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https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Le_Chateliers_Principle/The_Contact_Process |
The Contact Process is used in the manufacture of sulfuric acid. This Modules explain the reasons for the conditions used in the process by considering the effect of proportions, temperature, pressure and catalyst on the composition of the equilibrium mixture, the rate of the reaction and the economics of the process. The Contact Process: \[ S_{(s)} + O_2 (g) \rightarrow SO_{2\; (g)} \label{1}\] \[4FeS_2 (s) + 11O_2 (g) \rightarrow 2Fe_2O_3 (s) + 8SO_2 (g) \label{2}\] \[ H_2SO_{4 (l)} + SO_{3(g)} \rightarrow H_2S_2O_{7 (l)} \label{4}\] e fuming sulfuric acid. \[ H_2S_2O_{7 (l)} + H_2O_{(l)} \rightarrow 2H_2SO_{4 (l)} \label{5}\] \[2SO_{2 (g)} + O_{2(g)} \rightleftharpoons 2SO_{3 (g)} \;\;\; \Delta{H}=-196\;kJ/mol\] \[2SO_{3 (g)} + O_{2 (g)} \rightleftharpoons 2SO_{3(g)} \;\;\; \Delta{H} = -196\, kJ/mol\] \[ 2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)} \;\;\; \Delta H = -196\;kJ/mol\] â Jim Clark ( ) | 933 | 1,798 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/14%3A_Organohalogen_and_Organometallic_Compounds/14.12%3A_Organomagnesium_Compounds |
For many years the most important organometallic compounds for synthetic purposes have been the organomagnesium halides, or . They are named after Victor Grignard, who discovered them and developed their use as synthetic reagents, for which he received a Nobel Prize in 1912. As already mentioned, these substances customarily are prepared in dry ether solution from magnesium turnings and an organic halide: Chlorides often react sluggishly and, in addition, may give an unwelcome precipitate of magnesium chloride, which, unlike magnesium bromide and iodide, is only very slightly soluble in ether. Organomagnesium fluorides eluded preparation until quite recently. Although we usually write the structure of a Grignard reagent as \(\ce{RMgX}\), in which \(\ce{X}\) is a halogen, the structure of the reagent in ether solution is more complex. There is a rapidly established equilibrium between the organomagnesium halide \(\left( \ce{RMgX} \right)\) and the corresponding dialkylmagnesium \(\left( \ce{RMgR} \right)\): \[2 \ce{RMgX} \rightleftharpoons \ce{R_2Mg} + \ce{MgX_2}\] Both of these species, \(\ce{RMgX}\) and \(\ce{R_2Mg}\), are reactive, and in ether solvents are solvated by coordination of the ether oxygen to magnesium. They further associate as dimers or higher polymers in solution. Although it is an oversimplification to regard a Grignard reagent as \(\ce{RMgX}\), most of the reactions can be rationalized easily by this simple structure. and (1977) | 1,491 | 1,799 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Heterogeneous_Equilibria/Special_Equilibria |
Chemical Equilibrium: Previous page: Mass Action Law We treat and as special chemical equilibria. The expressions for equilibrium constants of these systems are discussed. The mass action law is valid for the case when the solvent is a reactant or product. However, due to the large amount of solvent present, the equilibrium constant expression can be simplified. Special consideration also applies to heterogeneous equilibria, in which solids or liquids are involved. Phases such as liquid and solid are not sensitive to pressure, and their "concentrations" are constant as long as these phases are present. Their "concentrations" are not defined, and that is why we use a " " to mark the special meaning of these phrases. Perhaps "activity" or "property" is a better term to use than "concentration" in these cases. As a solvent in the system, the concentration of the solvent is so large that its concentration rarely changes during the course of the reaction. For example, when water is used as a solvent, the concentration of water is almost always a constant. Its concentration can be calculated to be 55.6 M. (1 L of water has 1000 g, and molecular weight 18 g/mol; and thus \(\mathrm{[H_2O] = \dfrac{1000}{18} = 55.6\: mol/L}\)) For example, consider the following reaction, \[\ce{CH3COOH + C2H5OH \rightleftharpoons CH3COOC2H5 + H2O}\] According to the Mass Action Law, we can use either \[K' = \mathrm{\dfrac{[CH_3COOC_2H_5] {\color{Red} [H_2O]}}{[CH_3COOH] [C_2H_5OH]}}\] or \[K = \ce{\dfrac{[CH3COOC2H5]}{[CH3COOH] [C2H5OH]}}\] Of course, \[K = \mathrm{\dfrac{\mathit K\,' }{ {\color{Red} [H_2O]}}} = \dfrac{K' }{ {\color{Red} 55.6}}\] Heterogeneous reactions involve at least two phases. That is two of gas, liquid, and solid are present as reactants or products. In heterogeneous equilibria, the activities (or concentrations) of solid and liquid but not gas are always a constant. In these cases, their activities or concentrations are omitted in the expression of Q or K. Let us look at these expressions for the equilibria: \(\begin{align}
\ce{CaCO_{3\large{(s)}} &\rightleftharpoons CaO_{\large{(s)}} + CO_{2\large{(g)}}} &&K_{\ce p} = \ce{P(CO2)}\\
\ce{2 H2O_{\large{(l)}} &\rightleftharpoons 2 H2O_{\large{(g)}}} &&K_{\ce p} = \ce{P^2(H2O)}\\
\ce{2 H2O_{\large{(l)}} &\rightleftharpoons 2 H_{2\large{(g)}} + O_{2\large{(g)}}} &&K_{\ce p} = \ce{P^2(H2) P(O2)}\\
\mathrm{AgCl_{\large{(s)}}} &\mathrm{\rightleftharpoons \sideset{ }{_{\large{(aq)}}^{+}}{Ag} + \sideset{ }{_{\large{(aq)}}^{-}}{Cl}} &&K = \ce{[Ag^+] [Cl^- ]}
\end{align}\) Heterogeneous equilibria discusses this type of system in more detail. \(\ce{HgO_{\large{(s)}} \rightleftharpoons Hg_{\large{(s)}} + O_{2\large{(g)}}}\) \(\ce{H2O_{\large{(l)}} \rightleftharpoons H2O_{\large{(g)}}}\), the vapor pressure of water depends on temperature. At 298 K, the saturated vapor pressure is 23.8 torr. What is the equilibrium constant based on pressure in atm? (1 atm = 760 torr) \(\ce{CH3COOC2H5_{\large{(aq)}} + H2O_{\large{(l)}} \rightleftharpoons CH3COOH_{\large{(aq)}} + C2H5OH_{\large{(aq)}}}\) has an equilibrium constant of 0.10. If \(\mathrm{[CH_3COOC_2H_5] = 0.90\: M}\) in a system that is at equilibrium, what is \(\ce{[C2H5OH]}\)? \(\ce{NH4Cl_{\large{(s)}} \rightleftharpoons NH_{3\large{(g)}} + HCl_{\large{(g)}}}\). \(\begin{alignat}{2}
\ce{&CH3COOC2H5_{\large{(aq)}} + H2O_{\large{(l)}} \rightleftharpoons \; &&CH3COOH_{\large{(aq)}} +\; &&C2H5OH_{\large{(aq)}}}\\
&\hspace{55px}0.9 &&\hspace{40px}x &&\hspace{30px}x
\end{alignat}\) \(\dfrac{x^2}{0.90} = K = 0.1\); \(x =\: ?\) What was the original \(\ce{[CH3COOC2H5]}\) before any hydration reaction takes place? (Answer, 1.2) Consider...
\(\ce{P(HCl)} = \sqrt{K_{\ce p}}\) \(\begin{alignat}{2}
\ce{NH4Cl_{\large{(s)}} \rightleftharpoons \; &NH_{3\large{(g)}} + \; &&HCl_{\large{(g)}}}\\
&\:\:\:x &&\:\:\:x
\end{alignat}\) \(K_{\ce p} = x^2\) | 3,953 | 1,800 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Calorimetry/Constant_Volume_Calorimetry |
Make sure you thoroughly understand the following essential concept: Constant Volume Calorimetry, also know as bomb calorimetry, is used to measure the heat of a reaction while holding volume constant and resisting large amounts of pressure. Although these two aspects of bomb calorimetry make for accurate results, they also contribute to the difficulty of bomb calorimetry. In this module, the basic assembly of a bomb calorimeter will be addressed, as well as how bomb calorimetry relates to the heat of reaction and heat capacity and the calculations involved in regards to these two topics. is used to measure quantities of heat, and can be used to determine the heat of a reaction through experiments. Usually a is used since it is simpler than a bomb calorimeter, but to measure the heat evolved in a combustion reaction, constant volume or bomb calorimetry is ideal. A constant volume calorimeter is also more accurate than a coffee-cup calorimeter, but it is more difficult to use since it requires a well-built reaction container that is able to withstand large amounts of pressure changes that happen in many chemical reactions. Most serious calorimetry carried out in research laboratories involves the determination of \(\Delta H_{combustion}\), since these are essential to the determination of standard enthalpies of formation of the thousands of new compounds that are prepared and characterized each month. In a constant volume calorimeter, the system is sealed or isolated from its surroundings, which accounts for why its volume is fixed and there is no volume-pressure work done. A bomb calorimeter structure consists of the following: Since the process takes place at constant volume, the reaction vessel must be constructed to withstand the high pressure resulting from the combustion process, which amounts to a confined explosion. The vessel is usually called a “bomb”, and the technique is known as . The reaction is initiated by discharging a capacitor through a thin wire which ignites the mixture. Another consequence of the constant-volume condition is that the heat released corresponds to \(q_v\), and thus to the internal energy change \(ΔU\) rather than to \(ΔH\). The enthalpy change is calculated according to the formula \[ΔH = q_v + Δn_gRT\] where \(Δn_g\) is the change in the number of moles of gases in the reaction. A sample of biphenyl (\(\ce{(C6H5)2}\)) weighing 0.526 g was ignited in a bomb calorimeter initially at 25°C, producing a temperature rise of 1.91 K. In a separate calibration experiment, a sample of benzoic acid (\(\ce{C6H5COOH}\)) weighing 0.825 g was ignited under identical conditions and produced a temperature rise of 1.94 K. For benzoic acid, the heat of combustion at constant volume is known to be 3,226 kJ mol (that is, Δ = –3,226 kJ mol .) Use this information to determine the standard enthalpy of combustion of biphenyl. Begin by working out the calorimeter constant: \[\dfrac{0.825 g}{122.1 \;g/mol} = 0.00676\; mol \nonumber\] \[(0.00676\; mol) \times (3226\; kJ/mol) = 21.80\; kJ \nonumber\] \[\dfrac{21.80\; kJ}{1.94\; K} = 11.24\; kJ/K \nonumber\] Now determine \(ΔU_{combustion}\) of the biphenyl ("BP"): \[\dfrac{0.526\; g}{154.12\; g/mol} = 0.00341 \; mol \nonumber\] \[(1.91\; K) \times (11.24\; kJ/K) = 21.46\; kJ \nonumber\] \[\dfrac{21.46\; kJ}{0.00341\; mol} = 6,293\; kJ/mol \nonumber\] \[ΔU_{combustion} (BP) = –6,293\; kJ/mol \nonumber\] This is the heat change at constant volume, \(q_v\); the negative sign indicates that the reaction is exothermic, as all combustion reactions are. From the balanced reaction equation \[\ce{(C6H5)2(s) + 29/2 O2(g) \rightarrow 12 CO2(g) + 5 H2O(l)} \nonumber\] we can calculate the change in the moles of gasses for this reaction \[Δn_g = 12 - \frac{29}{2} = \frac{-5}{2} \nonumber\] Thus the volume of the system when the reaction takes place. Converting to \(ΔH\), we can write the following equation. Additionally, recall that at constant volume, \(ΔU = q_V\). \[ \begin{align*} ΔH &= q_V + Δn_gRT \\[4pt] &= ΔU -\left( \dfrac{5}{2}\right) (8.314\; J\; mol^{-1}\; K^{-1}) (298 \;K) \\[4pt] &= (-6,293 \; kJ/mol)–(6,194\; J/mol) \\[4pt] &= (-6,293-6.2)\;kJ/mol \\[4pt] &= -6299 \; kJ/mol \end{align*} \] The amount of heat that the system gives up to its surroundings so that it can return to its initial temperature is the . The heat of reaction is just the negative of the thermal energy gained by the calorimeter and its contents (\(q_{calorimeter}\)) through the combustion reaction. \[q_{rxn} = -q_{calorimeter} \label{2A}\] where \[q_{calorimeter} = q_{bomb} + q_{water} \label{3A}\] If the constant volume calorimeter is set up the same way as before, (same steel bomb, same amount of water, etc.) then the heat capacity of the calorimeter can be measured using the following formula: \[q_{calorimeter} = \text{( heat capacity of calorimeter)} \times \Delta{T} \label{4A}\] Heat capacity is defined as the amount of heat needed to increase the temperature of the entire calorimeter by 1 °C. The equation above can also be used to calculate \(q_{rxn}\) from \(q_{calorimeter}\) calculated by Equation \ref{2A}. The heat capacity of the calorimeter can be determined by conducting an experiment. 1.150 g of sucrose goes through combustion in a bomb calorimeter. If the temperature rose from 23.42 °C to 27.64 °C and the heat capacity of the calorimeter is 4.90 kJ/°C, then determine the heat of combustion of sucrose, \(\ce{C12H22O11}\) (in kJ per mole of \(\ce{C12H22O11}\)). Given: Using Equation \ref{4A} to calculate \(q_{calorimeter}\): \[\begin{align*} q_{calorimeter} &= (4.90\; kJ/°C) \times (27.64 - 23.42)°C \\[4pt] &= (4.90 \times 4.22) \;kJ = 20.7\; kJ \end{align*}\] Plug into Equation \ref{2A}: \[\begin{align*} q_{rxn} &= -q_{calorimeter} \\[4pt] &= -20.7 \; kJ \; \end{align*}\] But the question asks for kJ/mol \(\ce{C12H22O11}\), so this needs to be converted: \[\begin{align*}q_{rxn} &= \dfrac{-20.7 \; kJ}{1.150 \; g \; C_{12}H_{22}O_{11}} \\[4pt] &= \dfrac{-18.0 \; kJ}{g\; C_{12}H_{22}O_{11}} \end{align*}\] Convert to per Mole \(\ce{C12H22O11}\): \[\begin{align*}q_{rxn} &= \dfrac{-18.0 \; kJ}{\cancel{g \; \ce{C12H22O11}}} \times \dfrac{342.3 \; \cancel{ g \; \ce{C12H22O11}}}{1 \; mol \; \ce{C12H22O11}} \\[4pt] &= \dfrac{-6.16 \times 10^3 \; kJ \;}{mol \; \ce{C12H22O11}} \end{align*}\] Although calorimetry is simple in principle, its practice is a highly exacting art, especially when applied to processes that take place slowly or involve very small heat changes, such as the germination of seeds. Calorimeters can be as simple as a foam plastic coffee cup, which is often used in student laboratories. Research-grade calorimeters, able to detect minute temperature changes, are more likely to occupy table tops, or even entire rooms: The is an important tool for measuring the heat capacities of liquids and solids, as well as the heats of certain reactions. This simple yet ingenious apparatus is essentially a device for measuring the change in volume due to melting of ice. To measure a heat capacity, a warm sample is placed in the inner compartment, which is surrounded by a mixture of ice and water. The heat withdrawn from the sample as it cools causes some of the ice to melt. Since ice is less dense than water, the volume of water in the insulated chamber decreases. This causes an equivalent volume of mercury to be sucked into the inner reservoir from the outside container. The loss in weight of this container gives the decrease in volume of the water, and thus the mass of ice melted. This, combined with the heat of fusion of ice, gives the quantity of heat lost by the sample as it cools to 0°C. | 7,702 | 1,801 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Solubilty/Temperature_Effects_on_Solubility |
The solubility of solutes is dependent on temperature. When a solid dissolves in a liquid, a change in the physical state of the solid analogous to melting takes place. Heat is required to break the bonds holding the molecules in the solid together. At the same time, heat is given off during the formation of new solute -- solvent bonds. : Temperature dependent solubilities of three salts in water. The use of first-aid instant cold packs is an application of this solubility principle. A salt such as ammonium nitrate is dissolved in water after a sharp blow breaks the containers for each. \[NH_4NO_{3(s)} \rightarrow NH_{4(aq)}^+ + NO^-_{3(aq)}\] The dissolving reaction is endothermic and requires heat. Therefore the heat is drawn from the surroundings and the pack feels cold. | 805 | 1,802 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Complex_Ion_Chemistry/Ligand_Exchange_Reactions_(Introduction) |
This page describes some common ligand exchange (substitution) reactions involving complex metal ions. It assumes that you are familiar with basic ideas about complex ions. A involves the substitution of one or more ligands in a complex ion with one or more different ligands. If you add concentrated hydrochloric acid to a solution containing hexaaquacobalt(II) ions (for example, cobalt(II) chloride solution), the solution turns from its original pink color to a dark rich blue. The six aqua molecules are replaced by four chloro ions. The reaction taking place is reversible. \[ \ce{[Co(H2O)6]^{2+} + 4Cl^{-} <=> [CoCl4]^{2-} + 6H2O} \nonumber \] Concentrated hydrochloric acid is used as the source of chloride ions because it provides a very high concentration compared to what is possible with, say, sodium chloride solution. Concentrated hydrochloric acid has a chloride ion concentration of approximately 10 mol dm . The high chloride ion concentration pushes the position of the equilibrium to the right according to Le Chatelier's Principle. Notice the change in the co-ordination of the cobalt. Chloride ions are bigger than water molecules, and there isn't room to fit six of them around the central cobalt ion. This reaction can be easily reversed by adding water to the solution. Adding water to the right-hand side of the equilibrium has the effect of moving the position of equilibrium to the left. The pink color of the hexaaquacobalt(II) ion is produced again (only paler, of course, because it is more dilute). In terms of the chemistry, this is exactly the same as the last example - all that differs are the colors. Unfortunately, these are not quite so straightforward. The color of the tetrachlorocuprate(II) ion is almost always seen mixed with that of the original hexaaqua ion. What you normally see is: The reaction taking place is reversible, and you get a mixture of colors due to both of the complex ions. \[\ce{[Cu(H2O)6]^{2+} <=> [CuCl_4]^{2-} + 6H2O}\nonumber \] You may find the color of the tetrachlorocuprate(II) ion variously described as olive-green or yellow. Adding water to the green solution, replaces the chloro ligands by water molecules again, and returns the solution to blue. Water molecules and ammonia molecules are very similar in size, and so there is no change in co-ordination this time. Unfortunately, the reactions aren't quite so straightforward to describe. Ammonia solution can react with hexaaqua metal ions in two quite distinct ways, because it can act as a base as well as a ligand. If you add a small amount of ammonia solution you get precipitates of the metal hydroxide - the ammonia is acting as a base. In some cases, these precipitates redissolve when you add more ammonia to give solutions in which a ligand exchange reaction has occurred. In the diagrams below, both steps are shown, but we are only going to consider the chemistry of the overall ligand exchange reaction. The precipitates dissolve because of a complicated series of equilibrium shifts, and we shan't worry about that for the moment. This is a slightly untypical case, because only four of the six water molecules get replaced to give the tetraamminediaquacopper(II) ion, [Cu(NH ) (H O) ] . Notice that the four ammonias all lie in one plane, with the water molecules above and below. What you see in a test tube is: The main equilibrium involved in the ligand exchange reaction is: \[\ce{[Cu(H2O)6]^{2+} + 4NH3 <=> [Cu(NH3)4(H2O)2]^{2+} + 4H2O}\nonumber \] The color of the deep blue complex is so strong that this reaction is used as a sensitive test for copper(II) ions in solution. Even if you try to reverse the change by adding large amounts of water to the equilibrium, the strength of the deep blue (even highly diluted) always masks the pale blue of the aqua ion. This time, all the water molecules get replaced. \[\ce{[Co(H2O)6]^{2+} + 6NH3 <=> [Co(NH3)6]^{2+} + 6H2O}\nonumber \] The straw colored solution formed changes color very rapidly on standing to a deep reddish brown. The hexaamminecobalt(II) ions are oxidised by the air to hexaamminecobalt(III) ions. However, that is a quite separate reaction, and is not a part of the ligand exchange reaction. Again, all the agua ligands are replaced by ammine ligands. The difference this time is that the reaction is incomplete. The precipitate has to be left to stand in the presence of excess concentrated ammonia solution for some time to get the fully substituted ammine complex. Even so, you still get left with some unreacted precipitate. \[ \ce{ [Cr(H2O)6 ]^{3+} + 6NH3 <=> [Cr(NH3)6]^{3+} + 6H2O}\nonumber \] The color of the hexaaquachromium(III) ion has been shown as a "difficult to describe" violet-blue-grey in all the diagrams above. In practice, when it is produced during a reaction in a test tube, it is often green. A typical example of this is the use of acidified potassium dichromate(VI) as an oxidising agent. Whenever this is used, the orange solution turns green and we nearly always describe the green ion as being Cr - implying the hexaaquachromium(III) ion. That's actually an over-simplification. What happens is that one or more of the ligand water molecules get replaced by a negative ion in the solution - typically sulfate or chloride. You can do this simply by warming some chromium(III) sulfate solution. One of the water molecules is replaced by a sulfate ion. Notice the change in the charge on the ion. Two of the positive charges are canceled by the presence of the two negative charges on the sulfate ion. In the presence of chloride ions (for example with chromium(III) chloride), the most commonly observed color is green. This happens when two of the water molecules are replaced by chloride ions to give the tetraaquadichlorochromium(III) ion - [Cr(H O) Cl ] . Once again, notice that replacing water molecules by chloride ions changes the charge on the ion. This provides an extremely sensitive test for iron(III) ions in solution. If you add thiocyanate ions, SCN , (from, say, sodium or potassium or ammonium thiocyanate solution) to a solution containing iron(III) ions, you get an intense blood red solution containing the ion \(\ce{[Fe(SCN)(H2O)5]^{2+}}\). Jim Clark ( ) | 6,235 | 1,803 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Concepts_in_Biophysical_Chemistry_(Tokmakoff)/06%3A_Dynamics_and_Kinetics/21%3A_Binding_and_Association/21.05%3A_Diffusion-Limited_Reactions |
The diffusion-limited association rate is typically approximated from the expression for the relative diffusion of A and B with an effective diffusion constant D = D + D to within a critical encounter radius R = R + R , as described earlier. \[ k_a = 4\pi R_0 f(D_A + D_B) \] One can approximate association rates between two diffusing partners using the Stokes–Einstein expression: \(D_A = k_BT/6\pi \eta R_A \). For two identical spheres (i.e., dimerization) in water at T = 300 K, where η ~ 1 cP = 100 kg m s , \[ k_a = \dfrac{8k_BT}{3\eta} = 6.6 \times 10^9 M^{-1}s^{-1} \] Note that this model predicts that the association rate is not dependent on the size or mass of the object. For bimolecular reactions, the diffusion may also include those orientational factors that bring two binding sites into proximity. Several studies have investigated these geometric effects. During diffusive encounter in dilute solution, once two partners collide but do not react, there is a high probability of re-colliding with the same partner before diffusing over a longer range to a new partner. Depending on concentration and the presence of interaction potentials, there may be 5–50 microcollisions with the same partner before encountering a new partner. For the limit where associations are weak, k and k are fast and in equilibrium, and the dissociation is diffusion limited. Then we can calculate k \[A+B \underset{k_{-1}}{\stackrel{k_{1}}{\rightleftharpoons}} AB \] Now we consider boundary conditions for flux moving away from a sphere such that \[\begin{aligned} C_B(\infty) &= 0\\ C_B(R_0) &=\left( \dfrac{4}{3} \pi R^3_0 \right)^{-1} \end{aligned} \] The boundary condition for concentration at the surface of the sphere is written so that the number density is one molecule per sphere. The steady state distribution of B is found to be \[ C_B (r) = \dfrac{3}{4\pi R^2_0 r} \] The dissociation flux at the surface is \[ J = -D_B \left( \dfrac{\partial C_B}{\partial r} \right)_{r=R_0} = \dfrac{3D_B}{4\pi R_0^4} \] and the dissociation frequency is \[ \dfrac{J}{4\pi R_0^2} = \dfrac{3D_B}{R_0^2} \] When we also consider the dissociative flux for the other partner in the association reaction, \[ k_{-1} = k_d = 3(D_A +D_B) R_0^{-2} \] Written in a more general way for a system that may have an interaction potential \[k_d = \dfrac{4\pi De^{U(R_0)/ kT}}{\dfrac{4}{3}\pi R_0^3 \int^{\infty}_{R_0}e^{U(r)kT}r^{-2} dr} = 3DR^*R_0^{-3} \] Note that equilibrium constants do not depend on D for diffusion-limited association/dissociation \[ K_D = \dfrac{k_D}{k_A} = \dfrac{3DR_0^{-2}}{4\pi R_0D} = \dfrac{3}{4\pi R_0^3} \] Note this is the inverse of the volume of a sphere. | 2,693 | 1,804 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/28%3A_Photochemistry/28.03%3A_Organic_Photochemistry |
An extraordinary variety of reactions of organic compounds are known to occur under the influence of visible and ultraviolet light. Some of these, such as the photochemical halogenation of alkanes and photosynthesis in green plants, already have been discussed (see and ). It is not our purpose here to review organic photochemistry in detail - rather, we shall mention a few types of important photochemical reactions and show how these can be explained by the principles discussed in the preceding section. Compounds have very different chemical behavior in their excited states compared to their ground states. Not only is the energy much higher, but the molecular geometry and electronic configurations are different. Intuitively, we expect that excited states of molecules, in which two electrons occupy orbitals, would have substantial diradical character. This is the case, especially for triplet states, as we shall see. We have mentioned how chlorine molecules dissociate to chlorine atoms on absorption of near-ultraviolet light and thereby cause radical-chain chlorination of saturated hydrocarbons ( ). Photochemical chlorination is an example of a photochemical reaction that can have a high - that is, many molecules of chlorination product can be generated per quantum of light absorbed. The quantum yield of a reaction is said to be unity when \(1 \: \text{mol}\) of reactant is converted to product(s) per einstein\(^1\) of light absorbed. The symbol for quantum yield is usually \(\Phi\). 2-Propanone (acetone) vapor undergoes a photodissociation reaction with \(313\)-\(\text{nm}\) light with \(\Phi\) somewhat less than unity. Absorption of light by 2-propanone results in the formation of an excited state that has sufficient energy to undergo cleavage of a \(\ce{C-C}\) bond (the weakest bond in the molecule) and form a methyl radical and an ethanoyl radical. This is a photochemical reaction: \(\tag{28-1}\) The subsequent steps are dark reactions. At temperatures much above room temperature, the ethanoyl radical breaks down to give another methyl radical and carbon monoxide: \(\tag{28-2}\) If this reaction goes to completion, the principal reaction products are ethane and carbon monoxide: \[2 \ce{CH_3} \cdot \rightarrow \ce{CH_3-CH_3} \tag{28-3}\] If the ethanoyl radical does not decompose completely, then some 2,3-butanedione also is formed. This reaction is quite important at room temperature or below: \(\tag{28-4}\) Lesser amounts of methane and ketene also are formed as the result of disproportionation reactions involving hydrogen-atom transfers of the types we have encountered previously in radical reactions (see ): \(\tag{28-5}\) The product-forming reactions, Equations 28-2 through 28-5, all depend on the primary photochemical event, Equation 28-1, which breaks the \(\ce{C-C}\) bond to the carbonyl group. This cleavage has been termed a after the eminent photochemist, R. G. W. Norrish:\(^2\) Another photochemical reaction is important for ketones that have at least one \(\gamma\) hydrogen on a chain connected to the carbonyl group, as in In this pathway ( ), cleavage occurs at the \(\ce{C}_\alpha \ce{-C}_\beta\) bond to give, as the major product, a ketone of shorter chain length and an alkene. Thus for 2-pentanone: This reaction occurs in an interesting way. Whatever the nature of then \(n \rightarrow \pi^*\) excited state, \(S_1\) or \(T_1\), the primary photochemical reaction is the abstraction of a hydrogen atom from the \(\gamma\) carbon by the carbonyl oxygen to give the diradical, \(1\): The subsequent dark reactions readily are understood as typical of diradicals. Cleavage of \(1\) at \(\ce{C}_\alpha \ce{-C}_\beta\) gives ethene and an enol, which rearranges to the ketone. Alternatively, \(1\) can cyclize to a cyclobutanol: A variety of photodissociation reactions have been found to take place with ketones, but the products almost always can be explained as the result of Norrish type I and/or II cleavage. Examples are: Diaryl ketones do not undergo photodissociation in the same way as alkyl ketones, probably because cleavage to phenyl and other aryl radicals is unfavorable (Table 4-6). Nevertheless, aromatic ketones are photochemically reactive in the presence of compounds that can donate a hydrogen atom, with the result that the carbonyl group is reduced. Indeed, one of the classic photochemical reactions of organic chemistry is the formation of 1,1,2,2-tetraphenyl-1,2-ethanediol (\(3\), benzopinacol) by the action of light on a solution of diphenylmethanone (\(2\), benzophenone) in isopropyl alcohol. The yield is quantitative. The light is absorbed by \(2\) and the resulting activated ketone, \(2^*\), removes a hydrogen from isopropyl alcohol: Benzopinacol results from dimerization of the radicals, \(4\): Since the quantum yields of 2-propanone and benzopinacol both are nearly unity when the light intensity is not high, it is clear that two of the radicals, \(4\), must be formed for each molecule of \(2\) that becomes activated by light. This is possible if the 2-hydroxy-2-propyl radical formed by Equation 28-7 reacts with \(2\) to give a second diphenylhydroxymethyl radical: This reaction is energetically favorable because of the greater possibility for delocalization of the odd electron in \(4\) than in the 2-hydroxy-2-propyl radical. Photochemical formation of \(3\) also can be achieved from diphenylmethanone, \(2\), and diphenylmethanol, \(5\): The mechanism is similar to that for isopropyl alcohol as the reducing agent: This reduction is believed to involve the triplet state of \(2\) by the following argument: Formation of \(3\) is reasonably efficient even when the concentration of the alcohol, \(5\) is low; therefore, whatever the excited state of the ketone, \(2^*\), that accepts a hydrogen atom from \(5\), it must be a fairly long-lived one. Because solutions of \(2\) show no visible fluorescence, they must be converted rapidly to another state of longer life than the singlet \(\left( S_1 \right)\). The long-lived state is then most reasonably a triplet state. In fact, if naphthalene is added to the reaction mixture, formation of benzopinacol, \(3\) is drastically inhibited because the benzophenone triplet transfers energy to naphthalene more rapidly than it reacts with the alcohol, \(5\) (see ). An important problem in many syntheses is to produce the desired isomer of a cis-trans pair of alkenes. The problem would not arise if it were possible to isomerize the undesired isomer to the desired isomer. In many cases such isomerizations can be carried out photochemically. A typical example is afforded by and 1,2-diphenylethene (stilbene): Here the trans form is easily available by a variety of reactions and is more stable than the cis isomer because it is less sterically hindered. However, it is possible to produce a mixture containing mostly cis isomer by irradiating a solution of the trans isomer in the presence of a suitable photosensitizer. This process in no way contravenes the laws of thermodynamics because the input of radiant energy permits the equilibrium point to be shifted from what it would be normally. Isomerization appears to occur by the following sequence: The sensitizer, usually a ketone such as benzophenone or 1-(2-naphthyl)ethanone, is raised by an \(n \rightarrow \pi^*\) transition from the singlet ground state \(\left( S_0 \right)\) to an excited state \(\left( S_1 \right)\) by absorption of light. Intersystem crossing then occurs rapidly to give the triplet state \(\left( T_1 \right)\) of the sensitizer: The next step is excitation of the alkene by energy transfer from the triplet state of the sensitizer. Remember, the net electron spin is conserved during energy transfer, which means that the alkene will be excited to the triplet state: \[^3\text{Sens}^* \left( \uparrow \uparrow \right) + ^1\text{Alkene} \left( \uparrow \downarrow \right) \rightarrow ^1\text{Sens} \left( \uparrow \downarrow \right) + ^3\text{Alkene}^* \left( \uparrow \uparrow \right)\] The triplet state of the alkene is most stable when the \(p\) orbitals, which make up the normal \(\pi\) system of the double bond, are not parallel to one another (Figure 6-17). Therefore, if the energy-transfer process leads initially to a planar triplet, this is converted rapidly to the more stable nonplanar form. The excitation of either the cis or the trans isomer of the alkene appears to lead to a common triplet state, as shown in Figure 28-4. The final step in the isomerization is decay of the alkene triplet to the ground state. This happens either by emission of light (phosphorescence) or by having the triplet energy converted to thermal energy without emission of light. Either way, the cis or trans isomer could be formed and, as can be seen from Figure 28-4, the ratio of isomers produced depends on the relative rates of decay of the alkene triplet to the ground-state isomers, \(k_c/k_t\). This ratio turns out to favor formation of the isomer. Therefore, provided both isomers can be photosensitized efficiently, sensitized irradiation of either one will lead ultimately to a mixture of both, in which the thermochemically less stable isomer predominates. The sensitizer must have a triplet energy in excess of the triplet energy of the alkene for energy transfer to occur, and the or equilibrium point is independent of the nature of the sensitizer when the latter transfers energy efficiently to cis and trans isomers. In the practical use of the sensitized photochemical equilibrium of cis and trans isomers, it is normally necessary to carry out pilot experiments to determine what sensitizers are useful. Another example of how photochemical isomerization can be used is provided by the equilibration of the \(E\) and \(Z\) form of 1-bromo-2-phenyl-1-propene: The \(E\) isomer is formed to the extent of \(95\%\) in the dehydrohalogenation of 1,2-dibromo-2-phenylpropane: Photoisomerization of the elimination product with 1-(2-naphthyl)ethanone as sensitizer produce a mixture containing \(85\%\) of the \(Z\) isomer. One may well ask why the isomerization of alkenes discussed in the preceding section requires a sensitizer. Why cannot the same result be achieved by direct irradiation? One reason is that a \(\pi \rightarrow \pi^*\) singlet excited state \(\left( S_1 \right)\) produced by direct irradiation of an alkene or arene crosses over to the triplet state \(\left( T_1 \right)\) inefficiently (compared to \(n \rightarrow \pi^*\) excitation of ketones). Also, the \(S_1\) state leads to reactions beside isomerization which, in the case of 1,2-diphenylethene and other conjugated hydrocarbons, produce cyclic products. For example, -1,2-diphenylethene irradiated in the presence of oxygen gives phenanthrene by the sequence of Equation 28-8. The primary photoreaction is cyclization to a dihydrophenanthrene intermediate, \(6\), which, in the presence of oxygen, is converted to phenanthrene: The cyclization step of Equation 28-8 is a photochemical counterpart of the electrocyclic reactions discussed in . Many similar photochemical reactions of conjugated dienes and trienes are known, and they are of great interest because, like their thermal relatives, they often are stereospecific but tend to exhibit stereochemistry opposite to what is observed for formally similar thermal reactions. For example, These reactions are \(4n\)-electron concerted processes controlled by the symmetry of the reacting orbitals. The thermal reaction is most favorable with a Mobius transition state (achieved by conrotation), whereas the photochemical reaction is most favorable with a Huckel transition state (achieved by disrotation). Conjugated dienes also undergo photochemical cycloaddition reactions. Related thermal cycloadditions of alkadienes have been discussed in , , and , but the thermal and photochemical reactions frequently give different cyclic products. Butadiene provides an excellent example of the differences: In the thermal reaction the [4 + 2] or Diels-Alder adduct is the major product, whereas in the photochemical reaction [2 + 2] cycloadditions dominate. Because the photochemical additions are sensitized by a ketone, \(\ce{C_6H_5COCH_3}\), these cycloadditions occur through the triplet state of 1,3-butadiene and, as a result, it is not surprising that these cycloadditions are stepwise, nonstereospecific, and involve diradical intermediates. Direct irradiation of 1,3-butadiene with \(254\)-\(\text{nm}\) light produces cyclobutene and small amounts of bicyclo[1.1.0]butane along with dimers. In contrast to conjugated dienes, simple alkenes such as 2-butene do not react easily by photosensitized cycloaddition. But they will form [2 + 2] cycloadducts on direct irradiation. These additions occur by way of a singlet excited state and are stereospecific: A related reaction, which has no precedent in thermal chemistry, is the cycloaddition of an alkene and an aldehyde or ketone to form an oxacyclobutane: In this kind of addition the ground-state alkene \(\left( S_0 \right)\) reacts with an excited state (usually \(T_1\)) of the carbonyl compound by way of a diradical intermediate: The ground state of molecular oxygen is unusual because it is a triplet state. Two electrons of parallel spin occupy separate \(\pi\) orbitals of equal energy (degenerate), as shown schematically in Figure 28-5,\(^3\) The next two higher electronic states both are singlet states and lie respectively \(24\) and \(37 \: \text{kcal mol}^{-1}\) above the ground state. From this we can understand why ordinary oxygen has the properties of a diradical and reacts rapidly with many radicals, as in the radical-chain oxidation of hydrocarbons (autoxidation; and ): Oxygen also efficiently quenches excited triplet states of other molecules \(\left( ^3A^* \right)\) and, in accepting triplet energy, is itself promoted to an excited state. Notice that the total spin orientation is conserved: Singlet oxygen is highly reactive toward many organic molecules and will form oxygenated addition or substitution products. As one example, conjugated dienes react with singlet oxygen to give peroxides by [4 + 2] cycloaddition. Because only singlet states are involved, this addition is quite analogous to thermal Diels-Alder reactions ( ): If the alkene or alkadiene has at least one hydrogen on the carbon adjacent to the double bond, reaction with singlet oxygen may give hydroperoxides. The mechanism of this reaction is related to [4 + 2] cycloadditions and is presumed to occur through a Huckel pericyclic transition state (see ): Many reactions of this type can be achieved by allowing the hydrocarbon to react with oxygen in the presence of a sensitizing dye that strongly absorbs visible light. The dyes most commonly used for this purpose include fluorescein (and its chlorinated analogs, eosin and rose bengal), methylene blue, and porphyrin pigments (such as chlorophyll). The overall process of photosensitized oxygenation of a substrate \(\left( \ce{A} \right)\) proceeds by the following steps: \[^1\text{Sens} \overset{h \nu}{\longrightarrow} \: ^1\text{Sens}^* \longrightarrow ^3\text{Sens}^*\] \[^3\text{Sens}^* + ^3\ce{O_2} \longrightarrow ^1\text{Sens} + ^1\ce{O_2}*\] \[^1\ce{O_2}* + \ce{A} \longrightarrow \ce{AO_2}\] Singlet oxygen can be produced chemically as well as by photochemical sensitization. There are several chemical methods available, one of the best known being the reaction of sodium hypochlorite with peroxide: \[\ce{NaOCl} + \ce{H_2O_2} \rightarrow \ce{NaCl} + \ce{H_2O} + ^1\ce{O_2} \tag{28-9}\] An alternative method of formation, which can be used in organic solvents at low temperatures, involves the thermal decomposition of triethyl phosphite ozonide (Equation 28-10): Regardless of whether singlet oxygen is formed chemically or photochemically, it gives similar products in reactions with alkenes. Photosensitized reactions of oxygen are largely damaging to living organisms. Indeed, singlet oxygen reacts destructively with amino acids, proteins, and nucleic acids. How does an organism protect itself against the damaging effects of oxygen? There are no simple answers, but green plants provide a clue. Chlorophyll is an excellent sensitizing dye for singlet oxygen; yet green plants evidently are not harmed because of it. A reason may be that singlet oxygen is quenched very efficiently by other plant pigments, especially the carotenoid pigments such as \(\beta\)-carotene ( ). That this is the case is indicated by the fact that mutant plants unable to synthesize carotene are killed rapidly by oxygen and light. That direct irradiation with ultraviolet light is damaging to single-cell organisms is well known. It also is known that the nucleic acids, DNA and RNA, are the important targets of photochemical damage, and this knowledge has stimulated much research in the field of photobiology in the hope of unravelling the chemistry involved. An interesting and significant outcome is the finding that the pyrimidine bases of nucleic acids (uracil, thymine, and cytosine) are photoreactive and undergo on irradiation with ultraviolet light. Thymine, for example, gives a dimer of structure \(7\): Comparable experiments with the nucleic acids have confirmed that cycloaddition of their pyrimidine bases also occurs with ultraviolet light and effectively cross-links the chains, a process obviously quite inimical to the functioning of the DNA ( ). A remarkable and not well understood aspect of photobiology is the repair and defense mechanism both plants and animals possess to minimize the damaging effects of radiation. On the positive side, there are photochemical reactions that are essential for human health. One of these is the formation of vitamin D (the antirachitic vitamin) by irradiation of ergosterol. This photochemical reaction is an electrocyclic ring opening of the cyclohexadiene ring of ergosterol of the type described in . The product, previtamin D , subsequently rearranges thermally to vitamin D : \(^1\)The einstein unit is defined in . \(^2\)Recipient with G. Porter of the Nobel Prize in chemistry in 1967 for work on photochemical reactions. \(^3\)For a more detailed account of the electronic configuration of molecular oxygen, see M. Orchin and H. H. Jaffe, , Houghton Mifflin Co., Boston, 1967; or H. B. Gray, , W. A. Benjamin, Inc., Menlo Park, Calif., 1973. and (1977) | 18,493 | 1,805 |
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The energies of electrons in molecular orbitals can be observed directly by measuring the ionization energy. This is the energy required to remove an electron, in this case, from a molecule: \[H_2 (g) \rightarrow H_2^+(g) + e^-(g)\] The measured ionization energy of H is 1488 kJ mol . This number is primarily important in comparison to the ionization energy of a hydrogen atom, which is 1312 kJ mol . Therefore, it requires more energy to remove an electron from the hydrogen molecule than from the hydrogen atom; the electron therefore has a lower energy in the molecule. To pull the atoms apart, the energy of the electron must be increased. Hence, energy is required to break the bond, and the molecule is bound. A bond is formed when the energy of the electrons in the molecule is lower than the energy of the electrons in the separated atoms. This conclusion is consistent with the view of shared electrons in bonding molecular orbitals. As a second example, consider the nitrogen molecule, \(N_2\). The ionization energy of molecular nitrogen is 1503 kJ mol , and that of atomic nitrogen is 1402 kJ mol . Once again, the energy of the electrons in molecular nitrogen is lower than that of the electrons in the separated atoms, so the molecule is bound. As a third example, consider fluorine, \(F_2\). In this case, find that the ionization energy of molecular fluorine is 1515 kJ mol , which is smaller than the ionization energy of a fluorine atom, 1681 kJ mol . This seems inconsistent with the bonding orbital concept developed above, which states that the electrons in the bond have a lower energy than in the separated atoms. If the electron being ionized has a higher energy in \(F_2\) than in F, why is \(F_2\) a stable molecule? A more complete description of the molecular orbital concept of chemical bonding is required. Bond energies are next considered. Recall that the bond energy (or bond strength) is the energy required to separate the bonded atoms. The bond energy of \(N_2\) is 956 kJ mol ; this is much larger than the bond energy of H , 458 kJ mol , and of \(F_2\), which is 160 kJ mol . The unusually strong bond in nitrogen can be explained using both the valence shell electron pair sharing model and electron orbital descriptions. A nitrogen atom has three unpaired electrons in its valence shell; the three 2p electrons distribute themselves over the three 2p orbitals, each oriented along a different axis. Each of these unpaired electrons is available for sharing with a second nitrogen atom. The result, from valence shell electron pair sharing concepts, is that three pairs of electrons are shared between two nitrogen atoms in a “triple bond.” Intuition suggests that the triple bond in \(N_2\) should be much stronger than the single bond in H or in \(F_2\). Now consider the molecular orbital description of bonding in \(N_2\). Each of the three 2p atomic orbitals in each nitrogen atom must overlap to form a bonding molecular orbital, to accommodate three electron pairs. Each 2p orbital is oriented along a single axis. One 2p orbital from each atom is oriented in the direction of the other atom, that is, along the bond axis. When these two atomic orbitals overlap, they form a molecular orbital with cylindrical symmetry and is therefore a σ orbital (a σ* orbital is also formed). The two electrons are paired in the bonding orbital. This effect is depicted in Fig. 5 in a molecular orbital energy diagram. Each pair of atomic orbitals, one from each atom, overlaps to form a bonding and an anti-bonding orbital. The three 2p orbitals from each atom form one σ and σ* pair and two π and pairs. The lowering of the energies of the electrons in the σand π orbitals is apparent. The ten n=2 electrons from the nitrogen atoms are then placed pairwise, in order of increasing energy, into these molecular orbitals. Note that, in agreement with the , each pair in a single orbital consists of one spin up and one spin down electron. Recall now that the discussion of bonding was begun with \(N_2\) because of the curious result that the ionization energy of an electron in \(F_2\) is less than that of an electron in an F atom. Comparing the molecular orbital energy level diagrams for \(N_2\) and \(F_2\) allows an explanation for this puzzle. There are five p electrons in each fluorine atom. These ten electrons must be distributed over the molecular orbitals whose energies are shown in Fig. 6. (Note that the ordering of the bonding 2p orbitals differ between \(N_2\) and \(F_2\).) Two electrons are placed in the σ orbital, four more in the two π orbitals, and four more in the two orbitals. Overall, there are six electrons in bonding orbitals and four in anti-bonding orbitals. Because \(F_2\) is a stable molecule, we must conclude that the lowering of energy for the electrons in the bonding orbitals is greater than the raising of energy for the electrons in the antibonding orbitals. Overall, this distribution of electrons is, net, equivalent to having two electrons paired in a single bonding orbital. This also explains why the ionization energy of \(F_2\) is less than that of an F atom. The electron with the highest energy requires the least energy to remove from the molecule or atom. The molecular orbital energy diagram in Fig. 6 clearly shows that the highest energy electrons in \(F_2\) are in anti-bonding orbitals. Therefore, one of these electrons is easier to remove than an electron in an atomic 2p orbital, because the energy of an antibonding orbital is higher than that of the atomic orbitals, and the system is stabilized by the electron's removal. Therefore, the ionization energy of molecular fluorine is less than that of atomic fluorine. This clearly demonstrates the physical reality and importance of the antibonding orbitals. A particularly interesting case is the oxygen molecule, \(O_2\). In completing the molecular orbital energy level diagram for oxygen, the last two electrons must either be paired in the same 2p orbital or separated into different 2p orbitals. To determine which, it is important to note that oxygen molecules are paramagnetic—they are strongly attracted to a magnetic field. To account for this paramagnetism, recall that electron spin is a magnetic property. In most molecules, all electrons are paired, so for each “spin up” electron there is a “spin down” electron and their magnetic fields cancel out. If all electrons are paired, the molecule is diamagnetic, meaning that it responds only weakly to a magnetic field. If the electrons are not paired, they can adopt the same spin in the presence of a magnetic field. This accounts for the attraction of the paramagnetic molecule to the magnetic field. Therefore, for a molecule to be paramagnetic, it must have unpaired electrons. The correct molecular orbital energy level diagram for an \(O_2\) molecule is shown in Fig. 7. In comparing these three diatomic molecules, recall that \(N_2\) has the strongest bond, followed by \(O_2\) and \(F_2\). The comparative bond strengths were previously accounted for with Lewis structures, showing that \(N_2\) is a triple bond, \(O_2\) is a double bond, and \(F_2\) is a single bond. The molecular orbital energy level diagrams in Figs. 5 to 7 add more perspective to this analysis. Note that, in each case, the number of bonding electrons in these molecules is eight. The difference in bonding is entirely due to the number of antibonding electrons: 2 for \(N_2\), 4 for \(O_2\), and six for \(F_2\). Thus, the strength of a bond must be related to the relative numbers of bonding and antibonding electrons in the molecule. Therefore, the is defined as the following: \[\text{Bond order} = \frac{1}{2} (\# \ of\ bonding\ electrons - \# \ of\ antibonding\ electrons) \] Note that, defined this way, the bond order for \(N_2\) is 3, for \(O_2\) is 2, and for \(F_2\) is 1, which agrees with the conclusions made from Lewis structures. In conclusion, the relative strengths of bond can be predicted by comparing bond orders. | 8,056 | 1,806 |
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The internal energy of a system is identified with the random, disordered motion of molecules; the total (internal) energy in a system includes potential and kinetic energy. This is contrast to external energy which is a function of the sample with respect to the outside environment (e.g. kinetic energy if the sample is moving or potential energy if the sample is at a height from the ground etc). The symbol for Internal Energy Change is\( ΔU\). Energy on a smaller scale One gram of water at zero °Celsius compared with one gram of copper at zero °Celsius do NOT have the same internal energy because even though their kinetic energies are equal, water has a much higher potential energy causing its internal energy to be much greater than the copper's internal energy. The where is heat and is work An cannot exchange heat or work with its surroundings making the change in internal energy equal to zero. Energy is Conserved = The signs of internal energy | 995 | 1,807 |
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Aleksander Jablonski was a Polish academic who devoted his life to the study of molecular absorbance and emission of light. He developed a written representation that generally shows a portion of the possible consequences of applying photons from the visible spectrum of light to a particular molecule. These schematics are referred to as Jablonski diagrams. A Jablonski diagram is basically an energy diagram, arranged with energy on a vertical axis. The energy levels can be quantitatively denoted, but most of these diagrams use energy levels schematically. The rest of the diagram is arranged into columns. Every column usually represents a specific spin multiplicity for a particular species. However, some diagrams divide energy levels within the same spin multiplicity into different columns. Within each column, horizontal lines represent eigenstates for that particular molecule. Bold horizontal lines are representations of the limits of electronic energy states. Within each electronic energy state are multiple vibronic energy states that may be coupled with the electronic state. Usually only a portion of these vibrational eigenstates are represented due to the massive number of possible vibrations in a molecule. Each of these vibrational energy states can be subdivided even further into rotational energy levels; however, typical Jablonski diagrams omit such intense levels of detail. As electronic energy states increase, the difference in energy becomes continually less, eventually becoming a continuum that can be approach with classical mechanics. Additionally, as the electronic energy levels get closer together, the overlap of vibronic energy levels increases. Through the use of straight and curved lines, these figures show transitions between eigenstates that occur from the exposure of a molecule to a particular wavelength of light. Straight lines show the conversion between a photon of light and the energy of an electron. Curved lines show transitions of electrons without any interaction with light. Within a Jablonski diagram several different pathways show how an electron may accept and then dissipate the energy from a photon of a particular wavelength. Thus, most diagrams start with arrows going from the ground electronic state and finish with arrows going to the ground electronic state. The first transition in most Jablonski diagrams is the absorbance of a photon of a particular energy by the molecule of interest. This is indicated by a straight arrow pointing up. Absorbance is the method by which an electron is excited from a lower energy level to a higher energy level. The energy of the photon is transferred to the particular electron. That electron then transitions to a different eigenstate corresponding to the amount of energy transferred. Only certain wavelengths of light are possible for absorbance, that is, wavelengths that have energies that correspond to the energy difference between two different eigenstates of the particular molecule. Absorbance is a very fast transition, on the order of 10 Once an electron is excited, there are a multitude of ways that energy may be dissipated. The first is through vibrational relaxation, a non-radiative process. This is indicated on the Jablonski diagram as a curved arrow between vibrational levels. Vibrational relaxation is where the energy deposited by the photon into the electron is given away to other vibrational modes as kinetic energy. This kinetic energy may stay within the same molecule, or it may be transferred to other molecules around the excited molecule, largely depending on the phase of the probed sample. This process is also very fast, between 10 However, if vibrational energy levels strongly overlap electronic energy levels, a possibility exists that the excited electron can transition from a vibration level in one electronic state to another vibration level in a lower electronic state. This process is called internal conversion and mechanistically is identical to vibrational relaxation. It is also indicated as a curved line on a Jablonski diagram, between two vibrational levels in different electronic states. Internal Conversion occurs because of the overlap of vibrational and electronic energy states. As energies increase, the manifold of vibrational and electronic eigenstates becomes ever closer distributed. At energy levels greater than the first excited state, the manifold of vibrational energy levels strongly overlap with the electronic levels. This overlap gives a higher degree of probability that the electron can transition between vibrational levels that will lower the electronic state. Internal conversion occurs in the same time frame as vibrational relaxation, therefore, is a very likely way for molecules to dissipate energy from light perturbation. However, due to a lack of vibrational and electronic energy state overlap and a large energy difference between the ground state and first excited state, internal conversion is very slow for an electron to return to the ground state. This slow return to the ground state lets other transitive processes compete with internal conversion at the first electronically excited state. Both vibrational relaxation and internal conversion occur in most perturbations, yet are seldom the final transition. Another pathway for molecules to deal with energy received from photons is to emit a photon. This is termed fluorescence. It is indicated on a Jablonski diagram as a straight line going down on the energy axis between electronic states. Fluorescence is a slow process on the order of 10 Yet another path a molecule may take in the dissipation of energy is called intersystem crossing. This where the electron changes spin multiplicity from an excited singlet state to an excited triplet state. It is indicated by a horizontal, curved arrow from one column to another. This is the slowest process in the Jablonski diagram, several orders of magnitude slower than fluorescence. This slow transition is a forbidden transition, that is, a transition that based strictly on electronic selection rules should not happen. However, by coupling vibrational factors into the selection rules, the transition become weakly allowed and able to compete with the time scale of fluorescence. Intersystem crossing leads to several interesting routes back to the ground electronic state. One direct transition is phosphorescence, where a radiative transition from an excited triplet state to a singlet ground state occurs.This is also a very slow, forbidden transition. Another possibility is delayed fluorescence, the transition back to the first excited singlet level, leading to the emitting transition to the ground electronic state. Other non-emitting transitions from excited state to ground state exist and account for the majority of molecules not exhibiting fluorescence or phosphorescent behavior. One process is the energy transfer between molecules through molecular collisions (e.g., external conversion). Another path is through quenching, energy transfer between molecules through overlap in absorption and fluorescence spectra. These are non-emitting processes that will compete with fluorescence as the molecule relaxes back down to the ground electronic state. In a Jablonski diagram, each of these processes are indicated with a curved line going down to on the energy scale. It is important to note that a Jablonski diagram shows what sorts of transitions that can possibly happen in a particular molecule. Each of these possibilities is dependent on the time scales of each transition. The faster the transition, the more likely it is to happen as determined by selection rules. Therefore, understanding the time scales each process can happen is imperative to understanding if the process may happen. Below is a table of average time scales for basic radiative and non-radiative processes. Each process outlined above can be combined into a single Jablonski diagram for a particular molecule to give a overall picture of possible results of perturbation of a molecule by light energy. Jablonski diagrams are used to easily visualize the complex inner workings of how electrons change eigenstates in different conditions. Through this simple model, specific quantum mechanical phenomena are easily communicated. | 8,344 | 1,808 |
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This page explains how to use redox potentials (electrode potentials) to predict the feasibility of redox reactions. It also looks at how you go about choosing a suitable oxidizing agent or reducing agent for a particular reaction. Standard electrode potentials (redox potentials) are one way of measuring how easily a substance loses electrons. In particular, they give a measure of relative positions of equilibrium in reactions such as: \[ Zn^{2+} + 2e^- \rightleftharpoons Zn (s)\] with \(E^o = -0.76 \,V\) and \[ Cu^{2+} + 2e^- \rightleftharpoons Cu (s)\] with \(E^o = +0.34 \,V\). The more negative the E° value, the further the position of equilibrium lies to the left. Remember that this is always relative to the hydrogen equilibrium - and not in absolute terms. The negative sign of the zinc E° value shows that it releases electrons more readily than hydrogen does. The positive sign of the copper E° value shows that it releases electrons less readily than hydrogen. The more negative the E° value, the further the position of equilibrium lies to the left of a standard reduction half reaction. Whenever you link two of these equilibria together (either via a bit of wire, or by allowing one of the substances to give electrons directly to another one in a test tube) electrons flow from one equilibrium to the other. That upsets the equilibria and Le Chatelier's principle applies to figure out how new equilibria are established. The positions of equilibrium move - and keep on moving if the electrons continue to be transferred. The two equilibria essentially turn into two one-way reactions: Will magnesium react with dilute sulfuric acid? The relevant reduction reactions and associated potentials (via Table P2) for this system are \[ Mg^{2+} (aq) + 2e^- \rightleftharpoons Mg (s)\] with \(E^o = -2.37 \,V\) and \[S_2O_8^{2−} + 2e^− \rightleftharpoons 2SO_4^{2−}\] with \(E^o = +1.96 \,V\) and since we are in an aqueous solvent \[ 2H^{+} (aq) + 2e^- \rightleftharpoons H_2 (g)\] with \(E^o = 0 \,V\). The sulfate ions are spectator ions and play no part in the reaction; you are essentially starting with magnesium metal and hydrogen ions in the acid. Is there anything to stop the sort of movements we have suggested? No! The magnesium can freely turn into magnesium ions and give electrons to the hydrogen ions producing hydrogen gas. Now for a reaction which turns out not to be feasible . . Will copper react with dilute sulfuric acid? You probably know that the answer is that it will not. How do the E° values predict this? The relevant reduction reactions and associated potentials (via Table P2) for this system are \[ Cu^{2+} (aq) + 2e^- \rightleftharpoons Cu (s)\] with \(E^o = +0.34 \,V\) and since we are in an aqueous solvent \[ 2H^{+} (aq) + 2e^- \rightleftharpoons H_2 (g)\] with \(E^o = 0 \,V\). Doing the same sort of thinking as in Example \(\Page {1}\): The diagram shows the way that the E° values are telling us that the equilibria will tend to move. Is this possible? No! Will oxygen oxidize iron(II) hydroxide to iron(III) hydroxide under alkaline conditions? The relevant reduction reactions and associated potentials (via Table P2) for this system are \[ Fe(OH)_3 (s) + e^- \rightleftharpoons Fe(OH)_2 (s) + OH^- (aq)\] with \(E^o = -0.56 \,V\) and \[ O_2 (g) + 2H_2O (l) \rightleftharpoons 4 OH^- (aq)\] with \(E^o = 0.40 \,V\). Think about this before you read on. Remember that the equilibrium with the more negative E° value will tend to move to the left. The other one tends to move to the right. Is that possible? Yes, it is possible. Given what we are starting with, both of these equilibria can move in the directions required by the E° values. Will chlorine oxidize manganese (II) ions to manganate (VII) ions in acidic solutions? The relevant reduction reactions and associated potentials (via Table P2) for this system are \[ MnO_4^- (aq) + 8H^+(aq) + 5e^- \rightleftharpoons Mn^{2+} (aq) + 4H_2O (l) \] with \(E^o = +1.51 \,V\) and \[ Cl_2 (g) + 2e^- \rightleftharpoons 2Cl^- (aq) \] with \(E^o = +1.36 \,V\). Again, think about this before you read on. Given what you are starting from, these equilibrium shifts are impossible. The manganese equilibrium has the more positive E° value and so will tend to move to the right. However, because we are starting from manganese(II) ions, it is already as far to the right as possible. In order to get any reaction, the equilibrium would have to move to the left. That is against what the E° values are saying. Will dilute nitric acid react with copper? This is going to be more complicated because there are two different ways in which dilute nitric acid might possibly react with copper. The copper might react with the hydrogen ions or with the nitrate ions. Nitric acid reactions are always more complex than the simpler acids like sulfuric or hydrochloric acid because of this problem. The relevant reduction reactions and associated potentials (via Table P2) for this system are \[ Cu^{2+}(aq) + 2e^- \rightleftharpoons Cu (aq) \] with \(E^o = +0.34 \,V\) and \[ 2H^{+} (aq) + 2e^- \rightleftharpoons H_2 (g)\] with \(E^o = 0 \,V\) and \[ NO_3^{-}(aq) + 4H^+ + 3e^- \rightleftharpoons NO(g) + 2H_2O(l) \] with \(E^o = +0.96 \,V\). We have already discussed the possibility of copper reacting with hydrogen ions in Example \(\Page {2}\). Go back and look at it again if you need to, but the argument (briefly) goes like this: The copper equilibrium has a more positive E° value than the hydrogen one. That means that the copper equilibrium will tend to move to the right and the hydrogen one to the left. However, if we start from copper and hydrogen ions, the equilibria are already as far that way as possible. Any reaction would need them to move in the opposite direction to what the E° values want. The reaction isn't feasible. What about a reaction between the copper and the nitrate ions? This is feasible. The nitrate ion equilibrium has the more positive E° value and will move to the right. The copper E° value is less positive. That equilibrium will move to the left. The movements that the E° values suggest are possible, and so a reaction is feasible. Copper (II) ions are produced together with nitrogen monoxide gas. It sometimes happens that E° values suggest that a reaction ought to happen, but it does not. Occasionally, a reaction happens although the E° values seem to be the wrong way around. These next two examples explain how that can happen. By coincidence, both involve the dichromate(VI) ion in potassium dichromate(VI). Will acidified potassium dichromate(VI) oxidize water? The relevant reduction reactions and associated potentials (via Table P2) for this system are \[Cr_2O_7^{2-} (aq) + 14H^+(aq) + 6e^- \rightleftharpoons 2Cr^{3+} (aq) + 7H_2O (l) \] with \(E^o = +1.33 \,V\) and \[ O_2 (g) + 4H^+(aq) + 4e^- \rightleftharpoons 2H_2O (l) \] with \(E^o = +1.23 \,V\). The relative sizes of the E° values show that the reaction is feasible: However, in the test tube nothing happens however long you wait. An acidified solution of potassium dichromate(VI) with the water that it is dissolved in. So what is wrong with the argument? In fact, there is nothing wrong with the argument. The problem is that all the E° values show is that a reaction is possible. They do not tell you that it will actually happen as there may be very large activation barriers to the reaction which prevent it from taking place. Hence, always treat what E° values tell you with some caution. All they tell you is whether a reaction is feasible - they tell you nothing about how fast the reaction will happen; that is the subject of kinetics. The E° values show is that a reaction is possible, but they do not tell you that it will actually happen (or on what timescale). Will acidified potassium dichromate(VI) oxidize chloride ions to chlorine? The relevant reduction reactions and associated potentials (via Table P2) for this system are \[Cr_2O_7^{2-} (aq) + 14H^+(aq) + 6e^- \rightleftharpoons 2Cr^{3+} (aq) + 7H_2O (l) \] with \(E^o = +1.33 \,V\) and \[ Cl_2 (g) + 2e^- \rightleftharpoons 2Cl^- (aq) \] with \(E^o = +1.36 \,V\). Because the chlorine E° value is slightly greater than the dichromate(VI) one, there be any reaction. For a reaction to occur, the equilibria would have to move in the wrong directions. Unfortunately, in the test tube, potassium dichromate(VI) solution does oxidize concentrated hydrochloric acid to chlorine. The hydrochloric acid serves as the source of the hydrogen ions in the dichromate(VI) equilibrium and of the chloride ions. The problem here is that E° only apply under . If you change the conditions you will change the position of an equilibrium - and that will change its E value. The standard condition for concentration is 1 mol dm but concentrated hydrochloric acid is approximately 10 mol dm . The concentrations of the hydrogen ions and chloride ions are far in excess of standard. What effect does that have on the two positions of equilibrium? Because the E° values are so similar, you do not have to change them very much to make the dichromate(VI) one the more positive. As soon as that happens, it will react with the chloride ions to produce chlorine. In most cases, there is enough difference between E° values that you can ignore the fact that you are not doing a reaction under strictly standard conditions. But sometimes it does make a difference. Be careful! Remember that oxidation is loss of electrons and an oxidizing agent oxidizes something by removing electrons from it. That means that the oxidizing agent gains electrons. It is easier to explain this with a specific example. What could you use to oxidize iron(II) ions to iron(III) ions? The E° value for this reaction is: \[ Fe^{3+}(aq) + e^- \rightleftharpoons Fe^{2+} (aq) \] with \(E^o = +0.77 \,V\). To change iron(II) ions into iron(III) ions, you need to persuade this equilibrium to move to the left. That means that when you couple it to a second equilibrium, this iron E° value must be the (or less positive) one. An experimentally could use anything which has a more positive E° value. For example, you could use dilute nitric acid: \[ NO_3^{-}(aq) + 3e^- \rightleftharpoons NO(g) + 2H_2O (l) \] with \(E^o = +0.96 \,V\) or acidified potassium dichromate(VI): \[Cr_2O_7^{2-} (aq) + 14H^+(aq) + 6e^- \rightleftharpoons 2Cr^{3+} (aq) + 7H_2O (l) \] with \(E^o = +1.33 \,V\) or chlorine \[ Cl_2 (g) + 2e^- \rightleftharpoons 2Cl^- (aq) \] with \(E^o = +1.36 \,V\) or acidified potassium manganate(VII): \[ MnO_4^- (aq) + 8H^+(aq) + 5e^- \rightleftharpoons Mn^{2+} (aq) + 4H_2O (l) \] with \(E^o = +1.51 \,V\). Remember, reduction is gain of electrons and a reducing agent reduces something by giving electrons to it. That means that the reducing agent loses electrons. You have to be a little bit more careful this time, because the substance losing electrons is found on the right-hand side of one of these redox equilibria. Again, a specific example makes it clearer. For example, what could you use to reduce chromium(III) ions to chromium(II) ions? The E° value is: \[Cr^{3+}(aq) + e^- \rightleftharpoons Cr^{2+}(aq)\] with \(E^o = -0.41 \,V\). You need this equilibrium to move to the right. That means that when you couple it with a second equilibrium, this chromium E° value must be the most positive (least negative). In principle, you could choose anything with a E° value - for example, zinc: \[Zn^{2+} (aq) + 2e^- \rightleftharpoons Zn (s)\] with \(E^o = -0.76 \,V\). You would have to remember to start from metallic zinc, and not zinc ions. You need this second equilibrium to be able to move to the left to provide the electrons. If you started with zinc ions, it would already be on the left - and would have no electrons to give away. Nothing could possibly happen if you mixed chromium(III) ions and zinc ions. That is fairly obvious in this simple case. If you were dealing with a more complicated equilibrium, you would have to be careful to think it through properly. Jim Clark ( ) | 12,084 | 1,809 |
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Chemical reactions involve the making and breaking of chemical bonds ( and ) and the chemical energy of a system is the energy released or absorbed due to the making and breaking of these bonds. Breaking bonds requires energy, forming bonds releases energy, and the overall reaction can be either (\(\Delta G<0\)) or (\(\Delta G > 0\)) based on the overall changes in stability from reactants to products. Simply put, chemical energy is the potential of a chemical system to undergo a transformation from one system to another and to impart a transformation on another system (this may be chemical, but can also involve other energy-requiring processes like electron current or pressure-volume work). Chemical energy is a concept that is related to every single process of life on earth and powers the cars that we drive. Chemical energy plays a crucial role into each and every one of our every day lives. Through simple reactions and redox chemistry, the breaking and forming of bonds, energy can be extracted and harnessed into a usable fashion. | 1,070 | 1,810 |
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The traditional experimental methods described above all assume the possibility of following the reaction after its components have combined into a homogeneous mixture of known concentrations. But what can be done if the time required to complete the mixing process is comparable to or greater than the time needed for the reaction to run to completion? Flow instruments are a rapid mixing devices used to study the chemical kinetics of fast reactions in solution. There are different flavors that can be implement depending on the nature of the reaction as discussed below. For reactions that take place in milliseconds, the standard approach since the 1950s has been to employ a flow technique of some kind. An early example was used to study fast gas-phase reactions in which one of the reactants is a free radical such as OH that can be produced by an intense microwave discharge acting on a suitable source gas mixture. This gas, along with the other reactant being investigated, is made to flow through a narrow tube at a known velocity. If the distance between the point at which the reaction is initiated and the product detector is known, then the time interval can be found from the flow rate. By varying this distance, the time required to obtain the maximum yield can then be determined. Although this method is very simple in principle, it can be complicated in practice. Owing to the rather large volumes required, continuous flow method is more practical for the study of gas-phase reactions than for solutions, for which the stopped-flow method described below is generally preferred. These are by far the most common means of studying fast solution-phase reactions over time intervals of down to a fraction of a millisecond. The use of reasonably simple devices is now practical even in student laboratory experiments. These techniques make it possible to follow not only changes in the concentrations of reactants and products, but also the buildup and decay of reaction intermediates. The basic stopped-flow apparatus consists of two or more coupled syringes that rapidly inject the reactants into a small mixing chamber and then through an observation cell that can be coupled to instruments that measure absorption, fluorescence, light scattering, or other optical or electrical properties of the solution. As the solution flows through the cell, it empties into a stopping syringe that, when filled, strikes a backstop that abruptly stops the flow. The volume that the stopping syringe can accept is adjusted so that the mixture in the cell has just become uniform and has reached a steady state; at this point, recording of the cell measurement begins and its change is followed. In a quenched-flow instrument, the reaction is stopped after a certain amount of time has passed after mixing. The stopping of the reaction is called quenching and it can be achieved by various means, for example by mixing with another solution, which stops the reaction (chemical quenching), quickly lowering the temperature (freeze quenching) or even by exposing the sample to light of a certain wavelength (optical quenching). Of course, there are many reactions that cannot be followed by changes in light absorption or other physical properties that are conveniently monitored. In such cases, it is often practical to (stop) the reaction after a desired interval by adding an appropriate quenching agent. For example, an enzyme-catalyzed reaction can be stopped by adding an acid, base, or salt solution that denatures (destroys the activity of) the protein enzyme. Once the reaction has been stopped, the mixture is withdrawn and analyzed in an appropriate manner. works something like the stopped-flow method described above, with a slightly altered plumbing arrangement. The reactants A and B are mixed and fed directly through the diverter valve to the measuring cell, which is not shown in this diagram. After a set interval that can vary from a few milliseconds to 200 sec or more, the controller activates the quenching syringe and diverter valve, flooding the cell with the quenching solution. To investigate reactions that are complete in less than a millisecond, one can start with a pre-mixed sample in which one of active reactants is generated . Alternatively, a rapid change in pressure or temperature can alter the composition of a reaction that has already achieved equilibrium. Many reactions are known which do not take place without light of wavelength sufficiently short to supply the activation energy needed to break a bond, often leading to the creation of a highly reactive radical. A good example is the combination of gaseous Cl with H , which proceeds explosively when the system is illuminated with visible light. In , a short pulse of light is used to initiate a reaction whose progress can be observed by optical or other means. refers to the use of light to decompose a molecule into simpler units, often ions or free radicals. In contrast to (decomposition induced by high temperature), photolysis is able to inject energy into a molecule almost instantaneously and can be much "cleaner," meaning that there are fewer side reactions that often lead to complex mixtures of products. Photolysis can also be highly ; the wavelength of the light that triggers the reaction can often be adjusted to activate one particular kind of molecule without affecting others that might be present. All this had been known for a very long time, but until the mid-1940's there was no practical way of studying the kinetics of the reactions involving the highly reactive species produced by photolysis. In 1945, Ronald Norrish of Cambridge University and his graduate student George Porter conceived the idea of using a short-duration flash lamp to generate gas-phase CH radicals, and then following the progress of the reaction of these radicals with other species by means of absorption spectroscopy. In a flash photolysis experiment, recording of the absorbance of the sample cell contents is timed to follow the flash by an interval that can be varied in order to capture the effects produced by the product or intermediate as it is formed or decays. Norrish and Porter shared the 1967 Nobel Prize in Chemistry for this work. Many reactions, especially those that take place in solution, occur too rapidly to follow by flow techniques, and can therefore only be observed when they are already at equilibrium. The classical examples of such reactions are two of the fastest ones ever observed, the dissociation of water \[ 2 H_2O \rightarrow H_3O^+ + OH^- \] and the formation of the triiodide ion in aqueous solution \[ I^– + I_2 \rightarrow I_3^–\] Reactions of these kinds could not be studied until the mid-1950s when techniques were developed to shift the equilibrium by imposing an abrupt physical change on the system. The rate constants of reversible reactions can be measured using a relaxation method. In this method, the concentrations of reactants and products are allowed to achieve equilibrium at a specific temperature. Once equilibrium has been achieved, the temperature is rapidly changed, and then the time needed to achieve the new equilibrium concentrations of reactants and products is measured. For example, if the reaction \[\text{A} \overset{k_1}{\underset{k_{-1}}{\rightleftharpoons}} \text{B} \label{relax}\] is endothermic, then according to the , subjecting the system to a rapid jump in temperature will shift the equilibrium state to one in which the product B has a higher concentration. The composition of the system will than begin to shift toward the new equilibrium composition at a rate determined by the kinetics of the process. For the general case illustrated here, the quantity " " being plotted is a measurable quantity such as light absorption or electrical conductivity that varies linearly with the composition of the system. , will vary with time according to \[ x_t = x_o e^{-kt} \] After the abrupt perturbation at time , the relaxation time is defined as the half-time for the return to equilibrium — that is, as the time required for \(x_o\) to decrease by \(Δ /e = Δ /2.718\). The derivation of and the relations highlighted in yellow can be found in most standard kinetics textbooks. are likely most commonly used. The rate law for the reversible reaction in Equation \(\ref{relax}\) can be written as \[\dfrac{d \left[ \text{B} \right]}{dt} = k_1 \left[ \text{A} \right] - k_{-1} \left[ \text{B} \right] \label{19.33}\] Consider a system comprising \(\text{A}\) and \(\text{B}\) that is allowed to achieve equilibrium concentrations at a temperature, \(T_1\). After equilibrium is achieved, the temperature of the system is instantaneously lowered to \(T_2\) and the system is allowed to achieve new equilibrium concentrations of \(\text{A}\) and \(\text{B}\), \(\left[ \text{A} \right]_{\text{eq},2}\) and \(\left[ \text{B} \right]_{\text{eq}, 2}\). During the transition time from the first equilibrium state to the second equilibrium state, we can write the instantaneous concentration of \(\text{A}\) as \[\left[ \text{A} \right] = \left[ \text{B} \right]_{\text{eq}, 1} - \left[ \text{B} \right] \label{19.34}\] The rate of change of species \(\text{B}\) can then be written as \[\dfrac{d \left[ \text{B} \right]}{dt} = k_1 \left( \left[ \text{B} \right]_{\text{eq}, 1} - \left[ \text{B} \right] \right) - k_{-1} \left[ \text{B} \right] = k_1 \left[ \text{B} \right]_{\text{eq}, 1} - \left( k_1 + k_{-1} \right) \left[ \text{B} \right] \label{19.35}\] At equilibrium, \(d \left[ \text{B} \right]/dt = 0\) and \(\left[ \text{B} \right] = \left[ \text{B} \right]_{\text{eq}, 2}\), allowing us to write \[k_1 \left[ \text{B} \right]_{\text{eq}, 1} = \left( k_1 + k_{-1} \right) \left[ \text{B} \right]_{\text{eq}, 2} \label{19.36}\] Using the above equation, we can rewrite the rate equation as \[\dfrac{ d \text{B}}{\left( \left[ \text{B} \right]_{\text{eq}, 2} - \left[ \text{B} \right] \right)} = \left( k_1 + k_{-1} \right) dt \label{19.37}\] Integrating yields \[-\text{ln} \left( \left[ \text{B} \right] - \left[ \text{B} \right]_{\text{eq}, 2} \right) = -\left( k_1 + k_{-1} \right) t + C \label{19.38}\] We can rearrange the above equation in terms of \(\text{B}\) \[\left[ \text{B} \right] = Ce^{- \left( k_1 + k_{-1} \right) t} + \left[ \text{B} \right]_{\text{eq}, 2} \label{19.39}\] At \(t = 0\), \(\left[ \text{B} \right] = \left[ \text{B} \right]_{\text{eq}, 1}\), so \(C = \left[ \text{B} \right]_{\text{eq}, 1} - \left[ \text{B} \right]_{\text{eq}, 2}\). Plugging the the value of \(C\), we arrive at \[\left[ \text{B} \right] - \left[ \text{B} \right]_{\text{eq}, 2} = \left( \left[ \text{B} \right]_{\text{eq}, 1} - \left[ \text{B} \right]_{\text{eq}, 2} \right) e^{-\left( k_1 + k_{-1} \right) t} \label{19.40}\] which can also be expressed as \[\Delta \left[ \text{B} \right] = \Delta \left[ \text{B} \right]_0 e^{-\left( k_1 + k_{-1} \right) t} = \Delta \left[ \text{B} \right]_0 e^{-t/\tau} \label{19.41}\] where \(\Delta \left[ \text{B} \right]\) is the difference in the concentration of \(\text{B}\) from the final equilibrium concentration after the perturbation, and \(\tau\) is the . A plot of \(\text{ln} \left( \Delta \left[ \text{B} \right]/\Delta \left[ \text{B} \right]_0 \right)\) versus \(t\) will be linear with a slope of \(-\left( k_1 + k_{-1} \right)\), where \(k_1\) and \(k_{-1}\) are the rate constants at temperature, \(T_2\). ( ) ) | 11,493 | 1,812 |
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are simplified Bohr diagrams which only display electrons in the outermost energy level. The omitted electrons are those in filled energy levels, which do not contribute to the chemical properties of the species in question. Therefore Lewis Symbols are useful for studying elemental properties and reactions. A Lewis Symbol is constructed by placing dots representing electrons in the outer energy around the symbol for the element. For many common elements, the number of dots corresponds to the element's group number. Below are Lewis Symbols for various elements. Notice the correspondence to each element's group number. Molecules can be depicted by Lewis Diagrams by placing dots or lines around the constituent elemental symbols. Once again only valence electrons are shown. Lines denote bonded electron pairs, whereas dots are reserved for unbounded electrons. The following algorithm can be used to construct Lewis diagrams of most molecules. | 968 | 1,813 |
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A steam distillation apparatus is shown in Figure 5.58 that uses boiling water in the distilling flask. An apparatus using a steam line is shown in Figure 5.59. It is assumed that readers have previously performed a , so in this section are described differences between simple and steam distillations. Place large amount of plant material in a large round bottomed flask (no more than half full), and just cover with water. Always use a Claisen adapter, as there is often turbulence in the flask. Two variations are common: Collect the distillate at a rate of . The distillate may appear cloudy, or a second layer may form on the top. If milky, the distillation can be ceased when the distillate is clear. If a steam line was used, be sure to drain the liquid from the steam trap before turning off the steam, to prevent back suction. Separated oil can be pipetted (then dried with \(\ce{Na_2SO_4}\)), while milky distillates need to be extracted. | 962 | 1,814 |
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The macroscopic discussion of kinetics discussed in previous sections can be now expanded into a more microscopic picture in terms of molecular level properties (e..g, mass and velocities) involving two important theories: (1) collision theory and (2) transition-state theory. If two molecules need to collide in order for a reaction to take place, then factors that influence the ease of collisions will be important. The more energy there is available to the molecules, the faster they will move around, and the more likely they are to bump into each other. Higher temperatures ought to lead to more collisions and a greater frequency of reactions between molecules. In the drawing below, the cold, sluggish molecules on the left are not likely to collide, but the energetic molecules on the right are due to collide at any time. The rate at which molecules collide which is the frequency of collisions is called the , \(Z\), which has units of collisions per unit of time. Given a container of molecules \(A\) and \(B\), the collision frequency between \(A\) and \(B\) is defined by: \[Z=N_{A}N_{B}\sigma_{AB}\sqrt{\dfrac{8k_{B}T}{\pi\mu_{AB}}}\] where: \[ \sqrt{\dfrac{8k_{B}T}{\pi\mu_{AB}}} \] \[\mu = \dfrac{m_\text{A} m_\text{B}}{m_\text{A} + m_\text{B}}\] The concepts of collision frequency can be applied in the laboratory: (1) The temperature of the environment affects the average speed of molecules. Thus, reactions are heated to increase the reaction rate. (2) The initial concentration of reactants is directly proportional to the collision frequency; increasing the initial concentration will speed up the reaction. For a successful collision to occur, the reactant molecules must collide with enough kinetic energy to break original bonds and form new bonds to become the product molecules. This energy is called the activation energy for the reaction; it is also often referred to as the energy barrier. The fraction of collisions with enough energy to overcome the activation barrier is given by: where: The fraction of successful collisions is directly proportional to the temperature and inversely proportional to the activation energy. The fraction of successful collisions is directly proportional to the temperature and inversely proportional to the activation energy. The more complicated the structures of the reactants, the more likely that the value of the rate constant will depend on the trajectories at which the reactants approach each other. This kind of is well-known to all students of organic chemistry. Consider the addition of a hydrogen halide such as HCl to the double bond of an alkene, converting it to a chloroalkane. Experiments have shown that the reaction only takes place when the HCl molecule approaches the alkene with its hydrogen-end, and in a direction that is approximately perpendicular to the double bond, as shown at below. The reason for this becomes apparent when we recall that HCl is highly polar owing to the high electronegativity of chlorine, so that the hydrogen end of the molecule is slightly positive. The steric factor, \(\rho\) is then introduced to represent is the probability of the reactant molecules colliding with the right orientation and positioning to achieve a product with the desirable geometry and stereospecificity. Values of \(\rho\) are generally very difficult to assess and range from 0 to 1, but are sometime estimated by comparing the observed rate constant with the one in which the preexponential constant \(A\) is assumed to be the same as \(Z\). The lesson you should take from this example is that once you start combining a variety of chemical principles, you gradually develop what might be called "chemical intuition" which you can apply to a wide variety of problems. This is far more important than memorizing specific examples. The rate constant of the gas-phase reaction is proportional to the product of the and the . As stated above, sufficient kinetic energy is required for a successful reaction; however, they must also collide properly. Compare the following equation to the : where Although the collision theory deals with gas-phase reactions, its concepts can also be applied to reactions that take place in solvents; however, the properties of the solvents (for example: solvent cage) will affect the rate of reactions. Ultimately, collision theory illustrates how reactions occur; it can be used to approximate the rate constants of reactions, and its concepts can be directly applied in the laboratory. Read this for a more detailed discussion of Collision Theory. Transition state theory (TST) provides a more accurate alternative to the previously used and the collision theory. The transition state theory attempts to provide a greater understanding of activation energy, \(E_a\), and the thermodynamic properties involving the transition state. Collision theory of reaction rate, although intuitive, lacks an accurate method to predict the probability factor for the reaction. The theory assumes that reactants are hard spheres rather than molecules with specific structures. In 1935, Henry Eyring helped develop a new theory called the transition state theory to provide a more accurate alternative to the previously used Arrhenius equation and the collision theory. The Eyring equation involves the statistical frequency factory, v, which is fundamental to the theory. According to TST, between the state where molecules are reactants and the state where molecules are products, there is a state known as the transition state. In the transition state, the reactants are combined in a species called the activated complex. The theory suggests that there are three major factors that determine whether a reaction will occur: Collision theory proposes that not all reactants that combine undergo a reaction. However, assuming the stipulations of the collision theory are met and a successful collision occurs between the molecules, transition state theory allows one of two outcomes: a return to the reactants, or a rearranging of bonds to form the products. Consider a bimolecular reaction: \[A~+B~\rightarrow~C \label{2}\] \[K = \dfrac{[C]}{[A,B]} \label{3}\] where \(K\) is the equilibrium constant. In the transition state model, the activated complex AB is formed: \[A~+~B~\rightleftharpoons ~AB^\ddagger~\rightarrow ~C \label{4}\] \[K^\ddagger=\dfrac{[AB]^\ddagger}{[A,B]} \label{5}\] There is an energy barrier, called activation energy, in the reaction pathway. A certain amount of energy is required for the reaction to occur. The transition state, \(AB^\ddagger\), is formed at maximum energy. This high-energy complex represents an unstable intermediate. Once the energy barrier is overcome, the reaction is able to proceed and product formation occurs. The rate of a reaction is equal to the number of activated complexes decomposing to form products. Hence, it is the concentration of the high-energy complex multiplied by the frequency of it surmounting the barrier. \[\begin{eqnarray} rate~&=&~v[AB^\ddagger] \label{6} \\ &=&~v[A,B]K^\ddagger \label{7} \end{eqnarray} \] The rate can be rewritten: \[rate~=~k[A,B] \label{8}\] Combining Equations \(\ref{8}\) and \(\ref{7}\) gives: \[ \begin{eqnarray} k[A,B]~&=&~v[A,B]K^\ddagger \label{9} \\ k~&=&~vK^\ddagger \label{10} \end{eqnarray} \] where Statistical mechanics (not shown) provides that the frequency, v, is equivalent to the thermal energy, k T, divided by Planck's constant, h. \[v~=~\dfrac{k_BT}{h} \label{11}\] where Substituting Equation \(\ref{11}\) into Equation \(\ref{10}\) : \[k~=~\dfrac{k_BT}{h}K^\ddagger \label{12}\] Equation \({ref}\) is often tagged with another term \((M^{1-m})\) that makes the units equal with \(M\) is the molarity and \(m\) is the molecularly of the reaction. \[k~=~\dfrac{k_BT}{h}K^\ddagger (M^{1-m}) \label{E12}\] It is important to note here that the equilibrium constant \(K^\ddagger \) can be calculated by absolute, fundamental properties such as bond length, atomic mass, and vibration frequency. This gives the transition rate theory the alternative name absolute rate theory, because the rate constant, k, can be calculated from fundamental properties. To reveal the thermodynamics of the theory, \(K^\ddagger\) must be expressed in terms of \(\Delta {G^{\ddagger}}\). \(\Delta {G^{\ddagger}}\) is simply, \[\Delta{G}^{o\ddagger} = G^o (transitionstate) - G^o (reactants)\] By definition, at equilibrium, \(\Delta {G^{\ddagger}}\) can be expressed as: \[\Delta {G^{\ddagger}} = -RT \ln K^{\ddagger}\] Rearrangement gives: \[[K]^\ddagger = e^{ -\frac{\Delta{G}^\ddagger}{RT} }\] From Equation \(\ref{E12}\) \[ \color{red} k = ve^{ -\frac{\Delta{G}^{\ddagger}}{RT}} (M^{1-m})\] It is also possible to obtain terms for the change in enthalpy and entropy for the transition state. Because \[\Delta{G^\ddagger} = \Delta{H^\ddagger} - T\Delta{S^\ddagger}\] it follows that the derived equation becomes, \[k = \dfrac{k_BT}{h} e^{\Delta{S}^\ddagger /R} e^{-\Delta H^\ddagger/RT} M^{1-m} \label{Eyring}\] Equation \(\ref{Eyring}\) is known as the Eyring Equation and was developed by Henry Eyring in 1935, is based on transition state theory and is used to describe the relationship between reaction rate and temperature. It is similar to the , which also describes the temperature dependence of reaction rates. The linear form of the Eyring Equation is given below: \[\ln{\dfrac{k}{T}}~=~\dfrac{-\Delta H^\dagger}{R}\dfrac{1}{T}~+~\ln{\dfrac{k_B}{h}}~+~\dfrac{\Delta S^\ddagger}{R} \label{17}\] The values for \(\Delta H^\ddagger\) and \(\Delta S^\ddagger\) can be determined from kinetic data obtained from a \(\ln{\dfrac{k}{T}}\) vs. \(\dfrac{1}{T}\) plot. The Equation is a straight line with negative slope, \(\dfrac{-\Delta H^\ddagger}{R}\), and a y-intercept, \(\dfrac{\Delta S^\ddagger}{R}+\ln{\dfrac{k_B}{h}}\). In this article, the complete thermodynamic formulation of the transition state theory was derived. This equation is more reliable than either the and the equation for the . However, it has its limitations, especially when considering the concepts of quantum mechanics. Quantum mechanics implies that can occur, such that particles can bypass the energy barrier created by the transition state. This can especially occur with low activation energies, because the probability of tunneling increases when the barrier height is lowered. In addition, transition state theory assumes that an equilibrium exists between the reactants and the transition state phase. However, in solution non-equilibrium situations can arise, upsetting the theory. Several more complex theories have been presented to correct for these and other discrepancies. This theory still remains largely useful in calculating the thermodynamic properties of the transition state from the overall reaction rate. This presents immense usefulness in medicinal chemistry, in which the study of transition state analogs is widely implemented. | 10,986 | 1,815 |
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In general, it is necessary to experimentally measure the concentrations of species over time in order to determine the apparent rate law governing the reaction. If the reactions are elementary reactions, (i.e. they cannot be expressed as a series of simpler reactions), then we can directly define the rate law based on the chemical equation. For example, an elementary reaction in which a single reactant transforms into a single product, is unimolecular reaction. These reactions follow \(1^{st}\) order rate kinetics. An example of this type of reaction would be the isomerization of butane: \[nC_4 H_{10} \longrightarrow iC_4 H_{10}\] From the chemical reaction equation, we can directly write the rate law as \[\dfrac{d \left[ nC_4 H_{10} \right]}{dt} = -k \left[ nC_4 H_{10} \right] \label{19.18}\] without the need to carry out experiments. Elementary bimolecular reactions that involve two molecules interacting to form one or more products follow second order rate kinetics. An example would be the following reaction between a nitrate molecule and carbon monoxide to form nitrogen dioxide and carbon dioxide: \[NO_3 + CO \longrightarrow NO_2 + CO_2\] For the above elementary reaction, we can directly write the rate law as: \[\dfrac{d \left[ NO_3 \right]}{dt} = -k \left[ NO_3 \right] \left[ CO \right] \label{19.19}\] Trimolecular elementary reactions involving three reactant molecules to form one or more products are rare due to the low probability of three molecules simultaneously colliding with one another. | 1,544 | 1,816 |
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The Lindemann mechanism, sometimes called the Lindemann-Hinshelwood mechanism, is a schematic reaction mechanism. Frederick Lindemann the concept in 1921 and Cyril Hinshelwood developed it. It breaks down a stepwise reaction into two or more elementary steps, then it gives a rate constant for each elementary step. The rate law and rate equation for the entire reaction can be derived from this information. Lindemann mechanisms have been used to model gas phase decomposition reactions. Although the net formula for a decomposition may appear to be first-order (unimolecular) in the reactant, a Lindemann mechanism may show that the reaction is actually second-order (bimolecular). A Lindemann mechanism typically includes an activated reaction intermediate, labeled A* (where A can be any element or compound). The activated intermediate is produced from the reactants only after a sufficient activation energy is applied. It then either deactivates from A* back to A, or reacts with another (dis)similar reagent to produce yet another reaction intermediate or the final product. The schematic reaction \(A + M \rightarrow P\) is assumed to consist of two elementary steps: \[A + M \rightleftharpoons A^* + M \label{1} \] with \[A^* \overset{k_2}{\rightarrow} P \label{2} \] with Assuming that the concentration of intermediate \(A^*\) is held constant according to the quasi steady-state approximation, what is the rate of formation of product \(P\)? First, find the rates of production and consumption of intermediate \(A^*\). The rate of production of \(A^*\) in the first elementary step (Equation \(\ref{1}\)) and \(A^*\) is consumed both in the reverse first step and in the forward second step. The respective rates of consumption of \(A^*\) are: \[\dfrac{d[A^*]}{dt} = \underset{\text{(forward first step)}}{k_1 [A] [M]} - \underset{\text{(reverse first step)}}{k_{-1} [A^*] [M]} - \underset{\text{(forward second step)}}{k_2 [A^*]} \label{3} \] According to the steady-state approximation, \[\dfrac{d[A^*]}{dt} \approx 0 \label{4} \] Therefore the rate of production of \(A^*\) (first term in Equation \(\ref{3}\)) equals the rate of consumption (second and third terms in Equation \(\ref{3}\)): \[k_1 [A] [M] = k_{-1} [A^*] [M] + k_2 [A^*] \label{6} \] Solving for \([A^*]\), it is found that \[[A^*] = \dfrac{k_1 [A] [M]}{k_{-1} [M] + k_2} \label{7} \] The overall reaction rate is (Equation \(\ref{2}\)) \[\dfrac{d[P]}{dt} = k_2 [A^*] \label{8} \] Now, by substituting the calculated value for \([A^*]\) (Equation \(\ref{7}\) into Equation \(\ref{8}\)), the overall reaction rate can be expressed in terms of the original reactants \(A\) and \(M\) as follows: \[\dfrac{d[P]}{dt} = \dfrac{k_1k_2 [A] [M]}{k_{-1} [M] + k_2} \label{9} \] The rate law for the Lindemann mechanism is not a simple first or second order reaction. However, under certain conditions (discussed below), Equation \(\ref{9}\) can be simplified. The decomposition of dinitrogen pentoxide to nitrogen dioxide and nitrogen trioxide \[N_2O_5 \rightarrow NO_2 + NO_3 \nonumber \] is postulated to take place via two elementary steps, which are similar in form to the schematic example given above: Using the quasi steady-state approximation soluioten (Equation 9) with \([M]=[N_2O_5]\), then rate equation is: \[\text{Rate} = k_2 [N_2O_5]^* = \dfrac{k_1k_2 [N_2O_5]^2}{k_{-1}[N_2O_5] + k_2} \nonumber \] Experiment has shown that the rate is observed as first-order in the original concentration of \(N_2O_5\) sometimes, and second order at other times. The following first order rate constants for the gas phase decomposition of \(N_2O_5\) have been obtained as a function of number density at 298 K. Confirm that these data are consistent with the Lindemann mechanism and derive a rate constant and a ratio of two rate constants for elementary reactions in the mechanism. What are the units of the two quantities. Consider the isomerization of methylisonitrile gas, \(CH_3 NC\), to acetonitrile gas, \(CH_3 CN\): \[CH_3 NC \overset{k}{\longrightarrow} CH_3 CN \nonumber \] If the isomerization is a unimolecular elementary reaction, we should expect to see \(1^{st}\) order rate kinetics. Experimentally, however, \(1^{st}\) order rate kinetics are only observed at high pressures. At low pressures, the reaction kinetics follow a \(2^{nd}\) order rate law: \[\dfrac{d \left[ CH_3 NC \right]}{dt} = -k \left[ CH_3 NC \right]^2 \label{21.30} \] To explain this observation, J.A. Christiansen and F.A. Lindemann proposed that gas molecules first need to be energized via intermolecular collisions before undergoing an isomerization reaction. The reaction mechanism can be expressed as the following two elementary reactions \[\begin{align} \text{A} + \text{M} &\overset{k_1}{\underset{k_{-1}}{\rightleftharpoons}} \text{A}^* + \text{M} \\ \text{A}^* &\overset{k_2}{\rightarrow} \text{B} \end{align} \nonumber \] where \(\text{M}\) can be a reactant molecule, a product molecule or another inert molecule present in the reactor. Assuming that the concentration of \(\text{A}^*\) is small, or \(k_1 \ll k_2 + k_{-1}\), we can use a steady-state approximation to solve for the concentration profile of species \(\text{B}\) with time: \[\dfrac{d \left[ \text{A}^* \right]}{dt} = k_1 \left[ \text{A} \right] \left[ \text{M} \right] - k_{-1} \left[ \text{A}^* \right]_{ss} \left[ \text{M} \right] - k_2 \left[ \text{A}^* \right]_{ss} \approx 0 \label{21.31} \] Solving for \(\left[ \text{A}^* \right]\), \[\left[ \text{A}^* \right] = \dfrac{k_1 \left[ \text{M} \right] \left[ \text{A} \right]}{k_2 + k_{-1} \left[ \text{M} \right]} \label{21.32} \] The reaction rates of species \(\text{A}\) and \(\text{B}\) can be written as \[-\dfrac{d \left[ \text{A} \right]}{dt} = \dfrac{d \left[ \text{B} \right]}{dt} = k_2 \left[ \text{A}^* \right] = \dfrac{k_1 k_2 \left[ \text{M} \right] \left[ \text{A} \right]}{k_2 + k_{-1} \left[ \text{M} \right]} = k_\text{obs} \left[ \text{A} \right] \label{21.33} \] where \[k_\text{obs} = \dfrac{k_1 k_2 \left[ \text{M} \right]}{k_2 + k_{-1} \left[ \text{M} \right]} \label{21.34} \] At high pressures, we can expect collisions to occur frequently, such that \(k_{-1} \left[ \text{M} \right] \gg k_2\). Equation \(\ref{21.33}\) then becomes \[-\dfrac{d \left[ \text{A} \right]}{dt} = \dfrac{k_1 k_2}{k_{-1}} \left[ \text{A} \right] \label{21.35} \] which follows \(1^{st}\) order rate kinetics. At low pressures, we can expect collisions to occurs infrequently, such that \(k_{-1} \left[ \text{M} \right] \ll k_2\). In this scenario, Equation \(\ref{21.33}\) becomes \[-\dfrac{d \left[ \text{A} \right]}{dt} = k_1 \left[ \text{A} \right] \left[ \text{M} \right] \label{21.36} \] which follows second order rate kinetics, consistent with experimental observations. | 6,869 | 1,817 |
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Although a balanced chemical equation for a reaction describes the quantitative relationships between the amounts of reactants present and the amounts of products that can be formed, it gives us no information about whether or how fast a given reaction will occur. This information is obtained by studying the chemical kinetics of a reaction, which depend on various factors: reactant concentrations, temperature, physical states and surface areas of reactants, and solvent and catalyst properties if either are present. By studying the kinetics of a reaction, chemists gain insights into how to control reaction conditions to achieve a desired outcome. Two substances cannot possibly react with each other unless their constituent particles (molecules, atoms, or ions) come into contact. If there is no contact, the reaction rate will be zero. Conversely, the more reactant particles that collide per unit time, the more often a reaction between them can occur. Consequently, the reaction rate usually increases as the concentration of the reactants increases. One example of this effect is the reaction of sucrose (table sugar) with sulfuric acid, which is shown in . Mixing sucrose with sulfuric acid in a beaker (a, right) produces a simple solution. Mixing the same amount of sucrose with sulfuric acid (a, left) results in a dramatic reaction (b) that eventually produces a column of black porous graphite (c) and an intense smell of burning sugar. You learned in that increasing the temperature of a system increases the average kinetic energy of its constituent particles. As the average kinetic energy increases, the particles move faster, so they collide more frequently per unit time and possess greater energy when they collide. Both of these factors increase the reaction rate. Hence the reaction rate of virtually all reactions increases with increasing temperature. Conversely, the reaction rate of virtually all reactions decreases with decreasing temperature. For example, refrigeration retards the rate of growth of bacteria in foods by decreasing the reaction rates of biochemical reactions that enable bacteria to reproduce. shows how temperature affects the light emitted by two chemiluminescent light sticks. At temperature, the reaction that produces light in a chemiluminescent light stick occurs more rapidly, producing more photons of light per unit time. Consequently, the light glows brighter in hot water (left) than in ice water (right). In systems where more than one reaction is possible, the same reactants can produce different products under different reaction conditions. For example, in the presence of dilute sulfuric acid and at temperatures around 100°C, ethanol is converted to diethyl ether:
\( 2CH_{3}CH_{2}OH \overset{H_{2}SO_{4}}{\longrightarrow} CH_{3}CH_{2}OCH_{2}CH_{3} + H_{2}O \tag{14.1.1} \) At 180°C, however, a completely different reaction occurs, which produces ethylene as the major product:
\( CH_{3}CH_{2}OH \overset{H_{2}SO_{4}}{\longrightarrow} CH_{2}CH_{2} + H_{2}O \tag{14.1.2} \) When two reactants are in the same fluid phase, their particles collide more frequently than when one or both reactants are solids (or when they are in different fluids that do not mix). If the reactants are uniformly dispersed in a single homogeneous solution, then the number of collisions per unit time depends on concentration and temperature, as we have just seen. If the reaction is heterogeneous, however, the reactants are in two different phases, and collisions between the reactants can occur only at interfaces between phases. The number of collisions between reactants per unit time is substantially reduced relative to the homogeneous case, and, hence, so is the reaction rate. The reaction rate of a heterogeneous reaction depends on the surface area of the more condensed phase. Automobile engines use surface area effects to increase reaction rates. Gasoline is injected into each cylinder, where it combusts on ignition by a spark from the spark plug. The gasoline is injected in the form of microscopic droplets because in that form it has a much larger surface area and can burn much more rapidly than if it were fed into the cylinder as a stream. Similarly, a pile of finely divided flour burns slowly (or not at all), but spraying finely divided flour into a flame produces a vigorous reaction ( ). Similar phenomena are partially responsible for dust explosions that occasionally destroy grain elevators or coal mines. < | 4,511 | 1,818 |
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Aldehydes and ketones are carbonyl compounds in which the carbonyl carbon is connected only to carbons atoms (in the case of a ketone) or to one carbon and one hydrogen atom (in the case of an aldehyde). Carboxyloids, or carboxylic acid derivatives, are carbonyl compounds in which the carbonyl carbon is attached to one carbon and one heteroatom. Most commonly, the heteroatom is an oxygen, nitrogen, sulfur or chlorine Carboxylic acid derivatives were historically thought of as being made from carboxylic acids; hence the name. That name is a mouthful; we will use the term "carboxyloids", a term coined by the 20th century physical organic chemist, Christopher Ingold. The reactivity of carboxyloids is typically different from Because of this difference it is useful to study these compounds separately from the simple carbonyls. What are some things that the heteroatoms involved in carboxyloids have in common? Like simple carbonyls, carboxyloids react with nucleophiles Just like simple carbonyls, the LUMO of a carboxyloid is usually the C=O pi antibonding orbital (the π*) Populating this orbital with a pair of electrons from the nucleophile results in breaking the C=O pi bond, leaving only a C-O sigma bond. Remember that this addition to the C=O bond is often reversible In the case of carboxyloids, however, re-forming the pi bond can go through two pathways In one pathway, the nucleophile can be displaced, returning to starting materials In another pathway, the group attached to the carbonyl can be displaced instead In that pathway, a new product results. , | 1,592 | 1,819 |
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Chemists use various strategies to increase the yield of the desired products of reactions. When synthesizing an ester, for example, how can a chemist control the reaction conditions to obtain the maximum amount of the desired product? Only three types of stresses can change the composition of an equilibrium mixture: (1) a change in the concentrations (or partial pressures) of the components by adding or removing reactants or products, (2) a change in the total pressure or volume, and (3) a change in the temperature of the system. In this section, we explore how changes in reaction conditions can affect the equilibrium composition of a system. We will explore each of these possibilities in turn. If we add a small volume of carbon tetrachloride (CCl ) solvent to a flask containing crystals of iodine, we obtain a saturated solution of I in CCl , along with undissolved crystals:
\[I_{2(s)} \rightleftharpoons I_{2(soln)}\tag{15.5.1}\] The system reaches equilibrium, with = [I ]. If we add more CCl , thereby diluting the solution, is now less than . Le Chatelier’s principle tells us that the system will react to relieve the stress—but how? Adding solvent stressed the system by decreasing the concentration of dissolved I . Hence more crystals will dissolve, thereby increasing the concentration of dissolved I until the system again reaches equilibrium if enough solid I is available ( ). By adding solvent, we drove the reaction shown in to the right as written. We encounter a more complex system in the reaction of hydrogen and nitrogen to form ammonia:
\[N_{2(g)}+3H_{2(g)} \rightleftharpoons 2NH_{3(g)} \tag{15.5.2}\] The for this reaction is 2.14 × 10 at about 540 K. Under one set of equilibrium conditions, the partial pressure of ammonia is = 0.454 atm, that of hydrogen is = 2.319 atm, and that of nitrogen is = 0.773 atm. If an additional 1 atm of hydrogen is added to the reactor to give = 3.319 atm, how will the system respond? Because the stress is an increase in the system must respond in some way that decreases the partial pressure of hydrogen to counteract the stress. The reaction will therefore proceed to the right as written, consuming H and N and forming additional NH . Initially, the partial pressures of H and N will decrease, and the partial pressure of NH will increase until the system eventually reaches a new equilibrium composition, which will have a net increase in We can confirm that this is indeed what will happen by evaluating under the new conditions and comparing its value with . The equations used to evaluate and have the same form: substituting the values after adding hydrogen into the expression for results in the following: \[Q_p=\dfrac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3}=\dfrac{(0.454)^2}{(0.773)(2.319+1.00)^3}=7.29 \times 10^{-3}\] Thus < , which tells us that the ratio of products to reactants is less than at equilibrium. To reach equilibrium, the reaction must proceed to the right as written: the partial pressures of the products will increase, and the partial pressures of the reactants will decrease. will thereby increase until it equals , and the system will once again be at equilibrium. Changes in the partial pressures of the various substances in the reaction mixture ( ) as a function of time are shown in . We can force a reaction to go essentially to completion, regardless of the magnitude of , by continually removing one of the products from the reaction mixture. Consider, for example, the reaction, in which hydrogen reacts with carbon monoxide to form methane and water:
\[CO_{(g)}+3H_{2(g)} \rightleftharpoons CH_{4(g)}+H_2O_{(g)} \tag{15.5.3}\] This reaction is used for the industrial production of methane, whereas the reverse reaction is used for the production of H (Example 14). The expression for has the following form:
\[Q=\dfrac{[CH_4,H_2O]}{[CO,H_2]^3} \tag{15.5.4}\] Regardless of the magnitude of , if either H O or CH can be removed from the reaction mixture so that [H O] or [CH ] is approximately zero, then ≈ 0. In other words, when product is removed, the system is stressed ( << ), and more product will form to counter the stress. Because water (bp = 100°C) is much less volatile than methane, hydrogen, or carbon monoxide (all of which have boiling points below −100°C), passing the gaseous reaction mixture through a cold coil will cause the water vapor to condense to a liquid that can be drawn off. Continuing to remove water from the system forces the reaction to the right as the system attempts to equilibrate, thus enriching the reaction mixture in methane. This technique, referred to as , can be used to force a reaction to completion even if is relatively small. For example, esters are usually synthesized by removing water. The products of the condensation reaction are shown here. In , we will describe the thermodynamic basis for the change in the equilibrium position caused by changes in the concentrations of reaction components. For each equilibrium system, predict the effect of the indicated stress on the specified quantity. balanced chemical equations and changes effects of indicated stresses Use and to predict the effect of the stress on each reaction. Exercise For each equilibrium system, predict the effect that the indicated stress will have on the specified quantity. Because liquids are relatively incompressible, changing the pressure above a liquid solution has little effect on the concentrations of dissolved substances. Consequently, changes in external pressure have very little effect on equilibrium systems that contain only solids or liquids. In contrast, because gases are highly compressible, their concentrations vary dramatically with pressure. From the ideal gas law, = , described in , the concentration ( ) of a gas is related to its pressure as follows:
\[C=\dfrac{n}{V}=\dfrac{P}{RT} \tag{15.5.5}\] Hence the concentration of any gaseous reactant or product is directly proportional to the applied pressure ( ) and inversely proportional to the total volume ( ). Consequently, the equilibrium compositions of systems that contain gaseous substances are quite sensitive to changes in pressure, volume, and temperature. These principles can be illustrated using the reversible dissociation of gaseous N O to gaseous NO ( ). \[N_2O_4 \rightleftharpoons 2 NO_2 \tag{15.1.1} \] The syringe shown in initially contains an equilibrium mixture of colorless N O and red-brown NO . Decreasing the volume by 50% causes the mixture to become darker because all concentrations have doubled. Decreasing the volume also constitutes a stress, however, as we can see by examining the effect of a change in volume on . At equilibrium, \[Q = K = \dfrac{[NO_2]^2}{[N_2O_4]} \tag{15.5.6a}\] If the volume is decreased by half, the concentrations of the substances in the mixture are doubled, so the new reaction quotient is as follows:
\[Q=\dfrac{(2[NO_2]_i)^2}{2[N_2O_4]_i}=\dfrac{4([NO_2]_i)^2}{2[N_2O_4]_i}=2K \tag{15.5.6b}\] Because is now greater than , the system is no longer at equilibrium. The stress can be relieved if the reaction proceeds to the left, consuming 2 mol of NO for every 1 mol of N O produced. This will decrease the concentration of NO and increase the concentration of N O , causing to decrease until it once again equals . Thus, as shown in part (c) in , the intensity of the brown color due to NO decreases with time following the change in volume. (a) The syringe with a total volume of 15 mL contains an equilibrium mixture of N O and NO ; the red-brown color is proportional to the NO concentration. (b) If the volume is rapidly decreased by a factor of 2 to 7.5 mL, the initial effect is to double the concentrations of all species present, including NO . Hence the color becomes more intense. (c) With time, the system adjusts its composition in response to the stress as predicted by Le Chatelier’s principle, forming colorless N O at the expense of red-brown NO , which decreases the intensity of the color of the mixture. Increasing the pressure of a system (or decreasing the volume) favors the side of the reaction that has fewer gaseous molecules and vice versa. In general, . Increasing the pressure on a system (or decreasing the volume) will favor the side of the reaction that has fewer gaseous molecules and vice versa. For each equilibrium system, write the reaction quotient for the system if the pressure is decreased by a factor of 2 (i.e., if the volume is doubled) at constant temperature and then predict the direction of the reaction. balanced chemical equations direction of reaction if pressure is halved Use Le Chatelier’s principle to predict the effect of the stress. \[Q=\dfrac{[1/2\;NH_3]^2}{[1/2\;N_2,1/2H_2]^3}=\dfrac{1/4\;[NH_3]^2}{1/16\;[N_2,H_2]^3}=4K \notag \] \[Q=\dfrac{[C_2H_4]^2}{[C_2H_2,C_2H_6]}=\dfrac{[1/2C_2H_4]^2}{[1/2C_2H_2,1/2C_2H_6]}=\dfrac{1/4[C_2H_4]^2}{1/4[C_2H_2,C_2H_6]}=K \notag \] \[Q=\dfrac{[1/2NO]^2[1/2O_2]}{[1/2 NO_2]^2}=\dfrac{1/8[NO]^2[O_2]}{1/4[NO_2]^2}=1/2K \notag \] Exercise For each equilibrium system, write a new reaction quotient for the system if the pressure is increased by a factor of 2 (i.e., if the volume is halved) at constant temperature and then predict the direction in which the reaction will shift. In all the cases we have considered so far, the magnitude of the equilibrium constant, or , was constant. Changes in temperature can, however, change the value of the equilibrium constant without immediately affecting the reaction quotient ( ≠ ). In this case, the system is no longer at equilibrium; the composition of the system will change until equals at the new temperature. To predict how an equilibrium system will respond to a change in temperature, we must know something about the enthalpy change of the reaction (Δ ). As you learned in , heat is released to the surroundings in an exothermic reaction (Δ < 0), and heat is absorbed from the surroundings in an endothermic reaction (Δ > 0). We can express these changes in the following way:
Thus heat can be thought of as a product in an exothermic reaction and as a reactant in an endothermic reaction. Increasing the temperature of a system corresponds to adding heat. Le Chatelier’s principle predicts that an exothermic reaction will shift to the left (toward the reactants) if the temperature of the system is increased (heat is added). Conversely, an endothermic reaction will shift to the right (toward the products) if the temperature of the system is increased. If a reaction is thermochemically neutral (Δ = 0), then a change in temperature will not affect the equilibrium composition. We can examine the effects of temperature on the dissociation of N O to NO , for which Δ = +58 kJ/mol. This reaction can be written as follows:
\[58\; kJ+N_2O_{4(g)} \rightleftharpoons 2 NO_{2(g)} \tag{15.5.9}\] Increasing the temperature (adding heat to the system) is a stress that will drive the reaction to the right, as illustrated in . Thus increasing the temperature increases the ratio of NO to N O at equilibrium, which increases . The effect of increasing the temperature on a system at equilibrium can be summarized as follows: . shows the temperature dependence of the equilibrium constants for the synthesis of ammonia from hydrogen and nitrogen, which is an exothermic reaction with Δ ° = −91.8 kJ/mol. The values of both and decrease dramatically with increasing temperature, as predicted for an exothermic reaction. Temperature Dependence of for Increasing the temperature causes endothermic reactions to favor products and exothermic reactions to favor reactants. For each equilibrium reaction, predict the effect of decreasing the temperature: balanced chemical equations and values of Δ effects of decreasing temperature Use Le Chatelier’s principle to predict the effect of decreasing the temperature on each reaction. If the temperature of the mixture is decreased, heat (one of the reactants) is being removed from the system, which causes the equilibrium to shift to the left. Hence the thermal decomposition of calcium carbonate is less favored at lower temperatures. Exercise For each equilibrium system, predict the effect of increasing the temperature on the reaction mixture: Three types of stresses can alter the composition of an equilibrium system: adding or removing reactants or products, changing the total pressure or volume, and changing the temperature of the system. A reaction with an unfavorable equilibrium constant can be driven to completion by continually removing one of the products of the reaction. Equilibriums that contain different numbers of gaseous reactant and product molecules are sensitive to changes in volume or pressure; higher pressures favor the side with fewer gaseous molecules. Removing heat from an exothermic reaction favors the formation of products, whereas removing heat from an endothermic reaction favors the formation of reactants. If an equilibrium reaction is endothermic in the forward direction, what is the expected change in the concentration of each component of the system if the temperature of the reaction is increased? If the temperature is decreased? Write the equilibrium equation for the following system: \[4NH_{3(g)} +5O_{2(g)} \rightleftharpoons 4NO_{(g)} + 6H_2O_{(g)}\ Would you expect the equilibrium to shift toward the products or reactants with an increase in pressure? Why? The reaction rate approximately doubles for every 10°C rise in temperature. What happens to ? The formation of A B (g) via the equilibrium reaction is exothermic. What happens to the ratio / if the temperature is increased? If both temperature and pressure are increased? In each system, predict the effect that the indicated change will have on the specified quantity at equilibrium: H is removed; what is the effect on If Br is removed; what is the effect on ? If CO is removed; what is the effect on What effect will the indicated change have on the specified quantity at equilibrium? If NH Cl is increased; what is the effect on ? If O is added; what is the effect on If Cl is removed; what is the effect on For each equilibrium reaction, describe how and change when the pressure is increased, the temperature is increased, the volume of the system is increased, and the concentration(s) of the reactant(s) is increased. For each equilibrium reaction, describe how and change when the pressure is decreased, the temperature is increased, the volume of the system is decreased, and the concentration(s) of the reactant(s) is increased. Le Chatelier’s principle states that a system will change its composition to counteract stress. For the system write the equilibrium constant expression . What changes in the values of and would you anticipate when (a) the volume is doubled, (b) the pressure is increased by a factor of 2, and (c) COCl is removed from the system? For the equilibrium system Δ ° = 284 kJ, write the equilibrium constant expression . What happens to the values of and if the reaction temperature is increased? What happens to these values if both the temperature and pressure are increased? Carbon and oxygen react to form CO gas via for which = 1.2 × 10 . Would you expect to increase or decrease if the volume of the system were tripled? Why? The reaction has = 2.2 × 10 at 100°C. Starting with an initial ) of 1.0 atm, you determine the following values of at three successive time intervals: 6.32 × 10 atm, 1.78 × 10 atm, and 1.02 × 10 atm. Based on these data, in which direction will the reaction proceed after each measurement? If chlorine gas is added to the system, what will be the effect on ? The following table lists experimentally determined partial pressures at three temperatures for the reaction Is this an endothermic or an exothermic reaction? Explain your reasoning. The dissociation of water vapor proceeds according to the following reaction: At 1300 K, there is 0.0027% dissociation, whereas at 2155 K, the dissociation is 1.18%. Calculate and . Is this an endothermic reaction or an exothermic reaction? How do the magnitudes of the two equilibriums compare? Would increasing the pressure improve the yield of H gas at either temperature? (Hint: assume that the system initially contains 1.00 mol of H O in a 1.00 L container.) When 1.33 mol of CO and 1.33 mol of H are mixed in a 0.750 L container and heated to 395°C, they react according to the following equation: If = 0.802, what are the equilibrium concentrations of each component of the equilibrium mixture? What happens to if H O is removed during the course of the reaction? The equilibrium reaction has = 2.2 × 10 at 298 K. If you begin with 2.0 mol of Br and 2.0 mol of H in a 5.0 L container, what is the partial pressure of HBr at equilibrium? What is the partial pressure of H at equilibrium? If H is removed from the system, what is the effect on the partial pressure of Br ? Iron(II) oxide reacts with carbon monoxide according to the following equation: At 800°C, = 0.34; at 1000°C, = 0.40. The equilibrium constant for the reaction is 1.9 at 1000 K and 0.133 at 298 K. Data for the oxidation of methane, in a closed 5.0 L vessel are listed in the following table. Fill in the blanks and determine the missing values of and (indicated by ?) as the reaction is driven to completion. None of the changes would affect ; (a) doubles; (b) is halved; decreases. would not change; it does not depend on volume. [CO] = [H O] = 0.839 M, [CO ] = [H ] = 0.930 M; no effect on | 17,714 | 1,820 |
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This problem is answered through consideration of the . from easiest to hardest to replace: NO Cl H O NH PPh CO Associative Rate Law: Rate = [MLn,X], if MLn is the complex and X is the new ligand. Because Rate = k[MLn,L] and [MLn] is held constant while [L] is varied, then the slope of the line is k [MLn]. Since you would know the value of [MLn], you could obtain the rate constant from the quantity (slope/ [MLn]). Dissociative Rate Law: Rate = [MLn], if MLn is the complex. There is no dependence on [X], if X is the new ligand. The metal centre is becoming more crowded as the new ligand arrives, so an increase in energy owing to steric hindrance may also play a role in the transition state energetics. a) The new ligand, B, is arriving at the same time as the old ligand, A, is departing. We might also describe it as new ligand B pushing old ligand A out of the complex. b) Rate = k[ML5A,B], which looks like an associative rate law. c) This is a thought-provoking question without a definite answer. Associative mechanisms typically have lower activation enthalpy than dissociative mechanisms, because there has also been some bond-making prior to the bond-breaking in the rate determining step. The associative interchange would be a little more like the associative mechanism than dissociative. The mix of bond-making and bond-breaking at the transition state would make the enthalpy of activation relatively low. Associative mechanisms have negative activation entropies, whereas dissociative mechanisms have positive activation entropies. The associative interchange could be in between the two, given that the elementary step would be close to entropically neutral overall. What happens at the transition state is a little harder to imagine, but it might reflect the small changes in entropy through the course of the reaction, producing a small entropy of activation. On the other hand, if the incoming ligand is forced to adopt some specific approach as it comes into the molecule (to stay out of the way of the departing ligand, for example) then that restriction could show up as a small negative activation entropy. a) The lone pair donation from one ligand appears to push another ligand out. b) The first step is probably rate determining, because of the bond breaking involved. c) If the first step is rate determining, Rate = k [ML5A]. d) Another question without a very clear answer. Compared with an associative mechanism, the activation entropy is probably much more positive, because additional degrees of freedom are being gained as the molecule heads over the activation barrier and one of the ligands separates to be on its own. However, the activation entropy may be less positive than in a regular dissociation, because in this case the breaking of one bond has to be coordinated with the formation of another. The enthalpy of activation has both a bond-making and bond-breaking component, a little like in an associative mechanism. However, the amount of bond making here is probably less important, because pi bonds are typically not as strong as sigma bonds. The activation enthalpy is probably higher than an associative pathway but not as high as a dissociative one. e) The donor ligand must have a lone pair. Oxygen donors would be good candidates, because even if one lone pair is already donating in a sigma bond, an additional lone pair may be available for pi donation. The same thing is true for halogen donors. It would also be true for anionic nitrogen donors but not for neutral nitrogen donors, because a neutral nitrogen has only one lone pair. , | 3,613 | 1,821 |
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How do alkali metals differ from alkaline earth metals in atomic structure and general properties? The alkali metals all have a single electron in their outermost shell. In contrast, the alkaline earth metals have a completed subshell in their outermost shell. In general, the alkali metals react faster and are more reactive than the corresponding alkaline earth metals in the same period. Why does the reactivity of the alkali metals decrease from cesium to lithium? Predict the formulas for the nine compounds that may form when each species in column 1 of reacts with each species in column 2. \[\ce{Na + I2 ⟶ 2NaI\\
2Na + Se ⟶ Na2Se\\
2Na + O2 ⟶ Na2O2}\] \[\ce{Sr + I2⟶SrI2\\
Sr + Se⟶SeSe\\
2Sr + O2⟶2SrO}\] \[\ce{2Al + 3I2⟶2AlI3\\
2Al + 3Se⟶Al2Se3\\
4Al + 3O2⟶2Al2O3}\] Predict the best choice in each of the following. You may wish to review the chapter on electronic structure for relevant examples. Sodium chloride and strontium chloride are both white solids. How could you distinguish one from the other? The possible ways of distinguishing between the two include infrared spectroscopy by comparison of known compounds, a flame test that gives the characteristic yellow color for sodium (strontium has a red flame), or comparison of their solubilities in water. At 20 °C, NaCl dissolves to the extent of \(\mathrm{\dfrac{35.7\: g}{100\: mL}}\) compared with \(\mathrm{\dfrac{53.8\: g}{100\: mL}}\) for SrCl . Heating to 100 °C provides an easy test, since the solubility of NaCl is \(\mathrm{\dfrac{39.12\: g}{100\: mL}}\), but that of SrCl is \(\mathrm{\dfrac{100.8\: g}{100\: mL}}\). Density determination on a solid is sometimes difficult, but there is enough difference (2.165 g/mL NaCl and 3.052 g/mL SrCl ) that this method would be viable and perhaps the easiest and least expensive test to perform. The reaction of quicklime, CaO, with water produces slaked lime, Ca(OH) , which is widely used in the construction industry to make mortar and plaster. The reaction of quicklime and water is highly exothermic: \[\ce{CaO}(s)+\ce{H2O}(l)⟶\ce{Ca(OH)2}(s) \hspace{20px} ΔH=\mathrm{−350\: kJ\:mol^{−1}}\] Write a balanced equation for the reaction of elemental strontium with each of the following: (a) \(\ce{2Sr}(s)+\ce{O2}(g)⟶\ce{2SrO}(s)\); (b) \(\ce{Sr}(s)+\ce{2HBr}(g)⟶\ce{SrBr2}(s)+\ce{H2}(g)\); (c) \(\ce{Sr}(s)+\ce{H2}(g)⟶\ce{SrH2}(s)\); (d) \(\ce{6Sr}(s)+\ce{P4}(s)⟶\ce{2Sr3P2}(s)\); (e) \(\ce{Sr}(s)+\ce{2H2O}(l)⟶\ce{Sr(OH)2}(aq)+\ce{H2}(g)\) How many moles of ionic species are present in 1.0 L of a solution marked 1.0 mercury(I) nitrate? What is the mass of fish, in kilograms, that one would have to consume to obtain a fatal dose of mercury, if the fish contains 30 parts per million of mercury by weight? (Assume that all the mercury from the fish ends up as mercury(II) chloride in the body and that a fatal dose is 0.20 g of HgCl .) How many pounds of fish is this? 11 lb The elements sodium, aluminum, and chlorine are in the same period. Does metallic tin react with HCl? Yes, tin reacts with hydrochloric acid to produce hydrogen gas. What is tin pest, also known as tin disease? Compare the nature of the bonds in PbCl to that of the bonds in PbCl . In PbCl , the bonding is ionic, as indicated by its melting point of 501 °C. In PbCl , the bonding is covalent, as evidenced by it being an unstable liquid at room temperature. Is the reaction of rubidium with water more or less vigorous than that of sodium? How does the rate of reaction of magnesium compare? Write an equation for the reduction of cesium chloride by elemental calcium at high temperature. \[\ce{2CsCl}(l)+\ce{Ca}(g)\:\mathrm{\overset{countercurrent \\ fractionating \\ tower}{\xrightarrow{\hspace{40px}}}}\:\ce{2Cs}(g)+\ce{CaCl2}(l)\] Why is it necessary to keep the chlorine and sodium, resulting from the electrolysis of sodium chloride, separate during the production of sodium metal? Give balanced equations for the overall reaction in the electrolysis of molten lithium chloride and for the reactions occurring at the electrodes. You may wish to review the chapter on electrochemistry for relevant examples. Cathode (reduction): \(\ce{2Li+} + \ce{2e-}⟶\ce{2Li}(l)\); Anode (oxidation): \(\ce{2Cl-}⟶\ce{Cl2}(g)+\ce{2e-}\); Overall reaction: \(\ce{2Li+}+\ce{2Cl-}⟶\ce{2Li}(l)+\ce{Cl2}(g)\) The electrolysis of molten sodium chloride or of aqueous sodium chloride produces chlorine. Calculate the mass of chlorine produced from 3.00 kg sodium chloride in each case. You may wish to review the chapter on electrochemistry for relevant examples. What mass, in grams, of hydrogen gas forms during the complete reaction of 10.01 g of calcium with water? 0.5035 g H How many grams of oxygen gas are necessary to react completely with 3.01 × 10 atoms of magnesium to yield magnesium oxide? Magnesium is an active metal; it burns in the form of powder, ribbons, and filaments to provide flashes of brilliant light. Why is it possible to use magnesium in construction? Despite its reactivity, magnesium can be used in construction even when the magnesium is going to come in contact with a flame because a protective oxide coating is formed, preventing gross oxidation. Only if the metal is finely subdivided or present in a thin sheet will a high-intensity flame cause its rapid burning. Why is it possible for an active metal like aluminum to be useful as a structural metal? Describe the production of metallic aluminum by electrolytic reduction. Extract from ore: \(\ce{AlO(OH)}(s)+\ce{NaOH}(aq)+\ce{H2O}(l)⟶\ce{Na[Al(OH)4]}(aq)\) Recover: \(\ce{2Na[Al(OH)4]}(s)+\ce{H2SO4}(aq)⟶\ce{2Al(OH)3}(s)+\ce{Na2SO4}(aq)+\ce{2H2O}(l)\) Sinter: \(\ce{2Al(OH)3}(s)⟶\ce{Al2O3}(s)+\ce{3H2O}(g)\) Dissolve in Na AlF ( ) and electrolyze: \(\ce{Al^3+}+\ce{3e-}⟶\ce{Al}(s)\) What is the common ore of tin and how is tin separated from it? A chemist dissolves a 1.497-g sample of a type of metal (an alloy of Sn, Pb, Sb, and Cu) in nitric acid, and metastannic acid, H SnO , is precipitated. She heats the precipitate to drive off the water, which leaves 0.4909 g of tin(IV) oxide. What was the percentage of tin in the original sample? 25.83% Consider the production of 100 kg of sodium metal using a current of 50,000 A, assuming a 100% yield. (a) How long will it take to produce the 100 kg of sodium metal? (b) What volume of chlorine at 25 °C and 1.00 atm forms? What mass of magnesium forms when 100,000 A is passed through a MgCl melt for 1.00 h if the yield of magnesium is 85% of the theoretical yield? 39 kg Give the hybridization of the metalloid and the molecular geometry for each of the following compounds or ions. You may wish to review the chapters on chemical bonding and advanced covalent bonding for relevant examples. Write a Lewis structure for each of the following molecules or ions. You may wish to review the chapter on chemical bonding. (a) H BPH : ; (b) \(\ce{BF4-}\): ; (c) BBr : ; (d) B(CH ) : ; (e) B(OH) : Describe the hybridization of boron and the molecular structure about the boron in each of the following: Using only the periodic table, write the complete electron configuration for silicon, including any empty orbitals in the valence shell. You may wish to review the chapter on electronic structure. 1 2 2 3 3 3 . Write a Lewis structure for each of the following molecules and ions: Describe the hybridization of silicon and the molecular structure of the following molecules and ions: (a) (CH ) SiH: bonding about Si; the structure is tetrahedral; (b) \(\ce{SiO4^4-}\): sp bonding about Si; the structure is tetrahedral; (c) Si H : bonding about each Si; the structure is linear along the Si-Si bond; (d) Si(OH) : bonding about Si; the structure is tetrahedral; (e) \(\ce{SiF6^2-}\): sp bonding about Si; the structure is octahedral Describe the hybridization and the bonding of a silicon atom in elemental silicon. Classify each of the following molecules as polar or nonpolar. You may wish to review the chapter on chemical bonding. (a) SiH (b) Si H (c) SiCl H (d) SiF (e) SiCl F (a) nonpolar; (b) nonpolar; (c) polar; (d) nonpolar; (e) polar Silicon reacts with sulfur at elevated temperatures. If 0.0923 g of silicon reacts with sulfur to give 0.3030 g of silicon sulfide, determine the empirical formula of silicon sulfide. Name each of the following compounds: (a) tellurium dioxide or tellurium(IV) oxide; (b) antimony(III) sulfide; (c) germanium(IV) fluoride; (d) silane or silicon(IV) hydride; (e) germanium(IV) hydride Write a balanced equation for the reaction of elemental boron with each of the following (most of these reactions require high temperature): Why is boron limited to a maximum coordination number of four in its compounds? Boron has only and orbitals available, which can accommodate a maximum of four electron pairs. Unlike silicon, no orbitals are available in boron. Write a formula for each of the following compounds: From the data given in , determine the standard enthalpy change and the standard free energy change for each of the following reactions: (a) Δ ° = 87 kJ; Δ ° = 44 kJ; (b) Δ ° = −109.9 kJ; Δ = −154.7 kJ; (c) Δ ° = −510 kJ; Δ ° = −601.5 kJ A hydride of silicon prepared by the reaction of Mg Si with acid exerted a pressure of 306 torr at 26 °C in a bulb with a volume of 57.0 mL. If the mass of the hydride was 0.0861 g, what is its molecular mass? What is the molecular formula for the hydride? Suppose you discovered a diamond completely encased in a silicate rock. How would you chemically free the diamond without harming it? A mild solution of hydrofluoric acid would dissolve the silicate and would not harm the diamond. Carbon forms a number of allotropes, two of which are graphite and diamond. Silicon has a diamond structure. Why is there no allotrope of silicon with a graphite structure? Nitrogen in the atmosphere exists as very stable diatomic molecules. Why does phosphorus form less stable P molecules instead of P molecules? In the N molecule, the nitrogen atoms have an σ bond and two π bonds holding the two atoms together. The presence of three strong bonds makes N a very stable molecule. Phosphorus is a third-period element, and as such, does not form π bonds efficiently; therefore, it must fulfill its bonding requirement by forming three σ bonds. Write balanced chemical equations for the reaction of the following acid anhydrides with water: Determine the oxidation number of each element in each of the following compounds: (a) H = 1+, C = 2+, and N = 3−; (b) O = 2+ and F = 1−; (c) As = 3+ and Cl = 1− Determine the oxidation state of sulfur in each of the following: Arrange the following in order of increasing electronegativity: F; Cl; O; and S. S < Cl < O < F Why does white phosphorus consist of tetrahedral P molecules while nitrogen consists of diatomic N molecules? Why does hydrogen not exhibit an oxidation state of 1− when bonded to nonmetals? The electronegativity of the nonmetals is greater than that of hydrogen. Thus, the negative charge is better represented on the nonmetal, which has the greater tendency to attract electrons in the bond to itself. The reaction of calcium hydride, CaH , with water can be characterized as a Lewis acid-base reaction: \[\ce{CaH2}(s)+\ce{2H2O}(l)⟶\ce{Ca(OH)2}(aq)+\ce{2H2}(g)\] Identify the Lewis acid and the Lewis base among the reactants. The reaction is also an oxidation-reduction reaction. Identify the oxidizing agent, the reducing agent, and the changes in oxidation number that occur in the reaction. In drawing Lewis structures, we learn that a hydrogen atom forms only one bond in a covalent compound. Why? Hydrogen has only one orbital with which to bond to other atoms. Consequently, only one two-electron bond can form. What mass of CaH is necessary to react with water to provide enough hydrogen gas to fill a balloon at 20 °C and 0.8 atm pressure with a volume of 4.5 L? The balanced equation is: \[\ce{CaH2}(s)+\ce{2H2O}(l)⟶\ce{Ca(OH)2}(aq)+\ce{2H2}(g)\] What mass of hydrogen gas results from the reaction of 8.5 g of KH with water? \[\ce{KH + H2O ⟶ KOH + H2}\] 0.43 g H Carbon forms the \(\ce{CO3^2-}\) ion, yet silicon does not form an analogous \(\ce{SiO3^2-}\) ion. Why? Complete and balance the following chemical equations: (a) hardening of plaster containing slaked lime \[\ce{Ca(OH)2 + CO2 ⟶}\] (b) removal of sulfur dioxide from the flue gas of power plants \[\ce{CaO + SO2 ⟶}\] (c) the reaction of baking powder that produces carbon dioxide gas and causes bread to rise \[\ce{NaHCO3 + NaH2PO4 ⟶}\] (a) \(\ce{Ca(OH)2}(aq)+\ce{CO2}(g)⟶\ce{CaCO3}(s)+\ce{H2O}(l)\); (b) \(\ce{CaO}(s)+\ce{SO2}(g)⟶\ce{CaSO3}(s)\); (c) \(\ce{2NaHCO3}(s)+\ce{NaH2PO4}(aq)⟶\ce{Na3PO4}(aq)+\ce{2CO2}(g)+\ce{2H2O}(l)\) Heating a sample of Na CO ⋅ H O weighing 4.640 g until the removal of the water of hydration leaves 1.720 g of anhydrous Na CO . What is the formula of the hydrated compound? Write the Lewis structures for each of the following: (a) NH : For each of the following, indicate the hybridization of the nitrogen atom (for \(\ce{N3-}\), the central nitrogen). Explain how ammonia can function both as a Brønsted base and as a Lewis base. Ammonia acts as a Brønsted base because it readily accepts protons and as a Lewis base in that it has an electron pair to donate. Determine the oxidation state of nitrogen in each of the following. You may wish to review the chapter on chemical bonding for relevant examples. For each of the following, draw the Lewis structure, predict the ONO bond angle, and give the hybridization of the nitrogen. You may wish to review the chapters on chemical bonding and advanced theories of covalent bonding for relevant examples. (a) NO (b) \(\ce{NO2-}\) (c) \(\ce{NO2+}\) (a) NO : Nitrogen is hybridized. The molecule has a linear geometry with an ONO bond angle of 180°. How many grams of gaseous ammonia will the reaction of 3.0 g hydrogen gas and 3.0 g of nitrogen gas produce? Although PF and AsF are stable, nitrogen does not form NF molecules. Explain this difference among members of the same group. Nitrogen cannot form a NF molecule because it does not have orbitals to bond with the additional two fluorine atoms. The equivalence point for the titration of a 25.00-mL sample of CsOH solution with 0.1062 HNO is at 35.27 mL. What is the concentration of the CsOH solution? Write the Lewis structure for each of the following. You may wish to review the chapter on chemical bonding and molecular geometry. (a) ; (b) ; (c) ; (d) ; (e) Describe the molecular structure of each of the following molecules or ions listed. You may wish to review the chapter on chemical bonding and molecular geometry. Complete and balance each of the following chemical equations. (In some cases, there may be more than one correct answer.) (a) \(\ce{P4}(s)+\ce{4Al}(s)⟶\ce{4AlP}(s)\); (b) \(\ce{P4}(s)+\ce{12Na}(s)⟶\ce{4Na3P}(s)\); (c) \(\ce{P4}(s)+\ce{10F2}(g)⟶\ce{4PF5}(l)\); (d) \(\ce{P4}(s)+\ce{6Cl2}(g)⟶\ce{4PCl3}(l)\) or \(\ce{P4}(s)+\ce{10Cl2}(g)⟶\ce{4PCl5}(l)\); (e) \(\ce{P4}(s)+\ce{3O2}(g)⟶\ce{P4O6}(s)\) or \(\ce{P4}(s)+\ce{5O2}(g)⟶\ce{P4O10}(s)\); (f) \(\ce{P4O6}(s)+\ce{2O2}(g)⟶\ce{P4O10}(s)\) Describe the hybridization of phosphorus in each of the following compounds: P O , P O , PH I (an ionic compound), PBr , H PO , H PO , PH , and P H . You may wish to review the chapter on advanced theories of covalent bonding. What volume of 0.200 NaOH is necessary to neutralize the solution produced by dissolving 2.00 g of PCl is an excess of water? Note that when H PO is titrated under these conditions, only one proton of the acid molecule reacts. 291 mL How much POCl can form from 25.0 g of PCl and the appropriate amount of H O? How many tons of Ca (PO ) are necessary to prepare 5.0 tons of phosphorus if the yield is 90%? 28 tons Write equations showing the stepwise ionization of phosphorous acid. Draw the Lewis structures and describe the geometry for the following: (a) ; (b) ; (c) ; (d) Why does phosphorous acid form only two series of salts, even though the molecule contains three hydrogen atoms? Assign an oxidation state to phosphorus in each of the following: (a) P = 3+; (b) P = 5+; (c) P = 3+; (d) P = 5+; (e) P = 3−; (f) P = 5+ Phosphoric acid, one of the acids used in some cola drinks, is produced by the reaction of phosphorus(V) oxide, an acidic oxide, with water. Phosphorus(V) oxide is prepared by the combustion of phosphorus. Predict the product of burning francium in air. FrO Using equations, describe the reaction of water with potassium and with potassium oxide. Write balanced chemical equations for the following reactions: (a) \(\ce{2Zn}(s)+\ce{O2}(g)⟶\ce{2ZnO}(s)\); (b) \(\ce{ZnCO3}(s)⟶\ce{ZnO}(s)+\ce{CO2}(g)\); (c) \(\ce{ZnCO3}(s)+\ce{2CH3COOH}(aq)⟶\ce{Zn(CH3COO)2}(aq)+\ce{CO2}(g)+\ce{H2O}(l)\); (d) \(\ce{Zn}(s)+\ce{2HBr}(aq)⟶\ce{ZnBr2}(aq)+\ce{H2}(g)\) Write balanced chemical equations for the following reactions: Illustrate the amphoteric nature of aluminum hydroxide by citing suitable equations. \(\ce{Al(OH)3}(s)+\ce{3H+}(aq)⟶\ce{Al^3+}+\ce{3H2O}(l)\); \(\ce{Al(OH)3}(s)+\ce{OH-}⟶\ce{[Al(OH)4]-}(aq)\) Write balanced chemical equations for the following reactions: Write balanced chemical equations for the following reactions: (a) \(\ce{Na2O}(s)+\ce{H2O}(l)⟶\ce{2NaOH}(aq)\); (b) \(\ce{Cs2CO3}(s)+\ce{2HF}(aq)⟶\ce{2CsF}(aq)+\ce{CO2}(g)+\ce{H2O}(l)\); (c) \(\ce{Al2O3}(s)+\ce{6HClO4}(aq)⟶\ce{2Al(ClO4)3}(aq)+\ce{3H2O}(l)\); (d) \(\ce{Na2CO3}(aq)+\ce{Ba(NO3)2}(aq)⟶\ce{2NaNO3}(aq)+\ce{BaCO3}(s)\); (e) \(\ce{TiCl4}(l)+\ce{4Na}(s)⟶\ce{Ti}(s)+\ce{4NaCl}(s)\) What volume of 0.250 H SO solution is required to neutralize a solution that contains 5.00 g of CaCO ? Which is the stronger acid, HClO or HBrO ? Why? HClO is the stronger acid because, in a series of oxyacids with similar formulas, the higher the electronegativity of the central atom, the stronger is the attraction of the central atom for the electrons of the oxygen(s). The stronger attraction of the oxygen electron results in a stronger attraction of oxygen for the electrons in the O-H bond, making the hydrogen more easily released. The weaker this bond, the stronger the acid. Write a balanced chemical equation for the reaction of an excess of oxygen with each of the following. Remember that oxygen is a strong oxidizing agent and tends to oxidize an element to its maximum oxidation state. Which is the stronger acid, H SO or H SeO ? Why? You may wish to review the chapter on acid-base equilibria. As H SO and H SeO are both oxyacids and their central atoms both have the same oxidation number, the acid strength depends on the relative electronegativity of the central atom. As sulfur is more electronegative than selenium, H SO is the stronger acid. Explain why hydrogen sulfide is a gas at room temperature, whereas water, which has a lower molecular mass, is a liquid. Give the hybridization and oxidation state for sulfur in SO , in SO , and in H SO . SO , 4+; SO , , 6+; H SO , , 6+ Which is the stronger acid, NaHSO or NaHSO ? Determine the oxidation state of sulfur in SF , SO F , and KHS. SF : S = 6+; SO F : S = 6+; KHS: S = 2− Which is a stronger acid, sulfurous acid or sulfuric acid? Why? Oxygen forms double bonds in O , but sulfur forms single bonds in S . Why? Sulfur is able to form double bonds only at high temperatures (substantially endothermic conditions), which is not the case for oxygen. Give the Lewis structure of each of the following: Write two balanced chemical equations in which sulfuric acid acts as an oxidizing agent. There are many possible answers including: \[\ce{Cu}(s)+\ce{2H2SO4}(l)⟶\ce{CuSO4}(aq)+\ce{SO2}(g)+\ce{2H2O}(l)\] \[\ce{C}(s)+\ce{2H2SO4}(l)⟶\ce{CO2}(g)+\ce{2SO2}(g)+\ce{2H2O}(l)\] Explain why sulfuric acid, H SO , which is a covalent molecule, dissolves in water and produces a solution that contains ions. How many grams of Epsom salts (MgSO ⋅7H O) will form from 5.0 kg of magnesium? 5.1 × 10 g What does it mean to say that mercury(II) halides are weak electrolytes? Why is SnCl4 not classified as a salt? SnCl4 is not a salt because it is covalently bonded. A salt must have ionic bonds. The following reactions are all similar to those of the industrial chemicals. Complete and balance the equations for these reactions: (a) reaction of a weak base and a strong acid \[\ce{NH3 + HClO4⟶}\] (b) preparation of a soluble silver salt for silver plating \[\ce{Ag2CO3 + HNO3⟶}\] (c) preparation of strontium hydroxide by electrolysis of a solution of strontium chloride \[\ce{SrCl2}(aq)+\ce{H2O}(l)\xrightarrow{\ce{electrolysis}}\] Which is the stronger acid, HClO3 or HBrO3? Why? In oxyacids with similar formulas, the acid strength increases as the electronegativity of the central atom increases. HClO3 is stronger than HBrO3; Cl is more electronegative than Br. What is the hybridization of iodine in IF3 and IF5? Predict the molecular geometries and draw Lewis structures for each of the following. You may wish to review the chapter on chemical bonding and molecular geometry. (a) IF (b) \(\ce{I3-}\) (c) PCl (d) SeF (e) ClF (a) ; (b) ; (c) ; (d) ; (e) Which halogen has the highest ionization energy? Is this what you would predict based on what you have learned about periodic properties? Name each of the following compounds: (a) BrF (b) NaBrO (c) PBr (d) NaClO (e) KClO (a) bromine trifluoride; (b) sodium bromate; (c) phosphorus pentabromide; (d) sodium perchlorate; (e) potassium hypochlorite Explain why, at room temperature, fluorine and chlorine are gases, bromine is a liquid, and iodine is a solid. What is the oxidation state of the halogen in each of the following? (a) H IO (b) \(\ce{IO4-}\) (c) ClO (d) ICl (e) F (a) I: 7+; (b) I: 7+; (c) Cl: 4+; (d) I: 3+; Cl: 1−; (e) F: 0 Physiological saline concentration—that is, the sodium chloride concentration in our bodies—is approximately 0.16 . A saline solution for contact lenses is prepared to match the physiological concentration. If you purchase 25 mL of contact lens saline solution, how many grams of sodium chloride have you bought? Give the hybridization of xenon in each of the following. You may wish to review the chapter on the advanced theories of covalent bonding. (a) hybridized; (b) hybridized; (c) hybridized; (d) hybridized; (e) hybridized; What is the molecular structure of each of the following molecules? You may wish to review the chapter on chemical bonding and molecular geometry. Indicate whether each of the following molecules is polar or nonpolar. You may wish to review the chapter on chemical bonding and molecular geometry. (a) nonpolar; (b) nonpolar; (c) polar; (d) nonpolar; (e) polar What is the oxidation state of the noble gas in each of the following? You may wish to review the chapter on chemical bonding and molecular geometry. A mixture of xenon and fluorine was heated. A sample of the white solid that formed reacted with hydrogen to yield 81 mL of xenon (at STP) and hydrogen fluoride, which was collected in water, giving a solution of hydrofluoric acid. The hydrofluoric acid solution was titrated, and 68.43 mL of 0.3172 sodium hydroxide was required to reach the equivalence point. Determine the empirical formula for the white solid and write balanced chemical equations for the reactions involving xenon. The empirical formula is XeF , and the balanced reactions are: \[\ce{Xe}(g)+\ce{3F2}(g)\xrightarrow{Δ}\ce{XeF6}(s)\]
\[\ce{XeF6}(s)+\ce{3H2}(g)⟶\ce{6HF}(g)+\ce{Xe}(g)\] Basic solutions of Na XeO are powerful oxidants. What mass of Mn(NO ) •6H O reacts with 125.0 mL of a 0.1717 basic solution of Na XeO that contains an excess of sodium hydroxide if the products include Xe and solution of sodium permanganate? | 23,711 | 1,822 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Material_Properties/MSDS |
A Material Safety Data Sheet (MSDS) is designed to provide both workers and emergency personnel with the proper procedures for handling or working with a particular substance. MSDS's include information such as physical data (melting point, boiling point, flash point etc.), toxicity, health effects, first aid, reactivity, storage, disposal, protective equipment, and spill/leak procedures. These are of particular use if a spill or other accident occurs. The MSDS's are available from many sources such as the chemistry store, health and safety offices, material producing companies etc. This is one of the sources which points to many other sources of MSDS. It also contains information regarding the meaning of all the terms used in MSDS as well as the meaning of the data. For example, you may click see a MSDS for in this link. This site also contains tutorials as to how to read MSDS. I also suggest that you look up the MSDS's for ethanol, methanol, and ethyl ether and suggest a process of their purifications. These questions are suggested by Professor Please try to answer them according to the MSDS of a substance that interest you. A few of you may also work as a team to write specifics for the following using a MSDS for one substance that interest you. These activities were suggested by Michael Doherty | 1,344 | 1,824 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Book3A_Bioinorganic_Chemistry_(Bertini_et_al.)/04%3A_Biological_and_Synthetic_Dioxygen_Carriers/4.01%3A_Biological_Dioxygen_Transport_Systems |
Most organisms require molecular oxygen in order to survive. The dioxygen is used in a host of biochemical transformations, although most is consumed in the reaction \[O_{2} + 4 H^{+} + 4 e^{-} \rightarrow 2H_{2}O \tag{4.1}\] that is the terminal (or primary) step of oxidative phosphorylation (Chapters 5 and 6). For some small animals and for plants, where the surface-to-volume ratio is large, an adequate supply of dioxygen can be obtained from simple diffusion across cell membranes. The dioxygen may be extracted from air or water; for plants that produce dioxygen in photosynthesis, it is also available endogenously. For other organisms, particularly those with non-passive lifestyles, from scorpions to whales, diffusion does not supply sufficient dioxygen for respiration. An elegant three-component system has evolved to transport dioxygen from regions of high abundance—water (at least if free of pollutant reductants) and air—to regions of relatively low abundance and high demand—the interior cells of the organism. This process is illustrated in Figure 4.1. The central component is a dioxygen-carrier protein. In the three chemically distinct carriers that have evolved and are found today, the dioxygen-binding site in the protein, that is, the so-called "active site," is a complex either of copper or of iron. For hemoglobins, the most widely distributed family of dioxygen carriers, the active site has long been known to consist of an iron porphyrin (heme) group embedded in the protein. Almost all hemoglobins share the basic structure illustrated in Figure 4.2. Hemocyanin and hemerythrin, the other two biological dioxygen carriers, feature pairs of copper atoms and iron atoms, respectively, at the active sites.* Some basic properties of these metalloproteins are summarized in Table 4.1. * The use of the prefix - is confusing. In this context connotes blood. Thus, since hemocyanin and hemerythrin lack a heme group [an iron(II) porphyrin], they are nonheme metalloproteins. The second component of the dioxygen-transport system facilitates the sequestration of dioxygen by the dioxygen-carrier protein. Specialized organs, such as lungs in air-breathing creatures or gills in fish, offer a very large surface area to the outside environment in order to facilitate diffusion. The third component is the delivery system. The oxygen carrier is dissolved or suspended in a fluid, called blood plasma or hemolymph, that is pumped throughout the animal by another specialized organ, the heart, through a network of tubes, the blood vessels. In many organisms an additional dioxygen-binding protein, which stores dioxygen, is located in tissues that are subject to sudden and high dioxygen demand, such as muscles. These dioxygen-storage proteins are prefixed - (from the Greek root for muscle). Thus for the dioxygen-transport protein hemerythrin there exists a chemically similar dioxygen-storage protein myohemerythrin. For the hemoglobin family the corresponding storage protein is called myoglobin. Interestingly, some organisms that use hemocyanin as the dioxygen- protein use myoglobin as the dioxygen- protein. At the center of biological dioxygen transport are transition-metal complexes of iron or copper. To model such systems, chemists have prepared several synthetic oxygen carriers, especially of iron and cobalt porphyrins. In this chapter the structures and properties of biological and nonbiological oxygen carriers are described, with particular attention to the hemoglobin family. This family has been studied in more detail than any other group of proteins, and as a result a deeper understanding of the relationships among structure, properties, and biological function (i.e., physiology) exists. The central focus of this chapter is to delineate chemical features that determine the affinity of an active site, especially an iron porphyrin, for molecular oxygen. In order to develop this theme, macroscopic (thermodynamic and kinetic) factors associated with dioxygen binding and release are summarized first. The nonbiological chemistry of iron and copper in the presence of dioxygen is described briefly to elucidate the key role that the protein plays in supporting oxygen transport by preventing irreversible oxidation of the binding site or of its ligands. The macroscopic behavior of the biological systems is related to the microscopic picture that has been developed over the last 30 years from x-ray crystallographic studies and a miscellany of spectroscopic probes of the oxygen-binding site. Relationships between the geometry and charge distribution in the metal-dioxygen moiety and the nature of the interactions between this moiety and its surroundings are examined. Nonbiological dioxygen carriers have proved particularly useful in providing precise and accurate structural information as well as thermodynamic and kinetic data against which the corresponding data from biological oxygen carriers can be contrasted. The bioinorganic chemistry of the hemoglobin family of oxygen binders is particularly amenable to study by means of small-molecule model systems: four of the five ligands that make up the active site are provided by a square-planar tetradentate ligand, the protoporphyrin IX dianion (Figure 4.2). One axial ligand in hemoglobin, imidazole from a histidine residue, is provided by the protein, and the remaining sixth coordination site is available for the exogenous ligand, e.g., dioxygen or carbon monoxide. Thus a model system that approximates the stereochemistry of the active site in hemoglobin may be assembled from an iron(II) porphyrin and a ligand, such as imidazole or pyridine. On the other hand, in hemocyanin and hemerythrin most of the ligands are supplied by the protein. Thus the assembly of a model system that provides appropriate ligands correctly disposed around the pair of metal atoms poses a major synthetic challenge, especially for hemocyanin, where details on the number, type, and arrangement of ligands have been difficult to establish. Many aspects of the physical, inorganic, and structural chemistry underlying biological oxygen transport and utilization (Chapter 5) have been clarified through model systems. | 6,241 | 1,825 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_v2.0_(Soderberg)/INTERCHAPTER%3A_Retrosynthetic_analysis_and_metabolic_pathway_prediction |
Imagine that you are a biological chemist doing research on bacterial metabolism. You and your colleagues isolate an interesting biomolecule from a bacterial culture, then use mass spectrometry, NMR, and other analytical techniques to determine its structure. Using your 'toolbox' of known organic reaction types - nucleophilic substitution, phosphorylation, aldol additions, and so forth - can you figure out a chemically reasonable pathway by which your compound might be enzymatically synthesized from simple metabolic precursors? In other words, can you fill in the missing biochemical steps (or at least some of them) to come up with a potential new metabolic pathway, which can then be used a hypothesis for future experimental work to try to find and study the actual enzymes involved? An actual example approximating this scenario is shown below. A complete biosynthetic pathway for isopentenyl diphosphate (IPP), the building block molecule for all d compounds, has been known since the 1960's. This pathway, which begins with acetyl-CoA, was shown to be active in yeast, plants, and many other species including humans. However, researchers in the late 1980s uncovered evidence indicating that the known pathway is present in bacteria, although they clearly use IPP as a building block molecule just as other forms of life do. Over the next several years, the researchers conducted a number of experiments in which bacteria were grown on a medium containing glucose 'labeled' with the C isotope. With the results from these experiments, combined with their knowledge of common biological organic reaction types, the researchers were able to correctly predict that the bacterial pathway starts with precursor molecules (pyruvate and glyceraldehyde phosphate instead of acetyl CoA) and they also correctly predicted the first two enzymatic steps of the newly discovered bacterial pathway. This accomplishment eventually led to elucidation of every step in the pathway, and isolation of the enzymes catalyzing them. ( . , , 517; , , 2564; , , 1095) Why weren't they able to predict the whole pathway? It turns out that several of the later steps were somewhat unusual, unfamiliar reaction types - but discovery of these reactions hinged upon the correct prediction of the more familiar first two steps. Multi-step transformation problems of this type offer an unparalleled opportunity to use our knowledge of biological organic chemistry combined with creative reasoning to solve challenging, relevant scientific puzzles. At this point in your organic chemistry career, you have not yet accumulated quite enough tools in your reaction toolbox to tackle most real-life biochemical pathway problems such as the one addressed above - but by the time we finish with oxidation and reduction chemistry in chapter 15, you will be able to recognize most of the reaction types that you will encounter in real metabolism, and will be challenged to predict some real pathways in the end-of-chapter problems. You do, however, have enough of a bioorganic repertoire to begin to learn how multi-step pathway problems can be approached, using for practice some generalized, hypothetical examples in which the reaction types involved are limited to those with which you are already familiar. Imagine that you want to figure out how an old-fashioned mechanical clock is put together. One way to do this is to start with a working clock, and take it apart piece-by-piece. Alternatively, one could start with all of the disassembled pieces, plus a lot of other small parts from different clocks, and try to figure out how to put together the specific clock you are interested in. Which approach is easier? The answer is intuitively obvious - it's usually easier to take things apart than to put them back together. The same holds true for molecules. If we want to figure out the biosynthetic pathway by which a large, complex biomolecule might be made in a cell, it makes sense to start with the finished product and then mentally work backwards, taking it apart step-by-step using known, familiar reactions, until we get to simpler precursor molecules. Starting with a large collection of potential precursor molecules and trying to put the right ones together to make the target product would be a formidable task. - the concept of mentally dismantling a molecule step by step all the way back to smaller, simpler precursors using known reactions - is a powerful and widely-used intellectual tool first developed by synthetic organic chemists. The approach has also been adapted for use by biological chemists in efforts to predict pathways by which known biomolecules could be synthesized (or degraded) in living things. In retrosynthesis, we think about a series of reactions in reverse. A backwards (retro) chemical step is symbolized by a 'thick' arrow, commonly referred to as a , and visually conveys the phrase ' '. Consider a simple, hypothetical example: starting with the target molecule below, can we come up with a chemically reasonable pathway starting from the precursors indicated?
A first step is to identify the relevant : a key bond (usually a carbon-carbon bond) that must be formed to make the target product from smaller precursors. We search our mental 'toolbox' of common biochemical reaction types, and remember that the only way we know of (so far!) to make a new carbon-carbon bond is through an aldol addition reaction, which takes place at an alpha-carbon. Therefore, we can make a likely disconnection next to the alpha-carbon in the target molecule. Next, we need to recognize that the aldol addition reaction results in a beta-hydroxy ketone. But our target molecule is beta- ketone! Working backwards, we realize that the beta-methoxy group could be formed from beta-hydroxy group by a SAM methylation reaction. This is our first retrosynthetic (backwards) step. The second retro step (aldol) accounts for the disconnection we recognized earlier, and leads to the two precursor molecules. Now, consider the more involved (but still hypothetical) biochemical transformation below: Often the best thing to do first in this type of problem is to count the carbons in the precursor compounds and product - this allows us to recognize when extra carbons on either side must at some point be accounted for in our solution. In this case, one carbon (labeled 'f'') has been in the form of a methyl ether in the product. This is easy to account for: we know that the coenzyme -adenosyl methionine (SAM) often serves as the methyl group donor in enzymatic - or -methylation reactions. So, we can propose our first backwards (retro) step: the product as shown could be derived from SAM-dependent methylation of an alcohol group on a proposed intermediate I. How do we know that the methylation step occurs last? We don't - remember, we are proposing a pathway, so the best we can do is propose steps that make chemical sense, and which hopefully can be confirmed or invalidated later through actual experimentation. For now, we'll stick with our initial choice to make the methylation step the last one. Now that we have accounted for the extra carbon, a key thing to recognize regarding the transformation in question is that two linear molecules are combining to form a cyclic product. Thus, connections need to be made between reactants A and B, one to join the two, the other to close the circle. Our primary job in the retro direction, then, is to establish in the product the two points of : in other words, to find the two bonds in the product that need to be taken apart in our retrosynthetic analysis. Look closely at the product: what functional groups do you see? Hopefully, you can identify two alcohol groups, a methyl ether, and (critically) a . We've already accounted for the methyl ether. Identifying the cyclic hemiketal is important because it allows us to make our next 'disconnection': , so we can mentally work backwards and predict the open-chain intermediate II that could cyclize to form our product. Now, starting with the R group and working along the carbon chain, we can account for carbons a-e on the two precursors. Thus, Here's where our mastery of biological organic reactivity really comes into play: the OH at carbon c of intermediate II is in the beta position relative to carbonyl carbon a. reactions result in beta-hydroxy ketones or aldehydes. Therefore, we can work backward one more step and predict that our intermediate II was formed from an aldol addition reaction between intermediate III (as the nucleophile) and precursor molecule A (as the electrophile). We are most of the way home - we have successfully accounted for given precursor A. Intermediate III, however, is precursor B. What is different? Both III and B have a carbonyl and two alcohol groups, but the positioning is different: III is an aldehyde, while B is a ketone. Think back to earlier in this chapter: intermediate III could form from in compound B. We have now accounted for our second precursor - we are done! In the forward direction, a complete pathway diagram can be written as follows: A full 'retrosynthesis' diagram for this problem looks like this: | 9,289 | 1,827 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Acid/Lewis_Concept_of_Acids_and_Bases |
Acids and bases are an important part of chemistry. One of the most applicable theories is the Lewis acid/base motif that extends the definition of an acid and base beyond H and OH ions as described by . The Brø nsted acid-base theory has been used throughout the history of acid and base chemistry. However, this theory is very restrictive and focuses primarily on acids and bases acting as proton donors and acceptors. Sometimes conditions arise where the theory does not necessarily fit, such as in solids and gases. In 1923, G.N. Lewis from UC Berkeley proposed an alternate theory to describe acids and bases. His theory gave a generalized explanation of acids and bases based on structure and bonding. Through the use of the Lewis definition of acids and bases, chemists are now able to predict a wider variety of acid-base reactions. Lewis' theory used electrons instead of proton transfer and specifically stated that an acid is a species that accepts an electron pair while a base donates an electron pair. The reaction of a Lewis acid and a Lewis base will produce a coordinate covalent bond (Figure \(\Page {1}\)). A coordinate covalent bond is just a type of covalent bond in which one reactant gives it electron pair to another reactant. In this case the lewis base donates its electrons to the Lewis acid. When they do react this way the resulting product is called an addition compound, or more commonly an adduct. Lewis acids accept an electron pair. Lewis Acids are meaning that they are electron attracting. When bonding with a base the acid uses its lowest unoccupied molecular orbital or LUMO (Figure 2). Lewis Bases donate an electron pair. Lewis Bases are meaning that they “attack” a positive charge with their lone pair. They utilize the highest occupied molecular orbital or HOMO (Figure 2). An atom, ion, or molecule with a lone-pair of electrons can thus be a Lewis base. Each of the following anions can "give up" their electrons to an acid, e.g., \(OH^-\), \(CN^-\), \(CH_3COO^-\), \(:NH_3\), \(H_2O:\), \(CO:\). Lewis base's HOMO (highest occupied molecular orbital) interacts with the Lewis acid's LUMO (lowest unoccupied molecular orbital) to create bonded . Both contain HOMO and LUMOs but only the HOMO is considered for Bases and only the LUMO is considered for Acids (Figure \(\Page {2}\)). Complex ions are polyatomic ions, which are formed from a central metal ion that has other smaller ions joined around it. While Brønsted theory can't explain this reaction Lewis acid-base theory can help. A Lewis Base is often the ligand of a coordination compound with the metal acting as the Lewis Acid (see ). \[Al^{3+} + 6 H_2O \rightleftharpoons [Al(H_2O)_6]^{3+} \label{1}\] The aluminum ion is the metal and is a cation with an unfilled valence shell, and it is a Lewis Acid. Water has lone-pair electrons and is an anion, thus it is a Lewis Base. The Lewis Acid accepts the electrons from the Lewis Base which donates the electrons. Another case where Lewis acid-base theory can explain the resulting compound is the reaction of ammonia with Zn . \[ Zn^{2+} + 4NH_3 \rightarrow [Zn(NH_3)_4]^{4+} \label{2}\] Similarly, the Lewis Acid is the zinc Ion and the Lewis Base is NH . Note how Brønsted Theory of Acids and Bases will not be able to explain how this reaction occurs because there are no \(H^+\) or \(OH^-\) ions involved. Thus, Lewis Acid and Base Theory allows us to explain the formation of other species and complex ions which do not ordinarily contain hydronium or hydroxide ions. One is able to expand the definition of an acid and a base via the Lewis Acid and Base Theory. The lack of \(H^+\) or \(OH^-\) ions in many complex ions can make it harder to identify which species is an acid and which is a base. Therefore, by defining a species that donates an electron pair and a species that accepts an electron pair, the definition of a acid and base is expanded. As of now you should know that acids and bases are distinguished as two separate things however some substances can be both an acid and a base. You may have noticed this with water, which can act as both an acid or a base. This ability of water to do this makes it an amphoteric molecule. Water can act as an acid by donating its proton to the base and thus becoming its conjugate acid, OH-. However, water can also act as a base by accepting a proton from an acid to become its conjugate base, H O . \[H_2O + NH_3 \rightarrow NH_4^+ + OH^- \label{3}\] \[H_2O + HCl \rightarrow Cl^- + H_3O^+ \label{4}\] You may have noticed that the degree to which a molecule acts depends on the medium in which the molecule has been placed in. Water does not act as an acid in an acid medium and does not act as a base in a basic medium. Thus, the medium which a molecule is placed in has an effect on the properties of that molecule. Other molecules can also act as either an acid or a base. For example, \[Al(OH)_3 + 3H^+ \rightarrow Al^{3+} + 3H_2O \label{5}\] \[Al(OH)_3 + OH^- \rightarrow Al(OH)_4^- \label{6}\] Note how the amphoteric properties of the Al(OH) depends on what type of environment that molecule has been placed in. | 5,162 | 1,828 |
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Oxidation-Reduction or "redox" reactions occur when elements in a chemical reaction gain or lose electrons, causing an increase or decrease in oxidation numbers. The Half Equation Method is used to balance these reactions. In a redox reaction, one or more element becomes oxidized, and one or more element becomes reduced. Oxidation is the loss of electrons whereas reduction is the gain of electrons. An easy way to remember this is to think of the charges: an element's charge is reduced if it gains electrons (an acronym to remember the difference is LEO = Lose Electron Oxidation & GER = Gain Electron Reduction). Redox reactions usually occur in one of two environments: acidic or basic. In order to balance redox equations, understanding is necessary. Balance this reaction \[\ce{MnO4^{-} + I^{-} -> I2 + Mn^{2+}} \nonumber\] Steps to balance: Step 1: Separate the half-reactions that undergo oxidation and reduction. \[\ce{ I^{-} -> I2} \nonumber \] This is the oxidation half because the oxidation state changes from -1 on the left side to 0 on the right side. This indicates a gain in electrons. \[ \ce{ MnO4^{-} -> Mn^{2+}} \nonumber\] This is the reduction half because the oxidation state changes from +7 on the left side to +2 on the right side. This indicates a reduction in electrons. Step 2: In order to balance this half reaction we must start by balancing all atoms other than any Hydrogen or Oxygen atoms. : \[\ce{2I^{-} -> I2} \nonumber\] In order to balance the oxidation half of the reaction you must first add a 2 in front of the \(\ce{I}\) on the left hand side so there is an equal number of atoms on both sides. : \[ \ce{MnO4^{-} -> Mn^{2+}} \nonumber\] For the reduction half of the reaction, you can notice that all atoms other than Hydrogen and Oxygen are already balanced because there is one manganese atom on both sides of the half reaction. Step 3: Balance Oxygen atoms by adding \(\ce{H2O}\) to the side of the equation that needs Oxygen. Once you have completed this step add \(\ce{H^{+}}\) to the side of the equation that lacks \(\ce{H}\) atoms necessary to be balanced. \[\ce{2 I^{-} -> I2} \nonumber\] Because there are no Oxygen or Hydrogen atoms in this half of the reaction, it is not required to perform any kind of balancing. : \[ \ce{ MnO4^{-} -> Mn^{2+} + 4 H2O} \nonumber\] The first step in balancing this reaction using step 3 is to add4 H O atoms in order to balance the Oxygen atoms with the 4 on the other side of MnO4- : \[ \ce{MnO4^{-} + 8 H^{+} -> Mn^{2+} + 4 H2O} \nonumber\] Now that the Oxygen atoms have been balanced you can see that there are 8 \( \ce{H}\) atoms on the right hand side of the equation and none on the left. Therefore, you must add 8 \(\ce{H^{+}}\) atoms to the left hand side of the equation to make it balanced. Step 4: Now that the two half reactions have been balanced correctly one must balance the charges in each half reaction so that both the reduction and oxidation halves of the reaction consume the same number of electrons. : \[ \ce{2 I^{-} -> I2 + 2e^{-} } \nonumber\] Because of the fact that there are two \(\ce{I}\)'s on the left hand side of the equation which a charge of -1 we can state that the left hand side has an overall charge of -2. The I on the left side of the equation has an overall charge of 0. Therefore to balance the charges of this reaction we must add 2 electrons to the right side of the equation so that both sides of the equation have equal charges of -2. : \[ \ce{5 e^{-} + 8 H^{+} + MnO4^{-} -> Mn^{2+} + 4 H2O} \nonumber\] Looking at the left hand side of the equation you can notice that there are 8 Hydrogen atoms with a +1 charge. There is also a \(\ce{MnO4^{-}}\) ion that has a charge of -1. When we add these two charges up we can calculate that the left hand side of the equation has an overall charge of +7. The right hand side has an \(\ce{Mn}\) atom with a charge of +2 and then 4 water molecules that have charges of 0. Therefore, the overall charge of the right side is +2. We must add 5 electrons to the left side of the equation to make sure that both sides of the equation have equal charges of +2. Step 5: Multiply both sides of both reactions by the least common multiple that will allow the half-reactions to have the same number of electrons and cancel each other out. : \( 10I^- \rightarrow 5I_2 +10e^- \) We multiply this half reaction by 5 to come up with the following result above. Reduction: \(10e^- + 16H^+ + 2MnO_4^- \rightarrow 2Mn^{2+} + 8H_2O \) We multiply the reduction half of the reaction by 2 and arrive at the answer above. By multiplying the oxidation half by 5 and the reduction half by 2 we are able to observe that both half-reactions have 10 electrons and are therefore are able to cancel each other out. Step 6: Add the two half reactions in order to obtain the overall equation by canceling out the electrons and any \(\ce{H2O}\) and \(\ce{H^{+}}\) ions that exist on both sides of the equation. : \[\ce{10 I^{-} + 16 H^{+} + 2 MnO4^{-} -> 5 I2 + 2 Mn^{2+} + 8 H_2O} \nonumber\] In this problem, there is not anything that exists on both halves of the equation that can be cancelled out other than the electrons. Finally, double check your work to make sure that the mass and charge are both balanced. To double check this equation you can notice that everything is balanced because both sides of the equation have an overall charge of +4. The balancing procedure in basic solution differs slightly because \(\ce{OH^{-}}\) ions must be used instead of \(\ce{H^{+}}\) ions when balancing hydrogen atoms. To give the previous reaction under basic conditions, sixteen \(\ce{OH^{-}}\) ions can be added to both sides. on the left the \(\ce{OH^{-}}\) and the \(\ce{H^{+}}\) ions will react to form water, which will cancel out with some of the \(\ce{H2O}\) on the right. \[\ce{10I^{-} (aq) + 2MnO4^{-} (aq) + 16H^{+} (aq) + 16OH^{-} (aq) -> 5I2 (s) + 2Mn^{2+} (aq) + 8H2O (l) + 16OH^{-} (aq)} \nonumber\] On the left side the OH- and the \(\ce{H^{+}}\) ions will react to form water, which will cancel out with some of the \(\ce{H2O}\) on the right: \[\ce{10I^{-} (aq) + 2MnO4^{-} (aq) + 16H2O (l) -> 5I2 (s) + 2Mn^{2+} (aq) + 8H2O (l) + 16OH^{-} (aq)} \nonumber\] Eight water molecules can be canceled, leaving eight on the reactant side: \[\ce{10I^{-} (aq) + 2MnO4^{-} (aq) + 8H2O (l) -> 5I2 (s) + 2Mn^{2+} (aq) + 16OH^{-} (aq)} \nonumber\] This is the balanced reaction in basic solution. Balance the following in an acidic solution. \[\ce{SO3^{2-} (aq) + MnO4^{-} (aq) \rightarrow SO4^{2-} (aq) + Mn^{2+} (aq)} \nonumber\] Step 1: Split into two half reaction equations: Oxidation and Reduction Step 2: Balance each of the half equations in this order: The \(\ce{S}\) and \(\ce{Mn}\) atoms are already balanced, Balancing \(\ce{O}\) atoms \[\begin{align*} &\text{Oxidation}: \quad \ce{SO3^{2-}(aq)} + \ce{H2O(l)} \rightarrow \ce{SO4^{2-} (aq)} \\[4pt] &\text{Reduction}: \quad \ce{MnO4^{-} (aq)} \rightarrow \ce{Mn^{2+}(aq)} + \ce{4H2O(l)} \end{align*}\] Then balance out \(\ce{H}\) atoms on each side \[\begin{align*} &\text{Oxidation}: \quad \ce{SO3^{2-}(aq)} + \ce{H2O(l)} \rightarrow \ce{SO4^{2-} (aq)} + \ce{2H^{+}(aq)}\\[4pt] &\text{Reduction}: \quad \ce{MnO4^{-} (aq)} + 8 \ce{H^{+}} \rightarrow \ce{Mn^{2+}(aq)} + \ce{4H2O(l)} \end{align*}\] Step 3: Balance the charges of the half reactions by adding electrons Step 4: Obtain the overall redox equation by combining the half reaction, but multiply entire equation by number of electrons in oxidation with reduction equation, and number of electrons in reduction with oxidation equation. Overall Reaction: \[\begin{align*} &\text{Oxidation}: \quad \ce{5SO3^{2-}(aq)} + \ce{5H2O(l)} \rightarrow \ce{5 SO4^{2-} (aq)} + \ce{10 H^{+}(aq)} + \ce{10e^{-}} \\[4pt] &\text{Reduction}: \quad \ce{2 MnO4^{-} (aq)} + \ce{16H^{+}} + \ce{10e^{-}} \rightarrow \ce{2Mn^{2+}(aq)} + \ce{8H2O(l)} \\[4pt] \hline &\text{total}: \quad \ce{5SO3^{2-}(aq)} + \ce{5H2O(l)} + \ce{2 MnO4^{-} (aq)} + \ce{16H^{+}} + \ce{10e^{-}} \rightarrow \ce{5 SO4^{2-} (aq)} + \ce{10 H^{+}(aq)} + \ce{2Mn^{2+}(aq)} + \ce{8H2O(l)} + \ce{10e^{-}} \end{align*}\] Step 5: Simplify and cancel out similar terms on both sides \[\ce{5SO3^{2-}(aq)} + \cancel{\ce{5H2O(l)}} + \ce{2 MnO4^{-} (aq)} + \ce{\cancelto{6}{16}H^{+}} + \cancel{\ce{10e^{-}}} \rightarrow \ce{5 SO4^{2-} (aq)} + \cancel{\ce{10 H^{+}(aq)}} + \ce{2Mn^{2+}(aq)} + \ce{\cancelto{3}{8}H2O(l)} + \cancel{\ce{10e^{-}}} \nonumber\] To get \[\ce{5SO3^{2-}(aq)} + \ce{2 MnO4^{-} (aq)} + \ce{6H^{+}} \rightarrow \ce{5 SO4^{2-} (aq)} + \ce{2Mn^{2+}(aq)} + \ce{3H2O(l)} \nonumber\] Balance this reaction in both acidic and basic aqueous solutions \[\ce{MnO4^{-}(aq) + SO3^{2-}(aq) -> MnO2(s) + SO4^{2-}(aq)} \nonumber\] \[\ce{MnO4^{-}(aq) -> MnO2(s)} \nonumber\] \[\ce{SO3^{2-}(aq) -> SO4^{2-}(aq)} \nonumber\] \[\ce{MnO4^{-}(aq) -> MnO2(s) + 2H2O(l)} \nonumber \] \[\ce{H2O(l) + SO3^{2-}(aq) -> SO4^{2-}(aq)} \nonumber\] \[\ce{4H^{+} + MnO4^{-}(aq) -> MnO2(s) + 2H2O(l)} \nonumber\] \[\ce{H2O(l) + SO3^{2-}(aq) -> SO4^{2-}(aq) + 2H^{+}} \nonumber\] \[\ce{3e- + 4H^{+} + MnO4^{-}(aq) -> MnO2(s) + 2H2O(l)} \nonumber \] \[\ce{H2O(l) + SO3^{2-}(aq) --> SO4^{2-}(aq) + 2H^{+} + 2e^{-}} \nonumber \] \[\ce{2(3e^{-} + 4H^{+} + MnO4^{-}(aq) -> MnO2(s) + 2H2O(l))} \nonumber \] \[\ce{3(H2O(l) + SO3^{2-}(aq) -> SO4^{2-}(aq) + 2H^{+} + 2e^{-})} \nonumber \] \[\ce{6e^{-} + 8H^{+} + 2MnO4^{-}(aq) -> 2MnO2(s) + 4H2O(l)} \nonumber \] \[\ce{3H2O(l) + 3SO3^{2-}(aq) -> 3SO4^{2-}(aq) + 6H^{+} + 6e^{-}} \nonumber \] \(\ce{2H^{+}}\). The d of the \(\ce{3H2O(l)}\) on the bo \(\ce{H2O(l)}\) on the t \[\ce{2MnO4^{-}(aq) + 2H^{+} + 3SO3^{2-}(aq) -> H2O(l) + 2MnO2(s) + 3SO4^{2-}(aq)} \nonumber \] add \(\ce{OH^{-}}\) to each side to neutralize the \(\ce{H^{+}}\) into \[\ce{2MnO4^{-}(aq) + 2H2O + 3SO3^{2-}(aq) -> H2O(l) + 2MnO2(s) + 3SO4^{2-}(aq) + 2OH^{-}} \nonumber \] and then cancel the water molecules \[\ce{2MnO4^{-}(aq) + H2O + 3SO3^{2-}(aq) -> + 2MnO2(s) + 3SO4^{2-}(aq) + 2OH^{-}} \nonumber\] Balance this reaction in acidic solution \[\ce{Fe(OH)3 + OCl^{-} \rightarrow FeO4^{2-} + Cl^{-}} \nonumber \] Step 1: Steps 2 and 3: Step 4: \[\begin{align*} 3 \times \big[ \ce{ 2H^{+} + OCl^{-} + 2e^{-}} &\ce{ -> Cl^{-} + H2O} \big] \\[4pt] \ce{ 6H^{+} + 3OCl^{-} + 6e^{-}} &\ce{ -> 3Cl^{-} + 3H2O} \end{align*}\] and \[\begin{align*} 2 \times \big[ \ce{Fe(OH)3 + H2O} & \ce{-> FeO4^{2-} + 3e- + 5H^{+}} \big] \\[4pt] \ce{2Fe(OH)3 + 2H2O} & \ce{-> 2FeO4^{2-} + 6e- + 10H^{+}} \end{align*}\] Adding these together results Step 5: \[\ce{3OCl^{-} + 2Fe(OH)3 \rightarrow 3Cl^{-} + H2O + 2FeO4^{2-} + 4H^{+}} \nonumber\] Balance this equation in acidic aqueous solution \[\ce{VO4^{3-} + Fe^{2+} -> VO^{2+} + Fe^{3+}} \nonumber\] Step 1: Step 2/3: Step 4: Overall Reaction: \[\ce{Fe^{2+} -> Fe^{3+} + e^{-}} \nonumber\] + \[\ce{6H^{+} + VO4^{3-} + e^{-} -> VO^{2+} + 3H2O} \nonumber\] \[\ce{Fe^{2+} + 6H^{+} + VO4^{3-}} + \cancel{e^{-}} \ce{ \rightarrow Fe^{3+}} + \cancel{e^{-}} \ce{ + VO^{2+} + 3H2O} \nonumber\] Simplify: \[\ce{Fe^{2+} + 6H^{+} + VO4^{3-} \rightarrow Fe^{3+} + VO^{2+} + 3H2O} \nonumber\] \[\ce{Cr2O7^{2-}(aq) + C2H5OH(l) -> Cr^{3+}(aq) + CO2(g)} \nonumber\] In acidic aqueous solution: \[\ce{2Cr2O7^{-}(aq) + 16H^{+}(aq) + C2H5OH(l) -> 4Cr^{3+}(aq) + 2CO2(g) + 11H2O(l)} \nonumber\] In basic aqueous solution: \[\ce{2Cr2O7^{-}(aq) + 5H2O(l) + C2H5OH(l) -> 4Cr^{3+}(aq) + 2CO2(g) + 16OH^{-}(aq) } \nonumber\] \[\ce{Fe^{2+}(aq) + MnO4^{-}(aq) -> Fe^{3+}(aq) + Mn^{2+}(aq)} \nonumber\] In acidic aqueous solution: \[\ce{ MnO4^{-}(aq) + 5Fe^{2+}(aq) + 8H^{+}(aq) -> Mn^{2+}(aq) + 5Fe^{3+}(aq) + 4H2O(l)} \nonumber\] \[\ce{MnO4^{-}(aq) + 5Fe^{2+}(aq) + 4H2O(l) -> Mn^{2+}(aq) + 5Fe^{3+}(aq) + 8OH^{-}(aq) } \nonumber\] In a redox reaction, also known as an oxidation-reduction reaction, it is a must for oxidation and reduction to occur simultaneously. In the oxidation half of the reaction, an element gains electrons. A species loses electrons in the reduction half of the reaction. These reactions can take place in either acidic or basic solutions. | 12,115 | 1,832 |
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The 's are equilibrium constants in hetergeneous equilibria. 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: \(\mathrm{[Na^+] + [K^+] = [Cl^-]}\) 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. \[\mathrm{NaCl \rightleftharpoons Na^+ + {\color{Green} Cl^-}}\] \[\mathrm{KCl \rightleftharpoons K^+ + {\color{Green} Cl^-}}\] \[\mathrm{CaCl_2 \rightleftharpoons Ca^{2+} + {\color{Green} 2 Cl^-}}\] \[\mathrm{AlCl_3 \rightleftharpoons Al^{3+} + {\color{Green} 3 Cl^-}}\] \[\mathrm{AgCl \rightleftharpoons Ag^+ + {\color{Green} Cl^-}}\] 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 \(\mathrm{[Na^+] = [Ca^{2+}] = [H^+] = 0.10\: \ce M}\). but \(\begin{alignat}{3}
&\ce{[Cl- ]} &&= && && \:\textrm{0.10 (due to NaCl)}\\
& && && + &&\mathrm{\:0.20\: (due\: to\: CaCl_2)}\\
& && && + &&\mathrm{\:0.10\: (due\: to\: HCl)}\\
& &&= && &&\mathrm{\:0.40\: M}
\end{alignat}\) 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? \(\mathrm{[Cl^-] = \dfrac{0.1\: M\times 10\: mL+0.2\: M\times 5.0\: mL}{100.0\: mL} = 0.020\: M}\) What is the solubility (in moles per liter) of \(\ce{AgCl}\) in a solution that is 0.05 M in \(\ce{KCl}\)? for \(\ce{AgCl}\) is 1.0E-10. Assume \(x = \ce{[Ag+]}\) to be the molar solubility, then we can write the equilibrium concentration below the formula of the equilibrium. \(\begin{array}{cccccl}
\ce{AgCl &\rightarrow &Ag+ &+ &Cl- &}\\
&&x &&0.05+x &\leftarrow \ce{equilibrium\:\, concentration}
\end{array}\) By definition of the , we have \(x (0.05 + x) = K_{\ce{sp}}\); Thus, \(x = \dfrac{\textrm{1.0e-10}}{0.05} = \textrm{2e-9 M}\) Note that 0.05 + x is approximately 0.05, and you may use this approximation for the calculation. What is the solubility of \(\ce{Li2CO3}\) in a solution also containing 0.10 M of \(\ce{K2CO3}\)? The solubility product for \(\ce{Li2CO3}\) is 0.0017. In this case, \(\ce{CO3^2-}\) is the common ion between the two salts. \(\ce{K2CO3 \rightleftharpoons 2 K+ + CO3^2-}\)
\(\ce{Li2CO3 \rightleftharpoons 2 Li+ + CO3^2-}\) This question implies that \(\ce{K2CO3}\) is soluble. Thus, the only heterogeneous equilibrium to be considered is \(\ce{K2CO3 \rightleftharpoons 2 K+ + CO3^2-}\) Let \(\ce{[Li+]} = 2 x\), then \(\ce{[CO3^2- ]} = 0.10 + x\). Thus, we can write down the equilibrium equation again, and write the concentrations below the chemical formula: \[\begin{array}{ccccc}
\ce{Li2CO3 &\rightleftharpoons &2 Li+ &+ &CO3^2-}\\
&&2 x &&0.10+x
\end{array}\] \[\ce{[Li+]^2 [CO3^2- ]} = (2 x)^2 (0.10+x) = 0.0017\] Expanding the formula results in \[x^3 + 0.10 x^2 - 0.00043 = 0\.] A general method to solve this equation is to use the Newton's method. For which, we assume that \[y = x^3 + 0.10 x^2 - 0.00043 = 0.\] We assume the value of \(x = 0.06\), and substitute in the equation to calculate y. \[y = 0.06^3 + 0.10\times 0.06^2 - 0.00043 = 0.000146\] We now assume \(x = 0.05\), and substitute in the equation to evaluate y, \[y = 0.05^3 + 0.10\times 0.05^2 - 0.00043 = -0.000055\] Since the values for y calculated for x = 0.05 and 0.06 have different sign, the x value should lie between 0.05 and 0.06. We further assume the value for x = 0.053, and we obtain a y value \[y = 0.053^3 + 0.10\times 0.053^2 - 0.00043 = -2.2\times 10^{-7}\] Thus, the value of x should be greater than 0.053. We can increase x to 0.0531 or 0.0532, and evaluate y again: \[y = 0.0531^3 + 0.10\times 0.0531^2 - 0.00043 = 1.7\times 10^{-6}\] Thus, x value should lie between 0.0531 and 0.0530. By this method, we have evaluated x values to three significant figures. We only need two significant figures due to the nature of the data in the problem. Thus we have \[\ce{[Li+]} = 2 x = \textrm{0.106 \;M}\] The molar solubility of \(\ce{Li2CO3}\) in a solution also containing 0.10 M of \(\ce{K2CO3}\) is 0.053 mole per liter, but \[\ce{[Li+]} = 2 x = \textrm{0.106 M}\] This example illustrates the Newton's method for solving cubic equations. Note that if the is small, then x is a very small value. In this case, \(0.10 + x \approx 0.10\). You do not need to use the Newton's method in this case. What is the solubility of \(\ce{Li2CO3}\) in a solution which is 0.20 M in \(\ce{Na2CO3}\)? The solubility product of \(\ce{Li2CO3}\) is 1.7e-3 M . Since the solution contains 0.20 M \(\ce{Na2CO3}\), \(\ce{[CO3^2- ]} = \textrm{0.20 M}\). Assume the solubility to be M of \(\ce{Li2CO3}\), then we have \(\begin{array}{ccccc}
\ce{Li2CO3 &\rightleftharpoons &2 Li+ &+ &CO3^2-}\\
&&2 x &&x + 0.20\:\ce M
\end{array}\) Thus, \(\begin{align}
(2 x)^2 (x+0.20) &= \textrm{1.7e-3 M}^3\\
4x^3 + 0.20 x^2 &= \textrm{1.7e-3 M}^3
\end{align}\)
\(x^3 + 0.20 x^2 - 0.00043 = 0\) There is no definite way to solve this equation, but the Newton's method is very useful. In this method, we let \(y = x^3 + 0.20 x^2 - 0.00043\) We guesstimate and evaluate using the above expression. The solution is a value for so that =0. We refine values between two values that give positive and negative values in progression as shown in the Table below. This is the same type of question as . As the carbonate concentration increases from 0.10 to 0.20 M, the solubility of lithium carbonate reduces to 0.042 M from 0.053 M. Note that the lithium ion concentration is 0.084 M in this case. Note that when used to treat depression, lithium carbonate is usually called lithium in the health care profession. | 6,688 | 1,833 |
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This page introduces lattice enthalpies (lattice energies) and Born-Haber cycles. There are two different ways of defining lattice enthalpy which directly contradict each other, and you will find both in common use. In fact, there is a simple way of sorting this out, but many sources do not use it. Lattice enthalpy is a measure of the strength of the forces between the ions in an ionic solid. The greater the lattice enthalpy, the stronger the forces. Those forces are only completely broken when the ions are present as gaseous ions, scattered so far apart that there is negligible attraction between them. You can show this on a simple enthalpy diagram. For sodium chloride, the solid is more stable than the gaseous ions by 787 kJ mol , and that is a measure of the strength of the attractions between the ions in the solid. Remember that energy (in this case heat energy) is released when bonds are made, and is required to break bonds. So lattice enthalpy could be described in either of two ways. Both refer to the same enthalpy diagram, but one looks at it from the point of view of making the lattice, and the other from the point of view of breaking it up. Unfortunately, both of these are often described as "lattice enthalpy". This is an absurdly confusing situation which is easily resolved by never using the term "lattice enthalpy" without qualifying it. That immediately removes any possibility of confusion. The two main factors affecting lattice enthalpy are Sodium chloride and magnesium oxide have exactly the same arrangements of ions in the crystal lattice, but the lattice enthalpies are very different. You can see that the lattice enthalpy of magnesium oxide is much greater than that of sodium chloride. That's because in magnesium oxide, 2+ ions are attracting 2- ions; in sodium chloride, the attraction is only between 1+ and 1- ions. The lattice enthalpy of magnesium oxide is also increased relative to sodium chloride because magnesium ions are smaller than sodium ions, and oxide ions are smaller than chloride ions. That means that the ions are closer together in the lattice, and that increases the strength of the attractions. This effect of ion size on lattice enthalpy is clearly observed as you go down a Group in the Periodic Table. For example, as you go down Group 7 of the Periodic Table from fluorine to iodine, you would expect the lattice enthalpies of their sodium salts to fall as the negative ions get bigger - and that is the case: Attractions are governed by the distances between the centers of the oppositely charged ions, and that distance is obviously greater as the negative ion gets bigger. And you can see exactly the same effect if as you go down Group 1. The next bar chart shows the lattice enthalpies of the chlorides. It is impossible to measure the enthalpy change starting from a solid crystal and converting it into its scattered gaseous ions. It is even more difficult to imagine how you could do the reverse - start with scattered gaseous ions and measure the enthalpy change when these convert to a solid crystal. Instead, lattice enthalpies always have to be calculated, and there are two entirely different ways in which this can be done. Before we start talking about Born-Haber cycles, we need to define the , \(\Delta H^o_a\). The standard atomization enthalpy is the enthalpy change when 1 mole of gaseous atoms is formed from the element in its standard state. Enthalpy change of atomization is always positive. You are always going to have to supply energy to break an element into its separate gaseous atoms. All of the following equations represent changes involving atomization enthalpy: \[ \dfrac{1}{2} Cl_2 (g) \rightarrow Cl(g) \;\;\;\; \Delta H^o_a=+122\, kJ\,mol^{-1}\] \[ \dfrac{1}{2} Br_2 (l) \rightarrow Br(g) \;\;\;\; \Delta H^o_a=+122\, kJ\,mol^{-1}\] \[ Na (s) \rightarrow Na(g) \;\;\;\; \Delta H^o_a=+107\, kJ\,mol^{-1}\] Notice particularly that the "mol " is per mole of atoms - NOT per mole of element that you start with. You will quite commonly have to write fractions into the left-hand side of the equation. Getting this wrong is a common mistake. Consider a Born-Haber cycle for sodium chloride, and then talk it through carefully afterwards. You will see that I have arbitrarily decided to draw this for lattice formation enthalpy. If you wanted to draw it for lattice dissociation enthalpy, the red arrow would be reversed - pointing upwards. Focus to start with on the higher of the two thicker horizontal lines. We are starting here with the elements sodium and chlorine in their standard states. Notice that we only need half a mole of chlorine gas in order to end up with 1 mole of NaCl. The arrow pointing down from this to the lower thick line represents the enthalpy change of formation of sodium chloride. The Born-Haber cycle now imagines this formation of sodium chloride as happening in a whole set of small changes, most of which we know the enthalpy changes for - except, of course, for the lattice enthalpy that we want to calculate. Now we can use Hess' Law and find two different routes around the diagram which we can equate. As drawn, the two routes are obvious. The diagram is set up to provide two different routes between the thick lines. So, from the cycle we get the calculations directly underneath it . . . -411 = +107 + 496 + 122 - 349 + LE LE = -411 - 107 - 496 - 122 + 349 LE = -787 kJ mol How would this be different if you had drawn a lattice dissociation enthalpy in your diagram? Your diagram would now look like this: The only difference in the diagram is the direction the lattice enthalpy arrow is pointing. It does, of course, mean that you have to find two new routes. You cannot use the original one, because that would go against the flow of the lattice enthalpy arrow. This time both routes would start from the elements in their standard states, and finish at the gaseous ions. -411 + LE = +107 + 496 + 122 - 349 LE = +107 + 496 + 122 - 349 + 411 LE = +787 kJ mol Once again, the cycle sorts out the sign of the lattice enthalpy. Let's assume that a compound is fully ionic. Let's also assume that the ions are point charges - in other words that the charge is concentrated at the center of the ion. By doing physics-style calculations, it is possible to calculate a theoretical value for what you would expect the lattice energy to be. Calculations of this sort end up with values of , and not . If you know how to do it, you can then fairly easily convert between the two. There are several different equations, of various degrees of complication, for calculating lattice energy in this way. There are two possibilities: The explanation is that silver chloride actually has a significant amount of covalent bonding between the silver and the chlorine, because there is not enough electronegativity difference between the two to allow for complete transfer of an electron from the silver to the chlorine. Comparing experimental ( ) and theoretical values for lattice enthalpy is a good way of judging how purely ionic a crystal is. Jim Clark ( ) | 7,110 | 1,834 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/08%3A_Solids_Liquids_and_Gases/8.S%3A_Solids_Liquids_and_Gases_(Summary) |
A is a certain form of matter that has the same physical properties throughout. Three phases are common: the solid, the liquid, and the gas phase. What determines the phase of a substance? Generally, the strength of the determines whether a substance is a solid, liquid, or gas under any particular conditions. is a very strong form of intermolecular interaction. Diamond is one example of a substance that has this intermolecular interaction. , the forces of attraction due to oppositely charged ions, are also relatively strong. Covalent bonds are another type of interaction within molecules, but if the bonds are , then the unequal sharing of electrons can cause charge imbalances within molecules that cause interactions between molecules. These molecules are described as , and these interactions are called . A certain rather strong type of dipole-dipole interaction, involving a hydrogen atom, is called . On the other hand, equal sharing of electrons forms , and the interactions between different molecules is less because the molecules are nonpolar. All substances have very weak (also called ) caused by the movement of electrons within the bonds themselves. In the solid phase, intermolecular interactions are so strong that they hold the individual atoms or molecules in place. In many solids, the regular three-dimensional arrangement of particles makes a . In other solids, the irregular arrangement of particles makes an solid. In liquids, the intermolecular interactions are strong enough to keep the particles of substance together but not in place. Thus, the particles are free to move over each other but still remain in contact. In gases, the intermolecular interactions are weak enough that the individual particles are separated from each other in space. The is a collection of statements that describe the fundamental behavior of all gases. Among other properties, gases exert a on their container. Pressure is measured using units like , , , or (also called a ). There are several simple relationships between the variables used to describe a quantity of gas. These relationships are called . relates the pressure and volume of a gas, while relates the volume and absolute temperature of a gas. The relates the volume, pressure, and absolute temperature of a gas sample. All of these gas laws allow us to understand the changing conditions of a gas. The relates the pressure, volume, amount, and absolute temperature of a gas under any conditions. These four variables are related to the , which is the proportionality constant used to calculate the conditions of a gas. Because the conditions of a gas can change, a set of benchmark conditions called is defined. Standard temperature is 0ºC, and standard pressure is 1.00 atm. | 2,805 | 1,836 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Electrophilic_Addition_Reactions/Reactions_of_Alkenes_with_Sulfuric_Acid |
This page gives you the facts and a simple, uncluttered mechanism for the electrophilic addition reactions between sulfuric acid and alkenes like ethene and cyclohexene. Alkenes react with concentrated sulfuric acid in the cold to produce alkyl hydrogensulphates. For example, ethene reacts to give ethyl hydrogensulphate. \[ \ce{CH_2=CH_2 + H_2SO_4 \rightarrow CH_3CH_2OSO_2OH}\] The structure of the product molecule is sometimes written as \(CH_3CH_2HSO_4\), b ut the version in the equation is better because it shows how all the atoms are linked up. You may also find it written as \(CH_3CH_2OSO_3H\). All you need to do is to learn the structure of sulfuric acid, and after that the mechanism is exactly the same as the one with hydrogen bromide. As you will find out, the formula of the product follows from the mechanism in an inevitable way. This time we are going straight for the mechanism without producing an initial equation. This is to show that you can work out the structure of obscure products provided you can write the mechanism. Having worked out the structure of the product, you could then write a simple equation for the reaction if you wanted to. Jim Clark ( ) | 1,197 | 1,837 |
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The reaction of and ketones with ammonia or 1º-amines forms imine derivatives, also known as Schiff bases (compounds having a C=N function). Water is eliminated in the reaction, which is acid-catalyzed and reversible in the same sense as acetal formation. The pH for reactions which form imine compounds must be carefully controlled. The rate at which these imine compounds are formed is generally greatest near a pH of 5, and drops at higher and lower pH's. At high pH there will not be enough acid to protonate the OH in the intermediate to allow for removal as H O. At low pH most of the amine reactant will be tied up as its ammonium conjugate acid and will become non-nucleophilic. Converting reactants to products simply 1) Nucleophilic attack 2) Proton transfer 3) Protonation of OH 4) Removal of water 5) Deprotonation Imines can be hydrolyzed back to the corresponding primary amine under acidic conditons. Imines are sometimes difficult to isolate and purify due to their sensitivity to hydrolysis. Consequently, other reagents of the type Y–NH have been studied, and found to give stable products (R C=N–Y) useful in characterizing the aldehydes and ketones from which they are prepared. Some of these reagents are listed in the following table, together with the structures and names of their carbonyl reaction products. Hydrazones are used as part of the Wolff-Kishner reduction and will be discussed in more detail in another module. With the exception of unsubstituted hydrazones, these derivatives are easily prepared and are often crystalline solids - even when the parent aldehyde or ketone is a liquid. Since melting points can be determined more quickly and precisely than boiling points, derivatives such as these are useful for comparison and identification of carbonyl compounds. It should be noted that although semicarbazide has two amino groups (–NH ) only one of them is a reactive amine. The other is amide-like and is deactivated by the adjacent carbonyl group. 1)Please draw the products of the following reactions.
2) Please draw the structure of the reactant needed to produce the indicated product. 3) Please draw the products of the following reactions. 1) 2) 3) ) ), | 2,228 | 1,838 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/06%3A_Bonding_in_Organic_Molecules/6.06%3A_Resonance |
Until now, we have discussed bonding only in terms of electron pair associated with two nuclei. These we may call electrons. In fact, bonding electrons can be associated with more than two nuclei, and there is a measure of stability to be gained by this because the degree of bonding increases when the electrons can distribute themselves over a greater volume. This effect often is called or . It is important only if the component atomic orbitals overlap significantly, and this will depend in large part on the molecular geometry. The classic example of resonance is provided by the \(\pi\) bonding of benzene. This compound was shown in Chapter 1 to have the molecular formula \(C_6H_6\), to be planar, and hexagonal with bond angles of \(120^\text{o}\), and to possess six equivalent \(C-C\) bonds and six equivalent \(C-H\) bonds. Benzene usually is written with a structural formula proposed by Kekulé: That benzene is more stable than a single Kekulé, or 1,3,5-cyclohexatriene, structure can be gauged by comparing the experimental heat of combustion of benzene with the calculated value based on the average bond energies of Table 4-3: \[\ce{C6H6(g) + 15/2 O2 -> 6CO2(g) + 3H2O(g)}\] with \[\begin{aligned}
&\Delta H_{\exp }^{0}=-789 \mathrm{kcal} \\
&\Delta H_{\text {calc }}^{0}=-827 \mathrm{kcal}
\end{aligned}\] About \(38 \: \text{kcal}\) energy is released on combustion than calculated. Benzene, therefore, is \(38 \: \text{kcal mol}^{-1}\) than the cyclohexatriene structure predicts. Atomic-orbital models, like that shown for benzene, are useful descriptions of bonding from which to evaluate the potential for electron delocalization. But they are cumbersome to draw routinely. We need a simpler representation of electron delocalization. The method that commonly is used is to draw a set of structures, each of which represents a reasonable way in which the electrons (usually in \(p\) orbitals) could be paired. If more than one such structure can be written, the actual molecule, ion, or radical will have properties corresponding to some hybrid of these structures. A double-headed arrow \(\leftrightarrow\) is written between the structures that we consider to contribute to the hybrid. For example, the two Kekulé forms are two possible electron-pairing schemes or that could contribute to the resonance hybrid of benzene: It is very important to know what attributes a reasonable set of valence-bond structures has to have to contribute to a hybrid structure. It is equally important to understand what is and what is not implied in writing a set of structures. Therefore we shall emphasize the main points to remember in the rest of this section. 1. The members of a set of structures, as the two Kekulé structures for benzene, have no individual reality. They are hypothetical structures representing different electron-pairing schemes. We are not to think of benzene as a 50:50 mixture of equilibrating Kekulé forms. 2. To be reasonable, all structures in a set representing a resonance hybrid must have exactly the same locations of the atoms in space. For example, formula \(7\) does represent a valid member of the set of valence-bond structures of benzene, because the atoms of \(7\) have different positions from those of benzene (e.g., \(7\) is not planar): Structure \(7\) actually represents a known \(C_6H_6\) isomer that has a very different chemistry from that of benzene. 3. All members of the set must have the same number of paired or unpaired electrons. For the normal state of benzene, the six \(\pi\) electrons have three of one spin and three of the other. Structures such as \(8\), with four electrons of one spin and two of the other, are not valid contributors to the ground state of benzene: 4. The importance of resonance in any given case will depend on the energies of the contributing structures. The lower and more nearly equivalent the members of the set are in energy, the more important resonance becomes. That is to say, (as for the two Kekulé structures of benzene). As a corollary, the structure of a molecule is least likely to be satisfactorily represented by a conventional structural formula when two (or more) energetically equivalent, low-energy structures may be written. 5. If there is only low-energy structure in the set then, to a first approximation, the resonance hybrid may be assigned properties like those expected for that structure. As an example, we show three possible pairing schemes for ethene, \(9\), \(10\), and \(11\): Although \(10\) and \(11\) are equivalent, they are much higher in energy than \(9\) (see discussion in ). Therefore they do not contribute substantially to the structure of ethene that is best represented by \(9\). Resonance is by no means restricted to organic molecules. The following sets of valence-bond structures represent the hybrid structures of nitrate ion, \(NO_3^\ominus\), carbonate ion \(CO_3^{2 \ominus}\), and nitrous oxide, \(N_2O\). These are only representative examples. We suggest that you check these structures carefully to verify that each member of a set conforms to the general rules for resonance summarized above. A shorthand notation of hybrid structures frequently is used in which the delocalized \(\pi\)-bonding is shown as a broken line. For benzene, an inscribed circle also is used to indicate continuous \(\pi\) bonding: Electron delocalization is an important factor in the reactivity (or lack of it) of organic molecules. As an example, recall from Chapter 4 that the bond energies of various types of \(C-H\) bonds differ considerably (see Table 4-6). In particular, the methyl \(C-H\) bond in propene is about \(9 \: \text{kcal}\) than the methyl \(C-H\) bond of ethane or propane, and this difference can be explained by the use of the resonance concept. The following bond dissociations are involved: delocalization is possible for the propyl radical, propane, or propene. Accordingly, the methyl \(C-H\) bond strength in propene is less than in propane because of stabilization of the 2-propenyl radical. The foregoing discussion adds further to our understanding of the selectivity observed in the halogenation reactions discussed in Chapter 4. When propene is chlorinated in sunlight, the product is 3-chloropropene, and we may explain this on the basis that the radical-chain reaction involves propagation steps in which a chlorine atom attacks the hydrogen corresponding to the \(C-H\) bond: The resonance theory is very useful in accounting for, and in many cases predicting, the behavior of substances with \(\pi\) bonds. However, it is not omnipotent. One example where it fails is cyclobutadiene, for which we can write two equivalent valence-bond structures corresponding to the Kekulé structures for benzene: Despite this, cyclobutadiene is an extremely unstable substance, reacting with itself almost instantly at temperatures above \(-250^\text{o}\). For better understanding of this and some related problems, we provide a more detailed discussion of electron delocalization in Chapter 21. and (1977) | 7,108 | 1,839 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/14%3A_Organohalogen_and_Organometallic_Compounds/14.03%3A_Spectroscopic_Properties |
Organohalogen compounds give rise to strong absorptions in the arising from stretching vibrations of the carbon-halogen bond. The frequency of absorption decreases as the mass of the halogen increases. For monohaloalkanes the absorptions useful for identification are those of \(\ce{C-F}\) at \(1100\)-\(1000 \: \text{cm}^{-1}\) and \(\ce{C-Cl}\) at \(850\)-\(550 \: \text{cm}^{-1}\). The \(\ce{C-Br}\) and \(\ce{C-I}\) absorptions are below \(690 \: \text{cm}^{-1}\) and therefore are out of range of most commercial spectrophotometers. Because these bands are in the fingerprint region or far infrared, it is difficult to infer the presence of halogen in a molecule solely from its infrared spectrum. Apart from fluorine, the magnetic properties of halogen nuclei do not complicate proton or \(\ce{^{13}C}\) of organohalogen compounds. But fluorine \(\left( \ce{^{19}F} \right)\) has a spin of 1/2 and causes spin-spin splitting of the resonances of neighboring magnetic nuclei (\(\ce{^{13}C}\), \(\ce{^1H}\), and other \(\ce{^{19}F}\) nuclei). Proton chemical shifts are influenced strongly by the presence of halogen, which serves to deshield neighboring protons by electronegativity effects (see ). The of chlorine- and bromine-containing compounds clearly show the abundance ratios of the stable isotopes \(\ce{^{35}Cl}\):\(\ce{^{37}Cl} =\) 3:1 and \(\ce{^{79}Br}\):\(\ce{^{81}Br} =\) 1:1 in the molecular ions and those ionic fragments which contain halogens ( ). and (1977) | 1,506 | 1,840 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Energies_and_Potentials/THERMAL_ENERGY |
Thermal Energy, also known as random or internal Kinetic Energy, due to the random motion of molecules in a system. Kinetic Energy is seen in three forms: vibrational, rotational, and translational. Vibrational is the energy caused by an object or molecule moving in a vibrating motion, rotational is the energy caused by rotating motion, and translational is the energy caused by the movement of one molecule to to another location. Thermal energy is directly proportional to the temperature within a given (recall that a system is the subject of interest while the surroundings are located outside of the systems and the two interact via energy and matter exchange.) As a result of this relationship between thermal energy and the temperature of the system, the following applies:The more molecules present, the greater the movement of molecules within a given system, the greater the temperature and the greater the thermal energy + molecules = +movement = + temperature = + thermal energy As previously demonstrated, the thermal energy of a system is dependent on the temperature of a system which is dependent on the motion of the molecules of the system. As a result, the more molecules that are present, the greater the amount of movement within a given system which raises the temperature and thermal energy. Because of this, at a temperature of 0?C, the thermal energy within a given system is also zero. This means that a relatively small sample at a somewhat high temperature such as a cup of tea at its boiling temperature could have less thermal energy than a larger sample such as a pool that's at a lower temperature. If the cup of boiling tea is placed next to the freezing pool, the cup of tea will freeze first because it has less thermal energy than the pool. To keep definitions straight, remember the following Matter exists in three states: solid, liquid, or gas. When a given piece of matter undergoes a state change, thermal energy is either added or removed but the temperature remains constant. When a solid is melted, for example, thermal energy is what causes the bonds within the solid to break apart. Heat can be given off in three different processes: conduction, convection, or radiation. occurs when thermal energy is transferred through the interaction of solid particles. This process often occurs when cooking: the boiling of water in a metal pan causes the metal pan to warm as well. usually takes place in gases or liquids (whereas conduction most often takes place in solids) in which the transfer of thermal energy is based on differences in heat. Using the example of the boiling pot of of water, convection occurs as the bubbles rise to the surface and, in doing so, transfer heat from the bottom to the top. is the transfer of thermal energy through space and is responsible for the sunlight that fuels the earth. Thermal energy is a concept applicable in everyday life. For example, engines, such as those in cars or trains, do work by converting thermal energy into mechanical energy. Also, refrigerators remove thermal energy from a cool region into a warm region. On a larger scale, recent scientific research has been aiming to convert solar energy to thermal energy in order to create head and electricity. For example,scientific research centers such as NASA explore the uses and applications of thermal energy in order to provide for more efficient energy production. In 1990, for example, NASA extensively researched and explored the potentials of a hybrid power system which made use of Thermal Energy Storage (TES) devices. This power system would convert solar energy into thermal energy which would then be used to produce electrical power and heat. However, converting solar energy to thermal energy has been found to be much easier and much more feasible when systems are not in a state of thermodynamic equilibrium. Rather, scientists have proposed, a moving object or a running fluid can allow the energy to be converted into thermal energy. The ? states that whenever work is performed, the amount of entropy in the atmosphere is increased. Thus the flow of thermal energy is constantly increasing entropy. | 4,195 | 1,841 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Properties_of_Aldehydes_and_Ketones/Natural_Occurrence_of_Aldehydes_and_Ketones |
Aldehydes and ketones are widespread in nature and are often combined with other functional groups. Examples of naturally occurring molecules which contain a aldehyde or ketone functional group are shown in the following two figures. The compounds in the figure 1 are found chiefly in plants or microorganisms and those in the figure 2 have animal origins. Many of these molecular structures are chiral. When chiral compounds are found in nature they are usually enantiomerically pure, although different sources may yield different enantiomers. For example, carvone is found as its levorotatory (R)-enantiomer in spearmint oil, whereas, caraway seeds contain the dextrorotatory (S)-enantiomer. In this case the change of the stereochemistry causes a drastic change in the perceived scent. Aldehydes and ketones are known for their sweet and sometimes pungent odors. The odor from vanilla extract comes from the molecule vanillin. Likewise, benzaldehyde provides a strong scent of almonds and is this author’s favorite chemical smell. Because of their pleasant fragrances aldehyde and ketone containing molecules are often found in perfumes. However, not all of the fragrances are pleasing. In particular, 2-Heptanone provides part of the sharp scent from blue cheese and (R)-Muscone is part of the musky smell from the Himalayan musk deer. Lastly, ketones show up in many important hormones such as progesterone (a female sex hormone) and testosterone (a male sex hormone). Notice how subtle differences in structure can cause drastic changes in biological activity. The ketone functionality also shows up in the anti-inflammatory steroid, Cortisone. Figure 1. Aldehyde and ketone containing molecules isolated from plant sources. Figure 2. Aldehyde and ketone containing molecules isolated from animal sources. ) ), | 1,840 | 1,846 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/10%3A_Enzyme_Kinetics/10.03%3A_Chymotrypsin-_A_Case_Study |
Chymotrypsin is a digestive enzyme belonging to a super family of enzymes called serine proteases. It uses an active serine residue to perform hydrolysis on the C-terminus of the aromatic amino acids of other proteins. Chymotrypsin is a protease enzyme that cleaves on the C-terminal phenylalanine (F), tryptophan (W), and tyrosine (Y) on peptide chains. It shows specificity for aromatic amino acids because of its hydrophobic pocket. Chymotrypsin is one of the most studied enzymes due to its two phase kinetics: pre-steady-state and steady state. The study of these two kinetic states gives evidence of the "Ping-Pong" mechanism, the formation of covalent complexes leading to covalent hydrolysis reactions, and the rate of the catalyzed reactions. Synthesis of chymotrypsin occurs primarily in the pancreas. Instead of the active form, however, it is produced as an inactive zymogen called chymotrypsinogen to prevent its protease activity from digesting the pancreas. Upon secretion into the lumen of the small intestine, it is converted to its active form by another enzyme called trypsin. This dependence of a different enzyme for the activation of a protease is a common way for the body to prevent the digestion of organs and other harmful enzymatic side-effects. Chymotrypsin operates through a general mechanism known as the ping-pong mechanism (Figure \(\Page {1}\)) whereby the enzyme reacts with a substrate to form an enzyme intermediate. This intermediate has different properties than the initial enzyme, so to regenerate the initial enzymatic activity, it must react with a secondary substrate. This process is illustrated below: More specifically, chymotrypsin operates through a particular type of ping-pong mechanism called covalent hydrolysis. This means that the enzyme first forms a covalent bond with the target substrate, displacing the more stable moiety into solution. This enzyme-substrate complex is called the enzyme intermediate. The intermediate then reacts with water, which displaces the remaining part of the initial substrate and reforms the initial enzyme.
Chymotrypsin, like most enzymes, is specific in the types of substrates with which it reacts. As a protease, it cleaves polypeptides, and its inherent specificity allows it to act only on the carboxy-terminal of aromatic residues. It is a somewhat complicated mechanism, and is best explained in a series of steps.
Step 1: The target enters the active site of chymotrypsin, and it is held there by hydrophobic interactions between exposed non-polar groups of enzyme residues and the non-polar aromatic side-chain of the substrate. It is important to note the hydrogen bond between the Schiff nitrogen on histidine-57 and the oxygen side-chain of serine-195. Step 2: Aided by the histidine-serine hydrogen bonding, the hydroxyl group on serine-195 performs a nucleophilic attack on the carbonyl carbon of an aromatic amino acid while simultaneously transferring the hydroxyl hydrogen to the histidine Schiff nitrogen. This attack pushes the pi carbonyl electrons onto the carbonyl oxygen, forming a short-lived intermediate consisting of a c-terminal carbon with four single bonds: an oxygen anion, the beta-carbon of the aromatic amino acid, the n-terminus of the subsequent amino acid of the substrate protein, and the serine-195 side-chain oxygen. Step 3: This intermediate is short-lived, as the oxyanion electrons reform the pi bond with the c-terminus of the aromatic amino acid. The bond between the carboxy-terminus of the aromatic amino acid and the n-terminus of the subsequent residue is cleaved, and its electrons are used to extract the hydrogen of the protonated Schiff nitrogen on histidine-57. The bonds between the carbonyl carbon and the serine-195 oxygen remain in an ester configuration. This is called the acyl-enzyme intermediate. The c-terminal side of the polypeptide is now free to dissociate from the active site of the enzyme.
Step 4: Water molecules are now able to enter and bind to active site through hydrogen bonding between the hydrogen atoms of water and the histidine-57 Schiff nitrogen.
Step 5: The water oxygen now makes a nucleophilic attack on the carbonyl carbon of the acyl-enzyme intermediate, pushing the carbonyl’s pi electrons onto the carbonyl carbon as histidine-57 extracts one proton from water. This forms another quaternary carbon covalently bonded with serine, a hydroxyl, an oxyanion, and the aromatic amino acid. The proton on the recently protonated histidine-57 is now able to make a hydrogen bond with the serine oxygen.
Step 6: The oxyanion electrons reform the carbonyl pi bond, cleaving the bond between the carbonyl carbon and the serine hydroxyl. The electrons in this bond are used by the serine oxygen to deprotonate the histidine Schiff nitrogen and reform the original enzyme. The substrate no longer has affinity for the active site, and it soon dissociates from the complex.
Experiments were conducted in 1953 by B.S. Hartley and B.A. Kilby to investigate the kinetics of chymotrypsin-catalyzed hydrolysis. Instead of using a poly-peptide chain as a substrate, they used a nitro-phenyl ester, p-nitrophenyl acetate, that resembles an aromatic amino acid. Hydrolysis of this compound by chymotrypsin at the carbonyl group yields acetate and nitrophenolate, the latter of which absorbs near 400 nm light and its concentration can thus be measured by spectrophotometry (Figure \(\Page {2}\)). Spectrophotometric analysis of chymotrypsin acting on nitrophenylacetate showed that nitrophenolate was produced at a rate independent of substrate concentration, proving that the only factor contributing to the rate of product formation is the concentration of enzyme; this is typical for enzyme-substrate kinetics. However, when the slope of the 0-order absorbance plot was traced back to the starting point (time = 0), it was found that the initial concentration of nitrophenolate was not 0. In fact, it showed a 1:1 stoichiometric ratio with the amount of chymotrypsin used in the assay. This can only be explained by the fact that hydrolysis by chymotrypsin is biphasic in nature, meaning that it proceeds in two distinct steps. To analytically determine the rate of catalysis, all substrates, products, and intermediates must be defined. Refer to the figure below:
Using these abbreviations, kinetic analysis becomes less cumbersome. 1. The initial amount of enzyme can be represented as the sum of the free enzyme, the bound enzyme, and the inactive intermediate. \[[E]_o = [ES] + [^*ES] + [E] \nonumber \] 2. Assuming the initial step is the faster than the subsequent steps, the rate of nitrophenolate production can be described as: \[\dfrac{d[P_1]}{dt}=k_2[ES] \nonumber \] 3. Likewise, the rate of acetate formation can be represented by the equation: \[\dfrac{d[P_2]}{dt}=k_3[^*ES] \nonumber \] 4. Therefore, the net change in concentration of the inactive intermediate can be deduced: \[\dfrac{d[*ES]}{dt}=k_2[ES]-k_3[^*ES] \nonumber \] 5. The last inference that can be made from analysis of the measured kinetics data (Figure \(\Page {2}\)) is that the first step of the reaction equilibrates rapidly, and thus the change in bound substrate can be described in the following equation. This is a principal tenet in analyzing catalysis. \[\dfrac{d[ES]}{dt}=k_1[E,S]-k_{-1}[ES]=0 \nonumber \] 6. Where: \[\dfrac{k_{-1}}{k_1} = K_s = \dfrac{[E,S]}{[ES]} \nonumber \] 7. Combining all of these quantities, we can deduce the catalytic rate constant as: \[ k_{cat} = \dfrac{k_2k_3}{k_2+k_3} \nonumber \] 8. In ester hydrolysis, \(k_3 >> k_2\), so the resultant catalytic rate constant simplifies to: \[k_{cat}=k_2 \nonumber \] which is in agreement with the observed zeroth-order kinetics of Figure \(\Page {2}\). Speculate on how the catalytic rate constant can be determined from the spectrophotogram. The catalytic rate constant can be deduced from the graph by simply determining the slope of the line where the reaction demonstrates 0-order kinetics (the linear part) How can product be consistently produced if the rate of change of the ES complex is 0? This is pre-equilibrium kinetics in action. The ES complex is formed from E and S at a faster rate than any other step in the reaction. As soon as ES is converted to *ES, another mole of ES is produced from an infinite supply of E + S. This means that the amount of ES and E + S is constantly at equilibrium, and thus the change of either with respect to time is 0. How would the rate of product formation change if: Explain the role of hydrogen bonding in protein hydrolysis catalyzed by chymotrypsin. Initially, hydrogen bonding between the enzymes histidine and serine side chains weakens the bond of serine’s O-H. This allows for a facilitated nucleophilic attack of the hydroxyl Oxygen on the substrates carbonyl group. Conversely, in the final step of the reaction, the bound serine oxygen forms a hydrogen bond with a protonated histidine, which allows for easier cleavage from the substrate. What would the spectrophotogram look like if the reaction proceeded via a steady-state mechanism instead of pre-equilibrium. The graph would show similar 0-order kinetics, but the line would intercept the Y-axis at an absorbance of 0 instead of the 1:1 mole ratio of nitrophenolate to enzyme. | 9,348 | 1,847 |
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An atom is the smallest unit of an element that can exist. Every atom is made up of protons, neutrons, and electrons. These particles define a nuclide and its chemical properties and were discovered in the early 20 century and are described by modern . Nuclides are specific types of atoms or nuclei. Every nuclide has a chemical element symbol (E) as well as an atomic number (Z) , the number of protons in the nucleus, and a mass number (A), the total number of protons and neutrons in the nucleus. The symbol for the element is as shown below: \[^A_{Z}E\] An example is neon, which has the element symbol Ne, atomic number 10 and mass number 20. \[^{20}_{10}Ne\] A nuclide has a measurable amount of energy and lasts for a measurable amount of time. Stable nuclides can exist in the same state indefinitely, but unstable nuclides are radioactive and decay over time. Some unstable nuclides occur in nature, but others are synthesized artificially through nuclear reactions.They emit energy (\(\alpha\), \(\beta\), or \(\gamma\) emissions) until they reach stability. Every element has a defining atomic number, with the symbol "Z". If an atom is neutrally charged, it has the same number of protons and electrons. If it is charged, there may be more protons than electrons or vice versa, but the atomic number remains the same. In the element symbol, the charge goes on the right side of the element. For instance, O is an oxygen anion. O still has an atomic number of 8, corresponding to the 8 protons, but it has 10 electrons. Every element has a different atomic number, ranging from 1 to over 100. On the periodic table, the elements are arranged in the order of atomic number across a period. The atomic number is usually located above the element symbol. For example, hydrogen has one proton and one electron, so it has an atomic number of 1. Copper has the atomic number of 29 for its 29 protons. The atomic number defines an element's chemical properties. The number of electrons in an atom determines bonding and other chemical properties. In a neutral atom, the atomic number, Z, is also the number of electrons. These electrons are found in a cloud surrounding the nucleus, located by probability in electron shells or orbitals. The shell farthest from the nucleus is the valence shell. The electrons in this valence shell are involved in chemical bonding and show the behavior of the atom. The bonding electrons influence the molecular geometry and structure of the atom. They interact with each other and with other atoms in chemical reactions. The atomic number is unique to each atom and defines its characteristics of bonding or behavior or reactivity. Therefore, every atom, with a different atomic number, acts in a different manner. The mass of an atom is mostly localized to the nucleus. Because an electron has negligible mass relative to that of a proton or a neutron, the mass number is calculated by the sum of the number of protons and neutrons. Each proton and neutron's mass is approximately one atomic mass unit (AMU). The two added together results in the mass number: \[A=p^+ + n\] Elements can also have with the same atomic number, but different numbers of neutrons. There may be a few more or a few less neutrons, and so the mass is increased or decreased. On the periodic table, the mass number is usually located below the element symbol. The mass number listed is the average mass of all of the element's isotopes. Each isotope has a certain percentage abundance found in nature, and these are added and averaged to obtain the average mass number. For example, He has a mass number of 4. Its atomic number is 2, which is not always included in the notation because He is defined by the atomic number 2. | 3,764 | 1,848 |
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A is a special solution that stops massive changes in pH levels. Every buffer that is made has a certain buffer capacity, and buffer range. The buffer capacity is the amount of acid or base that can be added before the begins to change significantly. It can be also defined as the quantity of strong acid or base that must be added to change the pH of one liter of solution by one pH unit. The buffer range is the pH range where a buffer effectively neutralizes added acids and bases, while maintaining a relatively constant pH. The equation for pH also shows why pH does not change by much in buffers. \[K_a = \dfrac{[H^+,A^-]}{[HA]} \tag{1} \] \[pH = pK_a + \log\dfrac{[A^-]}{[HA]} \tag{2} \] Where, When the ratio between the conjugate base/ acid is equal to 1, the pH = pK . If the ratio between the two is 0.10, the pH drops by 1 unit from pK since log (0.10) = -1. If a ratio increases to a value of 10, then the pH increases by 1 unit since \log (10) = 1. The buffer capacity has a range of about 2. This means when a buffer is created, the pH can be changed by -1 by acid or +1 by base before the pH begins to change substantially. After the addition of base to raise the pH by 1 or more, most of the conjugate acid will have been depleted to try to maintain a certain pH, so the pH will be free to increase faster without the restraint of the conjugate acid. The same goes for the addition of acid, once the conjugate base has been used up, the pH will drop faster since most of the conjugate base has been used up. What is the effect on the pH of adding 0.006 mol HCL to 0.3L of a buffer solution that is 0.250M HC H O and 0.560 M NaC H O ? pK = 4.74 \[pH = 4.74 + \log\dfrac{0.560}{0.250} = 4.74 + 0.35 = 5.09 \nonumber \] \[C_2H_3O_2^- + H_3O^+ \rightleftharpoons HC_2H_3O_2 + H_2O \nonumber \] Calculate the starting amount of C H O \[0.300 \; L \times 0.560 \; M = 0.168 \; mol \; C_2H_3O_2^- \nonumber \] Calculate the starting amount of HC H O \[0.300 \; L \times 0.250 \; M = 0.075 \; mol \; HC_2H_3O_2 \nonumber \] Now calculate the new concentrations of C H O and HC H O : \[\dfrac{0.162 \; mol}{0.300 \; L} = 0.540 \; M \; C_2H_3O_2^- \nonumber \] \[\dfrac{0.081 \; mol}{0.300 \; L} = 0.540 \; M \; HC_2H_3O_2 \nonumber \] Using the new concentrations, we can calculate the new pH: \[pH = 4.74 + \log\dfrac{0.540}{0.270} = 4.74 + 0.30 = 5.04 \nonumber \] Calculate the pH change: \[pH_{final} - pH_{initial} = 5.04 - 5.09 = -0.05 \nonumber \] Therefore, the pH dropped by 0.05 pH units. What is the effect on the pH of adding 0.006 mol NaOH to 0.3L of a buffer solution that is 0.250M HC H O and 0.560 M NaC H O ? pK = 4.74 \[pH = 4.74 + \log\dfrac{0.560}{0.250} = 4.74 + 0.35 = 5.09 \nonumber \] \[HC_2H_3O_2 + OH^- \rightleftharpoons C_2H_3O_2^- + H_2O \nonumber \] Calculate the starting amount of HC H O \[0.300 \; L \times 0.250 \; M = 0.075 \; mol \; HC_2H_3O_2 \nonumber \] Calculate the starting amount of C H O \[ 0.300 \; L \times 0.560 \; M = 0.168 \; mol \; C_2H_3O_2^- \nonumber \] Now calculate the new concentrations of HC H O and C H O -: \[\dfrac{0.069 \; mol}{0.300 \; L} = 0.230 \; M \; HC_2H_3O_2 \nonumber \] \[\dfrac{0.174 \; mol}{0.300 \; L} = 0.580 \; M \; C_2H_3O_2^- \nonumber \] Using the new concentrations, we can calculate the new pH: \[pH = 4.74 + \log\dfrac{0.580}{0.230} = 4.74 + 0.40 = 5.14 \nonumber \] Calculate the pH change: \[pH_{final} - pH_{initial} = 5.14 - 5.09 = +0.05 \nonumber \] Therefore, the pH increased by 0.05 pH units. Blood contains large amounts of carbonic acid, a weak acid, and bicarbonate, a base. Together they help maintain the bloods pH at 7.4. If blood pH falls below 6.8 or rises above 7.8, one can become sick or die. The bicarbonate neutralizes excess acids in the blood while the carbonic acid neutralizes excess bases. Another example is when we consume antacids or milk of magnesia. After eating a meal with rich foods such as sea food, the stomach has to produce gastric acid to digest the food. Some of the acid can splash up the lower end of the esophagus causing a burning sensation. To relieve this burning, one would take an antacid, which when dissolved the bases buffer the excess acid by binding to them. | 4,225 | 1,851 |
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Ultra high vacuum is required for most surface science experiments for two principal reasons: To put these points in context we shall now look at the variation of various parameters with pressure The gas density \(\rho\) is easily estimated from the : \[\rho = \dfrac{N}{V} = \dfrac{P}{k T} \label{1}\] with The average distance that a particle (atom, electron, molecule ..) travels in the gas phase between collisions can be determined from a simple hard-sphere collision model - this quantity is known as the of the particle, here denoted by \(\lambda\), and for neutral molecules is given by the equation: \[\lambda = \dfrac{kT}{\sqrt{2} P \sigma} \label{2}\] with One of the crucial factors in determining how long a surface can be maintained clean (or, alternatively, how long it takes to build-up a certain surface concentration of adsorbed species) is the number of gas molecules impacting on the surface from the gas phase. The incident is the number of incident molecules per unit time per unit area of surface. The flux takes no account of the angle of incidence, it is merely a summation of all the arriving molecules over all possible incident angles For a given set of conditions ( etc.) the flux is readily calculated using a combination of the ideas of statistical physics, the ideal gas equation and the . : It can be readily shown that the incident flux, \(F\), is related to the gas density above the surface by \[F = \frac{1}{2} n \overline{c} \label{3}\] with : the molecular gas density is given by the ideal gas equation, namely \[n = \dfrac{N}{V}=\dfrac{P}{kT} \;\; (molecules \; m^{-3})\] : the mean molecular speed is obtained from the Maxwell-Boltzmann distribution of gas velocities by integration, yielding \[ \bar{c} = \sqrt{\dfrac{8kT}{m \pi}} \;\;\; ( m\,s^{-1})\] where Step 4: combining the equations from the first three steps gives the Hertz-Knudsen formula for the incident flux \[ F = \dfrac{P}{\sqrt{2 \pi m k T}} \;\;\; [ molecules \;m^{-2} s^{-1}]\] Note The is measure of the amount of gas which a surface has been subjected to. It is numerically quantified by taking the product of the pressure of the gas above the surface and the time of exposure (if the pressure is constant, or more generally by calculating the integral of pressure over the period of time of concern). Although the exposure may be given in the SI units of Pa s (Pascal seconds), the normal and far more convenient unit for exposure is the , where 1 L = 10 Torr s . i.e. \[(Exposure/L) = 10^6 \times (Pressure/Torr) \times (Time/s)\] The , , is a measure of the fraction of incident molecules which adsorb upon the surface i.e. it is a probability and lies in the range 0 - 1, where the limits correspond to no adsorption and complete adsorption of all incident molecules respectively. In general, depends upon many variables i.e. \[S = f ( surface coverage, temperature, crystal face .... )\] The of an adsorbed species may itself, however, be specified in a number of ways: \[\theta=\frac{\text { No. of adsorbed species per unit area of surface }}{\text { No. of surface substrate atoms per unit area }}\] where \(θ_{max}\) is usually less than one, but can for an adsorbate such as H occasionally exceed one. Note: How long will it take for a clean surface to become covered with a complete monolayer of adsorbate? This is dependent upon the flux of gas phase molecules incident upon the surface, the actual coverage corresponding to the monolayer and the coverage-dependent sticking probability and other aspect. however, it is possible to get a minimum estimate of the time required by assuming a unit sticking probability (i.e. = 1) and noting that monolayer coverages are generally of the order of 10 per cm or 10 per m . Then [ s ] All values given below are approximate and are generally dependent on factors such as temperature and molecular mass. Degree of Vacuum Pressure
(Torr) Gas Density
(molecules m ) Mean Free Path
(m) Time / ML
(s) Atmospheric 760 2 x 10 7 x 10 10 Low 1 3 x 10 5 x 10 10 Medium 10 3 x 10 5 x 10 10 High 10 3 x 10 50 1 UltraHigh 10 3 x 10 5 x 10 10 We can therefore conclude that the following requirements exist for: Collision Free Conditions ⇒ P < 10 Torr Maintenance of a Clean Surface ⇒ P < 10 Torr For most surface science experiments there are a number of factors necessitating a high vacuum environment: It is clear therefore that it is the last factor that usually determines the need for a very good vacuum in order to carry out reliable surface science experiments. | 4,614 | 1,852 |
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Although focused exclusively on acid–base equilibriums in aqueous solutions, equilibrium concepts can also be applied to many other kinds of reactions that occur in aqueous solution. In this chapter, we describe the equilibriums involved in the solubility of ionic compounds and the formation of complex ions. Solubility equilibriums involving ionic compounds are important in fields as diverse as medicine, biology, geology, and industrial chemistry. Carefully controlled precipitation reactions of calcium salts, for example, are used by many organisms to produce structural materials, such as bone and the shells that surround mollusks and bird eggs. In contrast, uncontrolled precipitation reactions of calcium salts are partially or wholly responsible for the formation of scale in coffee makers and boilers, “bathtub rings,” and kidney stones, which can be excruciatingly painful. The principles discussed in this chapter will enable you to understand how these apparently diverse phenomena are related. Solubility equilibriums are also responsible for the formation of caves and their striking features, such as stalactites and stalagmites, through a long process involving the repeated dissolution and precipitation of calcium carbonate. In addition to all of these phenomena, by the end of this chapter you will understand why barium sulfate is ideally suited for x-ray imaging of the digestive tract, and why soluble complexes of gadolinium can be used for imaging soft tissue and blood vessels using magnetic resonance imaging (MRI), even though most simple salts of both metals are toxic to humans. Hard water is a solution that consists largely of calcium and magnesium carbonate in CO -rich water. When the water is heated, CO gas is released, and the carbonate salts precipitate from solution and produce a solid called . | 1,855 | 1,853 |
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A potential energy surface (PES) describes the potential energy of a system, especially a collection of atoms, in terms of certain parameters, normally the positions of the atoms. The surface might define the energy as a function of one or more coordinates; if there is only one coordinate, the surface is called a potential energy curve or energy profile. It is helpful to use the analogy of a landscape: for a system with two degrees of freedom (e.g. two bond lengths), the value of the energy (analogy: the height of the land) is a function of two bond lengths (analogy: the coordinates of the position on the ground). The Potential Energy Surface represents the concepts that each geometry (both external and internal) of the atoms of the molecules in a chemical reaction is associated with it a unique potential energy. This creates a smooth energy “landscape” and chemistry can be viewed from a topology perspective (of particles evolving over "valleys""and passes"). The PES is the energy of a molecule as a function of the positions of its nuclei \(r\). This energy of a system of two atoms depends on the distance between them. At large distances the energy is zero, meaning “no interaction”. At distances of several atomic diameters attractive forces dominate, whereas at very close approaches the force is repulsive, causing the energy to rise. The attractive and repulsive effects are balanced at the minimum point in the curve. Plots that illustrate this relationship are quite useful in defining certain properties of a chemical bond. The internuclear distance at which the potential energy minimum occurs defines the . This is more correctly known as the bond length, because thermal motion causes the two atoms to vibrate about this distance. In general, the stronger the bond, the smaller will be the bond length. Attractive forces operate between all atoms, but unless the potential energy minimum is at least of the order of , the two atoms will not be able to withstand the disruptive influence of thermal energy long enough to result in an identifiable molecule. Thus we can say that a chemical bond exists between the two atoms in H . The weak attraction between argon atoms does not allow Ar to exist as a molecule, but it does give rise to the that holds argon atoms together in its liquid and solid forms. Potential energy and kinetic energy Quantum theory tells us that an electron in an atom possesses kinetic energy \(K\) as well as potential energy \(V\), so the total energy \(E\) is always the sum of the two: \(E = V + K\). The relation between them is surprisingly simple: \(K = –0.5 V\). This means that when a chemical bond forms (an exothermic process with \(ΔE < 0\)), the decrease in potential energy is accompanied by an increase in the kinetic energy (embodied in the momentum of the bonding electrons), but the magnitude of the latter change is only half as much, so the change in potential energy always dominates. The bond energy \(–ΔE\) has half the magnitude of the fall in potential energy. The geometry of a set of atoms can be described by a vector, r, whose elements represent the atom positions. The vector \(r\) could be the set of the Cartesian coordinates of the atoms, or could also be a set of inter-atomic distances and angles. Given \(r\), the energy as a function of the positions, \(V(r)\), is the value of \(V(r)\) for all values of \(r\) of interest. Using the landscape analogy from the introduction, \(V(r)\) gives the height on the "energy landscape" so that the concept of a potential energy surface arises. An example is the PES for water molecule (Figure \(\Page {1}\)) that show the energy minimum corresponding to optimized molecular structure for water- O-H bond length of 0.0958 nm and H-O-H bond angle of 104.5° To define an atom’s location in 3-dimensional space requires three coordinates (e.g., \(x\), \(y\),and \(z\) or \(r\), \(\theta\) and \(phi\) in Cartesian and Spherical coordinates) or . However, a reaction and hence the corresponding PESs do not depend of the absolute position of the reaction, only the relative positions (internal degrees). Hence both translation and rotation of the entire system can be removed (each with 3 degree of freedom, assuming non-linear geometries). So the dimensionality of a PES is \[3N-6\] where \(N\) is the number of atoms involves in the reaction, i.e., the number of atoms in each reactants). The PES is a hypersurface with many degrees of freedom and typically only a few are plotted at any one time for understanding. See to get a more details picture of how this applies to calculating the number of vibrations in a molecule To study a chemical reaction using the PES as a function of atomic positions, it is necessary to calculate the energy for arrangement of interest. Methods of calculating the energy of a particular atomic arrangement of atoms are well described in the computational chemistry article, and the emphasis here will be on finding approximations of \((V(r)\) to yield fine-grained energy-position information. For very simple chemical systems or when simplifying approximations are made about inter-atomic interactions, it is sometimes possible to use an analytically derived expression for the energy as a function of the atomic positions. An example is \[H + H_2 \rightarrow H_2 + H\] system as a function of the H-H distances. For more complicated systems, calculation of the energy of a particular arrangement of atoms is often too computationally expensive for large scale representations of the surface to be feasible. A PES is a conceptual tool for aiding the analysis of molecular geometry and chemical reaction dynamics. Once the necessary points are evaluated on a PES, the points can be classified according to the first and second derivatives of the energy with respect to position, which respectively are the gradient and the curvature. Stationary points (or points with a zero gradient) have physical meaning: energy minima correspond to physically stable chemical species and saddle points correspond to transition states, the highest energy point on the reaction coordinate (which is the lowest energy pathway connecting a chemical reactant to a chemical product). Three The PES concept finds application in fields such as chemistry and physics, especially in the theoretical sub-branches of these subjects. It can be used to theoretically explore properties of structures composed of atoms, for example, finding the minimum energy shape of a molecule or computing the rates of a chemical reaction. ) | 6,595 | 1,855 |
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If one needs to know how the Helmholtz function changes with changing volume at constant temperature, the following expression can be used: \[ \Delta A = \int_{V_1}^{V_2} \left( \dfrac{\partial A}{\partial V} \right)_T dV \label{eq1} \] But how does one derive an expression for the partial derivative in Equation \ref{eq1}? This is a fairly straight forward process that begins with the definition of \(A\): \[A = U - TS \nonumber \] Differentiating (and using the chain rule to evaluate \(d(TS)\) yields \[dA = dU - TdS - SdT \label{eq4} \] Now, it is convenient to use the combined first and second laws \[dU = TdS - pdV \label{eq5} \] which assumes: Substituting Equation \ref{eq5} into Equation \ref{eq4} yields \[dA = \cancel{TdS} - pdV - \cancel{TdS} - SdT \label{eq6} \] Canceling the \(TdS\) terms gives the important result \[dA = - pdV - SdT \label{eq6.5} \] The natural variables of \(A\) are therefore \(V\) and \(T\)! So the total differential of \(A\) is conveniently expressed as \[ dA = \left( \dfrac{\partial A}{\partial V} \right)_T dV + \left( \dfrac{\partial A}{\partial T} \right)_V dT \label{Total2} \] and by simple comparison of Equations \ref{eq6.5} and \ref{Total2}, it is clear that \[ \left( \dfrac{\partial A}{\partial V} \right)_T = - p \nonumber \] \[\left( \dfrac{\partial A}{\partial T} \right)_V = - S \nonumber \] And so, one can evaluate Equation \ref{eq1} as \[ \Delta A = - \int_{V_1}^{V_2} p\, dV \nonumber \] If the pressure is independent of the temperature, it can be pulled out of the integral. \[ \Delta A = - p \int_{V_1}^{V_2} dV = -p (V_2-V_1) \nonumber \] Otherwise, the temperature dependence of the pressure must be included. \[ \Delta A = - \int_{V_1}^{V_2} p(V)\, dV \nonumber \] Fortunately, this is easy if the substance is an ideal gas (or if some other equation of state can be used, such as the van der Waals equation.) Calculate \(\Delta A\) for the isothermal expansion of 1.00 mol of an ideal gas from 10.0 L to 25.0 L at 298 K. For an ideal gas, \[p =\dfrac{nRT}{V} \nonumber \] So \[\left( \dfrac{\partial A}{\partial V} \right)_T = -p \nonumber \] becomes \[\left( \dfrac{\partial A}{\partial V} \right)_T = -\dfrac{nRT}{V} \nonumber \] And so (Equation \ref{eq1}) \[ \Delta A = \int_{V_1}^{V_2} \left( \dfrac{\partial A}{\partial V} \right)_T dV \nonumber \] becomes \[ \Delta A = -nRT \int_{V_1}^{V_2} \dfrac{dV}{V} dT \nonumber \] or \[ \Delta A = -nRT \ln \left( \dfrac{V_2}{V_1} \right) \nonumber \] Substituting the values from the problem \[ \Delta A = -(1.00\,mol)(8.314 \, J/(mol\,K))(298\,K) \ln \left( \dfrac{25.0\,L}{10.0\,L} \right) \nonumber \] But further, it is easy to show that the Maxwell relation that arises from the simplified expression for the total differential of \(A\) is \[ \left( \dfrac{\partial p}{\partial T} \right)_V = \left( \dfrac{\partial S}{\partial V} \right)_T \nonumber \] This particular Maxwell relation is exceedingly useful since one of the terms depends only on \(p\), \(V\), and \(T\). As such it can be expressed in terms of our old friends, \(\alpha\) and \(\kappa_T\)! \[\left( \dfrac{\partial p}{\partial T} \right)_V = \dfrac{\alpha}{\kappa_T} \nonumber \] | 3,201 | 1,856 |
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As chemists, we are concerned with chemical changes and reactions. The thermodynamics of chemical reactions can be very important in terms of controlling the production of desired products and preventing safety hazards such as explosions. As such, measuring and understanding the thermochemistry of chemical reactions is not only useful, but essential! The techniques of can be used to measure q for a chemical reaction directly. The enthalpy change for a chemical reaction is of significant interest to chemists. An exothermic reaction will release heat (\(q_{reaction} < 0\), \(q_{surroundings} > 0\)) causing the temperature of the surrounding to increase. Conversely, an endothermic reaction (\(q_{reaction} > 0\), \(q_{surroundings} < 0\)) will draw heat from the surroundings, causing the temperature of the surrounding to drop. Measuring the temperature change in the surroundings allows for the determination of how much heat was released or absorbed in the reaction. Bomb calorimetry is used predominantly to measure the heat evolved in combustion reactions, but can be used for a wide variety of reactions. A typical bomb calorimetry set up is shown here. The reaction is contained in a heavy metallic container (the bomb) forcing the reaction to occur at constant volume. As such, the heat evolved (or absorbed) by the reaction is equal to the change in internal energy ( U ). The bomb is then submerged in a reproducible quantity of water, the temperature of which is monitored with a high-precision thermometer. For combustion reactions, the bomb will be loaded with a small sample of the compound to be combusted, and then the bomb is filled with a high pressure (typically about 10 atm) of O . The reaction is initiated by supplying heat using a short piece of resistive wire carrying an electrical current. The calorimeter must be calibrated by carrying out a reaction for which \(\Delta U_{rxn}\) is well known, so that the resulting temperature change can be related to the amount of heat released or absorbed. A commonly used reaction is the combustion of benzoic acid. This makes a good choice since benzoic acid reacts reliably and reproducibly under normal bomb calorimetry conditions. The “water equivalent” of the calorimeter can then be calculated from the temperature change using the following relationship: \[W = \dfrac{n\Delta U_c +e_{wrire}+e_{other}}{\Delta T} \nonumber \] where n is the number of moles of benzoic acid used, \(\Delta U_c\) is the internal energy of combustion for benzoic acid (3225.7 kJ mol at 25 C), \(e_{wire}\) accounts for the energy released in the combustion of the fuse wire, e account for any other corrections (such as heat released due to the combustion of residual nitrogen in the bomb), and T is the measured temperature change in the surrounding water bath. Once the “water equivalent” is determined for a calorimeter, the temperature change can be used to find \(\Delta U_c\) for an unknown compound from the temperature change created upon combustion of a known quantity of the substance. \[ \Delta U_c = \dfrac{W \Delta T - e_{wire} - e_{other}}{n_{sample}} \nonumber \] The experiment above is known as “isothermal bomb calorimetry” as the entire assembly sits in a constant temperature laboratory. Another approach is to employ “adiabatic bomb calorimetry” in which the assembly sits inside of a water jacket, the temperature of which is controlled to match the temperature of the water inside the insulated container. By matching this temperature, there is no thermal gradient, and thus no heat leaks into or out of the assembly during an experiment (and hence the experiment is effectively “adiabatic”). The can be calculated from the internal energy change if the balanced chemical reaction is known. Recall from the definition of enthalpy \[\Delta H = \Delta U + \Delta (pV) \nonumber \] and if the gas-phase reactants and products can be treated as ideal gases (\(pV = nRT\)) \[\Delta H = \Delta U + RT \Delta n_{gas} \nonumber \] at constant temperature. For the combustion of benzoic acid at 25 C \[\ce{C6H5COOH (s) + 15/2 O_2(g) -> 7 CO2(g) + 3 H2O(l)} \nonumber \] it can be seen that \(\Delta n_{gas}\) is -0.5 mol of gas for every mole of benzoic acid reacted. A student burned a 0.7842 g sample of benzoic acid (\(\ce{C7H6O2}\)) in a bomb calorimeter initially at 25.0 C and saw a temperature increase of 2.02 C. She then burned a 0.5348 g sample of naphthalene (\(\ce{C10H8}\)) (again from an initial temperature of 25 C) and saw a temperature increase of 2.24 C. From this data, calculate \(\Delta H_c\) for naphthalene (assuming and are unimportant.) First, the water equivalent: \[W = \dfrac{\left[ (0.7841\,g) \left(\frac{1\,mol}{122.124 \, g} \right)\right] (3225.7 \,kJ/mol)}{2.02 \,°C} = 10.254 \, kJ/°C \nonumber \] Then \(\Delta U_c\) for the sample: \[\Delta U_c = \dfrac{(10.254\, kJ/\,°C)(2.24\,°C )}{(0.5308 \,g)\left(\frac{1\,mol}{128.174 \, g} \right) } = 5546.4 \, kJ/°C \nonumber \] \(\Delta H_c\) is then given by \[\Delta H_c = \Delta U_c + RT \Delta n_{gas} \nonumber \] The reaction for the combustion of naphthalene at 25 C is: \[\ce{ C10H8(s) + 12 O2(g) -> 10 CO2(g) + 4 H2O(l)} \nonumber \] with \(\Delta n_{gas} = -2\). So \[ \Delta H_c = 5546.4 \,kJ/mol + \left( \dfrac{8.314}{1000} kJ/(mol \, K) \right) (298 \,L) (-2) = 5541\, kJ/mol \nonumber \] The literature value (Balcan, Arzik, & Altunata, 1996) is 5150.09 kJ/mol. So that’s not too far off! | 5,493 | 1,857 |
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Alkenes can donate their electrons to strong electrophiles and the resulting carbocations combine with the counterion of the electrophile to undergo an overall addition reaction. However, there may be some cases in which the counterion does not combine with the carbocation. Hydrobrominations of the type we have looked at only occur under certain conditions. Other conditions can lead to other products. For example, a solvent such as water can also participate in the reaction. The oxygen-based cation (or oxonium ion) that results can easily lose its charge through loss of a proton. As a result, a molecule of water adds to the alkene overall. The alkene becomes an alcohol. This reaction is called an "acid-catalyzed hdration" of an alkene. Explain how the hydration of an alkene in the presence of acid is a catalytic reaction. In many cases, equilibrium mixtures of multiple products may result from the addition of acids to alkenes. Show mechanisms, with curved arrows, for the following reactions. On the other hand, in the absence of any solvent, the bromide ion might still have some competition in the second step. The neat reaction (neat means "without solvent" of an alkene with a small amount of acid can result in polymerization. The alkene, which acted as a nucleophile in the first step, can also act as a nucleophile in the second step. It is important to remember that in any reaction, millions of molecules are involved. Even if one alkene molecule reacts with acid in the first step of a reaction, there are still plenty of other alkene molecules around to act as nucleophile in the second step. Provide a mechanism for the polymerization shown above. Assume there are four 2-methylpropene molecules and one hydrogen bromide molecule to begin. Chain reactions involve an , in which a reactive species is generated; , in which the reactive species reacts to make a new reactive species; and a in which the reactive species reacts to make a stable molecule. Label each of the steps in your mechanism from the previous question. , | 2,063 | 1,858 |
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Oxidative addition is a general term for the insertion of a metal between two atoms that were previously bonded together. Figure OA2.1. Our general scheme for an oxidative addition of hydrogen to a metal. Oxidative addition is a reaction type rather than mechanism. Several different mechanisms are possible, including polar reactions, non-polar / concerted reactions and radical reactions. For example the formation of Grignard nucleophiles by treatment of alkyl halides with magnesium could be described as an oxidative addition. Figure OA2.2. Formation of a Grignard reagent could be described as an oxidative addition, although the mechanism does not resemble the ones that we will look at here. However, Grignard formation is believed to occur via a radical reaction, and has very little relationship to the addition of hydrogen to Vaska's complex, for instance. In this section, we will look at polar oxidative additions and concerted oxidative additions. Radical oxidative additions will be left for a later chapter on radical chemistry. Where do the terms "oxidative addition" and "reductive elimination" come from? Think back to how we learned to count valence electrons in transition metal complexes. One of the first things we did was remove the ligands from the complex to see whether there would be a charge on the metal without the ligands. Figure OA2.3. Deconstruction of a metal complex to discern the charge on the metal. If each H ligand in Figure OA2.2 is a hydride (reasonable because H is more electronegative than the metal), then removing them would leave the metal with a 2+ charge. That means the metal is in the (II) oxidation state, even if it doesn't formally have a charge on it in the complex. Addition of H to a metal atom, M, is accompanied by increase in formal oxidation state at the metal (by +2). Reductive Elimination is microscopic reverse of oxidative addition. Figure OA2.3. Deconstruction of a metal complex to discern the charge on the metal. Two atoms that were bonded to one metal atom become bonded to each other, instead. Keep in mind the formal oxidation state of the metal. This elimination is accompanied by decrease in formal oxidation state at the metal (by -2). Remember, oxidation state can often be determined by giving each ligand the pair of electrons it shares with the metal. Sort out the formal charge on the donor atom in the ligand, and you will know the charge or oxidation state of the metal. For example, in the example of MH in Figure OA2.3., assume each bond between M and H is a pair of electrons that belongs with the hydrogen. A hydrogen with two electrons is an anion. As a result, the metal must have an oxidation state of +2 (usually written with Roman numeral II). Determine the oxidation state on the metal before and after each of the following reactions. Propose a reason why the addition of a ligand such as a phosphine can sometimes result in reductive elimination from a coordination complex. , | 2,988 | 1,859 |
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For an atom or an ion with only a single electron, we can calculate the potential energy by considering only the electrostatic attraction between the positively charged nucleus and the negatively charged electron. When more than one electron is present, however, the total energy of the atom or the ion depends not only on attractive electron-nucleus interactions but also on repulsive electron-electron interactions. When there are two electrons, the repulsive interactions depend on the positions of electrons at a given instant, but because we cannot specify the exact positions of the electrons, it is impossible to exactly calculate the repulsive interactions. Consequently, we must use approximate methods to deal with the effect of electron-electron repulsions on orbital energies. These effects are the underlying basis for the periodic trends in elemental properties that we will explore in this chapter. If an electron is far from the nucleus (i.e., if the distance \(r\) between the nucleus and the electron is large), then at any given moment, many of the other electrons will be that electron and the nucleus (Figure \(\Page {1}\)). Hence the electrons will cancel a portion of the positive charge of the nucleus and thereby decrease the attractive interaction between it and the electron farther away. As a result, the electron farther away experiences an effective nuclear charge (\(Z_{eff}\)) that is less than the actual nuclear charge \(Z\). This effect is called . As the distance between an electron and the nucleus approaches infinity, \(Z_{eff}\) approaches a value of 1 because all the other (\(Z − 1\)) electrons in the neutral atom are, on the average, between it and the nucleus. If, on the other hand, an electron is very close to the nucleus, then at any given moment most of the other electrons are farther from the nucleus and do not shield the nuclear charge. At \(r ≈ 0\), the positive charge experienced by an electron is approximately the full nuclear charge, or \(Z_{eff} ≈ Z\). At intermediate values of \(r\), the effective nuclear charge is somewhere between 1 and \(Z\): \[1 ≤ Z_{eff} ≤ Z. \nonumber \] Notice that \(Z_{eff} = Z\) only for hydrogen (Figure \(\Page {2}\)). Shielding refers to the core electrons repelling the outer electrons, which lowers the effective charge of the nucleus on the outer electrons. Hence, the nucleus has "less grip" on the outer electrons insofar as it is shielded from them. \(Z_{eff}\) can be calculated by subtracting the magnitude of shielding from the total nuclear charge and the effective nuclear charge of an atom is given by the equation: \[ Z_{eff}=Z-S \label{4} \] where \(Z\) is the atomic number (number of protons in nucleus) and \(S\) is the shielding constant and is approximated by number of electrons between the nucleus and the electron in question ( ). The value of \(Z_{eff}\) will provide information on how much of a charge an electron actually experiences. We can see from Equation \ref{4} that the effective nuclear charge of an atom increases as the number of protons in an atom increases (Figure \(\Page {2}\)). As we will discuss later on in the chapter, this phenomenon can explain the decrease in atomic radii we see as we go across the periodic table as electrons are held closer to the nucleus due to increase in number of protons and increase in effective nuclear charge. The shielding constant can be estimated by totaling the screening by (\(n\)) except the one in question. \[ S = \sum_{i}^{n-1} S_i \label{2.6.0} \] where \(S_i\) is the shielding of the i electron. Electrons that are shielded from the full charge of the nucleus experience an \(Z_{eff}\)) of the nucleus, which is some degree less than the full nuclear charge an electron would feel in a hydrogen atom or hydrogenlike ion. From Equations \ref{4} and \ref{2.6.0}, \(Z_{eff}\) for a specific electron can be estimated is the shielding constants for that electron of all other electrons in species is known. A simple approximation is that all other non-valence electrons shield equally and fully: \[S_i=1 \label{simple} \] This crude approximation is demonstrated in Example \(\Page {1}\). What is the effective attraction \(Z_{eff}\) experienced by the valence electrons in the three isoelectronic species: the fluorine anion, the neutral neon atom, and sodium cation? Each species has 10 electrons, and the number of nonvalence electrons is 2 (10 total electrons - 8 valence), but the effective nuclear charge varies because each has a different atomic number \(A\). This is an application of Equations \ref{4} and \ref{2.6.0}. We use the simple assumption that all electrons shield equally and fully the valence electrons (Equation \ref{simple}). The charge \(Z\) of the nucleus of a fluorine atom is 9, but the valence electrons are screened appreciably by the core electrons (four electrons from the 1s and 2s orbitals) and partially by the 7 electrons in the 2p orbitals. So the sodium cation has the greatest effective nuclear charge. This also suggests that \(\mathrm{Na}^+\) has the smallest radius of these species and that is correct. What is the effective attraction \(Z_{eff}\) experienced by the valence electrons in the magnesium anion, the neutral magnesium atom, and magnesium cation? Use the simple approximation for shielding constants. Compare your result for the magnesium atom to the more accurate value in Figure \(\Page {2}\) and proposed an origin for the difference. Remember that the simple approximations in Equations \ref{2.6.0} and \ref{simple} suggest that valence electrons other valence electrons. Therefore, each of these species has the same number of non-valence electrons and Equation \ref{4} suggests the effective charge on each valence electron is identical for each of the three species. This is not correct and a more complex model is needed to predict the experimental observed \(Z_{eff}\) value. The ability of valence electrons to shield other valence electrons or in partial amounts (e.g., \(S_i \neq 1\)) is in violation of Equations \ref{2.6.0} and \ref{simple}. That fact that these approximations are poor is suggested by the experimental \(Z_{eff}\) value shown in Figure \(\Page {2}\) for \(\ce{Mg}\) of 3.2+. This is appreciably larger than the +2 estimated above, which means these simple approximations overestimate the total shielding constant \(S\). A more sophisticated model is needed. The approximation in Equation \ref{simple} is a good first order description of electron shielding, but the actual \(Z_{eff}\) experienced by an electron in a given orbital depends not only on the spatial distribution of the electron in that orbital but also on the distribution of all the other electrons present. This leads to large differences in \(Z_{eff}\) for different elements, as shown in Figure \(\Page {2}\) for the elements of the first three rows of the periodic table. Penetration describes the proximity to which an electron can approach to the nucleus. In a multi-electron system, electron penetration is defined by an electron's relative electron density (probability density) near the nucleus of an atom (Figure \(\Page {3}\)). Electrons in different orbitals have different electron densities around the nucleus. In other words, penetration depends on the shell (\(n\)) and subshell (\(l\)). For example, a 1s electron (Figure \(\Page {3}\); purple curve) has greater electron density near the nucleus than a 2p electron (Figure \(\Page {3}\); red curve) and has a greater penetration. This related to the shielding constants since the 1s electrons are closer to the nucleus than a 2p electron, hence the 1s screens a 2p electron almost perfectly (\(S=1\). However, the 2s electron has a lower shielding constant (\(S<1\) because it can penetrate close to the nucleus in the small area of electron density within the first spherical node (Figure \(\Page {3}\); green curve). In this way the 2s electron can "avoid" some of the shielding effect of the inner 1s electron. For the same shell value (\(n\)) the penetrating power of an electron follows this trend in subshells (Figure \(\Page {3}\)): \[s > p > d \approx f. \label{better1} \] for different values of shell (n) and subshell (l), penetrating power of an electron follows this trend: \[\ce{1s > 2s > 2p > 3s > 3p > 4s > 3d > 4p > 5s > 4d > 5p > 6s > 4f ...} \label{better2} \] Penetration describes the proximity of electrons in an orbital to the nucleus. Electrons that have greater penetration can get closer to the nucleus and effectively block out the charge from electrons that have less proximity. Because of the effects of shielding and the different radial distributions of orbitals with the same value of but different values of , the different subshells are not degenerate in a multielectron atom. For a given value of , the orbital is always lower in energy than the orbitals, which are lower in energy than the orbitals, and so forth. As a result, some subshells with higher principal quantum numbers are actually lower in energy than subshells with a lower value of ; for example, the 4 orbital is lower in energy than the 3 orbitals for most atoms. The concepts of electron shielding, orbital penetration and effective nuclear charge were introduced above, but we did so in a qualitative manner (e.g., Equations \ref{better1} and \ref{better2}). A more accurate model for estimating electron shielding and corresponding effective nuclear charge experienced is . However, the application of these rules is outside the scope of this text. Zeff and Electron Shielding: The calculation of orbital energies in atoms or ions with more than one electron (multielectron atoms or ions) is complicated by repulsive interactions between the electrons. The concept of , in which intervening electrons act to reduce the positive nuclear charge experienced by an electron, allows the use of hydrogen-like orbitals and an (\(Z_{eff}\)) to describe electron distributions in more complex atoms or ions. The degree to which orbitals with different values of and the same value of overlap or penetrate filled inner shells results in slightly different energies for different subshells in the same principal shell in most atoms. | 10,267 | 1,860 |
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The equation that describes the equilibrium between solid calcium carbonate and its solvated ions is: \[\ce{CaCO3}(s)⇌\ce{Ca^2+}(aq)+\ce{CO3^2-}(aq)\] We can establish this equilibrium either by adding solid calcium carbonate to water or by mixing a solution that contains calcium ions with a solution that contains carbonate ions. If we add calcium carbonate to water, the solid will dissolve until the concentrations are such that the value of the reaction quotient \(\ce{(Q=[Ca^2+,CO3^2- ])}\) is equal to the solubility product ( = 4.8 × 10 ). If we mix a solution of calcium nitrate, which contains Ca ions, with a solution of sodium carbonate, which contains \(\ce{CO3^2-}\) ions, the slightly soluble ionic solid CaCO will precipitate, provided that the concentrations of Ca and \(\ce{CO3^2-}\) ions are such that is greater than for the mixture. The reaction shifts to the left and the concentrations of the ions are reduced by formation of the solid until the value of equals . A saturated solution in equilibrium with the undissolved solid will result. If the concentrations are such that is less than , then the solution is not saturated and no precipitate will form. We can compare numerical values of with to predict whether precipitation will occur, as Example \(\Page {7}\) shows. (Note: Since all forms of equilibrium constants are temperature dependent, we will assume a room temperature environment going forward in this chapter unless a different temperature value is explicitly specified.) The first step in the preparation of magnesium metal is the precipitation of Mg(OH) from sea water by the addition of lime, Ca(OH) , a readily available inexpensive source of OH ion: \[\ce{Mg(OH)2}(s)⇌\ce{Mg^2+}(aq)+\ce{2OH-}(aq) \hspace{20px} K_\ce{sp}=2.1×10^{−13}\] The concentration of Mg ( ) in sea water is 0.0537 . Will Mg(OH) precipitate when enough Ca(OH) is added to give a [OH ] of 0.0010 ? This problem asks whether the reaction: \[\ce{Mg(OH)2}(s)⇌\ce{Mg^2+}(aq)+\ce{2OH-}(aq)\] shifts to the left and forms solid Mg(OH) when [Mg ] = 0.0537 and [OH ] = 0.0010 . The reaction shifts to the left if is greater than . Calculation of the reaction quotient under these conditions is shown here: \[\mathrm{Q=[Mg^{2+},OH^-]^2=(0.0537)(0.0010)^2=5.4×10^{−8}}\] Because is greater than ( = 5.4 × 10 is larger than = 2.1 × 10 ), we can expect the reaction to shift to the left and form solid magnesium hydroxide. Mg(OH) ( ) forms until the concentrations of magnesium ion and hydroxide ion are reduced sufficiently so that the value of is equal to . Use the solubility products in Appendix J to determine whether CaHPO will precipitate from a solution with [Ca ] = 0.0001 and \(\ce{[HPO4^2- ]}\) = 0.001 . No precipitation of CaHPO ; = 1 × 10 , which is less than Does silver chloride precipitate when equal volumes of a 2.0 × 10 - solution of AgNO and a 2.0 × 10 - solution of NaCl are mixed? (Note: The solution also contains Na and \(\ce{NO3-}\) ions, but when referring to solubility rules, one can see that sodium nitrate is very soluble and cannot form a precipitate.) The equation for the equilibrium between solid silver chloride, silver ion, and chloride ion is: The solubility product is 1.8 × 10 (see Appendix J). AgCl will precipitate if the reaction quotient calculated from the concentrations in the mixture of AgNO and NaCl is greater than . The volume doubles when we mix equal volumes of AgNO and NaCl solutions, so each concentration is reduced to half its initial value. Consequently, immediately upon mixing, [Ag ] and [Cl ] are both equal to: The reaction quotient, , is greater than for AgCl, so a supersaturated solution is formed: Since supersaturated solutions are unstable, AgCl will precipitate from the mixture until the solution returns to equilibrium, with equal to . Will KClO precipitate when 20 mL of a 0.050- solution of K is added to 80 mL of a 0.50- solution of \(\ce{ClO4-}\)? (Remember to calculate the new concentration of each ion after mixing the solutions before plugging into the reaction quotient expression.) No, = 4.0 × 10 , which is less than = 1.07 × 10 In the previous two examples, we have seen that Mg(OH) or AgCl precipitate when is greater than . In general, when a solution of a soluble salt of the M ion is mixed with a solution of a soluble salt of the X ion, the solid, M X precipitates if the value of for the mixture of M and X is greater than for M X . Thus, if we know the concentration of one of the ions of a slightly soluble ionic solid and the value for the solubility product of the solid, then we can calculate the concentration that the other ion must exceed for precipitation to begin. To simplify the calculation, we will assume that precipitation begins when the reaction quotient becomes equal to the solubility product constant. Blood will not clot if calcium ions are removed from its plasma. Some blood collection tubes contain salts of the oxalate ion, \(\ce{C2O4^2-}\), for this purpose (Figure \(\Page {4}\)). At sufficiently high concentrations, the calcium and oxalate ions form solid, CaC O •H O (which also contains water bound in the solid). The concentration of Ca in a sample of blood serum is 2.2 × 10 . What concentration of \(\ce{C2O4^2-}\) ion must be established before CaC O •H O begins to precipitate? The equilibrium expression is: \[\ce{CaC2O4}(s)⇌\ce{Ca^2+}(aq)+\ce{C2O4^2-}(aq)\] For this reaction: \[K_\ce{sp}=\ce{[Ca^2+,C2O4^2- ]}=2.27×10^{−9}\] (see Appendix J) CaC O does not appear in this expression because it is a solid. Water does not appear because it is the solvent. Solid CaC O does not begin to form until equals . Because we know and [Ca ], we can solve for the concentration of \(\ce{C2O4^2-}\) that is necessary to produce the first trace of solid: A concentration of \(\ce{[C2O4^2- ]}\) = 1.0 × 10 is necessary to initiate the precipitation of CaC O under these conditions. If a solution contains 0.0020 mol of \(\ce{CrO4^2-}\) per liter, what concentration of Ag ion must be reached by adding solid AgNO before Ag CrO begins to precipitate? Neglect any increase in volume upon adding the solid silver nitrate. 7.0 × 10 It is sometimes useful to know the concentration of an ion that remains in solution after precipitation. We can use the solubility product for this calculation too: If we know the value of and the concentration of one ion in solution, we can calculate the concentration of the second ion remaining in solution. The calculation is of the same type as that in Example—calculation of the concentration of a species in an equilibrium mixture from the concentrations of the other species and the equilibrium constant. However, the concentrations are different; we are calculating concentrations after precipitation is complete, rather than at the start of precipitation. Clothing washed in water that has a manganese [Mn ( )] concentration exceeding 0.1 mg/L (1.8 × 10 ) may be stained by the manganese upon oxidation, but the amount of Mn in the water can be reduced by adding a base. If a person doing laundry wishes to add a buffer to keep the pH high enough to precipitate the manganese as the hydroxide, Mn(OH) , what pH is required to keep [Mn ] equal to 1.8 × 10 ? The dissolution of Mn(OH) is described by the equation: \[\ce{Mn(OH)2}(s)⇌\ce{Mn^2+}(aq)+\ce{2OH-}(aq) \hspace{20px} K_\ce{sp}=4.5×10^{−14}\] We need to calculate the concentration of OH when the concentration of Mn is 1.8 × 10 . From that, we calculate the pH. At equilibrium: \[K_\ce{sp}=\ce{[Mn^2+,OH- ]^2}\] or \[ (1.8×10^{−6})\ce{[OH- ]^2}=4.5×10^{−14}\] so \[\ce{[OH- ]}=1.6×10^{−4}\:M\] Now we calculate the pH from the pOH: If the person doing laundry adds a base, such as the sodium silicate (Na SiO ) in some detergents, to the wash water until the pH is raised to 10.20, the manganese ion will be reduced to a concentration of 1.8 × 10 ; at that concentration or less, the ion will not stain clothing. The first step in the preparation of magnesium metal is the precipitation of Mg(OH) from sea water by the addition of Ca(OH) . The concentration of Mg ( ) in sea water is 5.37 × 10 . Calculate the pH at which [Mg ] is diminished to 1.0 × 10 by the addition of Ca(OH) . 11.09 Due to their light sensitivity, mixtures of silver halides are used in fiber optics for medical lasers, in photochromic eyeglass lenses (glass lenses that automatically darken when exposed to sunlight), and—before the advent of digital photography—in photographic film. Even though AgCl ( = 1.6 × 10 ), AgBr ( = 7.7 × 10 ), and AgI ( = 8.3 × 10 ) are each quite insoluble, we cannot prepare a homogeneous solid mixture of them by adding Ag to a solution of Cl , Br , and I ; essentially all of the AgI will precipitate before any of the other solid halides form because of its smaller value for . However, we can prepare a homogeneous mixture of the solids by slowly adding a solution of Cl , Br , and I to a solution of Ag . When two anions form slightly soluble compounds with the same cation, or when two cations form slightly soluble compounds with the same anion, the less soluble compound (usually, the compound with the smaller ) generally precipitates first when we add a precipitating agent to a solution containing both anions (or both cations). When the values of the two compounds differ by two orders of magnitude or more (e.g., 10 vs. 10 ), almost all of the less soluble compound precipitates before any of the more soluble one does. This is an example of , where a reagent is added to a solution of dissolved ions causing one of the ions to precipitate out before the rest. Determining if a Precipitate forms (The Ion Product): A slightly soluble electrolyte begins to precipitate when the magnitude of the reaction quotient for the dissolution reaction exceeds the magnitude of the solubility product. Precipitation continues until the reaction quotient equals the solubility product. A reagent can be added to a solution of ions to allow one ion to selectively precipitate out of solution. The common ion effect can also play a role in precipitation reactions. In the presence of an ion in common with one of the ions in the solution, Le Chatelier’s principle applies and more precipitate comes out of solution so that the molar solubility is reduced. ). | 10,436 | 1,861 |
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To balance a redox equation using the , we conceptually separate the overall reaction into two parts: an oxidation—in which the atoms of one element lose electrons—and a reduction—in which the atoms of one element gain electrons. Consider, for example, the reaction of Cr (aq) with manganese dioxide (MnO ) in the presence of dilute acid. is the net ionic equation for this reaction before balancing; the oxidation state of each element in each species has been assigned using the procedure described in : \( \begin{matrix} +2 & +4 & +1 & +3 & +2 & +1\; \; \\ Cr^{2+}\left ( aq \right ) &+\; \; \; \; \; \; \; \;MnO_{2}\left ( aq \right ) &+\; \; H^{+}\left ( aq \right ) &\rightarrow\; Cr^{3+}\left ( aq \right ) &+\;Mn^{2+} \left ( aq \right ) &+\; \; \; H_{2}O\left ( l \right ) \\ & \; \; \; \; \; \; \;-2 & & & & \; \; \; \; -2 \end{matrix} \tag{8.8.1} \) Notice that chromium is oxidized from the +2 to the +3 oxidation state, while manganese is reduced from the +4 to the +2 oxidation state. We can write an equation for this reaction that shows only the atoms that are oxidized and reduced: \(Cr^{2+} + Mn^{4+} \rightarrow Cr^{3+} + Mn^{2+} \tag{8.8.2}\) The oxidation can be written as \(Cr^{2+} \rightarrow Cr^{3+} + e^- \tag{8.8.3}\) and the reduction as \(Mn^{4+} + 2e^- \rightarrow Mn^{2+} \tag{8.8.4}\) For the overall chemical equation to be balanced, the number of electrons lost by the reductant must equal the number gained by the oxidant. We must therefore multiply the oxidation and the reduction equations by appropriate coefficients to give us the same number of electrons in both. In this example, we must multiply the oxidation equation by 2 to give \(2Cr^{2+} \rightarrow 2Cr^{3+} + 2e^- \tag{8.8.5}\) In a balanced redox reaction, the number of electrons lost by the reductant equals the number of electrons gained by the oxidant. The number of electrons lost in the oxidation now equals the number of electrons gained in the reduction: \(2Cr^{2+} \rightarrow 2Cr^{3+} + 2e^- \tag{8.8.6}\) \( Mn^{4+} + 2e^- \rightarrow Mn^{2+} \) We then add the equations for the oxidation and the reduction and cancel the electrons on both sides of the equation, using the actual chemical forms of the reactants and products: \( \begin{matrix} 2Cr^{2+}\left ( aq \right ) & \rightarrow & 2Cr^{3+}\left ( aq \right )+\cancel{2e^{-}}\\ & & & \\ 2Cr^{2+}\left ( aq \right )+\cancel{2e^{-}}& \rightarrow & Mn^{2+}\left ( aq \right )\\ & & & \\ 2Cr^{2+}\left ( aq \right )+2Cr^{2+}\left ( aq \right )&\rightarrow & 2Cr^{3+}\left ( aq \right ) + Mn^{2+}\left ( aq \right ) \end{matrix} \tag{8.8.7}\) Although the electrons cancel and the metal atoms are balanced, the total charge on the left side of the equation (+4) does not equal the charge on the right side (+8). Because the reaction is carried out in the presence of aqueous acid, we can add H as necessary to either side of the equation to balance the charge. By the same token, if the reaction were carried out in the presence of aqueous base, we could balance the charge by adding OH as necessary to either side of the equation to balance the charges. In this case, adding four H ions to the left side of the equation gives Although the charges are now balanced, we have two oxygen atoms on the left side of the equation and none on the right. We can balance the oxygen atoms without affecting the overall charge balance by adding H O as necessary to either side of the equation. Here, we need to add two H O molecules to the right side: Although we did not explicitly balance the hydrogen atoms, we can see by inspection that the overall chemical equation is now balanced. All that remains is to check to make sure that we have not made a mistake. This procedure for balancing reactions is summarized in and illustrated in Example 17. Procedure for Balancing Oxidation–Reduction Reactions by the Oxidation State Method Arsenic acid (H AsO ) is a highly poisonous substance that was once used as a pesticide. The reaction of elemental zinc with arsenic acid in acidic solution yields arsine (AsH , a highly toxic and unstable gas) and Zn (aq). Balance the equation for this reaction using oxidation states: \(H_3AsO_4(aq) + Zn(s) \rightarrow AsH_3(g) + Zn^{2+}(aq) \) reactants and products in acidic solution balanced chemical equation using oxidation states Follow the procedure given in for balancing a redox equation using oxidation states. When you are done, be certain to check that the equation is balanced. The oxidation state of arsenic in arsenic acid is +6, and the oxidation state of arsenic in arsine is −3. Conversely, the oxidation state of zinc in elemental zinc is 0, and the oxidation state of zinc in Zn (aq) is +2: The arsenic atom in H AsO is reduced from the +5 to the −3 oxidation state, which requires the addition of eight electrons: \( Reduction \textrm: \: As^{5+} + 8e^- \rightarrow \underset{-3}{As ^{3-}}\) Each zinc atom in elemental zinc is oxidized from 0 to +2, which requires the loss of two electrons per zinc atom: \( Oxidation \textrm:\: Zn \rightarrow Zn^{2+} + 2e^- \) The reduction equation has eight electrons, and the oxidation equation has two electrons, so we need to multiply the oxidation equation by 4 to obtain Inserting the actual chemical forms of arsenic and zinc and adjusting the coefficients gives The sum of the two equations in step 5 is which then yields Because the reaction is carried out in acidic solution, we can add H ions to whichever side of the equation requires them to balance the charge. The overall charge on the left side is zero, and the total charge on the right side is 4 × (+2) = +8. Adding eight H ions to the left side gives a charge of +8 on both sides of the equation: There are 4 O atoms on the left side of the equation. Adding 4 H O molecules to the right side balances the O atoms: Although we have not explicitly balanced H atoms, each side of the equation has 11 H atoms. \(\begin{align} & Atoms \textrm : \: 1As + 4Zn + 4O + 11H = 1As + 4Zn + 4O + 11H \\ & Total\: charge \textrm : \: 8(+1) = 4(+2) = +8 \end{align}\) The balanced chemical equation for the reaction is therefore: \(H_3AsO_4(aq) + 4Zn(s) + 8H^+(aq) \rightarrow AsH_3(g) + 4Zn^{2+}(aq) + 4H_2O(l) \) Exercise Copper commonly occurs as the sulfide mineral CuS. The first step in extracting copper from CuS is to dissolve the mineral in nitric acid, which oxidizes the sulfide to sulfate and reduces nitric acid to NO. Balance the equation for this reaction using oxidation states: \(CuS(s) + H^+(aq) + NO_3^-(aq) \rightarrow Cu^{2+}(aq) + NO(g) + SO_4^{2-}(aq) \) \(3CuS(s) + 8H^+(aq) + 8NO_3^-(aq) \rightarrow 3Cu^{2+}(aq) + 8NO(g) + 3SO_4^{2-}(aq) + 4H_2O(l)\) Reactions in basic solutions are balanced in exactly the same manner. To make sure you understand the procedure, consider Example 18. The commercial solid drain cleaner, Drano, contains a mixture of sodium hydroxide and powdered aluminum. The sodium hydroxide dissolves in standing water to form a strongly basic solution, capable of slowly dissolving organic substances, such as hair, that may be clogging the drain. The aluminum dissolves in the strongly basic solution to produce bubbles of hydrogen gas that agitate the solution to help break up the clogs. The reaction is as follows: \(Al(s) + H_2O(aq) \rightarrow [Al(OH)_4]^-(aq) + H_2(g) \) Balance this equation using oxidation states. reactants and products in a basic solution balanced chemical equation Follow the procedure given in for balancing a redox reaction using oxidation states. When you are done, be certain to check that the equation is balanced. We will apply the same procedure used in Example 17 but in a more abbreviated form. The oxidation state of aluminum changes from 0 in metallic Al to +3 in [Al(OH) ] . The oxidation state of hydrogen changes from +1 in H O to 0 in H . Aluminum is oxidized, while hydrogen is reduced: \( \overset{0}{Al} (s) + \overset{+1}{H}_2 O(aq) \rightarrow [ \overset{+3}{Al} (OH)_4 ]^- (aq) + \overset{0}{H_2} (g) \) Multiply the reduction equation by 3 to obtain an equation with the same number of electrons as the oxidation equation: Insert the actual chemical forms of the reactants and products, adjusting the coefficients as necessary to obtain the correct numbers of atoms as in step 4. Because a molecule of H O contains two protons, in this case, 3H corresponds to 3/2H O. Similarly, each molecule of hydrogen gas contains two H atoms, so 3H corresponds to 3/2H . \( \begin{align} & Reduction\textrm : \: \dfrac{3}{2} H_2 O + 3e^- \rightarrow \dfrac{3}{2} H_2 \\ & Oxidation \textrm : \: Al \rightarrow [ Al ( OH )_4 ]^- + 3e^- \end{align}\) Adding the equations and canceling the electrons gives To remove the fractional coefficients, multiply both sides of the equation by 2: The right side of the equation has a total charge of −2, whereas the left side has a total charge of 0. Because the reaction is carried out in basic solution, we can balance the charge by adding two OH ions to the left side: The left side of the equation contains five O atoms, and the right side contains eight O atoms. We can balance the O atoms by adding three H O molecules to the left side: Be sure the equation is balanced: The balanced chemical equation is therefore Thus 3 mol of H gas are produced for every 2 mol of Al. Exercise The permanganate ion reacts with nitrite ion in basic solution to produce manganese(IV) oxide and nitrate ion. Write a balanced chemical equation for the reaction. \(2MnO_4^-(aq) + 3NO_2^-(aq) + H_2O(l) \rightarrow 2MnO_2(s) + 3NO_3^-(aq) + 2OH^-(aq)\) As suggested in Example 17 and Example 18, a wide variety of redox reactions are possible in aqueous solutions. The identity of the products obtained from a given set of reactants often depends on both the ratio of oxidant to reductant and whether the reaction is carried out in acidic or basic solution, which is one reason it can be difficult to predict the outcome of a reaction. Because oxidation–reduction reactions in solution are so common and so important, however, chemists have developed two general guidelines for predicting whether a redox reaction will occur and the identity of the products: Species in high oxidation states act as oxidants, whereas species in low oxidation states act as reductants. When an aqueous solution of a compound that contains an element in a high oxidation state is mixed with an aqueous solution of a compound that contains an element in a low oxidation state, an oxidation–reduction reaction is likely to occur. A widely encountered class of oxidation–reduction reactions is the reaction of aqueous solutions of acids or metal salts with solid metals. An example is the corrosion of metal objects, such as the rusting of an automobile ( ). Rust is formed from a complex oxidation–reduction reaction involving dilute acid solutions that contain Cl ions (effectively, dilute HCl), iron metal, and oxygen. When an object rusts, iron metal reacts with HCl(aq) to produce iron(II) chloride and hydrogen gas: \(Fe(s) + 2HCl(aq) \rightarrow FeCl_2(aq) + H_2(g) \tag{8.8.10}\) In subsequent steps, FeCl undergoes oxidation to form a reddish-brown precipitate of Fe(OH) . Rust Formation. Many metals dissolve through reactions of this type, which have the general form \(metal + acid \rightarrow salt + hydrogen \tag{8.8.11}\) 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: \( Pb(s) + 2H^+(aq) \rightarrow Pb^{2+}(aq) + H_2(g) \tag{8.8.12}\) 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 , 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 ( ) and the reduction of silver salts by copper ( and ): \( Zn(s) + Fe^{2+}(aq) \rightarrow Zn^{2+}(aq) + Fe(s) \tag{8.8.13}\) \( Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s) \tag{8.8.14}\) The reaction in 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. The Single-Displacement Reaction of Metallic Copper with a Solution of Silver Nitrate. Figure used with permission of In oxidation–reduction reactions, electrons are transferred from one substance or atom to another. We can balance oxidation–reduction reactions in solution using the ( ), in which the overall reaction is separated into an oxidation equation and a reduction equation. Which elements in the periodic table tend to be good oxidants? Which tend to be good reductants? If two compounds are mixed, one containing an element that is a poor oxidant and one with an element that is a poor reductant, do you expect a redox reaction to occur? Explain your answer. What do you predict if one is a strong oxidant and the other is a weak reductant? Why? In each redox reaction, determine which species is oxidized and which is reduced: Single-displacement reactions are a subset of redox reactions. In this subset, what is oxidized and what is reduced? Give an example of a redox reaction that is a single-displacement reaction. Balance each redox reaction under the conditions indicated. Balance each redox reaction under the conditions indicated. Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation; then write the complete ionic equation for the reaction. Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation; then write the complete ionic equation for the reaction. Dentists occasionally use metallic mixtures called for fillings. If an amalgam contains zinc, however, water can contaminate the amalgam as it is being manipulated, producing hydrogen gas under basic conditions. As the filling hardens, the gas can be released, causing pain and cracking the tooth. Write a balanced chemical equation for this reaction. Copper metal readily dissolves in dilute aqueous nitric acid to form blue Cu (aq) and nitric oxide gas. Classify each reaction as an acid–base reaction, a precipitation reaction, or a redox reaction, or state if there is no reaction; then complete and balance the chemical equation: Classify each reaction as an acid–base reaction, a precipitation reaction, or a redox reaction, or state if there is no reaction; then complete and balance the chemical equation: | 15,306 | 1,863 |
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The atoms, molecules, or ions that compose a chemical system can undergo several types of molecular motion, including translation, rotation, and vibration (Figure \(\Page {1}\)). The greater the molecular motion of a system, the greater the number of possible microstates and the higher the entropy. A perfectly ordered system with only a single microstate available to it would have an entropy of zero. The only system that meets this criterion is a perfect crystal at a temperature of absolute zero (0 K), in which each component atom, molecule, or ion is fixed in place within a crystal lattice and exhibits no motion (ignoring quantum effects). Such a state of perfect order (or, conversely, zero disorder) corresponds to zero entropy. In practice, absolute zero is an ideal temperature that is unobtainable, and a perfect single crystal is also an ideal that cannot be achieved. Nonetheless, the combination of these two ideals constitutes the basis for the third law of thermodynamics: the entropy of any perfectly ordered, crystalline substance at absolute zero is zero. The entropy of any perfectly ordered, crystalline substance at absolute zero is . The third law of thermodynamics has two important consequences: it defines the sign of the entropy of any substance at temperatures above absolute zero as positive, and it provides a fixed reference point that allows us to measure the absolute entropy of any substance at any temperature.In practice, chemists determine the absolute entropy of a substance by measuring the molar heat capacity (C ) as a function of temperature and then plotting the quantity C /T versus T. The area under the curve between 0 K and any temperature T is the absolute entropy of the substance at T. In contrast, other thermodynamic properties, such as internal energy and enthalpy, can be evaluated in only relative terms, not absolute terms. In this section, we examine two different ways to calculate ΔS for a reaction or a physical change. The first, based on the definition of absolute entropy provided by the third law of thermodynamics, uses tabulated values of absolute entropies of substances. The second, based on the fact that entropy is a state function, uses a thermodynamic cycle similar to those discussed previously. One way of calculating ΔS for a reaction is to use tabulated values of the standard molar entropy (S°), which is the entropy of 1 mol of a substance at a standard temperature of 298 K; the units of S° are J/(mol•K). Unlike enthalpy or internal energy, it is possible to obtain absolute entropy values by measuring the entropy change that occurs between the reference point of 0 K [corresponding to S = 0 J/(mol•K)] and 298 K. As shown in Table \(\Page {1}\), for substances with approximately the same molar mass and number of atoms, S° values fall in the order S°(gas) > S°(liquid) > S°(solid). For instance, S° for liquid water is 70.0 J/(mol•K), whereas S° for water vapor is 188.8 J/(mol•K). Likewise, S° is 260.7 J/(mol•K) for gaseous I and 116.1 J/(mol•K) for solid I2. This order makes qualitative sense based on the kinds and extents of motion available to atoms and molecules in the three phases. The correlation between physical state and absolute entropy is illustrated in Figure \(\Page {2}\), which is a generalized plot of the entropy of a substance versus temperature. Entropy increases with softer, less rigid solids, solids that contain larger atoms, and solids with complex molecular structures. A closer examination of Table \(\Page {1}\) also reveals that substances with similar molecular structures tend to have similar S° values. Among crystalline materials, those with the lowest entropies tend to be rigid crystals composed of small atoms linked by strong, highly directional bonds, such as diamond [S° = 2.4 J/(mol•K)]. In contrast, graphite, the softer, less rigid allotrope of carbon, has a higher S° [5.7 J/(mol•K)] due to more disorder in the crystal. Soft crystalline substances and those with larger atoms tend to have higher entropies because of increased molecular motion and disorder. Similarly, the absolute entropy of a substance tends to increase with increasing molecular complexity because the number of available microstates increases with molecular complexity. For example, compare the S° values for CH OH(l) and CH CH OH(l). Finally, substances with strong hydrogen bonds have lower values of S°, which reflects a more ordered structure. ΔS° for a reaction can be calculated from absolute entropy values using the same “products minus reactants” rule used to calculate ΔH°. To calculate ΔS° for a chemical reaction from standard molar entropies, we use the familiar “products minus reactants” rule, in which the absolute entropy of each reactant and product is multiplied by its stoichiometric coefficient in the balanced chemical equation. Example \(\Page {1}\) illustrates this procedure for the combustion of the liquid hydrocarbon isooctane (C H ; 2,2,4-trimethylpentane). Use the data in Table \(\Page {1}\) to calculate ΔS° for the reaction of liquid isooctane with O (g) to give CO (g) and H O(g) at 298 K. : standard molar entropies, reactants, and products : ΔS° : Write the balanced chemical equation for the reaction and identify the appropriate quantities in Table \(\Page {1}\). Subtract the sum of the absolute entropies of the reactants from the sum of the absolute entropies of the products, each multiplied by their appropriate stoichiometric coefficients, to obtain ΔS° for the reaction. : The balanced chemical equation for the complete combustion of isooctane (C H ) is as follows: We calculate ΔS° for the reaction using the “products minus reactants” rule, where m and n are the stoichiometric coefficients of each product and each reactant: \begin{align}\Delta S^\circ_{\textrm{rxn}}&=\sum mS^\circ(\textrm{products})-\sum nS^\circ(\textrm{reactants})
\\ &=[8S^\circ(\mathrm{CO_2})+9S^\circ(\mathrm{H_2O})]-[S^\circ(\mathrm{C_8H_{18}})+\dfrac{25}{2}S^\circ(\mathrm{O_2})]
\\ &=\left \{ [8\textrm{ mol }\mathrm{CO_2}\times213.8\;\mathrm{J/(mol\cdot K)}]+[9\textrm{ mol }\mathrm{H_2O}\times188.8\;\mathrm{J/(mol\cdot K)}] \right \}
\\ &-\left \{[1\textrm{ mol }\mathrm{C_8H_{18}}\times329.3\;\mathrm{J/(mol\cdot K)}]+\left [\dfrac{25}{2}\textrm{ mol }\mathrm{O_2}\times205.2\textrm{ J}/(\mathrm{mol\cdot K})\right ] \right \}
\\ &=515.3\;\mathrm{J/K}\end{align} ΔS° is positive, as expected for a combustion reaction in which one large hydrocarbon molecule is converted to many molecules of gaseous products. Use the data in Table \(\Page {1}\) to calculate ΔS° for the reaction of H (g) with liquid benzene (C H ) to give cyclohexane (C H ). : −361.1 J/K We can also calculate a change in entropy using a thermodynamic cycle. As you learned previously, the (C ) is the amount of heat needed to raise the temperature of 1 mol of a substance by 1°C at constant pressure. Similarly, C is the amount of heat needed to raise the temperature of 1 mol of a substance by 1°C at constant volume. The increase in entropy with increasing temperature in Figure \(\Page {2}\) is approximately proportional to the heat capacity of the substance. Recall that the entropy change (ΔS) is related to heat flow (q ) by ΔS = q /T. Because q = nC ΔT at constant pressure or nC ΔT at constant volume, where n is the number of moles of substance present, the change in entropy for a substance whose temperature changes from T to T is as follows: \[\Delta S=\dfrac{q_{\textrm{rev}}}{T}=nC_\textrm p\dfrac{\Delta T}{T}\hspace{4mm}(\textrm{constant pressure})\] As you will discover in more advanced math courses than is required here, it can be shown that this is equal to the following:For a review of natural logarithms, see Essential Skills 6 in Chapter 11 "Liquids". \[\Delta S=nC_\textrm p\ln\dfrac{T_2}{T_1}\hspace{4mm}(\textrm{constant pressure}) \label{18.20}\] Similarly, \[\Delta S=nC_{\textrm v}\ln\dfrac{T_2}{T_1}\hspace{4mm}(\textrm{constant volume}) \label{18.21}\] Thus we can use a combination of heat capacity measurements (Equation 18.20 or Equation 18.21) and experimentally measured values of enthalpies of fusion or vaporization if a phase change is involved (Equation 18.18) to calculate the entropy change corresponding to a change in the temperature of a sample. We can use a thermodynamic cycle to calculate the entropy change when the phase change for a substance such as sulfur cannot be measured directly. As noted in the exercise in Example 6, elemental sulfur exists in two forms (part (a) in Figure \(\Page {3}\)): an orthorhombic form with a highly ordered structure (S ) and a less-ordered monoclinic form (S ). The orthorhombic (α) form is more stable at room temperature but undergoes a phase transition to the monoclinic (β) form at temperatures greater than 95.3°C (368.5 K). The transition from S to S can be described by the thermodynamic cycle shown in part (b) in Figure \(\Page {3}\), in which liquid sulfur is an intermediate. The change in entropy that accompanies the conversion of liquid sulfur to S (−ΔS = ΔS in the cycle) cannot be measured directly. Because entropy is a state function, however, ΔS can be calculated from the overall entropy change (ΔS ) for the S –S transition, which equals the sum of the ΔS values for the steps in the thermodynamic cycle, using Equation 18.20 and tabulated thermodynamic parameters (the heat capacities of S and S , ΔH , and the melting point of S .) If we know the melting point of S (T = 115.2°C = 388.4 K) and ΔS for the overall phase transition [calculated to be 1.09 J/(mol•K) in the exercise in Example 6], we can calculate ΔS from the values given in part (b) in Figure \(\Page {3}\) where C = 22.70 J/mol•K and C = 24.77 J/mol•K (subscripts on ΔS refer to steps in the cycle): \(\begin{align}\Delta S_{\textrm t}&=\Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4
\\ 1.09\;\mathrm{J/(mol\cdot K)}&=C_{\textrm p({\alpha})}\ln\left(\dfrac{T_2}{T_1}\right)+\dfrac{\Delta H_{\textrm{fus}}}{T_{\textrm m}}+\Delta S_3+C_{\textrm p(\beta)}\ln\left(\dfrac{T_4}{T_3}\right)
\\ &=22.70\;\mathrm{J/(mol\cdot K)}\ln\left(\dfrac{388.4}{368.5}\right)+\left(\dfrac{1.722\;\mathrm{kJ/mol}}{\textrm{388.4 K}}\times1000\textrm{ J/kJ}\right)
\\ &+\Delta S_3+24.77\;\mathrm{J/(mol\cdot K)}\ln\left(\dfrac{368.5}{388.4}\right)
\\ &=[1.194\;\mathrm{J/(mol\cdot K)}]+[4.434\;\mathrm{J/(mol\cdot K)}]+\Delta S_3+[-1.303\;\mathrm{J/(mol\cdot K)}]\end{align}\) Solving for ΔS gives a value of −3.24 J/(mol•K). As expected for the conversion of a less ordered state (a liquid) to a more ordered one (a crystal), ΔS is negative. The of a substance at any temperature above 0 K must be determined by calculating the increments of heat required to bring the substance from 0 K to the temperature of interest, and then summing the ratios / . Two kinds of experimental measurements are needed: \[ S_{0 \rightarrow T^o} = \int _{0}^{T^o} \dfrac{C_p}{T} dt \] Because the heat capacity is itself slightly temperature dependent, the most precise determinations of absolute entropies require that the functional dependence of on be used in the above integral in place of a constant . \[ S_{0 \rightarrow T^o} = \int _{0}^{T^o} \dfrac{C_p(T)}{T} dt \] When this is not known, one can take a series of heat capacity measurements over narrow temperature increments Δ and measure the area under each section of the curve. The area under each section of the plot represents the entropy change associated with heating the substance through an interval Δ . To this must be added the enthalpies of melting, vaporization, and of any solid-solid phase changes. Values of for temperatures near zero are not measured directly, but can be estimated from quantum theory. The cumulative areas from 0 K to any given temperature (taken from the experimental plot on the left) are then plotted as a function of , and any phase-change entropies such as S = H / T are added to obtain the absolute entropy at temperature . The third law of thermodynamics states that the entropy of any perfectly ordered, crystalline substance at absolute zero is zero. At temperatures greater than absolute zero, entropy has a positive value, which allows us to measure the absolute entropy of a substance. Measurements of the heat capacity of a substance and the enthalpies of fusion or vaporization can be used to calculate the changes in entropy that accompany a physical change. The entropy of 1 mol of a substance at a standard temperature of 298 K is its standard molar entropy (S°). We can use the “products minus reactants” rule to calculate the standard entropy change (ΔS°) for a reaction using tabulated values of S° for the reactants and the products. | 12,758 | 1,864 |
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\( \newcommand{\irr}{\subs{irr}} % irreversible\)
\( \newcommand{\fric}{\subs{fric}} % friction\)
\( \newcommand{\diss}{\subs{diss}} % dissipation\)
\( \newcommand{\el}{\subs{el}} % electrical\)
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\( \newcommand{\E}{^\mathsf{E}} % excess quantity (superscript)\)
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\( \newcommand{\gpht}{\small\gph} % gamma phase tiny superscript\) \( \newcommand{\upOmega}{\Omega}\) \( \newcommand{\dif}{\mathop{}\!\mathrm{d}} % roman d in math mode, preceded by space\)
\( \newcommand{\Dif}{\mathop{}\!\mathrm{D}} % roman D in math mode, preceded by space\)
\( \newcommand{\df}{\dif\hspace{0.05em} f} % df\) \(\newcommand{\dBar}{\mathop{}\!\mathrm{d}\hspace-.3em\raise1.05ex{\Rule{.8ex}{.125ex}{0ex}}} % inexact differential \)
\( \newcommand{\dq}{\dBar q} % heat differential\)
\( \newcommand{\dw}{\dBar w} % work differential\)
\( \newcommand{\dQ}{\dBar Q} % infinitesimal charge\)
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\( \newcommand{\fug}{f} % fugacity\)
\( \newcommand{\g}{\gamma} % solute activity coefficient, or gamma in general\)
\( \newcommand{\G}{\varGamma} % activity coefficient of a reference state (pressure factor)\)
\( \newcommand{\ecp}{\widetilde{\mu}} % electrochemical or total potential\)
\( \newcommand{\Eeq}{E\subs{cell, eq}} % equilibrium cell potential\)
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\( \newcommand{\R}{8.3145\units{J$\,$K$\per\,$mol$\per$}} % gas constant value\)
\( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \)
\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) An of a species is a kind of adjustment factor that relates the actual behavior to ideal behavior at the same temperature and pressure. The ideal behavior is based on a for the species. We begin by describing reference states for nonelectrolytes. The thermodynamic behavior of an electrolyte solution is more complicated than that of a mixture of nonelectrolytes, and will be discussed in the next chapter. A of a constituent of a mixture has the same temperature and pressure as the mixture. When species \(i\) is in its reference state, its chemical potential \(\mu_i\rf\) depends only on the temperature and pressure of the mixture. If the pressure is the standard pressure \(p\st\), the reference state of species \(i\) becomes its . In the standard state, the chemical potential is the \(\mu_i\st\), which is a function only of temperature. Reference states are useful for derivations involving processes taking place at constant \(T\) and \(p\) when the pressure is not necessarily the standard pressure. Table 9.3 describes the reference states of nonelectrolytes used in this e-book, and lists symbols for chemical potentials of substances in these states. The symbols for solutes include \(x\), \(c\), or \(m\) in the subscript to indicate the basis of the reference state. Since the activity coefficient of a species relates its actual behavior to its ideal behavior at the same \(T\) and \(p\), let us begin by examining behavior in ideal mixtures. Consider first an ideal gas mixture at pressure \(p\). The chemical potential of substance \(i\) in this ideal gas mixture is given by Eq. 9.3.5 (the superscript “id” stands for ideal): \begin{equation} \mu_i\id\gas = \mu_i\st\gas + RT\ln\frac{p_i}{p\st} \tag{9.5.1} \end{equation} The reference state of gaseous substance \(i\) is pure \(i\) acting as an ideal gas at pressure \(p\). Its chemical potential is given by \begin{equation} \mu_i\rf\gas = \mu_i\st\gas + RT\ln\frac{p}{p\st} \tag{9.5.2} \end{equation} Subtracting Eq. 9.5.2 from Eq. 9.5.1, we obtain \begin{equation} \mu_i\id\gas - \mu_i\rf\gas = RT\ln\frac{p_i}{p} \tag{9.5.3} \end{equation} Consider the following expressions for chemical potentials in ideal mixtures and ideal-dilute solutions of nonelectrolytes. The first equation is a rearrangement of Eq. 9.5.3, and the others are from earlier sections of this chapter (in order of occurrence, Eqs. 9.4.8, 9.4.35, 9.4.24, 9.4.27, and 9.4.28). \begin{equation} \tx{Constituent of an ideal gas mixture} \quad \mu_i\id\gas = \s{\mu_i\rf\gas + RT\ln\frac{p_i}{p}} \tag{9.5.4} \end{equation} \begin{equation} \tx{Constituent of an ideal liquid or solid mixture} \quad \mu_i\id = \mu_i^* + RT\ln x_i \tag{9.5.5} \end{equation} \begin{equation} \tx{Solvent of an ideal-dilute solution} \quad \mu\A\id = \mu\A^* + RT\ln x\A \tag{9.5.6} \end{equation} \begin{equation} \tx{Solute, ideal-dilute solution, mole fraction basis} \quad \mu\B\id = \mu\xbB\rf + RT\ln x\B \tag{9.5.7} \end{equation} \begin{equation} \tx{Solute, ideal-dilute solution, concentration basis} \quad \mu\B\id = \mu\cbB\rf + RT\ln\frac{c\B}{c\st} \tag{9.5.8} \end{equation} \begin{equation} \tx{Solute, ideal-dilute solution, molality basis} \quad \mu\B\id = \mu\mbB\rf + RT\ln\frac{m\B}{m\st} \tag{9.5.9} \end{equation} Note that the equations for the condensed phases have the general form \begin{equation} \mu_i\id = \mu_i\rf + RT\ln \left( \frac{\tx{composition variable}} {\tx{standard composition}} \right) \tag{9.5.10} \end{equation} where \(\mu_i\rf\) is the chemical potential of component \(i\) in an appropriate reference state. (The standard composition on a mole fraction basis is \(x\st{=}1\).) If a mixture is ideal, we can write an expression for the chemical potential of each component that includes an . The expression is like one of those for the ideal case (Eqs. 9.5.4–9.5.9) with the activity coefficient multiplying the quantity within the logarithm. Consider constituent \(i\) of a gas mixture. If we eliminate \(\mu_i\st\gas\) from Eqs. 9.3.12 and 9.5.2, we obtain \begin{equation} \begin{split} \mu_i & = \mu_i\rf\gas + RT\ln\frac{\fug_i}{p} \cr & = \mu_i\rf\gas + RT\ln\frac{\phi_i p_i}{p} \end{split} \tag{9.5.11} \end{equation} where \(\fug_i\) is the fugacity of constituent \(i\) and \(\phi_i\) is its fugacity coefficient. Here the activity coefficient is the fugacity coefficient \(\phi_i\). For components of a condensed-phase mixture, we write expressions for the chemical potential having a form similar to that in Eq. 9.5.10, with the composition variable now multiplied by an activity coefficient: \begin{equation} \mu_i = \mu_i\rf + RT\ln \left[ (\tx{activity coefficient of \(i\)}) \times \left( \frac{\tx{composition variable}} {\tx{standard composition}} \right) \right] \tag{9.5.12} \end{equation} The activity coefficient of a species is a dimensionless quantity whose value depends on the temperature, the pressure, the mixture composition, and the choice of the reference state for the species. Under conditions in which the mixture behaves ideally, the activity coefficient is unity and the chemical potential is given by one of the expressions of Eqs. 9.5.4–9.5.9; otherwise, the activity coefficient has the value that gives the actual chemical potential. This e-book will use various symbols for activity coefficients, as indicated in the following list of expressions for the chemical potentials of nonelectrolytes: \begin{equation} \tx{Constituent of a gas mixture} \quad \mu_i = \mu_i\rf\gas + RT\ln\left(\phi_i\frac{p_i}{p}\right) \tag{9.5.13} \end{equation} \begin{equation} \tx{Constituent of a liquid or solid mixture} \quad \mu_i = \mu_i^* + RT\ln\left(\g_i x_i\right) \tag{9.5.14} \end{equation} \begin{equation} \tx{Solvent of a solution} \quad \mu\A = \mu\A^* + RT\ln\left(\g\A x\A\right) \tag{9.5.15} \end{equation} \begin{equation} \tx{Solute of a solution, mole fraction basis} \quad \mu\B = \mu\xbB\rf + RT\ln\left(\g\xbB x\B\right) \tag{9.5.16} \end{equation} \begin{equation} \tx{Solute of a solution, concentration basis} \quad \mu\B = \mu\cbB\rf + RT\ln\left(\g\cbB\frac{c\B}{c\st}\right) \tag{9.5.17} \end{equation} \begin{equation} \tx{Solute of a solution, molality basis} \quad \mu\B = \mu\mbB\rf + RT\ln\left(\g\mbB\frac{m\B}{m\st}\right) \tag{9.5.18} \end{equation} Equation 9.5.14 refers to a component of a liquid or solid mixture of substances that mix in all proportions. Equation 9.5.15 refers to the solvent of a solution. The reference states of these components are the pure liquid or solid at the temperature and pressure of the mixture. For the activity coefficients of these components, this e-book uses the symbols \(\g_i\) and \(\g\A\). The IUPAC Green Book (E. Richard Cohen et al, , 3rd edition, RSC Publishing, Cambridge, 2007, p. 59) recommends the symbol \(f_i\) for the activity coefficient of component \(i\) when the reference state is the pure liquid or solid. This e-book instead uses symbols such as \(\g_i\) and \(\g\A\), in order to avoid confusion with the symbol usually used for fugacity, \(\fug_i\). In Eqs. 9.5.16–9.5.18, the symbols \(\g\xbB\), \(\g\cbB\), and \(\g\mbB\) for activity coefficients of a nonelectrolyte solute include \(x\), \(c\), or \(m\) in the subscript to indicate the choice of the solute reference state. Although three different expressions for \(\mu\B\) are shown, for a given solution composition they must all represent the same of \(\mu\B\), equal to the rate at which the Gibbs energy increases with the amount of substance B added to the solution at constant \(T\) and \(p\). The value of a solute activity coefficient, on the other hand, depends on the choice of the solute reference state. You may find it helpful to interpret products appearing on the right sides of Eqs. 9.5.13–9.5.18 as follows. A change in pressure or composition that causes a mixture to approach the behavior of an ideal mixture or ideal-dilute solution must cause the activity coefficient of each mixture constituent to approach unity: How would we expect the activity coefficient of a nonelectrolyte solute to behave in a dilute solution as the solute mole fraction increases beyond the range of ideal-dilute solution behavior? The following argument is based on molecular properties at constant \(T\) and \(p\). We focus our attention on a single solute molecule. This molecule has interactions with nearby solute molecules. Each interaction depends on the intermolecular distance and causes a change in the internal energy compared to the interaction of the solute molecule with solvent at the same distance. In Sec. 11.1.5, it will be shown that roughly speaking the internal energy change is negative if the average of the attractive forces between two solute molecules and two solvent molecules is greater than the attractive force between a solute molecule and a solvent molecule at the same distance, and is positive for the opposite situation. The number of solute molecules in a volume element at a given distance from the solute molecule we are focusing on is proportional to the local solute concentration. If the solution is dilute and the interactions weak, we expect the local solute concentration to be proportional to the macroscopic solute mole fraction. Thus, the partial molar quantities \(U\B\) and \(V\B\) of the solute should be approximately linear functions of \(x\B\) in a dilute solution at constant \(T\) and \(p\). From Eqs. 9.2.46 and 9.2.50, the solute chemical potential is given by \(\mu\B=U\B+pV\B-TS\B\). In the dilute solution, we assume \(U\B\) and \(V\B\) are linear functions of \(x\B\) as explained above. We also assume the dependence of \(S\B\) on \(x\B\) is approximately the same as in an ideal mixture; this is a prediction from statistical mechanics for a mixture in which all molecules have similar sizes and shapes. Thus we expect the deviation of the chemical potential from ideal-dilute behavior, \(\mu\B = \mu\xbB\rf + RT\ln x\B\), can be described by adding a term proportional to \(x\B\): \(\mu\B = \mu\xbB\rf + RT\ln x\B + k_x x\B\), where \(k_x\) is a positive or negative constant related to solute-solute interactions. If we equate this expression for \(\mu\B\) with the one that defines the activity coefficient, \(\mu\B = \mu\xbB\rf + RT\ln(\g\xbB x\B)\) (Eq. 9.5.16), and solve for the activity coefficient, we obtain the relation \(\g\xbB = \exp (k_x x\B/RT)\). (This is essentially the result of the McMillan–Mayer solution theory from statistical mechanics.) An expansion of the exponential in powers of \(x\B\) converts this to \begin{equation} \g\xbB = 1 + (k_x/RT)x\B + \cdots \tag{9.5.25} \end{equation} Thus we predict that at constant \(T\) and \(p\), \(\g\xbB\) is a linear function of \(x\B\) at low \(x\B\). An ideal-dilute solution, then, is one in which \(x\B\) is much smaller than \(RT/k_x\) so that \(\g\xbB\) is approximately 1. An ideal mixture requires the interaction constant \(k_x\) to be zero. By similar reasoning, we reach analogous conclusions for solute activity coefficients on a concentration or molality basis. For instance, at low \(m\B\) the chemical potential of B should be approximately \(\mu\mbB\rf + RT\ln (m\B/m\st) + k_m m\B\), where \(k_m\) is a constant at a given \(T\) and \(p\); then the activity coefficient at low \(m\B\) is given by \begin{equation} \g\mbB = \exp (k_m m\B/RT) = 1 + (k_m/RT)m\B + \cdots \tag{9.5.26} \end{equation} The prediction from the theoretical argument above, that a solute activity coefficient in a dilute solution is a linear function of the composition variable, is borne out experimentally as illustrated in Fig. 9.10. This prediction applies only to a nonelectrolyte solute; for an electrolyte, the slope of activity coefficient versus molality approaches \(-\infty\) at low molality. | 19,244 | 1,866 |
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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 containing at least one volatile liquid can be described using \[P_j = x_jP^*_j \nonumber \] 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 diagram for an ideal mixture of two compounds, water and ethanol at 75 °C. At this temperature, water a pure vapor pressure of 384 Torr and ethanol has a pure vapor pressure of 945 Torr. In Figure 24.4.1, the composition of the liquid phase is represented by the solid line and the composition of the vapor phase is represented by the dashed line. Figure 24.4.1: The composition of the liquid phase (solid line) and the vapor phase (dashed lined) as a function of mole fraction according to Raoult's law. The solution is a mixture of water and ethanol. \(P_A\) is the vapor pressure of pure water and \(P_B\) is the vapor pressure of pure ethanol at 75 °C Often, 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 24.4.2 for an ideal mixture of two volatile liquids. In this diagram, \(T_A\) and \(T_B\) 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 re-vaporized, 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. When we add \(n_A\) moles of component \(A\) and \(n_B\) moles of component \(B\) to form an ideal liquid solution, this is generally a spontaneous process. Let us consider the Gibbs free energy change of that process: \[ \Delta_{mix}G = n_1\mu_1^{sln}+n_2\mu_2^{sln} - (n_1\mu_1^* + n_2\mu_2^*) \] Using: \[ \mu_i^{sln} \equiv \mu_i^* + nRTx_i\ln{x_i} \] his expression simplifies to: \[ \Delta_{mix} G = nRTx_A\ln{x_A} + nRTx_B\ln{x_B} \] where \(n\) is the total moles. Mole fraction, \(x_i\), is always less than one, so the Gibbs energy of mixing is always negative; mixing is always spontaneous. We can generalize this to mixtures with more than two components: \[ \Delta_{mix} G = nRT\sum_i{x_i\ln{x_i}} \] This expression looks suspiciously familiar. Apart from a factor of \(-T\), it is just like the : \[ \Delta_{mix} S = -nR\sum_i{x_i\ln{x_i}} \] Recalling the relationship between Gibbs energy and entropy: \[ \Delta_{mix} G = \Delta_{mix}H-T\Delta_{mix}S \] This leaves no room at all for an enthalpy effect: \[ \Delta_{mix} H = 0 \] Even though there are strong interactions between neighboring particles in liquids, there is no enthalpy change. This implies that it does not matter what the neighboring molecules are. If we represent the average interaction energy between molecule \(i\) and \(j\) by \(U_{ij}\), we are as \(U_{ij}\) is always the same. In practice, this is the case. It usually does matter and then the enthalpy term is not zero. As this affects the thermodynamics of the liquid solution, it should also affect the vapor pressures that are in equilibrium with it. From the change of \(G\) in its natural variables, we know that: \[ \left (\dfrac{\partial G}{\partial P} \right)_T =V \nonumber \] This means that if we take \[ \left (\dfrac{\partial \Delta G}{\partial P} \right)_T =\Delta V_{mix} \nonumber \] In the ideal case we get: \[ \left (\dfrac{\partial \Delta G^{ideal}}{\partial P} \right)_T =\Delta V_{mix}^{ideal} \nonumber \] \[ \left (\dfrac{\partial RT ( n_1 \ln x_1 + n_2\ln x_2)}{\partial P} \right)_T =\Delta V_{mix}^{ideal} =0 \nonumber \] In the ideal case, volumes are and we need not worry about how the partial molar volumes change with composition. | 5,134 | 1,868 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/08%3A_Substitution_Reactions_of_Alkyl_Halides/8.09%3A_The_Role_of_the_Solvent_in_(S_N2)_and_(S_N1)_Reactions |
Recall the definitions of electrophile and nucleophile: : An electron deficient atom, ion or molecule that has an affinity for an electron pair, and will bond to a base or nucleophile.
: An atom, ion or molecule that has an electron pair that may be donated in forming a covalent bond to an electrophile (or Lewis acid). If we use a common alkyl halide, such as methyl bromide, and a common solvent, ethanol, we can examine the rate at which various nucleophiles substitute the methyl carbon. is thereby related to the relative rate of substitution reactions at the halogen-bearing carbon atom of the reference alkyl halide. The most reactive nucleophiles are said to be more nucleophilic than less reactive members of the group. The nucleophilicities of some common Nu: reactants vary as shown in the following CH CO < Cl < Br < N < CH O < CN ≈ SCN < I < CH S The reactivity range encompassed by these reagents is over 5,000 fold, thiolate being the most reactive. Note that by using methyl bromide as the reference substrate, the complication of competing elimination reactions is avoided. The nucleophiles used in this study were all anions, but this is not a necessary requirement for these substitution reactions. Indeed reactions , presented at the beginning of this section, are examples of neutral nucleophiles participating in substitution reactions. The cumulative results of studies of this kind has led to useful empirical rules pertaining to nucleophilicity: of nucleophilic anions markedly influences their reactivity. The nucleophilicities cited above were obtained from reactions in methanol solution. Polar, protic solvents such as water and alcohols solvate anions by hydrogen bonding interactions, as shown in the diagram below. These solvated species are more stable and less reactive than the unsolvated "naked" anions. Polar, aprotic solvents such as DMSO (dimethyl sulfoxide), DMF (dimethylformamide) and acetonitrile do not solvate anions nearly as well as methanol, but provide good solvation of the accompanying cations. Consequently, most of the nucleophiles discussed here react more rapidly in solutions prepared from these solvents. These solvent effects are more pronounced for small basic anions than for large weakly basic anions. Thus, for reaction in DMSO solution we observe the following reactivity order: I < SCN < Br < Cl ≈ N < CH CO < CN ≈ CH S < CH O Note that this order is roughly the order of increasing basicity. | 2,502 | 1,872 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/02%3A_Properties_of_Gases/2.06%3A_Kinetic_Theory_of_Gases |
The kinetic theory describes a gas as a large number of submicroscopic particles (atoms or molecules), all of which are in constant, random motion. The rapidly moving particles constantly collide with each other and with the walls of the container. Kinetic theory explains macroscopic properties of gases, such as pressure, temperature, viscosity, thermal conductivity, and volume, by considering their molecular composition and motion. The theory posits that gas pressure is due to the impacts, on the walls of a container, of molecules or atoms moving at different velocities. The five basic tenets of the kinetic-molecular theory are as follows: If gases do in fact consist of widely-separated particles, then the observable properties of gases must be explainable in terms of the simple mechanics that govern the motions of the individual molecules. The kinetic molecular theory makes it easy to see why a gas should exert a pressure on the walls of a container. Any surface in contact with the gas is constantly bombarded by the molecules. Figure 2.6.1: When a molecule collides with a rigid wall, the component of its momentum perpendicular to the wall is reversed. A force is thus exerted on the wall, creating pressure. Image used with permisison from OpenSTAX At each collision, a molecule moving with momentum strikes the surface. Since the collisions are elastic, the molecule bounces back with the same velocity in the opposite direction. This change in velocity Δ is equivalent to a o ; according to Newton's second law, a is thus exerted on the surface of area exerting a pressure . According to the kinetic molecular theory, the average kinetic energy of an ideal gas is directly proportional to the absolute temperature. Kinetic energy is the energy a body has by virtue of its motion: \[ KE = \dfrac{mv^2}{2}\] As the temperature of a gas rises, the average velocity of the molecules will increase; a doubling of the temperature will increase this velocity by a factor of four. Collisions with the walls of the container will transfer more momentum, and thus more kinetic energy, to the walls. If the walls are cooler than the gas, they will get warmer, returning less kinetic energy to the gas, and causing it to cool until thermal equilibrium is reached. Because temperature depends on the kinetic energy, the concept of temperature only applies to a statistically meaningful sample of molecules. We will have more to say about molecular velocities and kinetic energies farther on. : Boyle's law is easily explained by the kinetic molecular theory. The pressure of a gas depends on the number of times per second that the molecules strike the surface of the container. If we compress the gas to a smaller volume, the same number of molecules are now acting against a smaller surface area, so the number striking per unit of area, and thus the pressure, is now greater. Kinetic molecular theory states that an increase in temperature raises the average kinetic energy of the molecules. If the molecules are moving more rapidly but the pressure remains the same, then the molecules must stay farther apart, so that the increase in the rate at which molecules collide with the surface of the container is compensated for by a corresponding increase in the area of this surface as the gas expands. : If we increase the number of gas molecules in a closed container, more of them will collide with the walls per unit time. If the pressure is to remain constant, the volume must increase in proportion, so that the molecules strike the walls less frequently, and over a larger surface area. : "Every gas is a vacuum to every other gas". This is the way Dalton stated what we now know as his law of partial pressures. It simply means that each gas present in a mixture of gases acts independently of the others. This makes sense because of one of the fundamental tenets of KMT theory that gas molecules have negligible volumes. So Gas A in mixture of A and B acts as if Gas B were not there at all. Each contributes its own pressure to the total pressure within the container, in proportion to the fraction of the molecules it represents. One of the triumphs of the kinetic molecular theory was the derivation of the ideal gas law from simple mechanics in the late nineteenth century. This is a beautiful example of how the principles of elementary mechanics can be applied to a simple model to develop a useful description of the behavior of macroscopic matter. We begin by recalling that the pressure of a gas arises from the force exerted when molecules collide with the walls of the container. This force can be found from Newton's law \[f = ma = m\dfrac{dv}{dt} \label{2.1}\] in which \(v\) is the velocity component of the molecule in the direction perpendicular to the wall and \(m\) is its mass. To evaluate the derivative, which is the velocity change per unit time, consider a single molecule of a gas contained in a cubic box of length . For simplicity, assume that the molecule is moving along the -axis which is perpendicular to a pair of walls, so that it is continually bouncing back and forth between the same pair of walls. When the molecule of mass strikes the wall at velocity (and thus with a momentum ) it will rebound elastically and end up moving in the opposite direction with . The total change in velocity per collision is thus 2 and the change in momentum is \(2mv\). After the collision the molecule must travel a distance to the opposite wall, and then back across this same distance before colliding again with the wall in question. This determines the time between successive collisions with a given wall; the number of collisions per second will be \(v/2l\). The \(F\) exerted on the wall is the rate of change of the momentum, given by the product of the momentum change per collision and the collision frequency: \[F = \dfrac{d(mv_x)}{dt} = (2mv_x) \times \left( \dfrac{v_x}{2l} \right) = \dfrac{m v_x^2}{l} \label{2-2}\] Pressure is force per unit area, so the pressure \(P\) exerted by the molecule on the wall of cross-section \(l^2\) becomes \[ P = \dfrac{mv^2}{l^3} = \dfrac{mv^2}{V} \label{2-3}\] in which \(V\) is the volume of the box. As noted near the beginning of this unit, any given molecule will make about the same number of moves in the positive and negative directions, so taking a simple average would yield zero. To avoid this embarrassment, we square the velocities before averaging them \[\bar{v^2} = \dfrac{v_1^2 + v_2^2 + v_3^2 + v_4^2 .. . v_N^2 }{N}= \dfrac{\sum_i v_i^2}{N} \] and take the square root of the average. This result is known as the (rms) velocity. \[v_{rms} = \sqrt{ \bar{v^2}}\] We have calculated the pressure due to a single molecule moving at a constant velocity in a direction perpendicular to a wall. If we now introduce more molecules, we must interpret \(v^2\) as an average value which we will denote by \(\bar{v^2}\). Also, since the molecules are moving randomly in all directions, only one-third of their total velocity will be directed along any one Cartesian axis, so the total pressure exerted by \(N\) molecules becomes \[ P=\dfrac{N}{3}\dfrac{m \bar{\nu}^2}{V} \label{2.4}\] Recalling that \(m\bar{v^2}/2\) is the average translational kinetic energy \(\epsilon\), we can rewrite the above expression as \[PV = \dfrac{1}{3} N m \bar{v^2} = \dfrac{2}{3} N \epsilon \label{2-5}\] The 2/3 factor in the proportionality reflects the fact that velocity components in each of the three directions contributes ½ to the kinetic energy of the particle. The average translational kinetic energy is directly proportional to temperature: \[\color{red} \epsilon = \dfrac{3}{2} kT \label{2.6}\] in which the proportionality constant \(k\) is known as the . Substituting Equation \(\ref{2.6}\) into Equation \(\ref{2-5}\) yields \[ PV = \left( \dfrac{2}{3}N \right) \left( \dfrac{3}{2}kT \right) =NkT \label{2.7}\] The Boltzmann constant is just the gas constant per molecule. For moles of particles, the Equation \(\ref{2.7}\) becomes \[ PV = nRT \label{2.8}\] which is the Ideal Gas law. ) | 8,131 | 1,873 |
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: Ad Ragas : Martien Janssen After this module, you should be able to: Risk, hazard, tiering, problem definition, exposure assessment, effect assessment, risk characterization We assess risks on a daily basis, although we may not always be aware of it. For example, when we cross the street, we - often implicitly - assess the benefits of crossing and weigh these against the risks of getting hit by a vehicle. If the risks are considered too high, we may decide not to cross the street, or to walk a bit further and cross at a safer spot with traffic lights. Risk assessment is common practice for a wide range of activities in society, for example for building bridges, protection against floods, insurance against theft and accidents, and the construction of a new industrial plant. The principle is always the same: we use the available knowledge to assess the probability of potential adverse effects of an activity as good as we can. And if these risks are considered too high, we consider options to reduce or avoid the risk. Risk assessment of chemicals aims to describe the risks resulting from the use of chemicals in our society. In chemical risk assessment, risk is commonly defined as "the probability of an adverse effect after exposure to a chemical". This is a very practical definition that provides natural scientists and engineers the opportunity to quantify risk using "objective" scientific methods, e.g. by quantifying exposure and the likelihood of adverse effects. However, it should be noted that this definition ignores more subjective aspects of risk, typically studied by social scientists, e.g. the perceptions of people and (dealing with) knowledge gaps. This subjective dimension can be important for risk management. For example, risk managers may decide to take action if a risk is perceived as high by a substantial part of the population, even if the associated health risks have been assessed as negligible by natural scientists and engineers. Next to the term "risk", the term "hazard" is often used. The difference between both terms is subtle, but important. A hazard is defined as the inherent capacity of a chemical (or agent/activity) to cause adverse effects. The labelling of a substance as "carcinogenic" is an example of a hazard-based action. The inherent capacity of the substance to trigger cancer, as for example demonstrated in an assay or an experiment with rats or mice, can be sufficient reason to label a substance as "carcinogenic". Hazard is thus independent of the actual exposure level of a chemical, whereas risk is not. Risk assessment is closely related to risk management, i.e. the process of dealing with risks in society. Decisions to accept or reduce risks belong to the risk management domain and involve consideration of the socio-economic implications of the risks as well as the risk management options. Whereas risk assessment is typically performed by natural scientists and engineers, often referred to as "risk assessors", risk management is performed by policy makers, often referred to as "risk managers". Risk assessment and risk management are often depicted as sequential processes, where assessment precedes management. However, strict separation of both processes is not always possible and management decisions may be needed before risks are assessed. For example, risk assessment requires political agreement on what should be protected and at what level, which is a risk management issue (see Section on ). Similarly, the identification, description and assessment of uncertainties in the assessment is an activity that involves risk assessors as well as risk managers. Finally, it is often more efficient to define alternative management options before performing a risk assessment. This enables the assessment of the current situation and alternative management scenarios (i.e., potential solutions) in one round. The scenario with the maximum risk reduction that is also feasible in practice would then be the preferred management option. This mapping of solutions and concurrent assessment of the associated risks is also known as solution-focused risk assessment. Chemical risk assessment is typically organized in a limited number of steps, which may vary depending on the regulatory context. Here, we distinguish four steps (Figure1): The four risk assessment steps are explained in more detail below. The four steps are often repeated multiple times before a final conclusion on the acceptability of the risk is reached. This repetition is called tiering (Figure 2). It typically starts with a simple, conservative assessment and then, in subsequent tiers, more data are added to the assessment resulting in less conservative assumptions and risk estimates. Tiering is used to focus the available time and resources for assessing risks on those chemicals that potentially lead to unacceptable risks. Detailed data are gathered only for chemicals showing potential risk in the lower, more conservative tiers. The order of the exposure and effect assessment steps has been a topic of debate among risk assessors and managers. Some argue that effect assessment should precede exposure assessment because effect information is independent of the exposure scenario and can be used to decide how exposure should be determined, e.g., information on toxicokinetics can be relevant to determine the exposure duration of interest. Others argue that exposure should precede effect assessment since assessing effects is expensive and unnecessary if exposure is negligible. The current consensus is that the preferred order should be determined on a case-by-case basis with parallel assessment of exposure and effects and exchange of information between the two steps as the preferred option. The scope of the assessment is determined during the problem definition phase. Questions typically answered in the problem definition include: Problem definition is not a task for risk assessors only, but should preferably be performed in a collaborative effort between risk managers, risk assessors and stakeholders. The problem definition should try to capture the worries of stakeholders as good as possible. This is not always an easy task as these worries may be very broad and sometimes also poorly articulated. Risk assessors need a clearly demarcated problem and they can only assess those aspects for which assessment methods are available. The dialogue should make transparent which aspects of the stakeholder concerns will be assessed and which not. Being transparent about this can avoid disappointments later in the process, e.g. if aspects considered important by stakeholders were not accounted for because suitable risk assessment methods were lacking. For example, if stakeholders are worried about the acute and chronic impacts of pesticide exposure, but only the acute impacts will be addressed, this should be made clear at the beginning of the assessment. The problem definition phase results in a risk assessment plan detailing how the risks will be assessed given the available resources and within the available timeframe. An important aspect of exposure assessment is the determination of an exposure scenario. An exposure scenario describes the situation for which the exposure is being assessed. In some cases, this exposure situation may be evident, e.g. soil organisms living a contaminated site. However, especially when we want to assess potential risks of future substance applications, we have to come up a typical exposure scenario. Such scenarios are for example defined before a substance is allowed to be used as a food additive or before a new pesticide is allowed on the market. Exposure scenarios are often conservative, meaning that the resulting exposure estimate will be higher than the expected average exposure. The exposure metric used to assess the risk depends on the protection target. For ecosystems, a medium concentration is often used such as the water concentration for aquatic systems, the sediment concentration for benthic systems and the soil concentration for terrestrial systems. These concentrations can either be measured or predicted using a fate model (see ) and may or may not take into account bioavailability (see ). For human risk assessment, the exposure metric depends on the exposure route. An air concentration is often used to cover inhalation, the average daily intake from food and water to cover oral exposure, and uptake through skin for dermal exposure. Uptake through multiple routes can also be combined in a dose metric for internal exposure, such as Area Under the Curve (AUC) in blood (see ). Exposure metrics for specific wildlife species (e.g. top predators) and farm animals are often similar as those for humans. Measuring and modelling route-specific exposures is generally more complex than quantifying a simple medium concentration, because it does not only require the quantification of the substance concentration in the contact medium (e.g., concentration in drinking water), but also quantification of the contact intensity (e.g., how much water is consumed per day). Especially oral exposure can be difficult to quantify because it covers a wide range of different contact media (e.g. food products) and intensities varying from organism to organism. The aim of the effect assessment is to estimate a reference exposure level, typically an exposure level which is expected to cause no or very limited adverse effects. There are many different types of reference levels in chemical risk assessment; each used in a different context. The most common reference level for ecological risk assessment is the Predicted No Effect Concentration (PNEC). This is the water, soil, sediment or air concentration at which no adverse effects at the ecosystem level are being expected. In human risk assessment, a myriad of different reference levels are being used, e.g. the Acceptable Daily Intake (ADI), the oral and inhalatory Reference Dose (RfD), the Derived No Effect Level (DNEL), the Point of Departure (PoD) and the Virtually Safe Dose (VSD). Each of these reference levels is used in a specific context, e.g. for addressing a specific exposure route (ADI is oral), regulatory domain (the DNEL is used in the EU for REACH, whereas the RfD is used in the USA), substance type (the VSD is typical for genotoxic carcinogens) or risk assessment method (the PoD is typical for the Margin of Safety approach). What all reference levels have in common, is that they reflect a certain level of protection for a specific protection goal. In ecological risk assessment, the protection goal typically is the ecosystem, but it can also be a specific species or even an organism. In human risk assessment, the protection goal typically comprises all individuals of the human population. The definition of protection goals is a normative issue and it therefore is not a task of risk assessors, but of politicians. The protection levels defined by politicians typically involve a high level of abstraction, e.g. "the entire ecosystem and all individuals of the human population should be protected". Such abstract protection goals do not always match with the methods used to assess the risks. For example, if one assumes that one molecule of a genotoxic carcinogen can trigger a deathly tumour, 100% protection for all individuals of the human population is feasible only by banning all genotoxic carcinogens (reference level = 0). Likewise, the safe concentration for an ecosystem is infinitely small if one assumes that the sensitivity of the species in the system follows a lognormal distribution which asymptotically approaches the x-axis. Hence, the abstract protection goals have to be operationalized, i.e. defined in more practical terms and matching the methods used for assessing effects. This is often done in a dialogue between scientific experts and risk managers. An example is the " " which is used by many (inter)national organizations as a basis for setting reference levels for genotoxic carcinogens. Likewise, the concentration at which the no observed effect concentration (NOEC) for only 5% of the species is being exceeded is often used as a basis for deriving a PNEC. Once a protection goal has been operationalized, it must be translated into a corresponding exposure level, i.e. the reference level. This is typically done using the outcomes of (eco)toxicity tests, i.e. tests with laboratory animals such as rats, mice and dogs for human reference levels and with primary consumers, invertebrates and vertebrates for ecological reference levels. Often, the toxicity data are plotted in a graph with the exposure level on the x-axis and the effect or response level on the y-axis. A mathematical function is then fitted to the data; the so-called dose-response relationship. This dose-response relationship is subsequently used to derive an exposure level that corresponds to a predefined effect or response level. Finally, this exposure level is extrapolated to the ultimate protection goal, accounting for phenomena such as differences in sensitivity between laboratory and field conditions, between tested species and the species to be protected, and the (often very large) variability in sensitivity in the human population or the ecosystem. This extrapolation is done by dividing the exposure level that corresponds to a predefined effect or response level by one or more assessment or safety factors. These assessment factors do not have a pure scientific basis in the sense that they account for physiological differences which have actually been proven to exist. These factors also account for uncertainties in the assessment and should make sure that the derived reference level is a conservative estimate. The determination of reference levels is an art in itself and is further explained in sections 6.3.1 for and 6.3.2 for . The aim of risk characterization is to come up with a risk estimate, including associated uncertainties. A comparison of the actual exposure level with the reference level provides an indication of the risk: If the reference level reflects the maximum safe exposure level, then the risk indicator should be below unity (1.0). A risk indicator higher than 1.0 indicates a potential risk. It is a "potential risk" because many conservative assumptions may have been made in the exposure and effect assessments. A risk indicator above 1.0 can thus lead to two different management actions: (1) if available resources (time, money) allow and the assessment was conservative, additional data may be gathered and a higher tier assessment may be performed, or (2) consideration of mitigation options to reduce the risk. Assessment of the uncertainties is very important in this phase, as it reveals how conservative the assessment was and how it can be improved by gathering additional data or applying more advanced risk assessment tools. Risks can also be estimated using a margin-of-safety approach. In this approach, the reference level used has not yet been extrapolated from the tested species to the protection goal, e.g. by applying assessment factors for interspecies and interindividual differences in sensitivity. As such, the reference level is not a conservative estimate. In this case, the risk indicator reflects the "margin of safety" between actual exposure and the non-extrapolated reference level. Depending on the situation at hand, the margin-of-safety typically should be 100 or higher. The main difference between the traditional and the margin-of-safety approach in risk assessment is the timing for addressing the uncertainties in the effect assessment. Figure 3 illustrates the risk assessment paradigm using the DPSIR chain ( ). It illustrates how reference exposure levels are being derived from protection goals, i.e. the maximum level of impact that we consider acceptable. The actual exposure level is either measured or predicted using estimated emission levels and dispersion models. When measured exposure levels are used, this is called retrospective or diagnostic risk assessment: the environmental is already polluted and the assessor wants to know whether the risk is acceptable and which substances are contributing to it. When the environment is not yet polluted, predictive tools can be used. This is called prospective risk assessment: the assessor wants to know whether a projected activity will result in unacceptable risks. Even if the environment is already polluted, the risk assessor may still decide to prefer predicted over measured exposure levels, e.g. if measurements are too expensive. This is possible only if the pollution sources are well-characterized. Retrospective (diagnostic) and prospective risk assessments can differ substantially in terms of problem definitions and methods used, and are therefore discussed in separate sections in this online book. Figure 3 can also be used to illustrate some important criticism on the current risk assessment paradigm, i.e. the comparison between the actual exposure level and a reference level. In current assessments, only one point of the dose-response relationship is being used to assess risk, i.e. the reference level. Critics argue that this is suboptimal and a waste of resources because the dose-response information is not used to assess the actual risk. A risk indicator with a value of 2.0 implies that the exposure is twice as high as the reference level but this does not give an indication of how many individuals or species are being affected or of the intensity of the effect. If the dose-response relationship would be used to determine the risk, this would result in a better-informed risk estimate. A final critical remark that should be made, is the fact that risk assessment is often performed on a substance-by-substance basis. Dealing with mixtures of chemicals is difficult because each mixture has a unique composition in terms of compounds and concentration ratios between compounds. This makes it difficult to determine a reference level for mixtures. Mixture toxicology is slowly progressing and several methods are now available to address mixtures, i.e. whole mixture methods and compound-based approaches ( ). Another promising development are effect-based methods ( ). These methods do not assess risk based on chemical concentration, but on the toxicity measured in an environmental sample. In terms of DPSIR, these methods are assessing risks on the level of impacts rather than the level of state or pressures. Imagine the herbicide glyphosate would be banned based on its carcinogenic properties. Would this intervention be risk-based or hazard-based? Indicate whether the following activities should involve risk assessors, risk managers/politicians and/or stakeholders: Indicate whether the following risk assessments are retrospective or prospective: A risk assessment was performed for two different substances, i.e. A and B. The risk indicator value of substance A was 1.5 and that of substance B was 2.0. A risk manager proposes to first address substance B and subsequently substance A. Do you agree? Motivate your answer. | 19,192 | 1,874 |
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The elements in group 18 are the noble gases (helium, neon, argon, krypton, xenon, and radon). They earned the name “noble” because they were assumed to be nonreactive since they have filled valence shells. In 1962, Dr. Neil at the University of British Columbia proved this assumption to be false. These elements are present in the atmosphere in small amounts. Some natural gas contains 1–2% helium by mass. Helium is isolated from natural gas by liquefying the condensable components, leaving only helium as a gas. The United States possesses most of the world’s commercial supply of this element in its helium-bearing gas fields. Argon, neon, krypton, and xenon come from the fractional distillation of liquid air. Radon comes from other radioactive elements. More recently, it was observed that this radioactive gas is present in very small amounts in soils and minerals. Its accumulation in well-insulated, tightly sealed buildings, however, constitutes a health hazard, primarily lung cancer. The boiling points and melting points of the noble gases are extremely low relative to those of other substances of comparable atomic or molecular masses. This is because only weak London dispersion forces are present, and these forces can hold the atoms together only when molecular motion is very slight, as it is at very low temperatures. Helium is the only substance known that does not solidify on cooling at normal pressure. It remains liquid close to absolute zero (0.001 K) at ordinary pressures, but it solidifies under elevated pressure. Helium is used for filling balloons and lighter-than-air craft because it does not burn, making it safer to use than hydrogen. Helium at high pressures is not a narcotic like nitrogen. Thus, mixtures of oxygen and helium are important for divers working under high pressures. Using a helium-oxygen mixture avoids the disoriented mental state known as nitrogen narcosis, the so-called rapture of the deep. Helium is important as an inert atmosphere for the melting and welding of easily oxidizable metals and for many chemical processes that are sensitive to air. Liquid helium (boiling point, 4.2 K) is an important coolant to reach the low temperatures necessary for cryogenic research, and it is essential for achieving the low temperatures necessary to produce superconduction in traditional superconducting materials used in powerful magnets and other devices. This cooling ability is necessary for the magnets used for magnetic resonance imaging, a common medical diagnostic procedure. The other common coolant is liquid nitrogen (boiling point, 77 K), which is significantly cheaper. Neon is a component of neon lamps and signs. Passing an electric spark through a tube containing neon at low pressure generates the familiar red glow of neon. It is possible to change the color of the light by mixing argon or mercury vapor with the neon or by utilizing glass tubes of a special color. Argon was useful in the manufacture of gas-filled electric light bulbs, where its lower heat conductivity and chemical inertness made it preferable to nitrogen for inhibiting the vaporization of the tungsten filament and prolonging the life of the bulb. Fluorescent tubes commonly contain a mixture of argon and mercury vapor. Argon is the third most abundant gas in dry air. Krypton-xenon flash tubes are used to take high-speed photographs. An electric discharge through such a tube gives a very intense light that lasts only \(\dfrac{1}{50,000}\) of a second. Krypton forms a difluoride, KrF , which is thermally unstable at room temperature. Stable compounds of xenon form when xenon reacts with fluorine. Xenon difluoride, XeF , forms after heating an excess of xenon gas with fluorine gas and then cooling. The material forms colorless crystals, which are stable at room temperature in a dry atmosphere. Xenon tetrafluoride, XeF , (Figure \(\Page {1}\)) and xenon hexafluoride, XeF , are prepared in an analogous manner, with a stoichiometric amount of fluorine and an excess of fluorine, respectively. Compounds with oxygen are prepared by replacing fluorine atoms in the xenon fluorides with oxygen. When XeF reacts with water, a solution of XeO results and the xenon remains in the 6+-oxidation state: \[\ce{XeF6}(s)+\ce{3H2O}(l)⟶\ce{XeO3}(aq)+\ce{6HF}(aq) \nonumber \] Dry, solid xenon trioxide, XeO , is extremely explosive—it will spontaneously detonate. Both XeF and XeO disproportionate in basic solution, producing xenon, oxygen, and salts of the perxenate ion, \(\ce{XeO6^4-}\), in which xenon reaches its maximum oxidation sate of 8+. Radon apparently forms RnF —evidence of this compound comes from radiochemical tracer techniques. Unstable compounds of argon form at low temperatures, but stable compounds of helium and neon are not known. The most significant property of the noble gases (group 18) is their inactivity. They occur in low concentrations in the atmosphere. They find uses as inert atmospheres, neon signs, and as coolants. The three heaviest noble gases react with fluorine to form fluorides. The xenon fluorides are the best characterized as the starting materials for a few other noble gas compounds. | 5,194 | 1,875 |
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The reaction quotient (\(Q\)) measures the relative amounts of products and reactants present during a reaction at a particular point in time. The reaction quotient aids in figuring out which direction a reaction is likely to proceed, given either the pressures or the concentrations of the reactants and the products. The \(Q\) value can be compared to the , \(K\), to determine the direction of the reaction that is taking place. The main difference between \(K\) and \(Q\) is that \(K\) describes a reaction that is at equilibrium, whereas \(Q\) describes a reaction that is not at equilibrium. To determine \(Q\), the concentrations of the reactants and products must be known. For a given general chemical equation: \[aA + bB \rightleftharpoons cC + dD \tag{1}\nonumber \] the Q equation is written by multiplying the (which are approximated by concentrations) for the species of the products and dividing by the activities of the reactants. If any component in the reaction has a coefficient, indicated above with lower case letters, the concentration is raised to the power of the coefficient. \(Q\) for the above equation is therefore: \[Q_c = \dfrac{[C]^c[D]^d}{[A]^a[B]^b} \tag{2}\nonumber \] This equation only shows components in the gaseous or aqueous states. Each pure liquid or solid has an activity of one and can be functionally omitted. Equilibrium constants really contain a ratio of concentrations (actual concentration divided by the reference concentration that defines the standard state). Because the standard state for concentrations is usually chosen to be 1 mol/L, it is not written out in practical applications. Hence, the ratio does not contain units. A comparison of \(Q\) with \(K\) indicates which way the reaction shifts and which side of the reaction is favored: Another important concept that is used in the calculation of the reaction quotient is called an . For example, consider the \(Q\) equation for this acid/base reaction: \[\ce{CH_3CH_2CO_2H(aq) + H_2O(l) <=> H_3O^{+}(aq) + CH_3CH_2CO_2^{-}(aq)} \nonumber \] The \(Q\) equation is written as the concentrations of the products divided by the concentrations of the reactants, but only including components in the gaseous or aqueous states and omitting pure liquid or solid states. The \(Q\) equation for this example is the following: \[Q = \dfrac{[\ce{H3O^{+}(aq)},\ce{CH3CH2CO2^{-}(aq)}]}{[\ce{CH3CH2CO2H(aq)}]} \nonumber \] What is the \(Q\) value for this equation? Which direction will the reaction shift if \(K_c\) = 1.0? \[\ce{CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)} \nonumber\] Step 1: Write the \(Q\) formula: \[Q_c = \dfrac{[CO_2,H_2]}{[CO,H_2O]} \nonumber\] Step 2: Plug in given concentration values: \(\begin{align*} Q_c &= \dfrac{(2.0)(2.0)}{(1.0)(1.0)} \\[4pt] &= 4.0 \end{align*}\] Step 3: Compare \(Q\) to K: Because \(4.0 > 1.0\), then \(Q > K\) and the reaction shifts left toward the reactants. Q= 4.0 and the reaction shifts left. Find the value of \(Q\) and determine which side of the reaction is favored with \(K=0.5\). \[\ce{HCl(g) + NaOH(aq) \rightleftharpoons NaCl(aq) + H_2O(l)} \nonumber\] with Step 1: Write the \(Q\) formula. Because the activity of a liquid is 1, we can omit the water component in the equation. \(Q_c = \dfrac{[NaCl{(aq)}]}{[HCl{(g)},NaOH{(aq)}]}\) Step 2: Plug in given concentrations into the \(Q\) formula: \(Q_c = \dfrac{[6]}{[3.2,4.3]}\) Step 3: Calculate using the given concentrations: \(Q = 0.436\) Step 4: Compare Q to K. The \(Q\) value, 0.436, is less than the given \(K\) value of 0.5, so \(Q < K\). Because \(Q\) < K, the reaction is not at equilibrium and proceeds to the products side to reach dynamic equilibrium once again. Answer: Q= 0.436 and the reaction favors the products. Given the equation with \(K= 0.040\). Find \(Q\) and determine which direction the reaction will shift to reach equilibrium. \[\ce{N_2(g) + 3H_2(aq) \rightleftharpoons 2NH_3(g)} \nonumber \] with Step 1: Write the \(Q\) formula: \[Q_c = \dfrac{[NH_3{(g)}]^2}{[N_2{(g)},H_2{(g)}]^3}\nonumber \] Step 2: Plug in values. Because the concentrations for \(N_2\) and \(H_2\) were given, they can be inserted directly into the equation. However, no concentration value was given for NH so the concentration is assumed to be 0. \[Q_c = \dfrac{(0)^2}{(0.04)(0.09)^3}\nonumber \] Step 3: Solve for Q: \[Q=0\nonumber \] Step 4: Compare \(Q\) to K. Because \(K=0.04\) and \(Q=0\), \(K > Q\) and the reaction will shift right to regain equilibrium. Answer: \(Q=0\), the reaction shifts right. | 4,557 | 1,876 |
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The conditions used for substitution reactions by the \(\text{S}_\text{N}2\) mechanism very often lead to elimination. The reaction of 2-bromopropane with sodium ethoxide in ethanol provides a good example: Elimination to give propene competes with substitution to give ethyl isopropyl ether. Furthermore, the rate of elimination, like the rate of substitution, is proportional to the concentrations of 2-bromopropane and ethoxide ion. Thus elimination here is a second-order reaction (it may be helpful to review Section 8-4 at this point): rate of substitution \(= k_S \left[ R Br \right] \left[ \overset{\ominus}{O} C_2H_5 \right]\) rate of elimination \(= k_E \left[ R Br \right] \left[ \overset{\ominus}{O} C_2H_5 \right]\) As to the of this kind of elimination, the attacking base, \(^\ominus OC_2H_5\), removes a proton from the \(\beta\) carbon more or less simultaneously with the formation of the double bond and the loss of bromide ion from the neighboring carbon: The abbreviation for this mechanism is , \(\text{E}\) for elimination and \(2\) for bimolecular, there being two reactants involved in the transition state. Structural influences on \(\text{E}2\) reactions have been studied in some detail. Like the competing \(\text{S}_\text{N}2\) process, a good leaving group is necessary and of these, the most commonly used are the halides, \(Cl\), \(Br\), and \(I\); sulfonate esters, \(RS \left( O_2 \right) O-\); and -onium ions such as ammonium, \(\overset{\oplus}{N} R_4\), and sulfonium, \(\overset{\oplus}{S} R_3\): Rather strong bases generally are required to bring about the \(\text{E}2\) reaction. The effectiveness of a series of bases generally parallels their base strengths, and the order \(\overset{\ominus}{N} H_2\) \(>\) \(\overset{\ominus}{O} C_2H_5\) \(>\) \(\overset{\ominus}{O} H\) \(>\) \(\overset{\ominus}{O} CCH_3\) is observed for \(\text{E}2\) reactions. This fact is important in planning practical syntheses, because the \(\text{E}2\) reaction tends to predominate with strongly basic, slightly polarizable reagents such as amide ion, \(\overset{\ominus}{N} H_2\), or ethoxide ion, \(\overset{\ominus}{O} C_2H_5\). In contrast, \(\text{S}_\text{N}2\) reactions tend to be favored with weakly basic nucleophiles such as iodide ion or ethanoate ion (unless dipolar aprotic solvents are used, which may markedly change the reactivity of anionic nucleophiles). As for the alkyl group, there are two important structural effects to notice. First, at least one \(C-H\) bond adjacent (\(\beta\)) to the leaving group is required. Second, the ease of \(\text{E}2\) elimination follows the order \(R\) \(>\) \(R\) \(>\) \(R\). Unlike \(\text{S}_\text{N}2\) reactions, which are observed for tertiary alkyl compounds because of steric hindrance to the approach of the nucleophile to carbon, the related \(\text{E}2\) reaction usually occurs readily with tertiary \(RX\) compounds. The reason is that little or no steric hindrance is likely for the approach of a base to a hydrogen unless the base is exceptionally bulky: The reactivity order also appears to correlate with the \(C-X\) bond energy, inasmuch as the tertiary alkyl halides both are more reactive and have weaker carbon-halogen bonds than either primary or secondary halides (see Table 4-6). In fact, elimination of \(HX\) from haloalkenes or haloarenes with relatively strong \(C-X\) bonds, such as chloroethene or chlorobenzene, is much less facile than for haloalkanes. Nonetheless, elimination does occur under the right conditions and constitutes one of the most useful general methods for the synthesis of alkynes. For example, The conditions and reagents used for \(\text{E}2\) and \(\text{S}_\text{N}2\) reactions are similar enough that it is difficult to have one occur without the other. However, \(\text{E}2\) elimination is favored over \(\text{S}_\text{N}2\) substitution by (a) strongly basic nucleophiles, (b) bulky nucleophiles, and (c) increasing alkyl substitution at the \(\alpha\) carbon. It also is observed that increasing the reaction temperature generally leads to an increase in elimination at the expense of substitution. In fact, surprisingly good yields of alkene or alkyne can be obtained by adding a halogen compound directly to molten or very hot \(KOH\) with no solvent present, whereupon the product is formed rapidly and distills immediately from the hot reaction mixture: With halides having unsymmetrical \(R\) groups, such as 2-chloro-2-methylbutane, it is possible to form two or more different alkenes, the proportion depending on the relative rates at which the different \(\beta\) hydrogens are removed. Most \(\text{E}2\) eliminations of alkyl halides with common bases, such as \(HO^\ominus\), \(C_2H_5O^\ominus\), and \(NH_2^\ominus\), tend to give mixtures of alkenes with a preference for the most stable alkene, which usually is the one with the hydrogens or alkyl groups attached to the carbons of the double bond. Thus However, the precise distribution of alkenes formed is found to vary enough with the nature of the leaving group, or the base used, so either product will predominated with some combination of reagents or conditions. For example, a change in the base alone can be decisive: The \(\text{E}2\) reaction occurs most easily if the molecule undergoing reaction can assume a conformation, \(2\), in which the leaving groups, \(H\) and \(X\), are trans to each other and the atoms \(H-C_\beta-C_\alpha-X\) lie in one plane. Elimination then proceeds from opposite sides of the incipient double bond to give an alkene of structure \(3\). We shall call this mode of elimination to distinguish it from another possible mode of elimination that is called . (See Figure 8-6).\(^8\) The transition state for conversion of \(2\) to \(3\) is particularly reasonable because it combines some of the geometry of both the reactants and the products and therefore gives the best overlap of the reacting orbitals necessary for the formation of the \(\pi\) bond. This is shown more explicitly below.\(^9\) As an illustration of the stereospecificity of eliminations, the meso compound \(4\) gives the -alkene \(5\), whereas the \(D,L\) isomers \(6\) give the -alkene \(7\) with ethoxide. Both reactions clearly proceed by antarafacial elimination: When antarafacial elimination is rendered difficult by the inability of the reacting groups to acquire the desired trans arrangement, then suprafacial elimination can occur, although less readily. An example is chlorocyclopentane, in which \(H\) and \(X\) cannot assume a trans configuration without very considerable strain but which does undergo suprafacial elimination at a reasonable rate: \(^9\)Persuasive arguments have been made that many \(\text{E}2\) reactions proceed by the sequence If this is so, antarafacial elimination still is predicted to be favored. and (1977) | 6,910 | 1,877 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Statistical_Mechanics/Boltzmann_Average/Fluctuations |
The methods developed allow us to calculate thermodynamic averages. The deviation of a mechanical variable from its mean value is called a fluctuation. The theory of fluctuations is useful for understanding how the different ensembles (NVT, NPT etc.). Fluctuations are also important in theories of light scattering and in the study of transport processes. We will show that the fluctuations about the mean energy are very small for macroscopic systems. Given the fact that the number of particles is much smaller the relative magnitude of fluctuations in a MD simulation are much larger. In part, we study fluctuations to have a criterion for determining the quality of an average property that we calculate in a computer simulation. The average of a property is also called the first moment of a distribution. The second moment of the distribution is called the variance \[ \langle (x- \langle x \rangle)^2 \rangle = \langle x^2 \rangle - \langle x \rangle^2 \label{var}\] The variance is a measure of spread of the probability distribution about the mean value. The mean value is \[ \langle x \rangle = \int_{-\infty}^{\infty} x P(x)dx\] or if descritized data are involved \[ \langle x \rangle = \sum_i x_i P_i\] where \(P(x)\) is a normalized probability distribution or \(P_i\) is a discrete probability distribution. The mean square value is \[ \langle x^2 \rangle = \int_{-\infty}^{\infty} x^2 P(x)dx\] or if descritized data are involved \[ \langle x^2 \rangle = \sum_i x_i^2 P_i\] We consider the fluctuations in the NVT or canonical ensemble. The number, volume, and temperature are fixed, and we can calculate the fluctuations in the energy in this ensemble. The variance in the energy (Equation \ref{var}) is \[ \begin{align} \sigma_E^2 & = \langle (E- \langle E \rangle)^2 \rangle = \langle E^2 \rangle - \langle E \rangle^2 \\ & = \sum_i E^2_iP_i - \left( \sum_i E^2_i P_i \right)^2 \end{align}\] where We need to evaluate the mean-square energy term Thus, The spread in energies about the mean is given by the ratio of the square root of the variance relative to the energy For an ideal gas C = 3/2Nk and E = 3/2NkT The fluctuations are proportional to 1/ N which is an extremely small number for macroscopic systems where N is of the order of Avagadro's number. | 2,293 | 1,878 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Le_Chateliers_Principle/The_Haber_Process |
The Haber Process is used in the manufacturing of ammonia from nitrogen and hydrogen, and then goes on to explain the reasons for the conditions used in the process. The process combines nitrogen from the air with hydrogen derived mainly from natural gas (methane) into ammonia. The reaction is reversible and the production of ammonia is exothermic. \[\ce{ N2(g) + 3H2(g) <=> 2NH3 (g)} \label{eq1}\] with \(ΔH=-92.4 kJ/mol\). A flow scheme for the Haber Process looks like this: The proportions of nitrogen and hydrogen: The mixture of nitrogen and hydrogen going into the reactor is in the ratio of 1 volume of nitrogen to 3 volumes of hydrogen. Avogadro's Law says that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. That means that the gases are going into the reactor in the ratio of 1 molecule of nitrogen to 3 of hydrogen. That is the proportion demanded by the equation. In some reactions you might choose to use an excess of one of the reactants. You would do this if it is particularly important to use up as much as possible of the other reactant - if, for example, it was much more expensive. That does not apply in this case. There is always a down-side to using anything other than the equation proportions. If you have an excess of one reactant there will be molecules passing through the reactor which cannot possibly react because there is not anything for them to react with. This wastes reactor space - particularly space on the surface of the catalyst. Notice that there are 4 molecules on the left-hand side of Equation \(\ref{eq1}\), but only 2 on the right. According to Le Chatelier's Principle, if you increase the pressure the system will respond by favoring the reaction which produces fewer molecules. That will cause the pressure to fall again. In order to get as much ammonia as possible in the equilibrium mixture, you need as high a pressure as possible. 200 atmospheres is a high pressure, but not amazingly high. By mixing one part ammonia to nine parts air with the use of a catalyst, the ammonia will get oxidized to . \[\begin{align*} \ce{4 NH_3} + \ce{5 O_2} &\rightarrow \ce{4 NO} + \ce{6 H_2O} \\[4pt] \ce{2 NO} + \ce{O_2} &\rightarrow \ce{2 NO_2} \\[4pt] \ce{2 NO_2} + \ce{2 H_2O} &\rightarrow \ce{2 HNO_3} + \ce{H_2} \end{align*}\] | 2,335 | 1,879 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Electrophilic_Addition_Reactions/The_Generalized_Electrophilic_Addition |
Electrophilic addition is a reaction between an electrophile and nucleophile, adding to double or triple bonds. An electrophile is defined by a molecule with a tendency to react with other molecules containing a donatable pair of electrons. Thus, it is an "electron lover." A nucleophile is one that possesses a lone pair of electrons that can be easily shared. In essence, all nucleophiles are Lewis bases that attack nonhydrogen atoms ( ). In a general electrophilic addition reaction, one of the pi bonds is removed and creates two new sigma bonds. Another electrophilic addition reaction known as halohydrination includes bromoalcohol, more commonly known as bromohydrin. There are a variety of electrophilic reactions, and therefore a variety of different products that are very useful. It simply depends on which reagents are used to determine the final product. In the first step there is an electrophilic addition of bromine to the cyclopentene, forming a cyclic bromonium ion or also known as an open-chained carbenium ion (Troll, T). For more information about the cyclic Bromonium Ion please look at Electrophilic Addition of Halogens to Alkenes. Now the nucleophilic water molecule attacks the back of the more substituted carbon and pushes the bromonium ion onto the less substituted carbon. In general the regiochemistry of this reaction follows (Troll, T). The stereochemistry of the reaction is anti-addition because of better orbital overlap from backside attack, which means that the Br and the H O are on opposite sides of the double bond. The electrophilic bromide in the product becomes linked to the less substituted carbon. The nucleophile attacks the more substituted carbon, because the carbon is more positively polarized than the other carbon. The negatively charged bromonium ion that was not used in the reaction is still floating freely in the water. The bromine atom attacks one of the H’s located on the water molecule. The Hydrogen drops off its electrons on the oxygen molecule making the oxygen neutral. Finally we are left with the trans-2-Bromocyclopentanol and hydrobromination. Here's the entire mechanism in gumdrop form. (Orange = C, Yellow = H, Red = Br, White = O) Just for some extra practice try and answer the following questions 1. What is the product for the following reaction 2. What is the product for the following reaction 3. Write the products (hint what acts as an electrophile and what acts as the nucleophile) 4. What is the stereochemistry and the Regiochemistry of Bromoalcohol? 5. In Markovnikov Addition the electrophile attacks the more substituted carbon and the nucleophile attacks the less substituted carbon? A. True B. False 1. 2. 3. 4. Stereochemistry: Anti-Addition & Regiochemistry: Markovnikov 5.B. False | 2,795 | 1,881 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/09._The_Hydrogen_Atom/Radial_Nodes |
A radial node is a sphere (rather than an angular node which is a flat plane) that occurs when the radial wavefunction for an atomic orbital is equal to zero or changes sign. There are two types of nodes within an atom: angular and radial. Angular nodes are or will be discussed in another section; this section is dedicated to the latter. Radial nodes, as one could guess, are determined radially. Using the radial probability density function, places without electrons, or radial nodes, can be found. A quick comparison of the two types of nodes can be seen in the diagram above. Angular nodes are either x, y, and z planes where electrons aren’t present while radial nodes are sections of these axes that are closed off to electrons. For atomic orbitals, the wavefunction can be separated into a radial part and an angular part so that it has the form \[Ψ(r,θ,ϕ)=R(r)Y(θ,ϕ)\] where \(R(r)\) is the radial component which depends only on the distance from the nucleus and Y(θ,ϕ) is the angular component. The radial nodes consist of spheres whereas the angular nodes consist of planes (or cones). A radial node will occur where the radial wavefunction, \(R(r)\), equals zero. At a node the probability of finding an electron is zero; which means that we will find an electron at a node. To solve for the number of radial nodes, the following simple equation can be used. The ‘n’ accounts for the total amount of nodes present. The ‘-1’ portion accounts for the node that exists at the ends. (A half of one node exists at one end and since there are two ends, there’s a total of one node located at the ends.) The azimuthal quantum number determines the shape of the orbital and how many angular nodes there are. The remaining number, which currently doesn’t have a symbol, is the amount of radial nodes which are present. Here’s a quick example: Radial nodes occur as the principle quantum number (n) increases and the number of radial nodes depends on the principle quantum number (n) and the number of angular nodes (l). The total number of nodes is found using From knowing the total nodes we can find the number of radial nodes by using which is just the total nodes minus the angular nodes. first shell (n=1) number of nodes= n-1=0 so there aren't any nodes second shell (n=2) number of nodes=n-1=1 total nodes for 2s orbital l=0 so there are 0 angular nodes and 1 radial node for 2p orbital l=1 so there is 1 angular node and 0 radial nodes third shell (n=3) number of nodes=n-1=2 total nodes for 3s orbital l=0 so there are 0 angular nodes and 2 radial nodes for 3p orbital l=1 so there is 1 angular node and 1 radial node for 3d orbital l=2 so there are 2 angular nodes and 0 radial nodes Find the radial nodes in a 3p orbital. For the 3p orbital, the ‘3’ means that ‘n’ = 3 and ‘p’ shows that ‘ℓ’ = 1. ‘ℓ’ also equals the number of angular nodes which means there is one angular node present. Using the equation for radial nodes, n - ℓ - 1 = 3 - 1 - 1 = 1. Thus there is one radial nodes. The following section will show how to determine radial nodes in a more complex way. We can calculate how many nodes there will be based off the equation above, however we can also see this from the wavefunction. For example the wavefunction for the Hydrogen atom 3d orbital: From the equation above we can see that the number of total nodes is n-1=2 and the number of angular nodes (l)=2 so the number of radial nodes is 0. From the wavefunction for the \(3dz^2\) orbital, we can see that (excluding r=0 and as r goes to infinity) the radial wavefunction will never equal to zero so there are 0 radial nodes for this orbital. For the angular wavefunction, we see there will be an angular node when \(3cos^2θ-1=0\); which corresponds to the 2 solutions θ=54.7º and 125.3º. As stated above, we know that at a node the probability of finding an electron is zero. The diagram below shows that as n increases, the number of radial nodes increases. From Figure 2 we can see that for the 1s orbital there are not any nodes (the curve for the 1s orbital doesn't equal zero probability other than at r=0 and as r goes to infinity). This is expected since n-l-1 for the 1s orbital is 1-0-1=0 radial nodes. For the 2s orbital, the curve has zero probability at 1 point (again other than r=0 and as r goes to infinity); which is consistent with the n-l-1 for the 2s orbital 2-0-1=1 radial node. For the 3s orbital, the curve has zero probability at 2 points; which is consistent with the n-l-1 for the 3s orbital 3-0-1=2 radial nodes. But what about for molecular orbitals? To separate the wavefunction into a radial part and an angular part, the system needs to be spherically symmetric. For an atom this is the case but a molecule can never be. Thus radial nodes do not exist for molecular orbitals. | 4,809 | 1,883 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Dissociation_Constant |
The dissociation constant specifies the tendency of a substance M N to reversibly dissociate (separate) in a solution (often aqueous) into smaller components M and N: \[ M_xN_{y(aq)} + H_2O_{(l)} \rightleftharpoons xM_{(aq)} + yN_{(aq)} \tag{1}\] The dissociation constant is denoted K and is calculated by \[ K_d =\dfrac{a_M^x·a_N^y}{a_{M_xN_y}·a_{H_2O}} \approx \dfrac{[M]^x[N]^y}{[M_xN_y](1)} \tag{2}\] where A represents the activity of a species, and [M], [N], and [M N ] are the molar concentrations of the entities M, N, and M N . Because water is the solvent, and the solution is assumed to be dilute, the water is assumed to be pure, and the activity of pure water is defined as 1. The activities of the solutes are approximated with molarities. The dissociation constant is an immediate consequence of the which describes equilibria in a more general way. The dissociation constant is also sometimes called ionization constant when applied to salts. The inverse of the dissociation constant is called association constant. Formally, the dissociation (autoprotolysis) of water follows the following equation: \[H_2O_{(l)} \rightleftharpoons H_{(aq)}^+ + OH_{(aq)}^- \tag{3}\] or \[H_2O_{(l)} + H_2O_{(l)} \rightleftharpoons H_3O_{(aq)}^+ + OH_{(aq)}^- \tag{4}\] These two reaction equations are thermodynamically equivalent, thus the law of mass action for the two equations must be equivalent. Their equivalency is shown by \[K_d = \dfrac{a_{H^+}·a_{OH^-}}{a_{H_2O}} = \dfrac{a_{H_3O^+}·a_{OH^-}}{a_{H_2O}^2}\approx\dfrac{[H^+,OH^-]}{1} =\dfrac{[H_3O^+,OH^-]}{[1]^2} = [H^+,OH^-]=[H_3O^+,OH^-] =1.00\times 10^{-14}=K_w\tag{5}\] In these reactions (and equations), the activity of water, as the solvent in a dilute solution, is approximated as the activity of pure water, which has a defined value of 1. The activity of the ions, as solutes, can be approximated as the molarity of the ions. Thus, equation 3 and equation 4 both have the same law of mass action and the same \(K_d\), which is commonly written as \(K_w\). The value of K changes considerably with temperature. Consequently this variation must be taken into account when making precise measurements (i.e. when determining the pH). The dissociation constant can also be applied to the reaction of acids with water, sometimes called the deprotonation of acids. These reactions can be written in two, equivalent forms, depending on whether water is shown as the solvent, or not: \[HA_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + A^-_{(aq)} \tag{6}\] or \[HA_{(aq)} \rightleftharpoons H^+_{(aq)} + A^-_{(aq)} \tag{7}\] In either case, the dissociation constant is denoted as K , and because the activity of water has a value of 1, there is no net difference in whether water is formally included in the reaction or the law of mass action equation. The greater the dissociation constant of an acid the stronger the acid. Polyprotic acids (e.g. carbonic acid or phosphoric acid) show several dissociation constants, because more than one proton can be separated (one after the other): A list of acid dissociation constants can be found here. The concept of the dissociation constant is applied in various fields of chemistry and pharmacology. In protein-ligand binding the dissociation constant describes the affinity between a protein and a ligand. A small dissociation constant indicates a more tightly bound ligand. In the case of antibody-antigen binding the inverted dissociation constant is used and is called affinity constant, which, confusingly, is also written as \(K_a\). {{template. Tom Neils (Grand Rapids Community College) | 3,638 | 1,884 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/03%3A_First_Law_of_Thermodynamics/3.07%3A_Lattice_Energy_and_the_Born-Haber_Cycle |
An important enthalpy change is the Lattice Energy, which is the energy required to take one mole of a crystalline solid to ions in the gas phase. For \(\ce{NaCl(s)}\), the lattice energy is defined as the enthalpy of the reaction \[ \ce{ NaCl(s) \rightarrow Na^{+}(g) + Cl^{-}(g) } \nonumber \] h \(\Delta H\) called the lattice energy (\(\Delta H_{Lat}\)). A very handy construct in thermodynamics is that of the thermodynamic cycle. This can be represented graphically to help to visualize how all of the pieces of the cycle add together. A very good example of this is the , describing the formation of an ionic solid. Two pathways can be envisioned for the formation. Added together, the two pathways form a cycle. In one pathway, the ionic solid if formed directly from elements in their standard states. \[ \ce{Na(s) + 1/2 Cl_2 \rightarrow NaCl(s)} \nonumber \] with \(\Delta H_f(NaCl)\). The other pathway involves a series of steps that take the elements from neutral species in their standard states to ions in the gas phase. \[Na(s) \rightarrow Na(g) \nonumber \] with \(\Delta H_{sub}(Na)\) \[Na(g) \rightarrow Na^+(g) + e^- \nonumber \] with \(1^{st}\, IP(Na)\) \[½ Cl_2(g) \rightarrow Cl(g) \nonumber \] with \(½ D(Cl-Cl)\) \[Cl(g) + e^- \rightarrow Cl^-(g) \nonumber \] with \(1^{st} EA(Cl)\) \[Na^+(g) + Cl^-(g) \rightarrow NaCl(s) \nonumber \] with \(\Delta H_{Lat}(NaCl)\) It should be clear that when added (after proper manipulation if needed), the second set of reactions yield the first reaction. Because of this, the total enthalpy changes must all add. \[\Delta H_{sub}(Na) + 1^{st} IP(Na) + ½ D(Cl-Cl) + 1^{st} EA(Cl) + \Delta H_{lat}(NaCl) = \Delta H_f(NaCl) \nonumber \] This can be depicted graphically, the advantage being that arrows can be used to indicate endothermic or exothermic changes. An example of the Born-Haber Cycle for NaCl is shown below. In many applications, all but one leg of the cycle is known, and the job is to determine the magnitude of the missing leg. Find \(\Delta H_f\) for KBr given the following data. \[\ce{K(s) \rightarrow K(g)} \nonumber \] with \(\Delta H_{sub} = 89\, kJ/mol\) \[\ce{Br_2(l) \rightarrow Br_2(g) } \nonumber \] with \(\Delta H_{vap} = 31\, kJ/mol\) \[\ce{Br_2(g) \rightarrow 2 Br(g)} \nonumber \] with \(D(Br-Br) = 193\, kJ/mol\) \[\ce{K(g) \rightarrow K^+(g) + e^- } \nonumber \] with \(1^{st} IP(K) = 419 kJ/mol\) \[\ce{Br(g) + e^- \rightarrow Br^-(g) } \nonumber \] with \(1^{st} EA(Br) = 194 kJ/mol\) \[\ce{K^+(g) + Br^-(g) \rightarrow KBr(s)} \nonumber \] with \(\Delta H_{Lat} = 672 kJ/mol\) \(\Delta H_f = -246 \,kJ/mol\) : This cycle required the extra leg of the vaporization of Br . Many cycles involve ions with greater than unit charge and may require extra ionization steps as well! | 2,796 | 1,885 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Surface_Science_(Nix)/04%3A_UHV_and_Effects_of_Gas_Pressure/4.E%3A_UHV_and_Effects_of_Gas_Pressure_(Exercises) |
This section provides a limited number of examples of the application of the formulae given in the previous section to determine the: If you have not already been through then I would suggest that you stop now and return to this page only after you have done so Calculate the molecular gas density for an ideal gas at 300 K, under the following conditions (giving your answer in molecules m ): At a pressure of 10 Torr At a pressure of 10 Torr Calculate the mean free path of CO molecules in a vessel at the indicated pressure and temperature, using a value for the collision cross section of CO of 0.42 nm . P = 10 Torr, at 300 K P = 10 Torr, at 300 K Calculate the flux of molecules incident upon a solid surface under the following conditions:
[Note - 1 u = 1.66 x 10 kg: atomic masses ; (O) =16.0 u, (H) = 1.0 u] Oxygen gas ( P = 1 Torr ) at 300 K Oxygen gas ( P = 10 Torr ) at 300 K Hydrogen gas ( P = 10 Torr ) at 300 K Hydrogen gas ( P = 10 Torr ) at 1000 K The rate of adsorption of molecules onto a surface can be determined from the flux of molecules incident on the surface and the sticking probability pertaining at that instant in time (note that in general the sticking probability itself will be dependent upon a number of factors including the existing coverage of adsorbed species). In the following examples we will assume that the surface is initially clean (i.e. the initial coverage is zero), and that there is no desorption of the molecules once they have adsorbed. You should determine coverages as the ratio of the adsorbate concentration to the density of surface substrate atoms ( ). In the first two questions we will assume that the sticking probability is constant over the coverage range concerned. Calculate the surface coverage obtained after exposure to a pressure of 10 Torr of CO for 20 s at 300 K - you may take the sticking probability of CO on this surface to have a constant value of 0.9 up to the coverage concerned.
Calculate the surface coverage of atomic nitrogen obtained by dissociative adsorption after exposure to a pressure of 10 Torr of nitrogen gas for 20 s at 300 K - you may take the dissociative sticking probability of molecular nitrogen on this surface to be constant and equal to 0.1 In general, the sticking probability varies with coverage - most obviously, the sticking probability must tend to zero as the coverage approaches its saturation value. These calculations are not quite so easy Calculate the surface coverage obtained after exposure to a pressure of 10 Torr of CO for 200 s at 300 K - the sticking probability of CO in this case should be taken to vary linearly with coverage between a value of unity at zero coverage and a value of zero at saturation coverage (which you should take to be 6.5 x 10 molecules m ). | 2,826 | 1,886 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/11%3A_Reactions_and_Other_Chemical_Processes/11.05%3A_Reaction_Calorimetry |
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\( \newcommand{\mA}{_{\text{m},\text{A}}} % subscript m,A (m=molar)\)
\( \newcommand{\mB}{_{\text{m},\text{B}}} % subscript m,B (m=molar)\)
\( \newcommand{\mi}{_{\text{m},i}} % subscript m,i (m=molar)\)
\( \newcommand{\fA}{_{\text{f},\text{A}}} % subscript f,A (for fr. pt.)\)
\( \newcommand{\fB}{_{\text{f},\text{B}}} % subscript f,B (for fr. pt.)\)
\( \newcommand{\xbB}{_{x,\text{B}}} % x basis, B\)
\( \newcommand{\xbC}{_{x,\text{C}}} % x basis, C\)
\( \newcommand{\cbB}{_{c,\text{B}}} % c basis, B\)
\( \newcommand{\mbB}{_{m,\text{B}}} % m basis, B\)
\( \newcommand{\kHi}{k_{\text{H},i}} % Henry's law constant, x basis, i\)
\( \newcommand{\kHB}{k_{\text{H,B}}} % Henry's law constant, x basis, B\)
\( \newcommand{\arrow}{\,\rightarrow\,} % right arrow with extra spaces\)
\( \newcommand{\arrows}{\,\rightleftharpoons\,} % double arrows with extra spaces\)
\( \newcommand{\ra}{\rightarrow} % right arrow (can be used in text mode)\)
\( \newcommand{\eq}{\subs{eq}} % equilibrium state\)
\( \newcommand{\onehalf}{\textstyle\frac{1}{2}\D} % small 1/2 for display equation\)
\( \newcommand{\sys}{\subs{sys}} % system property\)
\( \newcommand{\sur}{\sups{sur}} % surroundings\)
\( \renewcommand{\in}{\sups{int}} % internal\)
\( \newcommand{\lab}{\subs{lab}} % lab frame\)
\( \newcommand{\cm}{\subs{cm}} % center of mass\)
\( \newcommand{\rev}{\subs{rev}} % reversible\)
\( \newcommand{\irr}{\subs{irr}} % irreversible\)
\( \newcommand{\fric}{\subs{fric}} % friction\)
\( \newcommand{\diss}{\subs{diss}} % dissipation\)
\( \newcommand{\el}{\subs{el}} % electrical\)
\( \newcommand{\cell}{\subs{cell}} % cell\)
\( \newcommand{\As}{A\subs{s}} % surface area\)
\( \newcommand{\E}{^\mathsf{E}} % excess quantity (superscript)\)
\( \newcommand{\allni}{\{n_i \}} % set of all n_i\)
\( \newcommand{\sol}{\hspace{-.1em}\tx{(sol)}}\)
\( \newcommand{\solmB}{\tx{(sol,$\,$$m\B$)}}\)
\( \newcommand{\dil}{\tx{(dil)}}\)
\( \newcommand{\sln}{\tx{(sln)}}\)
\( \newcommand{\mix}{\tx{(mix)}}\)
\( \newcommand{\rxn}{\tx{(rxn)}}\)
\( \newcommand{\expt}{\tx{(expt)}}\)
\( \newcommand{\solid}{\tx{(s)}}\)
\( \newcommand{\liquid}{\tx{(l)}}\)
\( \newcommand{\gas}{\tx{(g)}}\)
\( \newcommand{\pha}{\alpha} % phase alpha\)
\( \newcommand{\phb}{\beta} % phase beta\)
\( \newcommand{\phg}{\gamma} % phase gamma\)
\( \newcommand{\aph}{^{\alpha}} % alpha phase superscript\)
\( \newcommand{\bph}{^{\beta}} % beta phase superscript\)
\( \newcommand{\gph}{^{\gamma}} % gamma phase superscript\)
\( \newcommand{\aphp}{^{\alpha'}} % alpha prime phase superscript\)
\( \newcommand{\bphp}{^{\beta'}} % beta prime phase superscript\)
\( \newcommand{\gphp}{^{\gamma'}} % gamma prime phase superscript\)
\( \newcommand{\apht}{\small\aph} % alpha phase tiny superscript\)
\( \newcommand{\bpht}{\small\bph} % beta phase tiny superscript\)
\( \newcommand{\gpht}{\small\gph} % gamma phase tiny superscript\) \( \newcommand{\upOmega}{\Omega}\) \( \newcommand{\dif}{\mathop{}\!\mathrm{d}} % roman d in math mode, preceded by space\)
\( \newcommand{\Dif}{\mathop{}\!\mathrm{D}} % roman D in math mode, preceded by space\)
\( \newcommand{\df}{\dif\hspace{0.05em} f} % df\) \(\newcommand{\dBar}{\mathop{}\!\mathrm{d}\hspace-.3em\raise1.05ex{\Rule{.8ex}{.125ex}{0ex}}} % inexact differential \)
\( \newcommand{\dq}{\dBar q} % heat differential\)
\( \newcommand{\dw}{\dBar w} % work differential\)
\( \newcommand{\dQ}{\dBar Q} % infinitesimal charge\)
\( \newcommand{\dx}{\dif\hspace{0.05em} x} % dx\)
\( \newcommand{\dt}{\dif\hspace{0.05em} t} % dt\)
\( \newcommand{\difp}{\dif\hspace{0.05em} p} % dp\)
\( \newcommand{\Del}{\Delta}\)
\( \newcommand{\Delsub}[1]{\Delta_{\text{#1}}}\)
\( \newcommand{\pd}[3]{(\partial #1 / \partial #2 )_{#3}} % \pd{}{}{} - partial derivative, one line\)
\( \newcommand{\Pd}[3]{\left( \dfrac {\partial #1} {\partial #2}\right)_{#3}} % Pd{}{}{} - Partial derivative, built-up\)
\( \newcommand{\bpd}[3]{[ \partial #1 / \partial #2 ]_{#3}}\)
\( \newcommand{\bPd}[3]{\left[ \dfrac {\partial #1} {\partial #2}\right]_{#3}}\)
\( \newcommand{\dotprod}{\small\bullet}\)
\( \newcommand{\fug}{f} % fugacity\)
\( \newcommand{\g}{\gamma} % solute activity coefficient, or gamma in general\)
\( \newcommand{\G}{\varGamma} % activity coefficient of a reference state (pressure factor)\)
\( \newcommand{\ecp}{\widetilde{\mu}} % electrochemical or total potential\)
\( \newcommand{\Eeq}{E\subs{cell, eq}} % equilibrium cell potential\)
\( \newcommand{\Ej}{E\subs{j}} % liquid junction potential\)
\( \newcommand{\mue}{\mu\subs{e}} % electron chemical potential\)
\( \newcommand{\defn}{\,\stackrel{\mathrm{def}}{=}\,} % "equal by definition" symbol\) \( \newcommand{\D}{\displaystyle} % for a line in built-up\)
\( \newcommand{\s}{\smash[b]} % use in equations with conditions of validity\)
\( \newcommand{\cond}[1]{\\[-2.5pt]{}\tag*{#1}}\)
\( \newcommand{\nextcond}[1]{\\[-5pt]{}\tag*{#1}}\)
\( \newcommand{\R}{8.3145\units{J$\,$K$\per\,$mol$\per$}} % gas constant value\)
\( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \)
\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) Reaction calorimetry is used to evaluate the molar integral reaction enthalpy \(\Del H\m\rxn\) of a reaction or other chemical process at constant temperature and pressure. The measurement actually made, however, is a temperature change. Sections 11.5.1 and 11.5.2 will describe two common types of calorimeters designed for reactions taking place at either constant pressure or constant volume. The constant-pressure type is usually called a , and the constant-volume type is known as a or . In either type of calorimeter, the chemical process takes place in a reaction vessel surrounded by an outer jacket. The jacket may be of either the adiabatic type or the isothermal-jacket type described in Sec. 7.3.2 in connection with heat capacity measurements. A temperature-measuring device is immersed either in the vessel or in a phase in thermal contact with it. The measured temperature change is caused by the chemical process, instead of by electrical work as in the determination of heat capacity. One important way in which these calorimeters differ from ones used for heat capacity measurements is that work is kept deliberately small, in order to minimize changes of internal energy and enthalpy during the experimental process. The contents of a constant-pressure calorimeter are usually open to the atmosphere, so this type of calorimeter is unsuitable for processes involving gases. It is, however, a convenient apparatus in which to study a liquid-phase chemical reaction, the dissolution of a solid or liquid solute in a liquid solvent, or the dilution of a solution with solvent. The process is initiated in the calorimeter by allowing the reactants to come into contact. The temperature in the reaction vessel is measured over a period of time starting before the process initiation and ending after the advancement has reached a final value with no further change. The heating or cooling curve (temperature as a function of time) is observed over a period of time that includes the period during which the advancement \(\xi\) changes. For an exothermic reaction occurring in an adiabatic calorimeter, the heating curve may resemble that shown in Fig. 7.3, and the heating curve in an isothermal-jacket calorimeter may resemble that shown in Fig. 7.4. Two points are designated on the heating or cooling curve: one at temperature \(T_1\), before the reaction is initiated, and the other at \(T_2\), after \(\xi\) has reached its final value. These points are indicated by open circles in Figs. 7.3 and 7.4. The relations derived here parallel those of Sec. 11.5.1 for a constant-pressure calorimeter. The three paths depicted in Fig. 11.13 are similar to those in Fig. 11.11, except that instead of being at constant pressure they are at constant volume. We shall assume the combustion reaction is exothermic, with \(T_2\) being greater than \(T_1\). The internal energy change of the experimental process that actually occurs in the calorimeter between times \(t_1\) and \(t_2\) is denoted \(\Del U\expt\) in the figure. Conceptually, the overall change of state during this process would be duplicated by a path in which the temperature of the system with the reactants present increases from \(T_1\) to \(T_2\), followed by the isothermal bomb process at temperature \(T_2\). (When one investigates a combustion reaction, the path in which temperature changes without reaction is best taken with reactants rather than products present because the reactants are more easily characterized.) In the figure these paths are labeled with the internal energy changes \(\Del U(\tx{R})\) and \(\Del U(\tx{IBP},T_2)\), and we can write \begin{equation} \Del U\expt = \Del U(\tx{R}) + \Del U(\tx{IBP},T_2) \tag{11.5.4} \end{equation} To evaluate \(\Del U(\tx{R})\), we can use the energy equivalent \(\epsilon\subs{R}\) of the calorimeter with reactants present in the bomb vessel. \(\epsilon\subs{R}\) is the average heat capacity of the system between \(T_1\) and \(T_2\)—that is, the ratio \(q/(T_2 - T_1)\), where \(q\) is the heat that would be needed to change the temperature from \(T_1\) to \(T_2\). From the first law, with expansion work assumed negligible, the internal energy change equals this heat, giving us the relation \begin{equation} \Del U(\tx{R}) = \epsilon\subs{R}(T_2 - T_1) \tag{11.5.5} \end{equation} The initial and final states of the path are assumed to be equilibrium states, and there may be some transfer of reactants or H\(_2\)O from one phase to another within the bomb vessel during the heating process. The value of \(\epsilon\subs{R}\) is obtained in a separate calibration experiment. The calibration is usually carried out with the combustion of a reference substance, such as benzoic acid, whose internal energy of combustion under controlled conditions is precisely known from standardization based on electrical work. If the bomb vessel is immersed in the same mass of water in both experiments and other conditions are similar, the difference in the values of \(\epsilon\subs{R}\) in the two experiments is equal to the known difference in the heat capacities of the initial contents (reactants, water, etc.) of the bomb vessel in the two experiments. The internal energy change we wish to find is \(\Del U(\tx{IBP},T_2)\), that of the isothermal bomb process in which reactants change to products at temperature \(T_2\), accompanied perhaps by some further transfer of substances between phases. From Eqs. 11.5.4 and 11.5.5, we obtain \begin{equation} \Del U(\tx{IBP},T_2) = -\epsilon (T_2 - T_1)+\Del U\expt \tag{11.5.6} \end{equation} The value of \(\Del U\expt\) is small. To evaluate it, we must look in detail at the possible sources of energy transfer between the system and the surroundings during the experimental process. These sources are We want to obtain the value of \(\Delsub{c}U\st(T\subs{ref})\), the molar internal energy change for the main combustion reaction at the reference temperature under standard-state conditions. Once we have this value, it is an easy matter to find the molar change under standard-state conditions, our ultimate goal. Consider a hypothetical process with the following three isothermal steps carried out at the reference temperature \(T\subs{ref}\): Thus, we calculate the standard internal energy change of the main combustion reaction at temperature \(T\subs{ref}\) from \begin{equation} \Del U\st(\tx{cmb},T\subs{ref}) = \Del U(\tx{IBP},T\subs{ref}) + \tx{(Washburn corrections)} - \sum_i \Del\xi_i \Delsub{r}U\st(i) \tag{11.5.9} \end{equation} where the sum over \(i\) is for side reactions and auxiliary reactions if present. Finally, we calculate the standard internal energy of combustion from \begin{equation} \Delsub{c}U\st(T\subs{ref}) = \frac{\Del U\st(\tx{cmb},T\subs{ref})}{\Del\xi\subs{c}} \tag{11.5.10} \end{equation} where \(\Del\xi\subs{c}\) is the advancement of the main combustion reaction in the bomb vessel. The Washburn corrections needed in Eq. 11.5.9 are internal energy changes for certain hypothetical physical processes occurring at the reference temperature \(T\subs{ref}\) involving the substances present in the bomb vessel. In these processes, substances change from their standard states to the initial state of the isothermal bomb process, or change from the final state of the isothermal bomb process to their standard states. For example, consider the complete combustion of a solid or liquid compound of carbon, hydrogen, and oxygen in which the combustion products are CO\(_2\) and H\(_2\)O and there are no side reactions or auxiliary reactions. In the initial state of the isothermal bomb process, the bomb vessel contains the pure reactant, liquid water with O\(_2\) dissolved in it, and a gaseous mixture of O\(_2\) and H\(_2\)O, all at a high pressure \(p_1\). In the final state, the bomb vessel contains liquid water with O\(_2\) and CO\(_2\) dissolved in it and a gaseous mixture of O\(_2\), H\(_2\)O, and CO\(_2\), all at pressure \(p_2\). In addition, the bomb vessel contains internal parts of constant mass such as the sample holder and ignition wires. In making Washburn corrections, we must use a single standard state for each substance in order for Eq. 11.5.9 to correctly give the standard internal energy of combustion. In the present example we choose the following standard states: pure solid or liquid for the reactant compound, pure liquid for the H\(_2\)O, and pure ideal gases for the O\(_2\) and CO\(_2\), each at pressure \(p\st=1\br\). We can calculate the amount of each substance in each phase, in both the initial state and final state of the isothermal bomb process, from the following information: the internal volume of the bomb vessel; the mass of solid or liquid reactant initially placed in the vessel; the initial amount of H\(_2\)O; the initial O\(_2\) pressure; the water vapor pressure; the solubilities (estimated from Henry’s law constants) of O\(_2\) and CO\(_2\) in the water; and the stoichiometry of the combustion reaction. Problem 11.7 guides you through these calculations. Experimenters have used great ingenuity in designing calorimeters to measure reaction enthalpies and to improve their precision. In addition to the constant-pressure reaction calorimeter and bomb calorimeter described above, three additional types will be briefly mentioned. A has two coexisting phases of a pure substance in thermal contact with the reaction vessel and an adiabatic outer jacket. The two coexisting phases constitute a univariant subsystem that at constant pressure is at the fixed temperature of the equilibrium phase transition. The thermal energy released or absorbed by the reaction, instead of changing the temperature, is transferred isothermally to or from the coexisting phases and can be measured by the volume change of the phase transition. A reaction enthalpy, of course, can only be measured by this method at the temperature of the equilibrium phase transition. The well-known Bunsen ice calorimeter uses the ice–water transition at \(0\units{\(\degC\)}\). The solid–liquid transition of diphenyl ether has a relatively large volume change and is useful for measurements at \(26.9\units{\(\degC\)}\). Phase-transition calorimeters are especially useful for slow reactions. A is a variation of an isothermal-jacket calorimeter. It uses a thermopile (Fig. 2.7) to continuously measure the temperature difference between the reaction vessel and an outer jacket acting as a constant-temperature heat sink. The heat transfer takes place mostly through the thermocouple wires, and to a high degree of accuracy is proportional to the temperature difference integrated over time. This is the best method for an extremely slow reaction, and it can also be used for rapid reactions. A is a flow system in which oxygen, fluorine, or another gaseous oxidant reacts with a gaseous fuel. The heat transfer between the flow tube and a heat sink can be measured with a thermopile, as in a heat-flow calorimeter. | 18,440 | 1,887 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Surface_Science_(Nix)/02%3A_Adsorption_of_Molecules_on_Surfaces/2.04%3A_PE_Curves_and_Energetics_of_Adsorption |
In this section, we will consider both the energetics of adsorption and factors which influence the kinetics of adsorption by looking at the "potential energy diagram/curve" for the adsorption process. The potential energy curve for the adsorption process is a representation of the variation of the energy (PE or E) of the system as a function of the distance (\(d\)) of an adsorbate from a surface. Within this simple one-dimensional (1D) model, the only variable is the distance (\(d\)) of the adsorbing molecule from the substrate surface. Thus, the energy of the system is a function only of this variable i.e. \[E = E(d)\] It should be remembered that this is a very simplistic model which neglects many other parameters which influence the energy of the system (a single molecule approaching a clean surface), including for example The interaction of a molecule with a given surface will also clearly be dependent upon the presence of any existing adsorbed species, whether these be surface impurities or simply pre-adsorbed molecules of the same type (in the latter case we are starting to consider the effect of surface coverage on the adsorption characteristics). Nevertheless, it is useful to first consider the interaction of an isolated molecule with a clean surface using the simple 1D model. For the purposes of this Module, we will also not be overly concerned whether the "energy" being referred to should strictly be the internal energy, the enthalpy or free energy of the system. In the case of pure physisorption, e.g., Ar/metals, the only attraction between the adsorbing species and the surface arises from weak, . As illustrated below, these forces give rise to a shallow minimum in the PE curve at a relatively large distance from the surface (typically \(d > 0.3\, nm\)) before the strong repulsive forces arising from electron density overlap cause a rapid increase in the total energy. There is no barrier to prevent the atom or molecule which is approaching the surface from entering this physisorption well, i.e. the process is not activated and the kinetics of physisorption are invariably fast. The weak physical adsorption forces and associated long-range attraction will be present to varying degrees in all adsorbate / substrate systems. However, in cases where chemical bond formation between the adsorbate and substrate can also occur, the PE curve is dominated by a much deeper chemisorption minimum at shorter values of . The graph above shows the PE curves due to physisorption and chemisorption separately - in practice, the PE curve for any real molecule capable of undergoing chemisorption is best described by a combination of the two curves, with a curve crossing at the point at which chemisorption forces begin to dominate over those arising from physisorption alone. The minimum energy pathway obtained by combining the two PE curves is now highlighted in red. Any perturbation of the combined PE curve from the original, separate curves is most likely to be evident close to the highlighted crossing point. For clarity, we will now consider only the overall PE curve: The depth of the chemisorption well is a measure of the strength of binding to the surface - in fact it is a direct representation of the energy of adsorption, whilst the location of the global minimum on the horizontal axis corresponds to the equilibrium bond distance ( ) for the adsorbed molecule on this surface. The energy of adsorption is , and since it corresponds to the energy upon adsorption it is better represented as or . However, you will also often find the depth of this well associated with the enthalpy of adsorption, Δ (ads). The " ", , is taken to be a positive quantity equal in magnitude to the enthalpy of adsorption ; i.e. = -Δ (ads) ) In this particular case, there is clearly no barrier to be overcome in the adsorption process and there is no activation energy of adsorption (i.e. = 0, but do remember the previously mentioned limitations of this simple 1D model). There is of course a significant barrier to the reverse, desorption process - the red arrow in the diagram below represents the activation energy for desorption. Clearly in this particular case, the magnitudes of the energy of adsorption and the activation energy for desorption can also be equated i.e. \[E_a^{des} = ΔE (ads)\] or \[ E_a^{des} \approx -ΔH (ads)\] In this case the main differences arise from the substantial changes in the PE curve for the chemisorption process. Again, we start off with the basic PE curve for the physisorption process which represents how the molecule can weakly interact with the surface: If we now consider a specific molecule such as H and initially treat it as being completely from the surface ( i.e. when the distance, , is very large ) then a substantial amount of energy has to be put into the system in order to cause dissociation of the molecule. \[\ce{H_2 → H + H}\] - this is the bond dissociation energy [ (H-H) ], some 435 kJ mol or 4.5 eV. The red dot in the diagram above thus represents two hydrogen atoms, equidistant (and a long distance) from the surface and also now well separated from each other. If these atoms are then allowed to approach the surface they may ultimately both form strong chemical bonds to the substrate .... this corresponds to the minimum in the red curve which represents the chemisorption PE curve for the two H atoms. In reality, of course, such a mechanism for dissociative hydrogen chemisorption is not practical - the energy downpayment associated with breaking the H-H bond is far too severe. Instead, a hydrogen molecule will initially approach the surface along the physisorption curve. If it has sufficient energy it may pass straight into the chemisorption well ( "direct chemisorption" ) .... or, alternatively, it may first undergo transient physisorption - a state from which it can then either desorb back as a molecule into the gas phase or cross over the barrier into the dissociated, chemisorptive state (as illustrated below). In this latter case, the molecule can be said to have undergone "precursor-mediated" chemisorption. The characteristics of this type of dissociative adsorption process are clearly going to be strongly influenced by the position of the crossing point of the two curves (molecular physisorption v's dissociative chemisorption) - relatively small shifts in the position of either of the two curves can significantly alter the size of any barrier to chemisorption. In the example immediately below there is no direct activation barrier to dissociative adsorption - the curve crossing is below the initial "zero energy" of the system. whilst, in this next case …. there is a substantial barrier to chemisorption. Such a barrier has a major influence on the of adsorption. At this point it is useful to return to consider the effect of such a barrier on the relationship between the activation energies for adsorption and desorption, and the energy (or enthalpy) of adsorption. Clearly, from the diagram \[E_a^{des} - E_a^{ads} = - ΔE_{ads}\] but, since the activation energy for adsorption is nearly always very much smaller than that for desorption, and the difference between the energy and enthalpy of adsorption is also very small, it is still quite common to see the relationship \[E_a^{des} \approx -ΔH_{ads}\] For a slightly more detailed treatment of the adsorption process, you are referred to the following examples of . | 7,497 | 1,888 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/10%3A_Electrolyte_Solutions/10.05%3A_Derivation_of_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$)}}\)
\( \newcommand{\dil}{\tx{(dil)}}\)
\( \newcommand{\sln}{\tx{(sln)}}\)
\( \newcommand{\mix}{\tx{(mix)}}\)
\( \newcommand{\rxn}{\tx{(rxn)}}\)
\( \newcommand{\expt}{\tx{(expt)}}\)
\( \newcommand{\solid}{\tx{(s)}}\)
\( \newcommand{\liquid}{\tx{(l)}}\)
\( \newcommand{\gas}{\tx{(g)}}\)
\( \newcommand{\pha}{\alpha} % phase alpha\)
\( \newcommand{\phb}{\beta} % phase beta\)
\( \newcommand{\phg}{\gamma} % phase gamma\)
\( \newcommand{\aph}{^{\alpha}} % alpha phase superscript\)
\( \newcommand{\bph}{^{\beta}} % beta phase superscript\)
\( \newcommand{\gph}{^{\gamma}} % gamma phase superscript\)
\( \newcommand{\aphp}{^{\alpha'}} % alpha prime phase superscript\)
\( \newcommand{\bphp}{^{\beta'}} % beta prime phase superscript\)
\( \newcommand{\gphp}{^{\gamma'}} % gamma prime phase superscript\)
\( \newcommand{\apht}{\small\aph} % alpha phase tiny superscript\)
\( \newcommand{\bpht}{\small\bph} % beta phase tiny superscript\)
\( \newcommand{\gpht}{\small\gph} % gamma phase tiny superscript\) \( \newcommand{\upOmega}{\Omega}\) \( \newcommand{\dif}{\mathop{}\!\mathrm{d}} % roman d in math mode, preceded by space\)
\( \newcommand{\Dif}{\mathop{}\!\mathrm{D}} % roman D in math mode, preceded by space\)
\( \newcommand{\df}{\dif\hspace{0.05em} f} % df\) \(\newcommand{\dBar}{\mathop{}\!\mathrm{d}\hspace-.3em\raise1.05ex{\Rule{.8ex}{.125ex}{0ex}}} % inexact differential \)
\( \newcommand{\dq}{\dBar q} % heat differential\)
\( \newcommand{\dw}{\dBar w} % work differential\)
\( \newcommand{\dQ}{\dBar Q} % infinitesimal charge\)
\( \newcommand{\dx}{\dif\hspace{0.05em} x} % dx\)
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\( \newcommand{\R}{8.3145\units{J$\,$K$\per\,$mol$\per$}} % gas constant value\)
\( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \)
\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) Debye and Hückel derived Eq. 10.4.1 using a combination of electrostatic theory, statistical mechanical theory, and thermodynamics. This section gives a brief outline of their derivation. The derivation starts by focusing on an individual ion of species \(i\) as it moves through the solution; call it the central ion. Around this central ion, the time-average spatial distribution of any ion species \(j\) is not random, on account of the interaction of these ions of species \(j\) with the central ion. (Species \(i\) and \(j\) may be the same or different.) The distribution, whatever it is, must be spherically symmetric about the central ion; that is, a function only of the distance \(r\) from the center of the ion. The local concentration, \(c'_j\), of the ions of species \(j\) at a given value of \(r\) depends on the ion charge \(z_j e\) and the electric potential \(\phi\) at that position. The time-average electric potential in turn depends on the distribution of all ions and is symmetric about the central ion, so expressions must be found for \(c'_j\) and \(\phi\) as functions of \(r\) that are mutually consistent. Debye and Hückel assumed that \(c'_j\) is given by the Boltzmann distribution \begin{equation} c'_j = c_j e^{-z_j e \phi/ kT} \tag{10.5.1} \end{equation} where \(z_j e\phi\) is the electrostatic energy of an ion of species \(j\), and \(k\) is the Boltzmann constant (\(k = R/N\subs{A}\)). As \(r\) becomes large, \(\phi\) approaches zero and \(c'_j\) approaches the macroscopic concentration \(c_j\). As \(T\) increases, \(c'_j\) at a fixed value of \(r\) approaches \(c_j\) because of the randomizing effect of thermal energy. Debye and Hückel expanded the exponential function in powers of \(1/T\) and retained only the first two terms: \(c'_j \approx c_j(1 - z_j e\phi/kT)\). The distribution of each ion species is assumed to follow this relation. The electric potential function consistent with this distribution and with the electroneutrality of the solution as a whole is \begin{equation} \phi = (z_i e / 4\pi\epsilon\subs{r}\epsilon_0 r) e^{\kappa(a-r)} / (1 + \kappa a) \tag{10.5.2} \end{equation} Here \(\kappa\) is defined by \(\kappa^2 = 2N\subs{A}^2 e^2 I_c/\epsilon_r \epsilon_0 RT\), where \(I_c\) is the defined by \(I_c = (1/2)\sum_i c_i z_i^2\). The electric potential \(\phi\) at a point is assumed to be a sum of two contributions: the electric potential the central ion would cause at infinite dilution, \(z_i e/4\pi \epsilon_r \epsilon_0 r\), and the electric potential due to all other ions, \(\phi'\). Thus, \(\phi'\) is equal to \(\phi - z_i e/4\pi \epsilon_r \epsilon_0 r\), or \begin{equation} \phi' = (z_i e/4\pi\epsilon\subs{r}\epsilon_0 r) [e^{\kappa(a-r)}/(1+\kappa a)-1] \tag{10.5.3} \end{equation} This expression for \(\phi'\) is valid for distances from the center of the central ion down to \(a\), the distance of closest approach of other ions. At smaller values of \(r\), \(\phi'\) is constant and equal to the value at \(r = a\), which is \(\phi'(a) = -(z_i e/4\pi \epsilon_r \epsilon_0)\kappa/(1 + \kappa a)\). The interaction energy between the central ion and the surrounding ions (the ion atmosphere) is the product of the central ion charge and \(\phi'(a)\). The last step of the derivation is the calculation of the work of a hypothetical reversible process in which the surrounding ions stay in their final distribution, and the charge of the central ion gradually increases from zero to its actual value \(z_i e\). Let \(\alpha z_i e\) be the charge at each stage of the process, where \(\alpha\) is a fractional advancement that changes from \(0\) to \(1\). Then the work \(w'\) due to the interaction of the central ion with its ion atmosphere is \(\phi'(a)\) integrated over the charge: \begin{equation} \begin{split} w' & = -\int_{\alpha=0}^{\alpha=1} [(\alpha z_i e / 4\pi\epsilon\subs{r}\epsilon_0)\kappa /(1+\kappa a)]\dif(\alpha z_i \epsilon) \cr & = -(z_i^2 e^2/8\pi\epsilon\subs{r}\epsilon_0) \kappa/(1+\kappa a) \end{split} \tag{10.5.4} \end{equation} Since the infinitesimal Gibbs energy change in a reversible process is given by \(\dif G = -S\dif T + V\difp + \dw'\) (Eq. 5.8.6), this reversible nonexpansion work at constant \(T\) and \(p\) is equal to the Gibbs energy change. The Gibbs energy change per amount of species \(i\) is \(w'N\subs{A} = -(z_i^2 e^2 N\subs{A}/8\pi \epsilon\subs{r}\epsilon_0)\kappa/(1 + \kappa a)\). This quantity is \(\Del G/n_i\) for the process in which a solution of fixed composition changes from a hypothetical state lacking ion–ion interactions to the real state with ion–ion interactions present. \(\Del G/n_i\) may be equated to the difference of the chemical potentials of \(i\) in the final and initial states. If the chemical potential without ion–ion interactions is taken to be that for ideal-dilute behavior on a molality basis, \(\mu_i=\mu_{m,i}\rf + RT\ln(m_i/m\st)\), then \(-(z_i^2 e^2 N\subs{A}/8\pi \epsilon\subs{r}\epsilon_0)\kappa/(1 + \kappa a)\) is equal to \(\mu_i - [\mu_{m,i}\rf + RT\ln(m_i/m\st)] = RT\ln\g_{m,i}\). In a dilute solution, \(c_i\) can with little error be set equal to \(\rho\A^* m_i\), and \(I_c\) to \(\rho\A^*I_m\). Equation 10.4.1 follows. | 12,569 | 1,889 |
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 | 1,891 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/10%3A_Electrolyte_Solutions/10.02%3A_Solution_of_a_Symmetrical_Electrolyte |
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\( \newcommand{\R}{8.3145\units{J$\,$K$\per\,$mol$\per$}} % gas constant value\)
\( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \)
\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) Let us consider properties of an electrolyte solute as a whole. The simplest case is that of a binary solution in which the solute is a symmetrical strong electrolyte—a substance whose formula unit has one cation and one anion that dissociate completely. This condition will be indicated by \(\nu = 2\), where \(\nu\) is the number of ions per formula unit. In an aqueous solution, the solute with \(\nu\) equal to 2 might be a 1:1 salt such as NaCl, a 2:2 salt such as MgSO\(_4\), or a strong monoprotic acid such as HCl. In this binary solution, the chemical potential of the solute as a whole is defined in the usual way as the partial molar Gibbs energy \begin{equation} \mu\B \defn \Pd{G}{n\B}{T,p,n\A} \tag{10.2.1} \end{equation} and is a function of \(T\), \(p\), and the solute molality \(m\B\). Although \(\mu\B\) under given conditions must in principle have a definite value, we are not able to actually evaluate it because we have no way to measure precisely the energy brought into the system by the solute. This energy contributes to the internal energy and thus to \(G\). We can, however, evaluate the differences \(\mu\B - \mu\mbB\rf\) and \(\mu\B - \mu\mbB\st\). We can write the additivity rule (Eq. 9.2.25) for \(G\) as either \begin{equation} G = n\A\mu\A + n\B\mu\B \tag{10.2.2} \end{equation} or \begin{equation} G = n\A\mu\A + n_{+}\mu_{+}+n_{-}\mu_{-} \tag{10.2.3} \end{equation} A comparison of these equations for a symmetrical electrolyte (\(n\B = n_+ = n_-\)) gives us the relation \begin{gather} \s{ \mu\B = \mu_{+} + \mu_{-} } \tag{10.2.4} \cond{(\(\nu{=}2\))} \end{gather} We see that the solute chemical potential in this case is the of the single-ion chemical potentials. The solution is a phase of electric potential \(\phi\). From Eqs. 10.1.4 and 10.1.5, the sum \(\mu_{+}+\mu_{-}\) appearing in Eq. 10.2.4 is \begin{equation} \mu_+(\phi) + \mu_-(\phi)=\mu_+(0) + \mu_-(0) + (z_+ + z_-)F\phi \tag{10.2.5} \end{equation} For the symmetrical electrolyte, the sum \((z_+ + z_-)\) is zero, so that \(\mu\B\) is equal to \(\mu_+(0) + \mu_-(0)\). We substitute the expressions of Eq. 10.1.10, use the relation \(\mu\mbB\rf=\mu_+\rf+\mu_-\rf\) with reference states at \(\phi{=}0\), set the ion molalities \(m_+\) and \(m_-\) equal to \(m\B\), and obtain \begin{gather} \s{ \mu\B=\mu\mbB\rf +RT\ln\left[\g_+\g_-\left(\frac{m\B}{m\st}\right)^2\right] } \tag{10.2.6} \cond{(\(\nu{=}2\))} \end{gather} The important feature of this relation is the appearance of the power of \(m\B/m\st\), instead of the first power as in the case of a nonelectrolyte. Also note that \(\mu\B\) does not depend on \(\phi\), unlike \(\mu_+\) and \(\mu_-\). Although we cannot evaluate \(\g_+\) or \(\g_-\) individually, we can evaluate the product \(\g_+\g_-\). This product is the square of the \(\g_{\pm}\), defined for a symmetrical electrolyte by \begin{gather} \s{ \g_{\pm} \defn \sqrt{\g_{+}\g_{-}} } \tag{10.2.7} \cond{(\(\nu{=}2\))} \end{gather} With this definition, Eq. 10.2.6 becomes \begin{gather} \s{ \mu\B=\mu\mbB\rf +RT\ln\left[\left(\g_{\pm}\right)^2 \left(\frac{m\B}{m\st}\right)^2\right] } \tag{10.2.8} \cond{(\(\nu{=}2\))} \end{gather} Since it is possible to determine the value of \(\mu\B-\mu\mbB\rf\) for a solution of known molality, \(\g_{\pm}\) is a measurable quantity. If the electrolyte (e.g., HCl) is sufficiently volatile, its mean ionic activity coefficient in a solution can be evaluated from partial pressure measurements of an equilibrated gas phase. Section 10.6 will describe a general method by which \(\g_{\pm}\) can be found from osmotic coefficients. Section 14.5 describes how, in favorable cases, it is possible to evaluate \(\g_{\pm}\) from the equilibrium cell potential of a galvanic cell. The activity \(a\mbB\) of a solute substance on a molality basis is defined by Eq. 9.7.8: \begin{equation} \mu\B = \mu\mbB\st + RT\ln a\mbB \tag{10.2.9} \end{equation} Here \(\mu\mbB\st\) is the chemical potential of the solute in its standard state, which is the solute reference state at the standard pressure. By equating the expressions for \(\mu\B\) given by Eqs. 10.2.8 and 10.2.9 and solving for the activity, we obtain \begin{gather} \s{ a\mbB = \G\mbB \left(\g_{\pm}\right)^2 \left(\frac{m\B}{m\st}\right)^2 } \tag{10.2.10} \cond{(\(\nu{=}2\))} \end{gather} where \(\G\mbB\) is the pressure factor defined by \begin{equation} \G\mbB \defn \exp\left(\frac{\mu\mbB\rf-\mu\mbB\st}{RT}\right) \tag{10.2.11} \end{equation} We can use the appropriate expression in Table 9.6 to evaluate \(\G\mbB\) at an arbitrary pressure \(p'\): \begin{equation} \G\mbB(p')=\exp\left(\int_{p\st}^{p'}\frac{V\B^{\infty}}{RT}\difp\right) \approx \exp\left[\frac{V\B^{\infty}(p'-p\st)}{RT}\right] \tag{10.2.12} \end{equation} The value of \(\G\mbB\) is \(1\) at the standard pressure, and close to \(1\) at any reasonably low pressure. For this reason it is common to see Eq. 10.2.10 written as \(a\mbB=\g_{\pm}^2(m\B/m\st)^2\), with \(\G\mbB\) omitted. Equation 10.2.10 predicts that the activity of HCl in aqueous solutions is proportional, in the limit of infinite dilution, to the of the HCl molality. In contrast, the activity of a electrolyte solute is proportional to the power of the molality in this limit. This predicted behavior of aqueous HCl is consistent with the data plotted in Fig. 10.1, and is confirmed by the data for dilute HCl solutions shown in Fig. 10.2(a). The dashed line in Fig. 10.2(a) is the extrapolation of the ideal-dilute behavior given by \(a\mbB=(m\B/m\st)^2\). The extension of this line to \(m\B = m\st\) establishes the hypothetical solute reference state based on molality, indicated by a filled circle in Fig. 10.2(b). (Since the data are for solutions at the standard pressure of \(1\br\), the solute reference state shown in the figure is also the solute standard state.) The solid curve of Fig. 10.2(c) shows how the mean ionic activity coefficient of HCl varies with molality in approximately the same range of molalities as the data shown in Fig. 10.2(b). In the limit of infinite dilution, \(\g_{\pm}\) approaches unity. The slope of the curve approaches \(-\infty\) in this limit, quite unlike the behavior described in Sec. 9.5.4 for the activity coefficient of a nonelectrolyte solute. For a symmetrical strong electrolyte, \(\g_{\pm}\) is the geometric average of the single-ion activity coefficients \(\g_+\) and \(\g_-\). We have no way of evaluating \(\g_+\) or \(\g_-\) individually, even if we know the value of \(\g_{\pm}\). For instance, we cannot assume that \(\g_+\) and \(\g_-\) are equal. | 13,976 | 1,892 |
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The concepts of VSEPR are illustrated with NH and NI . NI actually has a central N surrounded by four very large I atoms , which sterically repel, leading to the sensitivity of NI (better, NI *NH ) to explosive detonation. Prepare NI according to : "Specifically, 0.2-0.3 g of iodine are placed in a 30-ki beaker with 5 mL of concentrated aqueous ammonia and stirred briefly. The mixture is allowed to to stand for 5 min, and the supernatant liquid is decanted from the brown solid. It is then washed 5 times with water that is decanted off each time after allowing most of the brown solid to settle. The brown solid is then scraped onto a few pieces of filter paper and patted to absorb most of the water, then scraped onto a new filter paper. In our hands the nitrogen triiodide always has exploded totally 45 min later when touched with a long pole." | 870 | 1,893 |
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The bulk of the discussion in this chapter dealt with ideal solutions. However, real solutions will deviate from this kind of behavior. So much as in the case of gases, where fugacity was introduced to allow us to use the ideal models, is used to allow for the deviation of real solutes from limiting ideal behavior. The activity of a solute is related to its concentration by \[ a_B=\gamma \dfrac{m_B}{m^o} \nonumber \] where \(\gamma \) is the , \(m_B\) is the molaliy of the solute, and \(m^o\) is unit molality. The activity coefficient is unitless in this definition, and so the activity itself is also unitless. Furthermore, the activity coefficient approaches unity as the molality of the solute approaches zero, insuring that dilute solutions behave ideally. The use of activity to describe the solute allows us to use the simple model for chemical potential by inserting the activity of a solute in place of its mole fraction: \[ \mu_B =\mu_B^o + RT \ln a_B \nonumber \] The problem that then remains is the measurement of the activity coefficients themselves, which may depend on temperature, pressure, and even concentration. For an ionic substance that dissociates upon dissolving \[ MX(s) \rightarrow M^+(aq) + X^-(aq) \nonumber \] the chemical potential of the cation can be denoted \(\mu_+\) and that of the anion as \(\mu_-\). For a solution, the total molar Gibbs function of the solutes is given by \[G = \mu_+ + \mu_- \nonumber \] where \[ \mu = \mu^* + RT \ln a \nonumber \] where \(\ denotes the chemical potential of an ideal solution, and \(a\) is the activity of the solute. Substituting his into the above relationship yields \[G = \mu^*_+ + RT \ln a_+ + \mu_-^* + RT \ln a_- \nonumber \] Using a molal definition for the activity coefficient \[a_i = \gamma_im_i \nonumber \] The expression for the total molar Gibbs function of the solutes becomes \[G = \mu_+^* + RT \ln \gamma_+ m_+ + \mu_-^* + RT \ln \gamma_- m_- \nonumber \] This expression can be rearranged to yield \[ G = \mu_+^* + \mu_-^* + RT \ln m_+m_- + RT \ln \gamma_+\gamma _- \nonumber \] where all of the deviation from ideal behavior comes from the last term. Unfortunately, it impossible to experimentally deconvolute the term into the specific contributions of the two ions. So instead, we use a geometric average to define the , \(\gamma _\pm\). \[\gamma_{\pm} = \sqrt{\gamma_+\gamma_-} \nonumber \] For a substance that dissociates according to the general process \[ M_xX_y(s) \rightarrow x M^{y+} (aq) + yX^{x-} (aq) \nonumber \] the expression for the mean activity coefficient is given by \[ \gamma _{\pm} = (\gamma_+^x \gamma_-^y)^{1/x+y} \nonumber \] In 1923, Debeye and Hückel (Debye & Hückel, 1923) suggested a means of calculating the mean activity coefficients from experimental data. Briefly, they suggest that \[ \log _{10} \gamma_{\pm} = \dfrac{1.824 \times 10^6}{(\epsilon T)^{3/2}} |z_++z_- | \sqrt{I} \nonumber \] where \(\epsilon\) is the dielectric constant of the solvent, \(T\) is the temperature in K, \(z_+\) and \(z_-\) are the charges on the ions, and \(I\) is the of the solution. \(I\) is given by \[ I = \dfrac{1}{2} \dfrac{m_+ z_+^2 + m_-z_-^2}{m^o} \nonumber \] For a solution in water at 25 C, As seen before activities are a way to account for deviation from ideal behavior while still keeping the formulism for the ideal case intact. For example in a ideal solution we have: \[μ^{sln} = μ^* + RT \ln x_i \nonumber \] is replaced by \[μ^{sln} = μ^* + RT \ln a_i \nonumber \] The relationship between \(a_i\) and \(x_i\) is often written using an activity coefficient \(γ\): \[a_i= γ_ix_i \nonumber \] Implicitly we have made use of Raoult's law here because we originally used \[x_i = \dfrac{P_i}{P^*_i} \nonumber \] In the case of a solvent this makes sense because Raoult's law is still valid in the limiting case, but for the solute it would make more sense to use Henry's law as a basis for the definition of activity: \[a_{solute,H} ≡ \dfrac{P_{solute}}{K_{x,H}} \nonumber \] This does mean that the \(μ^*\) now becomes a \(μ^{*Henry}\) because the extrapolation of the Henry law all the way to the other side of the diagram where \(x_{solute}=1\) points to a point that is not the equilibrium vapor pressure of this component. In fact it represents a virtual state of the system that cannot be realized. This however does not affect the usefulness of the convention. The subscript X was added to the K value because we are still using mole fractions. However Henry's law is often used with other concentration measures. The most important are: Both the numerical values and the dimensions of K will differ depending on which concentration measure is used. In addition the pressure units can differ. For example for oxygen in water we have: As you can see K is simply 1/K , both conventions are used.. Note that in this case a choice based on Raoult is really not feasible. At room temperature we are far above the critical point of oxygen which make the equilibrium vapor pressure a non-existent entity. Returning to activities we could use each of the versions of K as a basis for the activity definition. This means that when using activities it must be specified what scale we are using. Activities and Henry coefficients of dissolved gases in water (both fresh and salt) are quite important in geochemistry, environmental chemistry etc. A special case arises if the vapor pressure of a solute is negligible. For example if we dissolve sucrose in water. In that case we can still use the Henry based definition \[a_{solute,H} ≡ \dfrac{P_{solute}}{K_{x,H}} \nonumber \] Even though both \(K\) and \(P\) will be exceedingly small their ratio is still finite. However how do we determine either? The answer lies in the solvent. Even if the vapor pressure of sucrose is immeasurably small, the water vapor pressure above the solution can be measured. The Gibbs-Duhem equation can then be used to translate one into the other. We can use Raoult Law to define the activity of the solvent: \[a_1 = \dfrac{P_1}{P^*_1} \nonumber \] We can measure the pressures as a function of the solute concentrations. At low concentrations \[\ln a_1 \ln x-1 ≈ -x-2 \nonumber \] At higher concentrations we will get deviations, we can write: \[\ln \dfrac{P_1}{P^*_1}=\ln a_1 ≈ -x_2φ \nonumber \] The 'fudge factor' \(φ\) is known as the and can thus be determined as a function of the solute concentration from the pressure data. What we are really interested in is \(a_2\), not \(a_1\): \[a_2= γ_2x_2 \nonumber \] Using Gibbs-Duhem we can convert \(φ\) into \(γ_2\). Usually this is done in terms of molalities rather than mole fractions and it leads to this integral: \[\ln γ_{2,m} = φ – 1 + \int_{m'=0}^m \dfrac{φ – 1}{m'} dm' \nonumber \] | 6,785 | 1,894 |
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The equipartition theorem, also known as the law of equipartition, equipartition of energy or simply equipartition, states that every degree of freedom that appears only quadratically in the total energy has an average energy of \(½k_BT\) in thermal equilibrium and contributes \(½k_B\) to the system's heat capacity. Here, \(k_B\) is the Boltzmann constant, and T is the temperature in Kelvin. The law of equipartition of energy states that each quadratic term in the classical expression for the energy contributes ½kBT to the average energy. For instance, the motion of an atom has three degrees of freedom (number of ways of absorbing energy), corresponding to the x, y and z components of its momentum. Since these momenta appear quadratically in the kinetic energy, every atom has an average kinetic energy of \(3/2k_BT\) in thermal equilibrium. The number of degrees of freedom of a polyatomic gas molecule is \(3N\) where \(N\) is the number of atoms in the molecule. This is equal to number of coordinates for the system; e.g. for two atoms you would have x, y, z for each atom. The translational contribution to the average energy is derived in terms of the derivative of the translational partition: \[ \langle E_{trans} \rangle = - \dfrac{1}{q_{trans}} \dfrac{\partial q_{trans}}{\partial \beta} \label{Eq1} \] Introducing the translational partition function derived earlier, Equation \(\ref{Eq1}\) becomes \[= - \dfrac{\Lambda^3}{V} \dfrac{\partial }{\partial \beta} \dfrac{V}{\Lambda^3} = - \dfrac{3}{\Lambda} \dfrac{\partial \Lambda}{\partial \beta} = \dfrac{3}{2} k_BT \nonumber \] Thus, the three translational degrees of freedom in three dimensions satisfy the equipartition theorem with each translational degree providing \( ½ k_BT\) of energy. Consider the molecular partition functions. The average rotational energy to the average energy is derived in terms of the derivative of the translational partition: \[ \langle E_{rot} \rangle = - \dfrac{1}{q_{rot}} \dfrac{\partial q_{rot}}{\partial \beta} \label{Eq2} \] which when you introduce the rotational partition function, Equation \(\ref{Eq2}\) becomes \[ \langle E_{rot} \rangle = -\sigma \beta \tilde{B} \dfrac{1}{\sigma \tilde{B}} \dfrac{\partial}{\partial \beta} \dfrac{1}{\beta} = \dfrac{1}{\beta} = k_BT \nonumber \] The classical expression for the rotational energy of a diatomic molecule is \[ E_{rot}^{(classical)}= \dfrac{1}{2} I (\omega_x^2 + \omega_y^2) \nonumber \] where \(I\) is the moment of inertia and \( \omega _x\) and \( \omega _y\) are the angular velocities in the \(x\) and \(y\) directions. The rotation along the molecular axis (the \(z\) axis here) has no meaning in quantum mechanics because the rotations along the molecular axis lead to configurations which are indistinguishable from the original configuration. The two rotational degrees of freedom have thus given a value of \(kT\) with with each rotational degree providing \( ½ k_BT\) of energy. Consider vibrational motions. The average vibrational energy to the average energy is derived in terms of the derivative of the translational partition: \[ \langle E_{vib} \rangle = - \dfrac{1}{q_{vib}} \dfrac{\partial q_{vib}}{\partial \beta} \label{Eq3} \] which when you introduce the partition function for vibration, Equation \(\ref{Eq3}\) becomes \[ \langle E_{vib} \rangle = \dfrac{-1}{q_{vib}} \left( -hc\tilde{\nu}\dfrac{ e^{-hc \tilde{\nu}/k_BT}}{(1-e^{-hc \tilde{\nu}/k_BT})^2 } \right) = hc\tilde{\nu} \dfrac{ e^{-hc \tilde{\nu}/k_BT}}{\left(1-e^{-hc \tilde{\nu}/k_BT}\right) } \label{18.1.7} \] This can be simplified by dividing both numerator and denominator of Equation \(\ref{18.1.7}\) by \(e^{-hc \tilde{\nu}/k_BT}\) \[\langle E_{vib} \rangle = hc \tilde{\nu} \left( \dfrac{ 1 }{e^{hc \tilde{\nu}/k_BT} -1} \right) \label{18.1.7B} \] Equation \(\ref{18.1.7B}\) is applicable at all temperatures, but if \( hc \tilde{\nu}/k_BT \ll 1\) (i.e., the high temperature limit), then the exponential in the denominator can be expanded \[ e^{hc \tilde{\nu}/k_BT} -1 \approx 1 + hc \tilde{\nu}/k_BT -1 = hc \tilde{\nu}/k_BT \label{expansion} \] and Equation \(\ref{18.1.7B}\) becomes \[ \langle E_{vib} \rangle \approx \cancel{hc \tilde{\nu}} \left( \dfrac{1}{\cancel{ hc \tilde{\nu}}/k_BT}\right) \nonumber \] \[ \langle E_{vib} \rangle \approx k_BT \label{18.1.10} \] with each vibrational degree providing \( k_BT\) of energy (since there ar two quadratic terms in the Hamiltonian for a harmonic oscillator (kinetics energy and potential energy). Compare Equation \(\ref{18.1.10}\) with the classical expression for the vibrational energy \[ E_{vib}^{(classical)} = ½ kx^2 + ½ μv_x^2 \nonumber \] At high temperature the equipartition theorem is valid, but at low temperature, the expansion in Equation \(\ref{expansion}\) fails (or more terms are required). In this case, only a few vibrational states are occupied and the equipartition principle is not typically applicable. Heat capacity at constant volume \(C_v\), is defined as \[ C_v = \left(\dfrac{\partial U}{\partial T} \right)_v \nonumber \] The equipartition theorem requires that each degree of freedom that appears only quadratically in the total energy has an average energy of ½k T in thermal equilibrium and, thus, contributes ½kB to the system's heat capacity. Thus the three translational degrees of freedom each contribute ½R to (3/2 R). The contribution of rotational kinetic energy will be R for the linear, and 3/2R for the nonlinear molecules. For the vibration, an oscillator has quadratic kinetic and potential terms, making the contribution of each vibrational mode R. However, k T has to be much greater than the spacing between the quantum energy levels. If this is not satisfied, the heat capacity will be reduced and which drop to zero at low temperatures. The corresponding degree of freedom is said to be ; this is the situation for the vibrational degrees of freedom at room temperature and that is why the usual assumption is that they will not contribute. For comparing the molar heat capacities of nitrogen dioxide and carbon dioxide at constant volume (at room temperature), let us use the law of equipartition and assume the vibrations to be frozen out at room temperature. The predicted molar for the linear \(CO_2\) (with three translational and two rotational degrees of freedom) is \(5/2R\) \(20.8\, JK^{-1}mo'^{-1})\). The estimated molar for \(NO_2\) (a bent molecule, with three translational and three rotational degrees of freedom) is \(3R \,(25.0\, JK^{-1}mol^{-1})\). These estimations are close to the experimental values: Especially for \(CO_2\), the deviation is significant. This suggests that, although not all vibrational degrees of freedom are available, they cannot be totally ignored. The bigger deviation in the prediction of molar heat capacities is probably due to the existence of the lower frequency-bending vibration in carbon dioxide. | 6,996 | 1,895 |
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After completing this section, you should be able to Make certain that you can define, and use in context, the key terms below. Recall the definitions of primary, secondary and tertiary hydrogen atoms given in Section 3.3. It follows that a “primary carbocation” is a carbocation in which the carbon atom carrying the positive charge is bonded to only one other carbon atom, a “secondary carbocation” is one in which the carbon atom carrying the positive charge is bonded to two other carbon atoms, and so on. If more than one reaction could occur between a set of reactants under the same conditions giving products that are constitutional isomers and if one product forms in greater amounts than the others, the overall reaction is said to be regioselective. Say three reactions could occur between the hypothetical reactants and under the same conditions giving the constitutionally isomeric products , , and . There are two possibilities: 1. The three products form in equal amounts, i.e., of the total product 33% is , another 33% , the remaining 33% . (These percentages are called relative yields of the products.) If this is what is observed, the overall reaction between and is not regioselective. 2. One product forms in greater amounts than the others. Say, for example, the relative yields of , , and are 25%, 50%, and 25%, respectively. If this is what is observed, the overall reaction between and is regioselective. eg: Experimentally, is the major product; is the minor product. Thus, the overall reaction between and HBr is regioselective toward . If more than one reaction could occur between a set of reactants under the same conditions giving products that are constitutional isomers and if only one product is observed, the overall reaction is said to be 100% regioselective or regiospecific. eg: The only observed product is . (Relative yields of and are 100% and 0%, respectively.) Thus the overall reaction between and HBr is regiospecific toward . Regiospecificity is merely the limiting case of regioselectivity. All regiospecific reactions are regioselective, but not all regioselective reactions are regiospecific. During the electrophilic addition of HX to an alkene, the halide (X) could attach to either carbon in the double bond producing two different isomers as products. But, when an unsymmertrically alkyl substituted alkene undergoes an electrophillic addition with HX a single isomer is typically produced. For example, if propene were reacted with HBr, two products could possibly form: 2-bromolpropane and 1-bromopropane. However, 2-bromopropane is produced as the reaction's only product. Reactions are called regiospecific when only one of multiple possible isomers is exclusively formed. If HCl adds to an unsymmetrical alkene like propene what will the major product be? The regiospecificity of electrophilic additions to alkenes is commonly known as , after the Russian chemist Vladimir Markovnikov who proposed it in 1869. The electrophilic addition of HX to an alkene is said to follow Markovnikov's rule. : During the electrophilic addition of HX to an alkene, the H adds to the carbon of the double bond with the fewest number of alkyl substitutent. The halide (X) adds to the double bond carbon with the most alkyl substituents. Although Markovnikov's rule has been specifically stated for the electrophilic addition of HX, later in this chapter many more reaction will be shown to also follow Markovnikov's rule. It is important to point out that Markovnikov's rule only truley applies when there is a difference between the number of alky groups attached to each carbon in the double bond. When both carbon of the double bond have the same degree of alkyl substitution, Markovnikovs's rule becomes void and a mixture of both possible isomers is produced. To consider an explanation for why Markovnikov's rule holds true, the mechanism of the reaction needs to be considered. As seen in the previous section, the mechanism starts with the addition of H to a carbon in the double bond. This in turn caused the other double bond carbon to become a carbocation intermediate. In the second step of the mechanism the halide ion attacks the carbocation to form a C-X. Because Markovnikov's rule says that the halogen adds to the carbon in the double bond with the most alky substituents, it can be said the carbocation also prefers to form on the carbon with the most alkyl substituents. The reason for this holds the explaination for Markovnikov's rule. To simplify this idea, Markovnikov's rule can be restated in a different form. : During the electrophilic addition of HX to an alkene, the carbocation intermediate forms on the double bond carbon with the greatest number of alkyl substitutents. Overall, during the electrophilic addition of HX to an alkene there are two major changes in the bonding. First, the pi bond of the alkene is broken. Second, a single bond is formed on each carbon that was originally in the double bond. The two single bonds will become attached to and H and an X. If the alkene is unsymmertrically alkyl substituted, Markovnikov's rule is followed and the X will be bonded to the more alky substituted carbon and the H to the less substiuted. If the alkene is symmertrically alkyl substituted a mixture of isomers will be produced in the product. Please draw the product of the following reaction: In answering these types of questions it is always important to first determine which reaction is occurring. Because an alkene is the reactant and HBr is the product this reaction is an electrophilic addition. Overall, the double bond will be broken as H and Br are added. The next step is to determine if Markovnikov's rule needs to be applied. In the reactant's double bond the upper carbon has two alkyl substituents and the lower carbon has only two. Markovnikov's rule says that Br will attach to the upper carbon and H to the lower. Understanding the starting material and reaction required for the synthesis a specific target molecule is an important concept in organic chemistry. The preferred method for answering these types of questions is to work backwards from the target molecule. Often there will be multiple variations of reactions and starting materials capable synthesizing the target molecule. Further analysis of these variations can draw out strengths and weaknesses and lead to the pathway with the best chance of success. To create a possible starting material for a given alkyl halide simply reverse the bonding changes expected during an electrophilic addition. Remove the C-X single bond and a C-H bond from an adjacent carbon. Then connect these two carbons with a double bond. What alkene would be required to make the following 2-bromopentane using an electrophilic addition? Possible starting materials can be made by first removing the C-Br bond and an adjacent C-H bond. Then connecting the two carbons with a double bond. Because there are two different locations where an adjacent C-H bond can be removed there will be two possible starting material. When comparing the the possible starting materials, #1 will follow Markovnikov's rule and produce 2-bromopentane as the sole product. Starting material #2 is symmertrically alkyl substituted, so Markovnikov's rule does not apply. A mixture of 2-bromopentane and 3-bromopentane would be produced. Between the two possible starting material #1 would be preferred because only one product is produced. Predict the product(s) for the following reactions: Give the IUPAC name for the product of the following reaction. Draw the reaction mechanism of the previous problem Identify the products of the following reactions. | 7,734 | 1,896 |
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Oxidative addition and reductive elimination are key steps in industrial catalysis. For example, both steps are featured in palladium-catalyzed cross-coupling reactions, the subject of the 2010 Nobel Prize in Chemistry. The prize was awarded to Richard Heck of the University of Delaware, Ei-Ichi Negishi of Purdue University and Akira Suzuki of Hokkaido University. With these reactions, workers in a variety of fields can make molecules that otherwise would be quite difficult to make. These molecules in turn may be important pharmaceuticals or useful compounds for electronic displays in computers and other devices, to name just a couple. Figure OA5.2. Catalytic cycle for the Negishi cross-coupling reaction. Oxidative addition is used to activate substrates. Substrates that normally might not react get ready to react with something. Then they sit on the metal atom and wait for something else to come along and react with them. Figure OA1.1. A generalized oxidative addition. Bonds can be broken via oxidative addition that cannot easily be broken via other reactions. For example, although the H-H bond is very strong, it can be cleaved in the presence of a variety of metal atoms and ions. Transition metals possess the right tools to coax the two hydrogen atoms apart from each other. Figure OA1.2. A generalized oxidative addition of dihydrogen. Figure OA1.3. A specific case of oxidative addition of dihydrogen. When hydrogen adds to Vaska's complex, the oxidation state of iridium changes from +1 to +3. Reductive elimination, in turn, is used to couple different groups together to form useful products. Once two groups are sitting beside each other on a transition metal atom or ion, they can bond to each other rather than the metal and go off together as a new molecule. Figure OA1.5. A general scheme for reductive elimination. Clearly, a reductive elimination is just an oxidative addition in reverse. The reaction can go in either direction. That means it can, conceivably, occur in equilibrium. At this level, you would not be expected to know which direction would be favoured for a particular reaction. However, you might be able to predict which direction a reaction would proceed based on factors such as le Chatelier's principle. For example, adding more reactant to the reaction shifts equilibrium to the right. More product is made, and some of the extra reactant is used up, so that the system can come back to its natural equilibrium. If products are somehow removed from the system, the reaction will also shift to the right, using up reactants and replacing the missing product. If the reaction is exothermic (produces heat) and more heat is added to the system, the reaction would shift to the left, using up some products and making more reactants in order to remove excess heat. The reversibility of oxidative addition / reductive elimination actually serves very well in catalytic processes. For example, one of the most important catalytic processes in the world is catalytic hydrogenation, in which two hydrogen atoms are added across a double bond (usually a C=C bond, but sometimes a C=O or C=N bond). The process requires oxidative addition of hydrogen to a metal, but it also requires reductive elimination of an alkyl and a hydride to form the final product, forming a hydrocarbon. Figure OA1.5. A reductive elimination of a methyl and a hydride to form methane. The addition of dihydrogen to Vaska's complex and other transition metals is a reversible reaction. The hydrogen can be released again if the reaction moves to the left in a reductive elimination. That reversibility makes transition metal compounds useful for hydrogen storage. Hydrogen gas is voluminous, flammable and generally dangerous. By cleaving H and binding hydrogen to metal atoms, hydrogen can be more safely stored and released again under the right conditions. In reductive elimination, bonds can be made that cannot be formed via other reactions. That makes it a useful part of strategies to make commodity chemicals and complex organic molecules such as pharmaceuticals. Based on le Chatelier's principle, propose conditions under which: Draw products of oxidative addition of the following compounds to (PPh ) Pd. a) HBr b) H c) I d) CH -Br , | 4,282 | 1,897 |
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The compression factor (\(Z\)) for O at 200 K is measured to have the following values: Using numerical integration, calculate the fugacity constant for O at 200 K from these data. The normal boiling point of ethanol is 78.4 C. Its enthalpy of vaporization is 38.6 kJ/mol. Estimate the vapor pressure of ethanol at 24.4 C. When 20.0 grams of an unknown nonelectrolyte compound are dissolved in 500.0 grams of benzene, the freezing point of the resulting solution is 3.77 °C. The freezing point of pure benzene is 5.444 °C and the cryoscopic constant (\(K_f\)) for benzene is 5.12 °C/m. What is the molar mass of the unknown compound? Consider a mixture of two volatile liquids, A and B. The vapor pressure of pure liquid A is 324.3 Torr and that of pure liquid B is 502.3 Torr. What is the total vapor pressure over a mixture of the two liquids for which x = 0.675? Consider the following expression for osmotic pressure \[\pi V = \chi_BRT\] where \(\pi\) is the osmotic pressure, \(V\) is the molar volume of the solvent, \(\chi_B\) is the mole fraction of the solute, \(R\) is the gas law constant, and \(T\) is the temperature (in Kelvin). The molar volume of a particular solvent is 0.0180 L/mol. 0.200 g of a solute (B) is dissolved in 1.00 mol of the solvent. The osmotic pressure of the solvent is then measured to be 0.640 atm at 298 K. Calculate the molar mass of the solute. At 300 K, the vapor pressure of HCl(g) over a solution of \(HCl\) in \(GeCl_4\) are summarized in the following table. Calculate the Henry’s Law constant for HCl based on these data. Consider the mixing of 1.00 mol of hexane (C H ) with 1.00 mole of benzene (C H ). Calculate \(\ H\), \(\ S\), and \(\ G\) of mixing, of the mixing occurs ideally at 298 K. | 1,766 | 1,898 |
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Ligand substitution refers to the replacement of one ligand in a coordination complex with another ligand. Figure LS1.1. Substitution of one ligand for another in a coordination complex. Remember, a ligand in coordination chemistry is just a Lewis base that binds to a metal atom or ion. It does so by donating a lone pair (or other pair of electrons). Generallly, this donation is reversible. The donor can always take its electrons back. Typically, there may be some balance between the metal's need for more electrons and the donor's attraction for its own electrons; donor atoms are frequently more electronegative than the metal. Figure LS1.2. An example of ligand substitution. THF replaces a carbon monoxide in this molybdenum complex. Even though the reaction is pretty simple, it can occur in different ways. That is, the elementary steps involved in the reaction can occur in different orders. The elementary reactions are the individual bond-making or bond-breaking events that lead to an overall change. Sometimes the order of steps is referred to as the mechanism or the mechanistic pathway. You may have seen reaction mechanisms before. For example, carbonyl addition chemistry can involve lengthy mechanisms, in which a number of proton transfers and other bond-making and bond-breaking steps must occur to get from one state to another. Because ligand substitution is simpler than that, it is a good place to study mechanism in a little more depth, without getting overwhelmed by the details. The sequence of steps in the mechanism influences how different factors will impact the reaction. For example, changing concentrations of different components in a reaction mixture can affect the time it takes for a reaction to finish. These kinds of considerations have a dramatic impact on industrial processes such as pharmaceutical production. In that setting, chemical engineers need to make decisions about how much of each reactant must be admitted to a reaction mixture and how long they should be allowed to react together. If they allow the reaction to proceed for too, long, there may be “side-reactions” that start to occur, interfering with the quality of the product, and they will waste valuable time in the production pipeline. If they don’t allow it to react long enough, the reaction may not finish, and the product will be contaminated with leftover starting materials. In this chapter, we will look at how this simple reaction can occur in different ways. We will see some different methods that are used to tell which way the reaction occurs (i.e. evidence of what is really happening). We will also look at some different factors that may influence whether the reaction is likely to occur one way or the other (i.e. reasons it is happening that way, or reasons we expect it will happen that way). Some kind of substitution occurs in each of the following reactions: an atom or group replaces another. In each case, identify what is being replaced, and what replaces it. , | 3,015 | 1,899 |
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Crystalline solids fall into one of four categories. All four categories involve packing discrete molecules or atoms into a lattice or repeating array, though are a special case. The categories are distinguished by the nature of the interactions holding the discrete molecules or atoms together. Based on the nature of the forces that hold the component atoms, molecules, or ions together, solids may be formally classified as ionic, molecular, covalent (network), or metallic. The variation in the relative strengths of these four types of interactions correlates nicely with their wide variation in properties. In ionic and molecular solids, there are no chemical bonds between the molecules, atoms, or ions. The solid consists of discrete chemical species held together by intermolecular forces that are or in nature. This behavior is most obvious for an ionic solid such as \(NaCl\), where the positively charged Na ions are attracted to the negatively charged \(Cl^-\) ions. Even in the absence of ions, however, electrostatic forces are operational. For polar molecules such as \(CH_2Cl_2\), the positively charged region of one molecular is attracted to the negatively charged region of another molecule ( ). For a nonpolar molecule such as \(CO_2\), which has no permanent dipole moment, the random motion of electrons gives rise to temporary polarity (a temporary dipole moment). Electrostatic attractions between two temporarily polarized molecules are called . is a term describing an attractive interaction between a hydrogen atom from a molecule or a molecular fragment X–H in which X is more electronegative than H, and an atom or a group of atoms in the same or a different molecule, in which there is evidence of bond formation. (See the Provisional Recommendation on the definition of a hydrogen bond.) Dots are employed to indicate the presence of a hydrogen bond: X–H•••Y. The attractive interaction in a hydrogen bond typically has a strong electrostatic contribution, but dispersion forces and weak covalent bonding are also present. In metallic solids and network solids, however, chemical bonds hold the individual chemical subunits together. The crystal is essential a single, macroscopic molecule with continuous chemical bonding throughout the entire structure. In metallic solids, the valence electrons are no longer exclusively associated with a single atom. Instead these electrons exist in molecular orbitals that are delocalized over many atoms, producing an electronic band structure. The metallic crystal essentially consists of a set of metal cations in a sea of electrons. This type of chemical bonding is called . You learned previously that an ionic solid consists of positively and negatively charged ions held together by electrostatic forces. The strength of the attractive forces depends on the charge and size of the ions that compose the lattice and determines many of the physical properties of the crystal. The (i.e., the energy required to separate 1 mol of a crystalline ionic solid into its component ions in the gas phase) is directly proportional to the product of the ionic charges and inversely proportional to the sum of the radii of the ions. For example, NaF and CaO both crystallize in the face-centered cubic (fcc) sodium chloride structure, and the sizes of their component ions are about the same: Na (102 pm) versus Ca (100 pm), and F (133 pm) versus O (140 pm). Because of the higher charge on the ions in CaO, however, the lattice energy of CaO is almost four times greater than that of NaF (3401 kJ/mol versus 923 kJ/mol). The forces that hold Ca and O together in CaO are much stronger than those that hold Na and F together in NaF, so the heat of fusion of CaO is almost twice that of NaF (59 kJ/mol versus 33.4 kJ/mol), and the melting point of CaO is 2927°C versus 996°C for NaF. In both cases, however, the values are large; that is, simple ionic compounds have high melting points and are relatively hard (and brittle) solids. Molecular solids consist of atoms or molecules held to each other by dipole–dipole interactions, London dispersion forces, or hydrogen bonds, or any combination of these. The arrangement of the molecules in solid benzene is as follows: Because the intermolecular interactions in a molecular solid are relatively weak compared with ionic and covalent bonds, molecular solids tend to be soft, low melting, and easily vaporized (\(ΔH_{fus}\) and \(ΔH_{vap}\) are low). For similar substances, the strength of the London dispersion forces increases smoothly with increasing molecular mass. For example, the melting points of benzene (C H ), naphthalene (C H ), and anthracene (C H ), with one, two, and three fused aromatic rings, are 5.5°C, 80.2°C, and 215°C, respectively. The enthalpies of fusion also increase smoothly within the series: benzene (9.95 kJ/mol) < naphthalene (19.1 kJ/mol) < anthracene (28.8 kJ/mol). If the molecules have shapes that cannot pack together efficiently in the crystal, however, then the melting points and the enthalpies of fusion tend to be unexpectedly low because the molecules are unable to arrange themselves to optimize intermolecular interactions. Thus toluene (C H CH ) and m-xylene [m-C H (CH ) ] have melting points of −95°C and −48°C, respectively, which are significantly lower than the melting point of the lighter but more symmetrical analog, benzene. Self-healing rubber is an example of a molecular solid with the potential for significant commercial applications. The material can stretch, but when snapped into pieces it can bond back together again through reestablishment of its hydrogen-bonding network without showing any sign of weakness. Among other applications, it is being studied for its use in adhesives and bicycle tires that will self-heal. Covalent solids are formed by networks or chains of atoms or molecules held together by covalent bonds. A perfect single crystal of a covalent solid is therefore a single giant molecule. For example, the structure of diamond, shown in part (a) in Figure \(\Page {1}\), consists of sp3 hybridized carbon atoms, each bonded to four other carbon atoms in a tetrahedral array to create a giant network. The carbon atoms form six-membered rings. The unit cell of diamond can be described as an fcc array of carbon atoms with four additional carbon atoms inserted into four of the tetrahedral holes. It thus has the zinc blende structure described in Section 12.3, except that in zinc blende the atoms that compose the fcc array are sulfur and the atoms in the tetrahedral holes are zinc. Elemental silicon has the same structure, as does silicon carbide (SiC), which has alternating C and Si atoms. The structure of crystalline quartz (SiO ), shown in Section 12.1, can be viewed as being derived from the structure of silicon by inserting an oxygen atom between each pair of silicon atoms. All compounds with the diamond and related structures are hard, high-melting-point solids that are not easily deformed. Instead, they tend to shatter when subjected to large stresses, and they usually do not conduct electricity very well. In fact, diamond (melting point = 3500°C at 63.5 atm) is one of the hardest substances known, and silicon carbide (melting point = 2986°C) is used commercially as an abrasive in sandpaper and grinding wheels. It is difficult to deform or melt these and related compounds because strong covalent (C–C or Si–Si) or polar covalent (Si–C or Si–O) bonds must be broken, which requires a large input of energy. Other covalent solids have very different structures. For example, graphite, the other common allotrope of carbon, has the structure shown in part (b) in Figure \(\Page {1}\). It contains planar networks of six-membered rings of sp2 hybridized carbon atoms in which each carbon is bonded to three others. This leaves a single electron in an unhybridized 2pz orbital that can be used to form C=C double bonds, resulting in a ring with alternating double and single bonds. Because of its resonance structures, the bonding in graphite is best viewed as consisting of a network of C–C single bonds with one-third of a π bond holding the carbons together, similar to the bonding in benzene. To completely describe the bonding in graphite, we need a molecular orbital approach similar to the one used for benzene in Chapter 9. In fact, the C–C distance in graphite (141.5 pm) is slightly longer than the distance in benzene (139.5 pm), consistent with a net carbon–carbon bond order of 1.33. In graphite, the two-dimensional planes of carbon atoms are stacked to form a three-dimensional solid; only London dispersion forces hold the layers together. As a result, graphite exhibits properties typical of both covalent and molecular solids. Due to strong covalent bonding within the layers, graphite has a very high melting point, as expected for a covalent solid (it actually sublimes at about 3915°C). It is also very soft; the layers can easily slide past one another because of the weak interlayer interactions. Consequently, graphite is used as a lubricant and as the “lead” in pencils; the friction between graphite and a piece of paper is sufficient to leave a thin layer of carbon on the paper. Graphite is unusual among covalent solids in that its electrical conductivity is very high parallel to the planes of carbon atoms because of delocalized C–C π bonding. Finally, graphite is black because it contains an immense number of alternating double bonds, which results in a very small energy difference between the individual molecular orbitals. Thus light of virtually all wavelengths is absorbed. Diamond, on the other hand, is colorless when pure because it has no delocalized electrons. Table \(\Page {2}\) compares the strengths of the intermolecular and intramolecular interactions for three covalent solids, showing the comparative weakness of the interlayer interactions. In network solids, conventional chemical bonds hold the chemical subunits together. The bonding between chemical subunits, however, is identical to that within the subunits, resulting in a continuous network of chemical bonds. One common examples of network solids are diamond (a form of pure carbon) Carbon exists as a pure element at room temperature in three different forms: graphite (the most stable form), diamond, and fullerene. The structure of diamond is shown at the right in a "ball-and-stick" format. The balls represent the carbon atoms and the sticks represent a covalent bond. Be aware that in the "ball-and-stick" representation the size of the balls do not accurately represent the size of carbon atoms. In addition, a single stick is drawn to represent a covalent bond irrespective of whether the bond is a single, double, or triple bond or requires resonance structures to represent. In the diamond structure, all bonds are single covalent bonds (\(\sigma\) bonds). The "space-filling" format is an alternate representation that displays atoms as spheres with a radius equal to the van der Waals radius, thus providing a better sense of the size of the atoms. Notice that diamond is a network solid. The entire solid is an "endless" repetition of carbon atoms bonded to each other by covalent bonds. (In the display at the right, the structure is truncated to fit in the display area.) The most stable form of carbon is graphite. Graphite consists of sheets of carbon atoms covalently bonded together. These sheets are then stacked to form graphite. Figure \(\Page {3}\) shows a ball-and-stick representation of graphite with sheets that extended "indefinitely" in the xy plane, but the structure has been truncated for display purposed. Graphite may also be regarded as a network solid, even though there is no bonding in the z direction. Each layer, however, is an "endless" bonded network of carbon atoms. Until the mid 1980's, pure carbon was thought to exist in two forms: graphite and diamond. The discovery of C molecules in interstellar dust in 1985 added a third form to this list. The existence of C , which resembles a soccer ball, had been hypothesized by theoreticians for many years. In the late 1980's synthetic methods were developed for the synthesis of C , and the ready availability of this form of carbon led to extensive research into its properties. The C molecule (Figure \(\Page {4}\); left), is called buckminsterfullerene, though the shorter name fullerene is often used. The name is a tribute to the American architect R. Buckminster Fuller, who is famous for designing and constructing geodesic domes which bear a close similarity to the structure of C . As is evident from the display, C is a sphere composed of six-member and five-member carbon rings. These balls are sometimes fondly referred to as "Bucky balls". It should be noted that fullerenes are an entire class of pure carbon compounds rather than a single compound. A distorted sphere containing more than 60 carbon atoms have also been found, and it is also possible to create long tubes (Figure \(\Page {4}\); right). All of these substances are pure carbon. such as crystals of copper, aluminum, and iron are formed by metal atoms Figure \(\Page {5}\). The structure of metallic crystals is often described as a uniform distribution of atomic nuclei within a “sea” of delocalized electrons. The atoms within such a metallic solid are held together by a unique force known as that gives rise to many useful and varied bulk properties. All exhibit high thermal and electrical conductivity, metallic luster, and malleability. Many are very hard and quite strong. Because of their malleability (the ability to deform under pressure or hammering), they do not shatter and, therefore, make useful construction materials. Metals are characterized by their ability to reflect light, called luster, their high electrical and thermal conductivity, their high heat capacity, and their malleability and ductility. Every lattice point in a pure metallic element is occupied by an atom of the same metal. The packing efficiency in metallic crystals tends to be high, so the resulting metallic solids are dense, with each atom having as many as 12 nearest neighbors. Bonding in metallic solids is quite different from the bonding in the other kinds of solids we have discussed. Because all the atoms are the same, there can be no ionic bonding, yet metals always contain too few electrons or valence orbitals to form covalent bonds with each of their neighbors. Instead, the valence electrons are delocalized throughout the crystal, providing a strong cohesive force that holds the metal atoms together. Valence electrons in a metallic solid are delocalized, providing a strong cohesive force that holds the atoms together. The strength of metallic bonds varies dramatically. For example, cesium melts at 28.4°C, and mercury is a liquid at room temperature, whereas tungsten melts at 3680°C. Metallic bonds tend to be weakest for elements that have nearly empty (as in Cs) or nearly full (Hg) valence subshells, and strongest for elements with approximately half-filled valence shells (as in W). As a result, the melting points of the metals increase to a maximum around group 6 and then decrease again from left to right across the d block. Other properties related to the strength of metallic bonds, such as enthalpies of fusion, boiling points, and hardness, have similar periodic trends. A somewhat oversimplified way to describe the bonding in a metallic crystal is to depict the crystal as consisting of positively charged nuclei in an electron sea (Figure \(\Page {6}\)). In this model, the valence electrons are not tightly bound to any one atom but are distributed uniformly throughout the structure. Very little energy is needed to remove electrons from a solid metal because they are not bound to a single nucleus. When an electrical potential is applied, the electrons can migrate through the solid toward the positive electrode, thus producing high electrical conductivity. The ease with which metals can be deformed under pressure is attributed to the ability of the metal ions to change positions within the electron sea without breaking any specific bonds. The transfer of energy through the solid by successive collisions between the metal ions also explains the high thermal conductivity of metals. This model does not, however, explain many of the other properties of metals, such as their metallic luster and the observed trends in bond strength as reflected in melting points or enthalpies of fusion. Some general properties of the four major classes of solids are summarized in Table \(\Page {2}\). The general order of increasing strength of interactions in a solid is: Classify Ge, RbI, C (CH ) , and Zn as ionic, molecular, covalent, or metallic solids and arrange them in order of increasing melting points. : compounds : classification and order of melting points : : Germanium lies in the p block just under Si, along the diagonal line of semimetallic elements, which suggests that elemental Ge is likely to have the same structure as Si (the diamond structure). Thus Ge is probably a covalent solid. RbI contains a metal from group 1 and a nonmetal from group 17, so it is an ionic solid containing Rb and I ions. The compound C6(CH3)6 is a hydrocarbon (hexamethylbenzene), which consists of isolated molecules that stack to form a molecular solid with no covalent bonds between them. Zn is a d-block element, so it is a metallic solid. Arranging these substances in order of increasing melting points is straightforward, with one exception. We expect C (CH ) to have the lowest melting point and Ge to have the highest melting point, with RbI somewhere in between. The melting points of metals, however, are difficult to predict based on the models presented thus far. Because Zn has a filled valence shell, it should not have a particularly high melting point, so a reasonable guess is C (CH ) < Zn ~ RbI < Ge. The actual melting points are C (CH ) , 166°C; Zn, 419°C; RbI, 642°C; and Ge, 938°C. This agrees with our prediction. Classify C , BaBr , GaAs, and AgZn as ionic, covalent, molecular, or metallic solids and then arrange them in order of increasing melting points. C (molecular) < AgZn (metallic) ~ BaBr (ionic) < GaAs (covalent). The actual melting points are C60, about 300°C; AgZn, about 700°C; BaBr , 856°C; and GaAs, 1238°C. The major types of solids are ionic, molecular, covalent, and metallic. Ionic solids consist of positively and negatively charged ions held together by electrostatic forces; the strength of the bonding is reflected in the lattice energy. Ionic solids tend to have high melting points and are rather hard. Molecular solids are held together by relatively weak forces, such as dipole–dipole interactions, hydrogen bonds, and London dispersion forces. As a result, they tend to be rather soft and have low melting points, which depend on their molecular structure. Covalent solids consist of two- or three-dimensional networks of atoms held together by covalent bonds; they tend to be very hard and have high melting points. Metallic solids have unusual properties: in addition to having high thermal and electrical conductivity and being malleable and ductile, they exhibit luster, a shiny surface that reflects light. An alloy is a mixture of metals that has bulk metallic properties different from those of its constituent elements. Alloys can be formed by substituting one metal atom for another of similar size in the lattice (substitutional alloys), by inserting smaller atoms into holes in the metal lattice (interstitial alloys), or by a combination of both. Although the elemental composition of most alloys can vary over wide ranges, certain metals combine in only fixed proportions to form intermetallic compounds with unique properties. ( ) ( ) ). | 19,893 | 1,900 |
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The common-ion effect is used to describe the effect on an equilibrium involving a substance that adds an ion that is a part of the equilibrium. The solubility products 's are equilibrium constants in heterogeneous 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: \[\mathrm{[Na^+] + [K^+] = [Cl^-]} \label{1} \] 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. \[\mathrm{NaCl \rightleftharpoons Na^+ + {\color{Green} Cl^-}} \nonumber \] \[\mathrm{KCl \rightleftharpoons K^+ + {\color{Green} Cl^-}} \nonumber \] \[\mathrm{CaCl_2 \rightleftharpoons Ca^{2+} + {\color{Green} 2 Cl^-}} \nonumber \] \[\mathrm{AlCl_3 \rightleftharpoons Al^{3+} + {\color{Green} 3 Cl^-}} \nonumber \] \[\mathrm{AgCl \rightleftharpoons Ag^+ + {\color{Green} Cl^-}} \nonumber \] 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 \[\mathrm{[Na^+] = [Ca^{2+}] = [H^+] = 0.10\: \ce M} \nonumber \] but \(\begin{alignat}{3}
\nonumber &\ce{[Cl- ]} &&= && && \:\textrm{0.10 (due to NaCl)}\\
\nonumber & && && + &&\mathrm{\:0.20\: (due\: to\: CaCl_2)}\\
\nonumber & && && + &&\mathrm{\:0.10\: (due\: to\: HCl)}\\
\nonumber & &&= && &&\mathrm{\:0.40\: M}
\nonumber \end{alignat}\) 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? \[\mathrm{[Cl^-] = \dfrac{0.1\: M\times 10\: mL+0.2\: M\times 5.0\: mL}{100.0\: mL} = 0.020\: M} \nonumber \] 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 PbCl . The balanced reaction is \[ PbCl_{2 (s)} \rightleftharpoons Pb^{2+} _{(aq)} + 2Cl^-_{(aq)} \nonumber \] 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 \(s\): \[\begin{eqnarray} K_{sp} &=& [Pb^{2+}] [Cl^-]^2 \\ &=& s \times (2s)^2 \\ 1.7 \times 10^{-5} &=& 4s^3 \\ s^3 &=& \frac{1.7 \times 10^{-5}}{4} \\ &=& 4.25 \times 10^{-6} \\ s &=& \sqrt[3]{4.25 \times 10^{-6}} \\ &=& 1.62 \times 10^{-2}\ mol\ dm^{-3} \end{eqnarray} \nonumber \]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 again: \[ PbCl_2 \; (s) \rightleftharpoons Pb^{2+} \; (aq) + 2Cl^- \; (aq) \nonumber \] What happens to that equilibrium if extra chloride ions are added? According to 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. \[[Pb^{2+}] = s \label{2} \] 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: \[[Cl^- ] = 0.100\; M \label{3} \] The rest of the mathematics looks like this: \begin{equation} \begin{split} K_{sp}& = [Pb^{2+},Cl^-]^2 \\ & = s \times (0.100)^2 \\ 1.7 \times 10^{-5} & = s \times 0.00100 \end{split} \nonumber \end{equation} therefore: \begin{equation} \begin{split} s & = \dfrac{1.7 \times 10^{-5}}{0.0100} \\ & = 1.7 \times 10^{-3} \, \text{M} \end{split} \label{4} \end{equation} Finally, compare that value with the simple saturated solution: Original solution: \[[Pb^{2+}] = 0.0162 \, M \label{5} \] Solution in 0.100 M NaCl solution: \[ [Pb^{2+}] = 0.0017 \, M \label{6} \] 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. A Discussing Finding the Solubility of a Salt: 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. Adding a common ion to a system at equilibrium affects the equilibrium composition, but not the ionization constant. The common ion effect of H O 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. Now 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 ), 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. \[Q_a = \frac{\ce{[NH_4^{+},OH^{-}]}}{\ce{[NH3]}} \nonumber \] 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. 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: \[\ce{Ca3(PO4)2(s) <=> 3Ca^{2+}(aq) + 2PO^{3−}4(aq)} \label{17.4.1} \] As you will discover 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 \[K=\dfrac{[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2}{[\mathrm{Ca_3(PO_4)_2}]} \label{17.4.2a} \] \[[\mathrm{Ca_3(PO_4)_2}]K=K_{\textrm{sp}}=[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2 \label{17.4.2b} \] At 25°C and pH 7.00, \(K_{sp}\) 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 vary dramatically for different compounds (Table E3). Although \(K_{sp}\) is not a function of pH in , changes in pH can affect the solubility of a compound. The solubility product expression tells us that the equilibrium concentrations of the cation and the anion are inversely related. That is, as the concentration of the anion increases, the maximum concentration of the cation needed for precipitation to occur decreases—and vice versa—so that \(K_{sp}\) is constant. This dependency is another example of the common ion effect where adding a common cation or anion shifts a solubility equilibrium in the direction predicted by Le Chatelier’s principle. As a result, the solubility of any sparingly soluble salt is almost always decreased by the presence of a soluble salt that contains a common ion. Consider, for example, the effect of adding a soluble salt, such as CaCl , to a saturated solution of calcium phosphate [Ca (PO ) ]. We have seen that the solubility of Ca (PO ) in water at 25°C is 1.14 × 10 M ( = 2.07 × 10 ). Thus a saturated solution of Ca (PO ) in water contains according to the stoichiometry shown in (neglecting hydrolysis to form HPO ). If CaCl is added to a saturated solution of Ca (PO ) , the Ca ion concentration will increase such that [Ca ] > 3.42 × 10 M, making > . The only way the system can return to equilibrium is for the reaction in to proceed to the left, resulting in precipitation of Ca (PO ) . This will decrease the concentration of both Ca and PO until = . 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: \[ PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq) \nonumber \] What happens to the solubility of PbCl (s) when 0.1 M NaCl is added? \[K_{sp}=1.7 \times 10^{-5} \nonumber \] \[Q_{sp}= 1.8 \times 10^{-5}\nonumber \] Identify the common ion: Cl Notice: Q > K The addition of 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 PbCl is equivalent to the concentration of Pb produced because they are in a 1:1 ratio. Because for the reaction is 1.7×10 , the overall reaction would be (s)(2s) = 1.7×10 . Solving the equation for s gives s= 1.62×10 M. The coefficient on Cl is 2, so it is assumed that twice as much Cl is produced as Pb , 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 Na and Cl are in a 1:1 ration in the ionic salt, NaCl. Therefore, the overall molarity of Cl would be 2s + 0.1, with 2s referring to the contribution of the chloride ion from the dissociation of lead chloride. \[\begin{eqnarray} Q_{sp} &=& [Pb^{2+},Cl^-]^2 \\ 1.8 \times 10^{-5} &=& (s)(2s + 0.1)^2 \\ s &=& [Pb^{2+}] \\ &=& 1.8 \times 10^{-3} M \\ 2s &=& [Cl^-] \\ &\approx & 0.1 M \end{eqnarray} \nonumber \] Notice that the molarity of Pb is lower when 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 [Cl ] is approximately 0.1M since the formation of the chloride ion from the dissociation of lead chloride is so small. The reaction quotient for PbCl is greater than the equilibrium constant because of the added 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. Calculate the solubility of silver carbonate in a 0.25 M solution of sodium carbonate. The solubility of silver carbonate in pure water is 8.45 × 10 at 25°C. 2.9 × 10 M (versus 1.3 × 10 M in pure water) A Discussing the Common Ion Effect in Solubility Products: | 13,214 | 1,901 |
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Artdej, Romklao; Thongpanchang, Tienthong. J. Chem. Educ. 2008, 85, 1382. Clyde R. Dillard; J. Chem. Educ., 1972, 49 (12), p A694. forums.jce.divched.org: Equal volumes of vinegar or other acid are added to a series of volumetric flasks, and increasing masses of NaHCO are added to balloons. The balloons are attached to the necks of the flasks, then lifted to dump the NaHCO into the vinegar. After the vinegar becomes limiting, the size of the CO -filled balloons no longer increases. Use bolts and a mismatched number of nuts (and washers) to model limiting reactants. Craig Blankenship, J. Chem. Educ., 1987, 64 (2), p 134. forums.jce.divched.org: The "nylon rope trick" can be used as a conceptual demonstration of limiting reagents, especially if different numbers of red and blue paper strips (of different lengths) are used to model the acid and amine. This can be related to amino acid nutrition and protein synthesis. Ed Vitz;J. Chem. Educ., 2005, 82 (7), p 1013l | 987 | 1,902 |
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In , you learned that in an acid–base titration, a buret is used to deliver measured volumes of an acid or a base solution of known concentration (the ) to a flask that contains a solution of a base or an acid, respectively, of unknown concentration (the ). If the concentration of the titrant is known, then the concentration of the unknown can be determined. The following discussion focuses on the pH changes that occur during an acid–base titration. Plotting the pH of the solution in the flask against the amount of acid or base added produces a titration curve . The shape of the curve provides important information about what is occurring in solution during the titration. Part (a) of shows a plot of the pH as 0.20 M HCl is gradually added to 50.00 mL of pure water. The pH of the sample in the flask is initially 7.00 (as expected for pure water), but it drops very rapidly as HCl is added. Eventually the pH becomes constant at 0.70—a point well beyond its value of 1.00 with the addition of 50.0 mL of HCl (0.70 is the pH of 0.20 M HCl). In contrast, when 0.20 M NaOH is added to 50.00 mL of distilled water, the pH (initially 7.00) climbs very rapidly at first but then more gradually, eventually approaching a limit of 13.30 (the pH of 0.20 M NaOH), again well beyond its value of 13.00 with the addition of 50.0 mL of NaOH as shown in part (b) in . As you can see from these plots, the titration curve for adding a base is the mirror image of the curve for adding an acid. Suppose that we now add 0.20 M NaOH to 50.0 mL of a 0.10 M solution of HCl. Because HCl is a strong acid that is completely ionized in water, the initial [H ] is 0.10 M, and the initial pH is 1.00. Adding NaOH decreases the concentration of H because of the neutralization reaction: (in part (a) in ). Thus the pH of the solution increases gradually. Near the equivalence point , however, the point at which the number of moles of base (or acid) added equals the number of moles of acid (or base) originally present in the solution, the pH increases much more rapidly because most of the H ions originally present have been consumed. (For more information on titrations and the equivalence point, see .) For the titration of a monoprotic strong acid (HCl) with a strong base (NaOH), we can calculate the volume of base needed to reach the equivalence point from the following relationship:
\[moles\;of \;base=(volume)_b(molarity)_bV_bM_b= moles \;of \;acid=(volume)_a(molarity)_a=V_aM_a \tag{16.5.1}\] If 0.20 M NaOH is added to 50.0 mL of a 0.10 M solution of HCl, we solve for : At the equivalence point (when 25.0 mL of NaOH solution has been added), the neutralization is complete: only a salt remains in solution (NaCl), and the pH of the solution is 7.00. Adding more NaOH produces a rapid increase in pH, but eventually the pH levels off at a value of about 13.30, the pH of 0.20 M NaOH. As shown in part (b) in , the titration of 50.0 mL of a 0.10 M solution of NaOH with 0.20 M HCl produces a titration curve that is nearly the mirror image of the titration curve in part (a) in . The pH is initially 13.00, and it slowly decreases as HCl is added. As the equivalence point is approached, the pH drops rapidly before leveling off at a value of about 0.70, the pH of 0.20 M HCl. The titration of either a strong acid with a strong base or a strong base with a strong acid produces an S-shaped curve. The curve is somewhat asymmetrical because the steady increase in the volume of the solution during the titration causes the solution to become more dilute. Due to the leveling effect, the shape of the curve for a titration involving a strong acid and a strong base depends on the concentrations of the acid and base, their identities. The shape of the titration curve involving a strong acid and a strong base depends only on their concentrations, not their identities. Calculate the pH of the solution after 24.90 mL of 0.200 M NaOH has been added to 50.00 mL of 0.100 M HCl. volumes and concentrations of strong base and acid pH Calculate the number of millimoles of H and OH to determine which, if either, is in excess after the neutralization reaction has occurred. If one species is in excess, calculate the amount that remains after the neutralization reaction. Determine the final volume of the solution. Calculate the concentration of the species in excess and convert this value to pH. Because 0.100 mol/L is equivalent to 0.100 mmol/mL, the number of millimoles of H in 50.00 mL of 0.100 M HCl can be calculated as follows: \[ 50.00 \cancel{mL} \left ( \dfrac{0.100 \;mmol \;HCl}{\cancel{mL}} \right )= 5.00 \;mmol \;HCl=5.00 \;mmol \;H^{+} \notag \] The number of millimoles of NaOH added is as follows: Thus H is in excess. To completely neutralize the acid requires the addition of 5.00 mmol of OH to the HCl solution. Because only 4.98 mmol of OH has been added, the amount of excess H is 5.00 mmol − 4.98 mmol = 0.02 mmol of H . The final volume of the solution is 50.00 mL + 24.90 mL = 74.90 mL, so the final concentration of H is as follows: The pH is −log[H ] = −log(3 × 10 ) = 3.5, which is significantly less than the pH of 7.00 for a neutral solution. Exercise Calculate the pH of a solution prepared by adding 40.00 mL of 0.237 M HCl to 75.00 mL of a 0.133 M solution of NaOH. 11.6 In contrast to strong acids and bases, the shape of the titration curve for a weak acid or a weak base depends dramatically on the identity of the acid or the base and the corresponding or . As we shall see, the pH also changes much more gradually around the equivalence point in the titration of a weak acid or a weak base. As you learned in , [H ] of a solution of a weak acid (HA) is equal to the concentration of the acid but depends on both its p and its concentration. Because only a fraction of a weak acid dissociates, [H ] is less than [HA]. Thus the pH of a solution of a weak acid is greater than the pH of a solution of a strong acid of the same concentration. Part (a) in shows the titration curve for 50.0 mL of a 0.100 M solution of acetic acid with 0.200 M NaOH superimposed on the curve for the titration of 0.100 M HCl shown in part (a) in . Below the equivalence point, the two curves are very different. Before any base is added, the pH of the acetic acid solution is greater than the pH of the HCl solution, and the pH changes more rapidly during the first part of the titration. Note also that the pH of the acetic acid solution at the equivalence point is greater than 7.00. That is, at the equivalence point, the solution is basic. In addition, the change in pH around the equivalence point is only about half as large as for the HCl titration; the magnitude of the pH change at the equivalence point depends on the p of the acid being titrated. Above the equivalence point, however, the two curves are identical. Once the acid has been neutralized, the pH of the solution is controlled only by the amount of excess NaOH present, regardless of whether the acid is weak or strong. The shape of the titration curve of a weak acid or weak base depends heavily on their identities and the or . The titration curve in part (a) in was created by calculating the starting pH of the acetic acid solution before any NaOH is added and then calculating the pH of the solution after adding increasing volumes of NaOH. The procedure is illustrated in the following subsection and Example 12 for three points on the titration curve, using the p of acetic acid (4.76 at 25°C; = 1.7 × 10 ). As explained , if we know or and the initial concentration of a weak acid or a weak base, we can calculate the pH of a solution of a weak acid or a weak base by setting up a table of initial concentrations, changes in concentrations, and final concentrations. In this situation, the initial concentration of acetic acid is 0.100 M. If we define as [H ] due to the dissociation of the acid, then the table of concentrations for the ionization of 0.100 M acetic acid is as follows: In this and all subsequent examples, we will ignore [H ] and [OH ] due to the autoionization of water when calculating the final concentration. However, you should use and to check that this assumption is justified. Inserting the expressions for the final concentrations into the equilibrium equation (and using approximations), \[K_a=\dfrac{[H^+,CH_3CO_2^-]}{[CH_3CO_2H]}=\dfrac{(x)(x)}{0.100 - x} \approx \dfrac{x^2}{0.100}=1.74 \times 10^{-5} \notag \] Solving this equation gives \(x = [H^+] = 1.32 \times 10^{-3}\; M\). Thus the pH of a 0.100 M solution of acetic acid is as follows: \[pH = −\log(1.32 \times 10^{-3}) = 2.879 \notag \] Now consider what happens when we add 5.00 mL of 0.200 M NaOH to 50.00 mL of 0.100 M CH CO H (part (a) in ). Because the neutralization reaction proceeds to completion, all of the OH ions added will react with the acetic acid to generate acetate ion and water:
\[ CH_3CO_2H_{(aq)} + OH^-_{(aq)} \rightarrow CH_3CO^-_{2\;(aq)} + H_2O_{(l)} \tag{16.5.2}\] All problems of this type must be solved in two steps: a stoichiometric calculation followed by an equilibrium calculation. In the first step, we use the stoichiometry of the neutralization reaction to calculate the of acid and conjugate base present in solution after the neutralization reaction has occurred. In the second step, we use the equilibrium equation ( ) to determine [H ] of the resulting solution. To determine the amount of acid and conjugate base in solution after the neutralization reaction, we calculate the amount of CH CO H in the original solution and the amount of OH in the NaOH solution that was added. The acetic acid solution contained The NaOH solution contained Comparing the amounts shows that CH CO H is in excess. Because OH reacts with CH CO H in a 1:1 stoichiometry, the amount of excess CH CO H is as follows: Each 1 mmol of OH reacts to produce 1 mmol of acetate ion, so the final amount of CH CO is 1.00 mmol. The stoichiometry of the reaction is summarized in the following table, which shows the numbers of moles of the various species, their concentrations. This table gives the initial amount of acetate and the final amount of OH ions as 0. Because an aqueous solution of acetic acid always contains at least a small amount of acetate ion in equilibrium with acetic acid, however, the initial acetate concentration is not actually 0. The value can be ignored in this calculation because the amount of CH CO in equilibrium is insignificant compared to the amount of OH added. Moreover, due to the autoionization of water, no aqueous solution can contain 0 mmol of OH , but the amount of OH due to the autoionization of water is insignificant compared to the amount of OH added. We use the initial amounts of the reactants to determine the stoichiometry of the reaction and defer a consideration of the equilibrium until the second half of the problem. To calculate [H ] at equilibrium following the addition of NaOH, we must first calculate [CH CO H] and [CH CO ] using the number of millimoles of each and the total volume of the solution at this point in the titration: Knowing the concentrations of acetic acid and acetate ion at equilibrium and for acetic acid (1.74 × 10 ), we can use to calculate [H ] at equilibrium: Calculating −log[H ] gives pH = −log(6.95 × 10 ) = 4.158. Comparing the titration curves for HCl and acetic acid in part (a) in , we see that adding the same amount (5.00 mL) of 0.200 M NaOH to 50 mL of a 0.100 M solution of both acids causes a much smaller pH change for HCl (from 1.00 to 1.14) than for acetic acid (2.88 to 4.16). This is consistent with the qualitative description of the shapes of the titration curves at the beginning of this section. In Example 16.5.2, we calculate another point for constructing the titration curve of acetic acid. What is the pH of the solution after 25.00 mL of 0.200 M NaOH is added to 50.00 mL of 0.100 M acetic acid? volume and molarity of base and acid pH Ignoring the spectator ion (Na ), the equation for this reaction is as follows: \[CH_3CO_2H_{ (aq)} + OH^-(aq) \rightarrow CH_3CO_2^-(aq) + H_2O(l) \notag \] The initial numbers of millimoles of OH and CH CO H are as follows: The number of millimoles of OH equals the number of millimoles of CH CO H, so neither species is present in excess. Because the number of millimoles of OH added corresponds to the number of millimoles of acetic acid in solution, this is the equivalence point. The results of the neutralization reaction can be summarized in tabular form. Because the product of the neutralization reaction is a weak base, we must consider the reaction of the weak base to calculate [H ] at equilibrium and thus the final pH of the solution. The initial concentration of acetate is obtained from the neutralization reaction: The equilibrium reaction of acetate with water is as follows: \[CH_3CO^-_{2(aq)}+H_2O_{(l)} \rightleftharpoons CH_3CO_2H_{(aq)}+OH^-_{(aq)} \notag \] The equilibrium constant for this reaction is = / , where is the acid ionization constant of acetic acid. We therefore define as [OH ] produced by the reaction of acetate with water. Here is the completed table of concentrations: Substituting the expressions for the final values from this table into , We can obtain by rearranging and substituting the known values: \[ K_{b}= \dfrac{K_w}{K_a} =\dfrac{1.01 \times 10^{-14}}{1.74 \times 10^{-5}} = 5.80 \times 10^{-10}=\dfrac{x^{2}}{0.0667} \notag \] Exercise Calculate the pH of a solution prepared by adding 45.0 mL of a 0.213 M HCl solution to 125.0 mL of a 0.150 M solution of ammonia. The p of ammonia is 4.75 at 25°C. 9.23 As shown in part (b) in , the titration curve for NH , a weak base, is the reverse of the titration curve for acetic acid. In particular, the pH at the equivalence point in the titration of a weak base is than 7.00 because the titration produces an acid. The identity of the weak acid or weak base being titrated strongly affects the shape of the titration curve. illustrates the shape of titration curves as a function of the p or the p . As the acid or the base being titrated becomes weaker (its p or p becomes larger), the pH change around the equivalence point decreases significantly. With very dilute solutions, the curve becomes so shallow that it can no longer be used to determine the equivalence point. One point in the titration of a weak acid or a weak base is particularly important: the midpoint of a titration is defined as the point at which exactly enough acid (or base) has been added to neutralize one-half of the acid (or the base) originally present and occurs halfway to the equivalence point. The midpoint is indicated in part (a) in and part (b) in for the two shallowest curves. By definition, at the midpoint of the titration of an acid, [HA] = [A ]. Recall from that the ionization constant for a weak acid is as follows: \[ K_{a}=\dfrac{\left [ H_{3}O^{+} \right ]\left [ A^{-} \right ]}{\left [ HA \right ]} \notag \] If [HA] = [A ], this reduces to = [H O ]. Taking the negative logarithm of both sides, \[ log \;K_{a}= -log \left [ H_{3}O^{+} \right ] \notag \] From the definitions of p and pH, we see that this is identical to
\[ pK_{a}= pH \tag{16.5.3} \notag \] Thus , as indicated in part (a) in for the weakest acid where we see that the midpoint for p = 10 occurs at pH = 10. Titration methods can therefore be used to determine both the concentration the p (or the p ) of a weak acid (or a weak base). The pH at the midpoint of the titration of a weak acid is equal to the p of the weak acid. When a strong base is added to a solution of a polyprotic acid, the neutralization reaction occurs in stages. The most acidic group is titrated first, followed by the next most acidic, and so forth. If the p values are separated by at least three p units, then the overall titration curve shows well-resolved “steps” corresponding to the titration of each proton. A titration of the triprotic acid H PO with NaOH is illustrated in and shows two well-defined steps: the first midpoint corresponds to p , and the second midpoint corresponds to p . Because HPO is such a weak acid, p has such a high value that the third step cannot be resolved using 0.100 M NaOH as the titrant. The titration curve for the reaction of a polyprotic base with a strong acid is the mirror image of the curve shown in The initial pH is high, but as acid is added, the pH decreases in steps if the successive p values are well separated. lists the ionization constants and p values for some common polyprotic acids and bases. Calculate the pH of a solution prepared by adding 55.0 mL of a 0.120 M NaOH solution to 100.0 mL of a 0.0510 M solution of oxalic acid (HO CCO H), a diprotic acid (abbreviated as H ox). Oxalic acid, the simplest dicarboxylic acid, is found in rhubarb and many other plants. Rhubarb leaves are toxic because they contain the calcium salt of the fully deprotonated form of oxalic acid, the oxalate ion ( O CCO , abbreviated ox ). volume and concentration of acid and base pH Calculate the initial millimoles of the acid and the base. Use a tabular format to determine the amounts of all the species in solution. Calculate the concentrations of all the species in the final solution. Use to determine [H ] and convert this value to pH. gives the p values of oxalic acid as 1.25 and 3.81. Again we proceed by determining the millimoles of acid and base initially present: \[ 100.00 \cancel{mL} \left ( \dfrac{0.510 \;mmol \;H_{2}ox}{\cancel{mL}} \right )= 5.10 \;mmol \;H_{2}ox \notag \] \[ 55.00 \cancel{mL} \left ( \dfrac{0.120 \;mmol \;NaOH}{\cancel{mL}} \right )= 6.60 \;mmol \;NaOH \notag \] The strongest acid (H ox) reacts with the base first. This leaves (6.60 − 5.10) = 1.50 mmol of OH to react with Hox , forming ox and H O. The reactions can be written as follows: \[ \underset{5.10\;mmol}{H_{2}ox}+\underset{6.60\;mmol}{OH^{-}} \rightarrow \underset{5.10\;mmol}{Hox^{-}}+ \underset{5.10\;mmol}{H_{2}O} \notag \] \[ \underset{5.10\;mmol}{Hox^{-}}+\underset{1.50\;mmol}{OH^{-}} \rightarrow \underset{1.50\;mmol}{ox^{2-}}+ \underset{1.50\;mmol}{H_{2}O} \notag \] In tabular form, The equilibrium between the weak acid (Hox ) and its conjugate base (ox ) in the final solution is determined by the magnitude of the second ionization constant, = 10 = 1.6 × 10 . To calculate the pH of the solution, we need to know [H ], which is determined using exactly the same method as in the acetic acid titration in Example 12: Thus the concentrations of Hox and ox are as follows: \[ \left [ Hox^{-} \right ] = \dfrac{3.60 \; mmol \; Hox^{-}}{155.0 \; mL} = 2.32 \times 10^{-2} \;M \notag \] \[ \left [ ox^{2-} \right ] = \dfrac{1.50 \; mmol \; ox^{2-}}{155.0 \; mL} = 9.68 \times 10^{-3} \;M \notag \] We can now calculate [H ] at equilibrium using the following equation: \[ K_{a2}= =\dfrac{\left [ ox^{2-} \right ]\left [ H^{+} \right ] }{\left [ Hox^{-} \right ]} \notag \] Rearranging this equation and substituting the values for the concentrations of Hox and ox , \[ \left [ H^{+} \right ] =\dfrac{K_{a2}\left [ Hox^{-} \right ]}{\left [ ox^{2-} \right ]} = \dfrac{\left ( 1.6\times 10^{-4} \right ) \left ( 2.32\times 10^{-2} \right )}{\left ( 9.68\times 10^{-3} \right )}=3.7\times 10^{-4} \; M \notag \] So \[ pH = -log\left [ H^{+} \right ]= -log\left ( 3.7 \times 10^{-4} \right )= 3.43 \notag \] This answer makes chemical sense because the pH is between the first and second p values of oxalic acid, as it must be. We added enough hydroxide ion to completely titrate the first, more acidic proton (which should give us a pH greater than p ), but we added only enough to titrate less than half of the second, less acidic proton, with p . If we had added exactly enough hydroxide to completely titrate the first proton plus half of the second, we would be at the midpoint of the second step in the titration, and the pH would be 3.81, equal to p . Exercise Piperazine is a diprotic base used to control intestinal parasites (“worms”) in pets and humans. A dog is given 500 mg (5.80 mmol) of piperazine (p = 4.27, p = 8.67). If the dog’s stomach initially contains 100 mL of 0.10 M HCl (pH = 1.00), calculate the pH of the stomach contents after ingestion of the piperazine. 4.9 ( ) In practice, most acid–base titrations are not monitored by recording the pH as a function of the amount of the strong acid or base solution used as the titrant. Instead, an acid–base indicator is often used that, if carefully selected, undergoes a dramatic color change at the pH corresponding to the equivalence point of the titration. Indicators are weak acids or bases that exhibit intense colors that vary with pH. The conjugate acid and conjugate base of a good indicator have very different colors so that they can be distinguished easily. Some indicators are colorless in the conjugate acid form but intensely colored when deprotonated (phenolphthalein, for example), which makes them particularly useful. We can describe the chemistry of indicators by the following general equation: \[ HIn\left ( aq \right ) \rightleftharpoons H^{+}\left ( aq \right ) + In^{-}\left ( aq \right ) \notag \] where the protonated form is designated by HIn and the conjugate base by In . The ionization constant for the deprotonation of indicator HIn is as follows:
\[ K_{In} =\dfrac{\left [ H^{+} \right ]\left [ In^{-} \right ]}{HIn} \tag{16.5.4}\] The p (its p ) determines the pH at which the indicator changes color. Many different substances can be used as indicators, depending on the particular reaction to be monitored. For example, red cabbage juice contains a mixture of colored substances that change from deep red at low pH to light blue at intermediate pH to yellow at high pH ( ). In all cases, though, a good indicator must have the following properties: Red cabbage juice contains a mixture of substances whose color depends on the pH. Each test tube contains a solution of red cabbage juice in water, but the pH of the solutions varies from pH = 2.0 (far left) to pH = 11.0 (far right). At pH = 7.0, the solution is blue. (From ) Synthetic indicators have been developed that meet these criteria and cover virtually the entire pH range. shows the approximate pH range over which some common indicators change color and their change in color. In addition, some indicators (such as thymol blue) are polyprotic acids or bases, which change color twice at widely separated pH values. It is important to be aware that an indicator does not change color abruptly at a particular pH value; instead, it actually undergoes a pH titration just like any other acid or base. As the concentration of HIn decreases and the concentration of In increases, the color of the solution slowly changes from the characteristic color of HIn to that of In . As we will see in , the [In ]/[HIn] ratio changes from 0.1 at a pH one unit p to 10 at a pH one unit p . Thus most indicators change color over a pH range of about two pH units. We have stated that a good indicator should have a p value that is close to the expected pH at the equivalence point. For a strong acid–strong base titration, the choice of the indicator is not especially critical due to the very large change in pH that occurs around the equivalence point. In contrast, using the wrong indicator for a titration of a weak acid or a weak base can result in relatively large errors, as illustrated in . This figure shows plots of pH versus volume of base added for the titration of 50.0 mL of a 0.100 M solution of a strong acid (HCl) and a weak acid (acetic acid) with 0.100 M NaOH. The pH ranges over which two common indicators (methyl red, p = 5.0, and phenolphthalein, p = 9.5) change color are also shown. The horizontal bars indicate the pH ranges over which both indicators change color cross the HCl titration curve, where it is almost vertical. Hence both indicators change color when essentially the same volume of NaOH has been added (about 50 mL), which corresponds to the equivalence point. In contrast, the titration of acetic acid will give very different results depending on whether methyl red or phenolphthalein is used as the indicator. Although the pH range over which phenolphthalein changes color is slightly greater than the pH at the equivalence point of the strong acid titration, the error will be negligible due to the slope of this portion of the titration curve. Just as with the HCl titration, the phenolphthalein indicator will turn pink when about 50 mL of NaOH has been added to the acetic acid solution. In contrast, methyl red begins to change from red to yellow around pH 5, which is near the of the acetic acid titration, not the equivalence point. Adding only about 25–30 mL of NaOH will therefore cause the methyl red indicator to change color, resulting in a huge error. In general, for titrations of strong acids with strong bases (and vice versa), any indicator with a p between about 4.0 and 10.0 will do. For the titration of a weak acid, however, the pH at the equivalence point is greater than 7.0, so an indicator such as phenolphthalein or thymol blue, with p > 7.0, should be used. Conversely, for the titration of a weak base, where the pH at the equivalence point is less than 7.0, an indicator such as methyl red or bromocresol blue, with p < 7.0, should be used. The existence of many different indicators with different colors and p values also provides a convenient way to estimate the pH of a solution without using an expensive electronic pH meter and a fragile pH electrode. Paper or plastic strips impregnated with combinations of indicators are used as “pH paper,” which allows you to estimate the pH of a solution by simply dipping a piece of pH paper into it and comparing the resulting color with the standards printed on the container (Figure 16.25). The shape of a , a plot of pH versus the amount of acid or base added, provides important information about what is occurring in solution during a titration. The shapes of titration curves for weak acids and bases depend dramatically on the identity of the compound. The of an acid–base titration is the point at which exactly enough acid or base has been added to react completely with the other component. The equivalence point in the titration of a strong acid or a strong base occurs at pH 7.0. In titrations of weak acids or weak bases, however, the pH at the equivalence point is greater or less than 7.0, respectively. The pH tends to change more slowly before the equivalence point is reached in titrations of weak acids and weak bases than in titrations of strong acids and strong bases. The pH at the , the point halfway on the titration curve to the equivalence point, is equal to the p of the weak acid or the p of the weak base. Thus titration methods can be used to determine both the concentration the p (or the p ) of a weak acid (or a weak base). are compounds that change color at a particular pH. They are typically weak acids or bases whose changes in color correspond to deprotonation or protonation of the indicator itself. Why is the portion of the titration curve that lies below the equivalence point of a solution of a weak acid displaced upward relative to the titration curve of a strong acid? How are the slopes of the curves different at the equivalence point? Why? Predict whether each solution will be neutral, basic, or acidic at the equivalence point of each titration. The p values of phenol red, bromophenol blue, and phenolphthalein are 7.4, 4.1, and 9.5, respectively. Which indicator is best suited for each acid–base titration? For the titration of any strong acid with any strong base, the pH at the equivalence point is 7.0. Why is this not usually the case in titrations of weak acids or weak bases? Why are the titration curves for a strong acid with a strong base and a weak acid with a strong base identical in shape above the equivalence points but not below? Describe what is occurring on a molecular level during the titration of a weak acid, such as acetic acid, with a strong base, such as NaOH, at the following points along the titration curve. Which of these points corresponds to pH = p ? On a molecular level, describe what is happening during the titration of a weak base, such as ammonia, with a strong acid, such as HCl, at the following points along the titration curve. Which of these points corresponds to pOH = p ? For the titration of a weak acid with a strong base, use the expression to show that pH = p at the midpoint of the titration. Chemical indicators can be used to monitor pH rapidly and inexpensively. Nevertheless, electronic methods are generally preferred. Why? Why does adding ammonium chloride to a solution of ammonia in water decrease the pH of the solution? Given the equilibrium system CH CO H(aq) ⇌ CH CO (aq) + H (aq), explain what happens to the position of the equilibrium and the pH in each case. Given the equilibrium system CH NH (aq) + H O(l) ⇌ CH NH (aq) + OH (aq), explain what happens to the position of the equilibrium and the pH in each case. Calculate the pH of each solution. What is the pH of a solution prepared by mixing 50.0 mL of 0.225 M HCl with 100.0 mL of a 0.184 M solution of NaOH? What volume of 0.50 M HCl is needed to completely neutralize 25.00 mL of 0.86 M NaOH? Calculate the final pH when each pair of solutions is mixed. Calculate the final pH when each pair of solutions is mixed. Calcium carbonate is a major contributor to the “hardness” of water. The amount of CaCO in a water sample can be determined by titrating the sample with an acid, such as HCl, which produces water and CO . Write a balanced chemical equation for this reaction. Generate a plot of solution pH versus volume of 0.100 M HCl added for the titration of a solution of 250 mg of CaCO in 200.0 mL of water with 0.100 M HCl; assume that the HCl solution is added in 5.00 mL increments. What volume of HCl corresponds to the equivalence point? For a titration of 50.0 mL of 0.288 M NaOH, you would like to prepare a 0.200 M HCl solution. The only HCl solution available to you, however, is 12.0 M. While titrating 50.0 mL of a 0.582 M solution of HCl with a solution labeled “0.500 M KOH,” you overshoot the endpoint. To correct the problem, you add 10.00 mL of the HCl solution to your flask and then carefully continue the titration. The total volume of titrant needed for neutralization is 71.9 mL. Complete the following table and generate a titration curve showing the pH versus volume of added base for the titration of 50.0 mL of 0.288 M HCl with 0.321 M NaOH. Clearly indicate the equivalence point. The following data were obtained while titrating 25.0 mL of 0.156 M NaOH with a solution labeled “0.202 M HCl.” Plot the pH versus volume of titrant added. Then determine the equivalence point from your graph and calculate the exact concentration of your HCl solution. Fill in the data for the titration of 50.0 mL of 0.241 M formic acid with 0.0982 M KOH. The p of formic acid is 3.75. What is the pH of the solution at the equivalence point? Glycine hydrochloride, which contains the fully protonated form of the amino acid glycine, has the following structure: It is a strong electrolyte that completely dissociates in water. Titration with base gives two equivalence points: the first corresponds to the deprotonation of the carboxylic acid group and the second to loss of the proton from the ammonium group. The corresponding equilibrium equations are as follows: \[ ^{+}NH_{3}-CH_{2}-CO_{2}H\left ( aq \right ) \rightleftharpoons \;\;\;\;\; pK_{a1}=2.3 \notag \] \[ ^{+}NH_{3}-CH_{2}-CO_{2}\left ( aq \right )+ H^{+} \] \[ ^{+}NH_{3}-CH_{2}-CO_{2}\left ( aq \right ) \rightleftharpoons \;\;\;\;\; pK_{a2}=9.6 \notag \] \[ NH_{2}-CH_{2}-COO\left ( aq \right )+ H^{+} \notag \] What is the pH of a solution prepared by adding 38.2 mL of 0.197 M HCl to 150.0 mL of 0.242 M pyridine? The p of pyridine is 8.77. What is the pH of a solution prepared by adding 40.3 mL of 0.289 M NaOH to 150.0 mL of 0.564 M succinic acid (HO CCH CH CO H)? (For succinic acid, p = 4.21 and p = 5.64). Calculate the pH of a 0.15 M solution of malonic acid (HO CCH CO H), whose p values are as follows: p = 2.85 and p = 5.70. 43 mL pH at equivalence point = 8.28 1.85 | 32,473 | 1,903 |
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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, coherent, 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. 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). The reaction is as follows: \[ Cu_2O (s) + H_2 (g) \rightarrow 2Cu (s) + H_2O (g) \label{3.4.1}\] Oxidation and reduction reactions are now defined as reactions that exhibit a change in the oxidation states of one or more elements in the reactants, which follows the mnemonic oxidation is loss reduction is gain, or oil rig. The oxidation state 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 \[ 4 Al (s) + 3O_2 \rightarrow 2Al_2O_3 (s) \label{3.4.2} \] 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 superscript 0 emphasizes the oxidation state of the elements): \[ 4 Al^0 + 3 O_2^0 \rightarrow 4 Al^{3+} + 6 O^{2-} \label{3.4.3}\] 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: \[ electrons \, lost = 4 \, Al \, atoms \times {3 \, e^- \, lost \over Al \, atom } = 12 \, e^- \, lost \label{3.4.4a}\] \[ electrons \, gained = 6 \, O \, atoms \times {2 \, e^- \, gained \over O \, atom} = 12 \, e^- \, gained \label{3.4.4a}\] 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 oxygen in air, 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. Previosuly, 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. 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. . 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 3.4.5 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: \[ \overset {+1}{Cu_2} \underset {-2}{O} (s) + \overset {0}{H_2} (g) \rightarrow 2 \overset {0}{Cu} (s) + \overset {+1}{H_2} \underset {-2}{O} (g) \label{3.4.5} \] Assigning oxidation states allows us to see that there has been a net transfer of electrons from hydrogen (0 → +1) to copper (+1 → 0). So 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: \[ electrons \, lost = 2 \, H \, atoms \times {1 \, e^- \, lost \over H \, atom } = 2 \, e^- \, lost \label{3.4.6a}\] \[ electrons \, gained = 2 \, Cu \, atoms \times {1 \, e^- \, gained \over Cu \, atom} = 2 \, e^- \, gained \label{3.4.6b}\] 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 \[ \underset {-3}{C} \overset {+1}{H_3} \overset {+3}{C} \underset {-2}{O_2} \overset {+1}{H} \] Thus the sum of the oxidation states of the two carbon atoms is indeed zero. Assign oxidation states to all atoms in each compound. : A widely encountered class of oxidation–reduction reactions is the reaction of aqueous solutions of acids or metal salts with solid metals. An example is the corrosion of metal objects, such as the rusting of an automobile (Figure \(\Page {1}\)). Rust is formed from a complex oxidation–reduction reaction involving dilute acid solutions that contain Cl ions (effectively, dilute HCl), iron metal, and oxygen. When an object rusts, iron metal reacts with HCl(aq) to produce iron(II) chloride and hydrogen gas: \(Fe(s) + 2HCl(aq) \rightarrow FeCl_2(aq) + H_2(g) \label{3.4.5}\) In subsequent steps, FeCl undergoes oxidation to form a reddish-brown precipitate of Fe(OH) . Many metals dissolve through reactions of this type, which have the general form \[metal + acid \rightarrow salt + hydrogen \label{3.4.6}\] 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: \[ Pb(s) + 2H^+(aq) \rightarrow Pb^{2+}(aq) + H_2(g) \label{3.4.7}\] 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 3.4.8) and the reduction of silver salts by copper (Equation 3.4.9 and Figure \(\Page {2}\)): \[ Zn(s) + Fe^{2+}(aq) \rightarrow Zn^{2+}(aq) + Fe(s) \label{3.4.8}\] \[ Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s) \label{3.4.9}\] The reaction in Equation 3.4.8 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 Zn . 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: \[ Zn(s) + Mg^{2+}(aq) \cancel{\rightarrow} Zn^{2+}(aq) + Mg(s) \label{3.4.10}\] \[ Mg(s) + Zn^{2+}(aq) \rightarrow Mg^{2+}(aq) + Zn(s) \label{3.4.11}\] 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 , , and . 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 \[ Al(s) + 3Ag^+(aq) \rightarrow Al^{3+}(aq) + 3Ag(s) \] 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. \[ Pb(s) + 2H^+(aq) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + H_2(g) \] 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. 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. 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 \(\Page {3}\)), 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. | 17,833 | 1,904 |
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Standard molar entropy (\(S_m^o\)) is one of the most useful types of thermodynamic data for a substance. But how does one measure this property? Unfortunately, there is no instrument that directly measures entropy. Instead, values are determined indirectly through the measurement of constant pressure heat capacities (\(C_p\)).
To understand how this works, we start with the Clausius Inequality \[ dS \ge \dfrac{đq}{T} \label{1}\] where the equality in Equation \ref{1} only applies to a reversible process. The slow heating of a substance at constant pressure so that the sample is always in quasi-static thermal equilibrium with the surroundings is approximately reversible. In such a case, Equation \ref{1} becomes \[ dS = \dfrac{C_p dT}{T} \label{2}\] where we have rewritten the differential heat transfer in terms of the constant pressure heat capacity times the differential temperature change. Imagine slowly heating a pure solid substance at constant pressure from absolute zero up to some arbitrary temperature, \(T\). Integration of Equation \ref{2} over the full temperature range will yield the absolute entropy of the substance relative to absolute zero, \[ S(T) = S(T=0) + \int_0^T \dfrac{Cp}{T} dt \label{3}\] The 3rd Law of Thermodynamics defines entropy to be zero for any pure crystalline solid at absolute zero, so the first term can be dropped from Equation \ref{3}. Whenever a substance in thermal equilibrium with its surroundings undergoes a phase transition at constant pressure, the corresponding entropy change given by \[ \Delta S_{trans} = \dfrac{\Delta H_{trans}}{T_{trans}} \label{4}\] If any phase transitions occur as we are heating a substance from absolute zero, each will contribute to the overall absolute entropy of the substance. A refinement of Equation \ref{3} that accounts for melting and vaporization is given by \[ S(T ) = \int_0^{T_{fus}} \dfrac{C_p(s)}{T} dt + \dfrac{\Delta H_{fus}}{T_{fus}} + \int_{T_{fus}}^{T_{vap}} \dfrac{C_p(l)}{T} dt + \dfrac{\Delta H_{vap}}{T_{vap}} + \int_{T_{vap}}^T \dfrac{C_p(g)}{T} dt \label{5}\] Equation \ref{5} provides a means of determining the standard molar entropy of most substances. The terms corresponding to heating the solid, liquid, and gas phases, described by the 1st, 3rd, and 5th terms on the right-hand-side of Equation \(5\), respectively, can be evaluated experimentally by measuring the heat capacity of the three phases over the appropriate temperature ranges, plotting these data as \(C_p/T\) verses \(T\), and measuring the area under the curve. The laboratory measurement of heat capacities is normally accomplished through the use of a constant pressure calorimeter that can be operated over a wide temperature range (nearly down to absolute zero). While these calorimeters are often custom designed, one can use a commercial instrument called a Differential Scanning Calorimeter (DSC). This device contains two sample compartments; a substance that has a well characterized heat capacity (a known) is placed in one compartment and the unknown is placed in the other. Starting at the low temperature limit, the known and unknown samples are heated using separate electrical heaters. Since the two samples likely have different heat capacities, more (or less) heat may be required by one sample in comparison to the other to achieve the same final temperature. The instrument measures this differential electrical heat at several points over the full temperature range. This ‘differential heat curve’ is subsequently analyzed to determine the heat capacity of the unknown substance over the same temperature range. Any phase transitions that occur as the sample is heated will yield peaks in the differential heat curve. The area under the peak is directly proportional to \(ΔH\) for the corresponding phase transition. In this fashion, each of the terms on the right-hand-side of Equation \ref{5} can be measured piecemeal. Designing a calorimeter that can measure heat capacities near absolute zero is problematic. A typical commercial DSC used in conjunction with a Helium cold-cycle refrigerator might have a low temperature limit of 15 Kelvin. Consequently, the entropy contribution for heating the sample from \(T = 0\) to \(15\, K\) has to be accounted for in some other fashion. Theoretical arguments have shown that the heat capacities of most substances at low temperatures are well described by the expression \[ C_p = a T^3 \label{6}\] where \(a\) is an empirical constant. Using Equation \ref{6} to describe the heat capacity of a substance at low temperatures is known as the 'Debye extrapolation.' If the constant \(a\) is known for a substance, then the absolute entropy at low temperatures is given by \[ S (T_{low}) = \int_0^{T_{low}} \dfrac{aT^3}{T} dT = \dfrac{aT_{low}^3}{3} \label{7}\] In this exercise, you will determine the standard molar entropy of solid aluminum oxide (\(\ce{Al2O3}\)) by analyzing a set of \(C_p\) versus \(T\) data and also applying the Debye extrapolation (Equation \ref{6}). on the following link and save the x-y data to an appropriate folder. The data was obtained from Furukawa, G.T., Douglas, T.B., McCoskey, R.E., and Ginnings, D.C. Journal of Research of the National Bureau of Standards, 57, 67-82 (1956). The first column of numbers is temperature in degrees Kelvin and the second column of numbers is standard \(C_{p,m}\) in units of J/(K mol). The data ranges from 15 K up to 300 K.
\(C_{p,m}\) verses T for \(\ce{Al2O3}\) | 5,518 | 1,905 |
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Sometimes, elimination reactions may lead to multiple stereoisomers; that is, they could lead to either the or the isomer, or in more complicated structures, either the or the isomer. Of course, if there were some inherent stability difference between these isomers, that could be a factor that plays a role in influencing the outcome. Elimination reactions aren't generally reversible, so products are not directly determined by alkene isomer stabilities. Nevertheless, sometimes the barrier leading to a more stable product is a little lower than the barrier leading to a less stable product. We do know that in simple cases, the isomer is generally lower in energy because of fewer steric interactions between the substituents on the double bond. In the absence of other information, we could take that as a starting point. Let's see whether elimination reactions generally lead to isomers. It turns out that sometimes this is true: eliminations often lead to the more stable product. Sometimes it isn't true, though. The answer depends on the mechanism. Instead, in an E2 reaction, stereochemistry of the double bond -- that is, whether the or isomer results -- is dictated by the stereochemistry of the starting material, if it is diastereomeric. In other words, if the carbon with the hydrogen and the carbon with the halogen are both chiral, then one diastereomer will lead to one product, and the other diastereomer will lead to the other product. The following reactions of potassium ethoxide with dibromostilbene (1,2-dibromo-1,2-diphenylethane) both occurred via an E2 mechanism. Two different diastereomers were used. Two different stereoisomers ( vs. ) resulted. Provide the stereochemical configurations of the following compounds from the above reactions: In E2 eliminations, the spatial relationship between the proton and leaving group determines the product stereochemistry. That's because pi bond formation happens at the same time that the halide leaves and at the same time that the base removes the proton. All of these events have to be coordinated together. The central, tricky event is the pi bond formation. The leaving group can leave in any direction, and the base can approach from many directions, but unless the pi bond is ready to form, nothing else happens. Let's slow the reaction down and imagine it takes place in slightly different stages. As the leaving group leaves, it takes its electrons with it. It begins to leave a positive charge behind. That positive charge will be centered on the carbon from which the halide is departing. That carbocation, if it fully formed, would have only three neighbours to bond with. It would be trigonal planar. It would have an unoccupied, non-bonding p orbital. As the base takes the proton, the hydrogen leaves behind the electrons from the C-H bond that held it in place. These electrons stay behind on the carbon atom. They are left in a non-bonding carbon valence orbital, a p orbital or something quite like it. Now we have a filled p orbital next to an empty p orbital. They overlap to form a pi bond. Of course, in an E2 reaction, things don't happen in stages. Everything happens at once. That means that, as the base removes the proton, the pi bond must already start forming. Because a pi bond requires parallel alignment of two p orbitals, and the p orbitals are forming from the C-H and C-LGp bonds, then those bonds must line up in order for the elimination to occur. So let's look again at that dibromostilbene example. In the first case, we need to spin the molecule so that we can see how the H on one carbon and the Br on the other are aligned and ready to eliminate via an E2 reaction. The substituents coming towards us in the reactant will still be coming towards us in the product. The substituents pointing away from us in the reactant will still be pointing away from us in the product. So the relationships between the substituents on the nascent double bond are determined by their relationship once the reactant is aligned for the E2 reaction. In the second case, we can spin the molecule but quickly realize the C-H and C-Br bonds are not lined up in this conformer. We need a bond rotation. Once we have made a conformational change, the C-H and C-Br bonds line up. It doesn't matter if this conformer is not favoured; if there is going to be any E2 reaction at all, this is the conformer it will have to go through. Again, the relationships between on the new double bond are determined by their relationship once the reactant is aligned for the E2 reaction. Sometimes it is easier to see the relationships between substituents by using a Newman projection. Draw Newman projections showing how the two isomers above proceed to different products in an E2 reaction. Predict the product of each of the following E2 reactions. Note that the compounds differ in the incorporation of a H isotope (deuterium, or D) in place of a regular H isotope (protium, or H). Conformational analysis of cyclohexanes requires the use of diamond lattice projections ("chairs"). Show why elimination can't proceed through the E2 mechanism in the following compound. Predict the E2 elimination products from the following compounds. In contrast to E2 reactions, E1 reactions do not occur in one step. That means there is time for reorganization in the intermediate. Once the leaving group leaves, the cation can sort itself into the most stable conformer. When the proton is taken, generally the most stable stereoisomer results because it comes from the most stable conformer of the cation. Any steric interactions in the alkene would also have occurred in the cation, so this interaction would have been sorted out at that point. Thus, in the case of the dibromostilbenes examined before, E1 elimination would result in the same product in either case. , | 5,872 | 1,906 |
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It is often required to know thermodynamic functions (such as enthalpy) at temperatures other than those available from tabulated data. Fortunately, the conversion to other temperatures is not difficult. At constant pressure \[ dH = C_p \,dT \nonumber \] And so for a temperature change from \(T_1\) to \(T_2\) \[ \Delta H = \int_{T_2}^{T_2} C_p\, dT \label{EQ1} \] Equation \ref{EQ1} is often referred to as . If \(C_p\) is independent of temperature, then \[\Delta H = C_p \,\Delta T \label{intH} \] If the temperature dependence of the heat capacity is known, it can be incorporated into the integral in Equation \ref{EQ1}. A common model used to fit heat capacities over broad temperature ranges is \[C_p(T) = a+ bT + \dfrac{c}{T^2} \label{EQ15} \] After combining Equations \ref{EQ15} and \ref{EQ1}, the enthalpy change for the temperature change can be found obtained by a simple integration \[ \Delta H = \int_{T_1}^{T_2} \left(a+ bT + \dfrac{c}{T^2} \right) dT \label{EQ2} \] Solving the definite integral yields \[ \begin{align} \Delta H &= \left[ aT + \dfrac{b}{2} T^2 - \dfrac{c}{T} \right]_{T_1}^{T_2} \\ &= a(T_2-T_1) + \dfrac{b}{2}(T_2^2-T_1^2) - c \left( \dfrac{1}{T_2} - \dfrac{1}{T_1} \right) \label{ineq} \end{align} \] This expression can then be used with experimentally determined values of \(a\), \(b\), and \(c\), some of which are shown in the following table. What is the molar enthalpy change for a temperature increase from 273 K to 353 K for Pb(s)? The enthalpy change is given by Equation \ref{EQ1} with a temperature dependence \(C_p\) given by Equation \ref{EQ1} using the parameters in Table \(\Page {1}\). This results in the integral form (Equation \ref{ineq}): \[ \Delta H = a(T_2-T_1) + \dfrac{b}{2}(T_2^2-T_1^2) - c \left( \dfrac{1}{T_2} - \dfrac{1}{T_1} \right) \nonumber \] when substituted with the relevant parameters of Pb(s) from Table \(\Page {1}\). \[ \begin{align*} \Delta H = \,& (22.14\, \dfrac{J}{mol\,K} ( 353\,K - 273\,K) \\ & + \dfrac{1.172 \times 10^{-2} \frac{J}{mol\,K^2}}{2} \left( (353\,K)^2 - (273\,K)^2 \right) \\ &- 9.6 \times 10^4 \dfrac{J\,K}{mol} \left( \dfrac{1}{(353\,K)} - \dfrac{1}{(273\,K)} \right) \\ \Delta H = \, & 1770.4 \, \dfrac{J}{mol}+ 295.5\, \dfrac{J}{mol}+ 470.5 \, \dfrac{J}{mol} \\ = & 2534.4 \,\dfrac{J}{mol} \end {align*} \] For chemical reactions, the reaction enthalpy at differing temperatures can be calculated from \[\Delta H_{rxn}(T_2) = \Delta H_{rxn}(T_1) + \int_{T_1}^{T_2} \Delta C_p \Delta T \nonumber \] The enthalpy of formation of NH (g) is -46.11 kJ/mol at 25 C. Calculate the enthalpy of formation at 100 C. \[\ce{N2(g) + 3 H2(g) \rightleftharpoons 2 NH3(g)} \nonumber \] with \(\Delta H \,(298\, K) = -46.11\, kJ/mol\) \[ \begin{align*} \Delta H (373\,K) & = \Delta H (298\,K) + \Delta C_p\Delta T \\ & = -46110 +\dfrac{J}{mol} \left[ 2 \left(35.06 \dfrac{J}{mol\,K}\right) - \left(29.12\, \dfrac{J}{mol\,K}\right) - 3\left(28.82\, \dfrac{J}{mol\,K}\right) \right] (373\,K -298\,K) \\ & = -49.5\, \dfrac{kJ}{mol} \end{align*} \] | 3,059 | 1,907 |
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Metal-containing monooxygenase enzymes are known that contain heme iron, nonheme iron, or copper at their active sites. For most of these enzymes, there is only limited information about the nature of the active site and the mode of interaction with dioxygen or substrates. But there are three monooxygenase enzymes that strongly resemble well-characterized reversible dioxygen-carrying proteins (see preceding chapter), suggesting that dioxygen binding to the metalloenzyme in its reduced state is an essential first step in the enzymatic mechanisms, presumably followed by other steps that result in oxygenation of substrates. The enzymes are: In addition to these three, there are also monooxygenase enzymes containing single nonheme iron or copper ions, or nonheme iron plus an organic cofactor such as a reduced pterin at their active sites. Just as with the dioxygenase enzymes, we do not know how similar the mechanisms of the different metal-containing monooxygenase enzymes are to one another. The enzyme for which we have the most information is cytochrome P-450, and we will therefore focus our discussion on that system. Speculations about the mechanisms for the other systems are discussed at the end of this section. Cytochrome P-450 enzymes are a group of monooxygenase enzymes that oxygenate a wide variety of substrates. Examples of such reactions are: \(\tag{5.59}\) \(\tag{5.60}\) \(\tag{5.61}\) \(\tag{5.62}\) \(\tag{5.63}\) \[Ph—O-CH_{3} \longrightarrow Ph—OH + HCHO \tag{5.64}\] Some of these reactions have great physiological significance, because they represent key transformations in metabolism, as in lipid metabolism and biosynthesis of corticosteroids, for example. Cytochrome P-450 is also known to catalyze the transformation of certain precarcinogens such as benzpyrene into their carcinogenic forms. Many of the P-450 enzymes have been difficult to characterize, because they are membrane-bound and consequently relatively insoluble in aqueous solution. However, cytochrome P-450 , which is a component of the camphor 5-monooxygenase system isolated from the bacterium , is soluble and has been particularly useful as the subject of numerous spectroscopic and mechanistic studies, as well as several x-ray crystallographic structure determinations. This enzyme consists of a single polypeptide chain, mainly \(\alpha\)-helical, with a heme b group (Fe-protoporphyrin IX) sandwiched in between two helices, with no covalent attachments between the porphyrin ring and the protein. One axial ligand complexed to iron is a cysteinyl thiolate. In the resting state, the iron is predominantly low-spin Fe , probably with a water as the other axial ligand. When substrate binds to the resting enzyme, the spin state changes to high-spin, and the non-cysteine axial ligand is displaced. The enzyme can be reduced to an Fe state, which is high-spin, and resembles deoxyhemoglobin or myoglobin in many of its spectroscopic properties. This ferrous form binds dioxygen to make an oxy form or carbon monoxide to make a carbonyl form. The CO derivative has a Soret band (high-energy \(\pi\)-\(\pi\)* transition of the porphyrin ring) at 450 nm, unusually low energy for a carbonyl derivative of a heme protein because of the presence of the axial thiolate ligand. This spectroscopic feature aids in the isolation of the enzyme and is responsible for its name. Camphor 5-monooxygenase is a three-component system, consisting of cytochrome P-450 and two electron-transfer proteins, a flavoprotein, and an iron-sulfur protein (see Chapters 6 and 7). The role of the electron-transfer proteins is to deliver electrons to the P-450 enzyme, but these may be replaced by other reducing agents. The reaction sequence is in Figure 5.10. For cytochrome P-450, the question that is possibly of greatest current interest to the bioinorganic chemist is just what mechanism enables activation of dioxygen and its reaction with substrate. It seems clear that dioxygen binds to the ferrous state of the enzyme-substrate complex, and that the resulting oxy ligand, which presumably is similar to the oxy ligand in oxyhemoglobin and oxymyoglobin, is not sufficiently reactive to attack the bound substrate. The oxy form is then reduced and the active oxidant is generated, but the nature of the active oxidant has not been deduced from studies of the enzyme itself, nor has it been possible to observe and characterize intermediates that occur between the time of the reduction and the release of product. Three species are potential candidates for "active oxygen," the oxygen-containing species that attacks the substrate, in cytochrome P-450. They are: The hydroxyl radical, HO•, although highly reactive and capable of attacking P-450 substrates, is considered to be an unlikely candidate for "active oxygen" because of the indiscriminate character of its reactivity. \[Fe^{II}P + O_{2} \rightarrow FePO_{2} \xrightarrow{e^{-}} [Fe^{III}P(O_{2}^{2-})]^{-} \xrightarrow{H^{+}} Fe^{III}P(O_{2}H^{-}) \tag{5.65}\] \[\qquad \qquad \qquad \qquad \qquad \bf{1a} \qquad \qquad \qquad \qquad \bf{1b}\] \[Fe^{III}P(O_{2}H^{-}) \rightarrow Fe^{IV}P(O) + HO \cdotp \tag{5.66}\] \[\bf{1b} \qquad \qquad \qquad \bf{2}\] \[Fe^{III}P(O_{2}H^{-}) \rightarrow [Fe^{V}(P^{2-})(O)^{+} \leftrightarrow Fe^{IV}(P^{-})(O)^{+}] +HO^{-} \tag{5.67}\] \[\bf{1b} \qquad \qquad \qquad\qquad \qquad \qquad \bf{3} \qquad \qquad \qquad \] (P = porphyrin ligand; P = one-electron oxidized porphyrin ligand) An iron(V) oxo complex (or a related species at the same oxidation level), , formed via Reaction (5.67), is the favored candidate for "active oxygen" in cytochrome P-450. This conclusion was initially drawn from studies of reactions of the enzyme with alkylhydroperoxides and single-oxygen-atom donors. Single-oxygen-atom donors are reagents such as iodosylbenzene, OIPh, and periodate, IO ,capable of donating a neutral oxygen atom to an acceptor, forming a stable product in the process (here, iodobenzene, IPh, and iodate, IO ). It was discovered that ferric cytochrome P-450 could catalyze oxygenation reactions using organic peroxides or single-oxygen-atom donors in place of dioxygen and reducing agents. Usually the same substrates would give the identical oxygenated product. This reaction pathway was referred to as the "peroxide shunt" (see Figure 5.10). The implication of this discovery was that the same form of "active oxygen" was generated in each reaction, and the fact that single-oxygen-atom donors could drive this reaction implied that this species contained only one oxygen atom, i.e., was generated subsequent to O—O bond cleavage. The mechanism suggested for this reaction was Reactions (5.68) and (5.69). \[Fe(III)P^{+} + OX \rightarrow \textbf{3} + X \tag{5.68}\] \[\textbf{3} + substrate \rightarrow Fe(III)P^{+} + substrate(O) \tag{5.69}\] Studies of the reactivities of synthetic metalloporphyrin complexes in oxygen-transfer reactions and characterization of intermediate species observed during the course of such reactions have been invaluable in evaluating potential intermediates and reaction pathways for cytochrome P-450. Logically, it would be most desirable if one could mimic the enzymatic oxygenation reactions of substrates using iron porphyrins, dioxygen, and reducing agents. However, studies of such iron-porphyrin-catalyzed reactions have failed to produce meaningful results that could be related back to the P-450 mechanism. This is perhaps not surprising, since the enzyme system is designed to funnel electrons into the iron-dioxygen-substrate complex, and thus to generate the active oxidant within the confines of the enzyme active site in the immediate proximity of the bound substrate. Without the constraints imposed by the enzyme, however, iron porphyrins generally will either (1) catalyze the oxidation of the reducing agent by dioxygen, leaving the substrate untouched, or (2) initiate free-radical autoxidation reactions (see Section II.C). A different approach was suggested by the observation of the peroxide shunt reaction (Reactions 5.68 and 5.69) using organic peroxides or single-oxygen-atom donors, and the earliest successful studies demonstrated that Fe(TPP)Cl (TPP = tetraphenylporphyrin) would catalyze the epoxidation of olefins and the hydroxylation of aliphatic hydrocarbons by iodosylbenzene (Reactions 5.70 and 5.71). \(\tag{5.70}\) \(\tag{5.71}\) Reactions (5.70) and (5.71) were postulated to occur via an iron-bound oxidant such as in Reaction (5.67). This hypothesis was tested by studying the reaction of dioctyl Fe(PPIX)CI with iodosylbenzene, which resulted in 60 percent hydroxylation at positions 4 and 5 on the hydrocarbon tail (see 5.72), positions for which there is no reason to expect increased reactivity except for the fact that those particular locations are predicted from molecular models to come closest to the iron center when the tail wraps around the porphyrin molecule. Visible absorption spectra of porphyrin complexes are due largely to \(\pi\)-\(\pi\)* transitions of the porphyrin ligand. The bright green color is unusual for iron-porphyrin complexes, which are usually red or purple. (However, this green color has been seen for compound I of catalase and peroxidases; see Section VI below.) The unusually long-wavelength visible absorption bands that account for the green color result from the fact that the porphyrin ring has been oxidized by one electron. Similar visible absorption bands can be seen, for example, in other oxidized porphyrin complexes, such as Co (P ) , formed by two-electron oxidation of Co (P )(see 5.73). \(\tag{5.73}\) Oxidized porphyrin ligands also give characteristic proton NMR spectra, which are seen for the green porphyrin complex as well. Magnetic measurements indicate that the green porphyrin complex contains three unpaired electrons. Detailed analysis of the Mössbauer spectra has indicated that the two unpaired electrons on the Fe ion are strongly ferromagnetically coupled to the unpaired electron on the porphyrin, accounting for the resulting S = \(\frac{5}{3}\) state. \(\tag{5.74}\) Studies of the reactions of this species with P-450-type substrates demonstrate that this species is reactive enough to make it an attractive candidate for" active oxygen" in the enzymatic mechanism. Synthetic analogues for two of the other candidates for "active oxygen" have also been synthesized and their reactivities assessed. For example, Fe and Mn -porphyrin peroxo complexes analogous to in Reaction (5.65) have been synthesized. The x-ray crystal structure of the Mn complex shows that the peroxo ligand is bound to the metal in a triangular, side-on fashion (see 5.75). The Fe complex is believed to have a similar structure. \(\tag{5.75}\) Studies of this species indicate that in Reaction (5.65) would not have the requisite reactivity to be a candidate for "active oxygen" in the cytochrome P-450 mechanism, since it will not even oxidize triphenylphosphine, PPh , to triphenylphosphine oxide, OPPh , one of the more facile oxygenation reactions known. Attempts to examine the protonated form, in Reaction (5.65), however, indicate that it is highly unstable, and its reactivity has not yet been thoroughly examined. Fe -oxo-porphyrin complexes analogous to in Reaction (5.66) have also been prepared in solution and characterized by NMR. Such complexes will react with PPh to give OPPh , but are relatively unreactive with olefins and totally unreactive with saturated hydrocarbons. Thus is also ruled out as a candidate for "active oxygen" in P-450 mechanisms. These reactivity studies, and the observation of the peroxide shunt described above, indicate that Fe (P )(O) or Fe (P )(O) is the most likely candidate for "active oxygen." These two formulations are, of course, isoelectronic, and it is tempting to conclude that the latter is the more likely formulation of the enzymatic intermediate. However, it is important to remember that the model systems lack the axial cysteinylligand present in cytochrome P-450. The effect of the relatively easily oxidized sulfur ligand on the electron distribution within that intermediate is not known, since model systems for high-valent iron-oxo complexes containing axial thiolate ligands have not been synthesized. The mechanism of reactions of the high-valent oxo complex in Reaction (5.67) with a variety of substrates is an area of active interest. Such studies are generally carried out by generation of the species from the reaction of a ferric porphyrin with a single-oxygen-atom donor, such as a peracid or iodosylbenzene. In hydroxylation reactions of aliphatic hydrocarbons, the initial step appears to be abstraction of a hydrogen atom from the substrate to form a substrate radical and an Fe hydroxide complex held together in a cage created by the enzyme active site so that they cannot diffuse away from each other (Reaction 5.76). This step is then followed by recombination of the OH fragment with the substrate radical to make the hydroxylated product (Reaction 5.77). This mechanism is referred to as the "oxygen rebound mechanism." \(\tag{5.76}\) \(\tag{5.77}\) The radical character of the intermediates formed in this reaction is supported by the observation that such reactions carried out using synthetic porphyrins and single-oxygen-atom donors in the presence of BrCCl give substantial amounts of alkyl bromides as products, a result that is consistent with radical intermediates and inconsistent with either carbanion or carbonium-ion intermediates. In the enzymatic reactions themselves, there is also strong evidence to support a stepwise mechanism involving free-radical intermediates. For example, cytochrome P-450 gives hydroxylation of d-camphor only in the 5- position, but deuterium-labeling studies show that either the 5- or the 5- hydrogen is lost (Reaction 5.78). \(\tag{5.78}\) Such results are obviously inconsistent with a concerted mechanism in which the oxygen atom would be inserted into the 5- C—H bond in one step; so there would be no chance for the hydrogens in the two positions to exchange. (Remember that alcohol protons exchange rapidly with water and therefore are not expected to remain deuterated when the reaction is carried out in H O.) The crystal structure of reduced cytochrome P-450 with CO bound to the iron and the substrate camphor bound adjacent to it has been examined and compared with the crystal structure of the oxidized enzyme with camphor bound. The former is expected to be similar in structure to the less-stable oxy complex. The comparison shows that the substrate camphor is closer to the iron center in the oxidized enzyme. It is therefore possible that a similar movement of the substrate occurs during the catalytic reaction after either a 5- or a 5- hydrogen is abstracted, and that the new position of the camphor molecule then restricts the hydroxylation step to the 5- position. It is interesting to note that the 5- position on the camphor that is hydroxylated is held in very close proximity to the Fe center, and therefore to the presumed location of the oxo ligand in the high-valent oxo intermediate in the structure of the ferric enzyme plus camphor derivative (Figure 5.12). Crystal structures of the ferric form of cytochrome P-450 with norcamphor and adamantanone bound in place of camphor have also been determined. These alternative substrates are smaller than camphor, and appear to fit more loosely than camphor. It is therefore reasonable to assume that they "rattle around" to a certain extent in the substrate binding site, which probably accounts for the less-specific pattern of hydroxylation observed for these alternative substrates. Mechanisms for olefin epoxidations catalyzed either by the enzyme or by model porphyrin complexes are not as well understood as those for hydroxylation of aliphatic hydrocarbons. Some of the possibilities that have been proposed are represented schematically in Figure 5.13. The evidence is persuasive that the "active oxygen" species that attacks substrate in cytochrome P-450 is a high-valent iron-oxo complex. However, the mechanism of formation of that species in the catalytic reaction with dioxygen is less well-understood. Heterolytic O—O bond cleavage of a ferric porphyrin hydroperoxide complex, (Reaction 5.67), is the logical and anticipated route, but it has not yet been unequivocally demonstrated in a model complex. The catalase and peroxidase enzymes catalyze heterolytic O—O bond cleavage in reactions of hydrogen peroxide, but in them the active sites contain amino-acid side chains situated to facilitate the developing charge separation that occurs in heterolytic cleavage (see Section VI). The crystal structure of cytochrome P-450 shows no such groups in the activesite cavity, nor does it give any clue to the source of a proton to protonate the peroxide ligand when it is produced. Also, we have little experimental evidence concerning possible roles that the cysteinyl sulfur axial ligand might play in facilitating O—O bond cleavage. These issues remain areas of active interest for researchers interested in cytochrome P-450 mechanisms. As mentioned above, much less is known about the structural characteristics and mechanisms of the nonheme metal-containing monooxygenase enzymes. From the similarities of the overall stoichiometries of the reactions and the resemblance of some of the enzymes to dioxygen-binding proteins, it is likely that the initial steps are the same as those for cytochrome P-450, i.e., dioxygen binding followed by reduction to form metal-peroxide or hydroperoxide complexes. It is not obvious that the next step is the same, however (i.e., O—O bond cleavage to form a high-valent metal-oxo complex prior to attack on substrate). The problem is that such a mechanism would generate metal-oxo complexes that appear to contain metal ions in chemically unreasonable high-oxidation states, e.g., Fe , Cu , or Cu (Reactions 5.79-5.81). \[(Fe^{III} - OOH)^{2+} \rightarrow (Fe^{V}O)^{3+} + OH^{-} \tag{5.79}\] \[(Cu^{II} - OOH)^{+} \rightarrow (Cu^{IV}O)^{2+} +OH^{-} \tag{5.80}\] \[(Cu^{II} - OO - Cu^{II})^{2+} \rightarrow 2 (Cu^{III}O)^{+} \tag{5.81}\] An alternative mechanism is for the peroxide or hydroperoxide ligand to attack the substrate directly; i.e., O—O bond cleavage could be concerted with attack on substrate. Another possibility is that the oxygen atom is inserted in a metalligand bond prior to transfer to the substrate. Neither of these alternative mechanisms has been demonstrated experimentally. These various possibilities remain to be considered as more information about the monooxygenase enzymes becomes available. | 18,758 | 1,908 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Nucleophilic_Substitution_at_Tetrahedral_Carbon/NS13._Regiochem._in_Elimination |
Sometimes, an elimination reaction could lead to formation of a double bond in more than one place. If the halide is on one carbon and there are protons that could be removed on either side, then taking one proton or the other might lead to two different products. This reaction could have different regiochemical outcomes, meaning it could happen at two different places in the molecule. What factors might influence which product forms? We might think about product stability, in case there are corresponding differences in barriers leading to those products. We already know about stereochemical effects in alkene stability, but what about other effects? It is well-established that alkene stability is influenced by degree of substitution of the double bond. The greater the number of carbons attached to the double bond, the more stable it is. That effect is related to hyperconjugation. Specifically, it's an interaction between bonding orbitals and antibonding orbitals on neighbouring carbons. The interaction allows the bonding electrons to drop a little lower in energy through delocalization. The interaction also pushes the antibonding orbitals a little higher in energy, but since they have no electrons, they don't contribute to the real energy of the molecule. Overall, the molecules goes down in energy. However, alkene stability isn't the only factor that plays a role in elimination. Steric hindrance can play a role, too. In a case in which there are two different hydrogens from which to select, the one leading to the more-substituted double bond is sometimes a little bit crowded. That leaves the base with fewer viable pathways to approach the proton. On the other hand, the removal of a proton leading to the less stable alkene is often less crowded, allowing the base to approach much more easily from a number of angles. , | 1,861 | 1,910 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Ligand_Substitution_in_Coordination_Complexes/LS3._Kinetics_of_Assoc._Mechanism |
We can measure the rate of an associative reaction and make changes in the reaction conditions to see how the rate is affected. For example, we could easily change the concentrations of the two reactants. All we have to do is change the amount of reactant we dissolve in the solution. If we did that, we would find a linear relationship between each concentration and rate. If we double the concentration of new ligand, the rate of reaction doubles. If we triple it, the rate triples. Also, if we double the amount of metal complex, the rate doubles and so on. We can write the following expression, called the rate law, to describe this relationship: \[ \text {Rate Law: Rate} = {-d[MLn]\over dt } = k[MLn,L']\] This type of reaction is sometimes called a second order reaction. That term just refers to the mathematical form of the rate law, which depends on concentration times concentration, or concentration squared. The "order" of the reaction is the number of concentrations multiplied together in the rate law. Why does the associative mechanism depend on concentrations in this specific way? This is a case of two molecules coming together. If both compounds are dissolved in solution, they must “swim around” or travel through the solution until they bump into each other and react. The more concentrated the solution is, or the more crowded it is with molecules, the more likely are the reactants to bump into each other. If we double the amount of new ligand in solution, an encounter between ligand and complex becomes twice as likely. If we double the amount of metal complex in solution, an encounter also becomes twice as likely. Figure LS3.1. The effect of concentration on collision probability. In the first beaker, there is a chance that a black molecule and white molecule will meet and react together. The chance of a meeting is much higher in both the second beaker, where there are lots more black molecules, and in the third beaker, where there are many more white molecules. Given the associative rate law above, what would happen to the reaction rate for an associative substitution in the following cases? Plot graphs of initial rate vs. concentration to show what you would see in associative substitution. In the previous problem, the experiment was run in a particular way for particular reasons. Given the following sets of initial rate data (rates measured at the beginning of a reaction), determine whether each case represents an associative substitution. a) b) c) d) What information can be gained from the slopes of lines in Problem LS3.4.? , | 2,592 | 1,911 |
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The series of straight-chain alkanes, in which \(n\) is the number of carbons in the chain, shows a remarkably smooth gradation of physical properties (see Table 4-1 and Figure 4-1). As \(n\) increases, each additional \(CH_2\) group contributes a fairly constant increment to the boiling point and density, and to a lesser extent to the melting point. This makes it possible to estimate the properties of an unknown member of the series from those of its neighbors. For example, the boiling points of hexane and heptane are \(69^\text{o}\) and \(98^\text{o}\), respectively. Thus a difference in structure of one \(CH_2\) group for these compounds makes a difference in boiling point of \(29^\text{o}\); we would predict the boiling point of the next higher member, octane, to be \(98^\text{o} + 29^\text{o} = 127^\text{o}\), which is close to the actual boiling point of \(126^\text{o}\). Members of a group of compounds, such as the alkanes, that have similar chemical structures and graded physical properties, and which differ from one another by the number of atoms in the structural backbone, are said to constitute a . When used to forecast the properties of unknown members of the series, the concept of works most satisfactorily for the higher-molecular-weight members because the introduction of additional \(CH_2\) groups makes a smaller relative change in the overall composition of such molecules. This is better seen from Figure 4-2, which shows how \(\Delta T\), the differences in boiling points and melting points between consecutive members of the homologous series of continuous-chain alkanes, changes with the number of carbons, \(n\). Branched-chain alkanes do not exhibit the same smooth gradation of physical properties as do the continuous-chain alkanes. Usually there is too great a variation in molecular structure for regularities to be apparent. Nevertheless, in any one set of isomeric hydrocarbons, volatility increases with increased branching. This can be seen from the data in Table 4-2, which lists the physical properties of the five hexane isomers. The most striking feature of the data is the \(19^\text{o}\) difference between the boiling points of hexane and 2,2-dimethylbutane. and (1977) | 2,251 | 1,916 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/01%3A_Introduction_to_Organic_Chemistry/1.04%3A_Why_Is_Organic_Chemistry_Special |
Let us consider some of the factors that make so much of chemistry center on a single element, carbon. One very important feature is that carbon-carbon bonds are strong, so long chains or rings of carbon atoms bonded to one another are possible. Diamond and graphite are two familiar examples, the diamond lattice being a three-dimensional network of carbon atoms, whereas graphite actually more closely resembles a planar network. The lubricating properties of graphite actually are related to its structure, which permits the planes to slide one past the other. But carbon is not unique in forming bonds to itself because other elements such as boron, silicon, and phosphorus form strong bonds . The uniqueness of carbon stems more from the fact that it forms strong carbon-carbon bonds that also are strong when in combination with other elements. For example, the combination of hydrogen with carbon affords a remarkable variety of carbon hydrides, or as they usually are called. In contrast, none of the other second-row elements except boron gives a very extensive system of stable hydrides, and most of the boron hydrides are much more reactive than hydrocarbons, especially to water and air. Carbon forms bonds not only with itself and with hydrogen but also with many other elements, including strongly electron-attracting elements such as fluorine and strongly electropositive metals such as lithium: Why is carbon so versatile in its ability to bond to very different kinds of elements? The special properties of carbon can be attributed to its being a relatively small atom with four valence electrons. To form simple saltlike compounds such as sodium chloride, \(Na^\oplus Cl^\ominus\), carbon would have to either lose the four valence electrons to an element such as fluorine and be converted to a quadripositive ion, \(C^{4 \oplus}\), or acquire four electrons from an element such as lithium and form a quadrinegative ion, \(C^{4 \ominus}\). Gain of four electrons would be energetically very unfavorable because of mutual repulsion between the electrons. Customarily, carbon completes its valence-shell octet by electrons with other atoms. In compounds with shared electron bonds (or covalent bonds) such as methane, ethane, or tetrafluoromethane, each of the bonded atoms including carbon has its valence shell filled, as shown in the following electron-pair or Lewis\(^6\) structures: In this way, repulsions between electrons associated with completion of the valence shell of carbon are compensated by the electron-attracting powers of the positively charged nuclei of the atoms to which the carbon is bonded. However, the electrons of a covalent bond are not necessarily shared equally by the bonded atoms, especially when the affinities of the atoms for electrons are very different. Thus, carbon-fluorine and carbon-lithium bonds, although they are not ionic, are polarized such that the electrons are associated more with the atom of higher electron affinity. This is usually the atom with the higher effective nuclear charge. \(\overset{\delta \oplus}{C} \: \: \: \: \: : \overset{\delta \ominus}{F} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \overset{\delta \ominus}{C:} \: \: \: \: \: \overset{\delta \oplus}{Li} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \scriptsize{\left( \delta \oplus, \: \delta \ominus \: \text{denote partial ionic bonds} \right)}\) We see then a gradation from purely ionic to purely covalent bonding in different molecules, and this is manifest in their chemical and physical properties. Consider, for instance, the hydrides of the elements in the second horizontal row of the periodic table. Their melting and boiling points,\(^7\) where known, are given below. Lithium hydride can be regarded as a saltlike compound, \(\overset{\oplus}{Li} \: \: \: : \overset{\ominus}{H}\). Electrostatic attractions between oppositely charged ions in the crystal lattice are strong, thereby causing lithium hydride to be a high-melting, nonvolatile solid like sodium chloride, lithium fluoride, and so on. Methane, \(CH_4\), is at the other extreme. It boils at \(-161^\text{o}\), which is about \(800^\text{o}\) lower even than the melting point of lithium hydride. Because carbon and hydrogen have about the same electron-attracting power, \(C-H\) bonds have little ionic character, and methane may be characterized as a substance. As a result, there is relatively little electrostatic attraction between methane molecules and this allows them to "escape" more easily from each other as gaseous molecules - hence the low boiling point. Hydrogen fluoride has a boiling point some \(200^\text{o}\) higher than that of methane. The bonding electron pair of \(HF\) is drawn more toward fluorine than to hydrogen so the bond may be formulated as \(\overset{\delta \oplus}{H}\) ---- \(\overset{\delta \ominus}{F}\). In liquid hydrogen fluoride, the molecules tend to aggregate through what is called in chains and rings arranged so the positive hydrogen on one molecule attracts a negative fluorine on the next: When liquid hydrogen fluoride is vaporized, the temperature must be raised sufficiently to overcome these intermolecular electrostatic attractions; hence the boiling point is high compared to liquid methane. Hydrogen fluoride is best characterized as a , but not ionic, substance. Although the \(O-H\) and \(N-H\) bonds of water and ammonia have somewhat less ionic character than the \(H-F\) bonds of hydrogen fluoride, these substances also are relatively polar in nature and also associate through hydrogen bonding in the same way as does hydrogen fluoride. The chemical properties of lithium hydride, methane, and hydrogen fluoride are in accord with the above formulations. Thus, when the bond to the hydrogen is broken, we might expect it to break in the sense \(\begin{array}{c:c} Li^\oplus & :H^\ominus \end{array}\) for lithium hydride, and \(\begin{array}{c:c} \overset{\delta \oplus}{H} & : \underset{\cdot \cdot}{\ddot{F}}:^{\delta \ominus} \end{array}\) for hydrogen fluoride so that the electron pair goes with the atom of highest electron affinity. This is indeed the case as the following reaction indicates: \(\begin{array}{c:c} Li^\oplus & :H^\ominus \end{array} + \begin{array}{c:c} H & : \underset{\cdot \cdot}{\ddot{F}} : \end{array} \longrightarrow Li^\oplus : \underset{\cdot \cdot}{\ddot{F}}:^\oplus + \: H : H\) Methane, with its relatively nonpolar bonds, is inert to almost all reagents that could remove hydrogen as \(H^\oplus\) or \(H:^\ominus\) except under anything but extreme conditions. As would be expected, methyl cations \(CH_3^\oplus\) and methyl anions \(CH_3:^\ominus\) are very difficult to generate and are extremely reactive. For this reason, the following reactions are not observed: From the foregoing you may anticipate that the chemistry of carbon compounds will be largely the chemistry of covalent compounds and will not at all resemble the chemistry of inorganic salts such as sodium chloride. You also may anticipate that the major differences in chemical and physical properties of organic compounds will arise from the nature of the elements bonded to carbon. Thus methane is not expected to, nor does it have, the same chemistry as other one-carbon compounds such as methyllithium, \(CH_3Li\), or methyl fluoride, \(CH_3F\). \(^6\)G. N. Lewis (1876-1946), the renowned U.S. chemist, was the first to grasp the significance of the electron-pair in molecular structure. He laid the foundation for modern theory of structure and bonding in his treatise on (1923). \(^7\)Throughout this text all temperatures not otherwise designated should be understood to be in \(^\text{o}C\); absolute temperatures will be shown as \(^\text{o}K\). and (1977) | 7,786 | 1,917 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Concepts_in_Biophysical_Chemistry_(Tokmakoff)/06%3A_Dynamics_and_Kinetics/22%3A_Biophysical_Reaction_Dynamics/22.02%3A_Computing_Dynamics |
There are a number of ways of computationally modeling time-dependent processes in molecular biophysics. These methods integrate equations of motion for the molecular degrees of freedom evolving under a classical force–field interaction potential, a quantum mechanical Hamiltonian, or an energy landscape that could be phenomenological or atomistically detailed. Examples include using classical force fields to propagate Newton’s equation of motion, integrating the Schrödinger equation, or integrating the Langevin equation on a potential of mean force. Since our interest is more on the description of computational or experimental data, this will just be a brief overview. An overview of how to integrate Newton’s equation of motion, leaving out many important details. This scheme, often used in MD simulations, is commonly called a Verlet integration. Building on our discussion of Brownian motion, the Langevin equation is an equation of motion for a particle acting under the influence of a fixed potential U, friction, and a time-dependent random force. Writing it in one dimension: \[ ma= f_{\textrm{potential}}+f_{\textrm{friction}}+f{\textrm{random}}(t) \] \[ m \frac{\partial^{2} x}{\partial t^{2}}=-\frac{\partial U}{\partial x}-\zeta \frac{\partial x}{\partial t}+f_{r}(t) \] The random force reflects the equilibrium thermal fluctuations acting on the particle, and is the source of the friction on the particle. In the Markovian limit, the friction coefficient ζ and the random force f (t) are related through a fluctuation–dissipation relationship: \[\langle f_r(t)\rangle = 0 \] \[\langle f_r(t)f_r(t_0)\rangle = 2ζ k_BT \delta (t-t_0) \] Also, the diffusion constant is = /ζ, and the time scale for loss of velocity correlations is \( \tau_c = \gamma-1 =\) m/ζ. The Langevin equation has high and low friction limits. In the low friction limit (ζ→0), the influence of friction and random force is minimal, and the behavior is dominated by the inertial motion of the particle. In the high friction limit, the particle’s behavior, being dominated by ζ, is diffusive. The limit is defined by any two of the following four linearly related variables: ζ, D, T, and \(\langle f_r^2 \rangle \). The high and low friction limit are also referred to as the low and high temperature limits: \( \langle f_r^2 \rangle / 2ζ = k_BT \). Example: Trajectory for a particle on a bistable potential from Langevin dynamics | 2,438 | 1,918 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Molecular_Orbital_Theory/MO_for_HF |
A simple approach to molecular orbital (MO) theory for heterogeneous diatomic molecules is to show the energy level diagram. The MO energy levels can be worked out following these steps: Recall that the energy \(E_n\) for the quantum number n is for an element with atomic Z is approximately \[E_n = 13.6 \dfrac{Z_{Eff}^2}{n^2} eV\] We use \(Z_{eff}\) instead of Z to mean that we have to modify the atomic number to get an effective atomic charge for the nucleus. Since we are dealing with approximate values, one may use Z directly. The 1s orbital energy level is -13.6 eV for hydrogen atoms, measured as the ionization energy of H. Thus, for the quantum number n = 1, the energy level for 1s of He is approximately - 54 eV. Similarly, the 1s energy level for F is - 1101 eV. The 2s and 2p energy levels for He is approximately - 13.6 eV, which is simlar to that of 1s orbital of H. Thus, the 2s energy level for Li is approximately -6 eV. However, for multi-electron atoms, the p-subshell and s-subshell have different energies due to penetration. At this level, we cannot be precise about it, but simply think that the 2p orbitals are at higher energy than the 2s orbital. Usually, atomic orbitals with energy levels similar to each other will overlap to form molecular orbitals. Thus, we match the energy levels of atomic orbitals, and then make bonding and anti-bonding MOs of them. However, in case the atomic orbital energy level is very different, we use atomic orbitals of the incomplete subshell to form MOs. Interaction occurs between the 1s orbital on hydrogen and the 2p orbital in fluorine causing the formation of a sigma-bonding and a sigma-antibonding molecular orbital, as shown below. | 1,725 | 1,920 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Acid-Base_Equilibria/7._Buffer_Solutions |
This page describes simple acidic and alkaline buffer solutions and explains how they work. A buffer solution is one which resists changes in pH when small quantities of an acid or an alkali are added to it. A buffer solution has to contain things which will remove any hydrogen ions or hydroxide ions that you might add to it - otherwise the pH will change. Acidic and alkaline buffer solutions achieve this in different ways. We'll take a mixture of ethanoic acid and sodium ethanoate as typical. Ethanoic acid is a weak acid, and the position of this equilibrium will be well to the left: Adding sodium ethanoate to this adds lots of extra ethanoate ions. According to Le Chatelier's Principle, that will tip the position of the equilibrium even further to the left. The solution will therefore contain these important things: Other things (like water and sodium ions) which are present aren't important to the argument. The buffer solution must remove most of the new hydrogen ions otherwise the pH would drop markedly. Hydrogen ions combine with the ethanoate ions to make ethanoic acid. Although the reaction is reversible, since the ethanoic acid is a weak acid, most of the new hydrogen ions are removed in this way. Since most of the new hydrogen ions are removed, the pH won't change very much - but because of the equilibria involved, it will fall a little bit. Alkaline solutions contain hydroxide ions and the buffer solution removes most of these. This time the situation is a bit more complicated because there are two processes which can remove hydroxide ions. Removal by reacting with ethanoic acid The most likely acidic substance which a hydroxide ion is going to collide with is an ethanoic acid molecule. They will react to form ethanoate ions and water. Because most of the new hydroxide ions are removed, the pH doesn't increase very much. Removal of the hydroxide ions by reacting with hydrogen ions Remember that there are some hydrogen ions present from the ionization of the ethanoic acid. Hydroxide ions can combine with these to make water. As soon as this happens, the equilibrium tips to replace them. This keeps on happening until most of the hydroxide ions are removed. Again, because you have equilibria involved, not all of the hydroxide ions are removed - just most of them. The water formed re-ionizes to a very small extent to give a few hydrogen ions and hydroxide ions. We'll take a mixture of ammonia and ammonium chloride solutions as typical. Ammonia is a weak base, and the position of this equilibrium will be well to the left: Adding ammonium chloride to this adds lots of extra ammonium ions. According to Le Chatelier's Principle, that will tip the position of the equilibrium even further to the left. The solution will therefore contain these important things: There are two processes which can remove the hydrogen ions that you are adding. The most likely basic substance which a hydrogen ion is going to collide with is an ammonia molecule. They will react to form ammonium ions. Most, but not all, of the hydrogen ions will be removed. The ammonium ion is weakly acidic, and so some of the hydrogen ions will be released again. Remember that there are some hydroxide ions present from the reaction between the ammonia and the water. Hydrogen ions can combine with these hydroxide ions to make water. As soon as this happens, the equilibrium tips to replace the hydroxide ions. This keeps on happening until most of the hydrogen ions are removed. Again, because you have equilibria involved, not all of the hydrogen ions are removed - just most of them. The hydroxide ions from the alkali are removed by a simple reaction with ammonium ions. Because the ammonia formed is a weak base, it can react with the water - and so the reaction is slightly reversible. That means that, again, most (but not all) of the the hydroxide ions are removed from the solution. This is easier to see with a specific example. Remember that an acid buffer can be made from a weak acid and one of its salts. Let's suppose that you had a buffer solution containing 0.10 mol dm of ethanoic acid and 0.20 mol dm of sodium ethanoate. How do you calculate its pH? In any solution containing a weak acid, there is an equilibrium between the un-ionized acid and its ions. So for ethanoic acid, you have the equilibrium: The presence of the ethanoate ions from the sodium ethanoate will have moved the equilibrium to the left, but the equilibrium still exists. That means that you can write the equilibrium constant, K , for it: Where you have done calculations using this equation previously with a weak acid, you will have assumed that the concentrations of the hydrogen ions and ethanoate ions were the same. Every molecule of ethanoic acid that splits up gives one of each sort of ion. That's no longer true for a buffer solution: If the equilibrium has been pushed even further to the left, the number of ethanoate ions coming from the ethanoic acid will be completely negligible compared to those from the sodium ethanoate. We therefore assume that the ethanoate ion concentration is the same as the concentration of the sodium ethanoate - in this case, 0.20 mol dm . In a weak acid calculation, we normally assume that so little of the acid has ionized that the concentration of the acid at equilibrium is the same as the concentration of the acid we used. That is even more true now that the equilibrium has been moved even further to the left. So the assumptions we make for a buffer solution are: Now, if we know the value for K , we can calculate the hydrogen ion concentration and therefore the pH. K for ethanoic acid is 1.74 x 10 mol dm . Remember that we want to calculate the pH of a buffer solution containing 0.10 mol dm of ethanoic acid and 0.20 mol dm of sodium ethanoate. Then all you have to do is to find the pH using the expression pH = -log [H ] You will still have the value for the hydrogen ion concentration on your calculator, so press the log button and ignore the negative sign (to allow for the minus sign in the pH expression). You should get an answer of 5.1 to two significant figures. You can't be more accurate than this, because your concentrations were only given to two figures. You could, of course, be asked to reverse this and calculate in what proportions you would have to mix ethanoic acid and sodium ethanoate to get a buffer solution of some desired pH. It is no more difficult than the calculation we have just looked at. Suppose you wanted a buffer with a pH of 4.46. If you un-log this to find the hydrogen ion concentration you need, you will find it is 3.47 x 10 mol dm . Feed that into the K expression. All this means is that to get a solution of pH 4.46, the concentration of the ethanoate ions (from the sodium ethanoate) in the solution has to be 0.5 times that of the concentration of the acid. All that matters is that ratio. In other words, the concentration of the ethanoate has to be half that of the ethanoic acid. One way of getting this, for example, would be to mix together 10 cm of 1.0 mol dm sodium ethanoate solution with 20 cm of 1.0 mol dm ethanoic acid. Or 10 cm of 1.0 mol dm sodium ethanoate solution with 10 cm of 2.0 mol dm ethanoic acid. And there are all sorts of other possibilities. We are talking here about a mixture of a weak base and one of its salts - for example, a solution containing ammonia and ammonium chloride. The modern, and easy, way of doing these calculations is to re-think them from the point of view of the ammonium ion rather than of the ammonia solution. Once you have taken this slightly different view-point, everything becomes much the same as before. So how would you find the pH of a solution containing 0.100 mol dm of ammonia and 0.0500 mol dm of ammonium chloride? The mixture will contain lots of unreacted ammonia molecules and lots of ammonium ions as the essential ingredients. The ammonium ions are weakly acidic, and this equilibrium is set up whenever they are in solution in water: \[ NH^+_{4 (aq)} \rightleftharpoons NH_{3(aq)} + H^+_{(aq)}\] You can write a K expression for the ammonium ion, and make the same sort of assumptions as we did in the previous case: The presence of the ammonia in the mixture forces the equilibrium far to the left. That means that you can assume that the ammonium ion concentration is what you started off with in the ammonium chloride, and that the ammonia concentration is all due to the added ammonia solution. The value for K for the ammonium ion is 5.62 x 10 mol dm . Remember that we want to calculate the pH of a buffer solution containing 0.100 mol dm of ammonia and 0.0500 mol dm of ammonium chloride. Just put all these numbers in the K expression, and do the sum: | 8,786 | 1,921 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Heterogeneous_Equilibria/Insoluble_Salts |
Phase transitions such as sublimation, deposition, melting, solidification, vaporization, and condensation are heterogeneous equilibria; so are the formation of crystals from a saturated solution, because a solid and its solution are separated phases. The picture shown here is the formation of solid from a gaseous solution. The equilibrium constants for saturated solution and solid formation (precipitate) are called . For unsaturated and supersaturated solutions, the system is not at equilibrium, and ion products, , which have the same expression as , are used. An oversaturated solution becomes a saturated solution by forming a solid to reduce the dissolved material. The crystals formed are called a . Often, however, a precipitate is formed when two clear solutions are mixed. For example, when a silver nitrate solution and sodium chloride solution are mixed, silver chloride crystals \(\ce{AgCl_{\large{(s)}}}\) (a precipitate) are formed. \(\ce{Na+}\)and \(\mathrm{\sideset{ }{_{3}^{-}}{NO}}\) are . \[\mathrm{Ag^+_{\large{(aq)}} + \sideset{ }{_{\large{(aq)}}^{-}}{Cl} \rightarrow AgCl_{\large{(s)}}} \hspace{10px} \ce{(precipitate)}\] Silver chloride is one of the few chlorides that has a limited solubility. A precipitate is also formed when sodium carbonate is added to a sample of hard water, \[\mathrm{Ca^{2+}_{\large{(aq)}} + CO^{2-}_{3\large{(aq)}} \rightarrow CaCO_{3\large{(s)}}} \hspace{10px} \ce{(precipitate)}\] When a solid dissolves, we have the reverse reaction: \[\mathrm{AgCl_{\large{(s)}} \rightarrow Ag^+_{\large{(aq)}} + \sideset{ }{_{\large{(aq)}}^{-}}{Cl}} \hspace{10px} \ce{(dissolving)}\] \[\ce{CaCO_{3\large{(s)}} \rightarrow Ca^2+_{\large{(aq)}} + CO^{2-}_{3\large{(aq)}}} \hspace{10px} \ce{(dissolving)}\] Formations of precipitates are chemical equilibria phenomena, and we usually write these in the following manner, and call the equilibrium constants : The solubility product, , is a special type of equilibrium constant given to a solution containing sparingly soluble salts. The symbols (aq) indicate that these ions are surrounded by water molecules; these ions are in the solution. Note the expression for the solubility product given above, please. These are special equilibrium constants, because the solid present has a constant tendency of being dissolved. Therefore, their role in \(K_{sp}\) is a constant. They do not appear in the \(K_{sp}\) expression. If the solution is not saturated, no precipitate will form. In this case, the product is called the , which is the product of the concentrations of the ions at any moment in time. and is equal to the solubility product for the salt when at equilibrium. The Table of solubility product is given as Salt, \(K_{sp}\) in the Handbook Section. In this table, the salts are divided into For some substances, formation of a solid or crystallization does not occur automatically whenever a solution is saturated. These substances have a tendency to form oversaturated solutions. For example, syrup and honey are oversaturated sugar solutions, containing other substances such as citric acids. For oversatureated solutions, is greater than . When a seed crystal is provided or formed, a precipitate will form immediately due to equilibrium of requiring to approach . Sodium acetate trihydrate, \(\ce{NaCH3COO\cdot 3H2O}\), when heated to 370 K will become a liquid. The sodium acetate is said to be dissolved in its own water of crystallization. The substance stays as a liquid when cooled to room temperature or even below 273 K. As soon as a seed crystal is present, crystallization occurs rapidly. In such a process, heat is released, and the liquid feels warm. Thus, the relationship among , \(K_{sp}\) and saturation is given below: Solubility products, , of salts are indirect indications of their solubilities expressed in mol/L (called ). However, the solubility products are more useful than molar solubility. The molar solubilities are affected when there are common ions present in the solution. We need to employ the solubility products to estimate the molar solubilities in these cases. When a salt is dissolved in pure water, solubility products and molar solubilities are related. This is illustrated using calcium carbonate. If is the concentration of \(\ce{Ca^2+}\) (= \(\ce{[CO3^2-]}\)) in the saturated solution, then \[K_{\ce{sp}} = x^2\] In this case, is also called the molar solubility of \(\ce{CaCO3}\). The following examples illustrate the relationship between solubility products, , and molar solubilities. The \(K_{sp}\) for \(\ce{AgCl}\) is 1.8e-10 M . What is the molar solubility of \(\ce{AgCl}\) in pure water?
Let be the molar solubility, then \(\begin{array}{ccccc}
\ce{AgCl &\rightleftharpoons &Ag+ &+ &Cl-}\\
&&x &&x
\end{array}\) \(\begin{align}
x &= (\textrm{1.8e-10 M}^2)^{1/2}\\
&= \textrm{1.3e-5 M}
\end{align}\) The solubility product, \(K_{sp}\) is a better indicator than the usual solubility specification of g per 100 mL of solvent or moles per unit volume of solvent. For the \(\ce{AgCl}\) case, when the cation concentration is not the same as the anion concentration (\(\ce{[Ag+] \neq [Cl- ]}\)), solubility of \(\ce{AgCl}\) can not be defined in terms of moles per L. In this case, the system can be divided into three zones. The condition \(\ce{[Ag+] [Cl- ]} = K_{\ce{sp}}\), is represented by a line which divides the plane into two zones. When \(\ce{[Ag+] [Cl- ]} < K_{\ce{sp}}\), no precipitate will be formed. When \(\ce{[Ag+] [Cl- ]} > K_{\ce{sp}}\), a precipitate will be formed. When \(\ce{AgCl}\) and \(\ce{NaCl}\) dissolve in a solution, both salts give \(\ce{Cl-}\) ions. The effect of \(\ce{[Cl- ]}\) on the solubility of \(\ce{AgCl}\) is called the . The \(K_{sp}\) for \(\ce{Ag2CrO4}\) is 9e-12 M . What is the molar solubility of \(\ce{Ag2CrO4}\) in pure water?
Let be the molar solubility of \(\ce{Ag2CrO4}\), then \(\begin{array}{ccccc}
\ce{Ag2CrO4 &\rightleftharpoons &2 Ag+ &+ &CrO4^2-}\\
&&2 x &&x
\end{array}\) \((2 x)^2 (x) = K_{\ce{sp}}\) \(\begin{align}
x &= \left (\dfrac{\textrm{9e-12}}{4} \right )^{1/3}\\
&= \textrm{1.3e-4 M}
\end{align}\) \(\ce{[Ag+]} = \textrm{2.6e-4 M}\) and the molar solubility is 1.3e-4 M. A similar diagram to the one given for \(\ce{AgCl}\) can be drawn, but the shape of the curve representing the \(K_{sp}\) is different. The \(K_{sp}\) for \(\ce{Cr(OH)3}\) is 1.2e-15 M . What is the molar solubility of \(\ce{Cr(OH)3}\) in pure water? Let be the molar solubility of \(\ce{Cr(OH)3}\), then you have \(\begin{array}{ccccc}
\ce{Cr(OH)3 &\rightleftharpoons &Cr^3+ &+ &3 OH-}\\
&&x &&3 x
\end{array}\) Thus, \(\begin{align}
x (3 x)^3 &= \textrm{1.2e-15}\\
x &= \left (\dfrac{\textrm{1.2e-15}}{27} \right )^{1/4}\\
&= \textrm{8.2e-5 M}
\end{align}\) Very careful experiment indicates that the molar solubility of \(\ce{Bi2S3}\) is 1.8e-15 M; what value of \(K_{sp}\) does this compound have? If the molar solubility of \(\ce{Bi2S3}\) is 1.8e-15, then \(\begin{array}{ccccc}
\ce{Bi2S3 &\rightleftharpoons &2 Bi^3+ &+ &3 S^2-}\\
&&\textrm{3.6e-15} &&\textrm{5.4e-15}
\end{array}\) \(\begin{align}
K_{\ce{sp}} &= (\textrm{3.6e-15})^2 (\textrm{5.4e-15})^3\\
&= \textrm{2.0e-72 M}^5
\end{align}\)
You should be able to calculate the \(K_{sp}\) if you know the molar solubility. \(\begin{array}{cccccc}
\ce{CuCl &\rightleftharpoons &Cu+ &+ &Cl- &}\\
&&x &&x &\hspace{10px}(x = \textrm{1.1e-3})
\end{array}\) \(K_{\ce{sp}} = \ce{[Cu+,Cl- ]} = (\textrm{1.1e-3})^2 =\: ?\: \ce M^2\)
Let be the molar solubility, then \(\ce{[Hg2^2+]} = x\); and \(\ce{[I- ]} = 2 x\); \(\begin{array}{ccccc}
\ce{Hg2I2 &\rightleftharpoons &Hg2^2+ &+ &2 I-}\\
x &&x &&2 x
\end{array}\) Molar solubility = \(\ce{[Hg2^2+]}\). \(\ce{[I- ]} = \ce{2 [Hg2^2+]}\). \(K_{\ce{sp}} = x\, (2 x)^2 = \textrm{4.5e-29}\); \(x =\: ?\) \(\begin{array}{ccccc}
\ce{Ca(OH)2 &\rightleftharpoons &Ca^2+ &+ &2 OH-}\\
&&x &&2 x
\end{array}\) \(x = \left (\dfrac{\textrm{5.5e-6}}{4} \right )^{1/3}\) \(\ce{[OH- ]} = 2 x\). | 8,102 | 1,922 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Addition_to_Carbonyls/CO11._Addition_of_Neutral%2C_Protic_Nucleophiles |
Nucleophiles do not have to be ionic, or even "semi-anionic." The basic requirement for a nucleophile is a lone pair. If a nucleophile has a lone pair, it can donate the lone pair to an electrophile such as a carbonyl. By donating a lone pair to a carbonyl, it can form a bond. However, donation of a lone pair to a carbonyl is reversible. If there is something about the nucleophile/electrophile adduct that isn't very stable, the reaction may revert to reactants again. In the case of neutral nucleophiles, charge separation may destabilize the first-formed products of the reaction. Let's think about addition of water to propanone. Two neutral molecules, water and propanone, come together. The water donates a lone pair to the carbonyl carbon in propanone. That leaves the oxygen atom from the water with a positive charge, and the oxygen atom from the propanone with a negative charge. One easy way to get rid of the charge separation is for the water to leave again. That step would just be the reverse of the first one. On the other hand, another way to solve the charge problem is to move a proton (H+) from the positively charged oxygen to the negatively charged one. That turns out to be pretty easy to do. If that happens, a "hydrate" or a "geminal diol" forms. A geminal diol, or twin diol, has two hydroxy groups on one carbon. In the following cases, a nucleophile donates to the carbonyl, followed by a proton transfer. Show mechanisms, with curved arrows, for each of the following reactions. The immediate products that form from the addition of neutral nucleophiles to carbonyls turn out to be a little unstable. That's partly because they can easily revert back to reactants. It's also because there are other processes that carry the reaction away from the first-formed products and turn them into other things. For example, in most cases the reaction does not involve the simple addition of one nucleophile, but involves a second molecule of the nucleophile as well. We will keep looking at these other processes and see where these reactions lead. In the meantime, it can be useful to know the patterns that different types of nucleophiles will usually follow. For example, addition of alcohols to aldehydes or ketones leads to the formation of ketals or acetals. Ketals or acetals have the specific chain of atoms C-O-C-O-C, in which each carbon is tetrahedral or sp . On the other hand, the addition of amines leads to a very different kind of structure. Instead of adding two amine molecules into the final structure, only one amine is incorporated, and a double bond appears again. The C=O of the aldehyde or ketone is replaced with the C=N of an imine. Imines form important linkages in biological chemistry, especially in connecting small molecules to lysine side chains in proteins. If the amine is "secondary", meaning the nitrogen is connected to two carbons instead of just one, the double bond ends up one position over. It is between the former carbonyl carbon and the one next to it, which is called the alpha position. Note that all three of these reactions involve ultimate loss of water from the original carbonyl compound. That's a very important feature of reactions with neutral nucleophiles. , | 3,249 | 1,923 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumentation_and_Analysis/Mass_Spectrometry/Introductory_Mass_Spectrometry/Isotopes%3A_13C |
Isotopomers or isotopic isomers are isomers with isotopic atoms, having the same number of each isotope of each element but differing in their positions. If you look closely at the mass spectrum of an organic compound, 2-butanone, you see a line at m/z 72, which corresponds to 4 carbons, an oxygen and 8 hydrogens. In addition, there are a number of other lines at lower values of m/z; these correspond to the masses of smaller pieces of those 2-butanone molecules that fall apart during the experiment. We won't look too closely at how those arise until we get to radical reactions later in the course. However, we will look at some factors that make cations stable later in this chapter. If you look closely at the mass spectrum of 2-butanone, you'll also see another little peak at m/z 73. This is referred to as the M+1 peak (one greater than the molecular ion), and it arises because of C. This compound is referred to as an isotopomer; that means the same compound with a different isotope. The chance that a molecule in a sample contains a C atom is related to the number of carbons present. If there is just one carbon atom in the molecule, it has a 1% chance of being a C. That means the M+1 peak would be only 1/100th as tall as M+, the peak for the molecular ion. , | 1,293 | 1,924 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Fundamentals/Fluorescence_Resonance_Energy_Transfer |
(FRET) is a special technique to gauge the distance between two chromophores, called a donor-acceptor pair. The limitation of FRET is that this transfer process is effective only when the separating distance of donor-acceptor pair is smaller than 10 nanometers. However, FRET is a highly distance-dependent phenomenon and thus has become a popular tool to measure the dynamic activities of biological molecules within nanoscale. FRET is the acronym of the Förster (Flourescence) Resonance Energy Transfer. The Förster energy transfer is the phenomenon that an excited donor transfers energy (not an electron) to an acceptor group through a non-radiative process. This process is highly distance-dependent, thus allowing one to probe biological structures. One common application is simply to measure the distance between two positions of interest on a big molecule, generally a biological macromolecule, by attaching appropriate donor-acceptor groups to the big one. If the big molecule only involves one donor and one acceptor group, the distance between the donor and the acceptor can be easily measured if there is no conformational change within this process. Besides, if the molecule has a huge conformational change, one may also measure the dynamical activities between two sites on this macromolecule such as protein interactions. Today, this technique is widely applied in many fields such as single-molecule experiments, molecular motors, biosensors and DNA mechanical movements. FRET is also called the "Spectroscopic Ruler" because of its intrinsic convenience. The theoretical analysis was well developed by Theodor Förster. This non-radiative transfer mechanism is schematically capsuled in Figure \(\Page {1}\). A donor group (D) is excited by a photon and then relaxes to the lowest excited singlet state, S (by Kasha’s rule). If the acceptor group is not too far, the energy released when the electron returns to the ground state (S ) may simultaneously excite the donor group. This non-radiative process is referred to as “resonance”. After excitation, the excited acceptor emits a photon and returns to the ground state, if the other quenching states do not exist. The resonance mechanism is associated with the Coulombic interaction between electrons. Thus, the relative distance of Coulombic interaction between the donor-acceptor pair could be longer than the electron exchange energy transfer which needs the overlap of wavefunctions, namely the . The Coulombic interaction only needs the overlap of the spectrum which means that the identity of resonance energy. Figure \(\Page {2}\) here shall give a sense about what the resonance mechanism is. (Note that the HOMO-LUMO gap does not equal the energy difference between the ground state and the lowest S excited state of the molecule.) The FRET efficiency (\(E\)) is the quantum yield of the energy transfer transition; i.e., the fraction of energy transfer event occurring per donor excitation event. The FR \[\ E = \dfrac{k_{ET}}{k_f + k_{ET}+\sum k_i} \label{1}\] where \(k_{ET}\) is the rate of FRET, \(k_f\) is the rate of radiative relaxation (i.e., fluorescence), and \(k_i\) are the non-radiative relaxation rates (e.g., internal conversion, intersystem crossing, external conversion etc). Within a point dipole-dipole approximation, the FRET efficiency can be related to the donar-acceptor distance via \[\ E = \dfrac{1}{1+\left( \dfrac{r}{R_0} \right)^6} \label{2}\] where To enhance the FRET efficiency, the donor group should have good abilities to absorb photons and emit photons. That means the donor group should have a high extinction coefficient and a high quantum yield. The overlap of emission spectrum of the donor and absorption spectrum of the acceptor means that the energy lost from excited donor to ground state could excite the acceptor group. The energy matching is called the resonance phenomenon. Thus, the more overlap of spectra, the better a donor can transfer energy to the acceptor. The overlap integral, \(J(λ)\), between the donor and the acceptor stands for the overlap of spectra, as shown in Figure \(\Page {3}\). The overlap integral is given by \[ J = \int_o^{\infty} F_D(\lambda)\epsilon_A(\lambda) \lambda^4 \,d\lambda \label{3}\] where \(F_D(λ)\) is the normalized emission spectrum of the donor, \(\epsilon_{A}\) standards for the molar absorption coefficient of the acceptor, and \(λ\) is the wavelength. The resonance energy transfer mechanism is also affected by the orientations of the emission transition dipole of the donor and the absorption dipole of the acceptor. The orientation parameter \(κ^2\) gives the quantitative value of interaction between two dipole moments. \(κ^2\) can theoretically be values from 0 (when dipoles are perpendicular to each other) to 4 (when dipoles are collinear). \[\begin{aligned}
\kappa^{2} &=\left(\cos \theta_{\tau} \cdot 3 \cos \theta_{0} \cos \theta_{A}\right)^{2} \\[4pt]
&=\left(\sin \theta_{0} \sin \theta_{A} \cos \phi \cdot 2 \cos \theta_{0} \sin \theta_{A}\right)^{2}
\end{aligned}\] \(κ^2\) is equal to 1 when these two transition dipoles are parallel. The orientation of transition dipoles is shown in Figure \(\Page {4}\) For a freely rotational donor and acceptor group, the average \(κ^2\) is treated 2/3. A long \(R_0\) can cause a high FRET efficiency. Based on Förster's analysis, \(R_0\) is a function of quantum yield of the donor chromophore \(\Phi_{D}\), spectral overlap of donor and acceptor \(J(λ)\), directional relationship of transition dipoles \(κ^2\) and the refractive index of the medium \[ (R_0)^6 \propto \kappa^2 \Phi_{D} J(\lambda) n^{-4} \label{4}\] Förster distance (\(R_0\), nm) FRET provides an efficient way to measure the distance between a donor and an acceptor chromophore. The energy transfer efficiency is highly influenced by the ratio of \(R\) and \(R_0\) because of the exponent 6. Thus, by measuring the FRET efficiency, one can easily get the precise distance between the donor and the acceptor. If choosing the donor and acceptor properly, this experiment can also be carried out . However, the FRET only gives the information about distances. If a dramatic conformational change happens, such as lengthening or kink, it is unable to know the exact movement of donor and the acceptor. Besides, attaching the chromophores to precise sites of a macromolecule is also important, both in quantity of chromophores and in position of a macromolecule, or the FRET might produce noise signals. (Please refer to question 5) The F-actin filament is composed by G-actin monomers. By attaching either a donor (D) or an acceptor (A) choromophore to the G-actin monomer and measuring the energy transfer efficiency to gauge the average distance between G-actin monomers in a F-actin filment (assuming that the monomers are well arranged in DADADADA....sequence), and one finds that the average energy transfer efficiencies is 23% If the \(R_0\) i 5.5 nm. Based on the question 1, if the filament sequence is composed by 8 monomers in the order of D-A-D-A-D-A-D-A, how many kinds of efficiencies might be detected if the filament does not bend and the \(R_0\) is bi ? 4 kinds of efficiencies will be detected. The cy3-dornor and cy5-acceptor pair is attached onto the theminals of a DNA sequence. If ignoring the orientations of transition dipoles of donor and acceptor, please plot the relationship between the ratio (\(R/R_0\)) and the energy transfer efficiency. Please refer to Professor Taekjip Ha's website. netfiles.uiuc.edu/tjha/www/newTechnique.html Use the example in question 3, and now consider the orientations of transition dipoles. Please plot the relationship between the separating distance of donor-acceptor pair and the energy transfer efficiency. (Remember that when elongating the DNA by adding base pairs, the orientations of donor and acceptor chromophores will change) Please refer to PNAS, 2008 Aug, 105(32), 11176-11181, doi:10.1073/pnas.0801707105 One of the most challenging problems within ion channels field is to observe how the channel works and the conformational change of the channel which is embedded on the cell membrane . If a scientist wants to investigate the movement of the ion channel gate by using FRET, what kinds of factors should be considered? For example, how to attached a proper donor and acceptor to precise positions of the channel gate. An open question. Please refer to this introductory paper about using FRET to investigate the ion channel movements. Biophysical Journal, 2003 Jan., 84(1), 1-2, doi:10.1016/S0006-3495(03)74827-9 | 8,616 | 1,925 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Acid-Base_Equilibria/pH_Titration_Curves |
This page describes how pH changes during various acid-base titrations. When you carry out a simple acid-base titration, you use an indicator to tell you when you have the acid and alkali mixed in exactly the right proportions to "neutralise" each other. When the indicator changes color, this is often described as the end point of the titration. In an ideal world, the color change would happen when you mix the two solutions together in exactly equation proportions. That particular mixture is known as the equivalence point. For example, if you were titrating sodium hydroxide solution with hydrochloric acid, both with a concentration of 1 mol dm , 25 cm of sodium hydroxide solution would need exactly the same volume of the acid - because they react 1 : 1 according to the equation. In this particular instance, this would also be the neutral point of the titration, because sodium chloride solution has a pH of 7. But that isn't necessarily true of all the salts you might get formed. For example, if you titrate ammonia solution with hydrochloric acid, you would get ammonium chloride formed. The ammonium ion is slightly acidic, and so pure ammonium chloride has a slightly acidic pH. That means that at the equivalence point (where you had mixed the solutions in the correct proportions according to the equation), the solution wouldn't actually be neutral. To use the term "neutral point" in this context would be misleading. Similarly, if you titrate sodium hydroxide solution with ethanoic acid, at the equivalence point the pure sodium ethanoate formed has a slightly alkaline pH because the ethanoate ion is slightly basic. To summarize: All the following titration curves are based on both acid and alkali having a concentration of 1 mol dm . In each case, you start with 25 cm of one of the solutions in the flask, and the other one in a burette. Although you normally run the acid from a burette into the alkali in a flask, you may need to know about the titration curve for adding it the other way around as well. Alternative versions of the curves have been described in most cases. We'll take hydrochloric acid and sodium hydroxide as typical of a strong acid and a strong base. You can see that the pH only falls a very small amount until quite near the equivalence point. Then there is a really steep plunge. If you calculate the values, the pH falls all the way from 11.3 when you have added 24.9 cm to 2.7 when you have added 25.1 cm . This is very similar to the previous curve except, of course, that the pH starts off low and increases as you add more sodium hydroxide solution. Again, the pH doesn't change very much until you get close to the equivalence point. Then it surges upwards very steeply. This time we are going to use hydrochloric acid as the strong acid and ammonia solution as the weak base. Because you have got a weak base, the beginning of the curve is obviously going to be different. However, once you have got an excess of acid, the curve is essentially the same as before. At the very beginning of the curve, the pH starts by falling quite quickly as the acid is added, but the curve very soon gets less steep. This is because a buffer solution is being set up - composed of the excess ammonia and the ammonium chloride being formed. Notice that the equivalence point is now somewhat acidic ( a bit less than pH 5), because pure ammonium chloride isn't neutral. However, the equivalence point still falls on the steepest bit of the curve. That will turn out to be important in choosing a suitable indicator for the titration. At the beginning of this titration, you have an excess of hydrochloric acid. The shape of the curve will be the same as when you had an excess of acid at the start of a titration running sodium hydroxide solution into the acid. It is only after the equivalence point that things become different. A buffer solution is formed containing excess ammonia and ammonium chloride. This resists any large increase in pH - not that you would expect a very large increase anyway, because ammonia is only a weak base. We'll take ethanoic acid and sodium hydroxide as typical of a weak acid and a strong base. For the first part of the graph, you have an excess of sodium hydroxide. The curve will be exactly the same as when you add hydrochloric acid to sodium hydroxide. Once the acid is in excess, there will be a difference. Past the equivalence point you have a buffer solution containing sodium ethanoate and ethanoic acid. This resists any large fall in pH. The start of the graph shows a relatively rapid rise in pH but this slows down as a buffer solution containing ethanoic acid and sodium ethanoate is produced. Beyond the equivalence point (when the sodium hydroxide is in excess) the curve is just the same as that end of the HCl - NaOH graph. The common example of this would be ethanoic acid and ammonia. It so happens that these two are both about equally weak - in that case, the equivalence point is approximately pH 7. This is really just a combination of graphs you have already seen. Up to the equivalence point it is similar to the ammonia - HCl case. After the equivalence point it is like the end of the ethanoic acid - NaOH curve. Notice that there isn't any steep bit on this graph. Instead, there is just what is known as a "point of inflexion". That lack of a steep bit means that it is difficult to do a titration of a weak acid against a weak base. The way you normally carry out a titration involves adding the acid to the alkali. Here are reduced versions of the graphs described above so that you can see them all together. The overall equation for the reaction between sodium carbonate solution and dilute hydrochloric acid is: If you had the two solutions of the same concentration, you would have to use twice the volume of hydrochloric acid to reach the equivalence point - because of the 1 : 2 ratio in the equation. Suppose you start with 25 cm of sodium carbonate solution, and that both solutions have the same concentration of 1 mol dm . That means that you would expect the steep drop in the titration curve to come after you had added 50 cm of acid. The actual graph looks like this: The graph is more complicated than you might think - and curious things happen during the titration. You expect carbonates to produce carbon dioxide when you add acids to them, but in the early stages of this titration, no carbon dioxide is given off at all. Then - as soon as you get past the half-way point in the titration - lots of carbon dioxide is suddenly released. The graph is showing two end points - one at a pH of 8.3 (little more than a point of inflection), and a second at about pH 3.7. The reaction is obviously happening in two distinct parts. In the first part, complete at in the diagram, the sodium carbonate is reacting with the acid to produce sodium hydrogencarbonate: You can see that the reaction doesn't produce any carbon dioxide. In the second part, the sodium hydrogencarbonate produced goes on to react with more acid - giving off lots of CO . That reaction is finished at on the graph. It is possible to pick up both of these end points by careful choice of indicator. That is explained on the separate page on indicators. Ethanedioic acid was previously known as oxalic acid. It is a diprotic acid, which means that it can give away 2 protons (hydrogen ions) to a base. Something which can only give away one (like HCl) is known as a monoprotic acid. The reaction with sodium hydroxide takes place in two stages because one of the hydrogens is easier to remove than the other. The two successive reactions are: If you run sodium hydroxide solution into ethanedioic acid solution, the pH curve shows the end points for both of these reactions. The curve is for the reaction between sodium hydroxide and ethanedioic acid solutions of equal concentrations. | 7,900 | 1,926 |
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