 mg/mL, 0.40 mg/mL, 0.60 mg/mL, 0.80 mg/mL, and 1.00 mg/mL, respectively. Determine the concentration of analyte in the sample by (a) ignoring the internal standards and creating an external standards calibration curve, and by (b) creating an internal standard calibration curve. For each approach, report the analyte’s concentration and the 95% confidence interval. Use peak heights instead of peak areas. The following table summarizes my measurements of the peak heights for each standard and the sample, and their ratio (although your absolute values for peak heights will differ from mine, depending on the size of your monitor or printout, your relative peak height ratios should be similar to mine). 0.20 35 Figure (a) shows the calibration curve and the calibration equation when we ignore the internal standard. Substituting the sample’s peak height into the calibration equation gives the analyte’s concentration in the sample as 0.49 mg/mL. The 95% confidence interval is ±0.24 mg/mL. The calibration curve shows quite a bit of scatter in the data because of uncertainty in the injection volumes. Figure (b) shows the calibration curve and the calibration equation when we include the internal standard. Substituting the sample’s peak height ratio into the calibration equation gives the analyte’s concentration in the sample as 0.54 mg/mL. The 95% confidence interval is ±0.04 mg/mL. To review the use of Excel or R for regression calculations and confidence intervals, see Chapter 5.5. The data for this exercise were created so that the analyte’s actual concentration is 0.55 mg/mL. Given the resolution of my ruler’s scale, my answer is pretty reasonable. Your measurements may be slightly different, but your answers should be close to the actual values. In addition to a quantitative analysis, we also can use chromatography to identify the components of a mixture. As noted earlier, when using an FT–IR or a mass spectrometer as the detector we have access to the eluent’s full spectrum for any retention time. By interpreting the spectrum or by searching against a library of spectra, we can identify the analyte responsible for each chromatographic peak. In addition to identifying the component responsible for a particular chromatographic peak, we also can use the saved spectra to evaluate peak purity. If only one component is responsible for a chromatographic peak, then the spectra should be identical throughout the peak’s elution. If a spectrum at the beginning of the peak’s elution is different from a spectrum taken near the end of the peak’s elution, then at least two components are co-eluting. When using a nonspectroscopic detector, such as a flame ionization detector, we must find another approach if we wish to identify the components of a mixture. One approach is to spike a sample with the suspected compound and look for an increase in peak height. We also can compare a peak’s retention time to the retention time for a known compound if we use identical operating conditions. Because a compound’s retention times on two identical columns are not likely to be the same—differences in packing efficiency, for example, will affect a solute’s retention time on a packed column—creating a table of standard retention times is not possible. provides one solution to the problem of matching retention times. Under isothermal conditions, the adjusted retention times for normal alkanes increase logarithmically. Kovat defined the retention index, , for a normal alkane as 100 times the number of carbon atoms. For example, the retention index is 400 for butane, C H , and 500 for pentane, C H . To determine the a compound’s retention index, , we use the following formula \[I_{cpd} = 100 \times \frac {\log t_{r,cpd}^{\prime} - \log t_{r,x}^{\prime}} {\log t_{r, x+1}^{\prime} - \log t_{r,x}^{\prime}} + I_x \label{12.1}\] where \(t_{r,cpd}^{\prime}\) is the compound’s adjusted retention time, \(t_{r,x}^{\prime}\) and \(t_{r,x+1}^{\prime}\) are the adjusted retention times for the normal alkanes that elute immediately before the compound and immediately after the compound, respectively, and is the retention index for the normal alkane that elutes immediately before the compound. A compound’s retention index for a particular set of chromatographic conditions—stationary phase, mobile phase, column type, column length, temperature, etc.—is reasonably consistent from day- to-day and between different columns and instruments. Tables of Kovat’s retention indices are available; see, for example, the . A search for toluene returns 341 values of for over 20 different stationary phases, and for both packed columns and capillary columns. In a separation of a mixture of hydrocarbons the following adjusted retention times are measured: 2.23 min for propane, 5.71 min for isobutane, and 6.67 min for butane. What is the Kovat’s retention index for each of these hydrocarbons? Kovat’s retention index for a normal alkane is 100 times the number of carbons; thus, for propane, = 300 and for butane, = 400. To find Kovat’s retention index for isobutane we use Equation \ref{12.1}. \[I_\text{isobutane} =100 \times \frac{\log (5.71)-\log (2.23)}{\log (6.67)-\log (2.23)}+300=386 \nonumber\] When using a column with the same stationary phase as in Example 12.4.2
, you find that the retention times for propane and butane are 4.78 min and 6.86 min, respectively. What is the expected retention time for isobutane? Because we are using the same column we can assume that isobutane’s retention index of 386 remains unchanged. Using Equation \ref{12.1}, we have \[386=100 \times \frac{\log x-\log (4.78)}{\log (6.86)-\log (4.78)}+300 \nonumber\] where is the retention time for isobutane. Solving for , we find that \[0.86=\frac{\log x-\log (4.78)}{\log (6.86)-\log (4.78)} \nonumber\] \[0.135=\log x-0.679 \nonumber\] \[0.814=\log x \nonumber\] \[x=6.52 \nonumber\] the retention time for isobutane is 6.5 min. The best way to appreciate the theoretical and the practical details discussed in this section is to carefully examine a typical analytical method. Although each method is unique, the following description of the determination of trihalomethanes in drinking water provides an instructive example of a typical procedure. The description here is based on a Method 6232B in , 20th Ed., American Public Health Association: Washing- ton, DC, 1998. Trihalomethanes, such as chloroform, CHCl , and bromoform, CHBr , are found in most chlorinated waters. Because chloroform is a suspected carcinogen, the determination of trihalomethanes in public drinking water supplies is of considerable importance. In this method the trihalomethanes CHCl , CHBrCl , CHBr Cl, and CHBr are isolated using a liquid–liquid extraction with pentane and determined using a gas chromatograph equipped with an electron capture detector. Collect the sample in a 40-mL glass vial equipped with a screw-cap lined with a TFE-faced septum. Fill the vial until it overflows, ensuring that there are no air bubbles. Add 25 mg of ascorbic acid as a reducing agent to quench the further production of trihalomethanes. Seal the vial and store the sample at 4 C for no longer than 14 days. Prepare a standard stock solution for each trihalomethane by placing 9.8 mL of methanol in a 10-mL volumetric flask. Let the flask stand for 10 min, or until all surfaces wetted with methanol are dry. Weigh the flask to the nearest ±0.1 mg. Using a 100-μL syringe, add 2 or more drops of trihalomethane to the volumetric flask, allowing each drop to fall directly into the methanol. Reweigh the flask before diluting to volume and mixing. Transfer the solution to a 40-mL glass vial equipped with a TFE-lined screw-top and report the concentration in μg/mL. Store the stock solutions at –10 to –20 C and away from the light. Prepare a multicomponent working standard from the stock standards by making appropriate dilutions of the stock solution with methanol in a volumetric flask. Choose concentrations so that calibration standards (see below) require no more than 20 μL of working standard per 100 mL of water. Using the multicomponent working standard, prepare at least three, but preferably 5–7 calibration standards. At least one standard must be near the detection limit and the standards must bracket the expected concentration of trihalomethanes in the samples. Using an appropriate volumetric flask, prepare the standards by injecting at least 10 μL of the working standard below the surface of the water and dilute to volume. Gently mix each standard three times only. Discard the solution in the neck of the volumetric flask and then transfer the remaining solution to a 40-mL glass vial with a TFE-lined screw-top. If the standard has a headspace, it must be analyzed within 1 hr; standards without a headspace may be held for up to 24 hr. Prepare an internal standard by dissolving 1,2-dibromopentane in hexane. Add a sufficient amount of this solution to pentane to give a final concentration of 30 μg 1,2-dibromopentane/L. To prepare the calibration standards and samples for analysis, open the screw top vial and remove 5 mL of the solution. Recap the vial and weigh to the nearest ±0.1 mg. Add 2.00 mL of pentane (with the internal standard) to each vial and shake vigorously for 1 min. Allow the two phases to separate for 2 min and then use a glass pipet to transfer at least 1 mL of the pentane (the upper phase) to a 1.8-mL screw top sample vial equipped with a TFE septum, and store at 4 C until you are ready to inject them into the GC. After emptying, rinsing, and drying the sample’s original vial, weigh it to the nearest ±0.1 mg and calculate the sample’s weight to ±0.1 g. If the density is 1.0 g/mL, then the sample’s weight is equivalent to its volume. Inject a 1–5 μL aliquot of the pentane extracts into a GC equipped with a 2-mm ID, 2-m long glass column packed with a stationary phase of 10% squalane on a packing material of 80/100 mesh Chromosorb WAW. Operate the column at 67 C and a flow rate of 25 mL/min. A variety of other columns can be used. Another option, for example, is a 30-m fused silica column with an internal diameter of 0.32 mm and a 1 µm coating of the stationary phase DB-1. A linear flow rate of 20 cm/s is used with the following temperature program: hold for 5 min at 35 C; increase to 70 C at 10 C/min; increase to 200 C at 20 C/min. 1. A simple liquid–liquid extraction rarely extracts 100% of the analyte. How does this method account for incomplete extractions? Because we use the same extraction procedure for the samples and the standards, we reasonably expect that the extraction efficiency is the same for all samples and standards; thus, the relative amount of analyte in any two samples or standards is unaffected by an incomplete extraction. 2. Water samples are likely to contain trace amounts of other organic compounds, many of which will extract into pentane along with the trihalomethanes. A short, packed column, such as the one used in this method, generally does not do a particularly good job of resolving chromatographic peaks. Why do we not need to worry about these other compounds? An electron capture detector responds only to compounds, such as the trihalomethanes, that have electronegative functional groups. Because an electron capture detector will not respond to most of the potential interfering compounds, the chromatogram will have relatively few peaks other than those for the trihalomethanes and the internal standard. 3. Predict the order in which the four analytes elute from the GC column. Retention time should follow the compound’s boiling points, eluting from the lowest boiling point to the highest boiling points. The expected elution order is CHCl (61.2 C), CHCl Br (90 C), CHClBr (119 C), and CHBr (149.1 C). 4. Although chloroform is an analyte, it also is an interferent because it is present at trace levels in the air. Any chloroform present in the laboratory air, for example, may enter the sample by diffusing through the sample vial’s silicon septum. How can we determine whether samples are contaminated in this manner? A sample blank of trihalomethane-free water is kept with the samples at all times. If the sample blank shows no evidence for chloroform, then we can safely assume that the samples also are free from contamination. 5. Why is it necessary to collect samples without a headspace (a layer of air that overlays the liquid) in the sample vial? Because trihalomethanes are volatile, the presence of a headspace allows for the loss of analyte from the sample to the headspace, resulting in a negative determinate error. 6. In preparing the stock solution for each trihalomethane, the procedure specifies that we add two or more drops of the pure compound by dropping them into a volumetric flask that contains methanol. When preparing the calibration standards, however, the working standard must be injected below the surface of the methanol. Explain the reason for this difference. When preparing a stock solution, the potential loss of the volatile trihalomethane is unimportant because we determine its concentration by weight after adding it to the methanol and diluting to volume. When we prepare the calibration standard, however, we must ensure that the addition of trihalomethane is quantitative; thus, we inject it below the surface to avoid the potential loss of analyte. Gas chromatography is used to analyze analytes present at levels ranging from major to ultratrace components. Depending on the detector, samples with major and minor analytes may need to be diluted before analysis. The thermal conductivity and flame ionization detectors can handle larger amounts of analyte; other detectors, such as an electron capture detector or a mass spectrometer, require substantially smaller amounts of analyte. Although the injection volume for gas chromatography is quite small—typically about a microliter—the amount of available sample must be sufficient that the injection is a representative subsample. For a trace analyte, the actual amount of injected analyte is often in the picogram range. Using as an example, a 3.0-μL injection of 1 μg/L CHCl is equivalent to 15 pg of CHCl , assuming a 100% extraction efficiency. The accuracy of a gas chromatographic method varies substantially from sample-to-sample. For routine samples, accuracies of 1–5% are common. For analytes present at very low concentration levels, for samples with complex matrices, or for samples that require significant processing before analysis, accuracy may be substantially poorer. In the analysis for trihalomethanes described in , for example, determinate errors as large as ±25% are possible. The precision of a gas chromatographic analysis includes contributions from sampling, sample preparation, and the instrument. The relative standard deviation due to the instrument typically is 1–5%, although it can be significantly higher. The principal limitations are detector noise, which affects the determination of peak area, and the reproducibility of injection volumes. In quantitative work, the use of an internal standard compensates for any variability in injection volumes. In a gas chromatographic analysis, sensitivity is determined by the detector’s characteristics. Of particular importance for quantitative work is the detector’s linear range; that is, the range of concentrations over which a calibration curve is linear. Detectors with a wide linear range, such as the thermal conductivity detector and the flame ionization detector, can be used to analyze samples over a wide range of concentrations without adjusting operating conditions. Other detectors, such as the electron capture detector, have a much narrower linear range. Because it combines separation with analysis, chromatographic methods provide excellent selectivity. By adjusting conditions it usually is possible to design a separation so that the analytes elute by themselves, even when the mixture is complex. Additional selectivity is obtained by using a detector, such as the electron capture detector, that does not respond to all compounds. Analysis time can vary from several minutes for samples that contain only a few constituents, to more than an hour for more complex samples. Preliminary sample preparation may substantially increase the analysis time. Instrumentation for gas chromatography ranges in price from inexpensive (a few thousand dollars) to expensive (>$50,000). The more expensive models are designed for capillary columns, include a variety of injection options, and use more sophisticated detectors, such as a mass spectrometer, or include multiple detectors. Packed columns typically cost <$200, and the cost of a capillary column is typically $300–$1000. | 44,192 | 2,500 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/04%3A_Reactions_in_Aqueous_Solution/4.04%3A_Oxidation-Reduction_Reactions |
The term oxidation was first used to describe reactions in which metals react with oxygen in air to produce metal oxides. When iron is exposed to air in the presence of water, for example, the iron turns to rust—an iron oxide. When exposed to air, aluminum metal develops a continuous, transparent layer of aluminum oxide on its surface. In both cases, the metal acquires a positive charge by transferring electrons to the neutral oxygen atoms of an oxygen molecule. As a result, the oxygen atoms acquire a negative charge and form oxide ions (O ). Because the metals have lost electrons to oxygen, they have been oxidized; oxidation is therefore the loss of electrons. Conversely, because the oxygen atoms have gained electrons, they have been reduced, so reduction is the gain of electrons. For every oxidation, there must be an associated reduction. Therefore, these reactions are known as oxidation-reduction reactions, or "redox" reactions for short. Any oxidation must be accompanied by a reduction and vice versa. Originally, the term reduction referred to the decrease in mass observed when a metal oxide was heated with carbon monoxide, a reaction that was widely used to extract metals from their ores. When solid copper(I) oxide is heated with hydrogen, for example, its mass decreases because the formation of pure copper is accompanied by the loss of oxygen atoms as a volatile product (water vapor). The reaction is as follows: \[ \ce{Cu_2O (s) + H_2 (g) \rightarrow 2Cu (s) + H_2O (g)} \label{4.4.1} \] Oxidation-reduction reactions are now defined as reactions that exhibit a change in the oxidation states of one or more elements in the reactants by a transfer of electrons, which follows the mnemonic "oxidation is loss, reduction is gain", or . The of each atom in a compound is the charge an atom would have if all its bonding electrons were transferred to the atom with the greater attraction for electrons. Atoms in their elemental form, such as O or H , are assigned an oxidation state of zero. For example, the reaction of aluminum with oxygen to produce aluminum oxide is \[\ce{ 4 Al (s) + 3O_2 \rightarrow 2Al_2O_3 (s)} \label{4.4.2} \] Each neutral oxygen atom gains two electrons and becomes negatively charged, forming an oxide ion; thus, oxygen has an oxidation state of −2 in the product and has been reduced. Each neutral aluminum atom loses three electrons to produce an aluminum ion with an oxidation state of +3 in the product, so aluminum has been oxidized. In the formation of Al O , electrons are transferred as follows (the small overset number emphasizes the oxidation state of the elements): \[ 4 \overset{0}{\ce{Al}} + 3 \overset{0}{\ce{O2}} \rightarrow \ce{4 Al^{3+} + 6 O^{2-} }\label{4.4.3} \] and are examples of oxidation–reduction (redox) reactions. In redox reactions, there is a net transfer of electrons from one reactant to another. In any redox reaction, the total number of electrons lost must equal the total of electrons gained to preserve electrical neutrality. In , for example, the total number of electrons lost by aluminum is equal to the total number gained by oxygen: \[ \begin{align*} \text{electrons lost} &= \ce{4 Al} \, \text{atoms} \times {3 \, e^- \, \text{lost} \over \ce{Al} \, \text{atom} } \\[4pt] &= 12 \, e^- \, \text{lost} \label{4.4.4a} \end{align*} \] \[ \begin{align*} \text{electrons gained} &= \ce{6 O} \, \text{atoms} \times {2 \, e^- \, \text{gained} \over \ce{O} \, \text{atom}} \\[4pt] &= 12 \, e^- \, \text{gained} \label{4.4.4b}\end{align*} \] The same pattern is seen in all oxidation–reduction reactions: the number of electrons lost must equal the number of electrons gained. An additional example of a redox reaction, the reaction of sodium metal with chlorine is illustrated in . In all oxidation–reduction (redox) reactions, the number of electrons lost equals the number of electrons gained. Assigning oxidation states to the elements in binary ionic compounds is straightforward: the oxidation states of the elements are identical to the charges on the monatomic ions. Previously, you learned how to predict the formulas of simple ionic compounds based on the sign and magnitude of the charge on monatomic ions formed by the neutral elements. Examples of such compounds are sodium chloride (NaCl; ), magnesium oxide (MgO), and calcium chloride (CaCl ). In covalent compounds, in contrast, atoms share electrons. However, we can still assign oxidation states to the elements involved by treating them as if they were ionic (that is, as if all the bonding electrons were transferred to the more attractive element). Oxidation states in covalent compounds are somewhat arbitrary, but they are useful bookkeeping devices to help you understand and predict many reactions. A set of rules for assigning oxidation states to atoms in chemical compounds follows. Nonintegral (fractional) oxidation states are encountered occasionally. They are usually due to the presence of two or more atoms of the same element with different oxidation states. In any chemical reaction, the net charge must be conserved; that is, in a chemical reaction, the total number of electrons is constant, just like the total number of atoms. Consistent with this, rule 1 states that the sum of the individual oxidation states of the atoms in a molecule or ion must equal the net charge on that molecule or ion. In NaCl, for example, Na has an oxidation state of +1 and Cl is −1. The net charge is zero, as it must be for any compound. Rule 3 is required because fluorine attracts electrons more strongly than any other element, for reasons you will discover in . Hence fluorine provides a reference for calculating the oxidation states of other atoms in chemical compounds. Rule 4 reflects the difference in chemistry observed for compounds of hydrogen with nonmetals (such as chlorine) as opposed to compounds of hydrogen with metals (such as sodium). For example, NaH contains the H ion, whereas HCl forms H and Cl ions when dissolved in water. Rule 5 is necessary because fluorine has a greater attraction for electrons than oxygen does; this rule also prevents violations of rule 2. So the oxidation state of oxygen is +2 in OF but −½ in KO . Note that an oxidation state of −½ for O in KO is perfectly acceptable. The reduction of copper(I) oxide shown in Equation \(\ref{4.4.5}\) demonstrates how to apply these rules. Rule 1 states that atoms in their elemental form have an oxidation state of zero, which applies to H and Cu. From rule 4, hydrogen in H O has an oxidation state of +1, and from rule 5, oxygen in both Cu O and H O has an oxidation state of −2. Rule 6 states that the sum of the oxidation states in a molecule or formula unit must equal the net charge on that compound. This means that each Cu atom in Cu O must have a charge of +1: 2(+1) + (−2) = 0. So the oxidation states are as follows: \[ \overset {\color{ref}{+1}}{\ce{Cu_2}} \overset {\color{ref}-2}{\ce{O}} (s) + \overset {\color{ref}0}{\ce{H_2}} (g) \rightarrow 2 \overset {\color{ref}0}{\ce{Cu}} (s) + \overset {\color{ref}+1}{\ce{H}}_2 \overset {\color{ref}-2}{\ce{O}} (g) \label{4.4.5} \] Assigning oxidation states allows us to see that there has been a net transfer of electrons from hydrogen (0 → +1) to copper (+1 → 0). Thus, this is a redox reaction. Once again, the number of electrons lost equals the number of electrons gained, and there is a net conservation of charge: \[ \text{electrons lost} = 2 \, H \, \text{atoms} \times {1 \, e^- \, \text{lost} \over H \, \text{atom} } = 2 \, e^- \, \text{lost} \label{4.4.6a} \] \[ \text{electrons gained} = 2 \, Cu \, \text{atoms} \times {1 \, e^- \, \text{gained} \over Cu \, \text{atom}} = 2 \, e^- \, \text{gained} \label{4.4.6b} \] Remember that oxidation states are useful for visualizing the transfer of electrons in oxidation–reduction reactions, but the oxidation state of an atom and its actual charge are the same only for simple ionic compounds. Oxidation states are a convenient way of assigning electrons to atoms, and they are useful for predicting the types of reactions that substances undergo. Assign oxidation states to all atoms in each compound. : molecular or empirical formula : oxidation states : Begin with atoms whose oxidation states can be determined unambiguously from the rules presented (such as fluorine, other halogens, oxygen, and monatomic ions). Then determine the oxidation states of other atoms present according to rule 1. : a. We know from rule 3 that fluorine always has an oxidation state of −1 in its compounds. The six fluorine atoms in sulfur hexafluoride give a total negative charge of −6. Because rule 1 requires that the sum of the oxidation states of all atoms be zero in a neutral molecule (here SF ), the oxidation state of sulfur must be +6: [(6 F atoms)(−1)] + [(1 S atom) (+6)] = 0 b. According to rules 4 and 5, hydrogen and oxygen have oxidation states of +1 and −2, respectively. Because methanol has no net charge, carbon must have an oxidation state of −2: [(4 H atoms)(+1)] + [(1 O atom)(−2)] + [(1 C atom)(−2)] = 0 c. Note that (NH ) SO is an ionic compound that consists of both a polyatomic cation (NH ) and a polyatomic anion (SO ) (see ). We assign oxidation states to the atoms in each polyatomic ion separately. For NH , hydrogen has an oxidation state of +1 (rule 4), so nitrogen must have an oxidation state of −3: [(4 H atoms)(+1)] + [(1 N atom)(−3)] = +1, the charge on the NH ion For SO42−, oxygen has an oxidation state of −2 (rule 5), so sulfur must have an oxidation state of +6: [(4 O atoms) (−2)] + [(1 S atom)(+6)] = −2, the charge on the sulfate ion d. Oxygen has an oxidation state of −2 (rule 5), giving an overall charge of −8 per formula unit. This must be balanced by the positive charge on three iron atoms, giving an oxidation state of +8/3 for iron: Fractional oxidation states are allowed because oxidation states are a somewhat arbitrary way of keeping track of electrons. In fact, Fe O can be viewed as having two Fe ions and one Fe ion per formula unit, giving a net positive charge of +8 per formula unit. Fe O is a magnetic iron ore commonly called magnetite. In ancient times, magnetite was known as lodestone because it could be used to make primitive compasses that pointed toward Polaris (the North Star), which was called the “lodestar.” e. Initially, we assign oxidation states to the components of CH CO H in the same way as any other compound. Hydrogen and oxygen have oxidation states of +1 and −2 (rules 4 and 5, respectively), resulting in a total charge for hydrogen and oxygen of [(4 H atoms)(+1)] + [(2 O atoms)(−2)] = 0 So the oxidation state of carbon must also be zero (rule 6). This is, however, an average oxidation state for the two carbon atoms present. Because each carbon atom has a different set of atoms bonded to it, they are likely to have different oxidation states. To determine the oxidation states of the individual carbon atoms, we use the same rules as before but with the additional assumption that bonds between atoms of the same element do not affect the oxidation states of those atoms. The carbon atom of the methyl group (−CH ) is bonded to three hydrogen atoms and one carbon atom. We know from rule 4 that hydrogen has an oxidation state of +1, and we have just said that the carbon–carbon bond can be ignored in calculating the oxidation state of the carbon atom. For the methyl group to be electrically neutral, its carbon atom must have an oxidation state of −3. Similarly, the carbon atom of the carboxylic acid group (−CO H) is bonded to one carbon atom and two oxygen atoms. Again ignoring the bonded carbon atom, we assign oxidation states of −2 and +1 to the oxygen and hydrogen atoms, respectively, leading to a net charge of [(2 O atoms)(−2)] + [(1 H atom)(+1)] = −3 To obtain an electrically neutral carboxylic acid group, the charge on this carbon must be +3. The oxidation states of the individual atoms in acetic acid are thus \[ \underset {-3}{C} \overset {+1}{H_3} \overset {+3}{C} \underset {-2}{O_2} \overset {+1}{H} \nonumber \] Thus the sum of the oxidation states of the two carbon atoms is indeed zero. Assign oxidation states to all atoms in each compound. Ba, +2; F, −1 C, 0; H, +1; O, −2 K, +1; Cr, +6; O, −2 Cs, +1; O, −½ C, −3; H, +1; C, −1; H, +1; O, −2; H, +1 Many types of chemical reactions are classified as redox reactions, and it would be impossible to memorize all of them. However, there are a few important types of redox reactions that you are likely to encounter and should be familiar with. These include: The following sections describe another important class of redox reactions: single-displacement reactions of metals in solution. \[\ce{ Fe(s) + 2HCl(aq) \rightarrow FeCl_2(aq) + H_2(g)} \label{4.4.81} \] In subsequent steps, \(\ce{FeCl2}\) undergoes oxidation to form a reddish-brown precipitate of \(\ce{Fe(OH)3}\). Many metals dissolve through reactions of this type, which have the general form \[\text{metal} + \text{acid} \rightarrow \text{salt} + \text{hydrogen} \label{4.4.82} \] Some of these reactions have important consequences. For example, it has been proposed that one factor that contributed to the fall of the Roman Empire was the widespread use of lead in cooking utensils and pipes that carried water. Rainwater, as we have seen, is slightly acidic, and foods such as fruits, wine, and vinegar contain organic acids. In the presence of these acids, lead dissolves: \[ \ce{Pb(s) + 2H^+(aq) \rightarrow Pb^{2+}(aq) + H_2(g) } \label{4.4.83} \] Consequently, it has been speculated that both the water and the food consumed by Romans contained toxic levels of lead, which resulted in widespread lead poisoning and eventual madness. Perhaps this explains why the Roman Emperor Caligula appointed his favorite horse as consul! Certain metals are oxidized by aqueous acid, whereas others are oxidized by aqueous solutions of various metal salts. Both types of reactions are called single-displacement reactions, in which the ion in solution is displaced through oxidation of the metal. Two examples of single-displacement reactions are the reduction of iron salts by zinc (Equation \(\ref{4.4.84}\)) and the reduction of silver salts by copper (Equation \(\ref{4.4.85}\) and Figure \(\Page {3}\)): \[ \ce{Zn(s) + Fe^{2+}(aq) \rightarrow Zn^{2+}(aq) + Fe(s)} \label{4.4.84} \] \[ \ce{ Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)} \label{4.4.85} \] The reaction in Equation \(\ref{4.4.84}\) is widely used to prevent (or at least postpone) the corrosion of iron or steel objects, such as nails and sheet metal. The process of “galvanizing” consists of applying a thin coating of zinc to the iron or steel, thus protecting it from oxidation as long as zinc remains on the object. By observing what happens when samples of various metals are placed in contact with solutions of other metals, chemists have arranged the metals according to the relative ease or difficulty with which they can be oxidized in a single-displacement reaction. For example, metallic zinc reacts with iron salts, and metallic copper reacts with silver salts. Experimentally, it is found that zinc reacts with both copper salts and silver salts, producing \(\ce{Zn2+}\). Zinc therefore has a greater tendency to be oxidized than does iron, copper, or silver. Although zinc will not react with magnesium salts to give magnesium metal, magnesium metal will react with zinc salts to give zinc metal: \[ \ce{Zn(s) + Mg^{2+}(aq) \xcancel{\rightarrow} Zn^{2+}(aq) + Mg(s)} \label{4.4.10} \] \[ \ce{Mg(s) + Zn^{2+}(aq) \rightarrow Mg^{2+}(aq) + Zn(s)} \label{4.4.11} \] Magnesium has a greater tendency to be oxidized than zinc does. Pairwise reactions of this sort are the basis of the activity series (Figure \(\Page {4}\)), which lists metals and hydrogen in order of their relative tendency to be oxidized. The metals at the top of the series, which have the greatest tendency to lose electrons, are the alkali metals (group 1), the alkaline earth metals (group 2), and Al (group 13). In contrast, the metals at the bottom of the series, which have the lowest tendency to be oxidized, are the precious metals or coinage metals—platinum, gold, silver, and copper, and mercury, which are located in the lower right portion of the metals in the periodic table. You should be generally familiar with which kinds of metals are active metals, which have the greatest tendency to be oxidized. (located at the top of the series) and which are inert metals, which have the least tendency to be oxidized. (at the bottom of the series). When using the activity series to predict the outcome of a reaction, keep in mind that . Because magnesium is above zinc in Figure \(\Page {4}\), magnesium metal will reduce zinc salts but not vice versa. Similarly, the precious metals are at the bottom of the activity series, so virtually any other metal will reduce precious metal salts to the pure precious metals. Hydrogen is included in the series, and the tendency of a metal to react with an acid is indicated by its position relative to hydrogen in the activity series. Because the precious metals lie below hydrogen, they do not dissolve in dilute acid and therefore do not corrode readily. Example \(\Page {2}\) demonstrates how a familiarity with the activity series allows you to predict the products of many single-displacement reactions. Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation. reactants overall reaction and net ionic equation \[ \ce{ Al(s) + 3Ag^+(aq) \rightarrow Al^{3+}(aq) + 3Ag(s)} \nonumber \] Recall from our discussion of solubilities that most nitrate salts are soluble. In this case, the nitrate ions are spectator ions and are not involved in the reaction. \[ \ce{Pb(s) + 2H^+(aq) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + H_2(g) } \nonumber \] Lead(II) sulfate is the white solid that forms on corroded battery terminals. Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation. \(no\: reaction\) \(3Zn(s) + 2Cr^{3+}(aq) \rightarrow 3Zn^{2+}(aq) + 2Cr(s)\) \(2Al(s) + 6CH_3CO_2H(aq) \rightarrow 2Al^{3+}(aq) + 6CH_3CO_2^-(aq) + 3H_2(g)\ Oxidation–reduction reactions are balanced by separating the overall chemical equation into an oxidation equation and a reduction equation. In oxidation–reduction reactions, electrons are transferred from one substance or atom to another. We can balance oxidation–reduction reactions in solution using the (Table \(\Page {1}\)), in which the overall reaction is separated into an oxidation equation and a reduction equation. There are many types of redox reactions. are reactions of metals with either acids or another metal salt that result in dissolution of the first metal and precipitation of a second (or evolution of hydrogen gas). The outcome of these reactions can be predicted using the (Figure \(\Page {4}\)), which arranges metals and H in decreasing order of their tendency to be oxidized. Any metal will reduce metal ions below it in the activity series. lie at the top of the activity series, whereas are at the bottom of the activity series. | 19,266 | 2,501 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Solubilty/Solubility_Rules |
When a substance is mixed with a solvent, there are several possible results. The determining factor for the result is the solubility of the substance, which is defined as the maximum possible concentration of the solute. The solubility rules help determine which substances are soluble, and to what extent. Depending on the solubility of a solute, there are three possible results: 1) if the solution has less solute than the maximum amount that it is able to dissolve The following are the solubility rules for common ionic solids. If there two rules appear to contradict each other, the preceding rule takes precedence. VJoKQ3ULCVs 1. Is FeCO soluble? According to Rule #5, carbonates tend to be insoluble. Therefore, . 2. Does ClO tend to form a precipitate? This is perchlorate, which according to Rule #2 is likely to be soluble. Therefore, 3. Which of these substances is likely to form a precipitate? a) CaSO b) table salt c) AgBr . Concerning a) CaSO , although sulfates tend to be soluble, Rule #5 indicates that calcium sulfate is an important exception to this rule. For b), Rule #1 indicates that table salt (NaCl) is soluble because it is a salt of an alkali metal. c) is an example of two rules contradicting each other. Rule #4 states that bromides are usually soluble, but Rule #3 states that salts of silver are insoluble. Because Rule #3 precedes Rule #4, the compound is insoluble and will form a precipitate. 4. Predict whether a precipitate will form as a result of this reaction: \[2AgNO_3 + Na_2S \rightarrow Ag_2S + 2NaNO_3 \] The products of the reaction must be examined; if either of the substances formed in the reaction is insoluble, a precipitate will form. Considering NaNO , Rule #3 states that nitrates tend to be soluble. A precipitate of this compound will not form. Next, consider Ag S. According to Rule #5, that sulfides tend to be insoluble. Therefore, because of this compound, . 5. Predict if a precipitate will form as a result of this reaction: \[2NaOH + K_2CrO_4 \rightarrow KOH + Na_2CrO_4 \] Consider again the products of the reaction: if either is insoluble, a precipitate will form. The first product, KOH, is an example of two rules contradicting each other. Although Rule #5 says that hydroxides tend to be insoluble, Rule #1 states that salts of alkali metal cations tend to be soluble, and Rule #1 precedes Rule #5. Therefore, this compound will not contribute to any precipitation being formed. The second product, Na CrO , also adheres to Rule #1, which states that salts of alkali metals tend to be soluble. Because both products are soluble, | 2,624 | 2,502 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/27%3A_Reactions_of_Organic_Compounds/27.02%3A_Introduction_to_Substitution_Reactions |
In many ways, the proton transfer process in a Brønsted-Lowry acid-base reaction can be thought of as simply a special kind of nucleophilic substitution reaction, one in which the electrophile is a hydrogen rather than a carbon. In both reaction types, we are looking at very similar players: an electron-rich species (the nucleophile/base) attacks an electron-poor species (the electrophile/proton), driving off the leaving group/conjugate base.In the next few sections, we are going to be discussing some general aspects of nucleophilic substitution reactions, and in doing so it will simplify things greatly if we can use some abbreviations and generalizations before we dive into real examples. Instead of showing a specific nucleophile like hydroxide, we will simply refer to the nucleophilic reactant as 'Nu'. In a similar fashion, we will call the leaving group 'X'. We will see as we study actual reactions that leaving groups are sometimes negatively charged, sometimes neutral, and sometimes positively charged. We will also see some examples of nucleophiles that are negatively charged and some that are neutral. Therefore, in this general picture we will not include a charge designation on the 'X' or 'Nu' species. In the same way, we will see later that nucleophiles and leaving groups are sometimes protonated and sometimes not, so for now, for the sake of simplicity, we will not include protons on 'Nu' or 'X'. We will generalize the three other groups bonded on the electrophilic central carbon as R , R , and R : these symbols could represent hydrogens as well as alkyl groups. Finally, in order to keep figures from becoming too crowded, we will use in most cases the line structure convention in which the central, electrophilic carbon is not drawn out as a 'C'. Here, then, is the generalized picture of a concerted (single-step) nucleophilic substitution reaction: There are two mechanistic models for how a nucleophilic substitution reaction can proceed. In the first picture, the reaction takes place in a single step, and bond-forming and bond-breaking occur simultaneously. This is called an ' , or ' mechanism. In the term S 2, S stands for 'substitution', the subscript N stands for 'nucleophilic', and the number 2 refers to the fact that this is a : the overall rate depends on a step in which two separate molecules (the nucleophile and the electrophile) collide. A potential energy diagram for this reaction shows the transition state (TS) as the highest point on the pathway from reactants to products. If you look carefully at the progress of the S 2 reaction, you will realize something very important about the outcome. The nucleophile, being an electron-rich species, must attack the electrophilic carbon from the relative to the location of the leaving group. Approach from the front side simply doesn't work: the leaving group - which is also an electron-rich group - blocks the way. The result of this backside attack is that the stereochemical configuration at the central carbon as the reaction proceeds. In a sense, the molecule is turned inside out. At the transition state, the electrophilic carbon and the three 'R' substituents all lie on the same plane. What this means is that S 2 reactions whether enzyme catalyzed or not, are inherently stereoselective: when the substitution takes place at a stereocenter, we can confidently predict the stereochemical configuration of the product. A second model for a nucleophilic substitution reaction is called the ' , or ' mechanism: in this picture, the C-X bond breaks , before the nucleophile approaches: This results in the formation of a carbocation: because the central carbon has only three bonds, it bears a formal charge of +1. Recall that a carbocation should be pictured as hybridized, with trigonal planar geometry. Perpendicular to the plane formed by the three hybrid orbitals is an empty, unhybridized orbital. In the second step of this two-step reaction, the nucleophile attacks the empty, 'electron hungry' orbital of the carbocation to form a new bond and return the carbon to tetrahedral geometry. We saw that S 2 reactions result specifically in inversion of stereochemistry at the electrophilic carbon center. What about the stereochemical outcome of S 1 reactions? In the model S 1 reaction shown above, the leaving group dissociates completely from the vicinity of the reaction before the nucleophile begins its attack. Because the leaving group is no longer in the picture, the nucleophile is free to attack from either side of the planar, -hybriduzed carbocation electrophile. This means that about half the time the product has the same stereochemical configuration as the starting material (retention of configuration), and about half the time the stereochemistry has been inverted. In other words, has occurred at the carbon center. As an example, the tertiary alkyl bromide below would be expected to form a racemic mix of and alcohols after an S 1 reaction with water as the incoming nucleophile. While most nonenzymatic S 1 reactions are not stereoselective, we will see later that enzyme-catalyzed nucleophilic substitution reactions - whether S 1 or S 2 - almost always stereoselective. The direct result of an enzymatic nucleophilic substitution reaction is more often than not of configuration - this is because the leaving group usually remains bound in the enzyme's active site long enough to block a nucleophilic attack from that side. This does not mean, however, that enzymes can only catalyze substitution reactions with inversion of configuration: we will see in the next chapter ( ) an example of an enzymatic nucleophilic substitution reaction in which the overall result is 100% of configuration. In the S 1 reaction we see an example of a reaction intermediate, a very important concept in the study of organic reaction mechanisms that was first introduced in Chapter 6. Recall that many important organic reactions do not occur in a single step; rather, they are the sum of two or more discreet bond-forming / bond-breaking steps, and involve transient intermediate species that go on to react very quickly. In the S 1 reaction, the carbocation species is a reaction intermediate. A potential energy diagram for an S 1 reaction shows that the carbocation intermediate can be visualized as a kind of valley in the path of the reaction, higher in energy than both the reactant and product but lower in energy than the two transition states. Recall ( ) that the first step of the reaction above, in which two charged species are formed from a neutral molecule, is much the slower of the two steps, and is therefore rate-determining. This is illustrated by the energy diagram, where the activation energy for the first step is higher than that for the second step. Also recall that an S 1 reaction has kinetics, because the rate determining step involves one molecule splitting apart, not two molecules colliding. Consider two nucleophilic substitutions that occur uncatalyzed in solution. Assume that reaction A is S 2, and reaction B is S 1. Predict, in each case, what would happen to the rate of the reaction if the concentration of the nucleophile were doubled, while all other conditions remained constant. Many S 1 reactions are of a class that are referred to as , where a solvent molecule participates in the reaction as a nucleophile. The S 1 reaction of allyl bromide in methanol is an example of what we would call , while if water is the solvent the reaction would be called : Because water and alcohols are relatively weak nucleophiles, they are unlikely to react in an S 2 fashion. One more important point must be made before continuing: nucleophilic substitutions as a rule occur at sp -hybridized carbons, and where the leaving group is attached to an sp -hybridized carbon:: Bonds on sp -hybridized carbons are inherently shorter and stronger than bonds on sp -hybridized carbons (section 1.5C), meaning that it is harder to break the C-X bond in these substrates. S 2 reactions of this type are unlikely also because the (hypothetical) electrophilic carbon is protected from nucleophilic attack by electron density in the p bond. S 1 reactions are highly unlikely, because the resulting carbocation intermediate, which would be sp-hybridized, would be very unstable (we’ll discuss the relative stability of carbocation intermediates in ). Before we look at some real-life nucleophilic substitution reactions in the next chapter, we will spend some time in the remainder of this chapter focusing more closely on the three principal partners in the nucleophilic substitution reaction: the nucleophile, the electrophile, and the leaving group. As a general rule, nucleophile substitution reactions that involve powerful nucleophiles tend to occur with S 2 mechanisms, while reactions with weaker nucleophiles tend to be S 1. But what makes a group a strong or weak nucleophile? Nucleophilic functional groups are those which have electron-rich atoms able to donate a pair of electrons to form a new covalent bond. In both laboratory and biological organic chemistry, the most relevant nucleophilic atoms are oxygen, nitrogen, and sulfur, and the most common nucleophilic functional groups are water, alcohols, phenols, amines, thiols, and occasionally carboxylates. Of course, carbons can also be nucleophiles - otherwise how could new carbon-carbon bonds be formed in the synthesis of large organic molecules like DNA or fatty acids? When thinking about nucleophiles, the first thing to recognize is that, for the most part, the same quality of 'electron-richness' that makes a something nucleophilic also makes it basic: . It should not be surprising, then, that most of the trends in basicity that we have already discussed also apply to nucleophilicity. The protonation state of a nucleophilic atom has a very large effect on its nucleophilicity. This is an idea that makes intuitive sense: a hydroxide ion is much more nucleophilic (and basic) than a water molecule, because the negatively charged oxygen on the hydroxide ion carries greater electron density than the oxygen atom of a neutral water molecule. In practical terms, this means that a hydroxide nucleophile will react in an S 2 reaction with methyl bromide much faster ( about 10,000 times faster) than a water nucleophile. A neutral amine is nucleophilic, whereas a protonated ammonium cation is not. This is why enzymes which have evolved to catalyze nucleophilic reactions often have a basic amino acid side chain poised in position to accept a proton from the nucleophilic atom as the nucleophilic attack occurs. Depending on the specific reaction being discussed, deprotonation of the nucleophile might occur before, during, or after the actual nucleophilic attack. In general for enzymatic reactions, however, it is most accurate to depict the proton abstraction and nucleophilic attack occurring simultaneously. The basic enzymatic group could be a histidine, a neutral (deprotonated) arginine or lysine, or a negatively-charged (deprotonated) aspartate, glutamate, or tyrosine. For example, a more complete picture of the DNA methylation reaction we saw in shows an aspartate from the enzyme's active site accepting a proton from the nucleophilic amine as it attacks the carbon electrophile. As it is deprotonated by the aspartate, the amine nitrogen becomes more electron-rich, and therefore more nucleophilic. There are predictable periodic trends in nucleophilicity. Moving horizontally across the second row of the table, the trend in nucleophilicity parallels the trend in basicity: The reasoning behind the horizontal nucleophilicity trend is the same as the reasoning behind the basicity trend: more electronegative elements hold their electrons more tightly, and are less able to donate them to form a new bond. This horizontal trends also tells us that amines are more nucleophilic than alcohols, although both groups commonly act as nucleophiles in both laboratory and biochemical reactions. Recall from that the basicity of atoms decreases as we move vertically down a column on the periodic table: thiolate ions are less basic than alkoxide ions, for example, and bromide ion is less basic than chloride ion, which in turn is less basic than fluoride ion. Recall also that this trend can be explained by considering the increasing size of the 'electron cloud' around the larger ions: the electron density inherent in the negative charge is spread around a larger area, which tends to increase stability (and thus reduce basicity). The vertical periodic trend for nucleophilicity is somewhat more complicated that that for basicity: depending on the solvent that the reaction is taking place in, the nucleophilicity trend can go in either direction. Let's take the simple example of the S 2 reaction below: . . .where Nu is one of the halide ions: fluoride, chloride, bromide, or iodide, and the leaving group I* is a radioactive isotope of iodine (which allows us to distinguish the leaving group from the nucleophile in that case where both are iodide). If this reaction is occurring in a (that is, a solvent that has a hydrogen bonded to an oxygen or nitrogen - water, methanol and ethanol are the most important examples), then the reaction will go fastest when iodide is the nucleophile, and slowest when fluoride is the nucleophile, reflecting the relative strength of the nucleophile. This of course, is opposite that of the vertical periodic trend for basicity, where iodide is the basic (you may want to review the reasoning for this trend in ). What is going on here? Shouldn't the stronger base, with its more reactive unbonded valence electrons, also be the stronger nucleophile? As mentioned above, it all has to do with the solvent. Remember, we are talking now about the reaction running in a solvent like ethanol. Protic solvent molecules form very strong ion-dipole interactions with the negatively-charged nucleophile, essentially creating a 'solvent cage' around the nucleophile: In order for the nucleophile to attack the electrophile, it must break free, at least in part, from its solvent cage. The lone pair electrons on the larger, less basic iodide ion interact less tightly with the protons on the protic solvent molecules - thus the iodide nucleophile is better able to break free from its solvent cage compared the smaller, more basic fluoride ion, whose lone pair electrons are bound more tightly to the protons of the cage. The picture changes if we switch to a , such as acetone, in which there is a molecular dipole but . Now, fluoride is the best nucleophile, and iodide the weakest. The reason for the reversal is that, with an aprotic solvent, the ion-dipole interactions between solvent and nucleophile are much weaker: the positive end of the solvent's dipole is hidden in the interior of the molecule, and thus it is shielded from the negative charge of the nucleophile. A weaker solvent-nucleophile interaction means a weaker solvent cage for the nucleophile to break through, so the solvent effect is much less important, and the more basic fluoride ion is also the better nucleophile. Why not use a completely nonpolar solvent, such as hexane, for this reaction, so that the solvent cage is eliminated completely? The answer to this is simple - the nucleophile needs to be in solution in order to react at an appreciable rate with the electrophile, and a solvent such as hexane will not solvate an a charged (or highly polar) nucleophile at all. That is why chemists use polar aprotic solvents for nucleophilic substitution reactions in the laboratory: they are polar enough to solvate the nucleophile, but not so polar as to lock it away in an impenetrable solvent cage. In addition to acetone, three other commonly used polar aprotic solvents are acetonitrile, dimethylformamide (DMF), and dimethyl sulfoxide (DMSO). In biological chemistry, where the solvent is protic (water), the most important implication of the periodic trends in nucleophilicity is that thiols are more powerful nucleophiles than alcohols. The thiol group in a , for example, is a powerful nucleophile and often acts as a nucleophile in enzymatic reactions, and of course negatively-charged thiolates (RS ) are even more nucleophilic. This is not to say that the hydroxyl groups on serine, threonine, and tyrosine do not also act as nucleophiles - they do. Resonance effects also come into play when comparing the inherent nucleophilicity of different molecules. The reasoning involved is the same as that which we used to understand resonance effects on basicity (see ). If the electron lone pair on a heteroatom is delocalized by resonance, it is inherently less reactive - meaning less nucleophilic, and also less basic. An alkoxide ion, for example, is more nucleophilic and more basic than a carboxylate group, even though in both cases the nucleophilic atom is a negatively charged oxygen. In the alkoxide, the negative charge is localized on a single oxygen, while in the carboxylate the charge is delocalized over two oxygen atoms by resonance. The nitrogen atom on an amide is less nucleophilic than the nitrogen of an amine, due to the resonance stabilization of the nitrogen lone pair provided by the amide carbonyl group. Steric hindrance is an important consideration when evaluating nucleophility. For example, -butanol is less potent as a nucleophile than methanol. This is because the comparatively bulky methyl groups on the tertiary alcohol effectively block the route of attack by the nucleophilic oxygen, slowing the reaction down considerably (imagine trying to walk through a narrow doorway while carrying three large suitcases!). It is not surprising that it is more common to observe serines acting as nucleophiles in enzymatic reactions compared to threonines - the former is a primary alcohol, while the latter is a secondary alcohol. by (University of Minnesota, Morris) | 18,076 | 2,503 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/01%3A_Introduction_-_Matter_and_Measurement/1.06%3A_Dimensional_Analysis |
Dimensional analysis is amongst the most valuable tools physical scientists use. Simply put, it is the conversion between an amount in one unit to the corresponding amount in a desired unit using various conversion factors. This is valuable because certain measurements are more accurate or easier to find than others. If you have every planned a party, you have used dimensional analysis. The amount of beer and munchies you will need depends on the number of people you expect. For example, if you are planning a Friday night party and expect 30 people you might estimate you need to go out and buy 120 bottles of sodas and 10 large pizza's. How did you arrive at these numbers? The following indicates the type of dimensional analysis solution to party problem: \[(30 \; \cancel{humans}) \times \left( \dfrac{\text{4 sodas}}{1 \; \cancel{human}} \right) = 120 \; \text{sodas} \label{Eq1} \] \[(30 \; \cancel{humans}) \times \left( \dfrac{\text{0.333 pizzas}}{1 \; \cancel{human}} \right) = 10 \; \text{pizzas} \label{Eq2} \] Notice that the units that canceled out are lined out and only the desired units are left (discussed more below). Finally, in going to buy the soda, you perform another dimensional analysis: should you buy the sodas in six-packs or in cases? \[(120\; { sodas}) \times \left( \dfrac{\text{1 six pack}}{6\; {sodas}} \right) = 20 \; \text{six packs} \label{Eq3} \] \[(120\; {sodas}) \times \left( \dfrac{\text{1 case }}{24\; {sodas}} \right) = 5 \; \text{cases} \label{Eq4} \] Realizing that carrying around 20 six packs is a real headache, you get 5 cases of soda instead. In this party problem, we have used dimensional analysis in two different ways: Consider the conversion in Equation \(\ref{Eq3}\): \[(120\; {sodas}) \times \left( \dfrac{\text{1 six pack}}{6\; {sodas}} \right) = 20 \; \text{six packs} \label{Eq3a} \] If we ignore the numbers for a moment, and just look at the units (i.e. ), we have: \[\text{soda} \times \left(\dfrac{\text{six pack}}{\text{sodas}}\right) \nonumber \] We can treat the dimensions in a similar fashion as other numerical analyses (i.e. any number divided by itself is 1). Therefore: \[\text{soda} \times \left(\dfrac{\text{six pack}}{\text{sodas}}\right) = \cancel{\text{soda}} \times \left(\dfrac{\text{six pack}}{\cancel{\text{sodas}}}\right) \nonumber \] So, the dimensions of the numerical answer will be "six packs". How can we use dimensional analysis to be sure we have set up our equation correctly? Consider the following alternative way to set up the above unit conversion analysis: \[ 120 \cancel{\text{soda}} \times \left(\dfrac{\text{6 sodas}}{\cancel{\text{six pack}}}\right) = 720 \; \dfrac{\text{sodas}^2}{\text{1 six pack}} \nonumber \] In the above case it was relatively straightforward keeping track of units during the calculation. What if the calculation involves powers, etc? For example, the equation relating kinetic energy to mass and velocity is: \[E_{kinetics} = \dfrac{1}{2} \text{mass} \times \text{velocity}^2 \label{KE} \] An example of units of mass is kilograms (kg) and velocity might be in meters/second (m/s). What are the dimensions of \(E_{kinetic}\)? \[(kg) \times \left( \dfrac{m}{s} \right)^2 = \dfrac{kg \; m^2}{s^2} \nonumber \] The \(\frac{1}{2}\) factor in Equation \ref{KE} is neglected since pure numbers have no units. Since the velocity is squared in Equation \ref{KE}, the associated with the numerical value of the velocity are also squared. We can double check this by knowing the the Joule (\(J\)) is a measure of energy, and as a composite unit can be decomposed thusly: \[1\; J = kg \dfrac{m^2}{s^2} \nonumber \] Pressure ( ) is a measure of the Force ( ) per unit area ( ): \[ P =\dfrac{F}{A} \nonumber \] Force, in turn, is a measure of the acceleration (\(a\)) on a mass (\(m\)): \[ F= m \times a \nonumber \] Thus, pressure (\(P\)) can be written as: \[ P= \dfrac{m \times a}{A} \nonumber \] What are the units of pressure from this relationship? ( ) \[ P =\dfrac{kg \times \frac{\cancel{m}}{s^2}}{m^{\cancel{2}}} \nonumber \] We can simplify this description of the units of Pressure by dividing numerator and denominator by \(m\): \[ P =\dfrac{\frac{kg}{s^2}}{m}=\dfrac{kg}{m\; s^2} \nonumber \] In fact, these are the units of a the composite ( ) unit and is the of pressure. The use of units in a calculation to ensure that we obtain the final proper units is called . For example, if we observe experimentally that an object’s potential energy is related to its mass, its height from the ground, and to a gravitational force, then when multiplied, the units of mass, height, and the force of gravity must give us units corresponding to those of energy. Energy is typically measured in joules, calories, or electron volts (eV), defined by the following expressions: Performing dimensional analysis begins with finding the appropriate . Then, you simply multiply the values together such that the units cancel by having equal units in the numerator and the denominator. To understand this process, let us walk through a few examples. Imagine that a chemist wants to measure out 0.214 mL of benzene, but lacks the equipment to accurately measure such a small volume. The chemist, however, is equipped with an analytical balance capable of measuring to \(\pm 0.0001 \;g\). Looking in a reference table, the chemist learns the density of benzene (\(\rho=0.8765 \;g/mL\)). How many grams of benzene should the chemist use? \[0.214 \; \cancel{mL} \left( \dfrac{0.8765\; g}{1\;\cancel{mL}}\right)= 0.187571\; g \nonumber \] Notice that the mL are being divided by mL, an equivalent unit. We can cancel these our, which results with the 0.187571 g. However, this is not our final answer, since this result has too many and must be rounded down to three significant digits. This is because 0.214 mL has three significant digits and the conversion factor had four significant digits. Since 5 is greater than or equal to 5, we must round the preceding 7 up to 8. Hence, the chemist should weigh out 0.188 g of benzene to have 0.214 mL of benzene. To illustrate the use of dimensional analysis to solve energy problems, let us calculate the kinetic energy in joules of a 320 g object traveling at 123 cm/s. To obtain an answer in joules, we must convert grams to kilograms and centimeters to meters. Using Equation \ref{KE}, the calculation may be set up as follows: \[ \begin{align*} &=\dfrac{1}{2}mv^2=\dfrac{1}{2}(g) \left(\dfrac{kg}{g}\right) \left[\left(\dfrac{cm}{s}\right)\left(\dfrac{m}{cm}\right) \right]^2 \\[4pt] &= (\cancel{g})\left(\dfrac{kg}{\cancel{g}}\right) \left(\dfrac{\cancel{m^2}}{s^2}\right) \left(\dfrac{m^2}{\cancel{cm^2}}\right) = \dfrac{kg⋅m^2}{s^2} \\[4pt] &=\dfrac{1}{2}320\; \cancel{g} \left( \dfrac{1\; kg}{1000\;\cancel{g}}\right) \left[\left(\dfrac{123\;\cancel{cm}}{1 \;s}\right) \left(\dfrac{1 \;m}{100\; \cancel{cm}}\right) \right]^2=\dfrac{0.320\; kg}{2}\left[\dfrac{123 m}{s(100)}\right]^2 \\[4pt] &=\dfrac{1}{2} 0.320\; kg \left[ \dfrac{(123)^2 m^2}{s^2(100)^2} \right]= 0.242 \dfrac{kg⋅m^2}{s^2} = 0.242\; J \end{align*} \nonumber \] Alternatively, the conversions may be carried out in a stepwise manner: Step 1: convert \(g\) to \(kg\) \[320\; \cancel{g} \left( \dfrac{1\; kg}{1000\;\cancel{g}}\right) = 0.320 \; kg \nonumber \] Step 2: convert \(cm\) to \(m\) \[123\;\cancel{cm} \left(\dfrac{1 \;m}{100\; \cancel{cm}}\right) = 1.23\ m \nonumber \] Now the natural units for calculating joules is used to get final results \[ \begin{align*} KE &=\dfrac{1}{2} 0.320\; kg \left(1.23 \;ms\right)^2 \\[4pt] &=\dfrac{1}{2} 0.320\; kg \left(1.513 \dfrac{m^2}{s^2}\right)= 0.242\; \dfrac{kg⋅m^2}{s^2}= 0.242\; J \end{align*} \nonumber \] Of course, steps 1 and 2 can be done in the opposite order with no effect on the final results. However, this second method involves an additional step. Now suppose you wish to report the number of kilocalories of energy contained in a 7.00 oz piece of chocolate in units of kilojoules per gram. To obtain an answer in kilojoules, we must convert 7.00 oz to grams and kilocalories to kilojoules. Food reported to contain a value in Calories actually contains that same value in kilocalories. If the chocolate wrapper lists the caloric content as 120 Calories, the chocolate contains 120 kcal of energy. If we choose to use multiple steps to obtain our answer, we can begin with the conversion of kilocalories to kilojoules: \[120 \cancel{kcal} \left(\dfrac{1000 \;\cancel{cal}}{\cancel{kcal}}\right)\left(\dfrac{4.184 \;\cancel{J}}{1 \cancel{cal}}\right)\left(\dfrac{1 \;kJ}{1000 \cancel{J}}\right)= 502\; kJ \nonumber \] We next convert the 7.00 oz of chocolate to grams: \[7.00\;\cancel{oz} \left(\dfrac{28.35\; g}{1\; \cancel{oz}}\right)= 199\; g \nonumber \] The number of kilojoules per gram is therefore \[\dfrac{ 502 \;kJ}{199\; g}= 2.52\; kJ/g \nonumber \] Alternatively, we could solve the problem in one step with all the conversions included: \[\left(\dfrac{120\; \cancel{kcal}}{7.00\; \cancel{oz}}\right)\left(\dfrac{1000 \;\cancel{cal}}{1 \;\cancel{kcal}}\right)\left(\dfrac{4.184 \;\cancel{J}}{1 \; \cancel{cal}}\right)\left(\dfrac{1 \;kJ}{1000 \;\cancel{J}}\right)\left(\dfrac{1 \;\cancel{oz}}{28.35\; g}\right)= 2.53 \; kJ/g \nonumber \] The discrepancy between the two answers is attributable to rounding to the correct number of significant figures for each step when carrying out the calculation in a stepwise manner. Recall that all digits in the calculator should be carried forward when carrying out a calculation using multiple steps. In this problem, we first converted kilocalories to kilojoules and then converted ounces to grams. Converting Between Units: Dimensional analysis is used in numerical calculations, and in converting units. It can help us identify whether an equation is set up correctly (i.e. the resulting units should be as expected). Units are treated similarly to the associated numerical values, i.e., if a variable in an equation is supposed to be squared, then the associated dimensions are squared, etc. ( ) | 10,080 | 2,504 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/13%3A_Properties_of_Solutions/13.E%3A_Properties_of_Solutions_(Exercises) |
. In addition to these individual basis; please contact 7. London dispersion forces increase with increasing atomic mass. Iodine is a solid while bromine is a liquid due to the greater intermolecular interactions between the heavier iodine atoms. Iodine is less soluble than bromine in virtually all solvents because it requires more energy to separate \(I_2\) molecules than \(Br_2\) molecules. 11. 15. In dental amalgam, the mercury atoms are locked in a solid phase that does not undergo corrosion under physiological conditions; hence, the mercury atoms cannot readily diffuse to the surface where they could vaporize or undergo chemical reaction. 21. Dissolve the mixture of A and B in a solvent in which they are both soluble when hot and relatively insoluble when cold, filter off any undissolved B, and cool slowly. Pure A should crystallize, while B stays in solution. If B were less soluble, it would be impossible to obtain pure A by this method in a single step, because some of the less soluble compound (B) will always be present in the solid that crystallizes from solution. hexane and methanol 1. \(6.7 \times 10^4\; amu\) 3. 9.24 atm 5. The \(CaCl_2\) solution will have a lower vapor pressure, because it contains three times as many particles as the glucose solution. 7. 0.36 m \(NaCl\), 2.6 g \(NaCl\) 9. 60 g NaBr 11. 700 g \(NaCl\) 13. \(MgCl_2\) produces three particles in solution versus two for \(NaCl\), so the same molal concentration of \(MgCl_2\) will produce a 50% greater freezing point depression than for \(NaCl\). Nonetheless, the molar mass of \(MgCl_2\) is 95.3 g/mol versus 48.45 g/mol for \(NaCl\). Consequently, a solution containing 1 g \(NaCl\) per 1000 g \(H_2O\) will produce a freezing point depression of 0.064°C versus 0.059°C for a solution containing 1 g \(MgCl_2\) per 1000 g \(H_2O\). Thus, given equal cost per gram, \(NaCl\) is more effective. Yes, \(MgCl_2\) would be effective at −8°C; a 1.43 m solution (136 g per 1000 g H2O) would be required. 16. k = 1.81(°C•kg)/mol, molecular mass of urea = 60.0 g/mol | 2,095 | 2,505 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Isomerism_in_Organic_Compounds/Structural_Isomerism_in_Organic_Molecules |
This page explains what structural isomerism is, and looks at some of the various ways that structural isomers can arise. Isomers are molecules that have the same molecular formula, but have a different arrangement of the atoms in space. That excludes any different arrangements which are simply due to the molecule rotating as a whole, or rotating about particular bonds. For example, both of the following are the same molecule. They are not isomers. Both are butane. There are also endless other possible ways that this molecule could twist itself. There is completely free rotation around all the carbon-carbon single bonds. If you had a model of a molecule in front of you, you would have to take it to pieces and rebuild it if you wanted to make an isomer of that molecule. If you can make an apparently different molecule just by rotating single bonds, it's not different - it's still the same molecule. In structural isomerism, the atoms are arranged in a completely different order. This is easier to see with specific examples. What follows looks at some of the ways that structural isomers can arise. The names of the various forms of structural isomerism probably don't matter all that much, but you must be aware of the different possibilities when you come to draw isomers. These isomers arise because of the possibility of branching in carbon chains. For example, there are two isomers of butane, \(C_4H_{10}\). In one of them, the carbon atoms lie in a "straight chain" whereas in the other the chain is branched. Be careful not to draw "false" isomers which are just twisted versions of the original molecule. For example, this structure is just the straight chain version of butane rotated about the central carbon-carbon bond. You could easily see this with a model. This is the example we've already used at the top of this page. Pentane, C H , has three chain isomers. If you think you can find any others, they are simply twisted versions of the ones below. If in doubt make some models. In position isomerism, the basic carbon skeleton remains unchanged, but important groups are moved around on that skeleton. For example, there are two structural isomers with the molecular formula C3H7Br. In one of them the bromine atom is on the end of the chain, whereas in the other it's attached in the middle. If you made a model, there is no way that you could twist one molecule to turn it into the other one. You would have to break the bromine off the end and re-attach it in the middle. At the same time, you would have to move a hydrogen from the middle to the end. Another similar example occurs in alcohols such as \(C_4H_9OH\) These are the only two possibilities provided you keep to a four carbon chain, but there is no reason why you should do that. You can easily have a mixture of chain isomerism and position isomerism - you aren't restricted to one or the other. So two other isomers of butanol are: You can also get position isomers on benzene rings. Consider the molecular formula \(C_7H_7Cl\). There are four different isomers you could make depending on the position of the chlorine atom. In one case it is attached to the side-group carbon atom, and then there are three other possible positions it could have around the ring - next to the \(CH_3\) group, next-but-one to the \(CH_3\) group, or opposite the \(CH_3\) group. In this variety of structural isomerism, the isomers contain different functional groups - that is, they belong to different families of compounds (different homologous series). A molecular formula \(C_3H_6O\) could be either propanal (an aldehyde) or propanone (a ketone). There are other possibilities as well for this same molecular formula - for example, you could have a carbon-carbon double bond (an alkene) and an -OH group (an alcohol) in the same molecule. Another common example is illustrated by the molecular formula \(C_3H_6O_2\). Amongst the several structural isomers of this are propanoic acid (a carboxylic acid) and methyl ethanoate (an ester). Jim Clark ( ) | 4,048 | 2,506 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/01%3A_General_Techniques/1.02%3A_Glassware_and_Equipment/1.2B%3A_Ground_Glass_Joints |
Most organic glassware uses "ground glass joints," which have a frosted appearance. They are precisely ground to a certain size (which makes them expensive) and have outer (female) and inner (male) joints so that pieces can be connected together with a tight fit (Figure 1.1a). Common joint sizes are 14/20, 19/22, and 24/40. The first number refers to the inner diameter (in millimeters) of a female joint or outer diameter of a male joint. The second number refers to the length of the joint (Figure 1.1b). It is best if ground glass joints are free of chemicals when pieces are connected, or else the compounds may undergo reactions that cause the joints to "freeze" together, or become inseparable. Solid in the joint can also compromise the seal between the pieces. If chemical residue were to get on the joint during transfer (Figure 1.1c), the joint should be wiped clean with a (lint-free tissue, Figure 1.2a) before connecting with another piece. Spillage on the joint can be minimized by using a funnel. Figures 1.2 b+c shows a "frozen" joint (notice the residue on the frosted joint), where benzaldehyde crept into the joint during storage and probably oxidized to seal the round bottomed flask and stopper together. To separate a frozen joint, first try to gently twist the two pieces apart from one another. If that fails, gently tap on the joint with a spatula or other piece of equipment (Figure 1.2c). If that fails, next try heating the joint in a hot water bath (heat may cause expansion of the outer joint), or sonicating the flask if a sonicator is available. As a last resort, see your instructor, and they may heat the joint briefly with a heat gun. The frozen joint in Figure 1.2 had to be heated to separate the pieces. | 1,756 | 2,507 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/01%3A_Introduction_-_Matter_and_Measurement/1.03%3A_Properties_of_Matter |
All matter has physical and chemical properties. are characteristics that scientists can measure without changing the composition of the sample under study, such as mass, color, and volume (the amount of space occupied by a sample). describe the characteristic ability of a substance to react to form new substances; they include its flammability and susceptibility to corrosion. All samples of a pure substance have the same chemical and physical properties. For example, pure copper is always a reddish-brown solid (a physical property) and always dissolves in dilute nitric acid to produce a blue solution and a brown gas (a chemical property). Physical properties can be extensive or intensive. vary with the amount of the substance and include mass, weight, and volume. , in contrast, do not depend on the amount of the substance; they include color, melting point, boiling point, electrical conductivity, and physical state at a given temperature. For example, elemental sulfur is a yellow crystalline solid that does not conduct electricity and has a melting point of 115.2 °C, no matter what amount is examined ( ). Scientists commonly measure intensive properties to determine a substance’s identity, whereas extensive properties convey information about the amount of the substance in a sample. Although mass and volume are both extensive properties, their ratio is an important intensive property called (\(\rho\)). Density is defined as mass per unit volume and is usually expressed in grams per cubic centimeter (g/cm ). As mass increases in a given volume, density also increases. For example, lead, with its greater mass, has a far greater density than the same volume of air, just as a brick has a greater density than the same volume of Styrofoam. At a given temperature and pressure, the density of a pure substance is a constant: \[\begin{align*} \text{density} &={\text{mass} \over \text{volume}} \\[4pt] \rho &={m \over V} \label{Eq1} \end{align*} \] Pure water, for example, has a density of 0.998 g/cm at 25 °C. The average densities of some common substances are in Notice that corn oil has a lower mass to volume ratio than water. This means that when added to water, corn oil will “float” (Figure \(\Page {2}\)). are changes in which no chemical bonds are broken or formed. This means that the same types of compounds or elements that were there at the beginning of the change are there at the end of the change. Because the ending materials are the same as the beginning materials, the properties (such as color, boiling point, etc) will also be the same. Physical changes involve moving molecules around, but not changing them. Some types of physical changes include: As an ice cube melts, its shape changes as it acquires the ability to flow. However, its composition does not change. is an example of a (Figure \(\Page {3}\)), since some properties of the material change, but the identity of the matter does not. Physical changes can further be classified as reversible or irreversible. The melted ice cube may be refrozen, so melting is a reversible physical change. Physical changes that involve a change of state are all reversible. Other changes of state include (liquid to gas), (liquid to solid), and (gas to liquid). Dissolving is also a reversible physical change. When salt is dissolved into water, the salt is said to have entered the aqueous state. The salt may be regained by boiling off the water, leaving the salt behind. Figure \(\Page {3}\): Ice Melting is a physical change. When solid water (\(\ce{H_2O}\)) as ice melts into a liquid (water), it appears changed. However, this change is only physical as the the composition of the constituent molecules is the same: 11.19% hydrogen and 88.81% oxygen by mass. occur when bonds are broken and/or formed between molecules or atoms. This means that one substance with a certain set of properties (such as melting point, color, taste, etc) is turned into a different substance with different properties. Chemical changes are frequently harder to reverse than physical changes. One good example of a chemical change is burning paper. In contrast to the act of ripping paper, the act of burning paper actually results in the formation of new chemicals (carbon dioxide and water, to be exact). Another example of chemical change occurs when water is formed. Each molecule contains two atoms of hydrogen and one atom of oxygen chemically bonded. Another example of a chemical change is what occurs when natural gas is burned in your furnace. This time, before the reaction we have a molecule of methane, \(\ce{CH_4}\), and two molecules of oxygen, \(\ce{O_2}\), while after the reaction we have two molecules of water, \(\ce{H_2O}\), and one molecule of carbon dioxide, \(\ce{CO_2}\). In this case, not only has the appearance changed, but the structure of the molecules has also changed. The new substances do not have the same chemical properties as the original ones. Therefore, this is a chemical change. The combustion of magnesium metal is also chemical change (Magnesium + Oxygen → Magnesium Oxide): \[\ce{2 Mg + O_2 \rightarrow 2 MgO } \nonumber \] as is the rusting of iron (Iron + Oxygen → Iron Oxide/ Rust): \[\ce{4 Fe + 3O_2 \rightarrow 2 Fe_2O_3} \nonumber \] Using the components of composition and properties, we have the ability to distinguish one sample of matter from the others. Different Definitions of Changes: Different Definitions of Properties: | 5,503 | 2,509 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/09%3A_Food_to_energy_metabolic_pathways/9.06%3A_Oxidation_of_fatty_acids |
Although glucose is a quick source of energy that animals need, fatty acids have higher energy per unit mass. Therefore, fats are the main energy storage compounds in animals. Fatty acids (\(\ce{R-COOH}\) in fats are a major source of energy for animals. Hydrolysis of fats, i.e., the 1st stage of catabolism of fats, begins in the digestive tract and completes in the cytosol of the cells. The second stage of fatty acid catabolism begins with the conversion of (\(\ce{R-COOH}\) to the thioester group of coenzyme A, ((\(\ce{R-CO-S-CoA}\), i.e., an acyl-\(\ce{CoA}\), at the cost of energy of \(\ce{ATP}\) to \(\ce{AMP}\) conversion. Some of the important terms used in this section are illustrated in the figure below. Acyl-\(\ce{CoA}\) is transported from the cytosol into mitochondria for the second stage of its catabolism. In the second stage, the Acyl-\(\ce{CoA}\) is fragmented into two \(\ce{C}\) fragments in the form of acetyl-\(\ce{CoA}\). The activated fatty acid goes through a set of four reactions in which the \(\beta\ce{C}\) is oxidized to a carbonyl \(\ce{C=O}\) group and then an acetyl-\(\ce{CoA}\) group is split off leaving behind an acyl-\(\ce{CoA}\) with (\ce{2C}\) less than the initial acyl-\(\ce{CoA}\). This process is called . The process repeats on the fragment acyl-\(\ce{CoA}\) again and again until the last acyl-\(\ce{CoA}\) fragment left is acetyl-\(\ce{CoA}\). Then acetyl-\(\ce{CoA}\) enters the citric acid cycle which is the third stage of the catabolism as in the case of the catabolism of glucose. These reactions are explained next. The activation of a fatty acid begins with an SN reaction of acylate anion as nucleophile with \{\ce{ATP}\) producing acyl-adenylate and pyrophosphate, as shown below. Pyrophosphate is removed by hydrolysis making the reaction irreversible: \(\ce{PP_i + H2O -> 2P_i + 2H^{+}}\). Acyl-adenylate then goes through a nucleophilic acyl substitution reaction with \(\ce{HS-CoA) attacking as a nucleophile an AMP as leaving group. Acyl-CoA is transported from cytosol into mitochondria for the \(\beta\)-oxidation process. The \(\alpha\) and \(\beta\) \(\ce{C's}\) of Acyl-CoA are activated for the reactions, as described next. Reaction 1 is the dehydrogenation of of acyl-\(\ce{CoA}\) by enzyme acyl-\(\ce{CoA}\) dehydrogenase that removes \(\ce{H}\) from \(\alpha\) and \(\beta\) \(\ce{C's}\) generation a trans-\(\ce{C=C}\) bond. An \(\ce{FAD}\) is reduced to \(\ce{FADH2}\) as the acyl-\(\ce{CoA}\) is oxidized in this reaction. Resonance stabilization of the \(\ce{C=C}\) bond by conjugated \(\ce{C=O}\) group makes this reaction easier to happen. The \(\ce{C=C}\) is hydrated by addition of \(\ce{H2O}\) that is a selective reaction that installs an \(\ce{-OH}\) group at \(\beta\)-\(\ce{C}\) creating a new chiral center in L-configuration. This is an electrophilic addition reaction. Again, the \(\ce{C=O}\) makes the \(\beta\)-\(\ce{C}\) more nucleophilic by withdrawing electrons from it by resonance. The secondary \(\ce{-OH}\) group is oxidized to a ketone (\(\ce{C=O}\) group at the expense of reduction of \(\ce{NAD^{+}}\) to \(\ce{NADH}\) by the following reaction. The fourth reaction is a nucleophilic acyl substitution reaction in which an anion of coenzyme A (\(\ce{^{-}S-CoA}\)) acts as a nucleophile, the ketone (\(\ce{C=O}\) as electrophile and \(\ce{^{-}CH2-CO-S-CoA}\) as the leaving group. \(\ce{^{-}CH2-CO-S-CoA}\) is a good leaving group because the negative charge is resonance stabilized and mainly resides on \(\ce{O}\) of carbonyl (\(ce{C=O}\) group. the negative charge is later neutralized by a proton addition that coverts the leaving group to acetyl-\(\ce{CoA}\). The second product of the above reaction is an acyl-\(\ce{CoA}\) with two less \(\ce{C's}\) than the initial acyl-\(\ce{CoA}\). The product acyl-\(\ce{CoA}\) of the first \(\beta\)-oxidation cycle goes through the cycle of four reactions again and again, until the last acyl-\(\ce{CoA}\) product is acetyl-\(\ce{CoA}\). The process is illustrated in the Figure \(\Page {1}\), with the help of an example of \(\beta\)-oxidation of stearic acid which is a typical \(\ce{18C}\) saturated fatty acid. An acetyl-\(\ce{CoA}\) product at the end of each \(\beta\)-oxidation cycle splits off \(\ce{2C's}\) from the initial acyl-\(\ce{CoA}\). So, if the starting fatty acid has \(\ce{nC's}\), there are: \(\frac{n}{2}\) acetyl-\(\ce{CoA}\) produced. Since the last cycle produced two acetyl-\(\ce{CoA}\), the number of cycles for a fatty acid containing \(\ce{nC's}\) is: \(\frac{n}{2}-1\). For example, stearic acid has \(\ce{18C's}\) and it goes through: \(\frac{18}{2}-1 = 8~\beta\text{-oxidation cycles}\). Each \(\beta\)-oxidation cycle of a fatty acid yields one \(\ce{FADH2}\), one \(\ce{NADH}\) and one acetyl-\(\ce{CoA}\). These molecules produce \(\ce{ATP's}\) when they enter the citric acid cycle and oxidative phosphorylation, as described earlier in the catabolism of glucose. One \(\ce{FADH2}\) ≈ 1.5 \(\ce{ATP}\), one \(\ce{NADH}\) ≈ 2.5 \(\ce{ATP}\), and one acetyl-\(\ce{CoA}\) ≈ 10 \(\ce{ATP}\) that makes 14 \(\ce{ATP}\) per \(\beta\)-oxidation cycle. Since there are (\(\frac{n}{2}-1\)) or (\(0.5\times{n}-1\)) cycles for a fatty acid containing \(\ce{nC's}\), there are: (\((0.5\times{n}-1)\times{14}))~\ce{ATP's}\) produced. The last cycle yields two acetyl-\(\ce{CoA}\), so there are addition -\(\ce{10ATP's}\) from the the last acetyl-\(\ce{CoA}\). The activation step converts one \(\ce{ATP}\) to one \(\ce{AMP}\) which is equivalent to conversion of \(\ce{2ATP}\) to \(\ce{2ADP}\). So, after adding 10 from the last acetyl-\(\ce{CoA}\) to the formula ( \((0.5\times{n}-1)\times{14})~\ce{ATP's}\) and subtracting 2 for the \(\ce{2ATP's}\) consumed in the activation step, the formula for the number of \(\ce{ATP's}\) produced per fatty acid containing \(\ce{nC's}\) become the following: \(\ce{ATP's}~\text{produced per fatty acid containing}~\ce{nC's} = ((0.5\times{n}-1)\times{14} + 10 -2)~\ce{ATP's}\). It simplifies to: \(\ce{ATP's}~\text{produced per fatty acid containing}~\ce{nC's} = (7\times{n}-6)~\ce{ATP's}\). For example, stearic acid contains \(\ce{18C's}\) and produces \((7\times18-6) = 120~\ce{ATP's}\). So, one mole of stearic acid (284.48 g/mol) produces 120 mol \(\ce{ATP's}\) which is significantly higher than 32 mol \(\ce{ATP's}\) produced by one mole of glucose (180.156 g/mol). When converted to moles of \(\ce{ATP's}\) produced per unit mass, stearic acid produces 0.42 mol \(\ce{ATP's}\)/g which is more than two times higher than 0.18 mol \(\ce{ATP's}\)/g of glucose. This is because the average oxidation state of glucose \(\ce{C's}\) is higher than those of fatty acid \(\ce{C's}\). Glucose is used as a quick energy source, but, due to higher energy density, fats are used as relatively longer-term energy storage molecules by animals. Unsaturated fatty acids have \(\ce{C=C}\) bond in their alkyl chain. \(\beta\)-Oxidation of unsaturated fatty acids happens the same way as for saturated fatty acids except for the following changes. If \(\ce{C=C}\) bond is a -\(\ce{C=C}\) bond between \(\alpha\) and \(\beta\) \(\ce{C'}\), reaction 1 in the \(\beta\)-oxidation is not needed and \(\ce{FADH2}\) from reaction 1 is not produced. If \(\ce{C=C}\) bond is a cis-\(\ce{C=C}\) bond or it is between \(\beta\) and \(\gamma\) \(\ce{C'}\), It goes through isomerization reactions catalyzed by isomerase enzymes to convert the \(\ce{C=C}\) bond into a -\(\ce{C=C}\) bond between \(\alpha\) and \(\beta\) \(\ce{C'}\) before reaction 3 happens on it. Unsaturated fatty acids yield a little less \(\ce{ATP's}\) due to one \(\ce{FADH2}\) less produced pre \(\ce{C=C}\) bond, but the difference is small and the formula: (7\times{n}-6)~\ce{ATP's}\), still gives a reasonably accurate estimate of \(\ce{ATP}\) yield. Acetyl-\(\ce{CoA}\) produced during the \(\beta\)-oxidation of fatty acids enter the citric acid cycle. When a large number of fatty acids degrade, the citric acid cycle can not take in all of the Acetyl-\(\ce{CoA}\). The excess Acetyl-\(\ce{CoA}\) accumulate in the liver and converts into ketone bodies by a pathway known as ketogenesis, as shown in the figure on the right (Copyright; Sav vas, CC0, via Wikimedia Commons). In this figure, the enzymes are colored red and the substrates/products are colored blue. The three , i.e., acetoacetate, acetone, and \(\beta\)-hydroxybutyrate are marked within an orange box. Ketone bodies are transported from the liver to other tissues where acetoacetate and \(\beta\)-hydroxybutyrate are converted back to acetyl-\(\ce{CoA}\) and enter the citric acid cycle to produce energy. Normally there is constant production and consumption of acetone bodies maintaining ~1 mg/dL concentration in blood. Acetone cannot be converted back into acetyl-\(\ce{CoA}\) except in the liver where it can be converted into lactate and then to pyruvate. When the rate of synthesis of ketone bodies exceeds their rate of consumption, the ketone bodies start accumulating and may start excreting in urine and via breathing -a condition called ketosis. Since two of the three ketone bodies are acids, their accumulation in the blood lowers the blood pH -a condition called ketoacidosis. The smell of acetone, which is fruity or like nail polish remover smaller, is detectable in the breath of persons suffering from ketosis. Ketone bodies can be tested in blood, urine, or breath, e.g. by using test strips with color chart for reading the results shown in figure on the left. In ketosis, the ketone bodies in the blood are between 0.5 to 3.0 millimole/L (mmol/L). In ketoacidosis, the concentration may go above 10 mmol/L. Ketosis can happen during fasting, starvation, prolonged low carbohydrate diets, prolonged intense exercises, alcoholism, or due to uncontrolled diabetes. Insulin is a hormone that regulates blood glucose levels. In diabetes, there is either insufficient insulin or it is not functioning properly. Therefore inappropriately higher levels of glucose in the blood and acetyl-\(\ce{CoA}\) produced from it causes the acetyl-\(\ce{CoA}\) from \(\beta\)-oxidation of fatty acids to covert into ketone bodies resulting in ketosis or ketoacidosis. Low pH of blood plasma triggers kidneys to excrete urine with high acid levels. Glucose and ketone bodies also spill into the urine. High glucose concentration in the blood causes higher osmotic pressure that results in increased urine output and dehydration. The symptoms include frequent urination and excessive thrust. These conditions can be treated with low carbohydrate diets, insulin therapy, or other medications to control diabetes. Glycerol which is a part of triglycerides (fats) accounts for about 5% of their energy. Glycerol is first converted into glycerol-3-phosphate by an SN2 reaction of a primary alcohol group of glycerol acting as a nucleophile, and phosphorus in the phosphate of ATP as nucleophile, and ADP as a leaving group. Than the secondary alcohol of glycerol is oxidized to a ketone group at the expense of reduction of a \(\ce{NAD^{+}}\) coenzyme into \(\ce{NADH}\) as shown below. Dihydroxyacetone phosphate produce of the second reaction is an intermediate in the glycolysis of glucose and enters the same catabolic pathway. | 11,293 | 2,510 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Hydration_of_Alkenes/Formation_of_alcohols_from_alkenes |
Alkenes can be converted to alcohols by the net addition of water across the double bond. There are multiple ways that are commonly used to do this transformation. The net addition of water to alkenes is known as hydration. The result involves breaking the pi bond in the alkene and an OH bond in water and the formation of a C-H bond and a C-OH bond. The reaction is typically by 10 - 15 kcal/mol, but has an change of -35 - -40 cal/mol K. Consequently, the net change for the process tends to close to 0, and the equilibrium constant for the direct addition is close to 1. Nonetheless, there are multiple approaches that allow this transformation to be carried out to completion. The direct addtion of water to an alkene is too slow to be of any significance. However, the addition can be catayzed by Lewis or Bronsted acids. The mechanism of hydration involves electrophlic addition of the proton (or acid) to the double bond to form a carbocation intermediate. Addition of water in the second step results in formation of an oxonium ion, which, upon deprotonation, gives the alcohol. The proton in the oxonium intermediate can be deprotonated by any base present, including the conjugate base of the acid used as a catalyst, or even by another alkene molecule, which would generate another carbocation intermediate and propagate the The extent of reaction of the acid catalyzed hydration of alkenes is determined by the equilibrium constant, which, as noted above, is near unity. Consequently, the reaction can be carried out by adding excess water to increase the yield of products, an application of . Alternatively, the reverse reaction, acid-catalyzed to form the alkene can promoted by removing water from the reaction by using a Dean-Stark trap. Protonation of the alkene in the first step of the reaction can occur at either carbon. However, the more stable carbocation is preferably formed. The order of stability of alkyl cations is 3 > 2 > 1 Therefore, protonation will occur at the , to create the more substituted carbocation, where the water adds. The addition of a proton at the less substituted carbon and the -OH to the more substituted carbon is known as Markovnikov's Rule. The carbocation formed in the reaction is prone to rearrangement, if possible. Therefore, hydration of an olefiin next to a branched aliphatic center will result in the alcohol forming in a position that was not part of the original double bond. Transition metals can also be used as the acids for hydration reactions. In some cases, coordination of the alkene to a metal leaves it susceptible to reaction with a nucleophile such as water. The classic case of nucleophilic donation to a coordinated alkene occurs with mercury (II) salts such as mercuric chloride, \(HgCl_2\), or mercuric acetate, \(Hg(OAc)_2\). The reaction, or rather the sequence of reactions, is called oxymercuration - demercuration or oxymercuration - reduction. We will break the two different reactions in this sequence apart and focus only on the first one: oxymercuration. This reaction qualifies as an electrophilic addition because, as in the previous cases, it begins with donation of a π-bonding pair to an electrophile. In this case, we will consider the electrophile to be aqueous \(Hg^{2+}\) ion. That electrophilic addition (from the alkene's perspective) results in the formation of an alkene complex. In reality, the mercury ion is also coordinated by several water molecules, but we will ignore them for simplicity. The complex formed by addition of mercury is not a localized carbocation, as is formed by protonation, but is better considered a bridged or cyclic structure, which results from addition of d electrons to the empty orbital in the cation. This situation is something like formation of a cyclic bromonium ion formed in the bromination of alkenes. As with bromination, the cyclic intermediate can be opened by attack of a nucleophile, in this case, water. The final part of the reaction sequence is displacement of mercury from the hydroxyalkylmercury complex, effected through addition of sodium borohydride. The details of the reaction are usually dismissed in textbooks because they have little to do with electrophilic addition, the topic we are focusing on. However, the result is that the mercury is replaced by a hydrogen atom. The metal is converted to silvery, liquid, elemental mercury. As shown above, oxymercuration leads to Markovnikov addition of water to the double bond. The reason for this can be seen by considering the electronic structure of the cyclic cation. The electronic structure of the cyclic intermediate can be deduced by using resonance structures as shown below. By considering these resonance structures, it can be seen that the postive charge is distributed over both of the carbons of the olefin. However, because the resonance structure on the left has the positive charge on a secondary carbon, it is energetically more favorable than the structure in the middle, where the charge is a primary cation. Consequently, although the charge is distributed over both carbons, the more substituted carbon has more positive charge density in the overall resonance hybrid. Therefore, it is more electrophilic, and is more susceptibleable to nucleophlic attack. An important advantage of oxymercuration over simple acid catalysis is that the cyclic structure is not prone to rearrangement, and can therefore is amenable to hydration of alkenes with branched substituents. 1. | 5,525 | 2,511 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/08%3A_Conjugated_compounds_and_aromaticity |
: Conjugation is the term we use to describe an arrangement of alternating single and double bonds. To explain how conjugated systems behave differently from non-conjugated systems we will compare 1,3-pentadiene (which is conjugated) and 1,4 pentadiene (which is not conjugated). To recognize the differences between the two, let us look at the orbitals that are involved in the \(\pi\) system bonding. Recall that \(\pi\) bonds can be considered as the side-to-side overlap of p orbitals, so that the electron density lies above and below the plane of the rest of the molecule. Consider the orbitals in 1,3-pentadiene: there is overlap of \(\mathrm{p}\) orbitals that result in a continuous π orbital system over carbons 1-4. The consequence of this is that there is some partial double-bond character between carbons 2 and 3. In 1,4-pentadiene there is no possibility of overlap between the two separate \(\pi\) bonds. Note that in the non-conjugated system, there is an \(\mathrm{sp}^{3}\) hybridized carbon between the two sets of \(\mathrm{sp}^{2}\) hybridized carbon-carbon double bonds which prevents any overlap of (\(\mathrm{p}\)) orbitals on carbons 2 and 4. One way to indicate and predict where this partial double bond character can occur is to use resonance structures. We can write resonance structures for the conjugated system which have double bond character between \(\mathrm{C}-2\) and \(\mathrm{C}-3\). Note that since resonance contributors A and B are equivalent, there is no actual charge separation in this molecule. It is not possible (without breaking a sigma bond) to write resonance structures like this for 1,4-pentadiene (try and convince yourself that this is true ). Another model that can be used to describe the bonding is to consider the π system in terms of molecular orbital theory. Molecular orbital theory considers the bonding orbitals as extending over the whole molecule. The number of molecular orbitals is equal to the sum of all the atomic orbitals. In practice, this approach is far too complex for even the smallest of organic molecules, which is why we usually use the simpler valence bond model in which we consider bonds as being located between two atoms. In the case of conjugated systems, it is often helpful to use the valence bond approach for the sigma (single bonds) framework and then consider the conjugated π system using molecular orbital theory. In this case, if we have a conjugated system of two π bonds, then four p atomic orbitals are involved in forming the four molecular orbitals (\(\mathrm{MO}\)s). Recall that when molecular orbitals are formed from atomic orbitals (\(\mathrm{AO}\)s), if the (quantum mechanical) wave functions add in phase, the resulting energy of the \(\mathrm{MO}\) is lower—that is, the interaction is stabilizing and the \(\mathrm{MO}\)s are bonding \(\mathrm{MO}\)s. If the \(\mathrm{AO}\)s add out of phase, the interaction is destabilizing and the result is antibonding \(\mathrm{MO}\)s. Note that, in the diagram (\(\downarrow\)), only the lowest energy \(\mathrm{MO}\) has electron density between carbons 2 and 3. All the other \(\mathrm{MO}\)s have a node (no electron density) at this position. Overall, we see that there is more \(\pi\) electron density between \(\mathrm{C}-1\) and \(\mathrm{C}-2\), and between \(\mathrm{C}-3\) and \(\mathrm{C}-4\). Since there are only 4 \(\pi\) electrons, only the two bonding \(\mathrm{MO}\)s are occupied while the two antibonding \(\mathrm{MO}\)s are unoccupied. As we will see, if we consider reactions using \(\mathrm{MO}\) theory, the Highest Occupied \(\mathrm{MO}\) (\(\mathrm{HOMO}\)) and the Lowest Unoccupied \(\mathrm{MO}\) (\(\mathrm{LUMO}\)) are the orbitals that participate in new bonding interactions. In general, we use the simplest model that allows us to predict and explain the outcome of reactions, which is usually valence bond theory, but we will call on \(\mathrm{MO}\) theory when necessary—and discussions of conjugated systems sometimes require \(\mathrm{MO}\) theory to explain phenomena. As we have seen before, one way to identify the thermodynamic stability of alkenes is to reduce them to the corresponding alkane by adding \(\mathrm{H}_{2}\) (hydrogens) across the double bond and determine the enthalpy change. In the case of 1,3- and 1,4-pentadiene, we can compare their heats of hydrogenation to produce the corresponding pentane; we find that the conjugated diene is about \(25 \mathrm{kJ/mol}\) more stable. This is a general finding: the more conjugated a system, the more stable it is and the less reactive it is.
| 4,615 | 2,512 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Lattices/Thermodynamics_of_Lattices/The_Born-Lande'_equation |
The Born-Landé equation is a concept originally formulated in 1918 by the scientists Born and Landé and is used to calculate the (measure of the strength of bonds) of a compound. This expression takes into account both the Born interactions as well as the Coulomb attractions. Due to its high simplicity and ease, the Born-Landé equation is commonly used by chemists when solving for lattice energy. This equation proposed by Max Born and Alfred Landé states that lattice energy can be derived from ionic lattice based on electrostatic potential and the potential energy due to repulsion. To solve for the Born-Landé equation, you must have a basic understanding of lattice energy: The Born-Landé equation was derived from these two following equations. the first is the electrostatic potential energy: \[ \Delta U = - \dfrac{N_A M\left | Z^+ \right | \left | Z^- \right |e^2}{4\pi\epsilon_o r} \label{1} \] with The second equation is the repulsive interaction: \[ \Delta U = \dfrac{N_A B}{r^n} \label{2}\] with These equations combine to form: \[ \Delta U (0K) = \dfrac{N_A M\left | Z^+ \right | \left | Z^- \right |e^2}{4\pi\epsilon_or_o} \left ( 1- \dfrac{1}{n} \right) \label{3}\] with Lattice energy, based on the equation from above, is dependent on multiple factors. We see that the charge of ions is proportional to the increase in lattice energy. In addition, as ions come into closer contact, lattice energy also increases. Which compound has the greatest lattice energy? This question requires basic knowledge of lattice energy. Since F gives the compound a +3 positive charge and the Al gives the compound a -1 negative charge, the compound has large electrostatic attraction. The bigger the electrostatic attraction, the greater the lattice energy. What is the lattice energy of NaCl? (Hint: you must look up the values for the constants for this compound) -756 kJ/mol (again, this value is found in a table of constants) Calculate the lattice energy of NaCl. | 1,994 | 2,513 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/01%3A_Introduction_-_Matter_and_Measurement/1.04%3A_Units_of_Measurement |
Have you ever estimated a distance by “stepping it off”— that is, by counting the number of steps required to take you a certain distance? Or perhaps you have used the width of your hand, or the distance from your elbow to a fingertip to compare two dimensions. If so, you have engaged in what is probably the first kind of measurement ever undertaken by primitive mankind. The results of a measurement are always expressed on some kind of a that is defined in terms of a particular kind of . The first scales of distance were likely related to the human body, either directly (the length of a limb) or indirectly (the distance a man could walk in a day). As civilization developed, a wide variety of measuring scales came into existence, many for the same quantity (such as length), but adapted to particular activities or trades. Eventually, it became apparent that in order for trade and commerce to be possible, these scales had to be defined in terms of standards that would allow measures to be verified, and, when expressed in different units (bushels and pecks, for example), to be correlated or converted. Over the centuries, hundreds of measurement units and scales have developed in the many civilizations that achieved some literate means of recording them. Some, such as those used by the Aztecs, fell out of use and were largely forgotten as these civilizations died out. Other units, such as the various systems of measurement that developed in England, achieved prominence through extension of the Empire and widespread trade; many of these were confined to specific trades or industries. The examples shown here are only some of those that have been used to measure length or distance. The history of measuring units provides a fascinating reflection on the history of industrial development. The most influential event in the history of measurement was undoubtedly the French Revolution and the Age of Rationality that followed. This led directly to the metric system that attempted to do away with the confusing multiplicity of measurement scales by reducing them to a few fundamental ones that could be combined in order to express any kind of quantity. The metric system spread rapidly over much of the world, and eventually even to England and the rest of the U.K. when that country established closer economic ties with Europe in the latter part of the 20th Century. The United States is presently the only major country in which “metrication” has made little progress within its own society, probably because of its relative geographical isolation and its vibrant internal economy. Science, being a truly international endeavor, adopted metric measurement very early on; engineering and related technologies have been slower to make this change, but are gradually doing so. Even the within the metric system, however, a variety of units were employed to measure the same fundamental quantity; for example, energy could be expressed within the metric system in units of ergs, electron-volts, joules, and two kinds of calories. This led, in the mid-1960s, to the adoption of a more basic set of units, the ( ) units that are now recognized as the standard for science and, increasingly, for technology of all kinds. In principle, any physical quantity can be expressed in terms of only seven base units (Table \(\Page {1}\)), with each base unit defined by a standard described in the . A few special points about some of these units are worth noting: Owing to the wide range of values that quantities can have, it has long been the practice to employ prefixes such as milli and mega to indicate decimal fractions and multiples of metric units. As part of the SI standard, this system has been extended and formalized (Table \(\Page {2}\)). There is a category of units that are “honorary” members of the SI in the sense that it is acceptable to use them along with the base units defined above. These include such mundane units as the hour, minute, and degree (of angle), etc., but the three shown here are of particular interest to chemistry, and you will need to know them. Most of the physical quantities we actually deal with in science and also in our daily lives, have units of their own: volume, pressure, energy and electrical resistance are only a few of hundreds of possible examples. It is important to understand, however, that all of these can be expressed in terms of the SI base units; they are consequently known as . In fact, most physical quantities can be expressed in terms of one or more of the following five fundamental units: Consider, for example, the unit of , which we denote as V. To measure the volume of a rectangular box, we need to multiply the lengths as measured along the three coordinates: \[V = x · y · z \nonumber \] We say, therefore, that volume has the dimensions of length-cubed: \[dim\{V\} = L^3 \nonumber \] Thus the units of volume will be m (in the SI) or cm , ft (English), etc. Moreover, any formula that calculates a volume must contain within it the L dimension; thus the volume of a sphere is \(4/3 πr^3\). The of a unit are the powers which M, L, t, Q and must be given in order to express the unit. Thus, \[dim\{V\} = M^0L^3T^0Q^0 Θ^0 \nonumber \] as given above. There are several reasons why it is worthwhile to consider the dimensions of a unit. The formal, detailed study of dimensions is known as and is a topic in any basic physics course. Find the dimensions of energy. When mechanical work is performed on a body, its energy increases by the amount of work done, so the two quantities are equivalent and we can concentrate on work. The latter is the product of the force applied to the object and the distance it is displaced. From Newton’s law, force is the product of mass and acceleration, and the latter is the rate of change of velocity, typically expressed in meters per second per second. Combining these quantities and their dimensions yields the result shown in Table \(\Page {1}\). Dimensional analysis is widely employed when it is necessary to convert one kind of unit into another, and chemistry students often use it in "chemical arithmetic" calculations, in which context it is also known as the "Factor-Label" method. In this section, we will look at some of the quantities that are widely encountered in Chemistry, and at the units in which they are commonly expressed. In doing so, we will also consider the actual range of values these quantities can assume, both in nature in general, and also within the subset of nature that chemistry normally addresses. In looking over the various units of measure, it is interesting to note that their unit values are set close to those encountered in everyday human experience These two quantities are widely confused. Although they are often used synonymously in informal speech and writing, they have different dimensions: is the exerted on a mass by the local gravational field: \[f = m a = m g \label{Eq1} \] where is the acceleration of gravity. While the nominal value of the latter quantity is 9.80 m s at the Earth’s surface, its exact value varies locally. Because it is a force, the SI unit of weight is properly the , but it is common practice (except in physics classes!) to use the terms "weight" and "mass" interchangeably, so the units and are acceptable in almost all ordinary laboratory contexts. The range of masses spans 90 orders of magnitude, more than any other unit. The range that chemistry ordinarily deals with has greatly expanded since the days when a microgram was an almost inconceivably small amount of material to handle in the laboratory; this lower limit has now fallen to the atomic level with the development of tools for directly manipulating these particles. The upper level reflects the largest masses that are handled in industrial operations, but in the recently developed fields of geochemistry and enivonmental chemistry, the range can be extended indefinitely. Flows of elements between the various regions of the environment (atmosphere to oceans, for example) are often quoted in teragrams. Chemists tend to work mostly in the moderately-small part of the distance range. Those who live in the lilliputian world of crystal- and molecular structures and atomic radii find the a convenient currency, but one still sees the older non-SI unit called the used in this context; 1Å = 10 m = 100pm. Nanotechnology, the rage of the present era, also resides in this realm. The largest polymeric molecules and colloids define the top end of the particulate range; beyond that, in the normal world of doing things in the lab, the and occasionally the commonly rule. For humans, time moves by the heartbeat; beyond that, it is the motions of our planet that count out the hours, days, and years that eventually define our lifetimes. Beyond the few thousands of years of history behind us, those years-to-the-powers-of-tens that are the fare for such fields as evolutionary biology, geology, and cosmology, cease to convey any real meaning for us. Perhaps this is why so many people are not very inclined to accept their validity. Most of what actually takes place in the chemist’s test tube operates on a far shorter time scale, although there is no limit to how slow a reaction can be; the upper limits of those we can directly study in the lab are in part determined by how long a graduate student can wait around before moving on to gainful employment. Looking at the microscopic world of atoms and molecules themselves, the time scale again shifts us into an unreal world where numbers tend to lose their meaning. You can gain some appreciation of the duration of a nanosecond by noting that this is about how long it takes a beam of light to travel between your two outstretched hands. In a sense, the material foundations of chemistry itself are defined by time: neither a new element nor a molecule can be recognized as such unless it lasts long enough to have its “picture” taken through measurement of its distinguishing properties. Temperature, the measure of thermal intensity, spans the narrowest range of any of the base units of the chemist’s measurement toolbox. The reason for this is tied into temperature’s meaning as a measure of the intensity of thermal kinetic energy. Chemical change occurs when atoms are jostled into new arrangements, and the weakness of these motions brings most chemistry to a halt as absolute zero is approached. At the upper end of the scale, thermal motions become sufficiently vigorous to shake molecules into atoms, and eventually, as in stars, strip off the electrons, leaving an essentially reaction-less gaseous fluid, or plasma, of bare nuclei (ions) and electrons. The degree is really an of temperature, a fixed fraction of the distance between two defined reference points on a . Although rough means of estimating and comparing temperatures have been around since 170, the first mercury thermometer and temperature scale were introduced in Holland in 1714 by Gabriel Daniel Fahrenheit. Fahrenheit established three fixed points on his thermometer. Zero degrees was the temperature of an ice, water, and salt mixture, which was about the coldest temperature that could be reproduced in a laboratory of the time. When he omitted salt from the slurry, he reached his second fixed point when the water-ice combination stabilized at "the thirty-second degree." His third fixed point was "found at the ninety-sixth degree, and the spirit expands to this degree when the thermometer is held in the mouth or under the armpit of a living man in good health." After Fahrenheit died in 1736, his thermometer was recalibrated using 212 degrees, the temperature at which water boils, as the upper fixed point. Normal human body temperature registered 98.6 rather than 96. In 1743, the Swedish astronomer Anders Celsius devised the aptly-named scale that places exactly 100 degrees between the two reference points defined by the freezing and boiling points of water. When we say that the temperature is so many degrees, we must specify the particular scale on which we are expressing that temperature. A temperature scale has two defining characteristics, both of which can be chosen arbitrarily: To express a temperature given on one scale in terms of another, it is necessary to take both of these factors into account. The key to temperature conversions is easy if you bear in mind that between the so-called ice- and steam-points of water there are 180 Fahrenheit degrees, but only 100 Celsius degrees, making the F° 100/180 = 5/9 the magnitude of the C°. Note the distinction between “°C” (a ) and “C°” (a temperature ). Because the ice point is at 32°F, the two scales are offset by this amount. If you remember this, there is no need to memorize a conversion formula; you can work it out whenever you need it. Near the end of the 19th Century when the physical significance of temperature began to be understood, the need was felt for a temperature scale whose zero really means zero— that is, the complete absence of thermal motion. This gave rise to the whose zero point is –273.15 °C, but which retains the same degree magnitude as the Celsius scale. This eventually got renamed after Lord Kelvin (William Thompson); thus the Celsius degree became the . Thus we can now express an increment such as five C° as “five kelvins” In 1859 the Scottish engineer and physicist William J. M. Rankine proposed an absolute temperature scale based on the Fahrenheit degree. Absolute zero (0° Ra) corresponds to –459.67°F. The Rankine scale has been used extensively by those same American and English engineers who delight in expressing heat capacities in units of BTUs per pound per F°. The importance of absolute temperature scales is that absolute temperatures can be entered directly in all the fundamental formulas of physics and chemistry in which temperature is a variable. Units of Temperature: is the measure of the exerted on a unit area of surface. Its SI units are therefore newtons per square meter, but we make such frequent use of pressure that a derived SI unit, the , is commonly used: \[1\; Pa = 1\; N \;m^{–2} \nonumber \] The concept of pressure first developed in connection with studies relating to the atmosphere and vacuum that were carried out in the 17th century. Atmospheric pressure is caused by the weight of the column of air molecules in the atmosphere above an object, such as the tanker car below. At sea level, this pressure is roughly the same as that exerted by a full-grown African elephant standing on a doormat, or a typical bowling ball resting on your thumbnail. These may seem like huge amounts, and they are, but life on earth has evolved under such atmospheric pressure. If you actually perch a bowling ball on your thumbnail, the pressure experienced is twice the usual pressure, and the sensation is unpleasant. The molecules of a gas are in a state of constant thermal motion, moving in straight lines until experiencing a collision that exchanges momentum between pairs of molecules and sends them bouncing off in other directions. This leads to a completely random distribution of the molecular velocities both in speed and direction— or it would in the absence of the Earth’s gravitational field which exerts a tiny downward force on each molecule, giving motions in that direction a very slight advantage. In an ordinary container this effect is too small to be noticeable, but in a very tall column of air the effect adds up: the molecules in each vertical layer experience more downward-directed hits from those above it. The resulting force is quickly randomized, resulting in an increased pressure in that layer which is then propagated downward into the layers below. At sea level, the total mass of the sea of air pressing down on each 1-cm of surface is about 1034 g, or 10340 kg m . The force (weight) that the Earth’s gravitional acceleration g exerts on this mass is \[f = ma = mg = (10340 \;kg)(9.81\; m\; s^{–2}) = 1.013 \times 10^5 \;kg \;m \;s^{–2} = 1.013 \times 10^5\; N \nonumber \] resulting in a pressure of 1.013 × 10 n m = 1.013 × 10 Pa. The actual pressure at sea level varies with atmospheric conditions, so it is customary to define standard atmospheric pressure as 1 atm = 1.01325 x 10 Pa or 101.325 kPa. Although the standard atmosphere is not an SI unit, it is still widely employed. In meteorology, the , exactly 1.000 × 10 = 0.967 atm, is often used. In the early 17th century, the Italian physicist and mathematician Evangalisto Torricelli invented a device to measure atmospheric pressure. The Torricellian consists of a vertical glass tube closed at the top and open at the bottom. It is filled with a liquid, traditionally mercury, and is then inverted, with its open end immersed in the container of the same liquid. The liquid level in the tube will fall under its own weight until the downward force is balanced by the vertical force transmitted hydrostatically to the column by the downward force of the atmosphere acting on the liquid surface in the open container. Torricelli was also the first to recognize that the space above the mercury constituted a vacuum, and is credited with being the first to create a vacuum. One will support a column of mercury that is 760 mm high, so the “millimeter of mercury”, now more commonly known as the , has long been a common pressure unit in the sciences: . International System of Units (SI Units): ) ). | 17,561 | 2,514 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Complex_Ion_Equilibria/Complex-Ion_Equilibria |
When a metal ion reacts with a in solution a is formed. This reaction can be described in terms of . A complex ion is comprised of two important parts: the central atom and its surrounding ligands. The central atom can be any metallic ion (usually a transition metal). The ligands are any combination of anions that can donate an electron pair, effectively meaning they are all Lewis bases. When combined they form . In general, chemical equilibrium is reached when the forward reaction rate is equal to the reverse reaction rate and can be described using an equilibrium constant, \(K\). Complex ion equilibria are no exception to this and have their own unique equilibrium constant. This , \(K_f\), describes the formation of a complex ion from its central ion and attached ligands. This constant may be called a or ; the units depend on the specific reaction it is describing. Common metal/ligand \(K_f\) values are tabulated in . At its most basic level, \(K_f\) can be explained as the following, where \(M\) is a metal ion, \(L\) is a ligand, and \(x\) and \(y\) are coefficients: \[K_f = \dfrac{[M_xL_y]} {[M]^x[L]^y}\] \(K_f\) values are very large in magnitude for complex ion formation reactions heavily favor the product and very small for poorly forming complex ions. For example, observe the formation reaction of the dicyanoargentate(I) complex ion \[\ce{Ag^{+}(aq) + 2CN^{-} (aq) <=> Ag(CN)2^{-}(aq)} \nonumber\] and the resulting formation constant expression: \[ K_f = \dfrac{[Ag(CN)_2^-]}{[Ag^{+},CN^{-}]^{2}} = 5.6 \times 10^{18}\] The larger the \(K_f\) value of a complex ion, the more stable it is. Due to how large formation constants often are it is not uncommon to see them listed as logarithms in the form \(\log K_f\). You may also see them in the form of a , \(K_d\) (which should not be confused with ). This is merely the inverse value of the formation constant and as such describes the instability of a complex ion. It may help to think of \(K_f\) and \(K_d\) as stability and instability constants, respectively. Using the above example, the dissociation constant expression would be: \(K_f = \dfrac{1}{K_d}\) \(K_{d} = \dfrac{1}{K_f} = \dfrac{[Ag^{+},CN^{-}]^{2}}{[Ag(CN)_2]^{-}} = 1.8 \times 10^{-19}\) In logarithm form \(K_f\)and \(K_d\) would be: \(\log K_{f} = \log(5.6 \times 10^{18}) = 18.7\) \(\log K_{d} = \log(1.8 \times 10^{-19}) = -18.7\) One might assume that the formation of complex ions is a one step process, however this is not always the case. As an example let's look at the formation of tetraamminecopper(II) ion in solution: \[\ce{[Cu(H2O)4]^{2+}(aq) + 4NH3 (aq) <=> [Cu(NH3)4]^{2+}(aq) + 4H2O(l)} \nonumber\] The reason for using the hydrated form of the copper ion is that cations in solution usually exist in their hydrated form. It is because of this that the creation of tetraamminecopper(II) ion is a multi-step process, the ammine ligands have to displace the existing aqua ligands. In calculations however, the concentration of water and the number of aqua ligands are effectively constant meaning that they can be safely ignored. Each step of this process has its own formation constant, and as with other chemical equilibria, the overall reaction constant is the product of these stepwise formation constants. \[[Cu(H_2O)_4]^{2+}_{(aq)} + NH_{3 \; (aq)} \rightleftharpoons [Cu(H_2O)_3(NH_3)]^{2+}_{(aq)} + H_2O_{(l)} \tag{step 1}\] \[K_{1} = \dfrac{[[Cu(H_2O)_3(NH_3)]^{2+}]}{[[Cu(H_2O)_4]^{2+},NH_3]} = 1.9 \times 10^4 \nonumber\] \[[Cu(H_2O)_3(NH_3)]^{2+}_{(aq)} + NH_{3 \; (aq)} \rightleftharpoons [Cu(H_2O)_2(NH_3)_2]^{2+}_{(aq)} + H_2O_{(l)} \tag{step 2}\] \[K_{2} = \dfrac{[[Cu(H_2O)_2(NH_3)_2]^{2+}]}{[[Cu(H_2O)_3(NH_3)]^{2+},NH_3]} = 3.9 \times 10^3 \nonumber\] \[[Cu(H_2O)_2(NH_3)_2]^{2+}_{(aq)} + NH_{3 \; (aq)} \rightleftharpoons [Cu(H_2O)(NH_3)_3]^{2+}_{(aq)} + H_2O_{(l)} \tag{step 3}\] \[K_{3} = \dfrac{[[Cu(H_2O)(NH_3)_3]^{2+}]}{[[Cu(H_2O)_2(NH_3)_2]^{2+},NH_3]} = 1.0 \times 10^3 \nonumber\] \[[Cu(H_2O)(NH_3)_3]^{2+}_{(aq)} + NH_{3 \; (aq)} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)} + H_2O_{(l)} \tag{step 4}\] \[K_{4} = \dfrac{[[Cu(NH_3)_4]^{2+}]}{[[Cu(H_2O)(NH_3)_3]^{2+},NH_3]} = 1.5 \times 10^2 \nonumber\] \[\begin{align*} K_f &= K_{1} \times K_{2} \times K_{3} \times K_{4} \\[4pt] &= \dfrac{[[Cu(H_2O)_3(NH_3)]^{2+}]}{[[Cu(H_2O)_4]^{2+},NH_3]} \times \dfrac{[[Cu(H_2O)_2(NH_3)_2]^{2+}]}{[[Cu(H_2O)_3(NH_3)]^{2+},NH_3]} \times \dfrac{[[Cu(H_2O)(NH_3)_3]^{2+}]}{[[Cu(H_2O)_2(NH_3)_2]^{2+},NH_3]} \times \dfrac{[[Cu(NH_3)_4]^{2+}]}{[[Cu(H_2O)(NH_3)_3]^{2+},NH_3]} \\[4pt] &= \dfrac{[[Cu(NH_3)_4]^{2+}]}{[[Cu(H_2O)_4]^{2+},NH_3]^{4}} \\[4pt] &= 1.1 \times 10^{13} \end{align*}\] If you wanted to find the formation constant of one of the intermediate steps, you would simply take the product of the \(K\) values up until that point. For example the formation constant of \([Cu(H_2O)_2(NH_3)_2]^{2+}\) would be \(\beta_2 = K_{1} \times K_{2} = 7.4 \times 10^7\). As seen here and above the standard notation for these values is a lowercase beta. Based on the given values, rank the following complex ions from most to least stable. The key to solving this problem is to convert all of the given values into the same form. In this case we will convert all of them into the form \(K_f\) though any other form could be used as long as you are consistent. After converting it is easy to rate the ions in terms of stability. From most to least stable: \[[Co(NH_3)_6]^{3+} > [Ni(CN)_4]^{2-} > [Cu(CN)_4]^{3-} > [Cr(OH)_4]^{-}. \nonumber\] You may notice that each stepwise formation constant is smaller than the one before it. This decreasing trend is due to the effects of entropy, causing each step to be progressively less likely to occur. You can think of this in the following way, continuing with the previous example: When the first ammine ligand goes to displace an aqua ligand it has four sites from which to choose from, making it "easier" to displace one. Yet with every step the number of sites decreases making it increasingly more difficult. As always though, there are exceptions to this rule. If the values do not continually decrease then the structure of the complex ion likely changed during one of the steps. Generally, complex ions with polydentate ligands have much higher formation constants than those with monodentate ligands. This additional stability is known as the . For example, the formation of the bis(ethylenediamine)copper(II) ion is identical to the tetraamminecopper(II) ion above except the ethylenediamine ligands displace two aqua ligands at a time. Observe the differences in formation constant ( : Of \(\ce{[Ni(en)3]^{2+}}\), \(\ce{[Ni(EDTA)]^{2-}}\), and \(\ce{[Ni(NH3)6]^{2+}}\) which would you expect to have the largest \(K_f\) value? The smallest? Explain. This is a question involving the effects of chelation. When the complex ion with EDTA is formed, the EDTA ligand displaces six aqua ligands at once. Ethylenediamine ligands displace two at a time and ammine ligands displace one at a time. Based on this we would expect \(\ce{[Ni(EDTA)]^{2-}}\) to be the most stable and \(\ce{[Ni(NH3)6]^{2+}}\) to be the least stable. The formation of complex ions can sometimes explain the solubility of certain compounds in solution. For example, let's say we have a saturated solution of silver bromide (\(AgBr\)) which also contains additional undissolved solid silver bromide. If high enough concentrations of ammonia are added to the solution, the solid silver bromide will dissolve despite the solution previously having been saturated. This is because of the formation diammineargentate(I) ion, illustrated by the following chemical equation: \[\ce{AgBr(s) + 2NH3 (aq) \rightarrow [Ag(NH3)2]^{+}(aq) + Br^{-}(l)} \nonumber\] The creation of \(\ce{[Ni(en)_3]^{2+}}\) is a three step process with the following stepwise formation constants \(K_1 = 3.3 \times 10^{7}\), \(K_2 = 1.9 \times 10^{6}\), and \(K_3 = 1.8 \times 10^{4}\). Using this information find the overall formation constant and the formation constant for each of the intermediate steps. Take the product of the stepwise formation constants up until the step you are trying to find. \([Ni(H_2O)_4(en)]^{2+}\) \(\beta_1 = K_1 = 3.3 \times 10^{7}\) \([Ni(H_2O)_2(en)_2]^{2+}\) \(\beta_2 = K_1 \times K_2 = 6.8 \times 10^{13}\) \([Ni(en)_3]^{2+}\) \(\beta_3 = K_f = K_1 \times K_2 \times K_3 = 1.1 \times 10^{18}\) | 8,435 | 2,515 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Concepts_in_Biophysical_Chemistry_(Tokmakoff)/04%3A_Transport/17%3A_Directed_and_Active_Transport/17.03%3A_Brownian_Ratchet |
The refers to a class of models for directed transport using that is rectified through the input of energy. For a diffusing particle, the energy is used to switch between two states that differ in their diffusive transport processes. This behavior results in biased diffusion. It is broadly applied for processive molecular motors stepping between discrete states, and it therefore particularly useful for understanding translational and rotational motor proteins. One common observation we find is that directed motion requires the object to switch between two states that are coupled to its motion, and for which the exchange is driven by input energy. Switching between states results in biased diffusion. The interpretation of real systems within the context of this model can vary. Some people consider this cycle as deterministic, whereas others consider it quite random and noisy, however, in either case, Brownian motion is exploited to an advantage in moving the particle. We will consider an example relevant to the ATP-fueled stepping of cytoskeletal motors along a filament. The motor cycles between two states: (1) a bound state (B), for which the protein binds to a particular site on the filament upon itself binding ATP, and (2) a free state (F) for which the protein freely diffuses along the filament upon ATP hydrolysis and release of ADP + P . The bound state is described by a periodic, spatially asymmetric energy profile \(U_B(x)\), for which the protein localizes to a particular energy minimum along the filament. Key characteristics of this potential are a series of sites separated by a barrier \(ΔU > k_BT\), and an asymmetry in each well that biases the system toward a local minimum in the direction of travel. In the free state, there are no barriers to motion and the protein diffuses freely. When the free protein binds another ATP, it returns to \(U_B(x)\) and relaxes to the nearest energy minimum. Let’s investigate the factors governing the motion of the particle in this Brownian ratchet, using the perspective of a biased random walk. The important parameters for our model are: \[x_f+x_r = \Delta x\] The asymmetry of \(U_B\) is described by \[ \alpha =(x_f-x_r)/\Delta x \] \[ \Delta t = \tau_F+\tau_B \nonumber \] \[ \ell_0(\tau_F)=\sqrt{4D\tau_F} \nonumber \] Let’s consider the conditions to maximize the velocity of the Brownian ratchet. \[ \ell_0=\sqrt{4D\tau_F} \nonumber \] \[ x_r > l_0 > x_F \nonumber \] Using the average value as a target: \[ \begin{aligned} \ell_0 &\approx \dfrac{x_r+x_F}{2}= \dfrac{\Delta x}{2} \\ \tau_F &\approx \dfrac{\Delta x^2}{16D} \end{aligned} \] 2. While in B: After the binding ATP, we would like the particle to stay with ATP bound long enough to relax to the minimum of the asymmetric energy landscape. Competing with this consideration, we do not want it to stay bound any longer than necessary if speed is the issue. We can calculate the time needed to relax from the barrier at x forward to the potential minimum, if we know the drift velocity v of this particle under the influence of the potential. \[\tau_B \approx x_r/ \nu_d \nonumber \] The drift velocity is related to the force on the particle through the friction coefficient, \(\nu_d = f/\zeta \), and we can obtain the magnitude of the force from the slope of the potential: \[ |f| = \dfrac{\Delta U}{x_r} \nonumber \] So the drift velocity is \(\nu_d = \dfrac{fD}{k_BT}= \dfrac{\Delta UD}{x_rk_BT} \) and the optimal bound time is \[\tau_B \approx \dfrac{x_r^2k_BT}{\Delta UD} \nonumber \] Now let’s look at this a bit more carefully. We can now calculate the probability of diffusing forward over the barrier during the free interval by integrating over the fraction of the population that has diffused beyond x during τ . Using the diffusive probability distribution with x0→0, \[ \begin{aligned} P_+ &= \dfrac{1}{\sqrt{4\pi D\tau_F}} \int_{x_f}^{\infty} e^{-x^2/4D\tau_F} dx \\ &=\dfrac{1}{2} erfc\left( \dfrac{x_f}{\ell_0} \right) \end{aligned}\] Similarly, the probability for diffusing backward over the barrier at x = ‒x is \[ P_- = \dfrac{1}{2} erfc \left( \dfrac{x_r}{\ell_0} \right) \nonumber \] Now we can determine the average velocity of the protein by calculating the average displacement in a given time step. The average displacement is the difference in probability for taking a forward versus a reverse step, times the step size. This displacement occurs during the time interval Δt. Therefore, \[ \begin{aligned} \nu &= \dfrac{\Delta P \Delta x}{\Delta t}\\ &=\dfrac{(P_+-P_-)\Delta x}{(\tau_B + \tau_F)} \\ &=\dfrac{\Delta x}{2\Delta t} \left[ erf\left( \dfrac{x_r}{\ell_0 (\tau_F)}\right) - erf \left( \dfrac{x_f}{\ell_0 (\tau_F)} \right) \end{aligned} \] It is clear from this expression that the velocity is zero when the asymmetry of the potential is zero. For asymmetric potentials, P and P are dependent on τ , with one rising in time faster than the other. As a result, the velocity, which depends on the difference of these reaches a maximum in the vicinity of \(\tau_F=x^2_f/D \). So how does the ATP hydrolysis influence the free energy gradient? Here free energy gradient is \[ \dfrac{\Delta G_{Hyd.}}{\Delta x} \nonumber \] \(k_+ = A_+e^{-(\Delta G_{barrier}-\Delta G_{hydrolysis})/kT} \) \(k_- = A_-e^{-(\Delta G_{barrier})/kT} \) \( \nu = (k_+-k_-) \Delta x \) ___________________________________________ | 5,428 | 2,516 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Addition_to_Carbonyls/CO7._What_is_a_Good_Nucleophile%3F |
Lots of things can be nucleophiles. In principle, a nucleophile only needs a lone pair. However, some nucleophiles are better than others. You already know something about nucleophiles if you know something about acidity and basicity. Nucleophiles are really Lewis bases. Some of the factors that account for basicity also account for nucleophilicity. Halides are not very good nucleophiles for carbonyls. The negative charge on a halide is pretty stable, either because of electronegativity or polarizability. If a halide donates to a carbonyl, producing an oxygen anion, the reaction is uphill. Hydroxide and alkoxide anions (such as CH O ) are more reactive than halides. They are better nucleophiles. The sulfur analogues are similarly good nucleophiles (such as CH S ). In addition, water, alcohols and thiols are nucleophilic, because they all have lone pairs that could be donated to an electrophile. Nitrogen also has a lone pair in most compounds. That means amines are good nucleophiles, too. Carbon does not normally have a lone pair, unless it is a carbanion. Carbanions are usually not very stable. As a result, they are not very common, except for cyanide (CN ) and acetylides (RCC , in which R is a hydrogen or an alkyl group). Carbanions such as CH (methyl anion) are very unstable and highly reactive. Explain why the following anions are more stable than a methyl anion. Nucleophilicity is the degree of attraction of a nucleophile to a positive charge (or partial positive charge). It is related to basicity. Choose the most nucleophilic item from each of the following pairs, and explain your answer. Carbonyl compounds such as aldehydes and ketones contain a very slightly acidic hydrogen next to the carbonyl. Some nucleophiles are basic enough to remove that proton instead of donating to the carbonyl. Show why the resulting anion is stable, using cyclopentanone as an example. Accidental deprotonation (proton removal) alpha to a carbonyl (one carbon away from the carbonyl) can occur when a nucleophile is added to a ketone. In the following cases, explain which nucleophile is more likely to add to the carbonyl in cyclohexanone and which is more likely to deprotonate it. , | 2,215 | 2,517 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/25%3A_Amino_Acids_Peptides_and_Proteins/25.12%3A_Enzyme_Technology |
Because enzymes function nearly to perfection in living systems, there is great interest in how they might be harnessed to carry on desired reactions of practical value outside of living systems. The potential value in the use of enzymes (separate from the organisms that synthesize them) is undeniable, but how to realize this potential is another matter. Practical use of separated enzymes is not new. Hydrolytic enzymes isolated from bacteria were widely used for a brief period to assist in removing food stains from clothing, but many people suffer allergic reactions to enzymes used in this way, and the practice was stopped. A major objective in enzyme technology is to develop an enzymatic process for the hydrolysis of cellulose to glucose ( ). Some microorganisms do possess the requisite enzymes to catalyze the hydrolysis of the \(\beta\)-1,4 glucoside links in cellulose. If these enzymes could be harnessed for industrial production of glucose from cellulose, this could be an important supplementary food source. Technology already is available to convert glucose into ethanol and ethanoic acid, and from there to many chemicals now derived from petroleum. A difficult problem in utilizing enzymes as catalysts for reactions in a noncellular environment is their instability. Most enzymes readily denature and become inactive on heating, exposure to air, or in organic solvents. An expensive catalyst that can be used only for one batch is not likely to be economical in an industrial process. Ideally, a catalyst, be it an enzyme or other, should be easily separable from the reaction mixtures and indefinitely reusable. A promising approach to the separation problem is to use the technique of . This means that the enzyme is modified by making it insoluble in the reaction medium. If the enzyme is insoluble and still able to manifest its catalytic activity, it can be separated from the reaction medium with minimum loss and reused. Immobilization can be achieved by linking the enzyme covalently to a polymer matrix in the same general manner as is used in solid-phase peptide synthesis ( ). Enzymes also have possible applications in organic synthesis. But there is another problem in addition to difficulties with enzyme stability. Enzymes that achieve carbon-carbon bond formation, the , normally require cofactors such as ATP. How to supply ATP in a commercial process and regenerate it continuously from ADP or AMP is a technical problem that has to be solved if the synthetases are to be economically useful. This is a challenging field of biological engineering. and (1977) | 2,621 | 2,518 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Basics_of_Electrochemistry/Electrochemistry/Half-Cell_Reaction |
A half cell is one of the two electrodes in a galvanic cell or simple battery. For example, in the \(\ce{Zn-Cu}\) battery, the two half cells make an oxidizing-reducing couple. Placing a piece of reactant in an electrolyte solution makes a half cell. Unless it is connected to another half cell via an electric conductor and salt bridge, no reaction will take place in a half cell. On the , reduction takes place. On the , oxidation takes place. A battery requires at least two electrodes, the anode at which oxidation occurs, and the cathode at which reduction occurs. Reduction and oxidation are always required in any battery setup. A battery operation requires (if the salt bridge is not there already). These are the key elements of a battery. Write the anode and cathode reactions for a galvanic cell that utilizes the reaction \(\ce{Ni_{\large{(s)}} + 2 Fe^3+ \rightarrow Ni^2+ + 2 Fe^2+}\) Oxidation takes place at the anode, and the electrode must be \(\ce{Ni\, |\, Ni^2+}\), \(\ce{Ni_{\large{(s)}} \rightarrow Ni^2+_{\large{(aq)}} + 2 e^-}\) and the reduction occurs at the cathode: \(\ce{Fe^3+,\: Fe^2+}\): \(\ce{2 Fe^3+ + 2 e^- \rightarrow 2 Fe^2+}\) For every \(\ce{Ni}\) atom oxidized, two \(\ce{Fe^3+}\) ions are reduced. The electrons from the \(\ce{Ni}\) metal will flow from the anode, pass the load, and then carry out the reduction at the surface of the cathode to reduce the ferric (\(\ce{Fe^3+}\)) ions to ferrous ions. In the meantime the ions in the solution move accordingly to keep the charges balanced. The galvanic cell is: \(\ce{Ni_{\large{(s)}}\, |\, Ni^2+_{\large{(aq)}}\, ||\, Fe^3+_{\large{(aq)}},\: Fe^2+_{\large{(aq)}}\, |\, Pt_{\large{(s)}}}\) where "\(\ce{Fe^3+_{\large{(aq)}},\: Fe^2+_{\large{(aq)}}}\)" represents a solution containing two types of ions. An inert \(\ce{Pt}\) electrode is placed in the solution to provide electrons for the reduction. The charge on an electron is 1.602x10 C (coulomb). What is the charge on 1 mole of electrons? The charge on one mole (Avogadro's number of) electrons is called a Faraday ( ). \(\begin{align}
F &= (6.022045\times10^{23} / \ce{mol}) \times (1.602\times10^{-19}\: \ce C)\\
&= \textrm{96485 C/mol}
\end{align}\) The chemical history involving the determination of Avogadro's number, and the charge on an electron, and how the two values agree with each other is very interesting. Who determined the charge on a single electron? Robert Millikan was awarded with the Nobel Prize for his determination of electron charge at University of Chicago. If 96485 C of charge is required to deposit 107.9 g of silver, what is the charge of an electron? A galvanic cell with a voltage of 1.1 V utilizes the reaction \(\ce{Zn + Cu^2+ \rightarrow Cu + Zn^2+}\) as a source of energy. If 6.3 g of \(\ce{Cu}\) and 11 g \(\ce{Zn}\) are used, what is the maximum usable energy in this battery? The 6.3 g \(\ce{Cu}\) and 11 g \(\ce{Zn}\) correspond to 0.10 and 0.17 mol of \(\ce{Cu}\) and \(\ce{Zn}\) respectively. Thus, \(\ce{Cu}\) is the limiting reagent, and 0.10 mol corresponds to a charge of 2×96485×0.10 C (2 significant figures). The maximum available energy is then \(\begin{align}
\textrm{Max. Energy} &= \mathrm{(1.1\: V)(96485\: C)(2)(0.10)}\\
&= \mathrm{22000\: J \hspace{15px} (1\: J = 1\: VC)}
\end{align}\) This energy corresponds to 2500 cal, which is enough to bring 25 g water from 273 K to its boiling point (373 K). Another way of looking at it: 22000 J is enough energy to send a 20-gram rocket to a height of 56 m. If the galvanic cell of Example 3 is used to power a calculator, which consumes 1 mW, how long theoretically will the battery last in continuous operation? Power consumption of 1 mW is equivalent to 0.001 J/sec. \(\begin{align}
\mathrm{\dfrac{22000\: J}{0.001\: J/sec}} &= \textrm{2.2E7 sec}\\
&= \textrm{6200 hrs}\\
&= \textrm{254 days}
\end{align}\) This is a realistic example. Most recent calculators use very little power. I noted that a SHARP programmable calculator uses 15 mW, a Casio calculator uses 0.5 mW, and an HP 25 uses 500 mW. A half cell consists of an electrode and the species to be oxidized or reduced. If the material conducts electricity, it may be used as an electrode. The hydrogen electrode consists of a \(\ce{Pt}\) electrode, \(\ce{H2}\) gas and \(\ce{H+}\). This half cell is represented by: \(\ce{Pt_{\large{(s)}}\, |\, H_{2\large{(g)}}\, |\, H+_{\large{(aq)}}}\) where the vertical bars represent the phase boundaries. Conventionally, the cell potential for the hydrogen electrode is defined to be exactly zero if it has the condition as given below: \(\ce{Pt\, |\, H_{2\: \large{(g,\: 1\: atm)}}\, |\, H+_{\large{(aq)}},\: 1\: M}\) The notations for half cells are not rigid, but a simplified way to represent a rather complicated setup. The tendency for a reduction reaction is measured by its . The reduction potential is a quantity measured by comparison. As mentioned earlier, the reduction potential of the standard hydrogen electrode (SHE) is arbitrarily defined to be zero as a reference point for comparison. When a half cell \(\ce{Cu^2+ || Cu}\) for the reaction \(\ce{Cu^2+ + 2 e^- \rightarrow Cu}\) is coupled with the Standard Hydrogen Electrode (SHE), the copper electrode is a cathode, where reduction takes place. The potential across the cell \(\ce{Pt\, |\, H_{2\: \large{(g,\: 1\: atm)}}\, |\, H+_{\large{(aq)}},\: 1\: M\, ||\, Cu^2+ \, |\, Cu}\) has been measured to be 0.339 V. This indicates that \(\ce{Cu^2+}\) ions are easier to reduce than the hydrogen ions, and we usually represent it by \(\ce{Cu^2+ + 2 e^- \rightarrow Cu} \hspace{15px} E^{\ce o} = 0.339\: \ce V\) A positive cell potential indicates a spontaneous reaction. When the cell \(\ce{Zn\, |\, Zn^2+}\) is coupled with the SHE, \(\ce{Zn\, |\, Zn^2+_{\large{(aq)}}\: 1\: M\, ||\, H+_{\large{(aq)}},\: 1\: M\, |\, H_{2\: \large{(g,\: 1\: atm)}}\, |\, Pt}\) The potential has been measured to be 0.76 V. However, in this cell, \(\ce{Zn}\) is oxidized, and its electrode is the anode. Therefore, the reduction potential has a negative value for the reduction reaction \(\ce{Zn^2+ + 2 e^- \rightarrow Zn} \hspace{15px} E^{\ce o} = - 0.76\: \ce V\) This means that \(\ce{Zn^2+}\) ions are less ready to accept electrons than hydrogen ions. Ideally, for every redox couple, there is a reduction potential. Reduction potentials of standard cells have been measured against the SHE or other standards; their potentials are measured. These values are usually tabulated in handbooks. A short Standard Reduction Potentials table is available from the HandbookMenu, but you may also click the live link to see one. \(\mathrm{2 H^+_{\large{(aq,\: 1.00\: F)}} + 2 e^- \rightarrow H_{2\:\large{(g,\: 1\:atm)}}}\) A.
No cell potential is ABSOLUTELY zero. \(\ce{H2}\) is reactive. This is not the reason at all. A.
The vertical bar | is used to indicate a boundary between two phases.
\(\ce{Pt\, |\, H2\, |\, H+}\: (1.0\: \ce M)\) represents the hydrogen half cell. C.
Only | and || are used among the four notations. Two vertical bars, ||, represent a salt bridge. | 7,145 | 2,519 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Reactivity_of_Arenes/Electrophilic_Substitution_of_Disubstituted_Benzene_Rings |
When a benzene ring has two substituent groups, each exerts an influence on subsequent substitution reactions. The activation or deactivation of the ring can be predicted more or less by the sum of the of these substituents. The site at which a new substituent is introduced depends on the orientation of the existing groups and their individual directing effects. We can identify two general behavior categories, as shown in the following table. Thus, the groups may be oriented in such a manner that their directing influences act in concert, reinforcing the outcome; or are opposed (antagonistic) to each other. Note that the orientations in each category change depending on whether the groups have similar or opposite individual directing effects. The products from substitution reactions of compounds having a reinforcing orientation of substituents are easier to predict than those having antagonistic substituents. For example, the six equations shown below are all examples of reinforcing or cooperative directing effects operating in the expected manner. Symmetry, as in the first two cases, makes it easy to predict the site at which substitution is likely to occur. Note that if two different sites are favored, substitution will usually occur at the one that is least hindered by ortho groups. The first three examples have two similar directing groups in a meta-relationship to each other. In examples 4 through 6, oppositely directing groups have an ortho or para-relationship. The major products of electrophilic substitution, as shown, are the sum of the individual group effects. The strongly activating hydroxyl (–OH) and amino (–NH ) substituents favor dihalogenation in examples 5 and six. Substitution reactions of compounds having an antagonistic orientation of substituents require a more careful analysis. If the substituents are identical, as in example 1 below, the symmetry of the molecule will again simplify the decision. When one substituent has a pair of non-bonding electrons available for adjacent charge stabilization, it will normally exert the product determining influence, examples 2, 4 & 5, even though it may be overall deactivating (case 2). Case 3 reflects a combination of steric hindrance and the superior innate stabilizing ability of methyl groups relative to other alkyl substituents. Example 6 is interesting in that it demonstrates the conversion of an activating ortho/para-directing group into a deactivating meta-directing "onium" cation [–NH(CH ) ] in a strong acid environment. ), | 2,553 | 2,520 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Surface_Science_(Nix)/04%3A_UHV_and_Effects_of_Gas_Pressure/4.01%3A_What_is_UltraHigh_Vacuum%3F |
Vacuum technology has advanced considerably over the last 25 years and very low pressures are now routinely obtainable. Firstly, we need to remind ourselves of the units of pressure: While the mbar is often used as a unit of pressure for describing the level of vacuum, the most commonly employed unit is still the Torr (the , the Pa, is almost never used). Classification of the degree of vacuum is hardly an exact science - it much depends upon who you are talking to - but as a rough guideline: Virtually all surface studies are carried out under UHV conditions - the question is ? This is the question that we will address in . | 647 | 2,521 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/08%3A_Phase_Equilibrium/8.07%3A_Liquid-Vapor_Systems_-_Raoults_Law |
Liquids tend to be volatile, and as such will enter the vapor phase when the temperature is increased to a high enough value (provided they do not decompose first!) A volatile liquid is one that has an appreciable vapor pressure at the specified temperature. An ideal mixture continuing at least one volatile liquid can be described using Raoult’s Law. Raoult’s law can be used to predict the total vapor pressure above a mixture of two volatile liquids. As it turns out, the composition of the vapor will be different than that of the two liquids, with the more volatile compound having a larger mole fraction in the vapor phase than in the liquid phase. This is summarized in the following theoretical diagram for an ideal mixture of two compounds, one having a pure vapor pressure of \(p_B^o = 450\, Torr\) and the other having a pure vapor pressure of \(p_B^o = 350\, Torr\). In Figure \(\Page {1}\), the liquid phase is represented at the top of the graph where the pressure is higher. Oftentimes, it is desirable to depict the phase diagram at a single pressure so that temperature and composition are the variables included in the graphical representation. In such a diagram, the vapor, which exists at higher temperatures) is indicated at the top of the diagram, while the liquid is at the bottom. A typical temperature vs. composition diagram is depicted in Figure \(\Page {2}\) for an ideal mixture of two volatile liquids. In this diagram, \(T_A^o\) and \(T_B^o\) represent the boiling points of pure compounds \(A\) and \(B\). If a system having the composition indicated by \(\chi_B^c\) has its temperature increased to that indicated by point c, The system will consist of two phases, a liquid phase, with a composition indicated by \(\chi_B^d\) and a vapor phase indicated with a composition indicated by \(\chi_B^b\). The relative amounts of material in each phase can be described by the lever rule, as described previously. Further, if the vapor with composition \(\chi_B^b\) is condensed (the temperature is lowered to that indicated by point b') and revaporized, the new vapor will have the composition consistent with \(\chi_B^{a}\). This demonstrates how the more volatile liquid (the one with the lower boiling temperature, which is A in the case of the above diagram) can be purified from the mixture by collecting and re-evaporating fractions of the vapor. If the liquid was the desired product, one would collect fractions of the residual liquid to achieve the desired result. This process is known as distillation. | 2,553 | 2,522 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/25%3A_Amino_Acids_Peptides_and_Proteins/25.11%3A_Enzyme_Regulation |
You may have wondered how the proteolytic enzymes such as trypsin, pepsin, chymotrypsin, carboxypeptidase, and others keep from self-destructing by catalyzing their own hydrolysis or by hydrolyzing each other. An interesting feature of the digestive enzymes is that they are produced in an inactive form in the stomach or the pancreas - presumably to protect the different kinds of proteolytic enzymes from attacking each other or other proteins. The inactive precursors are called trypinogen, pepsinogen, chymotrypsinogen, and procarboxypeptidase. These precursors are converted to the active enzymes by hydrolytic cleavage of a few specific peptide bonds under the influence of other enzymes (trypsin, for example, converts chymotrypsinogen to chymotrypsin). The digestive enzymes do not appear to self-destruct, probably because they are so constructed that it is sterically impossible to fit a part of one enzyme molecule into the active site of another. In this connection, it is significant that chymotrypsin attacks denatured proteins more rapidly than natural proteins with their compact structures of precisely folded chains. Presumably all enzymes must have some regulatory mechanism that turns them on and off as needed. Less is known about regulation mechanisms than about the enzymatic reactions themselves, but one type of control has been recognized. This occurs when a reaction product inhibits one of the reaction steps producing it by tying up the enzyme as a nonreactive complex ( ). As the simplest example, suppose that the product \(\left( \ce{P} \right)\) as well as the substrate \(\left( \ce{S} \right)\) complexes with the enzyme \(\left( \ce{E} \right)\); then we can write the following set of equilibria for the net reaction: \[\ce{E} + \ce{S} \rightleftharpoons \ce{ES} \rightarrow \ce{E} + \ce{P}\] \[\ce{E} + \ce{P} \rightleftharpoons \ce{EP}\] Clearly, a reaction of this type will decrease in rate as the product accumulates. It may stop altogether if the active sites are saturated with the product, and it will start again only on removal of the product. and (1977) | 2,120 | 2,523 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/03%3A_Organic_Nomenclature/3.01%3A_Prelude_to_Organic_Nomenclature |
Organic chemists, regardless of what languages they speak, can communicate with one another about their chemical work simply by writing equations and structural formulas. But this is a slow process if the molecules are complicated, and is not well suited for conversation (try describing a structural formula of a complex molecule to someone). For more rapid and efficient communication we need to have names for compounds, and we should have every reason to hope that, after 100 years, the names now in use would be clear, unambiguous, easy to pronounce, easy to spell and to remember, as well as being amenable to arrangement in alphabetical order. But, even more, we should hope that the names of organic compounds would contain enough information so we could generate the proper structures from them, and conversely, if we know the structures then the system would have simple enough rules that we could construct universally recognized and accepted names. Unfortunately, these splendid ideals have not yet been realized. A good part of the problem is that people are resistant to change and especially resistant to changes in . To give an example, the carboxylic acid, \(\ce{CH_3-CO_2H}\), commonly is known as "acetic acid". The name arises from the Latin word , for sour wine or vinegar, and acetic acid is the principal constituent, besides, water, of vinegar. A similarly common compound is called "acetone" and, in the ideal world, acetone should be structurally related to acetic acid. But acetone is \(\ce{CH_3-CO-CH_3}\), and the name arises only because acetone is formed by strong heating of the calcium salt of acetic acid, \(\ce{(CH_3-CO_2)_2Ca} \rightarrow \ce{CH_3-CO-CH_3} + \ce{CaCO_3}\), a reaction that is of no current importance whatsoever. Better nomenclature systems use names based on the . On this basis, \(\ce{CH_3CO_2H}\) is related to and is called , whereas \(\ce{CH_3-CO-CH_3}\) with three carbons is related to and called 2- . As far as possible, we shall use these names as our first choices, because organic chemistry is growing too fast to sustain the present chaos of nonsystematic nomenclatures in current use. You might well ask why nonsystematic names persist for so long. The reasons are complex and variable. Alchemists intentionally used abstruse names and symbolism to disguise what they really were working with. Chemical industry, especially in the drug area, has practiced much the same thing in using unintelligible trade names and codes for proprietary products. Obviously, everyone who handles or sells chemicals is not a chemist, and to the nonchemist, a short nonsystematic name will make more sense than a longer systematic name. A salesman who markets tons of "acrolein", \(\ce{CH_2=CH-CH=O}\), has little reason to adopt the systematic name, 2-propenal. People probably persist in using nicknames for chemicals for much the same reason that they use nicknames for people. Nicknames are less formal, usually shorter, and imply familiarity with the subject. Another cogent reason to resist dramatic changes in chemical nomenclature is that it would make the current and earlier literature archaic or even unintelligible. Universal adoption tomorrow of a nomenclature system different from the one we use here would render this book instantly obsolete. As a result, changes usually are made in small steps and may not be really effective until a generation or more passes. (Consider in this context the efforts to convert monetary systems and weights and measures to the decimal system.) Ideally, every organic substance should have a completely descriptive, systematic name to permit only one structural formula to be written for it. This ideal has been approached closely in some of the current nomenclature systems but, unfortunately, truly systematic nomenclature for very complicated compounds is often hopeless for conversational or routine scriptorial purposes. As a result, we will at times resort to using (common) trivial names, especially if it is impractical to do otherwise. Clearly, the description has phonetic disadvantages as a handy name for vitamin A: A very important consideration for becoming more familiar with the systematic names is their increasing use in indexing systems. When organic chemists dealt with relatively few compounds, it was possible to accommodate a wide variety of special nomenclature customs. However, the rapid growth of knowledge in the past twenty years, which probably has doubled the number of organic compounds, has also enormously increased the burden on those who dedicate themselves to making this knowledge easily available to others by indexing the current literature. A natural reaction is to discard common names in favor of more systematic ones and to develop numerical designations suitable for computer processing. The difference in sizes of the \(^1\) indexes for the years 1907-1916 and for the current year should be convincing as to the need for having systematic names become more widely used and important. But the fact remains that the naming systems used in indexing are not always the same as those used in practice, and we are left with the necessity of having to know both. Learning the nomenclature of organic compounds has many of the elements of learning a language, be it Latin or Fortran. Fortunately, like a language it does not have to be learned all at once. One can become familiar with naming of simple hydrocarbons, then study their chemistry (avoiding that part which involves compounds with as yet unlearned names), proceed to the naming of alkenes, study their chemistry, and so on. This is a very simple and natural way but can be inconvenient in a textbook if one wants to review the nomenclature of more than one class of compounds at a time. In this chapter, we consolidate the nomenclature of a number of classes of compounds - an undertaking that may not seem very logical to someone who will soon be troubled enough with the chemistry of these compounds let alone their names. We recommend, however, a thorough study now of alkane and haloalkane nomenclature (Section 3-1) followed by a more cursory examination of the rest of the chapter. Then, as unfamiliar names arise, you can quickly review the basic rules for alkanes and proceed to the new class you have encountered. The idea is to have many of the important rules in one place. Nomenclature rules for other types of compounds are given in Chapter 7. and (1977) | 6,496 | 2,525 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/06%3A_Putting_the_Second_Law_to_Work/6.01%3A_Free_Energy_Functions |
In the previous chapter, we saw that for a spontaneous process, \(ΔS_{universe} > 0\). While this is a useful criterion for determining whether or not a process is spontaneous, it is rather cumbersome, as it requires one to calculate not only the entropy change for the system, but also that of the surroundings. It would be much more convenient if there was a single criterion that would do the job and focus only on the system. As it turns out, there is. Since we know that \[ \Delta S_{univ} \ge 0 \nonumber \] for any natural process, and \[ \Delta S_{univ} = \Delta S_{sys} + \Delta S_{surr} \nonumber \] all we need to do is to find an expression for \(\Delta S_{sys}\) that can be determined by the changes in the system itself. Fortunately, we have already done that! Recalling that at constant temperature \[\Delta S = -\dfrac{q_{rev}}{T} \nonumber \] and at constant pressure \[\Delta H = q_p \nonumber \] it follows that at constant temperature and pressure \[\Delta S_{surr} = -\dfrac{\Delta H_{sys}}{T} \nonumber \] Substitution into the above equations yields an expression for the criterion of spontaneity that depends only on variables describing the changes in the system! \[ \Delta S_{univ} \ge \Delta S_{sys} -\dfrac{\Delta H_{sys}}{T} \nonumber \] so \[\Delta S_{sys}-\dfrac{\Delta H_{sys}}{T} \ge 0 \nonumber \] Multiplying both sides by \(-T\) yields \[\Delta H - T\Delta S \le 0 \label{chem} \] A similar derivation for constant volume processes results in the expression (at constant volume and temperature) \[\Delta U - T\Delta S \le 0 \label{geo} \] Equation \ref{chem} is of great use to chemists, as most of chemistry occurs at constant pressure. For geologists, however, who are interested in processes that occur at very high pressures (say, under the weight of an entire mountain) and expansion is not a possibility, the constant volume expression of Equation \ref{chem} may be of greater interest. All of the above arguments can be made for systems in which the temperature is not constant by considering infinitesimal changes. The resulting expressions are \[dH -TdS \le 0 \label{chem1} \] and \[dU -TdS \le 0 \label{chem2} \] Equation \ref{chem1} suggests a very convenient thermodynamic function to help keep track of both the effects of entropy and enthalpy changes. This function, the (or ) is defined by \[G \equiv H -TS \nonumber \] A change in the Gibbs function can be expressed \[\Delta G = \Delta H -\Delta (TS) \nonumber \] Or at constant temperature \[\Delta G = \Delta H -T \Delta S \nonumber \] And the criterion for a process to be spontaneous is the G < 0. As such, it should be clear spontaneity is not merely a function the enthalpy change (although exothermic processes tend to be spontaneous) but also a function of the entropy change, weighted by the temperature. Going back to an earlier example, \[ NaOH(s) \rightarrow Na^+ (aq) + OH^- (aq) \nonumber \] with \(\Delta H < 0\). and \[ NaHCO_3(s) \rightarrow Na^+ (aq) + HCO_3^- (aq) \nonumber \] with \(\Delta H > 0\). It is easy to see why both processes are spontaneous. In the first case, the process is exothermic (favorable) and proceeds with an increase in entropy (also favorable) due to the formation of fragments in the liquid phase (more chaotic) from a very ordered solid (more ordered). The second reaction is endothermic (unfavorable) but proceeds with an increase in entropy (favorable). So, so long as the temperature is high enough, the entropy term will overwhelm the enthalpy term and cause the process to be spontaneous. The conditions for spontaneous processes at constant temperature and pressure can be summarized in Table 6.1.1. Similarly to the Gibbs function, the is defined by \[A \equiv U -TS \nonumber \] and provides another important criterion for spontaneous processes at constant value and temperature. , the Helmholtz function can be expressed by \[\Delta A \equiv \Delta U -T\Delta S \nonumber \] Based on similar arguments used for the Gibbs function, the Helmholtz function also can be used to predict which processes will be spontaneous at constant volume and temperature according to Table 6.1.2. Much like in the case of enthalpy (and unlike entropy), free energy functions do not have an unambiguous zero to the energy scale. So, just like in the case of enthalpies of formation, by convention, the standard free energy of formation (\(\Delta G_f^o\)) for elements in their standard states is defined as zero. This allows for two important things to happen. First, \(\Delta G_f^o\) can be measured and tabulated for any substance (in principle, at least.) \(\Delta G_f^o\) is determined to be \(\Delta G_{rxn}^o\) for the reaction that forms one mole of a compound from elements in their standard states (similarly to how \(\Delta H_f^o\) is defined.) Secondly, tabulated (\(\Delta G_f^o\)) can be used to calculate standard reaction free energies (\(\Delta G_{rxn}^o\)) in much the same way as \(\Delta H_f^o\) is used for reaction enthalpies. Given the following data at 298 K, calculate \(\Delta G^o\) at 298 K for the following reaction: \[C_2H_4(g) + H_2(g) \rightarrow C_2H_6(g) \nonumber \] The \(\Delta G_f^o\) values can be used to calculate \(\Delta G^o\) for the reaction in exactly the same method as \(\Delta H_f^o\) can be used to calculate a reaction enthalpy. \[\Delta G^o = (1 \,mol)(-32.0\, kJ/mol) - (1\, mol)(68.4\, kJ/mol) \nonumber \] \[\Delta G^o= 100.4\, kJ \nonumber \] : \(H_2(g)\) is not included in the calculation since \(\Delta G_f^o\) for \(H_2(g)\) is 0 since it is an element in its standard state. | 5,602 | 2,526 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/08%3A_Phase_Equilibrium/8.06%3A_Phase_Diagrams_for_Binary_Mixtures |
As suggested by the Gibbs Phase Rule, the most important variables describing a mixture are pressure, temperature and composition. In the case of single component systems, composition is not important so only pressure and temperature are typically depicted on a phase diagram. However, for mixtures with two components, the composition is of vital important, so there is generally a choice that must be made as to whether the other variable to be depicted is temperature or pressure. Temperature-composition diagrams are very useful in the description of binary systems, many of which will for two-phase compositions at a variety of temperatures and compositions. In this section, we will consider several types of cases where the composition of binary mixtures are conveniently depicted using these kind of phase diagrams. A pair of liquids is considered partially miscible if there is a set of compositions over which the liquids will form a two-phase liquid system. This is a common situation and is the general case for a pair of liquids where one is polar and the other non-polar (such as water and vegetable oil.) Another case that is commonly used in the organic chemistry laboratory is the combination of diethyl ether and water. In this case, the differential solubility in the immiscible solvents allows the two-phase liquid system to be used to separate solutes using a separatory funnel method. As is the case for most solutes, their solubility is dependent on temperature. For many binary mixtures of immiscible liquids, miscibility increases with increasing temperature. And then at some temperature (known as the upper critical temperature), the liquids become miscible in all compositions. An example of a phase diagram that demonstrates this behavior is shown in Figure \(\Page {1}\). An example of a binary combination that shows this kind of behavior is that of methyl acetate and carbon disufide, for which the critical temperature is approximately 230 K at one atmosphere (Ferloni & Spinolo, 1974). Similar behavior is seen for hexane/nitrobenzene mixtures, for which the critical temperature is 293 K. Another condition that can occur is for the two immiscible liquids to become completely miscible below a certain temperature, or to have a lower critical temperature. An example of a pair of compounds that show this behavior is water and trimethylamine. A typical phase diagram for such a mixture is shown in Figure \(\Page {2}\). Some combinations of substances show both an upper and lower critical temperature, forming two-phase liquid systems at temperatures between these two temperatures. An example of a combination of substances that demonstrate the behavior is nicotine and water. The composition and amount of material in each phase of a two phase liquid can be determined using the . This rule can be explained using the following diagram. Suppose that the temperature and composition of the mixture is given by point b in the above diagram. The horizontal line segment that passes through point b, is terminated at points a and c, which indicate the compositions of the two liquid phases. Point a indicates the mole faction of compound B ( ) in the layer that is predominantly A, whereas the point c indicates the composition ( )of the layer that is predominantly compound B. The relative amounts of material in the two layers is then inversely proportional to the length of the tie-lines a-b and b-c, which are given by \(l_A\) and \(l_B\) respectively. In terms of mole fractions, \[ l_A = \chi_B - \chi_B^A \nonumber \] and \[ l_A = \chi_B^B - \chi_B \nonumber \] The number of moles of material in the A layer (\(n_A\)) and the number of moles in the B layer (\(n_B\)) are inversely proportional to the lengths of the two lines \(l_A\) and \(l_B\). \[ n_A l_A = n_B l_B \nonumber \] Or, substituting the above definitions of the lengths \(l_A\) and \(l_B\), the ratio of these two lengths gives the ratio of moles in the two phases. \[ \dfrac{n_A}{n_B} = \dfrac{l_B}{l_A} = \dfrac{ \chi_B^B - \chi_B}{\chi_B - \chi_B^A} \nonumber \] | 4,086 | 2,527 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/08%3A_Solutions/8.10%3A_Ions_and_Electrolytes/8.10.9E%3A_8.10.9E%3A_Some_applications_of_electrolytic_conduction |
From the chemist's standpoint, the most important examples of conduction are in connection with electrochemical cells, electrolysis and batteries. Owing to their high sensitivity, conductivity measurements are well adapted to the measurement of equilibrium constants for processes that involve very small ion concentrations. These include As long as the ion concentrations are so low, their values can be taken as activities, and limiting equivalent conductivities Λ can be used directly. The very small conductivity of pure water makes it rather difficult to obtain a precise value for ; better values are obtained by measuring the potential of an appropriate galvanic cell. But the principle illustrated here might be applicable to other autoprotolytic solvents such as H SO . Use the appropriate limiting molar ionic conductivities to estimate the autoprotolysis constant of water at 25° C. Use the reaction equation 2 H O → H O + OH . The data we need are λ = 349.6 and λ = 199.1 S cm mol . The conductivity of water is κ = [H ] λ + [OH ] λ whose units work out to (mol cm ) (S cm mol ). In order to express the ionic concentrations in molarities, we multiply the cm term by (1 L / 1000 cm ), yielding 1000 κ = [H ] λ + [OH ] λ with units S cm L . Recalling that in pure water, [H ] = [OH ] = , we obtain 1000 κ = ( )(λ + λ ) = ( )(548.7 S cm mol ). Solving for : = (1000 κ / 548.7 S cm mol ) Substituting Kohlrausch's water conductivity value of 0.043 × 10 S cm ) for κ gives =(1000 × 0.043 × 10 S cm / 548.7 S cm mol ) = 0.614 × 10 mol cm (i.e., mol L ). The accepted value for the autoprotolysis constant of water at 25° C is = 1.008 × 10 mol L . A saturated solution of silver chloride AgCl has a conductance 2.28 x 10 S cm at 25°C. The conductance of the water used to make up this solution is 0.116 x 10 S cm . The limiting ionic conductivities of the two ions are λ = 61.9 and λ = 76.3 S cm mol . Use this information to estimate the molar solubility of AgCl. The limiting molar conductivity of the solution is Λ = λ + λ = 138.2 S cm . The conductance due to the dissociated salt alone is the difference (2.785 – 0.043) = 2.16 x 10 S cm . Substituting into the expression yields A chemical reaction in which there is a significant change in the number or mobilities of ionic species can be followed by monitoring the change in conductance. Many acid-base reactions fall into this category. In , conductometry is employed to detect the end-point of a titration. Consider, for example, the titration of the strong acid HCl by the strong base NaOH. In ionic terms, the process can be represented as + Cl + Na + → + Na + Cl At the end point, only two ionic species remain, compared to the four during the initial stages of the titration, so the conductivity will be at a minimum. Beyond the end point, continued addition of base causes the conductivity to rise again. The very large mobilities of the H and OH ions cause the conductivity to rise very sharply on either side of the end point, making it quite easy to locate. The plot on the left depicts the conductivities due to the individual ionic species. But of course the conductivity we measure is just the sum of all of these (Kohlrausch's law of independent migration), so the plot on the right corresponds to what you actually see when the titration is carried out. The factor ( + )/ compensates for the dilution of the solution as more base is added. For most ordinary acid-base titrations, conductometry rarely offers any special advantage over regular volumetric analysis or potentiometry. But in some special cases such as those illustrated here, it is the only method capable of yielding useful results. Most people think of electrolytic conduction as something that takes place mainly in batteries, electrochemical plants and in laboratories, but by far the biggest and most important electrolytic system is the earth itself, or at least the thin veneer of soil sediments that coat much of its surface. Soils are composed of sand and humic solids within which are embedded pore spaces containing gases (air and microbial respiration products) and water. This water, both that which fills the pores as well as that which is adsorbed to soil solids, contains numerous dissolved ions on which soil conductivity mainly depends. These ions include exchangeable cations present on the surfaces of the clay components of soils. Although these electrolyte channels are small and tortuously connected, they are present in huge numbers and offer an abundance of parallel conduction paths, so the ground itself acts as a surprisingly efficient conductor. There is no better illustration of this than the use of a ground path to transmit a current of up to 1800 amperes along the 1360-km path of the Pacific DC Intertie that extends from the Celilo Converter Station (view below) in northern Oregon to Los Angeles. This system normally employs a two-conductor power line that operates at ±500,000 volts DC, but when one of these conductors is out of service, the ground path acts as a substitute conductor. This alternative path is said to have a lower resistance than the normal metallic conductor! From an electrochemical standpoint, the most interesting aspect of this system is the manner in which the current flows into or out of the ground. The grounding system at Celilo is composed of over 1000 cast-iron electrodes buried in a circular 3.3-km trench of petroleum coke which acts as the working electrode. At the Los Angeles end, grounding is achieved by means of 24 silicon-iron alloy electrodes submerged in the nearby Pacific Ocean. On a much smaller scale, single-wire earth return systems are often employed to supply regular single-phase ac power to remote regions, or as the return path for high-voltage DC submarine cables. For direct current submarine systems, a copper cable laid on the bottom is suitable for the cathode. The anodes are normally graphite rods surrounded by coke. You may have noticed that the pole-mounted step-down transformers used to distribute single-phase ac power in residential neighborhoods are connected to the high-voltage (10 Kv or so) primary line by only a single wire. The return circuit back to the local substation is completed by a ground connection which runs down the pole to a buried electrode. The Celilo Converter Station is located at The Dalles, Oregon | 6,445 | 2,528 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/12%3A_Chemical_Kinetics_II/12.05%3A_The_Steady-State_Approximation |
One of the most commonly used and most attractive approximations is the . This approximation can be applied to the rate of change of concentration of a highly reactive (short lived) intermediate that holds a constant value over a long period of time. The advantage here is that for such an intermediate (\(I\)), \[\dfrac{d[I]}{dt} = 0 \nonumber \] So long as one can write an expression for the rate of change of the concentration of the intermediate \(I\), the steady state approximation allows one to solve for its constant concentration. For example, if the reaction \[A +B \rightarrow C \label{total} \] is proposed to follow the mechanism \[\begin{align} A + A &\xrightarrow{k_1} A_2 \\[4pt] A_2 + B &\xrightarrow{k_2} C + A \end{align} \nonumber \] The time-rate of change of the concentration of the intermediate \(A_2\) can be written as \[ \dfrac{d[A_2]}{dt} = k_1[A]^2 - k_2[A_2,B] \nonumber \] In the limit that the steady state approximation can be applied to \(A_2\) \[ \dfrac{d[A_2]}{dt} = k_1[A]^2 - k_2[A_2,B] \approx 0 \nonumber \] or \[ [A_2] \approx \dfrac{k_1[A]^2}{k_2[B]} \nonumber \] So if the rate of the overall reaction is expressed as the rate of formation of the product \(C\), \[ \dfrac{d[C]}{dt} = k_2[A_2,B] \nonumber \] the above expression for \([A_2]\) can be substituted \[ \dfrac{d[C]}{dt} = k_2 \left ( \dfrac{k_1[A]^2}{k_2[B]} \right) [B] \nonumber \] of \[ \dfrac{d[C]}{dt} = k_1[A]^2 \nonumber \] and the reaction is predicted to be second order in \([A]\). Alternatively, if the mechanism for Equation \ref{total} is proposed to be \[\begin{align} A &\xrightarrow{k_1} A^* \\[4pt] A^* + B &\xrightarrow{k_2} C \end{align} \nonumber \] then the rate of change of the concentration of \(A^*\) is \[\dfrac{[A^*]}{dt} = k_1[A] - k_2[A^*,B] \nonumber \] And if the steady state approximation holds, then \[[A^*] \approx \dfrac{k_1[A]}{k_2[B]} \nonumber \] So the rate of production of \(C\) is \[\begin{align} \dfrac{d[C]}{dt} &= k_2[A^*,B] \\[4pt] &= \bcancel{k_2} \left( \dfrac{k_1[A]}{\bcancel{k_2} \cancel{[B]}} \right) \cancel{[B]} \end{align} \nonumber \] or \[\dfrac{d[C]}{dt} = k_1[A] \nonumber \] and the rate law is predicted to be first order in \(A\). In this manner, the plausibility of either of the two reaction mechanisms is easily deduced by comparing the predicted rate law to that which is observed. If the prediction cannot be reconciled with observation, then the scientific method eliminates that mechanism from consideration. | 2,500 | 2,529 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/08%3A_Properties_of_Organic_Compounds/8.17%3A_Carboxylic_Acids/8.17.01%3A_Biology-_Entomology |
Almost everyone has experienced the stinging of an ant or bee - for some, the reaction is not only mildly painful, but allergic. What causes the stinging sensation, though, is not the initial bite or sting of the insect; a group of organic compounds called carboxylic acids is responsible for the lasting effect. The Southern wood ant ( ) can actually shoot a simple carboxylic acid, formic acid, at large distances in self-defense - no stinger needed. Formic acid, like all carboxylic acids, ionize to release H ions in solution. Why does the (-RCOOH) group define such an acidic compound, when other such functional groups also release H ions? Why do Southern wood ants not shoot alcohol instead of formic acid? This is because their defining functional group is the carboxyl group, (-RCOOH), in which the electronegativity of the two oxygen atoms and of the resulting (-RCOO) anion contributes to deprotanation. In higher concentrations, formic acid can cause liver and kidney damage; however, ant venom is dilute enough to be eventually metabolized by the body, causing the sting to wear off. The structure of the carboxyl group effects not only the acidity of carboxylic acids, but a number of other physical and chemical properties. For example, hydrogen bonding raises the boiling point of carboxylic acids compared to other functional groups: The extended array of carboxylic acid properties can be made more visible by this JmoL of acetic acid. In the general menu to the left, click on partial charges. Each atom in the molecule will be assigned a partial charge. It is clear that the oxygen atoms are sharing electrons unequally and causing other parts of the molecule to gain a partial positive charge in the carboxyl carbon and hydrogen. Further, this induces a partial negative charge on the methyl carbon, leading to positive charges on the methyl hydrogen atoms. An even better way to view the electron distribution is with the Molecular Electrostatic Potential (MEP) Surface options. One can look at "MEP on isopotential surface", which show surfaces where electrostatic potential is the same, but the most informative option here is the "MEP on Van der Waals Surface" radio button. This shows the potential along the van der Waals surface of the molecule. The closer to red on the color spectrum, the more negative the potential at that surface is, the closer to blue, the more positive. One can see that both oxygen atoms are centers of partial negative charge, while the acidic hydrogen atom has a substantial partial positve charge, and the methyl group is also has a partial positive charge. One more way to look at the molecule, is to use the "MEP on a plane" button. Choose the XY plane, and then click "Set Plane Equation." This will show the electrostatic potential along the axis of symmetry for the molecule. While two hydrogen atoms on the methyl group are out of the plane, this view still allows one to see how partial charge is distributed along the backbone of the molecule in a way the van der Waals surface does not. From this modeling of the acetic acid molecule, hopefully it is becoming clear how the macroscopic properties we discussed arise. In entomology, carboxylic acids are not only used for self-defense. Pantothenic acid, C H NO , is a vital component of royal jelly, a rich honey secreted by worker bees to feed their queen . This compound is not exclusive to bees; pantothenic acid is recognized as vitamin B and is found in many foods, especially in healthily-touted whole grains. | 3,546 | 2,530 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/12%3A_Chemical_Kinetics_II/12.08%3A_The_Michaelis-Menten_Mechanism |
The mechanism (Michaelis & Menten, 1913) is one which many enzyme mitigated reactions follow. The basic mechanism involves an enzyme (\(E\), a biological catalyst) and a substrate (\(S\)) which must connect to form an enzyme-substrate complex (\(ES\)) in order for the substrate to be degraded (or augmented) to form a product (\(P\)). The overall reaction is \[S \rightarrow P \nonumber \] And the simple two-step mechanism is given by \[E + S \ce{<=>[k_1,k_{-1}]} ES \label{step1} \] \[ ES \xrightarrow{k_2} P \label{step2} \] Notice that the enzyme is necessary for the reaction to proceed, but is not part of the overall stoichiometry (as is the case for any catalyst!). Applying the equilibrium approximation to the first step \[ k_1[E,S] \approx k_{-1}[ES] \label{equil} \] And using a mass conservation relationship on the enzyme (noting that the enzyme must be either in its bare form (\(E\)) or complexed with a substrate (\(ES\))): \[ [E]_o = [E] + [ES] \nonumber \] or \[[E] = [E]_o - [ES] \nonumber \] Substituting this into the equilibrium expression (Equation \ref{equil}) yields \[ k_1([E]_o - [ES])[S] = k_{-1}[ES] \nonumber \] Solving this expression for \([ES]\) stepwise reveals \[ k_1[E]_o[S] - k_1[ES,S] = k_{-1}[ES] \nonumber \] \[ k_1[E]_o[S] = k_{-1}[ES] + k_1[ES,S] \nonumber \] \[= (k_{-1} + k_1 ) [ES] \nonumber \] \[\dfrac{k_1[E]_o[S]}{k_1[S] + k_{-1}} = [ES] \nonumber \] Substituting this into the expression for the rate of production of the product \(P\) \[\dfrac{d[P]}{dt} = k_2[ES] \label{step3} \] yields \[\dfrac{d[P]}{dt} = \dfrac{k_2 k_1 [E]_o [S]}{k_1[S] + k_{-1}} \nonumber \] Multiplying the top and bottom of the expression on the right hand side by 1/k gives the result \[\dfrac{d[P]}{dt} = \dfrac{k_2[E]_o[S]}{[S] + \frac{k_1}{k_{-1}}} \nonumber \] , \(K_d\) in Equation \ref{step1}. that \(k_2[E]_0\) giv m rate (\(V_{max}\)) hat \(\dfrac{d[P]}{dt}\) i \[\text{rate} = \dfrac{V_{max}[S]}{K_d+[S]} \nonumber \] This is because the maximum reaction rate is achieved when [ES] is low. As [ES] increases, the likelihood of the complex decomposing to reform reactants is higher, slowing the conversion. [ES] will be low if the concentration of the enzyme is much larger than that of the substrate, so there is never a shortage of enzyme available to form the complex with the substrate. However, if the substrate concentration is higher, the lack of available enzyme active sites will slow the reaction and cause it to become 0 order. In the limit that the substrate concentration is large compared to \(K_d\) (i.e., \(K_d + [S] \approx [S]\)), the reaction ends up zeroth order in substrate. \[\text{rate} = \dfrac{V_{max}[S]}{K_d+[S]} \approx \dfrac{V_{max}\cancel{[S]}}{\cancel{[S]}} = V_{max} \nonumber \] Hence, adding more substrate to the system under this limiting condition will have no effect on the observed rate. This is characteristic of a bottleneck in the mechanism, which would happen if there is a shortage of enzyme sites to which the substrate can attach. In the other extreme, in which \(K_d\) is very large compared to the substrate concentration (i.e., \(K_d + [S] \approx K_d \)), the reaction become first order in substrate. \[ \text{rate} = \dfrac{V_{max}[S]}{K_d+[S]} \approx \dfrac{V_{max}[S]}{K_d} = \dfrac{V_{max}}{K_d}[S] \nonumber \] In an alternate derivation (Briggs & Haldane, 1925) using the steady state approximation applied to the enzyme-substrate complex Solving for \([ES]\) gives the result \[ES] = \dfrac{k_1[E,S]}{k_{-1} + k_2} \nonumber \] or \[[ES] = \dfrac{[E,S]}{K_m} \nonumber \] where \[K_m = \dfrac{k_{-1}+K_2}{k_1} \nonumber \] \(K_M\) is the , which is affected by a number of factors, including pH, temperature, and the nature of the substrate itself. Proceeding as before, though the conservation of mass relationship and substitution into the expression for rate (Equation \ref{step3}) results in \[\dfrac{d[P]}{dt} =\dfrac{V_{max}[S]}{K_m + [S]} \nonumber \] The advantage to this approach is that it accounts for the loss of \(ES\) complex due to the production of products as well as the decomposition to reform the reactants E and S. As before, in the limit that \([S] \gg K_M\), the reaction reaches its maximum rate (\(V_{max}\)) and becomes independent of any concentrations. However in the limit that \([S] \ll K_M\), the reaction becomes 1 order in \([S]\). The Michalis constant and V parameters can be extracted in a number of ways. In the Lineweaver-Burk (Lineweaver & Burk, 1934) method, the reciprocal of the rate law is used to create a linear relationship. \[ \dfrac{1}{\text{rate}} = \dfrac{K_m + [S]}{V_{max}[S]} \nonumber \] or \[ \dfrac{1}{\text{rate}} = \dfrac{K_m}{V_{max}} \dfrac{1}{[S]} + \dfrac{1}{V_{max}} \nonumber \] So a plot of \(1/rate\) as a function of \(1/[S]\) results in a straight line, the slope of which is equal to \(K_M/V_{max}\) and the intercept is \(1/V_{max}\). This is called a . | 4,950 | 2,532 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Elimination_Reactions/Elimination_Reactions/C._Elimination_vs._Substitution |
This page discusses the factors that decide whether halogenoalkanes undergo elimination reactions or nucleophilic substitution when they react with hydroxide ions from, say, sodium hydroxide or potassium hydroxide. Both reactions involve heating the halogenoalkane under reflux with sodium or potassium hydroxide solution. The hydroxide ions present are good nucleophiles, and one possibility is a replacement of the halogen atom by an -OH group to give an alcohol via a nucleophilic substitution reaction. In the example, 2-bromopropane is converted into propan-2-ol. Halogenoalkanes also undergo elimination reactions in the presence of sodium or potassium hydroxide. The 2-bromopropane has reacted to give an alkene - propene. Notice that a hydrogen atom has been removed from one of the end carbon atoms together with the bromine from the centre one. In all simple elimination reactions the things being removed are on adjacent carbon atoms, and a double bond is set up between those carbons. The reagents you are using are the same for both substitution or elimination - the halogenoalkane and either sodium or potassium hydroxide solution. In all cases, you will get a mixture of both reactions happening - some substitution and some elimination. What you get most of depends on a number of factors. This is the most important factor. For example, whatever you do with tertiary halogenoalkanes, you will tend to get mainly the elimination reaction, whereas with primary ones you will tend to get mainly substitution. However, you can influence things to some extent by changing the conditions. The proportion of water to ethanol in the solvent matters. Higher temperatures encourage elimination. Higher concentrations favor elimination. For a given halogenoalkane, to favour elimination rather than substitution, use: In the substitution reaction between a halogenoalkane and OH ions, the hydroxide ions are acting as nucleophiles. For example, one of the lone pairs on the oxygen can attack the slightly positive carbon. This leads on to the loss of the bromine as a bromide ion, and the -OH group becoming attached in its place. Hydroxide ions have a very strong tendency to combine with hydrogen ions to make water - in other words, the OH ion is a very strong base. In an elimination reaction, the hydroxide ion hits one of the hydrogen atoms in the CH group and pulls it off. This leads to a cascade of electron pair movements resulting in the formation of a carbon-carbon double bond, and the loss of the bromine as Br . Jim Clark ( ) | 2,565 | 2,533 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/08%3A_Properties_of_Organic_Compounds/8.06%3A_Alkanes/8.6.03%3A_Cultural_Connections-_Rockets |
For millennia the skies have fascinated humans. It seemed to be an unreachable place that was filled with mystery. Finally, in the heart of the Cold War we finally made it into space. The Russians were the first to do it, with the spaceship Vostok 1, piloted by Yuri Gagarin in 1961. The booster that brought him into space was called the Vostok-K, and held over 10,400 pounds of fuel. That is a huge amount of fuel for a simple mission that lasted only a little under 2 hours. The rocket fuel was comprised of ammonium perchlorate, aluminum powder, and a polymer of butadiene and was a solid. This fuel was powerful, but it was not renewable, limiting flight time. The fact that Russia got a man in space first scared the United States. This was the heart of the cold war, and the US believed that if the Russians "owned"space then they could fire missiles down on them from anywhere. Therefore, the United States needed to be the first on the moon to counteract the Soviet actions. The United States put the mission of reaching the moon into the hands of the Apollo 11 astronauts. They rode the Saturn V rocket into space and to the moon. This rocket held 262,000 pounds of fuel for liftoff and another 100,000 pounds of fuel to reach the moon. That is a huge amount of fuel, almost 30 times the amount of fuel that was aboard the Vostok-K rocket. The Saturn V rocket uses an LOX fuel. LOX is a liquid fuel comprised of liquid oxygen and liquid hydrogen. It is extremely powerful, and also very unstable when ignited. This is why many accidents have occurred while using this type of fuel. LOX fuel is also used on the current day space shuttle missions. The space race was in full swing in the 60's and 70's. It has died down since, but in the near future it will heat up again. This time however, the race will be to put man and machine on farther and more out of reach planets like mars or even planets out of our solar system. In order to do this a reliable and renewable source of fuel must be used. This fuel source could come in the form of the alkane methane. Methane is a readily renewable source in space. It is common on Mars, and very abundant on Saturn's Moon Titan. It is also known to be on some other planets outside of our solar system. When the spacecraft reaches its destination, or in some cases on the way (methane is available in space too, not just on planets), it could refuel. This way the spaceship only has to carry enough fuel to get to the destination, or until more can be made. This is a huge step. The rockets previously required the over 300,000 pounds of fuel because it needed a lot for a return trip. With methane as a fuel source the starting fuel amount could be cut down immensely. Also, the LOX fuel is very unstable. Methane however is very stable, and actually hard to ignite unless the right conditions are met. Since methane is so abundant, and easily made from the following reaction called the : \[{\ce {CO2{}+4H2->[{} \atop 400\ ^{\circ }{\ce {C}},{\ce {pressure}}]CH4{}+2H2O}} \nonumber \] It is also a lot less expensive than the LOX fuel. This would leave NASA the ability to research many other things with the extra money, instead of having to spend it on rocket fuel. Another advantage to methane is that it can be stored at -161 degrees Celsius. This is a lot lower than the -252 degrees that hydrogen has to be stored at. Therefore there would not be as much need to cool the tanks, or insulate them, saving space, weight, and money. Also, methane is denser than hydrogen, so a smaller space could be used to store the material, again saving space and weight. Methane's abundance in the outer solar system is crucial to its possible effectiveness. Methane is not toxic like LOX fuel, and is called the "green fuel" because it is environmentally friendly as well. Having the ability to grab elements to form methane, or methane itself from space can expand the realms of space travel immensely in the future. An example of a methane engine, with thrust of 7,500 pounds (needs 3,000,000 eventually, but its a start) shown on the NASA website science.nasa.gov/media/medial...testfiring.wmv A disadvantage of using methane is that it is hard to ignite. The previous LOX fuel was extremely easy to ignite, which is a good and a bad. To ignite methane you need an ignition source, not just an oxidizer that could ignite LOX. This ignition source could be very hard to find in space, especially because of the extremely low temperatures of space. The ignition source is a problem that NASA is tackling, and can be solved. With the only major disadvantage being that methane is hard to ignite, this fuel type will be extremely useful and efficient once the few issues are figured out. Methane has four C—H bonds arranged tetrahedrally around a single carbon atom. It's boiling point is -161.6 °C and it has a melting point of -182.5 °C. Methane is related to other straight chain alkanes such as ethane (2 carbons), and propane (3 carbons). The properties (table ) are shown below. 1. 2. | 5,051 | 2,534 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkyl_Halides/Properties_of_Alkyl_Halides/Introduction_to_Alkyl_Halides/Electrophiles |
In the vast majority of the nucleophilic substitution reactions you will see in this and other organic chemistry texts, the electrophilic atom is a carbon which is bonded to an electronegative atom, usually oxygen, nitrogen, sulfur, or a halogen. The concept of electrophilicity is relatively simple: an electron-poor atom is an attractive target for something that is electron-rich, . a nucleophile. However, we must also consider the effect of steric hindrance on electrophilicity. In addition, we must discuss how the nature of the electrophilic carbon, and more specifically the stability of a potential carbocationic intermediate, influences the S 1 vs. S 2 character of a nucleophilic substitution reaction. Consider two hypothetical S 2 reactions: one in which the electrophile is a methyl carbon and another in which it is tertiary carbon. Because the three substituents on the methyl carbon electrophile are tiny hydrogens, the nucleophile has a relatively clear path for backside attack. However, backside attack on the tertiary carbon is blocked by the bulkier methyl groups. Once again, steric hindrance - this time caused by bulky groups attached to the electrophile rather than to the nucleophile - hinders the progress of an associative nucleophilic (S 2) displacement. The factors discussed in the above paragraph, however, do not prevent a sterically-hindered carbon from being a good electrophile - they only make it less likely to be attacked in a . Nucleophilic substitution reactions in which the electrophilic carbon is sterically hindered are more likely to occur by a two-step, dissociative (S 1) mechanism. This makes perfect sense from a geometric point of view: the limitations imposed by sterics are significant mainly in an S 2 displacement, when the electrophile being attacked is a sp -hybridized tetrahedral carbon with its relatively ‘tight’ angles of 109.4 . Remember that in an S 1 mechanism, the nucleophile attacks an sp -hybridized carbocation intermediate, which has trigonal planar geometry with ‘open’ 120 angles. With this open geometry, the empty p orbital of the electrophilic carbocation is no longer significantly shielded from the approaching nucleophile by the bulky alkyl groups. A carbocation is a very potent electrophile, and the nucleophilic step occurs very rapidly compared to the first (ionization) step. | 2,380 | 2,535 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/23%3A_Phase_Equilibria/23.04%3A_The_Clausius-Clapeyron_Equation |
In Section 23.3, the Clapeyron Equation was derived for melting points. \[\dfrac{dP}{dT} = \dfrac{ΔH_{molar}}{TΔV_{molar}} \nonumber \] However, our argument is actually quite general and should hold for vapor equilibria as well. The only problem is that the molar volume of gases are by no means so nicely constant as they are for condensed phases. (i. e., for condenses phases, both \(α\) and \(κ\) are pretty small). We can write: \[\dfrac{dP}{dT} =\dfrac{ ΔH_{molar}}{TΔV_{molar}} = \dfrac{ΔH_{molar}}{T} \left[V_{molar}^{gas}-V_{molar}^{liquid} \right] \nonumber \] as \[V_{molar}^{gas} \gg V_{molar}^{liquid} \nonumber \] we can approximate \[V_{molar}^{gas}-V_{molar}^{liquid} \nonumber \] by just taking \(V_{molar}^{gas}\). Further more if the vapor is considered an ideal gas, then \[V_{molar}^{gas} = \dfrac{RT}{P} \nonumber \] We get \[\dfrac{1}{P} .\dfrac{dP}{dT} = \dfrac{d \ln P}{dT} = \dfrac{ΔH_{molar}^{vap}}{RT^2} \label{CCe} \] Equation \(\ref{CCe}\) is known as the equation. We can further work our the integration and find the how the equilibrium vapor pressure changes with temperature: \[\ln \left( \dfrac{P_2}{P_1} \right)= \dfrac{-ΔH_{molar}^{vap}}{R} \left[\dfrac{1}{T_2}-\dfrac{1}{T_1} \right] \nonumber \] Thus if we know the molar enthalpy of vaporization we can predict the vapor lines in the diagram. Of course the approximations made are likely to lead to deviations if the vapor is not ideal or very dense (e.g., approaching the critical point). The Clapeyron attempts to answer the question of what the shape of a two-phase coexistence line is. In the \(P-T\) plane, we see the a function \(P(T)\), which gives us the dependence of \(P\) on \(T\) along a coexistence curve. Consider two phases, denoted \(\alpha\) and \(\beta\), in equilibrium with each other. These could be solid and liquid, liquid and gas, solid and gas, two solid phases, et. Let \(\mu_\alpha (P, T)\) and \(\mu_\beta (P, T)\) be the chemical potentials of the two phases. We have just seen that \[\mu_\alpha (P, T) = \mu_\beta (P, T) \label{14.1} \] Next, suppose that the pressure and temperature are changed by \(dP\) and \(dT\). The changes in the chemical potentials of each phase are \[ d \mu_{\alpha} (P, T) = d \mu_{\beta} (P, T) \label{14.2a} \] \[\left( \dfrac{\partial \mu_{\alpha}}{\partial P} \right)_T dP + \left( \dfrac{\partial \mu_{\alpha}}{\partial T} \right)_P dT = \left( \dfrac{\partial \mu_{\beta}}{\partial P} \right)_T dP + \left( \dfrac{\partial \mu_{\beta}}{\partial T} \right)_P dT \label{14.2b} \] However, since \(G(n, P, T) = n \mu (P, T)\), the free energy \(\bar{G} (P, T)\), which is \(G(n, P, T)/n\), is also just equal to the chemical potential \[\bar{G} (P, T) = \dfrac{G(n, P, T)}{n} = \mu (P, T) \label{14.3} \] Moreover, the derivatives of \(\bar{G}\) are \[\left( \dfrac{\partial \bar{G}}{\partial P} \right)_T = \bar{V}, \: \: \: \: \: \: \: \left( \dfrac{\partial \bar{G}}{\partial T} \right)_P = -\bar{S} \label{14.4} \] Applying these results to the chemical potential condition in Equation \(\ref{14.2b}\), we obtain \[\begin{align} \left( \dfrac{\partial \bar{G}_\alpha}{\partial P} \right)_T dP + \left( \dfrac{\partial \bar{G}_\alpha}{\partial T} \right)_P dT &= \left( \dfrac{\partial \bar{G}_\beta}{\partial P} \right)_T dP + \left( \dfrac{\partial \bar{G}_\beta}{\partial T} \right)_P dT \\[5pt] \bar{V}_\alpha dP - \bar{S}_\alpha dT &= \bar{V}_\beta dP - \bar{S}_\beta dT \end{align} \label{14.5} \] Dividing through by \(dT\), we obtain \[\begin{align} \bar{V}_\alpha \dfrac{\partial P}{\partial T} - \bar{S}_\alpha &= \bar{V}_\beta \dfrac{\partial P}{\partial T} - \bar{S}_\beta \\[5pt] (\bar{V}_\alpha - \bar{V}_\beta) \dfrac{\partial P}{\partial T} &= \bar{S}_\alpha - \bar{S}_\beta \\[5pt] \dfrac{dP}{dT} &= \dfrac{\bar{S}_\alpha - \bar{S}_\beta}{\bar{V}_\alpha - \bar{V}_\beta} \end{align} \label{14.6} \] The importance of the quantity \(dP/dT\) is that is represents the slope of the coexistence curve on the phase diagram between the two phases. Now, in equilibrium \(dG = 0\), and since \(G = H - TS\), it follows that \(dH = T \: dS\) at fixed \(T\). In the narrow temperature range in which the two phases are in equilibrium, we can assume that \(H\) is independent of \(T\), hence, we can write \(S = H/T\). Consequently, we can write the molar entropy difference as \[\bar{S}_\alpha - \bar{S}_\beta = \dfrac{\bar{H}_\alpha - \bar{H}_\beta}{T} \label{14.7} \] and the pressure derivative \(dP/dT\) becomes \[\dfrac{dP}{dT} = \dfrac{\bar{H}_\alpha - \bar{H}_\beta}{T (\bar{V}_\alpha - \bar{V}_\beta)} = \dfrac{\Delta_{\alpha \beta} \bar{H}}{T \Delta_{\alpha \beta} \bar{V}} \label{14.8} \] a result known as the , which tells us that the slope of the coexistence curve is related to the ratio of the molar enthalpy between the phases to the change in the molar volume between the phases. If the phase equilibrium is between the solid and liquid phases, then \(\Delta_{\alpha \beta} \bar{H}\) and \(\Delta_{\alpha \beta} \bar{V}\) are \(\Delta \bar{H}_\text{fus}\) and \(\Delta \bar{V}_\text{fus}\), respectively. If the phase equilibrium is between the liquid and gas phases, then \(\Delta_{\alpha \beta} \bar{H}\) and \(\Delta_{\alpha \beta} \bar{V}\) are \(\Delta \bar{H}_\text{vap}\) and \(\Delta \bar{V}_\text{vap}\), respectively. For the liquid-gas equilibrium, some interesting approximations can be made in the use of the Clapeyron equation. For this equilibrium, Equation \(\ref{14.8}\) becomes \[\dfrac{dP}{dT} = \dfrac{\Delta \bar{H}_\text{vap}}{T (\bar{V}_g - \bar{V}_l)} \label{14.9} \] In this case, \(\bar{V}_g \gg \bar{V}_l\), and we can approximate Equation \(\ref{14.9}\) as \[\dfrac{dP}{dT} \approx \dfrac{\Delta \bar{H}_\text{vap}}{T \bar{V}_g} \label{14.10} \] Suppose that we can treat the vapor phase as an ideal gas. Certainly, this is not a good approximation so close to the vaporization point, but it leads to an example we can integrate. Since \(PV_g = nRT\), \(P \bar{V}_g = RT\), Equation \(\ref{14.10}\) becomes \[\begin{align} \dfrac{dP}{dT} &= \dfrac{\Delta \bar{H}_\text{vap} P}{RT^2} \\[5pt] \dfrac{1}{P} \dfrac{dP}{dT} &= \dfrac{\Delta \bar{H}_\text{vap}}{RT^2} \\[5pt] \dfrac{d \: \text{ln} \: P}{dT} &= \dfrac{\Delta \bar{H}_\text{vap}}{RT^2} \end{align} \label{14.11} \] which is called the . We now integrate both sides, which yields \[\text{ln} \: P = -\dfrac{\Delta \bar{H}_\text{vap}}{RT} + C \nonumber \] where \(C\) is a constant of integration. Exponentiating both sides, we find \[P(T) = C' e^{-\Delta \bar{H}_\text{vap}/RT} \nonumber \] which actually has the wrong curvature for large \(T\), but since the liquid-vapor coexistence line terminates in a critical point, as long as \(T\) is not too large, the approximation leading to the above expression is not that bad. If we, instead, integrate both sides, the left from \(P_1\) to \(P_2\), and the right from \(T_1\) to \(T_2\), we find \[\begin{align} \int_{P_1}^{P_2} d \: \text{ln} \: P &= \int_{T_1}^{T_2} \dfrac{\Delta \bar{H}_\text{vap}}{RT^2} dT \\[5pt] \text{ln} \: \left( \dfrac{P_2}{P_1} \right) &= -\dfrac{\Delta \bar{H}_\text{vap}}{R} \left( \dfrac{1}{T_2} - \dfrac{1}{T_1} \right) \\[5pt] &= \dfrac{\Delta \bar{H}_\text{vap}}{R} \left( \dfrac{T_1 - T_1}{T_1 T_2} \right) \end{align} \label{14.12} \] assuming that \(\Delta \bar{H}_\text{vap}\) is independent of \(T\). Here \(P_1\) is the pressure of the liquid phase, and \(P_2\) is the pressure of the vapor phase. Suppose we know \(P_2\) at a temperature \(T_2\), and we want to know \(P_3\) at another temperature \(T_3\). The above result can be written as \[\text{ln} \: \left( \dfrac{P_3}{P_1} \right) = -\dfrac{\Delta \bar{H}_\text{vap}}{R} \left( \dfrac{1}{T_3} - \dfrac{1}{T_1} \right) \label{14.13} \] Subtracting the two results, we obtain \[\text{ln} \: \left( \dfrac{P_2}{P_3} \right) = -\dfrac{\Delta \bar{H}_\text{vap}}{R} \left( \dfrac{1}{T_2} - \dfrac{1}{T_3} \right) \label{14.14} \] so that we can determine the vapor pressure at any temperature if it is known as one temperature. In order to illustrate the use of this result, consider the following example: At \(1 \: \text{bar}\), the boiling point of water is \(373 \: \text{K}\). At what pressure does water boil at \(473 \: \text{K}\)? Take the heat of vaporization of water to be \(40.65 \: \text{kJ/mol}\). Let \(P_1 = 1 \: \text{bar}\) and \(T_1 = 373 \: \text{K}\). Take \(T_2 = 473 \: \text{K}\), and we need to calculate \(P_2\). Substituting in the numbers, we find \[\begin{align} \text{ln} \: P_2(\text{bar}) &= -\dfrac{(40.65 \: \text{kJ/mol})(1000 \: \text{J/kJ})}{8.3145 \: \text{J/mol} \cdot \text{K}} \left( \dfrac{1}{473 \: \text{K}} - \dfrac{1}{373 \: \text{K}} \right) = 2.77 \\[5pt] P_2(\text{bar}) &= (1 \: \text{bar}) \: e^{2.77} = 16 \: \text{bar} \end{align} \nonumber \] The s of most liquids have similar shapes with the vapor pressure steadily increasing as the temperature increases (Figure \(\Page {1}\)). A good approach is to find a mathematical model for the pressure increase as a function of temperature. Experiments showed that the vapor pressure \(P\) and temperature \(T\) are related, \[P \propto \exp \left(- \dfrac{\Delta H_{vap}}{RT}\right) \ \label{1}\] where \(\Delta{H_{vap}}\) is the Enthalpy (heat) of Vaporization and \(R\) is the gas constant (8.3145 J mol K ). A simple relationship can be found by integrating Equation \ref{1} between two pressure-temperature endpoints: \[\ln \left( \dfrac{P_1}{P_2} \right) = \dfrac{\Delta H_{vap}}{R} \left( \dfrac{1}{T_2}- \dfrac{1}{T_1} \right) \label{2}\] where \(P_1\) and \(P_2\) are the vapor pressures at two temperatures \(T_1\) and \(T_2\). Equation \ref{2} is known as the and allows us to estimate the vapor pressure at another temperature, if the vapor pressure is known at some temperature, and if the enthalpy of vaporization is known. The order of the temperatures in Equation \ref{2} matters as the Clausius-Clapeyron Equation is sometimes written with a negative sign (and switched order of temperatures): \[\ln \left( \dfrac{P_1}{P_2} \right) = - \dfrac{\Delta H_{vap}}{R} \left( \dfrac{1}{T_1}- \dfrac{1}{T_2} \right) \label{2B} \] The vapor pressure of water is 1.0 atm at 373 K, and the enthalpy of vaporization is 40.7 kJ mol . Estimate the vapor pressure at temperature 363 and 383 K respectively. Using the Clausius-Clapeyron equation (Equation \(\ref{2B}\)), we have: \[\begin{align} P_{363} &= 1.0 \exp \left[- \left(\dfrac{40,700}{8.3145}\right) \left(\dfrac{1}{363\;K} -\dfrac{1}{373\; K}\right) \right] \nonumber \\[4pt] &= 0.697\; atm \nonumber \end{align} \nonumber\] \[\begin{align} P_{383} &= 1.0 \exp \left[- \left( \dfrac{40,700}{8.3145} \right)\left(\dfrac{1}{383\;K} - \dfrac{1}{373\;K} \right) \right] \nonumber \\[4pt] &= 1.409\; atm \nonumber \end{align} \nonumber\] Note that the increase in vapor pressure from 363 K to 373 K is 0.303 atm, but the increase from 373 to 383 K is 0.409 atm. The increase in vapor pressure is not a linear process. We can use the Clausius-Clapeyron equation to construct the entire vaporization curve. There is a deviation from experimental value, that is because the enthalpy of vaporization varies slightly with temperature. The Clausius-Clapeyron equation can be also applied to sublimation; the following example shows its application in estimating the heat of sublimation. The vapor pressures of ice at 268 K and 273 K are 2.965 and 4.560 torr respectively. Estimate the heat of sublimation of ice. The enthalpy of sublimation is \(\Delta{H}_{sub}\). Use a piece of paper and derive the Clausius-Clapeyron equation so that you can get the form: \[\begin{align} \Delta H_{sub} &= \dfrac{ R \ln \left(\dfrac{P_{273}}{P_{268}}\right)}{\dfrac{1}{268 \;K} - \dfrac{1}{273\;K}} \nonumber \\[4pt] &= \dfrac{8.3145 \ln \left(\dfrac{4.560}{2.965} \right)}{ \dfrac{1}{268\;K} - \dfrac{1}{273\;K} } \nonumber \\[4pt] &= 52,370\; J\; mol^{-1}\nonumber \end{align} \nonumber\] Note that the heat of sublimation is the sum of heat of melting (6,006 J/mol at 0°C and 101 kPa) and the heat of vaporization (45,051 J/mol at 0 °C). Show that the vapor pressure of ice at 274 K is higher than that of water at the same temperature. Note the curve of vaporization is also called the curve of evaporization. Calculate \(\Delta{H_{vap}}\) for ethanol, given vapor pressure at 40 C = 150 torr. The normal boiling point for ethanol is 78 C. Recognize that we have TWO sets of \((P,T)\) data: We then directly use these data in Equation \ref{2B} \[\begin{align*} \ln \left(\dfrac{150}{760} \right) &= \dfrac{-\Delta{H_{vap}}}{8.314} \left[ \dfrac{1}{313} - \dfrac{1}{351}\right] \\[4pt] \ln 150 -\ln 760 &= \dfrac{-\Delta{H_{vap}}}{8.314} \left[ \dfrac{1}{313} - \dfrac{1}{351}\right] \\[4pt] -1.623 &= \dfrac{-\Delta{H_{vap}}}{8.314} \left[ 0.0032 - 0.0028 \right] \end{align*}\] Then solving for \(\Delta{H_{vap}}\) \[\begin{align*} \Delta{H_{vap}} &= 3.90 \times 10^4 \text{ joule/mole} \\[4pt] &= 39.0 \text{ kJ/mole} \end{align*} \] It is important to not use the Clausius-Clapeyron equation for the solid to liquid transition. That requires the use of the more general \[\dfrac{dP}{dT} = \dfrac{\Delta \bar{H}}{T \Delta \bar{V}} \nonumber\] where \(\Delta \bar{H}\) and \(\Delta \bar{V}\) is the molar change in enthalpy (the enthalpy of fusion in this case) and volume respectively between the two phases in the transition. | 13,487 | 2,536 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/02%3A_Atoms_Molecules_and_Chemical_Reactions/2.11%3A_The_Molar_Mass/2.11.01%3A_Biology-_Water |
For example, photosynthesis involves the reaction of carbon dioxide and water to make sugars. Richard Feynman noted chemitry's contribution in a very poetic way: “The world looks so different after learning science. For example, trees are made of air, primarily. When they are burned, they go back to air, and in the flaming, heat is released, the flaming heat of the sun which was bound in to convert the air into tree. And in the ash is the small remnant of the part which did not come from air, that came from the solid earth, instead. Sugar (and other "carbohydrates" in plants) is made photosynthetically according to overall equations like \[{6 CO_2 + 6 H_2O + Light Energy \longrightarrow\ C_6H_{12}O_6 + 6 O_2} \nonumber \] SID 24715013 CID 24749 This explains why plants die without water. But how much water is required to make a pound (457 grams) of sugar? The equation tells us 6 water molecules make 6 glucose (sugar) molecules. How do we get useful, macroscopic answers, like how much water (and CO ) is needed for a pound of sugar? We start with molar masses. It is often convenient to express physical quantities per (per mole), because in this way equal numbers of atoms or molecules are being compared. Such often tell us something about the atoms or molecules themselves. For example, if the molar volume of one solid is larger than that of another, it is reasonable to assume that the molecules of the first substance are larger than those of the second. (Comparing the molar volumes of liquids, and especially gases, would not necessarily give the same information since the molecules would not be as tightly packed.) A molar quantity is one which has been divided by the amount of substance. For example, an extremely useful molar quantity is the molar mass : \[ {{Molar\ mass} =\tfrac {\text {mass} } {\text {amount of substance} } } \nonumber \] Obtain the molar mass of (a) C (carbon) and (b) H O. The atomic weight of carbon is 12.01, and so 1 mol C weighs 12.01 g. \[M_{\text{C}}=\tfrac{\text{m}_{\text{C}}}{n_{\text{C}}}=\tfrac{\text{12}\text{.01 g}}{\text{1 mol}}=\text{12}\text{.01 g mol}^{-1} \nonumber \] \[M_{\text{H}_{\text{2}}\text{O}}= \tfrac {\text{m}_{\text{H}_{\text{2}}\text{O}}} {n_{\text{H}_{\text{2}}\text{O}}}=\text{18}\text{.02 g mol}^{-1} \nonumber \] \[\text{Mass}\overset{\text{Molar mass}}{\longleftrightarrow}\text{amount of substance}m\overset{M}{\longleftrightarrow}n \nonumber \] The molar mass is easily obtained from atomic weights and may be used as a conversion factor, provided the units cancel. Calculate the amount of glucose (C H O ) in 457 g of this solid. Any problem involving interconversion of mass and amount of substance requires molar mass = (6 x 12.01 + 12 x 1.008 + 6 x 16) g mol = 180 g mol \[\textit{n} = \textit{m}\ \cdot\text{ conversion factor} = m\ \cdot \tfrac {\text{1}} {M} = \text{457 g}\cdot \tfrac {\text{1 mol}} {\text{180 g}} = 2.54\ \text{mol} \nonumber \] The chemical equation above tells us that 6 glucose molecules require 6 water molecules, so 6 moles of glucose require 6 moles of water, and since they're equal, the 2.54 moles of glucose above would require 2.54 moles of water. What mass of water is that? The mass of substance will be the amount times a conversion factor which permits cancellation of units: \[\textit{m} = \textit{n} \cdot\text{conversion factor} = \text{n} \cdot \text{M} = \text{2.54 mol} \cdot 18.02 \tfrac{\text{g}}{\text{mol}} = \text{45.7 g} \nonumber \] How many molecules would be present in 50 mL of pure water? In previous examples, we showed that the number of molecules may be obtained from the amount of substance by using the Avogadro constant. The amount of substance may be obtained from mass by using the molar mass, and mass from volume by means of density. A road map to the solution of this problem is \[\text{Volume}\xrightarrow{\text{density}}\text{mass }\overset{\text{Molar mass}}{\longleftrightarrow}\text{amount}\overset{\text{Avogadro constant}}{\longleftrightarrow}\text{number of molecules} \nonumber \] \[V\xrightarrow{\rho }m\xrightarrow{M}n\xrightarrow{\text{N}_{\text{A}}}N \nonumber \] \begin{align} & \text{N}=\text{50}\text{.0 cm}^{\text{3}}\cdot \tfrac{\text{1}\text{.00 g}}{\text{1 cm}^{\text{3}}}\cdot \tfrac{\text{1 mol}}{\text{18}\text{.02 g}}\cdot \tfrac{\text{6}\text{.022}\cdot \text{10}^{\text{23}}\text{ molecules}}{\text{1 mol}} \\ & \text{ }=\text{1}\text{.67}\cdot \text{10}^{\text{24}}\text{ molecules} \\ \end{align} Notice that in this problem we had to techniques from previous examples. To do this you must remember relationships among quantities. For example, a volume was given, and we knew it could be converted to the corresponding mass by means of density, and so we looked up the density in a table. By writing a road map, or at least seeing it in your mind’s eye, you can keep track of such relationships, determine what conversion factors are needed, and then use them to solve the problem. A student's body may contain 10 water molecules and would need to accumulate about 10 (ten million million million) water molecules over 18 years. a. \[ 150 lb \cdot \dfrac{\text{0.453 kg}}{\text{1 lb}} \cdot .70 = 47.6 kg water \nonumber \] \[\dfrac{\text{47,600 g water}}{18\tfrac{\text{g}}{\text{mol}}} = 2.6 \cdot 10^3 \text{mol} \nonumber \] \[2.6 \cdot 10^3 \text {mol} \cdot 6.02 \cdot 10^{23} \tfrac{\text{molecules}}{\text{mol}} = 1.6 \cdot 10^{27} \text{molecules} \nonumber \] b.\[ \dfrac{10^{27} \text {water molecules}} { \text{18y} \cdot \tfrac{\text{365d}}{\text{y}} \cdot \tfrac{\text {24h}}{\text{d}} \cdot \tfrac{\text{60 min}}{\text{h}} \cdot \tfrac{\text{60 sec}}{\text{min}} } = 4 \cdot 10^{18} \tfrac{\text {molecules}}{\text{sec}} \nonumber \] | 5,766 | 2,537 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Properties_of_Arenes/Aromaticity/Aromaticity/Aromatic_Ions_and_Antiaromaticity |
Carbanions and carbocations may also show aromatic stabilization. Some examples are: The three-membered ring cation has 2 π-electrons and is surprisingly stable, considering its ring strain. Cyclopentadiene is as acidic as ethanol, reflecting the stability of its 6 π-electron conjugate base. Salts of cycloheptatrienyl cation (tropylium ion) are stable in water solution, again reflecting the stability of this 6 π-electron cation. Conjugated ring systems having 4n π-electrons (e.g. 4, 8, 12 etc. electrons) not only fail to show any aromatic properties, but appear to be less stable and more reactive than expected. As noted above, 1,3,5,7-cyclooctatetraene is non-planar and adopts a tub-shaped conformation. The compound is readily prepared, and undergoes addition reactions typical of alkenes. Catalytic hydrogenation of this tetraene produces cyclooctane. Planar bridged annulenes having 4n π-electrons have proven to be relatively unstable. Examples of 8 and 12-π-electron systems are shown below, together with a similar 10 π-electron aromatic compound. The simple C H hydrocarbon pentalene does not exist as a stable compound, and its hexaphenyl derivative is air sensitive. The 12-π-electron analog heptalene has been prepared, but is also extremely reactive (more so than cyclooctatetraene). On the other hand, azulene is a stable 10-π-electron hydrocarbon that incorporates structural features of both pentalene and heptalene. Azulene is a stable blue crystalline solid that undergoes a number of typical aromatic substitution reactions. The unexpected instability of 4n π-electron annulenes has been termed " ". Other examples may be cited. Thus, all attempts to isolate 1,3-cyclobutadiene have yielded its dimer, or products from reactions with other compounds introduced into the reaction system. Similarly, cyclopentadienyl cation (4 π-electrons) and cycloheptatrienyl anion (8 π-electrons) show very high reactivity when forced to form. Cyclooctatetraene is a fascinating compound. To see more of its chemistry Here. | 2,049 | 2,538 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Lattices/Lattice_Basics/How_to_Decide_What_Type_of_Structure |
The best place to start is usually the physical state. Melting point isn't always a good guide to the size of the attractions between particles, because the attractive forces have only been loosened on melting - not broken entirely. Boiling point is a much better guide, because enough heat has now been supplied to break the attractive forces completely. The stronger the attractions, the higher the boiling point. That being said, melting points are often used to judge the size of attractive forces between particles in solids, but you will find the occasional oddity. Those oddities usually disappear if you consider boiling points instead. You would expect stronger metallic bonding in aluminum than in magnesium, because aluminum has 3 electrons to delocalize into the "sea of electrons" rather than magnesium's 2. The boiling points reflect this: Al 2470 °C vs. Mg 1110 °C. However, aluminum's melting point is only 10 °C higher than magnesium's: Al 660 °C vs. Mg 650 °C. So, if it is a gas, liquid or low melting point solid, it will consist of covalently bound molecules (except the noble gases which have molecules consisting of single atoms). The size of the melting point or boiling point gives a guide to the strength of the intermolecular forces. That is then the end of the problem. If it is a gas, liquid or low melting point solid then you are talking about a simple molecular substance. Full stop! If it is a high melting point solid, it will be a giant structure - either ionic, metallic or giant covalent. You now have to sort out which of these it is. Solubility of a solid in water (without reaction) suggests it is ionic. There are exceptions to this, of course. Sugar (sucrose) is soluble in water despite being a covalent molecule. It is capable of extensive hydrogen bonding with water molecules. And there are a lot of ionic compounds which are insoluble in water, of course. Solubility of a low melting point solid or a liquid in water (without it reacting) suggests a small molecule capable of hydrogen bonding - or, at least, a small very polar molecule. Conduction of electricity in the solid state suggests delocalized electrons, and therefore either a metal or graphite. The clue as to which you had would usually come from other data - appearance, malleability, etc (see below). Semiconductors like silicon - a giant covalent structure with the same arrangement of atoms as diamond - also conduct electricity. If a substance doesn't conduct electricity as a solid, but undergoes electrolysis when it is molten, that would confirm that it was ionic. Electrolysis is the splitting up of a compound using electricity. For example, molten sodium chloride conducts electricity and is split into sodium and chlorine in the process Don't forget about obvious things like the shiny appearance of most metals, and their ease of working. Metals are malleable (can be bent or beaten into shape easily) and ductile (can be pulled out into wires). By contrast, giant ionic or giant covalent structures tend to be brittle - shattering rather than bending. Jim Clark ( ) | 3,103 | 2,539 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/01%3A_General_Techniques/1.04%3A_Heating_and_Cooling_Methods/1.4D%3A_Bunsen_Burners |
Bunsen burners are generally used to rapidly heat high-boiling liquids with low flammability (such as water). It is important to know that they can reach temperatures of approximately \(1500^\text{o} \text{C}\),\(^5\) and can easily ignite most organic compounds. If an apparatus is improperly set up, or if there is a small gap that allows organic vapors to escape from an apparatus, these vapors can ignite with a burner. Therefore, it is generally recommended to use other heat sources to warm flammable organic liquids (for example in distillation or reflux). Bunsen burners should never be used with highly flammable solvents such as diethyl ether. However, burners do have their place in the organic lab. Burners are often used in (Figure 1.44a) as the vapors are generally not flammable. In this context, a wire mesh set atop a ring clamp is often used under the flask to dissipate the heat and avoid overheating one area. Burners are also used in the Beilsten test for halogens (Figure 1.44b), with in melting and boiling point determinations (Figure 1.44c), and for softening pipettes to create capillary TLC spotters (Figure 1.44d). They may also be used in sublimations. Burners come in several different forms. The common Bunsen burner is six inches tall and has two models differing in how the gas and air are adjusted (a Bunsen burner is in Figure 1.45a, and a Tirrill burner is in Figure 1.45b). Small burners (microburners, Figure 1.45c) and large burners (Meker burners, Figure 1.45d) are also sometimes used. \(^5\)As reported in the Fischer Scientific catalog. | 1,597 | 2,540 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/11%3A_Fluids/11.03%3A_Unique_Properties_of_Liquids |
Although you have been introduced to some of the interactions that hold molecules together in a liquid, we have not yet discussed the consequences of those interactions for the bulk properties of liquids. We now turn our attention to three unique properties of liquids that intimately depend on the nature of intermolecular interactions: surface tension, capillary action, and viscosity. We stated in that liquids tend to adopt the shapes of their containers. Why, then, do small amounts of water on a freshly waxed car form raised droplets instead of a thin, continuous film? The answer lies in a property called , which depends on intermolecular forces. presents a microscopic view of a liquid droplet. A typical molecule in the of the droplet is surrounded by other molecules that exert attractive forces from all directions. Consequently, there is no force on the molecule that would cause it to move in a particular direction. In contrast, a molecule on the experiences a net attraction toward the drop because there are no molecules on the outside to balance the forces exerted by adjacent molecules in the interior. Because a sphere has the smallest possible surface area for a given volume, intermolecular attractive interactions between water molecules cause the droplet to adopt a spherical shape. This maximizes the number of attractive interactions and minimizes the number of water molecules at the surface. Hence raindrops are almost spherical, and drops of water on a waxed (nonpolar) surface, which does not interact strongly with water, form round beads (see the chapter opener photo). A dirty car is covered with a mixture of substances, some of which are polar. Attractive interactions between the polar substances and water cause the water to spread out into a thin film instead of forming beads. The same phenomenon holds molecules together at the surface of a bulk sample of water, almost as if they formed a skin. When filling a glass with water, the glass can be overfilled so that the level of the liquid actually extends the rim. Similarly, a sewing needle or a paper clip can be placed on the surface of a glass of water where it “floats,” even though steel is much denser than water (part (a) in ). Many insects take advantage of this property to walk on the surface of puddles or ponds without sinking (part (b) in 3.2). (a) A paper clip can “float” on water because of surface tension (from . (b) Surface tension also allows insects such as this water strider to “walk on water.” Such phenomena are manifestations of surface tension , which is defined as the energy required to increase the surface area of a liquid by a specific amount. Surface tension is therefore measured as energy per unit area, such as joules per square meter (J/m ) or dyne per centimeter (dyn/cm), where 1 dyn = 1 × 10 N. The values of the surface tension of some representative liquids are listed in . Note the correlation between the surface tension of a liquid and the strength of the intermolecular forces: the stronger the intermolecular forces, the higher the surface tension. For example, water, with its strong intermolecular hydrogen bonding, has one of the highest surface tension values of any liquid, whereas low-boiling-point organic molecules, which have relatively weak intermolecular forces, have much lower surface tensions. Mercury is an apparent anomaly, but its very high surface tension is due to the presence of strong metallic bonding, which we will discuss in more detail in . Adding soaps and detergents that disrupt the intermolecular attractions between adjacent water molecules can reduce the surface tension of water. Because they affect the surface properties of a liquid, soaps and detergents are called surface-active agents, or surfactants . In the 1960s, US Navy researchers developed a method of fighting fires aboard aircraft carriers using “foams,” which are aqueous solutions of fluorinated surfactants. The surfactants reduce the surface tension of water below that of fuel, so the fluorinated solution is able to spread across the burning surface and extinguish the fire. Such foams are now used universally to fight large-scale fires of organic liquids. Intermolecular forces also cause a phenomenon called capillary action , which is the tendency of a polar liquid to rise against gravity into a small-diameter tube (a ), as shown in . When a glass capillary is put into a dish of water, water is drawn up into the tube. The height to which the water rises depends on the diameter of the tube and the temperature of the water but on the angle at which the tube enters the water. The smaller the diameter, the higher the liquid rises. When a capillary is placed in liquid ink the ink rises up into the capillary. The smaller the diameter of the capillary, the higher the water rises. The height of the water does depend on the angle at which the capillary is tilted. Of course, you can see the same phenomina in anything with narrow channels, like bricks Water rises through a brick in a time lapse video Capillary action is the net result of two opposing sets of forces: cohesive forces , which are the intermolecular forces that hold a liquid together, and adhesive forces , which are the attractive forces between a liquid and the substance that composes the capillary. Water has both strong adhesion to glass, which contains polar SiOH groups, and strong intermolecular cohesion. When a glass capillary is put into water, the surface tension due to cohesive forces constricts the surface area of water within the tube, while adhesion between the water and the glass creates an upward force that maximizes the amount of glass surface in contact with the water. If the adhesive forces are stronger than the cohesive forces, as is the case for water, then the liquid in the capillary rises to the level where the downward force of gravity exactly balances this upward force. If, however, the cohesive forces are stronger than the adhesive forces, as is the case for mercury and glass, the liquid pulls itself down into the capillary below the surface of the bulk liquid to minimize contact with the glass (part (a) in ). The upper surface of a liquid in a tube is called the meniscus , and the shape of the meniscus depends on the relative strengths of the cohesive and adhesive forces. In liquids such as water, the meniscus is concave; in liquids such as mercury, however, which have very strong cohesive forces and weak adhesion to glass, the meniscus is convex (part (b) in ). Polar substances are drawn up a glass capillary and generally have a concave meniscus. Fluids and nutrients are transported up the stems of plants or the trunks of trees by capillary action. Plants contain tiny rigid tubes composed of cellulose, to which water has strong adhesion. Because of the strong adhesive forces, nutrients can be transported from the roots to the tops of trees that are more than 50 m tall. Cotton towels are also made of cellulose; they absorb water because the tiny tubes act like capillaries and “wick” the water away from your skin. The moisture is absorbed by the entire fabric, not just the layer in contact with your body. Viscosity (η) is the resistance of a liquid to flow. Some liquids, such as gasoline, ethanol, and water, flow very readily and hence have a . Others, such as motor oil, molasses, and maple syrup, flow very slowly and have a . The two most common methods for evaluating the viscosity of a liquid are (1) to measure the time it takes for a quantity of liquid to flow through a narrow vertical tube and (2) to measure the time it takes steel balls to fall through a given volume of the liquid. The higher the viscosity, the slower the liquid flows through the tube and the steel balls fall. Viscosity is expressed in units of the poise (mPa·s); the higher the number, the higher the viscosity. The viscosities of some representative liquids are listed in 1 and show a correlation between viscosity and intermolecular forces. Because a liquid can flow only if the molecules can move past one another with minimal resistance, strong intermolecular attractive forces make it more difficult for molecules to move with respect to one another. The addition of a second hydroxyl group to ethanol, for example, which produces ethylene glycol (HOCH CH OH), increases the viscosity 15-fold. This effect is due to the increased number of hydrogen bonds that can form between hydroxyl groups in adjacent molecules, resulting in dramatically stronger intermolecular attractive forces. There is also a correlation between viscosity and molecular shape. Liquids consisting of long, flexible molecules tend to have higher viscosities than those composed of more spherical or shorter-chain molecules. The longer the molecules, the easier it is for them to become “tangled” with one another, making it more difficult for them to move past one another. London dispersion forces also increase with chain length. Due to a combination of these two effects, long-chain hydrocarbons (such as motor oils) are highly viscous. Viscosity increases as intermolecular interactions or molecular size increases. Motor oils and other lubricants demonstrate the practical importance of controlling viscosity. The oil in an automobile engine must effectively lubricate under a wide range of conditions, from subzero starting temperatures to the 200°C that oil can reach in an engine in the heat of the Mojave Desert in August. Viscosity decreases rapidly with increasing temperatures because the kinetic energy of the molecules increases, and higher kinetic energy enables the molecules to overcome the attractive forces that prevent the liquid from flowing. As a result, an oil that is thin enough to be a good lubricant in a cold engine will become too “thin” (have too low a viscosity) to be effective at high temperatures. The viscosity of motor oils is described by an SAE (Society of Automotive Engineers) rating ranging from SAE 5 to SAE 50 for engine oils: the lower the number, the lower the viscosity. So-called can cause major problems. If they are viscous enough to work at high operating temperatures (SAE 50, for example), then at low temperatures, they can be so viscous that a car is difficult to start or an engine is not properly lubricated. Consequently, most modern oils are , with designations such as SAE 20W/50 (a grade used in high-performance sports cars), in which case the oil has the viscosity of an SAE 20 oil at subzero temperatures (hence the W for winter) and the viscosity of an SAE 50 oil at high temperatures. These properties are achieved by a careful blend of additives that modulate the intermolecular interactions in the oil, thereby controlling the temperature dependence of the viscosity. Many of the commercially available oil additives “for improved engine performance” are highly viscous materials that increase the viscosity and effective SAE rating of the oil, but overusing these additives can cause the same problems experienced with highly viscous single-grade oils. Based on the nature and strength of the intermolecular cohesive forces and the probable nature of the liquid–glass adhesive forces, predict what will happen when a glass capillary is put into a beaker of SAE 20 motor oil. Will the oil be pulled up into the tube by capillary action or pushed down below the surface of the liquid in the beaker? What will be the shape of the meniscus (convex or concave)? (Hint: the surface of glass is lined with Si–OH groups.) substance and composition of the glass surface behavior of oil and the shape of meniscus Identify the cohesive forces in the motor oil. Determine whether the forces interact with the surface of glass. From the strength of this interaction, predict the behavior of the oil and the shape of the meniscus. Motor oil is a nonpolar liquid consisting largely of hydrocarbon chains. The cohesive forces responsible for its high boiling point are almost solely London dispersion forces between the hydrocarbon chains. Such a liquid cannot form strong interactions with the polar Si–OH groups of glass, so the surface of the oil inside the capillary will be lower than the level of the liquid in the beaker. The oil will have a convex meniscus similar to that of mercury. Exercise Predict what will happen when a glass capillary is put into a beaker of ethylene glycol. Will the ethylene glycol be pulled up into the tube by capillary action or pushed down below the surface of the liquid in the beaker? What will be the shape of the meniscus (convex or concave)? Capillary action will pull the ethylene glycol up into the capillary. The meniscus will be concave. is the energy required to increase the surface area of a liquid by a given amount. The stronger the intermolecular interactions, the greater the surface tension. are molecules, such as soaps and detergents, that reduce the surface tension of polar liquids like water. is the phenomenon in which liquids rise up into a narrow tube called a capillary. It results when , the intermolecular forces in the liquid, are weaker than , the attraction between a liquid and the surface of the capillary. The shape of the , the upper surface of a liquid in a tube, also reflects the balance between adhesive and cohesive forces. The of a liquid is its resistance to flow. Liquids that have strong intermolecular forces tend to have high viscosities. Why is a water droplet round? How is the environment of molecules on the surface of a liquid droplet different from that of molecules in the interior of the droplet? How is this difference related to the concept of surface tension? Explain the role of intermolecular and intramolecular forces in surface tension. A mosquito is able to walk across water without sinking, but if a few drops of detergent are added to the water, the insect will sink. Why? Explain how soaps or surfactants decrease the surface tension of a liquid. How does the meniscus of an aqueous solution in a capillary change if a surfactant is added? Illustrate your answer with a diagram. Of CH Cl , hexane, and ethanol, which has the lowest viscosity? Which has the highest surface tension? Explain your reasoning in each case. At 25°C, cyclohexanol has a surface tension of 32.92 mN/m , whereas the surface tension of cyclohexanone, which is very simila ly 25.45 mN/m . Why is the surface tension of cyclohexanone so much less than that of cyclohexanol? What is the relationship between Explain your answers in terms of a microscopic picture. What two opposing forces are responsible for capillary action? How do these forces determine the shape of the meniscus? Which of the following liquids will have a concave meniscus in a glass capillary? Explain your reasoning. How does viscosity depend on molecular shape? What molecular features make liquids highly viscous? Adding a soap or a surfactant to water disrupts the attractive intermolecular interactions between water molecules, thereby decreasing the surface tension. Because water is a polar molecule, one would expect that a soap or a surfactant would also disrupt the attractive interactions responsible for adhesion of water to the surface of a glass capillary. As shown in the sketch, this would decrease the height of the water column inside the capillary, as well as making the meniscus less concave. As the structures indicate, cyclohexanol is a polar substance that can engage in hydrogen bonding, much like methanol or ethanol; consequently, it is expected to have a higher surface tension due to stronger intermolecular interactions. Cohesive forces are the intermolecular forces that hold the molecules of the liquid together, while adhesive forces are the attractive forces between the molecules of the liquid and the walls of the capillary. If the adhesive forces are stronger than the cohesive forces, the liquid is pulled up into the capillary and the meniscus is concave. Conversely, if the cohesive forces are stronger than the adhesive forces, the level of the liquid inside the capillary will be lower than the level outside the capillary, and the meniscus will be convex. Viscous substances often consist of molecules that are much longer than they are wide and whose structures are often rather flexible. As a result, the molecules tend to become tangled with one another (much like overcooked spaghetti), which decreases the rate at which they can move through the liquid. The viscosities of five liquids at 25°C are given in the following table. Explain the observed trends in viscosity. The following table gives values for the viscosity, boiling point, and surface tension of four substances. Examine these data carefully to see whether the data for each compound are internally consistent and point out any obvious errors or inconsistencies. Explain your reasoning. Surface tension data (in dyn/cm) for propanoic acid (C H O ), and 2-propanol (C H O), as a function of temperature, are given in the following table. Plot the data for each compound and explain the differences between the two graphs. Based on these data, which molecule is more polar? The plots of surface tension versus temperature for propionic acid and isopropanol have essentially the same slope, but at all temperatures the surface tension of propionic acid is about 30% greater than for isopropanol. Because surface tension is a measure of the cohesive forces in a liquid, these data suggest that the cohesive forces for propionic acid are significantly greater than for isopropanol. Both substances consist of polar molecules with similar molecular masses, and the most important intermolecular interactions are likely to be dipole–dipole interactions. Consequently, these data suggest that propionic acid is more polar than isopropanol. Thumbnail from | 17,859 | 2,542 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/01%3A_Introduction_-_The_Ambit_of_Chemistry/1.09%3A_Density |
The terms and are commonly used in two different ways. We refer to weight when we say that an adult is heavier than a child. On the other hand, something else is alluded to when we say that oak is heavier than balsa wood. A small shaving of oak would obviously weigh less than a roomful of balsa wood, but oak is heavier in the sense that a piece of given size weighs more than the same-size piece of balsa. What we are actually comparing is the , that is, the . In order to determine these densities, we might weigh a cubic centimeter of each type of wood. If the oak sample weighed 0.71 g and the balsa 0.15 g, we could describe the density of oak as 0.71 g cm and that of balsa as 0.15 g cm . (Note that the negative exponent in the units cubic centimeters indicates a reciprocal. Thus 1 cm = 1/cm and the units for our densities could be written as \(\frac{\text{g}}{\text{cm}^\text{3}}\), g/cm , or g cm . In each case the units are read as grams per cubic centimeter, the indicating division.) We often abbreviate "cm " as "cc", and 1 cm = 1 mL exactly by definition. In general it is not necessary to weigh exactly 1 cm of a material in order to determine its density. We simply measure mass and volume and divide volume into mass: \[\text{Density} = \dfrac{\text{mass}}{\text{volume}} \nonumber \] or \[\rho = \dfrac{m}{V} \quad \label{1} \] where \(ρ\) is the density, \(m\) is the mass, and \(V\) volume. Calculate the density of Since the submerged metal displaces its own volume, \[ \begin{align*} \text{Density} &= \rho =\dfrac{m}{V} \\[4pt] &= \dfrac{\text{37.42 g}}{\text{13.9 ml}} \\[4pt] &= \text{2.69 g}/\text{ml or 2.69 g ml}^{-\text{1}} \end{align*} \nonumber \] The volume of the cylinder must be calculated first, using the formula \[ \begin{align*} \text{V} &= \pi r^\text{2} h\\[4pt] &= \text{3.142}\times\text{(0.750 cm)}^\text{2}\times\text{5.25 cm} \\[4pt] &= \text{9.278 718 8 cm}^\text{3} \end{align*} \nonumber \] Then \[ \rho = \dfrac{m}{V} = \dfrac{\text{25.07 g}}{\text{9.278 718 8 cm}^\text{3}} = \begin{cases} 2.70 \dfrac{\text{g}}{\text{cm}^\text{3}} \\ \text{2.70 g cm}^{-\text{3}} \\ \text{2.70 g}/ \text{cm}^\text{3} \end{cases} \nonumber \] which are all acceptable alternatives. Note that unlike mass or volume ( ), the density of a substance is independent of the size of the sample ( ). Thus density is a property by which one substance can be distinguished from another. A sample of pure aluminum can be trimmed to any desired volume or adjusted to have any mass we choose, but its density will always be 2.70 g/cm at 20°C. The densities of some common pure substances are listed below. Tables and graphs are designed to provide a maximum of information in a minimum of space. When a physical quantity (number × units) is involved, it is wasteful to keep repeating the same units. Therefore it is conventional to use pure numbers in a table or along the axes of a graph. A pure number can be obtained from a quantity if we divide by appropriate units. For example, when divided by the units gram per cubic centimeter, the density of aluminum becomes a pure number 2.70: \[\dfrac{\text{Density of aluminum}}{\text{1 g cm}^{-\text{3}}} = \dfrac{\text{2.70 g cm}^{-\text{3}}}{\text{1 g cm}^{-\text{3}}} = \text{2.70} \nonumber \] Therefore, a column in a table or the axis of a graph is conveniently labeled in the following form: Quantity/units This indicates the units that must be divided into the quantity to yield the pure number in the table or on the axis. This has been done in the second column of the table. In our exploration of , notice that chemists may express densities differently depending on the subject. The density of pure substances may be expressed in kg/m in some journals which insist on strict compliance with SI units; densities of soils may be expressed in lb/ft in some agricultural or geological tables; the density of a cell may be expressed in mg/µL; and other units are in common use. It is easy to transform densities from one set of units to another, by multiplying the original quantity by one or more : Convert the density of water, 1 g/cm to (a) lb/cm and (b) lb/ft The equality \(\text{454 g} = \text{1 lb}\) can be used to write two unity factors, \[ \dfrac{\text{454 g}}{\text{1 lb}}\nonumber \] or \[\dfrac{\text{1 lb}}{\text{454 g}} \nonumber \] The given density can be multiplied by one of the unity factors to get the desired result. The correct conversion factor is chosen so that the units cancel: \[ \text{1} \dfrac{\text{g}}{\text{cm}^\text{3}} \times \dfrac{\text{1 lb}}{\text{454 g}} = \text{0.002203} \dfrac{\text{lb}}{\text{cm}^\text{3}}\nonumber \] Similarly, the equalities \(\text{2.54 cm} = \text{1 inch}\), and \(\text{12 inches} = \text{1 ft}\) can be use to write the unity factors: \[ \dfrac{\text{2.54 cm}}{\text{1 in}} \text{, } \dfrac{\text{1 in}}{\text{2.54 cm}} \text{, } \dfrac{\text{12 in}}{\text{1 ft}} \text{ and } \dfrac{\text{1 ft}}{\text{12 in}} \nonumber \] In order to convert the cm in the denominator of 0.002203 to in , we need to multiply by the appropriate unity factor three times, or by the cube of the unity factor: \[ \text{0.002203} \dfrac{\text{g}}{\text{cm}^\text{3}} \times \dfrac{\text{2.54 cm}}{\text{1 in}} \times \dfrac{\text{2.54 cm}}{\text{1 in}} \times \dfrac{\text{2.54 cm}}{\text{1 in}}\nonumber \] or \[ \text{0.002203} \dfrac{\text{g}}{\text{cm}^\text{3}} \times \left(\dfrac{\text{2.54 cm}}{\text{1 in}}\right)^\text{3} = \text{0.0361 lb}/ \text{in}^\text{3}\nonumber \] This can then be converted to lb/ft : \[ \text{0.0361 lb}/ \text{in}^\text{3}\times \left(\dfrac{\text{12 in}}{\text{1 ft}}\right)^\text{3} = \text{62.4 lb}/\text{ft}^\text{3}\nonumber \] It is important to notice that we have used conversion factors to convert from one unit to another unit of the | 5,844 | 2,543 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/09%3A_Molecular_Geometry_and_Bonding_Theories/9.07%3A_Molecular_Orbitals |
delocalized Previously, we described the electrons in isolated atoms as having certain spatial distributions, called , each with a particular . Just as the positions and energies of electrons in can be described in terms of (AOs), the positions and energies of electrons in can be described in terms of molecular orbitals (MOs) —a spatial distribution of electrons that is associated with a particular orbital energy. As the name suggests, molecular orbitals are not localized on a single atom but extend over the entire molecule. Consequently, the molecular orbital approach, called molecular orbital theory is a approach to bonding. Although the molecular orbital theory is computationally demanding, the principles on which it is based are similar to those we used to determine electron configurations for atoms. The key difference is that in molecular orbitals, the electrons are allowed to interact with more than one atomic nucleus at a time. Just as with atomic orbitals, we create an energy-level diagram by listing the molecular orbitals in order of increasing energy. We then fill the orbitals with the required number of valence electrons according to the Pauli principle. This means that each molecular orbital can accommodate a maximum of two electrons with opposite spins. We begin our discussion of molecular orbitals with the simplest molecule, H , formed from two isolated hydrogen atoms, each with a 1 electron configuration. As we explained in Chapter 9, electrons can behave like waves. In the molecular orbital approach, the overlapping atomic orbitals are described by mathematical equations called . The 1 atomic orbitals on the two hydrogen atoms interact to form two new molecular orbitals, one produced by taking the of the two H 1 wave functions, and the other produced by taking their : \[ \begin{matrix}
MO(1)= & AO(atom\; A) & +& AO(atomB) \\
MO(1)= & AO(atom\; A) & -&AO(atomB)
\end{matrix} \label{9.7.1} \] The molecular orbitals created from Equation \(\ref{9.7.1}\) are called linear combinations of atomic orbitals (LCAOs) . A molecule must have as many molecular orbitals as there are atomic orbitals. A molecule must have as many molecular orbitals as there are atomic orbitals. Adding two atomic orbitals corresponds to interference between two waves, thus reinforcing their intensity; the internuclear electron probability density is . The molecular orbital corresponding to the sum of the two H 1 orbitals is called a σ combination (pronounced “sigma one ess”) (part (a) and part (b) in Figure \(\Page {1}\)). In a sigma (σ) orbital, (i.e., a ), the electron density along the internuclear axis and between the nuclei has cylindrical symmetry; that is, all cross-sections perpendicular to the internuclear axis are circles. The subscript 1 denotes the atomic orbitals from which the molecular orbital was derived: \[ \sigma _{1s} \approx 1s\left ( A \right ) + 1s\left ( B \right ) \label{9.7.2} \] Conversely, subtracting one atomic orbital from another corresponds to interference between two waves, which reduces their intensity and causes a in the internuclear electron probability density (part (c) and part (d) in Figure \(\Page {1}\)). The resulting pattern contains a where the electron density is zero. The molecular orbital corresponding to the difference is called \( \sigma _{1s}^{*} \) (“sigma one ess star”). In a sigma star (σ*) orbital , there is a region of zero electron probability, a nodal plane, perpendicular to the internuclear axis: \[ \sigma _{1s}^{\star } \approx 1s\left ( A \right ) - 1s\left ( B \right ) \label{9.7.3} \] The electron density in the σ molecular orbital is greatest between the two positively charged nuclei, and the resulting electron–nucleus electrostatic attractions reduce repulsions between the nuclei. Thus the σ orbital represents a bonding molecular orbital. . In contrast, electrons in the \( \sigma _{1s}^{\star } \) orbital are generally found in the space outside the internuclear region. Because this allows the positively charged nuclei to repel one another, the \( \sigma _{1s}^{\star } \) orbital is an antibonding molecular orbital (a . Antibonding orbitals contain a node perpendicular to the internuclear axis; bonding orbitals do not. Because electrons in the σ orbital interact simultaneously with both nuclei, they have a lower energy than electrons that interact with only one nucleus. This means that the σ molecular orbital has a energy than either of the hydrogen 1 atomic orbitals. Conversely, electrons in the \( \sigma _{1s}^{\star } \) orbital interact with only one hydrogen nucleus at a time. In addition, they are farther away from the nucleus than they were in the parent hydrogen 1 atomic orbitals. Consequently, the \( \sigma _{1s}^{\star } \) molecular orbital has a energy than either of the hydrogen 1 atomic orbitals. The σ (bonding) molecular orbital is relative to the 1 atomic orbitals, and the \( \sigma _{1s}^{\star } \) (antibonding) molecular orbital is . The relative energy levels of these orbitals are shown in the energy-level diagram in Figure \(\Page {2}\) A bonding molecular orbital is lower in energy (more stable) than the component atomic orbitals, whereas an antibonding molecular orbital is higher in energy (less stable). To describe the bonding in a homonuclear diatomic molecule such as H , we use molecular orbitals; that is, for a molecule in which two identical atoms interact, we insert the total number of valence electrons into the energy-level diagram (Figure \(\Page {2}\)). We fill the orbitals according to the and : each orbital can accommodate a maximum of two electrons with opposite spins, and the orbitals are filled in order of increasing energy. Because each H atom contributes one valence electron, the resulting two electrons are exactly enough to fill the σ bonding molecular orbital. The two electrons enter an orbital whose energy is lower than that of the parent atomic orbitals, so the H molecule is more stable than the two isolated hydrogen atoms. Thus molecular orbital theory correctly predicts that H is a stable molecule. Because bonds form when electrons are concentrated in the space between nuclei, this approach is also consistent with our earlier discussion of electron-pair bonds. In the Lewis electron structures, the number of electron pairs holding two atoms together was called the . In the molecular orbital approach, bond order is defined as one-half the number of bonding electrons: \[ bond\; order=\dfrac{number\; of \; bonding\; electrons-number\; of \; antibonding\; electrons}{2} \label{9.7.4} \] To calculate the bond order of H , we see from Figure \(\Page {2}\) that the σ (bonding) molecular orbital contains two electrons, while the \( \sigma _{1s}^{\star } \) (antibonding) molecular orbital is empty. The bond order of H is therefore \[ \dfrac{2-0}{2}=1 \label{9.7.5} \] This result corresponds to the single covalent bond predicted by Lewis dot symbols. Thus molecular orbital theory and the Lewis electron-pair approach agree that a single bond containing two electrons has a bond order of 1. Double and triple bonds contain four or six electrons, respectively, and correspond to bond orders of 2 and 3. We can use energy-level diagrams such as the one in Figure \(\Page {2}\) to describe the bonding in other pairs of atoms and ions where = 1, such as the H ion, the He ion, and the He molecule. Again, we fill the lowest-energy molecular orbitals first while being sure not to violate the or . Figure \(\Page {3a}\) shows the energy-level diagram for the H ion, which contains two protons and only one electron. The single electron occupies the σ bonding molecular orbital, giving a (σ ) electron configuration. The number of electrons in an orbital is indicated by a superscript. In this case, the bond order is \[\dfrac{1-0}{2}=1/2 \nonumber \] Because the bond order is greater than zero, the H ion should be more stable than an isolated H atom and a proton. We can therefore use a molecular orbital energy-level diagram and the calculated bond order to predict the relative stability of species such as H . With a bond order of only 1/2 the bond in H should be weaker than in the H molecule, and the H–H bond should be longer. As shown in Table \(\Page {1}\), these predictions agree with the experimental data. Figure \(\Page {3b}\) is the molecular orbital energy-level diagram for He . This ion has a total of three valence electrons. Because the first two electrons completely fill the σ molecular orbital, the Pauli principle states that the third electron must be in the \( \sigma _{1s}^{\star} \) antibonding orbital, giving a \( \left (\sigma _{1s} \right )^{2}\left (\sigma _{1s}^{\star } \right )^{1} \) electron configuration. This electron configuration gives a bond order of \[\dfrac{2-1}{2}=1/2 \nonumber \] As with H , the He ion should be stable, but the He–He bond should be weaker and longer than in H . In fact, the He ion can be prepared, and its properties are consistent with our predictions (Table \(\Page {1}\)). Finally, we examine the He molecule, formed from two He atoms with 1 electron configurations. Figure \(\Page {3c}\) is the molecular orbital energy-level diagram for He . With a total of four valence electrons, both the σ bonding and \( \sigma _{1s}^{\star } \) antibonding orbitals must contain two electrons. This gives a \( \left (\sigma _{1s} \right )^{2}\left (\sigma _{1s}^{\star } \right )^{2} \) electron configuration, with a predicted bond order of (2 − 2) ÷ 2 = 0, which indicates that the He molecule has no net bond and is not a stable species. Experiments show that the He molecule is actually stable than two isolated He atoms due to unfavorable electron–electron and nucleus–nucleus interactions. In molecular orbital theory, . Consequently, any system that has equal numbers of bonding and antibonding electrons will have a bond order of 0, and it is predicted to be unstable and therefore not to exist in nature. In contrast to Lewis electron structures and the valence bond approach, molecular orbital theory is able to accommodate systems with an odd number of electrons, such as the H ion. In contrast to Lewis electron structures and the valence bond approach, molecular orbital theory can accommodate systems with an odd number of electrons. Use a molecular orbital energy-level diagram, such as those in Figure \(\Page {2}\), to predict the bond order in the He ion. Is this a stable species? chemical species molecular orbital energy-level diagram, bond order, and stability Two He 1 atomic orbitals combine to give two molecular orbitals: a σ bonding orbital at lower energy than the atomic orbitals and a \( \sigma _{1s}^{\star } \) antibonding orbital at higher energy. The bonding in any diatomic molecule with two He atoms can be described using the following molecular orbital diagram: The He ion has only two valence electrons (two from each He atom minus two for the +2 charge). We can also view He as being formed from two He ions, each of which has a single valence electron in the 1 atomic orbital. We can now fill the molecular orbital diagram: The two electrons occupy the lowest-energy molecular orbital, which is the bonding (σ ) orbital, giving a (σ ) electron configuration. To avoid violating the Pauli principle, the electron spins must be paired. So the bond order is \[ \dfrac{2-0}{2} =1 \nonumber \] He is therefore predicted to contain a single He–He bond. Thus it should be a stable species. Use a molecular orbital energy-level diagram to predict the valence-electron configuration and bond order of the H ion. Is this a stable species? H has a valence electron configuration of \( \left (\sigma _{1s} \right )^{2}\left (\sigma _{1s}^{\star } \right )^{2} \) with a bond order of 0. It is therefore predicted to be unstable. So far, our discussion of molecular orbitals has been confined to the interaction of valence orbitals, which tend to lie farthest from the nucleus. When two atoms are close enough for their valence orbitals to overlap significantly, the filled inner electron shells are largely unperturbed; hence they do not need to be considered in a molecular orbital scheme. Also, when the inner orbitals are completely filled, they contain exactly enough electrons to completely fill both the bonding and antibonding molecular orbitals that arise from their interaction. Thus the interaction of filled shells always gives a bond order of 0, so filled shells are not a factor when predicting the stability of a species. This means that we can focus our attention on the molecular orbitals derived from valence atomic orbitals. A molecular orbital diagram that can be applied to any with two identical alkali metal atoms (Li and Cs , for example) is shown in part (a) in Figure \(\Page {4}\), where M represents the metal atom. Only two energy levels are important for describing the valence electron molecular orbitals of these species: a σ bonding molecular orbital and a σ antibonding molecular orbital. Because each alkali metal (M) has an valence electron configuration, the M molecule has two valence electrons that fill the σ bonding orbital. As a result, a bond order of 1 is predicted for all homonuclear diatomic species formed from the alkali metals (Li , Na , K , Rb , and Cs ). The general features of these M diagrams are identical to the diagram for the H molecule in Figure \(\Page {4}\). Experimentally, all are found to be stable in the gas phase, and some are even stable in solution. Similarly, the molecular orbital diagrams for homonuclear diatomic compounds of the alkaline earth metals (such as Be ), in which each metal atom has an valence electron configuration, resemble the diagram for the He molecule in part (c) in Figure \(\Page {2}\). As shown in part (b) in Figure \(\Page {4}\), this is indeed the case. All the homonuclear alkaline earth diatomic molecules have four valence electrons, which fill both the σ bonding orbital and the antibonding orbital and give a bond order of 0. Thus Be , Mg , Ca , Sr , and Ba are all expected to be unstable, in agreement with experimental data. \(Be_2\) is st Use a qualitative molecular orbital energy-level diagram to predict the valence electron configuration, bond order, and likely existence of the Na ion. chemical species molecular orbital energy-level diagram, valence electron configuration, bond order, and stability Because sodium has a [Ne]3 electron configuration, the molecular orbital energy-level diagram is qualitatively identical to the diagram for the interaction of two 1 atomic orbitals. The Na ion has a total of three valence electrons (one from each Na atom and one for the negative charge), resulting in a filled σ molecular orbital, a half-filled σ and a \( \left ( \sigma _{3s} \right )^{2}\left ( \sigma _{3s}^{\star } \right )^{1} \) electron configuration. The bond order is (2-1) 2=1/2 With a fractional bond order, we predict that the Na ion exists but is highly reactive. Use a qualitative molecular orbital energy-level diagram to predict the valence electron configuration, bond order, and likely existence of the Ca ion. Ca has a \( \left ( \sigma _{4s} \right )^{2}\left ( \sigma _{4s}^{\star } \right )^{1} \) electron configurations and a bond order of 1/2 and should exist. Atomic orbitals other than orbitals can also interact to form molecular orbitals. Because individual , , and orbitals are not spherically symmetrical, however, we need to define a coordinate system so we know which lobes are interacting in three-dimensional space. Recall that for each subshell, for example, there are , , and orbitals. All have the same energy and are therefore degenerate, but they have different spatial orientations. \[ \sigma _{np_{z}}=np_{z}\left ( A \right )-np_{z}\left ( B \right ) \label{9.7.6} \] Just as with orbitals, we can form molecular orbitals from orbitals by taking their mathematical sum and difference. When two positive lobes with the appropriate spatial orientation overlap, as illustrated for two atomic orbitals in part (a) in Figure \(\Page {5}\), it is the mathematical of their wave functions that results in interference, which in turn increases the electron probability density between the two atoms. The difference therefore corresponds to a molecular orbital called a \( \sigma _{np_{z}} \) because, just as with the σ orbitals discussed previously, it is symmetrical about the internuclear axis (in this case, the -axis): \[ \sigma _{np_{z}}=np_{z}\left ( A \right )-np_{z}\left ( B \right ) \label{9.7.7A} \] The other possible combination of the two orbitals is the mathematical sum: \[ \sigma _{np_{z}}=np_{z}\left ( A \right )+np_{z}\left ( B \right ) \label{9.7.7} \] In this combination, shown in part (b) in Figure \(\Page {5}\), the positive lobe of one atomic orbital overlaps the negative lobe of the other, leading to interference of the two waves and creating a node between the two atoms. Hence this is an antibonding molecular orbital. Because it, too, is symmetrical about the internuclear axis, this molecular orbital is called a \( \sigma _{np_{z}}=np_{z}\left ( A \right )-np_{z}\left ( B \right ) \) . Whenever orbitals combine, (more stable) than the atomic orbitals from which it was derived, and (less stable). Overlap of atomic orbital lobes with the produces a bonding molecular orbital, regardless of whether it corresponds to the sum or the difference of the atomic orbitals. The remaining orbitals on each of the two atoms, and , do not point directly toward each other. Instead, they are perpendicular to the internuclear axis. If we arbitrarily label the axes as shown in Figure \(\Page {6}\), we see that we have two pairs of orbitals: the two orbitals lying in the plane of the page, and two orbitals perpendicular to the plane. Although these two pairs are equivalent in energy, the orbital on one atom can interact with only the orbital on the other, and the orbital on one atom can interact with only the on the other. These interactions are side-to-side rather than the head-to-head interactions characteristic of σ orbitals. Each pair of overlapping atomic orbitals again forms two molecular orbitals: one corresponds to the arithmetic sum of the two atomic orbitals and one to the difference. The sum of these side-to-side interactions increases the electron probability in the region above and below a line connecting the nuclei, so it is a bonding molecular orbital that is called a pi (π) orbital . The difference results in the overlap of orbital lobes with opposite signs, which produces a nodal plane perpendicular to the internuclear axis; hence it is an antibonding molecular orbital, called a pi star (π*) orbital . \[ \pi _{np_{x}}=np_{x}\left ( A \right )+np_{x}\left ( B \right ) \label{9.7.8} \] \[ \pi ^{\star }_{np_{x}}=np_{x}\left ( A \right )-np_{x}\left ( B \right ) \label{9.7.9} \] The two orbitals can also combine using side-to-side interactions to produce a bonding \( \pi _{np_{y}} \) molecular orbital and an antibonding \( \pi _{np_{y}}^{\star } \) molecular orbital. Because the and atomic orbitals interact in the same way (side-to-side) and have the same energy, the \( \pi _{np_{x}} \) and \( \pi _{np_{y}} \)molecular orbitals are a degenerate pair, as are the \( \pi _{np_{x}}^{\star } \) and \( \pi _{np_{y}}^{\star } \) molecular orbitals. Figure \(\Page {7}\) is an energy-level diagram that can be applied to two identical interacting atoms that have three atomic orbitals each. There are six degenerate atomic orbitals (three from each atom) that combine to form six molecular orbitals, three bonding and three antibonding. The bonding molecular orbitals are lower in energy than the atomic orbitals because of the increased stability associated with the formation of a bond. Conversely, the antibonding molecular orbitals are higher in energy, as shown. The energy difference between the σ and σ* molecular orbitals is significantly greater than the difference between the two π and π* sets. The reason for this is that the atomic orbital overlap and thus the strength of the interaction are greater for a σ bond than a π bond, which means that the σ molecular orbital is more stable (lower in energy) than the π molecular orbitals. Although many combinations of atomic orbitals form molecular orbitals, we will discuss only one other interaction: an atomic orbital on one atom with an atomic orbital on another. As shown in Figure \(\Page {8}\), the sum of the two atomic wave functions ( + ) produces a σ bonding molecular orbital. Their difference ( − ) produces a σ* antibonding molecular orbital, which has a nodal plane of zero probability density perpendicular to the internuclear axis. Molecular orbital theory, a delocalized approach to bonding, can often explain a compound’s color, why a compound with unpaired electrons is stable, semiconductor behavior, and resonance, none of which can be explained using a localized approach. A is an allowed spatial distribution of electrons in a molecule that is associated with a particular orbital energy. Unlike an atomic orbital (AO), which is centered on a single atom, a molecular orbital extends over all the atoms in a molecule or ion. Hence the of bonding is a approach. Molecular orbitals are constructed using , which are usually the mathematical sums and differences of wave functions that describe overlapping atomic orbitals. Atomic orbitals interact to form three types of molecular orbitals. A completely bonding molecular orbital contains no nodes (regions of zero electron probability) perpendicular to the internuclear axis, whereas a completely contains at least one node perpendicular to the internuclear axis. A (bonding) or a (antibonding) is symmetrical about the internuclear axis. Hence all cross-sections perpendicular to that axis are circular. Both a (bonding) and a (antibonding) possess a nodal plane that contains the nuclei, with electron density localized on both sides of the plane. The energies of the molecular orbitals versus those of the parent atomic orbitals can be shown schematically in an . The electron configuration of a molecule is shown by placing the correct number of electrons in the appropriate energy-level diagram, starting with the lowest-energy orbital and obeying the Pauli principle; that is, placing only two electrons with opposite spin in each orbital. From the completed energy-level diagram, we can calculate the , defined as one-half the net number of bonding electrons. In bond orders, electrons in antibonding molecular orbitals cancel electrons in bonding molecular orbitals, while electrons in nonbonding orbitals have no effect and are not counted. Bond orders of 1, 2, and 3 correspond to single, double, and triple bonds, respectively. Molecules with predicted bond orders of 0 are generally less stable than the isolated atoms and do not normally exist. ( ) | 23,164 | 2,544 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkanes/Reactivity_of_Alkanes/Chlorination_of_Methane_and_the_Radical_Chain_Mechanism |
(the most basic of all organic compounds) undergo very few reactions. One of these reactions is halogenation, or the substitution of a single hydrogen on the alkane for a single halogen to form a . This reaction is very important in organic chemistry because it opens a gateway to further chemical reactions. While the reactions possible with alkanes are few, there are many reactions that involve . In order to better understand the (a detailed look at the step by step process through which a reaction occurs), we will closely examine the chlorination of methane. When methane (CH ) and chlorine (Cl ) are mixed together in the absence of light at room temperature nothing happens. However, if the conditions are changed, so that either the reaction is taking place at high temperatures (denoted by Δ) or there is ultra violet irradiation, a product is formed, chloromethane (CH Cl). Why does this reaction occur? Is the reaction favorable? A way to answer these questions is to look at the change in (\(\Delta{H}\)) that occurs when the reaction takes place. ΔH = (Energy put into reaction) – (Energy given off from reaction) If more energy is put into a reaction than is given off, the ΔH is positive, the reaction is endothermic and not energetically favorable. If more energy is given off in the reaction than was put in, the ΔH is negative, the reaction is said to be exothermic and is considered favorable. The figure below illustrates the difference between endothermic and exothermic reactions. ΔH can also be calculated using bond dissociation energies (ΔH°): \[\Delta{H} = \sum \Delta{H^°} \text{ of bonds broken} - \sum \Delta{H^°} \text{ of bonds formed}\] Let’s look at our specific example of the chlorination of methane to determine if it is endothermic or exothermic: Since, the ΔH for the chlorination of methane is negative, the reaction is exothermic. Energetically this reaction is favorable. In order to better understand this reaction we need to look at the mechanism ( a detailed step by step look at the reaction showing how it occurs) by which the reaction occurs. The reaction proceeds through the radical chain mechanism. The radical chain mechanism is characterized by three steps: , and . Initiation requires an input of energy but after that the reaction is self-sustaining. The first propagation step uses up one of the products from initiation, and the second propagation step makes another one, thus the cycle can continue until indefinitely. Initiation breaks the bond between the chlorine molecule (Cl ). For this step to occur energy must be put in, this step is not energetically favorable. After this step, the reaction can occur continuously (as long as reactants provide) without input of more energy. It is important to note that this part of the mechanism cannot occur without some external energy input, through light or heat. The next two steps in the mechanism are called propagation steps. In the first propagation step, a chlorine radical combines with a hydrogen on the methane. This gives hydrochloric acid (HCl, the inorganic product of this reaction) and the methyl radical. In the second propagation step more of the chlorine starting material (Cl ) is used, one of the chlorine atoms becomes a radical and the other combines with the methyl radical. The first propagation step is endothermic, meaning it takes in heat (requires 2 kcal/mol) and is not energetically favorable. In contrast the second propagation step is exothermic, releasing 27 kcal/mol. Since the second propagation step is so exothermic, it occurs very quickly. The second propagation step uses up a product from the first propagation step (the methyl radical) and following , when the product of the first step is removed the equilibrium is shifted towards it's products. This principle is what governs the unfavorable first propagation step's occurance. In the termination steps, all the remaining radicals combine (in all possible manners) to form more product (CH Cl), more reactant (Cl ) and even combinations of the two methyl radicals to form a side product of ethane (CH CH ). The chlorination of methane does not necessarily stop after one chlorination. It may actually be very hard to get a monosubstituted chloromethane. Instead di-, tri- and even tetra-chloromethanes are formed. One way to avoid this problem is to use a much higher concentration of methane in comparison to chloride. This reduces the chance of a chlorine radical running into a chloromethane and starting the mechanism over again to form a dichloromethane. Through this method of controlling product ratios one is able to have a relative amount of control over the product. Answers to these questions are in an attached slide | 4,757 | 2,546 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/18%3A_Solubility_and_Complex-Ion_Equilibria/18.1%3A_Solubility_Product_Constant_Ksp |
We begin our discussion of solubility and complexation equilibria—those associated with the formation of complex ions—by developing quantitative methods for describing dissolution and precipitation reactions of ionic compounds in aqueous solution. Just as with acid–base equilibria, we can describe the concentrations of ions in equilibrium with an ionic solid using an equilibrium constant expression. When a slightly soluble ionic compound is added to water, some of it dissolves to form a solution, establishing an equilibrium between the pure solid and a solution of its ions. For the dissolution of calcium phosphate, one of the two main components of kidney stones, the equilibrium can be written as follows, with the solid salt on the left: \[Ca_3(PO_4)_{2(s)} \rightleftharpoons 3Ca^{2+}_{(aq)} + 2PO^{3−}_{4(aq)} \label{Eq1}\] As you will discover in and in more advanced chemistry courses, basic anions, such as S , PO , and CO , react with water to produce OH and the corresponding protonated anion. Consequently, their calculated molarities, assuming no protonation in aqueous solution, are only approximate. The equilibrium constant for the dissolution of a sparingly soluble salt is the of the salt. Because the concentration of a pure solid such as Ca (PO ) is a constant, it does not appear explicitly in the equilibrium constant expression. The equilibrium constant expression for the dissolution of calcium phosphate is therefore \[K=\dfrac{[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2}{[\mathrm{Ca_3(PO_4)_2}]} \label{Eq2a}\] \[[\mathrm{Ca_3(PO_4)_2}]K=K_{\textrm{sp}}=[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2 \label{Eq2b}\] At 25°C and pH 7.00, Ksp for calcium phosphate is 2.07 × 10 , indicating that the concentrations of Ca and PO ions in solution that are in equilibrium with solid calcium phosphate are very low. The values of for some common salts are listed in Table \(\Page {1}\), which shows that the magnitude of K varies dramatically for different compounds. Although K is not a function of pH in \(\ref{Eq2b}\), changes in pH can affect the solubility of a compound as discussed later. As with , the concentration of a pure solid does not appear explicitly in . Solubility products are determined experimentally by directly measuring either the concentration of one of the component ions or the solubility of the compound in a given amount of water. However, whereas solubility is usually expressed in terms of mass of solute per 100 mL of solvent, \(K_{sp}\), like \(K\), is defined in terms of the molar concentrations of the component ions. Definition of a Solubility Product: Calcium oxalate monohydrate [Ca(O CCO )·H O, also written as CaC O ·H O] is a sparingly soluble salt that is the other major component of kidney stones [along with Ca (PO ) ]. Its solubility in water at 25°C is 7.36 × 10 g/100 mL. Calculate its . solubility in g/100 mL We need to write the solubility product expression in terms of the concentrations of the component ions. For calcium oxalate monohydrate, the balanced dissolution equilibrium and the solubility product expression (abbreviating oxalate as ox ) are as follows: \(\mathrm{Ca(O_2CCO_2)}\cdot\mathrm{H_2O(s)}\rightleftharpoons \mathrm{Ca^{2+}(aq)}+\mathrm{^-O_2CCO_2^-(aq)}+\mathrm{H_2O(l)}\hspace{5mm}K_{\textrm{sp}}=[\mathrm{Ca^{2+}},\mathrm{ox^{2-}}]\) Neither solid calcium oxalate monohydrate nor water appears in the solubility product expression because their concentrations are essentially constant. Next we need to determine [Ca ] and [ox ] at equilibrium. We can use the mass of calcium oxalate monohydrate that dissolves in 100 mL of water to calculate the number of moles that dissolve in 100 mL of water. From this we can determine the number of moles that dissolve in 1.00 L of water. For dilute solutions, the density of the solution is nearly the same as that of water, so dissolving the salt in 1.00 L of water gives essentially 1.00 L of solution. Because each 1 mol of dissolved calcium oxalate monohydrate dissociates to produce 1 mol of calcium ions and 1 mol of oxalate ions, we can obtain the equilibrium concentrations that must be inserted into the solubility product expression. The number of moles of calcium oxalate monohydrate that dissolve in 100 mL of water is as follows: The number of moles of calcium oxalate monohydrate that dissolve in 1.00 L of the saturated solution is as follows: Because of the stoichiometry of the reaction, the concentration of Ca and ox ions are both 5.04 × 10 M. Inserting these values into the solubility product expression, \[K_{sp} = [Ca^{2+},ox^{2−}] = (5.04 \times 10^{−5})(5.04 \times10^{−5}) = 2.54 \times 10^{−9}\] In our calculation, we have ignored the reaction of the weakly basic anion with water, which tends to make the actual solubility of many salts greater than the calculated value. One crystalline form of calcium carbonate (CaCO ) is the mineral sold as “calcite” in mineral and gem shops. The solubility of calcite in water is 0.67 mg/100 mL. Calculate its . 4.5 × 10 The reaction of weakly basic anions with H O tends to make the actual solubility of many salts higher than predicted. Image used with permisison from Calcite, a structural material for many organisms, is found in the teeth of sea urchins. The urchins create depressions in limestone that they can settle in by grinding the rock with their teeth. Limestone, however, also consists of calcite, so how can the urchins grind the rock without also grinding their teeth? Researchers have discovered that the teeth are shaped like needles and plates and contain magnesium. The concentration of magnesium increases toward the tip, which contributes to the hardness. Moreover, each tooth is composed of two blocks of the polycrystalline calcite matrix that are interleaved near the tip. This creates a corrugated surface that presumably increases grinding efficiency. Toolmakers are particularly interested in this approach to grinding. Tabulated values of can also be used to estimate the solubility of a salt with a procedure that is essentially the reverse of the one used in Example \(\Page {1}\). In this case, we treat the problem as a typical equilibrium problem and set up a table of initial concentrations, changes in concentration, and final concentrations ( ), remembering that the concentration of the pure solid is essentially constant. We saw that the for Ca (PO ) is 2.07 × 10 at 25°C. Calculate the aqueous solubility of Ca (PO ) in terms of the following: molar concentration and mass of salt that dissolves in 100 mL of water Although the amount of solid Ca (PO ) changes as some of it dissolves, its molar concentration does not change. We now insert the expressions for the equilibrium concentrations of the ions into the solubility product expression ( ): This is the molar solubility of calcium phosphate at 25°C. However, the molarity of the ions is 2 and 3 , which means that [PO ] = 2.28 × 10 and [Ca ] = 3.42 × 10 . The solubility product of silver carbonate (Ag CO ) is 8.46 × 10 at 25°C. Calculate the following: The of a salt is the product of the concentrations of the ions in solution raised to the same powers as in the solubility product expression. It is analogous to the reaction quotient ( ) discussed for gaseous equilibria. Whereas describes equilibrium concentrations, the ion product describes concentrations that are not necessarily equilibrium concentrations. The ion product is analogous to the reaction quotient for gaseous equilibria. As summarized in , there are three possible conditions for an aqueous solution of an ionic solid: The process of calculating the value of the ion product and comparing it with the magnitude of the solubility product is a straightforward way to determine whether a solution is unsaturated, saturated, or supersaturated. More important, the ion product tells chemists whether a precipitate will form when solutions of two soluble salts are mixed. We mentioned that barium sulfate is used in medical imaging of the gastrointestinal tract. Its solubility product is 1.08 × 10 at 25°C, so it is ideally suited for this purpose because of its low solubility when a “barium milkshake” is consumed by a patient. The pathway of the sparingly soluble salt can be easily monitored by x-rays. Will barium sulfate precipitate if 10.0 mL of 0.0020 M Na SO is added to 100 mL of 3.2 × 10 M BaCl ? Recall that NaCl is highly soluble in water. and volumes and concentrations of reactants whether precipitate will form The only slightly soluble salt that can be formed when these two solutions are mixed is BaSO because NaCl is highly soluble. The equation for the precipitation of BaSO is as follows: \[BaSO_{4(s)} \rightleftharpoons Ba^{2+}_{(aq)} + SO^{2−}_{4(aq)}\] The solubility product expression is as follows: To solve this problem, we must first calculate the ion product— = [Ba ,SO ]—using the concentrations of the ions that are present after the solutions are mixed and before any reaction occurs. The concentration of Ba when the solutions are mixed is the total number of moles of Ba in the original 100 mL of BaCl solution divided by the final volume (100 mL + 10.0 mL = 110 mL): Similarly, the concentration of SO after mixing is the total number of moles of SO in the original 10.0 mL of Na SO solution divided by the final volume (110 mL): We can now calculate : We now compare with the . If > , then BaSO will precipitate, but if < , it will not. Because > , we predict that BaSO will precipitate when the two solutions are mixed. In fact, BaSO will continue to precipitate until the system reaches equilibrium, which occurs when [Ba ,SO ] = = 1.08 × 10 . The solubility product of calcium fluoride (CaF ) is 3.45 × 10 . If 2.0 mL of a 0.10 M solution of NaF is added to 128 mL of a 2.0 × 10 M solution of Ca(NO ) , will CaF precipitate? yes ( = 4.7 × 10 > ) The solubility product ( ) is used to calculate equilibrium concentrations of the ions in solution, whereas the ion product ( ) describes concentrations that are not necessarily at equilibrium. The equilibrium constant for a dissolution reaction, called the solubility product ( ), is a measure of the solubility of a compound. Whereas solubility is usually expressed in terms of mass of solute per 100 mL of solvent, is defined in terms of the molar concentrations of the component ions. In contrast, the ion product ( ) describes concentrations that are not necessarily equilibrium concentrations. Comparing and enables us to determine whether a precipitate will form when solutions of two soluble salts are mixed. | 10,696 | 2,547 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/22%3A_Chemistry_of_The_Main-Group_Elements_II/22.4%3A_Group_16%3A_The_Oxygen_Family |
The chalcogens are the first group in the p block to have no stable metallic elements. All isotopes of polonium (Po), the only metal in group 16, are radioactive, and only one element in the group, tellurium (Te), can even be described as a semimetal. As in groups 14 and 15, the lightest element of group 16, oxygen, is found in nature as the free element. Of the group 16 elements, only sulfur was known in ancient times; the others were not discovered until the late 18th and 19th centuries. Sulfur is frequently found as yellow crystalline deposits of essentially pure S in areas of intense volcanic activity or around hot springs. As early as the 15th century BC, sulfur was used as a fumigant in Homeric Greece because, when burned, it produces SO fumes that are toxic to most organisms, including vermin hiding in the walls and under the floors of houses. Hence references to sulfur are common in ancient literature, frequently in the context of religious purification. In fact, the association of sulfur with the divine was so pervasive that the prefixes thio- (meaning “sulfur”) and theo- (meaning “god”) have the same root in ancient Greek. Though used primarily in the production of sulfuric acid, sulfur is also used to manufacture gunpowder and as a cross-linking agent for rubber, which enables rubber to hold its shape but retain its flexibility. Group 16 is the first group in the p block with no stable metallic elements. Oxygen was not discovered until 1771, when the Swedish pharmacist Carl Wilhelm Scheele found that heating compounds such as KNO , Ag CO , and HgO produced a colorless, odorless gas that supported combustion better than air. The results were not published immediately, however, so Scheele’s work remained unknown until 1777. Unfortunately, this was nearly two years after a paper by the English chemist Joseph Priestley had been published, describing the isolation of the same gas by using a magnifying glass to focus the sun’s rays on a sample of HgO. Oxygen is used primarily in the steel industry during the conversion of crude iron to steel using the Bessemer process. Another important industrial use of oxygen is in the production of TiO , which is commonly used as a white pigment in paints, paper, and plastics. Tellurium was discovered accidentally in 1782 by the Austrian chemist Franz Joseph Müller von Reichenstein, the chief surveyor of mines in Transylvania who was also responsible for the analysis of ore samples. The silvery-white metal had the same density as antimony but very different properties. Because it was difficult to analyze, Müller called it metallum problematicum (meaning “difficult metal”). The name tellurium (from the Latin tellus, meaning “earth”) was coined by another Austrian chemist, Martin Klaproth, who demonstrated in 1798 that Müller’s “difficult metal” was actually a new element. Tellurium is used to color glass and ceramics, in the manufacture of blasting caps, and in thermoelectric devices. Berzelius was born into a well-educated Swedish family, but both parents died when he was young. He studied medicine at the University of Uppsala, where his experiments with electroshock therapy caused his interests to turn to electrochemistry. Berzelius devised the system of chemical notation that we use today. In addition, he discovered six elements (cerium, thorium, selenium, silicon, titanium, and zirconium). The heaviest chalcogen, polonium, was isolated after an extraordinary effort by Marie Curie. Although she was never able to obtain macroscopic quantities of the element, which she named for her native country of Poland, she demonstrated that its chemistry required it to be assigned to group 16. Marie Curie was awarded a second Nobel Prize in Chemistry in 1911 for the discovery of radium and polonium. Oxygen is by far the most abundant element in Earth’s crust and in the hydrosphere (about 44% and 86% by mass, respectively). The same process that is used to obtain nitrogen from the atmosphere produces pure oxygen. Oxygen can also be obtained by the electrolysis of water, the decomposition of alkali metal or alkaline earth peroxides or superoxides, or the thermal decomposition of simple inorganic salts, such as potassium chlorate in the presence of a catalytic amount of MnO : \[\mathrm{2KClO_3(s)\overset{MnO_2(s)}{\underset{\Delta}\rightleftharpoons}2KCl(s)+3O_2(g)} \label{22.4.1}\] Unlike oxygen, sulfur is not very abundant, but it is found as elemental sulfur in rock formations overlying salt domes, which often accompany petroleum deposits (Figure \(\Page {1}\)). Sulfur is also recovered from H S and organosulfur compounds in crude oil and coal and from metal sulfide ores such as pyrite (FeS ). Because selenium and tellurium are chemically similar to sulfur, they are usually found as minor contaminants in metal sulfide ores and are typically recovered as by-products. Even so, they are as abundant in Earth’s crust as silver, palladium, and gold. One of the best sources of selenium and tellurium is the “slime” deposited during the electrolytic purification of copper. Both of these elements are notorious for the vile odors of many of their compounds. For example, when the body absorbs even trace amounts of tellurium, dimethyltellurium [(CH ) Te] is produced and slowly released in the breath and perspiration, resulting in an intense garlic-like smell that is commonly called “tellurium breath.” With their ns np electron configurations, the chalcogens are two electrons short of a filled valence shell. Thus in reactions with metals, they tend to acquire two additional electrons to form compounds in the −2 oxidation state. This tendency is greatest for oxygen, the chalcogen with the highest electronegativity. The heavier, less electronegative chalcogens can lose either four np electrons or four np and two ns electrons to form compounds in the +4 and +6 oxidation state, respectively, as shown in Table Figure \(\Page {1}\). As with the other groups, the lightest member in the group, in this case oxygen, differs greatly from the others in size, ionization energy, electronegativity, and electron affinity, so its chemistry is unique. Also as in the other groups, the second and third members (sulfur and selenium) have similar properties because of shielding effects. Only polonium is metallic, forming either the hydrated Po or Po ion in aqueous solution, depending on conditions. As in groups 14 and 15, the lightest group 16 member has the greatest tendency to form multiple bonds. Thus elemental oxygen is found in nature as a diatomic gas that contains a net double bond: O=O. As with nitrogen, electrostatic repulsion between lone pairs of electrons on adjacent atoms prevents oxygen from forming stable catenated compounds. In fact, except for O , all compounds that contain O–O bonds are potentially explosive. Ozone, peroxides, and superoxides are all potentially dangerous in pure form. Ozone (O ), one of the most powerful oxidants known, is used to purify drinking water because it does not produce the characteristic taste associated with chlorinated water. Hydrogen peroxide (H O ) is so thermodynamically unstable that it has a tendency to undergo explosive decomposition when impure: \[2H_2O_{2(l)} \rightarrow 2H_2O_{(l)} + O_{2(g)} \;\;\; ΔG^o = −119\; kJ/mol \label{1}\] Despite the strength of the O=O bond (\(D_\mathrm{O_2}\) = 494 kJ/mol), \(O_2\) is extremely reactive, reacting directly with nearly all other elements except the noble gases. Some properties of O and related species, such as the peroxide and superoxide ions, are in Table \(\Page {2}\). With few exceptions, the chemistry of oxygen is restricted to negative oxidation states because of its high electronegativity (χ = 3.4). Unlike the other chalcogens, oxygen does not form compounds in the +4 or +6 oxidation state. Oxygen is second only to fluorine in its ability to stabilize high oxidation states of metals in both ionic and covalent compounds. For example, AgO is a stable solid that contains silver in the unusual Ag(II) state, whereas OsO is a volatile solid that contains Os(VIII). Because oxygen is so electronegative, the O–H bond is highly polar, creating a large bond dipole moment that makes hydrogen bonding much more important for compounds of oxygen than for similar compounds of the other chalcogens. Metal oxides are usually basic, and nonmetal oxides are acidic, whereas oxides of elements that lie on or near the diagonal band of semimetals are generally amphoteric. A few oxides, such as CO and PbO , are neutral and do not react with water, aqueous acid, or aqueous base. Nonmetal oxides are typically covalent compounds in which the bonds between oxygen and the nonmetal are polarized (E –O ). Consequently, a lone pair of electrons on a water molecule can attack the partially positively charged E atom to eventually form an oxoacid. An example is reacting sulfur trioxide with water to form sulfuric acid: \[H_2O_{(l)} + SO_{3(g)} \rightarrow H_2SO_{4(aq)} \label{2}\] The oxides of the semimetals and of elements such as Al that lie near the metal/nonmetal dividing line are amphoteric, as we expect: \[Al_2O_{3(s)} + 6H^+_{(aq)} \rightarrow 2Al^{3+}_{(aq)} + 3H_2O_{(l)} \label{3}\] \[Al_2O_{3(s)} + 2OH^−_{(aq)} + 3H_2O_{(l)} \rightarrow 2Al(OH)^−_{4(aq)} \label{4}\] Oxides of metals tend to be basic, oxides of nonmetals tend to be acidic, and oxides of elements in or near the diagonal band of semimetals are generally amphoteric. For each reaction, explain why the given products form. balanced chemical equations why the given products form Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the observed reaction products form. Predict the product(s) of each reaction and write a balanced chemical equation for each reaction. Because most of the heavier chalcogens (group 16) and pnicogens (group 15) are nonmetals, they often form similar compounds. For example, both third-period elements of these groups (phosphorus and sulfur) form catenated compounds and form multiple allotropes. Consistent with periodic trends, the tendency to catenate decreases as we go down the column. Sulfur and selenium both form a fairly extensive series of catenated species. For example, elemental sulfur forms S rings packed together in a complex “crankshaft” arrangement (Figure \(\Page {2}\)), and molten sulfur contains long chains of sulfur atoms connected by S–S bonds. Moreover, both sulfur and selenium form polysulfides (S ) and polyselenides (Se ), with n ≤ 6. The only stable allotrope of tellurium is a silvery white substance whose properties and structure are similar to those of one of the selenium allotropes. Polonium, in contrast, shows no tendency to form catenated compounds. The striking decrease in structural complexity from sulfur to polonium is consistent with the decrease in the strength of single bonds and the increase in metallic character as we go down the group. As in group 15, the reactivity of elements in group 16 decreases from lightest to heaviest. For example, selenium and tellurium react with most elements but not as readily as sulfur does. As expected for nonmetals, sulfur, selenium, and tellurium do not react with water, aqueous acid, or aqueous base, but all dissolve in strongly oxidizing acids such as HNO to form oxoacids such as H SO . In contrast to the other chalcogens, polonium behaves like a metal, dissolving in dilute HCl to form solutions that contain the Po ion. Just as with the other groups, the tendency to catenate, the strength of single bonds, and reactivity decrease down the group. Fluorine reacts directly with all chalcogens except oxygen to produce the hexafluorides (YF ), which are extraordinarily stable and unreactive compounds. Four additional stable fluorides of sulfur are known; thus sulfur oxidation states range from +1 to +6 (Figure \(\Page {2}\)). In contrast, only four fluorides of selenium (SeF , SeF , FSeSeF, and SeSeF ) and only three of tellurium (TeF , TeF , and Te F ) are known. Direct reaction of the heavier chalcogens with oxygen at elevated temperatures gives the dioxides (YO ), which exhibit a dramatic range of structures and properties. The dioxides become increasingly metallic in character down the group, as expected, and the coordination number of the chalcogen steadily increases. Thus SO is a gas that contains V-shaped molecules (as predicted by the valence-shell electron-pair repulsion model), SeO is a white solid with an infinite chain structure (each Se is three coordinate), TeO is a light yellow solid with a network structure in which each Te atom is four coordinate, and PoO is a yellow ionic solid in which each Po ion is eight coordinate. The dioxides of sulfur, selenium, and tellurium react with water to produce the weak, diprotic oxoacids (H YO —sulfurous, selenous, and tellurous acid, respectively). Both sulfuric acid and selenic acid (H SeO ) are strong acids, but telluric acid [Te(OH) ] is quite different. Because tellurium is larger than either sulfur or selenium, it forms weaker π bonds to oxygen. As a result, the most stable structure for telluric acid is Te(OH) , with six Te–OH bonds rather than Te=O bonds. Telluric acid therefore behaves like a weak triprotic acid in aqueous solution, successively losing the hydrogen atoms bound to three of the oxygen atoms. As expected for compounds that contain elements in their highest accessible oxidation state (+6 in this case), sulfuric, selenic, and telluric acids are oxidants. Because the stability of the highest oxidation state decreases with increasing atomic number, telluric acid is a stronger oxidant than sulfuric acid. The stability of the highest oxidation state of the chalcogens decreases down the column. Sulfur and, to a lesser extent, selenium react with carbon to form an extensive series of compounds that are structurally similar to their oxygen analogues. For example, CS and CSe are both volatile liquids that contain C=S or C=Se bonds and have the same linear structure as CO . Because these double bonds are significantly weaker than the C=O bond, however, CS , CSe , and related compounds are less stable and more reactive than their oxygen analogues. The chalcogens also react directly with nearly all metals to form compounds with a wide range of stoichiometries and a variety of structures. Metal chalcogenides can contain either the simple chalcogenide ion (Y ), as in Na S and FeS, or polychalcogenide ions (Y ), as in FeS and Na S . The dioxides of the group 16 elements become increasingly basic, and the coordination number of the chalcogen steadily increases down the group. Ionic chalcogenides like Na S react with aqueous acid to produce binary hydrides such as hydrogen sulfide (H S). Because the strength of the Y–H bond decreases with increasing atomic radius, the stability of the binary hydrides decreases rapidly down the group. It is perhaps surprising that hydrogen sulfide, with its familiar rotten-egg smell, is much more toxic than hydrogen cyanide (HCN), the gas used to execute prisoners in the “gas chamber.” Hydrogen sulfide at relatively low concentrations deadens the olfactory receptors in the nose, which allows it to reach toxic levels without detection and makes it especially dangerous. For each reaction, explain why the given product forms or no reaction occurs. balanced chemical equations why the given products (or no products) form Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the reaction products form or why no reaction occurs. Predict the products of each reaction and write a balanced chemical equation for each reaction. The chalcogens have no stable metallic elements. The tendency to catenate, the strength of single bonds, and the reactivity all decrease moving down the group. Because the electronegativity of the chalcogens decreases down the group, so does their tendency to acquire two electrons to form compounds in the −2 oxidation state. The lightest member, oxygen, has the greatest tendency to form multiple bonds with other elements. It does not form stable catenated compounds, however, due to repulsions between lone pairs of electrons on adjacent atoms. Because of its high electronegativity, the chemistry of oxygen is generally restricted to compounds in which it has a negative oxidation state, and its bonds to other elements tend to be highly polar. Metal oxides are usually basic, and nonmetal oxides are acidic, whereas oxides of elements along the dividing line between metals and nonmetals are amphoteric. The reactivity, the strength of multiple bonds to oxygen, and the tendency to form catenated compounds all decrease down the group, whereas the maximum coordination numbers increase. Because Te=O bonds are comparatively weak, the most stable oxoacid of tellurium contains six Te–OH bonds. The stability of the highest oxidation state (+6) decreases down the group. Double bonds between S or Se and second-row atoms are weaker than the analogous C=O bonds because of reduced orbital overlap. The stability of the binary hydrides decreases down the group. | 17,256 | 2,548 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/10%3A_Gases/10.06%3A_Stoichiometry_Involving_Gases |
With the ideal gas law, we can use the relationship between the amounts of gases (in moles) and their volumes (in liters) to calculate the stoichiometry of reactions involving gases, if the pressure and temperature are known. This is important for several reasons. Many reactions that are carried out in the laboratory involve the formation or reaction of a gas, so chemists must be able to quantitatively treat gaseous products and reactants as readily as they quantitatively treat solids or solutions. Furthermore, many, if not most, industrially important reactions are carried out in the gas phase for practical reasons. Gases mix readily, are easily heated or cooled, and can be transferred from one place to another in a manufacturing facility via simple pumps and plumbing. As a chemical engineer said to one of the authors, “Gases always go where you want them to, liquids sometimes do, but solids almost never do.” Sulfuric acid, the industrial chemical produced in greatest quantity (almost 45 million tons per year in the United States alone), is prepared by the combustion of sulfur in air to give SO , followed by the reaction of SO with O in the presence of a catalyst to give SO , which reacts with water to give H SO . The overall chemical equation is as follows: \[\rm 2S(s)+3O_2(g)+2H_2O(l)\rightarrow 2H_2SO_4(aq) \notag \] What volume of O (in liters) at 22°C and 745 mmHg pressure is required to produce 1.00 ton of H SO ? reaction, temperature, pressure, and mass of one product volume of gaseous reactant Calculate the number of moles of H SO in 1.00 ton. From the stoichiometric coefficients in the balanced chemical equation, calculate the number of moles of O required. Use the ideal gas law to determine the volume of O required under the given conditions. Be sure that all quantities are expressed in the appropriate units. We can see from the stoichiometry of the reaction that mol of O is required to produce 1 mol of H SO . This is a standard stoichiometry problem of the type presented in , except this problem asks for the volume of one of the reactants (O ) rather than its mass. We proceed exactly as in , using the strategy We begin by calculating the number of moles of H SO in 1.00 tn: \[\rm\dfrac{907.18\times10^3\;g\;H_2SO_4}{(2\times1.008+32.06+4\times16.00)\;g/mol}=9250\;mol\;H_2SO_4 \notag \] We next calculate the number of moles of O required: \[\rm9250\;mol\;H_2SO_4\times\dfrac{3mol\; O_2}{2mol\;H_2SO_4}=1.389\times10^4\;mol\;O_2 \notag \] After converting all quantities to the appropriate units, we can use the ideal gas law to calculate the volume of O : \[V=\dfrac{nRT}{P}=\rm\dfrac{1.389\times10^4\;mol\times0.08206\dfrac{L\cdot atm}{mol\cdot K}\times(273+22)\;K}{745\;mmHg\times\dfrac{1\;atm}{760\;mmHg}}=3.43\times10^5\;L \notag \] The answer means that more than 300,000 L of oxygen gas are needed to produce 1 tn of sulfuric acid. These numbers may give you some appreciation for the magnitude of the engineering and plumbing problems faced in industrial chemistry. Exercise In Example 5, we saw that Charles used a balloon containing approximately 31,150 L of H for his initial flight in 1783. The hydrogen gas was produced by the reaction of metallic iron with dilute hydrochloric acid according to the following balanced chemical equation: How much iron (in kilograms) was needed to produce this volume of H if the temperature was 30°C and the atmospheric pressure was 745 mmHg? 68.6 kg of Fe (approximately 150 lb) Many of the advances made in chemistry during the 18th and 19th centuries were the result of careful experiments done to determine the identity and quantity of gases produced in chemical reactions. For example, in 1774, Joseph Priestley was able to isolate oxygen gas by the thermal decomposition of mercuric oxide (HgO). In the 1780s, Antoine Lavoisier conducted experiments that showed that combustion reactions, which require oxygen, produce what we now know to be carbon dioxide. Both sets of experiments required the scientists to collect and manipulate gases produced in chemical reactions, and both used a simple technique that is still used in chemical laboratories today: collecting a gas by the displacement of water. As shown in , the gas produced in a reaction can be channeled through a tube into inverted bottles filled with water. Because the gas is less dense than liquid water, it bubbles to the top of the bottle, displacing the water. Eventually, all the water is forced out and the bottle contains only gas. If a calibrated bottle is used (i.e., one with markings to indicate the volume of the gas) and the bottle is raised or lowered until the level of the water is the same both inside and outside, then the pressure within the bottle will exactly equal the atmospheric pressure measured separately with a barometer. \notag The only gases that cannot be collected using this technique are those that readily dissolve in water (e.g., NH , H S, and CO ) and those that react rapidly with water (such as F and NO ). Remember, however, when calculating the amount of gas formed in the reaction, the gas collected inside the bottle is pure. Instead, it is a mixture of the product gas and water vapor. As we will discuss in , all liquids (including water) have a measurable amount of vapor in equilibrium with the liquid because molecules of the liquid are continuously escaping from the liquid’s surface, while other molecules from the vapor phase collide with the surface and return to the liquid. The vapor thus exerts a pressure above the liquid, which is called the liquid’s . In the case shown in 0.6.1 the bottle is therefore actually filled with a mixture of O and water vapor, and the total pressure is, by Dalton’s law of partial pressures, the sum of the pressures of the two components: \[P_{\rm tot}=P_{\rm gas}+P_{\rm H_2O}=P_{\rm bar.} \tag{10.6.1}\] If we want to know the pressure of the gas generated in the reaction to calculate the amount of gas formed, we must first subtract the pressure due to water vapor from the total pressure. This is done by referring to tabulated values of the vapor pressure of water as a function of temperature ( ). As shown in , the vapor pressure of water increases rapidly with increasing temperature, and at the normal boiling point (100°C), the vapor pressure is exactly 1 atm. The methodology is illustrated in Example 14. Sodium azide (NaN ) decomposes to form sodium metal and nitrogen gas according to the following balanced chemical equation: \[ 2NaN_3 \rightarrow 2Na_{(s)} + 3N_{2\; (g)} \notag \] This reaction is used to inflate the air bags that cushion passengers during automobile collisions. The reaction is initiated in air bags by an electrical impulse and results in the rapid evolution of gas. If the N gas that results from the decomposition of a 5.00 g sample of NaN could be collected by displacing water from an inverted flask, as in , what volume of gas would be produced at 22°C and 762 mmHg? reaction, mass of compound, temperature, and pressure volume of nitrogen gas produced Calculate the number of moles of N gas produced. From the data in , determine the partial pressure of N gas in the flask. Use the ideal gas law to find the volume of N gas produced. Because we know the mass of the reactant and the stoichiometry of the reaction, our first step is to calculate the number of moles of N gas produced: [\rm\dfrac{5.00\;g\;NaN_3}{(22.99+3\times14.01)\;g/mol}\times\dfrac{3mol\;N_2}{2mol\;NaN_3}=0.115\;mol\; N_2 \notag \] The pressure given (762 mmHg) is the pressure in the flask, which is the sum of the pressures due to the N gas and the water vapor present. tells us that the vapor pressure of water is 19.84 mmHg at 22°C (295 K), so the partial pressure of the N gas in the flask is only 762 − 19.84 = 742 mmHg = 0.976 atm. Solving the ideal gas law for and substituting the other quantities (in the appropriate units), we get \[V=\dfrac{nRT}{P}=\rm\dfrac{0.115\;mol\times0.08206\dfrac{atm\cdot L}{mol\cdot K}\times294\;K}{0.978\;atm}=2.84\;L \notag \] Exercise A 1.00 g sample of zinc metal is added to a solution of dilute hydrochloric acid. It dissolves to produce H gas according to the equation Zn(s) + 2 HCl(aq) → H (g) + ZnCl (aq). The resulting H gas is collected in a water-filled bottle at 30°C and an atmospheric pressure of 760 mmHg. What volume does it occupy? 0.397 L The relationship between the amounts of products and reactants in a chemical reaction can be expressed in units of moles or masses of pure substances, of volumes of solutions, or of volumes of gaseous substances. The ideal gas law can be used to calculate the volume of gaseous products or reactants as needed. In the laboratory, gases produced in a reaction are often collected by the displacement of water from filled vessels; the amount of gas can then be calculated from the volume of water displaced and the atmospheric pressure. A gas collected in such a way is not pure, however, but contains a significant amount of water vapor. The measured pressure must therefore be corrected for the vapor pressure of water, which depends strongly on the temperature. Why are so many industrially important reactions carried out in the gas phase? The volume of gas produced during a chemical reaction can be measured by collecting the gas in an inverted container filled with water. The gas forces water out of the container, and the volume of liquid displaced is a measure of the volume of gas. What additional information must be considered to determine the number of moles of gas produced? The volume of some gases cannot be measured using this method. What property of a gas precludes the use of this method? Equal masses of two solid compounds (A and B) are placed in separate sealed flasks filled with air at 1 atm and heated to 50°C for 10 hours. After cooling to room temperature, the pressure in the flask containing A was 1.5 atm. In contrast, the pressure in the flask containing B was 0.87 atm. Suggest an explanation for these observations. Would the masses of samples A and B still be equal after the experiment? Why or why not? Balance each chemical equation and then determine the volume of the indicated reactant at STP that is required for complete reaction. Assuming complete reaction, what is the volume of the products? During the smelting of iron, carbon reacts with oxygen to produce carbon monoxide, which then reacts with iron(III) oxide to produce iron metal and carbon dioxide. If 1.82 L of CO at STP is produced, Complete decomposition of a sample of potassium chlorate produced 1.34 g of potassium chloride and oxygen gas. The combustion of a 100.0 mg sample of an herbicide in excess oxygen produced 83.16 mL of CO and 72.9 mL of H O vapor at STP. A separate analysis showed that the sample contained 16.44 mg of chlorine. If the sample is known to contain only C, H, Cl, and N, determine the percent composition and the empirical formula of the herbicide. The combustion of a 300.0 mg sample of an antidepressant in excess oxygen produced 326 mL of CO and 164 mL of H O vapor at STP. A separate analysis showed that the sample contained 23.28% oxygen. If the sample is known to contain only C, H, O, and N, determine the percent composition and the empirical formula of the antidepressant. Percent composition: 58.3% C, 4.93% H, 23.28% O, and 13.5% N; empirical formula: C H O N | 11,438 | 2,549 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/13%3A_Solutions_and_their_Physical_Properties/13.06%3A_Vapor_Pressures_of_Solutions |
Adding a nonvolatile solute, one whose vapor pressure is too low to measure readily, to a volatile solvent decreases the vapor pressure of the solvent. We can understand this phenomenon qualitatively by examining Figure \(\Page {1}\), which is a schematic diagram of the surface of a solution of glucose in water. In an aqueous solution of glucose, a portion of the surface area is occupied by nonvolatile glucose molecules rather than by volatile water molecules. As a result, fewer water molecules can enter the vapor phase per unit time, even though the surface water molecules have the same kinetic energy distribution as they would in pure water. At the same time, the rate at which water molecules in the vapor phase collide with the surface and reenter the solution is unaffected. The net effect is to shift the dynamic equilibrium between water in the vapor and the liquid phases, decreasing the vapor pressure of the solution compared with the vapor pressure of the pure solvent. Figure \(\Page {2}\) shows two beakers, one containing pure water and one containing an aqueous glucose solution, in a sealed chamber. We can view the system as having two competing equilibria: water vapor will condense in both beakers at the same rate, but water molecules will evaporate more slowly from the glucose solution because fewer water molecules are at the surface. Eventually all of the water will evaporate from the beaker containing the liquid with the higher vapor pressure (pure water) and condense in the beaker containing the liquid with the lower vapor pressure (the glucose solution). If the system consisted of only a beaker of water inside a sealed container, equilibrium between the liquid and vapor would be achieved rather rapidly, and the amount of liquid water in the beaker would remain constant. If the particles of a solute are essentially the same size as those of the solvent and both solute and solvent have roughly equal probabilities of being at the surface of the solution, then the effect of a solute on the vapor pressure of the solvent is proportional to the number of sites occupied by solute particles at the surface of the solution. Doubling the concentration of a given solute causes twice as many surface sites to be occupied by solute molecules, resulting in twice the decrease in vapor pressure. The relationship between solution composition and vapor pressure is therefore \[ \color{red} P_A=X_AP^0_A \label{13.6.1}\] where \(P_A\) is the vapor pressure of component A of the solution (in this case the solvent), XA is the mole fraction of A in solution, and \(P^0_A\) is the vapor pressure of pure A. Equation \ref{13.6.1} is known as , after the French chemist who developed it. If the solution contains only a single nonvolatile solute (B), then \(X_A + X_B = 1\), and we can substitute \(X_A = 1 − X_B\) to obtain \[\begin{align} P_A &=(1−X_B)P^0_A \\[4pt] &=P^0_A−X_BP^0_A \label{13.6.2} \end{align}\] Rearranging and defining \(ΔP_A=P^0_A−P_A\), we obtain a relationship between the decrease in vapor pressure and the mole fraction of nonvolatile solute: \[P^0_A−P_A=ΔP_A=X_BP^0_A \label{13.6.3}\] We can solve vapor pressure problems in either of two ways: by using Equation \ref{13.6.1} to calculate the actual vapor pressure above a solution of a nonvolatile solute, or by using Equation \ref{13.6.3} to calculate the decrease in vapor pressure caused by a specified amount of a nonvolatile solute. Ethylene glycol (\(\ce{HOCH_2CH_2OH}\)), the major ingredient in commercial automotive antifreeze, increases the boiling point of radiator fluid by lowering its vapor pressure. At 100°C, the vapor pressure of pure water is 760 mmHg. Calculate the vapor pressure of an aqueous solution containing 30.2% ethylene glycol by mass, a concentration commonly used in climates that do not get extremely cold in winter. : identity of solute, percentage by mass, and vapor pressure of pure solvent : vapor pressure of solution : : A 30.2% solution of ethylene glycol contains 302 g of ethylene glycol per kilogram of solution; the remainder (698 g) is water. To use Raoult’s law to calculate the vapor pressure of the solution, we must know the mole fraction of water. Thus we must first calculate the number of moles of both ethylene glycol (EG) and water present: \[moles \; EG=(302 \;\cancel{g}) \left( \dfrac{1\; mol}{62.07\; \cancel{g}} \right)=4.87\; mol\; EG \nonumber\] \[moles \; \ce{H2O}=(698 \;\cancel{g}) \left( \dfrac{1\; mol}{18.02\; \cancel{g}} \right)=38.7\; mol\; H_2O \nonumber\] The mole fraction of water is thus \[X_{\ce{H2O}}=\dfrac{38.7\; \cancel{mol} \; H_2O}{38.7\; \cancel{mol}\; H_2O +4.87 \cancel{mol}\; EG} =0.888 \nonumber\] From Raoult’s law (Equation \ref{13.6.1}), the vapor pressure of the solution is \[P_{\ce{H2O}}=(X_{H2_O})(P^0_{H2_O)}=(0.888)(760\; mmHg) =675 \;mmHg \nonumber\] Alternatively, we could solve this problem by calculating the mole fraction of ethylene glycol and then using Equation \ref{13.6.3} to calculate the resulting decrease in vapor pressure: \[X_{EG}=\dfrac{4.87\; mol\; EG}{4.87\; mol\; EG+38.7\; mol\; H_2O}=0.112 \nonumber\] \[ΔP_{\ce{H2O}}=(X_{EG})(P^0_{H_2O})=(0.112)(760\; mmHg)=85.1\; mmHg \nonumber\] \[P_{\ce{H2O}}=P^0H_2O−ΔP_{H_2O}=760\; mmHg−85.1\; mmHg=675\; mmHg \nonumber\] The same result is obtained using either method. Seawater is an approximately 3.0% aqueous solution of \(NaCl\) by mass with about 0.5% of other salts by mass. Calculate the decrease in the vapor pressure of water at 25°C caused by this concentration of \(NaCl\), remembering that 1 mol of \(NaCl\) produces 2 mol of solute particles. The vapor pressure of pure water at 25°C is 23.8 mmHg. 0.45 mmHg. This may seem like a small amount, but it constitutes about a 2% decrease in the vapor pressure of water and accounts in part for the higher humidity in the north-central United States near the Great Lakes, which are freshwater lakes. The decrease therefore has important implications for climate modeling. Finding the Vapor Pressure of a Solution (Ionic-Nonvolatile Solute): Even when a solute is volatile, meaning that it has a measurable vapor pressure, we can still use Raoult’s law. In this case, we calculate the vapor pressure of each component separately. The total vapor pressure of the solution (\(P_T\)) is the sum of the vapor pressures of the components: \[ \begin{align} P_T &=P_A+P_B \\[4pt] &=X_AP^0_A+X_BP^0_B \label{13.6.4} \end{align}\] Because \(X_B = 1 − X_A\) for a two-component system, \[ P_T=X_AP^0_A+(1−X_A)P^0_B \label{13.6.5}\] Thus we need to specify the mole fraction of only one of the components in a two-component system. Consider, for example, the vapor pressure of solutions of benzene and toluene of various compositions. At 20°C, the vapor pressures of pure benzene and toluene are 74.7 and 22.3 mmHg, respectively. The vapor pressure of benzene in a benzene–toluene solution is \[P_{C_6H_6}=X_{C_6H_6}P^0_{C_6H_6} \label{13.6.6}\] and the vapor pressure of toluene in the solution is \[P_{C_6H_5CH_3}=X_{C_6H_5CH_3}P^0_{C_6H_5CH3} \label{13.6.7}\] Equations \ref{13.6.6} and \ref{13.6.7} are both in the form of the equation for a straight line: \(y = mx + b\), where \(b = 0\). Plots of the vapor pressures of both components versus the mole fractions are therefore straight lines that pass through the origin, as shown in Figure Figure \(\Page {3}\). Furthermore, a plot of the total vapor pressure of the solution versus the mole fraction is a straight line that represents the sum of the vapor pressures of the pure components. Thus the vapor pressure of the solution is always greater than the vapor pressure of either component. A solution of two volatile components that behaves like the solution in Figure \(\Page {3}\), which is defined as a solution that obeys Raoult’s law. Like an ideal gas, an ideal solution is a hypothetical system whose properties can be described in terms of a simple model. Mixtures of benzene and toluene approximate an ideal solution because the intermolecular forces in the two pure liquids are almost identical in both kind and magnitude. Consequently, the change in enthalpy on solution formation is essentially zero (\(ΔH_{soln} ≈ 0\)), which is one of the defining characteristics of an ideal solution. Ideal solutions and ideal gases are both simple models that ignore intermolecular interactions. Most real solutions, however, do not obey Raoult’s law precisely, just as most real gases do not obey the ideal gas law exactly. Real solutions generally deviate from Raoult’s law because the intermolecular interactions between the two components A and B differ. We can distinguish between two general kinds of behavior, depending on whether the intermolecular interactions between molecules A and B are stronger or weaker than the A–A and B–B interactions in the pure components. If the A–B interactions are stronger than the A–A and B–B interactions, each component of the solution exhibits a lower vapor pressure than expected for an ideal solution, as does the solution as a whole. The favorable A–B interactions effectively stabilize the solution compared with the vapor. This kind of behavior is called a negative deviation from Raoult’s law. Systems stabilized by hydrogen bonding between two molecules, such as acetone and ethanol, exhibit negative deviations from Raoult’s law. Conversely, if the A–B interactions are weaker than the A–A and B–B interactions yet the entropy increase is enough to allow the solution to form, both A and B have an increased tendency to escape from the solution into the vapor phase. The result is a higher vapor pressure than expected for an ideal solution, producing a positive deviation from Raoult’s law. In a solution of \(CCl_4\) and methanol, for example, the nonpolar \(CCl_4\) molecules interrupt the extensive hydrogen bonding network in methanol, and the lighter methanol molecules have weaker London dispersion forces than the heavier \(CCl_4\) molecules. Consequently, solutions of \(CCl_4\) and methanol exhibit positive deviations from Raoult’s law. For each system, compare the intermolecular interactions in the pure liquids and in the solution to decide whether the vapor pressure will be greater than that predicted by Raoult’s law (positive deviation), approximately equal to that predicted by Raoult’s law (an ideal solution), or less than the pressure predicted by Raoult’s law (negative deviation). : identity of pure liquids : predicted deviation from Raoult’s law : Identify whether each liquid is polar or nonpolar, and then predict the type of intermolecular interactions that occur in solution. : For each system, compare the intermolecular interactions in the pure liquids with those in the solution to decide whether the vapor pressure will be greater than that predicted by Raoult’s law (positive deviation), approximately equal to that predicted by Raoult’s law (an ideal solution), or less than the pressure predicted by Raoult’s law (negative deviation): approximately equal positive deviation (vapor pressure greater than predicted) negative deviation (vapor pressure less than predicted) Finding Vapor Pressure of a Solution (Nonionic-Volatile Solute): The vapor pressure of the solution is proportional to the mole fraction of solvent in the solution, a relationship known as . Solutions that obey Raoult’s law are called ideal solutions. Most real solutions exhibit positive or negative deviations from Raoult’s law. ) | 11,485 | 2,551 |
Subsets and Splits
No saved queries yet
Save your SQL queries to embed, download, and access them later. Queries will appear here once saved.