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https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/Electrophilic_Substitution_Reactions/C._The_Friedel-Crafts_Acylation_of_Benzene |
This page gives you the facts and a simple, uncluttered mechanism for the electrophilic substitution reaction between benzene and ethanoyl chloride in the presence of an aluminium chloride catalyst. An acyl group is an alkyl group attached to a carbon-oxygen double bond. If "R" represents any alkyl group, then an acyl group has the formula RCO-. Acylation means substituting an acyl group into something - in this case, into a benzene ring. The most commonly used acyl group is CH CO-. This is called the ethanoyl group. In the example which follows we are substituting a CH CO- group into the ring, but you could equally well use any other alkyl group instead of the CH . The most reactive substance containing an acyl group is an acyl chloride (also known as an acid chloride). These have the general formula RCOCl. Benzene is treated with a mixture of ethanoyl chloride, CH COCl, and aluminium chloride as the catalyst. A ketone called phenylethanone is formed. or better: The aluminium chloride isn't written into these equations because it is acting as a catalyst. If you wanted to include it, you could write AlCl over the top of the arrow. The electrophile is CH CO . It is formed by reaction between the ethanoyl chloride and the aluminium chloride catalyst. The hydrogen is removed by the AlCl ion which was formed at the same time as the CH CO electrophile. The aluminium chloride catalyst is re-generated in this second stage. | 1,454 | 1,427 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Basics_Thermodynamics_(General_Chemistry)/Entropy |
Entropy is a chemical concept that is very difficult to explain, because a one-sentence definition will not lead to a comprehensive statement. Thus, few people understand what entropy really is. You are not alone if you have some difficulty with this concept. The word entropy is used in many other places and for many other aspects. We confine our discussion to thermodynamics (science dealing with heat and changes) and to chemical and physical processes. We have define energy as the driving force for changes, entropy is also a driving force for physical and chemical changes (reactions). Entropy, symbol S, is related to energy, but it a different aspect of energy. This concept was developed over a long period of time. Human experienced chemical and physical changes that cannot be explained by energy alone. A different concept is required to explain spontaneous changes such as the expansion of a gas into an abailable empty space (vacumm) and heat transfer from a hot body into a cold body. These changes cause an increase in entropy for the system under consideration, but energy is not transferred into or out of the system. Traditionally, the entropy concept is associated with the second and third laws of thermodynamics. Entropy is related to the energy distribution of energy states of a collection of molecules, and this aspect is usually discussed in statistical mechanics. When a system receives an amount of energy at a constant temperature, , the entropy increase is defined by the following equation. D = . Entropy is the amount of energy transferred divided by the temperature at which the process takes place. Thus, entropy has the units of energy unit per Kelvin, J K . If the process takes place over a range of temperature, the quantity can be evaluated by adding bits of entropies at various temperatures. This sum can take the form of integration if the temperature various contineously. You have learned the concept of integration in a calculus course. in that it depends only on the initial and final state of the system, regardless of the path by which the changes take place. However, the changes are supposedly take place over a long period of time, or in an or condition. If the change takes place quickly in an manner, the entropy is greater than what is evaluated, because the temperature increase is not uniform. Nature has a tendency for entropy to increase, and the system changes in response to this tendency. Such a change is called a Thus, . This statement is one of the acceptable statement of second law of thermodynamics. Sorry for being so formal, but just so that you know you know something classical. Calculate entropy change when 36.0 g of ice melts at 273 K and 1 atm. If you look up the enthalpy of fusion for ice in a table, you would get a molar enthalpy of 6.01 kJ/mol. d = (6.01 kJ mol-1)/272 K * (36 g)/(18 g mol-1)
= 1.22 kJ / K
Gibbs Paradox illustrates an interesting aspect of entropy. By definition, the change in entropy can be evaluated by measuring the amount of energy transferred. Entropy contained in a system, say in a mole of a pure substance, is a that takes account of all heat transferred to it since the lowest atainable temperature, 0 K. At absolute zero Kelvin, the substance contains no removable energy. A substance is in a completely ordered crystalline state, at which the moleules contains no removalbe vibrational, rotational, translational, or even thermal disorder energy. As energy is absorbed by a substance, its temperature increases by . If the heat capacity is , then The area below the curve of a plot of versus from 0 K to is then the of the system. Of course, the heat capacity may also vary with temperature. In a phase transition, heat is absorbed, but the temperature remain constant. The entropy increase is . Recall that thermodynamic values at standard condition are called values. Thus, the entropies so evaluated = 298 K are called . They have been carefully measured for many substances. For example, the standard molar entropy of some solids are given below: The standard molar entropies (standard entropy per mole) for gases are usually higher because heat of melting and heat of vaporization must be included. The standard molar entropies for noble gases are: Note that the entropy increase as the atomic mass increase. The same trends is also found for the halogens, but the entropies for these diatomic gases are much greater than those of monoatomic noble gases. Standard entropy of some compounds have also been measured. For example, From the variations of standard entropies of these substances, one can conclude that The relationship between entropy and disorder is studied in a discipline called statistical mechanics. Simple illustrations are used here to explain the relationship of entropy and disorder. Entropy is also related to probability, as a measure of randomness or disorder, and entropy is proportional to the logarithm of the probability. The formula given by (and inscribed on the tomstone of) Ludwig Boltzmann, = ln sugests that the entropy, , is proportional to the natural logarithm (ln) of the number of possible state, . The proportional constant (=1.38*10 J/K) is called the Boltzmann's constant. When you toss three coins, the possibility of having 2 heads and 1 tail or 2 tails and 1 head are much higher than having 3 heads or 3 tails, becuase the possibility of having the former is three times higher than the latter groupings. Similarly, when two gases are placed in a container, the chances of having them separated in the two halves are much less than having them mixed. The mixed state have a higher entropy than the unmixed state. Thus, the general rule says: . Standard entropies of reaction, D , equals the entropy of products minus the entropy of reactants. Since the entropies of most substances have been measured and tabulated in handbooks and data banks, standard entropies of reactions can be evaluated in a similar manner as enthalpies of reactions. Similarly, when you put a drop of ink into a cup of water, the ink disperse into the water, or water disperse into the ink. In this process, the entropy is higher, because an increase of disorder results in an increase of entropy. Evaluate the entropy change for the reaction: CO + 3 H -> CH + H O in which all reactants and products are gaseous.
Standard entropies of reaction, , equals the entropy of products minus the entropy of reactants. The standard entropies of the reactants and products have been given above, and for clarity, the entropies are given below the formula of the reactants and products: (data are given in 3 significant digits) = ((186 + 189) - (198 + 3*131)) J (K mol)
= -216 J (K mol)
There are four (4) moles of reactants and two (2) moles of products (all in gaseous state), hense the large negative change in entropy for the reaction. Is the entropy change for the electrolytic decomposition of water, H O(l) -> 2 H + O , positive or negative? Gibbs Paradox illustrates an interesting aspect of entropy. The description above only covered some useful aspect of this important concept of entropy . O -> 2 O
Evaluate entropy changes for phase transitions from the enthalpies of phase transition.
Evaluate entropy changes of reactions.
Give trends of variation of standard molar entropies based on molecular structures and molar mass.
Actually, the difference is not very large for these two similar compounds.
Calculate the entropy of phase transition. At standard condition, the stable phase for water is liquid. However, for standard molar entropy of water vapour is / higher than that of water.
This estimate is very crude, because the process takes place only at elivated temperatures. | 7,841 | 1,429 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Complex_Ion_Chemistry/Reactions_of_the_Hexaaqua_Ions_with_Carbonate_Ions |
This page describes and explains the reactions between complex ions of the type \(\ce{[M(H2O)6]^{n+}}\) and carbonate ions from, for example, sodium carbonate solution. There is a difference in the reactions depending on whether the metal at the center of the hexaaqua ion carries a 2+ or a 3+ charge. We need to look at each cases separately. The 3+ hexaaqua ions are sufficiently acidic to react with carbonate ions to release carbon dioxide gas and precipitate a metal hydroxide. These have the formula \(\ce{[M(H2O)6]^{3+}}\), and they are fairly acidic. They react with water molecules from the solution: \[ \ce{ [M(H2O)6]^{3+} (aq) + H2O <=> [M(H2O)6]^{2+} (aq) + H3O^{+} (aq) }\] They are acting as acids by donating hydrogen ions to water molecules in the solution. Because of the confusing presence of water from two different sources (the ligands and the solution), it is easier to simplify this: \[\ce{ [M(H2O)6]^{3+} (aq) <=> [M(H2O)6]^{2+} (aq) + H^{+} (aq)} \] Carbonate ions combine with hydrogen ions in two stages - first to make hydrogencarbonate ions, and then to give carbon dioxide and water. \[ \ce{CO3^{2-} (aq) + H^{+} (aq) <=> HCO3^{-} + H2O (l)}\] \[ \ce{HCO3^{-} (aq) + H^{+} (aq) <=> CO2 (g) + H2O (l)}\] Provided the proportions are right, the 3+ hexaaqua ions are sufficiently acidic for the reactions to go all the way to carbon dioxide. There are two possible reactions. You can also usefully write the complete change as an overall reaction. \[\ce{ 2 [M(H2O)6]^{3+} (aq) + 3CO3^{2-} -> 2[M(H2O)3(OH)3] (s) + 3CO2 (g) + 3H2O (l)} \] The 2+ hexaaqua ions aren't strongly acidic enough to release carbon dioxide from carbonates. In these cases, you still get a precipitate - but it is a precipitate of what is loosely described as the "metal carbonate". \[ \ce{M^{2+} (aq) + CO_3^{2-} (aq) \rightarrow MCO3 (s)}\] We'll look at the reactions of three 3+ ions and three 2+ ions. The important ones are the 3+ ions. Starting from a colourless solution, you get a white precipitate - but with bubbles of gas as well. The precipitate is identical to the one you get if you add small amounts of either sodium hydroxide or ammonia solutions to a solution of the hexaaquaaluminium ions. Again, the precipitate is just the same as if you had added small amounts of either sodium hydroxide or ammonia solution. . . . and again, exactly the same precipitate as if you had added any other base. In each case you get a precipitate of the neutral complex - the metal hydroxide. This is exactly the same precipitate that you get if you add small amounts of either sodium hydroxide solution or ammonia solution to solutions of these ions. Bubbles of carbon dioxide are also given off. No gas this time - just a precipitate of "cobalt(II) carbonate". Again, there isn't any carbon dioxide - just a precipitate of the "copper(II) carbonate". You get a precipitate of the "iron(II) carbonate", but no carbon dioxide. Hexaaqua ions with a 2+ charge aren't sufficiently acidic to liberate carbon dioxide from carbonate ions. Instead you get a precipitate which you can think of as being the metal carbonate. Jim Clark ( ) | 3,143 | 1,431 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/22%3A_Helmholtz_and_Gibbs_Energies/22.05%3A_Thermodynamic_Functions_have_Natural_Variables |
The fundamental thermodynamic equations follow from five primary thermodynamic definitions and describe internal energy, enthalpy, Helmholtz energy, and Gibbs energy in terms of their natural variables. Here they will be presented in their differential forms. The fundamental thermodynamic equations describe the thermodynamic quantities U, H, G, and A in terms of their natural variables. The term "natural variable" simply denotes a variable that is one of the convenient variables to describe U, H, G, or A. When considered as a whole, the four fundamental equations demonstrate how four important thermodynamic quantities depend on variables that can be controlled and measured experimentally. Thus, they are essentially equations of state, and using the fundamental equations, experimental data can be used to determine sought-after quantities like \(G\) or \(H\). The first law of thermodynamics is represented below in its differential form \[ dU = đq+đw \nonumber \] where The "đ" symbol represent and indicates that both \(q\) and \(w\) are path functions. Recall that \(U\) is a state function. The first law states that internal energy changes occur only as a result of heat flow and work done. It is assumed that w refers only to PV work, where \[ w = -\int{pdV} \nonumber \] The fundamental thermodynamic equation for internal energy follows directly from the first law and the principle of Clausius: \[ dU = đq + đw \nonumber \] \[ dS = \dfrac{\delta q_{rev}}{T} \nonumber \] we have \[ dU = TdS + \delta w \nonumber \] Since only \(PV\) work is performed, \[ dU = TdS - pdV \label{DefU} \] The above equation is the fundamental equation for \(U\) with natural variables of entropy \(S\) and volume\(V\). The states that the entropy change of a system is equal to the ratio of heat flow in a reversible process to the temperature at which the process occurs. Mathematically this is written as \[ dS = \dfrac{\delta q_{rev}}{T} \nonumber \] where Mathematically, enthalpy is defined as \[ H = U + pV \label{DefEnth} \] where \(H\) is enthalpy of the system, p is pressure, and V is volume. The fundamental thermodynamic equation for enthalpy follows directly from it deffinition (Equation \(\ref{DefEnth}\)) and the fundamental equation for internal energy (Equation \(\ref{DefU}\)) : \[ dH = dU + d(pV) \nonumber \] \[ = dU + pdV + VdP \nonumber \] \[ dU = TdS - pdV \nonumber \] \[ dH = TdS - pdV + pdV + Vdp \nonumber \] \[ dH = TdS + Vdp \nonumber \] The above equation is the fundamental equation for H. The natural variables of enthalpy are S and p, entropy and pressure. The mathematical description of Gibbs energy is as follows \[ G = U + pV - TS = H - TS \label{Defgibbs} \] where \(G\) is the Gibbs energy of the system. The fundamental thermodynamic equation for Gibbs Energy follows directly from its definition \(\ref{Defgibbs}\) and the fundamental equation for enthalpy \(\ref{DefEnth}\): \[ dG = dH - d(TS) \nonumber \] \[ = dH - TdS - SdT \nonumber \] Since \[ dH = TdS + VdP \nonumber \] \[ dG = TdS + VdP - TdS - SdT \nonumber \] \[ dG = VdP - SdT \label{EqGibbs1} \] The above equation is the fundamental equation for G. The natural variables of Gibbs energy are P and T. Mathematically, Helmholtz energy is defined as \[ A = U - TS \label{DefHelm} \] where \(A\) is the Helmholtz energy of the system, which is often written as the symbol \(F\). The fundamental thermodynamic equation for Helmholtz energy follows directly from its definition (Equation \(\ref{DefHelm}\)) and the fundamental equation for internal energy (Equation \(\ref{DefU}\)): \[ dA = dU - d(TS) \nonumber \] \[ = dU - TdS - SdT \nonumber \] Since \[ dU = TdS - pdV \nonumber \] \[ dA = TdS - pdV -TdS - SdT \nonumber \] \[ dA = -pdV - SdT \label{EqHelm1} \] Equation \(\ref{EqHelm1}\) is the fundamental equation for A with natural variables of \(V\) and \(T\). For the definitions to hold, it is assumed that PV work is done and that processes are used. These assumptions are required for the first law and the principle of Clausius to remain valid. Also, these equations do not account include n, the number of moles, as a variable. When \(n\) is included, the equations appear different, but the essence of their meaning is captured without including the n-dependence. The fundamental equations derived above were not dependent on changes in the amounts of species in the system. Below the n-dependent forms are presented . \[ dU = TdS - PdV + \sum_{i=1}^{N}\mu_idn_i \nonumber \] \[ dH = TdS + VdP + \sum_{i=1}^{N}\mu_idn_i \nonumber \] \[ dG = -SdT + Vdp + \sum_{i=1}^{N}\mu_idn_i \nonumber \] \[ dA = -SdT - PdV + \sum_{i=1}^{N}\mu_idn_i \nonumber \] where μ is the chemical potential of species i and dn is the change in number of moles of substance i. The differential fundamental equations describe U, H, G, and A in terms of their natural variables. The natural variables become useful in understanding not only how thermodynamic quantities are related to each other, but also in analyzing relationships between measurable quantities (i.e. P, V, T) in order to learn about the thermodynamics of a system. Below is a table summarizing the natural variables for U, H, G, and A: The fundamental thermodynamic equations are the means by which the Maxwell relations are derived . The Maxwell Relations can, in turn, be used to group thermodynamic functions and relations into more general "families" . As we said dA is an . Let's write is out in its natural variables (Equation \(\ref{EqHelm1}\)) and take a cross derivative. The dA expression in natural variables is \[dA = \left( \dfrac{\partial A}{\partial V} \right)_T dV + \left( \dfrac{\partial A}{\partial T} \right) _V dT \nonumber \] The partial derivatives of A of first order can already be quite interesting we see e.g. in step 2 that the first partial of A versus V (at T constant) is the negative of the pressure. \[ \left( \dfrac{\partial A}{\partial V} \right)_T = -P \nonumber \] Likewise we find the (isochoric) slope with temperature gives us the negative of the entropy. Thus entropy is one of the of A. \[\left( \dfrac{\partial A}{\partial T} \right) _V = -S \nonumber \] When we apply a cross derivative \[ \left( \dfrac{\partial^2 A}{\partial V \partial T} \right) = \left( \dfrac{\partial (-S)}{\partial V} \right) _T + \left( \dfrac{\partial (-P)}{\partial T} \right) _V \nonumber \] we get what is known as a : \[ \left( \dfrac{\partial P}{\partial T} \right) _V = \left( \dfrac{\partial S}{\partial V} \right) _T \nonumber \] What does Equation three mean for the heat capacity? A similar treatment of dG (Equation \(\ref{EqGibbs1}\) gives: \[\left( \dfrac{\partial G}{\partial T} \right) _P = -S \nonumber \] \[\left( \dfrac{\partial G}{\partial P} \right) _T = V \nonumber \] and another Maxwell relation \[ - \left( \dfrac{\partial S}{\partial P} \right) _T = \left( \dfrac{\partial V}{\partial T} \right) _P \nonumber \] \[ \left (\dfrac{\partial H}{\partial P} \right)_{T,n} = -T \left(\dfrac{\partial V}{\partial T} \right)_{P,n} +V \nonumber \] Then apply this equation to an ideal gas. Does the result seem reasonable?
5. Using the definition of Gibbs energy and the conditions observed at phase equilibria, derive the Clapeyron equation. | 7,307 | 1,432 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/12%3A_Equilibrium_Conditions_in_Multicomponent_Systems/12.01%3A_Effects_of_Temperature |
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\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) For some of the derivations in this chapter, we will need an expression for the rate at which the ratio \(\mu_i/T\) varies with temperature in a phase of fixed composition maintained at constant pressure. This expression leads, among other things, to an important relation between the temperature dependence of an equilibrium constant and the standard molar reaction enthalpy. In a phase containing species \(i\), either pure or in a mixture, the partial derivative of \(\mu_i/T\) with respect to \(T\) at constant \(p\) and a fixed amount of each species is given by \begin{equation} \bPd{\left( \mu_i/T \right)}{T}{p,\allni} =\frac{1}{T}\Pd{\mu_i}{T}{\!\!p,\allni} - \frac{\mu_i}{T^2} \tag{12.1.1} \end{equation} This equality comes from a purely mathematical operation; no thermodynamics is involved. The relation is obtained from the formula \(\dif (uv)/\dx = u(\dif v/\dx) + v(\dif u/\dx)\) (Appendix E), where \(u\) is \(1/T\), \(v\) is \(\mu_i\), and \(x\) is \(T\). The partial derivative \(\pd{\mu_i}{T}{p,\allni}\) is equal to \(-S_i\) (Eq. 9.2.48), so that Eq. 12.1.1 becomes \begin{equation} \bPd{\left( \mu_i/T \right)}{T}{p,\allni} = -\frac{S_i}{T} - \frac{\mu_i}{T^2} = -\frac{TS_i + \mu_i}{T^2} \tag{12.1.2} \end{equation} The further substitution \(\mu_i = H_i - TS_i\) (Eq. 9.2.46) gives finally \begin{equation} \bPd{\left( \mu_i/T \right)}{T}{p,\allni} = -\frac{H_i}{T^2} \tag{12.1.3} \end{equation} For a pure substance in a closed system, Eq. 12.1.3 when multiplied by the amount \(n\) becomes \begin{equation} \bPd{\left( G/T \right)}{T}{p} = -\frac{H}{T^2} \tag{12.1.4} \end{equation} This is the . If we make the substitution \(\mu_i = \mu_i\st + RT\ln a_i\) in Eq. 12.1.3 and rearrange, we obtain \begin{equation} \frac{\dif(\mu_i\st/T)}{\dif T} = -\frac{H_i}{T^2} - R\Pd{\ln a_i}{T}{\!\!p,\allni} \tag{12.1.5} \end{equation} Because \(\mu_i\st/T\) is a function only of \(T\), its derivative with respect to \(T\) is itself a function only of \(T\). We can therefore use any convenient combination of pressure and composition in the expression on the right side of Eq. 12.1.5 in order to evaluate \(\dif(\mu_i\st/T)/\dif T\) at a given temperature. If species \(i\) is a constituent of a gas mixture, we take a constant pressure of the gas that is low enough for the gas to behave ideally. Under these conditions \(H_i\) is the standard molar enthalpy \(H_i\st\) (Eq. 9.3.7). In the expression for activity, \(a_i\gas = \G_i\gas \phi_i p_i/p\) (Table 9.5), the pressure factor \(\G_i\gas\) is constant when \(p\) is constant, the fugacity coefficient \(\phi_i\) for the ideal gas is unity, and \(p_i/p=y_i\) is constant at constant \(\allni\), so that the partial derivative \(\bpd{\ln a_i\gas}{T}{p,\allni}\) is zero. For component \(i\) of a condensed-phase mixture, we take a constant pressure equal to the standard pressure \(p\st\), and a mixture composition in the limit given by Eqs. 9.5.20–9.5.24 in which the activity coefficient is unity. \(H_i\) is then the standard molar enthalpy \(H_i\st\), and the activity is given by an expression in Table 9.5 with the pressure factor and activity coefficient set equal to 1: \(a_i{=}x_i\), \(a\A{=}x\A\), \(a\xbB{=}x\B\), \(a\cbB{=}c\B/c\st\), or \(a\mbB{=}m\B/m\st\). With the exception of \(a\cbB\), these activities are constant as \(T\) changes at constant \(p\) and \(\allni\). If solute B is an electrolyte, \(a\mbB\) is given instead by Eq. 10.3.10; like \(a\mbB\) for a nonelectrolyte, it is constant as \(T\) changes at constant \(p\) and \(\allni\). Thus for a gas-phase species, or a species with a standard state based on mole fraction or molality, \(\bpd{\ln a_i\gas}{T}{p,\allni}\) is zero and Eq. 12.1.5 becomes \begin{gather} \s{ \frac{\dif(\mu_i\st/T)}{\dif T} = -\frac{H_i\st}{T^2} } \tag{12.1.6} \cond{(standard state not based} \nextcond{on concentration)} \end{gather} Equation 12.1.6, as the conditions of validity indicate, does not apply to a solute standard state based on concentration, except as an approximation. The reason is the volume change that accompanies an isobaric temperature change. We can treat this case by considering the following behavior of \(\ln(c\B/c\st)\): \begin{equation} \begin{split} \bPd{\ln(c\B/c\st)}{T}{p,\allni} & = \frac{1}{c\B}\Pd{c\B}{T}{\!\!p,\allni} = \frac{1}{n\B/V}\bPd{(n\B/V)}{T}{p,\allni} \cr & = V\bPd{(1/V)}{T}{p,\allni} = -\frac{1}{V}\Pd{V}{T}{\!\!p,\allni} \cr & = -\alpha \end{split} \tag{12.1.7} \end{equation} Here \(\alpha\) is the cubic expansion coefficient of the solution (Eq. 7.1.1). If the activity coefficient is to be unity, the solution must be an ideal-dilute solution, and \(\alpha\) is then \(\alpha\A^*\), the cubic expansion coefficient of the pure solvent. Eq. 12.1.5 for a nonelectrolyte becomes \begin{equation} \frac{\dif(\mu\cbB\st/T)}{\dif T} = -\frac{H\B\st}{T^2} + R\alpha\A^* \tag{12.1.8} \end{equation} The thermodynamic equilibrium constant \(K\), for a given reaction equation and a given choice of reactant and product standard states, is a function of \(T\) and of \(T\). By equating two expressions for the standard molar reaction Gibbs energy, \(\Delsub{r}G\st= \sum_i\!\nu_i \mu_i\st\) and \(\Delsub{r}G\st=-RT\ln K\) (Eqs. 11.8.3 and 11.8.10), we obtain \begin{equation} \ln K = -\frac{1}{RT} \sum_i\nu_i\mu_i\st \tag{12.1.9} \end{equation} The rate at which \(\ln K\) varies with \(T\) is then given by \begin{equation} \frac{\dif\ln K}{\dif T} = -\frac{1}{R} \sum_i\nu_i \frac{\dif(\mu_i\st/T)}{\dif T} \tag{12.1.10} \end{equation} Combining Eq. 12.1.10 with Eqs. 12.1.6 or 12.1.8, and recognizing that \(\sum_i\!\nu_i H_i\st\) is the standard molar reaction enthalpy \(\Delsub{r}H\st\), we obtain the final expression for the temperature dependence of \(\ln K\): \begin{equation} \frac{\dif\ln K}{\dif T} = \frac{\Delsub{r}H\st}{RT^2} - \alpha\A^* \!\!\!\! \sum_{\stackrel{\tx{ solutes,}} {\tx{ conc. basis}}} \!\!\!\! \nu_i \tag{12.1.11} \end{equation} The sum on the right side includes only solute species whose standard states are based on concentration. The expression is simpler if all solute standard states are based on mole fraction or molality: \begin{gather} \s{ \frac{\dif\ln K}{\dif T} = \frac{\Delsub{r}H\st}{RT^2} } \tag{12.1.12} \cond{(no solute standard states} \nextcond{based on concentration)} \end{gather} We can rearrange Eq. 12.1.12 to \begin{gather} \s{ \Delsub{r}H\st = RT^2\frac{\dif\ln K}{\dif T} } \tag{12.1.13} \cond{(no solute standard states} \nextcond{based on concentration)} \end{gather} We can convert this expression for \(\Delsub{r}H\st\) to an equivalent form by using the mathematical identity \(\dif (1/T) = -(1/T^2)\dif T\): \begin{gather} \s{ \Delsub{r}H\st = -R\frac{\dif\ln K}{\dif(1/T)} } \tag{12.1.14} \cond{(no solute standard states} \nextcond{based on concentration)} \end{gather} Equations 12.1.13 and 12.1.14 are two forms of the . They allow us to evaluate the standard molar reaction enthalpy of a reaction by a noncalorimetric method from the temperature dependence of \(\ln K\). For example, we can plot \(\ln K\) versus \(1/T\); then according to Eq. 12.1.14, the slope of the curve at any value of \(1/T\) is equal to \(-\Delsub{r}H\st/R\) at the corresponding temperature \(T\). A simple way to derive the equation for this last procedure is to substitute \(\Delsub{r}G\st=\Delsub{r}H\st-T\Delsub{r}S\st\) in \(\Delsub{r}G\st=-RT\ln K\) and rearrange to \begin{equation} \ln K = -\frac{\Delsub{r}H\st}{R}\left(\frac{1}{T}\right) + \frac{\Delsub{r}S\st}{R} \tag{12.1.15} \end{equation} Suppose we plot \(\ln K\) versus \(1/T\). In a small temperature interval in which \(\Delsub{r}H\st\) and \(\Delsub{r}S\st\) are practically constant, the curve will appear linear. According to Eq. 12.1.15, the curve in this interval has a slope of \(-\Delsub{r}H\st/R\), and the tangent to a point on the curve has its intercept at \(1/T{=}0\) equal to \(\Delsub{r}S\st/R\). When we apply Eq. 12.1.14 to the A(l)\(\ra\)A(g) of pure A, it resembles the Clausius–Clapeyron equation for the same process (Eq. 8.4.15). These equations are not exactly equivalent, however, as the comparison in Table 12.1 shows. | 15,536 | 1,433 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Book3A_Bioinorganic_Chemistry_(Bertini_et_al.)/05%3A_Dioxygen_Reactions/5.02%3A_Chemistry_of_Dioxygen |
The reduction potential for the four-electron reduction of dioxygen (Reaction 5.1) is a measure of the great oxidizing power of the dioxygen molecule. However, the reaction involves the transfer of four electrons, a process that rarely, if ever, occurs in one concerted step, as shown in Reaction (5.2). \[O_{2} \xrightarrow{e^{-}} O_{2}^{-} \xrightarrow{e^{-}, 2H^{+}} H_{2}O_{2} \xrightarrow{e^{-}, H^{+}} H_{2}O + OH \xrightarrow{e^{-}, H^{+}} 2 H_{2}O \tag{5.2}\] \[dioxygen \xrightarrow{e^{-}} superoxide \xrightarrow{e^{-}, 2H^{+}} hydrogen\; peroxide \xrightarrow{e^{-}, H^{+}} water + hyrdoxyl\; radical \xrightarrow{e^{-}, H^{+}} water\] Since most reducing agents can transfer at most one or two electrons at a time to an oxidizing agent, the thermodynamics of the one- and two-electron reductions of dioxygen must be considered in order to understand the overall mechanism. In aqueous solution, the most common pathway for dioxygen reduction in the absence of any catalyst is one-electron reduction to give superoxide. But this is the least favorable of the reaction steps that make up the full four-electron reduction (see Table 5.1) and requires a moderately strong reducing agent. Thus if only one-electron pathways are available for dioxygen reduction, the low reduction potential for one-electron reduction of O to O presents a barrier that protects vulnerable species from the full oxidizing power of dioxygen that comes from the subsequent steps. If superoxide is formed (Reaction 5.3), however, it disproportionates quite rapidly in aqueous solution (except at very high pH) to give hydrogen peroxide and dioxygen (Reaction 5.4). The stoichiometry of the overall reaction is therefore that of a net two-electron reduction (Reaction 5.5). It is thus impossible under normal conditions to distinguish one-electron and two-electron reaction pathways for the reduction of dioxygen in aqueous solution on the basis of stoichiometry alone. \[2O_{2} + 2e^{-} \rightarrow 2 O_{2}^{-} \tag{5.3}\] \[2O_{2}^{-} + 2 H^{+} \rightarrow H_{2}O_{2} + O_{2} \tag{5.4}\] \[O_{2} + 2 e^{-} + 2 H^{+} \rightarrow H_{2}O_{2} \tag{5.5}\] The thermodynamics of dioxygen reactions with organic substrates is also of importance in understanding dioxygen reactivity. The types of reactions that are of particular interest to us here are hydroxylation of aliphatic and aromatic C—H bonds and epoxidation of olefins, since these typical reactions of oxygenase enzymes are ones that investigators are trying to mimic using synthetic reagents. Some of the simpler examples of such reactions (plus the reaction of H for comparison) are given in the reactions in Table 5.2. It is apparent that all these reactions of dioxygen with various organic substrates in Table 5.2 are thermodynamically favorable. However, reactions of dioxygen with organic substrates in the absence of a catalyst are generally very slow, unless the substrate is a particularly good reducing agent. To understand the sluggishness of dioxygen reactions with organic substrates, we must consider the kinetic barriers to these reactions. The principal kinetic barrier to direct reaction of dioxygen with an organic substrate arises from the fact that the ground state of the dioxygen molecule is triplet, i.e., contains two unpaired electrons. Typical organic molecules that are representative of biological substrates have singlet ground states, i.e., contain no unpaired electrons, and the products resulting from their oxygenation also have singlet ground states. Reactions between molecules occur in shorter times than the time required for conversions from triplet to singlet spin. Therefore the number of unpaired electrons must remain the same before and after each elementary step of a chemical reaction. For these reasons, we know that it is impossible for Reaction (5.6) to go in one fast, concerted step. \[\frac{1}{2} \;^{3}O_{2} + \;^{1}X \rightarrow \;^{1}XO \tag{5.6}\] \[\qquad \uparrow \uparrow \qquad \downarrow \uparrow \quad \qquad \downarrow \uparrow\] The arrows represent electron spins: \(\downarrow \uparrow\) represents a singlet molecule with all electron spins paired; \(\uparrow \uparrow\) represents a triplet molecule with two unpaired electrons; and \(\uparrow\) (which we will see in Reaction 5.13) represents a doublet molecule, also referred to as a free radical, with one unpaired electron. The pathways that do not violate the spin restriction are all costly in energy, resulting in high activation barriers. For example, the reaction of ground-state triplet dioxygen, i.e., O , with a singlet substrate to give the excited triplet state of the oxygenated product (Reaction 5.7) is spin-allowed, and one could imagine a mechanism in which this process is followed by a slow spin conversion to a singlet product (Reaction 5.8). \[\frac{1}{2} \;^{3}O_{2} + \;^{1}X \rightarrow \;^{3}XO \tag{5.7}\] \[\qquad \uparrow \uparrow \qquad \downarrow \uparrow \qquad \quad \uparrow \uparrow\] \[ \;^{3}XO \xrightarrow{slow} \;^{1}XO \tag{5.38}\] \[ \; \uparrow \uparrow \qquad \qquad \downarrow \uparrow\] But such a reaction pathway would give a high activation barrier, because the excited triplet states of even unsaturated molecules are typically 40-70 kcal/mol less stable than the ground state, and those of saturated hydrocarbons are much higher. Likewise, a pathway in which O is excited to a singlet state that then reacts with the substrate would be spin-allowed (Reactions 5.9 and 5.10). The high reactivity of singlet dioxygen, generated by photochemical or chemical means, is well-documented. However, such a pathway for a reaction of dioxygen, which is initially in its ground triplet state, would also require a high activation energy, since the lowest-energy singlet excited state of dioxygen is 22.5 kcal/mol higher in energy than ground-state triplet dioxygen. \[\;^{3}O_{2} + 22.5\; kcal/mol \rightarrow \;^{1}O_{2} \tag{5.9}\] \[ \uparrow \uparrow \qquad \qquad \qquad \qquad \qquad \downarrow \uparrow\] \[\frac{1}{2} \;^{1}O_{2} + \;^{1}X \rightarrow \;^{1}XO \tag{5.10}\] \[\quad \downarrow \uparrow \qquad \downarrow \uparrow \quad \qquad \downarrow \uparrow\] Moreover, the products of typical reactions of singlet-state dioxygen with organic substrates (Reactions 5.11 and 5.12, for example) are quite different in character from the reactions of dioxygen with organic substrates catalyzed by oxygenase enzymes (see Section V): \(\tag{5.11}\) \(\tag{5.12}\) One pathway for a direct reaction of triplet ground-state dioxygen with a singlet ground-state organic substrate that can occur readily without a catalyst begins with the one-electron oxidation of the substrate by dioxygen. The products of such a reaction would be two doublets, i.e., superoxide and the oneelectron oxidized substrate, each having one unpaired electron (Reaction 5.13). These free radicals can diffuse apart and then recombine with their spins paired (Reaction 5. 14). \[\;^{3}O_{2} + \;^{1}X \rightarrow \;^{2}O_{2}^{-} + \;^{2}X^{+} \tag{5.13}\] \[\uparrow \uparrow \qquad \downarrow \uparrow \quad \quad \uparrow \qquad \uparrow\] \[\;^{2}O_{2}^{-} + \;^{2}X^{+} \rightarrow \;^{2}O_{2}^{-} + 2X^{+} \rightarrow \;^{1}XO_{2} \tag{5.14}\] \[\uparrow \qquad \uparrow \qquad \qquad \uparrow \qquad \downarrow \qquad \qquad \downarrow \uparrow\] Such a mechanism has been shown to occur for the reaction of dioxygen with reduced flavins shown in Reaction (5.15). \(\tag{5.15}\) However, this pathway requires that the substrate be able to reduce dioxygen to superoxide, a reaction that requires an unusually strong reducing agent (such as a reduced flavin), since dioxygen is not a particularly strong one-electron oxidizing agent (see Table 5.1 and discussion above). Typical organic substrates in enzymatic and nonenzymatic oxygenation reactions usually are not sufficiently strong reducing agents to reduce dioxygen to superoxide; so this pathway is not commonly observed. The result of these kinetic barriers to dioxygen reactions with most organic molecules is that uncatalyzed reactions of this type are usually quite slow. An exception to this rule is an oxidation pathway known as free-radical autoxidation. The term free-radical autoxidation describes a reaction pathway in which dioxygen reacts with an organic substrate to give an oxygenated product in a free-radical chain process that requires an initiator in order to get the chain reaction started. (A free-radical initiator is a compound that yields free radicals readily upon thermal or photochemical decomposition.) The mechanism of free radical autoxidation is as shown in Reactions (5.16) to (5.21). Initiation: $$X_{2} \rightarrow 2X \cdotp \tag{5.16}\] \[X \cdotp + RH \rightarrow XH + R \cdotp \tag{5.17}\] Propagation: $$R \cdotp + O_{2} \rightarrow ROO \cdotp \tag{5.18}\] \[\downarrow \qquad \uparrow \uparrow \qquad \quad \uparrow\] \[ROO \cdotp + RH \rightarrow ROOH + R \cdotp \tag{5.19}\] Termination: $$R \cdotp + ROO \cdotp \rightarrow ROOR \tag{5.20}\] \[2 ROO \cdotp \rightarrow ROOOOR \rightarrow O_{2} + ROOR \tag{5.21}\] (plus other oxidized products, such as ROOH, ROH, RC(O)R, RC(O)H). This reaction pathway results in oxygenation of a variety of organic substrates, and is not impeded by the spin restriction, because triplet ground-state dioxygen can react with the free radical R• to give a free-radical product ROO•, in a spin-allowed process (Reaction 5.18). It is a chain reaction, since R• is regenerated in Reaction (5.19), and it frequently occurs with long chain lengths prior to the termination steps, resulting in a very efficient pathway for oxygenation of some organic substrates, such as, for example, the oxidation of cumene to give phenol and acetone (Reaction 5.22). \(\tag{5.22}\) When free-radical autoxidation is used for synthetic purposes, initiators are intentionally added. Common initiators are peroxides and other compounds capable of fragmenting readily into free radicals. Free-radical autoxidation reactions are also frequently observed when no initiator has been intentionally added, because organic substrates frequently contain peroxidic impurities that may act as initiators. Investigators have sometimes been deceived into assuming that a metal-complex catalyzed reaction of dioxygen with an organic substrate occurred by a nonradical mechanism. In such instances, the reactions later proved, upon further study, to be free-radical autoxidations, the role of the metal complex having been to generate the initiating free radicals. Although often useful for synthesis of oxygenated derivatives of relatively simple hydrocarbons, free-radical autoxidation lacks selectivity and therefore, with more complex substrates, tends to give multiple products. In considering possible mechanisms for biological oxidation reactions used for biosynthesis or energy production, free-radical autoxidation is not an attractive possibility, because such a mechanism requires diffusion of highly reactive free radicals. Such radicals, produced in the cell, will react indiscriminately with vulnerable sites on enzymes, substrates, and other cell components, causing serious damage. In fact, free-radical autoxidation is believed to cause certain deleterious reactions of dioxygen in biological systems, for example the oxidation of lipids in membranes. It is also the process that causes fats and oils to become rancid (Reaction 5.23). \(\tag{5.23}\) We see then the reasons that uncatalyzed reactions of dioxygen are usually either slow or unselective. The functions of the metalloenzymes for which dioxygen is a substrate are, therefore, to overcome the kinetic barriers imposed by spin restrictions or unfavorable one-electron reduction pathways, and, for the oxygenase enzymes, to direct the reactions and make them highly specific. It is instructive to consider (1) how these metalloenzymes function to lower the kinetic barriers to dioxygen reactivity, and (2) how the oxygenase enzymes redirect the reactions along different pathways so that very different products are obtained. The first example given below is cytochrome c oxidase. This enzyme catalyzes the four-electron reduction of dioxygen. It overcomes the kinetic barriers to dioxygen reduction by binding dioxygen to two paramagnetic metal ions at the dioxygen binding site, thus overcoming the spin restriction, and by reducing dioxygen in a two-electron step to peroxide, thus bypassing the unfavorable one-electron reduction to form free superoxide. The reaction occurs in a very controlled fashion, so that the energy released by dioxygen reduction can be used to produce ATP. A second example is provided by the catechol dioxygenases, which appear to represent substrate rather than dioxygen activation, and in which dioxygen seems to react with the substrate while it is complexed to the paramagnetic iron center. Another example given below is the monooxygenase enzyme cytochrome PASO, which catalyzes the reaction of dioxygen with organic substrates. It binds dioxygen at the paramagnetic metal ion at its active site, thus overcoming the spin restriction, and then carries out what can be formally described as a multielectron reduction of dioxygen to give a highly reactive high-valent metal-oxo species that has reactivity like that of the hydroxyl radical. Unlike a free hydroxyl radical, however, which would be highly reactive but nonselective, the reaction that occurs at the active site of cytochrome P-450 can be highly selective and stereospecific, because the highly reactive metal-oxo moiety is generated close to a substrate that is bound to the enzyme in such a way that it directs the reactive oxygen atom to the correct position. Thus, metalloenzymes have evolved to bind dioxygen and to increase while controlling its reactivity. | 13,842 | 1,434 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/24%3A_Solutions_I_-_Volatile_Solutes/24.06%3A_Vapor_Pressures_of_Volatile_Binary_Solutions |
The behaviors of ideal solutions of volatile compounds follow Raoult’s Law. Henry’s Law can be used to describe the deviations from ideality. Henry's law states: \[ P_B = k_H P_B^o\] For which the Henry’s Law constant (\(k_H\)) is determined for the specific compound. Henry’s Law is often used to describe the solubilities of gases in liquids. The relationship to Raoult’s Law is summarized in Figure \(\Page {1}\). Henry’s Law is depicted by the upper straight line and Raoult’s Law by the lower. The solubility of \(CO_2(g)\) in water at 25 C is 3.32 x 10 M with a partial pressure of \(CO_2\) over the solution of 1 bar. Assuming the density of a saturated solution to be 1 kg/L, calculate the Henry’s Law constant for \(CO_2\). : In one L of solution, there is 1000 g of water (assuming the mass of CO dissolved is negligible.) \[ (1000 \,g) \left( \dfrac{1\, mol}{18.02\,g} \right) = 55\, mol\, H_2O\] The solubility of \(CO_2\) can be used to find the number of moles of \(CO_2\) dissolved in 1 L of solution also: \[ \dfrac{3.32 \times 10^{-2} mol}{L} \cdot 1 \,L = 3.32 \times 10^{-2} mol\, CO_2\] and so the mol fraction of \(CO_2\) is \[ \chi_b = \dfrac{3.32 \times 10^{-2} mol}{55.5 \, mol} = 5.98 \times 10^{-4}\] And so \[10^5\, Pa = 5.98 \times 10^{-4} k_H\] or \[ k_H = 1.67 \times 10^9\, Pa\] An azeotrope is defined as the common composition of vapor and liquid when they have the same composition. Azeotropes can be either maximum boiling or minimum boiling, as show in Figure \(\Page {2; left}\). Regardless, distillation cannot purify past the azeotrope point, since the vapor and the liquid phases have the same composition. If a system forms a minimum boiling azeotrope and has a range of compositions and temperatures at which two liquid phases exist, the phase diagram might look like Figure \(\Page {2; right}\): Another possibility that is common is for two substances to form a two-phase liquid, form a minimum boiling azeotrope, but for the azeotrope to boil at a temperature below which the two liquid phases become miscible. In this case, the phase diagram will look like Figure \(\Page {3}\). In the diagram, make up of a system in each region is summarized below the diagram. The point e indicates the azeotrope composition and boiling temperature. Within each two-phase region (III, IV, and the two-phase liquid region, the lever rule will apply to describe the composition of each phase present. So, for example, the system with the composition and temperature represented by point b (a single-phase liquid which is mostly compound A, designated by the composition at point a, and vapor with a composition designated by that at point c), will be described by the lever rule using the lengths of tie lines and . What happens when Raoult does hold over the whole range? Recall that in a gas: \[μ_j = μ_j^o + RT \ln \dfrac{P_j}{P^o} \label{B} \] or \[μ_j = μ_j^o + RT \ln P_j \nonumber \] after dropping \(P^o=1\; bar\) out of the notation. Note that numerically this does not matter, since \(P_j\) is now to be dimensionless. Let's consider \(dμ_1\) at constant temperature: \[dμ_1 = RT\left(\dfrac{\partial \ln P_1}{ \partial x_1}\right)dx_1 \nonumber \] likewise: \[dμ_2 = RT\left(\dfrac{\partial \ln P_2}{ \partial x_2}\right)dx_2 \nonumber \] If we substitute into the Gibbs-Duhem expression we get: \[x_1 \left(\dfrac{∂\ln P_1}{ ∂x_1}\right) dx_1+x_2 \left(\dfrac{∂\ln P_2}{∂x_2} \right) dx_2=0 \nonumber \] Because \(dx_1= -dx_2\): \[x_ 1 \left( \dfrac{∂\ln P_1}{ ∂x_1} \right) =x_2 \left( \dfrac{∂\ln P_2}{∂x_2} \right) \nonumber \] (This is an alternative way of writing Gibbs-Duhem). If in the limit for \(x_1 \rightarrow 1\) Raoult Law holds then \[P_1 \rightarrow x_1P^*_1 \nonumber \] Thus: \[ \dfrac{∂ \ln P_1}{∂x_1} = \dfrac{1}{x_1} \nonumber \] and \[\dfrac{x_1}{x_1}=x_2 \dfrac{∂ \ln P_2}{∂x_2} \nonumber \] \[1=x_2 \dfrac{∂ \ln P_2}{ ∂x_2} \nonumber \] \[\dfrac{1}{x_2}= \dfrac{∂ \ln P_2}{∂x_2} \label{EqA12} \] We can integrate Equation \(\ref{EqA12}\) to form a logarithmic impression, but it will have an integration constant: \[\ln P_2 =\ln x_2 + constant \nonumber \] This constant of integration can be folded into the logarithm as a multiplicative constant, \(K\) \[\ln P_2 = \ln \left(K x_2 \right) \nonumber \] So for \(x_1 \rightarrow 1\) (i.e., \(x_2 \rightarrow 0\)), we get that \[P_2=K x_2 \nonumber \] where \(K\) is constant, but not necessarily \(P^*\). What this shows is that when one component follows Raoult the other follow Henry and vice versa. (Note that the ideal case is a subset of this case, in that the value of \(K\) then becomes \(P^*\) and the linearity must hold over the whole range.) Of course a big drawback of the Henry law is that it only describes what happens at the two extremes of the phase diagram and not in the middle. In cases of moderate non-ideality, it is possible to describe the whole range (at least in good approximation) using a : \[P_1= \left(x_1P^*_1 \right)f_{Mar} \nonumber \] The function \(f_{Mar}\) has the shape: \[f_{Mar}= \text{exp} \left[ αx_2^2+βx_2^3+δx_2^3 + .... \right] \nonumber \] Notice that the Margules function involves the mole fraction of the component. It is an exponential with a series expansion. with the constant and linear term missing. As you can see the function has a number of parameters \(α\), \(β\), \(δ\) etc. that need to be determined by experiment. In general, the more the system diverges from ideality, the more parameters you need. Using Gibbs-Duhem is is possible to translate the expression for \(P_1\) into the corresponding one for \(P_2\). | 5,629 | 1,435 |
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Nearly all of us have heated a pan of water with the lid in place and shortly thereafter heard the sounds of the lid rattling and hot water spilling onto the stovetop. When a liquid is heated, its molecules obtain sufficient kinetic energy to overcome the forces holding them in the liquid and they escape into the gaseous phase. By doing so, they generate a population of molecules in the vapor phase above the liquid that produces a pressure—the vapor pressure of the liquid. In the situation we described, enough pressure was generated to move the lid, which allowed the vapor to escape. If the vapor is contained in a sealed vessel, however, such as an unvented flask, and the vapor pressure becomes too high, the flask will explode (as many students have unfortunately discovered). In this section, we describe vapor pressure in more detail and explain how to quantitatively determine the vapor pressure of a liquid. Because the molecules of a liquid are in constant motion, we can plot the fraction of molecules with a given kinetic energy ( ) against their kinetic energy to obtain the kinetic energy distribution of the molecules in the liquid (Figure \(\Page {1}\)), just as we did for a gas. As for gases, increasing the temperature increases both the average kinetic energy of the particles in a liquid and the range of kinetic energy of the individual molecules. If we assume that a minimum amount of energy (\(E_0\)) is needed to overcome the intermolecular attractive forces that hold a liquid together, then some fraction of molecules in the liquid always has a kinetic energy greater than \(E_0\). The fraction of molecules with a kinetic energy greater than this minimum value increases with increasing temperature. Any molecule with a kinetic energy greater than \(E_0\) has enough energy to overcome the forces holding it in the liquid and escape into the vapor phase. Before it can do so, however, a molecule must also be at the surface of the liquid, where it is physically possible for it to leave the liquid surface; that is, only molecules at the surface can undergo evaporation (or vaporization), where molecules gain sufficient energy to enter a gaseous state above a liquid’s surface, thereby creating a vapor pressure. To understand the causes of vapor pressure, consider the apparatus shown in Figure \(\Page {2}\). When a liquid is introduced into an evacuated chamber (part (a) in Figure \(\Page {2}\)), the initial pressure above the liquid is approximately zero because there are as yet no molecules in the vapor phase. Some molecules at the surface, however, will have sufficient kinetic energy to escape from the liquid and form a vapor, thus increasing the pressure inside the container. As long as the temperature of the liquid is held constant, the fraction of molecules with \(KE > E_0\) will not change, and the rate at which molecules escape from the liquid into the vapor phase will depend only on the surface area of the liquid phase. As soon as some vapor has formed, a fraction of the molecules in the vapor phase will collide with the surface of the liquid and reenter the liquid phase in a process known as condensation (part (b) in Figure \(\Page {2}\)). As the number of molecules in the vapor phase increases, the number of collisions between vapor-phase molecules and the surface will also increase. Eventually, a will be reached in which exactly as many molecules per unit time leave the surface of the liquid (vaporize) as collide with it (condense). At this point, the pressure over the liquid stops increasing and remains constant at a particular value that is characteristic of the liquid at a given temperature. The rates of evaporation and condensation over time for a system such as this are shown graphically in Figure \(\Page {3}\). Two opposing processes (such as evaporation and condensation) that occur at the same rate and thus produce no change in a system, constitute a dynamic equilibrium. In the case of a liquid enclosed in a chamber, the molecules continuously evaporate and condense, but the amounts of liquid and vapor do not change with time. The pressure exerted by a vapor in dynamic equilibrium with a liquid is the equilibrium vapor pressure of the liquid. If a liquid is in an container, however, most of the molecules that escape into the vapor phase will collide with the surface of the liquid and return to the liquid phase. Instead, they will diffuse through the gas phase away from the container, and an equilibrium will never be established. Under these conditions, the liquid will continue to evaporate until it has “disappeared.” The speed with which this occurs depends on the vapor pressure of the liquid and the temperature. Volatile liquids have relatively high vapor pressures and tend to evaporate readily; nonvolatile liquids have low vapor pressures and evaporate more slowly. Although the dividing line between volatile and nonvolatile liquids is not clear-cut, as a general guideline, we can say that substances with vapor pressures greater than that of water (Figure \(\Page {4}\)) are relatively volatile, whereas those with vapor pressures less than that of water are relatively nonvolatile. Thus diethyl ether (ethyl ether), acetone, and gasoline are volatile, but mercury, ethylene glycol, and motor oil are nonvolatile. The equilibrium vapor pressure of a substance at a particular temperature is a characteristic of the material, like its molecular mass, melting point, and boiling point. It does depend on the amount of liquid as long as at least a tiny amount of liquid is present in equilibrium with the vapor. The equilibrium vapor pressure does, however, depend very strongly on the temperature and the intermolecular forces present, as shown for several substances in Figure \(\Page {4}\). Molecules that can hydrogen bond, such as ethylene glycol, have a much lower equilibrium vapor pressure than those that cannot, such as octane. The nonlinear increase in vapor pressure with increasing temperature is steeper than the increase in pressure expected for an ideal gas over the corresponding temperature range. The temperature dependence is so strong because the vapor pressure depends on the fraction of molecules that have a kinetic energy greater than that needed to escape from the liquid, and this fraction increases exponentially with temperature. As a result, sealed containers of volatile liquids are potential bombs if subjected to large increases in temperature. The gas tanks on automobiles are vented, for example, so that a car won’t explode when parked in the sun. Similarly, the small cans (1–5 gallons) used to transport gasoline are required by law to have a pop-off pressure release. Volatile substances have low boiling points and relatively weak intermolecular interactions; nonvolatile substances have high boiling points and relatively strong intermolecular interactions. A Discussing Vapor Pressure and Boiling Points. The exponential rise in vapor pressure with increasing temperature in Figure \(\Page {4}\) allows us to use natural logarithms to express the nonlinear relationship as a linear one. \[ \boxed{\ln P =\dfrac{-\Delta H_{vap}}{R}\left ( \dfrac{1}{T} \right) + C} \label{Eq1} \] where Plotting \(\ln P\) versus the inverse of the absolute temperature (\(1/T\)) is a straight line with a slope of −Δ / . Equation \(\ref{Eq1}\), called the , can be used to calculate the \(ΔH_{vap}\) of a liquid from its measured vapor pressure at two or more temperatures. The simplest way to determine \(ΔH_{vap}\) is to measure the vapor pressure of a liquid at temperatures and insert the values of \(P\) and \(T\) for these points into Equation \(\ref{Eq2}\), which is derived from the Clausius–Clapeyron equation: \[ \ln\left ( \dfrac{P_{1}}{P_{2}} \right)=\dfrac{-\Delta H_{vap}}{R}\left ( \dfrac{1}{T_{1}}-\dfrac{1}{T_{2}} \right) \label{Eq2} \] Conversely, if we know Δ and the vapor pressure \(P_1\) at any temperature \(T_1\), we can use Equation \(\ref{Eq2}\) to calculate the vapor pressure \(P_2\) at any other temperature \(T_2\), as shown in Example \(\Page {1}\). A Discussing the Clausius-Clapeyron Equation. Link: The experimentally measured vapor pressures of liquid Hg at four temperatures are listed in the following table: From these data, calculate the enthalpy of vaporization (Δ ) of mercury and predict the vapor pressure of the liquid at 160°C. (Safety note: mercury is highly toxic; when it is spilled, its vapor pressure generates hazardous levels of mercury vapor.) vapor pressures at four temperatures Δ of mercury and vapor pressure at 160°C The table gives the measured vapor pressures of liquid Hg for four temperatures. Although one way to proceed would be to plot the data using Equation \(\ref{Eq1}\) and find the value of Δ from the slope of the line, an alternative approach is to use Equation \(\ref{Eq2}\) to obtain Δ directly from two pairs of values listed in the table, assuming no errors in our measurement. We therefore select two sets of values from the table and convert the temperatures from degrees Celsius to kelvin because the equation requires absolute temperatures. Substituting the values measured at 80.0°C ( ) and 120.0°C ( ) into Equation \(\ref{Eq2}\) gives \[\begin{align*} \ln \left ( \dfrac{0.7457 \; \cancel{Torr}}{0.0888 \; \cancel{Torr}} \right) &=\dfrac{-\Delta H_{vap}}{8.314 \; J/mol\cdot K}\left ( \dfrac{1}{\left ( 120+273 \right)K}-\dfrac{1}{\left ( 80.0+273 \right)K} \right) \\[4pt] \ln\left ( 8.398 \right) &=\dfrac{-\Delta H_{vap}}{8.314 \; J/mol\cdot \cancel{K}}\left ( -2.88\times 10^{-4} \; \cancel{K^{-1}} \right) \\[4pt] 2.13 &=-\Delta H_{vap} \left ( -3.46 \times 10^{-4} \right) J^{-1}\cdot mol \\[4pt] \Delta H_{vap} &=61,400 \; J/mol = 61.4 \; kJ/mol \end{align*} \nonumber \] We can now use this value of Δ to calculate the vapor pressure of the liquid ( ) at 160.0°C ( ): \[ \ln\left ( \dfrac{P_{2} }{0.0888 \; torr} \right)=\dfrac{-61,400 \; \cancel{J/mol}}{8.314 \; \cancel{J/mol} \; K^{-1}}\left ( \dfrac{1}{\left ( 160+273 \right)K}-\dfrac{1}{\left ( 80.0+273 \right) K} \right) \nonumber \] Using the relationship \(e^{\ln x} = x\), we have \[\begin{align*} \ln \left ( \dfrac{P_{2} }{0.0888 \; Torr} \right) &=3.86 \\[4pt] \dfrac{P_{2} }{0.0888 \; Torr} &=e^{3.86} = 47.5 \\[4pt] P_{2} &= 4.21 Torr \end{align*} \nonumber \] At 160°C, liquid Hg has a vapor pressure of 4.21 torr, substantially greater than the pressure at 80.0°C, as we would expect. The vapor pressure of liquid nickel at 1606°C is 0.100 torr, whereas at 1805°C, its vapor pressure is 1.000 torr. At what temperature does the liquid have a vapor pressure of 2.500 torr? 1896°C As the temperature of a liquid increases, the vapor pressure of the liquid increases until it equals the external pressure, or the atmospheric pressure in the case of an open container. Bubbles of vapor begin to form throughout the liquid, and the liquid begins to boil. The temperature at which a liquid boils at exactly 1 atm pressure is the normal boiling point of the liquid. For water, the normal boiling point is exactly 100°C. The normal boiling points of the other liquids in Figure \(\Page {4}\) are represented by the points at which the vapor pressure curves cross the line corresponding to a pressure of 1 atm. Although we usually cite the normal boiling point of a liquid, the boiling point depends on the pressure. At a pressure greater than 1 atm, water boils at a temperature greater than 100°C because the increased pressure forces vapor molecules above the surface to condense. Hence the molecules must have greater kinetic energy to escape from the surface. Conversely, at pressures less than 1 atm, water boils below 100°C. Typical variations in atmospheric pressure at sea level are relatively small, causing only minor changes in the boiling point of water. For example, the highest recorded atmospheric pressure at sea level is 813 mmHg, recorded during a Siberian winter; the lowest sea-level pressure ever measured was 658 mmHg in a Pacific typhoon. At these pressures, the boiling point of water changes minimally, to 102°C and 96°C, respectively. At high altitudes, on the other hand, the dependence of the boiling point of water on pressure becomes significant. Table \(\Page {1}\) lists the boiling points of water at several locations with different altitudes. At an elevation of only 5000 ft, for example, the boiling point of water is already lower than the lowest ever recorded at sea level. The lower boiling point of water has major consequences for cooking everything from soft-boiled eggs (a “three-minute egg” may well take four or more minutes in the Rockies and even longer in the Himalayas) to cakes (cake mixes are often sold with separate high-altitude instructions). Conversely, pressure cookers, which have a seal that allows the pressure inside them to exceed 1 atm, are used to cook food more rapidly by raising the boiling point of water and thus the temperature at which the food is being cooked. As pressure increases, the boiling point of a liquid increases and vice versa. Use Figure \(\Page {4}\) to estimate the following. Data in Figure \(\Page {4}\), pressure, and boiling point corresponding boiling point and pressure Ethylene glycol is an organic compound primarily used as a raw material in the manufacture of polyester fibers and fabric industry, and polyethylene terephthalate resins (PET) used in bottling. Use the data in Figure \(\Page {4}\) to estimate the following. 200°C 450 mmHg Because the molecules of a liquid are in constant motion and possess a wide range of kinetic energies, at any moment some fraction of them has enough energy to escape from the surface of the liquid to enter the gas or vapor phase. This process, called or , generates a above the liquid. Molecules in the gas phase can collide with the liquid surface and reenter the liquid via . Eventually, a is reached in which the number of molecules evaporating and condensing per unit time is the same, and the system is in a state of . Under these conditions, a liquid exhibits a characteristic that depends only on the temperature. We can express the nonlinear relationship between vapor pressure and temperature as a linear relationship using the . This equation can be used to calculate the enthalpy of vaporization of a liquid from its measured vapor pressure at two or more temperatures. are liquids with high vapor pressures, which tend to evaporate readily from an open container; have low vapor pressures. When the vapor pressure equals the external pressure, bubbles of vapor form within the liquid, and it boils. The temperature at which a substance boils at a pressure of 1 atm is its . ( ) | 14,787 | 1,436 |
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The principle of is formulated for kinetic systems which are decomposed into elementary processes (collisions, or steps, or elementary reactions): Lewis put forward this general principle in 1925: According to Ter Haar, the essence of the detailed balance is: \(A\) \(B\) \(A\) \(B\) The principle of detailed balance was explicitly introduced for collisions by Ludwig Boltzmann. In 1872, he proved his H-theorem using this principle. The arguments in favor of this property are founded upon microscopic reversibility. In 1901, R. Wegscheider introduced the principle of detailed balance for chemical kinetics. In particular, he demonstrated that the irreversible cycles \[ A \rightarrow B \rightarrow C ... Y \rightarrow Z \label{1} \] are impossible and found explicitly the relations between kinetic constants that follow from the principle of detailed balance. This system is more accurately described thusly: \[ A \rightleftharpoons B \rightleftharpoons C ... Y \rightleftharpoons Z \label{2} \] | 1,027 | 1,437 |
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The functional group of alkyl halides is a carbon-halogen bond, the common halogens being fluorine, chlorine, bromine and iodine. With the exception of iodine, these halogens have electronegativities significantly greater than carbon. Consequently, this functional group is polarized so that the carbon is electrophilic and the halogen is nucleophilic, as shown in the drawing below. Two characteristics other than electronegativity also have an important influence on the chemical behavior of these compounds. The first of these is . The strongest of the carbon-halogen covalent bonds is that to fluorine. Remarkably, this is the strongest common single bond to carbon, being roughly 30 kcal/mole stronger than a carbon-carbon bond and about 15 kcal/mole stronger than a carbon-hydrogen bond. Because of this, , and do not share any of the reactivity patterns shown by the other alkyl halides. The carbon-chlorine covalent bond is slightly weaker than a carbon-carbon bond, and the bonds to the other halogens are weaker still, the bond to iodine being about 33% weaker. The second factor to be considered is the relative , which is likely the form in which these electronegative atoms will be replaced. This stability may be estimated from the relative acidities of the H-X acids, assuming that the strongest acid releases the most stable conjugate base (halide anion). With the exception of HF (pK = 3.2), all the hydrohalic acids are very strong, small differences being in the direction HCl < HBr < HI. The characteristics noted above lead us to anticipate certain types of reactions that are likely to occur with alkyl halides. The following table summarizes the expected outcome of alkyl halide reactions with nucleophiles. It is assumed that the alkyl halides have one or more beta-hydrogens, making elimination possible; and that low dielectric solvents (e.g. acetone, ethanol, tetrahydrofuran & ethyl acetate) are used. When a high dielectric solvent would significantly influence the reaction this is noted in red. .
( Weak Bases: I , Br , SCN , N ,
CH CO , RS , CN etc. ) pK 's from -9 to 10 (left to right)
( Strong Bases: HO , RO ) pK 's > 15
( H O, ROH, RSH, R N ) pK 's ranging from -2 to 11 | 2,287 | 1,438 |
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Various formulations of gunpowder were apparently discovered and used before 1000 AD in China, and its military use is documented during the Jin Dynasty(1115–1234). Rockets, guns, cannons, grenades, and bombs were used against invading Mongols. Since the late 19th century, the original formulation has been called "black powder" to distinguish it from modern smokeless varieties . Knowledge of gunpowder formulations, and of the products of their explosions, is essential in gunshot residue (GSR) analysis. Solids and gases resulting from explosion of gunpowder A modern black powder substitute for muzzleloading rifles in FFG size Black powder is usually 75% potassium nitrate (KNO , known as saltpeter or saltpetre), 15% softwood charcoal , and 10% sulfur (elemental S). Charcoal is made by heating wood with limited air, and is mostly carbon (elemental C), but contains trace minerals (such as potassium carbonate, K CO ) and some partially decomposed wood chemicals like lignin C H O , cellulose (C H O ) . There is no simple equation for the combustion of black powder because the products, as well as the reactants, are numerous and varied, as shown in this table: The main products are K CO , CO , and N , so the equation for the combustion can be given as But it is often simplified to : Sometimes, formulas for charcoal (like C H O) that approximate it's composition, but don't represent any actual compound in the charcoal, are used in place of C: Example 4 from Equations and Mass Relationships we noted that one reactant in a chemical equation may be completely consumed without using up all of another. A mixture like gunpowder is formulated for "average conditions", and some portion of reactants may be left unchanged after the reaction. Conversely, at least one reactant is usually completely consumed. When it is gone, the other excess reactants have nothing to react with and they cannot be converted to products. The substance which is used up first is the . If 100 g of black powder is made from the general recipe above (75 g KNO , 15 g C, 10 g S), and the combustion reaction is given by equation (2), which is the limiting reagent? What mass of solid product will be formed? The balanced equation Let's see how many moles of each we actually have \(\begin{align}
& n_{\text{KNO}_{\text{3}}}=\text{75.0 g}\times \frac{\text{1 mol KNO}_{\text{3}}}{\text{101.1 g}}=\text{0.742 mol KNO}_{\text{3}} \\
& n_{\text{S}}=\text{10.0 g}\times \frac{\text{1 mol S}}{\text{32.1 g}}=\text{0.312 mol S} \\
& n_{\text{C}}=\text{15.0 g}\times \frac{\text{1 mol C}}{\text{12.01 g}}=\text{1.249 mol C} \\
\end{align}\) Now we can use stoichiometric ratios to determine how much C and S would be required to react with all of the KNO : \(n_{\text{S}}=\text{0.742 mol KNO}_{\text{3}}~\times~\frac{\text{ mol S}}{\text{2 mol KNO}_{\text{3}}}~=~\text{0.371 mol S}\) \(n_{\text{C}}=\text{0.742 mol KNO}_{\text{3}}~\times~\frac{\text{3 mol C}}{\text{2 mol KNO}_{\text{3}}}~=~\text{1.13 mol C}\) Since only 0.312 mol S are present, and 0.371 mol S would be required to react with all of the KNO , it is clear that this can't happen, and KNO must be present in excess. One of the other reactants must be limiting. We can use stoichiometric ratios to discover how much KNO and S would be required if all the C reacts: \(n_{\text{KNO}_{\text{3}}}=\text{1.25 mol C}~\times~\frac{\text{2 mol KNO}_{\text{3}}}{\text{3 mol C}}~=~\text{0.833 mol KNO}_{\text{3}}\) \(n_{\text{S}}=\text{1.25 mol C}~\times~\frac{\text{1 mol S}}{\text{3 mol C}}~=~\text{0.416 mol S}\) We see that C is also present in excess, so S must be the limiting reactant. We can prove it by using stoichiometric ratios to find out that there is plenty of C and KNO to react with all the S: \(n_{\text{C}}=\text{0.312 mol S}~\times~\frac{\text{3 mol C}}{\text{1 mol S}}~=~\text{0.936 mol C}\) \(n_{\text{KNO}_{\text{3}}}=\text{0.312 mol S}~\times~\frac{\text{2 mol KNO}_{\text{3}}}{\text{1 mol S}}~=~\text{0.624 mol KNO}_{\text{3}}\) These calculations can be organized as a table, with entries below the respective reactants and products in the chemical equation. We can calculate (hypothetically) how much of each reactant would be required if the other were completely consumed to demonstrate which is in excess, and which is limiting. We use the amount of limiting reagent to calculate the amount of product formed. \(n_{\text{S}}~\xrightarrow{S\text{(K}_{\text{2}}\text{S/S)}}~n_{\text{K}_{\text{2}}\text{S}}\xrightarrow{M_{\text{K}_{\text{2}}\text{S}}} ~ m_{\text{K}_{\text{2}}\text{S}}\) \(m_{\text{K}_{\text{2}}\text{S}} ~=~ \text{0.312 mol S} ~\times~ \frac{\text{1 mol K}_{\text{2}}\text{S}}{\text{1 mol S}} ~ \times ~ \frac{\text{110.3 g}}{\text{mol K}_{\text{2}}\text{S}}=\text{34.4 g K}_{\text{2}}\text{S} \) When the reaction ends, there will be (0.742 – 0.624) mol KNO = 0.118 mol KNO , or 11.9 g left over. There will also be (1.25 - 0.936) = 0.314 mol C, or 3.76 g left over. S is therefore the limiting reagent. The left over solids in GSR (gunshot residue) are detected by swiping areas with adhesive coated samplers, which are then viewed with a scanning electron microscope to indentify the particles. From this example you can begin to see what needs to be done to determine which of two reagents, X or Y, is limiting. We must compare the stoichiometric ratio S(X/Y) with the actual ratio of amounts of X and Y which were initially mixed together. In Example 1 this ratio of initial amounts \(\frac{n_{\text{S}}\text{(initial)}}{n_{\text{KNO}_{\text{3}}}\text{(initial)}}=\frac{\text{0.312 mol S}}{\text{0.742 mol KNO}_{\text{3}}}=\frac{\text{0.420 mol S}}{\text{1 mol KNO}_{\text{3}}}\) was less than the stoichiometric ratio \(\text{S}\left( \frac{\text{S}}{\text{KNO}_{\text{3}}} \right)=\frac{\text{1 mol S}}{\text{2 mol KNO}_{\text{3}}}~=~ 0.5\) This indicated that there was not enough S to react with all the KNO and sulfur was the limiting reagent. The corresponding general rule, for any reagents X and Y, is \(\begin{align}
& \text{If}~ \frac{n_{\text{X}}\text{(initial)}}{n_{\text{Y}}\text{(initial)}}~\text{is less than S}\left( \frac{\text{X}}{\text{Y}} \right)\text{, then X is limiting}\text{.} \\
& \\
& \text{If}~\frac{n_{\text{X}}\text{(initial)}}{n_{\text{Y}}\text{(initial)}}~\text{is greater than S}\left( \frac{\text{X}}{\text{Y}} \right)\text{, then Y is limiting}\text{.} \\
\end{align}\) (Of course, when the amounts of X and Y are in exactly the stoichiometric ratio, both reagents will be completely consumed at the same time, and neither is in excess.). This general rule for determining the limiting reagent is applied in the next example. Iron can be obtained by reacting the ore hematite (Fe O ) with coke (C). The latter is converted to CO . As manager of a blast furnace you are told that you have 20.5 Mg (megagrams) of Fe O and 2.84 Mg of coke on hand. (a) Which should you order first—another shipment of iron ore or one of coke? (b) How many megagrams of iron can you make with the materials you have? a) Write a balanced equation 2Fe O + 3C → 3CO + 4Fe The stoichiometric ratio connecting C and Fe O is \(\text{S}\left( \frac{\text{C}}{\text{Fe}_{\text{2}}\text{O}_{\text{3}}} \right)=\frac{\text{3 mol C}}{\text{2 mol Fe}_{\text{2}}\text{O}_{\text{3}}}=\frac{\text{1}\text{.5 mol C}}{\text{1 mol Fe}_{\text{2}}\text{O}_{\text{3}}}\) The initial amounts of C and Fe O are calculated using appropriate molar masses \(\begin{align}
& \text{ }n_{\text{C}}\text{(initial)}=\text{2}\text{.84}\times \text{10}^{\text{6}}\text{g}\times \frac{\text{1 mol C}}{\text{12}\text{.01 g}}=\text{2}\text{.36}\times \text{10}^{\text{5}}\text{mol C} \\
& \\
& n_{\text{Fe}_{\text{2}}\text{O}_{\text{3}}}\text{(initial)}=\text{20}\text{.5}\times \text{10}^{\text{6}}\text{g}\times \frac{\text{1 mol Fe}_{\text{2}}\text{O}_{\text{3}}}{\text{159}\text{.69 g}}=\text{1}\text{.28}\times \text{10}^{\text{5}}\text{mol Fe}_{\text{2}}\text{O}_{\text{3}} \\
\end{align}\) Their ratio is \(\frac{n_{\text{C}}\text{(initial)}}{n_{\text{Fe}_{\text{2}}\text{O}_{\text{3}}}\text{(initial)}}=\frac{\text{2}\text{.36}\times \text{10}^{\text{5}}\text{mol C}}{\text{1}\text{.28}\times \text{10}^{\text{5}}\text{mol Fe}_{\text{2}}\text{O}_{\text{3}}}=\frac{\text{1}\text{.84 mol C}}{\text{1 mol Fe}_{\text{2}}\text{O}_{\text{3}}}\) Since this ratio is larger than the stoichiometric ratio, you have more than enough C to react with all the Fe O . Fe O is the limiting reagent, and you will want to order more of it first since it will be consumed first. b) The amount of product formed in a reaction may be calculated via an appropriate stoichiometric ratio from the amount of a reactant which was . Some of the excess reactant C will be left over, but all the initial amount of Fe O will be consumed. Therefore we use (initial) to calculate how much Fe can be obtained \(n_{\text{Fe}_{\text{2}}\text{O}_{\text{3}}}\text{ }\xrightarrow{S\text{(Fe/Fe}_{\text{2}}\text{O}_{\text{3}}\text{)}}\text{ }n_{\text{Fe}}\xrightarrow{M_{\text{Fe}}}\text{ }m_{\text{Fe}}\) \(m_{\text{Fe}}=\text{1}\text{.28 }\times \text{ 10}^{\text{5}}\text{ mol Fe}_{\text{2}}\text{O}_{\text{3}}\text{ }\times \text{ }\frac{\text{4 mol Fe}}{\text{2 mol Fe}_{\text{2}}\text{O}_{\text{3}}}\text{ }\times \text{ }\frac{\text{55}\text{.85 g}}{\text{mol Fe}}=\text{1}\text{.43 }\times \text{ 10}^{\text{7}}\text{ g Fe}\) This is 1.43 × 10 g, or 14.3 Mg, Fe. As you can see from the example, in a case where there is a limiting reagent, . Using the initial amount of a reagent present in excess would be incorrect, because such a reagent is not entirely consumed. The concept of a limiting reagent was used by the nineteenth century German chemist Justus von Liebig (1807 to 1873) to derive an important biological and ecological law. states that the essential substance available in the smallest amount relative to some critical minimum will control growth and reproduction of any species of plant or animal life. When a group of organisms runs out of that essential limiting reagent, the chemical reactions needed for growth and reproduction must stop. Vitamins, protein, and other nutrients are essential for growth of the human body and of human populations. Similarly, the growth of algae in natural bodies of water such as Lake Erie can be inhibited by reducing the supply of nutrients such as phosphorus in the form of phosphates. It is for this reason that many states have regulated or banned the use of phosphates in detergents and are constructing treatment plants which can remove phosphates from municipal sewage before they enter lakes or streams. | 10,671 | 1,439 |
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We take advantage of changes between the gas, liquid, and solid states to cool a drink with ice cubes (solid to liquid), cool our bodies by perspiration (liquid to gas), and cool food inside a refrigerator (gas to liquid and vice versa). We use dry ice, which is solid CO , as a refrigerant (solid to gas), and we make artificial snow for skiing and snowboarding by transforming a liquid to a solid. In this section, we examine what happens when any of the three forms of matter is converted to either of the other two. These changes of state are often called phase changes. The six most common phase changes are shown in Figure \(\Page {1}\). Phase changes are accompanied by a change in the energy of a system. For example, converting a liquid, in which the molecules are close together, to a gas, in which the molecules are, on average, far apart, requires an input of energy (heat) to give the molecules enough kinetic energy to allow them to overcome the intermolecular attractive forces. The stronger the attractive forces, the more energy is needed to overcome them. Solids, which are highly ordered, have the strongest intermolecular interactions, whereas gases, which are very disordered, have the weakest. Thus any transition from a more ordered to a less ordered state (solid to liquid, liquid to gas, or solid to gas) requires an input of energy; it is . Conversely, any transition from a less ordered to a more ordered state (liquid to solid, gas to liquid, or gas to solid) releases energy; it is . The energy change associated with each common phase change is shown in Figure \(\Page {1}\). Δ is positive for any transition from a more ordered to a less ordered state and negative for a transition from a less ordered to a more ordered state. Previously, we defined the enthalpy changes associated with various chemical and physical processes. The melting points and molar (\(ΔH_{fus}\)), the energy required to convert from a solid to a liquid, a process known as fusion (or melting), as well as the normal boiling points and enthalpies of vaporization (\(ΔH_{vap}\)) of selected compounds are listed in Table \(\Page {1}\). The substances with the highest melting points usually have the highest enthalpies of fusion; they tend to be ionic compounds that are held together by very strong electrostatic interactions. Substances with high boiling points are those with strong intermolecular interactions that must be overcome to convert a liquid to a gas, resulting in high enthalpies of vaporization. The enthalpy of vaporization of a given substance is much greater than its enthalpy of fusion because it takes more energy to completely separate molecules (conversion from a liquid to a gas) than to enable them only to move past one another freely (conversion from a solid to a liquid). Less energy is needed to allow molecules to move past each other than to separate them totally. The direct conversion of a solid to a gas, without an intervening liquid phase, is called sublimation. The amount of energy required to sublime 1 mol of a pure solid is the enthalpy of sublimation (Δ ). Common substances that sublime at standard temperature and pressure (STP; 0°C, 1 atm) include CO (dry ice); iodine (Figure \(\Page {2}\)); naphthalene, a substance used to protect woolen clothing against moths; and 1,4-dichlorobenzene. As shown in Figure \(\Page {1}\), the enthalpy of sublimation of a substance is the sum of its enthalpies of fusion and vaporization provided all values are at the same ; this is an application of . \[ΔH_{sub} =ΔH_{fus} +ΔH_{vap} \label{Eq1} \] Fusion, vaporization, and sublimation are endothermic processes; they occur only with the absorption of heat. Anyone who has ever stepped out of a swimming pool on a cool, breezy day has felt the heat loss that accompanies the evaporation of water from the skin. Our bodies use this same phenomenon to maintain a constant temperature: we perspire continuously, even when at rest, losing about 600 mL of water daily by evaporation from the skin. We also lose about 400 mL of water as water vapor in the air we exhale, which also contributes to cooling. Refrigerators and air-conditioners operate on a similar principle: heat is absorbed from the object or area to be cooled and used to vaporize a low-boiling-point liquid, such as ammonia or the (CFCs) and the hydrofluorocarbons (HCFCs). The vapor is then transported to a different location and compressed, thus releasing and dissipating the heat. Likewise, ice cubes efficiently cool a drink not because of their low temperature but because heat is required to convert ice at 0°C to liquid water at 0°C. The processes on the right side of Figure \(\Page {1}\)—freezing, condensation, and deposition, which are the reverse of fusion, sublimation, and vaporization—are exothermic. Thus heat pumps that use refrigerants are essentially air-conditioners running in reverse. Heat from the environment is used to vaporize the refrigerant, which is then condensed to a liquid in coils within a house to provide heat. The energy changes that occur during phase changes can be quantified by using a heating or cooling curve. Figure \(\Page {3}\) shows a heating curve, a plot of temperature versus heating time, for a 75 g sample of water. The sample is initially ice at 1 atm and −23°C; as heat is added, the temperature of the ice increases linearly with time. The slope of the line depends on both the mass of the ice and the specific heat ( ) of ice, which is the number of joules required to raise the temperature of 1 g of ice by 1°C. As the temperature of the ice increases, the water molecules in the ice crystal absorb more and more energy and vibrate more vigorously. At the melting point, they have enough kinetic energy to overcome attractive forces and move with respect to one another. As more heat is added, the temperature of the system does increase further but remains constant at 0°C until all the ice has melted. Once all the ice has been converted to liquid water, the temperature of the water again begins to increase. Now, however, the temperature increases more slowly than before because the specific heat capacity of water is than that of ice. When the temperature of the water reaches 100°C, the water begins to boil. Here, too, the temperature remains constant at 100°C until all the water has been converted to steam. At this point, the temperature again begins to rise, but at a rate than seen in the other phases because the heat capacity of steam is than that of ice or water. Thus . In this example, as long as even a tiny amount of ice is present, the temperature of the system remains at 0°C during the melting process, and as long as even a small amount of liquid water is present, the temperature of the system remains at 100°C during the boiling process. The rate at which heat is added does affect the temperature of the ice/water or water/steam mixture because the added heat is being used exclusively to overcome the attractive forces that hold the more condensed phase together. Many cooks think that food will cook faster if the heat is turned up higher so that the water boils more rapidly. Instead, the pot of water will boil to dryness sooner, but the temperature of the water does not depend on how vigorously it boils. The temperature of a sample does not change during a phase change. If heat is added at a constant rate, as in Figure \(\Page {3}\), then the length of the horizontal lines, which represents the time during which the temperature does not change, is directly proportional to the magnitude of the enthalpies associated with the phase changes. In Figure \(\Page {3}\), the horizontal line at 100°C is much longer than the line at 0°C because the enthalpy of vaporization of water is several times greater than the enthalpy of fusion. A superheated liquid is a sample of a liquid at the temperature and pressure at which it should be a gas. Superheated liquids are not stable; the liquid will eventually boil, sometimes violently. The phenomenon of superheating causes “bumping” when a liquid is heated in the laboratory. When a test tube containing water is heated over a Bunsen burner, for example, one portion of the liquid can easily become too hot. When the superheated liquid converts to a gas, it can push or “bump” the rest of the liquid out of the test tube. Placing a stirring rod or a small piece of ceramic (a “boiling chip”) in the test tube allows bubbles of vapor to form on the surface of the object so the liquid boils instead of becoming superheated. Superheating is the reason a liquid heated in a smooth cup in a microwave oven may not boil until the cup is moved, when the motion of the cup allows bubbles to form. The cooling curve, a plot of temperature versus cooling time, in Figure \(\Page {4}\) plots temperature versus time as a 75 g sample of steam, initially at 1 atm and 200°C, is cooled. Although we might expect the cooling curve to be the mirror image of the heating curve in Figure \(\Page {3}\), the cooling curve is an identical mirror image. As heat is removed from the steam, the temperature falls until it reaches 100°C. At this temperature, the steam begins to condense to liquid water. No further temperature change occurs until all the steam is converted to the liquid; then the temperature again decreases as the water is cooled. We might expect to reach another plateau at 0°C, where the water is converted to ice; in reality, however, this does not always occur. Instead, the temperature often drops below the freezing point for some time, as shown by the little dip in the cooling curve below 0°C. This region corresponds to an unstable form of the liquid, a supercooled liquid. If the liquid is allowed to stand, if cooling is continued, or if a small crystal of the solid phase is added (a seed crystal), the supercooled liquid will convert to a solid, sometimes quite suddenly. As the water freezes, the temperature increases slightly due to the heat evolved during the freezing process and then holds constant at the melting point as the rest of the water freezes. Subsequently, the temperature of the ice decreases again as more heat is removed from the system. Supercooling effects have a huge impact on Earth’s climate. For example, supercooling of water droplets in clouds can prevent the clouds from releasing precipitation over regions that are persistently arid as a result. Clouds consist of tiny droplets of water, which in principle should be dense enough to fall as rain. In fact, however, the droplets must aggregate to reach a certain size before they can fall to the ground. Usually a small particle (a ) is required for the droplets to aggregate; the nucleus can be a dust particle, an ice crystal, or a particle of silver iodide dispersed in a cloud during (a method of inducing rain). Unfortunately, the small droplets of water generally remain as a supercooled liquid down to about −10°C, rather than freezing into ice crystals that are more suitable nuclei for raindrop formation. One approach to producing rainfall from an existing cloud is to cool the water droplets so that they crystallize to provide nuclei around which raindrops can grow. This is best done by dispersing small granules of solid CO (dry ice) into the cloud from an airplane. Solid CO sublimes directly to the gas at pressures of 1 atm or lower, and the enthalpy of sublimation is substantial (25.3 kJ/mol). As the CO sublimes, it absorbs heat from the cloud, often with the desired results. A Discussing the Thermodynamics of Phase Changes. If a 50.0 g ice cube at 0.0°C is added to 500 mL of tea at 20.0°C, what is the temperature of the tea when the ice cube has just melted? Assume that no heat is transferred to or from the surroundings. The density of water (and iced tea) is 1.00 g/mL over the range 0°C–20°C, the specific heats of liquid water and ice are 4.184 J/(g•°C) and 2.062 J/(g•°C), respectively, and the enthalpy of fusion of ice is 6.01 kJ/mol. mass, volume, initial temperature, density, specific heats, and \(ΔH_{fus}\) final temperature Substitute the given values into the general equation relating heat gained (by the ice) to heat lost (by the tea) to obtain the final temperature of the mixture. When two substances or objects at different temperatures are brought into contact, heat will flow from the warmer one to the cooler. The amount of heat that flows is given by \[q=mC_sΔT \nonumber \] where \(q\) is heat, \(m\) is mass, \(C_s\) is the specific heat, and \(ΔT\) is the temperature change. Eventually, the temperatures of the two substances will become equal at a value somewhere between their initial temperatures. Calculating the temperature of iced tea after adding an ice cube is slightly more complicated. The general equation relating heat gained and heat lost is still valid, but in this case we also have to take into account the amount of heat required to melt the ice cube from ice at 0.0°C to liquid water at 0.0°C. The amount of heat gained by the ice cube as it melts is determined by its enthalpy of fusion in kJ/mol: \[q=nΔH_{fus} \nonumber \] For our 50.0 g ice cube: \[\begin{align*} q_{ice} &= 50.0 g⋅\dfrac{1\: mol}{18.02\:g}⋅6.01\: kJ/mol \\[4pt] &= 16.7\, kJ \end{align*} \nonumber \] Thus, when the ice cube has just melted, it has absorbed 16.7 kJ of heat from the tea. We can then substitute this value into the first equation to determine the change in temperature of the tea: \[q_{tea} = - 16,700 J = 500 mL⋅\dfrac{1.00\: g}{1\: mL}⋅4.184 J/(g•°C) ΔT \nonumber \] \[ΔT = - 7.98 °C = T_f - T_i \nonumber \] \[T_f = 12.02 °C \nonumber \] This would be the temperature of the tea when the ice cube has just finished melting; however, this leaves the melted ice still at 0.0°C. We might more practically want to know what the final temperature of the mixture of tea will be once the melted ice has come to thermal equilibrium with the tea. To determine this, we can add one more step to the calculation by plugging in to the general equation relating heat gained and heat lost again: \[\begin{align*} q_{ice} &= - q_{tea} \\[4pt] q_{ice} &= m_{ice}C_sΔT = 50.0g⋅4.184 J/(g•°C)⋅(T_f - 0.0°C) \\[4pt] &= 209.2 J/°C⋅T_f \end{align*} \nonumber \] \[q_{tea} = m_{tea}C_sΔT = 500g⋅4.184 J/(g•°C)⋅(T_f - 12.02°C) = 2092 J/°C⋅T_f - 25,150 J \nonumber \] \[209.2 J/°C⋅T_f = - 2092 J/°C⋅T_f + 25,150 J \nonumber \] \[2301.2 J/°C⋅T_f = 25,150 J \nonumber \] \[T_f = 10.9 °C \nonumber \] The final temperature is in between the initial temperatures of the tea (12.02 °C) and the melted ice (0.0 °C), so this answer makes sense. In this example, the tea loses much more heat in melting the ice than in mixing with the cold water, showing the importance of accounting for the heat of phase changes! Suppose you are overtaken by a blizzard while ski touring and you take refuge in a tent. You are thirsty, but you forgot to bring liquid water. You have a choice of eating a few handfuls of snow (say 400 g) at −5.0°C immediately to quench your thirst or setting up your propane stove, melting the snow, and heating the water to body temperature before drinking it. You recall that the survival guide you leafed through at the hotel said something about not eating snow, but you cannot remember why—after all, it’s just frozen water. To understand the guide’s recommendation, calculate the amount of heat that your body will have to supply to bring 400 g of snow at −5.0°C to your body’s internal temperature of 37°C. Use the data in Example \(\Page {1}\) 200 kJ (4.1 kJ to bring the ice from −5.0°C to 0.0°C, 133.6 kJ to melt the ice at 0.0°C, and 61.9 kJ to bring the water from 0.0°C to 37°C), which is energy that would not have been expended had you first melted the snow. Fusion, vaporization, and sublimation are endothermic processes, whereas freezing, condensation, and deposition are exothermic processes. Changes of state are examples of , or . All phase changes are accompanied by changes in the energy of a system. Changes from a more-ordered state to a less-ordered state (such as a liquid to a gas) are . Changes from a less-ordered state to a more-ordered state (such as a liquid to a solid) are always . The conversion of a solid to a liquid is called . The energy required to melt 1 mol of a substance is its enthalpy of fusion (Δ ). The energy change required to vaporize 1 mol of a substance is the enthalpy of vaporization (Δ ). The direct conversion of a solid to a gas is . The amount of energy needed to sublime 1 mol of a substance is its and is the sum of the enthalpies of fusion and vaporization. Plots of the temperature of a substance versus heat added or versus heating time at a constant rate of heating are called . Heating curves relate temperature changes to phase transitions. A , a liquid at a temperature and pressure at which it should be a gas, is not stable. A is not exactly the reverse of the heating curve because many liquids do not freeze at the expected temperature. Instead, they form a , a metastable liquid phase that exists below the normal melting point. Supercooled liquids usually crystallize on standing, or adding a of the same or another substance can induce crystallization. | 17,271 | 1,440 |
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There are many demonstrations that purport to exemplify mass relationships and stoichiometry, but they deal with solutions or gases. When reactions are demonstrated with solutions , students usually must (1) trust that a reported molarity is correct, and (2) deal with a layer of abstraction (reactant masses must be calculated from volumes, using the reported concentrations and molar masses. If gases are used (with a eudiometer, etc.) , the layer of abstraction exists through the ideal gas law. In reactions between weighable reactants, (1) the product must not be lost as a smoke, etc., and (2) one reactant must be easily removed, so that the mass of a reactant can be determined from the mass of product less the mass of the other, initially weighed reactant. There are several possibilities, but none gives truly quantitative data, so they are best carried out (possibly with openly "idealized" data) to illustrate how a more carefully controlled experiment could yield more accurate results. FeS is nonstoichiometric, but close to 1:1 . | 1,060 | 1,441 |
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Meticulously dry or oxygen-free conditions are sometimes necessary when using reagents that react with water or oxygen in the air. To safely and effectively use these reagents, glassware should be oven or flame dried, then the air displaced with a dry, inert gas (often nitrogen or argon). This creates an " " inside an apparatus, one that will not react with the reagents. Inert gases can be delivered to a flask through gas lines and a gas manifold (in a research setting, Figure 1.28), or through a balloon of inert gas (more common in teaching labs, Figure 1.29). The techniques shown in this section use balloons of nitrogen gas to create inert atmospheric conditions in a round bottomed flask, and syringes to transfer liquids from dry reagent bottles. These techniques can be easily adapted to use with a gas manifold if available. 1. Prepare a needle attachment for a balloon: Cut the end off a plastic \(1 \: \text{mL}\) syringe and fit the barrel into a piece of thick rubber tubing. Attach a helium-quality balloon to the rubber tubing, and seal all joints with Parafilm. Alternatively, attach a balloon directly to a \(2\)-\(3 \: \text{mL}\) plastic syringe. 2. Fill the balloon by connecting to a hose on the regulator of a tank of inert gas (nitrogen or argon, Figure 1.30a). Open the gas regulator to fill the balloon to between 7" – 8" in diameter (Figure 1.30b).
[For use with very sensitive reagents, the gas should first be passed through a column of drying agent.] 3. While holding the balloon close to your body, twist the balloon to prevent gas from escaping. Then attach a green needle (#21 gauge, \(0.8 \: \text{mm}\) × \(25 \: \text{mm}\), ) securely to the end of the syringe (Figure 1.30c). 4. To prevent gas from escaping when the balloon is untwisted, insert the needle into a rubber stopper (Figure 1.30d). The balloon can now be set aside while other parts of the setup are prepared. 5. Remove the surface water from a reagent flask (with stir bar if applicable), by either flame drying the flask or placing it in a hot oven for several hours. flask will be Use thick gloves to handle the hot glass.
6. Immediately insert a rubber septum (Figure 1.31a) into the ground glass joint. Fold one side of the septum over the lip of the flask and hold it in place while folding the opposite sides over as well (Figures 1.31b–d). This can be difficult to do with thick gloves. An alternative is to hold the flask against your body with the thick gloves, and fold the septum flaps over while using your bare hands (or thinner gloves, Figures 1.32 a+b).
7. Immediately secure the reaction flask to a ring stand or latticework using an extension clamp and insert the needle of the inert gas balloon into the inner circle on the septum (Figure 1.32c, see Figure 1.31d for the circle on the septum).
8. Insert a single needle into the circle on the septum (called an " ") to "flush" the air from the reaction flask (Figure 1.32d). The goal is to use the pressure from the balloon to force inert gas into the reaction flask and displace the air in the flask out the exit needle.
9. Allow the system to flush for at least 5 minutes if using nitrogen gas and perhaps 1-2 minutes if using argon gas (argon is denser than air so will displace the air more easily than nitrogen). Then remove the exit needle and allow the flask to fully cool under the balloon of inert gas. 10. If a mass is required of the empty flask, remove the inert gas balloon (insert the needle into a rubber stopper) and obtain the mass of the cool, empty flask with septum. 11. Remove a long, flexible needle from a hot oven and immediately into the barrel of a plastic syringe, freshly opened from its packaging (Figure 1.33a). The syringe needs to be able to hold a volume larger than the volume of reagent intended to deliver in order to have enough flexibility to properly manipulate the reagent. For example, a \(10\)-\(\text{mL}\) syringe is too small to deliver \(10 \: \text{mL}\) of reagent, but could be used to deliver \(7 \: \text{mL}\) of reagent. Hold the syringe such that the volume markings are visible, and connect the bent needle pointed , so that when screwed on (which normally requires roughly a half turn) the bent needle points downwards with the numbers visible. With this approach, the volume markings can be seen while withdrawing liquid, instead of being inconveniently on the back face of the syringe (as in Figure 1.33d). Glass syringes are often used with air-sensitive reagents dissolved in nonpolar solvents (e.g. hexanes), and require some further considerations that are not described in this section. Consult with your instructor for further instructions if you are to use a glass syringe. 12. Wrap the joint between the needle and syringe with Teflon tape or Parafilm (Figure 1.33b). 13. Flush the needle with inert gas: Insert the needle into the septum of an empty, dry flask attached to a balloon of inert gas (Figure 1.33c), withdraw a full volume of inert gas (Figure 1.33d), then expunge it into the air. 14. Immediately insert the flushed syringe into the reagent flask septum if nearby , or into a rubber stopper until the syringe is to be used. 15. A balloon of inert gas must be inserted into the reagent bottle in order to equalize pressures during withdrawal of liquid. A platform (e.g. ring clamp/wire mesh) should also be used beneath the reagent bottle if positioned above the bench, to provide support in case the bottle slips from the grasp of the clamp. 16. Insert the needle of the flushed syringe into the septum of the air-sensitive reagent, and into the liquid (Figure 1.34a). 17. withdraw some liquid into the syringe. If the plunger is pulled back too quickly, the low pressure inside the syringe may cause air to seep through the joint between the needle and syringe (through or around the Teflon tape or Parafilm). 18. Inevitably a bubble will form in the syringe. Keeping the syringe upside down and vertical (Figure 1.34b), push on the plunger to force the gas pocket back into the bottle. 19. Slowly withdraw liquid to \(1\)-\(2 \: \text{mL}\) greater than the desired volume (Figure 1.34c), then keeping the syringe vertical, expunge liquid back to the desired volume (Figure 1.34d shows \(2.0 \: \text{mL}\) of liquid). Withdrawing greater than the desired volume at first allows you to be confident that no gas bubbles are in the needle, and that you have measured an accurate volume. 20. The needle should be full of the air-sensitive reagent at this point, and if it were removed from the bottle the reagent would come into contact with the atmosphere at the needle tip. This can have disastrous consequences if the reagent is quite reactive (smoking or potentially fire). It is therefore essential that a "buffer" of inert gas (Figure 1.36) is placed between the air-sensitive reagent and the atmosphere before removing the needle. 21. To create the " ": a. Place the needle into the headspace of the reagent bottle (Figures 1.35 a+b). b. Keeping the syringe upside down and vertical, gently pull back on the plunger until a bubble is seen in the barrel (approximately \(20\%\) of the syringe capacity, Figure 1.35c). Immediately insert the syringe into the reaction flask septum if nearby, or into a rubber stopper if the flask is a distance away (Figure 1.35d). 22. With an inert gas balloon inserted in the reaction flask, place the syringe with reagent into the reaction flask septum. Keeping the syringe vertical, push on the plunger to first deliver the inert gas buffer (Figure 1.37a), then slowly deliver reagent to the flask. 23. Stop delivering reagent when the rubber plunger of the syringe meets the end of the barrel (Figure 1.37b). invert the syringe and push out the residual liquid: this would result in delivering a larger volume of reagent than measured by the syringe. 24. The needle will still be full of the air-sensitive reagent, so with the needle tip still in the headspace of the reaction flask, withdraw an inert gas buffer into the syringe. Insert the needle tip into a rubber stopper if the cleaning station is not nearby. 25. The syringe and needle should be cleaned as soon as possible, as over time deposits may form in the needle creating a plug. To clean the syringe and needle: a. Withdraw into the syringe a few \(\text{mL}\) of clean solvent similar to the solvent used in the air-sensitive solution (Figure 1.37c). For example, the pictures in this section show transfer of a \(\ce{BH_3}\) reagent dissolved in THF. An ideal rinse solvent would then be THF. As THF was not available, diethyl ether was a good substituted as the two solvents are structurally similar (they are both ethers). b. Expunge the solvent into a waste beaker. Repeat with another solvent rinse, being sure to rinse the entire area in the syringe where the reagent touched. c. Rinse the syringe once with water to dissolve and remove any inorganic salts. d. Further rinse the syringe and needle twice with a few \(\text{mL}\) of acetone. e. Remove the needle from the syringe and retain for future use. The plastic syringe should not be reused, but instead thrown away: solvent present in many air-sensitive solutions degrade the rubber plunger on the syringe, causing them to swell and be ineffective after one use. Fill a balloon with an inert gas (nitrogen or argon) to 7-8 inches in diameter. Twist the balloon to prevent gas from escaping, then attach a needle and insert into a rubber stopper to plug. Flame or oven dry a reaction flask (with stir bar, and fold a rubber septum over the joint while wearing thick gloves. Clamp the hot flask to the ring stand or latticework and insert the balloon of inert gas. Insert an " " and allow the flask to flush for ~5 minutes (to displace the air). Remove the exit needle and allow the flask to cool. Obtain a needle from the hot oven and into the tip of a freshly opened plastic syringe. Wrap the needle/syringe joint with Teflon tape or Parafilm. Flush the syringe with inert gas by using an empty flask attached to a balloon of inert gas. Withdraw a volume of gas and expunge into the air. Insert the needle tip into a rubber stopper to plug. Use the prepared syringe to withdraw some reagent (or air may enter the syringe). Push out the gas bubble. Withdraw reagent to a slightly greater volume than needed, then expunge liquid to the exact volume. Place the needle in the headspace of the bottle and withdraw an " ", ~\(20\%\) of the volume of the syringe. Insert the needle into a rubber stopper for transport. Into the reaction flask (with inert gas balloon), deliver first the inert gas buffer, then the air-sensitive reagent. Don't push out the residual liquid. Withdraw an inert gas buffer into the mostly empty syringe. Clean by rinsing with two portions (few \(\text{mL}\) each) of solvent (similar to reagent solvent), then one portion of water and two portions of acetone. | 10,949 | 1,442 |
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The properties of a substance depends upon its physical state. Water vapor, liquid water and ice all have the same properties, but their properties are considerably different. In general determine: molecular shape, bond energies, properties, while (non-covalent bonds) influence the properties of liquids and solids. The kinetic molecular theory of gases gives a reasonably accurate description of the behavior of gases. A similar model can be applied to liquids, but it must take into account the nonzero volumes of particles and the presence of strong intermolecular attractive forces. The of a substance depends on the balance between the of the individual particles (molecules or atoms) and the . The kinetic energy keeps the molecules apart and moving around, and is a function of the temperature of the substance. The intermolecular forces are attractive forces that try to draw the particles together (Figure \(\Page {2}\)). A discussed previously, gasses are very sensitive to temperatures and pressure. However, these also affect liquids and solids too. Heating and cooling can change the of the particles in a substance, and so, we can change the physical state of a substance by heating or cooling it. Increasing the pressure on a substance forces the molecules closer together, which the strength of intermolecular forces. Below is an overview of the general properties of the three different phases of matter. Thus, liquids can be poured and assume the shape of their containers. Due to the strong intermolecular forces between neighboring molecules, solids are rigid. In a gas, the distance between molecules, whether monatomic or polyatomic, is very large compared with the size of the molecules; thus gases have a low density and are highly compressible. In contrast, the molecules in liquids are very close together, with essentially no empty space between them. As in gases, however, the molecules in liquids are in constant motion, and their kinetic energy (and hence their speed) depends on their temperature. We begin our discussion by examining some of the characteristic properties of liquids to see how each is consistent with a modified kinetic molecular description. The properties of liquids can be explained using a modified version of the described previously. This model explains the higher density, greater order, and lower compressibility of liquids versus gases; the thermal expansion of liquids; why they diffuse; and why they adopt the shape (but not the volume) of their containers. A kinetic molecular description of liquids must take into account both the nonzero volumes of particles and the presence of strong intermolecular attractive forces. Solids and liquids have particles that are fairly close to one another, and are thus called " " to distinguish them from gases ( ) | 2,858 | 1,443 |
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Because galvanic cells can be self-contained and portable, they can be used as batteries and fuel cells. A is a galvanic cell (or a series of galvanic cells) that contains all the reactants needed to produce electricity. In contrast, a is a galvanic cell that requires a constant external supply of one or more reactants to generate electricity. In this section, we describe the chemistry behind some of the more common types of batteries and fuel cells. There are two basic kinds of batteries: disposable, or primary, batteries, in which the electrode reactions are effectively irreversible and which cannot be recharged; and rechargeable, or secondary, batteries, which form an insoluble product that adheres to the electrodes. These batteries can be recharged by applying an electrical potential in the reverse direction. The recharging process temporarily converts a rechargeable battery from a galvanic cell to an electrolytic cell. Batteries are cleverly engineered devices that are based on the same fundamental laws as galvanic cells. The major difference between batteries and the galvanic cells we have previously described is that commercial batteries use solids or pastes rather than solutions as reactants to maximize the electrical output per unit mass. The use of highly concentrated or solid reactants has another beneficial effect: the concentrations of the reactants and the products do not change greatly as the battery is discharged; consequently, the output voltage remains remarkably constant during the discharge process. This behavior is in contrast to that of the Zn/Cu cell, whose output decreases logarithmically as the reaction proceeds (Figure \(\Page {1}\)). When a battery consists of more than one galvanic cell, the cells are usually connected in series—that is, with the positive (+) terminal of one cell connected to the negative (−) terminal of the next, and so forth. The overall voltage of the battery is therefore the sum of the voltages of the individual cells. The major difference between batteries and the galvanic cells is that commercial typically batteries use solids or pastes rather than solutions as reactants to maximize the electrical output per unit mass. An obvious exception is the standard car battery which used solution phase chemistry. The dry cell, by far the most common type of battery, is used in flashlights, electronic devices such as the Walkman and Game Boy, and many other devices. Although the dry cell was patented in 1866 by the French chemist Georges Leclanché and more than 5 billion such cells are sold every year, the details of its electrode chemistry are still not completely understood. In spite of its name, the is actually a “wet cell”: the electrolyte is an acidic water-based paste containing \(MnO_2\), \(NH_4Cl\), \(ZnCl_2\), graphite, and starch (part (a) in Figure \(\Page {1}\)). The half-reactions at the anode and the cathode can be summarized as follows: \[\ce{2MnO2(s) + 2NH^{+}4(aq) + 2e^{−} -> Mn2O3(s) + 2NH3(aq) + H2O(l)} \nonumber \] \[\ce{Zn(s) -> Zn^{2+}(aq) + 2e^{−}} \nonumber \] The \(\ce{Zn^{2+}}\) ions formed by the oxidation of \(\ce{Zn(s)}\) at the anode react with \(\ce{NH_3}\) formed at the cathode and \(\ce{Cl^{−}}\) ions present in solution, so the overall cell reaction is as follows: \[\ce{2MnO2(s) + 2NH4Cl(aq) + Zn(s) -> Mn2O3(s) + Zn(NH3)2Cl2(s) + H2O(l)} \label{Eq3} \] The dry cell produces about 1.55 V and is inexpensive to manufacture. It is not, however, very efficient in producing electrical energy because only the relatively small fraction of the \(\ce{MnO2}\) that is near the cathode is actually reduced and only a small fraction of the zinc cathode is actually consumed as the cell discharges. In addition, dry cells have a limited shelf life because the \(\ce{Zn}\) anode reacts spontaneously with \(\ce{NH4Cl}\) in the electrolyte, causing the case to corrode and allowing the contents to leak out. The is essentially a Leclanché cell adapted to operate under alkaline, or basic, conditions. The half-reactions that occur in an alkaline battery are as follows: \[\ce{2MnO2(s) + H2O(l) + 2e^{−} -> Mn2O3(s) + 2OH^{−}(aq)} \nonumber \] \[\ce{Zn(s) + 2OH^{−}(aq) -> ZnO(s) + H2O(l) + 2e^{−}} \nonumber \] \[\ce{Zn(s) + 2MnO2(s) -> ZnO(s) + Mn2O3(s)} \nonumber \] This battery also produces about 1.5 V, but it has a longer shelf life and more constant output voltage as the cell is discharged than the Leclanché dry cell. Although the alkaline battery is more expensive to produce than the Leclanché dry cell, the improved performance makes this battery more cost-effective. Although some of the small button batteries used to power watches, calculators, and cameras are miniature alkaline cells, most are based on a completely different chemistry. In these "button" batteries, the anode is a zinc–mercury amalgam rather than pure zinc, and the cathode uses either \(\ce{HgO}\) or \(\ce{Ag2O}\) as the oxidant rather than \(\ce{MnO2}\) in Figure \(\Page {1b}\)). The cathode, anode and overall reactions and cell output for these two types of button batteries are as follows (two half-reactions occur at the anode, but the overall oxidation half-reaction is shown): The major advantages of the mercury and silver cells are their reliability and their high output-to-mass ratio. These factors make them ideal for applications where small size is crucial, as in cameras and hearing aids. The disadvantages are the expense and the environmental problems caused by the disposal of heavy metals, such as \(\ce{Hg}\) and \(\ce{Ag}\). None of the batteries described above is actually “dry.” They all contain small amounts of liquid water, which adds significant mass and causes potential corrosion problems. Consequently, substantial effort has been expended to develop water-free batteries. One of the few commercially successful water-free batteries is the . The anode is lithium metal, and the cathode is a solid complex of \(I_2\). Separating them is a layer of solid \(LiI\), which acts as the electrolyte by allowing the diffusion of Li ions. The electrode reactions are as follows: \[I_{2(s)} + 2e^− \rightarrow {2I^-}_{(LiI)}\label{Eq11} \] \[2Li_{(s)} \rightarrow 2Li^+_{(LiI)} + 2e^− \label{Eq12} \] \[2Li_{(s)}+ I_{2(s)} \rightarrow 2LiI_{(s)} \label{Eq12a} \] with \(E_{cell} = 3.5 \, V\) As shown in part (c) in Figure \(\Page {1}\), a typical lithium–iodine battery consists of two cells separated by a nickel metal mesh that collects charge from the anode. Because of the high internal resistance caused by the solid electrolyte, only a low current can be drawn. Nonetheless, such batteries have proven to be long-lived (up to 10 yr) and reliable. They are therefore used in applications where frequent replacement is difficult or undesirable, such as in cardiac pacemakers and other medical implants and in computers for memory protection. These batteries are also used in security transmitters and smoke alarms. Other batteries based on lithium anodes and solid electrolytes are under development, using \(TiS_2\), for example, for the cathode. Dry cells, button batteries, and lithium–iodine batteries are disposable and cannot be recharged once they are discharged. Rechargeable batteries, in contrast, offer significant economic and environmental advantages because they can be recharged and discharged numerous times. As a result, manufacturing and disposal costs drop dramatically for a given number of hours of battery usage. Two common rechargeable batteries are the nickel–cadmium battery and the lead–acid battery, which we describe next. The , or NiCad, battery is used in small electrical appliances and devices like drills, portable vacuum cleaners, and AM/FM digital tuners. It is a water-based cell with a cadmium anode and a highly oxidized nickel cathode that is usually described as the nickel(III) oxo-hydroxide, NiO(OH). As shown in Figure \(\Page {2}\), the design maximizes the surface area of the electrodes and minimizes the distance between them, which decreases internal resistance and makes a rather high discharge current possible. The electrode reactions during the discharge of a \(NiCad\) battery are as follows: \[2NiO(OH)_{(s)} + 2H_2O_{(l)} + 2e^− \rightarrow 2Ni(OH)_{2(s)} + 2OH^-_{(aq)} \label{Eq13} \] \[Cd_{(s)} + 2OH^-_{(aq)} \rightarrow Cd(OH)_{2(s)} + 2e^- \label{Eq14} \] \[Cd_{(s)} + 2NiO(OH)_{(s)} + 2H_2O_{(l)} \rightarrow Cd(OH)_{2(s)} + 2Ni(OH)_{2(s)} \label{Eq15} \] \(E_{cell} = 1.4 V\) Because the products of the discharge half-reactions are solids that adhere to the electrodes [Cd(OH) and 2Ni(OH) ], the overall reaction is readily reversed when the cell is recharged. Although NiCad cells are lightweight, rechargeable, and high capacity, they have certain disadvantages. For example, they tend to lose capacity quickly if not allowed to discharge fully before recharging, they do not store well for long periods when fully charged, and they present significant environmental and disposal problems because of the toxicity of cadmium. A variation on the NiCad battery is the nickel–metal hydride battery (NiMH) used in hybrid automobiles, wireless communication devices, and mobile computing. The overall chemical equation for this type of battery is as follows: \[NiO(OH)_{(s)} + \rightarrow Ni(OH)_{2(s)} + M_{(s)} \label{Eq16} \] The NiMH battery has a 30%–40% improvement in capacity over the NiCad battery; it is more environmentally friendly so storage, transportation, and disposal are not subject to environmental control; and it is not as sensitive to recharging memory. It is, however, subject to a 50% greater self-discharge rate, a limited service life, and higher maintenance, and it is more expensive than the NiCad battery. Directive 2006/66/EC of the European Union prohibits the placing on the market of portable batteries that contain more than 0.002% of cadmium by weight. The aim of this directive was to improve "the environmental performance of batteries and accumulators" The is used to provide the starting power in virtually every automobile and marine engine on the market. Marine and car batteries typically consist of multiple cells connected in series. The total voltage generated by the battery is the potential per cell (E° ) times the number of cells. As shown in Figure \(\Page {3}\), the anode of each cell in a lead storage battery is a plate or grid of spongy lead metal, and the cathode is a similar grid containing powdered lead dioxide (\(PbO_2\)). The electrolyte is usually an approximately 37% solution (by mass) of sulfuric acid in water, with a density of 1.28 g/mL (about 4.5 M \(H_2SO_4\)). Because the redox active species are solids, there is no need to separate the electrodes. The electrode reactions in each cell during discharge are as follows: \[PbO_{2(s)} + ^−_{4(aq)} + 3H^+_{(aq)} + 2e^− \rightarrow PbSO_{4(s)} + 2H_2O_{(l)} \label{Eq17} \] with \(E^°_{cathode} = 1.685 \; V\) \[Pb_{(s)} + HSO^−_{4(aq)} \rightarrow PbSO_{4(s) }+ H^+_{(aq)} + 2e^−\label{Eq18} \] with \(E^°_{anode} = −0.356 \; V\) \[Pb_{(s)} + PbO_{2(s)} + 2HSO^−_{4(aq)} + 2H^+_{(aq)} \rightarrow 2PbSO_{4(s)} + 2H_2O_{(l)} \label{Eq19} \] and \(E^°_{cell} = 2.041 \; V\) As the cell is discharged, a powder of \(PbSO_4\) forms on the electrodes. Moreover, sulfuric acid is consumed and water is produced, decreasing the density of the electrolyte and providing a convenient way of monitoring the status of a battery by simply measuring the density of the electrolyte. This is often done with the use of a hydrometer. When an external voltage in excess of 2.04 V per cell is applied to a lead–acid battery, the electrode reactions reverse, and \(PbSO_4\) is converted back to metallic lead and \(PbO_2\). If the battery is recharged too vigorously, however, of water can occur: \[ 2H_2O_{(l)} \rightarrow 2H_{2(g)} +O_{2 (g)} \label{EqX} \] This results in the evolution of potentially explosive hydrogen gas. The gas bubbles formed in this way can dislodge some of the \(PbSO_4\) or \(PbO_2\) particles from the grids, allowing them to fall to the bottom of the cell, where they can build up and cause an internal short circuit. Thus the recharging process must be carefully monitored to optimize the life of the battery. With proper care, however, a lead–acid battery can be discharged and recharged thousands of times. In automobiles, the alternator supplies the electric current that causes the discharge reaction to reverse. A fuel cell is a galvanic cell that requires a constant external supply of reactants because the products of the reaction are continuously removed. Unlike a battery, it does not store chemical or electrical energy; a fuel cell allows electrical energy to be extracted directly from a chemical reaction. In principle, this should be a more efficient process than, for example, burning the fuel to drive an internal combustion engine that turns a generator, which is typically less than 40% efficient, and in fact, the efficiency of a fuel cell is generally between 40% and 60%. Unfortunately, significant cost and reliability problems have hindered the wide-scale adoption of fuel cells. In practice, their use has been restricted to applications in which mass may be a significant cost factor, such as manned space vehicles. These space vehicles use a hydrogen/oxygen fuel cell that requires a continuous input of H (g) and O (g), as illustrated in Figure \(\Page {4}\). The electrode reactions are as follows: \[O_{2(g)} + 4H^+ + 4e^− \rightarrow 2H_2O_{(g)} \label{Eq20} \] \[2H_{2(g)} \rightarrow 4H^+ + 4e^− \label{Eq21} \] \[2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(g)} \label{Eq22} \] The overall reaction represents an essentially pollution-free conversion of hydrogen and oxygen to water, which in space vehicles is then collected and used. Although this type of fuel cell should produce 1.23 V under standard conditions, in practice the device achieves only about 0.9 V. One of the major barriers to achieving greater efficiency is the fact that the four-electron reduction of \(O_2 (g)\) at the cathode is intrinsically rather slow, which limits current that can be achieved. All major automobile manufacturers have major research programs involving fuel cells: one of the most important goals is the development of a better catalyst for the reduction of \(O_2 (g)\). Commercial batteries are galvanic cells that use solids or pastes as reactants to maximize the electrical output per unit mass. A battery is a contained unit that produces electricity, whereas a fuel cell is a galvanic cell that requires a constant external supply of one or more reactants to generate electricity. One type of battery is the Leclanché dry cell, which contains an electrolyte in an acidic water-based paste. This battery is called an alkaline battery when adapted to operate under alkaline conditions. Button batteries have a high output-to-mass ratio; lithium–iodine batteries consist of a solid electrolyte; the nickel–cadmium (NiCad) battery is rechargeable; and the lead–acid battery, which is also rechargeable, does not require the electrodes to be in separate compartments. A fuel cell requires an external supply of reactants as the products of the reaction are continuously removed. In a fuel cell, energy is not stored; electrical energy is provided by a chemical reaction. | 15,362 | 1,444 |
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The functional group of alkyl halides is a carbon-halogen bond, the common halogens being fluorine, chlorine, bromine and iodine. With the exception of iodine, these halogens have electronegativities significantly greater than carbon. Consequently, this functional group is polarized so that the carbon is electrophilic and the halogen is nucleophilic, as shown in the drawing on the right. Two characteristics other than electronegativity also have an important influence on the chemical behavior of these compounds. The first of these is . The strongest of the carbon-halogen covalent bonds is that to fluorine. Remarkably, this is the strongest common single bond to carbon, being roughly 30 kcal/mole stronger than a carbon-carbon bond and about 15 kcal/mole stronger than a carbon-hydrogen bond. Because of this, , and do not share any of the reactivity patterns shown by the other alkyl halides. The carbon-chlorine covalent bond is slightly weaker than a carbon-carbon bond, and the bonds to the other halogens are weaker still, the bond to iodine being about 33% weaker. The second factor to be considered is the relative , which is likely the form in which these electronegative atoms will be replaced. This stability may be estimated from the relative acidities of the H-X acids, assuming that the strongest acid releases the most stable conjugate base (halide anion). With the exception of HF (pK = 3.2), all the hydrohalic acids are very strong, small differences being in the direction HCl < HBr < HI. Alkyl haides undergo both and reactions. In describing these two reaction pathways, it is useful to designate the halogen-bearing carbon as and the carbon atom(s) adjacent to it as , as noted in the first four equations shown below. Replacement or substitution of the halogen on the α-carbon (colored maroon) by a nucleophilic reagent is a commonly observed reaction, as shown in equations below. Also, since the electrophilic character introduced by the halogen extends to the β-carbons, and since nucleophiles are also bases, the possibility of base induced H-X elimination must also be considered, as illustrated by equation . Finally, there are some combinations of alkyl halides and nucleophiles that fail to show any reaction over a 24 hour period, such as the example in equation . For consistency, alkyl bromides have been used in these examples. Similar reactions occur when alkyl chlorides or iodides are used, but the speed of the reactions and the exact distribution of products will change. In order to understand why some combinations of alkyl halides and nucleophiles give a substitution reaction, whereas other combinations give elimination, and still others give no observable reaction, we must investigate systematically the way in which changes in reaction variables perturb the course of the reaction. The following general equation summarizes the factors that will be important in such an investigation. One conclusion, relating the structure of the R-group to possible products, should be immediately obvious. , unless a structural rearrangement occurs first. The first four halides shown on the left below do not give elimination reactions on treatment with base, because they have no β-hydrogens. The two halides on the right do not normally undergo such reactions because the potential elimination products have highly strained double or triple bonds.
It is also worth noting that sp hybridized C–X compounds, such as the three on the right, do not normally undergo nucleophilic substitution reactions, unless other functional groups perturb the double bond(s). Using the general reaction shown above as our reference, we can identify the following variables and observables. change α-carbon from 1º to 2º to 3º
if the α-carbon is a chiral center, set as ( ) or ( )
change from Cl to Br to I (F is relatively unreactive)
change from anion to neutral; change basicity; change polarizability
polar vs. non-polar; protic vs. non-protic substitution, elimination, no reaction.
if the α-carbon is a chiral center what happens to its configuration?
measure as a function of reactant concentration. When several reaction variables may be changed, it is important to isolate the effects of each during the course of study. In other words: , the others being held as constant as possible. For example, we can examine the effect of changing the halogen substituent from Cl to Br to I, using ethyl as a common R–group, cyanide anion as a common nucleophile, and ethanol as a common solvent. We would find a common substitution product, C H –CN, in all cases, but the speed or rate of the reaction would increase in the order: Cl < Br < I. This reactivity order reflects both the strength of the C–X bond, and the stability of X as a leaving group, and leads to the general conclusion that alkyl iodides are the most reactive members of this functional class. | 5,015 | 1,445 |
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Atoms bond together to form compounds because in doing so they attain lower energies than they possess as individual atoms. A quantity of energy, equal to the difference between the energies of the bonded atoms and the energies of the separated atoms, is released, usually as heat. That is, the bonded atoms have a lower energy than the individual atoms do. When a chemical reaction occurs, molecular bonds are broken and other bonds are formed to make different molecules. For example, the bonds of two water molecules are broken to form hydrogen and oxygen. \[ 2H_2O \rightarrow 2H_2 + O_2\] Energy is always required to break a bond, which is known as bond energy. While the concept may seem simple, bond energy serves a very important purpose in describing the structure and characteristics of a molecule. It can be used to determine which Lewis Dot Structure is most suitable when there are multiple Lewis Dot Structures. Energy is always required to break a bond. Energy is released when a bond is made. Although each molecule has its own characteristic bond energy, some generalizations are possible. For example, although the exact value of a C–H bond energy depends on the particular molecule, all C–H bonds have a bond energy of roughly the same value because they are all C–H bonds. It takes roughly 100 kcal of energy to break 1 mol of C–H bonds, so we speak of the bond energy of a C–H bond as being about 100 kcal/mol. A C–C bond has an approximate bond energy of 80 kcal/mol, while a C=C has a bond energy of about 145 kcal/mol. We can calculate a more general bond energy by finding the average of the bond energies of a specific bond in different molecules to get the average bond energy. When a bond is strong, there is a higher bond energy because it takes more energy to break a strong bond. This correlates with . When the Bond order is higher, bond length is shorter, and the shorter the bond length means a greater the Bond Energy because of increased electric attraction. In general, The average bond energies in are the averages of bond dissociation energies. For example the average bond energy of O-H in H O is 464 kJ/mol. This is due to the fact that the H-OH bond requires 498.7 kJ/mol to dissociate, while the O-H bond needs 428 kJ/mol. \[\dfrac{498.7\; kJ/mol + 428\; kJ/mol}{2}=464\; kJ/mol\] When more bond energies of the bond in different molecules that are taken into consideration, the average will be more accurate. However, When a chemical reaction occurs, the atoms in the reactants rearrange their chemical bonds to make products. The new arrangement of bonds does not have the same total energy as the bonds in the reactants. Therefore, when chemical reactions occur, . In some reactions, the energy of the products is lower than the energy of the reactants. Thus, in the course of the reaction, the substances lose energy to the surrounding environment. Such reactions are exothermic and can be represented by an in Figure 1 (left). In most cases, the energy is given off as heat (although a few reactions give off energy as light). In chemical reactions where the products have a higher energy than the reactants, the reactants must absorb energy from their environment to react. These reactions are endothermic and can be represented by an energy-level diagrams like Figure 1 (right). It is not uncommon that textbooks and instructors to consider heat as a independent "species" in a reaction. While this is rigorously incorrect because one cannot "add or remove heat" to a reaction as with species, it serves as a convenient mechanism to predict the shift of reactions with changing temperature. For example, if heat is a "reactant" (\(\Delta{H} > 0 \)), then the reaction favors the formation of products at elevated temperature. Similarly, if heat is a "product" (\(\Delta{H} < 0 \)), then the reaction favors the formation of reactants. A more accurate, and hence preferred, description is discussed below. Exothermic and endothermic reactions can be thought of as having energy as either a "product" of the reaction or a "reactant." Exothermic reactions releases energy, so energy is a product. Endothermic reactions require energy, so energy is a reactant. Is each chemical reaction exothermic or endothermic? No calculates are required to address this question. Just look at where the "heat" is in the chemical reaction. If the bond energy for H-Cl is 431 kJ/mol. What is the overall bond energy of 2 moles of HCl? Simply multiply the average bond energy of H-Cl by 2. This leaves you with 862 kJ/mol ( ). \[H_2(g)+I_2(g) \rightarrow 2HI(g)\] First look at the equation and identify which bonds exist on in the reactants. Now do the same for the products Then identify the bond energies of these bonds from the table above: The sum of enthalpies on the reaction side is: 436 kJ/mole + 151 kJ/mole = 587 kJ/mol. This is how much energy is needed to break the bonds on the reactant side. Then we look at the bond formation which is on the product side: The sum of enthalpies on the product side is: 2 x 297 kJ/mol= 594 kJ/mol This is how much energy is released when the bonds on the product side are formed. The net change of the reaction is therefore 587-594= -7 kJ/mol. Since this is negative, the reaction is . Using the bond energies given in the chart above, find the enthalpy change for the thermal decomposition of water: \[ 2H_2O (g) \rightarrow 2H_2 + O_2 (g) \] Is the reaction written above exothermic or endothermic? Explain. The enthalpy change deals with breaking two mole of O-H bonds and the formation of 1 mole of O-O bonds and two moles of H-H bonds ( ). which is an output (released) energy = 872.8 kJ/mol + 498.7 kJ/mol = 1371.5 kJ/mol. Total energy difference is 1840 kJ/mol – 1371.5 kJ/mol = 469 kJ/mol, which indicates that the reaction is and that 469 kJ of heat is needed to be supplied to carry out this reaction. This reaction is endothermic since it requires energy in order to create bonds. Energy is released to generate bonds, which is why the enthalpy change for breaking bonds is positive. Energy is required to break bonds. Atoms are much happier when they are "married" and release energy because it is easier and more stable to be in a relationship (e.g., to generate ). The enthalpy change is negative because the system is releasing energy when forming bond. | 6,376 | 1,446 |
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1. Philosophers argue about the feasibility of a private language, a language that is known by only one individual. The case against the possibility of a private language is based on the assumption that language comes into existence as a tool for communication in a society comprising two or more individuals. Solipsism is a philosophical conception in which your sensory perceptions are internally generated; they are not the result of your interactions with the world. Solipsism assumes that the world you think you perceive does not in fact exist. Since a solipsistic individual is the only being that exists, any language he uses is necessarily a private language. Evidently, the existence of a solipsistic individual who uses some language to think and the impossibility of a private language are mutually exclusive: If a private language is impossible, a solipsistic individual cannot use any language. Since you are reading this, you are using a language, and therefore you cannot be a solipsistic individual. Does this argument convince you that you are not a solipsistic individual; that is, does this argument convince you of the existence of a physical reality that is external to yourself? Why or why not? Can a solipsistic individual engage in scientific inquiry? 2. Many people find the theory of evolution deeply repugnant. Some argue that the theory of evolution has not been proved and that is an alternative scientific theory, where creationism is the Biblical description of God’s creation of the world in six days. (a) Is it valid to say, “The theory of evolution is unproved.”? (b) Comment very briefly on whether or not the theory of evolution meets each of our criteria for a scientific theory. (c) Comment very briefly on whether or not creationism meets each of our criteria for a scientific theory. 3. Use an ordinary English sentence to state the meaning of propositions (a) and (b): (a) \(\sim (p\ \mathrm{and}\ q)\) \(\mathrm{\Rightarrow }\) (\(\sim p\ \ \mathrm{or}\ \sim q)\) (b) \(\sim (p\ \mathrm{or}\ q)\) \(\mathrm{\Rightarrow }\) (\(\sim p\ \ \mathrm{and}\ \sim q)\) Are propositions (a) and (b) true or false? Using propositions (a) and (b), prove propositions (c) and (d): (c) [\((p\ \mathrm{and}\ q)\) \(\mathrm{\Rightarrow }\)\(r\) ] \(\mathrm{\Rightarrow }\) [\(\sim r\) \(\mathrm{\Rightarrow }\) (\(\sim p\ \ \mathrm{or}\ \sim q)\)] (d) [\((p\ \mathrm{or}\ q)\) \(\mathrm{\Rightarrow }\)\(r\) ] \(\mathrm{\Rightarrow }\) [\(\sim r\) \(\mathrm{\Rightarrow }\) (\(\sim p\ \ \mathrm{and}\ \sim q)\)] \({}^{1}\)\({}^{\ }\)R. Clausius, , translated by Walter R. Browne, Macmillan and Co., London, 1879, p. 76. \({}^{2}\)\({}^{\ }\)We use square brackets around the symbol for a chemical substance to denote the concentration of that substance in molarity (moles per liter of solution) units. \({}^{3}\)\({}^{\ }\) , Merriam-Webster, Inc., Springfield, Massachusetts, 1988. | 2,924 | 1,447 |
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The alkali metals (Li, Na, K etc.) and the alkaline earth metals (Mg and Ca, together with Zn) are good reducing agents, the former being stronger than the latter. Sodium, for example, reduces elemental chlorine to chloride anion (sodium is oxidized to its cation), as do the other metals under varying conditions. In a similar fashion these same metals reduce the carbon-halogen bonds of alkyl halides. The halogen is converted to halide anion, and the carbon bonds to the metal (the carbon has carbanionic character). Halide reactivity increases in the order: Cl < Br < I. The following equations illustrate these reactions for the commonly used metals lithium and magnesium (R may be hydrogen or alkyl groups in any combination). The alkyl magnesium halides described in the second reaction are called after the French chemist, Victor Grignard, who discovered them. The other metals mentioned above react in a similar manner, but the two shown here are the most widely used. Although the formulas drawn here for the alkyl lithium and Grignard reagents reflect the stoichiometry of the reactions and are widely used in the chemical literature, they do not accurately depict the structural nature of these remarkable substances. Mixtures of polymeric and other associated and complexed species are in equilibrium under the conditions normally used for their preparation. R C- + 2 → R C- + An Alkyl Lithium Reagent
R C- + → R C- A Grignard Reagent The metals referred to here are insoluble in most organic solvents, hence these reactions are clearly heterogeneous, i.e. take place on the metal surface. The conditions necessary to achieve a successful reaction are critical. These reactions are obviously substitution reactions, but they cannot be classified as nucleophilic substitutions, as were the earlier reactions of alkyl halides. Because the functional carbon atom has been reduced, the polarity of the resulting functional group is inverted (an originally electrophilic carbon becomes nucleophilic). This change, shown below, makes alkyl lithium and Grignard reagents unique and useful reactants in synthesis. Reactions of organolithium and Grignard reagents reflect the nucleophilic (and basic) character of the functional carbon in these compounds. Many examples of such reactions will be encountered in future discussions, and five simple examples are shown below. The first and third equations demonstrate the strongly basic nature of these compounds, which bond rapidly to the weakly acidic protons of water and methyl alcohol (colored blue). The nucleophilic carbon of these reagents also bonds readily with electrophiles such as iodine (second equation) and carbon dioxide (fifth equation). The polarity of the carbon-oxygen double bonds of CO makes the carbon atom electrophilic, shown by the formula in the shaded box, so the nucleophilic carbon of the Grignard reagent bonds to this site. As noted above, solutions of these reagents must also be protected from oxygen, since peroxides are formed (equation 4). Another important reaction exhibited by these organometallic reagents is . In the first example below, methyl lithium reacts with cuprous iodide to give a lithium dimethylcopper reagent, which is referred to as a . Other alkyl lithiums give similar Gilman reagents. A useful application of these reagents is their ability to couple with alkyl, vinyl and aryl iodides, as shown in the second equation. Later we shall find that Gilman reagents also display useful carbon-carbon bond forming reactions with conjugated enones and with acyl chlorides. 2 CH Li + CuI → (CH ) CuLi + LiI
(C H ) CuLi + C H I → C H -C H + LiI + C H Cu . Changes in carbon hybridization have little effect on the reaction, and 1º, 2º and 3º-alkyl halides all react in the same manner. One restriction, of course, is the necessary absence of incompatible functional groups elsewhere in the reactant molecule. For example, 5-bromo-1-pentanol fails to give a Grignard reagent (or a lithium reagent) because the hydroxyl group protonates this reactive function as soon as it is formed. CH CH CH CH CH + → [ CH CH CH CH CH ] → CH CH CH CH CH Exchange metalation is particularly useful when it can be directed to specific sites in a molecule. One such case is the of aromatic rings bearing a suitable directing group. | 4,404 | 1,449 |
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As was stated before, molecules containing 2 or more chiral centers may or may not be chiral themselves. Let’s consider the case of achiral molecules first. -disubstituted cyclohexanes are examples of meso forms. Meso forms can also be open chain, as illustrated below. Now let’s consider the case of molecules that contain two or more stereocenters. Such molecules can have enantiomers because they are not the same as their mirror images. -1,2-dichlorocyclohexane. The mirror images are different compounds. Chiral molecules with two or more chiral centers can also have . Such sets of stereoisomers are called In the following example the two molecules shown are stereoisomers (same connectivity but different spacial arrangement of the atoms) but they are not mirror images. Their relationship is one of diastereomers. - and -1,2-dichlorocyclohexane. are examples of As a corollary, we can state that . Notice that we have referred to such sets before as . Geometric isomers are in fact a subcategory of diastereomers. The following example is an illustration of open chain molecules with a diastereomeric relationship. Also notice that in the above examples one of the members in each pair is chiral and the other is not. Diastereomeric sets are frequently made up of molecules where one of the molecules is chiral and the other is not. Also notice that although they are not mirror images, part of their structures do mirror each other. It is frequently the case that one half of one molecule mirrors one half of the other one, but the other halves are identical. In this pair of cis/trans isomers, the top half of one molecule mirrors the top half of the other one (the chiral centers have opposite configurations), while the bottom halves are the same (the chiral centers have the same configuration). If n = number of chiral centers, the maximum possible number of stereoisomers is 2 The diagram below shows the possible combinations of configurations for molecules with 2 chiral centers. Since each chiral center can be (R) or (S), the possible combinations are (R,R), (S,S), (R,S), (S,R). If two molecules are mirror images, then their configurations are exactly opposite and they are enantiomers (E). If they are not mirror images but still they are stereoisomers then they are diastereomers (D). | 2,332 | 1,450 |
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One of the most confusing things about this is the way the words are used. These days, the term "bond enthalpy" is normally used, but you will also find it described as "bond energy" - sometimes in the same article. An even older term is "bond strength". So you can take all these terms as being interchangeable. A diatomic molecule is one that only contains two atoms. They could be the same (for example, Cl ) or different (for example, HCl). The bond dissociation enthalpy is the energy needed to break one mole of the bond to give separated atoms - everything being in the gas state. The point about everything being in the gas state is essential; you cannot use bond enthalpies to do calculations directly from substances starting in the liquid or solid state. As an example of bond dissociation enthalpy, to break up 1 mole of gaseous hydrogen chloride molecules into separate gaseous hydrogen and chlorine atoms takes 432 kJ. The bond dissociation enthalpy for the H-Cl bond is +432 kJ mol . What happens if the molecule has several bonds, rather than just 1? Consider methane, CH . It contains four identical C-H bonds, and it seems reasonable that they should all have the same bond enthalpy. However, if you took methane to pieces one hydrogen at a time, it needs a different amount of energy to break each of the four C-H bonds. Every time you break a hydrogen off the carbon, the environment of those left behind changes. And the strength of a bond is affected by what else is around it. In cases like this, the bond enthalpy quoted is an average value. In the methane case, you can work out how much energy is needed to break a mole of methane gas into gaseous carbon and hydrogen atoms. That comes to +1662 kJ and involves breaking 4 moles of C-H bonds. The average bond energy is therefore +1662/4 kJ, which is +415.5 kJ per mole of bonds. That means that many bond enthalpies are actually quoted as mean (or average) bond enthalpies, although it might not actually say so. Mean bond enthalpies are sometimes referred to as "bond enthalpy terms". In fact, tables of bond enthalpies give average values in another sense as well, particularly in organic chemistry. The bond enthalpy of, say, the C-H bond varies depending on what is around it in the molecule. So data tables use average values which will work well enough in most cases. That means that if you use the C-H value in some calculation, you can't be sure that it exactly fits the molecule you are working with. So don't expect calculations using mean bond enthalpies to give very reliable answers. You may well have to know the difference between a bond dissociation enthalpy and a mean bond enthalpy, and you should be aware that the word mean (or average) is used in two slightly different senses. But for calculation purposes, it isn't something you need to worry about. Just use the values you are given. Remember that you can only use bond enthalpies directly if everything you are working with is in the gas state. We are going to estimate the enthalpy change of reaction for the reaction between carbon monoxide and steam. This is a part of the manufacturing process for hydrogen. \[ CO(g) + H_2O(g) \rightarrow CO_2 (g) +H_2(g)\] The bond enthalpies are : So let's do the sum. Here is the cycle - make sure that you understand exactly why it is the way it is. And now equate the two routes, and solve the equation to find the enthalpy change of reaction. ΔH + 2(805) + 436 = 1077 + 2(464) ΔH = 1077 + 2(464) - 2(805) - 436 ΔH = -41 kJ mol You could do any bond enthalpy sum by the method above - taking the molecules completely to pieces and then remaking the bonds. If you are happy doing it that way, just go on doing it that way. However, if you are prepared to give it some thought, you can save a bit of time - although only in very simple cases where the changes in a molecule are very small. For example, chlorine reacts with ethane to give chloroethane and hydrogen chloride gases (a ll of these are gases). It is always a good idea to draw full structural formulae when you are doing bond enthalpy calculations. It makes it much easier to count up how many of each type of bond you have to break and make. If you look at the equation carefully, you can see what I mean by a "simple case". Hardly anything has changed in this reaction. You could work out how much energy is needed to break every bond, and how much is given out in making the new ones, but quite a lot of the time, you are just remaking the same bond. All that has actually changed is that you have broken a C-H bond and a Cl-Cl bond, and made a new C-Cl bond and a new H-Cl bond. So you can just work those out. Work out the energy needed to break C-H and Cl-Cl: +413 + 243 = +656 kJ mol Work out the energy released when you make C-Cl and H-Cl: -346 - 432 = -778 kJ mol So the net change is +656 - 778 = -122 kJ mol You can only use bond enthalpies directly if everything you are working with is in the gas state. If you have one or more liquids present, you need an extra energy term to work out the enthalpy change when you convert from liquid to gas, or vice versa. That term is the enthalpy change of vaporization, and is given the symbol ΔH or ΔH . This is the enthalpy change when 1 mole of the liquid converts to gas at its boiling point with a pressure of 1 bar (100 kPa) (older sources might quote 1 atmosphere rather than 1 bar.) For water, the enthalpy change of vaporization is +41 kJ mol . That means that it take 41 kJ to change 1 mole of water into steam. If 1 mole of steam condenses into water, the enthalpy change would be -41 kJ. Changing from liquid to gas needs heat; changing gas back to liquid releases exactly the same amount of heat. You can only use bond enthalpies directly if everything you are working with is in the gas state. To see how this fits into bond enthalpy calculations, we will estimate the enthalpy change of combustion of methane - in other words, the enthalpy change for this reaction: Notice that the product is liquid water. You cannot apply bond enthalpies to this. You must first convert it into steam. To do this you have to supply 41 kJ mol . The bond enthalpies you need are: The cycle looks like this: This obviously looks more confusing than the cycles we've looked at before, but apart from the extra enthalpy change of vaporization stage, it isn't really any more difficult. Before you go on, make sure that you can see why every single number and arrow on this diagram is there. In particular, make sure that you can see why the first 4 appears in the expression "4(+464)". That is an easy thing to get wrong. (In fact, when I first drew this diagram, I carelessly wrote 2 instead of 4 at that point!) That's the hard bit done - now the calculation: ΔH + 2(805) + 2(41) + 4(464) = 4(413) + 2(498) ΔH = 4(413) + 2(498) - 2(805) - 2(41) - 4(464) ΔH = -900 kJ mol The measured enthalpy change of combustion is -890 kJ mol , and so this answer agrees to within about 1%. As bond enthalpy calculations go, that's a pretty good estimate. Jim Clark ( ) | 7,073 | 1,452 |
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There are a number of functional groups that contain a carbon-oxygen double bond, which is commonly referred to as a . and are two closely related carbonyl-based functional groups that react in very similar ways. In a ketone, the carbon atom of a carbonyl is bonded to two other carbons. In an aldehyde, the carbonyl carbon is bonded on one side to a hydrogen, and on the other side to a carbon. The exception to this definition is formaldehyde, in which the carbonyl carbon has bonds to two hydrogens. Molecules with carbon-nitrogen double bonds are called , or . If a carbonyl carbon is bonded on one side to a carbon (or hydrogen) and on the other side to a (in organic chemistry, this term generally refers to oxygen, nitrogen, sulfur, or one of the halogens), the functional group is considered to be one of the ‘ , a designation that describes a grouping of several functional groups. The eponymous member of this grouping is the functional group, in which the carbonyl is bonded to a hydroxyl (OH) group. As the name implies, carboxylic acids are acidic, meaning that they are readily deprotonated to form the conjugate base form, called a (much more about carboxylic acids in the acid-base chapter!). In , the carbonyl carbon is bonded to a nitrogen. The nitrogen in an amide can be bonded either to hydrogens, to carbons, or to both. Another way of thinking of an amide is that it is a carbonyl bonded to an amine. In , the carbonyl carbon is bonded to an oxygen which is itself bonded to another carbon. Another way of thinking of an ester is that it is a carbonyl bonded to an alcohol. are similar to esters, except a sulfur is in place of the oxygen. In an , the carbonyl carbon is bonded to the oxygen of a phosphate, and in an , the carbonyl carbon is bonded to a chlorine. Finally, in a group, a carbon is triple-bonded to a nitrogen. Nitriles are also often referred to as groups. A single compound often contains several functional groups. The six-carbon sugar molecules glucose and fructose, for example, contain aldehyde and ketone groups, respectively, and both contain five alcohol groups (a compound with several alcohol groups is often referred to as a ‘ ). Capsaicin, the compound responsible for the heat in hot peppers, contains phenol, ether, amide, and alkene functional groups. The male sex hormone testosterone contains ketone, alkene, and secondary alcohol groups, while acetylsalicylic acid (aspirin) contains aromatic, carboxylic acid, and ester groups. While not in any way a complete list, this section has covered most of the important functional groups that we will encounter in biological and laboratory organic chemistry. The table on the inside back cover provides a summary of all of the groups listed in this section, plus a few more that will be introduced later in the text. | 2,859 | 1,453 |
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A is a collection of elementary processes or steps (also called elementary steps) that explains how the overall reaction proceeds. A mechanism is a proposal from which you can work out a rate law that agrees with the observed rate laws. The fact that a mechanism explains the experimental results is not a proof that the mechanism is correct. As a mechanism, no proof is required. Proposing a mechanism is an interesting academic exercise for a mature chemist. Students in will not be required to propose a mechanism, but they are required to derive the rate law from a proposed mechanism. A summary of the elementary process or steps is given in a table form here. This previous link provides details about these steps. In the table below, \(\ce{A}\), \(\ce{B}\), and \(\ce{C}\) represent reactants, intermediates, or products in the elementary process. You all have experienced that an accident or road construction on a free way slows all the traffic on the road, because limitations imposed by the accident and constructions apply to all cars on that road. The narrow passing stretch limits the speed of the traffic. If several steps are involved in an overall chemical reaction, the slowest step limits the rate of the reaction. Thus, a slow step is called a The following examples illustrate the method of deriving rate laws from the proposed mechanism. Please learn the technique. If the reaction \(\ce{2 NO2 + F2 \rightarrow 2 NO2F}\) follows the mechanism, what is the rate law?
Since step i. is the rate-determining step, the rate law is \(\mathrm{- \dfrac{1}{2}\dfrac{d[NO_2]}{dt} = \mathit{k} [NO_2] [F_2]}\) Since both \(\ce{NO2}\) and \(\ce{F2}\) are reactants, this is the rate law for the reaction.
Addition of i. and ii. gives the overall reaction, but step ii. does not affect the rate law. Note that the rate law is not derived from the overall equation either. For the reaction \(\ce{H2 + Br2 \rightarrow 2 HBr}\), the following mechanism has been proposed Derive the rate law that is consistent with this mechanism.
For a problem of this type, you should give the rate law according to the rate-determining (slow) elementary process. In this case, step ii. is the rate-determining step, and the rate law is \(\mathrm{\dfrac{1}{2}\dfrac{d[HBr]}{dt} = \mathit{k}_2 [H_2] [Br]}\) The factor 1/2 results from the \(\ce{2 HBr}\) formed every time, one in step ii. and one in step iii. Since \(\ce{[Br]}\) is not one of the reactants, its relationship with the concentration of the reactants must be sought. The rapid reaction in both directions of step i. implies the following relationship: \(k_1 [\ce{Br2}] = k_{-1} [\ce{Br}]^2\) or \(\ce{[Br]} = \left(\dfrac{k_1}{k_{-1}} [\ce{Br2}]\right)^{1/2}\) Substituting this in the rate expression results in \(\ce{rate} = k_2 \left(\dfrac{k_1}{k_{-1}}\right)^{1/2} \ce{[H2] [Br2]}^{1/2}\) The overall reaction order is 3/2, 1 with respect to \(\ce{[H2]}\) and 1/2 with respect to \(\ce{[Br2]}\).
The important point in this example is that the rapid equilibrium in step i. allows you to express the concentration of an intermediate (\(\ce{[Br]}\)) in terms of concentrations of reactants (\(\ce{[Br2]}\)) so that the rate law can be expressed by concentrations of the reactants. The ratio / is often written as , and it is called the for the reversible elementary steps. Derive the rate law that is consistent with the proposed mechanism in the formation of phosgene from \(\ce{Cl2}\) and \(\ce{CO}\). (\(K_1 = \dfrac{k_1}{k_{-1}}\) and \(K_2 = \dfrac{k_2}{k_{-2}}\) may be considered as equilibrium constants of the elementary processes, and \(\ce{M}\) is any inert molecule.) The overall reaction is \(\ce{Cl2 + CO \rightarrow Cl2CO}\)
From the rate-determining (slow) step, \(\mathrm{\dfrac{d[Cl_2CO]}{dt}} = k_3 \mathrm{[ClCO] [Cl_2]}\tag{1}\) You should express \(\ce{[ClCO]}\) in terms of concentrations of \(\ce{Cl2}\) and \(\ce{CO}\). This is done by considering step ii. \(\ce{[ClCO]} = K_2 \ce{[Cl] [CO]}\tag{2}\) You should express \(\ce{[Cl]}\) in terms of \(\ce{[Cl2]}\). For this, you may use step i. \(\ce{[Cl]} = K_1^{1/2} \ce{[Cl2]}^{1/2} \tag{3}\) Substituting (3) in (2) and then in (1) gives the Rate, \(\begin{align*}
\ce{Rate} &= k_3 K_1^{1/2} K_2 \ce{[CO] [Cl2]}^{3/2}\\
&= k \ce{[CO] [Cl2]}^{3/2}
\end{align*}\) where \(k = k_1 K_1^{1/2} K_2\), the observed rate constant. The overall order of the reaction is 5/2: strange, but that is the observed rate law.
This example shows how the concentrations of intermediates are related to those of the reactants in a two-step equilibrium. If the third step is the rate law will be different from the result derived above. Derive the rate law using the alternate step from the Discussion in Example 3. In an acid solution, the mechanism for the reaction \(\ce{NH4+ + HNO2 \rightarrow N2 + 2 H2O + H+ }\) is: Derive the rate law.
From the rate-determining step, you have \(\mathrm{\dfrac{d[NH_3NO^+ ]}{dt}} = k_3 \mathrm{[NO^+] [NH_3]} \tag{4}\) Neither \(\ce{NO+}\) nor \(\ce{NH3}\) is a reactant. You must express their concentrations in terms of \(\ce{[NH4+]}\) and \(\ce{[HNO2]}\) from elementary processes i. and ii. From i, \(\ce{[NO+]} = K_1 \ce{\dfrac{[HNO2] [H+]}{[H2O]}} \tag{5}\) From ii, \(\ce{[NH3]} = K_2 \ce{\dfrac{[NH4+]}{[H+]}} \tag{6}\) Substituting (6) and (5) in (4) gives, \(\begin{align*}
\ce{Rate} &= k_3 K_1 K_2 \ce{\dfrac{[HNO2] [NH4+]}{[H2O]}}\\ \\
&= k \ce{[HNO2] [NH4+ ]}
\end{align*}\) where \(k = \dfrac{k_3 K_1 K_2}{[\ce{H2O}]}\) is the overall rate constant.
Figure out the order from a given mechanism.
What is the overall order?
If you got \(\ce{rate} = k \mathrm{[A_2]^{1/2} [B_2]}\), congratulations.
Figure out the order from a given mechanism.
You should get: \(rate = k \ce{[NO]^2 [O2]}\) \(\ce{A + 4 C \rightarrow 2 F}\)
The rate determining step is ii. Express \(\ce{[B]}\) in terms of \(\ce{[A]}\) from i. | 5,948 | 1,455 |
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This page is a brief introduction to solubility product calculations. The solubilities of the ionic compounds in the examples below are given in mol dm . If it is given in g dm , or another concentration unit, it must first be converted into mol dm . The solubility of barium sulfate at 298 K is 1.05 x 10 mol dm . Calculate the solubility product. The equilibrium is given below: \[BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq) \nonumber \] Notice that each mole of barium sulfate dissolves to give 1 mole of barium ions and 1 mole of sulfate ions in solution. That means that: \[[Ba^{2+}] = 1.05 \times 10^{-5}\ mol\ dm^{-3} \nonumber \] \[[SO_4^{2-}] = 1.05 \times 10^{-5}\ mol\ dm^{-3} \nonumber \] \[ \begin{eqnarray} K_{sp} &=& [Ba^{2+},SO_4^{2-}] \\ &=& (1.05 \times 10^{-5}) \times (1.05 \times 10^{-5}) \\ &=& 1.10 \times 10^{-10}\ mol^2\ dm^{-6} \end{eqnarray} \nonumber \] Verify that the units are correct. These calculations are very simple if for compounds in which ratio of the numbers of positive and negative ions is 1:1. This next example illustrates the procedure for a different ratio. The solubility of magnesium hydroxide at 298 K is 1.71 x 10 mol dm . Calculate the solubility product. The equilibrium is: \[Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq) \nonumber \] For every mole of magnesium hydroxide that dissolves, one mole of magnesium ions is generated, but twice that number of hydroxide ions form. So the concentration of the dissolved magnesium ions is the same as the dissolved magnesium hydroxide: \[[Mg^{2+}] = 1.71 \times 10^{-4}\ mol\ dm^{-3} \nonumber \] The concentration of dissolved hydroxide ions is twice that: \[\begin{eqnarray} [OH^-] &=& 2 \times 1.71 \times 10^{-4} \\ &=& 3.42 \times 10^{-4}\ mol\ dm^{-3} \end{eqnarray} \nonumber \] Substitute these values into the solubility product expression as before: \[\begin{eqnarray} K_{sp} &=& [Mg^{2+},OH^-]^2 \\ &=& (1.71 \times 10^{-4}) \times (3.42 \times 10^{-4})^2 \\ &=& 2.00 \times 10^{-11}\ mol^3\ dm^{-9} \end{eqnarray} \nonumber \] | 2,070 | 1,456 |
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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 | 1,458 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/Gen_Chem_Quantum_Theory/Electromagnetic_Waves |
Electromagnetic radiations are usually treated as wave motions. The electronic and magnetic fields oscillate in directions perpendicular to each other and to the direction of motion of the wave. The wavelength, the frequency, and the speed of light obey the following relationship: \(\mathrm{wavelength \times frequency = speed\: of\: light}\). The speed of light is usually represented by , the wavelength by the lower case Greek letter lambda, \(\lambda\), and the frequency by lower case Greek letter nu \(\nu\). In these symbols, the above formula is: \(\lambda \nu = c\) The electromagnetic radiation is the foundation for radar, which is used for guidance and remote sensing for the study of the planet Earth. Wavelengths of the visible region of the spectrum range from 700 nm for red light to 400 nm for violet light. There is no need to memorize these numbers, but knowing that the visible region has such a narrow range of 400-700 nm is handy at times when referring to certain light. | 1,010 | 1,459 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Nuclear_Chemistry/Radioactivity/Purifying_Radioactive_Materials |
The production of everyday used materials such as oil and gas results in the buildup of radioactive materials in high concentrations. As a result of the commonality of this occurrence, a solution has been presented as to rid the world of this toxic waste. Purifying these radioactive materials with the use of various methods allows the pure substance to be reused and prevents the depletion of resources. Before we can understand how to purify these substances, it is important to understand the chemistry involved within to get a deeper understanding. Radioactivity was first discovered by Henri Becquerel in 1896 when he considered that the phosphorescent materials could be related to the glow emitted by X-rays. Rutherford further enhanced this discovery with his gold foil experiment to class the particles emitted by radioactive materials into classes based on their ability to penetrate through materials. Alpha particles are larger in size and therefore the least harmful because they cannot pass through something as thin as a sheet of paper. Through analyzing radioactive decay, it can be determined that alpha particles are positive. Beta particles are larger, carry a negative charge, and require a more dense substance to hinder their path. Gamma rays are neutral and by far the most dangerous of the three. These rays cannot be easily deflected and can even go through concrete. The term radioactive is defined as an unstable particle that releases subatomic particles. Examples include and then using a substance that will bind this reduced radioactive material will allow it to be separated from the remaining solution. This method has recently acquired a patent and is still undergoing experimental procedures but remains effective. In nature water generally contains a plethora of impurities. These impurities can include small microbes to something as dangerous as radioactive substances. Methods such as boiling, Chlorination (use of household chlorine bleach), and purification tablets remove microorganisms. More rigorous modes of purification are used to rid the water of other wastes including radioactive materials. Groundwater is a common example in which radium, a radioactive element, is mixed with the water. This way produces a black sludge of radioactive water which is unhealthy for consumer usage. The radium can be removed through ion exchange or the conditioning of water. Other unnatural occurrences of radioactive materials require more meticulous methods. Distillation removed salts, heavy metals, and radioactive fallout (since water itself cannot become radioactive, the radioactive components are referred to as radioactive fallout). Filtering the water will also remove the radioactive fallout. | 2,757 | 1,462 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/16%3A_The_Organic_Chemistry_of_Amino_Acids_Peptides_and_Proteins/16.11%3A_The_Tertiary_Structure_of_Proteins |
Secondary structure refers to the shape of a folding protein due exclusively to hydrogen bonding between its backbone amide and carbonyl groups. Secondary structure does not include bonding between the R-groups of amino acids, hydrophobic interactions, or other interactions associated with tertiary structure. The two most commonly encountered secondary structures of a polypeptide chain are α-helices and beta-pleated sheets. These structures are the first major steps in the folding of a polypeptide chain, and they establish important topological motifs that dictate subsequent tertiary structure and the ultimate function of the protein.
Thumbnail: Structure of human hemoglobin. The proteins α and β subunits are in red and blue, and the iron-containing heme groups in green. (CC BY- ; ). | 817 | 1,464 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Fundamentals/Ladder_Operators_(Creation_Annihilation_Operators) |
Ladder Operators are operators that increase or decrease eigenvalue of another operator. There are two types; raising operators and lowering operators. In quantum mechanics the raising operator is called the creation operator because it adds a quantum in the eigenvalue and the annihilation operators removes a quantum from the eigenvalue. For example, in quantum harmonic oscillator, creation operators adds a quantum of energy to the system and annihilation operators removes a quantum of energy from the system. While in potential system quantum field theories, creation operators add a particle to the system and annihilation operators remove a particle from the system. Ladder operators increase or decrease eginvaules by a quantum. An example of ladder operators are the a type of angular momentum operators. Creation operator: \[{L_+ Y_{\ell m}}={\hbar}{\sqrt{\ell(\ell+1)-m(m+1)}}{Y_{\ell(m+1)}}\] Annihilation operator: \[{L_- Y_{\ell m}}={\hbar}{\sqrt{\ell(\ell+1)-m(m-1)}} {Y_{\ell(m-1)}} \] The creation operators increases the value of m and the annihilation operators decreases the value of m, both without affecting the value of l. Below is a graphical representation of what ladder operators do when related to energy eigenvalue of the quantum harmonic oscillator. The Creation operators a increases the energy value by a quantum and the annihilation operator decreases the the energy value by a quantum. In angular momentum, the ladder operators are J+ and J-. \[J_-=J_x-iJ_y\] \[J_+=J_x+iJ_y\] Then with the operator J \[J_z J_\pm|j m \rangle =(J_\pm J_z +[J_z,J_\pm]) |jm \rangle \] \[J_z J_\pm|j m \rangle =(J_\pm J_z \pm \hbar J_\pm)|jm \rangle \] \[J_z J_\pm|j m \rangle = \hbar (m \pm 1) J_\pm |jm \rangle \] therefore \[J_z |j m \pm 1 \rangle = \hbar (m \pm 1)|jm \rangle \] so we can see that Creation Operator: \[J_+|j m \rangle =\alpha|jm+1 \rangle \] Annihilation Operator: \[J_-|j m \rangle =\beta|m-1 \rangle \] Where \[\alpha=\hbar \sqrt{(j-m)(j+m+1)}|jm+1 \rangle \] \[\beta=\hbar \sqrt{(j+m)(j-m+1)}|jm-1 \rangle \] This shows why ladder operators are called creation and annihilation operators. The operators relate one eigenvalue to the next one by lower or rising the a quantum number. Another example of ladder operators is for the quantum harmonic oscillator. The ladder operators for quantum harmonic oscillator rise or lower the energy of the system by a quantam. To see where the operators come from, we start with the Schrödinger equation: \[\left(-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+\frac{1}{2}m\omega^2x^2\right)\psi(x)=E\psi(x)\] Setting x as \[x=\sqrt{\frac{\hbar}{m\omega}}q\] The Schrödinger equation is now \[\frac{\hbar\omega}{2}\left(-\frac{d^2}{dq^2}+q^2\right) \psi(q)=E\psi(q)\] \[\hat H=-\frac{d^2}{dq^2}+q^2\] \[\hat H=\left(-\frac{d}{dq}+q\right)\left(\frac{d}{dq}+q\right)+1\] So now the Schrödinger equation becomes \[\hbar\omega \left[\frac{1}{\sqrt2} \left(-\frac{d}{dq}+q\right) \frac{1}{\sqrt2} \left(\frac{d}{dq}+q\right) + \frac{1}{2}\right] \psi(q)=E\psi(q)\] The ladder operators are then Creation Operator: \[a^t=\frac{1}{\sqrt 2} \left(-\frac{d}{dq}+q \right )\] Annihilation Operator: \[a=\frac{1}{\sqrt 2} \left (\frac{d}{dq}+q \right )\] From this we can get our equation (we won't do it here) for the energy of an our eigenvaules. \[E_n=(n+\frac{1}{2})\hbar\omega\] There are two kinds of ladder operators, creation and annihilation operators. Like the word ladder suggests, these operators move eigenvalues up or down. They are used in angular momentum to rise or lower quantum numbers and quantum harmonic oscillators to move between energy levels. | 3,642 | 1,465 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/12%3A_Equilibrium_Conditions_in_Multicomponent_Systems/12.08%3A_Liquid-Gas_Equilibria |
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\( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \)
\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) This section describes multicomponent systems in which a liquid phase is equilibrated with a gas phase. If we vary the pressure of a liquid mixture at constant temperature and composition, there is a small effect on the fugacity of each volatile component in an equilibrated gas phase. One way to vary the pressure at essentially constant liquid composition is to change the partial pressure of a component of the gas phase that has negligible solubility in the liquid. At transfer equilibrium, component \(i\) has the same chemical potential in both phases: \(\mu_i\liquid =\mu_i\gas\). Combining the relations \(\bpd{\mu_i\liquid}{p}{T,\allni}=V_i\liquid\) and \(\mu_i\gas=\mu_i\st\gas +RT\ln(\fug_i/p\st)\) (Eqs. 9.2.49 and 9.3.12), we obtain \begin{gather} \s{ \frac{\dif\ln(\fug_i/p\st)}{\difp}=\frac{V_i\liquid}{RT} } \tag{12.8.1} \cond{(equilibrated liquid and} \nextcond{gas mixtures, constant \(T\)} \nextcond{and liquid composition)} \end{gather} Equation 12.8.1 shows that an increase in pressure, at constant temperature and liquid composition, causes an increase in the fugacity of each component in the gas phase. Integration of Eq. 12.8.1 between pressures \(p_1\) and \(p_2\) yields \begin{gather} \s{ \fug_i(p_2)=\fug_i(p_1) \exp\left[\int_{p_1}^{p_2}\frac{V_i\liquid}{RT} \difp\right] } \tag{12.8.2} \cond{(equilibrated liquid and} \nextcond{gas mixtures, constant \(T\)} \nextcond{and liquid composition)} \end{gather} The exponential on the right side is called the . The integral in the Poynting factor is simplified if we make the approximation that \(V_i\liquid\) is independent of pressure. Then we obtain the approximate relation \begin{gather} \s{ \fug_i(p_2) \approx \fug_i(p_1)\exp\left[ \frac{V_i\liquid (p_2-p_1)}{RT} \right] } \tag{12.8.3} \cond{(equilibrated liquid and} \nextcond{gas mixtures, constant \(T\)} \nextcond{and liquid composition)} \end{gather} The effect of pressure on fugacity is usually small, and can often be neglected. For typical values of the partial molar volume \(V_i\liquid\), the exponential factor is close to unity unless \(|p_2{-}p_1|\) is very large. For instance, for \(V_i\liquid{=}100\units{cm\(^3\) mol\(^{-1}\)}\) and \(T{=}300\K\), we obtain a value for the ratio \(\fug_i(p_2)/\fug_i(p_1)\) of \(1.004\) if \(p_2{-}p_1\) is \(1\br\), \(1.04\) if \(p_2{-}p_1\) is \(10\br\), and \(1.5\) if \(p_2{-}p_1\) is \(100\br\). Thus, unless the pressure change is large, we can to a good approximation neglect the effect of total pressure on fugacity. This statement applies only to the fugacity of a substance in a gas phase that is equilibrated with a liquid phase of constant composition containing the same substance. If the liquid phase is absent, the fugacity of \(i\) in a gas phase of constant composition is of course approximately proportional to the total gas pressure. We can apply Eqs. 12.8.2 and 12.8.3 to liquid A, in which case \(V_i\liquid\) is the molar volume \(V\A^*\liquid\). Suppose we have pure liquid A in equilibrium with pure gaseous A at a certain temperature. This is a one-component, two-phase equilibrium system with one degree of freedom (Sec. 8.1.7), so that at the given temperature the value of the pressure is fixed. This pressure is the saturation vapor pressure of pure liquid A at this temperature. We can make the pressure \(p\) greater than the saturation vapor pressure by adding a second substance to the gas phase that is essentially insoluble in the liquid, without changing the temperature or volume. The fugacity \(\fug\A\) is greater at this higher pressure than it was at the saturation vapor pressure. The vapor pressure \(p\A\), which is approximately equal to \(\fug\A\), has now become greater than the saturation vapor pressure. It is, however, safe to say that the difference is negligible unless the difference between \(p\) and \(p\A\) is much greater than \(1\br\). As an application of these relations, consider the effect of the size of a liquid droplet on the equilibrium vapor pressure. The calculation of Prob. 12.8(b) shows that the fugacity of H\(_2\)O in a gas phase equilibrated with liquid water in a small droplet is slightly greater than when the liquid is in a bulk phase. The smaller the radius of the droplet, the greater is the fugacity and the vapor pressure. Consider system 1 in Fig. 9.5. A binary liquid mixture of two volatile components, A and B, is equilibrated with a gas mixture containing A, B, and a third gaseous component C of negligible solubility used to control the total pressure. In order for A and B to be in transfer equilibrium, their chemical potentials must be the same in both phases: \begin{equation} \mu\A\liquid =\mu\A\st\gas +RT\ln\frac{\fug\A}{p\st} \qquad \mu\B\liquid =\mu\B\st\gas +RT\ln\frac{\fug\B}{p\st} \tag{12.8.4} \end{equation} Suppose we make an infinitesimal change in the liquid composition at constant \(T\) and \(p\). This causes infinitesimal changes in the chemical potentials and fugacities: \begin{equation} \dif\mu\A\liquid=RT\frac{\dif\fug\A}{\fug\A} \qquad \dif\mu\B\liquid=RT\frac{\dif\fug\B}{\fug\B} \tag{12.8.5} \end{equation} By inserting these expressions in the Gibbs–Duhem equation \(x\A\dif\mu\A = - x\B\dif\mu\B\) (Eq. 9.2.43), we obtain \begin{gather} \s{ \frac{x\A}{\fug\A}\dif\fug\A=-\frac{x\B}{\fug\B}\dif\fug\B} \tag{12.8.6} \cond{(binary liquid mixture equilibrated} \nextcond{with gas, constant \(T\) and \(p\))} \end{gather} This equation is a relation between changes in gas-phase fugacities caused by a change in the liquid-phase composition. It shows that a composition change at constant \(T\) and \(p\) that increases the fugacity of A in the equilibrated gas phase must decrease the fugacity of B. Now let us treat the liquid mixture as a binary solution with component B as the solute. In the ideal-dilute region, at constant \(T\) and \(p\), the solute obeys Henry’s law for fugacity:\begin{equation} \fug\B = \kHB x\B \tag{12.8.7} \end{equation} For composition changes in the ideal-dilute region, we can write \begin{equation} \frac{\dif\fug\B}{\dx\B} = \kHB = \frac{\fug\B}{x\B} \tag{12.8.8} \end{equation} With the substitution \(\dx\B=-\dx\A\) and rearrangement, Eq. 12.8.8 becomes \begin{equation} -\frac{x\B}{\fug\B}\dif\fug\B = \dx\A \tag{12.8.9} \end{equation} Combined with Eq. 12.8.6, this is \((x\A/\fug\A)\dif\fug\A=\dx\A\), which we can rearrange and integrate as follows within the ideal-dilute region: \begin{equation} \int_{\fug\A^*}^{\fug'\A}\frac{\dif\fug\A}{\fug\A} = \int_{1}^{x'\A}\frac{\dx\A}{x\A} \qquad \ln\frac{\fug'\A}{\fug\A^*}=\ln x'\A \tag{12.8.10} \end{equation} The result is \begin{gather} \s{ \fug\A=x\A\fug\A^* } \tag{12.8.11} \cond{(ideal-dilute binary solution)} \end{gather} Here \(\fug\A^*\) is the fugacity of A in a gas phase equilibrated with pure liquid A at the same \(T\) and \(p\) as the mixture. Equation 12.8.11 is Raoult’s law for fugacity applied to component A. If component B obeys Henry’s law at all compositions, then the Henry’s law constant \(\kHB\) is equal to \(\fug\B^*\) and B obeys Raoult’s law, \(\fug\B=x\B\fug\B^*\), over the entire range of \(x\B\). We can draw two conclusions: To a good approximation, by assuming an ideal gas mixture and neglecting the effect of total pressure on fugacity, we can apply Eq. 12.8.20 to a liquid–gas system in which the total pressure is constant, but instead is the sum of \(p\A\) and \(p\B\). Under these conditions, we obtain the following expression for the rate at which the total pressure changes with the liquid composition at constant \(T\): \begin{equation} \begin{split} \frac{\difp}{\dx\A} & = \frac{\dif(p\A+p\B)}{\dx\A} = \frac{\difp\A}{\dx\A} -\frac{x\A p\B}{x\B p\A} \frac{\difp\A}{\dx\A} = \frac{\difp\A}{\dx\A}\left( 1 - \frac{x\A/x\B}{p\A/p\B} \right) \cr & = \frac{\difp\A}{\dx\A}\left( 1 - \frac{x\A/x\B}{y\A/y\B}\right) \end{split} \tag{12.8.21} \end{equation} Here \(y\A\) and \(y\B\) are the mole fractions of A and B in the gas phase given by \(y\A=p\A/p\) and \(y\B=p\B/p\). We can use Eq. 12.8.21 to make several predictions for a binary liquid–gas system at constant \(T\). The activity of B in the gas phase is given by \(a\B\gas =\fug\B/p\st\). If the solute is a nonelectrolyte and we choose a standard state based on mole fraction, the activity in the solution is \(a\B\sln=\G\xbB \g\xbB x\B\). The equilibrium constant is then given by \begin{equation} K = \frac{\G\xbB \g\xbB x\B}{\fug\B/p\st} \tag{12.8.22} \end{equation} and the solubility, expressed as the equilibrium mole fraction of solute in the solution, is given by \begin{gather} \s{ x\B = \frac{K\fug\B/p\st}{\G\xbB \g\xbB} } \tag{12.8.23} \cond{(nonelectrolyte solute in} \nextcond{equilibrium with gas)} \end{gather} At a fixed \(T\) and \(p\), the values of \(K\) and \(\G\xbB\) are constant. Therefore any change in the solution composition that increases the value of the activity coefficient \(\g\xbB\) will decrease the solubility for the same gas fugacity. This solubility decrease is often what happens when a salt is dissolved in an aqueous solution, and is known as the (Prob. 12.11). Unless the pressure is much greater than \(p\st\), we can with negligible error set the pressure factor \(\G\xbB\) equal to 1. When the gas solubility is low and the solution contains no other solutes, the activity coefficient \(\g\xbB\) is close to 1. If furthermore we assume ideal gas behavior, then Eq. 12.8.23 becomes \begin{gather} \s{ x\B=K\frac{p\B}{p\st} } \tag{12.8.24} \cond{(nonelectrolyte solute in equilibrium} \nextcond{with ideal gas, \(\G\xbB{=}1\), \(\g\xbB{=}1\))} \end{gather} The solubility is predicted to be proportional to the partial pressure. The solubility of a gas that dissociates into ions in solution has a quite different dependence on partial pressure. An example is the solubility of gaseous HCl in water to form an electrolyte solution, shown in Fig. 10.1. If the actual conditions are close to those assumed for Eq. 12.8.24, we can use Eq. 12.1.13 to derive an expression for the temperature dependence of the solubility for a fixed partial pressure of the gas: \begin{equation} \Pd{\ln x\B}{T}{\!\!p\B} = \frac{\dif\ln K}{\dif T} = \frac{\Delsub{sol,B}H\st}{RT^2} \tag{12.8.25} \end{equation} At the standard pressure, \(\Delsub{sol,B}H\st\) is the same as the molar enthalpy of solution at infinite dilution. Since the dissolution of a gas in a liquid is invariably an exothermic process, \(\Delsub{sol,B}H\st\) is negative, and Eq. 12.8.25 predicts the solubility decreases with increasing temperature. Note the similarity of Eq. 12.8.25 and the expressions derived previously for the temperature dependence of the solubilities of solids (Eq. 12.5.8) and liquids (Eq. 12.6.3). When we substitute the mathematical identity \(\dif T=-T^2\dif(1/T)\), Eq. 12.8.25 becomes \begin{equation} \bPd{\ln x\B}{(1/T)}{p\B} = -\frac{\Delsub{sol,B}H\st}{R} \tag{12.8.26} \end{equation} We can use this form to evaluate \(\Delsub{sol,B}H\st\) from a plot of \(\ln x\B\) versus \(1/T\). The of a gas is the solubility calculated on the assumption that the dissolved gas obeys Raoult’s law for partial pressure: \(p\B = x\B p\B^*\). The ideal solubility, expressed as a mole fraction, is then given as a function of partial pressure by \begin{gather} \s{ x\B = \frac{p\B}{p\B^*} } \tag{12.8.27} \cond{(ideal solubility of a gas)} \end{gather} Here \(p\B^*\) is the vapor pressure of pure liquid solute at the same temperature and total pressure as the solution. If the pressure is too low for pure B to exist as a liquid at this temperature, we can with little error replace \(p\B^*\) with the saturation vapor pressure of liquid B at the same temperature, because the effect of total pressure on the vapor pressure of a liquid is usually negligible (Sec. 12.8.1). If the temperature is above the critical temperature of pure B, we can estimate a hypothetical vapor pressure by extrapolating the liquid–vapor coexistence curve beyond the critical point. We can use Eq. 12.8.27 to make several predictions regarding the ideal solubility of a gas at a fixed value of \(p\B\). As an example of the general validity of prediction 1, Hildebrand and Scott ( , 3rd edition, Dover, New York, 1964, Chap. XV) list the following solubilities of gaseous Cl\(_2\) in several dissimilar solvents at \(0\units{\(\degC\)}\) and a partial pressure of \(1.01\br\): \(x\B=0.270\) in heptane, \(x\B=0.288\) in SiCl\(_4\), and \(x\B=0.298\) in CCl\(_4\). These values are similar to one another and close to the ideal value \(p\B/p\B^*=0.273\). At the standard pressure \(p\st=1\br\), the value of \(\G\xbB\) is unity, and Eqs. 12.1.13 and 12.1.14 then give the following expressions for the dependence of the dimensionless quantity \(\kHB/p\st\) on temperature: \begin{gather} \s{ \frac{\dif\ln(\kHB/p\st)}{\dif T} = -\frac{\dif\ln K}{\dif T} = -\frac{\Delsub{sol,B}H\st}{RT^2}} \tag{12.8.31} \cond{(\(p{=}p\st\))} \end{gather} \begin{gather} \s{ \frac{\dif\ln(\kHB/p\st)}{\dif(1/T)} = -\frac{\dif\ln K}{\dif(1/T)} = \frac{\Delsub{sol,B}H\st}{R}} \tag{12.8.32} \cond{(\(p{=}p\st\))} \end{gather} These expressions can be used with little error at any pressure that is not much greater than \(p\st\), say up to at least \(2\br\), because under these conditions \(\G\xbB\) does not differ appreciably from unity. To find the dependence of \(\kHB\) on pressure, we substitute \(\G\xbB\) in Eq. 12.8.30 with the expression for \(\G\xbB\) at pressure \(p'\) found in Table 9.6: \begin{equation} \kHB(p') = \frac{\G\xbB(p') p\st}{K} = \frac{p\st}{K} \exp\left(\int_{p\st}^{p'}\frac{V\B^{\infty}}{RT}\difp\right) \tag{12.8.33} \end{equation} We can use Eq. 12.8.33 to compare the values of \(\kHB\) at the same temperature and two different pressures, \(p_1\) and \(p_2\): \begin{equation} \kHB(p_2) = \kHB(p_1)\exp\left(\int_{p_1}^{p_2}\frac{V\B^{\infty}}{RT}\difp\right) \tag{12.8.34} \end{equation} An approximate version of this relation, found by treating \(V\B^{\infty}\) as independent of pressure, is \begin{equation} \kHB(p_2) \approx \kHB(p_1)\exp\left[\frac{V\B^{\infty}(p_2-p_1)}{RT}\right] \tag{12.8.35} \end{equation} Unless \(|p_2-p_1|\) is much greater than \(1\br\), the effect of pressure on \(\kHB\) is small; see Prob. 12.12 for an example. | 21,727 | 1,466 |
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We noted previously that chemist's may be a misleading measure of a fuel's heating value in real conditions. Standard enthalpies are precisely defined as the heat energy absorbed or released when a process occurs at 25 C to give substances in their most stable state at that temperature. For example, for propane, water is produced as a liquid:
C H ( ) + 5 O ( ) → 3 CO ( )+ 4 H O ( ) Δ = –2219.2 kJ (1) But boilers are almost always operated with flue temperatures near the combustion temperature, and maintained at over 100 C to prevent water from condensing to liquid. Since water is produced as a gas, the standard thermodynamic equation (1), does not apply. If it were formed, liquid water would dissolve acidic flue gases like HCl and SO to make acidic solutions which corrode the system. These gases are normally removed by limestone "scrubbers", leaving just the water vapor. The white plumes over smokestacks are water droplets forming by condensation of water vapor. They often disappear as the droplets evaporate. The precision of our definition of guarantees that our calculations of heat energy will be sound, but it requires that we each energy cost and source. We'll see that inattention to such details has led to misinformation and confusion. Since Higher Heating Value (HHV) may be used by some air quality management authorities, while Lower Heating Values are used by many engineers and European power facilities, it is important to know precisely what the terms mean and which is being used. The USDA reports the following heating costs apparently, but not explicitly, stated to be LHVs: Gross heating value Efficiency (%) Net heating value Fuel required for 1 million Btu of usable heat Average cost/unit Total annual fuel cost Based on 100 million Btu of energy for the heating season, a typical value for an average sized house. There are many variables that affect total heating costs (furnace efficiencies, local energy costs, etc.), but one variable that can potentially be understood is the heat available from fuel combustion. Without precise definitions, even this may be obscured. To illustrate this point, let's examine what are known as the Higher Heating Value (HHV) and Lower Heating Value (LHV) are calculated for propane. The standard heat of combustion of propane is C H ( ) + 5 O ( ) → 3 CO ( )+ 4 H O ( ) Δ = -2219.2 kJ = = Δ (1) This is the process which yields the HHV, because the fuel is burned and the combustion products are cooled to 25 C, removing all heat resulting from the condensation of water in the process. But we want the enthalpy for the reaction: C H ( ) + 5 O ( ) → 3 CO ( )+ 4 H O ( ) Δ = = Δ (2) In this case, the water is not condensed, so some of the energy is not recovered, so the Δ is the Lower Heating Value. We can imagine reaction (2) occurring in two steps. First, reaction (1), then H O( ) → H O( ) Δ = 44.0 kJ (3) We notice that equation (1) produces 4 mol of H O ( ), so we multiply equation (3) by 4, so that 4 moles of H O ( ) are consumed in equation (3a): 4 H O( ) → 4 H O( ) Δ = 4 x 44.0 kJ = 176.0 kJ (3a) If we add equations (1) and (3a) as below, canceling the 4 mol H O ( ) that are produced (appear on the right) with the 4 mol H O ( ) which are consumed (appear on the left), we get equation (2). C H ( ) + 5 O ( ) → 3 CO ( ) + Δ = –2219.2 kJ (1) → 4 H O( ) Δ = 4 mol x 44.0 kJ mol = 176.0 kJ (3a) ————————————————————————————————————————
C H ( ) + 5 O ( ) → 3 CO ( )+ 4 H O ( ) Δ = LHV = Δ =–2043.2 kJ (2) Experimentally it is found that the enthalpy change for reaction (2) is the of the enthalpy changes for reactions (1) and (3a): Δ = –2219.2 kJ + (4 x 44.0 kJ) = –2043.2 kJ = Δ + 4 x Δ This value of Δ should be the Lower Heating Value. Let's see if it matches the USDA value of 91,200 BTU/gal in the table above, given that a gallon of propane is about 4.23 lb \(\frac{\text{2043.2.2 kJ}}{mol}
~\times~\frac{\text{1 mol}}{\text{44.1 g}}
~\times~\frac{\text{1 BTU}}{\text{1.055 kJ}}
~\times~\frac{\text{453.6 g}}{\text{lb}}
~\times~\frac{\text{4.23 lb}}{\text{gal}}
~=~\text{84 300 BTU/gal}
\) Our value is not close to the USDA LHV, but it matches the Oak Ridge National Laboratory (ORNL) value for the HHV of 84 250 BTU/gal . The Higher Heating Value is the enthalpy change for reaction (1), which includes the heat released when 4 mol of gaseous water from the combustion cool to 25 C, so its value is more negative than the LHV by four times the heat of condensation of water (–2043.2 + 4 x (–-44) = –2219.2 kJ):
H O( ) → H O( ) Δ = –44 kJ (4) Repeating the calculation for the heat liberated in equation (1) in BTU/gal, we get 91 500 BTU/gal, which is the HHV reported by ORNL. The USDA table above apparently reports the HHV, perhaps unknowingly. The variety of values found on the web for both HHV and LHV attests to the fact that much confusion can result from not being careful with the meaning and application of a . Technically, the lower heating value of a fuel is defined as the amount of heat released in the combustion of the fuel to give products at 150°C. In the general case it is always true that . This principle is known as . If it were not true, it would be possible to think up a series of reactions in which energy would be created but which would end up with exactly the same substances we started with. This would contradict the law of conservation of energy. Hess’ law enables us to obtain Δ values for reactions which cannot be carried out experimentally, as the next example shows. Acetylene (C H ) cannot be prepared directly from its elements according to the equation 2C( ) + H ( ) → C H ( ) (1) Calculate Δ for this reaction from the following thermochemical equations, all of which can be determined experimentally: C( ) + O ( ) → CO ( ) Δ = –393.5 kJ (2 ) H ( ) + ½O ( ) → H O( ) Δ = –285.8 kJ (2 ) C H ( ) + \(\tfrac{\text{5}}{\text{2}}\)O ( ) → 2CO ( ) + H O( ) Δ = –1299.8 kJ (2 ) We use the following strategy to manipulate the three experimental equations so that when added they yield Eq. (1): Since Eq. (1) has 2 mol C on the left, we multiply Eq. (2 ) by 2. Since Eq. (1) has mol H on the left, we leave Eq. (2b) unchanged. Since Eq. (1) has mol C H on the , whereas there is mol C H on the of Eq. (2c) we write Eq. (2c) in reverse. We then have \(\begin{align}
\text{2 C(s) + O}_{\text{2}}\text{(g)}&\to \text{CO}_{\text{2}}\text{(g)}~~~~~~~~~~~~~~~~\Delta H_{\text{m}}=\text{ 2 (}\text{-393.5 kJ)} \\
\text{H}_{\text{2}}\text{(g) + }\tfrac{1}{2}\text{O}_{\text{2}}\text{(g)}&\to \text{H}_{\text{2}}\text{O}(l)~~~~~~~~~~~~~~~~\Delta H_{\text{m}}=\text{-285.8 kJ} \\
\underline{\text{2 CO}_{\text{2}}\text{(g) + H}_{\text{2}}\text{O}(l)}&\underline{\to \text{ C}_{\text{2}}\text{H}_{\text{2}}\text{(g) +}\tfrac{5}{2}\text{O}_{\text{2}}\text{(g)}}~~\Delta H_{\text{m}}=-\text{(}-\text{1299}\text{.8 kJ)} \\
\text{2 C(s) + H}_{\text{2}}\text{(g) + 2}\tfrac{1}{2}\text{O}_{\text{2}}\text{(g)}&\to \text{C}_{\text{2}}\text{H}_{\text{2}}\text{(g) +}\tfrac{5}{2}\text{O}_{\text{2}}\text{(g)} \\
\Delta H_{\text{m}}=\text{(}-\text{787}\text{.0}&-\text{285}\text{.8 + 1299}\text{.8) kJ}=\text{227}\text{.0 kJ} \\
\end{align}\) Thus the desired result is 2C( ) + H ( ) → C H ( ) Δ = 227.0 kJ | 7,314 | 1,467 |
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Only very massive nuclei with high neutron-to-proton ratios can undergo , in which the nucleus breaks into two pieces that have different atomic numbers and atomic masses. This process is most important for the transactinide elements, with ≥ 104. Spontaneous fission is invariably accompanied by the release of large amounts of energy, and it is usually accompanied by the emission of several neutrons as well. An example is the spontaneous fission of \(^{254}_{98}\textrm{Cf}\), which gives a distribution of fission products; one possible set of products is shown in the following equation: \[^{254}_{98}\textrm{Cf}\rightarrow \,^{118}_{46}\textrm{Pd}+\,^{132}_{52}\textrm{Te}+4^{1}_{0}\textrm{n}\label{5.2.16}\] Once again, the number of nucleons is conserved. Thus the sum of the mass numbers of the products (118 + 132 + 4 = 254) equals the mass number of the reactant. Similarly, the sum of the atomic numbers of the products [46 + 52 + (4 × 0) = 98] is the same as the atomic number of the parent nuclide. Write a balanced nuclear equation to describe each reaction. radioactive nuclide and mode of decay balanced nuclear equation Identify the reactants and the products from the information given. Use the values of and to identify any missing components needed to balance the equation. a. We know the identities of the reactant and one of the products (a β particle). We can therefore begin by writing an equation that shows the reactant and one of the products and indicates the unknown product as \(^{A}_{Z}\textrm{X}\): Because both protons and neutrons must be conserved in a nuclear reaction, the unknown product must have a mass number of = 35 − 0 = 35 and an atomic number of = 16 − (−1) = 17. The element with = 17 is chlorine, so the balanced nuclear equation is as follows: We know the identities of both reactants: \(^{201}_{80}\textrm{Hg}\) and an inner electron, \(^{0}_{-1}\textrm{e}\). The reaction is as follows: \(^{201}_{80}\textrm{Hg}+\,^{0}_{-1}\textrm e\rightarrow\,^{A}_{Z}\textrm{X}\) Both protons and neutrons are conserved, so the mass number of the product must be = 201 + 0 = 201, and the atomic number of the product must be = 80 + (−1) = 79, which corresponds to the element gold. The balanced nuclear equation is thus \(^{201}_{80}\textrm{Hg}+\,^{0}_{-1}\textrm e\rightarrow\,^{201}_{79}\textrm{Au}\) As in part (a), we are given the identities of the reactant and one of the products—in this case, a positron. The unbalanced nuclear equation is therefore \(^{30}_{15}\textrm{P}\rightarrow\,^{A}_{Z}\textrm{X}+\,^{0}_{+1}\beta\) The mass number of the second product is = 30 − 0 = 30, and its atomic number is = 15 − 1 = 14, which corresponds to silicon. The balanced nuclear equation for the reaction is as follows: \(^{30}_{15}\textrm{P}\rightarrow\,^{30}_{14}\textrm{Si}+\,^{0}_{+1}\beta\) Write a balanced nuclear equation to describe each reaction. Predict the kind of nuclear change each unstable nuclide undergoes when it decays. nuclide type of nuclear decay Based on the neutron-to-proton ratio and the value of , predict the type of nuclear decay reaction that will produce a more stable nuclide. Predict the kind of nuclear change each unstable nuclide undergoes when it decays. | 3,279 | 1,468 |
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Cinnamon has been known for thousands of years. One of the first times it is mentioned is in the Old Testament when Moses uses cinnamon in holy oil. Using cinnamon in holy oil shows how prized this spice was in ancient times. In fact, in Greece cinnamon was used as a sacrifice to the Gods like Apollo. Although cinnamon was used in the Mediterranean, it originated in Sri Lanka. The cinnamon traveled along the spice route, and the actual origins of cinnamon were held quiet. By not allowing other traders to know the source of the spice the growers were able to keep a monopoly and control the price (at a high one of course). In the sixteenth century Portugal finally made it to Sri Lanka. In 1518 a trading fort was set up there. The Dutch fought the Portuguese and eventually won control of the cinnamon trade in 1658. The British eventually took control of the island in 1796 but by then cinnamon wasn't as highly prized. Through the spice trade cinnamon made its way out of Sri Lanka and into the cuisines and cultures of the rest of the world. In Mexico especially cinnamon is a huge part of their cuisine. In fact, Mexico is the main importer of raw cinnamon. The majority of this cinnamon is used in the processing of chocolate. Also, cinnamon in Mexico is used in desserts, hot chocolate, and many other things. The spice is used in the same kind of dishes in the United States, as well as things such as cereal. In the Middle East cinnamon is used to flavor lamb as well as soups and even as a flavor in pickling. Cinnamon's unique taste and ability to be used as a spice is the reason why it has been incorporated into cultures all around the world. The component in cinnamon that gives it it's properties is the presence of cinnamaldehyde. The essential oil that makes up cinnamon is over 90% cinnamaldehyde. Cinnamaldehyde is an organic compound that can also be classified as an aldehyde. Cinnamaldehyde is unique in that it also contains a benzene ring and a double bond, as is seen in the structure in Figure 1. Cinnamaldehyde is also used in many other foods as a flavoring. While cimmaldehyde can be synthesized, the main source of it is from cinnamon bark. The functional group found in formaldehyde is called a . Two classes of compounds may be distinguished on the basis of the location of the carbonyl group. In aldehydes it is at the end of a carbon chain and has at least one hydrogen attached. In the carbonyl group is attached to two carbon atoms. Some examples are
cinnamaldehyde Chemical tests may be performed to determine the presence of a carbonyl group. In the video below, a solution of 2,4-dinitrophenylhydrazine is added to test tubes containing 2-propanol, an alcohol; 2-propanone (acetone), a ketone; and propionic acid a carboxylic acid. 2,4-dinitrophenylhydrazine only reacts with the carbonyl group of 2-propanone, forming an orange precipitate. The reaction that occurs is: Another test can distinguish between aldehydes and ketones. Here is a video of the Silver Mirror Tollens Test for Aldehydes: Tollens reagent is an aqueous solution of silver nitrate, sodium hydroxide, and a little ammonia. (In the video the ammonia comes from reaction of ammonium ions in ammonium nitrate and hydroxide ions from sodium hydroxide to form water and NH .) If an aldehyde, in this case, glucose, is added to the solution, the Ag is reduced by the aldehyde, and the aldehyde is oxidized into a carboxylic acid. This produces silver metal, which coats the flask and creates the mirror. A similar reaction does not occur for ketones, so only aldehydes produce the silver mirror. The equation for the reaction in the video is: The endings and signify dehyde and ket , respectively. The general formula for an aldehyde is , while for a ketone it is . Note that every ketone is isomeric with at least one aldehyde. Acetone, for example, has the same molecular formula (C H O) as propanal. Aldehyde and ketone molecules cannot hydrogen bond among themselves for the same reason that ethers cannot—they do not contain hydrogens attached to highly electronegative atoms. The carbonyl group is rather polar, however, since the difference between the electronegativities of carbon (2.5) and oxygen (3.5) is rather large, and there are usually no other dipoles in an aldehyde or ketone molecule to cancel the effect of C==O. Therefore the boiling points of aldehydes and ketones are intermediate between those of alkanes or ethers on the one hand and alcohols on the other. Acetaldehyde, CH CH CHO, boils at 20.8°C midway between propane (–42°C) and ethanol (78.5°C). The boiling points of propanal and acetone are compared with other organic compounds in the table of the boiling points of comparable organic compounds which shows the same trend. Of all the aldehydes and ketones, formaldehyde and acetone are of greatest commercial importance. Uses of formaldehyde have already been mentioned. Like other ketones, acetone is mainly useful as a solvent, and you may have used it for this purpose in the laboratory. Acetone and other ketones are somewhat toxic and should not be handled carelessly. From ChemPRIME: 8.15: Aldehydes and Ketones | 5,197 | 1,469 |
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The electron-transfer reactions that occur within and between proteins typically involve prosthetic groups separated by distances that are often greater than 10 Å. When we consider these distant electron transfers, an explicit expression for the electronic factor is required. In the nonadiabatic limit, the rate constant for reaction between a donor and acceptor held at fixed distance and orientation is: \[k_{et} = \bigg[ \frac{H_{AB} \;^{2}}{\hbar}\left(\dfrac{\pi}{\lambda RT}\right)^{1/2} \bigg]^{\frac{-(\lambda + \Delta G^{o})^{2}}{4 \lambda RT}}\ldotp \tag{6.27}\] The electronic (or tunneling) matrix element H is a measure of the electronic coupling between the reactants and the products at the transition state. The magnitude of H depends upon donor-acceptor separation, orientation, and the nature of the intervening medium. Various approaches have been used to test the validity of Equation (6.27) and to extract the parameters H and \(\lambda\). Driving-force studies have proven to be a reliable approach, and such studies have been emphasized by many workers. In the nonadiabatic limit, the probability is quite low that reactants will cross over to products at the transition-state configuration. This probability depends upon the electronic hopping frequency (determined by H ) and upon the frequency of motion along the reaction coordinate. In simple models, the electronic-coupling strength is predicted to decay exponentially with increasing donor-acceptor separation (Equation 6.28): \[H_{AB} = (H_{AB}^{o})^{\frac{- \beta}{2} (\textbf{d} - \textbf{d}^{o})} \tag{6.28} \] In Equation (6.28), H ° is the electronic coupling at close contact ( °), and \(\beta\) is the rate of decay of coupling with distance ( ). Studies of the distance dependence of electron-transfer rates in donor-acceptor complexes, and of randomly oriented donors and acceptors in rigid matrices, have suggested 0.8 \(\leq \beta \leq\) 1.2 Å . Analysis of a large number of intramolecular electron-transfer rates has suggested a \(\beta\) value of 1.4 Å for protein reactions (Figure 6.24). Assigning a single protein \(\beta\) implies that the intervening medium is homogenous. At best this is a rough approximation, because the medium separating two redox sites in a protein is a heterogenous array of bonded and nonbonded interactions. Beratan and Onuchic have developed a formalism that describes the medium in terms of "unit blocks" connected together to form a tunneling pathway. A unit block may be a covalent bond, a hydrogen bond, or a through-space jump, each with a corresponding decay factor. Dominant tunneling pathways in proteins are largely composed of bonded groups (e.g., peptide bonds), with less favorable through-space interactions becoming important when a through-bond pathway is prohibitively long (Figure 6.25). The tunneling pathway model has been used successfully in an analysis of the electron-transfer rates in modified cytochromes c (Section IV.D.1). Plastocyanin cycles between the Cu and Cu oxidation states, and transfers electrons from cytochrome f to the P component of photosystem I in the chloroplasts of higher plants and algae. The low molecular weight (10.5 kDa) and availability of detailed structural information have made this protein an attractive candidate for mechanistic studies, which, when taken together, point to two distinct surface binding sites (i.e., regions on the plastocyanin molecular surface at which electron transfer with a redox partner occurs). The first of these, the solvent-exposed edge of the Cu ligand His-87 (the adjacent site A in Figure 6.26), is ~6 Å from the copper atom and rather nonpolar. The second site (the remote site R in Figure 6.26) surrounds Tyr-83, and is much farther (~15 Å) from the copper center. Negatively charged carboxylates at positions 42-45 and 59-61 make this latter site an attractive one for positively charged redox reagents. Bimolecular electron-transfer reactions are typically run under pseudo-first-order conditions (e.g., with an inorganic redox reagent present in ~15-fold excess): \[Rate = k[plastocyanin,complex] = k_{obs}[plastocyanin] \ldotp \tag{6.29}\] For some reactions [e.g., Co(phen) oxidation of plastocyanin (Cu )] the expected linear plot of k vs. [complex] is not observed. Instead, the rate is observed to saturate (Figure 6.27). A "minimal" model used to explain this behavior involves the two pathways for electron transfer shown in Equation (6.30). \(\tag{6.30}\) Surprisingly, the rate ratio k /k is 7. Calculations indicate that, despite the significant differences in distances, H for the remote site is ~15 percent of H for the adjacent site. This figure is much higher than would be expected from distance alone, suggesting that the value of the decay parameter \(\beta\) in Equation (6.28) depends strongly on the structure of the intervening medium. Chemical modification of structurally characterized metalloproteins by transition-metal redox reagents has been employed to investigate the factors that control long-range electron-transfer reactions. In these semisynthetic multisite redox systems, the distance is fixed, and tunneling pathways between the donor and acceptor sites can be examined. Sperm-whale myoglobin can be reacted with (NH ) Ru(OH ) and then oxidized to produce a variety of ruthenated products, including a His-48 derivative whose Ru \(\leftrightarrow\) Fe tunneling pathway is depicted in Figure 6.28. Electrochemical data (Table 6.5) indicate that the (NH ) Ru group does not significantly perturb the heme center, and that equilibrium (i.e., k = k + k ) should be approached when a mixed-valent intermediate is produced by flash-photolysis techniques: \[(NH_{3})_{5}Ru^{3+}-Mb(Fe^{3+}) \xrightarrow[e^{-}]{fast} (NH_{3})_{5}Ru^{2+}-Mb(Fe^{3+}) \xrightleftharpoons[k_{-1}]{k_{1}} (NH_{3})_{5}Ru^{3+}-Mb(Fe^{2+}) \tag{6.31}\] This kinetic behavior was observed, and both the forward (k ) and reverse (k ) reactions were found to be markedly temperature-dependent: k = 0.019 s (25 °C), \(\Delta\)H = 7.4 kcal/mol, k = 0.041 s ) (25°C), \(\Delta\)H = 19.5 kcal/mol. X-ray crystallographic studies indicate that the axial water ligand dissociates upon reduction of the protein. This conformational change does not control the rates, since identical results were obtained when a second flash-photolysis technique was used to generate (NH ) Ru -Mb(Fe ) in order to approach the equilibrium from the other direction. Fe (NH ) Ru Cyanogen bromide has been used to modify the six-coordinate metmyoglobin heme site, causing the coordinated water ligand to dissociate. The CNBr-modified myoglobin heme site is thus five-coordinate in oxidation states. As expected, the self-exchange rate increased from ~1 M s to ~10 M s . Recent efforts in modeling biological electron transfers using chemically modified redox proteins point the way toward the design of semisynthetic redox enzymes for catalytic applications. An intriguing example, termed flavohemoglobin, was produced by reaction of hemoglobin with a flavin reagent designed to react with Cys-93 of the \(\beta\)-chain (i.e., the hemoglobin molecule was modified by two flavin moieties). The resulting derivative, unlike native hemoglobin, accepts electrons directly from NADPH and catalyzes the -hydroxylation of aniline in the presence of O and NADPH. In physiologically relevant precursor complexes, both redox centers are frequently buried in protein matrices. Characterization of such protein-protein complexes is clearly important, and several issues figure prominently: Most of our knowledge about the structures of protein-protein complexes comes from crystallographic studies of antigen-antibody complexes and multisubunit proteins; such systems generally exhibit a high degree of thermodynamic stability. On the other hand, complexes formed as a result of bimolecular collisions generally are much less stable, and tend to resist attempts to grow x-ray-quality crystals; the high salt conditions typically used in protein crystallizations often lead to dissociation of such complexes. One of the most widely studied protein-protein complexes is that formed between mammalian cytochrome b and cytochrome c. Using the known x-ray structures of both proteins, Salemme generated a static computer graphics model of this electron-transfer complex by docking the x-ray structures of the individual proteins. Two features of this model and its revision by molecular dynamics simulations (Figure 6.29 See color plate section, page C-12.) are noteworthy: (1) several Lys residues on cytochrome c and carboxylate-containing groups on cytochrome b form "salt bridges" (i.e., intermolecular hydrogen bonds); and (2) the hemes are nearly coplanar and are ~17 Å (Fe-Fe) apart. This distance was confirmed by an energy-transfer experiment in which the fluorescence of Zn-substituted cytochrome c was quenched by cytochrome b . Spectroscopic studies have verified the suggestion that these proteins form a 1:1 complex at low ionic strength (Figure 6.30). In addition, chemical modification and spectroscopic analyses are all in agreement with the suggestion that the complex is primarily stabilized by electrostatic interactions of the (-NH ••• O C—) type. The effect of ionic strength on the reduction of cytochrome c by cytochrome b is also in accord with this picture: lowering the ionic strength increases the reaction rate, as expected for oppositely charged molecules. A common experimental strategy for studying electron transfers proteins uses a metal-substituted heme protein as one of the reactants. In particular, the substitution of zinc for iron in one of the porphyrin redox centers allows facile initiation of electron transfer through photoexcitation of the zinc porphyrin (ZnP). The excited zinc porphyrin, ZnP* in Equation (6.32), may decay back (k ~ 10 s ) to the ground state or transfer an electron to an acceptor. \(\tag{6.32}\) The ZnP cation radical produced in the k step is a powerful oxidant; back electron transfer (k ) will thus occur and regenerate the starting material. The reactions shown in Equation (6.32) have been investigated in mixedmetal [Zn, Fe] hemoglobins. A hemoglobin molecule can be viewed as two independent electron-transfer complexes, each consisting of an \(\alpha_{1}\)-\(\beta_{2}\) subunit pair (Figure 6.31), since the \(\alpha_{1}\)-\(\alpha_{2}\), (\beta_{1}\)-(\beta_{2}\), and \(\alpha_{1}\)-(\beta_{1}\) distances are prohibitively long (> 30 Å). Both [\(\alpha\)(Zn), \(\beta\)(Fe)] and [\(\alpha\)(Fe), \(\beta\)(Zn)] hybrids have been studied. The ZnP and FeP are nearly parallel, as in the cytochrome b -cytochrome c model complex. Long-range electron transfer ( ZnP* → Fe ) between the \(\alpha_{1}\) and \(\beta_{2}\) subunits has been observed (the heme-edge/heme-edge distance is ~20 Å). The driving force for the forward electron-transfer step is ~0.8 eV, and k (see Equation 6.32) is ~100 s at room temperature, but decreases to ~9 s in the low-temperature region (Figure 6.32). Below 140-160 K the vibrations that induce electron transfer "freeze out"; nuclear tunneling is usually associated with such slow, temperature-independent rates. A complete analysis of the full temperature dependence of the rate requires a quantum-mechanical treatment of \(\lambda_{i}\) rather than that employed in the Marcus theory. It is interesting to note that the heme b vinyl groups (see Figure 6.6) for a given [\(\alpha_{1}\)(Fe), \(\beta_{2}\)Zn)] hybrid point toward each other and appear to facilitate electron transfer. | 11,639 | 1,471 |
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Biological oils are different from petroleum oils ("regular" diesel fuel) in molecular structure and properties. Common petroleum diesel fuel is a mixure of simple hydrocarbons, with the average chemical formula C H (shown below), but components may range approximately from C H (dodecane) to C H (pentadecane). By contrast, biological oils are "triglycerides" (classified as "esters") like the glyceryl trilinoleate shown below: Because of their large size and consequent large intermolecular attractions, the viscosity of biological oils is generally too high for use in conventional diesel engines. Biological oils also burn a little less readily, and with a sootier flame than petroleum diesel. Biological oils can be used in conventional diesel engines if they are preheated to reduce their viscosity, but this requires an auxiliary electrical heater until the engine warms up. For these resons, bilogical oils require processing for use as biodiesel. A biological oil is an ester, which is a type of organic compound having the atom linkage shown below. The ester linkage. R and R' represent chains of carbon and hydrogen atoms. R = -CH and R' = -C H O for the methyl stearate in Example 1 The ester linkage in biological oils is created when a glycerol molecule reacts with organic acids. The glycerol molecule has a chain of 3 carbon atoms, each with an -OH (alcohol) group on it. The figure below shows how an organic alcohol reacts with a organic acid. Organic chemists abbreviate molecular structures--the "zig-zag" lines in the figure represent carbon chains with a C atom at each "zig" or "zag". Each carbon has 4 bonds, and if fewer than 4 are shown, it's assumed that they go to H atoms. So the alcohol is C H OH (ethanol), and the acid is acetic acid (or ethanoic acid, CH COOH) in the Figure: Since glycerol has 3 -OH groups, 3 long chain organic "fatty acids" attach to make the bulky "triglyceride". But just as easily as esters can be made from alcohols and acids, they can switch alcohols or acids. In the presence of a strong base catalyst, like NaOH, a triglyceride can react with 3 small alcohol molecules, like methanol (CH OH), which replace the glycerol "backbone", making 3 separate esters of lower molecular weight Quite often a mixture of two or more products is formed. For example, when a vegetable oil reacts with methanol, only one or two of the acids may be displaced from the glycerine, producing only 1 or 2 FAMEs. Usually, a large excess of methanol and sodium hydroxide are added, so that the reaction produces the maximum amount of FAME. But in the case of a transesterification, even though none of the reactants is completely consumed, no further increase in the amounts of the products occurs. We say that such a reaction does not . When a mixture of products is produced or a reaction does not go to completion, the effectiveness of the reaction is usually evaluated in terms of of the desired product. A is calculated by assuming that all the limiting reagent is converted to product. The experimentally determined mass of product is then compared to the theoretical yield and expressed as a percentage: \(\text{Percent yield}=\frac{\text{actual yield}}{\text{theoretical yield}}\times \text{100 percent}\) When 100.0 g C H (C H O ) gas and 15.0 g CH OH are mixed at 55°C with NaOH catalyst, they react to form 90.96 g C H COOCH methyl stearate biodiesel. Calculate the percent yield. We must calculate the theoretical yield of NH , and to do this, we must first discover whether N or H is the limiting reagent. For the balanced equation The stoichiometric ratio of the reactants is \(\text{S}\left( \frac{\text{stearin}}{\text{CH}_{\text{3}}\text{OH}} \right)=\frac{\text{1 mol stearin}}{\text{3 mol CH}_{\text{3}}\text{OH}}\) Now, the initial amounts of the two reagents are and \(\begin{align}
& n_{\text{stearin}}\text{(initial)}=\text{100}\text{.0 g stearin}\times \frac{\text{1 mol stearin}}{\text{891}\text{.5 stearin}}=\text{0}\text{.1122 mol stearin} \\
& \\
& n_{\text{CH}_{\text{3}}\text{OH}}\text{(initial)}=\text{15}\text{.0 g CH}_{\text{3}}\text{OH}\times \frac{\text{1 mol CH}_{\text{3}}\text{OH}}{\text{32}\text{.04 g CH}_{\text{3}}\text{OH}}=\text{0}\text{.4682 mol CH}_{\text{3}}\text{OH} \\
\end{align}\) The ratio of initial amounts is thus \(\frac{n_{\text{stearin}}\text{(initial)}}{n_{\text{CH}_{\text{3}}\text{OH}}\text{(initial}}~=~ \frac{\text{0}\text{.1122 mol stearin}}{\text{0}\text{.4682 mol CH}_{\text{3}}\text{OH}}~=~\frac{\text{0}\text{.240 mol stearin}}{\text{1 mol CH}_{\text{3}}\text{OH}}\) Since this ratio is less than \(\text{S}\left( \frac{\text{stearin}}{\text{CH}_{\text{3}}\text{OH}} \right)~=~0.33\), there is an excess of CH OH. Stearin is the limiting reagent. Accordingly we must use 0.1122 mol stearin and 0.3366 mol CH OH (rather than 0.4682 mol CH OH) to calculate the theoretical yield of C H COOCH (methyl stearate). We then have \(n_{\text{methyl stearate}}\text{(theoretical)}=\text{0}\text{.1122 mol stearin}\times \frac{\text{3 mol methyl stearate}}{\text{1 mol stearin}}=\text{0}\text{.3365 mol methyl stearate}\) so that \(\text{m}_{\text{methyl stearate}}\text{(theoretical)}=\text{0}\text{.3365 mol methyl stearate}\times \frac{\text{298}\text{.51 g methyl stearate}}{\text{1 mol methyl stearate}}=\text{100}\text{.5 g methyl stearate}\) We can organize these calculations in a table: The percent yield is then \(\text{Percent yield}=\frac{\text{actual yield}}{\text{theoretical yield}}\times \text{100 percent }=\frac{\text{90}\text{.96 g}}{\text{100}\text{.5 g}}\times \text{100 percent}=\text{90}\text{.55 percent}\) Transesterification is a classic example of a reaction which does not go to completion. | 5,751 | 1,473 |
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In the mid 1920's the German physicist Werner Heisenberg showed that if we try to locate an electron within a region \(Δx\); e.g. by scattering light from it, some momentum is transferred to the electron, and it is not possible to determine exactly how much momentum is transferred, even in principle. Heisenberg showed that consequently there is a relationship between the uncertainty in position \(Δx\) and the uncertainty in momentum \(Δp\). \[\Delta p \Delta x \ge \frac {\hbar}{2} \label {5-22}\] You can see from Equation \(\ref{5-22}\) that as \(Δp\) approaches 0, \(Δx\) must approach ∞, which is the case of the free particle discussed . This uncertainty principle, which also is discussed in , is a consequence of the wave property of matter. A wave has some finite extent in space and generally is not localized at a point. Consequently there usually is significant uncertainty in the position of a quantum particle in space. Activity 1 at the end of this chapter illustrates that a reduction in the spatial extent of a wavefunction to reduce the uncertainty in the position of a particle increases the uncertainty in the momentum of the particle. This illustration is based on the ideas described in the next section. Compare the minimum uncertainty in the positions of a baseball (mass = 140 gm) and an electron, each with a speed of 91.3 miles per hour, which is characteristic of a reasonable fastball, if the standard deviation in the measurement of the speed is 0.1 mile per hour. Also compare the wavelengths associated with these two particles. Identify the insights that you gain from these comparisons. | 1,643 | 1,474 |
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In we suppose that we have a thermometer that we can use to measure the temperature of a gas. We suppose that this thermometer uses a liquid, and we define an increase in temperature by the increase in the volume of this liquid. Our statement of Charles’ law asserts that the volume of a gas is a linear function of the volume of the liquid in our thermometer, and that the same linear function is observed for any gas. As we note in , there is a problem with this statement. Careful experiments with such thermometers produce results that deviate from Charles’ law. With sufficiently accurate volume measurements, this occurs to some extent for any choice of the liquid in the thermometer. If we make sufficiently accurate measurements, the volume of a gas is not exactly proportional to the volume of any liquid (or solid) that we might choose as the working substance in our thermometer. That is, if we base our temperature scale on a liquid or solid substance, we observe deviations from Charles’ law. There is a further difficulty with using a liquid as the standard fluid on which to base our temperature measurements: temperatures outside the liquid range of the chosen substance have to be measured in some other way. Evidently, we can choose to use a gas as the working fluid in our thermometer. That is, our gas-volume measuring device is itself a thermometer. This fact proves to be very useful because of a further experimental observation. To a very good approximation, we find: If we keep the pressures in the thermometer and in some other gaseous system constant at low enough values, both gases behave as ideal gases, and we find that the volumes of the two gases are proportional to each other over any range of temperature. Moreover, this proportionality is observed for any choice of either gas. This means that we can define temperature in terms of the expansion of any constant-pressure gas that behaves ideally. In principle, we can measure the same temperature using any gas, so long as the constant operating pressure is low enough. When we do so, our device is called the . In so far as any gas behaves as an ideal gas at a sufficiently low pressure, any real gas can be used in an ideal gas thermometer and to measure any temperature accurately. Of course, practical problems emerge when we attempt to make such measurements at very high and very low temperatures. The (very nearly) direct proportionality of two low-pressure real gas volumes contrasts with what we observe for liquids and solids. In general, the volume of a given liquid (or solid) substance is not exactly proportional to the volume of a second liquid (or solid) substance over a wide range of temperatures. In practice, the ideal-gas thermometer is not as convenient to use as other thermometers—like the mercury-in-glass thermometer. However, the ideal-gas thermometer is used to calibrate other thermometers. Of course, we have to calibrate the ideal-gas thermometer itself before we can use it. We do this by assigning a temperature of 273.16 K to the triple point of water. (It turns out that the melting point of ice isn’t sufficiently reproducible for the most precise work. Recall that the triple point is the temperature and pressure at which all three phases of water are at equilibrium with one another, with no air or other substances present. The triple-point pressure is 611 Pa or \(\mathrm{6.03\times }{\mathrm{10}}^{\mathrm{-3\ }}\)atm. See .) From both theoretical considerations and experimental observations, we are confident that no system can attain a temperature below absolute zero. Thus, the size\({}^{3}\) of the kelvin (one degree on the Kelvin scale) is fixed by the difference in temperature between a system at the triple point of water and one at absolute zero. If our ideal gas thermometer has volume \(V\) at thermal equilibrium with some other constant-temperature system, the proportionality of \(V\) and \(T\) means that \[\frac{T}{V}=\frac{273.16}{V_{273.16}}\] With the triple point fixed at 273.16 K, experiments find the freezing point of air-saturated water to be 273.15 K when the system pressure is 1 atmosphere. (So the melting point of ice is 273.15 K, and the triple-point is 0.10 C. We will find two reasons for the fact that the melting point is lower than the triple point: In we find that the melting point of ice decreases as the pressure increases. In we find that solutes usually decrease the temperature at which the liquid and solid states of a substance are in phase equilibrium.) If we could use an ideal gas in our ideal-gas thermometer, we could be confident that we had a rigorous operational definition of temperature. However, we note in that any real gas will exhibit departures from ideal gas behavior if we make sufficiently accurate measurements. For extremely accurate work, we need a way to correct the temperature value that we associate with a given real-gas volume. The issue here is the value of the partial derivative \[{\left(\frac{\partial V}{\partial T}\right)}_P\] For one mole of an ideal gas, \[{\left(\frac{\partial V}{\partial T}\right)}_P=\frac{R}{P}=\frac{V}{T}\] is a constant. For a real gas, it is a function of temperature. Let us assume that we know this function. Let the molar volume of the real gas at the triple point of water be \(V_{273.16}\) and its volume at thermal equilibrium with a system whose true temperature is \(V\) be \(V_T\). We have \[ \int_{273.16}^T \left( \frac{ \partial V}{ \partial T} \right)_P dT = \int_{V_{273.16}}^{V_T} dV = V_T - V_{273.16}\] When we know the integrand on the left as a function of temperature, we can do the integration and find the temperature corresponding to any measured volume, \(V_T\). When the working fluid in our thermometer is a real gas we make measurements to find \({\left({\partial V}/{\partial T}\right)}_P\) as a function of temperature. Here we encounter a circularity: To find \({\left({\partial V}/{\partial T}\right)}_P\) from pressure-volume-temperature data we must have a way to measure temperature; however, this is the very thing that we are trying to find. In principle, we can surmount this difficulty by iteratively correcting the temperature that we associate with a given real-gas volume. As a first approximation, we use the temperatures that we measure with an uncorrected real-gas thermometer. These temperatures are a first approximation to the ideal-gas temperature scale. Using this scale, we make non-pressure-volume-temperature measurements that establish \({\left({\partial V}/{\partial T}\right)}_P\) as a function of temperature for the real gas. [This function is \[{\left(\frac{\partial V}{\partial T}\right)}_P=\frac{V+{\mu }_{JT}C_P}{T}\] where \(C_P\) is the and \({\mu }_{JT}\) is the . Both are functions of temperature. We introduce \(C_P\) in . We discuss the Joule-Thomson coefficient further in below, and in detail in . Typically \(V\gg C_P\), and the value of \({\left({\partial V}/{\partial T}\right)}_P\) is well approximated by \({V}/{T}={R}/{P}\). With \({\left({\partial V}/{\partial T}\right)}_P\) established using this scale, integration yields a second-approximation to the ideal-gas temperatures. We could repeat this process until successive temperature scales converge at the number of significant figures that our experimental accuracy can support. In practice, there are several kinds of ideal-gas thermometers, and numerous corrections are required for very accurate measurements. There are also numerous other ways to measure temperature, each of which has its own complications. Our development has considered some of the ideas that have given rise to the concept\({}^{4}\) that temperature is fundamental property of nature that can be measured using a thermodynamic-temperature scale on which values begin at zero and increase to arbitrarily high values. This thermodynamic temperature scale is a creature of theory, whose real-world counterpart would be the scale established by an ideal-gas thermometer whose gas actually obeyed \(PV=nRT\) at all conditions. We have seen that such an ideal-gas thermometer is itself a creature of theory. The current real-world standard temperature scale is the ( ). This defines temperature over a wide range in terms of the pressure-volume relationships of helium isotopes and the triple points of several selected elements. The triple points fix the temperature at each of several conditions up to 1357.77 K (the freezing point of copper). Needless to say, the temperatures assigned at the fixed points are the results of painstaking experiments designed to give the closest possible match to the thermodynamic scale. A variety of measuring devices—thermometers—can be used to interpolate temperature values between different pairs of fixed points. | 8,825 | 1,476 |
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Lewis structures provide no information about the shapes of molecules, being limited to which atoms are connected to each other and whether bonds are single, double or triple. A more sophisticated treatment of bonding is needed to begin to understand the shapes of molecules. In this section, we present a quantum mechanical description of bonding, in which bonding electrons are viewed as being localized between the nuclei of the bonded atoms. The overlap of bonding orbitals is substantially increased through a process called , which results in the formation of stronger bonds. Using this model we will be able to predict the shapes of the molecules. In , you learned that as two hydrogen atoms approach each other from an infinite distance, the energy of the system reaches a minimum. This region of minimum energy in the energy diagram corresponds to the formation of a covalent bond between the two atoms at an H–H distance of 74 pm ( ). According to quantum mechanics, bonds form between atoms because their atomic orbitals overlap, with each region of overlap accommodating a maximum of two electrons with opposite spin, in accordance with the Pauli principle. In this case, a bond forms between the two hydrogen atoms when the singly occupied 1 atomic orbital of one hydrogen atom overlaps with the singly occupied 1 atomic orbital of a second hydrogen atom. Electron density between the nuclei is increased because of this orbital overlap and results in a ( ). Although Lewis structures also contain localized electron-pair bonds, it does not use an atomic orbital approach to predict the stability and direction of the bond. Doing so forms the basis for a description of chemical bonding known as valence bond theory ( , which is built on two assumptions: shows an electron-pair bond formed by the overlap of two atomic orbitals, two atomic orbitals, and an and an orbital where = 2. Maximum overlap occurs between orbitals with the same spatial orientation and similar energies. Three Different Ways to Form an Electron-Pair Bond Let’s examine the bonds in BeH , for example. Its bonding can also be described using an atomic orbital approach. Beryllium has a 1 2 electron configuration, and each H atom has a 1 electron configuration. Because the Be atom has a filled 2 subshell, however, it has no singly occupied orbitals available to overlap with the singly occupied 1 orbitals on the H atoms. If a singly occupied 1 orbital on hydrogen were to overlap with a filled 2 orbital on beryllium, the resulting bonding orbital would contain electrons, but the maximum allowed by quantum mechanics is . How then is beryllium able to bond to two hydrogen atoms? One way would be to add enough energy to excite one of its 2 electrons into an empty 2 orbital and reverse its spin, in a process called promotion ( : In this excited state, the Be atom would have two singly occupied atomic orbitals (the 2 and one of the 2 orbitals), each of which could overlap with a singly occupied 1 orbital of an H atom to form an electron-pair bond. Although this would produce BeH , the two Be–H bonds would not be equivalent: the 1 orbital of one hydrogen atom would overlap with a Be 2 orbital, and the 1 orbital of the other hydrogen atom would overlap with an orbital of a different energy, a Be 2 orbital. Experimental evidence indicates, however, that the two Be–H bonds have identical energies. To resolve this discrepancy and explain how molecules such as BeH form, scientists developed the concept of hybridization. The localized bonding approach uses a process called hybridization ( , in which atomic orbitals that are similar in energy but not equivalent are combined mathematically to produce sets of equivalent orbitals that are properly oriented to form bonds. These new combinations are called hybrid atomic orbitals because they are produced by combining ( ) two or more atomic orbitals from the same atom. In BeH , we can generate two equivalent orbitals by combining the 2 orbital of beryllium and any one of the three degenerate 2 orbitals. By taking the sum and the difference of Be 2 and 2 atomic orbitals, for example, we produce two new orbitals with major and minor lobes oriented along the -axes, as shown in . This gives us , where the value \( 1/\sqrt{2} \) is needed mathematically to indicate that the 2 and 2 orbitals contribute equally to each hybrid orbital.
\( sp_{1}=\dfrac{1}{\sqrt{2}}\left ( 2s+2p_{z} \right )\; \; \; and\; \; \; sp_{2}=\dfrac{1}{\sqrt{2}}\left ( 2s-2p_{z} \right ) \tag{6.2.1} \) Generally we do not use the subsript labels on the two orbitals, but we have done so here to emphasize that there are two different orbitals. It is also important to remember that the two orbitals are built from orbitals of the same atom and centered on the nucleus of that atom. Later in this chapter we will meet molecular orbitals which are linear combinations of atomic orbitals (LCAO in theoretical chemistry speak) centered on atoms. It is often difficult for students to figure out the difference The nucleus resides just inside the minor lobe of each orbital. In this case, the new orbitals are called because they are formed from one and one orbital. The two new orbitals are equivalent in energy, and their energy is between the energy values associated with pure and orbitals, as illustrated in this diagram: This indicates that each hybrid orbital They can now form an electron-pair bond with the singly occupied 1 atomic orbital of one of the H atoms. As shown in , each orbital on Be has the correct orientation for the major lobes to overlap with the 1 atomic orbital of an H atom. The formation of two energetically equivalent Be–H bonds produces a linear BeH molecule. Thus valence bond theory does what the Lewis electron structurel is able to do; it explains why the bonds in BeH are equivalent in energy and why BeH has a linear geometry. Because both promotion and hybridization require an input of energy, the formation of a set of singly occupied hybrid atomic orbitals is energetically uphill. The overall process of forming a compound with hybrid orbitals will be energetically favorable if the amount of energy released by the formation of covalent bonds is greater than the amount of energy used to form the hybrid orbitals ( ). As we will see, some compounds are highly unstable or do not exist because the amount of energy required to form hybrid orbitals is greater than the amount of energy that would be released by the formation of additional bonds. The concept of hybridization also explains why boron, with a 2 2 valence electron configuration, forms three bonds with fluorine to produce BF The Lewis approach only works on the assumption that molecules with a central atom having LESS than an octet can exist. You may remember that this was one of the handwaves that we had to introduce in the previous chapter discussing Lewis structures. With only a single unpaired electron in its ground state, boron should form only a single covalent bond. By the promotion of one of its 2 electrons to an unoccupied 2 orbital, however, followed by the hybridization of the three singly occupied orbitals (the 2 and two 2 orbitals), boron acquires a set of three equivalent hybrid orbitals with one electron each, as shown here: The hybrid orbitals are degenerate and are oriented at 120° angles to each other ( ). Because the hybrid atomic orbitals are formed from one and two orbitals, boron is said to be (pronounced “s-p-two” or “s-p-squared”). The singly occupied hybrid atomic orbitals can overlap with the singly occupied orbitals on each of the three F atoms to form a trigonal planar structure with three energetically equivalent B–F bonds. Looking at the 2 2 valence electron configuration of carbon, we might expect carbon to use its two unpaired 2 electrons to form compounds with only two covalent bonds. We know, however, that carbon typically forms compounds with four covalent bonds. We can explain this apparent discrepancy by the hybridization of the 2 orbital and the three 2 orbitals on carbon to give a set of four degenerate (“s-p-three” or “s-p-cubed”) hybrid orbitals, each with a single electron: The large lobes of the hybridized orbitals are oriented toward the vertices of a tetrahedron, with 109.5° angles between them ( ). It is rather difficult to visualize a tetrahedron. A simple, though inelegant way of doing so is to stand with one foot forward and the other backwards. Now raise your hands from your sides to above your shoulders. Like all the hybridized orbitals discussed earlier, the hybrid atomic orbitals ( are predicted to be equal in energy. In addition to explaining why some elements form more bonds than would be expected based on their valence electron configurations, and why the bonds formed are equal in energy, valence bond theory explains why these compounds are so stable: the amount of energy released increases with the number of bonds formed. In the case of carbon, for example, much more energy is released in the formation of four bonds than two, so compounds of carbon with four bonds tend to be more stable than those with only two. Carbon does form compounds with only two covalent bonds (such as CH or CF ), but these species are highly reactive, unstable intermediates that form in only certain chemical reactions. Valence bond theory explains the number of bonds formed in a compound and the relative bond strengths. The bonding in molecules such as NH or H O, which have lone pairs on the central atom, can also be described in terms of hybrid atomic orbitals. In NH , for example, N, with a 2 2 valence electron configuration, can hybridize its 2 and 2 orbitals to produce four hybrid orbitals. Placing five valence electrons in the four hybrid orbitals, we obtain three that are singly occupied and one with a pair of electrons: The three singly occupied lobes can form bonds with three H atoms, while the fourth orbital accommodates the lone pair of electrons. Similarly, H O has an hybridized oxygen atom that uses two singly occupied lobes to bond to two H atoms, and two to accommodate the two lone pairs predicted by the VSEPR model. Such descriptions explain the approximately tetrahedral distribution of electron pairs on the central atom in NH and H O. Unfortunately, however, recent experimental evidence indicates that in CH and NH , the hybridized orbitals are entirely equivalent in energy, making this bonding model an active area of research. Predict the number of electron pairs and molecular geometry in each compound and then describe the hybridization and bonding of all atoms except hydrogen. two chemical compounds number of electron pairs and molecular geometry, hybridization, and bonding Using the approach from Example 1, determine the number of electron pairs and the molecular geometry of the molecule. From the valence electron configuration of the central atom, predict the number and type of hybrid orbitals that can be produced. Fill these hybrid orbitals with the total number of valence electrons around the central atom and describe the hybridization. Exercise Predict the number of electron pairs and molecular geometry in each compound and then describe the hybridization and bonding of all atoms except hydrogen. The number of hybrid orbitals used by the central atom is the same as the number of electron pairs around the central atom. Hybridization is not restricted to the and atomic orbitals. The bonding in compounds with central atoms in the period 3 and below can also be described using hybrid atomic orbitals. In these cases, the central atom can use its valence ( − 1) orbitals as well as its and orbitals to form hybrid atomic orbitals, which allows it to accommodate five or more bonded atoms (as in PF and SF ). Using the orbital, all three orbitals, and one ( − 1) orbital gives a set of five hybrid orbitals ( that point toward the vertices of a trigonal bipyramid (part (a) in ). In this case, the five hybrid orbitals are all equivalent: three form a triangular array oriented at 120° angles, and the other two are oriented at 90° to the first three and at 180° to each other. Similarly, the combination of the orbital, all three orbitals, and orbitals gives a set of six equivalent hybrid orbitals ( oriented toward the vertices of an octahedron (part (b) in ). In the VSEPR model, PF and SF are predicted to be trigonal bipyramidal and octahedral, respectively, which agrees with a valence bond description in which or hybrid orbitals are used for bonding. What is the hybridization of the central atom in each species? Describe the bonding in each species. three chemical species hybridization of the central atom Determine the geometry of the molecule using the strategy in Example 1. From the valence electron configuration of the central atom and the number of electron pairs, determine the hybridization. Place the total number of electrons around the central atom in the hybrid orbitals and describe the bonding. To accommodate five electron pairs, the sulfur atom must be hybridized. Filling these orbitals with 10 electrons gives four hybrid orbitals forming S–F bonds and one with a lone pair of electrons. Exercise What is the hybridization of the central atom in each species? Describe the bonding. Hybridization using orbitals allows chemists to explain the structures and properties of many molecules and ions. Like most such models, however, it is not universally accepted. Nonetheless, it does explain a fundamental difference between the chemistry of the elements in the period 2 (C, N, and O) and those in period 3 and below (such as Si, P, and S). Period 2 elements do not form compounds in which the central atom is covalently bonded to five or more atoms, although such compounds are common for the heavier elements. Thus whereas carbon and silicon both form tetrafluorides (CF and SiF ), only SiF reacts with F to give a stable hexafluoro dianion, SiF . Because there are no 2 atomic orbitals, the formation of octahedral CF would require hybrid orbitals created from 2 , 2 , and 3 atomic orbitals. The 3 orbitals of carbon are so high in energy that the amount of energy needed to form a set of hybrid orbitals cannot be equaled by the energy released in the formation of two additional C–F bonds. These additional bonds are expected to be weak because the carbon atom (and other atoms in period 2) is so small that it cannot accommodate five or six F atoms at normal C–F bond lengths due to repulsions between electrons on adjacent fluorine atoms. Perhaps not surprisingly, then, species such as CF have never been prepared. What is the hybridization of the oxygen atom in OF ? Is OF likely to exist? chemical compound hybridization and stability Predict the geometry of OF using the VSEPR model. From the number of electron pairs around O in OF , predict the hybridization of O. Compare the number of hybrid orbitals with the number of electron pairs to decide whether the molecule is likely to exist. T OF will have five electron pairs, resulting in a trigonal bipyramidal geometry with four bonding pairs and one lone pair. To accommodate five electron pairs, the O atom would have to be hybridized. The only orbital available for forming a set of hybrid orbitals is a 3 orbital, which is higher in energy than the 2 and 2 valence orbitals of oxygen. As a result, the OF molecule is unlikely to exist. In fact, it has not been detected. Exercise What is the hybridization of the boron atom in BF ? Is this ion likely to exist? hybridization; no The model (called ) assumes that covalent bonds are formed when atomic orbitals overlap and that the strength of a covalent bond is proportional to the amount of overlap. It also assumes that atoms use combinations of atomic orbitals ( ) to maximize the overlap with adjacent atoms. The formation of can be viewed as occurring via of an electron from a filled subshell to an empty or ( − 1) valence orbital, followed by , the combination of the orbitals to give a new set of (usually) equivalent orbitals that are oriented properly to form bonds. The combination of an and an orbital gives rise to two equivalent oriented at 180°, whereas the combination of an and two or three orbitals produces three equivalent or four equivalent , respectively. The bonding in molecules with more than an octet of electrons around a central atom can be explained by invoking the participation of one or two ( − 1) orbitals to give sets of five or six orbitals, capable of forming five or six bonds, respectively. The spatial orientation of the hybrid atomic orbitals is consistent with the geometries predicted using the VSEPR model. Arrange , , and in order of increasing strength of the bond formed to a hydrogen atom. Explain your reasoning. What atomic orbitals are combined to form , , , and ? What is the maximum number of electron-pair bonds that can be formed using each set of hybrid orbitals? Why is it incorrect to say that an atom with hybridization will form only three bonds? The carbon atom in the carbonate anion is hybridized. How many bonds to carbon are present in the carbonate ion? Which orbitals on carbon are used to form each bond? If hybridization did not occur, how many bonds would N, O, C, and B form in a neutral molecule, and what would be the approximate molecular geometry? How are hybridization and molecular geometry related? Which has a stronger correlation—molecular geometry and hybridization or Lewis structures and hybridization? In the valence bond approach to bonding in BeF , which step(s) require(s) an energy input, and which release(s) energy? The energies of hybrid orbitals are intermediate between the energies of the atomic orbitals from which they are formed. Why? How are lone pairs on the central atom treated using hybrid orbitals? Because nitrogen bonds to only three hydrogen atoms in ammonia, why doesn’t the nitrogen atom use hybrid orbitals instead of hybrids? Using arguments based on orbital hybridization, explain why the CCl ion does not exist. Species such as NF and OF are unknown. If 3 atomic orbitals were much lower energy, low enough to be involved in hybrid orbital formation, what effect would this have on the stability of such species? Why? What molecular geometry, electron-pair geometry, and hybridization would be expected for each molecule? Draw an energy-level diagram showing promotion and hybridization to describe the bonding in CH . How does your diagram compare with that for methane? What is the molecular geometry? Draw an energy-level diagram showing promotion and hybridization to describe the bonding in CH . How does your diagram compare with that for methane? What is the molecular geometry? Draw the molecular structure, including any lone pairs on the central atom, state the hybridization of the central atom, and determine the molecular geometry for each molecule. Draw the molecular structure, including any lone pairs on the central atom, state the hybridization of the central atom, and determine the molecular geometry for each species. What is the hybridization of the central atom in each of the following? What is the hybridization of the central atom in each of the following? What is the hybridization of the central atom in PF ? Is this ion likely to exist? Why or why not? What would be the shape of the molecule? What is the hybridization of the central atom in SF ? Is this ion likely to exist? Why or why not? What would be the shape of the molecule? The promotion and hybridization process is exactly the same as shown for CH in the chapter. The only difference is that the C atom uses the four singly occupied hybrid orbitals to form electron-pair bonds with only H atoms, and an electron is added to the fourth hybrid orbital to give a charge of 1–. The electron-pair geometry is tetrahedral, but the molecular geometry is pyramidal, as in NH . , trigonal planar , pyramidal , trigonal planar The central atoms in CF , CCl , IO , and SiH are all sp hybridized. The phosphorus atom in the PF ion is hybridized, and the ion is octahedral. The PF ion is isoelectronic with SF and has essentially the same structure. It should therefore be a stable species. | 20,483 | 1,477 |
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One of the more recent experimental methods of studying the nucleus of an atom is to probe the nucleus with very high energy electrons. Calculate the order of magnitude of the energy of an electron when it is bound inside a nucleus with a diameter 1 10 cm. Compare this value with the order of magnitude of the energy of an electron bound to an atom of diameter 1 10 cm. Nuclear particles, protons or neutrons have masses approximately 2 10 times the mass of an electron. Estimate the average energy of a nuclear particle bound in a nucleus and compare it with the order of magnitude energy for an electron bound to an atom. This result should indicate that chemical changes which involve changes in the electronic energies of the system do not affect the nucleus of an atom. ( / ) | 814 | 1,478 |
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\[\ce{N2O4(g) → NO2 + NO2} \qquad \Delta S_{m}°(298 \text{K})=+176 \text{JK}^{-1} \text{mol}^{-1} \nonumber \] \[\ce{2O3 → 3O2} \qquad \Delta S°_{m}(298 \text{K})=+137 \text{JK}^{-1} \text{mol}^{-1} \nonumber \] A further extension of this argument leads us to the general conclusion that in any reaction involving gases if the amount of substance in the gaseous phase increases, Δ will be positive, while if it decreases, so will Δ . For example, in the reaction \[\ce{2CO(g) + O2(g) → 2CO2(g)} \nonumber \] The amount of gas decreases from 3 to 2 mol (i.e., Δ = –1 mol). The entropy change should thus be negative for this reaction. From the Table of Standard Molar Entropies we can readily find that Δ °(298 K) has the value –173 J K mol . This table shows molar entropies for the standard conditions of 298.15 K (25°C) and 101.3 kPa. Such conditions need to be specificed, since entropy is propotional to substance amount, and dependent on temperature, pressure. Entropy is also dependent upon volume, but since the amount, , temperature, and pressure are given, volume is implicitly defined. This table is taken from CoreChem:Standard Molar Entropies, and is also used on CoreChem:Dependence of S on Molecular Structure as well as CoreChem:Some Trends In Entropy Values. | 1,292 | 1,479 |
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In this section, consider the differences between two types of changes in a system: Those that occur spontaneously and those that occur only with the continuous input of energy. In doing so, we’ll gain an understanding as to why some systems are naturally inclined to change in one direction under certain conditions. We’ll also gain insight into how the spontaneity of a process affects the distribution of energy and matter within the system. Processes have a natural tendency to occur in one direction under a given set of conditions. Water will naturally flow downhill, but uphill flow requires outside intervention such as the use of a pump. A is one that occurs naturally under certain conditions. A , on the other hand, will not take place unless it is “driven” by the continual input of energy from an external source. A process that is spontaneous in one direction under a particular set of conditions is nonspontaneous in the reverse direction. At room temperature and typical atmospheric pressure, for example, ice will spontaneously melt, but water will not spontaneously freeze. The spontaneity of a process is correlated to the speed of the process. A spontaneous change may be so rapid that it is essentially instantaneous or so slow that it cannot be observed over any practical period of time. To illustrate this concept, consider the decay of radioactive isotopes, a topic more thoroughly treated in the chapter on nuclear chemistry. Radioactive decay is by definition a spontaneous process in which the nuclei of unstable isotopes emit radiation as they are converted to more stable nuclei. All the decay processes occur spontaneously, but the rates at which different isotopes decay vary widely. Technetium-99m is a popular radioisotope for medical imaging studies that undergoes relatively rapid decay and exhibits a half-life of about six hours. Uranium-238 is the most abundant isotope of uranium, and its decay occurs much more slowly, exhibiting a half-life of more than four billion years (Figure \(\Page {1}\)). As another example, consider the conversion of diamond into graphite (Figure \(\Page {2}\)). \[\ce{C(s, diamond)}⟶\ce{C(s, graphite)} \label{Eq1} \] The phase diagram for carbon indicates that graphite is the stable form of this element under ambient atmospheric pressure, while diamond is the stable allotrope at very high pressures, such as those present during its geologic formation. Thermodynamic calculations of the sort described in the last section of this chapter indicate that the conversion of diamond to graphite at ambient pressure occurs spontaneously, yet diamonds are observed to exist, and persist, under these conditions. Though the process is spontaneous under typical ambient conditions, its rate is extremely slow, and so for all practical purposes diamonds are indeed “forever.” Situations such as these emphasize the important distinction between the thermodynamic and the kinetic aspects of a process. In this particular case, diamonds are said to be but under ambient conditions. As we extend our discussion of thermodynamic concepts toward the objective of predicting spontaneity, consider now an isolated system consisting of two flasks connected with a closed valve. Initially there is an ideal gas on the left and a vacuum on the right (Figure \(\Page {3}\)). When the valve is opened, the gas spontaneously expands to fill both flasks. Recalling the definition of pressure-volume work from the chapter on thermochemistry, note that no work has been done because the pressure in a vacuum is zero. \[ \begin{align} w&=−PΔV \\[4pt]&=0 \,\,\, \mathrm{(P=0\: in\: a\: vaccum)} \label{Eq2} \end{align} \] Note as well that since the system is isolated, no heat has been exchanged with the surroundings (q = 0). The first law of thermodynamics confirms that there has been no change in the system’s internal energy as a result of this process. \[ \begin{align} ΔU&=q+w \tag{First Law of Thermodynamics} \\[4pt] &=0+0=0 \label{Eq3}\end{align} \] The spontaneity of this process is therefore not a consequence of any change in energy that accompanies the process. Instead, the movement of the gas appears to be related to the greater, more that results when the gas is allowed to expand. Initially, the system was comprised of one flask containing matter and another flask containing nothing. After the spontaneous process took place, the matter was distributed both more widely (occupying twice its original volume) and more uniformly (present in equal amounts in each flask). Now consider two objects at different temperatures: object X at temperature and object Y at temperature , with > (Figure \(\Page {4}\)). When these objects come into contact, heat spontaneously flows from the hotter object (X) to the colder one (Y). This corresponds to a loss of thermal energy by X and a gain of thermal energy by Y. \[q_\ce{X}<0 \hspace{20px} \ce{and} \hspace{20px} q_\ce{Y}=−q_\ce{X}>0 \label{Eq4} \] From the perspective of this two-object system, there was no net gain or loss of thermal energy, rather the available thermal energy was redistributed among the two objects. This spontaneous process resulted in a . As illustrated by the two processes described, an important factor in determining the spontaneity of a process is the extent to which it changes the dispersal or distribution of matter and/or energy. In each case, a spontaneous process took place that resulted in a more uniform distribution of matter or energy. Describe how matter and energy are redistributed when the following spontaneous processes take place: Describe how matter and energy are redistributed when you empty a canister of compressed air into a room. This process entails both a greater and more uniform dispersal of matter as the compressed air in the canister is permitted to expand into the lower-pressure air of the room. The process also requires an input of energy to disrupt the intermolecular forces between the closely-spaced gas molecules that are originally compressed into the container. If you were to touch the nozzle of the canister, you would notice that it is cold because the exiting molecules are taking energy away from their surroundings, and the canister is part of the surroundings. Chemical and physical processes have a natural tendency to occur in one direction under certain conditions. A spontaneous process occurs without the need for a continual input of energy from some external source, while a nonspontaneous process requires such. Systems undergoing a spontaneous process may or may not experience a gain or loss of energy, but they will experience a change in the way matter and/or energy is distributed within the system. In this section we have only discussed nuclear decay, physical changes of pure substances, and macroscopic events such as water flowing downhill. In the following sections we will discuss mixtures and chemical reactions, situations in which the description of sponteneity becomes more challenging. | 7,030 | 1,480 |
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For hemoglobin, the majority of ligands around the iron center are provided by a fairly rigid macrocycle, the protoporphyrin IX dianion (Figure 4.1), that by itself enforces a square-planar stereochemistry. Thus, the task of assembling synthetic analogues of the active site in hemoglobin is simplified. Essentially, any square-planar, tetradentate ligand containing at least a couple of nitrogen atoms will suffice: to this end a variety of other porphyrins have been used, as well as Schiff-base and non-porphyrinato nitrogen-containing macrocycles that serve to delineate the role of the porphyrin in dioxygen binding. Tetraphenylporphyrin, in place of the naturally occurring porphyrins, has served as the basis of numerous model systems. It is easily synthesized and derivatized (see below). Its fourfold symmetry precludes formation of chemical isomers that may arise if substitution on the asymmetric, naturally occurring porphyrins is attempted. Moreover, its derivatives can be crystallized. Finally, with the porphyrin positions occupied by phenyl groups, the molecule is less susceptible to photoinduced oxidation. In order to obtain five-coordinate species, putative models for a deoxy hemoglobin, access to one side of the porphyrin must be blocked to the coordination of a second axial base (Reaction 4.33), but still must be accessible for the binding of small molecules (O , NO, CO, etc). Or second, an axial base may be attached covalently to the porphyrin, to give so-called "tail-under" or "chelated" porphyrins. Here the chelate effect ensures an effectively 100 percent five-coordinate complex with only a 1:1 stoichiometric ratio of axial base to porphyrin. These two approaches are illustrated in Figure 4.14. A third means is to incorporate a sterically bulky substituent in the 2-position of the base, such as a methyl group, to give 2-methylimidazole (4.34). The formation of the hemochrome Fe(porph)(2-MeIm) , where the iron atom is in the center of the porphyrin ring, is strongly disfavored relative to the 1-methylimidazole analogue, because the 2-methyl substituents clash with the porphyrin ring. The axial base needs to be a strong \(\sigma\) donor, such as imidazole or pyridine, in order to increase affinity at the iron (or cobalt) center for dioxygen (Figure 4.18). Steric hindrance on one side, or on both, provides a pocket for small molecules to bind and, for O , prevents the bimolecular contact of two iron(II)-porphyrinato species that would lead to irreversible oxidation (Reaction 4.29). A picturesque collection of substituted porphyrins has been synthesized. Some of these are illustrated in Figure 4.23. The only system that has led to crystalline dioxygen complexes stable at room temperature is the "picket-fence" porphyrin. * A derivative of this, the "pocket" porphyrin, and various "capped" porphyrins, provide binding sites with steric hindrance even to small diatomic molecules. In the next section the structures of various derivatives of hemoglobin and its models are presented, and the relationship of structure to ligand-binding properties is examined. Although there is now a wealth of thermodynamic data available from model systems, attention is focused primarily on those for which structural data are also available. * The pivalamido pickets (—NH—CO—C(CH ) ) of the picket-fence porphyrin are sufficiently bulky that their free rotation is sterically hindered. Thus the various , , , , where \(\alpha\) denotes picket "up" and \(\beta\) denotes picket "down"—can be separated chromatographically on the basis of their different polarities. The tail-under picket-fence porphyrin is derived from the atropisomer. | 3,702 | 1,481 |
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Consider a N-particle ensemble. The particles are not necessarily indistinguishable and possibly have mutual potential energy. Since this is a large system, there are many different ways to arrange its particles and yet yield the same thermodynamic state. Only one arrangement can occur at a time. The sum of the probabilities of each separate arrangement equals the total number of separate arrangements. Then the probability of a system is: \[p_N=W_N p_i \nonumber \] where \(p_N\) is the probability of the system, \(W_N\) is the total number of different possible arrangements of the N particles in the system, and \(p_i\) is the probability of each separate arrangement. Heisenberg's uncertainty principle states that it is impossible to simultaneously know the momentum and the position of an object with complete precision. In agreement with the uncertainty principle, the total possible number of combinations can be defined as the total number of distinguishable rearrangements of the N particles. The most practical ensemble is the canonical ensemble with \(N\), \(V\), and \(T\) fixed. We can imagine a collection of boxes with equal volumes and number of particles with the entire collection kept in thermal equilibrium. Based on the Boltzmann factor, we know that for a system that has states with energies \(e_1,e_2,e_3\)..., the probability \(p_j\) that the system will be in the state \(j\) with energy \(E_j\) is exponentially proportional to the energy of state \(j\). The partition functions of the state places a very important role in calculating the properties of a system, for example, it can be used to calculate the probability, as well as the energy, heat capacity, and pressure. We are ultimately interested in the probability that a given distribution will occur. The reason for this is that we must have this information in order to obtain useful thermodynamic averages. Let's consider an ensemble of \(A\) systems. We will define \(a_j\) as the number of systems in the ensemble that are in the quantum state \(j\). For example, \(a_1\) represents the number of systems in the quantum state 1. The total number of possible microstates is: \[W(a_1,a_2,...) = \frac{A!}{a_1!a_2!...} \nonumber \] The overall probability that \(P_j\) that a system is in the j quantum state is obtained by averaging \(a_j/A\) over all the allowed distributions. Thus, \(P_j\) is given by: \[ \begin{align*} P_j &= \dfrac{\langle a_j \rangle}{A} \\[4pt] &= \dfrac{1}{A} \dfrac{ \displaystyle \sum_a W(a) a_j(a)}{\displaystyle \sum_a W(a)} \end{align*} \] where the angle brackets indicate an ensemble average. Using this definition we can calculate any average property (i.e. any thermodynamic property): \[ \langle M \rangle = \sum_j M_j P_j \label{avg} \] The method of the most probable distribution is based on the idea that the average over \(P_j\) is to the most probable distribution. Physically, this results from the fact that we have so many particles in a typical system that the fluctuations from the mean are extremely (immeasurably) small. The equivalence of the average probability of an occupation number and the most probable distribution is expressed as follows: \[ P_j = \dfrac{\langle a_j \rangle}{A} = \dfrac{a_j}{A} \nonumber \] The probability function is subject to the following constraints: As we will learn in later chapters, the system will tend towards the distribution of \(a_j\) that maximizes the total number of microstates. This can be expressed as: \[\sum_j \left(\dfrac{\partial \ln W }{\partial a_j}\right) = 0 \nonumber \] Our constraints becomes: \[ \sum_j e_j da_j =0 \nonumber \] \[ \sum_j da_j =0 \nonumber \] The method of Lagrange multipliers (named after Joseph Louis Lagrange is a strategy for finding the local maxima and minima of a function subject to equality constraints. Using the method of LaGrange undetermined multipliers we have: \[ \sum_j \left[ \left(\dfrac{\partial \ln W }{\partial a_j}\right)da_j + \alpha da_j - \beta e_j da_j \right] = 0 \nonumber \] We can use Stirling's approximation: \[\ln x! \approx x\ln x – x \nonumber \] to evaluate: \[ \left(\dfrac{\partial \ln W }{\partial a_j}\right) \nonumber \] to get: \[ \left(\dfrac{\partial A! }{\partial a_j}\right) - \sum_i \left(\dfrac{\partial \ln a_i }{\partial a_j}\right) = 0 \nonumber \] as outlined below. First step is to note that: \[\ln W = \ln A! - \sum_j \ln a_j! a \approx A \ln A – A - \sum_j a_j \ln a_j - \sum_j a_j \nonumber \] Since (from Equation \(\ref{con2}\)): \[A = \sum_j a_j \nonumber \] these two cancel to give: \[\ln W = A \ln A - \sum_j a_j \ln a_j \nonumber \] The derivative is: \[ \left(\dfrac{\partial \ln W}{\partial a_j} \right) = \dfrac{\partial A \ln A}{\partial a_j} - \sum_i \dfrac{\partial a_i \ln a_i}{\partial a_j} \nonumber \] Therefore we have: \[\left(\dfrac{\partial A \ln A}{\partial a_j} \right) = \dfrac{\partial A }{\partial a_j} \ln A - \dfrac{\partial A }{\partial a_j} = \ln A -1 \nonumber \] \[\left(\dfrac{\partial a_i \ln a_i}{\partial a_j} \right) = \dfrac{\partial a_i }{\partial a_j} \ln a_i - \dfrac{\partial a_i }{\partial a_j} = \ln a_j +1 \nonumber \] These latter derivatives result from the fact that: \[\left( \dfrac{\partial a_i}{\partial a_i} \right) = 1 \nonumber \] \[\left( \dfrac{\partial a_j}{\partial a_i}\right)=0 \nonumber \] The simple expression that results from these manipulations is: \[ - \ln \left( \dfrac{a_j}{A} \right) + \alpha - \beta e_j =0 \nonumber \] The most probable distribution is: \[ \dfrac{a_j}{A} = e^a \sum_j e^{-\beta e_j} \label{Eq3} \] Now we need to find the undetermined multipliers \(\alpha\) and \(\beta\). The left hand side of Equation \(\ref{Eq3}\) is 1. Thus, we have: \[ P_j= \dfrac{a_j}{A} = \dfrac{ e^{-\beta e_j}} {\sum_j e^{-\beta e_j}} \nonumber \] This determines \(a\) and defines the Boltzmann distribution. We will show that \(\beta\) from the optimization procedure of method of Lagrange multipliers is: \[\beta=\dfrac{1}{kT} \nonumber \] This identification will show the importance of temperature in the Boltzmann distribution. The distribution represents a thermally equilibrated most probable distribution over all energy levels (Figure 17.2.1
). The Boltzmann distribution represents a thermally equilibrated distribution over all energy levels. There is always a population in a state of lower energy than in one of higher energy. Once we know the probability distribution for energy, we can calculate thermodynamic properties like the energy, entropy, free energies and heat capacities, which are all average quantities (Equation \(\ref{avg}\)). To calculate \(P_j\), we need the energy levels of a system (i.e., \(\{e_i\}\)). The energy ("levels") of a system can be built up from the quantum energy levels It must always be remembered that no matter how large the energy spacing is, there is always a non-zero probability of the upper level being populated. The only exception is a system that is at absolute zero. This situation is however hypothetical as absolute zero can be approached but not reached. The sum over all factors \( e^{-\beta e_j} \) is given a name. It is called the molecular partition function, \(q\): \[ q = \sum_j e^{-\beta e_j} \nonumber \] The molecular partition function \(q\) gives an indication of the average number of states that are thermally accessible to a molecule at the temperature of the system. The partition function is a sum over states (of course with the Boltzmann factor \(\beta\) multiplying the energy in the exponent) and is a number. Larger the value of \(q\), larger the number of states which are available for the molecular system to occupy (Figure 17.2.2
). We distinguish here between the partition function of the ensemble, \(Q\) and that of an individual molecule, \(q\). Since \(Q\) represents a sum over all states accessible to the system it can written as: \[ Q(N,V,T) = \sum_{i,j,k ...} e^{-\beta ( e_i + e_j +e_k ...)} \nonumber \] where the indices \(i,\,j,\,k\) represent energy levels of particles. Regardless of the type of particle the molecular partition function, \(q\) represents the energy levels of one individual molecule. We can rewrite the above sum as: \[Q = q_iq_jq_k… \nonumber \] or: \[Q = q^N \nonumber \] for \(N\) particles. Note that \(q_i\) means a sum over states or energy levels accessible to molecule \(i\) and \(q_j\) means the same for molecule \(j\). The molecular partition function, \(q\) counts the energy levels accessible to molecule \(i\) only. \(Q\) counts not only the states of all of the molecules, but all of the possible combinations of occupations of those states. However, if the particles are not distinguishable then we will have counted \(N!\) states too many. The factor of \(N!\) is exactly how many times we can swap the indices in \(Q(N,V,T)\) and get the same value (again provided that the particles are not distinguishable). See this for more information. | 8,949 | 1,482 |
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In this organic chemistry topic, we shall see how alcohols (R-OH) add to carbonyl groups. Carbonyl groups are characterized by a carbon-oxygen double bond. The two main functional groups that consist of this carbon-oxygen double bond are . It has been demonstrated that water adds rapidly to the carbonyl function of aldehydes and ketones to form geminal-diol. In a similar reaction alcohols add reversibly to aldehydes and ketones to form hemiacetals (h Greek, half). This reaction can continue by adding another alcohol to form an acetal. Hemiacetals and acetals are important functional groups because they appear in sugars. To achieve effective hemiacetal or acetal formation, two additional features must be implemented. First, an acid catalyst must be used because alcohol is a weak nucleophile; and second, the water produced with the acetal must be removed from the reaction by a process such as a molecular sieves or a . The latter is important, since acetal formation is reversible. Indeed, once pure hemiacetal or acetals are obtained they may be hydrolyzed back to their starting components by treatment with aqueous acid and an excess of water. Acetals are geminal-diether derivatives of aldehydes or ketones, formed by reaction with two equivalents (or an excess amount) of an alcohol and elimination of water. Ketone derivatives of this kind were once called ketals, but modern usage has dropped that term. It is important to note that a hemiacetal is formed as an intermediate during the formation of an acetal. The mechanism shown here applies to both acetal and hemiacetal formation 1) Protonation of the carbonyl 2) Nucleophilic attack by the alcohol 3) Deprotonation to form a hemiacetal 4) Protonation of the alcohol 5) Removal of water 6) Nucleophilic attack by the alcohol 7) Deprotonation by water Molecules which have an alcohol and a carbonyl can undergo an intramolecular reaction to form a cyclic hemiacetal. Intramolecular Hemiacetal formation is common in sugar chemistry. For example, the common sugar glucose exists in the cylcic manner more than 99% of the time in a mixture of aqueous solution. Carbonyls reacting with diol produce a cyclic acetal. A common diol used to form cyclic acetals is ethylene glycol. The importance of acetals as carbonyl derivatives lies chiefly in their stability and lack of reactivity in neutral to strongly basic environments. As long as they are not treated by acids, especially aqueous acid, acetals exhibit all the lack of reactivity associated with ethers in general. Among the most useful and characteristic reactions of aldehydes and ketones is their reactivity toward strongly nucleophilic (and basic) metallo-hydride, alkyl and aryl reagents. If the carbonyl functional group is converted to an acetal these powerful reagents have no effect; thus, acetals are excellent protective groups, when these irreversible addition reactions must be prevented. In the following example we would like a to react with the ester and not the ketone. This cannot be done without a protecting group because Grignard reagents react with esters and ketones. ) ), | 3,146 | 1,483 |
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This Module describes the Arrhenius, Brønsted-Lowry, and Lewis theories of acids and bases, and explains the relationships between them. It also explains the concept of a conjugate pair - an acid and its conjugate base, or a base and its conjugate acid. Neutralization happens because hydrogen ions and hydroxide ions react to produce water. \[ H^+_{(aq)} + OH^-{(aq)} \rightarrow H_2O_{(l)}\] Hydrochloric acid is neutralized by both sodium hydroxide solution and ammonia solution. In both cases, you get a colorless solution which you can crystallize to get a white salt - either sodium chloride or ammonium chloride. These are clearly very similar reactions. The full equations are: \[ NaOH_{(aq)} + HCl_{(aq)} \rightarrow NaCl_{(aq)} + H_2O_{(l)}\] \[ NH_{3(aq)} + HCl_{(aq)} \rightarrow NH_4Cl_{(aq)}\] In the sodium hydroxide case, hydrogen ions from the acid are reacting with hydroxide ions from the sodium hydroxide - in line with the Arrhenius theory. However, in the ammonia case, there do not appear to be any hydroxide ions! You can get around this by saying that the ammonia reacts with the water it is dissolved in to produce ammonium ions and hydroxide ions: \[ NH_{3(aq)} + H_2O_{(l)} \rightleftharpoons NH^+_{4(aq)} + OH_{(aq)}^-\] This is a reversible reaction, and in a typical dilute ammonia solution, about 99% of the ammonia remains as ammonia molecules. Nevertheless, there are hydroxide ions there, and we can squeeze this into the Arrhenius theory. However, this same reaction also happens between ammonia gas and hydrogen chloride gas. \[NH_{3(g)} + HCl_{(g)} \rightarrow NH_4Cl_{(s)}\] In this case, there are no hydrogen ions or hydroxide ions in solution - because there is not any solution. The Arrhenius theory would not count this as an acid-base reaction, despite the fact that it is producing the same product as when the two substances were in solution. The Brønsted-Lowry theory does not go against the Arrhenius theory in any way - it just adds to it. Hydroxide ions are still bases because they accept hydrogen ions from acids and form water. An acid produces hydrogen ions in solution because it reacts with the water molecules by giving a proton to them. When hydrogen chloride gas dissolves in water to produce hydrochloric acid, the hydrogen chloride molecule gives a proton (a hydrogen ion) to a water molecule. A is formed between one of the lone pairs on the oxygen and the hydrogen from the \(HCl\). , \(H_3O^+_{(aq)}\), are pr oduced. \[H_2O + HCl \rightarrow H_3O^+ +Cl^- \] When an acid in solution reacts with a base, what is actually functioning as the acid is the hydronium ion. For example, a proton is transferred from a hydronium ion to a hydroxide ion to make water. \[ H_3O^+_{(aq)} + OH^-_{(aq)} \rightarrow 2H_2O_{(l)}\] Showing the electrons, but leaving out the inner ones: It is important to realize that whenever you talk about hydrogen ions in solution, \(H^+_{(aq)}\), wha t you are actually talking about are hydronium ions. This is no longer a problem using the Brønsted-Lowry theory. Whether you are talking about the reaction in solution or in the gas state, ammonia is a base because it accepts a proton (a hydrogen ion). The hydrogen becomes attached to the lone pair on the nitrogen of the ammonia via a co-ordinate bond. If it is in solution, the ammonia accepts a proton from a hydronium ion: \[ NH_{3 (aq)} + H_3O_{(aq)}^+ \rightarrow NH^+_{4(aq)} + H_2O_{(l)}\] If the reaction occurs in the gas state, the ammonia accepts a proton directly from the hydrogen chloride: \[ NH_{3 (g)} + HCl_{(g)} \rightarrow NH_4Cl_{(s)}\] Either way, the ammonia acts as a base by accepting a hydrogen ion from an acid. When hydrogen chloride dissolves in water, almost 100% of it reacts with the water to produce hydronium ions and chloride ions. Hydrogen chloride is a strong acid, and we tend to write this as a one-way reaction: \[H_2O + HCl \rightarrow H_3O^+ + Cl^-\] In fact, the reaction between HCl and water is reversible, but only to a very minor extent. To generalize, consider an acid \(HA\), and think of the reaction as being reversible. \[HA + H_2O \rightleftharpoons H_3O^+ + A^- \] Thinking about the forward reaction: However, there is also a back reaction between the hydronium ion and the \(A^-\) i on: The reversible reaction contains two acids and two bases. We think of them in pairs, called . When the acid, \(HA\), loses a proton it forms a base, \(A^-\), which can a ccept a proton back again to refom the acid, \(HA\). These two are a . Members of a conjugate pair differ from each other by the presence or absence of the transferable hydrogen ion. The water and the hydronium ion are also a conjugate pair. Thinking of the water as a base, the hydronium ion is its conjugate acid because it has the extra hydrogen ion which it can give away again. Thinking about the hydronium ion as an acid, then water is its conjugate base. The water can accept a hydrogen ion back again to reform the hydronium ion. This is the reaction between ammonia and water that we looked at earlier: Think first about the forward reaction. Ammonia is a base because it is accepting hydrogen ions from the water. The ammonium ion is its conjugate acid - it can release that hydrogen ion again to reform the ammonia. The water is acting as an acid, and its conjugate base is the hydroxide ion. The hydroxide ion can accept a hydrogen ion to reform the water. Looking at it from the other side, the ammonium ion is an acid, and ammonia is its conjugate base. The hydroxide ion is a base and water is its conjugate acid. You may possibly have noticed (although probably not!) that in one of the last two examples, water was acting as a base, whereas in the other one it was acting as an acid. A substance which can act as either an acid or a base is described as being amphoteric. This theory extends well beyond the things you normally think of as acids and bases. It is easiest to see the relationship by looking at exactly what Brønsted-Lowry bases do when they accept hydrogen ions. Three Brønsted-Lowry bases we've looked at are hydroxide ions, ammonia and water, and they are typical of all the rest. The Brønsted-Lowry theory says that they are acting as bases because they are combining with hydrogen ions. The reason they are combining with hydrogen ions is that they have lone pairs of electrons - which is what the Lewis theory says. The two are entirely consistent. So how does this extend the concept of a base? At the moment it doesn't - it just looks at it from a different angle. But what about other similar reactions of ammonia or water, for example? On the Lewis theory, any reaction in which the ammonia or water used their lone pairs of electrons to form a co-ordinate bond would be counted as them acting as a base. Here is a reaction which you will find talked about on the page dealing with co-ordinate bonding. Ammonia reacts with BF by using its lone pair to form a co-ordinate bond with the empty orbital on the boron. As far as the ammonia is concerned, it is behaving exactly the same as when it reacts with a hydrogen ion - it is using its lone pair to form a co-ordinate bond. If you are going to describe it as a base in one case, it makes sense to describe it as one in the other case as well. Lewis acids are electron pair acceptors. In the above example, the BF is acting as the Lewis acid by accepting the nitrogen's lone pair. On the Brønsted-Lowry theory, the BF has nothing remotely acidic about it. This is an extension of the term acid well beyond any common use. What about more obviously acid-base reactions - like, for example, the reaction between ammonia and hydrogen chloride gas? \[ NH_{3(g)} + HCl_{g)} \rightarrow NH^+_{4(s)} + Cl^-_{(s)}\] What exactly is accepting the lone pair of electrons on the nitrogen. Textbooks often write this as if the ammonia is donating its lone pair to a hydrogen ion - a simple proton with no electrons around it. That is misleading! You don't usually get free hydrogen ions in chemical systems. They are so reactive that they are always attached to something else. There aren't any uncombined hydrogen ions in HCl. There isn't an empty orbital anywhere on the HCl which can accept a pair of electrons. Why, then, is the HCl a Lewis acid? Chlorine is more electronegative than hydrogen, and that means that the hydrogen chloride will be a polar molecule. The electrons in the hydrogen-chlorine bond will be attracted towards the chlorine end, leaving the hydrogen slightly positive and the chlorine slightly negative. The lone pair on the nitrogen of an ammonia molecule is attracted to the slightly positive hydrogen atom in the HCl. As it approaches it, the electrons in the hydrogen-chlorine bond are repelled still further towards the chlorine. Eventually, a is formed between the nitrogen and the hydrogen, and the chlorine breaks away as a chloride ion. This is best shown using the "curly arrow" notation commonly used in organic reaction mechanisms. The whole HCl molecule is acting as a Lewis acid. It is accepting a pair of electrons from the ammonia, and in the process it breaks up. Lewis acids don't necessarily have to have an existing empty orbital. All you need to remember for Lewis acids and bases is: For all general purposes, stick with the Brønsted-Lowry theory. | 9,349 | 1,486 |
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"For approximately one century from the early 1900s to the early 2000s, petroleum-based feedstocks gave rise to a vast petrochemicals industry that resulted in the production of synthetic rubber, plastics, polymers with a variety of properties, pesticides and literally hundreds of other products, many of which replaced materials biosynthesized in nature, especially by plants. Now with diminishing petroleum supplies, a new generation of materials made from biomaterials such as the lignocellulose that composes plant structural matter is developing that will take the place of many of the petroleum-based chemicals. This massive shift to renewable feedstocks is leading to a new age of green chemistry.”
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In the same way that every enzyme has an optimum temperature, so each enzyme also has an optimum pH at which it works best. For example, trypsin and pepsin are both enzymes in the digestive system which break protein chains in the food into smaller bits - either into smaller peptide chains or into individual amino acids. Pepsin works in the highly acidic conditions of the stomach. It has an optimum pH of about 1.5. On the other hand, trypsin works in the small intestine, parts of which have a pH of around 7.5. Trypsin's optimum pH is about 8. If you think about the structure of an enzyme molecule, and the sorts of bonds that it may form with its substrate, it isn't surprising that pH should matter. Suppose an enzyme has an optimum pH around 7. Imagine that at a pH of around 7, a substrate attaches itself to the enzyme via two ionic bonds. In the diagram below, the groups allowing ionic bonding are caused by the transfer of a hydrogen ion from a -COOH group in the side chain of one amino acid residue to an -NH group in the side chain of another. In this simplified example, that is equally true in both the substrate and the enzyme. Now think about what happens at a lower pH - in other words under acidic conditions. It won't affect the -NH group, but the -COO will pick up a hydrogen ion. What you will have will be this: You no longer have the ability to form ionic bonds between the substrate and the enzyme. If those bonds were necessary to attach the substrate and activate it in some way, then at this lower pH, the enzyme won't work. What if you have a pH higher than 7 - in other words under alkaline conditions. This time, the -COO group won't be affected, but the -NH group will lose a hydrogen ion. That leaves . . . Again, there is no possibility of forming ionic bonds, and so the enzyme probably won't work this time either. At extreme pH's, something more drastic can happen. Remember that the tertiary structure of the protein is in part held together by ionic bonds just like those we've looked at between the enzyme and its substrate. At very high or very low pH's, these bonds within the enzyme can be disrupted, and it can lose its shape. If it loses its shape, the active site will probably be lost completely. This is essentially the same as denaturing the protein by heating it too much. The rates of enzyme-catalysed reactions vary with pH and often pass through a maximum as the pH is varied. If the enzyme obeys Michaelis-Menten kinetics the kinetic parameters and often behave similarly. The pH at which the rate or a suitable parameter is a maximum is called the and the plot of rate or parameter against pH is called a Neither the pH optimum nor the pH profile of an enzyme has any absolute significance and both may vary according to which parameter is plotted and according to the conditions of the measurements. If the pH is changed and then brought back to its original value, the behavior is said to be if the original properties of the enzyme are restored; otherwise it is Reversible pH behavior may occur over a narrow range of pH, but effects of large changes in pH are in most cases irreversible. The diminution in rate as the pH is taken to the acid side of the optimum can be regarded as inhibition by hydrogen ions. The diminution in rate on the alkaline side can be regarded as inhibition by hydroxide ions. The equations describing pH effects are therefore analogous to inhibition equations. For single-substrate reactions the pH behavior of the parameters and can sometimes be represented by an equation of the form \[ k = \dfrac{k_{opt}}{1 + \dfrac{[H^+]}{K_1} + \dfrac{K_2}{[H^+]}} \label{eq1}\] in which k represents or , and \(k_{opt}\) is the value of the same parameter that would be observed if the enzyme existed entirely in the optimal state of protonation; it may be called the value of the parameter. The constants and can sometimes be identified as acid dissociation constants for the enzyme. substrates or other species in the reaction mixture. The identification is, however, never straight forward and has to be justified by independent evidence. The behavior is frequently much more complicated than represented by Equation \(\ref{eq1}\). It is not accidental that this section has referred exclusively to pH dependences of and . The pH dependence of the initial rate or, worse, the extent of reaction after a given time is rarely meaningful; the pH dependence of the Michaelis constant is often too complex to be readily interpretable. The pH dependence of the Michaelis constant is often too complex to be readily interpretable. In principle, we can use any of the range of possible analytical techniques to follow a reaction’s kinetics provided that the reaction does not proceed to any appreciable extent during the time it takes to make a measurement. As you might expect, this requirement places a serious limitation on kinetic methods of analysis. If the reaction’s kinetics are slow relative to the analysis time, then we can make our measurements without the analyte undergoing a significant change in concentration. When the reaction’s rate is too fast—which often is the case—then we introduce a significant error if our analysis time is too long. One solution to this problem is to stop, or the reaction by adjusting experimental conditions. For example, many reactions show a strong pH dependency, and may be quenched by adding a strong acid or a strong base. Figure 13.7 shows a typical example for the enzymatic analysis of -nitrophenylphosphate using the enzyme wheat germ acid phosphatase to hydrolyze the analyte to -nitrophenol. The reaction has a maximum rate at a pH of 5. Increasing the pH by adding NaOH quenches the reaction and converts the colorless -nitrophenol to the yellow-colored -nitrophenolate, which absorbs at 405 nm. | 5,879 | 1,490 |
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Organic compounds, molecules created around a chain of carbon atom (more commonly known as carbon backbone), play an essential role in the chemistry of life. These molecules derive their importance from the energy they carry, mainly in a form of potential energy between atomic molecules. Since such potential force can be widely affected due to changes in atomic placement, it is important to understand the concept of an isomer, a molecule sharing same atomic make up as another but differing in structural arrangement. Stereoisomers are isomers that differ in spatial arrangement of atoms, rather than order of atomic connectivity. One of their most interesting type of isomer is the mirror-image stereoisomers, a non-superimposable set of two molecules that are mirror image of one another. The existence of these molecules are determined by concept known as chirality. Molecules with the same connectivity but different arrangements of the atoms in space are called stereoisomers. There are two types of stereoisomers: geometric and optical. Geometric isomers differ in the relative position(s) of substituents in a rigid molecule. Simple rotation about a C–C σ bond in an alkene, for example, cannot occur because of the presence of the π bond. The substituents are therefore rigidly locked into a particular spatial arrangement. Thus a carbon–carbon multiple bond, or in some cases a ring, prevents one geometric isomer from being readily converted to the other. The members of an isomeric pair are identified as either cis or trans, and interconversion between the two forms requires breaking and reforming one or more bonds. Because their structural difference causes them to have different physical and chemical properties, cis and trans isomers are actually two distinct chemical compounds. Stereoisomers have the same connectivity, but different arrangements of atoms in space. Optical isomers are molecules whose structures are mirror images but cannot be superimposed on one another in any orientation. Optical isomers have identical physical properties, although their chemical properties may differ in asymmetric environments. Molecules that are nonsuperimposable mirror images of each other are said to be chiral (pronounced “ky-ral,” from the Greek cheir, meaning “hand”). Examples of some familiar chiral objects are your hands, feet, and ears. As shown in Figure \(\Page {1a}\), your left and right hands are nonsuperimposable mirror images. (Try putting your right shoe on your left foot—it just doesn’t work.) An achiral object is one that can be superimposed on its mirror image, as shown by the superimposed flasks in Figure \(\Page {1b}\). Most chiral organic molecules have at least one carbon atom that is bonded to four different groups, as occurs in the bromochlorofluoromethane molecule shown in part (a) in Figure \(\Page {2}\). This carbon, often designated by an asterisk in structural drawings, is called a chiral center or asymmetric carbon atom. If the bromine atom is replaced by another chlorine (Figure \(\Page {2b}\)), the molecule and its mirror image can now be superimposed by simple rotation. Thus the carbon is no longer a chiral center. Asymmetric carbon atoms are found in many naturally occurring molecules, such as lactic acid, which is present in milk and muscles, and nicotine, a component of tobacco. A molecule and its nonsuperimposable mirror image are called enantiomers (from the Greek enantiou, meaning “opposite”). In the 1960’s, a drug called thalidomide was widely prescribed in the Western Europe to alleviate morning sickness in pregnant women. Thalidomide had previously been used in other countries as an antidepressant, and was believed to be safe and effective for both purposes. The drug was not approved for use in the U.S.A. It was not long, however, before doctors realized that something had gone horribly wrong: many babies born to women who had taken thalidomide during pregnancy suffered from severe birth defects. Researchers later realized the that problem lay in the fact that thalidomide was being provided as a mixture of two different isomeric forms. One of the isomers is an effective medication, the other caused the side effects. Both isomeric forms have the same molecular formula and the same atom-to-atom connectivity, so they are not constitutional isomers. Where they differ is in the arrangement in three-dimensional space about one tetrahedral, sp -hybridized carbon. These two forms of thalidomide are . Looking for planes of symmetry in a molecule is useful, but often difficult in practice. In most cases, the easiest way to decide whether a molecule is chiral or achiral is to look for one or more stereocenters - with a few rare exceptions, the general rule is that molecules with at least one stereocenter are chiral, and molecules with no stereocenters are achiral. Carbon stereocenters are also referred to quite frequently as . When evaluating a molecule for chirality, it is important to recognize that the question of whether or not the dashed/solid wedge drawing convention is used is irrelevant. Chiral molecules are sometimes drawn without using wedges (although obviously this means that stereochemical information is being omitted). Conversely, wedges may be used on carbons that are not stereocenters – look, for example, at the drawings of glycine and citrate in the figure above. Just because you see dashed and solid wedges in a structure, do not automatically assume that you are looking at a stereocenter. | 5,537 | 1,491 |
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This page explains what is meant by the ionic product for water. It looks at how the ionic product varies with temperature, and how that determines the pH of pure water at different temperatures. Water molecules can function as both acids and bases. One water molecule (acting as a base) can accept a hydrogen ion from a second one (acting as an acid). This will be happening anywhere there is even a trace of water - it doesn't have to be pure. A hydroxonium ion and a hydroxide ion are formed. However, the hydroxonium ion is a very strong acid, and the hydroxide ion is a very strong base. As fast as they are formed, they react to poduce water again. The net effect is that an equilibrium is set up. At any one time, there are incredibly small numbers of hydroxonium ions and hydroxide ions present. Further down this page, we shall calculate the concentration of hydroxonium ions present in pure water. It turns out to be 1.00 x 10 mol dm at room temperature. You may well find this equilibrium written in a simplified form: This is OK provided you remember that H actually refers to a hydroxonium ion. K is essentially just an equilibrium constant for the reactions shown. You may meet it in two forms: Based on the fully written equilibrium . . . . . . or on the simplified equilibrium: You may find them written with or without the state symbols. Whatever version you come across, they all mean exactly the same thing! You may wonder why the water isn't written on the bottom of these equilibrium constant expressions. So little of the water is ionised at any one time, that its concentration remains virtually unchanged - a constant. K is defined to avoid making the expression unnecessarily complicated by including another constant in it. Like any other equilibrium constant, the value of K varies with temperature. Its value is usually taken to be 1.00 x 10 mol dm at room temperature. In fact, this is its value at a bit less than 25°C. The relationship between K and pK is exactly the same as that between K and pK , or [H ] and pH. The K value of 1.00 x 10 mol dm at room temperature gives you a pK value of 14. Try it on your calculator! Notice that pK doesn't have any units. That question is actually misleading! In fact, pure water only has a pH of 7 at a particular temperature - the temperature at which the K value is 1.00 x 10 mol dm . This is how it comes about: To find the pH you need first to find the hydrogen ion concentration (or hydroxonium ion concentration - it's the same thing). Then you convert it to pH. In pure water at room temperature the K value tells you that: [H ] [OH ] = 1.00 x 10 But in pure water, the hydrogen ion (hydroxonium ion) concentration must be equal to the hydroxide ion concentration. For every hydrogen ion formed, there is a hydroxide ion formed as well. That means that you can replace the [OH ] term in the K expression by another [H ]. [H ] = 1.00 x 10 Taking the square root of each side gives: [H ] = 1.00 x 10 mol dm Converting that into pH: pH = - log [H ] pH = 7 That's where the familiar value of 7 comes from. The formation of hydrogen ions (hydroxonium ions) and hydroxide ions from water is an endothermic process. Using the simpler version of the equilibrium: The forward reaction absorbs heat. According to Le Chatelier's Principle, if you make a change to the conditions of a reaction in dynamic equilibrium, the position of equilibrium moves to counter the change you have made. According to Le Chatelier, if you increase the temperature of the water, the equilibrium will move to lower the temperature again. It will do that by absorbing the extra heat. That means that the forward reaction will be favored, and more hydrogen ions and hydroxide ions will be formed. The effect of that is to increase the value of K as temperature increases. The table below shows the effect of temperature on K . For each value of K , a new pH has been calculated using the same method as above. It might be useful if you were to check these pH values yourself. You can see that the pH of pure water falls as the temperature increases. If the pH falls as temperature increases, does this mean that water becomes more acidic at higher temperatures? NO! A solution is acidic if there is an of hydrogen ions over hydroxide ions. In the case of pure water, there are always the same number of hydrogen ions and hydroxide ions. That means that the water remains neutral - even if its pH changes. The problem is that we are all so familiar with 7 being the pH of pure water, that anything else feels really strange. Remember that you calculate the neutral value of pH from K . If that changes, then the neutral value for pH changes as well. At 100°C, the pH of pure water is 6.14. That is the neutral point on the pH scale at this higher temperature. A solution with a pH of 7 at this temperature is slightly alkaline because its pH is a bit higher than the neutral value of 6.14. Similarly, you can argue that a solution with a pH of 7 at 0°C is slightly acidic, because its pH is a bit lower than the neutral value of 7.47 at this temperature. | 5,140 | 1,492 |
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Molecular geometry, also known as the molecular structure, is the three-dimensional structure or arrangement of atoms in a molecule. Understanding the molecular structure of a compound can help determine the polarity, reactivity, phase of matter, color, magnetism, as well as the biological activity. To determine the shapes of molecules, we must become acquainted with the Lewis electron dot structure. Although the Lewis theory does not determine the shapes of molecules, it is the first step in predicting shapes of molecules. The Lewis structure helps us identify the bond pairs and the lone pairs. Then, with the Lewis structure, we apply the to determine the molecular geometry and the electron-group geometry. To identify and have a complete description of the three-dimensional shape of a molecule, we need to know also learn about state the bond angle as well. Lewis Electron Dot Structures play crucial role in determining the geometry of molecules because it helps us identify the valence electrons. To learn how to draw a Lewis electron dot structure click the link above. Now that we have a background in the Lewis electron dot structure we can use it to locate the the valence electrons of the center atom. The states that electron pairs repel each other whether or not they are in bond pairs or in lone pairs. Thus, electron pairs will spread themselves as far from each other as possible to minimize repulsion. VSEPR focuses not only on electron pairs, but it also focus on electron groups as a whole. An can be an electron pair, a lone pair, a single unpaired electron, a double bond or a triple bond on the center atom. Using the VSEPR theory, the electron bond pairs and lone pairs on the center atom will help us predict the shape of a molecule. The shape of a molecule is determined by the location of the nuclei and its electrons. The electrons and the nuclei settle into positions that minimize repulsion and maximize attraction. Thus, the molecule's shape reflects its equilibrium state in which it has the lowest possible energy in the system. Although VSEPR theory predicts the distribution of the electrons, we have to take in consideration of the actual determinant of the molecular shape. We separate this into two categories, the and the . Electron-group geometry is determined by the number of electron groups. Molecular geometry, on the other hand, depends on not only on the number of electron groups, but also on the number of lone pairs. When the electron groups are all bond pairs, they are named exactly like the electron-group geometry. See the chart below for more information on how they are named depending on the number of lone pairs the molecule has. As stated above, molecular geometry and electron-group geometry are the same when there are no lone pairs. The VSEPR notation for these molecules are AX "A" represents the central atom and n represents the number of bonds with the central atom. When lone pairs are present, the letter E is added. The x represents the number of lone pairs present in the molecule. For example, a molecule with two bond pairs and two lone pairs would have this notation: AX E Tetrahedral octahedral ICl Lets try determining the geometric structures of H O and CO . So starting off by drawing the Lewis structure: H O: Water has four electron groups so it falls under tetrahedral for the electron-group geometry. The four electron groups are the 2 single bonds to Hydrogen and the 2 lone pairs of Oxygen. Since water has two lone pairs it's molecular shape is bent. According to the VSEPR theory, the electrons want to minimize repulsion, so as a result, the lone pairs are adjacent from each other. CO : Carbon dioxide has two electron groups and no lone pairs. Carbon dioxide is therefore linear in electron-group geometry and in molecular geometry. The shape of CO is linear because there are no lone pairs affecting the orientation of the molecule. Therefore, the linear orientation minimizes the repulsion forces. The VSEPR theory not only applies to one central atom, but it applies to molecules with more than one central atom. We take in account the geometric distribution of the terminal atoms around each central atom. For the final description, we combine the separate description of each atom. In other words, we take long chain molecules and break it down into pieces. Each piece will form a particular shape. Follow the example provided below: Butane is C H . C-C-C-C is the simplified structural formula where the Hydrogens (not shown) are implied to have single bonds to Carbon. You can view a better structural formula of butane at en. .org/wiki/File:Butane-2D-flat.png
If we break down each Carbon, the central atoms, into pieces, we can determine the relative shape of each section. Let's start with the leftmost side. We see that C has three single bonds to 2 Hydrogens and one single bond to Carbon. That means that we have 4 electron groups. By checking the geometry of molecules chart above, we have a tetrahedral shape. Now, we move on to the next Carbon. This Carbon has 2 single bonds to 2 Carbons and 2 single bonds to 2 Hydrogens. Again, we have 4 electron groups which result in a tetrahedral. Continuing this trend, we have another tetrahedral with single bonds attached to Hydrogen and Carbon atoms. As for the rightmost Carbon, we also have a tetrahedral where Carbon binds with one Carbon and 3 Hydrogens. Let me recap. We took a look at butane provided by the wonderful link. We, then, broke the molecule into parts. We did this by looking at a particular central atom. In this case, we have 4 central atoms, all Carbon. By breaking the molecule into 4 parts (each part looks at 1 of the 4 Carbons), we determine how many electron groups there are and find out the shapes. We aren't done, yet! We need to determine if there are any lone pairs because we only looked at bonds. Remember that electron groups include lone pairs! Butane doesn't have any lone pairs. Hence, we have 4 tetrahedrals. Now, what are we going to do with 4 tetrahedrals? Well, we want to optimize the bond angle of each central atom attached to each other. This is due to the electrons that are shared are more likely to repel each other. With 4 tetrahedrals, the shape of the molecule looks like this: en. .org/wiki/File:Butane-3D-balls.png. That means that if we look back at every individual tetrahedral, we match the central Carbon with the Carbon it's bonded to. Bond angles also contribute to the shape of a molecule. are the angles between adjacent lines representing bonds. The bond angle can help differentiate between linear, trigonal planar, tetraheral, trigonal-bipyramidal, and octahedral. The ideal bond angles are the angles that demonstrate the maximum angle where it would minimize repulsion, thus verifying the VSEPR theory. Essentially, bond angles is telling us that electrons don't like to be near each other. Electrons are negative. Two negatives don't attract. Let's create an analogy. Generally, a negative person is seen as bad or mean and you don't want to talk to a negative person. One negative person is bad enough, but if you have two put together...that's just horrible. The two negative people will be mean towards each other and they won't like each other. So, they will be far away from each other. We can apply this idea to electrons. Electrons are alike in charge and will repel each other. The farthest way they can get away from each other is through angles. Now, let's refer back to tetrahedrals. Why is it that 90 degrees does not work? Well, if we draw out a tetrahedral on a 2-D plane, then we get 90 degrees. However, we live in a 3-D world. To visualize this, think about movies. Movies in 3D pop out at us. Before, we see movies that are just on the screen and that's good. What's better? 3D or 2D? For bond angles, 3D is better. Therefore, tetrahedrals have a bond angle of 109.5 degrees. How scientists got that number was through experiments, but we don't need to know too much detail because that is not described in the textbook or lecture. Using the example above, we would add that H O has a bond angle of 109.5 and CO would have a bond angle of 180 . To sum up there are four simple steps to apply the VSEPR theory. A molecule is polar when the electrons are not distributed equally and the molecule has two poles. The more electronegative end of the molecule is the negative end and the less electronegative end is the positive end. A common example is HCl. Using the capital sigma + or - as a symbol to show the the positive end and the negative end we can draw the net dipole. So sigma + would be on the hydrogen atom and sigma - would be on the Chlorine atom. Using the cross bow arrow shown below we can show that it has a net dipole. The net dipole is the measurable, which is called the . Dipole moment is equal to the product of the partial charge and the distance. The equation for dipole moment is as follows. \[ \mu = \delta \times d\] with The units for dipole is expressed in which is also known as Coulombs x meter (C x m) Example of a Dipole On the cross-base arrow, the cross represents the positive charge and the arrow represents the negative charge. Here's another way to determine dipole moments. We need to comprehend electronegativity which is abbreviated EN. What is EN? Well, EN is how much an element really wants an electron. Think about basketball and how two players pass the ball to each other. Each player represent an element and the ball represents the electron. Let's say one player is a ball hog. The player that is the ball hog is more electronegative because he or she wants the ball more. Here is a link that has all the EN listed: www.green-planet-solar-energy...electroneg.gif What if we are not given EN? Luckily, there is a trend in the periodic table for EN. From bottom to the top, EN will increase. From left to right, EN will increase. The most electronegative element is Flourine with 4.0. Now, we are ready to apply EN to determine whether or not molecules are polar. We look back at the picture of H O above. The EN is given. What do we do with all the EN? We compare the EN between each bond. Oxygen has a greater EN than Hydrogen. Therefore, we can draw a cross bow arrow towards Oxygen. We have two arrows because Oxygen is bonded to two Hydrogens. Since both arrows point toward Oxygen, we can say that there is a net EN. We added the arrows that point to Oxygen and we end up with a new, bigger arrow. This is examplified in the picture above. If arrows are drawn away from each other like <--- and --->, then we are more likely to have no net EN because the molecule is symmetrical. Refer back to the Lewis dot diagram of CO . The shape is linear and the EN arrows point towards Oxygen. The arrows are opposite of each other and have the same EN difference. Therefore, we have no net charge and the molecule is non-polar. To recap, when a molecule is polar it means that the electron is not distributed evenly and there is a difference in the electronegativity of the atoms. If a molecule is polar, it means that it had a net dipole which results in having a dipole moment. Is it polar? There are three ways to go about determining whether a molecule is polar or not. A. If the molecule has a net dipole, then it is polar.
B. If the structure is symmetric, then it is non-polar
C. There are three rules to this part:
Draw the Lewis Structure and name the shape of each compound. Also determine the polarity and whether or not it has a dipole moment. Name the shape and determine whether they are polar or non-polar. 1. 2. 3. 4. 5. 1. electron group geometry: octahedral molecular geometry: square planar non polar because it is symmetrical 2. electron group geometry: octahedral molecular geometry: square planar polar because it is not symmetrical | 11,883 | 1,493 |
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Considering the relation between solubility and \(K_{sq}\) is important when describing the solubility of slightly ionic compounds. However, this article discusses ionic compounds that are difficult to dissolve; they are considered "slightly soluble" or "almost insoluble." Solubility product constants (\(K_{sq}\)) are given to those solutes, and these constants can be used to find the molar solubility of the compounds that make the solute. This relationship also facilitates finding the \(K_{sq}\) of a slightly soluble solute from its solubility. Solubility is the ability of a substance to dissolve. The two participants in the dissolution process are the solute and the solvent. The solute is the substance that is being dissolved, and the solvent is the substance that is doing the dissolving. For example, sugar is a solute and water is a solvent. Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium. Equilibrium is the state at which the concentrations of products and reactant are constant after the reaction has taken place. The solubility product constant (\(K_{sq}\)) describes the equilibrium between a solid and its constituent ions in a solution. The value of the constant identifies the degree to which the compound can dissociate in water. For example, the higher the \(K_{sq}\), the more soluble the compound is. \(K_{sq}\) is defined in terms of activity rather than concentration because it is a measure of a concentration that depends on certain conditions such as temperature, pressure, and composition. It is influenced by surroundings. \(K_{sq}\) is used to describe the saturated solution of ionic compounds. (A saturated solution is in a state of equilibrium between the dissolved, dissociated, undissolved solid, and the ionic compound.) Consider the compound barium carbonate BaCO (an ionic compound that is not very soluble): \[ \ce{BaCO_{3(s)} \rightleftharpoons Ba^{2+} (aq)+CO_3^{2-} (aq) } \nonumber\] First, write down the equilibrium constant expression: \[K_c = \dfrac{[\ce{Ba^{2+}},\ce{CO_3^{2-}}]}{[\ce{BaCO_3}]} \nonumber \] The activity of solid \(\ce{BaCO is 1, and considering that the concentrations of these ions are small, the activities of the ions are approximated to their molar concentrations. \(K_{sq}\) is therefore equal to the product of the ion concentrations: \[\begin{align*} K_{sp} &= [\ce{Ba^{2+}},\ce{CO_3^{2-}}] \\[4pt] &= 5.1 \times 10^{-9} \end{align*}\] The relation between solubility and the solubility product constants is that one can be used to derive the other. In other words, there is a relationship between the solute's molarity and the solubility of the ions because \(K_{sq}\) is the product of the solubility of each ion in moles per liter. For example, to find the \(K_{sq}\) of a slightly soluble compound from its solubility, the solubility of each ion must be converted from mass per volume to moles per liter to find the molarity of each ion. These numbers can then be substituted into the \(K_{sq}\) formula, which is the product of the solubility of each ion. An example of this process is given below: Suppose the aqueous solubility for compound PbI is 0.54 grams/100 ml at 25 °C and calculate the \(K_{sq}\) of PbI at 25°C. \[ PbI_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2I^-_{(aq)} \nonumber\] Convert 0.54 grams of \(PbI_2\) to moles: \[ 0.54\ grams \times \dfrac{1\ mol\ PbI_2}{461.0\ grams} = 0.001171\ mol\ PbI_2 \nonumber\] Convert ml to L: \[\dfrac{100\ mL}{1000\ L} = 0.100\ L \nonumber\] Find the molarity: \[ \dfrac{0.001171\ mol}{0.100\ L} = 0.01171\ M\ PbI_2 \nonumber\] Now find the molarity of each ion by using the stoichiometric ratio (remember there are two I ions for each Pb ion): \[ \begin{eqnarray} [Pb^{2+}] &=& \dfrac{0.01171\ M}{1\ L} \times \dfrac{1\ mol\ Pb}{1\ mol\ PbI_2} \\ &=& 0.011714\ M\ Pb^{2+} \\ [I^-] &=& \dfrac{0.01171\ M}{1\ L} \times \dfrac{2\ mol\ I^-}{1\ mol\ PbI_2} \\ &=& 0.23427\ M\ I^- \end{eqnarray} \] Finally, plug in the molarity to find \(K_{sq}\): \[\begin{eqnarray} K_{sp} &=& [Pb^{2+},I^-]^2 \\ &=& (0.011714\ M)(0.023427\ M)^2 \\ &=& 6.4 \times 10^{-6} \end{eqnarray} \] *This relation facilitates solving for the molar solubility of the ionic compounds when the \(K_{sq}\) is given to us. The process involves working backwards from \(K_{sq}\) to the molarity of the ionic compound.* Suppose the \(K_{sq}\) at 25 C is 8.5 x 10 for the compound AgI. What is the molar solubility? \[ AgI_{(s)} \rightleftharpoons Ag^+_{(aq)} + I^-_{(aq)}\nonumber \] Let "g" represent the number of moles: \[ \begin{eqnarray} K_{sp} &=& [Ag^{2+},I^-] \\ &=& g^2 \\ &=& 8.5 \times 10^{-17} \end{eqnarray} \] Solve for "g": \[ \begin{eqnarray} g^2 &=& 8.5 \times 10^{-17} \\ g &=& (8.5 \times 10^{-17})^{\dfrac{1}{2}} \\ &=& 9.0 \times 10^{-9} \end{eqnarray} \] | 4,869 | 1,494 |
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Addition of hydrogen to a carbon-carbon double bond is called . The overall effect of such an addition is the reductive removal of the double bond functional group. Regioselectivity is not an issue, since the same group (a hydrogen atom) is bonded to each of the double bond carbons. The simplest source of two hydrogen atoms is molecular hydrogen (H ), but mixing alkenes with hydrogen does not result in any discernible reaction. Although the overall hydrogenation reaction is exothermic, a high activation energy prevents it from taking place under normal conditions. This restriction may be circumvented by the use of a catalyst, as shown in the following diagram. Catalysts are substances that changes the rate (velocity) of a chemical reaction without being consumed or appearing as part of the product. Catalysts act by lowering the activation energy of reactions, but they do not change the relative potential energy of the reactants and products. Finely divided metals, such as platinum, palladium and nickel, are among the most widely used hydrogenation catalysts. Catalytic hydrogenation takes place in at least two stages, as depicted in the diagram. First, the alkene must be adsorbed on the surface of the catalyst along with some of the hydrogen. Next, two hydrogens shift from the metal surface to the carbons of the double bond, and the resulting saturated hydrocarbon, which is more weakly adsorbed, leaves the catalyst surface. The exact nature and timing of the last events is not well understood. As shown in the energy diagram, the hydrogenation of alkenes is exothermic, and heat is released corresponding to the ΔE (colored green) in the diagram. This heat of reaction can be used to evaluate the thermodynamic stability of alkenes having different numbers of alkyl substituents on the double bond. For example, the following table lists the heats of hydrogenation for three C H alkenes which give the same alkane product (2-methylbutane). Since a large heat of reaction indicates a high energy reactant, these heats are inversely proportional to the stabilities of the alkene isomers. To a rough approximation, we see that each alkyl substituent on a double bond stabilizes this functional group by a bit more than 1 kcal/mole. From the mechanism shown here we would expect the addition of hydrogen to occur with syn-stereoselectivity. This is often true, but the hydrogenation catalysts may also cause isomerization of the double bond prior to hydrogen addition, in which case stereoselectivity may be uncertain.
The formation of transition metal complexes with alkenes has been convincingly demonstrated by the isolation of stable platinum complexes such as Zeise's salt, K[PtCl (C H )].H O, and ethylenebis(triphenylphosphine)platinum, [(C H ) P] Pt(H C=CH ). In the latter, platinum is three-coordinate and zero-valent, whereas Zeise's salt is a derivative of platinum(II). A model of Zeise's salt and a discussion of the unusual bonding in such complexes may be viewed by clicking here. Similar complexes have been reported for nickel and palladium, metals which also function as catalysts for alkene hydrogenation. A non-catalytic procedure for the syn-addition of hydrogen makes use of the unstable compound diimide, N H . This reagent must be freshly generated in the reaction system, usually by oxidation of hydrazine, and the strongly exothermic reaction is favored by the elimination of nitrogen gas (a very stable compound). Diimide may exist as cis-trans isomers; only the cis-isomer serves as a reducing agent. Examples of alkene reductions by both procedures are shown below. | 3,630 | 1,495 |
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Koopmans' theorem states that the first ionization energy of a molecule is equal to the negative of the energy of the highest occupied molecular orbital (HOMO). Koopmans' theorem uses the Hartree-Fock method for approximation of orbital energy ε which is derived from the wavefunction of the spin orbital and the kinetic and nuclear attraction energies. This theorem applies when an electron is removed from a molecular orbital in order to form a positive ion. It was originally only used for ionization energies in a closed-shell system, but has been generalized to be used to calculate energy changes when electrons are added to or removed from a system. Based on this generalization, it is possible to use the same method to approximate the electron affinity. In this case, the molecular orbital energy would be the one associated with the orbital to which the electron is being added. Koopmans' theorem is useful because the use of this approximation means that it is not necessary to calculate the two separate energies of the original molecule and its ion in order to find the ionization energy and electron affinity. In order to understand Koopmans' theorem, we must first understand its background, which is based in Hartree-Fock method. Hartree-Fock method is used to approximate the wave function and energy of a multi-electron system. This method starts by expressing the wave function of the system as a Slater determinant of the wave function of each single-particle orbital. The Slater determinant for a system with N electons is \[{\displaystyle \Psi (\mathbf {x} _{1},\mathbf {x} _{2},\ldots ,\mathbf {x} _{N})={\frac {1}{\sqrt {N!}}}\left|{\begin{matrix}\chi _{1}(\mathbf {x} _{1})&\chi _{2}(\mathbf {x} _{1})&\cdots &\chi _{N}(\mathbf {x} _{1})\\\chi _{1}(\mathbf {x} _{2})&\chi _{2}(\mathbf {x} _{2})&\cdots &\chi _{N}(\mathbf {x} _{2})\\\vdots &\vdots &\ddots &\vdots \\\chi _{1}(\mathbf {x} _{N})&\chi _{2}(\mathbf {x} _{N})&\cdots &\chi _{N}(\mathbf {x} _{N})\end{matrix}}\right|\equiv \left|{\begin{matrix}\chi _{1}&\chi _{2}&\cdots &\chi _{N}\\\end{matrix}}\right|,}\] By breaking up the electrons into individual wave functions, we can create single-particle Hartree-Fock equations, which can be used as an operator whose eigenvalue is the energy of a particle in a particular orbital. \[ \hat {F} | \varphi _i \rangle = \epsilon _i| \varphi _i \rangle \] Where \(\hat {F}\) is the Fock operator corresponding to the Hartree-Fock equation, \(\epsilon _i\) is the energy of the particle in the orbital, and \(\varphi _i\) is the wave function of the particle. Using Koopmans' theorem, the equation from above and the wave function of the HOMO, we can approximate the first order ionization energy of a molecule. The Hartree-Fock equation for a particle takes the form \[-\dfrac{\hbar^{2}}{2m}\nabla^{2}\psi_{i}(\mathbf{r}) + V_{nucleus}(\mathbf{r})\psi_{i}(\mathbf{r}) + V_{electron}(\mathbf{r})\psi_{i}(\mathbf{r}) - \sum_{j} \int d\mathbf{r}^{\prime} \dfrac{\psi^{\star}_{j}(\mathbf{r}') \psi^{\star}_{i}(\mathbf{r}') \psi_{j}(\mathbf{r}) } {\left \vert \mathbf{r} - \mathbf{r}^{\prime} \right\vert} = \epsilon_{i}\psi_{i}(\mathbf{r})\] So, the Fock operator derived from this is \[\hat {F} = \hat {H} ^0 + \sum _{j=1}^N ( 2 \hat {J} _j - \hat {K} _j ) = -\dfrac {\hbar ^2}{2m} \nabla ^2 - \dfrac {Ze^2}{4 \pi \epsilon _0 r} + \sum _{j=1}^N (2\hat {J}_j - \hat {K} _j )\] This is the operator that is applied to the wave function, and gives the eigenvalues that describe the energy of a given orbital. Notice that the first two terms in this equation are the same as the operator which corresponds to the energy of a hydrogen atom. The third term accounts for the potential energy applied by the other electrons in the molecule. This method of calculating the potential energy assumes that the electron only interacts with the average charge of the electron cloud. Ionization energies and electron affinities are equal to the negative of the energy of the orbital which is added or removed, i.e. \[E_{N-1} - E_N = - \epsilon_k \] for ionization, and \[E_N - E_{N+1} = - \epsilon_k .\] for electron affinity. Koopmans' theorem also applies to the calculation of electron affinity. We will use Hartree-Fock equations to calculate the energy change when an electron is added to the N+1 orbital. The energy of the N-electron determinant with spin-orbitals \(\phi _1\) through \(\phi _N\) occupied is \[E_N = \sum_{i=1}^N \langle \phi_i | T + V | \phi_i \rangle + \sum_{i=1}^{N} [ J_{i,j} - K_{i,j} ] \nonumber\] or \[E_N = \sum_{i=1}^N \langle \phi_i | T + V | \phi_i \rangle + \frac{1}{2} \sum_{i,j=1}^{N} [ J_{i,j} - K_{i,j} ].\nonumber\] And the energy of the N+1 electron determinant is \[E_{N+1} = \sum_{i=1}^{N+1} \langle \phi_i | T + V | \phi_i \rangle + \frac{1}{2} \sum_{i,j=1}^{N+1} [ J_{i,j} - K_{i,j} ].\nonumber = \epsilon_{N+1}. \nonumber\] Since we are adding the N+1 orbital the electron affinity is equal to the negative of the energy of the N+1 electron determinant. \[EA = - \epsilon_{N+1}. \nonumber\] However, we must remember that Koopmans' theorem is merely an approximation of ionization energies and electron affinities. The more accurate method would be to calculate the separate energies of the parent and daughter molecules and subtract them to calculate the difference. This is because Koopmans' theorem does not account for orbital relaxation; it uses the orbitals of only one of the molecules to describe both of the species, which is not necessarily true. | 5,528 | 1,496 |
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The case of a quantum particle confined a one-dimensional ring is similar to the . Consider a variant of the one-dimensional particle in a box problem in which the x-axis is bent into a ring of radius \(R\). We can write the same Schrödinger equation There are no boundary conditions in this case since the x-axis closes upon itself. A more appropriate independent variable for this problem is the angular position on the ring given by, \( \phi = x {/} R \) Schrödinger \[ -\dfrac{\hbar^2}{2mR^2} \dfrac{d^2 \psi (\phi)} {d (\phi)^2} = E \psi (\phi) \label{2}\] The kinetic energy of a body rotating in the xy-plane can be expressed as \[ E = \dfrac{L_z^2}{2I} \label{3}\] where \(I = mR^2\) is the moment of inertia and \( L_z\), the -component of angular momentum. (Since \( L = r \times p\), if and lie in the -plane, points in the -direction.) The structure of Equation \ref{2} suggests that this angular-momentum operator is given by \[ \hat{L_z} = -{i} \hbar \dfrac{\partial}{\partial \phi} \label{4}\] This result will follow from a more general derivation elsewhere. The Schrödinger Equation (Equation \ref{2}) can now be written more compactly as \[ \psi \prime \prime \ (\phi) + m^2 \psi (\phi) = 0 \label{5}\] where \[m^2 \equiv 2IE/ \hbar^2\label{6}\] Do not confuse the variable \(m\) with the mass of the particle! Possible solutions to Equation \ref{5} are \[ \psi (\phi) = \text{const}\, e^{\pm{i}m\phi} \label{7}\] For this wavefunction to be physically acceptable, it must be . Since \(\phi \) increased by any multiple of 2\(\pi \) represents the same point on the ring, we must have \[ \psi (\phi + 2\pi ) = \psi (\phi) \label{8}\] and therefore \[e^{{i}m (\phi + 2\pi)} = e^{{i}m \phi} \label{9}\] This requires that \[e^{2\pi {i}m} = 1 \label{10}\] which is true only if m is an integer: \[ m = 0, \pm 1, \pm 2... \] Using Equation \ref{6}, this gives the quantized \[E_m = \dfrac{\hbar^2}{2I} m^2 \] In contrast to the particle in a box, the eigenfunctions corresponding to \(+m \) and \(-m \) (Equation \ref{7}) are , so both must be accepted. Therefore all eigenvalues, except \(E_0 \), are two-fold (or doubly) degenerate. The eigenfunctions can all be written in the form const \( e^{{i}m \phi} \), with \(m\) allowed to take either values (or 0), as in Equation \ref{10}. The normalized eigenfunctions are \[ {\psi _{m}} (\phi) = \dfrac{1}{\sqrt{2 \pi}} e^{im\phi} \] and can be verified to satisfy the normalization condition containing the complex conjugate \[ \int\limits_{0}^{2\pi} {\psi_{m}^*} (\phi) {\psi _{m}} (\phi) d\phi = 1 \] where we have noted that \( {\psi_{m}^*} (\phi) = (2\pi)^{-1/2} e^{-{i}m\phi} \). The mutual orthogonality of the functions also follows easily, for \[ \begin{align} \int\limits_{0}^{2\pi} {\psi_{m^\prime}^*} {\psi _{m}} (\phi) d\phi &= \dfrac{1}{2\pi} \int\limits_{0}^{2\pi} e^{{i}(m-m^\prime) \phi} d\phi \\[4pt] &= \dfrac{1}{2\pi} \int\limits_{0}^{2\pi} [\cos(m-m^\prime)\phi + {i} \sin(m-m^\prime)\phi] d\phi =0 \end{align}\] for \( m^\prime \neq m \). The solutions of Equation \ref{2} are also eigenfunctions of the angular momentum operator (Equation \ref{4}), with \[ \hat{L_z} \psi_{m} (\phi) = m\hbar \psi_{m} (\phi), m = 0, \pm 1, \pm 2...\] This is a instance of a fundamental result in quantum mechanics, that any measured component of orbital angular momentum is restricted to integral multiples of \( \hbar \). The of the hydrogen atom can be derived from this principle alone. The benzene molecule consists of a ring of six carbon atoms around which six delocalized -electrons can circulate. A variant of the FEM for rings predicts the ground-state electron configuration which we can write as \( 1\pi^{2} 2\pi^{4} \), as shown here: The enhanced stability the benzene molecule can be attributed to the complete shells of \(\pi \)-electron orbitals, analogous to the way that noble gas electron configurations configuration fulfill Hückel's 4 cyclobutadiene cyclooctatetraene they contain partially-filled \[ \dfrac{hc}{\lambda} = E_2 - E_1 = \dfrac{\hbar^2}{2mR^2} {(2^2 -1^2)} \] The ring radius can be approximated by the C-C distance in benzene, 1.39 Å. We predict \( \lambda \approx \) 210 nm, whereas the experimental absorption has \( \lambda_{max} \approx \) 268 nm. One type of rotational motion in quantum mechanics is a particle in a ring. An important aspect of this is the angular momentum which includes a vector with a direction that shows axis of rotation . The particle’s magnitude of angular momentum that is traveling along a circular path of radius \(r\) is classified as \(J=p \times r\) where \(p\) is the linear momentum at any moment. When the particle of mass travels in the horizontal radius \(r\), the particle has purely kinetic energy since potential energy is constant and is set to zero everywhere. The energy with regards of angular momentum can be expressed as: \[E = \dfrac{J^2_z}{2mr^2} \label{1C}\] A particle has a moment of inertia when traveling along a circular path. is defined by m (mass) multiplied by \(r^2\) (radius squared). The heavier particle in the top picture has a large moment of inertia on the central point while the lighter particle in the lower picture has a smaller moment of inertia while traveling on the path of the same radius . For the heavier particle, the is large and therefore, the particle's energy can be expressed by: \[E = \dfrac{J^2_z}{2I} \label{2C}\] We then use the to quantize the energy of rotation. This is done by expressing the angular momentum in wavelengths: \[J_z = pr = \dfrac{hr}{\lambda} \label{3C}\] where p is the linear momentum and h is Planck's constant (6.626 x 10- Js). It can also be written: \[\lambda = \dfrac{h}{p}\] With this equation, de Broglie postulated that there is a wave correlated with the electron via wavelength. He had done this to explain Bohr's model of the Hydrogen atom, in which the electron is only allowed permitted to orbit from the nucleus at certain distances. (Professor Emeritus of Chemistry and Physics at the , ) | 6,073 | 1,497 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Energies_and_Potentials/Enthalpy/Heat_of_Fusion |
Solids can be heated to the point where the molecules holding their bonds together break apart and form a liquid. The most common example is solid ice turning into liquid water. This process is better known as melting, or , and results in the molecules within the substance becoming less organized. When a substance converts from a solid state to a liquid state, the change in enthalpy (\(ΔH\)) is positive. However, if the substance is transforming from a liquid state to a solid state the change in enthalpy (ΔH) is negative. This process is commonly known as the , and results in the molecules within the substance becoming more ordered. Determining the heat of fusion is fairly straightforward. When a solid undergoes melting or freezing, the temperature stays at a constant rate until the entire phase change is complete. One can visualize this process by examining the heating/cooling chart. By drawing this chart before conducting a heat of fusion analysis, one can easily map out the required steps in completing the analysis. The equation for determining the enthalpy of fusion (\(ΔH\)) is listed below. \[\Delta{H}=n \,\Delta{H_{fus}}\] with Calculate the heat when 36.0 grams of water at 113 °C is cooled to 0 °C. Answer \[q =- 110.6\, kJ\] In some cases, the solid will bypass the liquid state and transition into the gaseous state. This direct transformation from solid to gas is called sublimation. The opposite reaction, when a gas directly transforms into a solid, is known as deposition. Therefore, these two processes can be summarized in the following equation: \[\Delta{H_{sub}}= \Delta{H_{fus}}+\Delta{H_{vap}}\] with The heat of fusion process can be seen in countless applications and evidenced in the creation of many common household items. As mentioned in the opening paragraph, the most common application of the heat of fusion is the melting of ice to water. The vast majority of examples where heat of fusion is commonplace can be seen in the manufacturing industry. The following examples have been used for hundreds of years and are still perfected to this day. The processes of coin making, glassblowing, forging metal objects, and transforming blow molded plastics into household products all require heat of fusion to become final product. The change in your wallet, the glass vase on your fireplace mantel, and the plastic soda bottle from the vending machine all went through a heat of fusion manufacturing process. In coin making, solid zinc and copper (metals in American pennies) are placed into a casting furnace and heated by the heat of fusion process until they reach the liquid phase. Once in the liquid phase, the molten zinc and copper are poured into a mold, and cast into long bars. In the casting process, the molten metal transforms from the liquid phase to the solid phase, becoming a solid bar. The long bars are flattened by heavy machinery and stamped into thousands of coins. Without the heat of fusion process, a monetary system would not exist in the United States. | 3,042 | 1,498 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Surface_Science_(Nix)/03%3A_The_Langmuir_Isotherm/3.02%3A_Langmuir_Isotherm_-_derivation_from_equilibrium_considerations |
We may derive the Langmuir isotherm by treating the adsorption process as we would any other equilibrium process - except in this case the equilibrium is between the gas phase molecules (\(M\)), together with vacant surface sites, and the species adsorbed on the surface. Thus, for a non-dissociative (molecular) adsorption process, we consider the adsorption to be represented by the following chemical equation: \[S - * + M_{(g)} \rightleftharpoons S - M \label{Eq1}\] where: In writing Equation \(\ref{Eq1}\) we are making an inherent assumption that there are a fixed number of localized surface sites present on the surface. This is the first major assumption of the Langmuir isotherm. We may now define an equilibrium constant (\(K\)) in terms of the concentrations of "reactants" and "products" \[ K = \dfrac{[S-M]}{[S-*,M]}\] We may also note that: Hence, it is also possible to define another equilibrium constant, , as given below: \[ b =\dfrac{\theta}{(1- \theta)P}\] Rearrangement then gives the following expression for the surface coverage \[ \theta =\dfrac{b P}{1 + bP}\] which is the usual form of expressing the Langmuir Isotherm. As with all chemical reactions, the equilibrium constant, \(b\), is both temperature-dependent and related to the Gibbs free energy and hence to the enthalpy change for the process. \(b\) is only a constant (independent of \(\theta\)) if the enthalpy of adsorption is independent of coverage. This is the second major assumption of the Langmuir Isotherm. | 1,515 | 1,499 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Statistical_Mechanics/Boltzmann_Average/The_Boltzmann_constant |
The ( or ) is the physical constant relating temperature to energy. It is named after the Austrian physicist Ludwig Eduard Boltzmann. Its experimentally determined value (in SI units, 2002 CODATA value) is: \[k_B =1.380 6505(24) \times 10^{-23}\left. JK^{-1}\right.\] Max Planck, Nobel Lecture, June 2, 1920 Boltzmann's constant can be obtained from the ratio of the molar gas constant to the Avogadro constant. The molar gas constant can be obtained via acoustic gas thermometry, and Avogadros constant from either the , or via the watt balance. Recently laser spectroscopy has been used to determine the constant (Refs. 3 and 4). Other techniques include Coulomb blockade thermometry (Refs. 5 and 6). | 725 | 1,500 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Liquid_Crystals |
A true liquid is , meaning that its properties are uniform in all directions— the result of its molecules being in constant random motion. Crystalline solids, in contrast, are ; optical- and other properties such as thermal and electrical conductivity vary with direction. A liquid crystal phase has many of the physical attributes of a liquid, but its molecular units are sufficiently ordered to give rise to some anisotropy, most notably in their optical properties. As with so many scientific discoveries, it all started with an unexpected observation. In 1888, an Austrian plant physiologist, working in Prague, attempted to measure the melting point of a cholesterol derivative that he had extracted from a plant. To his surprise, he found that this substance appeared to have melting points. At 145° C the crystalline solid first melted into a cloudy liquid, and then at 178° the cloudiness suddenly disappeared, leaving the clear, transparent liquid that one ordinarily expects after melting. A physicist who examined this material recognized that the cloudy liquid had a certain degree of order; he proposed that it was a hitherto unknown state of matter, and suggested the name "liquid crystal". But science was not quite ready to accept this concept, and despite a number of confirming experiments between 1910 and 1930, the field remained largely dormant until the mid-1960s when the French physicist Pierre-Gilles de Gennes (1932-2007) developed a thorough theoretical model for the properties of liquid crystals, particularly their ability to scatter light. For this and for related studies on polymers, de Gennes was awarded the 1991 Nobel Prize in Physics. The structural units capable of forming liquid crystals are always molecules, usually rather elongated organic ones that possess dissimilar local structural regions that can interact in an organized way with their neighbors. Over a certain range of temperatures, these attractive forces can lead to a degree of self-organization in which crystal-like order persists in some directions even though it is lost in other directions. Although a large variety of molecules are known to form liquid crystals, the simplest and most common structures can be represented by the following generic scheme: Liquid crystal phases are generally cloudy in appearance, which means that they scatter light in much the same way as colloids such as milk. This light scattering is a consequence of fluctuating regions of non-uniformity as small groups of molecules form and disperse. The anisotropy of liquid crystals causes them to exhibit . That is, light that enters the crystal is broken up into two oppositely-polarized rays that travel at different velocities. Observation of a birefringent material between crossed polarizing filters reveals striking patterns and color effects. The colors arise from interference between the and the ; the latter traverses a slightly longer path through the material, and thus emerges later (and out-of-phase) with the former. Liquid crystals, like all other kinds of matter, are subject to thermal expansion. As the temperature rises, the average spacing between the aligned molecules of a nematic phase (see below) increases, thus causing the e-ray to be increasingly retarded with respect to the o-ray. If a suitable liquid crystal mixture is painted onto the surface of a patient's body, it can often reveal the sites of infection or tumors, which cause increases or reductions in local blood flow giving rise to temperature anomalies. Inexpensive can be made by printing a succession of suitably formulated LC mixtures on a paper or plastic strip which is held in contact with the surface whose temperature is to be measured. The first working liquid crystal display ("LCD") was demonstrated at RCA in 1968. In the following year, James Fergason of Kent State University (OH) discovered the twisted nematic field effect which allowed a much higher-quality display and led to the first commercial LCD wristwatch in 1979. LCDs were first used in calculators in the late 1970s, but they are now widely encountered in computer- and television displays. The thickness of the chiral phase is such that polarized light passing through it is rotated by 90°, which corresponds to the orientation of the right-hand polarizing filter. So in this state, the light passes through and illuminates the pixel surface. When an electric field is imposed on the liquid crystal phase, the component molecules line up in the field and the chirality is lost. Light passing through the cell does not undergo rotation of its polarization plane, and is therefore stopped by the right polarizing filter, turning the display off. There are many classes and sub-classes of liquid crystals, but for the purposes we will divide them into the two kinds depicted on the right side of this Figure below which also compares them with the two conventional condensed phases of matter: In a (the term means "thread-like") the molecules are aligned in the same direction but are free to drift around randomly, very much as in an ordinary liquid. Owing to their polarity, the alignment of the rod-like molecules can be controlled by applying an electric field; this is the physical basis for liquid crystal displays and certain other electrooptic devices. In ("soap-like") phases the molecules are arranged in layers, with the long molecular axes approximately perpendicular to the laminar planes. The only long-range order extends along this axis, with the result that individual layers can slip over each other (hence the "soap-like" nature) in a manner similar to that observed in graphite. Within a layer there is a certain amount of short-range order. There are a large number of sub-categories of smectic phases which we will not go into here. Special cases of nematic and smectic phases are sometimes formed by molecules that display — that is, they can exist in either left- or right-handed forms that cannot be superposed on each other. In the resulting , successive molecules positioned along the long axis are rotated around this axis, giving rise to a periodicity that repeats itself at distances corresponding to a complete rotation. These twisted phases are able to rotate the plane of polarized light that passes along the axis. If the molecules are polar, this twisting can be turned off by imposing an external electric field at either end of the long axis. Besides the very important application of this property (known as ) to liquid crystal displays, these materials can be used to make electrooptic shutters which can be switched open and closed in microseconds. A typical chiral molecule capable of exhibiting ferroelectric behavior is shown below. The chiral part of the molecule is indicated by the asterisk. The chirality arises because this carbon atom is joined to four different groups. ) | 6,907 | 1,501 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumentation_and_Analysis/Mass_Spectrometry/Introductory_Mass_Spectrometry/Molecular_Ion_and_Nitrogen |
This phenomenon is a result of the fact that the most common elements in organic compounds, carbon and oxygen, have even atomic weights (12 and 16, respectively), so any number of carbons and oxygens will have even weights. The most common isotope of hydrogen has an odd molecular weight, but because carbon and oxygen both have even valences (carbon forms four bonds and oxygen forms two), there is always an even number of hydrogen atoms in an organic compound containing those elements, so they also add up to an even numbered weight. Nitrogen has an even atomic weight (14), so any number of nitrogen atoms will add up to an even molecular weight. Nitrogen, however, has an odd valence (it forms three bonds), and as a result there will be an odd number of hydrogens in a nitrogenous compound, and the molecular weight will be odd because of the presence of an extra hydrogen. Of course, if there are two nitrogens in a molecule, there will be two extra hydrogens, so the molecular weight will actually be even. That means the rule about molecular weight and nitrogen should really be expressed as: What about those other atoms that sometimes show up in organic chemistry, such as the halogens? Halogens all have odd molecular weights (19 amu for fluorine, 35 or 37 for chlorine, 79 or 81 for bromine, and X for iodine). However, halogens all have a valence of 1, just like hydrogen. As a result, to add a halogen to methane, we would need to erase one of the hydrogen atoms and replace it with the halogen. Since we are just substituting one odd numbered atomic weight for another, the total weight remains even. Calculate molecular weights for the following compounds. , | 1,689 | 1,502 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Stereoselectivity_in_Addition_Reactions_to_Double_Bonds/Oxidations/Hydroxylation |
Dihydroxylated products (glycols) are obtained by reaction with aqueous potassium permanganate (pH > 8) or osmium tetroxide in pyridine solution. Both reactions appear to proceed by the same mechanism (shown below); the metallocyclic intermediate may be isolated in the osmium reaction. In basic solution the purple permanganate anion is reduced to the green manganate ion, providing a nice color test for the double bond functional group. From the mechanism shown here we would expect syn-stereoselectivity in the bonding to oxygen, and regioselectivity is not an issue. When viewed in context with the previously discussed addition reactions, the hydroxylation reaction might seem implausible. Permanganate and osmium tetroxide have similar configurations, in which the metal atom occupies the center of a tetrahedral grouping of negatively charged oxygen atoms. How, then, would such a species interact with the nucleophilic pi-electrons of a double bond? A possible explanation is that an empty d-orbital of the electrophilic metal atom extends well beyond the surrounding oxygen atoms and initiates electron transfer from the double bond to the metal, in much the same fashion noted above for platinum. Back-bonding of the nucleophilic oxygens to the antibonding π*-orbital completes this interaction. The result is formation of a metallocyclic intermediate, as shown below. | 1,391 | 1,503 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/Electrophilic_Substitution_Reactions/The_Friedel-Crafts_Alkylation_of_Benzene |
This page gives you the facts and a simple, uncluttered mechanism for the electrophilic substitution reaction between benzene and chloromethane in the presence of an aluminium chloride catalyst. Any other chloroalkane would work similarly. Alkylation means substituting an alkyl group into something - in this case into a benzene ring. A hydrogen on the ring is replaced by a group like methyl or ethyl and so on. Benzene is treated with a chloroalkane (for example, chloromethane or chloroethane) in the presence of aluminum chloride as a catalyst. On this page, we will look at substituting a methyl group, but any other alkyl group could be used in the same way. Substituting a methyl group gives methylbenzene - once known as toluene. \[C_6H_6 + CH_3Cl \rightarrow C_6H_5CH_3 + HCl\] or better: The aluminium chloride isn't written into these equations because it is acting as a . If you wanted to include it, you could write AlCl over the top of the arrow. The electrophile is CH . It is formed by reaction between the chloromethane and the aluminum chloride catalyst. \[ CH_3Cl + AlCl_3 \rightarrow ^+CH_3 + AlCl_4^-\] Stage one Stage two The hydrogen is removed by the \(AlCl_4^-\) ion wh ich was formed at the same time as the \(CH_3^+\) ele ctrophile . The aluminum chloride catalyst is re-generated in this second stage. Jim Clark ( ) | 1,358 | 1,504 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Thermodynamic_Cycles/Hesss_Law |
(or just ) states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes. This law is a manifestation that enthalpy is a . Hess's Law is named after Russian Chemist and Doctor Germain Hess. Hess helped formulate the early principles of thermochemistry. His most famous paper, which was published in 1840, included his law on thermochemistry. Hess's law is due to enthalpy being a , which allows us to calculate the overall change in enthalpy by simply summing up the changes for each step of the way, until product is formed. All steps have to proceed at the same temperature and the equations for the individual steps must balance out. The principle underlying Hess's law does not just apply to Enthalpy and can be used to calculate other state functions like changes in and The heat of any reaction \(\Delta{H^°_f}\) for a specific reaction is equal to the sum of the heats of reaction for any set of reactions which in sum are equivalent to the overall reaction: (Although we have not considered the restriction, applicability of this law requires that all reactions considered proceed under similar conditions: we will consider all reactions to occur at constant pressure.) Hydrogen gas, which is of potential interest nationally as a clean fuel, can be generated by the reaction of carbon (coal) and water: \[C_{(s)} + 2 H_2O_{(g)} \rightarrow CO_{2\, (g)} + 2 H_{2\, (g)} \tag{2}\] Calorimetry reveals that this reaction requires the input of 90.1 kJ of heat for every mole of \(C_{(s)}\) consumed. By convention, when heat is absorbed during a reaction, we consider the quantity of heat to be a positive number: in chemical terms, \(q > 0\) for an endothermic reaction. When heat is evolved, the reaction is exothermic and \(q < 0\) by convention. It is interesting to ask where this input energy goes when the reaction occurs. One way to answer this question is to consider the fact that the reaction converts one fuel, \(C_{(s)}\), into another, \(H_{2(g)}\). To compare the energy available in each fuel, we can measure the heat evolved in the combustion of each fuel with one mole of oxygen gas. We observe that \[C_{(s)}+O_{2(g)} \rightarrow CO_{2(g)} \tag{3}\] produces \(393.5\, kJ\) for one mole of carbon burned; hence \(q=-393.5\, kJ\). The reaction \[2 H_{2(g)} + O_{2(g)} \rightarrow 2 H_2O_{(g)} \tag{4}\] produces 483.6 kJ for two moles of hydrogen gas burned, so q=-483.6 kJ. It is evident that more energy is available from combustion of the hydrogen fuel than from combustion of the carbon fuel, so it is not surprising that conversion of the carbon fuel to hydrogen fuel requires the input of energy. Of considerable importance is the observation that the heat input in equation [2], 90.1 kJ, is exactly equal to the difference between the heat evolved, -393.5 kJ, in the combustion of carbon and the heat evolved, -483.6 kJ, in the combustion of hydrogen. This is not a coincidence: if we take the combustion of carbon and add to it the reverse of the combustion of hydrogen, we get \[C_{(s)}+O_{2(g)} \rightarrow CO_{2(g)}\] \[2 H_2O_{(g)} \rightarrow 2 H_{2(g)} + O_{2(g)}\] \[C_{(s)} + O_{2(g)} + 2 H_2O_{(g)} \rightarrow CO_{2(g)} + 2 H_{2(g)} + O_{2(g)} \tag{5}\] Canceling the \(O_{2(g)}\) from both sides, since it is net neither a reactant nor product, equation [5] is equivalent to equation [2]. Thus, taking the combustion of carbon and "subtracting" the combustion of hydrogen (or more accurately, adding the reverse of the combustion of hydrogen) yields equation [2]. And, the heat of the combustion of carbon minus the heat of the combustion of hydrogen equals the heat of equation [2]. By studying many chemical reactions in this way, we discover that this result, known as Hess's Law, is general. A pictorial view of Hess's Law as applied to the heat of equation [2] is illustrative. In figure 1, the reactants C(s) + 2 H O(g) are placed together in a box, representing the state of the materials involved in the reaction prior to the reaction. The products CO (g) + 2 H (g) are placed together in a second box representing the state of the materials involved after the reaction. The reaction arrow connecting these boxes is labeled with the heat of this reaction. Now we take these same materials and place them in a third box containing C(s), O (g), and 2 H (g). This box is connected to the reactant and product boxes with reaction arrows, labeled by the heats of reaction in equation [3] and equation [4]. This picture of Hess's Law reveals that the heat of reaction along the "path" directly connecting the reactant state to the product state is exactly equal to the total heat of reaction along the alternative "path" connecting reactants to products via the intermediate state containing \(C_{(s)}\), \(O_{2(g)}\), and 2 \(H_{2(g)}\). A consequence of our observation of Hess's Law is therefore that the net heat evolved or absorbed during a reaction is independent of the path connecting the reactant to product (this statement is again subject to our restriction that all reactions in the alternative path must occur under constant pressure conditions). A slightly different view of figure 1 results from beginning at the reactant box and following a complete circuit through the other boxes leading back to the reactant box, summing the net heats of reaction as we go. We discover that the net heat transferred (again provided that all reactions occur under constant pressure) is exactly zero. This is a statement of the conservation of energy: the energy in the reactant state does not depend upon the processes which produced that state. Therefore, we cannot extract any energy from the reactants by a process which simply recreates the reactants. Were this not the case, we could endlessly produce unlimited quantities of energy by following the circuitous path which continually reproduces the initial reactants. By this reasoning, we can define an energy function whose value for the reactants is independent of how the reactant state was prepared. Likewise, the value of this energy function in the product state is independent of how the products are prepared. We choose this function, H, so that the change in the function, ΔH = H - H , is equal to the heat of reaction q under constant pressure conditions. H, which we call the enthalpy, is a state function, since its value depends only on the state of the materials under consideration, that is, the temperature, pressure and composition of these materials. The concept of a state function is somewhat analogous to the idea of elevation. Consider the difference in elevation between the first floor and the third floor of a building. This difference is independent of the path we choose to get from the first floor to the third floor. We can simply climb up two flights of stairs, or we can climb one flight of stairs, walk the length of the building, then walk a second flight of stairs. Or we can ride the elevator. We could even walk outside and have a crane lift us to the roof of the building, from which we climb down to the third floor. Each path produces exactly the same elevation gain, even though the distance traveled is significantly different from one path to the next. This is simply because the elevation is a "state function". Our elevation, standing on the third floor, is independent of how we got to the third floor, and the same is true of the first floor. Since the elevation thus a state function, the elevation gain is independent of the path. Now, the existence of an energy state function H is of considerable importance in calculating heats of reaction. Consider the prototypical reaction in subfigure 2.1, with reactants R being converted to products P. We wish to calculate the heat absorbed or released in this reaction, which is ΔH. Since H is a state function, we can follow any path from R to P and calculate ΔH along that path. In subfigure 2.2, we consider one such possible path, consisting of two reactions passing through an intermediate state containing all the atoms involved in the reaction, each in elemental form. This is a useful intermediate state since it can be used for any possible chemical reaction. For example, in figure 1, the atoms involved in the reaction are C, H, and O, each of which are represented in the intermediate state in elemental form. We can see in subfigure 2.2 that the ΔH for the overall reaction is now the difference between the ΔH in the formation of the products P from the elements and the ΔH in the formation of the reactants R from the elements. The ΔH values for formation of each material from the elements are thus of general utility in calculating ΔH for any reaction of interest. We therefore define the standard formation reaction for reactant R, as elements in standard state R and the heat involved in this reaction is the standard enthalpy of formation, designated by ΔH °. The subscript f, standing for "formation," indicates that the ΔH is for the reaction creating the material from the elements in standard state. The superscript ° indicates that the reactions occur under constant standard pressure conditions of 1 atm. From subfigure 2.2, we see that the heat of any reaction can be calculated from \[\Delta{H^°_f} = \Delta{H^°_{f,products}} -\Delta{H^°_{f,reactants}} \tag{6}\] Extensive tables of ΔH° values ( ) have been compiled that allows us to calculate with complete confidence the heat of reaction for any reaction of interest, even including hypothetical reactions which may be difficult to perform or impossibly slow to react. The enthalpy of a reaction does not depend on the elementary steps, but on the final state of the products and initial state of the reactants. Enthalpy is an extensive property and hence changes when the size of the sample changes. This means that the enthalpy of the reaction scales proportionally to the moles used in the reaction. For instance, in the following reaction, one can see that doubling the molar amounts simply doubles the enthalpy of the reaction. H (g) + 1/2O (g) → H O (g) ΔH° = -572 kJ 2H (g) + O (g) → 2H O (g) ΔH° = -1144kJ The sign of the reaction enthalpy changes when a process is reversed. H (g) + 1/2O (g) → H O (g) ΔH° = -572 kJ When switched: H O (g) → H (g) + 1/2O (g) ΔH° = +572 kJ Since enthalpy is a state function, it is path independent. Therefore, it does not matter what reactions one uses to obtain the final reaction. | 10,512 | 1,505 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/08%3A_Phase_Transitions_and_Equilibria_of_Pure_Substances/8.04%3A_Coexistence_Curves |
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\( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \)
\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) A coexistence curve on a pressure–temperature phase diagram shows the conditions under which two phases can coexist in equilibrium, as explained in Sec. 8.2.2. From the relation \(\pd{\mu}{T}{p}=-S\m\), we see that at constant \(p\) the slope of \(\mu\) versus \(T\) is negative since molar entropy is always positive. Furthermore, the magnitude of the slope increases on going from solid to liquid and from liquid to gas, because the molar entropies of sublimation and vaporization are positive. This difference in slope is illustrated by the curves for H\(_2\)O in Fig. 8.13(a). The triple-point pressure of H\(_2\)O is \(0.0062\br\). At a pressure of \(0.03\br\), greater than the triple-point pressure, the curves for solid and liquid intersect at a melting point (point A) and the curves for liquid and gas intersect at a boiling point (point B). From \(\pd{\mu}{p}{T}=V\m\), we see that a pressure reduction at constant temperature lowers the chemical potential of a phase. The result of a pressure reduction from \(0.03\br\) to \(0.003\br\) (below the triple-point pressure of H\(_2\)O) is a downward shift of each of the curves of Fig. 8.13(a) by a distance proportional to the molar volume of the phase. The shifts of the solid and liquid curves are too small to see (\(\Del \mu\) is only \(-0.002\units{kJ mol\(^{-1}\)}\)). Because the gas has a large molar volume, the gas curve shifts substantially to a position where it intersects with the solid curve at a sublimation point (point C). At \(0.003\br\), or any other pressure below the triple-point pressure, only a solid–gas equilibrium is possible for H\(_2\)O. The liquid phase is not stable at any pressure below the triple-point pressure, as shown by the pressure–temperature phase diagram of H\(_2\)O in Fig. 8.13(b). If we start with two coexisting phases, \(\pha\) and \(\phb\), of a pure substance and change the temperature of both phases equally without changing the pressure, the phases will no longer be in equilibrium, because their chemical potentials change unequally. In order for the phases to remain in equilibrium during the temperature change \(\dif T\) of both phases, there must be a certain simultaneous change \(\difp\) in the pressure of both phases. The changes \(\dif T\) and \(\difp\) must be such that the chemical potentials of both phases change equally so as to remain equal to one another: \(\dif\mu\aph = \dif\mu\bph\). The infinitesimal change of \(\mu\) in a phase is given by \(\dif \mu = -S\m\dif T + V\m\difp\) (Eq. 7.8.2). Thus, the two phases remain in equilibrium if \(\dif T\) and \(\difp\) satisfy the relation \begin{equation} -S\m\aph\dif T + V\m\aph\difp = -S\m\bph\dif T + V\m\bph\difp \tag{8.4.2} \end{equation} which we rearrange to \begin{equation} \frac{\difp}{\dif T} = \frac{S\m\bph-S\m\aph}{V\m\bph-V\m\aph} \tag{8.4.3} \end{equation} or \begin{gather} \s{ \frac{\difp}{\dif T} = \frac{\Delsub{trs}S}{\Delsub{trs}V} } \tag{8.4.4} \cond{(pure substance)} \end{gather} Equation 8.4.4 is one form of the , which contains no approximations. We find an alternative form by substituting \(\Delsub{trs}S = \Delsub{trs}H/T\subs{trs}\) (Eq. 8.3.5): \begin{gather} \s{ \frac{\difp}{\dif T} = \frac{\Delsub{trs}H}{T\Delsub{trs}V} } \tag{8.4.5} \cond{(pure substance)} \end{gather} Equations 8.4.4 and 8.4.5 give the slope of the coexistence curve, \(\difp/\dif T\), as a function of quantities that can be measured. For the sublimation and vaporization processes, both \(\Delsub{trs}H\) and \(\Delsub{trs}V\) are positive. Therefore, according to Eq. 8.4.5, the solid–gas and liquid–gas coexistence curves have positive slopes. For the fusion process, however, \(\Delsub{fus}H\) is positive, but \(\Delsub{fus}V\) may be positive or negative depending on the substance, so that the slope of the solid–liquid coexistence curve may be either positive or negative. The absolute value of \(\Delsub{fus}V\) is small, causing the solid–liquid coexistence curve to be relatively steep; see Fig. 8.13(b) for an example. Most substances on melting, making the slope of the solid–liquid coexistence curve positive. This is true of carbon dioxide, although in Fig. 8.2(c) the curve is so steep that it is difficult to see the slope is positive. Exceptions at ordinary pressures, substances that on melting, are H\(_2\)O, rubidium nitrate, and the elements antimony, bismuth, and gallium. The phase diagram for H\(_2\)O in Fig. 8.4 clearly shows that the coexistence curve for ice I and liquid has a negative slope due to ordinary ice being less dense than liquid water. The high-pressure forms of ice are more dense than the liquid, causing the slopes of the other solid–liquid coexistence curves to be positive. The ice VII–ice VIII coexistence curve is vertical, because these two forms of ice have identical crystal structures, except for the orientations of the H\(_2\)O molecule; therefore, within experimental uncertainty, the two forms have equal molar volumes. We may rearrange Eq. 8.4.5 to give the variation of \(p\) with \(T\) along the coexistence curve: \begin{equation} \difp = \frac{\Delsub{trs}H}{\Delsub{trs}V} \cdot \frac{\dif T}{T} \tag{8.4.6} \end{equation} Consider the transition from solid to liquid (fusion). Because of the fact that the cubic expansion coefficient and isothermal compressibility of a condensed phase are relatively small, \(\Delsub{fus}V\) is approximately constant for small changes of \(T\) and \(p\). If \(\Delsub{fus}H\) is also practically constant, integration of Eq. 8.4.6 yields the relation \begin{equation} p_2-p_1 \approx \frac{\Delsub{fus}H}{\Delsub{fus}V} \ln \frac{T_2}{T_1} \tag{8.4.7} \end{equation} or \begin{gather} \s{ T_2 \approx T_1 \exp\left[ \frac{\Delsub{fus}V(p_2-p_1)}{\Delsub{fus}H}\right] } \tag{8.4.8} \cond{(pure substance)} \end{gather} from which we may estimate the dependence of the melting point on pressure. When the gas phase of a substance coexists in equilibrium with the liquid or solid phase, and provided \(T\) and \(p\) are not close to the critical point, the molar volume of the gas is much greater than that of the condensed phase. Thus, we may write for the processes of vaporization and sublimation \begin{equation} \Delsub{vap}V = V\m\sups{g}-V\m\sups{l} \approx V\m\sups{g} \qquad \Delsub{sub}V = V\m\sups{g}-V\m\sups{s} \approx V\m\sups{g} \tag{8.4.9} \end{equation} The further approximation that the gas behaves as an ideal gas, \(V\m\sups{g} \approx RT/p\), then changes Eq. 8.4.5 to \begin{gather} \s{ \frac{\difp}{\dif T} \approx \frac{p\Delsub{trs}H}{RT^2} } \tag{8.4.10} \cond{(pure substance,} \nextcond{vaporization or sublimation)} \end{gather} Equation 8.4.10 is the . It gives an approximate expression for the slope of a liquid–gas or solid–gas coexistence curve. The expression is not valid for coexisting solid and liquid phases, or for coexisting liquid and gas phases close to the critical point. At the temperature and pressure of the triple point, it is possible to carry out all three equilibrium phase transitions of fusion, vaporization, and sublimation. When fusion is followed by vaporization, the net change is sublimation. Therefore, the molar transition enthalpies at the triple point are related by \begin{equation} \Delsub{fus}H + \Delsub{vap}H = \Delsub{sub}H \tag{8.4.11} \end{equation} Since all three of these transition enthalpies are positive, it follows that \(\Delsub{sub}H\) is greater than \(\Delsub{vap}H\) at the triple point. Therefore, according to Eq. 8.4.10, the slope of the solid–gas coexistence curve at the triple point is slightly greater than the slope of the liquid–gas coexistence curve. We divide both sides of Eq. 8.4.10 by \(p\st\) and rearrange to the form \begin{equation} \frac{\dif(p/p\st)}{p/p\st} \approx \frac{\Delsub{trs}H}{R}\cdot\frac{\dif T}{T^2} \tag{8.4.12} \end{equation} Then, using the mathematical identities \(\dif(p/p\st)/(p/p\st) = \dif\ln(p/p\st)\) and \(\dif T/T^2 = -\dif(1/T)\), we can write Eq. 8.4.12 in three alternative forms: \begin{gather} \s{ \frac{\dif\ln (p/p\st)}{\dif T} \approx \frac{\Delsub{trs}H}{RT^2} } \tag{8.4.13} \cond{(pure substance,} \nextcond{vaporization or sublimation)} \end{gather} \begin{gather} \s{ \dif\ln (p/p\st) \approx -\frac{\Delsub{trs}H}{R}\dif (1/T) } \tag{8.4.14} \cond{(pure substance,} \nextcond{vaporization or sublimation)} \end{gather} \begin{gather} \s{ \frac{\dif\ln (p/p\st)}{\dif (1/T)} \approx -\frac{\Delsub{trs}H}{R} } \tag{8.4.15} \cond{(pure substance,} \nextcond{vaporization or sublimation)} \end{gather} Equation 8.4.15 shows that the curve of a plot of \(\ln(p/p\st)\) versus \(1/T\) (where \(p\) is the vapor pressure of a pure liquid or solid) has a slope at each temperature equal, usually to a high degree of accuracy, to \(-\Delsub{vap}H/R\) or \(-\Delsub{sub}H/R\) at that temperature. This kind of plot provides an alternative to calorimetry for evaluating molar enthalpies of vaporization and sublimation. If we use the recommended standard pressure of \(1\br\), the ratio \(p/p\st\) appearing in these equations becomes \(p/\tx{bar}\). That is, \(p/p\st\) is simply the numerical value of \(p\) when \(p\) is expressed in bars. For the purpose of using Eq. 8.4.15 to evaluate \(\Delsub{trs}H\), we can replace \(p\st\) by any convenient value. Thus, the curves of plots of \(\ln(p/\tx{bar})\) versus \(1/T\), \(\ln(p/\tx{Pa})\) versus \(1/T\), and \(\ln(p/\tx{Torr})\) versus \(1/T\) using the same temperature and pressure data all have the same slope (but different intercepts) and yield the same value of \(\Delsub{trs}H\). If we assume \(\Delsub{vap}H\) or \(\Delsub{sub}H\) is essentially constant in a temperature range, we may integrate Eq. 8.4.14 from an initial to a final state along the coexistence curve to obtain \begin{gather} \s{ \ln\frac{p_2}{p_1} \approx -\frac{\Delsub{trs}H}{R} \left( \frac{1}{T_2}-\frac{1}{T_1} \right) } \tag{8.4.16} \cond{(pure substance,} \nextcond{vaporization or sublimation)} \end{gather} Equation 8.4.16 allows us to estimate any one of the quantities \(p_1\), \(p_2\), \(T_1\), \(T_2\), or \(\Delsub{trs}H\), given values of the other four. | 17,547 | 1,506 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/17%3A_Boltzmann_Factor_and_Partition_Functions/17.04%3A_Heat_Capacity_at_Constant_Volume |
The heat capacity at constant volume (\(C_V\)) is defined to be the change in internal energy with respect to temperature: \[C_V = \left( \dfrac{\partial U}{\partial T} \right)_{N, V} \label{Eq3.26} \] Since: \[E = -\dfrac{\partial \ln{Q(N, V, \beta)}}{\partial \beta} \label{Eq3.27} \] We see that: \[\begin{align} C_V &= \dfrac{\partial U}{\partial T} \nonumber \\[4pt] &= \dfrac{\partial U}{\partial \beta} \dfrac{\partial \beta}{\partial T} \nonumber \\[4pt] &= \dfrac{1}{kT^2} \dfrac{\partial^2}{\partial \beta^2} \: \ln Q(N, V, \beta) \nonumber\\[4pt] &= k \beta^2 \dfrac{\partial^2}{\partial \beta^2} \: \ln Q(N, V, \beta) \label{Eq3.28} \end{align} \] where \(k\) is the Boltzmann constant. Energy can be stored in materials/gases via populating the specific degrees of freedom that exist in the sample. Understanding how this occurs requires the usage of Quantum Mechanics. Pierre Louis Dulong and Alexis Thèrèse Petit conducted experiments in 1819 on three dimensional solid crystals to determine the heat capacities of a variety of these solids (Heat capacity is the solids ability to absorb and retain heat). Dulong and Petit discovered that all investigated solids had a heat capacity of approximately 2.49 x 10 J kilomole K at around 298 K or room temperature. The result from their experiment was explained by considering every atom inside the solid as an oscillator with six degrees of freedom (an oscillator can be thought of as a spring connecting all the atoms in the solid lattice). These springs extend into three dimensional space. The more energy that is added to the solid the more these springs vibrate. Each atom has an energy of \(\frac{1}{2}kT\), where \(k\) is the Boltzmann constant and \(T\) is the absolute temperature. Thus, \[ C_V=\dfrac{6R}{2}=3R \label{1}\tag{Dulong-Petit Law} \] The number 6 in this equation is the number of degrees of freedom for the molecule. Petit and Dulong suggested that these results supported their foundation for the heat capacity of solids. The explanation for Petit and Dulong's experiment was not sufficient when it was discovered that heat capacity decreased as temperature approached absolute zero. The degrees of freedom do not slow down or cease to move when the solid reaches a sufficiently cold temperature. An additional model was proposed to explain this deviance. Two main theories were developed to explain this puzzling deviance in the heat capacity experiments. The first was constructed by Einstein and the second was authored by Debye. Einstein assumed three things when he investigated the heat capacity of solids. First, he assumed that each solid was composed of a lattice structure consisting of \(N\) atoms. Each atom was treated as moving independently in three dimensions within the lattice (3 degrees of freedom). This meant that the entire lattice's vibrational motion could be described by a total of \(3N\) motions, or degrees of freedom. Secondly, he assumed that the atoms inside the solid lattice did not interact with each other and thirdly, all of the atoms inside the solid vibrated at the same frequency. The third point highlights the main difference in Einstein's and Debye's two models. Einstein's first point is accurate because the experimental data supported his hypothesis, however his second point is not because if atoms inside a solid could not interact, sound could not propagate through it. For example, a tuning fork's atoms, when struck, interact with one another to create sound which travels through air to the listener's ear. Atoms also interact in a solid when they are heated. Take for example a frying pan. If the pan is heated on one side, the heat transfers throughout the metal effectively warming the entire pan. Molecules that make up the frying pan interact to transfer heat. Much in the same way the oscillators in a solid interact when energy is added to the system. The extent of these interactions lead to the physically observed heat capacity. The heat capacity of a solid at a constant volume is \[ \begin{align*} C_V &= \left(\dfrac{\partial{U}}{\partial{T}}\right)_V \\[4pt] &=3Nk\left(\dfrac{\theta_E}{T}\right)^2 \label{2} \\[4pt] &= \dfrac{\exp \left(\dfrac{\theta_E}{T} \right)}{\left(\exp \dfrac{\theta_E}{T} -1\right)^2} \end{align*} \] where The Einstein temperature's accessibility of the vibrational energy inside of a solid molecule determines the heat capacity of that solid. The greater the accessibility the greater the heat capacity. If the vibrational energy is easily accessible the collisions in the molecule have a greater probability of exciting the atom into an upper vibrational level. This is displayed below. So the Einstein temperature specifically indicates the probability that a molecule has in its degrees of freedom to store energy in its atomic oscillators (or bonds). Comparing the Einstein temperature to the traditional classical values of Heat capacity will illustrate the differences the specific strengths (high temperature) and weaknesses (low temperature) of the Einstein model. Thus, in the high temperature limit (\(\dfrac{\theta_E}{T}\ll 1\)) (i.e., the temperature is very large compared to the Einstein temperature) then \[ C_V \approx 3Nk=3nR. \label{4} \] Einstein's model reveals the accuracy of the Petit and Dulong model and models high temperatures accurately. However, just as Petit and Dulong's model decreased in accuracy as the temperature decreased, so followed Einstein's. When examining the extremely low temperature limit: \(\dfrac{\theta_E}{T}\gg1\), it can be seen: \[ C_V=3Nk\left(\dfrac{\theta_E}{T}\right)^2e^\dfrac{-\theta_E}{T} \label{5} \] As temperature (\(T\)) goes to zero, the exponential portion of the above equation goes to zero and therefore \(C_V\) also approaches zero. This supports the experimental values as temperature approaches zero the heat capacity of the solid likewise decreases to zero. Einstein's theory also explains solids that exhibit a low heat capacity even at relatively high temperatures. An example of such a solid is diamond. The heat capacity of diamond approaches \(3Nk\) as temperature greatly increases. Einstein's model supports this through the definition of an Einstein temperature. As the Einstein temperature increases, \(\nu\) must increase likewise. This is the equivalent of each atom possessing more energy and therefore vibrating more rapidly within the solid itself. The oscillator frequency, \(\nu\), can be written as: \[ \nu=\dfrac{1}{2\pi}\sqrt{\dfrac{\kappa}{\mu}} \label{6} \] where \(\kappa\) is the force constant and \(\mu\) is the reduced mass. This formula better predicts solids with high force constants or low reduced masses. This corrects deviations from the Petit and Dulong model. Essentially the Einstein temperature allows for the heat capacity equation and the vibrational frequencies in the solid to change as the temperature changes. This effectively adjusts for the deviations seen in the Petit/Dulong model. As the temperature increases or decreases, the Einstein temperature increases or decreases likewise to mirror the actual physical activity within the solid. Einstein's model predicts relatively low temperatures well. However, when decreasing from approximately 15 K, Einstein's model deviates from experimental values. Also, this can be observed, although not as dramatically, for temperatures from 25 K to 30 K. Clearly a term or correction is still missing from Einstein's model to increase its accuracy. | 7,505 | 1,507 |
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Every molecule has a point group associated with it, which are assigned by a set for rules (explained by Group theory). The takes the point group and represents all of the symmetry that the molecule has. Symbols under the first column of the character tables anti-symmetric with respect to the inverse Symbols in the first row of the character tables The order is the number in front of the the classes. If there is not number then it is considered to be one. The number of classes is the representation of symmetries.The D h has six classes and an order of twelve. The identity does nothing to the matrix. [1 0 0] [X] [X] [0 1 0] [Y] = [Y] [0 0 1] [Z] [Z] σ the x and y stay positive, while z turns into a negative. [1 0 0] [X] [X] [0 1 0] [Y] = [Y] [0 0 -1] [Z] [-Z] Inversion (I) is when all of the matrix turns into a negative. [-1 0 0] [X] [-X] [0 -1 0] [Y] = [-Y] [0 0 -1] [Z] [-Z] C is when one would use cos and sin. for an example C [cos (2 /4 -sin (2 /4 0] [X] [] [sin (2 /4) cos (2 /4) 0] [Y] = [] [0 0 1] [Z] [] There are two columns on the far right. One is ir and the other is Raman. Try moving the molecule around using reflections and rotations. Remember when the positive side of the orbitals goes into the negative side, the number is negative (in the character tables). Also, remember if it is moved or reflected and there is no change then the number is positive (in the character tables). also, molecules are placed on a x, y, and z (three dimensional) graph. The highest C fold rotation is always on the z axis. | 1,554 | 1,509 |
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\[yY + cC \rightleftharpoons cC + zZ\] ts \(K_c\), \(K_{\chi}\), and \(K_p\) with res \[K_c = \dfrac{[Z]^z}{[Y]^y}\] \[K_{\chi} = \dfrac{\chi_Z^z}{\chi_Y^y}\] \[K_p = \dfrac{P_Z^z}{P_Y^y}\] \(K_p\) \(K_c\) \(K_{\chi}\) \[CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(s)}\] The decomposition of sodium chlorate is: \[2NaClO_{3(s)} \rightleftharpoons 2NaCl_{(s)} + 3O_{2(g)}\] Suppose 0.760 mole of sodium chlorate was to be decomposed in a 3.75 L container. At the temperature 336 C, the percent decomposition of NaClO is 3.5%. Find the Hint: Use partial pressure and look at a) number of moles of NaClO decomposed is: \[ \dfrac{ \# moles \, decomposed}{.760 moles} = 0.035 \Rightarrow \# mole\,decomposed = 0.0266 mole \,NaClO_{3} \] number of moles of O formed is: \[ 0.0266 mol \,NaClO_{3} \times \dfrac{3 mol O_{3}}{2 mol \,NaClO_{3}} = 0.0399 mol \,O_{2} \] \[ P_{O_{2}} V = nRT \Rightarrow P_{O_{2}} = \dfrac{nRT}{V} \] \[ P_{O_{2}} = \dfrac{(0.0399 mol)(0.08206 \dfrac{L \cdot K}{K \cdot mol})(336+273)K}{3.75 L} = {\color{red} 0.532 atm} \] b) \[K_{p} = \dfrac{P_{O_{2}}}{P^{\circ}} = \dfrac{0.532 atm \times \dfrac{1.013 bar}{1atm}}{1bar} = {\color{red} 0.539 } \] Using the decomposition reaction of hydrogen peroxide (H O ), assume that at 1048 °C and the pressure of the oxygen gas (O ) is 1.5 bar. \[\ce{2H2O2 (aq) \rightleftharpoons 2H_2O (l) + O2 (g) }\] a. b. In order to calculate the number of moles of H O that are decomposed, you first need to determine the moles of O formed by the reaction. Treat O as an ideal gas. c. If 1.3 moles of H O were used instead of 0.92 mole, the pressure of the O would not be affected and would remain at 1.5 bar. The number of moles of H O decomposed would still be 0.0802 mole, therefore, the fraction of H O decomposed would be as follows: d. The pressure of O must be greater than or equal to 1.5 bar in order for equilibrium to take place, therefore, the number of moles of H O cannot be less than 0.0802 mol. Find the value of K for cellular respiration if \(P_{O_{2}}=350torr\), with a 3:2 ratio to the pressure of CO \[C_6H_{12}O_6(s)+O_2(g)\rightleftharpoons CO_2(g)+H_2O(l)\] \[P_{CO_{2}}=(350torr)\dfrac{2}{3}=233.3torr\] \[K_p=\dfrac{\dfrac{233.3torr}{750torr}}{\dfrac{350torr}{750torr}}=0.666\] For the reaction of \[\ce{N2O4 → 2 NO2}\] with K = 0.167 at 300 K. If 1.0 g of N O is placed into a 250.0 mL container: Calculate \(\Delta_{vap}H\) for the evaporation of methanol when the temperature is raised from 20 °C to 100 °C and if the \(K_2/K_1\) ratio is 22.14? \[CH_3OH_{(l)} \rightarrow CH_3OH_{(g)}\] Using the decomposition reaction of coppr (II) oxide below: \[4CuO_{(s)} \rightleftharpoons 2Cu_2O_{(s)} + O_{2(g)}\] determine the standard enthalpy of the reaction. (Note: The equilibrium vapor pressures of O are 15.4 mmHg at 600°C and 927 mmHg at 850°C). In this case, we must use the : \[ \ln \dfrac{K_2}{K_1} = \dfrac{\Delta_ H^o}{R} \left( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right)\] in order to solve this particular problem. We can conclude that is proportional to due to the defined relationship of: . Therefore, A container of water at 20 C was placed in a freezer that was at a temperature of -5.0 C. The vapor pressure of water in the container went from 0.60 bar to 0.38 bar. Calculate the enthalpy of fusion for the reaction that took place. \[\ln\dfrac{k_2}{k_1}= \dfrac{\Delta_r H^\circ }{R}\dfrac{T_2-T_1}{T_2T_1}\] \[\rightarrow \Delta_r H^\circ = \ln\dfrac{k_2}{k_1}(R)\dfrac{T_2T_1}{T_2-T_1}\] \[ \ln\dfrac{0.38 bar}{0.60 bar} 8.314 \dfrac{J}{mol\cdot K} \dfrac{(293\; K)(268\; K)}{268\; K-293\; K}(\dfrac{1\;kJ}{1000\; J})= 11.9 \dfrac{kJ}{mol}\] The chemical responsible for the brown air throughout the Los Angeles area is NO (g). To learn more about NO (g), you decide to study this pollutant spectroscopically (by light absorption). You fill a gas cell with N O , equilibrate the temperature to 298.0 K and then open the stopcock on the cell to equilibrate the pressure to the barometric pressure (723.4 mm Hg) that day. You then reequilibrate the cell at 323.1 K, 348.0 K, and 372.9 K. The following data is obtained: Find K , Δ G, H, and S at each temperature for the reaction \[\ce{N2O4(g) ⇌ 2 NO2(g)}.\] What is the equilibrium constant and standard Gibbs energy change for: \[CO_{(g)} \rightleftharpoons CO_{(g)} + ½O_{2(g)}\] Assume that when \(CO_{(g)}\) dissociates into \(CO_{(g)}\) and \(O_{2(g)}\) at 300 K and 1.5 atm, the overall dissociation is one two fifths complete. The standard Gibbs energies of formation of proban-1-ol and proban-2-ol are -171.3 kJ mol and -180.3 kJ mol , respectively. Find the ratio of equilibrium vapor pressures of each isomer at 300 K. The ratio of equilibrium pressure is : 171.3/180.3 = 0.95 At what temperature does a particular reaction favor the formation of products at equilibrium if Δ H = 215.7 kJ mol and Δ S = 348.8 J mol K ? The reaction favors the formation of products at equilibrium when \[\Delta_{r} G^{\circ} = \Delta_{r} H^{\circ}-T\Delta_{r} S^{\circ}<0\] \[\Delta_{r} H^{\circ}-T\Delta_{r} S^{\circ} = 215.7 \ast 10^{3} J\ mol^{-1} - T(348.8 J\ mol^{-1}\ K^{-1})<0\] \[T>\dfrac{215.7 \ast 10^{3} J\ mol^{-1}}{348.8 J\ mol^{-1}\ K^{-1}}\] \[T > 616 K\] Using a table of thermodynamic data, find Ksp at 298.15K for: \[CO_{(g)} \rightarrow CO_{(g)} + ½O_{2(g)}\] (Hint, this problem can be solved using the van’t Hoff equation in its integrated form) Consider the the : \[N_{2(g)} + 3H_{2 (g)} \rightleftharpoons 2NH_{3 (g)}\] How does the equilibrium shift under the following changes Given the general reaction A( ) ⇔2B( ), calculate the degree of dissociation of A at 25 C and 7.00 bar if Δ G = 6.76 kJ mol . According to Le Chatelier's Principal, in what direction should this reaction proceed? \[\Delta_{r} G^{\circ} = -RTlnK_{P}\] \[K_{P} = e^{\dfrac{\Delta_{r} G^{\circ}}{-RT}} = e^{\dfrac{6.76\ast 10^{3}J\ mol^{-1}}{-(8.314J\ mol^{-1}\ K^{-1})298K}}\] \[K_{P} = 0.0653\] \[K_{P} = \dfrac{4\alpha ^{2}}{1-\alpha ^{2}}P\] \[\dfrac{4\alpha ^{2}}{1-\alpha ^{2}} = \dfrac{0.0653}{7}\] \[\alpha = .048\] Degree of dissociation α = 0.048, the reaction proceeds 4.8%. Le Chatelier's Principal says that reactions will move towards the side with fewer moles of gas at high pressures. How does Le Chatelier's Principle relate to the following equation? Le Chatelier's Principle basically states that a system will adjust itself in efforts to re-establish equilibrium when an outside stress is placed on it. Using the equation provided, we can conclude that raising the temperature causes the equilibrium to shift from left to right in an endothermic reaction. This suggests that it is favoring the formation of products. We can also conclude that the opposite is true. This conclusion reinforces Le Chatelier's Principle because the temperature acts as the external stress placed on the system in this case. H and H molecules are at equilibrium pressures of .35 bar and .30 bar, respectively. If the size of the container they are in is reduced by a factor of two, what will be the new partial pressures? \[H_2(g)\rightleftharpoons 2H(g)\] \[K_p=\dfrac{P_H^2}{P_{H_{2}}}=\dfrac{.30^2}{.35}=.675\] At new volume: \[P_H=.60\] \[P_{H_{2}}=.70\] Pressures will increase with decreased volume. Less molecules of gas will be favored. \[.675=\dfrac{.60-(2x^2)}{.70+x}\rightarrow 4x^2 +.675x+.113\rightarrow x=.272\] \[P_{H_{2}}=.70bar+.272bar=.972bar\] \[P_H=.60bar-(2).272bar=.056bar\] With Le Chatlier's principle in mind, does raising the temperature favor the forward reaction or the reverse? This reaction favors the reverse reaction. When a gas was heated at atmospheric pressure and 25°C, its color deepened. Heating above 150°C caused the color to fade, and at 550°C the color was barely detectable. At 550°C, however, the color was partially restored by increasing the pressure of the system. Which of the following scenarios best fits the above description? Justify your answer. (Hint: Gas B is bluish, and Gas Y is yellow. The other gases are colorless. Gas A and B are in their natural state, Gas AB has a Δ H° = -43.2 kJ / mol. The reaction of Gas X into Gas Y is endothermic.) Hydrogen gas and iodine react at equilibrium in a glass canister to form Hydrogen Iodide, a strong acid: \[H_{2(g)}+I_{2(g)}\rightleftharpoons HI_{(g)}\] Iodine gas is a deep purple color. Both Hydrogen Iodide and hydrogen gas are colorless. Assume that iodine sublimates readily at 37 C, and that the reaction is endothermic in the direction written. What color is the gas mixture in the canister in the following scenarios? Provide an explanation for each. Hint: At room temperature, the gas is colorless. Solid iodine must be in gas phase to react. a.) The color of the gas will be a deep purple. The iodine will have sublimated, but the temperature is not high enough to drive the reaction forward. b.) The gas will be colorless, or almost completely so. This is because an increase in the temperature in an endothermic reaction drives the equilibrium constant higher. This can be justified using , which states that added stress to an equilibrium will be offset by the system. \[H_{2(g)}+I_{2(g)}+heat\rightleftharpoons HI_{(g)}\] Thus, when heat is added, the system will compensate by driving the reactants of the reaction forward into products. Another way to assess this is by the altered form of the : \[lnK=-\dfrac{\Delta _{r}H^{\circ}}{RT}+\dfrac{\Delta _{r}S^{\circ}}{R}\] Assuming that neither change in entropy nor enthalpy changes due to change in temperature, and that in an endothermic reaction the enthalpy is positive, as the temperature increases, so does K. c.) The gas will become slightly purple. This is because a decrease in pressure will alter K and drive the reaction in the direction which produces more moles of gas. By this, Le Chatelier's principle once again holds. \[Mg + Pb^{2+} \rightarrow Mg^{2+}+Pb\] Write the two separate reactions happening here. Label which one is happening at the cathode end and which is happening at the anode end. Cathode: \[Pb^{2+}+2e^- \rightarrow Pb\] Anode: \[Mg \rightarrow Mg^{2+} + 2e^-\] The reaction of N (g) and H (g) gas produces NH (g). This reaction is exothermic explain what happens when you increase the temperature of the reaction. What happens when you increase the pressure? Since this reaction is exothermic heat is produced. Using Le Chatelier's principle we see that an increase in the temperature of the reaction will drive the reaction backward since heat is already in the product side. If we increase pressure we decrease the volume of the reaction therefore by Le Chatelier's principle an increase in the pressure should drive the reaction forward. At places such as high mountain, the air pressure is lower than 1 atm, resulting lower partial pressure of Oxygen. What would you expect for the concentration of hemoglobin for the people living at such places? Show the steps to get from \[Y = \dfrac{[PL]}{[L] + [P]}\] to \[\dfrac{1}{Y} = 1 + \dfrac{K_d}{L}\] After an experiment of protein-binding you find the data respectively Total \(\dfrac{Mg^{2+}}{\mu M}\): 60 120 180 240 480 \(\dfrac{Mg^{2+} bound to protein}{\mu M}\): 33.8 120 180 240 480 Determine the dissociation constant of \(Ca^{2+}\) graphically. The protein concentration was kept at 96 \(\mu M\) for each run. Solution: \[54.2\mu M\,pH=\,pKa\] \[54.2\mu M(\dfrac{1\times10^{6}M}{1\mu M})=5.42\times10^{-5}\] \[Ka=10^{-5.42E-5}\] \[Ka=.9998\] The dissociation constant for the following reaction is 3.2x10 . Dissolve 0.02M of C H O in water, calculate the molarity of reactants at equilibrium. C H O H O <---> H O + C H O The reaction \[ \begin{align} &Glyceraldehyde \ 3-phosphate + NAD^{+} + HPO_{4}^{2-} \rightarrow 1,3-Biphosphoglycerate + NADH + H^{+} \\ & \triangle_rG^{\circ'}=6.3 \ kJ \ mol^{-1} \end{align}\] is catalyzed by GAPDH (Glyceraldehyde 3-phosphate Dehydrogenase). At 298 K, predict whether or not the reaction will be spontaneous with the following information: Use the following equation. Calculate [H ] \[ [H^{+}] = 10^{-7.5} = 3.162 X 10^{-8} M \\ since \ pH=-log([H^{+}])\] Plug values into equation and solve. Note: Since there is no coefficient in front of H in the reaction, x=1. \[\triangle_rG = 6.3 \ kJ mol^{-1} + (\dfrac{8.3145}{1000}kJ/mol^{-1} K^{-1}) (298K) \ln\dfrac{(3 \cdot 10^{-3} \ M)(3.162 \cdot 10^{-8} \ M/(1 \cdot 10^{-7})]^{1} \left(1.0 \cdot 10^{-4} \ M \right )}{ \left(1.5 \cdot 10^{-5} \ M \right ) \left(1.2 \cdot 10^{-5} \ M \right ) \left(1.2 \cdot 10^{-5} \ M \right )}\] \[ \underline {\triangle_rG = 49.90 \ kJ \ mol^{-1}}\] for some more information on bioenergetics and free energy. The reaction: \[Glucose+ATP\rightleftharpoons Glucose 6-phosphate + ADP\] At 298K, the equilibrium constant for the reaction is \(3.7 \times 10^{-3}\).Will the reaction occur spontaneously if the reaction is at the following concentrations: [Glucose]=\(3.2 \times 10^{-4}M\), [ATP]=\(2.5 \times 10^{-3}M\), [G-6-P]=\(1.2 \times 10^{-5}M\), [ADP]=\(1.0 \times 10^{-5}M\). \(\begin{align*}\Delta G^\circ &=-RTln(K_{eq})=-(8.314J/K*mol)(298K)ln(3.7 \times 10^{-3})\\&=13872.97J/mol=13.87297kJ/mol\end{align*}\) \(\begin{align*}
\Delta G &=\Delta G^\circ +RTln(\dfrac{[Product]}{[Reactant]})\\&=13.87297kJ/mol + (8.314 \times 10^{-3}kJ/K*mol)(298K)ln(\dfrac{[1.2 \times 10^{-5},1.0 \times 10^{-5}]}{[3.2 \times 10^{-4},2.5 \times 10^{-3}]})\\&=-7.942kJ/mol
\end{align*}\) \(\Delta G\) is negative, so the reaction is spontaneous at the given concentrations. The established standard Gibbs energy for hydrolysis of ATP to ADP at 310K is \(-30.5kj\,mol^{-1}\). At \(-4.6^oC\), determine the \(\Delta rG^{o'}\) in the process of the muscle of a hippo. (Hint: \(\Delta rH^{o'}=-20.1kjmol^{-1}\) Step 1: \[\Delta rG^{o'}=\Delta rH^{o'}-T\Delta rS^{o'}\] Step 2: \[\Delta rS^{o'}=\dfrac{\Delta rH^{o'}-\Delta rG^{o'}}{T}$$
$$=\dfrac{(-20.1kj\,mol^{-1})-(-3.0kj\,mol^{-1})}{310K}=3.355\times10^{-2}kj\,K^{-1}\,mol^{-1}\] Step 3: \[\Delta rG^{o'}=\Delta rH^{o'}-T\Delta rS^{o'}\] \[=(-20.1kj\,mol^{-1})-(298.15K)(3.355\times10^{-2}kj\,K^{-1}\,mol^{-1})\] Which step of glycolysis would not occur spontaneously at standard-state conditions and why? Consider a hydrolysis of PEP, a phosphate compound. PEP + H O → pyruvate + P Δ G°' = -61.9 kJ/mol At the mperature of 288K, the following reaction took place and has a reaction Gibbs free energy of -49.5 kJ/mole. Find the concentration of PEP for the reaction if the other concentrations are: [Pi] = 3.54 x 10 M, [Pyruvate] = 1.85 x 10 M. Hint: Use equation Δ G° = -RTlnK \[\Delta _{r} G = \Delta _{r} G^{\circ '} + RTln\dfrac{[pyruvate,P_i]}{[PEP]}\] \[-49500 \dfrac{J}{mol} = -61900 \dfrac{J}{mol} + (8.314 \dfrac{J}{K \cdot mol})(288K)ln\dfrac{(1.85 \times 10^{-2}M)(3.54 \times 10^{-1}M)}{[PEP]}\] \[ {\color{red} [PEP] = 3.69 \times 10^{-5} M} \] For the following reaction: \[Fructose 1,6-bisphosphate\rightleftharpoons dihydroxyacetone phosphate + glyceraldehyde 3-phosphate\] The \(\Delta G°'=5.7kcal/mol\). Calculate the equilibrium constant and determine if the reaction is spontaneous or not at 310K. \(\Delta G°'=5.7kcal/mol=23.8488kJ/mol \) \(\Delta G°'=-RTln(k_{eq}')\\k_{eq}'=e^{\dfrac{\Delta G°'}{-RT}}=e^{\dfrac{23848.8J/mol}{(-8.314J/K*mol)(310K)}}=9.58 \times 10^{-5}\) Small \(k_{eq}'\) suggest that this reaction is not a spontaneous process under the given conditions. Consider the following reaction: Glucose + Fructose → Sucrose + H O Find: a) The standard free energy = (-1544.3)- (-908.9 + -875.9) = 240.2 kJ/mol b)delta G at 300 K = delta G + RTln Q = -34.6 kJ/mol delta G at 333 K = delta G + RTln Q = -26.8 kJ/mol Consider the formation of the dipeptide glycylclycine. Using the following information, calculate the \[ \begin{align} & 2 \ Glycine \rightarrow Glycylglycine + H_2O \\ & \triangle _{r}G^{\circ ' }=29.5 \ kJ \ mol^{-1} \\ & \left [ Glycine \right ]= 1.4M\\ & \left [ Glycylglycine \right ]= 0.7M \\ & \left [ H_2O \right ]= 1.0M \end{align}\] Use the following equation and plug in values. \[ \begin{align} & \triangle_rG^{\circ} = \triangle_rG^{\circ '} + RT\ln(K) \\ & \triangle_rG^{\circ} = \triangle_rG^{\circ '} + RT\ln\dfrac{[Glycylglycine]\cdot [H_2O]}{[Glycine]^{2}} \\ & \triangle_rG^{\circ} = 29.5 \ kJ \ mol^{-1} + (\dfrac{8.3145}{1000} \ kJ \ mol^{-1} \ K^{-1})(298K)\ln\dfrac{0.7M\cdot 1.0M}{1.4^{2} \ M} \\ & \underline{\triangle_rG^{\circ} = 26.7 \ kJ \ mol^{-1}} \end{align}\] for some more information on bioenergetics and free energy. Consider the following reaction: \[2NH_{3(g)}+H_{2}O_{(l)}\rightleftharpoons 2NH^{+}_{4(aq)}\] \[\Delta _{f}\bar{G}^{\circ}(NH_{3(g)})=-16.6 \dfrac{KJ}{mol}\] \[\Delta _{f}\bar{G}^{\circ}(H_{2}O_{(l)})=-237.2 \dfrac{KJ}{mol}\] \[\Delta _{f}\bar{G}^{\circ}(NH^{+}_{4(aq)})=-79.3 \dfrac{KJ}{mol}\] What is the equilibrium constant of this process? Is this process typically spontaneous process under standard conditions? To calculate change in Gibbs free energy for the equation, we use the standard molar Gibbs energies of formation for the reactants and products: \[\Delta _{r}G = 2\Delta _{f}\bar{G}^{\circ}(NH^{+}_{4(aq)})-\Delta _{f}\bar{G}^{\circ}(H_{2}O_{(l)})-2\Delta _{f}\bar{G}^{\circ}(NH_{3(g)})\] \[=2\left(-79.3\dfrac{kJ}{mol}\right)-\left(-267.2\dfrac{kJ}{mol}\right)-2\left(-16.6\dfrac{kJ}{mol}\right)\] \[=141.8 \dfrac{kJ}{mol}\] To calculate the equilibrium constant: \[\ln K=-\dfrac{\Delta _{r}G }{RT}=-\dfrac{141800\dfrac{J}{mol}}{(8.3145\dfrac{J}{K\cdot mol})(298K)}=-57.23\] \[K=e^{-57.23}=1.40 \times 10^{-25}\] Because the K value is so low for this process, the reaction does not occur spontaneously at 25 C Calculate the Δ G and equilibrium constant for the following reaction at 298K: 3-Phosphoglycerate → Phosphoenolpyruvate +H O Given that: 2-Phosphoglycerate → 3-Phosphoglycerate Δ G = -4.2 kJ mol 2-Phosphoglycerate → Phosphoenolpyruvate +H O Δ G = -16.4 kJ mol Δ G = 4.2 kJ mol - 16.4 kJ mol = 12.2 kJ mol \[\Delta _{r}G^{\circ} = -RTlnK\] \[lnK = \dfrac{12.2 \ast 10^{3} J\ mol^{-1}}{-(8.314J\ mol^{-1}\ K^{-1})298K}\] \[K = e^{.00492}\] \[K = 1.00\] Suppose the isomerization of DHAP to GAP in glycolysis has an enthalpy of -1.20 kJ/mol. At 25°C the Gibbs free energy of the reaction is 1.98 kJ/mol. Determine the equilibrium constant K of the isomerization at 25°C and at 35°C Hint: Use equation \[1980 \dfrac{J}{mol} = -(8.314 \dfrac{J}{K \cdot})(25+273)K (lnK_{25^{\circ}}) \Rightarrow {\color{red}K_{25^{\circ}} = 0.450}\] Steady state and equilibrium state have an important role in understanding enzyme kinetics. What are some significant differences between the two? In a steady state, there is no net change over time in concentrations of reactants and products of a reaction since they are being produced and consumed at constant rates. In this sense a steady state is a dynamic equilibrium. A steady state can be going in either the forward reaction or the backward reaction. A chemical equilibrium on the other hand is when a reaction goes in the forward and backward reaction at the same rate so there is no net change in the system. This is important because cells maintain steady states so that they are able to use particular reactions continuously. If a cell were at chemical equilibrium it would be dead because it would be at the point where all reactions are not going anymore, among other reasons. Also, it is important to note that in a steady state, reactions are reversible compared to chemical equilibrium where the reaction rate is zero. to see more information about steady states. Is the following an example of a , , or neither: 1.45 The reaction in question is H + I ⇌ 2HI P /(P P ), the activity equilibrium constant is given by γ /(γ γ ), and the thermodynamic equilibrium constant is given by the product of those two. K = (980 mmhg) /(812 mmhg * 587 mmhg) = 2.01 K = 1.23 /(1.45 0.844) = 1.24 K = 1.24 * 2.01 = 2.49 \[SO_2 + Cl_2 \rightarrow SO_2Cl_2 \] This reaction happens at 273K. You are given a Kp of 0.683. The pressure for SO is 0.58 bar for Cl is 0.93 bar, and SO Cl is 0.776 bar. From this information, determine (delta) G. =(-8.314Jmol K )(273K)ln(0.683) =865.4 J mol =1.69 kj mol Assuming oxygen binding to hemoglobin can be represented by the following reaction: \[Hb_{(aq)} + O_{2(g)} \rightarrow HbO_{2(aq)}\] If the value of Δ G° for the reaction is -11.2 kJ mol at 37°C, calculate the value of Δ G° for the reaction. At T=300K, given the mole ratio between 2 isomers Cis-2-butene and Trans-2-butene in an equilibrium mixture is 1:4. Evaluate Cis-2-butene <------> Trans-2-butene When discussing the reaction in biological cells, why would you use concentrations instead of activities? Concentrations are generally smaller and it's easier to compare the concentrations between two parts of a cell or between different molecules in a reaction whereas activity describes behavior. The following data shows the oxygen binding concentration in snails. The protein concentration is 15mM. Find \(n\) and \(K_d\) by using the \(y=mx+b\\
\dfrac{Y}{[L]}=(-\dfrac{1}{K_d})Y+\dfrac{n}{K_d}\\
Y=\dfrac{[O_2]_{bound}}{[P]}\\.[L]=[O_2]_{free}\) [O2]tot \(-\dfrac{1}{K_d}=-0.102\\
K_d=9.804\\
\dfrac{n}{K_d}=0.564\\
n=K_d*0.564=5.529\) | 21,313 | 1,510 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Energies_and_Potentials/Enthalpy/Kirchhoff_Law |
Kirchhoff's Law describes the enthalpy of a reaction's variation with temperature changes. In general, enthalpy of any substance increases with temperature, which means both the products and the reactants' enthalpies increase. The overall enthalpy of the reaction will change if the increase in the enthalpy of products and reactants is different. At constant pressure, the heat capacity is equal to change in enthalpy divided by the change in temperature. \[ c_p = \dfrac{\Delta H}{\Delta T} \label{1}\] Therefore, if the heat capacities do not vary with temperature then the change in enthalpy is a function of the difference in temperature and heat capacities. The amount that the enthalpy changes by is proportional to the product of temperature change and change in heat capacities of products and reactants. A weighted sum is used to calculate the change in heat capacity to incorporate the ratio of the molecules involved since all molecules have different heat capacities at different states. \[ H_{T_f}=H_{T_i}+\int_{T_i}^{T_f} c_{p} dT \label{2}\] If the heat capacity is temperature independent over the temperature range, then Equation \ref{1} can be approximated as \[ H_{T_f}=H_{T_i}+ c_{p} (T_{f}-T_{i}) \label{3}\] with Equation \ref{3} can only be applied to small temperature changes, (<100 K) because over a larger temperature change, the is not constant. There are many biochemical applications because it allows us to predict enthalpy changes at other temperatures by using standard enthalpy data. | 1,538 | 1,511 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/08%3A_Phase_Transitions_and_Equilibria_of_Pure_Substances/8.05%3A_Chapter_8_Problems |
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\( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \)
\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) An underlined problem number or problem-part letter indicates that the numerical answer appears in Appendix I. Consider the system described in Sec. 8.1.5 containing a spherical liquid droplet of radius \(r\) surrounded by pure vapor. Starting with Eq. 8.1.15, find an expression for the total differential of \(U\). Then impose conditions of isolation and show that the equilibrium conditions are \(T\sups{g} = T\sups{l}\), \(\mu\sups{g} = \mu\sups{l}\), and \(p\sups{l} = p\sups{g} + 2\g /r\), where \(\g\) is the surface tension. This problem concerns diethyl ether at \(T=298.15\K\). At this temperature, the standard molar entropy of the gas calculated from spectroscopic data is \(S\m\st\gas = 342.2\units{J K\(^{-1}\) mol\(^{-1}\)}\). The saturation vapor pressure of the liquid at this temperature is \(0.6691\br\), and the molar enthalpy of vaporization is \(\Delsub{vap}H = 27.10\units{kJ mol\(^{-1}\)}\). The second virial coefficient of the gas at this temperature has the value \(B=-1.227\timesten{-3}\units{m\(^3\) mol\(^{-1}\)}\), and its variation with temperature is given by \(\dif B/\dif T = 1.50\timesten{-5}\units{m\(^3\) K\(^{-1}\) mol\(^{-1}\)}\). Use these data to calculate the standard molar entropy of liquid diethyl ether at \(298.15\K\). A small pressure change has a negligible effect on the molar entropy of a liquid, so that it is a good approximation to equate \(S\m\st\liquid\) to \(S\m\liquid\) at the saturation vapor pressure. Calculate the standard molar entropy of vaporization and the standard molar enthalpy of vaporization of diethyl ether at \(298.15\K\). It is a good approximation to equate \(H\m\st\liquid\) to \(H\m\liquid\) at the saturation vapor pressure. Explain why the chemical potential surfaces shown in Fig. 8.12 are concave downward; that is, why \(\pd{\mu}{T}{p}\) becomes more negative with increasing \(T\) and \(\pd{\mu}{p}{T}\) becomes less positive with increasing \(p\). Potassium has a standard boiling point of \(773\units{\(\degC\)}\) and a molar enthalpy of vaporization \(\Delsub{vap}H = 84.9\units{kJ mol\(^{-1}\)}\). Estimate the saturation vapor pressure of liquid potassium at \(400.\units{\(\degC\)}\). Naphthalene has a melting point of \(78.2\units{\(\degC\)}\) at \(1\br\) and \(81.7\units{\(\degC\)}\) at \(100\br\). The molar volume change on melting is \(\Delsub{fus}V=0.019\units{cm\(^{3}\) mol\(^{-1}\)}\). Calculate the molar enthalpy of fusion to two significant figures. The dependence of the vapor pressure of a liquid on temperature, over a limited temperature range, is often represented by the , \(\log_{10}(p/\tx{Torr})=A-B/(t+C)\), where \(t\) is the Celsius temperature and \(A\), \(B\), and \(C\) are constants determined by experiment. A variation of this equation, using a natural logarithm and the thermodynamic temperature, is \[ \ln(p/\tx{bar}) = a - \frac{b}{T+c} \] The vapor pressure of liquid benzene at temperatures close to \(298\K\) is adequately represented by the preceding equation with the following values of the constants: \[ a = 9.25092 \qquad b = 2771.233\K \qquad c = -53.262\K \] Find the standard boiling point of benzene. Use the Clausius–Clapeyron equation to evaluate the molar enthalpy of vaporization of benzene at \(298.15\K\). At a pressure of one atmosphere, water and steam are in equilibrium at \(99.97\units{\(\degC\)}\) (the normal boiling point of water). At this pressure and temperature, the water density is \(0.958\units{g cm\(^{-3}\)}\), the steam density is \(5.98\timesten{-4}\units{g cm\(^{-3}\)}\), and the molar enthalpy of vaporization is \(40.66\units{kJ mol\(^{-1}\)}\). Use the Clapeyron equation to calculate the slope \(\difp/\dif T\) of the liquid–gas coexistence curve at this point. Repeat the calculation using the Clausius–Clapeyron equation. Use your results to estimate the standard boiling point of water. (Note: The experimental value is \(99.61\units{\(\degC\)}\).) At the standard pressure of \(1\br\), liquid and gaseous H\(_2\)O coexist in equilibrium at \(372.76\K\), the standard boiling point of water. (a) Do you expect the standard molar enthalpy of vaporization to have the same value as the molar enthalpy of vaporization at this temperature? Explain. The molar enthalpy of vaporization at \(372.76\K\) has the value \(\Delsub{vap}H=40.67\units{kJ mol\(^{-1}\)}\). Estimate the value of \(\Delsub{vap}H\st\) at this temperature with the help of Table 7.5 and the following data for the second virial coefficient of gaseous H\(_2\)O at \(372.76\K\): \[ B=-4.60\timesten{-4}\units{m\(^3\) mol\(^{-1}\)} \qquad \dif B/\dif T=3.4\timesten{-6}\units{m\(^3\) K\(^{-1}\) mol\(^{-1}\)} \] (c) Would you expect the values of \(\Delsub{fus}H\) and \(\Delsub{fus}H\st\) to be equal at the standard freezing point of water? Explain. The standard boiling point of H\(_2\)O is \(99.61\units{\(\degC\)}\). The molar enthalpy of vaporization at this temperature is \(\Delsub{vap}H=40.67\units{kJ mol\(^{-1}\)}\). The molar heat capacity of the liquid at temperatures close to this value is given by \begin{equation*} \Cpm=a+b(t-c) \end{equation*} where \(t\) is the Celsius temperature and the constants have the values \[ a=75.94\units{J K\(^{-1}\) mol\(^{-1}\)} \qquad b = 0.022\units{J K\(^{-2}\) mol\(^{-1}\)} \qquad c = 99.61\units{\(\degC\)} \] Suppose \(100.00\mol\) of liquid H\(_2\)O is placed in a container maintained at a constant pressure of \(1\br\), and is carefully heated to a temperature \(5.00\units{\(\degC\)}\) above the standard boiling point, resulting in an unstable phase of superheated water. If the container is enclosed with an adiabatic boundary and the system subsequently changes spontaneously to an equilibrium state, what amount of water will vaporize? (Hint: The temperature will drop to the standard boiling point, and the enthalpy change will be zero.) | 13,209 | 1,515 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Acid/Graphical_Treatment_of_Acid-Base_Systems |
Most students in General Chemistry are never taken beyond the traditional, algebraic treatment of acid-base equilibria. This is unfortunate for the following reasons: Yes, I still teach my students how to set up a quadratic equation to calculate the pH of an acetic acid solution, but I insist that they also be able to do the same thing graphically. The major advantages of the graphical approach: Log-C vs. pH graphs are plotted on coordinates of the form shown above.Be sure you understand the y-axis; "4", for example, corresponds to a concentration of 10 M for whatever species we plot. Note that concentrations increases with height on this graph, and that those less than about 10 M are usually negligible for most practical purposes. The two plots shown on this graph are no more than formal definitions of pH and pOH; for example, when the pH is 4, –log pH = 4 and [H ] = 10 M. (Notice that the ordinate is the negative of the log concentration, so the smaller numbers near the top of the scale refer to larger concentrations.) The above graph of is of no use by itself, but it forms the basis for the construction of other graphs specific to a given acid-base system. You should be able to draw this graph from memory. The H and OH log concentration lines are the same ones that we saw in Figure 1. The other two lines show how the concentrations of CH COOH and of the the acetate ion vary with the pH of the solution. How do we construct the plots for [HAc] and [Ac ]? If you look carefully at Fig 2, you will observe that each line is horizontal at the top, and then bends to become diagonal. There are thus three parameters that define these two lines: the location of their top, horizontal parts, their crossing points with the other lines, and the slopes of their diagonal parts. The horizontal sections of these lines are placed at 3 on the ordinate scale, corresponding to the nominal acid concentration of 10 M. This value corresponds to C = [HAc] + [Ac ] which you will recognize as the mass balance condition saying that "acetate" is conserved; C is the nominal "acid concentration" of the solution, and is to be distinguished from the concentration of the actual acidic species HAc. At low pH values (strongly acidic solution) the acetate species is completely protonated, so [HAc] = 10 M and [Ac ]=0. Similarly, at high pH, –log [Ac ]=3$ and [HAc]=0. If the solution had some other nominal concentration, such as 0.1 M or 10 , we would simply move the pair of lines up or down. The diagonal parts of the lines have slopes of equal magnitude but opposite sign. It can easily be shown that these slopes d(–log [HAc]}/d{pH} etc.are +-1, corresponding to the slopes of the [OH ] and [H ] lines. Using the latter as a guide, the diagonal portions of lines 3 and 4 Can easily be drawn. The crossing point of the plots for the acid and base forms corresponds to the condition [HAc]=[Ac ]. You already know that this condition holds when the pH is the same as the pK of the acid, so the the pH coordinate of the crossing point must be 4.75 for acetic acid. The vertical location of the crossing point is found as follows: When [HAc] = [Ac], the concentration of each species must be C /2 or in this case 0.0005 M. The logarithm of 1/2 is 0.3, so a 50% reduction in the concentration of a species will shift its location down on the log concentration scale by 0.3 unit. The crossing point therefore falls at a log-C value of (–3) – 0.3 = –3.3. Knowing the value of log 0.5 is one of the few new "facts" that must be learned in order to construct these graphs. Of special interest in acid-base chemistry are the pH values of a solution of an acid and of its conjugate base in pure water; as you know, these correspond to the beginning and equivalence points in the titration of an acid with a strong base. Except for the special cases of extremely dilute solutions or very weak acids in which the autoprotolysis of water is a major contributor to the hydrogen ion concentration, the pH of a solution of an acid in pure water will be determined largely by the extent of the reaction \[\ce{HAc + H2O -> H3O^{+} + Ac^{–}}\] so that at equilibrium, the approximate relation \(\ce{[H3O^{+}] \approx [Ac^{–}]}\) will hold. The equivalence of these two concentrations corresponds to the point labeled 1 in Fig 3; this occurs at a pH of about 3.7, and this is the pH of a 0.001M solution of acetic acid in pure water. Now consider a 0.001M solution of sodium acetate in pure water. This, you will recall, corresponds to the composition of a solution of acetic acid that has been titrated to its equivalence point with sodium hydroxide. The acetate ion, being the conjugate base of a weak acid, will undergo hydrolysis according to \[\ce{Ac– + H2O -> HAc + OH–}\] As long as we can neglect the contribution of OH from the autoprotolysis of the solvent, we can expect the relation [HAc]=[OH ] to be valid in such a solution. The equivalence of these two concentrations corresponds to the intersection point 3 in Fig 3; a 0.001M solution of sodium or potassium acetate should have a pH of about 8. The top part of this figure is the log-C plot for acetic acid that we saw previously, the blue dashed vertical lines now showing the pH of a 0.001M solution of HAc before and and after its titration with strong base. The extension of these lines (and the one representing the pK) down to the bottom part of the figure locates the three points that define the titration curve for the acid. Take a moment to verify that the lower figure is indeed a titration curve; that is, a plot of the pH as a function of the fraction of acid neutralized. These two plots nicely sum up the complete acid-base chemistry of a monoprotic system. Diprotic acids are no more complicated to treat graphically than monoprotic acids-- something that definitely cannot be said for exact algebraic treatment! The above example for oxalic acid (H A = HOOCCOOH) is just the superposition for separate plots of the two acid-base systems H A-HA and HA -A whose pKs are 1.2 and 4.2. The system points corresponding to 0.01M solutions of pure H A; and A are determined just as in the monoprotic case. For a solution of the ampholyte, several equilibria can be written, but the one that dominates is \[2 HA^– \rightleftharpoons H_2A + A^{2–}\] which leads to the approximation \[[HA^–] \approx [A^{2–}]\] Thus fixing the pH at about 2.7. This same result can be obtained from thewell known approximation \[pH = \dfrac{pK_1 + pK_2}{2}\] The quantitative treatment of a solution of a salt of a weak acid a weak base is algebraically complicated, even when done to the crudest approximation. Graphically, it is a piece of cake! As you can see above, we just construct plots of the two acid-base system on the same graph. The stoichiometry of the salt defines the system point at about pH 6.3, showing that hydrolyses of the two systems does not quite "cancel out". Of course, this is still an approximation that neglects ionization of either the anion or cation, but it is probably as valid as any calculation that is likely to be carried out ordinarily-available data. ... but why stop at diprotic acids? With the phosphate system we have three pKs, and thus two ampholytes H PO and HPO . A glance at the graph allows one to estimate the relative amounts of the two major species at any pH. The pH of a solution of the acid H PO or of the base PO is found from the system points 1 and 4 which are constructed in the same way as those for a monoprotic acid. System points 2 and 3 for solutions of the two ampholytes provide only approximate values pH values because of the effects of competing equilibria, but the accuracy is good enough for most applications. Here it is -- the most important of all acid-base systems in natural waters and physiology! Nothing new as far as theory goes -- see the explanation given for oxalic acid. What's different here is the two sts of plots. The lower (more dilute) one corresponds to the concentration of dissolved CO present in water in equlibrium with the atmosphere having the normal CO partial pressure of about .0005 atm. You can see from this that when pure water is exposed to the atmosphere, its pH will fall from 7 to about 5 (but see the last paragraph below); thus all rain is in a sense "acid rain". Note that around pH 8 (the pH of the ocean and of blood), over 99% of all carbonate is in the form of bicarbonate HCO . As a result, the ocean acts as a sink for atmospheric CO and contains about 50 times as much as does the atmosphere. Similarly, CO resulting from cellular respiration is converted to HCO for transport through the blood stream, and is converted back to CO in the lungs. The upper (.001M) set of graphs correspond to the approximate carbonate content of natural waters in contact with rocks and sediments (which normally contain carbonate components). It describes the situation in lakes, streams, ground waters and the ocean. Notice how the pH of this more concentrated CO solution is lower (about 2, compared to 4.3), which is just what you would expect. This is the reason the pH of an algae-containing pond rises during the day (when CO is being consumed by photosynthesis) and falls at night (CO restored by respiration.) The term "closed system" in the caption at the top of the graph means that C , the sum of the concentrations of all three carbonate species, is assumed constant here. For an open system the plots for the bicarbonate and carbonate concentrations would be dramatically different because an alkaline solution open to the atmosphere has access to an infinite supply of CO and will absorb it up to the solubility limit of sodium carbonate. (You may recall that a common test for CO is to watch for a precipitate to appear in a saturated solution of calcium hydroxide.) Thus the estimate of pH=5 for pure water exposed to the atmosphere is not quite correct; using an "open system" plot, a somewhat lower pH, around 4.5, will be found. It turns out, however, that for many practical situations, the relatively slow transport of CO between air and water makes the "closed system" predictions reasonably accurate within the time scale of interest. Glycine is the simplest of the amino acids, the building blocks of proteins. It is also a (the word comes from the German term for "hermaphrodite" ) which chemists use to denote a species that possesses both an acidic and a basic functional group, both of which can simultaneously exist in their ionized (conjugate) forms. Examination of this graph shows that the double-ionic form (the true zwitterion) reigns supreme between pH 3 and 9; outside this range, the glycinium anion or the glycinate cation prevails. Notice that the completely un-ionized form does not exist in solution. The log-C vs. pH diagram is constructed as s superposition of plots for each conjugate pair at its respective pKa. Note especially that the pH of a solution of glycine does lie exactly at the crossing point [Gly ] = [H ], but is slightly displaced from it according to the proton balance equation shown in the inset on the graph. The other important quantity shown on this graph is the , which is the pH at which the concentrations of the cationic and anionic forms are identical. It goes without saying that treating this system algebraically would be far more complicated than the results would warrant for most applications. Note especially that ) | 11,482 | 1,516 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/01._Waves_and_Particles/Blackbody_Radiation |
Blackbody radiation is a cornerstone in the study of quantum mechanics.This experiment is what led to the discovery of a field that would revolutionize physics and chemistry. Quantum mechanics gives a more complete understanding of the fundamental mechanisms at the sub-atomic level. The work done at the turn of the 20 century on blackbody radiation was the beginning of a totally new field of science. Blackbody radiation is a theoretical concept in quantum mechanics in which a material or substance completely absorbs all frequencies of light. Because of the laws of thermodynamics, this ideal body must also re-emit as much light as it absorbs.Although there is no material that can truly be a blackbody, some have come close.Carbon in its graphite form is about 96% efficient in its absorption of light. The concept of blackbody radiation is seen in many different places.The intensity of the energy coming from the radiator is a function only of temperature. A good example of this temperature dependence is a flame. The flame starts out with a low frequency emitting red light in the visible range, as the temperature increases the flame turns white and then blue as is moves across the visible spectrum with an increasing temperature. Also, with each temperature corresponds a new maximum radiance which can be emitted. As the temperature increases, the total radiation emitted also increases due to an increase in the area under the curve. Lord Rayleigh and J. H. Jeans developed an equation which explained blackbody radiation at low frequencies.The equation which seemed to express blackbody radiation was built upon all the known assumptions of physics at the time. The big assumption which Rayleigh and Jean implied was that infinitesimal amounts of energy were continuously added to the system when the frequency was increased. Classical physics assumed that energy emitted by atomic oscillations could have any continuous value. This was true for anything that had been studied up until that point, including things like acceleration, position, or energy.They came up with and the equation they derived was \[ d\rho \left( \nu ,T \right) = \rho_{\nu} \left( T \right) d \nu = \frac{8 \pi k_B T}{c^3} \nu^2 d\nu \] During the 19th century much attention was given to the study of heat properties of various objects. An idealised model that was considered was the Black Body, an object which absorbs all incident radiation and then re-emits all this energy again. We can think of the radiating energy as standing waves inside our blackbody cavity. The energy of the radiating waves at a given frequency , should be proportional to the number of modes at this frequency. Classical physics states that all these modes have the same energy (a result derived from classical thermodynamics) and as the number of modes is proportional to \(\nu^2\): \[E \propto \nu^2 kT \label{1.1.1}\] This implies that we would expect most of the energy at higher frequency, and this energy diverges with frequency. If we try and sum the energies at each frequency we find that there is an infinite energy in ths system! This paradox was called the . It was left to Planck to resolve this gaping paradox, but postulated that the energy of the modes could only come in discrete packets – quanta – of energy: \[E = h\nu, 2h\nu, 3h\nu, \ldots \qquad \Delta E = h\nu \label{1.1.2}\] Using statistical mechanics Planck found that the modes at higher frequency were less likely excited so the average energy of these modes would decrease with the frequency. The exact expression for the average energy of each mode is given by the Planck distribution: \[\langle E \rangle = \frac{h\nu}{\exp(\frac{h\nu}{KT}) - 1} \label{1.1.3}\] You can see that if the frequency is low then the average energy tends towards the classical result, and as frequency goes to infinity we get that the average energy goes to zero as expected. Max Planck was the first person to properly explain this experimental data. Rayleigh and Jean made the assumption that energy is continuous, but Planck took a slightly different approach. He said energy must come in certain unit intervals instead of being any random unit or number. He instead “quantized” energy in the form of \(E= nh\nu\) where \(n\) is an integer, \(h\) is a constant, and \(\nu\) is the frequency. This assumption proved to be the missing piece of the puzzle and Planck derived an expression which could explain the experimental data. \[ d\rho \left( \nu ,T \right) = \rho_{\nu} \left( T \right) d\nu = \dfrac{8 \pi k_B T}{c^3} \dfrac{nu^2}{e^{hv/K_bT}-1} d\nu \] | 4,617 | 1,517 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Chemical_Energetics/Energy%2C_Heat%2C_and_Work |
Energy is one of the most fundamental and universal concepts of physical science, but one that is remarkably difficult to define in a way that is meaningful to most people. This perhaps reflects the fact that energy is not a “thing” that exists by itself, but is rather an attribute of matter (and also of electromagnetic radiation) that can manifest itself in different ways. It can be observed and measured only indirectly through its effects on matter that acquires, loses, or possesses it. The concept that we call energy was very slow to develop; it took more than a hundred years just to get people to agree on the definitions of many of the terms we use to describe energy and the interconversion between its various forms. But even now, most people have some difficulty in explaining what it is; somehow, the definition we all learned in elementary science ("the capacity to do work") seems less than adequate to convey its meaning. Although the term "energy" was not used in science prior to 1802, it had long been suggested that certain properties related to the motions of objects exhibit an endurance which is incorporated into the modern concept of "conservation of energy". In the 17th Century, the great mathematician Gottfried Leibnitz (1646-1716) suggested the distinction between ("live force") and ("dead force"), which later became known as kinetic energy (1829) and potential energy (1853). Whatever energy may be, there are basically two kinds. is associated with the of an object, and its direct consequences are part of everyone's daily experience; the faster the ball you catch in your hand, and the heavier it is, the more you feel it. Quantitatively, a body with a mass and moving at a velocity possesses the kinetic energy mv /2. A rifle shoots a 4.25 g bullet at a velocity of 965 m s . What is its kinetic energy? The only additional information you need here is that 1 J = 1 kg m s : KE = ½ × (.00425 kg) (965 m s ) = 1980 J More generally, the restoring force comes from what we call a — a gravitational, electrostatic, or magnetic field. We observe the consequences of potential energy all the time, such as when we walk, but seldom give it any thought. If an object of mass is raised off the floor to a height , its potential energy increases by , where is a proportionality constant known as the ; its value at the earth's surface is 9.8 m s . Find the change in potential energy of a 2.6 kg textbook that falls from the 66-cm height of a table top onto the floor. PE = = (2.6 kg)(9.8 m s )(0.66 m) = 16.8 kg m s = 16.8 J Electrostatic potential energy plays a major role in chemistry; the potential energies of electrons in the force field created by atomic nuclei lie at the heart of the chemical behavior of atoms and molecules. "Chemical energy" usually refers to the energy that is stored in the chemical bonds of molecules. These bonds form when electrons are able to respond to the force fields created by two or more atomic nuclei, so they can be regarded as manifestations of electrostatic potential energy. In an chemical reaction, the electrons and nuclei within the reactants undergo rearrangement into products possessing lower energies, and the difference is released to the environment in the form of heat. Transitions between potential and kinetic energy are such an intimate part of our daily lives that we hardly give them a thought. It happens in walking as the body moves up and down. Our bodies utilize the chemical energy in glucose to keep us warm and to move our muscles. In fact, life itself depends on the conversion of chemical energy to other forms. it can neither be created nor destroyed. So when you go uphill, your kinetic energy is transformed into potential energy, which gets changed back into kinetic energy as you coast down the other side. And where did the kinetic energy you expended in peddling uphill come from? By conversion of some of the chemical potential energy in your breakfast cereal. Kinetic energy is associated with motion, but in two different ways. For a macroscopic object such as a book or a ball, or a parcel of flowing water, it is simply given by ½ . However, as we mentioned above, when an object is dropped onto the floor, or when an exothermic chemical reaction heats surrounding matter, the kinetic energy gets dispersed into the molecular units in the environment. This "microscopic" form of kinetic energy, unlike that of a speeding bullet, is completely random in the kinds of motions it exhibits and in its direction. We refer to this as "thermalized" kinetic energy, or more commonly simply as . We observe the effects of this as a rise in the temperature of the surroundings. The temperature of a body is direct measure of the quantity of thermal energy is contains. Once kinetic energy is thermalized, only a portion of it can be converted back into potential energy. The remainder simply gets dispersed and diluted into the environment, and is effectively lost. To summarize, then: You might at first think that a book sitting on the table has zero kinetic energy since it is not moving. But if you think about it, the earth itself is moving; it is spinning on its axis, it is orbiting the sun, and the sun itself is moving away from the other stars in the general expansion of the universe. Since these motions are normally of no interest to us, we are free to adopt an arbitrary scale in which the velocity of the book is measured with respect to the table; on this so-called , the kinetic energy of the book can be considered zero. We do the same thing with potential energy. If the book is on the table, its potential energy with respect to the surface of the table will be zero. If we adopt this as our zero of potential energy, and then push the book off the table, its potential energy will be negative after it reaches the floor. Energy is measured in terms of its ability to perform work or to transfer heat. Mechanical work is done when a force f displaces an object by a distance : \[w = f × d\] The basic unit of energy is the . One joule is the amount of work done when a force of 1 newton acts over a distance of 1 m; thus 1 J = 1 N-m. The newton is the amount of force required to accelerate a 1-kg mass by 1 m/sec , so the basic dimensions of the joule are kg m s . The other two units in wide use. the and the (British thermal unit) are defined in terms of the heating effect on water. Because of the many forms that energy can take, there are a correspondingly large number of units in which it can be expressed, a few of which are summarized below. 1 calorie will raise the temperature of 1 g of water by 1 C°. The “dietary” calorie is actually 1 kcal. An average young adult expends about 1800 kcal per day just to stay alive. (you should know this definition) 1 J = 10 ergs
1 erg = 1 d-cm = 1 g cm s 1 J = 2.78 × 10 watt-hr
1 w-h = 3.6 kJ 1 bboe = 6.1 GJ
1 cmge = 37-39 mJ
1 toce = 29 GJ Heat and work are both measured in energy units, so they must both represent energy. How do they differ from each other, and from just plain “energy” itself? In our daily language, we often say that "this object contains a lot of heat", but this is gibberish in thermodynamics terms, although it is ok to say that the object is "hot", indicating that its temperature is high. The term "heat" has a special meaning in thermodynamics: it is a in which a body (the contents of a tea kettle, for example) acquires or loses energy as a direct consequence of its having a than its surroundings. Hence, thermal energy can only flow from a higher temperature to a lower temperature. It is this flow that constitutes "heat". Use of the term "flow" of heat recalls the incorrect 18th-century notion that heat is an actual substance called “caloric” that could flow like a liquid. We often say that "this object contains a lot of heat," however, this makes no sense since heat represents an energy transfer. Transfer of thermal energy can be accomplished by bringing two bodies into physical contact (the kettle on top of the stove, or through an electric heating element inside the kettle). Another mechanism of thermal energy transfer is by radiation; a hot object will convey energy to any body in sight of it via electromagnetic radiation in the infrared part of the spectrum. In many cases, both modes will be active. Work refers to the transfer of energy some means that does not depend on temperature difference. Work, like energy, can take various forms, the most familiar being mechanical and electrical. A transfer of energy to or from a system by any means other than heat is called “work”. Work can be completely converted into heat (by friction, for example), but heat can only be partially converted to work. is accomplished by means of a , the most common example of which is an ordinary gasoline engine. The science of thermodynamics developed out of the need to understand the limitations of steam-driven heat engines at the beginning of the Industrial Age. The Second Law of Thermodynamics, states that the complete conversion of heat into work is impossible. Something to think about when you purchase fuel for your car! ) | 9,259 | 1,518 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Fundamentals_of_Chemical_Bonding/Ionic_Bonds |
Ionic bonding is the complete transfer of valence electron(s) between atoms and is a type of chemical bond that generates two oppositely charged ions. It is observed because metals with few electrons in its outer-most orbital. By losing those electrons, these metals can achieve noble-gas configuration and satisfy the . Similarly, nonmetals that have close to 8 electrons in its valence shell tend to readily accept electrons to achieve its noble gas configuration. In ionic bonding, electrons are transferred from one atom to another resulting in the formation of positive and negative ions. The electrostatic attractions between the positive and negative ions hold the compound together. The predicted overall energy of the ionic bonding process, which includes the of the metal and of the nonmetal, is usually positive, indicating that the reaction is endothermic and unfavorable. However, this reaction is highly favorable because of their electrostatic attraction. At the most ideal inter-atomic distance, attraction between these particles releases enough energy to facilitate the reaction. Most ionic compounds tend to dissociate in polar solvents because they are often polar. This phenomenon is due to the opposite charges on each ions. At a simple level, a lot of importance is attached to the electronic structures of noble gases like neon or argon which have eight electrons in their outer energy levels (or two in the case of helium). These noble gas structures are thought of as being in some way a "desirable" thing for an atom to have. One may well have been left with the strong impression that when other atoms react, they try to organize things such that their outer levels are either completely full or completely empty. Sodium Chloride: The answer is obvious. If a sodium atom gives an electron to a chlorine atom, both become more stable. The sodium has lost an electron, so it no longer has equal numbers of electrons and protons. Because it has one more proton than electron, it has a charge of 1+. If electrons are lost from an atom, positive ions are formed. Positive ions are sometimes called . The chlorine has gained an electron, so it now has one more electron than proton. It therefore has a charge of 1-. If electrons are gained by an atom, negative ions are formed. A negative ion is sometimes called an . The sodium ions and chloride ions are held together by the strong electrostatic attractions between the positive and negative charges. You need one sodium atom to provide the extra electron for one chlorine atom, so they combine together 1:1. The formula is therefore NaCl. Magnesium Oxide: Again, noble gas structures are formed, and the magnesium oxide is held together by very strong attractions between the ions. The ionic bonding is stronger than in sodium chloride because this time you have 2+ ions attracting 2- ions. The greater the charge, the greater the attraction. The formula of magnesium oxide is MgO. Calcium Chloride: This time you need two chlorines to use up the two outer electrons in the calcium. The formula of calcium chloride is therefore CaCl . Potassium Oxide: Again, noble gas structures are formed. It takes two potassiums to supply the electrons the oxygen needs. The formula of potassium oxide is K O. You may have come across some of the following ions, which are all perfectly stable, but not one of them has a noble gas structure. What needs modifying is the view that there is something magic about noble gas structures. There are far more ions which do not have noble gas structures than there are which do. If elements are not aiming for noble gas structures when they form ions, what decides how many electrons are transferred? The answer lies in the energetics of the process by which the compound is made. Jim Clark ( ) | 3,823 | 1,519 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Periodic_Trends_of_Elemental_Properties/Periodic_Trends_in_Ionic_Radii |
\(Z_{eff}\) \[Z_{eff} = Z - S\] Neutral atoms that have lost an electron exhibit a positive charge and are called cations. These cations are smaller than their respective atoms; this is because when an electron is lost, electron-electron repulsion (and therefore, shielding) decreases and the protons are better able to pull the remaining electrons towards the nucleus (in other words, \(Z_{eff}\)increases). A second lost electron further reduces the radius of the ion. For instance, the ionic radius of Fe is 76 pm, while that of Fe is 65 pm. If creation of an ion involves completely emptying an outer shell, then the decrease in radius is especially great. \(Z_{eff}\) Figure 2 shows an isoelectric series of atoms and ions (each has the same number of electrons, and thus the same degree of electron-electron repulsion and shielding) with differing numbers of protons (and thus different nuclear attraction), giving the relative ionic sizes of each atom or ion. Due to each atom’s unique ability to lose or gain an electron, periodic trends in ionic radii are not as ubiquitous as trends in atomic radii across the periodic table. Therefore, trends must be isolated to specific groups and considered for either cations or anions. Consider the s- and d-block elements. All metals can lose electrons and form cations. The alkali and alkali earth metals (groups 1 and 2) form cations which increase in size down each group; atomic radii behave the same way. Beginning in the d-block of the periodic table, the ionic radii of the cations do not significantly change across a period. However, the ionic radii do slightly decrease until group 12, after which the trend continues (Shannon 1976). It is important to note that metals, not including groups 1 and 2, can have different ionic states, or oxidation states, (e.g. Fe or Fe for iron) so caution must be employed when generalizing about trends in ionic radii across the periodic table. All non-metals (except for the noble gases which do not form ions) form anions which become larger down a group. For non-metals, a subtle trend of decreasing ionic radii is found across a pegroup theoryriod (Shannon 1976). Anions are almost always larger than cations, although there are some exceptions (i.e. fluorides of some alkali metals). The ionic radius of an atom is measured by calculating its spatial proportions in an ionic bond with another ion within a crystal lattice. However, it is to consistently and accurately determine the proportions of the ionic bonds. After comparing many compounds, chemist Linus Pauling assign a radius of 140 pm to O and use this as a reference point to determine the sizes of other Ionic Radii (Jensen 2010). Ionic radius is not a permanent trait of an ion, but changes depending on coordination number, spin state, and other variables (Shannon 1976). For a given ion, the ionic radius increases with increasing coordination number and is larger in a high-spin state than in a low-spin state. According to group theory, the idea of ionic radii as a measurement of spherical shapes only applies to ions that form highly-symmetric crystal lattices like Na and Cl . The point group symmetry of a lattice determines whether or not the ionic radii in that lattice can be accurately measured (Johnson 1973). For instance, lattices with O and T symmetries are considered to have high symmetry; thus the electron densities of the component ions occupy relatively-spherical regions and ionic radii can be measured fairly accurately. However, for less symmetrical and more polar lattices such as those with C , C , and C symmetries, significant changes in the electron density can occur, causing deviations from spherical shape; these deviations make ionic radii more difficult to measure. | 3,789 | 1,520 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Chemical_Reactions/Chemical_Reactions_Examples/Revealing_the_Dates_of_Buffalo_Nickels_(Demo) |
First, this caveat. Experienced coin collectors will tell you that chemical etching of coins almost never increases their value. It practically always decreases their value, so none of the following will be of any use if you collect coins as an investment. It is simply here to demonstrate a point of interest about revealing hidden dates of buffalo nickels using a gentle etching process. The U.S. "Buffalo" Nickel was minted between 1913 and 1938. The date was placed on a raised area beneath the image of a native American; the date was usually the first thing to disappear as a Buffalo nickel wore down in people's pockets, purses and being handled in transactions. During the 1950s I found the dates on most circulating buffalo nickels already to be worn beyond recognition. Those minted in the 30s often had readable dates, but those minted between 1913 and 1929 were practically always unreadable. Occasionally I was delighted to find one like that at the right where natural corrosion or some coating process gave a good contrast to a still-existing date. There were those rare finds of a nickel with a reasonably good date on a bright coin, like that one at the left. A buffalo nickel can be etched with vinegar so as to "develop" a latent image of the date of mint, owing presumably to the variable "etchability" of regions which had experienced high or low die pressures during the minting process. If a buffalo nickel is dropped into about 10 mL vinegar, date side up, and left for anywhere from one to eight weeks (frequent inspections are advised), a readable date develops in about a third of the cases tried. Here are three from my old coin collection. Most often one obtains marginal results, like the 1916 coin on the right. Occasionally one gets a clearly identifiable date, as in the case of the 1925 coin at the left. Here's an enlargement of the date from the coin above.
For some percentage of trials, less than 50% but frustratingly close to it, one gets a result which can be described only as wishful thinking, such as the 1913 coin at the right. However, 1913 nickels were the only minting in which there was "raised ground" under the buffalo. Notice that the "FIVE CENTS" partially emerged, but there was no mint mark under it, so I entered it as a 1913 Philadelphia, even though the date can't be seen. During a recent discussion on the Internet list of chemical educators, CHEMED-L, someone brought up the topic of etching buffalo nickels to produce latent images of the dates of mint. Local coin stores have large collections of unreadable buffalo nickels because some craftspeople make bracelets and necklaces from them. For 30 cents I bought three unreadable buffalo nickels in December, 1999, and dropped each one into 10 mL standard vinegar, 5% acetic acid. Here are some "before" (on the left) and "after" (on the right) pictures following an etching period of about eight weeks. The first pair offer the standard marginal success story. Is that a "9" or a "5" as the last digit? Is it 1919 or 1929? Here's an enlargement to emphasize the 5/9 uncertainty on the last digit and to suggest that it is probably 1925 or 1929 rather than 1915 or 1919 because of the suggested space between the 2nd and fourth digits. For my money (!) I'd say that I can actually see the date 1929. For those old enough to remember it, these pictures are a little reminiscent of the movie "Blowup," aren't they?
The second case is a stunning success. After deciding that I really COULD read the date, I handed this nickel to my wife who is near-sighted and asked her to tell me the date. She removed her glasses, took one look at it and without a pause said, "Oh, that's 1915." Voila! The image above and to the right appears marginal because the resolution of most screens is less than the image itself. The enlargement to the right shows the date much better. Finally the third case gives us another example of wishful thinking. Can you even begin to see the first two digits of the date, which you already know to be 19?
Here's an enlargement better to illustrate the problem. | 4,114 | 1,521 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/13%3A_Solutions/13.04%3A_Effects_of_Temperature_and_Pressure_on_Solubility |
Experimentally it is found that the solubility of most compounds depends strongly on temperature and, if a gas, on pressure as well. As we shall see, the ability to manipulate the solubility by changing the temperature and pressure has several important consequences. Figure \(\Page {1}\) shows plots of the solubilities of several organic and inorganic compounds in water as a function of temperature. Although the solubility of a solid generally increases with increasing temperature, there is no simple relationship between the structure of a substance and the temperature dependence of its solubility. Many compounds (such as glucose and \(\ce{CH_3CO_2Na}\)) exhibit a dramatic increase in solubility with increasing temperature. Others (such as \(\ce{NaCl}\) and \(\ce{K_2SO_4}\)) exhibit little variation, and still others (such as \(\ce{Li_2SO_4}\)) become less soluble with increasing temperature. Notice in particular the curves for \(\ce{NH4NO3}\) and \(\ce{CaCl2}\). The dissolution of ammonium nitrate in water is endothermic (\(ΔH_{soln} = +25.7\; kJ/mol\)), whereas the dissolution of calcium chloride is exothermic (\(ΔH_{soln} = −68.2 \;kJ/mol\)), yet Figure \(\Page {1}\) shows that the solubility of both compounds increases sharply with increasing temperature. In fact, the magnitudes of the changes in both enthalpy and entropy for dissolution are temperature dependent. Because the solubility of a compound is ultimately determined by relatively small differences between large numbers, there is generally no good way to predict how the solubility will vary with temperature. The variation of solubility with temperature has been measured for a wide range of compounds, and the results are published in many standard reference books. Chemists are often able to use this information to separate the components of a mixture by the separation of compounds on the basis of their solubilities in a given solvent. For example, if we have a mixture of 150 g of sodium acetate (\(\ce{CH_3CO_2Na}\)) and 50 g of \(\ce{KBr}\), we can separate the two compounds by dissolving the mixture in 100 g of water at 80°C and then cooling the solution slowly to 0°C. According to the temperature curves in Figure \(\Page {1}\), both compounds dissolve in water at 80°C, and all 50 g of \(\ce{KBr}\) remains in solution at 0°C. Only about 36 g of \(\ce{CH3CO2Na}\) are soluble in 100 g of water at 0°C, however, so approximately 114 g (150 g − 36 g) of \(\ce{CH_3CO_2Na}\) crystallizes out on cooling. The crystals can then be separated by filtration. Thus fractional crystallization allows us to recover about 75% of the original \(\ce{CH_3CO_2Na}\) in essentially pure form in only one step. Fractional crystallization is a common technique for purifying compounds as diverse as those shown in Figure \(\Page {1}\) and from antibiotics to enzymes. For the technique to work properly, the compound of interest must be more soluble at high temperature than at low temperature, so that lowering the temperature causes it to crystallize out of solution. In addition, the impurities must be more soluble than the compound of interest (as was \(\ce{KBr}\) in this example) and preferably present in relatively small amounts. The solubility of gases in liquids decreases with increasing temperature, as shown in Figure \(\Page {2}\). Attractive intermolecular interactions in the gas phase are essentially zero for most substances. When a gas dissolves, it does so because its molecules interact with solvent molecules. Because heat is released when these new attractive interactions form, dissolving most gases in liquids is an exothermic process (\(ΔH_{soln} < 0\)). Conversely, adding heat to the solution provides thermal energy that overcomes the attractive forces between the gas and the solvent molecules, thereby decreasing the solubility of the gas. The phenomenon is similar to that involved in the increase in the vapor pressure of a pure liquid with increasing temperature. In the case of vapor pressure, however, it is attractive forces between solvent molecules that are being overcome by the added thermal energy when the temperature is increased. The decrease in the solubilities of gases at higher temperatures has both practical and environmental implications. Anyone who routinely boils water in a teapot or electric kettle knows that a white or gray deposit builds up on the inside and must eventually be removed. The same phenomenon occurs on a much larger scale in the giant boilers used to supply hot water or steam for industrial applications, where it is called “boiler scale,” a deposit that can seriously decrease the capacity of hot water pipes (Figure \(\Page {3}\)). The problem is not a uniquely modern one: aqueducts that were built by the Romans 2000 years ago to carry cold water from alpine regions to warmer, drier regions in southern France were clogged by similar deposits. The chemistry behind the formation of these deposits is moderately complex and will be described elsewhere, but the driving force is the loss of dissolved \(\ce{CO2}\) from solution. Hard water contains dissolved \(\ce{Ca^{2+}}\) and \(\ce{HCO3^{-}}\) (bicarbonate) ions. Calcium bicarbonate (\(\ce{Ca(HCO3)2}\) is rather soluble in water, but calcium carbonate (\(\ce{CaCO3}\)) is quite insoluble. A solution of bicarbonate ions can react to form carbon dioxide, carbonate ion, and water: \[\ce{2HCO3^{-}(aq) -> CO3^{2-}(aq) + H2O(l) + CO2(aq)} \label{13.9} \] Heating the solution decreases the solubility of \(\ce{CO2}\), which escapes into the gas phase above the solution. In the presence of calcium ions, the carbonate ions precipitate as insoluble calcium carbonate, the major component of boiler scale. In , lake or river water that is used to cool an industrial reactor or a power plant is returned to the environment at a higher temperature than normal. Because of the reduced solubility of \(\ce{O2}\) at higher temperatures (Figure \(\Page {2}\)), the warmer water contains less dissolved oxygen than the water did when it entered the plant. Fish and other aquatic organisms that need dissolved oxygen to live can literally suffocate if the oxygen concentration of their habitat is too low. Because the warm, oxygen-depleted water is less dense, it tends to float on top of the cooler, denser, more oxygen-rich water in the lake or river, forming a barrier that prevents atmospheric oxygen from dissolving. Eventually even deep lakes can be suffocated if the problem is not corrected. Additionally, most fish and other nonmammalian aquatic organisms are cold-blooded, which means that their body temperature is the same as the temperature of their environment. Temperatures substantially greater than the normal range can lead to severe stress or even death. Cooling systems for power plants and other facilities must be designed to minimize any adverse effects on the temperatures of surrounding bodies of water. A similar effect is seen in the rising temperatures of bodies of water such as the k0oi89Chesapeake Bay, the largest estuary in North America, where \lobal warming has been implicated as the cause. For each 1.5°C that the bay’s water warms, the capacity of water to dissolve oxygen decreases by about 1.1%. Many marine species that are at the southern limit of their distributions have shifted their populations farther north. In 2005, the eelgrass, which forms an important nursery habitat for fish and shellfish, disappeared from much of the bay following record high water temperatures. Presumably, decreased oxygen levels decreased populations of clams and other filter feeders, which then decreased light transmission to allow the eelsgrass to grow. The complex relationships in ecosystems such as the Chesapeake Bay are especially sensitive to temperature fluctuations that cause a deterioration of habitat quality. External pressure has very little effect on the solubility of liquids and solids. In contrast, the solubility of gases increases as the partial pressure of the gas above a solution increases. This point is illustrated in Figure \(\Page {4}\), which shows the effect of increased pressure on the dynamic equilibrium that is established between the dissolved gas molecules in solution and the molecules in the gas phase above the solution. Because the concentration of molecules in the gas phase increases with increasing pressure, the concentration of dissolved gas molecules in the solution at equilibrium is also higher at higher pressures. The relationship between pressure and the solubility of a gas is described quantitatively by Henry’s law, which is named for its discoverer, the English physician and chemist, William Henry (1775–1836): \[C = kP \label{13.3.1} \] where Although the gas concentration may be expressed in any convenient units, we will use molarity exclusively. The units of the Henry’s law constant are therefore mol/(L·atm) = M/atm. Values of the Henry’s law constants for solutions of several gases in water at 20°C are listed in Table \(\Page {1}\). As the data in Table \(\Page {1}\) demonstrate, the concentration of a dissolved gas in water at a given pressure depends strongly on its physical properties. For a series of related substances, London dispersion forces increase as molecular mass increases. Thus among the elements, the Henry’s law constants increase smoothly from \(\ce{He}\) to \(\ce{Ne}\) to \(\ce{Ar}\). Nitrogen and oxygen are the two most prominent gases in the Earth’s atmosphere and they share many similar physical properties. However, as Table \(\Page {1}\) shows, \(\ce{O2}\) is twice as soluble in water as \(\ce{N2}\). Many factors contribute to solubility including the nature of the intermolecular forces at play. For a details discussion, see "The O /N Ratio Gas Solubility Mystery" by Rubin Battino and Paul G. Seybold ( ), Gases that react chemically with water, such as \(\ce{HCl}\) and the other hydrogen halides, \(\ce{H2S}\), and \(\ce{NH3}\), do obey Henry’s law; all of these gases are much more soluble than predicted by Henry’s law. For example, \(\ce{HCl}\) reacts with water to give \(\ce{H^{+}(aq)}\) and \(\ce{Cl^{-}(aq)}\), not dissolved \(\ce{HCl}\) molecules, and its dissociation into ions results in a much higher solubility than expected for a neutral molecule. Gases that with water do not obey Henry’s law. Henry’s law has important applications. For example, bubbles of \(\ce{CO2}\) form as soon as a carbonated beverage is opened because the drink was bottled under \(\ce{CO2}\) at a pressure greater than 1 atm. When the bottle is opened, the pressure of \(\ce{CO2}\) above the solution drops rapidly, and some of the dissolved gas escapes from the solution as bubbles. Henry’s law also explains why scuba divers have to be careful to ascend to the surface slowly after a dive if they are breathing compressed air. At the higher pressures under water, more N2 from the air dissolves in the diver’s internal fluids. If the diver ascends too quickly, the rapid pressure change causes small bubbles of \(\ce{N2}\) to form throughout the body, a condition known as “the bends.” These bubbles can block the flow of blood through the small blood vessels, causing great pain and even proving fatal in some cases. Due to the low Henry’s law constant for \(\ce{O2}\) in water, the levels of dissolved oxygen in water are too low to support the energy needs of multicellular organisms, including humans. To increase the \(\ce{O2}\) concentration in internal fluids, organisms synthesize highly soluble carrier molecules that bind \(\ce{O2}\) reversibly. For example, human red blood cells contain a protein called hemoglobin that specifically binds \(\ce{O2}\) and facilitates its transport from the lungs to the tissues, where it is used to oxidize food molecules to provide energy. The concentration of hemoglobin in normal blood is about 2.2 mM, and each hemoglobin molecule can bind four \(\ce{O2}\) molecules. Although the concentration of dissolved \(\ce{O2}\) in blood serum at 37°C (normal body temperature) is only 0.010 mM, the total dissolved \(\ce{O2}\) concentration is 8.8 mM, almost a thousand times greater than would be possible without hemoglobin. Synthetic oxygen carriers based on fluorinated alkanes have been developed for use as an emergency replacement for whole blood. Unlike donated blood, these “blood substitutes” do not require refrigeration and have a long shelf life. Their very high Henry’s law constants for \(\ce{O2}\) result in dissolved oxygen concentrations comparable to those in normal blood. A Discussing Henry's Law. Link: The Henry’s law constant for \(\ce{O2}\) in water at 25°C is \(1.27 \times 10^{-3} M/atm\), and the mole fraction of \(\ce{O2}\) in the atmosphere is 0.21. Calculate the solubility of \(\ce{O2}\) in water at 25°C at an atmospheric pressure of 1.00 atm. : Henry’s law constant, mole fraction of \(\ce{O2}\), and pressure : : A According to , the partial pressure of \(\ce{O2}\) is proportional to the mole fraction of \(\ce{O2}\): \[\begin{align*} P_A &= \chi_A P_t \\[4pt] &= (0.21)(1.00\; atm) \\[4pt] &= 0.21\; atm \end{align*} \nonumber \] B From Henry’s law, the concentration of dissolved oxygen under these conditions is \[\begin{align*} [\ce{CO2}] &= k P_{\ce{O2}} \\[4pt] &=(1.27 \times 10^{-3}\; M/\cancel{atm}) (0.21\; \cancel{atm}) \\[4pt] &=2.7 \times 10^{-4}\; M \end{align*} \nonumber \] To understand why soft drinks “fizz” and then go “flat” after being opened, calculate the concentration of dissolved \(\ce{CO2}\) in a soft drink: \(0.17 M\) \(1 \times 10^{-5} M\) The solubility of most substances depends strongly on the temperature and, in the case of gases, on the pressure. The solubility of most solid or liquid solutes increases with increasing temperature. The components of a mixture can often be separated using fractional crystallization, which separates compounds according to their solubilities. The solubility of a gas decreases with increasing temperature. Henry’s law describes the relationship between the pressure and the solubility of a gas. | 14,099 | 1,523 |
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We described the defining characteristics of oxidation–reduction, or redox, reactions in . Most of the reactions we considered there were relatively simple, and balancing them was straightforward. When oxidation–reduction reactions occur in aqueous solution, however, the equations are more complex and can be more difficult to balance by inspection. Because a balanced chemical equation is the most important prerequisite for solving any stoichiometry problem, we need a method for balancing oxidation–reduction reactions in aqueous solution that is generally applicable. One such method uses , and a second is referred to as the method. We show you how to balance redox equations using oxidation states in this section; the half-reaction method is discussed in the second semester To balance a redox equation using the oxidation state method , 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. Equation 4.9.1 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 Section 3.5:
\[ \begin{matrix}
+2 & +4 & +1 & +3 & +2 & +1\; \; \\
Cr^{2+}\left ( aq \right ) &+\; \; \; \; \; \; \; \;MnO_{2}\left ( aq \right ) &+\; \; H^{+}\left ( aq \right ) &\rightarrow\; Cr^{3+}\left ( aq \right ) &+\;Mn^{2+} \left ( aq \right ) &+\; \; \; H_{2}O\left ( l \right ) \\
& \; \; \; \; \; \; \;-2 & & & & \; \; \; \; -2
\end{matrix} \] 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:
\[ Cr^{2+} + Mn^{4+} \rightarrow Cr^{3+} + Mn^{2+} \] The oxidation can be written as
\[ Cr^{2+} \rightarrow Cr^{3+} + e^- \] and the reduction as
\[ Mn^{4+} + 2e^- \rightarrow Mn^{2+} \] 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
\[ 2Cr^{2+} \rightarrow 2Cr^{3+} + 2e^- \] 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:
\[ 2Cr^{2+} \rightarrow 2Cr^{3+} + 2e^- \] \[ Mn^{4+} + 2e^- \rightarrow Mn^{2+} \] 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:
\[ \begin{matrix}
2Cr^{2+}\left ( aq \right ) & \rightarrow & 2Cr^{3+}\left ( aq \right )+\cancel{2e^{-}}\\
& & & \\
2Cr^{2+}\left ( aq \right )+\cancel{2e^{-}}& \rightarrow & Mn^{2+}\left ( aq \right )\\
& & & \\
2Cr^{2+}\left ( aq \right )+2Cr^{2+}\left ( aq \right )&\rightarrow & 2Cr^{3+}\left ( aq \right ) + Mn^{2+}\left ( aq \right )
\end{matrix} \] 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 \[ 2Cr^{2+}(aq) + MnO_2(s) + 4H^+(aq) \rightarrow 2Cr^{3+}(aq) + Mn^{2+}(aq) \] 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: \[ 2Cr^{2+}(aq) + MnO_2(s) + 4H^+(aq) \rightarrow 2Cr^{3+}(aq) + Mn^{2+}(aq) + 2H_2O(l) \] 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 Table \(\Page {1}\) and illustrated in Example 17. \(\Page {1}\) 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: \(H_3AsO_4(aq) + Zn(s) \rightarrow AsH_3(g) + Zn^{2+}(aq) \notag \) reactants and products in acidic solution balanced chemical equation using oxidation states Follow the procedure given in Table \(\Page {1}\) for balancing a redox equation using oxidation states. When you are done, be certain to check that the equation is balanced. \( Reduction \textrm: \: As^{5+} + 8e^- \rightarrow \underset{-3}{As ^{3-}} \) Each zinc atom in elemental zinc is oxidized from 0 to +2, which requires the loss of two electrons per zinc atom: \( Oxidation \textrm:\: Zn \rightarrow Zn^{2+} + 2e^- \) which then yields Although we have not explicitly balanced H atoms, each side of the equation has 11 H atoms. \(\begin{align} & Atoms \textrm : \: 1As + 4Zn + 4O + 11H = 1As + 4Zn + 4O + 11H \\
& Total\: charge \textrm : \: 8(+1) = 4(+2) = +8 \notag \end{align}\) The balanced chemical equation for the reaction is therefore: \(H_3AsO_4(aq) + 4Zn(s) + 8H^+(aq) \rightarrow AsH_3(g) + 4Zn^{2+}(aq) + 4H_2O(l) \) 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 Table \(\Page {1}\) 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. \( \overset{0}{Al} (s) + \overset{+1}{H}_2 O(aq) \rightarrow [ \overset{+3}{Al} (OH)_4 ]^- (aq) + \overset{0}{H_2} (g) \notag \) \( \begin{align}
& Reduction\textrm : \: \dfrac{3}{2} H_2 O + 3e^- \rightarrow \dfrac{3}{2} H_2 \\
& Oxidation \textrm : \: Al \rightarrow [ Al ( OH )_4 ]^- + 3e^-
\end{align}\) To remove the fractional coefficients, multiply both sides of the equation by 2: Thus 3 mol of H gas are produced for every 2 mol of Al. 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: Compounds of elements in high oxidation states (such as ClO , NO , MnO , Cr O , and UF ) tend to act as and in chemical reactions. Compounds of elements in low oxidation states (such as CH , NH , H S, and HI) tend to act as and in chemical reactions. 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 (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:
\[ Fe(s) + 2HCl(aq) \rightarrow FeCl_2(aq) + H_2(g) \] In subsequent steps, FeCl undergoes oxidation to form a reddish-brown precipitate of Fe(OH) . \(\Page {1}\) Rust Formation Many metals dissolve through reactions of this type, which have the general form
\[metal + acid \rightarrow salt + hydrogen \] 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:
\[ Pb(s) + 2H^+(aq) \rightarrow Pb^{2+}(aq) + H_2(g) \] 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 \(\Page {13}\) ) and the reduction of silver salts by copper (Equation \(\Page {14}\) and Figure \(\Page {2}\) ):
\[ Zn(s) + Fe^{2+}(aq) \rightarrow Zn^{2+}(aq) + Fe(s) \]
\( Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s) \tag{8.10.14}\) The reaction in Equation \(\Page {13}\) 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. \(\Page {2}\) The Single-Displacement Reaction of Metallic Copper with a Solution of Silver Nitrate 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, we saw in and that 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:
\[ Zn(s) + Mg^{2+}(aq) \cancel{\rightarrow} Zn^{2+}(aq) + Mg(s) \]
\[ Mg(s) + Zn^{2+}(aq) \rightarrow Mg^{2+}(aq) + Zn(s) \] Magnesium has a greater tendency to be oxidized than zinc does. Pairwise reactions of this sort are the basis of the activity series (Figure \(\Page {3}\) ), which lists metals and hydrogen in order of their relative tendency to be oxidized. The metals at the top of the series, which have the greatest tendency to lose electrons, are the alkali metals (group 1), the alkaline earth metals (group 2), and Al (group 13). In contrast, the metals at the bottom of the series, which have the lowest tendency to be oxidized, are the precious metals or coinage metals—platinum, gold, silver, and copper, and mercury, which are located in the lower right portion of the metals in the periodic table. You should be generally familiar with which kinds of metals are active metals (located at the top of the series) and which are inert metals (at the bottom of the series). \(\Page {3}\) The Activity Series When using the activity series to predict the outcome of a reaction, keep in mind that . Because magnesium is above zinc in Figure \(\Page {3}\), 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 19 demonstrates how a familiarity with the activity series allows you to predict the products of many single-displacement reactions. We will return to the activity series when we discuss oxidation–reduction reactions in more detail in Chapter 19 . 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 Locate the reactants in the activity series in Figure \(\Page {3}\) and from their relative positions, predict whether a reaction will occur. If a reaction does occur, identify which metal is oxidized and which is reduced. Write the net ionic equation for the redox reaction. \( Al(s) + 3Ag^+(aq) \rightarrow Al^{3+}(aq) + 3Ag(s) \) 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. The white solid is lead(II) sulfate, formed from the reaction of solid lead with a solution of sulfuric acid. 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 8.9.1), 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. 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. Of the following elements, which would you expect to have the greatest tendency to be oxidized: Zn, Li, or S? Explain your reasoning. Of these elements, which would you expect to be easiest to reduce: Se, Sr, or Ni? Explain your reasoning. Which of these metals produces H in acidic solution? Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation. 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: ( ) | 19,019 | 1,525 |
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One of the most commonly used and most attractive approximations is the . This approximation can be applied to the rate of change of concentration of a highly reactive (short lived) intermediate that holds a constant value over a long period of time. The advantage here is that for such an intermediate (\(I\)), \[\dfrac{d[I]}{dt} = 0 \nonumber \] So long as one can write an expression for the rate of change of the concentration of the intermediate \(I\), the steady state approximation allows one to solve for its constant concentration. For example, if the reaction \[A +B \rightarrow C \label{total} \] is proposed to follow the mechanism \[\begin{align} A + A &\xrightarrow{k_1} A_2 \\[4pt] A_2 + B &\xrightarrow{k_2} C + A \end{align} \nonumber \] The time-rate of change of the concentration of the intermediate \(A_2\) can be written as \[ \dfrac{d[A_2]}{dt} = k_1[A]^2 - k_2[A_2,B] \nonumber \] In the limit that the steady state approximation can be applied to \(A_2\) \[ \dfrac{d[A_2]}{dt} = k_1[A]^2 - k_2[A_2,B] \approx 0 \nonumber \] or \[ [A_2] \approx \dfrac{k_1[A]^2}{k_2[B]} \nonumber \] So if the rate of the overall reaction is expressed as the rate of formation of the product \(C\), \[ \dfrac{d[C]}{dt} = k_2[A_2,B] \nonumber \] the above expression for \([A_2]\) can be substituted \[ \dfrac{d[C]}{dt} = k_2 \left ( \dfrac{k_1[A]^2}{k_2[B]} \right) [B] \nonumber \] of \[ \dfrac{d[C]}{dt} = k_1[A]^2 \nonumber \] and the reaction is predicted to be second order in \([A]\). Alternatively, if the mechanism for Equation \ref{total} is proposed to be \[\begin{align} A &\xrightarrow{k_1} A^* \\[4pt] A^* + B &\xrightarrow{k_2} C \end{align} \nonumber \] then the rate of change of the concentration of \(A^*\) is \[\dfrac{[A^*]}{dt} = k_1[A] - k_2[A^*,B] \nonumber \] And if the steady state approximation holds, then \[[A^*] \approx \dfrac{k_1[A]}{k_2[B]} \nonumber \] So the rate of production of \(C\) is \[\begin{align} \dfrac{d[C]}{dt} &= k_2[A^*,B] \\[4pt] &= \bcancel{k_2} \left( \dfrac{k_1[A]}{\bcancel{k_2} \cancel{[B]}} \right) \cancel{[B]} \end{align} \nonumber \] or \[\dfrac{d[C]}{dt} = k_1[A] \nonumber \] and the rate law is predicted to be first order in \(A\). In this manner, the plausibility of either of the two reaction mechanisms is easily deduced by comparing the predicted rate law to that which is observed. If the prediction cannot be reconciled with observation, then the scientific method eliminates that mechanism from consideration. | 2,500 | 1,528 |
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Adsorbed species on single crystal surfaces are frequently found to exhibit long-range ordering ; that is to say that the adsorbed species form a well-defined overlayer structure. Each particular structure may only exist over a limited coverage range of the adsorbate, and in some adsorbate/substrate systems a whole progression of adsorbate structures are formed as the surface coverage is gradually increased. This section deals with the classification of such ordered structures - in most cases this involves describing the overlayer structure in terms of the underlying structure of the substrate. There are two principal methods for specifying the structure: Before we start to discuss overlayer structures, however, we need to make sure that we can adequately describe the structure of the substrate ! The primitive unit cell is the simplest periodically repeating unit which can be identified in an ordered array - the array in this instance being the ordered arrangement of surface atoms. By repeated translation of a unit cell, the whole array can be constructed. Let us consider the clean surface structures of the low index surface planes of fcc metals. The (100) surface has 4-fold rotational symmetry ("square symmetry") - perhaps it should not surprise us therefore to find that the primitive unit cell for this surface is square in shape ! Two possible choices of unit cell are highlighted - it is clear that a unit cell of this size is indeed going to be the simplest possible repeating unit for this surface. The two alternatives drawn are in fact but two of an infinite number of possibilities; they have the same shape/symmetry, size and orientation, differing only in their translational position or "origin". Whichever we choose then it is clear that we can indeed generate the whole surface structure by repeated translation of the unit cell; for example .... In fact, I personally prefer the alternative choice of unit cell which has the corners of the unit cell coincident with the atomic centres. We now need to think how to define the unit cell shape, size and symmetry - this is best done using two vectors which have a common origin and define two sides of the unit cell ... For this (100) surface the two vectors which define the unit cell, conventionally called & , are: By convention, one also selects the vectors such that you go anticlockwise from to get to . The length of the vectors & is related to the bulk unit cell parameter, , by | | = | | = / √2 ] In the case of the (110) surface, which has 2-fold rotational symmetry, the unit cell is rectangular By convention, | | > | | - if we also recall the convention that one goes anticlockwise to get from to , then this leads to the choice of vectors shown. With the (111) surface we again have a situation where the length of the two vectors are the same i.e. | | = | | . We can either keep the angle between the vectors less than 90 degrees or let it be greater than 90 degrees. The normal convention is to choose the latter, i.e. the right hand cell of the two illustrated with an angle of 120 degrees between the two vectors. If we have an ordered overlayer of adsorbed species (atoms or molecules), then we can use the same basic ideas as outlined in the previous section to define the structure. The adsorbate unit cell is usually defined by the two vectors and . To avoid ambiguities, it again helps if we stick to a set of conventions in choosing the unit cell vectors. In this case: Once the unit cell vectors for substrate and adsorbate have been selected then it is a relatively simple matter to work out how to denote the structure. Wood's notation is the simplest and most frequently used method for describing a surface structure - it only works, however, if the two unit cells are of the same symmetry or closely-related symmetries (more specifically, the angle between & must be the same as that between & ). In essence, Wood's notation first involves specifying the lengths of the two overlayer vectors, & in terms of & respectively - this is then written in the format: \[( |b_1|/|a_1| \times |b_2|/|a_2| )\] i.e. a ( 2 x 2 ) structure has | | = 2| | and | | = 2| | . The following diagram shows a ( 2 x 2 ) adsorbate overlayer on an (100) surface in which the adsorbate is bonded terminally on-top of individual atoms of the substrate. The unit cells of the (100) substrate and the ( 2 x 2 ) overlayer are both highlighted. The next diagram shows another ( 2 x 2 ) structure, but in this case the adsorbate species is bonded in the four-fold hollows of the substrate surface. Of course, only a very limited section of the structure can be shown here - in practice, the unit cell shown would repeat to give a complete overlayer structure extending across the substrate surface. The highlighted unit cells of the adsorbate and substrate are identical in size, shape and orientation to those of the previously illustrated ( 2 x 2 ) structure. Both this and the previous structure are examples of ( 2 x 2 ), or p( 2 x 2 ), structures. That is to say that they are indeed the simplest unit cells that may be used to describe the overlayer structure, and contain only one "repeat unit". For the purposes of this tutorial I shall follow common practice and omit the preceding "p" - referring to such structures simply as ( 2 x 2 ) structures (in spoken language, " " structures). Such ( 2 x 2 ) structures are also found on other surfaces, but they may differ markedly in superficial appearance from the structure on the (100) surface. The following diagram, for example, shows a ( 2 x 2 ) structure on a (110) surface The adsorbate unit cell is again twice as large as that of the substrate in both dimensions - it retains the same aspect ratio as the rectangular substrate unit cell (1: 1.414) and does not exhibit any rotation with respect to the substrate cell. The following diagram shows yet another ( 2 x 2 ) structure, in this case on the fcc(111) surface ... Again, the adsorbate unit cell is of the same symmetry as the substrate cell but is scaled up by a factor of two in its linear dimensions (and corresponds to a surface area four times as large as that of the substrate unit cell). The next example is a surface structure which is closely related to the ( 2 x 2 ) structure: it differs in that there is an additional atom in the middle/centre of the ( 2 x 2 ) adsorbate unit cell. Since the middle atom is "crystallographically equivalent" to those at the corners (i.e. it is not distinguishable by means of different coordination to the underlying substrate or any other structural feature), then this is no longer a primitive ( 2 x 2 ) structure. Instead it may be classified in one of two ways: In using the latter Wood's notation we are stating that the adsorbate unit cell is a factor of 2 larger than the substrate unit cell in both directions and is also rotated by 45 degrees with respect to the substrate unit cell. If the "central" atom is not completely crystallographically equivalent, then the structure formally remains a p(2x2) unit cell but now has a basis of two adsorbate atoms per unit cell. In some instances it is possible to use a centred unit cell description for a structure for which the primitive unit cell cannot be described using Wood's notation - for example, the c( 2 x 2 ) structure on the (110) surface shown below. As a final example, the next diagram illustrates a commonly-observed structure on (111) surfaces which can be readily described using Wood's notation. (You should confirm for yourself that the adsorbate unit cell is indeed scaled up from the substrate cell by the factor given and rotated by 30 degrees ! ) This is a much more general system of describing surface structures which can be applied to all ordered overlayers: quite simply it relates the vectors & to the substrate vectors & using a simple matrix i.e. remember , , and are vectors. To illustrate the use of matrix notation we shall now consider two surface structures with which we are already familiar ... For the (2 x 2) structure we have: By contrast, for the c(2 x 2) structure: we have Ordered surface structures may be described by defining the adsorbate unit cell in terms of that of the underlying substrate using: | 8,320 | 1,529 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/12%3A_Equilibrium_Conditions_in_Multicomponent_Systems/12.09%3A_Reaction_Equilibria |
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\( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \)
\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) The definition of the thermodynamic equilibrium constant of a reaction or other chemical process is given by Eq. 11.8.9: \begin{equation} K = \prod_i (a_i)\eq^{\nu_i} \tag{12.9.1} \end{equation} The activity \(a_i\) of each reactant or product species is based on an appropriate standard state. We can replace each activity on the right side of Eq. 12.9.1 by an expression in Table 12.2. For example, consider the following heterogeneous equilibrium that is important in the formation of limestone caverns: \[ \ce{CaCO3}\tx{(cr, calcite)} + \ce{CO2}\tx{(g)} + \ce{H2O}\tx{(sln)} \arrows \ce{Ca^2+}\tx{(aq)} + \ce{2HCO3-}\tx{(aq)} \] If we treat H\(_2\)O as a solvent and Ca\(^{2+}\) and HCO\(_3\)\(^-\) as the solute species, then we write the thermodynamic equilibrium constant as follows: \begin{equation} K = \frac{a_+ a_-^2}{a\subs{CaCO\(_3\)} a\subs{CO\(_2\)} a\subs{H\(_2\)O}} = \G\subs{r} \frac{\g_+\g_-^2m_+m_-^2/(m\st)^3} {\left(\fug\subs{CO\(_2\)}/p\st\right)\g\subs{H\(_2\)O} x\subs{H\(_2\)O}} \tag{12.9.2} \end{equation} The subscripts \(+\) and \(-\) refer to the Ca\(^{2+}\) and HCO\(_3\)\(^-\) ions, and all quantities are for the system at reaction equilibrium. \(\G\subs{r}\) is the proper quotient of pressure factors, given for this reaction by \begin{equation} \G\subs{r} = \frac{\G_+\G_-^2}{\G\subs{CaCO\(_3\)}\G\subs{H\(_2\)O}} \tag{12.9.3} \end{equation} Unless the pressure is very high, we can with little error set the value of \(\G\subs{r}\) equal to unity. The product \(\G_+\G_-^2\) in the numerator of Eq. 12.9.3 is the pressure factor \(\G\mbB\) for the solute Ca(HCO\(_3\))\(_2\) (see Eq. 10.3.11). Equation 12.9.2 is an example of a “mixed” equilibrium constant—one using more than one kind of standard state. From the definition of the mean ionic activity coefficient (Eq. 10.3.7), we can replace the product \(\g_+\g_-^2\) by \(\g_{\pm}^3\), where \(\g_{\pm}\) is the mean ionic activity coefficient of aqueous Ca(HCO\(_3\))\(_2\): \begin{equation} K = \G\subs{r} \frac{\g_{\pm}^3 m_+m_-^2/(m\st)^3} {\left(\fug\subs{CO\(_2\)}/p\st\right)\g\subs{H\(_2\)O} x\subs{H\(_2\)O}} \tag{12.9.4} \end{equation} Instead of treating the aqueous Ca\(^{2+}\) and HCO\(_3\)\(^-\) ions as solute species, we can regard the dissolved Ca(HCO\(_3\))\(_2\) electrolyte as the solute and write \begin{equation} K = \frac{a\mbB}{a\subs{CaCO\(_3\)} a\subs{CO\(_2\)} a\subs{H\(_2\)O}} \tag{12.9.5} \end{equation} We then obtain Eq. 12.9.4 by replacing \(a\mbB\) with the expression in Table 12.2 for an electrolyte solute. The value of \(K\) depends only on \(T\), and the value of \(\G\subs{r}\) depends only on \(T\) and \(p\). Suppose we dissolve some NaCl in the aqueous phase while maintaining the system at constant \(T\) and \(p\). The increase in the ionic strength will alter \(\g_{\pm}\) and necessarily cause a compensating change in the solute molarity in order for the system to remain in reaction equilibrium. An example of a different kind of reaction equilibrium is the dissociation (ionization) of a weak monoprotic acid such as acetic acid \[ \ce{HA}\tx{(aq)} \arrows \ce{H+}\tx{(aq)} + \ce{A-}\tx{(aq)} \] for which the thermodynamic equilibrium constant (the ) is \begin{equation} K\subs{a} = \G\subs{r} \frac{\g_+\g_-m_+m_-}{\g\subs{\(m\),HA} m\subs{HA}m\st} = \G\subs{r} \frac{\g_{\pm}^2 m_+m_-}{\g\subs{\(m\),HA} m\subs{HA}m\st} \tag{12.9.6} \end{equation} Suppose the solution is prepared from water and the acid, and H\(^+\) from the dissociation of H\(_2\)O is negligible compared to H\(^+\) from the acid dissociation. We may then write \(m_+=m_-=\alpha m\B\), where \(\alpha\) is the degree of dissociation and \(m\B\) is the overall molality of the acid. The molality of the undissociated acid is \(m\subs{HA}=(1-\alpha)m\B\), and the dissociation constant can be written \begin{equation} K\subs{a}=\G\subs{r} \frac{\g_{\pm}^2 \alpha^2m\B/m\st} {\g\subs{\(m\),HA}(1-\alpha)} \tag{12.9.7} \end{equation} From this equation, we see that a change in the ionic strength that decreases \(\g_{\pm}\) when \(T\), \(p\), and \(m\B\) are held constant must increase the degree of dissociation (Prob. 12.17). | 11,501 | 1,530 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/IV._Nucleophilic_Substitution_Reactions/B._What_is_Nucleophilic_Substitution |
Halogenoalkanes (also known as haloalkanes or alkyl halides) are compounds containing a halogen atom (fluorine, chlorine, bromine or iodine) joined to one or more carbon atoms in a chain. The interesting thing about these compounds is the carbon-halogen bond, and all the nucleophilic substitution reactions of the halogenoalkanes involve breaking that bond. With the exception of iodine, all of the halogens are more electronegative than carbon. That means that the electron pair in the carbon-halogen bond will be dragged towards the halogen end, leaving the halogen slightly negative ( -) and the carbon slightly positive ( +) - except in the carbon-iodine case. Although the carbon-iodine bond doesn't have a permanent dipole, the bond is very easily polarized by anything approaching it. Imagine a negative ion approaching the bond from the far side of the carbon atom: The fairly small polarity of the carbon-bromine bond will be increased by the same effect. In all of these nucleophilic substitution reactions, the carbon-halogen bond has to be broken at some point during the reaction. In general, the harder it is to break, the slower the reaction will be. Look at the strengths of various bonds (all values in kJ mol ). The carbon-fluorine bond is very strong (stronger than C-H) and isn't easily broken. It doesn't matter that the carbon-fluorine bond has the greatest polarity - the strength of the bond is much more important in determining its reactivity. You might therefore expect fluoroalkanes to be very unreactive - and they are! We shall simply ignore them from now on. In the other halogenoalkanes, the bonds get weaker as you go from chlorine to bromine to iodine. That means that chloroalkanes react most slowly, bromoalkanes react faster, and iodoalkanes react faster still. Where "<" is read as "is less than" - or, in this instance, "is slower than", and represents any alkyl group. You will need to know about this if your syllabus talks about "primary halogenoalkanes" or about S 2 reactions. If the syllabus is vague, check recent exam papers and mark schemes, and compare them against what follows. A nucleophile is a species (an ion or a molecule) which is strongly attracted to a region of positive charge in something else. Nucleophiles are either fully negative ions, or else have a strongly - charge somewhere on a molecule. Common nucleophiles are hydroxide ions, cyanide ions, water and ammonia. Notice that each of these contains at least one lone pair of electrons, either on an atom carrying a full negative charge, or on a very electronegative atom carrying a substantial - charge. We'll talk this mechanism through using an ion as a nucleophile, because it's slightly easier. The water and ammonia mechanisms involve an extra step which you can read about on the pages describing those particular mechanisms. We'll take bromoethane as a typical primary halogenoalkane. The bromoethane has a polar bond between the carbon and the bromine. We'll look at its reaction with a general purpose nucleophilic ion which we'll call Nu . This will have at least one lone pair of electrons. Nu could, for example, be OH or CN . The lone pair on the Nu ion will be strongly attracted to the + carbon, and will move towards it, beginning to make a co-ordinate (dative covalent) bond. In the process the electrons in the C-Br bond will be pushed even closer towards the bromine, making it increasingly negative. The movement goes on until the -Nu is firmly attached to the carbon, and the bromine has been expelled as a Br ion. The Nu ion approaches the \(\delta^+\) carbon from the side away from the bromine atom. The large bromine atom hinders attack from its side and, being \(\delta^-\), would repel the incoming Nu anyway. This attack from the back is important if you need to understand why tertiary halogenoalkanes have a different mechanism. We'll discuss this later on this page. There is obviously a point in which the Nu is half attached to the carbon, and the C-Br bond is half way to being broken. This is called a transition state. It isn't an intermediate. You can't isolate it - even for a very short time. It's just the mid-point of a smooth attack by one group and the departure of another. A major factor affecting the S 2 reaction is strength of the nucleophile. Recall that a nucleophile is a molecule which is attracted to positive charge. We can describe several general trends which determine the strength of a nucleophile: In addition to the strength of the nucleophile, the structure of the nucleophile also influences the capacity of an S 2 reaction to proceed. As stated many times previously, steric effects are the most important factor in determining whether an S 2 reaction can proceed. We have discussed these effects with regard to the substrate, but they also apply to the nucleophile. A small, unbranched nucleophile will be more effective in an S 2 reaction than a large, branched nucleophile. Based on the factors described above, we can outline several trends that affect the likelihood of an S 2 reaction occurring. As described above, the most important factor in deciding whether an S 2 reaction occurs is steric effects. When we describe the nature of the molecule being attacked, called the , we can look at how many alkyl substituents are present on the molecule. For example, the molecule may contain a carbon atom attached to a leaving group and three hydrogen atoms. This would constitute a carbon atom, in analogy to methane. Given that such a molecule has almost no (substituents blocking access to the carbon atom), an S 2 reaction will be highly favored in a molecule with a methyl carbon. We call a carbon atom with one alkyl substituent a carbon atom. S 2 reactions will proceed easily with a primary carbon atom, though not as fast as with a methyl carbon. If there are two alkyl substituents, we call the carbon atom . It is possible for an S 2 reaction to proceed at a secondary carbon atom, but it is not as favorable as in the methyl or primary cases. Finally, if there are three alkyl substituents attached to the carbon atom, we have the case of a carbon atom. In this case, there is far too much steric hindrance, and the S 2 reaction cannot proceed. The principle that increasing substitution leads to decreasing reactivity is outlined in the following table: The degree of substitution of the molecule undergoing attack is not the only factor that influences the rate of an S 2 reaction. Recall that steric effects (the likelihood that attack by the nucleophile will be blocked) are of paramount importance in determining whether an S 2 reaction will take place. Degree of substitution is one factor in determining steric effects, but it is not the only one. We also need to consider the bulkiness of the substituents. The simplest way is: Technically, this is known as an S 2 reaction. S stands for substitution, N for nucleophilic, and the 2 is because the initial stage of the reaction involves two species - the bromoethane and the Nu ion. If your syllabus doesn't refer to S 2 reactions by name, you can just call it nucleophilic substitution. Some examiners like you to show the transition state in the mechanism, in which case you need to write it in a bit more detail - showing how everything is arranged in space. Be very careful when you draw the transition state to make a clear difference between the dotted lines showing the half-made and half-broken bonds, and those showing the bonds going back into the paper. Notice that the molecule has been inverted during the reaction - rather like an umbrella being blown inside-out. Remember that a tertiary halogenoalkane has three alkyl groups attached to the carbon with the halogen on it. These alkyl groups can be the same or different, but in this section, we shall just consider a simple one, (CH ) CBr - 2-bromo-2-methylpropane. Once again, we'll talk this mechanism through using an ion as a nucleophile, because it's slightly easier, and again we'll look at the reaction of a general purpose nucleophilic ion which we'll call Nu . This will have at least one lone pair of electrons. You will remember that when a nucleophile attacks a primary halogenoalkane, it approaches the + carbon atom from the side away from the halogen atom. With a tertiary halogenoalkane, this is impossible. The back of the molecule is completely cluttered with CH groups. Since any other approach is prevented by the bromine atom, the reaction has to go by an alternative mechanism. The reaction happens in two stages. In the first, a small proportion of the halogenoalkane ionises to give a carbocation and a bromide ion. This reaction is possible because tertiary carbocations are relatively stable compared with secondary or primary ones. Even so, the reaction is slow. Once the carbocation is formed, however, it would react immediately it came into contact with a nucleophile like Nu . The lone pair on the nucleophile is strongly attracted towards the positive carbon, and moves towards it to create a new bond. How fast the reaction happens is going to be governed by how fast the halogenoalkane ionises. Because this initial slow step only involves one species, the mechanism is described as S 1 - substitution, nucleophilic, one species taking part in the initial slow step. If a primary halogenoalkane did use this mechanism, the first step would be, for example: A primary carbocation would be formed, and this is much more energetically unstable than the tertiary one formed from tertiary halogenoalkanes - and therefore much more difficult to produce. This instability means that there will be a very high activation energy for the reaction involving a primary halogenoalkane. The activation energy is much less if it undergoes an S 2 reaction - and so that's what it does instead. There isn't anything new in this. Secondary halogenoalkanes will use both mechanisms - some molecules will react using the S 2 mechanism and others the S 1. The S 2 mechanism is possible because the back of the molecule isn't completely cluttered by alkyl groups and so the approaching nucleophile can still get at the + carbon atom. The S 1 mechanism is possible because the secondary carbocation formed in the slow step is more stable than a primary one. It isn't as stable as a tertiary one though, and so the S 1 route isn't as effective as it is with tertiary halogenoalkanes. Jim Clark ( ) (McGill University) | 10,490 | 1,531 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/12%3A_Equilibrium_Conditions_in_Multicomponent_Systems/12.07%3A_Membrane_Equilibria |
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\( \newcommand{\timesten}[1]{\mbox{$\,\times\,10^{#1}$}}\)
\( \newcommand{\per}{^{-1}} % minus one power\)
\( \newcommand{\m}{_{\text{m}}} % subscript m for molar quantity\)
\( \newcommand{\CVm}{C_{V,\text{m}}} % molar heat capacity at const.V\)
\( \newcommand{\Cpm}{C_{p,\text{m}}} % molar heat capacity at const.p\)
\( \newcommand{\kT}{\kappa_T} % isothermal compressibility\)
\( \newcommand{\A}{_{\text{A}}} % subscript A for solvent or state A\)
\( \newcommand{\B}{_{\text{B}}} % subscript B for solute or state B\)
\( \newcommand{\bd}{_{\text{b}}} % subscript b for boundary or boiling point\)
\( \newcommand{\C}{_{\text{C}}} % subscript C\)
\( \newcommand{\f}{_{\text{f}}} % subscript f for freezing point\)
\( \newcommand{\mA}{_{\text{m},\text{A}}} % subscript m,A (m=molar)\)
\( \newcommand{\mB}{_{\text{m},\text{B}}} % subscript m,B (m=molar)\)
\( \newcommand{\mi}{_{\text{m},i}} % subscript m,i (m=molar)\)
\( \newcommand{\fA}{_{\text{f},\text{A}}} % subscript f,A (for fr. pt.)\)
\( \newcommand{\fB}{_{\text{f},\text{B}}} % subscript f,B (for fr. pt.)\)
\( \newcommand{\xbB}{_{x,\text{B}}} % x basis, B\)
\( \newcommand{\xbC}{_{x,\text{C}}} % x basis, C\)
\( \newcommand{\cbB}{_{c,\text{B}}} % c basis, B\)
\( \newcommand{\mbB}{_{m,\text{B}}} % m basis, B\)
\( \newcommand{\kHi}{k_{\text{H},i}} % Henry's law constant, x basis, i\)
\( \newcommand{\kHB}{k_{\text{H,B}}} % Henry's law constant, x basis, B\)
\( \newcommand{\arrow}{\,\rightarrow\,} % right arrow with extra spaces\)
\( \newcommand{\arrows}{\,\rightleftharpoons\,} % double arrows with extra spaces\)
\( \newcommand{\ra}{\rightarrow} % right arrow (can be used in text mode)\)
\( \newcommand{\eq}{\subs{eq}} % equilibrium state\)
\( \newcommand{\onehalf}{\textstyle\frac{1}{2}\D} % small 1/2 for display equation\)
\( \newcommand{\sys}{\subs{sys}} % system property\)
\( \newcommand{\sur}{\sups{sur}} % surroundings\)
\( \renewcommand{\in}{\sups{int}} % internal\)
\( \newcommand{\lab}{\subs{lab}} % lab frame\)
\( \newcommand{\cm}{\subs{cm}} % center of mass\)
\( \newcommand{\rev}{\subs{rev}} % reversible\)
\( \newcommand{\irr}{\subs{irr}} % irreversible\)
\( \newcommand{\fric}{\subs{fric}} % friction\)
\( \newcommand{\diss}{\subs{diss}} % dissipation\)
\( \newcommand{\el}{\subs{el}} % electrical\)
\( \newcommand{\cell}{\subs{cell}} % cell\)
\( \newcommand{\As}{A\subs{s}} % surface area\)
\( \newcommand{\E}{^\mathsf{E}} % excess quantity (superscript)\)
\( \newcommand{\allni}{\{n_i \}} % set of all n_i\)
\( \newcommand{\sol}{\hspace{-.1em}\tx{(sol)}}\)
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\( \newcommand{\pha}{\alpha} % phase alpha\)
\( \newcommand{\phb}{\beta} % phase beta\)
\( \newcommand{\phg}{\gamma} % phase gamma\)
\( \newcommand{\aph}{^{\alpha}} % alpha phase superscript\)
\( \newcommand{\bph}{^{\beta}} % beta phase superscript\)
\( \newcommand{\gph}{^{\gamma}} % gamma phase superscript\)
\( \newcommand{\aphp}{^{\alpha'}} % alpha prime phase superscript\)
\( \newcommand{\bphp}{^{\beta'}} % beta prime phase superscript\)
\( \newcommand{\gphp}{^{\gamma'}} % gamma prime phase superscript\)
\( \newcommand{\apht}{\small\aph} % alpha phase tiny superscript\)
\( \newcommand{\bpht}{\small\bph} % beta phase tiny superscript\)
\( \newcommand{\gpht}{\small\gph} % gamma phase tiny superscript\) \( \newcommand{\upOmega}{\Omega}\) \( \newcommand{\dif}{\mathop{}\!\mathrm{d}} % roman d in math mode, preceded by space\)
\( \newcommand{\Dif}{\mathop{}\!\mathrm{D}} % roman D in math mode, preceded by space\)
\( \newcommand{\df}{\dif\hspace{0.05em} f} % df\) \(\newcommand{\dBar}{\mathop{}\!\mathrm{d}\hspace-.3em\raise1.05ex{\Rule{.8ex}{.125ex}{0ex}}} % inexact differential \)
\( \newcommand{\dq}{\dBar q} % heat differential\)
\( \newcommand{\dw}{\dBar w} % work differential\)
\( \newcommand{\dQ}{\dBar Q} % infinitesimal charge\)
\( \newcommand{\dx}{\dif\hspace{0.05em} x} % dx\)
\( \newcommand{\dt}{\dif\hspace{0.05em} t} % dt\)
\( \newcommand{\difp}{\dif\hspace{0.05em} p} % dp\)
\( \newcommand{\Del}{\Delta}\)
\( \newcommand{\Delsub}[1]{\Delta_{\text{#1}}}\)
\( \newcommand{\pd}[3]{(\partial #1 / \partial #2 )_{#3}} % \pd{}{}{} - partial derivative, one line\)
\( \newcommand{\Pd}[3]{\left( \dfrac {\partial #1} {\partial #2}\right)_{#3}} % Pd{}{}{} - Partial derivative, built-up\)
\( \newcommand{\bpd}[3]{[ \partial #1 / \partial #2 ]_{#3}}\)
\( \newcommand{\bPd}[3]{\left[ \dfrac {\partial #1} {\partial #2}\right]_{#3}}\)
\( \newcommand{\dotprod}{\small\bullet}\)
\( \newcommand{\fug}{f} % fugacity\)
\( \newcommand{\g}{\gamma} % solute activity coefficient, or gamma in general\)
\( \newcommand{\G}{\varGamma} % activity coefficient of a reference state (pressure factor)\)
\( \newcommand{\ecp}{\widetilde{\mu}} % electrochemical or total potential\)
\( \newcommand{\Eeq}{E\subs{cell, eq}} % equilibrium cell potential\)
\( \newcommand{\Ej}{E\subs{j}} % liquid junction potential\)
\( \newcommand{\mue}{\mu\subs{e}} % electron chemical potential\)
\( \newcommand{\defn}{\,\stackrel{\mathrm{def}}{=}\,} % "equal by definition" symbol\) \( \newcommand{\D}{\displaystyle} % for a line in built-up\)
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\( \newcommand{\R}{8.3145\units{J$\,$K$\per\,$mol$\per$}} % gas constant value\)
\( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \)
\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) A semipermeable membrane used to separate two liquid phases can, in principle, be permeable to certain species and impermeable to others. A membrane, however, may not be perfect in this respect over a long time period. We will assume that during the period of observation, those species to which the membrane is supposed to be permeable quickly achieve transfer equilibrium, and only negligible amounts of the other species are transferred across the membrane. Section 12.2.2 sketched a derivation of the conditions needed for equilibrium in a two-phase system in which a membrane permeable only to solvent separates a solution from pure solvent. We can generalize the results for any system with two liquid phases separated by a semipermeable membrane: in an equilibrium state, both phases must have the same temperature, and any species to which the membrane is permeable must have the same chemical potential in both phases. The two phases, however, need not and usually do not have the same pressure. An equilibrium state in a system with two solutions of the same solvent and different solute compositions, separated by a membrane permeable only to the solvent, is called an . We have already seen this kind of equilibrium in an apparatus that measures osmotic pressure (Fig. 12.2). Consider a system with transfer equilibrium of the solvent across a membrane separating phases \(\pha\) and \(\phb\). The phases have equal solvent chemical potentials but different pressures: \begin{equation} \mu\A\bph(p\bph)=\mu\A\aph(p\aph) \tag{12.7.1} \end{equation} The dependence of \(\mu\A\) on pressure in a phase of fixed temperature and composition is given by \(\pd{\mu\A}{p}{T,\allni}=V\A\) (from Eq. 9.2.49), where \(V\A\) is the partial molar volume of A in the phase. If we apply this relation to the solution of phase \(\phb\), treat the partial molar volume \(V\A\) as independent of pressure, and integrate at constant temperature and composition from the pressure of phase \(\pha\) to that of phase \(\phb\), we obtain \begin{equation} \mu\A\bph(p\bph)=\mu\A\bph(p\aph)+V\A\bph(p\bph-p\aph) \tag{12.7.2} \end{equation} By equating the two expressions for \(\mu\A\bph(p\bph)\) and rearranging, we obtain the following expression for the pressure difference needed to achieve transfer equilibrium: \begin{equation} p\bph-p\aph=\frac{\mu\A\aph(p\aph)-\mu\A\bph(p\aph)}{V\A\bph} \tag{12.7.3} \end{equation} The pressure difference can be related to the osmotic pressures of the two phases. From Eq. 12.2.11, the solvent chemical potential in a solution phase can be written \(\mu\A(p)=\mu\A^*(p)-V\A\varPi(p)\). Using this to substitute for \(\mu\A\aph(p\aph)\) and \(\mu\A\bph(p\aph)\) in Eq. 12.7.3, we obtain \begin{equation} p\bph-p\aph=\varPi\bph(p\aph)-\left( \frac{V\A\aph}{V\A\bph} \right) \varPi\aph(p\aph) \tag{12.7.4} \end{equation} Equilibrium dialysis is a useful technique for studying the binding of a small uncharged solute species (a ligand) to a macromolecule. The macromolecule solution is placed on one side of a membrane through which it cannot pass, with a solution without the macromolecule on the other side, and the ligand is allowed to come to transfer equilibrium across the membrane. If the same solute standard state is used for the ligand in both solutions, at equilibrium the unbound ligand must have the same activity in both solutions. Measurements of the total ligand molality in the macromolecule solution and the ligand molality in the other solution, combined with estimated values of the unbound ligand activity coefficients, allow the amount of ligand bound per macromolecule to be calculated. If one of the solutions in a two-phase membrane equilibrium contains certain solute species that are unable to pass through the membrane, whereas other ions can pass through, the situation is more complicated than the osmotic membrane equilibrium described in Sec. 12.7.1. Usually if the membrane is impermeable to one kind of ion, an ion species to which it is permeable achieves transfer equilibrium across the membrane only when the phases have different pressures and different electric potentials. The equilibrium state in this case is a , and the resulting electric potential difference across the membrane is called the . This phenomenon is related to the membrane potentials that are important in the functioning of nerve and muscle cells (although the cells of a living organism are not, of course, in equilibrium states). A Donnan potential can be measured electrically, with some uncertainty due to unknown liquid junction potentials, by connecting silver-silver chloride electrodes (described in Sec. 14.1) to both phases through salt bridges. Consider solution phases \(\pha\) and \(\phb\) separated by a semipermeable membrane. Both phases contain a dissolved salt, designated solute B, that has \(\nu_+\) cations and \(\nu_-\) anions in each formula unit. The membrane is permeable to these ions. Phase \(\phb\) also contains a protein or other polyelectrolyte with a net positive or negative charge, together with counterions of the opposite charge that are the same species as the cation or anion of the salt. The presence of the counterions in phase \(\phb\) prevents the cation and anion of the salt from being present in stoichiometric amounts in this phase. The membrane is impermeable to the polyelectrolyte, perhaps because the membrane pores are too small to allow the polyelectrolyte to pass through. The condition for transfer equilibrium of solute B is \(\mu\B\aph=\mu\B\bph\), or \begin{equation} (\mu\mbB\st)\aph+RT\ln a\mbB\aph=(\mu\mbB\st)\bph+RT\ln a\mbB\bph \tag{12.7.5} \end{equation} Solute B has the same standard state in the two phases, so that \((\mu\mbB\st)\aph\) and \((\mu\mbB\st)\bph\) are equal. The activities \(a\mbB\aph\) and \(a\mbB\bph\) are therefore equal at equilibrium. Using the expression for solute activity from Eq. 10.3.16, which is valid for a multisolute solution, we find that at transfer equilibrium the following relation must exist between the molalities of the salt ions in the two phases: \begin{equation} \G\mbB\aph\left(\g_{\pm}\aph\right)^{\nu} \left(m_+\aph\right)^{\nu_+} \left(m_-\aph\right)^{\nu_-} = \G\mbB\bph\left(\g_{\pm}\bph\right)^{\nu} \left(m_+\bph\right)^{\nu_+} \left(m_-\bph\right)^{\nu_-} \tag{12.7.6} \end{equation} To find an expression for the Donnan potential, we can equate the single-ion chemical potentials of the salt cation: \(\mu_+\aph(\phi\aph)=\mu_+\bph(\phi\bph)\). When we use the expression of Eq. 10.1.15 for \(\mu_+(\phi)\), we obtain \begin{gather} \s{ \phi\aph-\phi\bph = \frac{RT}{z_+F}\ln\frac{\G_+\bph \g_+\bph m_+\bph}{\G_+\aph \g_+\aph m_+\aph} } \tag{12.7.7} \cond{(Donnan potential)} \end{gather} The condition needed for an osmotic membrane equilibrium related to the solvent can be written \begin{equation} \mu\A\bph(p\bph) - \mu\A\aph(p\aph) = 0 \tag{12.7.8} \end{equation} The chemical potential of the solvent is \(\mu\A=\mu\A\st+RT\ln a\A=\mu\A\st+RT\ln(\G\A \g\A x\A)\). From Table 9.6, we have to a good approximation the expression \(RT\ln\G\A = V\A^*(p-p\st)\). With these substitutions, Eq. 12.7.8 becomes \begin{equation} RT\ln\frac{\g\A\bph x\A\bph}{\g\A\aph x\A\aph} + V\A^*\left(p\bph-p\aph\right) = 0 \tag{12.7.9} \end{equation} We can use this equation to estimate the pressure difference needed to maintain an equilibrium state. For dilute solutions, with \(\g\A\aph\) and \(\g\A\bph\) set equal to 1, the equation becomes \begin{equation} p\bph-p\aph \approx \frac{RT}{V\A^*}\ln\frac{x\A\aph}{x\A\bph} \tag{12.7.10} \end{equation} In the limit of infinite dilution, \(\ln x\A\) can be replaced by \(-M\A\sum_{i\ne\tx{A}}m_i\) (Eq. 9.6.12), giving the relation \begin{equation} p\bph-p\aph \approx \frac{M\A RT}{V\A^*}\sum_{i\ne\tx{A}}\left(m_i\bph-m_i\aph\right) = \rho\A^*RT\sum_{i\ne\tx{A}}\left(m_i\bph-m_i\aph\right) \tag{12.7.11} \end{equation} As a specific example of a Donnan membrane equilibrium, consider a system in which an aqueous solution of a polyelectrolyte with a net negative charge, together with a counterion M\(^+\) and a salt MX of the counterion, is equilibrated with an aqueous solution of the salt across a semipermeable membrane. The membrane is permeable to the H\(_2\)O solvent and to the ions M\(^+\) and X\(^-\), but is impermeable to the polyelectrolyte. The species in phase \(\pha\) are H\(_2\)O, M\(^+\), and X\(^-\); those in phase \(\phb\) are H\(_2\)O, M\(^+\), X\(^-\), and the polyelectrolyte. In an equilibrium state, the two phases have the same temperature but different compositions, electric potentials, and pressures. Because the polyelectrolyte in this example has a negative charge, the system has more M\(^+\) ions than X\(^-\) ions. Figure 12.9(a) is a schematic representation of an initial state of this kind of system. Phase \(\phb\) is shown as a solution confined to a closed dialysis bag immersed in phase \(\pha\). The number of cations and anions shown in each phase indicate the relative amounts of these ions. For simplicity, let us assume the two phases have equal masses of water, so that the molality of an ion is proportional to its amount by the same ratio in both phases. It is clear that in the initial state shown in the figure, the chemical potentials of both M\(^+\) and X\(^-\) are greater in phase \(\phb\) (greater amounts) than in phase \(\pha\), and this is a nonequilibrium state. A certain quantity of salt MX will therefore pass spontaneously through the membrane from phase \(\phb\) to phase \(\pha\) until equilibrium is attained. The equilibrium ion molalities must agree with Eq. 12.7.6. We make the approximation that the pressure factors and mean ionic activity coefficients are unity. Then for the present example, with \(\nu_+=\nu_-=1\), the equation becomes \begin{equation} m_+\aph m_-\aph \approx m_+\bph m_-\bph \tag{12.7.12} \end{equation} There is furthermore an electroneutrality condition for each phase: \begin{equation} m\aph_+=m\aph_- \qquad m\bph_+ = m\bph_- + |z\subs{P}|m\subs{P} \tag{12.7.13} \end{equation} Here \(z\subs{P}\) is the negative charge of the polyelectrolyte, and \(m\subs{P}\) is its molality. Substitution of these expressions into Eq. 12.7.12 gives the relation \begin{equation} \left(m\aph_-\right)^2 \approx \left(m\bph_- + |z\subs{P}|m\subs{P}\right)m\bph_- \tag{12.7.14} \end{equation} This shows that in the equilibrium state, \(m\aph_-\) is greater than \(m\bph_-\). Then Eq. 12.7.12 shows that \(m\aph_+\) is less than \(m\bph_+\). These equilibrium molalities are depicted in Fig. 12.9(b). The chemical potential of a cation, its activity, and the electric potential of the phase are related by Eq. 10.1.9: \(\mu_+=\mu_+\st + RT\ln a_+ + z_+F\phi\). In order for M\(^+\) to have the same chemical potential in both phases, despite its lower activity in phase \(\pha\), the electric potential of phase \(\pha\) must be greater than that of phase \(\phb\). Thus the Donnan potential \(\phi\aph-\phi\bph\) in the present example is positive. Its value can be estimated from Eq. 12.7.7 with the values of the single-ion pressure factors and activity coefficients approximated by 1 and with \(z_+\) for this example set equal to 1: \begin{equation} \phi\aph-\phi\bph \approx \frac{RT}{F}\ln\frac{m_+\bph}{m_+\aph} \tag{12.7.15} \end{equation} The existence of a Donnan potential in the equilibrium state is the result of a very small departure of the phases on both sides of the membrane from exact electroneutrality. In the example, phase \(\pha\) has a minute net positive charge and phase \(\phb\) has a net negative charge of equal magnitude. The amount of M\(^+\) ion transferred across the membrane to achieve equilibrium is slightly greater than the amount of X\(^-\) ion transferred; the difference between these two amounts is far too small to be measured chemically. At equilibrium, the excess charge on each side of the membrane is distributed over the boundary surface of the solution phase on that side, and is not part of the bulk phase composition. The pressure difference \(p\bph-p\aph\) at equilibrium can be estimated with Eq. 12.7.11, and for the present example is found to be positive. Without this pressure difference, the solution in phase \(\pha\) would move spontaneously through the membrane into phase \(\phb\) until phase \(\pha\) completely disappears. With phase \(\pha\) open to the atmosphere, as in Fig. 12.9, the volume of phase \(\phb\) must be constrained in order to allow its pressure to differ from atmospheric pressure. If the volume of phase \(\phb\) remains practically constant, the transfer of a minute quantity of solvent across the membrane is sufficient to cause the pressure difference. It should be clear that the existence of a Donnan membrane equilibrium introduces complications that would make it difficult to use a measured pressure difference to estimate the molar mass of the polyelectrolyte by the method of Sec. 12.4, or to study the binding of a charged ligand by equilibrium dialysis. | 20,471 | 1,532 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/25%3A_Solutions_II_-_Nonvolatile_Solutes/25.02%3A_The_Activities_of_Nonvolatile_Solutes |
For non-ideal gases, we introduced in chapter the concept of as an that accounts for non-ideal behavior. If we extend this concept to non-ideal solution, we can introduce the of a liquid or a solid, \(a\), as: \[ \mu_{\text{non-ideal}} = \mu^{- {\ominus} } + RT \ln a, \label{14.1.1} \] where \(\mu\) is the chemical potential of the substance or the mixture, and \(\mu^{-\ominus}\) is the chemical potential at standard state. Comparing this definition to , it is clear that the activity is equal to the fugacity for a non-ideal gas (which, in turn, is equal to the pressure for an ideal gas). However, for a liquid and a liquid mixture, it depends on the chemical potential at standard state. This means that the activity is not an absolute quantity, but rather a relative term describing how “active” a compound is compared to standard state conditions. The choice of the standard state is, in principle, arbitrary, but conventions are often chosen out of mathematical or experimental convenience. We already discussed the convention that standard state for a gas is at \(P^{-\ominus}=1\;\text{bar}\), so the activity is equal to the fugacity. The standard state for a component in a solution is the pure component at the temperature and pressure of the solution. This definition is equivalent to setting the activity of a pure component, \(i\), at \(a_i=1\). For a component in a solution we can use Equation 11.4.2 to write the chemical potential in the gas phase as: \[ \mu_i^{\text{vapor}} = \mu_i^{- {\ominus} } + RT \ln \dfrac{P_i}{P^{-\ominus}}. \label{14.1.2} \] If the gas phase is in equilibrium with the liquid solution, then: \[ \mu_i^{\text{solution}} = \mu_i^{\text{vapor}} = \mu_i^*, \label{14.1.3} \] where \(\mu_i^*\) is the chemical potential of the pure element. Subtracting Equation \ref{14.1.3} from Equation \ref{14.1.2}, we obtain: \[ \mu_i^{\text{solution}} = \mu_i^* + RT \ln \dfrac{P_i}{P^*_i}. \label{14.1.4} \] For an ideal solution, we can use Raoult’s law, Equation 13.1.1, to rewrite Equation \ref{14.1.4} as: \[ \mu_i^{\text{solution}} = \mu_i^* + RT \ln x_i, \label{14.1.5} \] which relates the chemical potential of a component in an ideal solution to the chemical potential of the pure liquid and its mole fraction in the solution. For a non-ideal solution, the partial pressure in Equation \ref{14.1.4} is either larger (positive deviation) or smaller (negative deviation) than the pressure calculated using Raoult’s law. The chemical potential of a component in the mixture is then calculated using: \[ \mu_i^{\text{solution}} = \mu_i^* + RT \ln \left(\gamma_i x_i\right), \label{14.1.6} \] where \(\gamma_i\) is a positive coefficient that accounts for deviations from ideality. This coefficient is either larger than one (for positive deviations), or smaller than one (for negative deviations). The activity of component \(i\) can be calculated as an , using: \[ a_i = \gamma_i x_i, \label{14.1.7} \] where \(\gamma_i\) is defined as the . The partial pressure of the component can then be related to its vapor pressure, using: \[ P_i = a_i P_i^*. \label{14.1.8} \] Comparing Equation \ref{14.1.8} with Raoult’s law, we can calculate the activity coefficient as: \[ \gamma_i = \dfrac{P_i}{x_i P_i^*} = \dfrac{P_i}{P_i^{\text{R}}}, \label{14.1.9} \] where \(P_i^{\text{R}}\) is the partial pressure calculated using Raoult’s law. This result also proves that for an ideal solution, \(\gamma=1\). Equation \ref{14.1.9} can also be used experimentally to obtain the activity coefficient from the phase diagram of the non-ideal solution. This is achieved by measuring the value of the partial pressure of the vapor of a non-ideal solution. Examples of this procedure are reported for both positive and negative deviations in Figure \(\Page {1}\). | 3,832 | 1,533 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/14%3A_Chemical_Kinetics/14.06%3A_Half_Lives_and_Radioactive_Decay_Kinetics |
Another approach to describing reaction rates is based on the time required for the concentration of a reactant to decrease to one-half its initial value. This period of time is called the half-life of the reaction, written as . Thus the half-life of a reaction is the time required for the reactant concentration to decrease from [A] to [A] /2. If two reactions have the same order, the faster reaction will have a shorter half-life, and the slower reaction will have a longer half-life. The half-life of a first-order reaction under a given set of reaction conditions is a constant. This is not true for zeroth- and second-order reactions. The half-life of a first-order reaction is . This becomes evident when we rearrange the integrated rate law for a first-order reaction ( ) to produce the following equation:
\( ln \dfrac{\left [ A \right ]_{0}}{\left [ A \right ]} = kt \tag{14.5.1} \) Substituting [A] /2 for [A] and for (to indicate a half-life) into gives \( ln \dfrac{\left [ A \right ]_{0}}{\left [ A \right ]_{0}/2} = ln 2=kt_{1/2} \tag{14.5.2} \) The natural logarithm of 2 (to three decimal places) is 0.693. Substituting this value into the equation, we obtain the expression for the half-life of a first-order reaction:
\( t_{1/2}=\dfrac{0.693}{k} \tag{14.5.3} \) Thus, for a first-order reaction, each successive half-life is the same length of time, as shown in , and is independent of [A]. If we know the rate constant for a first-order reaction, then we can use half-lives to predict how much time is needed for the reaction to reach a certain percent completion. As you can see from this table, the amount of reactant left after half-lives of a first-order reaction is (1/2) times the initial concentration. For a first-order reaction, the concentration of the reactant decreases by a constant with each half-life and is independent of [A]. The anticancer drug cisplatin hydrolyzes in water with a rate constant of 1.5 × 10 min at pH 7.0 and 25°C. Calculate the half-life for the hydrolysis reaction under these conditions. If a freshly prepared solution of cisplatin has a concentration of 0.053 M, what will be the concentration of cisplatin after 5 half-lives? after 10 half-lives? What is the percent completion of the reaction after 5 half-lives? after 10 half-lives? rate constant, initial concentration, and number of half-lives half-life, final concentrations, and percent completion Use to calculate the half-life of the reaction. Multiply the initial concentration by 1/2 to the power corresponding to the number of half-lives to obtain the remaining concentrations after those half-lives. Subtract the remaining concentration from the initial concentration. Then divide by the initial concentration, multiplying the fraction by 100 to obtain the percent completion. We can calculate the half-life of the reaction using : \( t_{1/2}=\dfrac{0.693}{k}=\dfrac{0.693}{1.5 \times 10^{-3}\;min^{-1}}=4.6 \times 10^{2}\;min \) Thus it takes almost 8 h for half of the cisplatin to hydrolyze. After 5 half-lives (about 38 h), the remaining concentration of cisplatin will be as follows: \( \dfrac{0.053\;M}{2^{5}}=\dfrac{0.053\;M}{32}=0.0017\;M \) After 10 half-lives (77 h), the remaining concentration of cisplatin will be as follows: \( \dfrac{0.053\;M}{2^{10}}=\dfrac{0.053\;M}{1024}=5.2 \times 10^{-5}\;M \) The percent completion after 5 half-lives will be as follows: \( percent\;completion=\dfrac{\left (0.053\;M-0.0017\;M \right )100}{0.53\;M}=97\% \) The percent completion after 10 half-lives will be as follows: \( percent\;completion=\dfrac{\left (0.053\;M-5.2 \times 10^{-5}\;M \right )100}{0.53\;M}=100\% \) Thus a first-order chemical reaction is 97% complete after 5 half-lives and 100% complete after 10 half-lives. Exercise In Example 4 you found that ethyl chloride decomposes to ethylene and HCl in a first-order reaction that has a rate constant of 1.6 × 10 s at 650°C. What is the half-life for the reaction under these conditions? If a flask that originally contains 0.077 M ethyl chloride is heated at 650°C, what is the concentration of ethyl chloride after 4 half-lives? 4.3 × 10 s = 120 h = 5.0 days; 4.8 × 10 M As you learned in radioactivity, or radioactive decay, is the emission of a particle or a photon that results from the spontaneous decomposition of the unstable nucleus of an atom. The rate of radioactive decay is an intrinsic property of each radioactive isotope that is independent of the chemical and physical form of the radioactive isotope. The rate is also independent of temperature. In this section, we will describe radioactive decay rates and how half-lives can be used to monitor radioactive decay processes. In any sample of a given radioactive substance, the number of atoms of the radioactive isotope must decrease with time as their nuclei decay to nuclei of a more stable isotope. Using to represent the number of atoms of the radioactive isotope, we can define the rate of decay of the sample, which is also called its activity ( ) as the decrease in the number of the radioisotope’s nuclei per unit time:
\( A=-\dfrac{\Delta N}{\Delta t} \tag{14.5.4}\) Activity is usually measured in . The activity of a sample is directly proportional to the number of atoms of the radioactive isotope in the sample:
\( A=kN \tag{14.5.5}\) Here, the symbol is the , which has units of inverse time (e.g., s , yr ) and a characteristic value for each radioactive isotope. If we combine and we obtain the relationship between the number of decays per unit time and the number of atoms of the isotope in a sample:
\( -\dfrac{\Delta N}{\Delta t}=kN \tag{14.5.6}\) is the same as the equation for the reaction rate of a first-order reaction ( ), except that it uses numbers of atoms instead of concentrations. In fact, radioactive decay a first-order process and can be described in terms of either the differential rate law ( ) or the integrated rate law: \( N=N_{0}e^{-kt} \tag{14.5.7}\) \( ln \dfrac{N}{N_{0}}=-kt \tag{14.5.8} \) Because radioactive decay is a first-order process, the time required for half of the nuclei in any sample of a radioactive isotope to decay is a constant, called the . The half-life tells us how radioactive an isotope is (the number of decays per unit time); thus it is the most commonly cited property of any radioisotope. For a given number of atoms, isotopes with shorter half-lives decay more rapidly, undergoing a greater number of radioactive decays per unit time than do isotopes with longer half-lives. The half-lives of several isotopes are listed in , along with some of their applications. Radioactive decay is a first-order process. In our earlier discussion, we used the half-life of a first-order reaction to calculate how long the reaction had been occurring. Because nuclear decay reactions follow first-order kinetics and have a rate constant that is independent of temperature and the chemical or physical environment, we can perform similar calculations using the half-lives of isotopes to estimate the ages of geological and archaeological artifacts. The techniques that have been developed for this application are known as techniques. The most common method for measuring the age of ancient objects is . The carbon-14 isotope, created continuously in the upper regions of Earth’s atmosphere, reacts with atmospheric oxygen or ozone to form CO . As a result, the CO that plants use as a carbon source for synthesizing organic compounds always includes a certain proportion of CO molecules as well as nonradioactive CO and CO . Any animal that eats a plant ingests a mixture of organic compounds that contains approximately the same proportions of carbon isotopes as those in the atmosphere. When the animal or plant dies, the carbon-14 nuclei in its tissues decay to nitrogen-14 nuclei by a radioactive process known as , which releases low-energy electrons (β particles) that can be detected and measured:
\( ^{14}C\rightarrow ^{14}N + \beta ^{-} \tag{14.5.8} \) The half-life for this reaction is 5700 ± 30 yr. The C/ C ratio in living organisms is 1.3 × 10 , with a decay rate of 15 dpm/g of carbon ( ). Comparing the disintegrations per minute per gram of carbon from an archaeological sample with those from a recently living sample enables scientists to estimate the age of the artifact, as illustrated in Example 11. In 1990, the remains of an apparently prehistoric man were found in a melting glacier in the Italian Alps. Analysis of the C content of samples of wood from his tools gave a decay rate of 8.0 dpm/g carbon. How long ago did the man die? isotope and final activity elapsed time Use to calculate / . Then substitute the value for the half-life of C into to find the rate constant for the reaction. Using the values obtained for / and the rate constant, solve to obtain the elapsed time. We know the initial activity from the isotope’s identity (15 dpm/g), the final activity (8.0 dpm/g), and the half-life, so we can use the integrated rate law for a first-order nuclear reaction ( ) to calculate the elapsed time (the amount of time elapsed since the wood for the tools was cut and began to decay). \( ln \dfrac{N}{N_{0}}=-kt \) \( \dfrac{ln \left (N_{0}/N \right )}{k}=t \) From .5, we know that = . We can therefore use the initial and final activities ( = 15 dpm and = 8.0 dpm) to calculate / : \( \dfrac{A_{0}}{A}= \dfrac{\cancel{k}N_{0}}{\cancel{k}N}=\dfrac{N_{0}}{N}=\dfrac{15.}{8.0} \) Now we need only calculate the rate constant for the reaction from its half-life (5730 yr) using : \( t_{1/2}=\dfrac{0.693}{k} \) This equation can be rearranged as follows: \( k=\dfrac{0.693}{t_{1/2}}=\dfrac{0.693}{5730\;yr}=1.22 \times 10^{-4} \; yr^{-1} \) Substituting into the equation for , \( t=\dfrac{ln\left ( N_{0}/N \right )}{k}=\dfrac{ln\left ( 15./8.0 \right )}{1.22 \times 10^{-4} \; yr^{-1}}=5.2 \times 10^{3} \; yr \) From our calculations, the man died 5200 yr ago. Exercise It is believed that humans first arrived in the Western Hemisphere during the last Ice Age, presumably by traveling over an exposed land bridge between Siberia and Alaska. Archaeologists have estimated that this occurred about 11,000 yr ago, but some argue that recent discoveries in several sites in North and South America suggest a much earlier arrival. Analysis of a sample of charcoal from a fire in one such site gave a C decay rate of 0.4 dpm/g of carbon. What is the approximate age of the sample? 30,000 yr The of a reaction is the time required for the reactant concentration to decrease to one-half its initial value. The half-life of a first-order reaction is a constant that is related to the rate constant for the reaction: t = 0.693/ . Radioactive decay reactions are first-order reactions. The , or , of a sample of a radioactive substance is the decrease in the number of radioactive nuclei per unit time. \( t_{1/2}=\dfrac{0.693}{k} \) .5: \( A=kN \) What do chemists mean by the of a reaction? If a sample of one isotope undergoes more disintegrations per second than the same number of atoms of another isotope, how do their half-lives compare? Half-lives for the reaction A + B → C were calculated at three values of [A] , and [B] was the same in all cases. The data are listed in the following table: Does this reaction follow first-order kinetics? On what do you base your answer? Ethyl-2-nitrobenzoate (NO C H CO C H ) hydrolyzes under basic conditions. A plot of [NO C H CO C H ] versus was used to calculate , with the following results: Is this a first-order reaction? Explain your reasoning. Azomethane (CH N CH ) decomposes at 600 K to C H and N . The decomposition is first order in azomethane. Calculate from the data in the following table: How long will it take for the decomposition to be 99.9% complete? The first-order decomposition of hydrogen peroxide has a half-life of 10.7 h at 20°C. What is the rate constant (expressed in s ) for this reaction? If you started with a solution that was 7.5 × 10 M H O , what would be the initial rate of decomposition (M/s)? What would be the concentration of H O after 3.3 h? No; the reaction is second order in A because the half-life decreases with increasing reactant concentration according to = 1/ [A ]. = 1.92 × 10 s or 1920 s; 19100 s or 5.32 hrs. | 12,387 | 1,534 |
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In any electrochemical process, electrons flow from one chemical substance to another, driven by an oxidation–reduction (redox) reaction. A redox reaction occurs when electrons are transferred from a substance that is oxidized to one that is being reduced. The is the substance that loses electrons and is oxidized in the process; the is the species that gains electrons and is reduced in the process. The associated potential energy is determined by the potential difference between the valence electrons in atoms of different elements. Because it is impossible to have a reduction without an oxidation and vice versa, a redox reaction can be described as two , one representing the oxidation process and one the reduction process. For the reaction of zinc with bromine, the overall chemical reaction is as follows: \[\ce{Zn(s) + Br2(aq) \rightarrow Zn^{2+} (aq) + 2Br^{−} (aq)} \nonumber \] The half-reactions are as follows: \[\ce{Br2 (aq) + 2e^{−} \rightarrow 2Br^{−} (aq)} \nonumber \] \[\ce{Zn (s) \rightarrow Zn^{2+} (aq) + 2e^{−} }\nonumber \] Each half-reaction is written to show what is actually occurring in the system; \(\ce{Zn}\) is the in this reaction (it loses electrons), and \(\ce{Br2}\) is the (it gains electrons). Adding the two half-reactions gives the overall chemical reaction (Equation \(\Page {1}\)). A redox reaction is balanced when the number of electrons lost by the reductant equals the number of electrons gained by the oxidant. Like any balanced chemical equation, the overall process is electrically neutral; that is, the net charge is the same on both sides of the equation. In any redox reaction, the number of electrons lost by the oxidation reaction(s) equals the number of electrons gained by the reduction reaction(s). In most of our discussions of chemical reactions, we have assumed that the reactants are in intimate physical contact with one another. Acid–base reactions, for example, are usually carried out with the acid and the base dispersed in a single phase, such as a liquid solution. With redox reactions, however, it is possible to physically separate the oxidation and reduction half-reactions in space, as long as there is a complete circuit, including an external electrical connection, such as a wire, between the two half-reactions. As the reaction progresses, the electrons flow from the reductant to the oxidant over this electrical connection, producing an electric current that can be used to do work. An apparatus that is used to generate electricity from a spontaneous redox reaction or, conversely, that uses electricity to drive a nonspontaneous redox reaction is called an . There are two types of electrochemical cells: galvanic cells and electrolytic cells. Galvanic cells are named for the Italian physicist and physician Luigi Galvani (1737–1798), who observed that dissected frog leg muscles twitched when a small electric shock was applied, demonstrating the electrical nature of nerve impulses. A uses the energy released during a spontaneous redox reaction (\(ΔG < 0\)) to generate electricity. This type of electrochemical cell is often called a voltaic cell after its inventor, the Italian physicist Alessandro Volta (1745–1827). In contrast, an consumes electrical energy from an external source, using it to cause a nonspontaneous redox reaction to occur (\(ΔG > 0\)). Both types contain two , which are solid metals connected to an external circuit that provides an electrical connection between the two parts of the system (Figure \(\Page {1}\)). The oxidation half-reaction occurs at one electrode (the ), and the reduction half-reaction occurs at the other (the ). When the circuit is closed, electrons flow from the anode to the cathode. The electrodes are also connected by an electrolyte, an ionic substance or solution that allows ions to transfer between the electrode compartments, thereby maintaining the system’s electrical neutrality. In this section, we focus on reactions that occur in galvanic cells. To illustrate the basic principles of a galvanic cell, let’s consider the reaction of metallic zinc with cupric ion (Cu ) to give copper metal and Zn ion. The balanced chemical equation is as follows: \[\ce{Zn (s) + Cu^{2+} (aq) \rightarrow Zn^{2+} (aq) + Cu(s)} \label{20.3.4} \] We can cause this reaction to occur by inserting a zinc rod into an aqueous solution of copper(II) sulfate. As the reaction proceeds, the zinc rod dissolves, and a mass of metallic copper forms. These changes occur spontaneously, but all the energy released is in the form of heat rather than in a form that can be used to do work. This same reaction can be carried out using the galvanic cell illustrated in Figure \(\Page {3a}\). To assemble the cell, a copper strip is inserted into a beaker that contains a 1 M solution of \(\ce{Cu^{2+}}\) ions, and a zinc strip is inserted into a different beaker that contains a 1 M solution of \(\ce{Zn^{2+}}\) ions. The two metal strips, which serve as electrodes, are connected by a wire, and the compartments are connected by a , a U-shaped tube inserted into both solutions that contains a concentrated liquid or gelled electrolyte. The ions in the salt bridge are selected so that they do not interfere with the electrochemical reaction by being oxidized or reduced themselves or by forming a precipitate or complex; commonly used cations and anions are \(\ce{Na^{+}}\) or \(\ce{K^{+}}\) and \(\ce{NO3^{−}}\) or \(\ce{SO4^{2−}}\), respectively. (The ions in the salt bridge do not have to be the same as those in the redox couple in either compartment.) When the circuit is closed, a spontaneous reaction occurs: zinc metal is oxidized to \(\ce{Zn^{2+}}\) ions at the zinc electrode (the anode), and \(\ce{Cu^{2+}}\) ions are reduced to \(\ce{Cu}\) metal at the copper electrode (the cathode). As the reaction progresses, the zinc strip dissolves, and the concentration of \(\ce{Zn^{2+}}\) ions in the solution increases; simultaneously, the copper strip gains mass, and the concentration of \(\ce{Cu^{2+}}\) ions in the solution decreases (Figure \(\Page {3b}\)). Thus we have carried out the same reaction as we did using a single beaker, but this time the oxidative and reductive half-reactions are physically separated from each other. The electrons that are released at the anode flow through the wire, producing an electric current. Galvanic cells therefore transform chemical energy into electrical energy that can then be used to do work. The electrolyte in the salt bridge serves two purposes: it completes the circuit by carrying electrical charge and maintains electrical neutrality in both solutions by allowing ions to migrate between them. The identity of the salt in a salt bridge is unimportant, as long as the component ions do not react or undergo a redox reaction under the operating conditions of the cell. Without such a connection, the total positive charge in the \(\ce{Zn^{2+}}\) solution would increase as the zinc metal dissolves, and the total positive charge in the \(\ce{Cu^{2+}}\) solution would decrease. The salt bridge allows charges to be neutralized by a flow of anions into the \(\ce{Zn^{2+}}\) solution and a flow of cations into the \(\ce{Cu^{2+}}\) solution. In the absence of a salt bridge or some other similar connection, the reaction would rapidly cease because electrical neutrality could not be maintained. A voltmeter can be used to measure the difference in electrical potential between the two compartments. Opening the switch that connects the wires to the anode and the cathode prevents a current from flowing, so no chemical reaction occurs. With the switch closed, however, the external circuit is closed, and an electric current can flow from the anode to the cathode. The (\(E_{cell}\)) of the cell, measured in volts, is the difference in electrical potential between the two half-reactions and is related to the energy needed to move a charged particle in an electric field. In the cell we have described, the voltmeter indicates a potential of 1.10 V (Figure \(\Page {3a}\)). Because electrons from the oxidation half-reaction are released at the anode, the anode in a galvanic cell is negatively charged. The cathode, which attracts electrons, is positively charged. Not all electrodes undergo a chemical transformation during a redox reaction. The electrode can be made from an inert, highly conducting metal such as platinum to prevent it from reacting during a redox process, where it does not appear in the overall electrochemical reaction. This phenomenon is illustrated in Example \(\Page {1}\). A galvanic (voltaic) cell converts the energy released by a chemical reaction to electrical energy. An electrolytic cell consumes electrical energy from an external source to drive a chemical reaction. A chemist has constructed a galvanic cell consisting of two beakers. One beaker contains a strip of tin immersed in aqueous sulfuric acid, and the other contains a platinum electrode immersed in aqueous nitric acid. The two solutions are connected by a salt bridge, and the electrodes are connected by a wire. Current begins to flow, and bubbles of a gas appear at the platinum electrode. The spontaneous redox reaction that occurs is described by the following balanced chemical equation: \[\ce{3Sn(s) + 2NO3^{-}(aq) + 8H^{+} (aq) \rightarrow 3Sn^{2+} (aq) + 2NO (g) + 4H2O (l)} \nonumber \] For this galvanic cell, galvanic cell and redox reaction half-reactions, identity of anode and cathode, and electrode assignment as positive or negative In the reduction half-reaction, nitrate is reduced to nitric oxide. (The nitric oxide would then react with oxygen in the air to form NO , with its characteristic red-brown color.) In the oxidation half-reaction, metallic tin is oxidized. The half-reactions corresponding to the actual reactions that occur in the system are as follows: : \[\ce{NO3^{−} (aq) + 4H^{+}(aq) + 3e^{−} → NO(g) + 2H2O(l)} \nonumber \] : \[\ce{Sn(s) → Sn^{2+}(aq) + 2e^{−}} \nonumber \] Thus nitrate is reduced to NO, while the tin electrode is oxidized to Sn . Because the reduction reaction occurs at the Pt electrode, it is the cathode. Conversely, the oxidation reaction occurs at the tin electrode, so it is the anode. Electrons flow from the tin electrode through the wire to the platinum electrode, where they transfer to nitrate. The electric circuit is completed by the salt bridge, which permits the diffusion of cations toward the cathode and anions toward the anode. Because electrons flow from the tin electrode, it must be electrically negative. In contrast, electrons flow toward the Pt electrode, so that electrode must be electrically positive. Consider a simple galvanic cell consisting of two beakers connected by a salt bridge. One beaker contains a solution of \(\ce{MnO_4^{−}}\) in dilute sulfuric acid and has a Pt electrode. The other beaker contains a solution of \(\ce{Sn^{2+}}\) in dilute sulfuric acid, also with a Pt electrode. When the two electrodes are connected by a wire, current flows and a spontaneous reaction occurs that is described by the following balanced chemical equation: \[\ce{2MnO^{−}4(aq) + 5Sn^{2+}(aq) + 16H^{+}(aq) \rightarrow 2Mn^{2+}(aq) + 5Sn^{4+}(aq) + 8H2O(l)} \nonumber \] For this galvanic cell, \[\begin{align*} \ce{MnO4^{−}(aq) + 8H^{+}(aq) + 5e^{−}} &→ \ce{Mn^{2+}(aq) + 4H2O(l)} \\[4pt] \ce{Sn^{2+}(aq)} &→ \ce{Sn^{4+}(aq) + 2e^{−}} \end{align*} \nonumber \] The Pt electrode in the permanganate solution is the cathode; the one in the tin solution is the anode. The cathode (electrode in beaker that contains the permanganate solution) is positive, and the anode (electrode in beaker that contains the tin solution) is negative. Electrochemical Cells: Because it is somewhat cumbersome to describe any given galvanic cell in words, a more convenient notation has been developed. In this line notation, called a cell diagram, the identity of the electrodes and the chemical contents of the compartments are indicated by their chemical formulas, with the anode written on the far left and the cathode on the far right. Phase boundaries are shown by single vertical lines, and the salt bridge, which has two phase boundaries, by a double vertical line. Thus the cell diagram for the \(\ce{Zn/Cu}\) cell shown in Figure \(\Page {3a}\) is written as follows: Galvanic cells can have arrangements other than the examples we have seen so far. For example, the voltage produced by a redox reaction can be measured more accurately using two electrodes immersed in a single beaker containing an electrolyte that completes the circuit. This arrangement reduces errors caused by resistance to the flow of charge at a boundary, called the . One example of this type of galvanic cell is as follows: \[\ce{Pt(s)\, | \, H2(g) | HCl(aq, \, 1\,M)\,|\, AgCl(s) \,Ag(s)} \nonumber \] This cell diagram does not include a double vertical line representing a salt bridge because there is no salt bridge providing a junction between two dissimilar solutions. Moreover, solution concentrations have not been specified, so they are not included in the cell diagram. The half-reactions and the overall reaction for this cell are as follows: cathode reaction: \[\ce{AgCl (s) + e^{−} \rightarrow Ag(s) + Cl^{−}(aq)} \nonumber \] anode reaction: \[\ce{ 1/2 H2(g) -> H^{+}(aq) + e^{-}} \nonumber \] overall: \[\ce{ AgCl(s) + 1/2H2(g) -> Ag(s) + Cl^{-} + H^{+}(aq)} \nonumber \] A single-compartment galvanic cell will initially exhibit the same voltage as a galvanic cell constructed using separate compartments, but it will discharge rapidly because of the direct reaction of the reactant at the anode with the oxidized member of the cathodic redox couple. Consequently, cells of this type are not particularly useful for producing electricity. Draw a cell diagram for the galvanic cell described in Example \(\Page {1}\). The balanced chemical reaction is as follows: \[\ce{3Sn(s) + 2NO^{−}3(aq) + 8H^{+}(aq) \rightarrow 3Sn^{2+}(aq) + 2NO(g) + 4H2O(l)} \nonumber \] galvanic cell and redox reaction cell diagram Using the symbols described, write the cell diagram beginning with the oxidation half-reaction on the left. The anode is the tin strip, and the cathode is the \(\ce{Pt}\) electrode. Beginning on the left with the anode, we indicate the phase boundary between the electrode and the tin solution by a vertical bar. The anode compartment is thus \(\ce{Sn(s)∣Sn^{2+}(aq)}\). We could include \(\ce{H2SO4(aq)}\) with the contents of the anode compartment, but the sulfate ion (as \(\ce{HSO4^{−}}\)) does not participate in the overall reaction, so it does not need to be specifically indicated. The cathode compartment contains aqueous nitric acid, which does participate in the overall reaction, together with the product of the reaction (\(\ce{NO}\)) and the \(\ce{Pt}\) electrode. These are written as \(\ce{HNO3(aq)∣NO(g)∣Pt(s)}\), with single vertical bars indicating the phase boundaries. Combining the two compartments and using a double vertical bar to indicate the salt bridge, \[\ce{Sn(s)\,|\,Sn^{2+}(aq)\,||\,HNO3(aq)\,|\,NO(g)\,|\,Pt_(s)} \nonumber \] The solution concentrations were not specified, so they are not included in this cell diagram. Draw the cell diagram for the following reaction, assuming the concentration of \(\ce{Ag^{+}}\) and \(\ce{Mg^{2+}}\) are each 1 M: \[\ce{Mg(s) + 2Ag^{+}(aq) \rightarrow Mg^{2+}(aq) + 2Ag(s)} \nonumber \] \[ \ce{ Mg(s) \,|\,Mg^{2+}(aq, \;1 \,M )\,||\,Ag^+(aq, \;1\, M)\,|\,Ag(s)} \nonumber \] Cell Diagrams: A galvanic (voltaic) cell uses the energy released during a spontaneous redox reaction to generate electricity, whereas an electrolytic cell consumes electrical energy from an external source to force a reaction to occur. Electrochemistry is the study of the relationship between electricity and chemical reactions. The oxidation–reduction reaction that occurs during an electrochemical process consists of two half-reactions, one representing the oxidation process and one the reduction process. The sum of the half-reactions gives the overall chemical reaction. The overall redox reaction is balanced when the number of electrons lost by the reductant equals the number of electrons gained by the oxidant. An electric current is produced from the flow of electrons from the reductant to the oxidant. An electrochemical cell can either generate electricity from a spontaneous redox reaction or consume electricity to drive a nonspontaneous reaction. In a galvanic (voltaic) cell, the energy from a spontaneous reaction generates electricity, whereas in an electrolytic cell, electrical energy is consumed to drive a nonspontaneous redox reaction. Both types of cells use two electrodes that provide an electrical connection between systems that are separated in space. The oxidative half-reaction occurs at the anode, and the reductive half-reaction occurs at the cathode. A salt bridge connects the separated solutions, allowing ions to migrate to either solution to ensure the system’s electrical neutrality. A voltmeter is a device that measures the flow of electric current between two half-reactions. The potential of a cell, measured in volts, is the energy needed to move a charged particle in an electric field. An electrochemical cell can be described using line notation called a cell diagram, in which vertical lines indicate phase boundaries and the location of the salt bridge. Resistance to the flow of charge at a boundary is called the junction potential. | 17,535 | 1,535 |
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\( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \)
\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) An underlined problem number or problem-part letter indicates that the numerical answer appears in Appendix I. Consider the heterogeneous equilibrium \(\ce{CaCO3}\tx{(s)} \arrows \ce{CaO}\tx{(s)} + \ce{CO2}\tx{(g)}\). Table 12.3 lists pressures measured over a range of temperatures for this system. The method described in Prob. 12.13 has been used to obtain high-precision values of the Henry’s law constant, \(\kHB\), for gaseous methane dissolved in water (Timothy R. Rettich, Y. Paul Handa, Rubin Battino, and Emmerich Wilhelm, , , 3230–3237, 1981). Table 12.6 lists values of \(\ln (\kHB/p\st)\) at eleven temperatures in the range \(275\K\)–\(328\K\) and at pressures close to \(1\br\). Use these data to evaluate \(\Delsub{sol,B}H\st\) and \(\Delsub{sol,B}C\st_p\) at \(T=298.15\K\). This can be done by a graphical method. Better precision will be obtained by making a least-squares fit of the data to the three-term polynomial \begin{equation*} \ln (\kHB/p\st) = a + b(1/T) + c(1/T)^2 \end{equation*} and using the values of the coefficients \(a\), \(b\), and \(c\) for the evaluations. Liquid water and liquid benzene have very small mutual solubilities. Equilibria in the binary water–benzene system were investigated by Tucker, Lane, and Christian ( , , 1–20, 1981) as follows. A known amount of distilled water was admitted to an evacuated, thermostatted vessel. Part of the water vaporized to form a vapor phase. Small, precisely measured volumes of liquid benzene were then added incrementally from the sample loop of a liquid-chromatography valve. The benzene distributed itself between the liquid and gaseous phases in the vessel. After each addition, the pressure was read with a precision pressure gauge. From the known amounts of water and benzene and the total pressure, the liquid composition and the partial pressure of the benzene were calculated. The fugacity of the benzene in the vapor phase was calculated from its partial pressure and the second virial coefficient. At a fixed temperature, for mole fractions \(x\B\) of benzene in the liquid phase up to about \(3\timesten{-4}\) (less than the solubility of benzene in water), the fugacity of the benzene in the equilibrated gas phase was found to have the following dependence on \(x\B\): \[ \frac{\fug\B}{x\B} = \kHB - Ax\B \] Here \(\kHB\) is the Henry’s law constant and \(A\) is a constant related to deviations from Henry’s law. At \(30\units{\(\degC\)}\), the measured values were \(\kHB=385.5\br\) and \(A=2.24\timesten{4}\br\). Treat benzene (B) as the solute and find its activity coefficient on a mole fraction basis, \(\g\xbB\), at \(30\units{\(\degC\)}\) in the solution of composition \(x\B=3.00\timesten{-4}\). The fugacity of benzene vapor in equilibrium with pure liquid benzene at \(30\units{\(\degC\)}\) is \(\fug\B^*=0.1576\br\). Estimate the mole fraction solubility of liquid benzene in water at this temperature. (c) The calculation of \(\g\xbB\) in part (a) treated the benzene as a single solute species with deviations from infinite-dilution behavior. Tucker et al suggested a dimerization model to explain the observed negative deviations from Henry’s law. (Classical thermodynamics, of course, cannot prove such a molecular interpretation of observed macroscopic behavior.) The model assumes that there are two solute species, a monomer (M) and a dimer (D), in reaction equilibrium: \(\ce{2M} \arrows \ce{D}\). Let \(n\B\) be the total amount of C\(_6\)H\(_6\) present in solution, and define the mole fractions \[ x\B \defn \frac{n\B}{n\A+n\B} \approx \frac{n\B}{n\A} \] \[ x\subs{M} \defn \frac{n\subs{M}}{n\A+n\subs{M}+n\subs{D}} \approx \frac{n\subs{M}}{n\A} \qquad x\subs{D} \defn \frac{n\subs{D}}{n\A+n\subs{M}+n\subs{D}} \approx \frac{n\subs{D}}{n\A} \] where the approximations are for dilute solution. In the model, the individual monomer and dimer particles behave as solutes in an ideal-dilute solution, with activity coefficients of unity. The monomer is in transfer equilibrium with the gas phase: \(x\subs{M}=\fug\B/\kHB\). The equilibrium constant expression (using a mole fraction basis for the solute standard states and setting pressure factors equal to 1) is \(K=x\subs{D}/x\subs{M}^2\). From the relation \(n\B=n\subs{M}+2n\subs{D}\), and because the solution is very dilute, the expression becomes \[ K = \frac{x\B-x\subs{M}}{2x\subs{M}^2} \] Make individual calculations of \(K\) from the values of \(\fug\B\) measured at \(x\B=1.00\timesten{-4}\), \(x\B=2.00\timesten{-4}\), and \(x\B=3.00\timesten{-4}\). Extrapolate the calculated values of \(K\) to \(x\B{=}0\) in order to eliminate nonideal effects such as higher aggregates. Finally, find the fraction of the benzene molecules present in the dimer form at \(x\B=3.00\timesten{-4}\) if this model is correct. Use data in Appendix H to evaluate the thermodynamic equilibrium constant at \(298.15\K\) for the limestone reaction \[ \ce{CaCO3}\tx{(cr, calcite)} + \ce{CO2}\tx{(g)} + \ce{H2O}\tx{(l)} \arrow \ce{Ca^2+}\tx{(aq)} + \ce{2HCO3-}\tx{(aq)} \] For the dissociation equilibrium of formic acid, \(\ce{HCO2H}\tx{(aq)} \arrows \ce{H+}\tx{(aq)} + \ce{HCO2-}\tx{(aq)}\), the acid dissociation constant at \(298.15\K\) has the value \(K\subs{a}=1.77\timesten{-4}\). Use Eq. 12.9.7 to find the degree of dissociation and the hydrogen ion molality in a 0.01000 molal formic acid solution. You can safely set \(\G\subs{r}\) and \(\g\subs{\(m\),HA}\) equal to \(1\), and use the Debye–Hückel limiting law (Eq. 10.4.8) to calculate \(\g_{\pm}\). You can do this calculation by iteration: Start with an initial estimate of the ionic strength (in this case 0), calculate \(\g_{\pm}\) and \(\alpha\), and repeat these steps until the value of \(\alpha\) no longer changes. Estimate the degree of dissociation of formic acid in a solution that is 0.01000 molal in both formic acid and sodium nitrate, again using the Debye–Hückel limiting law for \(\g_{\pm}\). Compare with the value in part (a). Use the following experimental information to evaluate the standard molar enthalpy of formation and the standard molar entropy of the aqueous chloride ion at \(298.15\K\), based on the conventions \(\Delsub{f}H\st(\tx{H\(^+\), aq})=0\) and \(S\m\st(\tx{H\(^+\), aq})=0\) (Secs. 11.3.2 and 11.8.4). (Your calculated values will be close to, but not exactly the same as, those listed in Appendix H, which are based on the same data combined with data of other workers.) From these data and the Debye–Hückel limiting law, calculate the solubility product \(K\subs{s}\) of AgCl at \(298.15\K\). The following reaction was carried out in an adiabatic solution calorimeter by Wagman and Kilday ( , , 569–579, 1973): \[ \tx{AgNO\(_3\)(s)} + \tx{KCl(aq, } m\B=0.101\units{mol kg\(^{-1}\)}) \arrow \tx{AgCl(s)} + \tx{KNO\(_3\)(aq)} \] The reaction can be assumed to go to completion, and the amount of KCl was in slight excess, so the amount of AgCl formed was equal to the initial amount of AgNO\(_3\). After correction for the enthalpies of diluting the solutes in the initial and final solutions to infinite dilution, the standard molar reaction enthalpy at \(298.15\K\) was found to be \(\Delsub{r}H\st=-43.042\units{kJ mol\(^{-1}\)}\). The same workers used solution calorimetry to obtain the molar enthalpy of solution at infinite dilution of crystalline AgNO\(_3\) at \(298.15\K\): \(\Delsub{sol,B}H^{\infty}=22.727\units{kJ mol\(^{-1}\)}\). Show that the difference of these two values is the standard molar reaction enthalpy for the precipitation reaction \[ \ce{Ag+}\tx{(aq)} + \ce{Cl-}\tx{(aq)} \arrow \ce{AgCl}\tx{(s)} \] and evaluate this quantity. Evaluate the standard molar enthalpy of formation of aqueous Ag\(^+\) ion at \(298.15\K\), using the results of part (a) and the values \(\Delsub{f}H\st(\tx{Cl\(^-\), aq})=-167.08\units{kJ mol\(^{-1}\)}\) and \(\Delsub{f}H\st(\tx{AgCl, s})=-127.01\units{kJ mol\(^{-1}\)}\) from Appendix H. (These values come from calculations similar to those in Probs. 12.18 and 14.4.) The calculated value will be close to, but not exactly the same as, the value listed in Appendix H, which is based on the same data combined with data of other workers. | 15,546 | 1,536 |
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Hernan Cortes (1485-1547) was a Spanish conquistador who was responsible for the fall of the Aztec empire. In 1518, Cortes left with a fleet of 11 ships and 500 men to explore and secure inner Mexico for Spain. Cortes entered Aztec territory and was peacefully received by the emperor Moctezuma II. Moctezuma offered the Spaniards gold and took them into the heart of the city. This enticed the Spanish even more to take over the territory for what they believed was vast quantities of gold. Many battles took place with the Aztecs, and over time Cortes needed more gun powder for his army. Spain, being across the ocean, would not be a quick source that he needed to win this war. As a result, he had his men make the gun powder. The main ingredients of gunpowder are Carbon and Sulfur. Carbon was easy to come by, but Sulfur was not. Cortes sent an expedition of men on a treacherous mission to the top of Popocatépetl, an active volcano. Volcanoes are a good source of sulfur because they emit sulfur dioxide. Sulfur has many similarities with the elements in its group. This group is VIA. Except polonium, which is radioactive and usually omitted from discussion, all members of the group form X ions when combined with highly electropositive metals. The tendency to be reduced to the –2 oxidation state decreases significantly from top to bottom of the group, however, and tellurium shows some metallic properties. The group VIA elements are called chalcogens because most ores of copper (Greek chalkos) are oxides or sulfides, and such ores contain traces of selenium and tellurium. Atomic properties of the chalcogens are summarized in the table. Electron Configuration Usual Oxidation State Radius/pm Ionic (X ) 140 184 198 Density/ g cm Electro- negativity Melting Point (in °C) -218 When heated to 96°C, solid rhombic sulfur changes very slowly into monoclinic sulfur, in which one-third of the S molecules are randomly oriented in the crystal lattice. When either form of sulfur melts, the liquid is at first pale yellow and flows readily, but above 160°C it becomes increasingly viscous. Only near the boiling point of 444.6°C does it thin out again. This unusual change in viscosity with temperature is attributed to opening of the eight-membered ring of S and formation of long chains of sulfur atoms. These intertwine and prevent the liquid from flowing. This explanation is supported by the fact that if the viscous liquid is cooled rapidly by pouring it into water, the amorphous sulfur produced can be shown experimentally to consist of long chains of sulfur atoms. Although this element is only sixteenth in abundance at the surface of the earth, it is one of the few that has been known and used throughout history. Deposits of elemental sulfur are not uncommon, and, because they were stones that would burn, were originally called brimstone. Burning sulfur produces sulfur dioxide, The natural mechanism for removal of sulfur oxides from the air is solution in raindrops, followed by precipitation. This makes the rainwater more acidic than it would otherwise be, and acid rain is now common in industrialized areas of the United States and Europe. Acid rain can slowly dissolve limestone and marble, both of which consist of CaCO : Despite the fact that a tremendous amount of sulfur is released to the environment by coal combustion and ore smelting, this element is not usually recovered from such processes. Instead it is obtained commercially from large deposits along the U.S. Gulf Coast and from refining of sour petroleum. Sour petroleum contains numerous sulfur compounds, including H S, which smells like rotten eggs. The deposits of elemental sulfur in Texas and Louisiana are mined by the . Water at 170°C is pumped down a pipe to melt the sulfur, and the latter is forced to the surface by compressed air. Most of the H S or S obtained from these sources is oxidized to SO , passed over a vanadium catalyst to make SO , and dissolved in water to make H SO . In 2005 an estimated 190 billion kg of H SO was produced in the world, making H SO one of the most important industrial chemicals. About half of it is used in phosphate fertilizer production . Pure H SO is a liquid at room temperature and has a great affinity for H O. This is apparently due to the reaction 2. en. .org/wiki/Hernán_Cortés Ed Vitz (Kutztown University), (University of | 4,414 | 1,537 |
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Demonstrate a model mass spectrometer built by affixing a magnet to the bottom at the center of a ~60 x 20 cm plexiglas ramp. Roll ball bearings of different size (from BB to 1 cm) down the ramp, and observe different deflections depending on mass. The top of the ramp is ~2 cm higher than the bottom, and a small dimple is drilled in the plexiglas near the top as a starting point. The bottom may rest on collection trays for different deflections. The differences between magnetic deflection of metal balls and charged particles should be emphasized: The force depends on the velocity and the charge for nuclei, and is in a direction determined by the right hand rule. The radius of curvature is given by: r = m (kg) v (m/s) / q (Coulombs) B (Tesla) Several simulations are available on the web, for example and . | 831 | 1,540 |
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The state exhibited by a given sample of matter depends on the identity, temperature, and pressure of the sample. A phase diagram is a graphic summary of the physical state of a substance as a function of temperature and pressure in a closed system. A typical phase diagram consists of discrete regions that represent the different phases exhibited by a substance (Figure \(\Page {1}\)). Each region corresponds to the range of combinations of temperature and pressure over which that phase is stable. The combination of high pressure and low temperature (upper left of Figure \(\Page {1}\)) corresponds to the solid phase, whereas the gas phase is favored at high temperature and low pressure (lower right). The combination of high temperature and high pressure (upper right) corresponds to a supercritical fluid. The solid phase is favored at low temperature and high pressure; the gas phase is favored at high temperature and low pressure. The lines in a phase diagram correspond to the combinations of temperature and pressure at which two phases can coexist in equilibrium. In Figure \(\Page {1}\), the line that connects points A and D separates the solid and liquid phases and shows how the melting point of a solid varies with pressure. The solid and liquid phases are in equilibrium all along this line; crossing the line horizontally corresponds to melting or freezing. The line that connects points A and B is the vapor pressure curve of the liquid, which we discussed in . It ends at the critical point, beyond which the substance exists as a supercritical fluid. The line that connects points A and C is the vapor pressure curve of the phase. Along this line, the solid is in equilibrium with the vapor phase through sublimation and deposition. Finally, point A, where the solid/liquid, liquid/gas, and solid/gas lines intersect, is the triple point; it is the combination of temperature and pressure at which all three phases (solid, liquid, and gas) are in equilibrium and can therefore exist simultaneously. Because no more than three phases can ever coexist, a phase diagram can never have more than three lines intersecting at a single point. Remember that a phase diagram, such as the one in Figure \(\Page {1}\), is for a single pure substance in a closed system, not for a liquid in an open beaker in contact with air at 1 atm pressure. In practice, however, the conclusions reached about the behavior of a substance in a closed system can usually be extrapolated to an open system without a great deal of error. Figure \(\Page {2}\) shows the phase diagram of water and illustrates that the triple point of water occurs at 0.01°C and 0.00604 atm (4.59 mmHg). Far more reproducible than the melting point of ice, which depends on the amount of dissolved air and the atmospheric pressure, the triple point (273.16 K) is used to define the absolute (Kelvin) temperature scale. The triple point also represents the lowest pressure at which a liquid phase can exist in equilibrium with the solid or vapor. At pressures less than 0.00604 atm, therefore, ice does not melt to a liquid as the temperature increases; the solid sublimes directly to water vapor. Sublimation of water at low temperature and pressure can be used to “freeze-dry” foods and beverages. The food or beverage is first cooled to subzero temperatures and placed in a container in which the pressure is maintained below 0.00604 atm. Then, as the temperature is increased, the water sublimes, leaving the dehydrated food (such as that used by backpackers or astronauts) or the powdered beverage (as with freeze-dried coffee). The phase diagram for water illustrated in Figure \(\Page {2b}\) shows the boundary between ice and water on an expanded scale. The melting curve of ice slopes up and slightly to the left rather than up and to the right as in Figure \(\Page {1}\); that is, the melting point of ice with increasing pressure; at 100 MPa (987 atm), ice melts at −9°C. Water behaves this way because it is one of the few known substances for which the crystalline solid is than the liquid (others include antimony and bismuth). Increasing the pressure of ice that is in equilibrium with water at 0°C and 1 atm tends to push some of the molecules closer together, thus decreasing the volume of the sample. The decrease in volume (and corresponding increase in density) is smaller for a solid or a liquid than for a gas, but it is sufficient to melt some of the ice. In Figure \(\Page {2b}\) point A is located at = 1 atm and = −1.0°C, within the solid (ice) region of the phase diagram. As the pressure increases to 150 atm while the temperature remains the same, the line from point A crosses the ice/water boundary to point B, which lies in the liquid water region. Consequently, applying a pressure of 150 atm will melt ice at −1.0°C. We have already indicated that the pressure dependence of the melting point of water is of vital importance. If the solid/liquid boundary in the phase diagram of water were to slant up and to the right rather than to the left, ice would be denser than water, ice cubes would sink, water pipes would not burst when they freeze, and antifreeze would be unnecessary in automobile engines. Until recently, many textbooks described ice skating as being possible because the pressure generated by the skater’s blade is high enough to melt the ice under the blade, thereby creating a lubricating layer of liquid water that enables the blade to slide across the ice. Although this explanation is intuitively satisfying, it is incorrect, as we can show by a simple calculation. Recall that pressure ( ) is the force ( ) applied per unit area ( ): \[P=\dfrac{F}{A} \nonumber \] To calculate the pressure an ice skater exerts on the ice, we need to calculate only the force exerted and the area of the skate blade. If we assume a 75.0 kg (165 lb) skater, then the force exerted by the skater on the ice due to gravity is \[ F = mg \nonumber \] where is the mass and is the acceleration due to Earth’s gravity (9.81 m/s ). Thus the force is \[F = (75.0\; kg)(9.81\; m/s^2) = 736\; (kg•m)/s^2 = 736 N \nonumber \] If we assume that the skate blades are 2.0 mm wide and 25 cm long, then the area of the bottom of each blade is \[ A = (2.0 \times 10^{−3}\; m)(25 \times 10^{−2}\; m) = 5.0 \times 10^{−4} m^2 \nonumber \] If the skater is gliding on one foot, the pressure exerted on the ice is \[ P= \dfrac{736\;N}{5.0 \times 10^{-4} \; m^2} = 1.5 \times 10^6 \; N/m^2 = 1.5 \times 10^6\; Pa =15 \; atm \nonumber \] The pressure is much lower than the pressure needed to decrease the melting point of ice by even 1°C, and experience indicates that it is possible to skate even when the temperature is well below freezing. Thus pressure-induced melting of the ice cannot explain the low friction that enables skaters (and hockey pucks) to glide. Recent research indicates that the surface of ice, where the ordered array of water molecules meets the air, consists of one or more layers of almost liquid water. These layers, together with melting induced by friction as a skater pushes forward, appear to account for both the ease with which a skater glides and the fact that skating becomes more difficult below about −7°C, when the number of lubricating surface water layers decreases. Referring to the phase diagram of water in Figure \(\Page {2}\): phase diagram, temperature, and pressure physical form and physical changes Referring to the phase diagram of water in Figure \(\Page {2}\), predict the physical form of a sample of water at −0.0050°C as the pressure is gradually increased from 1.0 mmHg to 218 atm. The sample is initially a gas, condenses to a solid as the pressure increases, and then melts when the pressure is increased further to give a liquid. In contrast to the phase diagram of water, the phase diagram of CO (Figure \(\Page {3}\)) has a more typical melting curve, sloping up and to the right. The triple point is −56.6°C and 5.11 atm, which means that liquid CO cannot exist at pressures lower than 5.11 atm. At 1 atm, therefore, solid CO sublimes directly to the vapor while maintaining a temperature of −78.5°C, the normal sublimation temperature. Solid CO is generally known as dry ice because it is a cold solid with no liquid phase observed when it is warmed. Also notice the critical point at 30.98°C and 72.79 atm. Supercritical carbon dioxide is emerging as a natural refrigerant, making it a low carbon (and thus a more environmentally friendly) solution for domestic heat pumps. As the phase diagrams above demonstrate, a combination of high pressure and low temperature allows gases to be liquefied. As we increase the temperature of a gas, liquefaction becomes more and more difficult because higher and higher pressures are required to overcome the increased kinetic energy of the molecules. In fact, for every substance, there is some temperature above which the gas can no longer be liquefied, regardless of pressure. This temperature is the critical temperature ( ), the highest temperature at which a substance can exist as a liquid. Above the critical temperature, the molecules have too much kinetic energy for the intermolecular attractive forces to hold them together in a separate liquid phase. Instead, the substance forms a single phase that completely occupies the volume of the container. Substances with strong intermolecular forces tend to form a liquid phase over a very large temperature range and therefore have high critical temperatures. Conversely, substances with weak intermolecular interactions have relatively low critical temperatures. Each substance also has a critical pressure ( ), the minimum pressure needed to liquefy it at the critical temperature. The combination of critical temperature and critical pressure is called the critical point. The critical temperatures and pressures of several common substances are listed in Figure \(\Page {1}\). High-boiling-point, nonvolatile liquids have high critical temperatures and vice versa. A Discussing Phase Diagrams. | 10,098 | 1,541 |
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states that atoms are the units of chemical changes. What this means can be seen in the macroscopic and microscopic views of a chemical change in the video and figure farther down this page. The video shows that when macroscopic quantities of mercury and bromine are mixed at room temperature, a chemical reaction occurs, and a new substance, Mercury(II) bromide (mercuric bromide), is produced. Mercury(II) bromide is a white solid, quite different in appearance from the two elements from which it was formed. A chemist’s sub-microscopic (nanoscale or atomic scale) theory about what is going on is shown below the video of the macroscopic reaction. Soon after the two liquids are mixed together, a rearrangement of atoms begins. The two bromine atoms of each Br molecule become separated and combine instead with mercury atoms. When the chemical reaction is complete, all that remains is a collection of mercury(II) bromide molecules, each of which contains one mercury atom and two bromine atoms. Notice that there are just as many mercury atoms after the reaction as there were before the reaction. The same applies to bromine atoms. Atoms were neither created, destroyed, divided into parts, or changed into other kinds of atoms during the chemical reaction. The view of solid mercury(II) bromide shown in part is our first sub-microscopic example of a . Each molecule of a compound is made up of two (or more) different of atoms. Since these atoms may be rearranged during a chemical reaction, the molecules may be broken apart and the compound can be decomposed into two (or more) different elements. The formula for a compound involves at least two chemical symbols—one for each element, present. In the case of mercury(II) bromide each molecule contains one mercury atom and two bromine atoms, and so the formula is HgBr . Both part of the figure and the formula tell you that any sample of pure mercury(II) bromide contains twice as many bromine atoms as mercury atoms. This 2:1 ratio agrees with Dalton’s fourth postulate that atoms combine in the ratio of small whole numbers. Although John Dalton originally used circular symbols like those in the figure to represent atoms in chemical reactions, a modern chemist would use chemical symbols and a chemical equation like \[\underset{\text{Reactants}}{\ce{Hg + Br2}} \rightarrow \underset{\text{Products}}{\ce{HgBr2}} \label{1} \] This equation may be interpreted at the sub-microscopic scale to mean that 1 mercury atom and 1 bromine molecule react to form 1 mercury(II) bromide molecule. It should also call to mind the macroscopic change shown in the video, in which a silvery liquid and a red-brown liquid change into a white solid. This macroscopic interpretation is often strengthened by specifying physical states of the reactants and products: \[\text{Hg}(l) + \text{Br}_2 (l) \rightarrow \text{HgBr}_2 (s) \label{2} \] Thus liquid mercury and liquid bromine react to form solid mercuric bromide. [If the bromine had been in gaseous form, Br ( ) might have been written as a product of the reaction. Occasionally ( ) may be used instead of ( ) to indicate a crystalline solid.] Chemical equations such as \(\ref{1}\) and \(\ref{2}\) summarize a great deal of information on both macroscopic and sub-microscopic levels, for those who know how to interpret them. | 3,351 | 1,542 |
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The heat capacity (\(C\)) of a substance is a measure of how much heat is required to raise the temperature of that substance by one degree Kelvin. For a simple molecular gas, the molecules can simultaneously store kinetic energy in the translational, vibrational, and rotational motions associated with the individual molecules. In this case, the heat capacity of the substance can be broken down into translational, vibrational, and rotational contributions; \[ C = C_{trans} + C_{vib} + C_{rot}\] Monoatomic crystalline solids represent a much simpler case. Einstein proposed a simple model for such substances whereby the atoms only have vibrational energy (each atom can vibrate in three perpendicular directions around its lattice position). Specifically, the ‘Einstein Solid Model’ assumes that the atoms act like three-dimensional harmonic oscillators (with the vibrational motion of each atom in each perpendicular dimension entirely independent). Statistical mechanics provides a relatively simple expression for the constant volume molar heat capacity (\(C_{V,m}\)) of a one-dimensional harmonic oscillator \[ C_{V,m}^{1-D} = R \left( \dfrac{\Theta_v }{T} \right)^2 \left( \dfrac{e^{-\Theta_v/2T} }{1- e^{-\Theta_v/T}} \right) ^2 \label{1}\] where \(R\) is the universal gas constant, \(T\) is absolute temperature, and \(Θ_v\) is called the ‘characteristic vibrational temperature’ of the oscillator and depends on the vibrational frequency (\(ν\)) according to \[ \Theta_v = \dfrac{hv}{k} \label{2}\] with \(h\) representing Plank’s constant and \(k\) representing Boltzmann’s constant. Since the vibrations in each dimension are assumed to be independent, the expression for the constant volume molar heat capacity of a 'three-dimensional' Einstein Solid is obtained by simply multiplying Equation \ref{1} by three; \[ C_{V,m}^{3-D} = 3R \left( \dfrac{\Theta_v }{T} \right)^2 \left( \dfrac{e^{-\Theta_v/2T} }{1- e^{-\Theta_v/T}} \right) ^2 \label{3}\] The temperature variation of the heat capacity of most metallic solids is well described by Equation \ref{3}. Furthermore, plots of Equation \ref{3} as a function of temperature for metals with widely varying vibrational frequencies reveal that the heat capacity always approaches the same asymptotic limit of \(3R\) at high temperatures. Stated another way, at high temperatures \[ \lim_{T \to \infty} \left[ \left( \dfrac{\Theta_v }{T} \right)^2 \left( \dfrac{e^{-\Theta_v/2T} }{1- e^{-\Theta_v/T}} \right) ^2 \right] = 1 \label{4}\] and Equation \ref{3} reduces to \[ \lim_{T \to \infty} \left[ C_{V,m}^{3D} \right] = 3R \label{5}\] (You will be asked to verify this result in the exercise below). According to Equation \ref{5}, the molar heat capacities of metallic solids should approach 24.9 J/(K mol) at high temperatures, regardless of the identity of the metal. The vibrational frequencies of most metallic solids are usually small enough so that \(Θ_v\) lies considerably below room temperature (\(Θ_v \ll 298\, K\)). For these substances, the limits implied by Equations \ref{4} and \ref{5} are well approximated even at room temperature, leading to the result that \(C_{v,m} = 24.9\, J/(K·mol)\) for most metals at room temperature. In the early 1800s, two French scientists by the names of Pierre Louis Dulong and Alexis Therese Petit empirically discovered the same remarkable result. The Dulong-Petit Law is normally expressed in terms of the specific heat capacity (\(C_s\)) and the molar mass (\(M\)) of the metal \[C_s M = C_{V,m} \approx 25 (J\, K^{-1} \, mol^{-1}) \label{6}\] where \(C_s\) represents how much heat is required to raise the temperature of 'one gram' of that substance by one degree Kelvin. Dulong and Petit, as well as other scientists of their time, used this famous relationship as a means of establishing more accurate values for the atomic weight of metallic elements (by instead measuring the specific heat capacity of the element and using the Dulong-Petit relationship, which is a relatively simple method of establishing weights in comparison to the more disputable gravimetric methods that were being used at the time to establish the equivalent weights of elements). In the exercise below, you will look up the specific heat capacities of a number of elements that exist as simple monoatomic solids at room temperature and assess the accuracy of the Dulong-Petit law. Consult the CRC Handbook of Chemistry and Physics (CRC Press: Boca Raton, FL) and compile a table of specific heat capacities for a large number of elements that are known to exist as monoatomic solids at room temperature. Also look up and record the molar mass of these elements. The elements that you consider should be restricted to those appearing in groups 1-14 of the periodic table. Make sure you generate a fairly large list which includes a number of elements that are normally considered as metallic in character (such as copper, iron, sodium, lithium, gold, platinum, barium, and aluminum), but also some non-metallic elements that are nonetheless monoatomic isotropic solids (such as carbon-diamond, beryllium, boron, and silicon). Heat capacities that are usually reported in the literature are not actual constant volume heat capacities (\(C_v\)), but are instead constant pressure heat capacities (\(C_p\)). Fortunately, \(C_p\) and \(C_v\) are essentially equal for simple solids (within the level of precision that we consider in this exercise), and you can assume that the values from the CRC Handbook represent \(C_s\). | 5,531 | 1,543 |
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To the student the subject of biological and synthetic molecular-oxygen carriers offers unusual insights into how bioinorganic chemistry works and what its aims and uses are. First, consider another bioinorganic problem, that of the nature of the blue copper proteins. When bioinorganic chemists entered the scene, the nature, function, and structure of copper blues were largely unknown. To take a Cu(lI) solution, add a nitrogen base, and obtain a spectrum that "resembled" that of the proteins was not a contribution to the solution of the biological puzzle, although it was the activity of some bioinorganic chemists. The difficulty here was that too little was known about the protein system. But the challenge did not diminish once the structure of a blue copper protein was known, for that structure allowed definition of the active site, one that contained (in the oxidized form) a Cu(lI) center surrounded by two imidazole, one cysteine, and one methionine residue. Now the bioinorganic chemist was faced with the formidable (and still incompletely solved) synthetic problem of designing a tetradentate ligand that (i) would present two N atoms, one thiolate S atom, and one thioether SR group to a Cu(lI) center and (ii) would remain intact if the Cu(lI) center were reduced to Cu(I). If we could prepare such complexes, we would be in a position to examine in some detail the effects on physical properties, such as redox potentials or spectra, of chemical substitution. In other words, we could learn about structure-function relationships in the copper blues. But the risks involved included the possibility that the specially designed ligand, even if it could be synthesized, might not bind Cu(Il) or Cu(l) in the desired manner. Contrast such a situation with that of the oxygen carriers. Hemoglobin and myoglobin were the first crystallized proteins to have their structures determined. Their functions were well-known. They had been studied by a wide variety of physical techniques, in part because their structures were known, and even before that because of their role in human health. The central tetradentate ligand of the heme group, namely, the porphyrin, was well-defined and much porphyrin chemistry was known. The structural puzzles that intrigued chemists and biologists were not answered in the initial, early structural studies of the proteins; for example, how is O bound and why is CO not bound more firmly? Model chemistry in this area looked as though it would be easy; after all, the metalloporphyrins were readily synthesized, and all one needed was an axial base, some spectroscopic equipment, perhaps some single crystals, and then the structure-function relationships in the biological oxygen carriers would be understood! Indeed, as often happens, the situation was more complicated than it appeared. The irreversible oxidation of iron porphyrins was a major stumbling block to simple modeling. This obstacle was overcome in solution studies through the use of low temperatures and aprotic solvents; some very useful measurements of O and CO binding were made on model systems in such solutions. But in order to isolate oxygen complexes so that they could be studied by diffraction methods, another approach, that of synthesizing elaborated porphyrins, such as those in Figure 4.23, was necessary. This task entailed difficult organic chemistry that ultimately led to successful models that proved to be stable under ambient conditions. From such models we have learned much about local stereochemistry and, through spectroscopic congruence, about the biological systems. In short, bioinorganic chemistry has made a major contribution to the understanding of biological molecular-oxygen carriers, primarily because knowledge of the biological systems was advanced, the systems "self-assemble," and the goals of the studies were well-defined. The complementarity of the two approaches continues. There are several unanswered questions, including: There remain many intriguing questions about biological molecular-oxygen carriers, questions that will be answered by complementary studies on the biological and model systems. To make and study such model systems is an example of the challenge and excitement of this aspect of bioinorganic chemistry. | 4,307 | 1,545 |
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Although the conversion of one element to another is the basis of natural radioactive decay, it is also possible to convert one element to another artificially. The conversion of one element to another is the process of transmutation. Between 1921 and 1924, Patrick Blackett conducted experiments in which he converted a stable isotope of nitrogen to a stable isotope of . By bombarding \(\ce{^{14}N}\) with \(\alpha\) particles he created \(\ce{^{17}O}\). Transmutation may also be accomplished by bombardment with neutrons. \[\ce{^{14}_7N + ^4_2He \rightarrow ^{17}_8O + ^1_1H}\] Historically, part of Alchemy was the study of methods of creating gold from base metals, such lead. Where the Alchemists failed in this quest, we can now succeed. Thus, bombardment of platinum-198 with a neutron creates an unstable isotope of platinum that undergoes beta decay to gold-199. Unfortunately, while we may succeed in making gold, the platinum we make it from is actually worth more than the gold making this particular transmutation economically non-viable! \[\ce{^{198}_{78}Pt + ^1_0n \rightarrow ^{199}_{78}Pt \rightarrow ^{199}_{79}Au + ^0_{-1}\beta}\] | 1,170 | 1,546 |
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Because of the low hydrogen to carbon ratio in aromatic compounds (note that the H:C ratio in an alkane is >2), chemists expected their structural formulas would contain a large number of double or triple bonds. Since double bonds are easily cleaved by oxidative reagents such as potassium permanganate or ozone, and rapidly add bromine and chlorine, these reactions were applied to these aromatic compounds. Surprisingly, products that appeared to retain many of the double bonds were obtained, and these compounds exhibited a high degree of chemical stability compared with known and cycloalkenes (aliphatic compounds). On treatment with hot permanganate solution, cinnamaldehyde gave a stable, crystalline C H O compound, now called benzoic acid. The H:C ratio in benzoic acid is <1, again suggesting the presence of several double bonds. Benzoic acid was eventually converted to the stable hydrocarbon benzene, C H , which also proved unreactive to common double bond transformations, as shown below. For comparison, reactions of cyclohexene, a typical , with these reagents are also shown (green box). As experimental evidence for a wide assortment of compounds was acquired, those incorporating this exceptionally stable six-carbon core came to be called "aromatic". If benzene is forced to react by increasing the temperature and/or by addition of a catalyst, It undergoes rather than the addition reactions that are typical of alkenes. This further confirms the previous indication that the six-carbon benzene core is unusually stable to chemical modification. The conceptual contradiction presented by a high degree of unsaturation (low H:C ratio) and high chemical stability for benzene and related compounds remained an unsolved puzzle for many years. Eventually, the presently accepted structure of a regular-hexagonal, planar ring of carbons was adopted, and the exceptional thermodynamic and chemical stability of this system was attributed to resonance stabilization of a conjugated cyclic triene. Benzene: Here, two structurally and energetically equivalent electronic structures for a stable compound are written, but no single structure provides an accurate or even an adequate representation of the true molecule. The six-membered ring in benzene is a perfect hexagon (all carbon-carbon bonds have an identical length of 1.40 Å). The cyclohexatriene contributors would be expected to show alternating bond lengths, the double bonds being shorter (1.34 Å) than the single bonds (1.54 Å). An alternative representation for benzene (circle within a hexagon) emphasizes the pi-electron delocalization in this molecule, and has the advantage of being a single diagram. In cases such as these, the electron delocalization described by resonance enhances the stability of the molecules, and compounds composed of such molecules often show exceptional stability and related properties. Evidence for the enhanced thermodynamic stability of benzene was obtained from measurements of the heat released when double bonds in a six-carbon ring are hydrogenated (hydrogen is added catalytically) to give cyclohexane as a common product. In the following diagram cyclohexane represents a low-energy reference point. Addition of hydrogen to cyclohexene produces cyclohexane and releases heat amounting to 28.6 kcal per mole. If we take this value to represent the energy cost of introducing one double bond into a six-carbon ring, we would expect a cyclohexadiene to release 57.2 kcal per mole on complete hydrogenation, and 1,3,5-cyclohexatriene to release 85.8 kcal per mole. These would reflect the relative thermodynamic stability of the compounds. In practice, 1,3-cyclohexadiene is slightly more stable than expected, by about 2 kcal, presumably due to conjugation of the double bonds. . This sort of stability enhancement is now accepted as a characteristic of all aromatic compounds. A molecular orbital description of benzene provides a more satisfying and more general treatment of "aromaticity". We know that benzene has a planar hexagonal structure in which all the carbon atoms are sp hybridized, and all the carbon-carbon bonds are equal in length. As shown below, the remaining cyclic array of six p-orbitals ( one on each carbon) overlap to generate six molecular orbitals, three bonding and three antibonding. The plus and minus signs shown in the diagram do not represent electrostatic charge, but refer to phase signs in the equations that describe these orbitals (in the diagram the phases are also color coded). When the phases correspond, the orbitals overlap to generate a common region of like phase, with those orbitals having the greatest overlap (e.g. π ) being lowest in energy. The remaining carbon valence electrons then occupy these molecular orbitals in pairs, resulting in a fully occupied (6 electrons) set of bonding molecular orbitals. It is this completely filled set of bonding orbitals, or , that gives the benzene ring its thermodynamic and chemical stability, just as a filled valence shell octet confers stability on the inert gases. ), | 5,112 | 1,547 |
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. In addition to these individual basis; please contact Plot molar volume versus molar mass for these substances. According to Meyer, which would be considered metals and which would be considered nonmetals? Meyer found that the alkali metals had the highest molar volumes, and that molar volumes decreased steadily with increasing atomic mass, then leveled off, and finally rose again. The elements located on the rising portion of a plot of molar volume versus molar mass were typically nonmetals. If we look at the plot of the data in the table, we can immediately identify those elements with the largest molar volumes (A, B, F) as metals located on the left side of the periodic table. The element with the smallest molar volume (E) is aluminum. The plot shows that the subsequent elements (C, D) have molar volumes that are larger than that of E, but smaller than those of A and B. Thus, C and D are most likely to be nonmetals (which is the case: C = sulfur, D = phosphorus). What happens to the energy of a given orbital as the nuclear charge of a species increases? In a multielectron atom and for a given nuclear charge, the experienced by an electron depends on its value of . Why? The electron density of a particular atom is divided into two general regions. Name these two regions and describe what each represents. As the principal quantum number increases, the energy difference between successive energy levels decreases. Why? What would happen to the electron configurations of the transition metals if this decrease did not occur? Describe the relationship between electron shielding and on the outermost electrons of an atom. Predict how chemical reactivity is affected by a decreased effective nuclear charge. If a given atom or ion has a single electron in each of the following subshells, which electron is easier to remove? The electrons of the 1 shell have a stronger electrostatic attraction to the nucleus than electrons in the 2 shell. Give two reasons for this. Predict whether Na or Cl has the more stable 1 shell and explain your rationale. Arrange K, F, Ba, Pb, B, and I in order of decreasing atomic radius. Arrange Ag, Pt, Mg, C, Cu, and Si in order of increasing atomic radius. Using the periodic table, arrange Li, Ga, Ba, Cl, and Ni in order of increasing atomic radius. Element M is a metal that forms compounds of the type MX , MX , and MX , where X is a halogen. What is the expected trend in the ionic radius of M in these compounds? Arrange these compounds in order of decreasing ionic radius of M. The atomic radii of Na and Cl are 190 and 79 pm, respectively, but the distance between sodium and chlorine in NaCl is 282 pm. Explain this discrepancy. Are shielding effects on the atomic radius more pronounced across a row or down a group? Why? What two factors influence the size of an ion relative to the size of its parent atom? Would you expect the ionic radius of S to be the same in both MgS and Na S? Why or why not? Arrange Br , Al , Sr , F , O , and I in order of increasing ionic radius. Arrange P , N , Cl , In , and S in order of decreasing ionic radius. How is an isoelectronic series different from a series of ions with the same charge? Do the cations in magnesium, strontium, and potassium sulfate form an isoelectronic series? Why or why not? What isoelectronic series arises from fluorine, nitrogen, magnesium, and carbon? Arrange the ions in this series by What would be the charge and electron configuration of an ion formed from calcium that is isoelectronic with The 1 shell is closer to the nucleus and therefore experiences a greater electrostatic attraction. In addition, the electrons in the 2 subshell are shielded by the filled 1 shell, which further decreases the electrostatic attraction to the nucleus. Ba > K > Pb > I > B > F The sum of the calculated atomic radii of sodium and chlorine is 253 pm. The sodium cation is significantly smaller than a neutral sodium atom (102 versus 154 pm), due to the loss of the single electron in the 3 orbital. Conversely, the chloride ion is much larger than a neutral chlorine atom (181 versus 99 pm), because the added electron results in greatly increased electron–electron repulsions within the filled = 3 principal shell. Thus, transferring an electron from sodium to chlorine decreases the radius of sodium by about 50%, but causes the radius of chlorine to almost double. The net effect is that the distance between a sodium ion and a chloride ion in NaCl is than the sum of the atomic radii of the neutral atoms. Plot the ionic charge versus ionic radius using the following data for Mo: Mo , 69 pm; Mo , 65 pm; and Mo , 61 pm. Then use this plot to predict the ionic radius of Mo . Is the observed trend consistent with the general trends discussed in the chapter? Why or why not? Internuclear distances for selected ionic compounds are given in the following table. If the ionic radius of Li is 76 pm, what is the ionic radius of each of the anions? What is the ionic radius of Na ? Arrange the gaseous species Mg , P , Br , S , F , and N in order of increasing radius and justify your decisions. Identify each statement as either true or false and explain your reasoning. Based on electronic configurations, explain why the first ionization energies of the group 16 elements are lower than those of the group 15 elements, which is contrary to the general trend. The first through third ionization energies do not vary greatly across the lanthanides. Why? How does the effective nuclear charge experienced by the n electron change when going from left to right (with increasing atomic number) in this series? Most of the first row transition metals can form at least two stable cations, for example iron(II) and iron(III). In contrast, scandium and zinc each form only a single cation, the Sc and Zn ions, respectively. Use the electron configuration of these elements to provide an explanation. Of the elements Nd, Al, and Ar, which will readily form(s) +3 ions? Why? Orbital energies can reverse when an element is ionized. Of the ions B , Ga , Pr , Cr , and As , in which would you expect this reversal to occur? Explain your reasoning. The periodic trends in electron affinities are not as regular as periodic trends in ionization energies, even though the processes are essentially the converse of one another. Why are there so many more exceptions to the trends in electron affinities compared to ionization energies? Elements lying on a lower right to upper left diagonal line cannot be arranged in order of increasing electronegativity according to where they occur in the periodic table. Why? Why do ionic compounds form, if energy is required to form gaseous cations? Why is Pauling’s definition of electronegativity considered to be somewhat limited? Based on their positions in the periodic table, arrange Sb, O, P, Mo, K, and H in order of increasing electronegativity. Based on their positions in the periodic table, arrange V, F, B, In, Na, and S in order of decreasing electronegativity. 5. Both Al and Nd will form a cation with a +3 charge. Aluminum is in Group 13, and loss of all three valence electrons will produce the Al ion with a noble gas configuration. Neodymium is a lanthanide, and all of the lanthanides tend to form +3 ions because the ionization potentials do not vary greatly across the row, and a +3 charge can be achieved with many oxidants. 11. K < Mo ≈ Sb < P ≈ H < O The following table gives values of the first and third ionization energies for selected elements: Plot the ionization energies versus the number of electrons. Explain why the slopes of the and plots are different, even though the species in each row of the table have the same electron configurations. Would you expect the third ionization energy of iron, corresponding to the removal of an electron from a gaseous Fe ion, to be larger or smaller than the fourth ionization energy, corresponding to the removal of an electron from a gaseous Fe ion? Why? How would these ionization energies compare to the first ionization energy of Ca? Which would you expect to have the highest first ionization energy: Mg, Al, or Si? Which would you expect to have the highest third ionization energy. Why? Use the values of the first ionization energies given in Figure 7.11 to construct plots of first ionization energy versus atomic number for (a) boron through oxygen in the second period; and (b) oxygen through tellurium in group 16. Which plot shows more variation? Explain the reason for the variation in first ionization energies for this group of elements. Arrange Ga, In, and Zn in order of increasing first ionization energies. Would the order be the same for second and third ionization energies? Explain your reasoning. Arrange each set of elements in order of increasing magnitude of electron affinity. Arrange each set of elements in order of decreasing magnitude of electron affinity. Of the species F, O , Al , and Li , which has the highest electron affinity? Explain your reasoning. Of the species O , N , Hg , and H , which has the highest electron affinity? Which has the lowest electron affinity? Justify your answers. The Mulliken electronegativity of element A is 542 kJ/mol. If the electron affinity of A is −72 kJ/mol, what is the first ionization energy of element A? Use the data in the following table as a guideline to decide if A is a metal, a nonmetal, or a semimetal. If 1 g of A contains 4.85 × 10 molecules, what is the identity of element A? Based on their valence electron configurations, classify the following elements as either electrical insulators, electrical conductors, or substances with intermediate conductivity: S, Ba, Fe, Al, Te, Be, O, C, P, Sc, W, Na, B, and Rb. Using the data in Problem 10, what conclusions can you draw with regard to the relationship between electronegativity and electrical properties? Estimate the approximate electronegativity of a pure element that is very dense, lustrous, and malleable. Of the elements Al, Mg, O , Ti, I , and H , which, if any, would you expect to be a good reductant? Explain your reasoning. Of the elements Zn, B, Li, Se, Co, and Br , which if any, would you expect to be a good oxidant? Explain your reasoning. Determine whether each species is a good oxidant, a good reductant, or neither. Determine whether each species is a good oxidant, a good reductant, or neither. Of the species I , O , Zn, Sn , and K , choose which you would expect to be a good oxidant. Then justify your answer. The general features of both plots are roughly the same, with a small peak at 12 electrons and an essentially level region from 15–16 electrons. The slope of the plot is about twice as large as the slope of the plot, however, because the values correspond to removing an electron from an ion with a +2 charge rather than a neutral atom. The greater charge increases the effect of the steady rise in effective nuclear charge across the row. Electron configurations: Mg, 1 2 2 3 ; Al, 1 2 2 3 3 ; Si, 1 2 2 3 3 ; First ionization energies increase across the row due to a steady increase in effective nuclear charge; thus, Si has the highest first ionization energy. The third ionization energy corresponds to removal of a 3 electron for Al and Si, but for Mg it involves removing a 2 electron from a filled inner shell; consequently, the third ionization energy of Mg is the highest. insulators: S, O, C (diamond), P; conductors: Ba, Fe, Al, C (graphite), Be, Sc, W, Na, Rb; Te and B are semimetals and semiconductors. Mg, Al, Ti, and H I is the best oxidant, with a moderately strong tendency to accept an electron to form the I ion, with a closed shell electron configuration. O would probably also be an oxidant, with a tendency to add an electron to form salts containing the oxide ion, O . Zn and Sn are all reductants, while K has no tendency to act as an oxidant or a reductant. see above question to tease out | 12,000 | 1,548 |
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As the idea of heat as a form of transferring energy was first being developed, a unit amount of heat was taken to be the amount that was needed to increase the temperature of a reference material by one degree. Water was the reference material of choice, and the calorie was defined as the quantity of heat that raised the temperature of one gram of water one degree kelvin. The amount of heat exchanged by a known amount of water could then be calculated from the amount by which the temperature of the water changed. If, for example, introducing 63.55 g (1 mole) of copper metal, initially at 274.0 K, into 100 g of water, initially at 373.0 K, resulted in thermal equilibrium at 288.5 K, the water surrendered \[\mathrm{100\ \ g}\ \times \mathrm{1\ cal\ }{\mathrm{g}}^{\mathrm{-1}}\mathrm{\ }{\mathrm{K}}^{\mathrm{-1}}\times 84.5\ \mathrm{K}=8450\mathrm{\ cal}\] This amount of heat was taken up by the copper, so that 0.092 cal was required to increase the temperature of one gram of copper by one degree K. Given this information, the amount of heat gained or lost by a known mass of copper in any subsequent experiment can be calculated from the change in its temperature. Joule developed the idea that mechanical work can be converted entirely into heat. The quantity of heat that could be produced from one unit of mechanical work was called the . Today we the unit of heat in mechanical units. That is, we define the unit of energy, the joule (\(\mathrm{J}\)), in terms of the mechanical units mass (\(\mathrm{kg}\)), distance (\(\mathrm{m}\)), and time (\(\mathrm{s}\)). One joule is one newton-meter or one \(\mathrm{kg\ }{\mathrm{m}}^{\mathrm{2}}\mathrm{\ }{\mathrm{s}}^{\mathrm{-2}}\). \(\boldsymbol{\mathrm{4}}.\boldsymbol{\mathrm{184}}\boldsymbol{\mathrm{\ }}\mathrm{J}\) This definition assumes that heat and work are both forms of energy. This assumption is an intrinsic element of the first law of thermodynamics. This aspect of the first law is, of course, just a restatement of Joule’s original idea. When we want to measure the heat added to a system, measuring the temperature increase that occurs is often the most convenient method. If we know the temperature increase in the system, and we know the temperature increase that accompanies the addition of one unit of heat, we can calculate the heat input to the system. Evidently, it is useful to know how much the temperature increases when one unit of heat is added to various substances. Let us consider a general procedure for accumulating such information. First, we need to choose some standard amount of the substance in question. After all, if we double the amount, it takes twice as much heat to effect the same temperature change. One mole is a natural choice for this standard amount. If we add small increments of heat to one mole of a pure substance, we can measure the temperature after each addition and plot heat temperature. Figure 5 shows such a plot. (In experiments like this, it is often convenient to introduce the heat by passing a known electrical current, \(I\), through a known resistance, \(R\), immersed in the substance. The rate at which heat is produced is \(I^{\mathrm{2}}R\). Except for the usually negligible amount that goes into warming the resistor, all of it is transferred to the substance.) At any particular temperature, the slope of the graph is the increment of heat input divided by the incremental temperature increase. This slope is so useful, it is given a name; it is the molar heat capacity of the substance, \(C\). Since this slope is also the derivative of the \(q\)- -\(T\) curve, we have \[C=\frac{dq}{dT}\] The temperature increase accompanying a given heat input varies with the particular conditions under which the experiment is done. In particular, the temperature increase will be less if some of the added heat is converted to work, as is the case if the volume of the system increases. If the volume increases, the system does work on the surroundings. For a given \(q\), \(\Delta T\) will be less when the system is allowed to expand, which means that \({q}/{\Delta T}\) will be greater. Heat capacity measurements are most conveniently done with the system at a constant pressure. However, the heat capacity at constant volume plays an important role in our theoretical development. The heat capacity is denoted \(C_P\) when the pressure is constant and \(C_V\) when the volume is constant. We have the important definitions \[C_P={\left(\frac{\partial q}{\partial T}\right)}_P\] and \[C_V={\left(\frac{\partial q}{\partial T}\right)}_V\] Since no pressure–volume work can be done when the volume is constant, less heat is required to effect a given temperature change, and we have \(C_P>C_V\), as a general result. (In 14, we consider this point further.) If the system contains a gas, the effect of the volume increase can be substantial. For a monatomic ideal gas, the temperature increase at constant pressure is only \(\mathrm{60\%}\) of the temperature increase at constant volume for the same input of heat. | 5,072 | 1,549 |
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Of the two methods for bringing the stationary phase and the mobile phases into contact, the most important is column chromatography. In this section we develop a general theory that we may apply to any form of column chromatography. Figure 12.2.1
provides a simple view of a liquid–solid column chromatography experiment. The sample is introduced as a narrow band at the top of the column. Ideally, the solute’s initial concentration profile is rectangular (Figure 12.2.2
a). As the sample moves down the column, the solutes begin to separate (Figure 12.2.1
b,c) and the individual solute bands begin to broaden and develop a Gaussian profile (Figure 12.2.2
b,c). If the strength of each solute’s interaction with the stationary phase is sufficiently different, then the solutes separate into individual bands (Figure 12.2.1
d and Figure 12.2.2
d).
. An alternative view of the separation in Figure 12.2.1
showing the concentration of each solute as a function of distance down the column. We can follow the progress of the separation by collecting fractions as they elute from the column (Figure 12.2.1
e,f), or by placing a suitable detector at the end of the column. A plot of the detector’s response as a function of elution time, or as a function of the volume of mobile phase, is known as a (Figure 12.2.3
), and consists of a peak for each solute. There are many possible detectors that we can use to monitor the separation. Later sections of this chapter describe some of the most popular. We can characterize a chromatographic peak’s properties in several ways, two of which are shown in Figure 12.2.4
. , , is the time between the sample’s injection and the maximum response for the solute’s peak. A chromatographic peak’s , , as shown in Figure 12.2.4
, is determined by extending tangent lines from the inflection points on either side of the peak through the baseline. Although usually we report and using units of time, we can report them using units of volume by multiplying each by the mobile phase’s velocity, or report them in linear units by measuring distances with a ruler. For example, a solute’s retention volume, , is \(t_\text{r} \times u\) where is the mobile phase’s velocity through the column. In addition to the solute’s peak, Figure 12.2.4
also shows a small peak that elutes shortly after the sample is injected into the mobile phase. This peak contains all , which move through the column at the same rate as the mobile phase. The time required to elute the nonretained solutes is called the column’s , . The goal of chromatography is to separate a mixture into a series of chromatographic peaks, each of which constitutes a single component of the mixture. The between two chromatographic peaks, , is a quantitative measure of their separation, and is defined as \[R_{A B}=\frac{t_{t, B}-t_{t,A}}{0.5\left(w_{B}+w_{A}\right)}=\frac{2 \Delta t_{r}}{w_{B}+w_{A}} \label{12.1}\] where is the later eluting of the two solutes. As shown in Figure 12.2.5
, the separation of two chromatographic peaks improves with an increase in . If the areas under the two peaks are identical—as is the case in Figure 12.2.5
—then a resolution of 1.50 corresponds to an overlap of only 0.13% for the two elution profiles. Because resolution is a quantitative measure of a separation’s success, it is a useful way to determine if a change in experimental conditions leads to a better separation. In a chromatographic analysis of lemon oil a peak for limonene has a retention time of 8.36 min with a baseline width of 0.96 min. \(\gamma\)-Terpinene elutes at 9.54 min with a baseline width of 0.64 min. What is the resolution between the two peaks? Using Equation \ref{12.1} we find that the resolution is \[R_{A B}=\frac{2 \Delta t_{r}}{w_{B}+w_{A}}=\frac{2(9.54 \text{ min}-8.36 \text{ min})}{0.64 \text{ min}+0.96 \text{ min}}=1.48 \nonumber\] Figure 12.2.6
shows the separation of a two-component mixture. What is the resolution between the two components? Use a ruler to measure \(\Delta t_\text{r}\), , and in millimeters. Because the relationship between elution time and distance is proportional, we can measure \(\Delta t_\text{r}\), , and using a ruler. My measurements are 8.5 mm for \(\Delta t_\text{r}\), and 12.0 mm each for and . Using these values, the resolution is \[R_{A B}=\frac{2 \Delta t_{t}}{w_{A}+w_{B}}=\frac{2(8.5 \text{ mm})}{12.0 \text{ mm}+12.0 \text{ mm}}=0.70 \nonumber\] Your measurements for \(\Delta t_\text{r}\), , and will depend on the relative size of your monitor or printout; however, your value for the resolution should be similar to the answer above. Equation \ref{12.1} suggests that we can improve resolution by increasing \(\Delta t_\text{r}\), or by decreasing and (Figure 12.2.7
). To increase \(\Delta t_\text{r}\) we can use one of two strategies. One approach is to adjust the separation conditions so that both solutes spend less time in the mobile phase—that is, we increase each —which provides more time to effect a separation. A second approach is to increase by adjusting conditions so that only one solute experiences a significant change in its retention time. The baseline width of a solute’s peak depends on the solutes movement within and between the mobile phase and the stationary phase, and is governed by several factors that collectively we call . We will consider each of these approaches for improving resolution in more detail, but first we must define some terms. Let’s assume we can describe a solute’s distribution between the mobile phase and stationary phase using the following equilibrium reaction \[S_{\text{m}} \rightleftharpoons S_{\text{s}} \nonumber\] where is the solute in the mobile phase and is the solute in the stationary phase. Following the same approach we used in for liquid–liquid extractions, the equilibrium constant for this reaction is an equilibrium partition coefficient, . \[K_{D}=\frac{\left[S_{\mathrm{s}}\right]}{\left[S_\text{m}\right]} \nonumber\] This is not a trivial assumption. In this section we are, in effect, treating the solute’s equilibrium between the mobile phase and the stationary phase as if it is identical to the equilibrium in a liquid–liquid extraction. You might question whether this is a reasonable assumption. There is an important difference between the two experiments that we need to consider. In a liquid–liquid extraction, which takes place in a separatory funnel, the two phases remain in contact with each other at all times, allowing for a true equilibrium. In chromatography, however, the mobile phase is in constant motion. A solute that moves into the stationary phase from the mobile phase will equilibrate back into a different portion of the mobile phase; this does not describe a true equilibrium. So, we ask again: Can we treat a solute’s distribution between the mobile phase and the stationary phase as an equilibrium process? The answer is yes, if the mobile phase velocity is slow relative to the kinetics of the solute’s movement back and forth between the two phase. In general, this is a reasonable assumption. In the absence of any additional equilibrium reactions in the mobile phase or the stationary phase, is equivalent to the distribution ratio, , \[D=\frac{\left[S_{0}\right]}{\left[S_\text{m}\right]}=\frac{(\operatorname{mol} \text{S})_\text{s} / V_\text{s}}{(\operatorname{mol} \text{S})_\text{m} / V_\text{m}}=K_{D} \label{12.2}\] where and are the volumes of the stationary phase and the mobile phase, respectively. A conservation of mass requires that the total moles of solute remain constant throughout the separation; thus, we know that the following equation is true. \[(\operatorname{mol} \text{S})_{\operatorname{tot}}=(\operatorname{mol} \text{S})_{\mathrm{m}}+(\operatorname{mol} \text{S})_\text{s} \label{12.3}\] Solving Equation \ref{12.3} for the moles of solute in the stationary phase and substituting into Equation \ref{12.2} leaves us with \[D = \frac{\left\{(\text{mol S})_{\text{tot}} - (\text{mol S})_\text{m}\right\} / V_{\mathrm{s}}}{(\text{mol S})_{\mathrm{m}} / V_{\mathrm{m}}} \nonumber\] Rearranging this equation and solving for the fraction of solute in the mobile phase, , gives \[f_\text{m} = \frac {(\text{mol S})_\text{m}} {(\text{mol S})_\text{tot}} = \frac {V_\text{m}} {DV_\text{s} + V_\text{m}} \label{12.4}\] which is identical to the result for a liquid-liquid extraction (see ). Because we may not know the exact volumes of the stationary phase and the mobile phase, we simplify Equation \ref{12.4} by dividing both the numerator and the denominator by ; thus \[f_\text{m} = \frac {V_\text{m}/V_\text{m}} {DV_\text{s}/V_\text{m} + V_\text{m}/V_\text{m}} = \frac {1} {DV_\text{s}/V_\text{m} + 1} = \frac {1} {1+k} \label{12.5}\] where \[k=D \times \frac{V_\text{s}}{V_\text{m}} \label{12.6}\] is the solute’s . Note that the larger the retention factor, the more the distribution ratio favors the stationary phase, leading to a more strongly retained solute and a longer retention time. Other (older) names for the retention factor are capacity factor, capacity ratio, and partition ratio, and it sometimes is given the symbol \(k^{\prime}\). Keep this in mind if you are using other resources. Retention factor is the approved name from the IUPAC Gold Book. We can determine a solute’s retention factor from a chromatogram by measuring the column’s void time, , and the solute’s retention time, (see ). Solving Equation \ref{12.5} for , we find that \[k=\frac{1-f_\text{m}}{f_\text{m}} \label{12.7}\] Earlier we defined as the fraction of solute in the mobile phase. Assuming a constant mobile phase velocity, we also can define as \[f_\text{m}=\frac{\text { time spent in the mobile phase }}{\text { time spent in the stationary phase }}=\frac{t_\text{m}}{t_\text{r}} \nonumber\] Substituting back into Equation \ref{12.7} and rearranging leaves us with \[k=\frac{1-\frac{t_{m}}{t_{t}}}{\frac{t_{\mathrm{m}}}{t_{\mathrm{r}}}}=\frac{t_{\mathrm{t}}-t_{\mathrm{m}}}{t_{\mathrm{m}}}=\frac{t_{\mathrm{r}}^{\prime}}{t_{\mathrm{m}}} \label{12.8}\] where \(t_\text{r}^{\prime}\) is the . In a chromatographic analysis of low molecular weight acids, butyric acid elutes with a retention time of 7.63 min. The column’s void time is 0.31 min. Calculate the retention factor for butyric acid. \[k_{\mathrm{but}}=\frac{t_{\mathrm{r}}-t_{\mathrm{m}}}{t_{\mathrm{m}}}=\frac{7.63 \text{ min}-0.31 \text{ min}}{0.31 \text{ min}}=23.6 \nonumber\] Figure 12.2.8
is the chromatogram for a two-component mixture. Determine the retention factor for each solute assuming the sample was injected at time = 0. Because the relationship between elution time and distance is proportional, we can measure , , and using a ruler. My measurements are 7.8 mm, 40.2 mm, and 51.5 mm, respectively. Using these values, the retention factors for solute A and solute B are \[k_{1}=\frac{t_{\mathrm{r} 1}-t_\text{m}}{t_\text{m}}=\frac{40.2 \text{ mm}-7.8 \text{ mm}}{7.8 \text{ mm}}=4.15 \nonumber\] \[k_{2}=\frac{t_{\mathrm{r} 2}-t_\text{m}}{t_\text{m}}=\frac{51.5 \text{ mm}-7.8 \text{ mm}}{7.8 \text{ mm}}=5.60 \nonumber\] Your measurements for , , and will depend on the relative size of your monitor or printout; however, your value for the resolution should be similar to the answer above. Selectivity is a relative measure of the retention of two solutes, which we define using a selectivity factor, \(\alpha\) \[\alpha=\frac{k_{B}}{k_{A}}=\frac{t_{r, B}-t_{\mathrm{m}}}{t_{r, A}-t_{\mathrm{m}}} \label{12.9}\] where solute has the smaller retention time. When two solutes elute with identical retention time, \(\alpha = 1.00\); for all other conditions \(\alpha > 1.00\). In the chromatographic analysis for low molecular weight acids described in , the retention time for isobutyric acid is 5.98 min. What is the selectivity factor for isobutyric acid and butyric acid? First we must calculate the retention factor for isobutyric acid. Using the void time from we have \[k_{\mathrm{iso}}=\frac{t_{\mathrm{r}}-t_{\mathrm{m}}}{t_{\mathrm{m}}}=\frac{5.98 \text{ min}-0.31 \text{ min}}{0.31 \text{ min}}=18.3 \nonumber\] The selectivity factor, therefore, is \[\alpha=\frac{k_{\text {but }}}{k_{\text {iso }}}=\frac{23.6}{18.3}=1.29 \nonumber\] Determine the selectivity factor for the chromatogram in . Using the results from , the selectivity factor is \[\alpha=\frac{k_{2}}{k_{1}}=\frac{5.60}{4.15}=1.35 \nonumber\] Your answer may differ slightly due to differences in your values for the two retention factors. Suppose we inject a sample that has a single component. At the moment we inject the sample it is a narrow band of finite width. As the sample passes through the column, the width of this band continually increases in a process we call . Column efficiency is a quantitative measure of the extent of band broadening. See and . When we inject the sample it has a uniform, or rectangular concentration profile with respect to distance down the column. As it passes through the column, the band broadens and takes on a Gaussian concentration profile. In their original theoretical model of chromatography, Martin and Synge divided the chromatographic column into discrete sections, which they called theoretical plates. Within each theoretical plate there is an equilibrium between the solute present in the stationary phase and the solute present in the mobile phase [Martin, A. J. P.; Synge, R. L. M. . , , 1358–1366]. They described column efficiency in terms of the number of , , \[N=\frac{L}{H} \label{12.10}\] where is the column’s length and is the height of a theoretical plate. For any given column, the column efficiency improves—and chromatographic peaks become narrower—when there are more theoretical plates. If we assume that a chromatographic peak has a Gaussian profile, then the extent of band broadening is given by the peak’s variance or standard deviation. The height of a theoretical plate is the peak’s variance per unit length of the column \[H=\frac{\sigma^{2}}{L} \label{12.11}\] where the standard deviation, \(\sigma\), has units of distance. Because retention times and peak widths usually are measured in seconds or minutes, it is more convenient to express the standard deviation in units of time, \(\tau\), by dividing \(\sigma\) by the solute’s average linear velocity, \(\overline{u}\), which is equivalent to dividing the distance it travels, , by its retention time, . \[\tau=\frac{\sigma}{\overline{u}}=\frac{\sigma t_{r}}{L} \label{12.12}\] For a Gaussian peak shape, the width at the baseline, , is four times its standard deviation, \(\tau\). \[w = 4 \tau \label{12.13}\] Combining Equation \ref{12.11}, Equation \ref{12.12}, and Equation \ref{12.13} defines the height of a theoretical plate in terms of the easily measured chromatographic parameters and . \[H=\frac{L w^{2}}{16 t_\text{r}^{2}} \label{12.14}\] Combing Equation \ref{12.14} and Equation \ref{12.10} gives the number of theoretical plates. \[N=16 \frac{t_{\mathrm{r}}^{2}}{w^{2}}=16\left(\frac{t_{\mathrm{r}}}{w}\right)^{2} \label{12.15}\] A chromatographic analysis for the chlorinated pesticide Dieldrin gives a peak with a retention time of 8.68 min and a baseline width of 0.29 min. Calculate the number of theoretical plates? Given that the column is 2.0 m long, what is the height of a theoretical plate in mm? Using Equation \ref{12.15}, the number of theoretical plates is \[N=16 \frac{t_{\mathrm{r}}^{2}}{w^{2}}=16 \times \frac{(8.68 \text{ min})^{2}}{(0.29 \text{ min})^{2}}=14300 \text{ plates} \nonumber\] Solving Equation \ref{12.10} for gives the average height of a theoretical plate as \[H=\frac{L}{N}=\frac{2.00 \text{ m}}{14300 \text{ plates}} \times \frac{1000 \text{ mm}}{\mathrm{m}}=0.14 \text{ mm} / \mathrm{plate} \nonumber\] For each solute in the chromatogram for , calculate the number of theoretical plates and the average height of a theoretical plate. The column is 0.5 m long. Because the relationship between elution time and distance is proportional, we can measure , , , and using a ruler. My measurements are 40.2 mm, 51.5 mm, 8.0 mm, and 13.5 mm, respectively. Using these values, the number of theoretical plates for each solute is \[N_{1}=16 \frac{t_{r,1}^{2}}{w_{1}^{2}}=16 \times \frac{(40.2 \text{ mm})^{2}}{(8.0 \text{ mm})^{2}}=400 \text { theoretical plates } \nonumber\] \[N_{2}=16 \frac{t_{r,2}^{2}}{w_{2}^{2}}=16 \times \frac{(51.5 \text{ mm})^{2}}{(13.5 \text{ mm})^{2}}=233 \text { theoretical plates } \nonumber\] The height of a theoretical plate for each solute is \[H_{1}=\frac{L}{N_{1}}=\frac{0.500 \text{ m}}{400 \text { plates }} \times \frac{1000 \text{ mm}}{\mathrm{m}}=1.2 \text{ mm} / \mathrm{plate} \nonumber\] \[H_{2}=\frac{L}{N_{2}}=\frac{0.500 \text{ m}}{233 \text { plates }} \times \frac{1000 \text{ mm}}{\mathrm{m}}=2.15 \text{ mm} / \mathrm{plate} \nonumber\] Your measurements for , , , and will depend on the relative size of your monitor or printout; however, your values for and for should be similar to the answer above. It is important to remember that a theoretical plate is an artificial construct and that a chromatographic column does not contain physical plates. In fact, the number of theoretical plates depends on both the properties of the column and the solute. As a result, the number of theoretical plates for a column may vary from solute to solute. One advantage of improving column efficiency is that we can separate more solutes with baseline resolution. One estimate of the number of solutes that we can separate is \[n_{c}=1+\frac{\sqrt{N}}{4} \ln \frac{V_{\max }}{V_{\min }} \label{12.16}\] where is the column’s , and and are the smallest and the largest volumes of mobile phase in which we can elute and detect a solute [Giddings, J. C. , Wiley-Interscience: New York, 1991]. A column with 10 000 theoretical plates, for example, can resolve no more than \[n_{c}=1+\frac{\sqrt{10000}}{4} \ln \frac{30 \mathrm{mL}}{1 \mathrm{mL}}=86 \text { solutes } \nonumber\] if and are 1 mL and 30 mL, respectively. This estimate provides an upper bound on the number of solutes and may help us exclude from consideration a column that does not have enough theoretical plates to separate a complex mixture. Just because a column’s theoretical peak capacity is larger than the number of solutes, however, does not mean that a separation is feasible. In most situations the practical peak capacity is less than the theoretical peak capacity because the retention characteristics of some solutes are so similar that a separation is impossible. Nevertheless, columns with more theoretical plates, or with a greater range of possible elution volumes, are more likely to separate a complex mixture. The smallest volume we can use is the column’s void volume. The largest volume is determined either by our patience—the maximum analysis time we can tolerate—or by our inability to detect solutes because there is too much band broadening. Our treatment of chromatography in this section assumes that a solute elutes as a symmetrical Gaussian peak, such as that shown in Figure 12.2.4
. This ideal behavior occurs when the solute’s partition coefficient, \[K_{\mathrm{D}}=\frac{[S_\text{s}]}{\left[S_\text{m}\right]} \nonumber\] is the same for all concentrations of solute. If this is not the case, then the chromatographic peak has an asymmetric peak shape similar to those shown in Figure 12.2.9
. The chromatographic peak in Figure 12.2.9
a is an example of , which occurs when some sites on the stationary phase retain the solute more strongly than other sites. Figure 12.2.9
b, which is an example of most often is the result of overloading the column with sample. As shown in Figure 12.2.9
a, we can report a peak’s asymmetry by drawing a horizontal line at 10% of the peak’s maximum height and measuring the distance from each side of the peak to a line drawn vertically through the peak’s maximum. The asymmetry factor, , is defined as \[T=\frac{b}{a} \nonumber\] The number of theoretical plates for an asymmetric peak shape is approximately \[N \approx \frac{41.7 \times \frac{t_{r}^{2}}{\left(w_{0.1}\right)^{2}}}{T+1.25}=\frac{41.7 \times \frac{t_{r}^{2}}{(a+b)^{2}}}{T+1.25} \nonumber\] where is the width at 10% of the peak’s height [Foley, J. P.; Dorsey, J. G. , , 730–737]. Asymmetric peaks have fewer theoretical plates, and the more asymmetric the peak the smaller the number of theoretical plates. For example, the following table gives values for for a solute eluting with a retention time of 10.0 min and a peak width of 1.00 min. 0.5 0.5 0.6 0.4 0.7 0.3 | 20,705 | 1,550 |
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Of the two methods for bringing the stationary phase and the mobile phases into contact, the most important is column chromatography. In this section we develop a general theory that we may apply to any form of column chromatography. Figure 12.2.1
provides a simple view of a liquid–solid column chromatography experiment. The sample is introduced as a narrow band at the top of the column. Ideally, the solute’s initial concentration profile is rectangular (Figure 12.2.2
a). As the sample moves down the column, the solutes begin to separate (Figure 12.2.1
b,c) and the individual solute bands begin to broaden and develop a Gaussian profile (Figure 12.2.2
b,c). If the strength of each solute’s interaction with the stationary phase is sufficiently different, then the solutes separate into individual bands (Figure 12.2.1
d and Figure 12.2.2
d).
. An alternative view of the separation in Figure 12.2.1
showing the concentration of each solute as a function of distance down the column. We can follow the progress of the separation by collecting fractions as they elute from the column (Figure 12.2.1
e,f), or by placing a suitable detector at the end of the column. A plot of the detector’s response as a function of elution time, or as a function of the volume of mobile phase, is known as a (Figure 12.2.3
), and consists of a peak for each solute. There are many possible detectors that we can use to monitor the separation. Later sections of this chapter describe some of the most popular. We can characterize a chromatographic peak’s properties in several ways, two of which are shown in Figure 12.2.4
. , , is the time between the sample’s injection and the maximum response for the solute’s peak. A chromatographic peak’s , , as shown in Figure 12.2.4
, is determined by extending tangent lines from the inflection points on either side of the peak through the baseline. Although usually we report and using units of time, we can report them using units of volume by multiplying each by the mobile phase’s velocity, or report them in linear units by measuring distances with a ruler. For example, a solute’s retention volume, , is \(t_\text{r} \times u\) where is the mobile phase’s velocity through the column. In addition to the solute’s peak, Figure 12.2.4
also shows a small peak that elutes shortly after the sample is injected into the mobile phase. This peak contains all , which move through the column at the same rate as the mobile phase. The time required to elute the nonretained solutes is called the column’s , . The goal of chromatography is to separate a mixture into a series of chromatographic peaks, each of which constitutes a single component of the mixture. The between two chromatographic peaks, , is a quantitative measure of their separation, and is defined as \[R_{A B}=\frac{t_{t, B}-t_{t,A}}{0.5\left(w_{B}+w_{A}\right)}=\frac{2 \Delta t_{r}}{w_{B}+w_{A}} \label{12.1}\] where is the later eluting of the two solutes. As shown in Figure 12.2.5
, the separation of two chromatographic peaks improves with an increase in . If the areas under the two peaks are identical—as is the case in Figure 12.2.5
—then a resolution of 1.50 corresponds to an overlap of only 0.13% for the two elution profiles. Because resolution is a quantitative measure of a separation’s success, it is a useful way to determine if a change in experimental conditions leads to a better separation. In a chromatographic analysis of lemon oil a peak for limonene has a retention time of 8.36 min with a baseline width of 0.96 min. \(\gamma\)-Terpinene elutes at 9.54 min with a baseline width of 0.64 min. What is the resolution between the two peaks? Using Equation \ref{12.1} we find that the resolution is \[R_{A B}=\frac{2 \Delta t_{r}}{w_{B}+w_{A}}=\frac{2(9.54 \text{ min}-8.36 \text{ min})}{0.64 \text{ min}+0.96 \text{ min}}=1.48 \nonumber\] Figure 12.2.6
shows the separation of a two-component mixture. What is the resolution between the two components? Use a ruler to measure \(\Delta t_\text{r}\), , and in millimeters. Because the relationship between elution time and distance is proportional, we can measure \(\Delta t_\text{r}\), , and using a ruler. My measurements are 8.5 mm for \(\Delta t_\text{r}\), and 12.0 mm each for and . Using these values, the resolution is \[R_{A B}=\frac{2 \Delta t_{t}}{w_{A}+w_{B}}=\frac{2(8.5 \text{ mm})}{12.0 \text{ mm}+12.0 \text{ mm}}=0.70 \nonumber\] Your measurements for \(\Delta t_\text{r}\), , and will depend on the relative size of your monitor or printout; however, your value for the resolution should be similar to the answer above. Equation \ref{12.1} suggests that we can improve resolution by increasing \(\Delta t_\text{r}\), or by decreasing and (Figure 12.2.7
). To increase \(\Delta t_\text{r}\) we can use one of two strategies. One approach is to adjust the separation conditions so that both solutes spend less time in the mobile phase—that is, we increase each —which provides more time to effect a separation. A second approach is to increase by adjusting conditions so that only one solute experiences a significant change in its retention time. The baseline width of a solute’s peak depends on the solutes movement within and between the mobile phase and the stationary phase, and is governed by several factors that collectively we call . We will consider each of these approaches for improving resolution in more detail, but first we must define some terms. Let’s assume we can describe a solute’s distribution between the mobile phase and stationary phase using the following equilibrium reaction \[S_{\text{m}} \rightleftharpoons S_{\text{s}} \nonumber\] where is the solute in the mobile phase and is the solute in the stationary phase. Following the same approach we used in for liquid–liquid extractions, the equilibrium constant for this reaction is an equilibrium partition coefficient, . \[K_{D}=\frac{\left[S_{\mathrm{s}}\right]}{\left[S_\text{m}\right]} \nonumber\] This is not a trivial assumption. In this section we are, in effect, treating the solute’s equilibrium between the mobile phase and the stationary phase as if it is identical to the equilibrium in a liquid–liquid extraction. You might question whether this is a reasonable assumption. There is an important difference between the two experiments that we need to consider. In a liquid–liquid extraction, which takes place in a separatory funnel, the two phases remain in contact with each other at all times, allowing for a true equilibrium. In chromatography, however, the mobile phase is in constant motion. A solute that moves into the stationary phase from the mobile phase will equilibrate back into a different portion of the mobile phase; this does not describe a true equilibrium. So, we ask again: Can we treat a solute’s distribution between the mobile phase and the stationary phase as an equilibrium process? The answer is yes, if the mobile phase velocity is slow relative to the kinetics of the solute’s movement back and forth between the two phase. In general, this is a reasonable assumption. In the absence of any additional equilibrium reactions in the mobile phase or the stationary phase, is equivalent to the distribution ratio, , \[D=\frac{\left[S_{0}\right]}{\left[S_\text{m}\right]}=\frac{(\operatorname{mol} \text{S})_\text{s} / V_\text{s}}{(\operatorname{mol} \text{S})_\text{m} / V_\text{m}}=K_{D} \label{12.2}\] where and are the volumes of the stationary phase and the mobile phase, respectively. A conservation of mass requires that the total moles of solute remain constant throughout the separation; thus, we know that the following equation is true. \[(\operatorname{mol} \text{S})_{\operatorname{tot}}=(\operatorname{mol} \text{S})_{\mathrm{m}}+(\operatorname{mol} \text{S})_\text{s} \label{12.3}\] Solving Equation \ref{12.3} for the moles of solute in the stationary phase and substituting into Equation \ref{12.2} leaves us with \[D = \frac{\left\{(\text{mol S})_{\text{tot}} - (\text{mol S})_\text{m}\right\} / V_{\mathrm{s}}}{(\text{mol S})_{\mathrm{m}} / V_{\mathrm{m}}} \nonumber\] Rearranging this equation and solving for the fraction of solute in the mobile phase, , gives \[f_\text{m} = \frac {(\text{mol S})_\text{m}} {(\text{mol S})_\text{tot}} = \frac {V_\text{m}} {DV_\text{s} + V_\text{m}} \label{12.4}\] which is identical to the result for a liquid-liquid extraction (see ). Because we may not know the exact volumes of the stationary phase and the mobile phase, we simplify Equation \ref{12.4} by dividing both the numerator and the denominator by ; thus \[f_\text{m} = \frac {V_\text{m}/V_\text{m}} {DV_\text{s}/V_\text{m} + V_\text{m}/V_\text{m}} = \frac {1} {DV_\text{s}/V_\text{m} + 1} = \frac {1} {1+k} \label{12.5}\] where \[k=D \times \frac{V_\text{s}}{V_\text{m}} \label{12.6}\] is the solute’s . Note that the larger the retention factor, the more the distribution ratio favors the stationary phase, leading to a more strongly retained solute and a longer retention time. Other (older) names for the retention factor are capacity factor, capacity ratio, and partition ratio, and it sometimes is given the symbol \(k^{\prime}\). Keep this in mind if you are using other resources. Retention factor is the approved name from the IUPAC Gold Book. We can determine a solute’s retention factor from a chromatogram by measuring the column’s void time, , and the solute’s retention time, (see ). Solving Equation \ref{12.5} for , we find that \[k=\frac{1-f_\text{m}}{f_\text{m}} \label{12.7}\] Earlier we defined as the fraction of solute in the mobile phase. Assuming a constant mobile phase velocity, we also can define as \[f_\text{m}=\frac{\text { time spent in the mobile phase }}{\text { time spent in the stationary phase }}=\frac{t_\text{m}}{t_\text{r}} \nonumber\] Substituting back into Equation \ref{12.7} and rearranging leaves us with \[k=\frac{1-\frac{t_{m}}{t_{t}}}{\frac{t_{\mathrm{m}}}{t_{\mathrm{r}}}}=\frac{t_{\mathrm{t}}-t_{\mathrm{m}}}{t_{\mathrm{m}}}=\frac{t_{\mathrm{r}}^{\prime}}{t_{\mathrm{m}}} \label{12.8}\] where \(t_\text{r}^{\prime}\) is the . In a chromatographic analysis of low molecular weight acids, butyric acid elutes with a retention time of 7.63 min. The column’s void time is 0.31 min. Calculate the retention factor for butyric acid. \[k_{\mathrm{but}}=\frac{t_{\mathrm{r}}-t_{\mathrm{m}}}{t_{\mathrm{m}}}=\frac{7.63 \text{ min}-0.31 \text{ min}}{0.31 \text{ min}}=23.6 \nonumber\] Figure 12.2.8
is the chromatogram for a two-component mixture. Determine the retention factor for each solute assuming the sample was injected at time = 0. Because the relationship between elution time and distance is proportional, we can measure , , and using a ruler. My measurements are 7.8 mm, 40.2 mm, and 51.5 mm, respectively. Using these values, the retention factors for solute A and solute B are \[k_{1}=\frac{t_{\mathrm{r} 1}-t_\text{m}}{t_\text{m}}=\frac{40.2 \text{ mm}-7.8 \text{ mm}}{7.8 \text{ mm}}=4.15 \nonumber\] \[k_{2}=\frac{t_{\mathrm{r} 2}-t_\text{m}}{t_\text{m}}=\frac{51.5 \text{ mm}-7.8 \text{ mm}}{7.8 \text{ mm}}=5.60 \nonumber\] Your measurements for , , and will depend on the relative size of your monitor or printout; however, your value for the resolution should be similar to the answer above. Selectivity is a relative measure of the retention of two solutes, which we define using a selectivity factor, \(\alpha\) \[\alpha=\frac{k_{B}}{k_{A}}=\frac{t_{r, B}-t_{\mathrm{m}}}{t_{r, A}-t_{\mathrm{m}}} \label{12.9}\] where solute has the smaller retention time. When two solutes elute with identical retention time, \(\alpha = 1.00\); for all other conditions \(\alpha > 1.00\). In the chromatographic analysis for low molecular weight acids described in , the retention time for isobutyric acid is 5.98 min. What is the selectivity factor for isobutyric acid and butyric acid? First we must calculate the retention factor for isobutyric acid. Using the void time from we have \[k_{\mathrm{iso}}=\frac{t_{\mathrm{r}}-t_{\mathrm{m}}}{t_{\mathrm{m}}}=\frac{5.98 \text{ min}-0.31 \text{ min}}{0.31 \text{ min}}=18.3 \nonumber\] The selectivity factor, therefore, is \[\alpha=\frac{k_{\text {but }}}{k_{\text {iso }}}=\frac{23.6}{18.3}=1.29 \nonumber\] Determine the selectivity factor for the chromatogram in . Using the results from , the selectivity factor is \[\alpha=\frac{k_{2}}{k_{1}}=\frac{5.60}{4.15}=1.35 \nonumber\] Your answer may differ slightly due to differences in your values for the two retention factors. Suppose we inject a sample that has a single component. At the moment we inject the sample it is a narrow band of finite width. As the sample passes through the column, the width of this band continually increases in a process we call . Column efficiency is a quantitative measure of the extent of band broadening. See and . When we inject the sample it has a uniform, or rectangular concentration profile with respect to distance down the column. As it passes through the column, the band broadens and takes on a Gaussian concentration profile. In their original theoretical model of chromatography, Martin and Synge divided the chromatographic column into discrete sections, which they called theoretical plates. Within each theoretical plate there is an equilibrium between the solute present in the stationary phase and the solute present in the mobile phase [Martin, A. J. P.; Synge, R. L. M. . , , 1358–1366]. They described column efficiency in terms of the number of , , \[N=\frac{L}{H} \label{12.10}\] where is the column’s length and is the height of a theoretical plate. For any given column, the column efficiency improves—and chromatographic peaks become narrower—when there are more theoretical plates. If we assume that a chromatographic peak has a Gaussian profile, then the extent of band broadening is given by the peak’s variance or standard deviation. The height of a theoretical plate is the peak’s variance per unit length of the column \[H=\frac{\sigma^{2}}{L} \label{12.11}\] where the standard deviation, \(\sigma\), has units of distance. Because retention times and peak widths usually are measured in seconds or minutes, it is more convenient to express the standard deviation in units of time, \(\tau\), by dividing \(\sigma\) by the solute’s average linear velocity, \(\overline{u}\), which is equivalent to dividing the distance it travels, , by its retention time, . \[\tau=\frac{\sigma}{\overline{u}}=\frac{\sigma t_{r}}{L} \label{12.12}\] For a Gaussian peak shape, the width at the baseline, , is four times its standard deviation, \(\tau\). \[w = 4 \tau \label{12.13}\] Combining Equation \ref{12.11}, Equation \ref{12.12}, and Equation \ref{12.13} defines the height of a theoretical plate in terms of the easily measured chromatographic parameters and . \[H=\frac{L w^{2}}{16 t_\text{r}^{2}} \label{12.14}\] Combing Equation \ref{12.14} and Equation \ref{12.10} gives the number of theoretical plates. \[N=16 \frac{t_{\mathrm{r}}^{2}}{w^{2}}=16\left(\frac{t_{\mathrm{r}}}{w}\right)^{2} \label{12.15}\] A chromatographic analysis for the chlorinated pesticide Dieldrin gives a peak with a retention time of 8.68 min and a baseline width of 0.29 min. Calculate the number of theoretical plates? Given that the column is 2.0 m long, what is the height of a theoretical plate in mm? Using Equation \ref{12.15}, the number of theoretical plates is \[N=16 \frac{t_{\mathrm{r}}^{2}}{w^{2}}=16 \times \frac{(8.68 \text{ min})^{2}}{(0.29 \text{ min})^{2}}=14300 \text{ plates} \nonumber\] Solving Equation \ref{12.10} for gives the average height of a theoretical plate as \[H=\frac{L}{N}=\frac{2.00 \text{ m}}{14300 \text{ plates}} \times \frac{1000 \text{ mm}}{\mathrm{m}}=0.14 \text{ mm} / \mathrm{plate} \nonumber\] For each solute in the chromatogram for , calculate the number of theoretical plates and the average height of a theoretical plate. The column is 0.5 m long. Because the relationship between elution time and distance is proportional, we can measure , , , and using a ruler. My measurements are 40.2 mm, 51.5 mm, 8.0 mm, and 13.5 mm, respectively. Using these values, the number of theoretical plates for each solute is \[N_{1}=16 \frac{t_{r,1}^{2}}{w_{1}^{2}}=16 \times \frac{(40.2 \text{ mm})^{2}}{(8.0 \text{ mm})^{2}}=400 \text { theoretical plates } \nonumber\] \[N_{2}=16 \frac{t_{r,2}^{2}}{w_{2}^{2}}=16 \times \frac{(51.5 \text{ mm})^{2}}{(13.5 \text{ mm})^{2}}=233 \text { theoretical plates } \nonumber\] The height of a theoretical plate for each solute is \[H_{1}=\frac{L}{N_{1}}=\frac{0.500 \text{ m}}{400 \text { plates }} \times \frac{1000 \text{ mm}}{\mathrm{m}}=1.2 \text{ mm} / \mathrm{plate} \nonumber\] \[H_{2}=\frac{L}{N_{2}}=\frac{0.500 \text{ m}}{233 \text { plates }} \times \frac{1000 \text{ mm}}{\mathrm{m}}=2.15 \text{ mm} / \mathrm{plate} \nonumber\] Your measurements for , , , and will depend on the relative size of your monitor or printout; however, your values for and for should be similar to the answer above. It is important to remember that a theoretical plate is an artificial construct and that a chromatographic column does not contain physical plates. In fact, the number of theoretical plates depends on both the properties of the column and the solute. As a result, the number of theoretical plates for a column may vary from solute to solute. One advantage of improving column efficiency is that we can separate more solutes with baseline resolution. One estimate of the number of solutes that we can separate is \[n_{c}=1+\frac{\sqrt{N}}{4} \ln \frac{V_{\max }}{V_{\min }} \label{12.16}\] where is the column’s , and and are the smallest and the largest volumes of mobile phase in which we can elute and detect a solute [Giddings, J. C. , Wiley-Interscience: New York, 1991]. A column with 10 000 theoretical plates, for example, can resolve no more than \[n_{c}=1+\frac{\sqrt{10000}}{4} \ln \frac{30 \mathrm{mL}}{1 \mathrm{mL}}=86 \text { solutes } \nonumber\] if and are 1 mL and 30 mL, respectively. This estimate provides an upper bound on the number of solutes and may help us exclude from consideration a column that does not have enough theoretical plates to separate a complex mixture. Just because a column’s theoretical peak capacity is larger than the number of solutes, however, does not mean that a separation is feasible. In most situations the practical peak capacity is less than the theoretical peak capacity because the retention characteristics of some solutes are so similar that a separation is impossible. Nevertheless, columns with more theoretical plates, or with a greater range of possible elution volumes, are more likely to separate a complex mixture. The smallest volume we can use is the column’s void volume. The largest volume is determined either by our patience—the maximum analysis time we can tolerate—or by our inability to detect solutes because there is too much band broadening. Our treatment of chromatography in this section assumes that a solute elutes as a symmetrical Gaussian peak, such as that shown in Figure 12.2.4
. This ideal behavior occurs when the solute’s partition coefficient, \[K_{\mathrm{D}}=\frac{[S_\text{s}]}{\left[S_\text{m}\right]} \nonumber\] is the same for all concentrations of solute. If this is not the case, then the chromatographic peak has an asymmetric peak shape similar to those shown in Figure 12.2.9
. The chromatographic peak in Figure 12.2.9
a is an example of , which occurs when some sites on the stationary phase retain the solute more strongly than other sites. Figure 12.2.9
b, which is an example of most often is the result of overloading the column with sample. As shown in Figure 12.2.9
a, we can report a peak’s asymmetry by drawing a horizontal line at 10% of the peak’s maximum height and measuring the distance from each side of the peak to a line drawn vertically through the peak’s maximum. The asymmetry factor, , is defined as \[T=\frac{b}{a} \nonumber\] The number of theoretical plates for an asymmetric peak shape is approximately \[N \approx \frac{41.7 \times \frac{t_{r}^{2}}{\left(w_{0.1}\right)^{2}}}{T+1.25}=\frac{41.7 \times \frac{t_{r}^{2}}{(a+b)^{2}}}{T+1.25} \nonumber\] where is the width at 10% of the peak’s height [Foley, J. P.; Dorsey, J. G. , , 730–737]. Asymmetric peaks have fewer theoretical plates, and the more asymmetric the peak the smaller the number of theoretical plates. For example, the following table gives values for for a solute eluting with a retention time of 10.0 min and a peak width of 1.00 min. 0.5 0.5 0.6 0.4 0.7 0.3 | 20,705 | 1,551 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/12%3A_Chromatographic_and_Electrophoretic_Methods/12.02%3A_General_Theory_of_Column_Chromatography |
Of the two methods for bringing the stationary phase and the mobile phases into contact, the most important is column chromatography. In this section we develop a general theory that we may apply to any form of column chromatography. Figure 12.2.1
provides a simple view of a liquid–solid column chromatography experiment. The sample is introduced as a narrow band at the top of the column. Ideally, the solute’s initial concentration profile is rectangular (Figure 12.2.2
a). As the sample moves down the column, the solutes begin to separate (Figure 12.2.1
b,c) and the individual solute bands begin to broaden and develop a Gaussian profile (Figure 12.2.2
b,c). If the strength of each solute’s interaction with the stationary phase is sufficiently different, then the solutes separate into individual bands (Figure 12.2.1
d and Figure 12.2.2
d).
. An alternative view of the separation in Figure 12.2.1
showing the concentration of each solute as a function of distance down the column. We can follow the progress of the separation by collecting fractions as they elute from the column (Figure 12.2.1
e,f), or by placing a suitable detector at the end of the column. A plot of the detector’s response as a function of elution time, or as a function of the volume of mobile phase, is known as a (Figure 12.2.3
), and consists of a peak for each solute. There are many possible detectors that we can use to monitor the separation. Later sections of this chapter describe some of the most popular. We can characterize a chromatographic peak’s properties in several ways, two of which are shown in Figure 12.2.4
. , , is the time between the sample’s injection and the maximum response for the solute’s peak. A chromatographic peak’s , , as shown in Figure 12.2.4
, is determined by extending tangent lines from the inflection points on either side of the peak through the baseline. Although usually we report and using units of time, we can report them using units of volume by multiplying each by the mobile phase’s velocity, or report them in linear units by measuring distances with a ruler. For example, a solute’s retention volume, , is \(t_\text{r} \times u\) where is the mobile phase’s velocity through the column. In addition to the solute’s peak, Figure 12.2.4
also shows a small peak that elutes shortly after the sample is injected into the mobile phase. This peak contains all , which move through the column at the same rate as the mobile phase. The time required to elute the nonretained solutes is called the column’s , . The goal of chromatography is to separate a mixture into a series of chromatographic peaks, each of which constitutes a single component of the mixture. The between two chromatographic peaks, , is a quantitative measure of their separation, and is defined as \[R_{A B}=\frac{t_{t, B}-t_{t,A}}{0.5\left(w_{B}+w_{A}\right)}=\frac{2 \Delta t_{r}}{w_{B}+w_{A}} \label{12.1}\] where is the later eluting of the two solutes. As shown in Figure 12.2.5
, the separation of two chromatographic peaks improves with an increase in . If the areas under the two peaks are identical—as is the case in Figure 12.2.5
—then a resolution of 1.50 corresponds to an overlap of only 0.13% for the two elution profiles. Because resolution is a quantitative measure of a separation’s success, it is a useful way to determine if a change in experimental conditions leads to a better separation. In a chromatographic analysis of lemon oil a peak for limonene has a retention time of 8.36 min with a baseline width of 0.96 min. \(\gamma\)-Terpinene elutes at 9.54 min with a baseline width of 0.64 min. What is the resolution between the two peaks? Using Equation \ref{12.1} we find that the resolution is \[R_{A B}=\frac{2 \Delta t_{r}}{w_{B}+w_{A}}=\frac{2(9.54 \text{ min}-8.36 \text{ min})}{0.64 \text{ min}+0.96 \text{ min}}=1.48 \nonumber\] Figure 12.2.6
shows the separation of a two-component mixture. What is the resolution between the two components? Use a ruler to measure \(\Delta t_\text{r}\), , and in millimeters. Because the relationship between elution time and distance is proportional, we can measure \(\Delta t_\text{r}\), , and using a ruler. My measurements are 8.5 mm for \(\Delta t_\text{r}\), and 12.0 mm each for and . Using these values, the resolution is \[R_{A B}=\frac{2 \Delta t_{t}}{w_{A}+w_{B}}=\frac{2(8.5 \text{ mm})}{12.0 \text{ mm}+12.0 \text{ mm}}=0.70 \nonumber\] Your measurements for \(\Delta t_\text{r}\), , and will depend on the relative size of your monitor or printout; however, your value for the resolution should be similar to the answer above. Equation \ref{12.1} suggests that we can improve resolution by increasing \(\Delta t_\text{r}\), or by decreasing and (Figure 12.2.7
). To increase \(\Delta t_\text{r}\) we can use one of two strategies. One approach is to adjust the separation conditions so that both solutes spend less time in the mobile phase—that is, we increase each —which provides more time to effect a separation. A second approach is to increase by adjusting conditions so that only one solute experiences a significant change in its retention time. The baseline width of a solute’s peak depends on the solutes movement within and between the mobile phase and the stationary phase, and is governed by several factors that collectively we call . We will consider each of these approaches for improving resolution in more detail, but first we must define some terms. Let’s assume we can describe a solute’s distribution between the mobile phase and stationary phase using the following equilibrium reaction \[S_{\text{m}} \rightleftharpoons S_{\text{s}} \nonumber\] where is the solute in the mobile phase and is the solute in the stationary phase. Following the same approach we used in for liquid–liquid extractions, the equilibrium constant for this reaction is an equilibrium partition coefficient, . \[K_{D}=\frac{\left[S_{\mathrm{s}}\right]}{\left[S_\text{m}\right]} \nonumber\] This is not a trivial assumption. In this section we are, in effect, treating the solute’s equilibrium between the mobile phase and the stationary phase as if it is identical to the equilibrium in a liquid–liquid extraction. You might question whether this is a reasonable assumption. There is an important difference between the two experiments that we need to consider. In a liquid–liquid extraction, which takes place in a separatory funnel, the two phases remain in contact with each other at all times, allowing for a true equilibrium. In chromatography, however, the mobile phase is in constant motion. A solute that moves into the stationary phase from the mobile phase will equilibrate back into a different portion of the mobile phase; this does not describe a true equilibrium. So, we ask again: Can we treat a solute’s distribution between the mobile phase and the stationary phase as an equilibrium process? The answer is yes, if the mobile phase velocity is slow relative to the kinetics of the solute’s movement back and forth between the two phase. In general, this is a reasonable assumption. In the absence of any additional equilibrium reactions in the mobile phase or the stationary phase, is equivalent to the distribution ratio, , \[D=\frac{\left[S_{0}\right]}{\left[S_\text{m}\right]}=\frac{(\operatorname{mol} \text{S})_\text{s} / V_\text{s}}{(\operatorname{mol} \text{S})_\text{m} / V_\text{m}}=K_{D} \label{12.2}\] where and are the volumes of the stationary phase and the mobile phase, respectively. A conservation of mass requires that the total moles of solute remain constant throughout the separation; thus, we know that the following equation is true. \[(\operatorname{mol} \text{S})_{\operatorname{tot}}=(\operatorname{mol} \text{S})_{\mathrm{m}}+(\operatorname{mol} \text{S})_\text{s} \label{12.3}\] Solving Equation \ref{12.3} for the moles of solute in the stationary phase and substituting into Equation \ref{12.2} leaves us with \[D = \frac{\left\{(\text{mol S})_{\text{tot}} - (\text{mol S})_\text{m}\right\} / V_{\mathrm{s}}}{(\text{mol S})_{\mathrm{m}} / V_{\mathrm{m}}} \nonumber\] Rearranging this equation and solving for the fraction of solute in the mobile phase, , gives \[f_\text{m} = \frac {(\text{mol S})_\text{m}} {(\text{mol S})_\text{tot}} = \frac {V_\text{m}} {DV_\text{s} + V_\text{m}} \label{12.4}\] which is identical to the result for a liquid-liquid extraction (see ). Because we may not know the exact volumes of the stationary phase and the mobile phase, we simplify Equation \ref{12.4} by dividing both the numerator and the denominator by ; thus \[f_\text{m} = \frac {V_\text{m}/V_\text{m}} {DV_\text{s}/V_\text{m} + V_\text{m}/V_\text{m}} = \frac {1} {DV_\text{s}/V_\text{m} + 1} = \frac {1} {1+k} \label{12.5}\] where \[k=D \times \frac{V_\text{s}}{V_\text{m}} \label{12.6}\] is the solute’s . Note that the larger the retention factor, the more the distribution ratio favors the stationary phase, leading to a more strongly retained solute and a longer retention time. Other (older) names for the retention factor are capacity factor, capacity ratio, and partition ratio, and it sometimes is given the symbol \(k^{\prime}\). Keep this in mind if you are using other resources. Retention factor is the approved name from the IUPAC Gold Book. We can determine a solute’s retention factor from a chromatogram by measuring the column’s void time, , and the solute’s retention time, (see ). Solving Equation \ref{12.5} for , we find that \[k=\frac{1-f_\text{m}}{f_\text{m}} \label{12.7}\] Earlier we defined as the fraction of solute in the mobile phase. Assuming a constant mobile phase velocity, we also can define as \[f_\text{m}=\frac{\text { time spent in the mobile phase }}{\text { time spent in the stationary phase }}=\frac{t_\text{m}}{t_\text{r}} \nonumber\] Substituting back into Equation \ref{12.7} and rearranging leaves us with \[k=\frac{1-\frac{t_{m}}{t_{t}}}{\frac{t_{\mathrm{m}}}{t_{\mathrm{r}}}}=\frac{t_{\mathrm{t}}-t_{\mathrm{m}}}{t_{\mathrm{m}}}=\frac{t_{\mathrm{r}}^{\prime}}{t_{\mathrm{m}}} \label{12.8}\] where \(t_\text{r}^{\prime}\) is the . In a chromatographic analysis of low molecular weight acids, butyric acid elutes with a retention time of 7.63 min. The column’s void time is 0.31 min. Calculate the retention factor for butyric acid. \[k_{\mathrm{but}}=\frac{t_{\mathrm{r}}-t_{\mathrm{m}}}{t_{\mathrm{m}}}=\frac{7.63 \text{ min}-0.31 \text{ min}}{0.31 \text{ min}}=23.6 \nonumber\] Figure 12.2.8
is the chromatogram for a two-component mixture. Determine the retention factor for each solute assuming the sample was injected at time = 0. Because the relationship between elution time and distance is proportional, we can measure , , and using a ruler. My measurements are 7.8 mm, 40.2 mm, and 51.5 mm, respectively. Using these values, the retention factors for solute A and solute B are \[k_{1}=\frac{t_{\mathrm{r} 1}-t_\text{m}}{t_\text{m}}=\frac{40.2 \text{ mm}-7.8 \text{ mm}}{7.8 \text{ mm}}=4.15 \nonumber\] \[k_{2}=\frac{t_{\mathrm{r} 2}-t_\text{m}}{t_\text{m}}=\frac{51.5 \text{ mm}-7.8 \text{ mm}}{7.8 \text{ mm}}=5.60 \nonumber\] Your measurements for , , and will depend on the relative size of your monitor or printout; however, your value for the resolution should be similar to the answer above. Selectivity is a relative measure of the retention of two solutes, which we define using a selectivity factor, \(\alpha\) \[\alpha=\frac{k_{B}}{k_{A}}=\frac{t_{r, B}-t_{\mathrm{m}}}{t_{r, A}-t_{\mathrm{m}}} \label{12.9}\] where solute has the smaller retention time. When two solutes elute with identical retention time, \(\alpha = 1.00\); for all other conditions \(\alpha > 1.00\). In the chromatographic analysis for low molecular weight acids described in , the retention time for isobutyric acid is 5.98 min. What is the selectivity factor for isobutyric acid and butyric acid? First we must calculate the retention factor for isobutyric acid. Using the void time from we have \[k_{\mathrm{iso}}=\frac{t_{\mathrm{r}}-t_{\mathrm{m}}}{t_{\mathrm{m}}}=\frac{5.98 \text{ min}-0.31 \text{ min}}{0.31 \text{ min}}=18.3 \nonumber\] The selectivity factor, therefore, is \[\alpha=\frac{k_{\text {but }}}{k_{\text {iso }}}=\frac{23.6}{18.3}=1.29 \nonumber\] Determine the selectivity factor for the chromatogram in . Using the results from , the selectivity factor is \[\alpha=\frac{k_{2}}{k_{1}}=\frac{5.60}{4.15}=1.35 \nonumber\] Your answer may differ slightly due to differences in your values for the two retention factors. Suppose we inject a sample that has a single component. At the moment we inject the sample it is a narrow band of finite width. As the sample passes through the column, the width of this band continually increases in a process we call . Column efficiency is a quantitative measure of the extent of band broadening. See and . When we inject the sample it has a uniform, or rectangular concentration profile with respect to distance down the column. As it passes through the column, the band broadens and takes on a Gaussian concentration profile. In their original theoretical model of chromatography, Martin and Synge divided the chromatographic column into discrete sections, which they called theoretical plates. Within each theoretical plate there is an equilibrium between the solute present in the stationary phase and the solute present in the mobile phase [Martin, A. J. P.; Synge, R. L. M. . , , 1358–1366]. They described column efficiency in terms of the number of , , \[N=\frac{L}{H} \label{12.10}\] where is the column’s length and is the height of a theoretical plate. For any given column, the column efficiency improves—and chromatographic peaks become narrower—when there are more theoretical plates. If we assume that a chromatographic peak has a Gaussian profile, then the extent of band broadening is given by the peak’s variance or standard deviation. The height of a theoretical plate is the peak’s variance per unit length of the column \[H=\frac{\sigma^{2}}{L} \label{12.11}\] where the standard deviation, \(\sigma\), has units of distance. Because retention times and peak widths usually are measured in seconds or minutes, it is more convenient to express the standard deviation in units of time, \(\tau\), by dividing \(\sigma\) by the solute’s average linear velocity, \(\overline{u}\), which is equivalent to dividing the distance it travels, , by its retention time, . \[\tau=\frac{\sigma}{\overline{u}}=\frac{\sigma t_{r}}{L} \label{12.12}\] For a Gaussian peak shape, the width at the baseline, , is four times its standard deviation, \(\tau\). \[w = 4 \tau \label{12.13}\] Combining Equation \ref{12.11}, Equation \ref{12.12}, and Equation \ref{12.13} defines the height of a theoretical plate in terms of the easily measured chromatographic parameters and . \[H=\frac{L w^{2}}{16 t_\text{r}^{2}} \label{12.14}\] Combing Equation \ref{12.14} and Equation \ref{12.10} gives the number of theoretical plates. \[N=16 \frac{t_{\mathrm{r}}^{2}}{w^{2}}=16\left(\frac{t_{\mathrm{r}}}{w}\right)^{2} \label{12.15}\] A chromatographic analysis for the chlorinated pesticide Dieldrin gives a peak with a retention time of 8.68 min and a baseline width of 0.29 min. Calculate the number of theoretical plates? Given that the column is 2.0 m long, what is the height of a theoretical plate in mm? Using Equation \ref{12.15}, the number of theoretical plates is \[N=16 \frac{t_{\mathrm{r}}^{2}}{w^{2}}=16 \times \frac{(8.68 \text{ min})^{2}}{(0.29 \text{ min})^{2}}=14300 \text{ plates} \nonumber\] Solving Equation \ref{12.10} for gives the average height of a theoretical plate as \[H=\frac{L}{N}=\frac{2.00 \text{ m}}{14300 \text{ plates}} \times \frac{1000 \text{ mm}}{\mathrm{m}}=0.14 \text{ mm} / \mathrm{plate} \nonumber\] For each solute in the chromatogram for , calculate the number of theoretical plates and the average height of a theoretical plate. The column is 0.5 m long. Because the relationship between elution time and distance is proportional, we can measure , , , and using a ruler. My measurements are 40.2 mm, 51.5 mm, 8.0 mm, and 13.5 mm, respectively. Using these values, the number of theoretical plates for each solute is \[N_{1}=16 \frac{t_{r,1}^{2}}{w_{1}^{2}}=16 \times \frac{(40.2 \text{ mm})^{2}}{(8.0 \text{ mm})^{2}}=400 \text { theoretical plates } \nonumber\] \[N_{2}=16 \frac{t_{r,2}^{2}}{w_{2}^{2}}=16 \times \frac{(51.5 \text{ mm})^{2}}{(13.5 \text{ mm})^{2}}=233 \text { theoretical plates } \nonumber\] The height of a theoretical plate for each solute is \[H_{1}=\frac{L}{N_{1}}=\frac{0.500 \text{ m}}{400 \text { plates }} \times \frac{1000 \text{ mm}}{\mathrm{m}}=1.2 \text{ mm} / \mathrm{plate} \nonumber\] \[H_{2}=\frac{L}{N_{2}}=\frac{0.500 \text{ m}}{233 \text { plates }} \times \frac{1000 \text{ mm}}{\mathrm{m}}=2.15 \text{ mm} / \mathrm{plate} \nonumber\] Your measurements for , , , and will depend on the relative size of your monitor or printout; however, your values for and for should be similar to the answer above. It is important to remember that a theoretical plate is an artificial construct and that a chromatographic column does not contain physical plates. In fact, the number of theoretical plates depends on both the properties of the column and the solute. As a result, the number of theoretical plates for a column may vary from solute to solute. One advantage of improving column efficiency is that we can separate more solutes with baseline resolution. One estimate of the number of solutes that we can separate is \[n_{c}=1+\frac{\sqrt{N}}{4} \ln \frac{V_{\max }}{V_{\min }} \label{12.16}\] where is the column’s , and and are the smallest and the largest volumes of mobile phase in which we can elute and detect a solute [Giddings, J. C. , Wiley-Interscience: New York, 1991]. A column with 10 000 theoretical plates, for example, can resolve no more than \[n_{c}=1+\frac{\sqrt{10000}}{4} \ln \frac{30 \mathrm{mL}}{1 \mathrm{mL}}=86 \text { solutes } \nonumber\] if and are 1 mL and 30 mL, respectively. This estimate provides an upper bound on the number of solutes and may help us exclude from consideration a column that does not have enough theoretical plates to separate a complex mixture. Just because a column’s theoretical peak capacity is larger than the number of solutes, however, does not mean that a separation is feasible. In most situations the practical peak capacity is less than the theoretical peak capacity because the retention characteristics of some solutes are so similar that a separation is impossible. Nevertheless, columns with more theoretical plates, or with a greater range of possible elution volumes, are more likely to separate a complex mixture. The smallest volume we can use is the column’s void volume. The largest volume is determined either by our patience—the maximum analysis time we can tolerate—or by our inability to detect solutes because there is too much band broadening. Our treatment of chromatography in this section assumes that a solute elutes as a symmetrical Gaussian peak, such as that shown in Figure 12.2.4
. This ideal behavior occurs when the solute’s partition coefficient, \[K_{\mathrm{D}}=\frac{[S_\text{s}]}{\left[S_\text{m}\right]} \nonumber\] is the same for all concentrations of solute. If this is not the case, then the chromatographic peak has an asymmetric peak shape similar to those shown in Figure 12.2.9
. The chromatographic peak in Figure 12.2.9
a is an example of , which occurs when some sites on the stationary phase retain the solute more strongly than other sites. Figure 12.2.9
b, which is an example of most often is the result of overloading the column with sample. As shown in Figure 12.2.9
a, we can report a peak’s asymmetry by drawing a horizontal line at 10% of the peak’s maximum height and measuring the distance from each side of the peak to a line drawn vertically through the peak’s maximum. The asymmetry factor, , is defined as \[T=\frac{b}{a} \nonumber\] The number of theoretical plates for an asymmetric peak shape is approximately \[N \approx \frac{41.7 \times \frac{t_{r}^{2}}{\left(w_{0.1}\right)^{2}}}{T+1.25}=\frac{41.7 \times \frac{t_{r}^{2}}{(a+b)^{2}}}{T+1.25} \nonumber\] where is the width at 10% of the peak’s height [Foley, J. P.; Dorsey, J. G. , , 730–737]. Asymmetric peaks have fewer theoretical plates, and the more asymmetric the peak the smaller the number of theoretical plates. For example, the following table gives values for for a solute eluting with a retention time of 10.0 min and a peak width of 1.00 min. 0.5 0.5 0.6 0.4 0.7 0.3 | 20,705 | 1,552 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/07%3A_Obtaining_and_Preparing_Samples_for_Analysis/7.04%3A_Separating_the_Analyte_From_Interferents |
When an analytical method is selective for the analyte, analyzing a sample is a relatively simple task. For example, a quantitative analysis for glucose in honey is relatively easy to accomplish if the method is selective for glucose, even in the presence of other reducing sugars, such as fructose. Unfortunately, few analytical methods are selective toward a single species. In the absence of an interferent, the relationship between the sample’s signal, , and the analyte’s concentration, , is \[S_{samp}=k_{A} C_{A} \label{7.1}\] where is the analyte’s sensitivity. In Equation \ref{7.1}, and the equations that follow, you can replace the analyte’s concentration, , with the moles of analyte, , when working with methods, such as gravimetry, that respond to the absolute amount of analyte in a sample. In this case the interferent also is expressed in terms of moles. If an interferent, is present, then Equation \ref{7.1} becomes \[S_{samp}=k_{A} C_{A}+k_{I} C_{I} \label{7.2}\] where and are, respectively, the interferent’s sensitivity and concentration. A method’s selectivity for the analyte is determined by the relative difference in its sensitivity toward the analyte and the interferent. If is greater than , then the method is more selective for the analyte. The method is more selective for the interferent if is greater than . Even if a method is more selective for an interferent, we can use it to determine if the interferent’s contribution to is insignificant. The , , which we introduced in , provides a way to characterize a method’s selectivity. \[K_{A, I}=\frac{k_{I}}{k_{A}} \label{7.3}\] Solving Equation \ref{7.3} for , substituting into Equation \ref{7.2}, and simplifying, gives \[S_{samp}=k_{A}\left(C_{A}+K_{A, I} \times C_{I}\right) \label{7.4}\] An interferent, therefore, does not pose a problem as long as the product of its concentration and its selectivity coefficient is significantly smaller than the analyte’s concentration. \[K_{A, I} \times C_{I}<<C_{A} \nonumber\] If we cannot ignore an interferent’s contribution to the signal, then we must begin our analysis by separating the analyte and the interferent. | 2,180 | 1,553 |
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