url
stringlengths 90
342
| html
stringlengths 602
98.8k
| text_length
int64 602
98.8k
| __index_level_0__
int64 0
5.02k
|
|---|---|---|---|
https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Environmental_Toxicology_(van_Gestel_et_al.)/05%3A_Population_Community_and_Ecosystem_Ecotoxicology/5.08%3A_Structure_versus_function_incl._ecosystem_services |
: Herman Eijsackers : Nico van den Brink, Kees van Gestel, Lorraine Maltby : You should be able to structural biodiversity, functional biodiversity, functional redundancy, food web interactions In ecology, biodiversity describes the richness of natural life at three levels: , (the most well-known) and . The most commonly used index, the Shannon Wiener index, expresses biodiversity in general terms as the number of species in relation to the number of individuals per species. Precisely, this index stands for the sum of the natural logarithm of the number of individuals per species present: -∑(p *ln(p )) with p = n /N in which n is the number of individuals of species i and N the total number of individuals of all species combined. In environmental toxicology, most attention is paid to species diversity. Genetic diversity plays a role in the assessment of more or less sensitive or resistant subspecies or local populations of a species, like in various mining areas. Landscape diversity is receiving attention only recently and aims primarily at the total load of e.g. pesticides applied in an agronomic landscape (see Section on ), although it should more logically focus on the interactions between the various ecosystems in a landscape, for instance a lake surrounded partly by a forest, partly by a grassland. In general, the various types of interactions between species do not play a major role in the study of biodiversity neither within ecology nor in environmental toxicology. The diversity in interactions described in the food web or food chain is not expressed in a term like the Shannon-Wiener index. However, in aquatic as well as soil ecological research, extensive, quantitative descriptions have been made of various ecosystems. These model descriptions, like the one for arable soil below, are partly based on the taxonomic background of species groups and partly on their functional role in the food web, expressed as their way of feeding (see for instance the phytophagous nematodes feeding from plants, the fungivorous nematodes eating fungi and the predaceous nematodes eating other nematodes). The scheme in Figure 1 shows a very general soils food web and the different trophic levels. Much more detailed soil food web descriptions also are available, that do not only link the different trophic groups but also describe the energy flows within the system and through these flows the intensity and thus strength of the interactions that together determine the stability of the system (see e.g. de Ruiter et al., 1998). This food web shown in Figure 1 illustrates that biodiversity not only has a structural side: the various types of species, but also a functional one: which species are involved in the execution of which process. At the species level this functional aspect has been further elaborated in specific feeding leagues. At the ecosystem level this functional aspect has clearly been recognized in the last decades and resulted in the development of the concept of ecosystem services (see Section on ). However, these do not trace back to the individual species level and as such not to the functional aspect of biodiversity. Another development to be mentioned is that of trait-based approaches, which attempt to group species according to certain traits that are linked not only to exposure and sensitivity but also to their functioning. With that the trait-based approach may enable linking structural and functional biodiversity (see Section on ). When effects of contaminants on species are compared to effects on processes, the species effects are mostly more distinct than the process effects. In other words: effects on structural diversity will be seen already at lower concentrations, and probably also sooner, than effects on functional diversity. This can be explained by the fact that processes are executed by more than one species. When with increasing chemical exposure levels the most sensitive species disappear, their role is taken over by less sensitive species. This reasoning has been generalized in the concept of " ", which postulates that not all species that can perform a specific process are always active, and thus necessary, in a specific situation. Consequently they are superfluous or redundant. When a sensitive species that can perform a similar function disappears, a redundant species may take over, so the function is still covered. It has to be realized, however, that in case this is relevant in situation A, that does not mean it is also relevant for situation B with different environmental conditions and another species composition. Another consequence of functional redundancy is that when functional biodiversity is affected, there is (severe) damage to structural biodiversity: most likely several important species will have gone extinct or are strongly inhibited. Redundant species are often less efficient in performing a certain function. Tyler (1984) observed in a gradient of Cu-contamination by a copper manufacturing plant in Gusum Sweden that specific enzyme functions as well as a general processes like mineralisation decreased faster than the total fungal biomass. (Figure 2b). The explanation was provided by experimental research by Ruhling et al. (1984) who selected a number of these micro-fungi in the field and tested them for their sensitivity for copper. The various species showed different concentration-effect relationships but all going to zero, except for two species which increased in abundance at the higher concentration so that the total biomass stayed more or less the same (Figure 2A). Another example of the importance of a combined approach to structural and functional diversity are the different ecological types of earthworms. According to their behaviour and role they are distinguished in the anecics (deep burrowing earthworms moving up and down from deeper soil layers to the soil surface and consuming leaf litter), the endogeics (active in the deeper laying mineral and humus soil layers and consuming fragmented litter material and humus), and the epigeics (active in the upper soil litter layer and consuming leaf litter). Adverse effects of contamination on the anecics will result in accumulation of litter at the soil surface, in reduced litter fragmentation by the epigeics and reduced humus-forming by the endogeics. In various studies it is shown that these earthworms have different sensitivities for different types of pesticides. However, so far the ranking of more or less sensitive species is different for different groups of pesticides. So, there is no general relation between the function of a species e.g. surface active earthworms (epigeics) and their exposure to and sensitivity for pesticides. Nevertheless, pesticide effects on anecics generally lead to reduced litter removal, effects on endogeics result in slower fragmentation, reduced humification etc., and an effect on earthworm communities in general may hamper soil aeration and lead to soil compaction. Another example of the impact of contaminants on functional diversity is from microbiological research on the impact of heavy metals by Doelman et al. (1994). They isolated fungi, bacteria and actinomycetes from various heavy metal contaminated and clean areas, tested these species for their sensitivity to zinc and cadmium, and divided them accordingly in a sensitive and resistant group. As a next step they measured to what extent both groups were able to degrade and mineralize a series of organic compounds. Figure 3 shows that the sensitive group is much more effective in degrading a variety of organic compounds, whereas the heavy metal resistant microbes are far less effective. This would indicate that although functional redundancy may alleviate some of the effects that contaminants have on ecosystem functioning, the overall performance of the community generally decreases upon contaminant exposure. The latter example also shows that genetic diversity, expressed as the numbers of sensitive and resistant species, plays a role in the functional stability and sustainability of microbial degradation processes in the soil. In conclusion, ecosystem services are worth studying in relation to contamination (Faber et al., 2019), but also more specific in relation to functional diversity at the species level. A promising field of research in this framework would include microorganisms in relation to the variety of degradation processes they are involved in. De Ruiter, J.C., Neutel, A-M., Moore, J.C. 1995. Energetics, patterns of interaction strenghts and stability in real ecosystems. Science 269, 1257-60. Doelman, P., Jansen, E., Michels, M., Van Til, M. (1994). Effects of heavy metals in soil on microbial diversity and activity as shown by the sensitivity-resistance index, an ecologically relevant parameter Biology and Fertility of Soils 17, 177-184. Faber, J.H., Marshall, S., Van den Brink, P.J., Maltby, L. (2019). Priorities and opportunities in the application of the ecosystem services concept in risk assessment for chemicals in the environment. Science of the Total Environment 651, 1067-1077. Rühling, Å., Bååth, E., Nordergren, A., Söderström, B. (1984) Fungi in a metal-contaminated soil near the Gusum brass mill, Sweden. Ambio 13, 34-36. Tyler, G. (1984) The impact of heavy metal pollution on forests: A case study of Gusum, Sweden. Ambio 13, 18-24. Describe the structural and functional diversity at the species and at the landscape level What is meant by redundancy? Does redundancy have an impact of the sensitivity of species (structural diversity) versus processes (functional diversity)? | 9,676 | 1,038 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/05.5%3A_Particle_in_Boxes/Particle_in_a_1-Dimensional_box |
A particle in a 1-dimensional box is a fundamental quantum mechanical approximation describing the translational motion of a single particle confined inside an infinitely deep well from which it escape. The particle in a box problem is a common application of a quantum mechanical model to a simplified system consisting of a particle moving horizontally within an infinitely deep well from which it cannot escape. The solutions to the problem give possible values of E and \(\psi\) that the particle can possess. E represents allowed energy values and \(\psi(x)\) is a wavefunction, which when squared gives us the probability of locating the particle at a certain position within the box at a given energy level. To solve the problem for a particle in a 1-dimensional box, we must follow our The potential energy is (V=0 for 0<x<L) and (V=∞ for x<0 or x>L). We assume the walls have infinite potential energy to ensure that the particle has zero probability of being at the walls or outside the box. Doing so significantly simplifies our later mathematical calculations as we employ these when solving the Schrödinger Equation. The time-independent Schrödinger equation for a particle of mass moving in one direction with energy is \[-\dfrac{\hbar^2}{2m} \dfrac{d^2 \psi(x)}{dx^2} + V(x)\psi(x) = E\psi(x) \label{5.5.1}\] with This equation can be modified for a particle of mass free to move parallel to the x-axis with zero potential energy (V = 0 everywhere) resulting in the quantum mechanical description of free motion in one dimension: \[ -\dfrac{\hbar^2}{2m} \dfrac{d^2\psi(x)}{dx^2} = E\psi(x) \label{5.5.2}\] This equation has been well studied and gives a general solution of: \[\psi(x) = A\sin(kx) + B\cos(kx) \label{5.5.3}\] where A, B, and k are constants. The solution to the Schrödinger equation we found above is the general solution for a 1-dimensional system. We now need to apply our to find the solution to our particular system. According to our boundary conditions, the probability of finding the particle at x=0 or x=L is zero. When \(x=0\), \(\sin(0)=0\), and \(\cos(0)=1\); therefore, to fulfill this boundary condition giving: \[\psi(x) = A\sin(kx) \label{5.5.4}\] We can now solve for our constants (A and k) systematically to define the wavefunction.
Differentiate the wavefunction with respect to x: \[\dfrac{d\psi}{dx} = kA\cos(kx) \label{5.5.5}\] \[\dfrac{d^{2}\psi}{dx^{2}} = -k^{2}A\sin(kx) \label{5.5.6}\] Since \(\psi(x) = Asin(kx)\), then \[\dfrac{d^{2}\psi}{dx^{2}} = -k^{2}\psi \label{5.5.7}\] If we then solve for k by comparing with the Schrödinger equation above, we find: \[k = \left( \dfrac{8\pi^2mE}{h^2} \right)^{1/2} \label{5.5.8}\] Now we plug k into our wavefunction: \[\psi = A\sin\left(\dfrac{8\pi^{2}mE}{h^{2}}\right)^{1/2}x \label{5.5.9}\]
To determine A, we have to apply the boundary conditions again. Recall that the When \(x = L\): \[0 = A\sin\left(\dfrac{8\pi^{2}mE}{h^{2}}\right)^{1/2}L \label{5.5.10}\] This is only true when \[\left(\dfrac{8\pi^{2}mE}{h^{2}}\right)^{1/2}L = n\pi \label{5.5.11}\] where n = 1,2,3… Plugging this back in gives us: \[\psi = A\sin{\dfrac{n\pi}{L}}x \label{5.5.12}\] To determine \(A\), recall that the total probability of finding the particle inside the box is 1, meaning there is no probability of it being outside the box. When we find the probability and set it equal to 1, we are the wavefunction. \[\int^{L}_{0}\psi^{2}dx = 1 \label{5.5.13}\] For our system, the normalization looks like: \[A^2 \int^{L}_{0}\sin^2\left(\dfrac{n\pi x}{L}\right) dx = 1 \label{5.5.14}\] Using the solution for this integral from an integral table, we find our normalization constant, \(A\): \[A = \sqrt{\dfrac{2}{L}} \label{5.5.15}\] Which results in the normalized wavefunction for a particle in a 1-dimensional box: \[\psi = \sqrt{\dfrac{2}{L}}\sin{\dfrac{n\pi}{L}}x \label{5.5.16}\] Solving for E results in the allowed energies for a particle in a box: \[E_n = \dfrac{n^{2}h^{2}}{8mL^{2}} \label{5.5.17}\] This is an important result that tells us: This is also consistent with the : if the particle had zero energy, we would know where it was in both space and time. The wavefunction for a particle in a box at the \(n=1\) and \(n=2\) energy levels look like this: The probability of finding a particle a certain spot in the box is determined by squaring \(\psi\). The probability distribution for a particle in a box at the \(n=1\) and \(n=2\) energy levels looks like this: Notice that the number of (places where the particle has zero probability of being located) increases with increasing energy n. Also note that as the energy of the particle becomes greater, the quantum mechanical model breaks down as the energy levels get closer together and overlap, forming a continuum. This continuum means the particle is free and can have any energy value. At such high energies, the classical mechanical model is applied as the particle behaves more like a continuous wave. Therefore, the particle in a box problem is an example of . | 5,063 | 1,039 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.02%3A_Equations_and_Mass_Relationships/3.2.05%3A_Foods-_Metabolism_of_Dietary_Sugar |
We eat a lot of sugars, and in the next few sections we'll explore some of the body chemistry that explains why they may (or may not) lead to weight gain, why they're a good source of energy, and why astronauts excrete more water than they drink. As we've seen, there are many sugars, but one of the most common is sucrose, C H O . A balanced overall chemical equation for the metabolism of sucrose will help us understand the questions above, and many others. A balanced chemical equation such as 2 C H O ( ) + 35 O ( ) → 12 CO ( ) + 11 H O( ) (1) not only tells how many molecules of each kind are involved in a reaction, it also indicates the of each substance that is involved. Equation (1) says that 2 C H O can react with 35 O to give 12 CO and 11 H O . It also says that 2 2C H O would react with 35 O yielding 12 CO and 11 H O. In other words, the equation indicates that exactly 35 mol O must react 2 mol C H O consumed. For the purpose of calculating how much O is required to react with a certain amount of C H O therefore, the significant information contained in Eq. (1) is the \(\frac{\text{35 mol O}_{\text{2}}}{\text{2 mol C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}\) We shall call such a ratio derived from a balanced chemical equation a and give it the symbol . Thus, for Eq. (1), \(\text{S}\left( \frac{\text{O}_{\text{2}}}{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}\right)=\frac{\text{35 mol O}_{\text{2}}}{\text{2 mol C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}\) The word comes from the Greek words , “element,“ and , “measure.“ Hence the stoichiometric ratio measures one element (or compound) against another. Derive all possible stoichiometric ratios from Eq. (1) Any ratio of amounts of substance given by coefficients in the equation may be used, so in addition to (2) above, we have: \(\text{S}\left( \frac{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}{\text{O}_{\text{2}}}\right)=\frac{\text{2 mol C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}{\text{35 mol O}_{\text{2}}}\) \(\text{S}\left( \frac{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}{\text{CO}_{\text{2}}}\right)=\frac{\text{2 mol C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}{\text{12 mol CO}_{\text{2}}}=\frac{\text{1 mol C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}{\text{6 mol CO}_{\text{2}}}\) \(\text{S}\left( \frac{\text{CO}_{\text{2}}}{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}\right)=\frac{\text{12 mol CO}_{\text{2}}}{\text{2 mol C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}\) \(\text{S}\left( \frac{\text{CO}_{2}}{\text{H}_{\text{2}}\text{O}} \right)=\frac{\text{12 mol CO}_{2}}{\text{11 mol H}_{\text{2}}\text{O}}\) \(\text{S}\left( \frac{\text{CO}_{2}}{\text{O}_{\text{2}}} \right)=\frac{\text{12 mol CO}_{2}}{\text{35 mol O}_{\text{2}}}\) \(\text{S}\left( \frac{\text{O}_{\text{2}}}{\text{H}_{\text{2}}\text{O}} \right)=\frac{\text{35 mol O}_{\text{2}}}{\text{11 mol H}_{\text{2}}\text{O}}\) \(\text{S}\left( \frac{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}{\text{H}_{\text{2}}\text{O}} \right)=\frac{\text{2 mol C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}{\text{11 mol H}_{\text{2}}\text{O}}\) There are six more stoichiometric ratios, each of which is the reciprocal of one of these. [Eq. (2) gives one of them.] . Using Eq. (1) as an example, this means that the ratio of the amount of O consumed to the amount of NH consumed must be the stoichiometric ratio S(O /NH ): \(\frac{n_{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}\text{ consumed}}}{n_{\text{O}_{\text{2}}\text{ consumed}}}=\text{S}\left( \frac{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}{\text{O}_{\text{2}}}\right)=\frac{\text{2 mol C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}{\text{35 mol O}_{\text{2}}}\) Similarly, the ratio of the amount of H O produced to the amount of NH consumed must be S(H O/C H O ): \(\frac{n_{\text{H}_{\text{2}}\text{O produced}}}{n_{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}\text{ consumed}}} = \text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}} \right) =\frac{\text{11 mol H}_{\text{2}}\text{O}}{\text{2 mol C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}\) In general we can say that \(\text{Stoichiometric ratio }\left( \frac{\text{X}}{\text{Y}} \right)=\frac{\text{amount of X consumed or produced}}{\text{amount of Y consumed or produced}}\text{ (3}\text{a)}\) or, in symbols, \(\text{S}\left( \frac{\text{X}}{\text{Y}} \right)=\frac{n_{\text{X consumed or produced}}}{n_{\text{Y consumed or produced}}}\text{ (3}\text{b)}\) Note that in the word Eq. (3a) and the symbolic Eq. (3b), X and Y may represent reactant or product in the balanced chemical equation from which the stoichiometric ratio was derived. No matter how much of each reactant we have, the amounts of reactants and the amounts of products will be in appropriate stoichiometric ratios. Find the amount of water produced when 1 cup (roughly 8 oz or 240 g) C H O is consumed according to Eq. (1). The amount of water produced must be in the stoichiometric ratio S(H O/C H O ) to the amount of sugar consumed, and the amount is n = m/M = 240 g /342.3 g mol = .70 mol. \(\frac{n_{\text{H}_{\text{2}}\text{O produced}}}{n_{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}\text{ consumed}}} = \text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}} \right) =\frac{\text{11 mol H}_{\text{2}}\text{O}}{\text{2 mol C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}\) Multiplying both sides by This calculation shows why an astronaut dirnks about 2 L of H per day, but excretes about 2.4 L of H per day! Think about it, and check your answer This is a typical illustration of the use of a stoichiometric ratio as a conversion factor. Example 2 is analogous to Examples 1 and 2 from Conversion Factors and Functions, where density was employed as a conversion factor between mass and volume. Example 2 is also analogous to Examples 2.4 and 2.6, in which the Avogadro constant and molar mass were used as conversion factors. As in these previous cases, there is no need to memorize or do algebraic manipulations with Eq. (3) when using the stoichiometric ratio. Simply remember that the coefficients in a balanced chemical equation give stoichiometric ratios, and that the proper choice results in cancellation of units. In road-map form or symbolically. When using stoichiometric ratios, be sure you indicate moles . You can only cancel moles of the same substance. In other words, 1 mol C H O cancels 1 mol C H O but does not cancel 1 mol H O. The next example shows that stoichiometric ratios are also useful in problems involving the mass of a reactant or product. It is estimated that each human exhales about 1 kg (2.2 lb) of carbon dioxide per day. If that came entirely from glucose, what mass of glucose must be metabolized according to the equation: The problem asks that we calculate the mass of C H O consumed. As we learned in Example 2 of The Molar Mass, the molar mass can be used to convert from the amount of CO to the mass of CO . Then we can calculate the amount of C H O consumed from the amount of CO produced with a stoichiometric ratio, just as in Example 2. Finally, we can convert the amount of glucose to its mass using the molar mass as a conversion factor. It requires the stoichiometric ratio. \(\text{S}\left( \frac{\text{C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}{\text{CO}_{\text{2}}} \right)=\frac{\text{1 mol C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}{\text{6 mol CO}_{\text{2}}}\) The of CO produced is n = m/M = 1000 g/44 g mol = 22.7 mol The of C H O consumed is then The of C H O is By similar calculations, we could show that to produce 6 mol of carbon dioxide, it takes the same of oxygen (22.7 mol) and the same amount (22.7 mol)of water will be produced. That's 22.7 mol x 32 g/mol = 726 g of oxygen consumed and 22.7 mol x 18 g/mol = 409 g of water produced. The reactants weigh 682 + 726 = 1408 g and the products weigh the same (within error), 1000 + 409 = 1409 g. We notice two things: First, both products are eliminated (CO in breath, water in urine, so how do we gain weight? Only sugar that isn't metabolized goes into weight gain, and we'll see soon how to calculate that. Second, this accounts for .409 kg of the total 2.4 L of water excreted per day. It's a rough estimate of the part that results from metabolism (not from ingested water). With practice this kind of problem can be solved in one step by concentrating on the units. The appropriate stoichiometric ratio will convert moles of CO to amount (in moles) of C H O and the molar mass will convert amount of C H O to mass (in grams) of SO . A schematic road map for the one-step calculation can be written as Thus The chemical reaction in this example is of environmental interest. If each person on Earth exhales 1000 g of CO /day, and there are 6.7 billion people, this accounts for 6.7 x 10 (6.7 quadrillion) g or 6.2 Tg (teragrams) of the greenhouse gas CO . How does this compare with the amount produced by burning fuels in its effect on the atmosphere? In 2000, fossil fuel burning released 7 x 10 tons of carbon dioxide . That's 7 x 10 tons x 907 kg/ton x 1000 g/kg or 6.3 x 10 . Our breath contributes about 0.1%. The mass of atmospheric carbon dioxide comes from calculations like the following: What mass of oxygen and carbon dioxide would be consumed when 3.3 × 10 g, 3.3 Pg (petagrams), of octane (C H ) is burned to produce CO and H O? First, write a balanced equation The problem gives the mass of C H burned and asks for the mass of O required to combine with it. Thinking the problem through before trying to solve it, we realize that the molar mass of octane could be used to calculate the amount of octane consumed. Then we need a stoichiometric ratio to get the amount of O consumed. Finally, the molar mass of O permits calculation of the mass of O . Symbolically Thus 12 Pg (petagrams) of O would be needed. By a similar calculation, we see that the amount of carbon dioxide would be 16/2 = 8 times the amount of octane consumed, or 2.3 x 10 mol, which, multiplied by the molar mass 44 g/mol, is 1 x 10 g or 10 Pg. The large mass of oxygen obtained in this example is an estimate of how much O is removed from the earth’s atmosphere each year by human activities. Octane, a component of gasoline, was chosen to represent coal, gas, and other fossil fuels. Fortunately, the total mass of oxygen in the air (1.2 × 10 g) is much larger than the yearly consumption. If we were to go on burning fuel at the present rate, it would take about 100 000 years to use up all the O . Actually we will consume the fossil fuels long before that! One of the least of our environmental worries is running out of atmospheric oxygen. | 11,209 | 1,040 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/24%3A_Chemistry_of_Life-_Organic_and_Biological_Chemistry/24.E%3A_Organic_and_Biological_Chemistry_(Exercises) |
. In addition to these individual basis; please contact Classify each compound as organic or inorganic. Which compound is likely organic and which is likely inorganic? Classify each compound as organic or inorganic. Classify each compound as organic or inorganic. Which member of each pair has a higher melting point? Which member of each pair has a higher melting point? What is the functional group of an alkene? An alkyne? Does CH CH CH CH CH CH CH CH CH CH CH have a functional group? Explain. carbon-to-carbon double bond; carbon-to-carbon triple bond No; it has nothing but carbon and hydrogen atoms and all single bonds. What is the functional group of 1-butanol (CH CH CH CH OH)? What is the functional group of butyl bromide, CH CH CH CH Br? OH | 795 | 1,042 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Energies_and_Potentials/Enthalpy/Enthalpy_Change_of_Solution |
This page looks at the relationship between enthalpies of solution, hydration enthalpies and lattice enthalpies. The enthalpy change of solution is the enthalpy change when 1 mole of an ionic substance dissolves in water to give a solution of infinite dilution. Enthalpies of solution may be either positive or negative - in other words, some ionic substances dissolved endothermically (for example, NaCl); others dissolve exothermically (for example NaOH). An infinitely dilute solution is one where there is a sufficiently large excess of water that adding any more does not cause any further heat to be absorbed or evolved. So, when 1 mole of sodium chloride crystals are dissolved in an excess of water, the enthalpy change of solution is found to be +3.9 kJ mol . The change is slightly endothermic, and so the temperature of the solution will be slightly lower than that of the original water. Why is heat sometimes evolved and sometimes absorbed when a substance dissolves in water? To answer that it is useful to think about the various enthalpy changes that are involved in the process. You can think of an imaginary process where the crystal lattice is first broken up into its separate gaseous ions, and then those ions have water molecules wrapped around them. That is how they exist in the final solution. Hydration enthalpy is a measure of the energy released when attractions are set up between positive or negative ions and water molecules. The size of the hydration enthalpy is governed by the amount of attraction between the ions and the water molecules. The hydration enthalpies for calcium and chloride ions are given by the equations: The following cycle is for calcium chloride, and includes a lattice dissociation enthalpy of +2258 kJ mol . We have to use double the hydration enthalpy of the chloride ion because we are hydrating 2 moles of chloride ions. Make sure you understand exactly how the cycle works. So . . . ΔH = +2258 - 1650 + 2(-364) ΔH = -120 kJ mol Whether an enthalpy of solution turns out to be negative or positive depends on the relative sizes of the lattice enthalpy and the hydration enthalpies. In this particular case, the negative hydration enthalpies more than made up for the positive lattice dissociation enthalpy. Jim Clark ( ) | 2,294 | 1,044 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/The_Four_Laws_of_Thermodynamics/Second_Law_of_Thermodynamics |
The Second Law of Thermodynamics states that the state of entropy of the entire universe, as an , will always increase over time. The second law also states that the changes in the entropy in the universe can never be negative. Why is it that when you leave an ice cube at room temperature, it begins to melt? Why do we get older and never younger? And, why is it whenever rooms are cleaned, they become messy again in the future? Certain things happen in one direction and not the other, this is called the "arrow of time" and it encompasses every area of science. The thermodynamic arrow of time (entropy) is the measurement of disorder within a system. Denoted as \(\Delta S\), the change of entropy suggests that time itself is asymmetric with respect to order of an isolated system, meaning: a system will become more disordered, as time increases. If a given state can be accomplished in more ways, then it is more probable than the state that can only be accomplished in a fewer/one way. Assume a box filled with jigsaw pieces were jumbled in its box, the probability that a jigsaw piece will land randomly, away from where it fits perfectly, is very high. Almost every jigsaw piece will land somewhere away from its ideal position. The probability of a jigsaw piece landing correctly in its position, is very low, as it can only happened one way. Thus, the misplaced jigsaw pieces have a much higher multiplicity than the correctly placed jigsaw piece, and we can correctly assume the misplaced jigsaw pieces represent a higher entropy. To understand why entropy increases and decreases, it is important to recognize that two changes in entropy have to considered at all times. The entropy change of the surroundings and the entropy change of the system itself. Given the entropy change of the universe is equivalent to the sums of the changes in entropy of the system and surroundings: \[\Delta S_{univ}=\Delta S_{sys}+\Delta S_{surr}=\dfrac{q_{sys}}{T}+\dfrac{q_{surr}}{T} \label{1}\] In an isothermal reversible expansion, the heat q absorbed by the system from the surroundings is \[q_{rev}=nRT\ln\frac{V_{2}}{V_{1}}\label{2}\] Since the heat absorbed by the system is the amount lost by the surroundings, \(q_{sys}=-q_{surr}\).Therefore, for a truly reversible process, the entropy change is \[\Delta S_{univ}=\dfrac{nRT\ln\frac{V_{2}}{V_{1}}}{T}+\dfrac{-nRT\ln\frac{V_{2}}{V_{1}}}{T}=0 \label{3}\] If the process is irreversible however, the entropy change is \[\Delta S_{univ}=\frac{nRT\ln \frac{V_{2}}{V_{1}}}{T}>0 \label{4}\] If we put the two equations for \(\Delta S_{univ}\)together for both types of processes, we are left with the second law of thermodynamics, \[\Delta S_{univ}=\Delta S_{sys}+\Delta S_{surr}\geq0 \label{5}\] where \(\Delta S_{univ}\) equals zero for a truly reversible process and is greater than zero for an irreversible process. In reality, however, truly reversible processes never happen (or will take an infinitely long time to happen), so it is safe to say all thermodynamic processes we encounter everyday are irreversible in the direction they occur. The second law of thermodynamics can also be stated that "all processes produce an in the entropy of the universe". Given another equation: \[\Delta S_{total}=\Delta S_{univ}=\Delta S_{surr}+\Delta S{sys} \label{6}\] The formula for the entropy change in the surroundings is \(\Delta S_{surr}=\Delta H_{sys}/T\). If this equation is replaced in the previous formula, and the equation is then multiplied by T and by -1 it results in the following formula. \[-T \, \Delta S_{univ}=\Delta H_{sys}-T\, \Delta S_{sys} \label{7}\] If the left side of the equation is replaced by \(G\), which is know as Gibbs energy or free energy, the equation becomes \[\Delta G_{}=\Delta H-T\Delta S \label{8}\] Now it is much simpler to conclude whether a system is spontaneous, non-spontaneous, or at equilibrium. According to the equation, when the entropy decreases and enthalpy increases the free energy change, \(\Delta G_{}\), is positive and not spontaneous, and it does not matter what the temperature of the system is. Temperature comes into play when the entropy and enthalpy both increase or both decrease. The reaction is not spontaneous when both entropy and enthalpy are positive and at low temperatures, and the reaction is spontaneous when both entropy and enthalpy are positive and at high temperatures. The reactions are spontaneous when the entropy and enthalpy are negative at low temperatures, and the reaction is not spontaneous when the entropy and enthalpy are negative at high temperatures. Because all spontaneous reactions increase entropy, one can determine if the entropy changes according to the spontaneous nature of the reaction (Equation \(\ref{8}). Lets start with an easy reaction: \[2 H_{2(g)}+O_{2(g)} \rightarrow 2 H_2O_{(g)}\] The enthalpy, \(\Delta H_{}\), for this reaction is -241.82 kJ, and the entropy, \(\Delta S_{}\), of this reaction is -233.7 J/K. If the temperature is at 25º C, then there is enough information to calculate the standard free energy change, \(\Delta G_{}\). The first step is to convert the temperature to Kelvin, so add 273.15 to 25 and the temperature is at 298.15 K. Next plug \(\Delta H_{}\), \(\Delta S_{}\), and the temperature into the \(\Delta G=\Delta H-T \Delta S_{}\). \(\Delta G\)= -241.8 kJ + (298.15 K)(-233.7 J/K) = -241.8 kJ + -69.68 kJ (Don't forget to convert Joules to Kilojoules) = -311.5 kJ Here is a little more complex reaction: \[2 ZnO_{(s)}+2 C_{(g)} \rightarrow 2 Zn_{(s)}+2 CO_{(g)}\] If this reaction occurs at room temperature (25º C) and the enthalpy, \(\Delta H_{}\), and standard free energy, \(\Delta G_{}\), is given at -957.8 kJ and -935.3 kJ, respectively. One must work backwards somewhat using the same equation from Example 1 for the free energy is given. -935.3 kJ = -957.8 kJ + (298.15 K) (\(\Delta S_{}\)) 22.47 kJ = (298.15 K) (\(\Delta S_{}\)) (Add -957.8 kJ to both sides) 0.07538 kJ/K = \(\Delta S_{}\) (Divide by 298.15 K to both sides) Multiply the entropy by 1000 to convert the answer to Joules, and the new answer is 75.38 J/K. For the following dissociation reaction \[O_{2(g)} \rightarrow 2 O_{(g)}\] under what temperature conditions will it occurs spontaneously? By simply viewing the reaction one can determine that the reaction increases in the number of moles, so the entropy increases. Now all one has to do is to figure out the enthalpy of the reaction. The enthalpy is positive, because covalent bonds are broken. When covalent bonds are broken energy is absorbed, which means that the enthalpy of the reaction is positive. Another way to determine if enthalpy is positive is to to use the formation data and subtract the enthalpy of the reactants from the enthalpy of the products to calculate the total enthalpy. So, if the temperature is low it is probable that \(\Delta H_{}\) is more than \(T*\Delta S_{}\), which means the reaction is not spontaneous. If the temperature is large then \(T*\Delta S_{}\) will be larger than the enthalpy, which means the reaction is spontaneous. The following reaction \[CO_{(g)} + H_2O_{(g)} \rightleftharpoons CO_{2(g)} + H_{2(g)}\] occurs spontaneously under what temperature conditions? The enthalpy of the reaction is -40 kJ. One may have to calculate the enthalpy of the reaction, but in this case it is given. If the enthalpy is negative then the reaction is exothermic. Now one must find if the entropy is greater than zero to answer the question. Using the entropy of formation data and the enthalpy of formation data, one can determine that the entropy of the reaction is -42.1 J/K and the enthalpy is -41.2 kJ. Because both enthalpy and entropy are negative, the spontaneous nature varies with the temperature of the reaction. The temperature would also determine the spontaneous nature of a reaction if both enthalpy and entropy were positive. When the reaction occurs at a low temperature the free energy change is also negative, which means the reaction is spontaneous. However, if the reaction occurs at high temperature the reaction becomes nonspontaneous, for the free energy change becomes positive when the high temperature is multiplied with a negative entropy as the enthalpy is not as large as the product. Under what temperature conditions does the following reaction occurs spontaneously ? \[H_{2(g)} + I_{(g)} \rightleftharpoons 2 HI_{(g)}\] Only after calculating the enthalpy and entropy of the reaction is it possible for one can answer the question. The enthalpy of the reaction is calculated to be -53.84 kJ, and the entropy of the reaction is 101.7 J/K. Unlike the previous two examples, the temperature has no affect on the spontaneous nature of the reaction. If the reaction occurs at a high temperature, the free energy change is still negative, and \(\Delta G_{}\) is still negative if the temperature is low. Looking at the formula for spontaneous change one can easily come to the same conclusion, for there is no possible way for the free energy change to be positive. Hence, the reaction is spontaneous at all temperatures. The second law occurs all around us all of the time, existing as the biggest, most powerful, general idea in all of science. When scientists were trying to determine the age of the Earth during 1800s they failed to even come close to the value accepted today. They also were incapable of understanding how the earth transformed. Lord Kelvin, who was mentioned earlier, first hypothesized that the earth's surface was extremely hot, similar to the surface of the sun. He believed that the earth was cooling at a slow pace. Using this information, Kelvin used thermodynamics to come to the conclusion that the earth was at least twenty million years, for it would take about that long for the earth to cool to its current state. Twenty million years was not even close to the actual age of the Earth, but this is because scientists during Kelvin's time were not aware of radioactivity. Even though Kelvin was incorrect about the age of the planet, his use of the second law allowed him to predict a more accurate value than the other scientists at the time. Some critics claim that evolution violates the Second Law of Thermodynamics, because organization and complexity increases in evolution. However, this law is referring to isolated systems only, and the earth is not an isolated system or closed system. This is evident for constant energy increases on earth due to the heat coming from the sun. So, order may be becoming more organized, the universe as a whole becomes more disorganized for the sun releases energy and becomes disordered. This connects to how the second law and cosmology are related, which is explained well in the video below. | 10,780 | 1,045 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Ionization_Constants/Calculating_Equilibrium_Concentrations |
\(K_a\) is an , also known as the . It describes the likelihood of the compounds and the ions to break apart from each other. As we already know, strong acids completely dissociate, whereas weak acids only partially dissociate. A big \(K_a\) value will indicate that you are dealing with a very strong acid and that it will completely dissociate into ions. A small \(K_a\) will indicate that you are working with a weak acid and that it will only partially dissociate into ions. Generally, the problem usually gives an initial acid concentration and a \(K_a\) value. From there you are expected to know: The general formula of an acid dissociating into ions is \[HA_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + A^-_{(aq)} \label{1}\] with By definition, the \(K_a\) formula is written as the products of the reaction divided by the reactants of the reaction \[K_a = \dfrac{[Products]}{[Reactants]} \label{2}\] Based off of this general template, we plug in our concentrations from the chemical equation. The concentrations on the right side of the arrow are the products and the concentrations on the left side are the reactants. Using this information, we now can plug the concentrations in to form the \(K_a\) equation. We then write: \[K_a = \dfrac{[H_3O^+,A^-]}{[HA]} \label{3}\] The concentration of the hydrogen ion (\([H^+]\)) is often used synonymously with the hydrated (\([H_3O^+]\)). To find a concentration of hydronium ions in solution from a pH, we use the formula: \[[H_3O^+]= 10^{-pH}\] This can be flipped to calculate pH from hydronium concentration: \[pH = -\log[H_3O^+]\] At 25 °C, we can correlate whether a solution is acidic, basic, or neutral based off of the measured pH of the solutions: However, these relationships are not valid at temperatures outside 25 °C. Calculate the pH of a weak acid solution of 0.2 M HOBr, given: \[HOBr + H_2O \rightleftharpoons H_3O^+ + OBr^-\] \[K_a = 2 \times 10^{-9}\] Step 1: The ICE Table Since we were given the initial concentration of HOBr in the equation, we can plug in that value into the Initial Concentration box of the ICE chart. Considering that no initial concentration values were given for H O and OBr-, we can assume that none was present initially, and we indicate this by placing a zero in the corresponding boxes. M stands for molarity. Because we started off without an initial concentration of H O and OBr , it has to come from somewhere. In the Change in Concentration box, we add a +x because while we do not know what the numerical value of the concentration is at the moment, we do know that it has to be added and not taken away. In contrast, since we did start off with a numerical value of the initial concentration, we know that it has to be taken away to reach equilibrium. Because of this, we add a -x in the HOBr box. Now its time to add it all together! Go from top to bottom and add the Initial concentration boxes to the Change in concentration boxes to get the Equilibrium concentration. Step 2: Create the \(K_a\) equation using this equation: \(K_a = \dfrac{[Products]}{[Reactants]}\) \(K_a = \dfrac{[H_3O^+,OBr-]}{[HOBr-]}\) Step 3: Plug in the information we found in the ICE table \(K_a = \dfrac{(x)(x)}{(0.2 - x)}\) Step 4: Set the new equation equal to the given Ka \[2 \times 10^{-9} = \dfrac{(x)(x)}{(0.2 - x)}\] Step 5: Solve for x \[x^2 + (2 \times 10^{-9})x - (4 \times 10^{-10}) = 0\] To solve for x, we use the quadratic formula \[x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}= \dfrac{-2 \times 10^{-9} \pm \sqrt{(2 \times10^{-9})^2 - 4(1)(-4 \times 10^{-10})}}{2(1)}\] Step 6: Plug x back into the ICE table to find the concentration \[x= [H_3O^+] = 2 \times 10^{-5} \; M\] Step 7: Use the formula using the concentration to find pH \[pH = -\log[H_3O^+] = -\log(2 \times 10^{-5}) = -(-4.69) = 4.69\] \[pH= 4.69\] For acetic acid, HC H O , the \(K_a\) value is \(1.8 \times 10^{-5}\). Calculate the concentration of H O in a 0.3 M solution of HC H O Step 1: The Since we were given the initial concentration of HC H O in the original equation, we can plug in that value into the Initial Concentration box of the ICE chart. Considering that no initial concentration values were given for \(H_3O^+\) and \(C_2H_3O_2^-\), we assume that none was present initially, and we indicate this by placing a zero in the corresponding boxes. Because we started off without any initial concentration of H O and C H O -, is has to come from somewhere. For the Change in Concentration box, we add a +x because while we do not know what the numerical value of the concentration is at the moment, we do know that it has to be added and not taken away. In contrast, since we did start off with a numerical value of the initial concentration, we know that it has to be taken away to reach equilibrium. Because of this, we add a -x in the \(HC_2H_3O_2\) box. Now its time to add it all together! Go from top to bottom and add the Initial concentration boxes to the Change in concentration boxes to get the Equilibrium concentration. Step 2: Create the \(K_a\) equation using this equation: \(K_a = \dfrac{[Products]}{[Reactants]}\) \(K_a = \dfrac{[H_3O^+,C_2H_3O_2]}{[HC_2H_3O_2]}\) Step 3: Plug in the information we found in the ICE table \[K_a = \dfrac{(x)(x)}{(0.3 - x)}\] Step 4: Set the new equation equal to the given K \[1.8 x 10^{-5} = \dfrac{(x)(x)}{(0.3 - x)}\] Step 5: Solve for x \[(x^2)+ (1.8 \times 10^{-5}x)-(5.4 \times 10^{-6})\] To solve for x, we use the quadratic formula \[x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}= \dfrac{-1.8 \times 10^{-5} \pm \sqrt{(1.8 \times10^{-5})^2 - 4(1)(-5.4 \times 10^{-6})}}{2(1)}\] Step 6: Plug x back into the ICE table to find the concentration \[x= [H_3O^+] = 0.0023\; M\] Find the equilibrium concentration of HC H O from a 0.43 M solution of Benzoic Acid, HC H O Given: \(K_a\) for HC H O = 6.4 x 10 Step 1: The ICE Table Step 2: Create the \(K_a\) equation using this equation :\(K_a = \dfrac{[Products]}{[Reactants]}\) \(K_a = \dfrac{[H_3O^+,C_7H_5O_2-]}{[HC_7H_5O_2]}\) Step 3: Plug in the information we found in the ICE table \(K_a = \dfrac{(x)(x)}{(0.43 - x)}\) Step 4: Set the new equation equal to the given Ka \(6.4 x 10^{-5} = \dfrac{(x)(x)}{(0.43 - x)}\) Step 5: Solve for x. Step 6: Plug x back into the ICE table to find the concentration [HC H O ]= (0.43-x)M [HC H O ]= (0.43-0.0052)M Answer: [HC H O ]= 0.425 M For a 0.2 M solution of Hypochlorous acid, calculate all equilibrium concentrations. Given: \(K_a = 3.5 \times 10^{-8}\) Step 1: The ICE Table Step 2: Create the \(K_a\) equation using this equation: \(K_a = \dfrac{[Products]}{[Reactants]}\) \(K_a = \dfrac{[H_3O^+,OCl-]}{[HOCl-]}\) Step 3: Plug in the information we found in the ICE table \(K_a = \dfrac{(x)(x)}{(0.2 - x)}\) Step 4: Set the new equation equal to the given Ka \(3.5 x 10^{-8} = \dfrac{(x)(x)}{(0.2 - x)}\) Step 5: Solve for x Step 6: Plug x back into the ICE table to find the concentration [HOCl]= [(.2)-(8.4 x 10 )]=.199 [H O+]=8.4 x 10 [OCl-]=8.4 x 10 Calculate the pH from the equilibrium concentrations of [H O ] in Example \(\Page {4}\). Given: [HOCl]=0.199 [H O ]=8.4 x 10 [OCl-]=8.4 x 10 Step 1: Use the formula using the concentration of [H O ] to find pH \[pH = -\log[H3O+] = -\log(8.4 x 10^{-5}) = 4.08\] | 7,283 | 1,046 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.03%3A_The_Limiting_Reagent/3.3.04%3A_Everyday_Life-_Sodium_Silicide_Fueled_Bicycles |
Previously, we explored the Stoichiometry of hydrogen powered bicycles that "run on water". They typically have a rechargeable battery pack and electric hub motor. The company that makes the electrical fuel cell won a Presidential Green Chemistry Award . Electric hub motors on front or back wheels Now we'll see how the water cannister and fuel canister need to be matched for maximum range. In the laboratory as well as the environment, inexpensive reagents like atmospheric O or water are often supplied in excess. Some portion of such a reagent will be left unchanged after the reaction. Conversely, at least one reagent 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 . A new electricity source combines a hydrogen fuel cell with a fuel cartridge, winner of a "Green Chemistry Challenge Award" . The sodium silicyde reacts with water to make the hydrogen fuel : The composition of sodium silicide may depend on the method of synthesis. Silicides can be made by the reaction of active metals (like Mg) with sand, or by heating sodium with silicon. Dye et al prepare sodium silicide by the reaction of sodium metal with silica gel, obtaining black powders of (hypothetically) Na Si nanoparticles. Sodium silicide/water bikes produce about 200W to run a bicycle for about 30 miles with a NaSi cartridge weighing about 1.5 lb . If a bicyclist wants to travel about 30 miles and brings 1 quart of water and one 1.5 lb cartridge, which is the limiting reagent? What mass of solid product will be formed? The balanced equation tells us that according to the atomic theory, 2 mol NaSi is required for each 5 moles of 5H O. That is, the stoichiometric ratio S(NaSi/H O) = 2 mol NaSi/ 5 mol H O. Let us calculate the amount in moles of each we actually have, assuming the density of water is 1.00 g/mL: \(\text{m}_{\text{H}_{2}\text{O}}~=~\) \(\text{1 quart}~\times~\frac{\text{1 L}}{\text{1.05668821 quart}}~\times~\frac{\text{1000 g}}{\text{1 L} }~ =~\text{1057 g}\) \(\text{m}_{\text{NaSi}}~=~
<math>\text{1.5 lb}~\times~\frac{\text{453.59 g}}{\text{ quart}}~ =~\text{680 g}\) \(\text{n}_{\text{NaSi}}=\text{680 g}\times \frac{\text{1 mol NaSi}}{\text{51.075 g}}=\text{13.3 mol NaSi}\) \(\text{n}_{\text{H}_{\text{2}}\text{O}}=\text{1057 g}\times \frac{\text{1 mol H}_{\text{2}}\text{O}}{\text{18.015 g}}=\text{58.57 mol H}_{\text{2}\text{O}}\) If all the H O were to react, the stoichiometric ratio allows us to calculate the amount of NaSi that would be required: \(\text{n}_{\text{NaSi}}~=~ \text{n}_{\text{H}_{2}\text{O}}~~\times~~\) \(~~\frac{\text{2 mol NaSi}}{\text{5 mol H}_{2}\text{O}}~~\) \(~=~\text{23.5 mol NaSi}\) This is more than the amount of NaSi present,so NaSi is the and H O is present in excess. If all the NaSi reacts, the stoichiometric ratio allows us to calculate the amount of H O that would be required: \(\text{n}_{\text{H}_{2}\text{O}}~=~ \text{n}_{\text{NaSi}}~~\times~~\) \(~~\frac{\text{5 mol H}_{2}\text{O}}{\text{2 mol NaSi}}~~\)
\(~=~\text{13.3 mol NaSi}\) \(~\times~\frac{\text{5 mol H}_{2}\text{O}}{\text{2 mol NaSi}}~=~\) We require less than the amount of H O present, so it is the . When the reaction ends, 13.3 mol of NaSi will have reacted with 33.3 mol H O and there will be (58.57 mol - 33.3 mol) = 25.4 mol H O left over. NaSi is therefore the limiting reagent. b. Since the water doesn't all react, we need to calculate the amount of solid product produced from the amount of NaSi consumed, by using the stoichiometric ratio: \(\text{n}_{\text{Na}_{2}\text{Si}_{2}\text{O}_{5}}~=~ \text{n}_{\text{NaSi}}~~\times~~\) \(~~\frac{\text{1 mol Na}_{2}\text{Si}_{2}\text{O}_{5}}{\text{2 mol NaSi}}~~\) \(~=~\text{6.65 mol Na}_{2}\text{Si}_{2}\text{O}_{5}\) The mass of water is then calculated by using the molar mass: \(\text{6.65 mol Na}_{2}\text{Si}_{2}\text{O}_{5}~\times~\) \(\frac{\text{182.148 g}}{\text{1 mol Na}_{2}\text{Si}_{2}\text{O}_{5}}\) \(~=~\text{1211 g Na}_{2}\text{Si}_{2}\text{O}_{5}\) c. About a pound (458 g) of water will react, and 1057-458 = 599 g of water will remain. That's enough for another 1.5 lb canister of NaSi. 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. In the end, 1057 - 599 = 458 g of H O will remain, along with 1211 g of sodium silicate. The 67.2 g of hydrogen produced gives a total of 1736 g. The mass of the reactants was also 1057 + 680 = 1737 g, equal within the error of measurement. 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. The 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. Suppose the sodium silicide made by a different method is Na Si , and it produces ordinary "water glass", or sodium silicate (Na SiO (s) ) rather than Na Si O (s) . In that case, if the bicyclist wants to travel about 30 miles and brings 1 quart of water and one 1.5 lb cartridge, which is the limiting reagent? The balanced equation is: Now the stoichiometric ratio S(NaSi/H O) = 1 mol Na Si/ 3 mol H O. Again, let's calculate the amount in moles of each we actually have, assuming the density of water is 1.00 g/mL: \(\text{m}_{\text{H}_{2}\text{O}}~=~\) \(\text{1 quart}~\times~\frac{\text{1 L}}{\text{1.05668821 quart}}~\times~\frac{\text{1000 g}}{\text{1 L} }~ =~\text{1057 g}\) \(\text{m}_{\text{NaSi}}~=~
<math>\text{1.5 lb}~\times~\frac{\text{453.59 g}}{\text{ quart}}~ =~\text{680 g}\) \(\text{n}_{\text{Na}_{2}\text{Si}}~=~\) \(\text{680 g}\times \frac{\text{1 mol Na}_{2}\text{Si}}{\text{74.065 g}}~=~\) \(\text{n}_{\text{H}_{\text{2}}\text{O}}=\text{1057 g}\times \frac{\text{1 mol H}_{\text{2}}\text{O}}{\text{18.015 g}}=\text{58.57 mol H}_{2}\text{O}\) If all the H O were to react, the stoichiometric ratio allows us to calculate the amount of NaSi that would be required: \(\text{n}_{\text{Na}_{2}\text{Si}}~=~\) \(\text{n}_{\text{H}_{2}\text{O}}~~\times~~\) \(~~\frac{\text{1 mol Na}_{2}\text{Si}}{\text{3 mol H}_{2}\text{O}}~~\) \(~=~\text{19.52 mol Na}_{2}\text{Si}\) This is more than the amount of Na Si present,so Na Si is the and H O is present in excess. If all the Na Si reacts, the stoichiometric ratio allows us to calculate the amount of H O that would be required: \(\text{n}_{\text{H}_{2}\text{O}}~=~\) \(\text{n}_{\text{Na}_{2}\text{Si}}~~\times~~\) \(~~\frac{\text{3 mol H}_{2}\text{O}}{\text{1 mol Na}_{2}\text{Si}}~~\)
\(~=~\text{13.3 mol Na}_{2}\text{Si}\) \(~\times~\frac{\text{3 mol H}_{2}\text{O}}{\text{1 mol Na}_{2}\text{Si}}~=~\) We require less than the amount of H O present, so it is the . When the reaction ends, 9.18 mol of Na Si will have reacted with 27.54 mol H O and there will be (58.57 mol - 27.54 mol) = 31.03 mol H O left over. Na Si is therefore the limiting reagent. These calculations can again 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. In the end, 1057 - 496 = 561 g of H O will remain, along with 1121 g of sodium silicate. The 55.5 g of hydrogen produced gives a total of 1176 g. The mass of the reactants was also 496 + 680 = 1176 g, equal within the error of measurement. This process seems less efficient than the one in Example 1, because only 55.5 g of hydrogen is produced, compared to 67.2 g in Ex. 1. Also, 496 g of water and 680 g of silicide are consumed (totalling 1176 g), while in Ex. 1, 599 g water and 680 g of silicide (1280 g total) were consumed, so we get 0.047 g hydrogen/g reactants in Ex. 2, while we get 0.053 g hydrogen/g reactants in Ex. 1. | 8,989 | 1,047 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/05%3A_Orbital_Picture_of_Bonding-_Orbital_Combinations_Hybridization_Theory_and_Molecular_Orbitals/5.02%3A_Orbital_Hybridization_Theory |
If we look at the valence shell configuration of carbon, we find two paired electrons in the 2 orbital, and two unpaired electrons in the 2 and 2 orbitals, one in each: In order to fulfill the octet rule, carbon must use its 4 valence electrons when bonding to other atoms. However, only unpaired electrons can bond. That means that Since the energy gap between the 2 and 2 orbitals is very small, one of the 2 electrons can be promoted to the empty 2 orbital, leading to the following situation Now the four electrons appear to be ready for bonding, but there is a problem. The 2 orbitals are known to be at right angles to each other. If bonding occurs in this state, the 3 equivalent electrons would form 3 equivalent bonds oriented at 90 to each other, and the s electron would form a bond of a different type and orientation from the other three. No such compound exists. The simplest hydrocarbon –methane (CH )– is known to have tetrahedral geometry, where the four C–H bonds are all equivalent and positioned at 109.5 angles to each other. In addition, there are some carbon compounds where the bond angles are 120 or even 180 . HYBRIDIZATION THEORY ATTEMPTS TO EXPLAIN THE ACTUAL SHAPES OF MOLECULES BY INVOKING THE FORMATION OF HYBRID ORBITALS DURING, OR PRIOR TO, THE BONDING PROCESS. Going back to the carbon model with four unpaired electrons in the valence shell, we can take it as a point of departure for formation of hybrid orbitals. The first step is to take either 2, 3, or all four of those orbitals and equalize their energies. Let’s say that we take all four of them and form 4 equivalent new orbitals. These orbitals are now of the same energy, which is intermediate between those of the original 2 and 2 orbitals. At the same time, we cannot name the new orbitals or , for they’re neither. We have to find a new name that reflects the fact that they were created from . We will call them . The process that leads to their formation is called All four orbitals that result from this process are That means that they have According to VSEPR (valence shell electron pair repulsion) theory, such orbitals will orient themselves in 3-D space The resulting shape is then a tetrahedron, where the carbon nucleus is at the center and the orbitals point to the corners of the tetrahedron. The ideal angle between orbitals is then 109.5 degrees. When an hybridized carbon bonds to hydrogen, it forms methane, whose geometry is known to be tetrahedral. A 3-D representation of methane. The single lines represent bonds that are positioned on the plane of the paper. The solid wedge represents a bond coming out of the plane of the paper towards the front. The broken wedge represents a bond going behind the plane of the paper towards the back. Hydrocarbons are substances containing only carbon and hydrogen. Hydrocarbons are classified into the following major categories: and In the following pages we will do an overview of the basic characteristics of the first three, but will postpone the study of aromatics until later. is the simplest alkane, followed by , etc. The carbon chain constitutes the basic skeleton of alkanes. Carbon chains with four or more atoms can be or . Some examples are shown below including Lewis, molecular, and condensed formulas. Refer to chapter 2 of the Wade textbook for additional examples. Some examples of branched alkanes are shown below. Notice that sometimes Lewis formulas become cumbersome and difficult to write without cluttering. Condensed formulas are more convenient to use in such situations. As can be seen from the above examples, where is the total number of carbon atoms. This holds regardless of whether the alkane is linear or branched. | 3,755 | 1,048 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/16%3A_Aqueous_AcidBase_Equilibriums/16.04%3A_Quantitative_Aspects_of_Acid-Base_Equilibria |
This section presents a quantitative approach to analyzing acid–base equilibriums. You will learn how to determine the values of and , how to use or to calculate the percent ionization and the pH of an aqueous solution of an acid or a base, and how to calculate the equilibrium constant for the reaction of an acid with a base from the and of the reactants. The ionization constants and are equilibrium constants that are calculated from experimentally measured concentrations, just like the equilibrium constants discussed in . Before proceeding further, it is important to understand exactly what is meant when we describe the concentration of an aqueous solution of a weak acid or a weak base. Suppose, for example, we have a bottle labeled 1.0 M acetic acid or 1.0 M ammonia. As you learned in , such a solution is usually prepared by dissolving 1.0 mol of acetic acid or ammonia in water and adding enough water to give a final volume of exactly 1.0 L. If, however, we were to list the actual concentrations of all the species present in either solution, we would find that of the values is exactly 1.0 M because a weak acid such as acetic acid or a weak base such as ammonia always reacts with water to some extent. The extent of the reaction depends on the or the , the concentration of the acid or the base, and the temperature. Consequently, only the of both the ionized and unionized species is equal to 1.0 M. The ( ) is defined as the concentration of all forms of an acid or a base that are present in solution, regardless of their state of protonation. Thus a “1.0 M” solution of acetic acid has an analytical concentration of 1.0 M, which is the sum of the concentrations of unionized acetic acid (CH CO H) and the ionized form (CH CO ):
\[C_{CH_3CO_2H}=[CH_3CO_2H] + [CH_3CO_2^−] \tag{16.4.1}\] As we shall see shortly, if we know the analytical concentration and the , we can calculate the actual values of [CH CO H] and [CH CO ]. The equilibrium equations for the reaction of acetic acid and ammonia with water are as follows:
\[K_a=\dfrac{[H^+,CH_3CO_2^−]}{[CH_3CO_2H]} \tag{16.4.2}\] \[ K_b=\dfrac{[NH_4^+,OH^−]}{[NH_3]} \tag{16.4.3}\] where and are the ionization constants for acetic acid and ammonia, respectively. In addition to the analytical concentration of the acid (or the base), we must have a way to measure the concentration of at least of the species in the equilibrium constant expression to determine the (or the ). There are two common ways to obtain the concentrations: (1) measure the electrical conductivity of the solution, which is related to the total concentration of ions present, and (2) measure the pH of the solution, which gives [H ] or [OH ]. Example 6 and Example 7 illustrate the procedure for determining for a weak acid and for a weak base. In both cases, we will follow the procedure developed in : the analytical concentration of the acid or the base is the concentration, and the stoichiometry of the reaction with water determines the in concentrations. The concentrations of all species are calculated from the initial concentrations and the changes in the concentrations. Inserting the final concentrations into the equilibrium constant expression enables us to calculate the or the . Electrical conductivity measurements indicate that 0.42% of the acetic acid molecules in a 1.00 M solution are ionized at 25°C. Calculate and p for acetic acid at this temperature. analytical concentration and percent ionization and p Write the balanced equilibrium equation for the reaction and derive the equilibrium constant expression. Use the data given and the stoichiometry of the reaction to construct a table showing the initial concentrations, the changes in concentrations, and the final concentrations for all species in the equilibrium constant expression. Substitute the final concentrations into the equilibrium constant expression and calculate the . Take the negative logarithm of to obtain the p . The balanced equilibrium equation for the dissociation of acetic acid is as follows: \[CH_3CO_2H_{(aq)} \rightleftharpoons H^+_{(aq)}+CH3CO^−_{2(aq)}\] and the equilibrium constant expression is as follows: \[K_a=\dfrac{[H^+,CH_3CO_2^−]}{[CH_3CO_2H]}\] To calculate the , we need to know the equilibrium concentrations of CH CO H, CH CO , and H . The most direct way to do this is to construct a table that lists the initial concentrations and the changes in concentrations that occur during the reaction to give the final concentrations, using the procedure introduced in . The initial concentration of unionized acetic acid ([CH CO H] ) is the analytical concentration, 1.00 M, and the initial acetate concentration ([CH CO ] ) is zero. The initial concentration of H is not zero, however; [H ] is 1.00 × 10 M due to the autoionization of water. The measured percent ionization tells us that 0.42% of the acetic acid molecules are ionized at equilibrium. Consequently, the change in the concentration of acetic acid is Δ[CH CO H] = −(4.2 × 10 )(1.00 M) = −0.0042 M. Conversely, the change in the acetate concentration is Δ[CH CO ] = +0.0042 M because every 1 mol of acetic acid that ionizes gives 1 mol of acetate. Because one proton is produced for each acetate ion formed, Δ[H ] = +0.0042 M as well. These results are summarized in the following table. The final concentrations of all species are therefore as follows: We can now calculate by inserting the final concentrations into the equilibrium constant expression: \[K_a=\dfrac{[H^+,CH_3CO_2^−]}{[CH_3CO_2H]}=\dfrac{(0.0042)(0.0042)}{1.00}=1.8 \times 10^{−5}\] The p is the negative logarithm of : p = −log = −log(1.8 × 10 ) = 4.74. Exercise Picric acid is the common name for 2,4,6-trinitrophenol, a derivative of phenol (C H OH) in which three H atoms are replaced by nitro (–NO ) groups. The presence of the nitro groups removes electron density from the phenyl ring, making picric acid a much stronger acid than phenol (p = 9.99). The nitro groups also make picric acid potentially explosive, as you might expect based on its chemical similarity to 2,4,6-trinitrotoluene, better known as TNT. A 0.20 M solution of picric acid is 73% ionized at 25°C. Calculate and p for picric acid. = 0.39; p = 0.41 A 1.0 M aqueous solution of ammonia has a pH of 11.63 at 25°C. Calculate and p for ammonia. analytical concentration and pH and p Write the balanced equilibrium equation for the reaction and derive the equilibrium constant expression. Use the data given and the stoichiometry of the reaction to construct a table showing the initial concentrations, the changes in concentrations, and the final concentrations for all species in the equilibrium constant expression. Substitute the final concentrations into the equilibrium constant expression and calculate the . Take the negative logarithm of to obtain the p . The balanced equilibrium equation for the reaction of ammonia with water is as follows: \[NH_{3(aq)}+H_2O_{(l)} \rightleftharpoons NH^+_{4(aq)}+OH^−_{(aq)} \notag \] and the equilibrium constant expression is as follows: \[K_b=\dfrac{[NH_4^+,OH^−]}{[NH_3]} \notag \] Remember that water does not appear in the equilibrium constant expression for . To calculate , we need to know the equilibrium concentrations of NH , NH , and OH . The initial concentration of NH is the analytical concentration, 1.0 M, and the initial concentrations of NH and OH are 0 M and 1.00 × 10 M, respectively. In this case, we are given the pH of the solution, which allows us to calculate the final concentration of one species (OH ) directly, rather than the change in concentration. Recall that p = pH + pOH = 14.00 at 25°C. Thus pOH = 14.00 − pH = 14.00 − 11.63 = 2.37, and [OH ] = 10 = 4.3 × 10 M. Our data thus far are listed in the following table. The final [OH ] is much greater than the initial [H ], so the change in [OH ] is as follows: The stoichiometry of the reaction tells us that 1 mol of NH is converted to NH for each 1 mol of OH formed, so We can now insert these values for the changes in concentrations into the table, which enables us to complete the table. Inserting the final concentrations into the equilibrium constant expression gives : \[K_b=\dfrac{[NH_4^+,OH^−]}{[NH_3]}=\dfrac{(4.3 \times 10^{−3})^2}{1.0}=1.8 \times 10^{−5} \notag \] and p = −log = 4.74. The and the p for ammonia are almost exactly the same as the and the p for acetic acid at 25°C. In other words, ammonia is almost exactly as strong a base as acetic acid is an acid. Consequently, the extent of the ionization reaction in an aqueous solution of ammonia at a given concentration is the same as in an aqueous solution of acetic acid at the same concentration. Exercise The pH of a 0.050 M solution of pyridine (C H N) is 8.96 at 25°C. Calculate and p for pyridine. = 1.7 × 10 ; p = 8.77 When carrying out a laboratory analysis, chemists frequently need to know the concentrations of all species in solution. Because the reactivity of a weak acid or a weak base is usually very different from the reactivity of its conjugate base or acid, we often need to know the percent ionization of a solution of an acid or a base to understand a chemical reaction. The percent ionization is defined as follows:
\[\text{percent ionization οf acid} =\dfrac{[H^+]}{C_{HA}}×100 \tag{16.4.4}\] \[\text{percent ionization οf base}=\dfrac{[OH^−]}{C_B}×100 \tag{16.4.5}\] One way to determine the concentrations of species in solutions of weak acids and bases is a variation of the tabular method we used previously to determine and values. As a demonstration, we will calculate the concentrations of all species and the percent ionization in a 0.150 M solution of formic acid at 25°C. The data in show that formic acid ( = 1.8 × 10 at 25°C) is a slightly stronger acid than acetic acid. The equilibrium equation for the ionization of formic acid in water is as follows:
\[HCO_2H_{(aq)} \rightleftharpoons H^+_{(aq)}+HCO^−_{2(aq)} \tag{16.4.6}\] and the equilibrium constant expression for this reaction is as follows:
\[K_a=\dfrac{[H^+,HCO_2^−]}{[HCO_2H]} \tag{16.4.7}\] We set the initial concentration of HCO H equal to 0.150 M, and that of HCO is 0 M. The initial concentration of H is 1.00 × 10 M due to the autoionization of water. Because the equilibrium constant for the ionization reaction is small, the equilibrium will lie to the left, favoring the unionized form of the acid. Hence we can define as the amount of formic acid that dissociates. If the change in [HCO H] is − , then the change in [H ] and [HCO ] is + . The final concentration of each species is the sum of its initial concentration and the change in concentration, as summarized in the following table. We can calculate by substituting the final concentrations from the table into the equilibrium constant expression: \[K_a=\dfrac{[H^+,HCO_2^−]}{[HCO_2H]}=\dfrac{(1.00 \times 10^{−7}+x) x}{0.150−x}\] Because the ionization constant is small, is likely to be small compared with the initial concentration of formic acid: (0.150 − ) M ≈ 0.150 M. Moreover, [H ] due to the autoionization of water (1.00 × 10 M) is likely to be negligible compared with [H ] due to the dissociation of formic acid: (1.00 × 10 + ) M ≈ M. Inserting these values into the equilibrium constant expression and solving for , \[K_a=\dfrac{x^2}{0.150} =1.8 \times 10^{−4}\] \[x=5.2 \times 10^{−3}\] We can now calculate the concentrations of the species present in a 0.150 M formic acid solution by inserting this value of into the expressions in the last line of the table: \[[HCO_2H]=(0.150−x)\; M=0.145 \;M\] \[[HCO_2]=x=5.2 \times 10^{−3}\; M\] \[[H^+]=(1.00×10−7+x) M=5.2 \times 10^{−3} M\] Thus the pH of the solution is –log(5.2 × 10 ) = 2.28. We can also use these concentrations to calculate the fraction of the original acid that is ionized. In this case, the percent ionization is the ratio of [H ] (or [HCO ]) to the analytical concentration, multiplied by 100 to give a percentage: \[\text{percent ionization}=\dfrac{[H^+]}{C_{HA}}×100=\dfrac{5.2 \times 10^{−3}\; M}{0.150} \times 100=3.5\%\] As a general rule of thumb, approximations such as those used here are valid if the quantity being neglected is no more than about 5% of the quantity to which it is being added or from which it is being subtracted. If the quantity that was neglected is much greater than about 5%, then the approximation is probably not valid, and you should go back and solve the problem using the quadratic formula. In the previous demonstration, both simplifying assumptions were justified: the percent ionization is only 3.5%, which is well below the approximately 5% limit, and the 1.00 × 10 M [H ] due to the autoionization of water is much, much less than the 5.2 × 10 M [H ] due to the ionization of formic acid. As a general rule, the [H ] contribution due to the autoionization of water can be ignored as long as the product of the acid or the base ionization constant and the analytical concentration of the acid or the base is at least 10 times greater than the [H ] or [OH ] from the autoionization of water—that is, if
\[K_aC_{HA} \ge 10(1.00 \times 10^{−7}) = 1.0 \times 10^{−6} \tag{16.4.8}\] or \[K_bC_B \ge 10(1.00 \times 10^{−7}) = 1.0 \times 10^{−6} \tag{16.4.9}\] By substituting the appropriate values for the formic acid solution into , we see that the simplifying assumption is valid in this case:
\[K_aC_{HA} = (1.8 \times 10^{−4})(0.150) = 2.7 \times 10^{−5} > 1.0 \times 10^{−6} \tag{16.4.10}\] Doing this simple calculation before solving this type of problem saves time and allows you to write simplified expressions for the final concentrations of the species present. In practice, it is necessary to include the [H ] contribution due to the autoionization of water only for extremely dilute solutions of very weak acids or bases. Example 8 illustrates how the procedure outlined previously can be used to calculate the pH of a solution of a weak base. Calculate the pH and percent ionization of a 0.225 M solution of ethylamine (CH CH NH ), which is used in the synthesis of some dyes and medicines. The p of ethylamine is 3.19 at 20°C. concentration and p pH and percent ionization Write the balanced equilibrium equation for the reaction and the equilibrium constant expression. Calculate from p . autoionization Use the relationship = [OH ,H ] to obtain [H ]. Then calculate the pH of the solution. We begin by writing the balanced equilibrium equation for the reaction: \[CH_3CH_2NH_{2(aq)}+H_2O_{(l)} \rightleftharpoons CH_3CH_2NH^+_{3(aq)}+OH^−_{(aq)} \notag \] The corresponding equilibrium constant expression is as follows: \[K_b=\dfrac{[CH_3CH_2NH_3^+,OH^−]}{[CH_3CH_2NH_2}] \notag \] From the \(pK_b\), we have \(K_b = 10−3.19 = 6.5 \times 10^{−4}\). To calculate the pH, we need to determine the H concentration. Unfortunately, H does not appear in either the chemical equation or the equilibrium constant expression. However, [H ] and [OH ] in an aqueous solution are related by = [H ,OH ]. Hence if we can determine [OH ], we can calculate [H ] and then the pH. The initial concentration of CH CH NH is 0.225 M, and the initial [OH ] is 1.00 × 10 M. Because ethylamine is a weak base, the extent of the reaction will be small, and it makes sense to let equal the amount of CH CH NH that reacts with water. The change in [CH CH NH ] is therefore − , and the change in both [CH CH NH ] and [OH ] is + . To see whether the autoionization of water can safely be ignored, we substitute and into : Thus the simplifying assumption is valid, and we will not include [OH ] due to the autoionization of water in our calculations. Substituting the quantities from the last line of the table into the equilibrium constant expression, \[K_b=\dfrac{[CH_3CH_2NH_3^+,OH^−]}{[CH_3CH_2NH_2]}=\dfrac{(x)(x)}{0.225−x}=6.5\times 10^{−4} \notag \] As before, we assume the amount of CH CH NH that ionizes is small compared with the initial concentration, so [CH CH NH ] = 0.225 − ≈ 0.225. With this assumption, we can simplify the equilibrium equation and solve for : \[K_b=\dfrac{x^2}{0.225} =6.5 \times 10^{−4} \notag \] \[x=0.012=[CH_3CH_2NH_3^+]_f=[OH^−]_f \notag \] The percent ionization is therefore \[\text{percent ionization}=\dfrac{[OH^−]}{C_B} \times 100=\dfrac{0.012\; M}{0.225\; M} \times 100=5.4\%\] which is at the upper limit of the approximately 5% range that can be ignored. The final hydroxide concentration is thus 0.012 M. We can now determine the [H ] using the expression for : \[K_w=[OH^−,H^+] \notag \] \[1.01 \times 10^{−14} =(0.012 \;M)[H^+] \notag \] \[8.4 \times 10^{−13}\; M=[H^+] \notag \] The pH of the solution is −log(8.4 × 10 ) = 12.08. Alternatively, we could have calculated pOH as −log(0.012) = 1.92 and determined the pH as follows: \[pH + pOH =pKw=14.00 \notag \] The two methods are equivalent. Exercise Aromatic amines, in which the nitrogen atom is bonded directly to a phenyl ring (−C H ) tend to be much weaker bases than simple alkylamines. For example, aniline (C H NH ) has a p of 9.13 at 25°C. What is the pH of a 0.050 M solution of aniline? 8.78 The previous examples illustrate a key difference between solutions of strong acids and bases and solutions of weak acids and bases. Because strong acids and bases ionize essentially completely in water, the percent ionization is always approximately 100%, regardless of the concentration. In contrast, the percent ionization in solutions of weak acids and bases is small and depends on the analytical concentration of the weak acid or base. As illustrated for benzoic acid in , the percent ionization of a weak acid or a weak base actually as its analytical concentration decreases. The percent ionization also increases as the magnitude of and increases. Unlike the or the , the percent ionization is not a constant for weak acids and bases but depends on the or the the analytical concentration. Consequently, the procedure in Example 8 must be used to calculate the percent ionization and pH for solutions of weak acids and bases. Example 9 and its corresponding exercise demonstrate that the combination of a dilute solution and a relatively large or can give a percent ionization much greater than 5%, making it necessary to use the quadratic equation to determine the concentrations of species in solution. The percent ionization in a solution of a weak acid or a weak base as the analytical concentration and as the or the . Benzoic acid (C H CO H) is used in the food industry as a preservative and medically as an antifungal agent. Its p at 25°C is 4.20, making it a somewhat stronger acid than acetic acid. Calculate the percentage of benzoic acid molecules that are ionized in each solution. concentrations and p percent ionization Write both the balanced equilibrium equation for the ionization reaction and the equilibrium equation ( ). Use to calculate the from the p . For both the concentrated solution and the dilute solution, use a tabular format to write expressions for the final concentrations of all species in solution. Substitute these values into the equilibrium equation and solve for [C H CO ] for each solution. Use the values of [C H CO ] and to calculate the percent ionization. If we abbreviate benzoic acid as PhCO H where Ph = –C H , the balanced equilibrium equation for the ionization reaction and the equilibrium equation can be written as follows: \[PhCO_2H_{(aq)} \rightleftharpoons H^+_{(aq)}+PhCO^−_{2(aq)} \notag \] \[K_a=\dfrac{[H^+,PhCO_2^−]}{[PhCO_2H]} \notag \] From the p , we have = 10 = 6.3 × 10 . Inserting the expressions for the final concentrations into the equilibrium equation and making our usual assumptions, that [PhCO ] and [H ] are negligible due to the autoionization of water, \[K_a=\dfrac{[H^+,PhCO_2^−]}{[PhCO_2H]}=\dfrac{(x)(x)}{0.0500−x}=\dfrac{x^2}{0.0500}=6.3 \times 10^{−5} \notag \] \[1.8 \times 10^{-3}=x \notag \] This value is less than 5% of 0.0500, so our simplifying assumption is justified, and [PhCO ] at equilibrium is 1.8 × 10 M. We reach the same conclusion using : = (6.3 × 10 )(0.0500) = 3.2 × 10 > 1.0 × 10 . The percent ionized is the ratio of the concentration of PhCO to the analytical concentration, multiplied by 100: Because only 3.6% of the benzoic acid molecules are ionized in a 0.0500 M solution, our simplifying assumptions are confirmed. Inserting the expressions for the final concentrations into the equilibrium equation and making our usual simplifying assumptions, \[K_a=\dfrac{[H^+,PhCO_2^−]}{[PhCO_2H]}=\dfrac{(x)(x)}{0.00500−x}=\dfrac{x^2}{0.00500}=6.3 \times 10^{−5} \notag \] \[5.6 \times 10^{−4}=x \notag \] Unfortunately, this number is greater than 10% of 0.00500, so our assumption that the fraction of benzoic acid that is ionized in this solution could be neglected and that (0.00500 − ) ≈ is not valid. Furthermore, we see that = (6.3 × 10 )(0.00500) = 3.2 × 10 < 1.0 × 10 . Thus the relevant equation is as follows: which must be solved using the quadratic formula. Multiplying out the quantities, Rearranging the equation to fit the standard quadratic equation format, This equation can be solved by using the quadratic formula: \[x=\dfrac{-b \pm \sqrt{b^2 -4ac}}{2a} \notag \] \[x=\dfrac{−(6.3 \times10^{−5}) \pm \sqrt{(6.3 \times 10^{−5})^2−4(1)(−3.2 \times 10^{−7})}}{2(1)} \notag \] \[x=\dfrac{−(6.3 \times 10^{−5}) \pm (1.1 \times 10^{−3})}{2} = 5.3 \times 10^{−4} \notag \] Because a negative value corresponds to a negative [PhCO ], which is not physically meaningful, we use the positive solution: = 5.3 × 10 . Thus [PhCO ] = 5.3 × 10 M. The percent ionized is therefore
In the more dilute solution ( = 0.00500 M), 11% of the benzoic acid molecules are ionized versus only 3.6% in the more concentrated solution ( = 0.0500 M). Decreasing the analytical concentration by a factor of 10 results in an approximately threefold increase in the percentage of benzoic acid molecules that are ionized. Exercise Lactic acid (CH CH(OH)CO H) is a weak acid with a p of 3.86 at 25°C. What percentage of the lactic acid is ionized in each solution? In , you learned how to use and values to qualitatively predict whether reactants or products are favored in an acid–base reaction. Tabulated values of (or p ) and (or p ), plus the , enable us to quantitatively determine the direction and extent of reaction for a weak acid and a weak base by calculating for the reaction. To illustrate how to do this, we begin by writing the dissociation equilibriums for a weak acid and a weak base and then summing them:
\[acid:\;\;HA \rightleftharpoons H^++A^− \;\;\; K_a \tag{16.4.11a}\] \[\;base:\;\;B + H_2O \rightleftharpoons HB^++OH^− \;\;\; K_b\tag{16.4.11b}\] \[sum:\;\;HA + B+ H_2O \rightleftharpoons H^++A^−+HB^++OH^− \;\;\; K_{sum}=K_aK_b \tag{16.4.11c}\] The overall reaction has H O on the left and H and OH on the right, which means it involves the autoionization of water (\(H_2O \rightleftharpoons H^++OH^−\)) in addition to the acid–base equilibrium in which we are interested. We can obtain an equation that includes only the acid–base equilibrium by simply adding the equation for the reverse of the autoionization of water (\(H^++OH^− \rightleftharpoons H_2O\)), for which = 1/ , to the overall equilibrium in and canceling:
\[HA + B+ \cancel{H_2O} \rightleftharpoons \cancel{H^+} + A^−+HB^++\cancel{OH^−} \;\;\; K_{sum}=K_aK_b \tag{16.4.12a}\] \[\cancel{ H^+} + \cancel{OH^−} \rightleftharpoons \cancel{H_2O} \;\;\; 1/K_w \tag{16.4.12b}\] \[HA + B \rightleftharpoons A^- + HB^+ \;\;\; K=(K_aK_b)/K_w \tag{16.4.12c}\] Thus the equilibrium constant for the reaction of a weak acid with a weak base is the product of the ionization constants of the acid and the base divided by . Example 10 illustrates how to calculate the equilibrium constant for the reaction of a weak acid with a weak base. Fish tend to spoil rapidly, even when refrigerated. The cause of the resulting “fishy” odor is a mixture of amines, particularly methylamine (CH NH ), a volatile weak base (p = 3.34). Fish is often served with a wedge of lemon because lemon juice contains citric acid, a triprotic acid with p values of 3.13, 4.76, and 6.40 that can neutralize amines. Calculate the equilibrium constant for the reaction of excess citric acid with methylamine, assuming that only the first dissociation constant of citric acid is important. p for base and p for acid Write the balanced equilibrium equation and the equilibrium constant expression for the reaction. Convert p and p to and and then use to calculate . If we abbreviate citric acid as H citrate, the equilibrium equation for its reaction with methylamine is as follows: \[CH_3NH_{2(aq)}+H_3citrate_{(aq)} \rightleftharpoons CH_3NH^+_{3(aq)}+H_2citrate^−_{(aq)} \notag \] The equilibrium constant expression for this reaction is as follows: \[K=\dfrac{[CH_3NH_3^+,H_2citrate^−]}{[CH_3NH_2,H_3citrate]} \notag \] is = ( )/ . Converting p and p to and gives = 10 = 7.4 × 10 for citric acid and = 10 = 4.6 × 10 for methylamine. Substituting these values into the equilibrium equation, \[K=\dfrac{K_aK_b}{K_w}=\dfrac{(7.4 \times 10^{−4})(4.6 \times 10^{−4})}{1.01 \times 10^{−14}}=3.4 \times 10^7 \notag \] The value of p can also be calculated directly by taking the negative logarithm of both sides of , which gives \(pK = pK_a + pK_b − pK_w = 3.13 + 3.34 − 14.00 = −7.53 \notag \] Thus = 10 = 3.4 × 10 , in agreement with the earlier value. In either case, the values show that the reaction of citric acid with the volatile, foul-smelling methylamine lies very far to the right, favoring the formation of a much less volatile salt with no odor. This is one reason a little lemon juice helps make less-than-fresh fish more appetizing. Exercise Dilute aqueous ammonia solution, often used as a cleaning agent, is also effective as a deodorizing agent. To see why, calculate the equilibrium constant for the reaction of aqueous ammonia with butyric acid (CH CH CH CO H), a particularly foul-smelling substance associated with the odor of rancid butter and smelly socks. The p of ammonia is 4.75, and the p of butyric acid is 4.83. 2.6 × 10 If the concentration of one or more of the species in a solution of an acid or a base is determined experimentally, and can be calculated, and , p , , and p can be used to quantitatively describe the composition of solutions of acids and bases. The concentrations of all species present in solution can be determined, as can the pH of the solution and the percentage of the acid or base that is ionized. The equilibrium constant for the reaction of a weak acid with a weak base can be calculated from (or p ), (or p ), and . : \(=\dfrac{[H^+]}{C_{HA}}×100 \) : \(=\dfrac{[OH^−]}{C_B}×100 \) : \(K=\dfrac{K_aK_b}{K_w} \) Explain why the analytical concentration ( ) of H SO is equal to [H SO ] + [HSO ] + [SO ]. Write an expression for the analytical concentration ( ) of H PO in terms of the concentrations of the species actually present in solution. For relatively dilute solutions of a weak acid such as acetic acid (CH CO H), the concentration of undissociated acetic acid in solution is often assumed to be the same as the analytical concentration. Explain why this is a valid practice. How does dilution affect the percent ionization of a weak acid or a weak base? What is the relationship between the of a weak acid and its percent ionization? Does a compound with a large p value have a higher or a lower percent ionization than a compound with a small p value (assuming the same analytical concentration in both cases)? Explain. For a dilute solution of a weak acid (HA), show that the pH of the solution can be approximated using the following equation (where is the analytical concentration of the weak acid): \[pH=−\log \sqrt{K_a C_{HA}}\] Under what conditions is this approximation valid? The p of NH is estimated to be 35. Its conjugate base, the amide ion (NH ), can be isolated as an alkali metal salt, such as sodium amide (NaNH ). Calculate the pH of a solution prepared by adding 0.100 mol of sodium amide to 1.00 L of water. Does the pH differ appreciably from the pH of a NaOH solution of the same concentration? Why or why not? Phenol is a topical anesthetic that has been used in throat lozenges to relieve sore throat pain. Describe in detail how you would prepare a 2.00 M solution of phenol (C H OH) in water; then write equations to show all the species present in the solution. What is the equilibrium constant expression for the reaction of phenol with water? Use the information in Table E1 and Table E2 to calculate the pH of the phenol solution. Describe in detail how you would prepare a 1.50 M solution of methylamine in water; then write equations to show all the species present in the solution. What is the equilibrium constant expression for the reaction of methylamine with water? Use the information in Table E1 and Table E2 to calculate the pH of the solution. A 0.200 M solution of diethylamine, a substance used in insecticides and fungicides, is only 3.9% ionized at 25°C. Write an equation showing the equilibrium reaction and then calculate the p of diethylamine. What is the p of its conjugate acid, the diethylammonium ion? What is the equilibrium constant expression for the reaction of diethylammonium chloride with water? A 1.00 M solution of fluoroacetic acid (FCH CO H) is 5% dissociated in water. What is the equilibrium constant expression for the dissociation reaction? Calculate the concentration of each species in solution and then calculate the p of FCH CO H. The p of 3-chlorobutanoic acid (CH CHClCH CO H) is 4.05. What percentage is dissociated in a 1.0 M solution? Do you expect the p of butanoic acid to be greater than or less than the p of 3-chlorobutanoic acid? Why? The p of the ethylammonium ion (C H NH ) is 10.64. What percentage of ethylamine is ionized in a 1.00 M solution of ethylamine? The p of Cl CCO H is 0.64. What is the pH of a 0.580 M solution? What percentage of the Cl CCO H is dissociated? The pH of a 0.150 M solution of aniline hydrochloride (C H NH Cl ) is 2.70. What is the p of the conjugate base, aniline (C H NH )? Do you expect the p of (CH ) CHNH to be greater than or less than the p of C H NH ? Why? What is the pH of a 0.620 M solution of CH NH Br if the p of CH NH is 10.62? The p of 4-hydroxypyridine is 10.80 at 25°C. What is the pH of a 0.0250 M solution? The p values of formic acid and the methylammonium ion are 3.75 and 10.62, respectively. Calculate for the following reaction: \[HCO^−_{2(aq)} + CH_3NH^+_{3(aq)} \rightleftharpoons HCO_2H_{(aq)}+CH_3NH_{2(aq)}\] The p values of butanoic acid and the ammonium ion are 4.82 and 9.24, respectively. Calculate for the following reaction: \[CH_3CH_2CH_2CO^−_{2(aq)}+NH^+_{4(aq)} \rightleftharpoons CH_3CH_2CH_2CO_2H_{(aq)}+NH_{3(aq)}\] Use the information in Tables E1 and E2 to calculate the pH of a 0.0968 M solution of calcium formate. Calculate the pH of a 0.24 M solution of sodium lactate. The p of lactic acid is 3.86. Use the information in Tables E1 and E2 to determine the pH of a solution prepared by dissolving 750.0 mg of methylammonium chloride (CH NH Cl ) in enough water to make 150.0 mL of solution. Use the information in Tables E1 and E2 to determine the pH of a solution prepared by dissolving 855 mg of sodium nitrite (NaNO ) in enough water to make 100.0 mL of solution. p = 9.43; (CH ) CHNH will be a stronger base and have a lower p ; aniline is a weaker base because the lone pair on the nitrogen atom can be delocalized on the aromatic ring. 3.8 × 10 8.18 | 31,882 | 1,050 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Balanced_Equations_and_Equilibrium_Constants |
In a , the total number of atoms of each element present is the same on both sides of the equation. are the coefficients required to balance a chemical equation. These are important because they relate the amounts of reactants used and products formed. The coefficients relate to the because they are used to calculate them. For this reason, it is important to understand how to balance an equation before using the equation to calculate equilibrium constants. There are several important rules for balancing an equation: \[H_2\; (g) + O_2 \; (g) \rightleftharpoons H_2O \; (l) \nonumber \] \[2H_2(g) + O_2(g) \rightleftharpoons 2H_2O(l)\nonumber \] \[Al \; (s) + MnSO_4 \; (aq) \rightleftharpoons Al_2(SO_4)_3 + Mn ; (s) \nonumber \] \[P_4S_3 + KClO_3 \rightleftharpoons P_2O_5 + KCl + SO_2 \nonumber \] Balanced chemical equations can now be applied to the concept of chemical equilibrium, the state in which the reactants and products experience no net change over time. This occurs when the forward and reverse reactions occur at equal rates. The equilibrium constant is used to determine the amount of each compound that present at equilibrium. Consider a chemical reaction of the following form: \[ aA + bB \rightleftharpoons cC + dD\nonumber \] For this equation, the equilibrium constant is defined as: \[ K_c = \dfrac{[C]^c [D]^d}{[A]^a [B]^b} \nonumber \] The activities of the products are in the numerator, and those of the reactants are in the denominator. For K , the activities are defined as the molar concentrations of the reactants and products ([A], [B] etc.). The lower case letters are the stoichiometric coefficients that balance the equation. An important aspect of this equation is that pure liquids and solids are not included. This is because their activities are defined as one, so plugging them into the equation has no impact. This is due to the fact that pure liquids and solids have no effect on the physical equilibrium; no matter how much is added, the system can only dissolve as much as the solubility allows. For example, if more sugar is added to a solution after the equilibrium has been reached, the extra sugar will not dissolve (assuming the solution is not heated, which would increase the solubility). Because adding more does not change the equilibrium, it is not accounted for in the expression. The following are concepts that apply when adjusting K in response to changes to the corresponding balanced equation: A balanced equation is very important in using the constant because the coefficients become the powers of the concentrations of products and reactants. If the equation is not balanced, then the constant is incorrect. For gas-phase equilibria, the equation is a function of the reactants' and products' partial pressures. The equilibrium constant is expressed as follows: \[ K_p = \dfrac{P_C^c P_D^d}{P_A^a P_B^b} \nonumber \] P represents partial pressure, usually in atmospheres. As before, pure solids and liquids are not accounted for in the equation. K and K are related by the following equation: \[ K_p = K_c(RT)^{\Delta n} \nonumber \] where \[ \Delta n = (c+d) - (a+b) \nonumber \] This represents the change in gas molecules. a,b,c and d are the stoichiometric coefficients of the gas molecules found in the balanced equation. Neither K nor K have units. This is due to their formal definitions in terms of activities. Their units cancel in the calculation, preventing problems with units in further calculations. \[ PbI_2 \rightleftharpoons Pb \; (aq) + I \; (aq) \nonumber \] First, balance the equation. Next, calculate find Kc. Use these concentrations: Pb 0.3 mol/L, I 0.2 mol/L, PbI - 0.5 mol/L \[ K_c = \dfrac{(0.3) * (0.2)^2}{(0.5)} \nonumber \] \[K_c= 0.024\nonumber \] : If the equation had not been balanced when the equilibrium constant was calculated, the concentration of I would not have been squared. This would have given an incorrect answer. utomatic number to work, you need to add the "AutoNum" template (preferably at the end) to the page. \[SO_2 \; (g) + O_2 \; (g) \rightleftharpoons SO_3 \; (g) \nonumber \] First, make sure the equation is balanced. Calculate K . The partial pressures are as follows: SO - 0.25 atm, O - 0.45 atm, SO - 0.3 atm \( K_p = \dfrac{(0.3)^2}{(0.25)^2 \times (0.45)} \) \( K_p= 3.2\) | 4,331 | 1,051 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Electronic_Structure_of_Atoms_and_Molecules/Bohr_Diagrams_of_Atoms_and_Ions |
Niels Bohr proposed an early model of the atom as a central nucleus containing protons and neutrons being orbited by electrons in shells. As previously discussed, there is a connection between the number of protons in an element, the atomic number that distinguishes one element from another, and the number of electrons it has. In all electrically-neutral atoms, the number of electrons is the same as the number of protons. Each element, when electrically neutral, has a number of electrons equal to its atomic number. An early model of the atom was developed in 1913 by Danish scientist Niels Bohr (1885–1962). The Bohr model shows the atom as a central nucleus containing protons and neutrons with the electrons in circular orbitals at specific distances from the nucleus (Figure \(\Page {1}\)). These orbits form electron shells or energy levels, which are a way of visualizing the number of electrons in the various shells. These energy levels are designated by a number and the symbol "n." For example, the 1n shell represents the first energy level located closest to the nucleus. An electron normally exists in the lowest energy shell available, which is the one closest to the nucleus. Energy from a photon of light can bump it up to a higher energy shell, but this situation is unstable and the electron quickly decays back to the ground state. Bohr diagrams show electrons orbiting the nucleus of an atom somewhat like planets orbit around the sun. In the Bohr model, electrons are pictured as traveling in circles at different shells, depending on which element you have. Figure \(\Page {2}\) contrast the Bohr diagrams for lithium, fluorine and aluminum atoms. The shell closest to the nucleus is called the K shell, next is the L shell, next is the M shell. Each shell can only hold certain number of electrons. K shell can have 2, L can have 8 , M can have 18 electrons and so on. Electrons fill orbit shells in a consistent order. Under standard conditions, atoms fill the inner shells (closer to the nucleus) first, often resulting in a variable number of electrons in the outermost shell. The innermost shell has a maximum of two electrons, but the next two electron shells can each have a maximum of eight electrons. This is known as the octet rule which states that, with the exception of the innermost shell, atoms are more stable energetically when they have eight electrons in their valence shell, the outermost electron shell. Examples of some neutral atoms and their electron configurations are shown in Figure \(\Page {3}\). As shown, helium has a complete outer electron shell, with two electrons filling its first and only shell. Similarly, neon has a complete outer 2n shell containing eight electrons. In contrast, chlorine and sodium have seven and one electrons in their outer shells, respectively. Theoretically, they would be more energetically stable if they followed the octet rule and had eight. Bohr diagrams indicate how many electrons fill each principal shell. Group 18 elements (helium, neon, and argon are shown) have a full outer, or valence, shell. A full valence shell is the most stable electron configuration. Elements in other groups have partially-filled valence shells and gain or lose electrons to achieve a stable electron configuration. An atom may gain or lose electrons to achieve a full valence shell, the most stable electron configuration. The periodic table is arranged in columns and rows based on the number of electrons and where these electrons are located, providing a tool to understand how electrons are distributed in the outer shell of an atom. As shown in , the group 18 atoms helium (He), neon (Ne), and argon (Ar) all have filled outer electron shells, making it unnecessary for them to gain or lose electrons to attain stability; they are highly stable as single atoms. Their non-reactivity has resulted in their being named the inert gases (or noble gases). In comparison, the group 1 elements, including hydrogen (H), lithium (Li), and sodium (Na), all have one electron in their outermost shells. This means that they can achieve a stable configuration and a filled outer shell by donating or losing an electron. As a result of losing a negatively-charged electron, they become positively-charged ions. When an atom loses an electron to become a positively-charged ion, this is indicated by a plus sign after the element symbol; for example, Na . Group 17 elements, including fluorine and chlorine, have seven electrons in their outermost shells; they tend to fill this shell by gaining an electron from other atoms, making them negatively-charged ions. When an atom gains an electron to become a negatively-charged ion this is indicated by a minus sign after the element symbol; for example, \(F^-\). Thus, the columns of the periodic table represent the potential shared state of these elements' outer electron shells that is responsible for their similar chemical characteristics. Lewis Symbols are simplified Bohr diagrams which only display electrons in the outermost energy level. Boundless (www.boundless.com) | 5,111 | 1,052 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Molecular_Orbital_Theory/00%3A_Front_Matter/02%3A_InfoPage |
Instructors can adopt existing LibreTexts texts or Remix them to quickly build course-specific resources to meet the needs of their students. Unlike traditional textbooks, LibreTexts’ web based origins allow powerful integration of advanced features and new access texts to improve postsecondary education at all levels of higher learning by developing an Open Access Resource environment. The project currently consists of 14 independently operating and interconnected libraries that are constantly being integrated. and are supported by the Department of Education Open Textbook Pilot 1525057, and 1413739. Unless otherwise noted, LibreTexts content is licensed by . Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not ). and are supported by the Department of Education Open Textbook Pilot Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. . . | 1,070 | 1,053 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.11%3A_Solution_Concentrations/3.11.01%3A_Biology-_Solution_Concentrations_and_Cells |
Solution concentration plays critical roles in biology: In the laboratory, in your body, and in the outside environment, the majority of chemical reactions take place in solutions. Macroscopically a is defined as a homogeneous mixture of two or more substances, that is, a mixture which appears to be uniform throughout. On the microscopic scale a solution involves the random arrangement of one kind of atom or molecule with respect to another. There are a number of reasons why solutions are so often encountered both in nature and in the laboratory. The most common type of solution involves a liquid which dissolves a solid . (The term solvent usually refers to the substance present in greatest amount. There may be more than one solute dissolved in it.) Because a liquid adopts the shape of its container but does not expand to fill all space available to it, liquid solutions are convenient to handle. You can easily pour them from one container to another, and their volumes are readily measured using graduated cylinders, pipets, burets, volumetric flasks, or other laboratory glass-ware. Moreover, atoms or molecules of solids dissolved in a liquid are close together but still able to move past one another. They contact each other more frequently than if two solids were placed next to each other. This “intimacy” in liquid solutions often facilitates chemical reactions. Since solutions offer a convenient medium for carrying out chemical reactions, it is often necessary to know how much of one solution will react with a given quantity of another. Examples in other sections have shown that the amount of substance is the quantity which determines how much of one material will react with another. The ease with which solution volumes may be measured suggests that it would be very convenient to know the amount of substance dissolved per unit volume of solution. Then by measuring a certain volume of solution, we would also be measuring a certain amount of substance. The of a substance in a solution (often called ) is the : \[\text{Concentration of solute, M}=\frac{\text{amount of solute, mol}}{\text{volume of solution, L}} \nonumber \] \[c_{\text{solute, M}}~=~\frac{n_{\text{solute, mol}}}{V_{\text{solution,L}}} \label{2} \] Usually the units moles per cubic decimeter (mol dm ) or moles per liter (mol liter ) are used to express concentration. If a pure substance is soluble in water, it is easy to prepare a solution of known concentration. A container with a sample of the substance is weighed accurately, and an appropriate mass of sample is poured through a funnel into a volumetric flask, as shown in Figure 3.3. The container is then reweighed. Any solid adhering to the funnel is rinsed into the flask, and water is added until the flask is about three-quarters full. After swirling the flask to dissolve the solid, water is added carefully until the bottom of the meniscus coincides with the calibration mark on the neck of the flash. Isotonic saline solution for medical applications is made according to the method above. The container and sodium chloride weigh 43.2874 g and the final mass after pouring was 41.0374 g. The flask has a volume of 250 mL. The concentration can be calculated by dividing the amount of solute by the volume of solution [Eq. (1)]: = 2.2500 g × \(\frac{\text{1 mol}}{\text{58}\text{.44 g}}\) = 3.850 × 10 mol Isotonic saline solutions have the same "osmolarity" as blood plasma, so they are isotonic with it and red blood cells. Isotonic intravenous saline is 9 g NaCl in 1000 mL water, or roughly 0.15 M NaCl, as we see above. Its is 0.30 Osm ("osmolar"), because each of the two ions, Na and Cl affect osmosis, and 0.15 M NaCl gives 0.30 M total concentration of ions. Note that the definition of concentration is entirely analogous to the definitions of density, molar mass, and stoichiometric ratio that we have previously encountered. Concentration will serve as a conversion factor relating the volume of solution to the amount of dissolved solute. \(\text{Volume of solution}\overset{concentration}{\longleftrightarrow}\text{amount of solute}\) \(V\overset{c}{\longleftrightarrow}n\) Because the volume of a liquid can be measured quickly and easily, concentration is a much-used quantity. The next two examples show how this conversion factor may be applied to commonly encountered solutions in which water is the solvent ( solutions). There are various IV fluids used in hospitals. "D5W" only contains glucose; NS only contains Sodium Chloride; LR Solution contains Sodium, Chloride, Lactate, Potassium, and Calcium. If D5W has a concentration of 0.2520 mol dm , what amount (in mol) of glucose is administered to a patient in 24.71 cm (24.71 ml) of this solution? Using concentration as a conversion factor, we have \(=\text{24}\text{.71 cm}^{\text{3}}\times \frac{\text{0}\text{.2520 mol}}{\text{1 dm}^{\text{3}}}\times \frac{\text{1 dm}^{\text{3}}}{\text{10}^{\text{3}}\text{ cm}^{\text{3}}}\) Alternatively, 24.71 mL x \(\frac{\text{1 liter}}{\text{1000 mL}} x \frac{\text{0.2520 mol}}{liter} = \text{0.006227 mol}\) The concentration units of moles per cubic decimeter are often abbreviated , pronounced . That is, a 0.1- (one-tenth molar) solution contains 0.1 mol solute per cubic decimeter of solution. This abbreviation is very convenient for labeling laboratory bottles and for writing textbook problems; however, when doing calculations, it is difficult to see that \(\text{1 dm}^{\text{3}}\times \text{1 }\text{M}=\text{1mol}\) Therefore we recommend that you when doing any calculations involving solution concentrations. That is, \(\text{1 dm}^{\text{3}}\times \text{1 }\dfrac{\text{mol}}{\text{dm}^{\text{3}}}=\text{1mol}\)
\(\text{1 L}\times \text{1 }\dfrac{\text{mol}}{\text{L}}=\text{1mol}\) Problems such as Example 2 are easier for some persons to solve if the solution concentration is expressed in millimoles per cubic centimeter (mmol cm ) instead of moles per cubic decimeter. Since the SI prefix m means 10 , 1 mmol = 10 mol, and \(\text{1 M} ~ = ~ \dfrac{\text{1 mol}}{\text{1 dm}^{\text{3}}} ~ \times ~ \dfrac{\text{1 dm}^{\text{3}}}{\text{1 L}} ~ = ~ \dfrac{\text{1 mol}}{\text{L}}\) \(\text{1 M} ~ = ~ \dfrac{\text{1 mol}}{\text{L}} ~ \times ~ \dfrac{\text{10}^{\text{-3}}\text{ L}}{\text{1 ml}} ~ \times ~ \dfrac{\text{1 mmol}}{\text{10}^{\text{-3}}\text{ mol}} ~ = ~ \dfrac{\text{1 mmol}}{\text{1 ml}}\) \(\text{1 M} ~ = ~ \dfrac{\text{1 mol}}{\text{1 dm}^{\text{3}}} \times \dfrac{\text{1 dm}^{\text{3}}}{\text{10}^{\text{3}}\text{ cm}^{\text{3}}} \times \dfrac{\text{1 mmol}}{\text{10}^{\text{-3}}\text{ mol}} ~ = ~ \dfrac{\text{1 mmol}}{\text{1 cm}^{\text{3}}} \) Thus a concentration of 0.1396 mol dm (0.1396 ) can also be expressed as 0.1396 mmol cm . Expressing the concentration this way is very convenient when dealing with laboratory glassware calibrated in milliliters or cubic centimeters. Exactly 250 ml of NS (Normal Saline) I.V. solution whose concentration is 0.154 in NaCl was delivered from a pipet. (a) What amount of NaCl was present? (b) What mass of NaCl would remain if all the water evaporated? Since 0.154 means 0.154 mol dm , or 0.154 mmol cm , we choose the latter, more convenient quantity as a conversion factor: From ChemPRIME: 3.10: Solution Concentrations | 7,340 | 1,054 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/09%3A_Separation_Purification_and_Identification_of_Organic_Compounds/9.06%3A_Energy_States_of_Molecules |
The energy states and spectra of molecules are much more complex than those of isolated atoms. In addition to the energies associated with molecular electronic states, there is kinetic energy associated with vibrational and rotational motions. The total energy, \(E\), of a molecule (apart from its translational\(^5\) and nuclear energy) can be expressed as the sum of three terms: \[E = E_{electronic} + E_{vibrational} + E_{rotational}\] Absorption of electromagnetic radiation by molecules occurs not only by electronic excitation of the type described for atoms, but also by changes in the vibrational and rotational energies. Both rotations and vibrations of molecules are quantized. This means that only particular values of rotational angular momentum or vibrational energy are possible. We speak of these permitted values of the energies as the vibrational and rotational energy levels. \(^5\)Translational energy is not very important in connection with spectroscopy and will not be considered here. and (1977) | 1,039 | 1,055 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/28%3A_Photochemistry/28.02%3A_Light_Absorption_Flourescence_and_Phosphorescence |
Electromagnetic radiation in the ultraviolet and visible region spans a wavelength range of about \(800\)-\(100 \: \text{nm}\) corresponding to energies of \(36\)-\(286 \: \text{kcal mol}^{-1}\). Absorption of such radiation by molecules is not to be regarded as equivalent to simple excitation by thermal energy of \(36\)-\(286 \: \text{kcal mol}^{-1}\). Instead, all the energy of the light quantum is taken up in excitation of an electron to a high-energy, usually antibonding, orbital ( ). An important point about such processes is that they occur more rapidly than the atoms vibrate in the bonds ( ). The short transition time of an electron between ground and excited states is in complete contrast to what happens during absorption of a quantum of radio-frequency energy in nmr spectroscopy, wherein the absorption process may be slow compared to chemical reactions ( ). Therefore an electronically excited molecule is, in the first instant that it is produced \(\left( <10^{-13} \: \text{sec} \right)\), just like the ground-state molecule as far as and of the atoms go, but has a very different electronic configuration. What happens at this point depends on several factors, some of which can be best illustrated by energy diagrams of the type used previously ( ). We shall consider diatomic molecules, but the argument can be extended to more complicated systems. Consider Figure 28-1, which shows schematic potential-energy curves for a molecule \(\ce{A-B}\) in the ground state \(\left( \ce{A-B} \right)\) and in excited electronic states \(\left( \ce{A-B} \right)^*\). We have noted previously ( ) that in the ground states of most molecules all electrons are ; excited states also can have all electrons paired. Such states with paired electrons are called states. But, because the bonding is in excited states, the average bond length \(r_e\) between the nuclei is in the excited state than in the ground state. For this reason the upper curve \(\left( S_1 \right)\) in Figure 28-1 is displaced toward a larger average bond length relative to the lower or ground-state curve \(\left( S_0 \right)\). Excited states also can have electrons. States with two unpaired electrons are called states \(\left( T \right)\) and normally are more stable than the corresponding singlet states because, by Hund's rule, less interelectronic repulsion is expected with unpaired than paired electrons (Sections and ). (For clarity, the potential-energy curve for the excited triplet state \(\left( T_1 \right)\) of \(\ce{A-B}\) is given an unrealistically long equilibrium bond distance, which puts it to the right of the curve for the \(S_1\) state in Figure 28-1.) The electronic configurations for ground singlet \(\left( S_0 \right)\), excited singlet \(\left( S_1 \right)\), and triplet \(\left( T_1 \right)\) states of the \(\sigma\) electrons of a diatomic molecule are shown in Figure 28-2. This diagram will be helpful in interpreting the transitions between \(S_0\), \(S_1\), and \(T_1\) states shown in Figure 28-1, and which we will now discuss in more detail. When a molecule absorbs sufficient radiant energy to cause electronic excitation, the of the excited electron remains in the transition. That is to say, ground-state molecules with paired electrons \(\left( S_0 \right)\) give excited states with paired electrons \(\left( S_1 \right)\), not triplet states \(\left( T_1 \right)\). The transition marked \(1\) in Figure 28-1 corresponds to a singlet-singlet \(\left( S_0 \rightarrow S_1 \right)\) transition from a relatively high vibrational level of \(\ce{A-B}\). The energy change occurs with no change in \(r\) (Franck-Condon principle), and the electronic energy of the \(\ce{A-B}^*\) molecule so produced is seen to be the level required for dissociation of \(\ce{A-B}^*\). The vibration of the excited molecule therefore has no restoring force and leads to dissociation to \(\ce{A}\) and \(\ce{B}\) atoms. In contrast, the transition marked \(2\) leads to an excited vibrational state of \(\ce{A-B}^*\), which is not expected to dissociate but can lose vibrational energy to the surroundings and come down to a lower vibrational state. This is called "vibrational relaxation" and usually requires about \(10^{-12} \: \text{sec}\). The vibrationally "relaxed" excited state can return to the ground state with emission of radiation (transition \(F\), \(S_1 \rightarrow S_0\)); this is known as , the wavelength of fluorescence being different from that of the original light absorbed. Normally, fluorescence, if it occurs at all, occurs in \(10^{-9}\) to \(10^{-7} \: \text{sec}\) after absorption of the original radiation. In many cases, the excited state \(\left( S_1 \right)\) can return to the ground state \(\left( S_0 \right)\) by processes. The most important processes are: 1. By chemical reaction, often with surrounding molecules. This process forms the basis of much organic photochemistry, which will be described in a later section.
2. By transfer of its excess electronic energy to other molecules. This kind of energy transfer also is a very important aspect of photochemistry, and we shall return to it shortly.
3. By decay through a lower energy state. If, for example, the potential-energy curves for the upper and lower singlet states were closer together than shown in Figure 28-1, they may actually cross at some point, thus providing a pathway for \(S_1\) to relax to \(S_0\) without fluorescing. But what about decay of \(S_1\) through the triplet state \(\left( T_1 \right)\)? Conversion of a singlet excited state to a triplet state \(\left( S_1 \rightarrow T_1 \right)\) is energetically favorable but usually occurs rather slowly, in accord with the spectroscopic selection rules, which predict that spontaneous changes of electron spin should have very low probabilities. Nonetheless, if the singlet state is sufficiently long-lived, the singlet-triplet change, \(S_1 \rightarrow T_1\), (often called ) may occur for a very considerable proportion of the excited singlet molecules. The triplet state, like the singlet state, can return to the ground state by nonradiative processes, but in many cases a radiative transition \(\left( T_1 \rightarrow S_0 \right)\) occurs, even though it has low probability. Such transitions result in emission of light of considerably longer wavelength than either that absorbed originally or resulting from fluorescence. This type of radiative transition is called (transition \(P\) in Figure 28-1). Because phosphorescence is a process with a low probability, the \(T_1\) state may persist from fractions of a second to many seconds. For benzene at \(-200^\text{o}\), the absorption of light at \(254 \: \text{nm}\) leads to fluorescence centered on \(290 \: \text{nm}\) and phosphorescence at \(340 \: \text{nm}\). The half-life of the triplet state of benzene at \(-200^\text{o}\) is \(7 \: \text{sec}\). In previous discussions of electronic absorption spectra ( ), we have identified two different kinds of transitions in the spectra of simple carbonyl compounds such as 2-propanone or methanal. One involves excitation of an electron in a nonbonding \(n\) orbital on oxygen to an antibonding \(\left( \pi^* \right)\) orbital of the carbon-oxygen double bond (an \(n \rightarrow \pi^*\) transition), and the other involves excitation of an electron in the bonding \(\left( \pi \right)\) orbital to the corresponding antibonding orbital (a \(\pi \rightarrow \pi^*\) transition). These changes are shown for methanal in Figure 28-3. Besides the transitions already discussed, methanal shows strong absorption at \(175 \: \text{nm}\), which possibly is \(n \rightarrow \sigma^*\), or else \(\sigma \rightarrow \sigma^*\). Although the \(n \rightarrow \pi^*\) and \(\pi \rightarrow \pi^*\) transitions of Figure 28-3 are singlet-singlet transitions, each of the two singlet excited states produced has a corresponding triplet state. Accordingly, there are easily accessible excited states of a carbonyl group - the \(n \rightarrow \pi^*\) singlet \(\left( S_1 \right)\), \(n \rightarrow \pi^*\) triplet \(\left( T_1 \right)\), \(\pi \rightarrow \pi^*\) singlet \(\left( S_2 \right)\), and \(\pi \rightarrow \pi^*\) triplet \(\left( T_2 \right)\). The energies of these electronic states for methanal decrease in the order \(S_2 > T_2 > S_1 > T_1\), although this ordering may not hold for all carbonyl compounds. As we shall see, \(n \rightarrow \pi^*\) singlet and triplet states of carbonyl compounds play an important role in photochemistry. Aldehydes and ketones display all the characteristics of absorption, fluorescence, phosphorescence, and intersystem crossing \(\left( S_1 \rightarrow T_1 \right)\) illustrated in Figure 28-1. Generally, they are more efficient at intersystem crossing than are unsaturated hydrocarbons, perhaps because the energies of the \(S\) and \(T\) states involved are not widely different. Besides the bond lengths being longer in excited states of molecules, the molecular shapes differ from those of the ground states. Although the Franck-Condon principle requires that absorption produce excited states with the same geometry as the ground states, the excited molecules thereafter can relax to more stable shapes, which may be nonplanar and twisted about the erstwhile \(\pi\) bonds. Methanal is planar with a \(\ce{C-O}\) bond length of \(1.21 \: \text{Å}\) in the ground state, but in the \(n \rightarrow \pi^*\) singlet \(\left( S_1 \right)\) state, methanal is pyramidal, with a \(\ce{C-O}\) bond length of \(1.32 \: \text{Å}\). Methanal is even more distorted in the \(n \rightarrow \pi^*\) triplet state, although the bond length remains about the same at \(1.32 \: \text{Å}\). It is possible to produce electronic excited states of molecules indirectly by way of energy transfer from other excited molecules. An example is provided by excitation of naphthalene as the result of energy transfer from excited benzophenone. Benzophenone, \(\ce{C_6H_5COC_6H_5}\), absorbs ultraviolet light with \(\lambda_\text{max} = 330 \: \text{nm}\) in an \(n \rightarrow \pi^*\) transition. Naphthalene does not absorb appreciably in this region. Yet irradiation of a mixture of benzophenone and naphthalene with \(330\)-\(\text{nm}\) light produces phosphorescent emission from naphthalene. Thus benzophenone absorbs the light and transfers its excess energy to naphthalene, which returns to the ground state by emission. Because the emission is from the triplet state of naphthalene, benzophenone must be involved in exciting the naphthalene to the triplet state. We may write the process as follows: Energy transfer does not involve a change in electron spin. For this to hold for excitation of naphthalene from \(S_0\) to \(T_1\), the energy transfer must come from triplet (not singlet) benzophenone). The process of producing excited states in this way is called . Singlet-singlet, as well as triplet-triplet, energy transfers are possible, but in all cases there is no net change in spin. Efficient energy transfer will only be possible if \(\Delta G^0\) for the transfer is small or negative. and (1977) | 11,186 | 1,056 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/09%3A_Separation_Purification_and_Identification_of_Organic_Compounds/9.04%3A_Why_Can't_We_See_Molecules |
The most straightforward way to determine the structures of molecules would be to "see" how the nuclei are arranged and how the electrons are distributed. This is not possible with visible light, because the wavelengths of visible light are very much longer than the usual molecular dimensions. A beam of electrons can have the requisite short wavelengths, but small organic molecules are destroyed rapidly by irradiation with electrons of the proper wavelengths. Nonetheless, is a valuable technique for the study of large molecules, such as DNA, which can be stained with heavy-metal atoms before viewing, or are themselves reasonably stable to an electron beam (Figures 9-4 and 9-5). Virtually all parts of the spectrum of electromagnetic radiation, from x rays lo radio waves, have some practical application for the study of organic molecules. The use of for determination of the structures of molecules in crystals is of particular value, and in the past ten years this technique has become almost routine. Figure 9-6 shows the detailed arrangement of the carbons, hydrogens, and bromines in 1,8-bis(bromomethyl)naphthalene, \(1\), as determined by x-ray diffraction. The apparatus and techniques used are highly complex and are not available yet to very many organic laboratories.\(^3\) Other diffraction methods include , which may be used to’determine the structures of gases or of volatile liquid substances that cannot be obtained as crystals suitable for x-ray diffraction, and , which has special application for crystals in which the exact location of hydrogens is desired. Hydrogen does not have sufficient scattering power for x rays to be located precisely by x-ray diffraction. The diffraction methods can be used to determine complete structures of organic molecules, but they are not sufficiently routine to be utilized generally in practical organic laboratory work. For this reason, in the remainder of this chapter we will emphasize those forms of spectroscopy that are generally available for routine laboratory use. As will be seen, these methods are used by organic chemists in more or less empirical ways. In general, spectroscopic methods depend on some form of excitation of molecules by absorption of electromagnetic radiation and, as we have said, virtually all parts of the electromagnetic spectrum have utility in this regard. The commonly used span of the electromagnetic spectrum is shown in Figure 9-7 along with a comparison of the various units that are employed to express energy or wavelength. The major kinds of spectroscopy used for structural analysis of organic compounds are listed in Table 9-1. The range of frequencies of the absorbed radiation are shown, as well as the effect produced by the radiation and specific kind of information that is utilized in structural analysis. After a brief account of the principles of spectroscopy, we will describe the methods that are of greatest utility to practical laboratory work. Nonetheless, it is very important to be aware of the other, less routine, methods that can be used to solve special problems, and some of these are discussed in this chapter and in Chapters 19 and 27. You may have problems with the relationships among the variety of wavelength and frequency units commonly used in spectroscopy. The relationship between wavelength, frequency, and velocity should become clear to you by considering yourself standing on a pier watching ocean waves going by. Assuming the waves are uniformly spaced, there will be a uniform distance between the crests, which is \(\lambda\), the wavelength. The wave crests will pass by at a certain number per minute, which is \(\nu\), the frequency. The velocity, \(c\), at which the crests move by you is related to \(\lambda\) and \(\nu\) by the relationship \(c = \lambda \nu\). This is not really very complicated and it applies equally well to water waves or electromagnetic radiation. What is almost needlessly complicated is the variety of units commonly used to express \(\lambda\) and \(\nu\) for electromagnetic radiation. One problem is tradition, the other is the desire to avoid very large or very small numbers. Thus, as Figure 9-7 shows, we may be interested in electromagnetic wavelengths that differ by as much as a factor of \(10^{16}\). Because the velocity of electromagnetic radiation in a vacuum is constant at \(3 \times 10^8 \: \text{m sec}^{-1}\), the frequencies will differ by the same factor. Units commonly used for are meters (\(\text{m}\)), centimeters (\(\text{cm}\)), nanometers (\(\text{nm}\)), and microns (\(\mu\)). In the past, angstroms (\(Å\)) and millimicrons (\(\text{m} \mu m\)) also were used rather widely. \[1 \: \text{m} = 10^2 \: \text{cm} = 10^9 \: \text{nm} = 10^6 \: \mu m\] \[10^{-2} \: \text{m} = 1 \: \text{cm} = 10^7 \: \text{nm} = 10^4 \: \mu m\] \[10^{-6} \: \text{m} = 10^{-4} \: \text{cm} = 10^3 \: \text{nm} = 1 \: \mu m\] \[10^{-9} \: \text{m} = 10^{-7} \: \text{cm} = 1 \: \text{nm} = 10^{-3} \: \mu m= 1 \: \text{m} \mu m= 10 \: \text{Å}\] Frequency units are in cycles per second (cps) or hertz (\(\text{Hz}\)), which are equivalent (radians per second are used widely by physicists). \[1 \: \text{Hz} = 10^{-6} \: \text{MHz} \cong 3.3 \times 10^{-11} \: \text{cm}^{-1}\] \[10^6 \: \text{Hz} = 1 \: \text{MHz} \cong 3.3 \times 10^{-5} \: \text{cm}^{-1}\] \[3 \times 10^{10} \: \text{Hz} = 3 \times 10^4 \: \text{MHz} \cong 1 \: \text{cm}^{-1}\] \(^3\)A useful description of how molecular structures can be determined by "x-ray vision" is given in Chapter XI of by M. Goodman and F. Morehouse, Gordon and Breach, New York, 1973. and (1977) | 5,670 | 1,057 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/06%3A_Structures_and_Energetics_of_Metallic_and_Ionic_solids/6.11%3A_Ionic_Lattices/6.11E%3A_Structure_-_Zinc_Blende_(ZnS) |
Summary: Zinc blend is a compound that comes in two forms: sphalerite and wurtzite. These are characterized by a 1:1 stoichiometric ratio of Zinc to Sulfur. It maintains a tetrahedral arrangement in both forms. Zinc sulfide (ZnS) is a unique compound that forms two types of crystalline structures. These two polymorphs are wurtzite and zincblende (also known as sphalerite). Wurtzite has a hexagonal structure, while zincblende is cubic. It is characterized by single bonds between each atom and maintenance of a 1:1 zinc to sulfur ratio. sphalerite: Since the number of atoms in a single unit cell of Zn and S is the same, it is consistent with the formula ZnS. The ionic radius for Zn is 74pm and for S is 190pm. Therefore the ratio between cationic and anionic radii in zinc blend is 0.39 (74pm/190 pm) .This suggests a tetrahedral ion arrangement and four nearest neighbors from standard crystal structure prediction tables. Therefore, four sulfur atoms surround each zinc atom and four zinc atoms surround each sulfur atom. The coordination number, the number of of electron pairs donated to a metal by its ligands, for both zinc an sulfur is four. The difference between wurtzite and zincblende lies in the different arrangements of layers of ions. Zincblende is characterized as a (ccp), also known as face-centered cubic, structure. This crystal lattice structure is shown in Figures 1 & 2 below. Fig. 1. A break down of cubic closest packing. (Author: Maghémite Date: May 5, 2008. d under the Creative Commons Attribution-Share Alike , , and license.) Fig. 2. A representation of ccp structure. (from Public Domain) Notice how only half of the tetrahedral sites are occupied. Density tends to decrease as temperature increases. In this case, since ccp structures are more dense than hcp structures, so a conversion from sphalerite to wurtzite occurs naturally over time at a rate similar to that of diamond to graphite. The sphalerite structure is favored at 298k by 13kJ/mol, but at 1296K the transition to wurtzite occurs. Wurtzite has a hexagonal closest packing structure (hcp), which is characterized by 12 ions in the corners of each unit that create a hexagonal prism (seen in Fig. 3). As discussed previously, zincblende slowly transforms to wurtzite due to thermodynamic stability. Fig 3. HCP structure of wurtzite. (Creator: Alexander Mann Date: 01/14/2006 d under the Creative Commons license) Density = Mass of unit cell / volume of unit cell. where: Mass of unit cell = Number of atoms in a unit cell x the mass of each atom volume of unit cell = a x 10 | 2,613 | 1,059 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Solutions_and_Mixtures/Colloid/Tyndall_Effect |
The Tyndall Effect is the effect of light scattering in colloidal dispersion, while showing no light in a true solution. This effect is used to determine whether a mixture is a true solution or a colloid. "To be classified , a material must have one or more of its dimensions (length, width, or thickness) in the approximate range of 1-1000 nm." Because a colloidal solution or substance (like fog) is made up of scattered particles (like dust and water in air), light cannot travel straight through. Rather, it collides with these micro-particles and scatters causing the effect of a visible light beam. This effect was observed and described by John Tyndall as the Tyndall Effect. The Tyndall effect is an easy way of determining whether a mixture is colloidal or not. When light is shined through a true solution, the light passes cleanly through the solution, however when light is passed through a colloidal solution, the substance in the dispersed phases scatters the light in all directions, making it readily seen. For example, light being shined through water and milk. The light is not reflected when passing through the water because it is not a colloid. It is however reflected in all directions when it passes through the milk, which is colloidal. A second example is shining a flashlight into fog or ; the beam of light can be easily seen because the fog is a colloid. | 1,404 | 1,060 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/4_f-Block_Elements/The_Lanthanides/Chemistry_of_Lutetium |
Lutetium ranks among the rare-earths in abundance only above thulium and promethium (and there's none of that anyway!). It official name comes from an ancient name for Paris, Lutecia, but it has had many names, most recently lutecium (only a change in official spelling). It was discovered independently by von Welsbach and Urbain in 1907-08. The refinement of ion exchange methods and their application to the separation of the rare-earths made the separation of lutetium from ytterbium possible. von Welsbach decided to rename ytterbium aldebaranium and picked cassiopium for element 71. Urbain preferred neoytterbium and lutecium. Urbain's choices eventually were accepted, altho ugh the prefix was dropped from ytterbium and the spelling of lutecium was eventually changed. The metal is the hardest and densest of the rare-earths and is the last of the lanthanides. | 889 | 1,061 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/4_f-Block_Elements/The_Lanthanides/Chemistry_of_Europium |
Europium looks and feels a lot like lead, although it is not as dense. It was discovered in 1896 and isolated in 1901 by Demarcay, working with samples of supposedly "pure" samarium. Named for the continent of Europe, the element ranks thirteenth in abundance among the rare earth metals, but there is more of it than silver and gold combined. It is the most reactive of the rare earth metals, behaving with water in a manner similar to calcium. Generally refined from monazite sand, the pure metal has few applications, but you would find it less interesting to read this without some of its compounds which are used as activators and red phosphors in color CRT screens for television and computers. All lanthanoids tend to group together in nature because all the lanthanoids are most stable in their +3 oxidation state, and their bonding is nearly exclusively ionic/electrostatic. While the lanthanides have valent 4f electrons, the 4f orbitals are radially constricted and cannot achieve appropriate overlap with ligand orbitals to engage in sigma- or pi-bonding interactions. So they can all easily incorporate into similar minerals. Similarly, Europium is most stable in the +3 oxidation state, but can be easily reduced to the +2 oxidation state. This change in Lewis acidity and causes it to out into other minerals that the relatively hard +3 Lewis acids do not go into. It is well known as the . | 1,428 | 1,062 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/04%3A_The_Second_Law_of_Thermodynamics/4.03%3A_The_Second_Law_of_Thermodynamics |
The Second Law of Thermodynamics states that the state of entropy of the entire universe, as an , will always increase over time. The second law also states that the changes in the entropy in the universe can never be negative. Why is it that when you leave an ice cube at room temperature, it begins to melt? Why do we get older and never younger? And, why is it whenever rooms are cleaned, they become messy again in the future? Certain things happen in one direction and not the other, this is called the "arrow of time" and it encompasses every area of science. The thermodynamic arrow of time (entropy) is the measurement of disorder within a system. Denoted as \(\Delta S\), the change of entropy suggests that time itself is asymmetric with respect to order of an isolated system, meaning: a system will become more disordered, as time increases. To understand why entropy increases and decreases, it is important to recognize that two changes in entropy have to considered at all times. The entropy change of the surroundings and the entropy change of the system itself. Given the entropy change of the universe is equivalent to the sums of the changes in entropy of the system and surroundings: \[\Delta S_{univ}=\Delta S_{sys}+\Delta S_{surr}=\dfrac{q_{sys}}{T}+\dfrac{q_{surr}}{T} \label{1}\] In an isothermal reversible expansion, the heat q absorbed by the system from the surroundings is \[q_{rev}=nRT\ln\frac{V_{2}}{V_{1}}\label{2}\] Since the heat absorbed by the system is the amount lost by the surroundings, \(q_{sys}=-q_{surr}\).Therefore, for a truly reversible process, the entropy change is \[\Delta S_{univ}=\dfrac{nRT\ln\frac{V_{2}}{V_{1}}}{T}+\dfrac{-nRT\ln\frac{V_{2}}{V_{1}}}{T}=0 \label{3}\] If the process is irreversible however, the entropy change is \[\Delta S_{univ}=\frac{nRT\ln \frac{V_{2}}{V_{1}}}{T}>0 \label{4}\] If we put the two equations for \(\Delta S_{univ}\)together for both types of processes, we are left with the second law of thermodynamics, \[\Delta S_{univ}=\Delta S_{sys}+\Delta S_{surr}\geq0 \label{5}\] where \(\Delta S_{univ}\) equals zero for a truly reversible process and is greater than zero for an irreversible process. In reality, however, truly reversible processes never happen (or will take an infinitely long time to happen), so it is safe to say all thermodynamic processes we encounter everyday are irreversible in the direction they occur. The second law of thermodynamics can also be stated that "all processes produce an in the entropy of the universe". | 2,545 | 1,064 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Electronic_Spectroscopy/Electronic_Spectroscopy%3A_Application |
Electronic Absorption and Fluorescence spectroscopy are both analytical methods that center around the idea that when one perturbs a known or unknown solution with a spectrum of energetic photons, those photons that have the correct energy to interact with the molecules in solution will do so, and those molecules under observation will always interact with photons of energies characteristic to that molecule. These methods have tremendously helped scientists elucidate and characterize the physical properties of a variety of molecules by giving a means to probe the fundamental electronic structure of those molecules. The theory behind Electronic Absorption and Fluorescence was described in a . If you are unfamiliar with electronic spectroscopy, browsing the theory might help paint a better picture of what will be discussed in this module.But even if you understand what a Jablonski diagram represents or the mathematical description behind the transition dipole moment operator, eventually you want to know how to apply the information you've learned to your world. What this means to you is that you now have the ability that allow you to find the reason why carrots are orange or why plants are green. You could take leaves and carrots, chop them into fine pieces, extract and separate the molecules inside of them, and use Electronic Absorption or Fluorescence spectroscopy to figure out that the orange color in carrots is due to a family of molecules known as the carotenoids and that the green color in the leaves is due to a variety of different chlorophyll molecules. So excited by this knowledge, you could take the carotenoids that you now know are responsible for the color orange and dye cloth to make yourself a carrot costume! But what if you wanted a different color? What if you wanted a lighter shade of orange? What if you wanted a very exact and very specific color? Before you can go explore the world (or lab) to find (or design) a molecule that has those properties you need to be able to understand the instruments that allow you to do the work so you can be certain you get the results you desire. The remainder of this wikitext describes the design and components of a photospectrometer to measure the the Ultra Violet through Visable (UV-Vis) region of the electromagnetic spectrum. Basic instrumental design and theory behind a spectrofluorometer will also be discussed along with the errors and limits known to plague these instruments. By the end of this text, you should be fully able to design a a photospectrometer in order to characterize the electronic behavior of molecules in order to prove to yourself that carotenoids are responsible for the color orange (building of a carrot costume is optional) As stated in the introduction, the purpose of a spectrophotometer is to characterize how molecules react with photons of varying wavelengths. This easiest way to do this is to design an instrument that can control what type of light is in contact with the sample and the amount of that light that makes it through. Shown below is a simple spectrophotometer that does just what was described. A light source produces light that can be separated and controlled by the monochrometer which then allows only the desired light to impinge upon the sample. The light that travels through the sample is detected and sent to a recorder. Keep this overall picture in mind when thinking about the components and what each of their parts are in whole experiment. As shown above, the first piece of equipment needed is a source of photons (light). Since we are interested in the UV-Vis region of the electromagnetic spectrum, we need to assure that this source is able to produce photons with the correct wavelengths (180-780 nm) that are characteristic to the spectrums of interest so that the experiment can be completed. Typical sources come in two forms, either in the form of a lamp, or in the form of a laser. Lamps can provide a wide spectrum of light that can be used for almost any purpose at a very reasonable cost, while lasers provide very intense monochromatic light and tend can be used for fluorescence and other specialized experiments due to their high costs. A common lamp source that might be familiar is the tungsten filament lamp or sometimes referred to as the incandescent light bulb. This type of lamp produces a spectrum of light perfect for the visible spectrum (320 nm - 3.5 µm) by applying a voltage through a thin filament of tungsten until it heats up enough to produce black body radiation. The Tungsten-Halogen lamp is a significant improvement over the normal tungsten filament lamp due to its longer lifetime and the wider range of light it produces. This is achieved by incasing the filament with quartz instead of silica so the filament can be forced to higher temperatures (thus increasing the range of light produced) and by trapping a halogen gas (normally Iodine) that can react with sublimated tungsten atoms to form WI2 that can then redeposit the tungsten by coming into contact with the filament (thus increasing the lifespan). This type of lamp is used to obtain a spectrum of light in the Ultra Violet region (190-400 nm). The spectrum is obtained by trapping Hydrogen at low pressures inside of a glass tube and creating an electrical arc inside of the cell. This electrical arc creates high energy hydrogen gas atoms that dissociate into two hydrogen atoms along with producing photons. The wavelength of the photons created from the dissociation is dependent on the resulting kinetic energies of the two hydrogen atoms after the dissociation. Since the kinetic energies of the dissociated atoms can vary from zero to the original energy of the excited atom, the photon produced can also vary which leads to a spectrum of photons. Deuterium is often used in lieu of hydrogen because it can produce more a more intense spectrum and has a longer lifetime. Both types of this lamp are housed within a quartz shell since silica absorbs in the Ultraviolet range. Tunable dye and fixed wavelength lasers that produce photons within the UV-Vis spectrum are commonly used for fluorescence experiments where the intensity of the resulting fluorescence is directly proportional to the intensity of the source. Lasers are inherently focused which provides the benefit of being able to work with very small or dilute samples. The caveats of using a laser are that the spectrum produced is physically limited and that care must also be taken to avoid photodegredation of your samples by attenuating the power of the laser. A description of the construction and theory behind lasers won't be discussed here, but can be found through a variety of sources Unless a laser is used, a device is needed to restrict and isolate the wavelengths that our sources provide in order to control the experiment. If a isolated wavelength is needed or isn't needed, interference or absorption filters can be used. If a spectrum of wavelengths needs to be passed through the sample, monochrometers provide the ability to separate and control the light passed through the sample while preserving the available spectrum. Intereference filters are designed to provide constructive or destructive interference of light by taking advantage of the refraction of light through different materials. As light passes from one medium to the other the direction and wavelength of light can be changed based on the index of refraction of both mediums involved and the angle of the incident and exiting light (For more look at Snell's Law). Due to this behavior, constructive and destructive interference can be controlled by varying the thickness (d) of a transparent dielectric material between two semi-reflective sheets and the angle the light is shined upon the surface. As light hits the first semi-reflective sheet, a portion is reflected, while the rest travels through the dielectric to be bent and reflected by the second semi-reflective sheet. If the conditions are correct, the reflected light and the initial incident light will be in phase and constructive interference occurs for only a particular wavelength. Another common filter is the Absorption Filter. Absorption filters work on the premise of being able to filter light by absorbing all other wavelengths that aren't of interest. They are normally constructed from a colored glass that absorbs over a wide range. Although normally cheaper than interference filters, absorbance filters tend to be less precise at filtering for a selected wavelength and also have the added penalty of absorbing some of the selected light thus lowering the intensity. Monochrometers, as previously mentioned, are used to control and separate light so that a sample can be subjected to a span of wavelengths. Light entering a monochrometer is filtered by a thin slit. The filtered light is then focused by a mirror onto a dispersing element that separates the light into its different wavelengths. The separated light is then focused again and angled toward an exit slit which then filters against all wavelengths except the desired one. The wavelength selected can then easily be changed by rotating the dispersing element to support the transmission of the new desired wavelength. Though monochrometers nearly all have the same practical design, the difference normally is determined by the type of dispersing element. One type of dispersing element is a prism which disperses light by refraction through two angled surfaces. In order to separate the light into different wavelengths, the prism needs to be made of material that has a change in the index of refraction with respect to wavelength so that each wavelength is bent at a different angle. The larger the difference in the index of refraction the better the separation is between wavelengths. For the UV-Vis range, a typical prism is cut from left handed quartz at a thirty degree angle and attached to another piece of quarts cut the same way except from right handed quartz to make a Conru Prism. The pitfall of prism based dispersing elements is the fact that the index of refraction for most materials varies nonlinearly when compared to wavelength which results in a smaller degree of separation at longer wavelengths. More commonly used due to less expensive fabrication costs and its ability to separate light in a linear fashion are grating dispersing elements. Gratings are designed to diffract light which is a separation based on the angle of the incident light to the grating normal and the spacing between groves. When beams of light impinge upon the grating some beams travel farther than others and this causes an effect akin to a interference filter which allows for constructive and destructive interference and provides the means to reflect a specific wavelength based on the angle of incidence. The most common type of grating is the echellette grating in which the grooves are angled in order to provide maximum reflection of the incident light into a single order of reflected light. Care must be taken when considering a sample container to avoid unwanted absorption in the range of interest. While glass might be perfect for the visible range, it absorbs in the UV range. Quartz can be used for both but most likely would cost considerably more. Disposable Plastic cuvets created from polystyrene or polymethyl methacrylate are in common use today as they are cheap to purchase and eliminate the need for cleaning cuvettes in order to analyze multiple samples. The shape of the sample holder is also very important as unwanted scattering of light should be minimized and pathlength through the cell should remain constant . Square (1 cm) cuvettes with frosted sides have been designed to minimize scattering and provide a surface to hold the cuvette, thereby reducing smudges or smears of the optical window all the while keeping a fixed path length. Cuvettes for fluorescent experiments cannot allow the frosted sides due to the 90° angle of most instruments and have to be carefully cleaned as bodily oils have the ability to fluoresce. Placement of the sample within either a absorption or fluorescence device is imperative and is normally fixed by a sample holder to insure reproducible results. After the light has passed through the sample, we want to be able to detect and measure the resulting light. These types of detectors come in the form of transducers that are able to take energy from light and convert it into an electrical signal that can be recorded, and if necessary, amplified.Photomultiplier tubes are a common example of a transducer that is used in a variety of devices. The idea is that when a photon hits the top of the tube electrons are released, which are pulled toward the other end of the tube by an electric field. The way the electrons are multiplied is due to the fact that along the length of the tube there are several dynodes that have a slightly less negative potential than surface before it which causes the electrons traveling down the tube to hit each surface which then in turn produces more electrons until at the very end there is a large amount of electrons (~1,000,000) representing the one photon that started the cascade. Another type of transducer is a Charge injection device (CID). This device works by having a p-type and n-type semiconductor next to each other in which the n-type material is separated from the anodes by a silica layer that acts as a capacitor. As a photon interacts with the n-type layer, an electron migrates to the p-type semiconductor and a positive charge is generated that migrates toward the silica capacitor. This process continues to happen until the potential is measured by means of comparing the more negative of the anodes to ground (V ) . The charge is then transferred to the other anode and measured (V ). The difference between V and V is directly proportional to the amount of photons that collided with the n-type layer. The positive charges are then repelled when the anodes switch momentarily to having a positive charge and then the cycle can be repeated. Thousands of these little detectors can be aligned and used to describe the light that interacts with it and can rival the performance of the photomultiplier tube. In the end of the experiment all of the data collected needs to be stored or written down. In the past printers have been used to record the transmittance as the experiment progresses. These days, computers are used to record, store, and even manipulate data along with controlling the devices within the spectrometer. This ability allowed by having a computer that can quickly integrate, compare, or annotate spectra dramatically decreases the learning curve needed to operate the instrument as well as reduces the labor associated with simple or even complex experiments. Instead of having to record the transmittance, convert it to absorbance and then subtract the background noise, the computer can accomplish all that for you in the matter of a few seconds . Now that we've covered the instrument and its components in some detail, there are some different modifications that can be made in order to tailor the device to the experiment it'll be performing. The device introduced in the beginning of this text is the basic single beam spectrometer. This device can measure the transmittance or absorbance of a particular analyte at a given wavelength provided by the source and monochrometer. Background subtraction in these machines needs to be done separately before the analyte is inserted and can be stored and subtracted by the signal processor attached. Although this is the simplest design, the cost of these types of instruments can vary greatly depending on the components that the machine is comprised of. For fluorescence spectroscopy the equivalent to a single beam spectrometer has a few slight modifications in that the detector is perpendicular to the source, and that there is an additional monochrometer that can be used to vary the wavelength detected. In this fashion, the source light is unable to interfere with the fluorescence light being measured, and the fluorescent light produced can be separated and described as the resulting wavelengths of fluorescence as a function of incident light. A double beam spectrometer takes a more analytical approach to the design. Because of fluctuations in the source intensity and inconsistencies in the transducer, the source light is split into two beams, one which travels through the sample, and another that is sent through a blank or standard solution. Both beams are then read by separate transducers and the difference between the two is recorded as the corrected transmittance. This allows for quick screening of analytes and negates the need for two separate scans to complete a background subtraction. The same idea can be applied to a fluorometer as well to obtain the same benefit The latest instrument designs use a multichannel detector, such as an array of CID's, that allow for the spectrum of an analyte to be gathered in seconds due to the fact that the light transmitted through the sample can be split and a spectrum of wavelengths can be monitored simultaniously instead of individually. Also fiber optic cables can be used to transfer the light from the source to the sample or from the sample to the dispersion grating and negate the need to consider the slit when thinking about resolution. As with any instrument, the measurements we make are limited by the tools we use. In spectrophotometers, the only measurement they are designed to monitor is the transmittance of light through the sample. Thus any noise associated with the transmittance will correlate with errors in our analyses. The analysis of the noise in spectrophotometers has been done and is outlined by Skoog and others in "Principles of Analytical Chemistry" . In the outline they summarize that the types of errors commonly experienced fall into three cases and are listed below. In the first case, the standard deviation of the transmittance is a constant. \[ S_T=k_1\] Errors associated with this situation are due to detectors with limited readout resolution and dark currant and amplifier noise. Limited readout resolution keeps the deviation constant because the measurements more precise than the readout can not be expressed or displayed by the machine. Dark Currant and Amplifier noise are only an issue when the lamp intensity and transducer sensitivity are low and random fluctuations in the currant become the dominant source of error. In the second case, the deviation in transmittance varies with the equation. \[ S_T=K_2 \sqrt{T^2+T} \] Errors associated with this equation are those related to shot noise or the transfer of electrons across any sort of junction or barrier like those found in the photomultiplier tube in the detector. Since the currant is dependant on this transfers, which happen randomly, the currant becomes a random distribution that centers around and average. This error becomes considerable when dealing with either really low or really high transmittances. The last situation deals with errors that relate proportionally to the transmittance. Errors related to this type of noise have to do with source flickering along with cell positioning. Both of these errors can be easily corrected either by attaching a constant voltage source on the lamp or by placing a fixed cuvette holde \[ S_T=k_3T\] Other big sources of error can be introduced from the entrance and exit slits not having the appropriate separation or if the experiment being conducted is outside or near the limit of the instrumentation. In older machines, a scan time that was faster than the recording printer could be a source of error as the printer would not be able to record the data as fast it was being given. | 19,890 | 1,065 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/16%3A_The_Organic_Chemistry_of_Amino_Acids_Peptides_and_Proteins/16.01%3A_Classification_and_Nomenclature_of_Amino_Acids |
Intermolecular forces are the attractive or repulsive forces between molecules. They are separated into two groups; short range and long range forces. Short range forces happen when the centers of the molecules are separated by three angstroms (10 cm) or less. Short range forces tend to be repulsive, where the long range forces that act outside the three angstroms range are attractive. Long range forces are also known as Van der Waals forces. They are responsible for surface tension, friction, viscosity and differences between actual behavior of gases and that predicted by the ideal gas law. Intermolecular forces are responsible for most properties of all the phases. The viscosity, diffusion, and surface tension are examples of physical properties of liquids that depend on intermolecular forces. Vapor pressure, critical point, and boiling point are examples of properties of gases. Melting and sublimation are examples of properties of solids that depend on intermolecular forces.
| 1,014 | 1,066 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/05%3A_The_Electronic_Structure_of_Atoms/5.08%3A_Orbitals/5.8.01%3A_Cultural_Connections-_Tones_on_a_Drum_and_Orbital_Wave_Functions |
A drummer can influence the pitch of a drum by touching its surface. A drum vibrates in a complicated way, but all the vibrations can be broken down into several of vibration (wavefunctions), like the ones we showed before and will further explain below. Earlier we suggested that two Dimensional Standing Waves on a drum might help understand the wave patterns for electrons constrained to a nucleus. We'll now show how that works. There are actual movies of the vibrations of a drum head on YouTube, but the cartoons below will make a number of modes easy to see. We'll label the first set of modes m , m , and m , or in general,m . Later we'll see that the first number corresponds to the shell number, n, or "principal quantum number" of an atom, and the second to the "angular quantum number". In this first set, all the ("el") values are 0. mode m mode m mode m Note that all of these vibrate up and down in the middle, although the last two have circular nodes where they don't vibrate at all. For m , if you start at "3:00--in the middle of the right side, and move the mouse pointer in about 7 graph markers from the right (about 1/3 to 1/2 of the way to the center), you'll find a node, where there is no vibration. If a drummer touched the drum here, it would not affect the second mode, but it would make the first and third modes impossible, because they don't have nodes here. The sound of the drum would change because two modes were eliminated. Note that the total number of nodes is n-1 (0 for m , 1 for m , and 2 for m ). In this case, all the nodes are mode m mode m mode m Here's another set of vibrational modes, m , m , m . Note that the second number, "el", has changed from 0 to 1, and it tells us the number of ( nodes in 3D). We'll see later that , corresponds to the for electrons. It may be difficult to see these node at first, but in mode m there is one linear node along a diagonal, so half the drum is displaced upwards, and half downwards. The other modes have one and two circular nodes, respectively, in addition to the diagonal linear node, because the total number of nodes must be n-1. Finally, the set of modes below has = 2, so that there are planar nodes, which are two diagonals, perpendicular to each other. Thus mode m divides the drum into four quadrants, with alternate quadrants vibrating up and down, respectively. The other two modes have additional circular nodes. mode m mode m mode m Now let's look at 3 dimensional wavefunctions that might be used to describe the electron, compared to the 3 D drum vibrations. In three dimensions, the circular nodes will become and the linear nodes will become . We also use a shorthand designation for the shapes of wavefunctions, in which letters which are used to designate the values of : So if n = 1 and = 0, it's a "1s orbital"; if n = 3 and = 2, it's a "3d orbital", and if n = 4 and = 3, it's a "4f orbital", etc. We'll start with the lowest energy wavefunction, the one with the fewest nodes. It has n = 1, = 0, so it's designated . We'll show the 2D drum vibration, the 3D orbital as a rotatable Jmol model, and a "surface plot" where the surface is drawn so that the electron would be within it (typically) 95% of the time. mode m We use the label "1 " to distinguish this pattern from other wave patterns the electron could possibly adopt if it moved about the nucleus with a higher energy. Each of these three-dimensional wave patterns is different in shape, size, or orientation from all the others and is called an . The word is used in order to make a distinction between these wave patterns and the circular or elliptical of the Bohr picture shown in The Wave Nature of the Electron. At ordinary temperatures, the electron in a hydrogen atom is almost invariably found to have the lowest energy available to it. That is, the electron occupies the 1 orbital. The electron cloud looks like the dot-density diagram shown in Figure 1 from Electron Waves in the Hydrogen Atom. This orbital is shown below as both a boundary surface diagram , a dot density diagram , and a rotatable Jmol. At a very high temperature, though, some collisions between the atoms are sufficiently hard to provide one of the electrons with enough energy so that it can occupy one of the other orbitals, say a 2 orbital, but this is unusual. Nevertheless a knowledge of these higher energy orbitals is necessary since electron clouds having the same shapes as for hydrogen are found to apply to all the other atoms in the periodic table as well. The 2 orbital is shown below, once again represented by a dot density diagram, a boundary surface diagram, and a rotatable Jmol. Notice how the dot density diagram reveals a feature about the 2 orbital that boundary surface does not: Since n = 2 and = 0, there must one "spherical" node. A node divides the 2 orbital in two, a portion of the electron cloud is near the center, while another portion lies beyond the node (the circular region with no dots). At the node the wave has no amplitude, its square is also zero, and there is zero probability of finding the electron. This additional node accounts, in part, for the higher energy of the 2s orbital. mode m In the case of a particle in a one-dimensional box or drum vibrations, the energy was determined by a positive whole number . The same situation prevails in the case of the hydrogen atom. An integer called the , also designated by the symbol , is used to label each orbital. The larger the value of , the greater the energy of the electron and the larger the average distance of the electron cloud from the nucleus. In the two orbitals already considered, = 1 for the 1 orbital, = 2 for the 2s orbital. For the second shell, n=2, the total number of nodes is 1, so may have values of 0 or 1 (since there may be 0 or 1 planar node). If = 1 we get a new set of orbitals, designate with one node. In the H atom, these orbitals have the same energy as the 2s orbitals because they have the same number of nodes, but when more electrons are added, they have different energies. Since there are three independent planes defined by 3D geometry (each slicing through the x, y or z axis), there must be three orbitals with 1 planar node. They are designated 2p , 2p , and 2p . This is analogous to the 2D drum vibrations, where there are 2 perpendicular defining the nodes. The boundary diagrams show why the mnemonic "propeller" might be appropriate. mode m nodal plane yz mode m nodal plane xz mode m nodal plane xy When equals , there are 2 nodes, and may have values of 0, 1, or 2 (since there may be 0, 1, or 2 planar nodes). The first two are familiar, the orbital and orbitals. Below are representations of the 3 orbital, and the 3 orbitals. As the 2 orbital was slightly different in shape from the 1 orbital due to the introduction of a node, so the 3 and 3 orbitals differ slightly in shape from the 2 and 2 orbitals. mode m mode m nodal plane yz In the third shell there are 2 nodes, and for the drum, they may both be , when = 2. In 3D, this gives rise to the "d orbitals", with 4-lobed "daisy" wavefunctions, each having 2 perpendicular nodes. In the case of the orbitals the subscripts are more difficult to follow. You can puzzle them out from the Jmols, the dot density diagrams and the orbital surface diagrams if you like, but analysis of these orbitals is usually considered beyond the scope of general chemistry. You should, however, be aware that there are possible orientations for orbitals. Below are representations of the orbitals. Since all have 2 nodes, they all have the same energy in the hydrogen atom, but different energies when more electrons are added. mode m nodal planes xz and yz The same pattern extends to = 4 where orbital types, namely, 4 , 4 , 4 and 4 , are found. While none of these orbitals will be shown, the patterns seen in moving from 1 to 2 or from 2 to 3 continue with the , , and orbitals. The new orbitals are even more complicated than the orbitals. For an understanding of general chemistry, it is important to know that there are different orientations for orbitals, since the number of orbitals of each type ( , etc.) is important in determining the shell structure of the atom. Orbitron: [1] D. Manthey's Orbital Viewer: [2] Falstad.com: [3] JCE: [4] Darmstadt: [5] [6] | 8,429 | 1,067 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/18%3A_Partition_Functions_and_Ideal_Gases/18.04%3A_Most_Molecules_are_in_the_Ground_Vibrational_State |
The vibrational energy levels of a diatomic are given by: \[E_v = (v +1/2) h \nu \] where is \(\nu\) the vibrational frequency and \(v\) is the vibrational quantum number. In this case, it is easy to sum the geometric series shown below: \[\begin{align*} q_\text{vib} &= \sum_{v=0}^{\infty} e^{-( v + 1/2) h\nu / k T} \\[4pt] &= e^{-h \nu/ 2kT} \left( 1 + e^{-h \nu/ 2kT} + e^{-2 h \nu/ 2kT} + ... \right) \end{align*} \nonumber \] or rewritten as: \[ q_\text{vib} = e^{-h \nu / 2kT} \left( 1 +x + x^2 +x^3 + ... \right) \label{eq0} \] where \(x = e^{-h \nu/kT} \). Given the following power series expansion: \[ \dfrac{1}{1 - x} = 1 + x + x^2 + x^3 + x^4 + .... \nonumber \] Equation \(\ref{eq0}\) can be rewritten as: \[ q_\text{vib} = e^{-h \nu / 2kT} \left( \dfrac{1}{1-x} \right) \nonumber \] or: \[q_\text{vib} = \dfrac{e^{-h \nu / 2 k T}} {1 - e^{-h \nu/ k T}} \label{eq1} \] If the zero of energy scale is at \(h \nu /2kT\), then Equation \(\ref{eq1}\) can be rewritten as: \[q_\text{vib} \approx \dfrac{1} {1 - e^{-h \nu / kT}} \label{VIBQ} \] A vibrational temperature \(Θ_\text{vib}\) may be defined as: \[ Θ_\text{vib}= \dfrac{ hc \tilde{\nu}}{k} \nonumber \] where \(\tilde{\nu}\) is the vibrational frequency in cm . \(Θ_\text{vib}\) is a good way to express the of the vibrating bond in units of the Boltzmann constant. Because the stiffness obviously depends on what bond your are talking about, this is a good way to do the same thing we did for the critical temperature of the non-ideal gases. The vibrational frequency of of \(I_2\) is \(214.57\; cm^{-1}\). Calculate the vibrational partition function of \(I_2\) at 300 K. Solution: \[\dfrac{h\nu}{kT} = \dfrac{ 214.57}{209.7} = 1.0232 \nonumber \] so \[e^{-h\nu/kT} = 0.3595 \nonumber \] and \[q_\text{vib} = \dfrac{1}{1-0.3595} = 1.561 \nonumber \nonumber \] This implies, as before, that very few vibrational states are accessible and much less than rotation states and many orders less than translation states. The vibrational energy is given by the above expression and the molar heat capacity at constant volume, \(\bar{C}_V\) is given by: \[ \bar{C}_V= \left(\dfrac{\partial E}{\partial T} \right)_V \nonumber \] We have: \[ \begin{align} \dfrac{\partial }{\partial T} &= \dfrac{\partial \beta}{\partial T} \dfrac{∂}{\partial \beta} \\[4pt] &= \dfrac{-1}{kT^2} \dfrac{\partial }{\partial \beta} \\[4pt] &= \left(-k \beta^2 \right) \left(\dfrac{\partial }{\partial \beta}\right) \label{3.54} \end{align} \] Therefore: \[\bar{C}_V = (-k \beta^2 ) \left(\dfrac{\partial ε_\text{vib}}{\partial \beta}\right) \nonumber \] and when the vibrational partition function (Equation \(\ref{VIBQ}\)) is introduced: \[\begin{align} q_\text{vib} &= -k \beta^2 \dfrac{ \left[ (1 - e^{-hc \nu / k T}) (-hc \tilde{\nu}) - e^{-hc \nu / k T} (+ hc\tilde{\nu}) \right] e^{-hc \nu / k T} }{ (1 - e^{-hc \nu / k T})^2 } hv \tilde{\nu} \\[4pt] &= k \left( \dfrac{ Θ_\text{vib} }{T} \right)^2 \dfrac{ e^{- Θ_\text{vib} /T} }{ \left( 1- e^{- Θ_\text{vib}/T} \right)^2 } \label{FinalQ} \end{align} \] For large \(T\), the \(\bar{C}_V\) becomes: \[N_A k = R \nonumber \] and for small T, \(\bar{C}_V\) goes to zero as demonstrated in Figure 18.4.1
. The vibrational heat capacity is shown as function of the reduced temperature \(T/Θ\) to get a general picture valid for all diatomic gases. Compare the parameters on Table 18.4.1
to see on what different absolute scales we have to think for different gases. Clearly the vibrational contribution to the heat capacity . For many molecules (especially light ones), the vibrational contribution only kicks in at quite high temperatures. The value of \(Θ_\text{vib}\) is determined mostly by Molecules with low \(Θ_\text{vib}\) often dissociate at lower temperatures, although the oscillator model is not sufficient to describe that phenomenon. We can calculate the fraction of molecules in each vibrational state. The fraction of molecules in the \(v\)th vibrational state is given by: \[ f_v = \frac{e^{-h\nu(v+\frac{1}{2})/kT}}{q_\text{vib}} \] Substituting in Equation \(\ref{eq1}\), we get: \[\begin{split} f_v &= \left(1-e^{-h\nu/kT}\right)e^{-h\nu v/kT} \\ &= \left(1-e^{-\Theta_\text{vib}/T}\right)e^{-\Theta_\text{vib} v/T} \end{split} \label{vib1} \] We can plot the fraction molecules in each vibrational state. From the figure, we can see that most of the Cl molecules are in the ground vibrational state at room temperature (300 K). This is true for most molecules. Only molecules with very weak bonds and low vibrational temperatures will populate a significant fraction of molecules in excited vibrational states. | 4,650 | 1,069 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Book3A_Bioinorganic_Chemistry_(Bertini_et_al.)/04%3A_Biological_and_Synthetic_Dioxygen_Carriers/4.14%3A_Requirements_for_Effective_Oxygen_Carriers |
In order for dioxygen transport to be more efficient than simple diffusion through cell membranes and fluids, it is not sufficient that a metalloprotein merely binds dioxygen. Not only is there an optimal affinity of the carrier for dioxygen, but also, and more importantly, the carrier must bind and release dioxygen at a rapid rate. These thermodynamic and kinetic aspects are illustrated in Figure 4.3, a general diagram of energy vs. reaction coordinate for the process \[M + O_{2} \xrightleftharpoons[k_{-1}]{k_{1}} MO_{2}\tag{4.2}\] where M is an oxygen carrier, for example hemocyanin or a simple nonbiological metal complex. Thermodynamic or equilibrium aspects are summarized by \(\Delta\)G in Figure 4.3. As illustrated there, \(\Delta\)G is negative, and thus the forward reaction, dioxygen binding, is spontaneous. The equilibrium constant (K) is given by \[K = \frac{a(MO_{2})}{a(M)a(O_{2})} , \tag{4.3}\] where \(a\) is the (i.e., effective concentration) of the component. The equilibrium constant is related to the change in free energy by \[\Delta G^{o} = -RT\ln K \ldotp \tag{4.4}\] The rate of the forward reaction (\(k_1\)) is related to \(\Delta G_1^*\); the rate of the reverse reaction (\(k_{-1}\)) is related to \(\Delta G^*_{-1}\). Provided that oxygen binding is effectively a single-step process, then \[K = \frac{k_{1}}{k_{-1}} \ldotp \tag{4.5}\] Usually the rates of the forward and reverse reactions are related by the empirical Arrhenius expression to quantities termed the activation energies (E and E ) of the reactions, where \[k_{1} = A_{1}^{\frac{-E_{1}}{RT}}\; and\; k_{-1} = A_{-1}^{\frac{-E_{-1}}{RT}} \ldotp \tag{4.6}\] These quantities are experimentally accessible through the change in rate as a function of temperature. It is of little benefit to the organism if its dioxygen carrier, such as hemoglobin, binds and releases O at such slow rates that O is not delivered faster than it would be by simple diffusive processes. Thus, a binding rate within a couple of orders of magnitude of the rate of diffusion, together with the high carrying capacity of O that high concentrations of oxygen carrier enable (noted ), and a pumping system ensure adequate O supplies under all but the most physiologically stressful conditions. Whereas measurements of equilibrium give little or no molecular information, rather more molecular information may be inferred from kinetic data. The processes of binding and release can be examined by a variety of techniques, with timescales down to the picosecond range. The temperature behavior of the rates gives information on the heights of energy barriers that are encountered as dioxygen molecules arrive at or depart from the binding site. The quantitative interpretation of kinetic data generally requires a molecular model of some sort. It is because of this multibarrier pathway that the equilibrium constant measured as k /k (Equation 4.5) may differ substantially from the thermodynamically measured value (Equation 4.3). The simple Adair scheme outlined above is readily adapted to cater to kinetic data. Most biological conversions involving dioxygen require enzymatic catalysis. It is reasonable then that metals found in the proteins involved in the transport and storage of O also frequently appear, with minor modification of ligands, in enzymes that incorporate oxygen from dioxygen into some substrate. Dioxygen, in this case, is not only coordinated, but also activated and made available to the substrate. In the family of proteins with heme groups, hemoglobin is a dioxygen carrier and cytochrome P-450 is an oxygenase. A similar differentiation in function is also found for hemocyanin and tyrosinase from the family of proteins with a dinuclear copper complex at the active site. Note that not all enzymes that mediate the incorporation of oxygen from O into some substrate coordinate and activate dioxygen. For example, lipoxygenase probably catalyzes the conversion of a 1,4-diene to a 1,3-diene-4-hydroxyperoxy species by activation of the organic substrate. The active site does not resemble that of any known oxygen-carrier protein. This topic is discussed more fully in . | 4,201 | 1,070 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/IV._Nucleophilic_Substitution_Reactions/C._Substitution_Reactions_Involving_Hydroxide_Ions |
This page gives you the facts and simple, uncluttered mechanisms for the nucleophilic substitution reactions between halogenoalkanes and hydroxide ions (from, for example, sodium hydroxide). If a halogenoalkane is heated under reflux with a solution of sodium or potassium hydroxide, the halogen is replaced by -OH and an alcohol is produced. Heating under reflux means heating with a condenser placed vertically in the flask to prevent loss of volatile substances from the mixture. The solvent is usually a 50/50 mixture of ethanol and water, because everything will dissolve in that. The halogenoalkane is insoluble in water. If you used water alone as the solvent, the halogenoalkane and the sodium hydroxide solution wouldn't mix and the reaction could only happen where the two layers met. For example, using 1-bromopropane as a typical primary halogenoalkane: You could write the full equation rather than the ionic one, but it slightly obscures what's going on: The bromine (or other halogen) in the halogenoalkane is simply replaced by an -OH group - hence a substitution reaction. In this example, propan-1-ol is formed. Here is the mechanism for the reaction involving bromoethane: This is an example of nucleophilic substitution. Because the mechanism involves collision between two species in the slow step (in this case, the only step) of the reaction, it is known as an S 2 reaction. If your examiners want you to show the transition state, draw the mechanism like this: The facts of the reaction are exactly the same as with primary halogenoalkanes. If the halogenoalkane is heated under reflux with a solution of sodium or potassium hydroxide in a mixture of ethanol and water, the halogen is replaced by -OH, and an alcohol is produced. For example: Or if you want the full equation rather than the ionic one: This mechanism involves an initial ionisation of the halogenoalkane: followed by a very rapid attack by the hydroxide ion on the carbocation (carbonium ion) formed: This is again an example of nucleophilic substitution. This time the slow step of the reaction only involves one species - the halogenoalkane. It is known as an S 1 reaction. The facts of the reaction are exactly the same as with primary or tertiary halogenoalkanes. The halogenoalkane is heated under reflux with a solution of sodium or potassium hydroxide in a mixture of ethanol and water. For example: Secondary halogenoalkanes use both S 2 and S 1 mechanisms. For example, the S 2 mechanism is: Should you need it, the two stages of the S 1 mechanism are: | 2,570 | 1,071 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/07%3A_Case_Studies-_Kinetics/7.03%3A_Depletion_of_the_Ozone_Layer |
The earth's stratospheric ozone layer plays a critical role in absorbing ultraviolet radiation emitted by the sun. In the last thirty years, it has been discovered that stratospheric ozone is depleting as a result of anthropogenic pollutants. There are a number of chemical reactions that can deplete stratospheric ozone; however, some of the most significant of these involves the catalytic destruction of ozone by halogen radicals such as chlorine and bromine. The atmosphere of the Earth is divided into five layers. In order of closest and thickest to farthest and thinnest the layers are listed as follows: troposphere, stratosphere, mesosphere, thermosphere and exosphere. The majority of the ozone in the atmosphere resides in the stratosphere, which extends from six miles above the Earth’s surface to 31 miles. Humans rely heavily on the absorption of ultraviolet B rays by the ozone layer because UV-B radiation causes skin cancer and can lead to genetic damage. The ozone layer has historically protected the Earth from the harmful UV rays, although in recent decades this protection has diminished due to stratospheric ozone depletion. Ozone depletion is largely a result of man-made substances. Humans have introduced gases and chemicals into the atmosphere that have rapidly depleted the ozone layer in the last century. This depletion makes humans more vulnerable to the UV-B rays which are known to cause skin cancer as well as other genetic deformities. The possibility of ozone depletion was first introduced by scientists in the late 1960's as dreams of super sonic transport began to become a reality. Scientists had long been aware that nitric oxide (NO) can catalytically react with ozone (\(O_3\)) to produce \(O_2\) molecules; however, \(NO\) molecules produced at ground level have a half life far too short to make it into the stratosphere. It was not until the advent of commercial super sonic jets (which fly in the stratosphere and at an altitude much higher then conventional jets) that the potential for \(NO\) to react with stratospheric ozone became a possibility. The threat of ozone depletion from commercial super sonic transport was so great that it is often cited as the main reason why the US federal government pulled support for its development in 1971. Fear of ozone depletion was abated until 1974 when Sherwood Rowland and Mario Molina discovered that chlorofluorocarbons could be photolyzed by high energy photons in the stratosphere. They discovered that this process could releasing chlorine radicals that would catalytically react with \(O_3\) and destroy the molecule. This process is called the Rowland-Molina theory of \(O_3\) depletion. The stratosphere is in a constant cycle with oxygen molecules and their interaction with ultraviolet rays. This process is considered a cycle because of its constant conversion between different molecules of oxygen. The ozone layer is created when ultraviolet rays react with oxygen molecules (O ) to create ozone (O ) and atomic oxygen (O). This process is called the . Step 1: An oxygen molecules is photolyzed by solar radiation, creating two oxygen radicals: \[ h\nu + O_2 \rightarrow 2O^. \nonumber \] Step 2: Oxygen radicals then react with molecular oxygen to produce ozone: \[O_2 + O^. \rightarrow O_3 \nonumber \] Step 3: Ozone then reacts with an additional oxygen radical to form molecular oxygen: \[O_3 + O^. \rightarrow 2O_2 \nonumber \] Step 4: Ozone can also be recycled into molecular oxygen by reacting with a photon: \[O_3 + h\nu \rightarrow O_2 + O^. \nonumber \] It is important to keep in mind that ozone is constantly being created and destroyed by the Chapman cycle and that these reactions are natural processes, which have been taking place for millions of years. Because of this, the thickness the ozone layer at any particular time can vary greatly. It is also important to know that O is constantly being introduced into the atmosphere through photosynthesis, so the ozone layer has the capability of regenerating itself. CFC molecules are made up of chlorine, fluorine and carbon atoms and are extremely stable. This extreme stability allows CFC's to slowly make their way into the stratosphere (most molecules decompose before they can cross into the stratosphere from the troposphere). This prolonged life in the atmosphere allows them to reach great altitudes where photons are more energetic. When the CFC's come into contact with these high energy photons, their individual components are freed from the whole. The following reaction displays how Cl atoms have an ozone destroying cycle: \[Cl + O_3 \rightarrow ClO + O_2 \tag{step 1} \] \[ClO + O^. \rightarrow Cl + O_2 \tag{step 2} \] \[O_3 + O^. \rightarrow 2O_2 \tag{Overall reaction} \] Chlorine is able to destroy so much of the ozone because it acts as a catalyst. Chlorine initiates the breakdown of ozone and combines with a freed oxygen to create two oxygen molecules. After each reaction, chlorine begins the destructive cycle again with another ozone molecule. One chlorine atom can thereby destroy thousands of ozone molecules. Because ozone molecules are being broken down they are unable to absorb any ultraviolet light so we experience more intense UV radiation at the earths surface. From 1985 to 1988, researchers studying atmospheric properties over the south pole continually noticed significantly reduced concentrations of ozone directly over the continent of Antarctica. For three years it was assumed that the ozone data was incorrect and was due to some type of instrument malfunction. In 1988, researchers finally realized their error and concluded that an enormous hole in the ozone layer had indeed developed over Antarctica. Examination of NASA satellite data later showed that the hole had begun to develop in the mid 1970's. The ozone hole over Antarctica is formed by a slew of unique atmospheric conditions over the continent that combine to create an ideal environment for ozone destruction. As spring comes to Antarctica, the PSC's melt in the stratosphere and release all of the halogenated compounds that were previously absorbed to the cloud. In the antarctic summer, high energy photons are able to photolyze the halogenated compounds, freeing halogen radicals that then catalytically destroy O . Because Antarctica is constantly surrounded by a polar vortex, radical halogens are not able to be diluted over the entire globe. The ozone hole develops as result of this process. Resent research suggests that the strength of the polar vortex from any given year is directly correlated to the size of the ozone hole. In years with a strong polar vortex, the ozone hole is seen to expand in diameter, whereas in years with a weaker polar vortex, the ozone hole is noted to shrink The following substances are listed as ozone depleting substances under Title VI of the United State Clean Air Act: | 6,928 | 1,072 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Atomic_Radii |
Atomic radii is useful for determining many aspects of chemistry such as various physical and chemical properties. The greatly assists in determining atomic radius and presents a number of trends. Atomic radius is generally stated as being the total distance from an atom’s nucleus to the outermost orbital of electron. In simpler terms, it can be defined as something similar to the radius of a circle, where the center of the circle is the nucleus and the outer edge of the circle is the outermost orbital of electron. As you begin to move across or down the periodic table, trends emerge that help explain how atomic radii change. The effective nuclear charge (\(Z_{eff}\)) of an atom is the net positive charge felt by the valence electron. Some positive charge is shielded by the core electrons therefore the total positive charge is not felt by the valence electron. A detailed description of shielding and effective nuclear charge can be found . \(Z_{eff}\) greatly affects the atomic size of an atom. So as the \(Z_{eff}\) decreases, the atomic radius will grow as a result because there is more screening of the electrons from the nucleus, which decreases the attraction between the nucleus and the electron. Since \(Z_{eff}\)decreases going down a group and right to left across the periodic table, the atomic radius will going down a group and right to left across the periodic table. Determining the atomic radii is rather difficult because there is an uncertainty in the position of the outermost electron – we do not know exactly where the electron is. This phenomenon can be explained by the . To get a precise measurement of the radius, but still not an entirely correct measurement, we determine the radius based on the distance between the nuclei of two bonded atoms. The radii of atoms are therefore determined by the bonds they form. An atom will have different radii depending on the bond it forms; so there is no fixed radius of an atom. When a is present between two atoms, the covalent radius can be determined. When two atoms of the same element are covalently bonded, the radius of each atom will be half the distance between the two nuclei because they equally attract the electrons. The distance between two nuclei will give the diameter of an atom, but you want the radius which is half the diameter. Covalent radii will increase in the same pattern as atomic radii. The reason for this trend is that the bigger the radii, the further the distance between the two nuclei. See explanation for \(Z_{eff}\) for more details. The covalent radius depicted below in Figure 1 will be the same for both atoms because they are of the same element as shown by X. The ionic radius is the radius of an atom forming or an ion. The radius of each atom in an ionic bond will be different than that in a covalent bond. This is an important concept. The reason for the variability in radius is due to the fact that the atoms in an ionic bond are of greatly different size. One of the atoms is a cation, which is smaller in size, and the other atom is an anion which is a lot larger in size. So in order to account for this difference, one most get the total distance between the two nuclei and divide the distance according to atomic size. The bigger the atomic size, the larger radius it will have. This is depicted in Figure 2 as shown below where the cation is displayed on the left as X+, and clearly has a smaller radius than the anion, which is depicted as Y- on the right. If we were able to determine the atomic radius of an atom from experimentation, say Se, which had an atomic radius of 178 pm, then we could determine the atomic radius of any other atom bonded to Se by subtracting the size of the atomic radius of Se from the total distance between the two nuclei. So, if we had the compound CaSe, which had a total distance of 278 pm between the nucleus of the Ca atom and Se atom, then the atomic radius of the Ca atom will be 278 pm (total distance) - 178 pm (distance of Se), or 100 pm. This process can be applied to other examples of ionic radius. Cations have ionic radii than their neutral atoms. In contrast, anions have ionic radii than their corresponding neutral atoms. A detailed explanation is given below: Figure 3 below depicts this process. A neutral atom X is shown here to have a bond length of 180 pm and then the cation X is smaller with a bond length of 100 pm. An , on the other hand, will be bigger in size than that of the atom it was made from because of a gain of an electron. This can be seen in the Figure 4 below. The gain of an electron adds more electrons to the outermost shell which the radius because there are now more electrons further away from the nucleus and there are more electrons to pull towards the nucleus so the pull becomes slightly weaker than of the neutral atom and causes an in atomic radius. The metallic radius is the radius of an atom joined by metallic bond. The metallic radius is half of the total distance between the nuclei of two adjacent atoms in a metallic cluster. Since a metal will be a group of atoms of the same element, the distance of each atom will be the same (Figure 5). The radius of atoms increases as you go down a certain group. The size of an atom will decrease as you move from left to the right of a period. Because the electrons added in the transition elements are added in the inner electron shell and at the same time, the outer shell remains constant, the nucleus attracts the electrons inward. The electron configuration of the transition metals explains this phenomenon. This is why Ga is the same size as its preceding atom and why Sb is slightly bigger than Sn. When solving a radius-bond problem, identify the bond first and then use the standard method of finding the radius for that particular bond. Also remember the trend for the atomic radii. | 5,898 | 1,073 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/15%3A_AcidBase_Equilibria/15.E%3A_Acid-Base_Equilibria_(Exercises) |
.
Vinegar contains acetic acid \(\mathrm{CH_3COOH}\). What species serves as a base when vinegar is mixed with baking soda, sodium bicarbonate, during the preparation of bread? \[NaHCO_3 + CH_3COOH \rightarrow NaCH_3COO + CO_2 + H_2O\nonumber \] \(\ce{NaHCO_3}\) serves as the base. \[\ce{CaO + CO2 \rightleftharpoons CaCO3 }\nonumber \] Al(III) oxide is amphorteric. What is the balanced chemical equation of Al(III) oxide react with aqueous \(\ce{H_2SO_4}\)? What is the balanced equation of it reacts with \(\ce{KOH}\)? \[\ce{Al2O3(s) + 3H2SO4(aq) \rightleftharpoons Al2(SO4)3(aq) + 3H2O(l)} \nonumber\] \[\ce{Al2O3 + 2KOH + 3H2O \rightleftharpoons 2KAl(OH)4 \nonumber \] The \(\ce{[H_3O^+]}\) concentration in a glass of orange juice is 3.96 x 10 M. What is the juice's pH? \[ pH = -\log[H_3O+] \nonumber \] Since the concentration of hydronium ions is given, the pH calculation is as follows: \[ pH = -\log[3.96 \times 10^{-5}] = 4.4023\nonumber \] pH is a measure of hydrogen ions in a solution. This concentration defines the acidity or alkalinity of a solution. The pK of an unknown salty water at 25 °C is 13.665. This differs from the usual K of 14.00 at this temperature because dissolved salts make this unknown salty water a non- ideal solution. If the pH in the salty water is 7.8, what are the concentrations of H O and OH in the salty water at 25 °C? \(\mathrm{pH = -log[H_3O^+]}\), hence, \(\mathrm{[H_3O^+] = 10^{-7.8} = 1.5849 \times 10^-8 \: M}\) Since \(\mathrm{pK_w = [OH^-,H_3O^+]}\), \(\mathrm{[OH^-] = \frac{10^{-13.665}}{1.5849 \times 10^{-8} \: M} = 1.3645 \times 10^{-6} M }\) When rubidium (Rb) solid is added to water, there is an instantaneous and vigorous reaction (i.e., an explosion) as this video demonstrates. Based on this information, which of these two equations is a more accurate representation the reaction? \[\ce{2Rb(s) + 2H2O(l) \rightarrow 2RbOH(aq) + H2 (aq)}\nonumber \] Acetic acid gives vinegar a sour taste and strong aroma. Its \(K_a\) value is 1.75 x 10 . What is the pH of the solution if 0.59 grams of acetic acid is dissolved in 40 mL of water? First, convert grams of acetic acid to moles. \[(0.59\; g\; CH_{3}COOH) \left(\dfrac{1\: mol}{60.05\: g\; CH_{3}COOH} \right) = 0.0098\; mol\nonumber \] Then, find the molarity of acetic acid by dividing the number of moles of acetic acid by the number of liters of water. \[ \dfrac{0.0098\: mol}{0.04\: {L\; water}} = 0.246\; M\nonumber \] Using the molarity and K , construct and solve an ICE table to find out how much the acetic acid dissociates. \(\ce{CH_3COOH_{(aq)} + H_2O_{(l)} \rightleftharpoons CH_3COO^{-}_{(aq)} + H_3O+_{(aq)}}\) The acid dissociation constant works in the below equation: \[ K_a = \dfrac{[H_3O^+,CH_3COO^-]}{[CH_3COOH]} \nonumber \] Plug in the final concentration values from the ICE table and solve for x. \[ 1.75 \times 10^{-5} = \dfrac{x^2}{0.246-x} \nonumber \] \[ x = 0.0021 \nonumber \] Use the calculated value of x to calculate the concentration of H O+, which in this case is equal to x. Then, plug this concentration into the equation for pH. \[ pH = -log[0.0021] \nonumber \] \[ pH = 2.6848 \nonumber \] Construct an ICE table based on the equation: \[\mathrm{HCOOH + H_2O \rightleftharpoons HCO_2^- + H_3O^+} \nonumber \] Ka can then be equated to an algebraic expression: \[\mathrm{Ka= \dfrac{x^2 }{0.6-x}}\nonumber \] \[\mathrm{1.8 \times 10^{-4}= \dfrac{x^2 }{0.6-x}}\nonumber \] \[\mathrm{x=0.0103026}\nonumber \] or \[\mathrm{x=-0.0104}\nonumber \] x must be postive. Thus, x = 0.0103026. \[\mathrm{[H_3O^+]}\nonumber \] \[\mathrm{pH = -log(0.0103026) = 1.987}\nonumber \] \[\mathrm{-log[H_3O^+] = 1.987}\nonumber \] \[\mathrm{[H_3O^+] = 10^{-1.987}}\nonumber \] \[\mathrm{[H_3O^+] = 0.0103026}\nonumber \] An ICE table can also be constructed for the reaction: \[CCl_3CO_2H + H_2O \rightleftharpoons CCl_3CO_2^- + H_3O^+\nonumber \] \[\mathrm{K_a = \dfrac{(0.010303)^2}{(x-0.010303)}}\nonumber \] \[\mathrm{2.2 \times 10^{-1} = \dfrac{(0.010303)^2}{(x-0.010303)}}\nonumber \] \[\mathrm{x = 0.0107855 M}\nonumber \] To calculate the mass of trichloroacetic acid, we can calculate the molarity of trichloroacetic acid by its molar mass. \[\mathrm{Mass \: of \: trichloroacetic \: acid= 0.0107855 \: M \times (35.5 \dfrac{g}{mol} \times 3 + 12 \dfrac{g}{mol} \times 2 + 16 \dfrac{g}{mol} \times 2 + 1 \dfrac{g}{mol})= 1.76 \: g}\nonumber \] Rank each of the 0.2 M solution below in an order of increasing pH: \(NH_4I\), \(KF\), \(HCl\), \(KCl\), \(KOH\). \[HCl< NH_{4}I<KF<KCl<KOH\nonumber \] 0.040 mol of pK =11.09 Because equivalence point is not reached yet, we can employ . \[\begin{align*} pOH &\approx pK_b + \log \frac{[BH^+]}{[B]} \\[5pt] pOH &\approx (11.09) + \log \frac{[C_4H_{12}N^+]}{[C_4H_{11}N]} \\[5pt] pOH &\approx (11.09) + \log \frac{0.015}{0.040-15} \\[5pt] pOH &\approx 10.87 \end{align*}\] At room temperature \(K_w = 14\) and \[\begin{align*} pH &= pK_w - pOH \\[5pt] &= 14.00 - 10.87 \\[5pt] &= 3.13 \end{align*}\] Prepare a Hypochlorous acid/Hypochlorite buffer at pH 7. Using the : \[\mathrm{pH \approx pK_a + \log{\dfrac{[ClO^-]_o}{[HClO]_o}}}\nonumber \] It's easy to re-arrange the equation to solve for \(\mathrm{[HClO]_\circ}\). Multiplying by 100 mL yields the moles of acid added. \[\mathrm{pH \approx pK_a + \log{\dfrac{[ClO^-]_o}{[HClO]_o}}}\nonumber \] \[\mathrm{7 \approx pK_a + \log{\dfrac{[ClO^-]_o}{[HClO]_o}}}\nonumber \] \[\mathrm{7 \approx 7.53 + \log{\dfrac{5}{[HClO]_o}}}\nonumber \] \[\mathrm{[HClO]_\circ \approx 16.94}\nonumber \] \[\mathrm{Moles \: HClO \approx 1.694}\nonumber \] A student is given 500 mL of a 0.500 M acetic acid solution and wants to create a pH 5.0 buffer. How many mL of 1 M \(\ce{NaOH}\) must be added to the original solution? Acetic acid has a pK of 4.756. We first use the to determine the required ratio of base to acid in the solution: \[ \begin{align*} pH&=pK_a+\log\left(\dfrac{[base]}{[acid]}\right) \\[5pt] 5.00 &=4.756+\log\left(\dfrac{[base]}{[acid]}\right) \\[5pt] 0.244 &=\log\left(\dfrac{[base]}{[acid]}\right) \\[5pt] 1.75 &=\dfrac{[base]}{[acid]} \end{align*}\nonumber \] Next, we will determine the molar amount of base required to get a pH of 5.00. In order to simplify things we will first solve for the molar quantities as of each using the simple ratio above in relation to 1.75: \[\begin{align*} \text{Acetic Acid in 500 mL solution} &= {(0.5 \; M) \times (0.5 \; L)} \\[5pt] &= 0.25 \; mol \end{align*}\nonumber \] \[NaOH\; in\; 500\; mL\; solution={(0.25 \; mol) \times (1.75)} +0.25=0.6875+ \; mol \nonumber \] Now we'll determine the volume of 1M NaOH needed to raise the pH: \[\left(\dfrac{1 \; mol}{1 \; L}\right)=\left(\dfrac{0.6875 \; mol}{x}\right)\nonumber \] \[s=0.6875\; L\; NaOH\;\nonumber \] Now we'll check to make sure everything is right: \[Acetic\; Acid\;=\left(\dfrac{0.25}{0.9375}\right)=0.2667M\nonumber \] \[NaOH\;=\left(\dfrac{0.6875 - 0.25}{0.9375}\right)=0.4667M\nonumber \] \[5.00=4.756+\log\left(\dfrac{[base]}{[acid]}\right)\nonumber \] \[pH=4.756+\log\left(\dfrac{[0.4667]}{[0.2667]}\right)\nonumber \] \[pH=4.756+0.243=4.999\nonumber \] We get a pH of 4.999 which is about 5.00 and isn't exactly 5.00 due to rounding early in the problem, so the problem was done correctly. 0.15M of HBr is added into 50mL of 0.1M Ca(OH) for the titration. \[2HBr+Ca(OH)_{2}\rightleftharpoons 2 H_{2}O+CaBr_{2}\] a) \(Ca(OH)_2\) is strong base. They dissociate completely. \[\mathrm{[OH^{-}]=2[Ca(OH)_{2}]=0.2M}\nonumber \] \[\mathrm{pH = pK_w - pOH}\nonumber \] \[\mathrm{pH = 14 + log(0.2 \; M)}\nonumber \] \[\mathrm{pH = 13.3}\nonumber \] b) Because Ca(OH) and HBr are both strong base and strong acid, at equilibrium, the pH is 7.00. \[\mathrm{mole_{Ca(OH)_{2}}=(Volume)(Molarity)}\] \[\mathrm{mole_{Ca(OH)_{2}}=(0.05L)(0.1M)}\] \[\mathrm{mole_{Ca(OH)_{2}}=0.005\ mol}\] @ equilibrium; \[\mathrm{mole_{HBr}=2\ mole_{Ca(OH)_{2}}}\] \[\mathrm{mole_{HBr}=2(0.005\ mol)=0.01\ mol}\] \[\mathrm{Volume_{HBr} = \frac{mole_{HBr}}{molarity_{HBr}}=\frac{0.01\ mol}{0.15M}=0.0667\ L=66.7\ mL}\] \[\mathrm{Total\ Volume\ at\ equilibrium=66.7mL + 50mL=116.7mL}\] \[\mathrm{Total\ Volume\ 1mL\ short\ of\ equilibrium=116.7mL-1mL=115.7mL}\] \[\mathrm{Volume_{HBr}\ 1mL\ short\ of\ equilibrium=66.7mL-1mL=65.7\ mL}\] \[\mathrm{mols \; of \; OH^- \; not \; neutralized \; by \; HBr = 0.01 \; mol - 0.0657 \; L \times (0.15 \;M) =1.45 \times 10^{-4} \; mol }\] \[\mathrm{pH = 14 + log(\dfrac{1.45^{-4} \; mol}{0.1157 \; L}) = 11.1}\] c) At 1mL after equilibrium, Ca(OH) has been neutralized by HBr. Only HBr exists in the solution. \[\mathrm{ mole_{HBr}=(0.001L)(0.15M)=1.5\times 10^{-4}\ mol}\] \[\mathrm{Volume=Total\ Volume+1mL=117.7mL=0.1177\ L}\] \[\mathrm{M_{HBr}\ in\ the\ solution=\frac{1.5\times 10^{-4}\ mol}{0.1177\ L}=1.274\times 10^{-3}M}\] \[\mathrm{M_{HBr}\ in\ the\ solution=M_{H^{+}}\ in\ the\ solution}\] \[\mathrm{pH=-log[H^{+}]=-log(1.274\times 10^{-3}M)=2.89}\] K at 25°C for Diethylamine (\(\ce{(C_2H_5)_2NH}\)) is \(1.3 \times 10^{-3}\). Consider the titration of 50.00 mL of a 0.1000 M solution of Diethylamine with 0.100 M \(\ce{HCl}\) added with the following volumes: 0, 10.00, 50.00 mL. Calculate pH for each solutions. At an unknown volume beyond 50.00 mL, the pH is 3.90. Find the corresponding amount volume of \(\ce{HCl}\) needed to obtain that pH. \[DiEtNH_{(aq)}+H_{2}O_{(l)}\rightleftharpoons DiEtNH_{2(aq)}^{+}+OH_{aq}^{-}\] \[\frac{[DiEtNH_{2}^{+},OH^{-}]}{[DiEtNH]}= 1.3\times 10^{-3}\] \[ [OH^-] = [DiEtNH_2^+] = y\nonumber \] \[ [DiEtNH] = 0.1000 - y \nonumber \] \[\frac{y^{2}}{0.1000-y}= 1.3\times 10^{-3}\] \[\mathrm{y = 0.01077M= [OH^-] }\nonumber \] \[\mathrm{pOH = 1.97}\nonumber \] \[\mathrm{pH = 12.03}\nonumber \] \[\mathrm{[DiEtNH_{2}^{+}]=\frac{(0.1000M)(0.01L)}{(0.050+0.010)L}= 0.0167 \; M}\] \[\mathrm{[DiEtNH]=\frac{(0.1000M(0.050L)-(0.1000M)(0.01L)}{(0.050+0.010)L}= 0.0667 \; M}\] Plug it back to Henerson Hasselbalch equation \[\mathrm{pOH = pK_b + \log(\dfrac{[BH^+]}{[B]})}\] \[\mathrm{pOH = -\log(1.3 \times 10^{-3}) +\log(\dfrac{[0.0167 \; M]}{[0.0667 \; M]})}$$ \[\mathrm{pH = pH - pOH = 14.00 - 2.28 = 11.72}\] The titration is at the equivalence point. At equivalence, the reaction consists of 100 mL of 0.050 mol \(\ce{DiEtNH_2^+}\) \[\mathrm{DiEtNH_{2(aq)}^{+}+H_{2}O_{(l)}\rightleftharpoons DiEtNH_{(aq)}+H_{3}O_{(aq)}^{+}}\] \[\mathrm{K_a= \dfrac{1.00 \times 10^{-14}}{1.3 \times 10^{-3}} = 7.69 \times 10^{-12}}\nonumber \] \[\mathrm{K_{a}= 7.69\times 10^{-12}=\frac{x^{2}}{(0.5000-x)};x=[H_{3}O^{+}]}\] \[\mathrm{x = 1.96 \times 10^{-6}; pH = 5.71}\nonumber \] Beyond the equivalence point, solution behaves like \(HCl\). Given pH = 3.90. \[\mathrm{10^{-3.90}=\frac{(z-0.050 \; L)}{0.100 \; L+z} \times 0.1000 \; M}\] \(\mathrm{z= 0.0502 \: L}\) Sodium Bicarbonate (\(\ce{NaHCO_3}\)) is a very weak base when dissolved in water. Some amount of sodium bicarbonate is dissolved in 125 mL of a 0.25 M solution of HNO The 168 mL of 0.15 M NaOH was used to titrate the solution. How many grams of sodium bicarbonate were added? We are titrating an acid with two bases so solve for the amount of acid the \(\ce{NaOH}\) neutralizes and the remaining moles of acid will be the number of moles of sodium bicarbonate. \[\mathrm{Moles \: HNO_3= \frac{0.25 \; moles}{1 \: L} \times 0.125 \: L = 0.03125 \: moles}\nonumber \] \[\mathrm{Moles \: NaOH= \frac{0.15 \: moles}{1 \: L} \times 0.168 \: L =0.0252 \: moles} \nonumber \] \[\mathrm{0.03125-0.0252= 0.00605 \; moles \; sodium \; bicarbonate}\nonumber \] Now just multiply by sodium bicarbonate's molar mass (84.007 \(\frac{g}{mol}\) ) to find the mass of sodium bicarbonate added \[\mathrm{0.00605 \times 84.007 = 0.51 g}\nonumber \] What is the molarity of a \(\ce{HNO3}\) water solution if it requires 31.80 mL of such solution to titrate 0.0662 g Aniline in 100 mL aqueous solution to equivalence point? What will the pH value be at the equivalence point if \(\mathrm{K_b(Aniline) = 3.8 \times 10^{-10}}\)? Aniline and \(\ce{HNO3}\) react with a one-to-one stoichiometry \[\ce{C6H5NH2(aq) + HNO3(aq) \rightleftharpoons C6H5NH3^{+}(aq) + NO3^{-} (aq) } \nonumber \] Therefore \[M(HNO_{3})=\dfrac{0.0662 \; g}{93.13 \; g/mol \;} \times \dfrac{1}{0.03180 \; L}=0.0224 \; M\nonumber \] Suppose at the equivalence point all Aniline is converted to its conjugate acid, then its concentration equals \[[\ce{C6H5NH3^{+}}]= \dfrac {0.0662\,g} {(93.13\,g/mol ) \; 0.1318\,L}=0.00539\, M \nonumber \] Also as some Aniline's conjugate acid reacts with water, \[K_a=\dfrac{K_w}{K_b}=\dfrac{1.0 \times 10^{-14}}{3.8\times 10^{-10}}=2.63 \times 10^{-5}=\dfrac{[H_3O^+,C_6H_5NH_2]}{[C_6H_5NH_3^+]} = \dfrac{x^2}{0.00539-x}\nonumber \] Therefore, \[x =[H_{3}O^{+}]=3.636\times 10^{-4}M\nonumber \] so \(\mathrm{pH=3.44}\). | 12,812 | 1,074 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/05%3A_Introduction_To_Reactions_In_Aqueous_Solutions/5.7%3A_Stoichiometry_of_Reactions_in_Aqueous_Solutions%3A_Titrations |
The Learning Objective of this Module is to use titration methods to analyze solutions quantitatively. To determine the amounts or concentrations of substances present in a sample, chemists use a combination of chemical reactions and stoichiometric calculations in a methodology called . Suppose, for example, we know the identity of a certain compound in a solution but not its concentration. If the compound reacts rapidly and completely with another reactant, we may be able to use the reaction to determine the concentration of the compound of interest. In a , a carefully measured volume of a solution of known concentration, called the , is added to a measured volume of a solution containing a compound whose concentration is to be determined (the ). The reaction used in a titration can be an acid–base reaction, a precipitation reaction, or an oxidation–reduction reaction. In all cases, the reaction chosen for the analysis must be fast, complete, and ; that is, the compound of interest should react with the titrant. The is reached when a stoichiometric amount of the titrant has been added—the amount required to react completely with the unknown. The chemical nature of the species present in the unknown dictates which type of reaction is most appropriate and also how to determine the equivalence point. The volume of titrant added, its concentration, and the coefficients from the balanced chemical equation for the reaction allow us to calculate the total number of moles of the unknown in the original solution. Because we have measured the volume of the solution that contains the unknown, we can calculate the molarity of the unknown substance. This procedure is summarized graphically here: The calcium salt of oxalic acid [Ca(O CCO )] is found in the sap and leaves of some vegetables, including spinach and rhubarb, and in many ornamental plants. Because oxalic acid and its salts are toxic, when a food such as rhubarb is processed commercially, the leaves must be removed, and the oxalate content carefully monitored. The reaction of MnO with oxalic acid (HO CCO H) in acidic aqueous solution produces Mn and CO : \( \begin{pmatrix} MnO_{4}^{-}\left ( aq \right )+HO_{2}CCO_{2}H\left ( aq \right ) &\rightarrow Mn^{2+}\left ( aq \right )&+CO_{2}\left ( g \right )+H_{2}O\left ( l \right ) \\ purple & colorless & \end{pmatrix} \) Because this reaction is rapid and goes to completion, potassium permanganate (KMnO ) is widely used as a reactant for determining the concentration of oxalic acid. The following video (chemistry Nakajima) demonstrates the reaction Figure 8.9.1 Suppose you stirred a 10.0 g sample of canned rhubarb with enough dilute H SO (aq) to obtain 127 mL of colorless solution. Because the added permanganate is rapidly consumed, adding small volumes of a 0.0247 M KMnO solution, which has a deep purple color, to the rhubarb extract does not initially change the color of the extract. When 15.4 mL of the permanganate solution have been added, however, the solution becomes a faint purple due to the presence of a slight excess of permanganate ( ). If we assume that oxalic acid is the only species in solution that reacts with permanganate, what percentage of the mass of the original sample was calcium oxalate? The video below demonstrates the titration when small, measured amounts of a known permaganate solution are added. At the endpoint, the number of moles of permaganage added equals the number of moles of oxalate in the solution, thus determining how many moles of oxalate we started with Titration of oxalate using a known permaganate solution equation, mass of sample, volume of solution, and molarity and volume of titrant mass percentage of unknown in sample As in all other problems of this type, the first requirement is a balanced chemical equation for the reaction. Using oxidation states gives \[2MnO_4^-(aq) + 5HO_2CCO_2H(aq) + 6H^+(aq) \rightarrow 2Mn^{2+}(aq) + 10CO_2(g) + 8H_2O(l)\] Thus each mole of MnO added consumes 2.5 mol of oxalic acid. Because we know the concentration of permanganate (0.0247 M) and the volume of permanganate solution that was needed to consume all the oxalic acid (15.4 mL), we can calculate the number of moles of MnO consumed. To do this we first convert the volume in mL to a volume in liters. Then simply multiplying the molarity of the solution by the volume in liters we find the number of moles of \[ 15.4\; \cancel{mL}\left ( \dfrac{1 \: \cancel{L}}{1000\; \cancel{mL}} \right )\left ( \dfrac{0.0247\; mol\; MnO_{4}^{-}}{1\; \cancel{L}} \right )=3.80\times 10^{4\;} mol\; MnO_{4}^{-} \] The number of moles of oxalic acid, and thus oxalate, present can be calculated from the mole ratio of the reactants in the balanced chemical equation. We can abbreviate the table needed to calculate the number of moles of oxalic acid in the \[\begin{align} moles\: HO_2 CCO_2 H & = 3 .80 \times 10^{-4}\: \cancel{mol\: MnO_4^-} \left( \dfrac{5\: mol\: HO_2 CCO_2 H} {2\:\cancel{mol\: MnO_4^-}} \right) \\ &= 9 .50 \times 10^{-4}\: mol\: HO_2 CCO_2 H \end{align}\] The problem asks for the of calcium oxalate by mass in the original 10.0 g sample of rhubarb, so we need to know the mass of calcium oxalate that produced 9.50 × 10 mol of oxalic acid. Because calcium oxalate is Ca(O CCO ), 1 mol of calcium oxalate gave 1 mol of oxalic acid in the initial acid extraction: \[Ca(O_2CCO_2)(s) + 2H^+(aq) \rightarrow Ca^{2+}(aq) + HO_2CCO_2H(aq) \] The mass of calcium oxalate originally present was thus \(\begin{align} mass\: of\: CaC_2 O_4 &= 9 .50 \times 10^{-4}\: \cancel{mol\: HO_2 CCO_2 H} \left( \dfrac{1\: \cancel{mol\: CaC_2 O_4}} {1\: \cancel{mol\: HO_2 CCO_2 H}} \right) \left( \dfrac{128 .10\: g\: CaC_2 O_4} {1\: \cancel{mol\: CaC_2 O_4}} \right) \\ &= 0 .122\: g\: CaC_2 O_4 \end{align}\) The original sample contained 0.122 g of calcium oxalate per 10.0 g of rhubarb. The percentage of calcium oxalate by mass was thus \[ \% CaC_2 O_4 = \dfrac{0 .122\: g} {10 .0\: g} \times 100 = 1 .22\% \] Because the problem asked for the percentage by mass of calcium oxalate in the original sample rather than for the of oxalic acid in the extract, we do not need to know the volume of the oxalic acid solution for this calculation. Glutathione is a low-molecular-weight compound found in living cells that is produced naturally by the liver. Health-care providers give glutathione intravenously to prevent side effects of chemotherapy and to prevent kidney problems after heart bypass surgery. Its structure is as follows: Glutathione is found in two forms: one abbreviated as (left) GSH (indicating the presence of an –SH group) and the other (right) as GSSG (the form, in which an S–S bond links two glutathione units). The GSH form is easily oxidized to GSSG with elemental iodine: \[2GSH(aq) + I_2(aq) \rightarrow GSSG(aq) + 2HI(aq) \] A small amount of soluble starch is added as an indicator. Because starch reacts with excess I to give an intense blue color, the appearance of a blue color indicates that the equivalence point of the reaction has been reached. Adding small volumes of a 0.0031 M aqueous solution of I to 194 mL of a solution that contains glutathione and a trace of soluble starch initially causes no change. After 16.3 mL of iodine solution have been added, however, a permanent pale blue color appears because of the formation of the starch-iodine complex. What is the concentration of glutathione in the original solution? 5.2 × 10 M In Example 20, the concentration of the titrant (I ) was accurately known. The accuracy of any titration analysis depends on an accurate knowledge of the concentration of the titrant. Most titrants are first ; that is, their concentration is measured by titration with a , which is a solution whose concentration is known precisely. Only pure crystalline compounds that do not react with water or carbon dioxide are suitable for use in preparing a standard solution. One such compound is potassium hydrogen phthalate (KHP), a weak monoprotic acid suitable for standardizing solutions of bases such as sodium hydroxide. The reaction of KHP with NaOH is a simple acid–base reaction. If the concentration of the KHP solution is known accurately and the titration of a NaOH solution with the KHP solution is carried out carefully, then the concentration of the NaOH solution can be calculated precisely. The standardized NaOH solution can then be used to titrate a solution of an acid whose concentration is unknown. As with all acid-base reactions, a salt is formed. Because most common acids and bases are not intensely colored, a small amount of an acid–base is usually added to detect the equivalence point in an acid–base titration. The point in the titration at which an indicator changes color is called the . The procedure is illustrated in Example 21. The structure of vitamin C (ascorbic acid, a monoprotic acid) is as follows: The upper figure shows the three-dimensional representation of ascorbic acid. Hatched lines indicate bonds that are behind the plane of the paper, and wedged lines indicate bonds that are out of the plane of the paper. An absence of vitamin C in the diet leads to the disease known as scurvy, a breakdown of connective tissue throughout the body and of dentin in the teeth. Because fresh fruits and vegetables rich in vitamin C are readily available in developed countries today, scurvy is not a major problem. In the days of slow voyages in wooden ships, however, scurvy was common. Ferdinand Magellan, the first person to sail around the world, lost more than 90% of his crew, many to scurvy. Although a diet rich in fruits and vegetables contains more than enough vitamin C to prevent scurvy, many people take supplemental doses of vitamin C, hoping that the extra amounts will help prevent colds and other illness. Suppose a tablet advertised as containing 500 mg of vitamin C is dissolved in 100.0 mL of distilled water that contains a small amount of the acid–base indicator bromothymol blue, an indicator that is yellow in acid solution and blue in basic solution, to give a yellow solution. The addition of 53.5 mL of a 0.0520 M solution of NaOH results in a change to green at the endpoint, due to a mixture of the blue and yellow forms of the indicator ( ). What is the actual mass of vitamin C in the tablet? (The molar mass of ascorbic acid is 176.13 g/mol.) : Bromothymol Blue is used as an indicator. If placed in a acidic solution it will turn the solution yellow, and if placed in a basic solution it will turn the solution blue. The equivalence point is seen when the solution turns green. pH is important for a variety of reasons. For example if you are trying to discard an acidic solution, you could add bromothymol blue to the solution and then titrate in some base until the solution turned green. This indicates that it is neutral reactant, volume of sample solution, and volume and molarity of titrant mass of unknown Write the balanced chemical equation for the reaction and calculate the number of moles of base needed to neutralize the ascorbic acid. Using mole ratios, determine the amount of ascorbic acid consumed. Calculate the mass of vitamin C by multiplying the number of moles of ascorbic acid by its molar mass. Because ascorbic acid acts as a monoprotic acid, we can write the balanced chemical equation for the reaction as \[HAsc(aq) + OH^-(aq) \rightarrow Asc^-(aq) + H_2O(l) \] where HAsc is ascorbic acid and Asc is ascorbate. The number of moles of OH ions needed to neutralize the ascorbic acid is \[ moles\: OH^- = 53 .5\: \cancel{mL} \left( \dfrac{\cancel{1\: L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .0520\: mol\: OH^-} {\cancel{1\: L}} \right) = 2 .78 \times 10^{-3}\: mol\: OH^- \] The mole ratio of the base added to the acid consumed is 1:1, so the number of moles of OH added equals the number of moles of ascorbic acid present in the tablet: \[ mass\: ascorbic\: acid = 2 .78 \times 10^{-3}\: \cancel{mol\: HAsc} \left( \dfrac{176 .13\: g\: HAsc} {1 \: \cancel{mol\: HAsc}} \right) = 0 .490\: g\: HAsc \] Because 0.490 g equals 490 mg, the tablet contains about 2% less vitamin C than advertised. Vinegar is essentially a dilute solution of acetic acid in water. Vinegar is usually produced in a concentrated form and then diluted with water to give a final concentration of 4%–7% acetic acid; that is, a 4% m/v solution contains 4.00 g of acetic acid per 100 mL of solution. If a drop of bromothymol blue indicator is added to 50.0 mL of concentrated vinegar stock and 31.0 mL of 2.51 M NaOH are needed to turn the solution from yellow to green, what is the percentage of acetic acid in the vinegar stock? (Assume that the density of the vinegar solution is 1.00 g/mL.) 9.35% The concentration of a species in solution can be determined by . One such method is a , in which a measured volume of a solution of one substance, the , is added to a solution of another substance to determine its concentration. The in a titration is the point at which exactly enough reactant has been added for the reaction to go to completion. A , a solution whose concentration is known precisely, is used to determine the concentration of the titrant. Many titrations, especially those that involve acid–base reactions, rely on an indicator. The point at which a color change is observed is the , which is close to the equivalence point if the indicator is chosen properly. The titration procedure is an application of the use of limiting reactants. Explain why this is so. Explain how to determine the concentration of a substance using a titration. Following are two graphs that illustrate how the pH of a solution varies during a titration. One graph corresponds to the titration of 100 mL 0.10 M acetic acid with 0.10 M NaOH, and the other corresponds to the titration of 100 mL 0.10 M NaOH with 0.10 M acetic acid. Which graph corresponds to which titration? Justify your answer. Following are two graphs that illustrate how the pH of a solution varies during a titration. One graph corresponds to the titration of 100 mL 0.10 M ammonia with 0.10 M HCl, and the other corresponds to the titration of 100 mL 0.10 M NH Cl with 0.10 M NaOH. Which graph corresponds to which titration? Justify your answer. Following are two graphs that illustrate how the electrical conductivity of a solution varies during a titration. One graph corresponds to the titration of 100 mL 0.10 M Ba(OH) with 0.10 M H SO , and the other corresponds to the titration of 100 mL of 0.10 M NaOH with 0.10 M H SO . Which graph corresponds to which titration? Justify your answer. A 10.00 mL sample of a 1.07 M solution of potassium hydrogen phthalate (KHP, formula mass = 204.22 g/mol) is diluted to 250.0 mL. What is the molarity of the final solution? How many grams of KHP are in the 10.00 mL sample? What volume of a 0.978 M solution of NaOH must be added to 25.0 mL of 0.583 M HCl to completely neutralize the acid? How many moles of NaOH are needed for the neutralization? A student was titrating 25.00 mL of a basic solution with an HCl solution that was 0.281 M. The student ran out of the HCl solution after having added 32.46 mL, so she borrowed an HCl solution that was labeled as 0.317 M. An additional 11.5 mL of the second solution was needed to complete the titration. What was the concentration of the basic solution? ( ) | 15,470 | 1,075 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Molecular_Orbital_Theory/Bonding_and_antibonding_orbitals |
Molecular orbital theory is concerned with the combination of atomic orbitals to form new molecular orbitals. These new orbitals arise from the linear combination of atomic orbitals to form bonding and antibonding orbitals. The bonding orbitals are at a lower energy than the antibonding orbitals, so they are the first to fill up. By figuring out the molecular orbitals, it is easy to calculate bond order. The valence bond theory is an extension of the Lewis Structures that considers the overlapping of orbitals to create bonds. The valence bond theory is only limited in its use because it does not explain the molecular geometry of molecules very well. This is where hybridization and the molecular orbital theory comes into place. Hybridization is a simple model that deals with mixing orbitals to from new, hybridized, orbitals. This is part of the valence bond theory and helps explain bonds formed, the length of bonds, and bond energies; however, this does not explain molecular geometry very well. If you add the exponents of the hybridized orbitals, you get the amount of sigma bonds associated with that bond. The sp hybridized orbital has one p orbitals that is not hybridized and so it can form a pi bond. This means that sp orbitals allow for the formation of a double bond. Also, sp hybridized orbitals form a triple bond. Electrons that spend most of their time between the nuclei of two atoms are placed into the bonding orbitals, and electrons that spend most of their time outside the nuclei of two atoms are placed into antibonding orbitals. This is because there is an increasing in electron density between the nuclei in bonding orbitals, and a decreasing in electron density in antibonding orbitals (Chang 459). Placing an electron in the bonding orbital stabilizes the molecule because it is in between the two nuclei. Conversely, placing electrons into the antibonding orbitals will decrease the stability of the molecule. Electrons will fill according to the energy levels of the orbitals. They will first fill the lower energy orbitals, and then they will fill the higher energy orbitals. If a bond order of zero is obtained, that means that the molecule is too unstable and so it will not exist. Below are a few examples of bonding and antibonding orbitals drawn out: Bond order is the amount of bonds formed between two atoms. For example, two bonds are formed between oxygen atoms, so the bond order is 2. The following is the equation to find bond order. 1/2(electrons in bonding molecular orbitals - electrons in antibonding molecular orbitals) Bond order gives information about bond length and strength. Generally, higher bond order correlates to a shorter bond length. This is due to the greater number of bonds between the atoms. In addition, because of the greater number of bonds between the atoms, the strength should also be greater as bond order increases. | 2,918 | 1,076 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_5%3A_Rates_of_Chemical_and_Physical_Processes/18%3A_Chemical_Kinetics/18.1%3A_Rates_of_Chemical_Reactions |
Reaction rates are usually expressed as the concentration of reactant consumed or the concentration of product formed per unit time. The units are thus moles per liter per unit time, written as M/s, M/min, or M/h. To measure reaction rates, chemists initiate the reaction, measure the concentration of the reactant or product at different times as the reaction progresses, perhaps plot the concentration as a function of time on a graph, and then calculate the change in the concentration per unit time. The progress of a simple reaction (A → B) is shown in the beakers are snapshots of the composition of the solution at 10 s intervals. The number of molecules of reactant (A) and product (B) are plotted as a function of time in the graph. Each point in the graph corresponds to one beaker in . The reaction rate is the change in the concentration of either the reactant or the product over a period of time. The concentration of A decreases with time, while the concentration of B increases with time. \[\textrm{rate}=\dfrac{\Delta [\textrm B]}{\Delta t}=-\dfrac{\Delta [\textrm A]}{\Delta t} \label{Eq1} \] Square brackets indicate molar concentrations, and the capital Greek delta (Δ) means “change in.” Because chemists follow the convention of expressing all reaction rates as positive numbers, however, a negative sign is inserted in front of Δ[A]/Δt to convert that expression to a positive number. The reaction rate calculated for the reaction A → B using Equation \(\ref{Eq1}\) is different for each interval (this is not true for every reaction, as shown below). A greater change occurs in [A] and [B] during the first 10 s interval, for example, than during the last, meaning that the reaction rate is greatest at first. Reaction rates generally decrease with time as reactant concentrations decrease. A Discussing Average Reaction Rates. Link: We can use Equation \(\ref{Eq1}\) to determine the reaction rate of hydrolysis of aspirin, probably the most commonly used drug in the world (more than 25,000,000 kg are produced annually worldwide). Aspirin (acetylsalicylic acid) reacts with water (such as water in body fluids) to give salicylic acid and acetic acid, as shown in Figure \(\Page {2}\). Because salicylic acid is the actual substance that relieves pain and reduces fever and inflammation, a great deal of research has focused on understanding this reaction and the factors that affect its rate. Data for the hydrolysis of a sample of aspirin are in Table \(\Page {1}\) and are shown in the graph in . The data in Table \(\Page {1}\) were obtained by removing samples of the reaction mixture at the indicated times and analyzing them for the concentrations of the reactant (aspirin) and one of the products (salicylic acid). The for a given time interval can be calculated from the concentrations of either the reactant or one of the products at the beginning of the interval (time = t ) and at the end of the interval (t ). Using salicylic acid, the reaction rate for the interval between t = 0 h and t = 2.0 h (recall that change is always calculated as final minus initial) is calculated as follows: The reaction rate can also be calculated from the concentrations of aspirin at the beginning and the end of the same interval, remembering to insert a negative sign, because its concentration decreases: If the reaction rate is calculated during the last interval given in Table \(\Page {1}\)(the interval between 200 h and 300 h after the start of the reaction), the reaction rate is significantly slower than it was during the first interval (t = 0–2.0 h): In the preceding example, the stoichiometric coefficients in the balanced chemical equation are the same for all reactants and products; that is, the reactants and products all have the coefficient 1. Consider a reaction in which the coefficients are not all the same, the fermentation of sucrose to ethanol and carbon dioxide: \[\underset{\textrm{sucrose}}{\mathrm{C_{12}H_{22}O_{11}(aq)}}+\mathrm{H_2O(l)}\rightarrow\mathrm{4C_2H_5OH(aq)}+4\mathrm{CO_2(g)} \label{Eq2} \] The coefficients indicate that the reaction produces four molecules of ethanol and four molecules of carbon dioxide for every one molecule of sucrose consumed. As before, the reaction rate can be found from the change in the concentration of any reactant or product. In this particular case, however, a chemist would probably use the concentration of either sucrose or ethanol because gases are usually measured as volumes and, as explained in , the volume of CO gas formed depends on the total volume of the solution being studied and the solubility of the gas in the solution, not just the concentration of sucrose. The coefficients in the balanced chemical equation tell us that the reaction rate at which ethanol is formed is always four times faster than the reaction rate at which sucrose is consumed: \[\dfrac{\Delta[\mathrm{C_2H_5OH}]}{\Delta t}=-\dfrac{4\Delta[\textrm{sucrose}]}{\Delta t} \label{Eq3} \] The concentration of the reactant—in this case sucrose— with time, so the value of Δ[sucrose] is negative. Consequently, a minus sign is inserted in front of Δ[sucrose] in so the rate of change of the sucrose concentration is expressed as a positive value. Conversely, the ethanol concentration with time, so its rate of change is automatically expressed as a positive value. Often the reaction rate is expressed in terms of the reactant or product with the smallest coefficient in the balanced chemical equation. The smallest coefficient in the sucrose fermentation reaction ( ) corresponds to sucrose, so the reaction rate is generally defined as follows: \[\textrm{rate}=-\dfrac{\Delta[\textrm{sucrose}]}{\Delta t}=\dfrac{1}{4}\left (\dfrac{\Delta[\mathrm{C_2H_5OH}]}{\Delta t} \right ) \label{Eq4} \] Consider the thermal decomposition of gaseous N O to NO and O via the following equation: Write expressions for the reaction rate in terms of the rates of change in the concentrations of the reactant and each product with time. balanced chemical equation reaction rate expressions Because O has the smallest coefficient in the balanced chemical equation for the reaction, define the reaction rate as the rate of change in the concentration of O and write that expression. The balanced chemical equation shows that 2 mol of N O must decompose for each 1 mol of O produced and that 4 mol of NO are produced for every 1 mol of O produced. The molar ratios of O to N O and to NO are thus 1:2 and 1:4, respectively. This means that the rate of change of [N O ] and [NO ] must be divided by its stoichiometric coefficient to obtain equivalent expressions for the reaction rate. For example, because NO is produced at four times the rate of O , the rate of production of NO is divided by 4. The reaction rate expressions are as follows: \(\textrm{rate}=\dfrac{\Delta[\mathrm O_2]}{\Delta t}=\dfrac{\Delta[\mathrm{NO_2}]}{4\Delta t}=-\dfrac{\Delta[\mathrm{N_2O_5}]}{2\Delta t}\) The is used in the manufacture of sulfuric acid. A key step in this process is the reaction of \(SO_2\) with \(O_2\) to produce \(SO_3\). \[2SO_{2(g)} + O_{2(g)} \rightarrow 2SO_{3(g)} \nonumber \] Write expressions for the reaction rate in terms of the rate of change of the concentration of each species. The of a reaction is the reaction rate at any given point in time. As the period of time used to calculate an average rate of a reaction becomes shorter and shorter, the average rate approaches the instantaneous rate. Comparing this to calculus, the instantaneous rate of a reaction at a given time corresponds to the slope of a line tangent to the concentration-versus-time curve at that point—that is, the derivative of concentration with respect to time. The distinction between the instantaneous and average rates of a reaction is similar to the distinction between the actual speed of a car at any given time on a trip and the average speed of the car for the entire trip. Although the car may travel for an extended period at 65 mph on an interstate highway during a long trip, there may be times when it travels only 25 mph in construction zones or 0 mph if you stop for meals or gas. The average speed on the trip may be only 50 mph, whereas the instantaneous speed on the interstate at a given moment may be 65 mph. Whether the car can be stopped in time to avoid an accident depends on its instantaneous speed, not its average speed. There are important differences between the speed of a car during a trip and the speed of a chemical reaction, however. The speed of a car may vary unpredictably over the length of a trip, and the initial part of a trip is often one of the slowest. In a chemical reaction, the initial interval typically has the fastest rate (though this is not always the case), and the reaction rate generally changes smoothly over time. Chemical kinetics generally focuses on one particular instantaneous rate, which is the initial reaction rate, t = 0. Initial rates are determined by measuring the reaction rate at various times and then extrapolating a plot of rate versus time to t = 0. Using the reaction shown in Example \(\Page {1}\), calculate the reaction rate from the following data taken at 56°C: \[2N_2O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)} \nonumber \] balanced chemical equation and concentrations at specific times reaction rate Calculate the reaction rate in the interval between t = 240 s and t = 600 s. From Example \(\Page {1}\), the reaction rate can be evaluated using any of three expressions: Subtracting the initial concentration from the final concentration of N O and inserting the corresponding time interval into the rate expression for N O , Substituting actual values into the expression, Similarly, NO can be used to calculate the reaction rate: Allowing for experimental error, this is the same rate obtained using the data for N O . The data for O can also be used: Again, this is the same value obtained from the N O and NO data. Thus, the reaction rate does not depend on which reactant or product is used to measure it. Using the data in the following table, calculate the reaction rate of \(SO_2(g)\) with \(O_2(g)\) to give \(SO_3(g)\). \[2SO_{2(g)} + O_{2(g)} \rightarrow 2SO_{3(g)} \nonumber \] In this Module, the quantitative determination of a reaction rate is demonstrated. Reaction rates can be determined over particular time intervals or at a given point in time. A rate law describes the relationship between reactant rates and reactant concentrations. Reaction rates are reported as either the average rate over a period of time or as the instantaneous rate at a single time. Reaction rates can be determined over particular time intervals or at a given point in time. | 10,775 | 1,077 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/07%3A_Alkenes-_Structure_and_Reactivity/7.12%3A_Evidence_for_the_Mechanism_of_Electrophilic_Additions_-_Carbocation_Rearrangements |
After completing this section, you should be able to explain the “unusual” products formed in certain reactions in terms of the rearrangement of an intermediate carbocation. Make certain that you can define, and use in context, the key terms below. Whenever possible, carbocations will rearrange from a less stable isomer to a more stable isomer. This rearrangement can be achieved by either a hydride shift, where a hydrogen atom migrates from one carbon atom to the next, taking a pair of electrons with it; or an alkyl shift, in which an alkyl group undergoes a similar migration, again taking a bonding pair of electrons with it. These migrations usually occur between neighbouring carbon atoms, and hence are termed 1,2-hydride shifts or 1,2-alkyl shifts. [A hydride ion consists of a proton and two electrons, that is, [H:] . Hydride ions exist in compounds such as sodium hydride, NaH, and calcium hydride, CaH .] An electrophilic reaction such as HX with an alkene will often yield more than one product. This is strong evidence that the mechanism includes intermediate rearrangement steps of the cation. Throughout this textbook many reaction mechanisms will be presented. It is impossible to know with absolute certainty that a mechanism is correct. At best a proposed mechanism can be shown to be consistent with existing experimental data. Virtually all of the mechanisms in this textbook have been carefully studied by experiments designed to test their validity although the details are not usually discussed. An excellent example of experimental evidence which supports the carbocation based mechanism for electrophilic addition, is that structural rearrangements often occur during the reaction. A 1,2-hydride shift is a carbocation rearrangement in which a hydrogen atom in a carbocation migrates to the carbon atom bearing the formal charge of +1 (carbon 2 in the example below) from an adjacent carbon (carbon 1). An example of this structural rearrangment occurs during the reaction of 3-methyl-1-butene with HBr. Markovnikov's rule predicts that the preferred product would be 2-bromo-3-methylbutane, however, very little of this product forms. The predominant product is actually 2-bromo-2-methylbutane. This result comes from a during the reaction mechanism. The mechanism begins with protonation of the alkene which places a positive charge on the more alkyl substituted double bond carbon resulting in a secondary carbocation. In step 2, The electrons in the C-H bond on carbon #3 are attracted by the positive charge on carbon #2, and they simply shift over to fill the carbocation's empty orbital, pulling the proton over with them. The process called a carbocation rearrangement, and more specifically, a hydride shift. A hydride ion (H: ) is a proton plus two electrons which not to be confused with H , which is just a proton without any electrons. Notice that the hydride, in shifting, is not acting as an actual leaving group - a hydride ion is a very strong base and a very poor leaving group. As the hydride shift proceeds, a new \(C-H\) \(\sigma \) bond is formed at carbon #2, and carbon #3 is left with an empty \(p\) orbital and a positive charge. What is the thermodynamic driving force for this process? Notice that the hydride shift results in the conversion of a secondary carbocation (on carbon 2) to a (more stable) tertiary carbocation (on carbon 3) - a thermodynamically downhill step. As it turns out, the shift occurs so quickly that it is accomplished before the bromide nucleophile has time to attack at carbon #2. Rather, the bromide will attack after the hydride shift (step 3) at carbon #3 to complete the addition. A 1,2-alkyl shift is a carbocation rearrangement in which an alkyl group migrates to the carbon atom bearing the formal charge of +1 (carbon 2) from an adjacent carbon atom (carbon 1), e.g. Consider another example. When HBr is added to 3,3-dimethyl-1-butene the preferred product is 2-bromo-2,3-dimethylbutane and not 3-bromo-2,2-dimethylbutane as predicted by Markovnikov's rule. Notice that in the observed product, the carbon framework has been rearranged: a methyl carbon has been shifted. This is an example of another type of carbocation rearrangement, called an or more specifically a . Below is the mechanism for the reaction. Once again a secondary carbocation intermediate is formed in step 1. In this case, there is no hydrogen on carbon #3 available to shift over create a more stable tertiary carbocation. Instead, it is a methyl group that does the shifting, as the electrons in the carbon-carbon \(\sigma \) bond shift over to fill the empty orbital on carbon #2 (step 2 below). The methyl shift results in the conversion of a secondary carbocation to a more stable tertiary carbocation. It is this tertiary carbocation which is attacked by the bromide nucleophile to make the rearranged end product. The end result is a rearrangement of the carbon framework of the molecule. Electrophilic addition with methyl shift: Carbocation shifts occur in many more reactions than just electrophilic additions as some of which will be discussed in subsequent chapters of this textbook. Whenever a carbocation is produced in a reaction's mechanism the possibility of rearrangements should be considered. As discussed in Section 7.9, there are multiple ways to stabilize a carbocation all of which could induce a rearrangement. The most common situation for a rearrangement to occur during electrophilic addition is: When considering the possibility of a carbocation rearrangement the most important factors are the designation of the carbocation formed and the designation of the alkyl groups attached to the carbocation. When a 2 carbocation has a 3 alkyl substituent a hydride shift will occur to create a more stable 3 carbocation. When a 2 carbocation has a 4 alkyl substituent an alkyl shift will occur to create a more stable 3 carbocation. First, draw the unrearranged product. Add HX to the double bond following Markovnikov's rule if necessary. Then determine if a hydride or alkyl shift is occurring by observing the designation of the alkyl substituent. The switch X ⇔H for a hydride shift and X ⇔ CH for an alkyl shift this will produce the rearranged product. Draw the expectred produts of the following reaction. Carbocation rearrangements are involved in many known biochemical reactions. Rearrangements are particularly important in carbocation-intermediate reactions in which isoprenoid molecules cyclize to form complex multi-ring structures. For example, one of the key steps in the biosynthesis of cholesterol is the electrophilic cyclization of oxidosqualene to form a steroid called lanosterol. This complex but fascinating reaction has two phases. The first phase is where the actual cyclization takes place, with the formation of four new carbon-carbon bonds and a carbocation intermediate. The second phase involves a series of hydride and methyl shifts culminating in a deprotonation. In the exercise below, you will have the opportunity to work through the second phase of the cyclase reaction mechanism. The second phase of the cyclase reaction mechanism involves multiple rearrangement steps and a deprotonation. Please supply the missing mechanistic arrows. 1) The following reaction shows a rearrangement within the mechanism. Propose a mechanism that shows this. 2) Propose a mechanism for the following reaction. It involves an electrophilic addition and the shift of a C-C and a C-H bond. 2) In most examples of carbocation rearrangements that you are likely to encounter, the shifting species is a hydride or methyl group. However, pretty much any alkyl group is capable of shifting. Sometimes, the entire side of a ring will shift over in a ring-expanding rearrangement. The first 1,2-alkyl shift is driven by the expansion of a five-membered ring to a six-membered ring, which has slightly less ring strain. A hydride shift then converts a secondary carbocation to a tertiary carbocation, which is the electrophile ultimately attacked by the bromide nucleophile. Once again, the driving force for this process is an increase in stability of the carbocation. Initially, there is a primary carbocation at C2, and this becomes a tertiary carbocation at C1 as a result of the (1,2)-methyl shift. | 8,373 | 1,078 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/14%3A_Chemical_Equilibrium/14.E%3A_Chemical_Equilibria_(Exercises) |
.
Write an expression for the equilibrium constant K for each reaction below. If a reaction \(\ce {aA + bB \rightleftharpoons cC + dD}\) occurs, then \[ K_c = \dfrac{[C]^c[D]^d}{[A]^a[B]^b} \nonumber \] Note that for part c, \(H_2O_{(l)}\) is not included because pure solids and liquids are not included in the equilibrium constant. Gaseous chlorine reacts with water vapor to form hydrogen chloride and oxygen gas. Write down the equilibrium expression for the reaction. First, a balanced reaction is constructed to ensure the equilibrium expression will be correctly states. \[\ce{2Cl_2 + 2H_2O \rightleftharpoons 4HCl + O2} \nonumber \] Equilibrium expression \(K_p\) is as follows: \[ K_p = \dfrac{p^4_{\ce{HCl}} p_{\ce{O2}}}{p^2_{\ce{Cl2}}p^2_{\ce{H2O}}}\nonumber \] For each thermodynamics equilibrium expressions outlined below, determine one possible chemical reaction related to each. Based on the following data, calculate the equilibrium constant and the value of \(ΔG\) at 273 K. a: \(\ce{CO2(g) + H2(g) \rightleftharpoons CO(g) + H2O(g)}\nonumber \) with b: \(\ce{2NO(g) + 2H2(g) \rightleftharpoons N2(g) + 2H2O(g)}\nonumber \) with a) K = [CO,H O] / [CO ,H ] K = [0.0046,0.0046] / [0.0954,0.0454] K = 4.9 x 10 ΔG = nRT ln(K) ΔG = (1 mol)(8.3145 J/K mol)(273 K) ln(4.9 x 10 ) ΔG = -12072.288 J b) K = [N ,H O] / [NO] [H ] K = [0.95,1.3] / [0.45] [0.63] K = 19.98 ΔG = nRT ln(K) ΔG = (1 mol)(8.3145 J/K mol)(273 K) ln(19.98) ΔG = 6797.62 J Calculate the appropriate \(K\) value for the following reactions using the information provided. Assume that they are preformed under standard conditions (and \(298.15K\)). The core of this problem is to find the K value and understanding what information it can describe about the reaction. Since the problem provides many, many \(\Delta{G_{f}}\), it is likely that to calculate the K value, a relationship needs to be drawn from \(\Delta{G}\) to \(K\). This relationship is as follows: \[\Delta{G_{rxn}^{\circ}}= -RT \ln\left(K_{p}\right)\nonumber \] Relevant Information obtained from the thermodynamic tables \(\Delta{G_f^{\circ}}\; NO_{2\, (g)} = 51.30 \dfrac{kJ}{mol}\) \(\Delta{G_f^{\circ}}\; H_{3}O_{(aq)}^{+} = -103.45 \dfrac{kJ}{mol}\) \(\Delta{G_f^{\circ}}\; NO_{(g)} = 86.57 \dfrac{kJ}{mol}\) \(\Delta{G_f^{\circ}}\; NH_{3\, (aq)} = -26.5 \dfrac{kJ}{mol}\) \(\Delta{G_f^{\circ}}\; O_{2\, (g)} = 0 \dfrac{kJ}{mol}\) \(\Delta{G_f^{\circ}}\; CH_{4\, (g)} = -50.84 \dfrac{kJ}{mol}\) \(\Delta{G_f^{\circ}}\; NH_{4\, (aq)}^{+} = -79.37 \dfrac{kJ}{mol}\) \(\Delta{G_f^{\circ}}\; CO_{(g)} = -137.28 \dfrac{kJ}{mol}\) \(\Delta{G_f^{\circ}}\; H_{2}O_{(l)} = -237.14 \dfrac{kJ}{mol}\) \(\Delta{G_f^{\circ}}\; H_{2}O_{(g)} = -228.61 \dfrac{kJ}{mol}\) Note that the relationship here is between \(\Delta{G}\) and \(K_{p}\), which is: \[K_{p}\approx\dfrac{P_{products}}{P_{reactants}}\nonumber \] Where \(P\) is partial pressure of the products or reactants. The reason this is an approximation is because the equilibrium actually measures the of the species, but by using just the partial pressures a good estimate is procured. Since the equation relating \(K_{p}\) and \(\Delta{G}\) only has these two variables as unknown, once \(\Delta{G}\) is found, algebra can find \(K_{p}\). The process for utilizing to find state variables is outlined extensively . The values below were found using the exact same process, now for \(\Delta{G}\) (Just in case, the Hess law formula is provided). \(\Delta{G^{\circ}}=\sum{n_{products}\times{\Delta{G_{f_{products}}^{\circ}}}} - \sum{n_{reactants}\times{\Delta{G_{f_{reactants}}^{\circ}}}}\) \(\Delta{G^{\circ}_{2 NO_{2\,(g)} \rightleftharpoons 2 NO_{(g)} + O_{2\, (g)}}} = 70.42\dfrac{kJ}{mol}\) \(\Delta{G^{\circ}_{NH_{4\,(aq)}^{+} + H_{2}O{(l)} \rightleftharpoons H_{3}O_{(aq)}^{+} + NH_{3\, (aq)}}}= -437.93 \dfrac{kJ}{mol}\) \(\Delta{G^{\circ}_{2 CH_{4\, (g)} + 3 O_{2\, (g)} \rightarrow 2 CO_{(g)} + 4 H_{2}O_{l}}}= -1121.44\dfrac{kJ}{mol}\) To find K, plug in the \(\Delta{G}\) into the equation above. (Note that the solution does NOT show the conversion of the gas constant from \(J\) to \(kJ\)!!) \(\Delta{G_{rxn}^{\circ}}= -\left(8.314\dfrac{J}{molK}\right)\left(298.15K\right)\times{ln\left(K_{p}\right)}\) \(70.54\dfrac{kJ}{mol}= -\left(8.314\dfrac{J}{molK}\right)\left(298.15K\right)\times{ln\left(K_{p}\right)}\) \(-\dfrac{70.54\dfrac{kJ}{mol}}{\left(8.314\dfrac{J}{molK}\right)\left(298.15K\right)}= ln\left(K_{p}\right)\) \(e^{-\dfrac{70.54\dfrac{kJ}{mol}}{\left(8.314\dfrac{J}{molK}\right)\left(298.15K\right)}} = K_{p}\) \(K_{p}\approx e^{-28.457}\approx4.3776\times10^{-13}\) Thus, the \(K_{p}\) value for \(2 NO_{2\,(g)} \rightleftharpoons 2 NO_{(g)} + O_{2\, (g)}\) is a very small number!. Since the equilibrium value depicts the ratio of products to reactants at the equilibrium point, this very tiny \(K_{p}\) value means that the reaction will heavily favor the reactants. The third reaction where Txeltoqlztop wants to poison his brother with \(CO\) also follows the same process, and the answer is such: \(K_{p}\approx{e^{452}}\) \(e^{452}\) is a big number! That means, while the previous reaction will stay mostly on the reactant sides, this reaction heavily favors the products side, meaning that pretty much all of the methane and oxygen that Txeltoqlztop uses will be transformed into carbon monoxide, which is all the better for him! The same method can be applied to the aqueous reaction, but the \(K\) value calculated is \(K_{c}\) rather than \(K_{p}\). \(K_{c}\approx{e^{-176.65}}\approx1.898\times10^{-77}\)* *These equilibrium constants are based solely off the calculated Gibbs energy, some of the reactions, therefore, may have actual K values much different from the ones calculated. : use of \(\Delta{G_{rxn}^{\circ}}= -RT\times{ln\left(K_{p}\right)}\) Use the thermodynamic data in and to calculate the equilibrium constants for the following equations at 25°C. In order to determine the value of G , the following the reaction is needed \[ \Delta G_{rxn} = \sum nG_{f}°(products) - \sum nG_{f}°(reactants) \nonumber \] By looking in the tables, the \( \Delta G\) of formation can be found and they are listed for each compound below Next, these values are plugged into the second equation to the the change in Gibbs energy for the reaction \[ \Delta G_{rxn} = -275.4\; J\; mol^{-1} - (203.3\; J\; mol^{-1} + 62.3\; J\; mol^{-1}) = -514\; J\; mol^{-1} \nonumber \] Using the the value for the Gibbs Energy, the value of K can be determined using the first equation \[ -541\; J\; mol^{-1} = -8.314\; J\; K^{-1}\; mol^{-1} \times 298K \times lnK \nonumber \] \[ \dfrac{-541\; J\; mol^{-1}}{-8.314\; J\; K^{-1}\; mol^{-1} \times 298K} = lnK \nonumber \] \[K = 1.25 \nonumber \] The \( \Delta G\) of formation for the compounds are FeCl = -334.0 J mol , FeCl = -302.3 J mol , Cl =105.3 J mol The calculations are \[ \Delta G_{rxn} = (-302.3\; J\; mol^{-1} + 105.3\; J\; mol^{-1}) + 334.0\; J\; mol^{-1} =137\; J\; mol^{-1} \nonumber \] \[ 137\; J\; mol^{-1} = -8.314\; J\; K^{-1}\; mol^{-1} \times 298K \times lnK \nonumber \] \[ \dfrac{137\; J\; mol^{-1}}{-8.314\; J\; K^{-1}\; mol^{-1} \times 298K} = lnK \nonumber \] \[ K = 0.945 \nonumber \] The \( \Delta G\) of formation for the compounds are CaSO = -1322.0 J/mol, Ca (aq) = -553.58 J/mol, SO (aq) = -744.53 J/mol The calculations are \[\Delta G_{rxn} = (-553.58\; J\; mol^{-1} - 744.53\; J\; mol^{-1}) + 1322.0\; J\; mol^{-1} =23.89\; J\; mol^{-1} \nonumber \] \[ 23.89\; J\; mol^{-1} = -8.314\; J\; K^{-1}\; mol^{-1} \times 298K \times lnK \nonumber \] \[ \dfrac{23.89\; J\; mol^{-1}}{-8.314\; J\; K^{-1}\; mol^{-1} \times 298K} = lnK \nonumber \] \[ K = 0.99 \nonumber \] To ignite this house on fire, you dowsed the house in some ethanol and used a combustion reaction: \[ \ce{C2H5OH(l) + O2(g) <=> H2O(l) + CO2 (g)}\nonumber \] This reaction has an equilibrium constant of K . If you multiply this equation by 5, will the equilibrium constant change? If so, write this new equilibrium constant, K , in terms of K . Because: \(K = \dfrac{[A]^{a}[B]^{b}}{[C]^{c}[D]^{d}}\) If we multiply a chemical equation by a constant, the exponents of K will be multiplied by that multiple as well. An unknown gas B D reacts at room temperature and becomes B D , the reaction is illustrated with the equation \[\ce{2B5D10(g) <=> B10D2(g)} \nonumber \] At room temperature the equilibrium partial pressure of B D is \(1.23 \times 10^{-4}atm\) and that of B D is \(3.14 \times 10^{-16}atm\). Given these constants, what is \(K_p\) of this reaction at room temperature. Given that we already know the partial pressures and this reaction is at equilibrium, we simply substitute the values into the equilibrium equation and solve for K . \[ K_p=\dfrac{[products]}{[reactants]}\nonumber \] \[ K_p=\dfrac{[3.14\times10^{-16}]}{[1.23\times10^{-4}]^2}=2.08\times10^{-8}\nonumber \] \[ K_p=2.08\times10^{-8}\nonumber \] \[\ce{X_{red} \rightleftharpoons X_{blue}}\nonumber \] \[\ce{X_{red} \rightleftharpoons X_{blue}}\nonumber \] \[ \ce{K_{eq}} = \frac{ \left[ X_{blue} \right] }{ \left[X_{red} \right] } \nonumber \] \[ \ce{K_{eq}} = \frac{ \left[ 0.17 \right] }{ \left[0.83 \right] } = 0.205\nonumber \] 1,2 dimethylcyclohexane can exist in both boat and chair conformation. There is equilibrium between the two forms. What is the equilibrium constant for the reaction of boat conformation to chair conformation if the molecules in the boat form are 68.7%? \[ Boat \rightarrow Chair\] Let B stand for Boat and C stand for Chair; \[K=\frac{[C]}{[B]}\] @ constant volume, the ratio of mole is equal to the ratio of Molarity, therefore: \[\frac{mole_{B}}{mole_{C}+mole_{B}}=\frac{[B]}{[C]+[B]}=68.7\%=0.687\] Let [B]=0.687, \[\frac{0.687}{0.687+[C]}=0.687\] \[0.687+[C]=1\] \[[C]=0.313\] \[K=\frac{0.313}{0.687}=0.456\] \(K_p\) fo A 100.0-mL glass bulb was filled which a weighed sample of solid \(XO_3\), where \(X\) is an unknown element. The bulb was then attached to a pressure gauge and heated to 325 K, at which, the pressure was read to be 0.891 atm. Given that all of the \(XO_3\) in the bulb at 325 K was in the gas phase, and it also partially dissociated into O and XO : $$XO_{3 (g)} \rightleftharpoons O_{2(g)} + XO_{(g)} \nonumber \] At 325 K, K = 3.14 for this reaction. Calculate the partial pressures of all three species in the bulb at equilibrium. \[XO_{3 (g)} \rightleftharpoons O_{2(g)} + XO_{(g)}\] Let Y be the initial partial pressure of X O , and x is the change in partial pressure of O : The total pressure is 0.891 atm, therefore: $$ (y - x) + x + x = 0.891 \; atm \] $$ y + x = 0.891 \; atm \] $$ y = 0.891 - x \] So the ICE table now would be: \[ k_{p} = 3.14 = \dfrac{P_{XO} \cdot P_{O_{2}}}{P_{XO_{3}}} = \dfrac {x^{2}}{0.891 - 2x}\] \[ x^{2} + 6.28x - 2.80 = 0\] \[ x = 0.418 \] \[ P_{O_{2}} = P_{XO} = 0.418 \; atm \] \[ P_{XO_{3}} = 0.891 - 2(0.418) = 0.055 atm\] The equilibrium constant for the reaction \[2 F(g) + H_2(g) \rightarrow 2 HF(g)\nonumber \] at 298 K is \(5.07 \times 10^4\). Hydrogen with a partial pressure of 0.03500 atm is mixed with fluorine with a partial pressure of 0.06800 atm, and allowed to reach the equilibrium. What is the partial pressure of each of the gasses at the equilibrium? We are looking for partial pressures of the reactants and we have starting partial pressures and a Kc value. First we need to convert Kc to Kp \[K_p = K_c(RT)^{\Delta{n_{gas}}}\nonumber \] Kp=(5.07*10 )(.0821*298) =2072.28 The equilibrium constant can be constructed from the equation given \[K_p = \dfrac{ [HF]^2}{[H_2,F]^2}\nonumber \] Use an ICE table to find the equilibrium partial pressures of the equilibrium \(2072.28 = \dfrac{ [2x]^2}{[.035-x,.068-2x]^2}\) Simplify the equation and solve as you wish. You can't use the quadratic formula because you end up with a degree three equation but you can put the equation into a graphing calculator and solve for x using the zero function. x=.027 atm At 25 °C, the equilibrium constant for the reaction below has the value of \(1.7 \times 10^{-13}\): \[\ce{ N2O(g) + 1/2 O2(g) <=> 2 NO(g)} \nonumber \] In a container where \(N_2O\) has an initial partial pressure of 0.62 atm, \(O_2\) has a pressure of 0.24 atm, and \(NO\) has an initial pressure of 0.08 atm, what will the partial pressure of the three gases be after reaching equilibrium at the same temperature? The reaction: \[\ce{2HI(g) <=> H_2(g) + I_2(g)} \nonumber \] has an equilibrium constant \(K_c= 1.82 \times 10^{-2}\) at 698K. If the reaction took place in a tank at 698 K and started off with HI having a partial pressure of 1 atm, what would the partial pressures of all the gases be at equilibrium? Firstly, we need to convert Kc into Kp with the equation Kp=Kc(RT) Kp= 1.82 x 10 (0.0821Latm/Kmol x 698K) Kp=1.82 x 10 An ICE table can also be constructed for the reaction: \[2HI_{(g)} \rightleftharpoons H_{2(g)} + I_{2(g)}\nonumber \] 1 0 0 -2x +x +x 1-2x x x \[0.82 \times 10^{-2} = \dfrac{x^2}{1-2x} \nonumber\] x = 1.82 x 10 -0.0368x x=0.118488 partial pressure of HI= 0.763 atm partial pressure of H = partial pressure of I = 0.1189 atm What is the concentration of a \(XY\) that will be found in equilibrium at room temperature, given that \(X_2\) has a concentration of \(3.14 \times 10{-15} M\), \(Y_2\) has a concentration of \(1.23 \times 10^{-4}\; M\), and the \(K_c\) of the reaction is \(2.22 \times 10^{-7} M\). First we must state what the formulas for both the reaction and K and then we can substitute the known values and solve for the concentration of XY as variable B. \[\ce{2XY}(g) \rightleftharpoons \ce{X2}(g)+\ce{Y2}(g)\nonumber \] \[ K_{\ce c}=\dfrac{[Products]}{[Reactants]}\nonumber \] \[ K_{\ce c}=\dfrac{[3.14\times10^{-15},1.23\times10^{-4}]}{[X]^{2}}=2.22\times10^{-7}\nonumber \] \[ K_{\ce c}=\dfrac{[3.86\times10^{-19}]}{[X]^{2}}=2.22\times10^{-7}\nonumber \] \[ K_{\ce c}=3.86\times10^{-19}=2.22\times10^{-7}X^{2}\nonumber \] \[ X=1.32\times10^{-6}\nonumber \] So the concentration of XY is 1.32E-6 M in equilibrium at 25ºC. Hydrogen and oxygen gas react with each other to form gaseous water with an equilibrium constant for the reaction is \(K_C = 1.33 \times 10^{20}\) at 1000 K. a)\[2H_{2}{(g)} +O_{2(g)} \rightleftharpoons 2H_{2}O_{(l)}\] \[K=3.4=\frac{1}{P_{H_{2}}^{2}P_{O_{2}}}\] \[P_{H_{2}}=1.2\ atm\] \[3.4=\frac{1}{(1.2)^{2}P_{O_{2}}}\] \[P_{O_{2}}=\frac{1}{(1.2)^{2}(3.4)}=0.204\ atm\] b) Excess H O is added drive the reaction to the left towards the reactant to reach the equilibrium; therefore, pressure of H and O increased. \[K=\frac{1}{(2x)^{2}(0.3+x)}=3.4\] \[3.4(0.3+x)(2x)^{2}=1\] \[x=0.339\ atm\] \[P_{O_{2}}=0.3+0.339=0.639\ atm\] \[P_{H_{2}}=2(0.339)=0.678\ atm\] \[\ce{ 2 NH4NO3(s) \rightleftharpoons 2 N2(g) + O2(g) + 4 H2O(g)} \nonumber \] (a) If there is no ammonium nitrate left, then the reaction was products favoring, meaning K is extremely high at 170°C. (b) There would have to be an enormous amount of Nitrogen to force the reaction backwards. Because 0.3 atm is likely not enough to push back an essentially irreversible reaction, it has no noticeable affect on the amount of end product Al Cl at a partial pressure of 0.600 atm is placed in a closed container at 454 K. Al Cl (partial pressure 1.98 X 10 atm) is also placed in it as well. Argon is added to raise the total pressure up to 1.00 atm. Find whether if there is going to be net production or consumption of Al Cl given K = 1.04 x 10 . Then find the final pressure of Al Cl . \[Q=\frac{(Al_{3}Cl_{9})^{2}}{(Al_{2}Cl_{6})^{3}}= \frac{(1.98\cdot 10^{-3})^{2}}{(0.600)^{3}}=1.82\cdot 10^{-5}<K= 1.04\cdot 10^{-4}\] There will be production of Al Cl . \[K= 1.04\cdot 10^{-4}=\frac{(Al_{3}Cl_{9})^{2}}{(Al_{2}Cl_{6})^{3}}= \frac{(1.98\cdot 10^{-3}+2y)^{2}}{(1-(1.98\cdot 10^{-3}+2y))^{3}}; y = 0.00403atm\] Al Cl final pressure = 0.01004atm ssures of \(\ce{NH3(g)}\) and \(\ce{HCl(g)}\) are A tube contains a mixture of NO and N O gas is set at 298K, in which the initial partial pressure NO is 0.38 atm, and the initial partial pressure of N O is 0.59 atm. NO is a brownish gas, while N O is a colorless gas. \[ \ce{2NO2(g) \rightleftharpoons N2O4(g)} \nonumber \] a) \[2NO_{2(g)} \rightleftharpoons N_{2}O_{4(g)}\] \[P_{N_{2}O_{4}} = 0.59 \; atm \; and \; P_{NO_{2}} = 0.38 \; atm \] \[ Q_{p} = \frac{P_{N_{2}O_{4}}}{(P_{NO_{2}})^{2}}\] \[ Q_{p} = \frac{0.59}{0.38^{2}} = 4.1\] b) Since the mixture becomes more brown as the reaction reaches equilibrium, more NO gas is being produced, therefore the reverse reaction is favored, so Q > k , in conclusion, K < 4.1. At 500 K the equilibrium constant for the reaction: \[\ce{ PCl3 (g) + Cl2 (g) <=> PCl5 (g)}\] is \(K_{eq}=2.1\). Suppose the equilibrium is disturbed and shifts, while temperature remains constant. Calculate the reaction quotient for each one of the possible changes in equilibrium concentration given below and evaluate in what direction the reaction will shift to reestablish equilibrium. NO has a brownish color. At elevated temperatures, NO reacts with CO \[\ce{NO2 (g) + CO (g) <=> NO(g) + CO2 (g)}\nonumber \] The other gases in this equation are colorless. When a gas mixture is prepared at 600K, in which 9.6 atm is the initial partial pressure of both NO and CO, and 5.4 atm is the partial pressure of both NO and CO , the brown color of the mixture observed begins to get stronger as the reaction progresses toward equilibrium. Give a condition that must be satisfied by the equilibrium constant K. The equilibrium constant (\(K_p\)) for the reaction \[ \ce{As4(g) \rightarrow 2 As2(g)} \nonumber\nonumber \] is \(5.5837 \times 10^{-4}\) at 1090 K. (a) First, because you are given the initial molarities of your products and reactants and a Kp instead of a Kc value, you have to convert Kp to Kc using the formula: \[K_p = K_c(RT)^{\Delta{n_{gas}}}\nonumber \] Plug in your given values 5.5837 x 10 =Kc(.0821*1090) 5.5837 x 10 =Kc(89.489) Kc=6.24 x 10 Set up a reaction quotient for the equation given Q = \(\frac{ [Sc_2]^2}{[Sc_4]}\) Q = \(\frac{ [.001]^2}{[2.3]}\) (b) Set up an ice chart to find the equilibrium molarities 6.24 x 10 = \(\frac{ [.001+2x]^2}{[2.3-x]}\) 4x +.00400624x-1.3352x10 =0 Use the quadratic equation to find x x= -.0024, .00139 .001-.0024 is negative, which can't happen in real life so we know x actually equals .00139 (c) Consider a reaction involving only gaseous compounds. The product yield at equilibrium decreases when the temperature and volume increase. Exothermic Decreases In exothermic reactions, product yield is indirectly related to temperature. In endothermic reactions, product yield is directly related to temperature. These properties can be further explained when heat is thought of as a reactant or a product. In exothermic reactions, heat is a product, and in endothermic reactions, heat is a reactant. When heat is added to a reaction, the equilibrium shifts to the opposite side to relieve stress on the system. The number of gas molecules is directly related to the volume of the container in which the reaction is happening. For example, if the volume of the container is decreased, the reaction will shift to whichever side has fewer moles of gas. Haber process, also called the Haber–Bosch process, is an artificial Because Haber process is an exothermic reaction, a lower temperature will maximize the yield of product. Also, according to the ideal gas law, because the moles of gas are proportional to the volume, and the volume is inverse proportional to the pressure; a higher pressure is required to maximize the yield of product. At room temperature, a vessel is filled with gaseous carbon dioxide. Some water is added at 2.00 atm. It is well-shaken, integrating carbon dioxide gas into the water. 0.5 kg of the solution is taken out and boiled to extract 2.50 L of carbon dioxide. The system is at 10 degree Celsius and 1.00 atm. What is the Henry’s Law constant for carbon dioxide in water? given that the extracted solution: \[PV=nRT\] \[n=\frac{PV}{RT}=\frac{1atm\times 2.5L}{0.082\frac{L\ atm}{mol\ K} \times 283.15L}=0.1077\ mol\] \[mass_{CO_{2}}=n_{CO_{2}}\times molar\ mass = 0.1077\ mol \times 44\ g/mol=4.7388\ g\] \[mass_{H_{2}O}=500\ g - 4.7388\ g = 495.29\ g\] \[n_{H_{2}O}=\frac{495.29\ g}{18\ g/mol}$$ = 27.5 mol H O \[Solubility\ of\ CO_{2}\ in\ water=\frac{0.1077\ mol}{0.5\ kg}=0.2154\ mol/kg\] Due to the Ideal gas law, the ratio of moles equals to the ratio of pressure pressure: pressure = mol:mol = 0.1077: 27.51 because the total pressure is 2 atm, partial pressure of CO = 0.0724 atm Due to Henry's law: \[k=\frac{c}{P}=\frac{0.2154mol/kg}{0.0724\ atm}=2.9751\frac{mol}{kg\ atm}\] To calculate the values of the remaining two unknown variables, a substitution can be made that expresses \(\ce{\Delta}\text{H}\) in terms of \(\ce{\Delta}\text{S}\). The equilibrium constant for the reaction \(N_{2}O_{4}\rightleftharpoons 2NO_{2}(g)\) is measured to be \(5.9\times10^{-3}\) at 298\;K and \(1.3\times10^{-6}\) at \(398\;K\) a.Use the equation \(\Delta\;G^{\circ}=-RT\ln K\) \(\Delta\;G^{\circ}=-RTlnK=(-8.3145\;J\;K\;mol^{-1})(298\;K)(ln5.9\times10^{-3})=-1.29\times10^{4}\;J\;mol^{-1}\) For one mole of the reaction as written, \(\Delta\;G^{\circ}\) is -1290 kJ. b. Use van't Hoff equation and the values of K at 298 K and at 398 K to obtain \(\Delta\;H^{\circ}\). Then get \(\Delta\;S^{\circ}\) from \(\Delta\;G^{\circ}\) and the equation \(\Delta\;G^{\circ}=-RTlnK\) \(ln(\frac{1.3\times10^{-6}}{5.9\times10^{-3}})=\frac{\Delta\;H^{\circ}}{8.3145\;J\;K^{-1}{mol^{-1}}}(\frac{1}{398\;K}-\frac{1}{298\;K})\) \(\Delta\;H^{\circ}=83.0\;kJ\) The answer equals both \(\Delta\;H^{\circ}\) at 298 K and 398 K because it is assumed in the derivation of the van't Hoff equation that \(\Delta\;H^{\circ}\) is independent of temperature. Next, \(\Delta\;S^{\circ}=\frac{\Delta\;H^{\circ}-\Delta\;G^{\circ}}{T}=\frac{83.0\times10^{3}\;J-(-1.29\times10^{4}\;J)}{298\;K}=323\;J\;K^{-1}\) For the reaction: \(\ce{ Br2(g) <=> 2Br(g)}\), the equilibrium constant is \(4.64 \times 10^{-29}\) at 25 C and \(1.1 \times 10^{-3} \) at 1280 K. The equilibrium constant for the following reaction was experientially calculated at different temperatures. \[\ce{N2 (g) + 3H2 (g) \rightleftharpoons 2NH3 (g)} \nonumber \] The graph bellow shows what the graph should look like: \[\ln K = - \dfrac{\Delta{H^o}}{R} \dfrac{1}{T} + \dfrac{\Delta{S^o}}{R}\label{20}\nonumber \] Writing the equation like this how it can be compared with y = mx + b. With this information we then know that the slope of the graphe is equal to -\(\Delta{H^o}\)/R. So: \(\Delta{H^o}\) = -mR were R = 8.3145 J K mol \(\Delta{H^o}\) = -104.9 kJ mol \[\ce{I_{2\, (aq)} \rightleftharpoons I_{2} (CCl_4)}\nonumber \] 5°C \[\ce{C2H2ClO2H(s)} \rightleftharpoons \ce{C2H2ClO2H}_{(aq)} \nonumber \] \[\ce{C2H2ClO2H(s)} \rightleftharpoons \ce{C2H2ClO2H}_{(octanol)} \nonumber \] \[\ce{C2H2ClO2H(aq)} \rightleftharpoons \ce{C2H2ClO2H}_{(octanol)} \nonumber \] First, we will need g/L to M so that we can calculate the equilibrium constant. \[\left(\dfrac{0.130\; \cancel{g\; C_{2}H_{3}ClO_{2}} }{1\;L}\right) \left(\dfrac{1\: mol}{94.494\: \cancel{ g\; C_{2}H_{3}ClO_{2}}} \right) = 0.0013757\; M\nonumber \] \[\left(\dfrac{0.2834\; \cancel{g\; C_{2}H_{3}ClO_{2}} }{1\;L}\right) \left(\dfrac{1\: mol}{94.494\: \cancel{ g\; C_{2}H_{3}ClO_{2}}} \right) = 3.00x10^{-3}\; M\nonumber \] \[K = \dfrac{K_{forminde}}{K_{aq}}=\dfrac{3.00 \times 10^{-3}}{0.001376} =2.18\nonumber \] \[\log_{10} K_{oct/wat} =0.338\nonumber \] \[2NOBr_{(g)} \rightleftharpoons 2NO_{(s)} + Br_{2(g)} \nonumber \] At 300 K and 1 atm, N O is partly dissociated into NO . The density of the equilibrium mixture was 2.076 g/L. What is the degree of dissociation of N O under these circumstances? \[N_2O_4(g) \rightarrow 2NO_2(g) \nonumber \] Therefore, the degree of dissociation of SO is 0.8. (Multiplying number of moles by molar mass to get the total mass) 92*(1- x) + [46*(2x)/ (1+x) ]= [ 2.076 g/L * 0.08206 (L*atm)/(mol*K) ] / 1 atm On solving the equation, we get the equation, x= 0.8 The decompressions of calcium carbonate (\(\ce{CaCO3}\)) occurs according to the following reaction: \[\ce{ CaCO3 (s) \rightleftharpoons CaO(s) + CO2 (g)}\nonumber \] 100 g of solid calcium carbonate (\(\ce{CaCO3}\)) and 50 g of calcium oxide (\(\ce{CaO}\)) are placed in an evacuated round bottom flask that is heated up to 35°C. At this temperature the pressure in the round bottom flask is found to be 0.5 atm. Let's say you have the reaction: \[A(s) + B(g) \leftrightharpoons C(s) + D(g)\nonumber \] At 25°C, this reaction has an equilibrium constant of 157.2. Find the partial pressures of \(B(g)\) and \(D(g)\) when the total pressure of the system is equal to 4.3 atm. First, the equilibrium expression of this reaction is \[K_c = \frac {D(g)}{B(g)} = 157.2\nonumber \] Note, the A(s) and C(s) are not included in this expression because they are solids. Solids have an activity equal to 1 which basically means they do not affect the equilibrium of the reaction. To calculate the partial pressure of our unknowns, we can use Dalton's Law of partial pressures: \[P_A = X_A P_{Total}\nonumber \] Since we are looking for the partial pressure of each of our gases, and we already have the total pressure of the system, we need to find the mole fraction of each of our gases. To find it, we can manipulate the above equilibrium expression to get: \[D(g) = 157.2 \; B(g)\nonumber \] Now that we have the ratio of molecules in this reaction, we can construct workable mole fractions. \(X_D = \frac {157.2}{1 + 157.2}\) \(X_B = \frac {1}{1 + 157.2}\) Finally, we can plug everything in to find our partial pressures. \(P_D = \frac {157.2}{158.2} \times (4.3 \; atm) = 4.27 \; atm\) \(P_B = \frac {1}{157.8} \times (4.3 \; atm) = 0.027 \; atm\) Given the reaction \[C_3H_5N(g) + 3 \; H_2(g) \leftrightharpoons C_5H_{11}N(g)\nonumber \] whose equilibrium constant is solved by this equation \[log_{10}(K) = -20.28 + \frac {10.56 \; K}{T}\nonumber \] Part a) \[log_{10}(K) = -20.28 + \frac {10.56 \; K}{200 \; K}\nonumber \] \(K = 10^{-20.23} = 5.89 \times 10^{-21}\) part b) Set up the equilibrium constant of the reaction, plug in your partial pressures, and solve for \(C_3H_5N(g)\) \(K = \frac {[C_5H_{11}N]}{[C_5H_5N] \; [H_2]^3}\) \(K = \frac {2.0 \times 10^{-5}}{[C_5H_5N] \; [1]}\) \([C_5H_5N] = 3.40 \times 10^{15}\) The chemical equation for this reaction is \[HCOOH_{(aq)}+H_{2}O_{l}\leftrightarrow H_{3}O^{+}_{(aq)}+HCOO^{-}_{(aq)}\nonumber \] This indicates that \[K_{a}=\frac{[HCOO^{-},H_{3}O^{+}]}{[HCOOH]}\nonumber \] Because the volumes are equal and additive, the concentrations of the two solutions become half of what they were initially. At equilibrium, \[[HCOO^{-}]=1+x\nonumber \] \[[H_{3}O^{+}]=x\nonumber \] \[[HCOOH]=1-x\nonumber \] Plugging in these values into the equilibrium expression gives \[1.77\bullet 10^{-4}=\frac{[1+x,x]}{[1-x]}\nonumber \] Solving the equation for x gives \[x= 1.77\bullet 10^{-4}M\nonumber \] The ratio of HCOOH and HCOO is \[\frac{1+x}{1-x}\nonumber \] and due to x being such a small value, the ratio between the amounts of HCOOH and HCOO is essentially 1. °C. PbO(s) decomposes according to the equation \[\ce{PbO_{(s)} <=> Pb (s) + 1/2 O2(g)}\nonumber \] Since G ° for O (g) and Pb (s) is 0, for the reaction \[\Delta G^{\circ}=(0+0)-(-188.95kJ mol^{-1})=188.95kJ mol^{-1}\nonumber \] The equilibrium constant of the reaction can be computed by \[ln K= -\frac{\Delta G^{\circ}}{RT}=-76.22\nonumber \] \[K=7.898\bullet 10^{-34}\nonumber \] From the chemical equation, we can tell \[K=(P_{O_{2}})^{1/2}\nonumber \] Therefore the equilibrium pressure of O °C \[P_{O_{2}}=6.2\bullet 10^{-67}atm\nonumber \] If that weren’t enough, Jim’s sister, ., knocked his grass juice (a mixture of 1-Hexanol, \(Cl_{2}\), \(CaCl_{2}\), and cyclohexane) into the family’s benzene-ethanol vat! Assume that the two solvents in the vat are immiscible with each other but are miscible with the grass juice components. Jim separates the benzene and ethanol, and drinks the one with the greater amount of 1-Hexanol. Which one does he drink? Which of the two solvents have the highest concentration of the other components of the grass juice? There are two immiscible solvents in the family vat. What this means is that they do not mix/ dissolve well in each other. Consequently, when . poured the grass juice into this vat, the solution technically contains two solvents. Evidently, what this means is that each of the solutes (which are the grass juice constituents mentioned in the question) dissolves to a different extent in each solvent. When the question asks which solvent has the greater amount of 1-Hexanol, it is the same as asking does 1-Hexanol dissolve “more” in either benzene or ethanol (the two solvents). The concept of “like dissolves like” is very important here, and to solve this problem first benzene must be identified as non-polar and driven by dispersion forces, while ethanol is polar. Since polar solvents dissolve polar solutes, then ethanol should be able to dissolve \(CaCl_{2}\) and 1-Hexanol, both polar solutes, much better, while benzene can dissolve \(Cl_{2}\) and cyclohexane, which are nonpolar, to a greater extent. The aptitude that each solvent has in dissolving the solutes directly correlate to the concentrations of solutes that are present in the solvent. Thus, Jim would drink the ethanol, since it has the highest concentration of the polar 1-Hexanol. Also, the benzene would have the highest concentration of \(Cl_{2}\) and cyclohexane, and the ethanol would also have the highest concentration of \(CaCl_{2}\), which is ionic and therefore “very polar”. : “Like dissolves Like”: polar solvents dissolve polar solutes, and nonpolar solvents dissolve polar solutes Calculate the pH of a 0.5 M \(H_3PO_4\) acid solution given that \(K_{a1}=7.1 \times 10^{-3}\), \(K_{a2}=6.3 \times 10^{-8}\) and \(K_{a3}=4.2 \times 10^{-13}\). 1.25 Review on how to calculate the pH of a polyprotic acids . \[7.1 \times 10^{-3}=\dfrac{x^2}{(0.5-x)}\nonumber \] \[x=0.0561\;M=[H^+]\nonumber \] \[pH=-\log[H^+]=1.25\nonumber \] We assume that all of the \(H^+\) in solution comes from the dissociation of the first H because the second and the third is too small to have a significant effect in the pH of the solution. The Haber-Bosch process is a method invented in the 20th century as a method for hydrogen fixation. It is widely used in industry today, and involves reacting nitrogen and hydrogen in the following manner: \[\ce{N2 (g) + 3H2 (g) \rightleftharpoons 2 NH3 (g)}\nonumber \] Calculate \(\Delta{G_{rxn}^{\circ}}\). If the pressure of the products and reaction are all kept constant at the same value, in which direction will the reaction go? : \(\Delta{G_f^{\circ}}\; NH_{2\, (g)} = 199.83\dfrac{kJ}{mol}\) The process of calculating \(\Delta{G}\) should be very familiar after reading Q15, so this solution will not elaborate much on this method. With that being said, the \(\Delta{G_{rxn}^{\circ}}\) is as follows: \(\Delta{G_{rxn}^{\circ}} = 399.66\dfrac{kJ}{mol}\) Both \(N_{2}\) and \(H_{2}\) have respective \(\Delta{G_{f}}\) of zero. Now, remember that if \(\Delta{G_{rxn}^{\circ}}\) is negative (\(\Delta{G_{rxn}^{\circ}}\)<0), then the reaction is spontaneous in the direction mentioned. However, in this case the \(\Delta{G_{rxn}^{\circ}}\) is definitely NOT less than zero, so it is not spontaneous in the direction \(N_{2\, (g)} + 3H_{2\, (g)} \rightarrow 2 NH_{3\, (g)}\). However, if the chemical formula is "flipped" so that it goes from \(N_{2\, (g)} + 3H_{2\, (g)} \leftarrow 2 NH_{3\, (g)}\), Hess's law states that the corresponding \(\Delta{G_{rxn}^{\circ}}\) is also "flipped" from positive to negative (or vice versa, depending on the value. For this situation, however, the \(\Delta{G_{rxn}^{\circ}}\) does turn negative). What information can be drawn from this is that, since a negative \(\Delta{G_{rxn}^{\circ}}\) means that the reaction will be spontaneous, then it can be inferred that the reaction will favor going from \(N_{2\, (g)} + 3H_{2\, (g)} \leftarrow 2 NH_{3\, (g)}\), rather than \(N_{2\, (g)} + 3H_{2\, (g)} \rightarrow 2 NH_{3\, (g)}\). Without even having to calculate the equilibrium constant, from this information alone, it can be inferred that, at equilibrium, the reactants are favored over the products. Therefore, if both gasses are kept at constant pressure, the reaction will tend to increase concentration of reactants. Thus, the reaction will go left. : Find the \(\Delta{G_{rxn}^{\circ}}\), understand what implications it has in regards to equilibrium, and apply this knowledge to infer that the reaction will favor the reactants. At 273 K, the equation: H (g) + I (g) → 2HI(g) has an equilibrium constant of 60.2 a) ΔG° = -RT ln(K) ΔG° = -(8.3145 J/K mol)(273 K) ln(60.2) ΔG° = -9301.1 J/mol b) ΔG = ΔG° + RT ln(Q) ΔG = -9301.1 J/mol + (8.3145 J/K mol)(273 K) ln(P / P P ) ΔG = -9301.1 J/mol + (8.3145 J/K mol)(273 K) ln(2 / (6x4)) ΔG = -14941.49 J/mol. c) This reaction proceeds spontaneously in the forward direction. At 773 K the equilibrium constant for the Haber-Bosch process is 1.45x10 . The chemical reaction involved in the Haber-Bosch process is: \[ N_{2\, (g)} + 3H_{2\, (g)} \rightleftharpoons 2NH_{3\: (g)} \nonumber \] The value of \(\Delta G°(773) \) can be found using the equation \(\Delta G° = -RT \ln K \) where T and K are already known to be 773 K and 1.45x10 respectively while R=8.134 J K mol . \[ \Delta G°= -8.314\; J\; K^{-1}\; mol^{-1} \times 773K \times ln(1.45 \times 10^{-5}) = 70052\; J \nonumber \] \[\Delta G°= 70052\; J \nonumber \] The equation needed to determine \(\Delta G\) is \( \Delta G = RTln(Q/K) \). The values of R, T, and K are known while Q is \[Q = \dfrac{P_{NH_{3}}^{2}/P_{ref}} {(P_{N_{2}}/P_{ref})(P_{H_{2}}^{3}/P_{ref})} \nonumber \] P =1 atm and the other pressures are stated in the question. So to find Q, the pressures have to be plugged into the equation \[Q = \dfrac{(1)^{2}} {(20)(20)^{3}} = \dfrac{1}{160000} = 6.25 \times 10^{-6}\nonumber \] Now that the value of Q is known, \(\Delta G \) can be found \[ \Delta G= -8.314\; J\; K^{-1}\; mol^{-1} \times 773K \times ln(\frac{6.25 \times 10^{-6}}{1.45 \times 10^{-5}}) = -5291\; J \nonumber \] \[ \Delta G = -5291\; J \nonumber \] The previous reaction would run towards the products spontaneously because \(\Delta G < 0 \). The following reaction is exothermic \[\ce{SnO2(s) + 2H2(g) → Sn(g) + 2H2O(l)} \nonumber\] Given the chemical reaction \[\ce{ SnO2(s) \rightleftharpoons Sn (s) + O2 (g)} \nonumber \] The activities of solids and liquids are always equal to 1. The activities of gases are equal to the partial pressure of the gas in bars. Therefore the equilibrium expression would be \[ K = P_{O_{2}} \nonumber \] \(P_{O_{2}} \) would not change because the activity of a liquid is always 1 meaning it has no effect on the equilibrium constant \(P_{O_{2}} \) would decrease because the extra gas would push the system out of equilibrium and in order for the system to return to equilibrium, some of the O gas would have to react with some of the Sn . This would decrease the amount of O present in the system and therefore decrease the the pressure of O . Arrange the following solutions in order of most acidic to basic: NaOH, NaCN, H SO , NH NO , NaCl \[\ce{H2SO4 > NH4NO3 > NaCl > NaCN > NaOH }\] Review acids bases Given that the \(K_b\) values for \(\ce{NH_3}\) and \(\ce{NH_2NH_2}\) are \(1.8 \times 10^{-5}\) and \(8.5 \times 10^{-7}\), respectively. Which is the stronger acid? \(\ce{NH2NH2}\) Calculate the \(K_a\) values first. Review on how to identify the strength of an acid from its \(K_a\) value For NH , \[K_a= \dfrac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}\nonumber \] For NH NH , \[K_a= \dfrac{(1.0 \times 10^{-14}}{8.5 \times 10^{-7}}= 1.18 \times 10^{-8}\nonumber \] Since the Ka value of NH NH is larger, NH NH is the stronger acid. Calculate the pH of the following solutions. Ka=(1*10 )/(1.8*10 )=5.56*10
5.56*10 =x /(0.01-x)
x=2.36*10
pH= -log(x)= 5.63 Review on the properties of salts \[\ce{CN (aq) + H O (l) <=> HCN (aq) + OH (aq)}\] Kb=(1*10 )/(6.2*10 )=1.62*10 \[ 1.62 \times 10^{-5} =\dfrac{x^2}{0.5-x}\] x=0.00283 pOH= -log(x)= 2.55 pH= 14-pOH= 11.5 Identify the stronger acid in each of the following combination. Explain. Identify the Lewis acid and Lewis base for each of the following reaction. Lewis acid Lewis base 1. NH 2. Al H O 3. CO H O Study the concept of electron pairs H is the electron-pair acceptor while NH is the electron-pair donor. Al is the electron-pair acceptor while H O is the electron-pair donor. CO is the electron-pair acceptor while H O is the electron-pair donor. Beryllium hydroxide is an amphoteric compound. It can react as an acid as well as a base. Please write the reactions of Be(OH) and describe its role base on Brønsted-Lowry and Lewis theories. Be(OH) (s)+2H (aq)-> Be (aq)+2H O(l) (Brønsted-Lowry base, 2H acceptor) Be(OH) (s)+2OH (aq)-> Be(OH) (aq) (Lewis acid, electron pair acceptor) Consider Brønsted-Lowry and Lewis theories. | 37,217 | 1,079 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Solutions_and_Mixtures/Case_Studies/Dialysis |
is the separation of from dissolved ions or molecules of small dimensions, or , in a solution. A is any substance that is made of particles that are of an extremely small size: larger than atoms but generally have the size of 10 cm ranging to 10 cm. A is a substance that has some or all of the properties of a crystal or a substance that forms a true solution and diffuses through a membrane by dialysis. Dialysis is a process that is like osmosis. Osmosis is the process in which there is a diffusion of a solvent through a semipermeable membrane. In 1861, chemist (how developed ) used the process of dialysis, a process used to separate colloidal particles from dissolved ions or molecules. Dialysis is possible because of the through a . A semipermeable membrane is a membrane that lets some molecules to pass through it while not letting others (Figure \(\Page {1}\)). Examples of semipermeable membranes include parchment and cellophane. Another way to think of a semipermeable membrane is to think of a net like object that traps larger objects, but lets smaller object pass through because they can pass through the holes in the net. When a colloidal mixture is places in a semipermeable membrane, which is then placed in an aqueous solution or pure water, dissolved ions and small molecules are allowed to pass through this membrane. This causes colloidal particles to stay in the membrane, because these particles are unable to pass through the small pores of the membrane. Dialysis is not a quick process; the rate of dialysis depends on the speed of the unequal diffusion rates between the crystalloids and the colloids and the differences in particle size. The rate of dialysis can be changed through , or if the crystalloids are charged, then , called . Electrodialysis is the type of dialysis in which electrodes are placed on the sides of the membrane. In this way, positive ions can pass through one side of this membrane while the negatively charged ions can pass through the other side of the membrane. This causes acceleration in the process of dialysis. is a method in which kidney failure is treated with the process of dialysis. In hemodialysis, blood is removed, purified through dialysis, and returned to the bloodstream. In kidney failure, there is a retention of salts and water, urea, and metabolic acids. The patient is then connected to a dialysis machine, which is also called a hemodialyzer. The blood flows through small channels made of semipermeable membranes (Figure \(\Page {2}\)). The dissolved substances like urea and salts pass through a sterile solution. Compounds like sugar and amino acids are added to the sterile solution. The dialysis solution is on the other side of the membranes, and the molecules flow through the membranes. The molecules diffuse from a higher concentration to low concentration area. The concentrations of molecules needed to be removed from the blood are zero in the dialysis fluid. The process of hemodialysis helps many patients who have kidney failure because a person who suffers from kidney failure are at great risk, because someone who has complete kidney failure will need a kidney transplant within two weeks, or else he/she will face death. Between the time that the person finds a suitable kidney to be transplanted, the hemodialyzer comes into great help in facing the fight against death. | 3,409 | 1,081 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Energies_and_Potentials/Free_Energy/Helmholtz_(Free)_Energy |
(Hermann Ludwig Ferdinand von Helmholtz) \(A\) (for ): \[\left.A\right.=U-TS\] where is the internal energy, is the temperature and is the entropy. is a . The differential of this function is \[\left.dA\right.=dU-TdS-SdT\] From the second law of thermodynamics one obtains \[\left.dA\right.=TdS -pdV -TdS-SdT\] thus one arrives at \[\left.dA\right.=-pdV-SdT.\] For one has the following \[dA=\left(\dfrac{\partial A}{\partial T}\right)_V dT + \left(\dfrac{\partial A}{\partial V}\right)_T dV\] The following equation provides a link between classical thermodynamics and statistical mechanics: \[\left.A\right.=-k_B T \ln Q_{NVT}\] where \(k_B\) is the Boltzmann constant, is the temperature, and \(Q_{NVT}\) is the canonical ensemble partition function. A quantum correction can be calculated by making use of the of the partition function, resulting in (Eq. 3.5 in ): \[\dfrac{A-A_{ {\mathrm{classical}} }}{N} = \dfrac{\hbar^2}{24m(k_BT)^2} \langle F^2 \rangle \] where \(\langle F^2 \rangle\) is the mean squared force on any one atom due to all the other atoms. | 1,101 | 1,082 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Physical_Equilibria/Liquid-Solid_Phase_Diagrams%3A_Tin_and_Lead |
Throughout the whole experiment, heat is being lost to the surroundings - and yet the temperature doesn't fall at all while the lead is freezing. This is because the freezing process liberates heat at exactly the same rate that it is being lost to the surroundings. Energy is released when new bonds form - in this case, the strong metallic bonds in the solid lead. If you repeated this process for pure liquid tin, the shape of the graph would be exactly the same, except that the freezing point would now be at 232°C (the graph for this is further down the page.) If you add some tin to the lead, the shape of the cooling curve changes. The next graph shows what happens if you cool a liquid mixture containing about 67% lead and 33% tin by mass. There are lots of things to look at: If you had less tin in the mixture, the overall shape of the curve stays much the same, but the point at which the lead first starts to freeze changes. The less tin there is, the smaller the drop in the freezing point of the lead. For a mixture containing only 20% of tin, the freezing point of the lead is about 275°C. That's where the graph would suddenly become less steep. BUT . . . you will still get the graph going horizontal (showing the freezing of both the tin and lead) at exactly the same temperature: 183°C. As you increase the proportion of tin, the first signs of solid lead appear at lower and lower temperatures, but the final freezing of the whole mixture still happens at 183°C. That continues until you have added enough tin that the mixture contains 62% tin and 38% lead. At that point, the graph changes. This particular mixture of lead and tin has a cooling curve which looks exactly like that of a pure substance rather than a mixture. There is just the single horizontal part of the graph where everything is freezing. However, it is still a (not a solution). If you use a microscope to look at the solid formed after freezing, you can see the individual crystals of tin and lead. This particular mixture is known as a . The word "eutectic" comes from Greek and means "easily melted". The eutectic mixture has the lowest melting point (which is, of course, the same as the freezing point) of any mixture of lead and tin. The temperature at which the eutectic mixture freezes or melts is known as the eutectic temperature. You can trace it through in exactly the same way, by imagining starting with pure tin and then adding lead to it. The cooling curve for pure liquid tin looks like this: It's just like the pure lead cooling curve except that tin's freezing point is lower. If you add small amounts of lead to the tin, so that you have perhaps 80% tin and 20% lead, you will get a curve like this: Notice the lowered freezing point of the tin. Notice also the final freezing of the whole mixture again takes place at 183°C. As you increase the amount of lead (or decrease the amount of tin - same thing!) until there is 62% of tin and 38% of lead, you will again get the eutectic mixture with the curve we've already looked at. You start from data obtained from the cooling curves. You draw a graph of the temperature at which freezing first starts against the proportion of tin and lead in the mixture. The only unusual thing is that you draw the temperature scale at each end of the diagram instead of only at the left-hand side. Notice that at the left-hand side and right-hand sides of the curves you have the freezing points (melting points) of the pure lead and tin. To finish off the phase diagram, all you have to do is draw a single horizontal line across at the eutectic temperature. Then you label each area of the diagram with what you would find under the various different conditions. Suppose you have a mixture of 67% lead and 33% tin. That's the mixture from the first cooling curve plotted above. Suppose it is at a temperature of 300°C. That corresponds to a set of conditions in the area of the phase diagram labeled as molten tin and lead. Now consider what happens if you cool that mixture. Eventually the temperature will drop to a point where it crosses the line into the next region of the diagram. At that point, the mixture will start to produce some solid lead - in other words, the lead (but not the tin) starts to freeze. That happens at a temperature of about 250°C. Now, it is the next bit that needs careful thinking about, because there are two different ways you can look at it. If you have been taught to do it one way, then stick with that - otherwise you risk getting very confused! When the first of the lead freezes, the composition of the remaining liquid changes. It obviously becomes proportionally richer in tin. That lowers the freezing point of the lead a bit more, and so the next bit of lead freezes at a slightly lower temperature - leaving a liquid still richer in tin. This process goes on. The liquid gets richer and richer in tin, and the temperature needed to freeze the next lot of lead continues to fall. The set of conditions of temperature and composition of the liquid essentially moves down the curve - until it reaches the eutectic point. Once it has reached the eutectic point, if the temperature continues to fall, you obviously just move into the region of a mixture of solid lead and solid tin - in other words, all the remaining liquid freezes. We've seen that as the liquid gradually freezes, its composition changes. But if you look at the system as a whole, obviously the proportions of lead and tin remain constant - you aren't taking anything away or adding anything. All that is happening is that things are changing from liquids to solids. So suppose we continue the cooling beyond the temperature that the first solid lead appears and the temperature drops to the point shown in the next diagram - a point clearly in the "solid lead and molten mixture" area. What would you see in the mixture? To find out, you draw a horizontal through that point, and then look at the ends of it. At the left-hand end, you have 100% lead. That represents the solid lead that has frozen from the mixture. At the right-hand end, you have the composition of the liquid mixture. This is now much richer in tin than the whole system is - because obviously a fair bit of solid lead has separated out. As the temperature continues to fall, the composition of the liquid mixture (as shown by the right-hand end of the tie line) will get closer and closer to the eutectic mixture. It will finally reach the eutectic composition when the temperature drops to the eutectic temperature - and the whole lot then freezes. At a temperature lower than the eutectic temperature, you are obviously in the solid lead plus solid tin region. If you cooled a liquid mixture on the right-hand side of the phase diagram (to the right of the eutectic mixture), everything would work exactly the same except that solid tin would be formed instead of solid lead. If you have understood what has gone before, it isn't at all difficult to work out what happens. Finally . . . what happens if you cool a liquid mixture which has exactly the eutectic composition? It simply stays as a liquid mixture until the temperature falls enough that it all solidifies. You never get into the awkward areas of the phase diagram. Traditionally, tin-lead mixtures have been used as solder, but these are being phased out because of health concerns over the lead. This is especially the case where the solder is used to join water pipes where the water is used for drinking. New non-lead solders have been developed as safer replacements. Typical old-fashioned solders include: Jim Clark ( ) | 7,635 | 1,083 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/The_Four_Laws_of_Thermodynamics/Third_Law_of_Thermodynamics |
The 3rd law of thermodynamics will essentially allow us to quantify the absolute amplitude of entropies. It says that when we are considering a totally perfect (100% pure) crystalline structure, at absolute zero (0 Kelvin), it will have no entropy (S). Note that if the structure in question were not , then although it would only have an extremely small disorder (entropy) in space, we could not precisely say it had no entropy. One more thing, we all know that at zero Kelvin, there will still be some atomic motion present, but to continue making sense of this world, we have to assume that at absolute Kelvin there is no entropy whatsoever. From physics we know that the change in entropy \( \Delta S \) equals to the area under the graph of heat capacity (C) versus some temperature range. We can now extend this reasoning when trying to make sense of absolute entropies as well. First off, since absolute entropy depends on pressure we must define a standard pressure. It is conventional to choose the standard pressure of just 1 bar. Also, from now on when you see "S" we mean the absolute molar entropy at one bar of pressure. We know that \( \Delta S = S_{T=final} - S_{T=0} \); however, by the 3rd law this equation becomes \( \Delta S = S_{T=final} \). Now note that we can calculate the absolute entropy simply by extrapolating (from the above graph) the heat capacities all the way down to zero Kelvin. Actually, it is not exactly zero, but as close as we can possible get. For several reasons, it is so hard to measure the heat capacities at such low temperatures (T=0) that we must reserve to a different approach, much simpler. Debye's 3 thermodynamic law says that the heat capacities for most substances (does not apply to metals) is: \(C = bT^3\). It's possible to find the constant \(b\) if you fit Debye's equation to some experimental measurements of heat capacities extremely close to absolute zero (T=0 K). Just remember that \(b\) depends on the type of substance. Debye's law can be used to calculate the molar entropy at values infinitely close to absolute Kelvin temperatures: S(T) = (1/3)C(T) Note that \(C\) is the molar and constant volume heat capacity. | 2,205 | 1,085 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/05%3A_Energy_Changes_in_Chemical_Reactions/5.01%3A_Energy_Changes_in_Chemical_Reactions |
Because energy takes many forms, only some of which can be seen or felt, it is defined by its effect on matter. For example, microwave ovens produce energy to cook food, but we cannot see that energy. In contrast, we can see the energy produced by a light bulb when we switch on a lamp. In this section, we describe the forms of energy and discuss the relationship between energy, heat, and work. The forms of energy include thermal energy, radiant energy, electrical energy, nuclear energy, and chemical energy (Figure \(\Page {1}\) ). Thermal energy results from atomic and molecular motion; the faster the motion, the greater the thermal energy. The temperature of an object is a measure of its thermal energy content. Radiant energy is the energy carried by light, microwaves, and radio waves. Objects left in bright sunshine or exposed to microwaves become warm because much of the radiant energy they absorb is converted to thermal energy. Electrical energy results from the flow of electrically charged particles. When the ground and a cloud develop a separation of charge, for example, the resulting flow of electrons from one to the other produces lightning, a natural form of electrical energy. Nuclear energy is stored in the nucleus of an atom, and chemical energy is stored within a chemical compound because of a particular arrangement of atoms. Electrical energy, nuclear energy, and chemical energy are different forms of potential energy ( ) , which is energy stored in an object because of the relative positions or orientations of its components. A brick lying on the windowsill of a 10th-floor office has a great deal of potential energy, but until its position changes by falling, the energy is contained. In contrast, kinetic energy ( ) is energy due to the of an object. When the brick falls, its potential energy is transformed to kinetic energy, which is then transferred to the object on the ground that it strikes. The electrostatic attraction between oppositely charged particles is a form of potential energy, which is converted to kinetic energy when the charged particles move toward each other. Energy can be converted from one form to another (Figure \(\Page {2}\) ) or, as we saw with the brick, transferred from one object to another. For example, when you climb a ladder to a high diving board, your body uses chemical energy produced by the combustion of organic molecules. As you climb, the chemical energy is converted to to overcome the force of gravity. When you stand on the end of the diving board, your potential energy is greater than it was before you climbed the ladder: the greater the distance from the water, the greater the potential energy. When you then dive into the water, your potential energy is converted to kinetic energy as you fall, and when you hit the surface, some of that energy is transferred to the water, causing it to splash into the air. Chemical energy can also be converted to radiant energy; one common example is the light emitted by fireflies, which is produced from a chemical reaction. Although energy can be converted from one form to another, . This is known as the law of conservation of energy . One definition of energy is the capacity to do work. The easiest form of work to visualize is mechanical work Figure \(\Page {3}\) \[ \begin{matrix}
work &=&force\times distance \\
w & = & Fd
\end{matrix} \]
Because the force ( ) that opposes the action is equal to the mass ( ) of the object times its acceleration ( ), we can also write as follows:
\[ \begin{matrix}
work &=&mass\times acceleration \times distance \\
w & = & mad
\end{matrix} \] work performed on or by the object, or some combination of heat and work. Consider, for example, the energy stored in a fully charged battery. As shown in Figure \(\Page {4}\), this energy can be used primarily to perform work (e.g., running an electric fan) or to generate light and heat (e.g., illuminating a light bulb). When the battery is fully discharged in either case, the total change in energy is the same, even though the fraction released as work or heat varies greatly. The sum of the heat produced and the work performed equals the change in energy (Δ ):
\[ \begin{matrix}
Energy Change (\Delta E) &=&work + heat \\
\Delta E & = & q + w
\end{matrix} \] Energy can be transferred only in the form of heat, work performed on or by an object, or some combination of heat and work. Energy is an extensive property of matter—for example, the amount of in an object is proportional to both its mass and its temperature. (For more information on the properties of matter, see Chapter 1 ) A water heater that holds 150 L of water at 50°C contains much more thermal energy than does a 1 L pan of water at 50°C. Similarly, a bomb contains much more chemical energy than does a firecracker. We now present a more detailed description of kinetic and potential energy. The kinetic energy of an object is related to its mass and velocity :
\[ KE = \frac{1}{2} mv^{2} \] For example, the kinetic energy of a 1360 kg (approximately 3000 lb) automobile traveling at a velocity of 26.8 m/s (approximately 60 mi/h) is
\[ KE = \frac{1}{2} mv^{2} = \frac{1}{2} \left (1360 \; kg \right ) \] Because all forms of energy can be interconverted, energy in any form can be expressed using the same units as kinetic energy. The SI unit of energy, the joule (J). is defined as 1 kilogram·meter /second (kg·m /s ). Because a joule is such a small quantity of energy, chemists usually express energy in kilojoules (1 kJ = 10 J). For example, the kinetic energy of the 1360 kg car traveling at 26.8 m/s is 4.88 × 10 J or 4.88 × 10 kJ. It is important to remember that , whether thermal, radiant, chemical, or any other form. Because heat and work result in changes in energy, their units must also be the same. To demonstrate, let’s calculate the potential energy of the same 1360 kg automobile if it were parked on the top level of a parking garage 36.6 m (120 ft) high. Its potential energy is equivalent to the amount of work required to raise the vehicle from street level to the top level of the parking garage, which is given by Equation \(\Page {1}\) ( = ). According to Equation \(\Page {5}\), the force ( ) exerted by gravity on any object is equal to its mass ( , in this case, 1360 kg) times the acceleration ( ) due to gravity ( , 9.81 m/s at Earth’s surface). The distance ( ) is the height ( ) above street level (in this case, 36.6 m). Thus the potential energy of the car is as follows:
\[ \begin{matrix}
PE = Fd=mad=mgh \\
PE = \left ( 1360 \; kg \right ) \left ( 9.81 \; m \; s^{-2} \right ) \left ( 36.6 \; m \right ) = 4.88\times 10^{5} kg \; m^{2} \; s^{-2} \\
= 4.88\times 10^{5} J = 488 \; kJ
\end{matrix} \] The units of potential energy are the same as the units of kinetic energy. Notice that in this case the potential energy of the stationary automobile at the top of a 36.6 m high parking garage is the same as its kinetic energy at 60 mi/h. If the vehicle fell from the roof of the parking garage, its potential energy would be converted to kinetic energy, and it is reasonable to infer that the vehicle would be traveling at 60 mi/h just before it hit the ground, neglecting air resistance. After the car hit the ground, its potential and kinetic energy would both be zero. Potential energy is usually defined relative to an arbitrary standard position (in this case, the street was assigned an elevation of zero). As a result, we usually calculate only differences in potential energy: in this case, the difference between the potential energy of the car on the top level of the parking garage and the potential energy of the same car on the street at the base of the garage. A recent and spectacular example of the conversion of potential energy to kinetic energy was seen by the earthquake near the east coast of Honshu, Japan, on March 11, 2011. The magnitude 9.0 earthquake occurred along the Japan Trench subduction zone, the interface boundary between the Pacific and North American geological plates. During its westward movement, the Pacific plate became trapped under the North American plate, and its further movement was prevented. When there was sufficient potential energy to allow the Pacific plate to break free, approximately 7.1 × 10 kJ of potential energy was released as kinetic energy, the equivalent of 4.75 × 10 tn of TNT (trinitrotoluene) or 25,003 nuclear bombs. The island of Japan experienced the worst devastation in its history from the earthquake, resulting tsunami, and aftershocks. Historical records indicate that an earthquake of such force occurs in some region of the globe approximately every 1000 years. One such earthquake and resulting tsunami is speculated to have caused the destruction of the lost city of Atlantis, referred to by the ancient Greek philosopher Plato. The units of energy are the same for all forms of energy. Energy can also be expressed in the non-SI units of calories (cal) , where 1 cal was originally defined as the amount of energy needed to raise the temperature of exactly 1 g of water from 14.5°C to 15.5°C. The name is derived from the Latin , meaning “heat.” Although energy may be expressed as either calories or joules, calories were defined in terms of heat, whereas joules were defined in terms of motion. Because calories and joules are both units of energy, however, the calorie is now defined in terms of the joule:
\( 1\;cal = 4.184\; J exactly \tag{9.1.6a} \) \( 1\;J = 0.2390 \; cal \tag{9.1.6b} \) In this text, we will use the SI units—joules (J) and kilojoules (kJ)—exclusively, except when we deal with nutritional information, addressed in Section 9.7 . mass and velocity or height kinetic and potential energy Use Equation \(\Page {4}\) to calculate the kinetic energy and Equation \(\Page {6}\) to calculate the potential energy, as appropriate. Exercise Equation \(\Page {1}\) : = Equation \(\Page {3}\) : Δ = + Equation \(\Page {4}\) : Equation \(\Page {5}\).: = is a branch of chemistry that qualitatively and quantitatively describes the energy changes that occur during chemical reactions. is the capacity to do work. is the amount of energy required to move an object a given distance when opposed by a force. is due to the random motions of atoms, molecules, or ions in a substance. The of an object is a measure of the amount of thermal energy it contains. is the transfer of thermal energy from a hotter object to a cooler one. Energy can take many forms; most are different varieties of , energy caused by the relative position or orientation of an object. is the energy an object possesses due to its motion. Energy can be converted from one form to another, but the states that energy can be neither created nor destroyed. The most common units of energy are the , defined as 1 (kg·m )/s , and the , defined as the amount of energy needed to raise the temperature of 1 g of water by 1°C (1 cal = 4.184 J). What is the relationship between mechanical work and energy? Does a person with a mass of 50 kg climbing a height of 15 m do work? Explain your answer. Does that same person do work while descending a mountain? If a person exerts a force on an immovable object, does that person do work? Explain your answer. Explain the differences between electrical energy, nuclear energy, and chemical energy. The chapter describes thermal energy, radiant energy, electrical energy, nuclear energy, and chemical energy. Which form(s) of energy are represented by each of the following? Describe the various forms of energy that are interconverted when a flashlight is switched on. Describe the forms of energy that are interconverted when the space shuttle lifts off. Categorize each of the following as representing kinetic energy or potential energy. Are the units for potential energy the same as the units for kinetic energy? Can an absolute value for potential energy be obtained? Explain your answer. Categorize each of the following as representing kinetic energy or potential energy. Why does hammering a piece of sheet metal cause the metal to heat up? Technically, the person is not doing any work, since the object does not move. The kinetic energy of the hammer is transferred to the metal. Describe the mathematical relationship between (a) the thermal energy stored in an object and that object’s mass and (b) the thermal energy stored in an object and that object’s temperature. How much energy (in kilojoules) is released or stored when each of the following occurs? Calculate how much energy (in kilojoules) is released or stored when each of the following occurs: A car weighing 1438 kg falls off a bridge that is 211 ft high. Ignoring air resistance, how much energy is released when the car hits the water? A 1 tn roller coaster filled with passengers reaches a height of 28 m before accelerating downhill. How much energy is released when the roller coaster reaches the bottom of the hill? Assume no energy is lost due to friction. 250 kJ released | 13,113 | 1,086 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Advanced_Thermodynamics/2._The_Postulates_of_Thermodynamics |
Thermodynamics has three important ingredients: fundamental relations, which describe the relationship between state functions for a particular system; postulates that make statements about the state functions from which all other thermodynamic statements can be derived; and a body of mathematical manipulations that allows one to derive theorems from the postulates, and manipulate the fundamental relations in order to obtain the desired result. It is worth noting at this stage that fundamental relations cannot be derived from the postulates. The postulates provide some constraint on what is a valid fundamental relation, but otherwise leave a lot of freedom. Fundamental relations must be determined either empirically, or through a model. Statistical mechanics provides a means of deriving fundamental relations from the most generally valid of models, dynamics, either classical or quantum. For now, we will assume that fundamental relations have been obtained via measurement or from a model, and we only manipulate them. We begin with the postulates of thermodynamics before considering their mathematical manipulation. The postulates are like axioms in mathematics, but with one important difference: they can actually be derived as special cases of the postulates of statistical mechanics, by letting the particle number go to infinity. The quantity \(U\) is conserved for a closed system. Notes: For a set of simple systems \(\{S_k\}\), there exist single-valued, continuous, and differentiable extensive state functions \(S_k(U_k, X_{ik})\), defined for stable equilibrium states, such that for a composite system \(\{S\}=\sum_k \oplus \{s_k\}\), the state functions \(U_k\) and \(X_{ik}\) take on those values that maximize the entropy \(S=\sum S_k\) of the composite system, subject to its internal constraints. Notes: Note: This will later be seen equivalent to the statement \[\lim_{t \rightarrow 0} S=0\] because \[ \left( \dfrac{\partial U}{\partial S} \right)_x = T\]. We now can outline a method for the general solution of thermodynamic problems: Consider the following example of this method: a closed box is partitioned into volumes \(V_1\) and \(V_2\) by an impermeable wall, each side of which is filled with \(n_1\) and \(n_2\) moles of a gas having the fundamental relation \(S_k = c + n_kR\ln V_k\). When equilibrium is reached, what is the relationship between volumes \(V_1\) and \(V_2\)? The main problem with this approach using the postulates directly is that the fundamental relations usually are unknown! Instead, partial information about the system in the form of equations of state such as \(PV=nRT\), \(U = 3/2 nRT\) is usually available, and one must see what information can be extracted from them subject to the known constraints. Note that thermodynamics provides no clue as to the functional nature of \(S(U, N_i)\), except that it must be compatible with the postulates. As we will see in the next chapter, the fundamental relations can be obtained if enough equations of state are known. | 3,056 | 1,088 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/06%3A_Structures_and_Energetics_of_Metallic_and_Ionic_solids/6.11%3A_Ionic_Lattices/6.11A%3A_Structure_-_Rock_Salt_(NaCl) |
Rock salt also known as NaCl is an ionic compound. It occurs naturally as white cubic crystals. The structure of NaCl is formed by repeating the unit cell. It has an organized structure and has a 1:1 ratio of Na:Cl. Rock salt (\(\ce{NaCl}\)) is an ionic compound that occurs naturally as white crystals. It is extracted from the mineral form halite or evaporation of seawater. The structure of NaCl is formed by repeating the face centered cubic unit cell. It has 1:1 stoichiometry ratio of Na:Cl with a molar mass of 58.4 g/mol. Compounds with the sodium chloride structure include alikali halides and metal oxides and transition-metal compounds. An important role to many important applications is structure and dynamics of water. Some applications include crystallization of proteins and conformational behavior of peptides and nucleic acids. Figure \(\Page {1}\) shows how the Na and Cl ions occupy the space. The smaller ions are the Na with has an atomic radius of 102 pm, and the larger ions are the Cl with an atomic radium of 181 pm. Since NaCl are one to one ratio as a compound, the coordination numbers of Na and Cl are equal. The larger green ions represent Cl and the smaller purple ions represent Na . However, the structure of this molecule allows their positions to be switched since the coordination numbers are equivalent. The unit cell of \(\ce{NaCl}\) consists of \(\ce{Na^{+}}\) ions and \(\ce{Cl^{-}}\) ions. There are four types of site: unique central position, face site, edge sites and corner site, which are used to determine the number of Na ions and Cl ions in the unit cell of NaCl. When counting the number of ions, a corner site would be shared by 7 other unit cells. Therefore, 1 corner would be 1/8 of an ion. A similar occurrence happens with the face site and the edge sites. For a face site, it is shared by 1 other unit cell and for an edge site, the ion is shared by 3 other unit cells. \(\ce{NaCl}\) is a face centered cubic unit cell which has four cations and four anions. This can be shown by counting the number of ions and multiplying them in relation to their position. Each ion in this lattice has six of the other kind of ion as its nearest neighbors, and twelve of the same kind of ions as its second nearest neighbors. There are many ionic compounds that assume this structure including all other halides of Na, Li, K and Rb. CsF, AgF, AgCl, BaO, CoO, and SrS are also among many that will form similar structures to NaCl. | 2,497 | 1,089 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/11%3A_Quantum_Mechanics_and_Atomic_Structure/11.01%3A_The_Wave_Theory_of_Light |
Scientists discovered much of what we know about the structure of the atom by observing the interaction of atoms with various forms of radiant, or transmitted, energy, such as the energy associated with the visible light we detect with our eyes, the infrared radiation we feel as heat, the ultraviolet light that causes sunburn, and the x-rays that produce images of our teeth or bones. All these forms of radiant energy should be familiar to you. We begin our discussion of the development of our current atomic model by describing the properties of waves and the various forms of electromagnetic radiation. . Anyone who has visited a beach or dropped a stone into a puddle has observed waves traveling through water (Figure 1.1.1). These waves are produced when wind, a stone, or some other disturbance, such as a passing boat, transfers energy to the water, causing the surface to oscillate up and down as the energy travels outward from its point of origin. As a wave passes a particular point on the surface of the water, anything floating there moves up and down. Waves have characteristic properties (Figure 1.1.2). As you may have noticed in Figure 1.1.1, waves are periodic, that is, they repeat regularly in both space and time. The distance between two corresponding points in a wave—between the midpoints of two peaks, for example, or two troughs—is the wavelength (λ), distance between two corresponding points in a wave—between the midpoints of two peaks or two troughs. \(\lambda\) is the lowercase Greek lambda, and \(\nu\) is the lowercase Greek nu. Wavelengths are described by a unit of distance, typically meters. The frequency (ν), the number of oscillations (i.e., of a wave) that pass a particular point in a given period of time. of a wave is the number of oscillations that pass a particular point in a given period of time. The usual units are oscillations persecond (1/s = s ), which in the SI system is called the hertz (Hz). (wavelength)(frequency) = speed \[ \lambda \nu =v \tag{1.1.1a}\] \[ \left ( \dfrac{meters}{\cancel{wave}} \right )\left ( \dfrac{\cancel{wave}}{second} \right )=\dfrac{meters}{second} \tag{1.1.1b}\] Be careful not to confuse the symbols for the speed, \(v\), with the frequency, \(\nu\). Water waves are slow compared to sound waves, which can travel through solids, liquids, and gases. Whereas water waves may travel a few meters per second, the speed of sound in dry air at 20°C is 343.5 m/s. Ultrasonic waves, which travel at an even higher speed (>1500 m/s) and have a greater frequency, are used in such diverse applications as locating underwater objects and the medical imaging of internal organs. Water waves transmit energy through space by the periodic oscillation of matter (the water). In contrast, energy that is transmitted, or radiated, through space in the form of periodic oscillations of electric and magnetic fields is known as electromagnetic radiation, which is energy that is transmitted, or radiated, through space in the form of periodic oscillations of electric and magnetic fields. (Figure 1.1.3). Some forms of electromagnetic radiation are shown in Figure 1.1.4. In a vacuum, all forms of electromagnetic radiation—whether microwaves, visible light, or gamma rays—travel at the speed of light (c), which is the speed with which all forms of electromagnetic , a fundamental physical constant with a value of 2.99792458 × 10 m/s (which is about 3.00 ×10 m/s or 1.86 × 10 mi/s). This is about a million times faster than the speed of sound. Because the various kinds of electromagnetic radiation all have the same speed ( ), they differ in only wavelength and frequency. As shown in Figure 1.1.4 and Table 1.1.1 , the wavelengths of familiar electromagnetic radiation range from 10 m for radio waves to 10 m for gamma rays, which are emitted by nuclear reactions. By replacing with in Equation 6.1.1, we can show that the frequency of electromagnetic radiation is inversely proportional to its wavelength: \[ \begin{array}{cc} c=\lambda \nu \\ \nu =\dfrac{c}{\lambda } \end{array} \tag{1.1.2} \] For example, the frequency of radio waves is about 10 Hz, whereas the frequency of gamma rays is about 10 Hz. Visible light, which is electromagnetic radiation that can be detected by the human eye, has wavelengths between about 7 × 10 m (700 nm, or 4.3 × 10 Hz) and 4 × 10 m (400 nm, or 7.5 × 10 Hz). Note that when frequency increases, wavelength decreases; c being a constant stays the same. Similarly when frequency decreases, the wavelength increases. Within this visible range our eyes perceive radiation of different wavelengths (or frequencies) as light of different colors, ranging from red to violet in order of decreasing wavelength. The components of white light—a mixture of all the frequencies of visible light—can be separated by a prism, as shown in part (b) in Figure 1.1.4. A similar phenomenon creates a rainbow, where water droplets suspended in the air act as tiny prisms. As you will soon see, the energy of electromagnetic radiation is directly proportional to its frequency and inversely proportional to its wavelength: \[ E\; \propto\; \nu \tag{1.1.3}\] \[ E\; \propto\; \dfrac{1}{\lambda } \tag{1.1.4}\] Whereas visible light is essentially harmless to our skin, ultraviolet light, with wavelengths of ≤ 400 nm, has enough energy to cause severe damage to our skin in the form of sunburn. Because the ozone layer absorbs sunlight with wavelengths less than 350 nm, it protects us from the damaging effects of highly energetic ultraviolet radiation. The energy of electromagnetic radiation with frequency and wavelength. Your favorite FM radio station, WXYZ, broadcasts at a frequency of 101.1 MHz. What is the wavelength of this radiation? frequency wavelength Substitute the value for the speed of light in meters per second into Equation 1.1.2 to calculate the wavelength in meters. From Equation 1.1.2 , we know that the product of the wavelength and the frequency is the speed of the wave, which for electromagnetic radiation is 2.998 × 10 m/s: \[λν = c = 2.998 \times 10^8 m/s \nonumber\] Thus the wavelength λ is given by \[ \lambda =\dfrac{c}{\nu }=\left ( \dfrac{2.988\times 10^{8}\; m/\cancel{s}}{101.1\; \cancel{MHz}} \right )\left ( \dfrac{1\; \cancel{MHz}}{10^{6}\; \cancel{s^{-1}}} \right )=2.965\; m \nonumber\] As the police officer was writing up your speeding ticket, she mentioned that she was using a state-of-the-art radar gun operating at 35.5 GHz. What is the wavelength of the radiation emitted by the radar gun? A basic knowledge of the electronic structure of atoms requires an understanding of the properties of waves and electromagnetic radiation. A wave is a periodic oscillation by which energy is transmitted through space. All waves are periodic, repeating regularly in both space and time. Waves are characterized by several interrelated properties: wavelength (λ), the distance between successive waves; frequency (ν), the number of waves that pass a fixed point per unit time; speed (v), the rate at which the wave propagates through space; and amplitude, the magnitude of the oscillation about the mean position. The speed of a wave is equal to the product of its wavelength and frequency. Electromagnetic radiation consists of two perpendicular waves, one electric and one magnetic, propagating at the speed of light (c). Electromagnetic radiation is radiant energy that includes radio waves, microwaves, visible light, x-rays, and gamma rays, which differ in their frequencies and wavelengths. ( ) | 7,578 | 1,094 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/4_f-Block_Elements/The_Actinides/Chemistry_of_Plutonium |
Discovered by Seaborg (see element 106) in 1941, plutonium was named for the 9th planet, completing the series of elements named for planets which begins with uranium. Evidence for the existence of plutonium was obtained from work with samples of neptunium and the eventual synthesis in fact results from the beta decay of neptunium-239. The longest-lived isotope of plutonium is Pu-244 with a half-life of 82 million years. However the isotope of chief interest is Pu-239 which, like U-235, is fissionable. Most of the nuclear weapons built by the "great powers" today are based on Pu-239 which is derived from U-238 in special "breeder" reactors. Pu-239 is also a by-product of normal fission power reactors and accounts for a good deal of the concern over nuclear waste by-products since it is both highly radioactive and exceptionally toxic. Also, since the critical mass of plutonium is only about one third that of U-235, the possibility for terrorist diversion of the material is considered a serious matter. A piece of plutonium about the size of a softball would feel hot to the touch because of the high level of alpha particle radiation given off. A somewhat larger piece of the metal would boil water within minutes. Plutonium is occasionally used in deep-space probes as a source of energy (too far from the sun for effective solar power), the heat being directly converted into electricity by a special device. | 1,444 | 1,095 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/11%3A_Quantum_Mechanics_and_Atomic_Structure/11.03%3A_The_Photoelectric_Effect |
Nature, it seemed, was quantized (non-continuous, or discrete). If this was so, how could Maxwell’s equations correctly predict the result of the blackbody radiator? Planck spent a good deal of time attempting to reconcile the behavior of electromagnetic waves with the discrete nature of the blackbody radiation, to no avail. It was not until 1905, with yet another paper published by Albert Einstein, that the wave nature of light was expanded to include the particle interpretation of light which adequately explained Planck’s equation. The photoelectric effect was first documented in 1887 by the German physicist Heinrich Hertz and is therefore sometimes referred to as the Hertz effect. While working with a spark-gap transmitter (a primitive radio-broadcasting device), Hertz discovered that upon absorption of certain frequencies of light, substances would give off a visible spark. In 1899, this spark was identified as light-excited electrons (called ) leaving the metal's surface by J.J. Thomson (Figure 1.3.1
). The classical picture underlying the photoelectron effect was that the atoms in the metal contained electrons, that were shaken and caused to vibrate by the oscillating electric field of the incident radiation. Eventually some of them would be shaken loose, and would be ejected from the cathode. It is worthwhile considering carefully how the and of electrons emitted would be expected to vary with the and of the incident radiation along with the time needed to observe the photoelectrons. In 1902, Hertz's student, Lenard, studied how the energy of the emitted photoelectrons varied with the intensity of the light. He used a carbon arc light and could increase the intensity a thousand-fold. The ejected electrons hit another metal plate, the collector, which was connected to the cathode by a wire with a sensitive ammeter, to measure the current produced by the illumination (Figure 1.3.2
). To measure the energy of the ejected electrons, Lenard charged the collector plate negatively, to repel the electrons coming towards it. Thus, only electrons ejected with enough kinetic energy to get up this potential hill would contribute to the current. Lenard discovered that there was a well defined minimum voltage that stopped any electrons getting through (\(V_{stop}\)). To Lenard's surprise, he found that \(V_{stop}\) did not depend at all on the intensity of the light! Doubling the light intensity doubled the of electrons emitted, but did not affect the of the emitted electrons. The more powerful oscillating field ejected more electrons, but the maximum individual energy of the ejected electrons was the same as for the weaker field (Figure 1.3.2
). The American experimental physicist Robert Millikan followed up on Lenard's experiments and using a powerful arc lamp, he was able to generate sufficient light intensity to separate out the colors and check the photoelectric effect using light of different colors. He found that the maximum energy of the ejected electrons depend on the color - the shorter wavelength, higher frequency light eject photoelectrons with greater kinetic energy (Figures 1.3.3
). As shown in Figure 1.3.4
, just the opposite behavior from classical is observed from Lenard's and Millikan's experiments. The intensity affects the number of electrons, and the frequency affects the kinetic energy of the emitted electrons. From these sketches, we see that Classical theory predicts that energy carried by light is proportional to its amplitude independent of its frequency, and this fails to correctly explain the observed wavelength dependence in Lenard's and Millikan's observations. As with most of the experimental results we discuss in this text, the behavior described above is a simplification of the true experimental results observed in the laboratory. A more complex description involves a greater introduction of more complex physics and instrumentation, which will be ignored for now. In 1905 Einstein gave a very simple interpretation of Lenard's results and borrowed Planck's hypothesis about the quantized energy from his blackbody research and assumed that the incoming radiation should be thought of as quanta of energy \(h\nu\), with \(\nu\) the frequency. In photoemission, one such quantum is absorbed by one electron. If the electron is some distance into the material of the cathode, some energy will be lost as it moves towards the surface. There will always be some electrostatic cost as the electron leaves the surface, which is the workfunction, \(\Phi\). The most energetic electrons emitted will be those very close to the surface, and they will leave the cathode with kinetic energy \[KE = h\nu - \Phi \label{Eq1} \] On cranking up the negative voltage on the collector plate until the current just stops, that is, to \(V_{stop}\), the highest kinetic energy electrons (\(KE_e\)) must have had energy \(eV_{stop}\) upon leaving the cathode. Thus, \[eV_{stop} = h\nu - \Phi \label{Eq2} \] Thus, Einstein's theory makes a very definite quantitative prediction: if the frequency of the incident light is varied, and \(V_{stop}\) plotted as a function of frequency, the slope of the line should be \(\frac{h}{e}\) (Figure \(\Page {4A}\)). It is also clear that there is a minimum light frequency for a given metal \(\nu_o\), that for which the quantum of energy is equal to \(\Phi\) (Equation \ref{Eq1}). Light below that frequency, no matter how bright, will not eject electrons. Since every photon of sufficient energy excites only one electron, increasing the light's intensity (i.e. the number of photons/sec) only increases the of released electrons and not their kinetic energy. In addition, no time is necessary for the atom to be heated to a critical temperature and therefore the release of the electron is nearly instantaneous upon absorption of the light. Finally, because the photons must be above a certain energy to satisfy the workfunction, a threshold frequency exists below which no photoelectrons are observed. This frequency is measured in units of Hertz (1/second) in honor of the discoverer of the photoelectric effect. Einstein's Equation \(\ref{Eq1}\) explains the properties of the photoelectric effect quantitatively. A strange implication of this experiment is that light can behave as a kind of massless "particle" now known as a whose energy \(E=h\nu\) can be transferred to an actual particle (an electron), imparting kinetic energy to it, just as in an elastic collision between to massive particles such as billiard balls. Robert Millikan initially did not accept Einstein's theory, which he saw as an attack on the wave theory of light, and worked for ten years until 1916, on the photoelectric effect. He even devised techniques for scraping clean the metal surfaces inside the vacuum tube. For all his efforts he found disappointing results: he confirmed Einstein's theory after ten years. In what he writes in his paper, Millikan is still desperately struggling to avoid this conclusion. However, by the time of his Nobel Prize acceptance speech, he has changed his mind rather drastically! Einstein's simple explanation (Equation \ref{Eq1}) completely accounted for the observed phenomena in Lenard's and Millikan's experiments (Figure 1.3.4
) and began an investigation into the field we now call . This new field seeks to provide a quantum explanation for classical mechanics and create a more unified theory of physics and thermodynamics. The study of the photoelectric effect has also lead to the creation of new field of photoelectron spectroscopy. Einstein's theory of the photoelectron presented a completely different way to measure Planck's constant than from black-body radiation. While the workfunctions and ionization energies appear as similar concepts, they are independent. The workfunction of a metal is the minimum amount of energy (\(\ce{E}\)) necessary to remove an electron from the surface of the bulk ( ) metal (sometimes referred to as ). \[\ce{M (s) + \Phi \rightarrow M^{+}(s) + e^{-}}(\text{free with no kinetic energy}) \nonumber \] The workfunction is qualitatively similar to ionization energy (\(\ce{IE}\)), which is the amount of energy required to remove an electron from an atom or molecule in the state. \[\ce{M (g) + IE \rightarrow M^{+}(g) + e^{-}} (\text{free with no kinetic energy}) \nonumber \] However, these two energies differ in magnitude (Table 1.3.1
). For instance, copper has a workfunction of about 4.7 eV, but has a higher ionization energy of 7.7 eV. Generally, the ionization energies for metals are greater than the corresponding workfunctions (i.e., the electrons are less tightly bound in bulk metal). To solve part (a), note that the energy of a photon is given by \(E=h\nu\). For part (b), once the energy of the photon is calculated, it is a straightforward application of Equation \ref{Eq1} to find the ejected electron’s maximum kinetic energy, since \(\Phi\) is given. Photon energy is given by \[E = h\nu \nonumber \] Since we are given the wavelength rather than the frequency, we solve the familiar relationship \(c=\nu\lambda\) for the frequency, yielding \[\nu=\dfrac{c}{\lambda} \nonumber \] Combining these two equations gives the useful relationship \[E=\dfrac{hc}{\lambda} \nonumber \] Now substituting known values yields \[\begin{align*} E &= \dfrac{(6.63 \times 10^{-34}\; J \cdot s)(3.00 \times 10^8 m/s)}{420 \times 10^{-9}\; m} \\[4pt] &= 4.74 \times 10^{-19}\; J \end{align*} \nonumber \] Converting to eV, the energy of the photon is \[\begin{align*} E&=(4.74 \times 10^{-19}\; J) \left( \dfrac{1 \;eV}{1.6 \times 10^{-19}\;J} \right) \\[4pt] &= 2.96\; eV. \nonumber \end{align*} \nonumber \] Finding the kinetic energy of the ejected electron is now a simple application of Equation \ref{Eq1}. Substituting the photon energy and binding energy yields \[\begin{align*} KE_e &=h\nu – \Phi \\[4pt] &= 2.96 \;eV – 2.71 \;eV \\[4pt] &= 0.246\; eV.\nonumber \end{align*} \nonumber \] The energy of this 420-nm photon of violet light is a tiny fraction of a joule, and so it is no wonder that a single photon would be difficult for us to sense directly—humans are more attuned to energies on the order of joules. But looking at the energy in electron volts, we can see that this photon has enough energy to affect atoms and molecules. A DNA molecule can be broken with about 1 eV of energy, for example, and typical atomic and molecular energies are on the order of eV, so that the UV photon in this example could have biological effects. The ejected electron (called a photoelectron) has a rather low energy, and it would not travel far, except in a vacuum. The electron would be stopped by a retarding potential of 0.26 eV. In fact, if the photon wavelength were longer and its energy less than 2.71 eV, then the formula would give a negative kinetic energy, an impossibility. This simply means that the 420-nm photons with their 2.96-eV energy are not much above the frequency threshold. You can show for yourself that the threshold wavelength is 459 nm (blue light). This means that if calcium metal is used in a light meter, the meter will be insensitive to wavelengths longer than those of blue light. Such a light meter would be insensitive to red light, for example. What is the longest-wavelength electromagnetic radiation that can eject a photoelectron from silver? Is this in the visible range? Given that the workfunction is 4.72 eV from Table 1.3.1
, then only photons with wavelengths lower than 263 nm will induce photoelectrons (calculated via \(E=h \nu\)). This is ultraviolet and not in the visible range. The workfunction of a metal refers to the minimum energy required to extract an electron from the surface of a ( ) metal by the absorption a photon of light. The workfunction will vary from metal to metal. In contrast, ionization energy is the energy needed to detach electrons from and also varies with each particular atom, with the valence electrons require less energy to extract than core electrons (i.e., from lower shells) that are more closely bound to the nuclei. The electrons in the metal lattice there less bound (i.e., free to move within the metal) and removing one of these electrons is much easier than removing an electron from an atom because the metallic bonds of the bulk metal reduces their binding energy. As we will show in subsequent chapters, the more delocalized a particle is, the lower its energy. Although Hertz discovered the photoelectron in 1887, it was not until 1905 that a theory was proposed that explained the effect completely. The theory was proposed by Einstein and it made the claim that electromagnetic radiation had to be thought of as a series of particles, called photons, which collide with the electrons on the surface and emit them. This theory ran contrary to the belief that electromagnetic radiation was a wave and thus it was not recognized as correct until 1916 when Robert Millikan experimentally confirmed the theory The photoelectric effect is the process in which electromagnetic radiation ejects electrons from a material. Einstein proposed photons to be quanta of electromagnetic radiation having energy \(E=h\nu\) is the frequency of the radiation. All electromagnetic radiation is composed of photons. As Einstein explained, all characteristics of the photoelectric effect are due to the interaction of individual photons with individual electrons. The maximum kinetic energy \(KE_e\) of ejected electrons (photoelectrons) is given by \(KE_e=h\nu – \Phi\), where \(h\nu\) is the photon energy and \(\Phi\) is the workfunction (or binding energy) of the electron to the particular material. (Beams Professor, , ( ) ") ). | 13,769 | 1,096 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/4_f-Block_Elements/The_Lanthanides/Chemistry_of_Yttrium |
Like the histories of the discoveries of many other rare-earths, the tale of erbium reads like a series of mistaken identities. These elements are generally found as oxides and most often together. Chemically, the oxides are very similar and at the time of their first examination were difficult to separate. Thus a sample of "lanthanum" might end up containing two additional elements that no one had bothered to look for. Many chemists thought that the oxides were elements themselves at one time. The oxide of yttrium (which along with scandium and lanthanum is generally included with the "rare-earths") known as yttria was eventually found to contain erbia and terbia as well, the oxides of, respectively, erbium and terbium. But the two are so similar that they were often confused in early work and what we now call erbium was originally terbium! In both cases, the credit for discovery goes to Mosander (1843 for erbium) and both elements were named for the Swedish town of Ytterby (which, by the way, also lends its name to Ytterbium and Yttrium----certainly some kind of record where naming elements is concerned!). Like most of the rare-earth metals, erbium is silvery and soft, tarnishing slightly in air. | 1,237 | 1,098 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/04%3A_The_Second_Law_of_Thermodynamics/4.04%3A_The_Third_Law_of_Thermodynamics |
The atoms, molecules, or ions that compose a chemical system can undergo several types of molecular motion, including translation, rotation, and vibration (Figure \(\Page {1}\)). The greater the molecular motion of a system, the greater the number of possible microstates and the higher the entropy. A perfectly ordered system with only a single microstate available to it would have an entropy of zero. The only system that meets this criterion is a perfect crystal at a temperature of absolute zero (0 K), in which each component atom, molecule, or ion is fixed in place within a crystal lattice and exhibits no motion (ignoring quantum ). This system may be described by a single microstate, as its purity, perfect crystallinity and complete lack of motion (at least classically, quantum mechanics argues for constant motion) means there is but one possible location for each identical atom or molecule comprising the crystal (\(\Omega = 1\)). According to the Boltzmann equation, the entropy of this system is zero. \[\begin{align*} S&=k\ln \Omega \\[4pt] &= k\ln(1) \\[4pt] &=0 \label{\(\Page {5}\)} \end{align*}\] In practice, absolute zero is an ideal temperature that is unobtainable, and a perfect single crystal is also an ideal that cannot be achieved. Nonetheless, the combination of these two ideals constitutes the basis for the third law of thermodynamics: the entropy of any perfectly ordered, crystalline substance at absolute zero is zero. The entropy of a pure, perfect crystalline substance at 0 K is zero. The third law of thermodynamics has two important consequences: it defines the sign of the entropy of any substance at temperatures above absolute zero as positive, and it provides a fixed reference point that allows us to measure the absolute entropy of any substance at any temperature. In this section, we examine two different ways to calculate ΔS for a reaction or a physical change. The first, based on the definition of absolute entropy provided by the third law of thermodynamics, uses tabulated values of absolute entropies of substances. The second, based on the fact that entropy is a state function, uses a thermodynamic cycle similar to those discussed previously. One way of calculating \(ΔS\) for a reaction is to use tabulated values of the standard molar entropy (\(S^o\)), which is the entropy of 1 mol of a substance under standard pressure (1 bar). Often the standard molar entropy is given at 298 K and is often demarked as \(ΔS^o_{298}\). The units of \(S^o\) are J/(mol•K). Unlike enthalpy or internal energy, it is possible to obtain absolute entropy values by measuring the entropy change that occurs between the reference point of 0 K (corresponding to \(S = 0\)) and 298 K ( and ). As shown in Table \(\Page {1}\), for substances with approximately the same molar mass and number of atoms, \(S^o\) values fall in the order \[S^o(\text{gas}) \gg S^o(\text{liquid}) > S^o(\text{solid}).\] For instance, \(S^o\) for liquid water is 70.0 J/(mol•K), whereas \(S^o\) for water vapor is 188.8 J/(mol•K). Likewise, \(S^o\) is 260.7 J/(mol•K) for gaseous \(\ce{I2}\) and 116.1 J/(mol•K) for solid \(\ce{I2}\). This order makes qualitative sense based on the kinds and extents of motion available to atoms and molecules in the three phases (Figure \(\Page {1}\)). The correlation between physical state and absolute entropy is illustrated in Figure \(\Page {2}\), which is a generalized plot of the entropy of a substance versus temperature. The of a substance at any temperature above 0 K must be determined by calculating the increments of heat \(q\) required to bring the substance from 0 K to the temperature of interest, and then summing the ratios \(q/T\). Two kinds of experimental measurements are needed: \[ S_{0 \rightarrow T} = \int _{0}^{T} \dfrac{C_p}{T} dt \label{eq20}\] Because the heat capacity is itself slightly temperature dependent, the most precise determinations of absolute entropies require that the functional dependence of \(C\) on \(T\) be used in the integral in Equation \ref{eq20}, i.e.,: \[ S_{0 \rightarrow T} = \int _{0}^{T} \dfrac{C_p(T)}{T} dt. \label{eq21}\] When this is not known, one can take a series of heat capacity measurements over narrow temperature increments \(ΔT\) and measure the area under each section of the curve. The area under each section of the plot represents the entropy change associated with heating the substance through an interval \(ΔT\). To this must be added the enthalpies of melting, vaporization, and of any solid-solid phase changes. Values of \(C_p\) for temperatures near zero are not measured directly, but can be estimated from quantum theory. The cumulative areas from 0 K to any given temperature (Figure \(\Page {3}\)) are then plotted as a function of \(T\), and any phase-change entropies such as \[S_{vap} = \dfrac{H_{vap}}{T_b}\] are added to obtain the absolute entropy at temperature \(T\). As shown in Figure \(\Page {2}\) above, the entropy of a substance increases with temperature, and it does so for two reasons: We can make careful calorimetric measurements to determine the temperature dependence of a substance’s entropy and to derive absolute entropy values under specific conditions. are given the label \(S^o_{298}\) for values determined for one mole of substance at a pressure of 1 bar and a temperature of 298 K. The for any process may be computed from the standard entropies of its reactant and product species like the following: \[ΔS^o=\sum νS^o_{298}(\ce{products})−\sum νS^o_{298}(\ce{reactants}) \label{\(\Page {6}\)}\] Here, \(ν\) represents stoichiometric coefficients in the balanced equation representing the process. For example, \(ΔS^o\) for the following reaction at room temperature is computed as the following: \[ΔS^o=[xS^o_{298}(\ce{C})+yS^o_{298}(\ce{D})]−[mS^o_{298}(\ce{A})+nS^o_{298}(\ce{B})] \label{\(\Page {8}\)}\] Table \(\Page {1}\) lists some standard entropies at 298.15 K. You can find additional standard entropies in and A closer examination of Table \(\Page {1}\) also reveals that substances with similar molecular structures tend to have similar \(S^o\) values. Among crystalline materials, those with the lowest entropies tend to be rigid crystals composed of small atoms linked by strong, highly directional bonds, such as diamond (\(S^o = 2.4 \,J/(mol•K)\)). In contrast, graphite, the softer, less rigid allotrope of carbon, has a higher \(S^o\) (5.7 J/(mol•K)) due to more (microstates) in the crystal. Soft crystalline substances and those with larger atoms tend to have higher entropies because of increased molecular motion and disorder. Similarly, the absolute entropy of a substance tends to increase with increasing molecular complexity because the number of available microstates increases with molecular complexity. For example, compare the \(S^o\) values for CH OH(l) and CH CH OH(l). Finally, substances with strong hydrogen bonds have lower values of \(S^o\), which reflects a more ordered structure. Entropy increases with softer, less rigid solids, solids that contain larger atoms, and solids with complex molecular structures. To calculate \(ΔS^o\) for a chemical reaction from standard molar entropies, we use the familiar “products minus reactants” rule, in which the absolute entropy of each reactant and product is multiplied by its stoichiometric coefficient in the balanced chemical equation. Example \(\Page {1}\) illustrates this procedure for the combustion of the liquid hydrocarbon isooctane (\(\ce{C8H18}\); 2,2,4-trimethylpentane). Use the data in Table \(\Page {1}\) to calculate \(ΔS^o\) for the reaction of liquid isooctane with \(\ce{O2(g)}\) to give \(\ce{CO2(g)}\) and \(\ce{H2O(g)}\) at 298 K. : standard molar entropies, reactants, and products : ΔS° : Write the balanced chemical equation for the reaction and identify the appropriate quantities in Table \(\Page {1}\). Subtract the sum of the absolute entropies of the reactants from the sum of the absolute entropies of the products, each multiplied by their appropriate stoichiometric coefficients, to obtain \(ΔS^o\) for the reaction. : The balanced chemical equation for the complete combustion of isooctane (\(\ce{C8H18}\)) is as follows: \[\ce{C8H18(l) + 25/2 O2(g) -> 8CO2(g) + 9H2O(g)} \nonumber\] We calculate \(ΔS^o\) for the reaction using the “products minus reactants” rule, where m and n are the stoichiometric coefficients of each product and each reactant: \[\begin{align*}\Delta S^o_{\textrm{rxn}}&=\sum mS^o(\textrm{products})-\sum nS^o(\textrm{reactants})
\\[4pt] &=[8S^o(\mathrm{CO_2})+9S^o(\mathrm{H_2O})]-[S^o(\mathrm{C_8H_{18}})+\dfrac{25}{2}S^o(\mathrm{O_2})]
\\[4pt] &=\left \{ [8\textrm{ mol }\mathrm{CO_2}\times213.8\;\mathrm{J/(mol\cdot K)}]+[9\textrm{ mol }\mathrm{H_2O}\times188.8\;\mathrm{J/(mol\cdot K)}] \right \}
\\[4pt] & \,\,\, -\left \{[1\textrm{ mol }\mathrm{C_8H_{18}}\times329.3\;\mathrm{J/(mol\cdot K)}]+\left [\dfrac{25}{2}\textrm{ mol }\mathrm{O_2}\times205.2\textrm{ J}/(\mathrm{mol\cdot K})\right ] \right \}
\\[4pt] &=515.3\;\mathrm{J/K}\end{align*}\] \(ΔS^o\) is positive, as expected for a combustion reaction in which one large hydrocarbon molecule is converted to many molecules of gaseous products. Use the data in Table \(\Page {1}\) to calculate \(ΔS^o\) for the reaction of \(\ce{H2(g)}\) with liquid benzene (\(\ce{C6H6}\)) to give cyclohexane (\(\ce{C6H12}\)) at 298 K. -361.1 J/K Calculate the standard entropy change for the following process at 298 K: \[\ce{H2O}(g)⟶\ce{H2O}(l)\nonumber\] The value of the standard entropy change at room temperature, \(ΔS^o_{298}\), is the difference between the standard entropy of the product, H O( ), and the standard entropy of the reactant, H O( ). \[\begin{align*} ΔS^o_{298} &=S^o_{298}(\ce{H2O (l)})−S^o_{298}(\ce{H2O(g)})\nonumber \\[4pt] &= (70.0\: J\:mol^{−1}K^{−1})−(188.8\: Jmol^{−1}K^{−1})\nonumber \\[4pt] &=−118.8\:J\:mol^{−1}K^{−1} \end{align*}\] The value for \(ΔS^o_{298}\) is negative, as expected for this phase transition (condensation), which the previous section discussed. Calculate the standard entropy change for the following process at 298 K: \[\ce{H2}(g)+\ce{C2H4}(g)⟶\ce{C2H6}(g)\nonumber\] −120.6 J mol K Calculate the standard entropy change for the combustion of methanol, CH OH at 298 K: \[\ce{2CH3OH}(l)+\ce{3O2}(g)⟶\ce{2CO2}(g)+\ce{4H2O}(l)\nonumber\] The value of the standard entropy change is equal to the difference between the standard entropies of the products and the entropies of the reactants scaled by their stoichiometric coefficients. The standard entropy of formations are found in Table \(\Page {1}\). \[\begin{align*} ΔS^o &=ΔS^o_{298} \\[4pt] &= ∑νS^o_{298}(\ce{products})−∑νS^o_{298} (\ce{reactants}) \\[4pt] & = 2S^o_{298}(\ce{CO2}(g))+4S^o_{298}(\ce{H2O}(l))]−[2S^o_{298}(\ce{CH3OH}(l))+3S^o_{298}(\ce{O2}(g))]\nonumber \\[4pt] &= [(2 \times 213.8) + (4×70.0)]−[ (2 \times 126.8) + (3 \times 205.03) ]\nonumber \\[4pt] &= −161.6 \:J/mol⋅K\nonumber \end{align*} \] Calculate the standard entropy change for the following reaction at 298 K: \[\ce{Ca(OH)2}(s)⟶\ce{CaO}(s)+\ce{H2O}(l)\nonumber\] 24.7 J/mol•K Energy values, as you know, are all relative, and must be defined on a scale that is completely arbitrary; there is no such thing as the absolute energy of a substance, so we can arbitrarily define the enthalpy or internal energy of an element in its most stable form at 298 K and 1 atm pressure as zero. The same is true of the entropy; since entropy is a measure of the “dilution” of thermal energy, it follows that the less thermal energy available to spread through a system (that is, the lower the temperature), the smaller will be its entropy. In other words, as the absolute temperature of a substance approaches zero, so does its entropy. This principle is the basis of the , which states that the entropy of a perfectly-ordered solid at 0 K is zero. In practice, chemists determine the absolute entropy of a substance by measuring the molar heat capacity (\(C_p\)) as a function of temperature and then plotting the quantity \(C_p/T\) versus \(T\). The area under the curve between 0 K and any temperature T is the absolute entropy of the substance at \(T\). In contrast, other thermodynamic properties, such as internal energy and enthalpy, can be evaluated in only relative terms, not absolute terms. The second law of thermodynamics states that a spontaneous process increases the entropy of the universe, > 0. If Δ < 0, the process is nonspontaneous, and if Δ = 0, the system is at equilibrium. The third law of thermodynamics establishes the zero for entropy as that of a perfect, pure crystalline solid at 0 K. With only one possible microstate, the entropy is zero. We may compute the standard entropy change for a process by using standard entropy values for the reactants and products involved in the process. ). ) | 12,907 | 1,100 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/20%3A_Periodic_Trends_and_the_s-Block_Elements/20.06%3A_The_s-Block_Elements_in_Biology |
The -block elements play important roles in biological systems. Covalent hydrides, for example, are the building blocks of organic compounds, and other compounds and ions containing -block elements are found in tissues and cellular fluids. In this section, we describe some ways in which biology depends on the properties of the group 1 and group 2 elements. There are three major classes of hydrides—covalent, ionic, and metallic—but only covalent hydrides occur in living cells and have any biochemical significance. As you learned in , carbon and hydrogen have similar electronegativities, and the C–H bonds in organic molecules are strong and essentially nonpolar. Little acid–base chemistry is involved in the cleavage or formation of these bonds. In contrast, because hydrogen is less electronegative than oxygen and nitrogen (symbolized by Z), the H–Z bond in the hydrides of these elements is polarized (H –Z ). Consequently, the hydrogen atoms in these H–Z bonds are relatively acidic. Moreover, S–H bonds are relatively weak due to poor orbital overlap, so they are readily cleaved to give a proton. Hydrides in which H is bonded to O, N, or S atoms are therefore polar, hydrophilic molecules that form hydrogen bonds. They also undergo acid–base reactions by transferring a proton. Covalent hydrides in which H is bonded to O, N, or S atoms are polar and hydrophilic, form hydrogen bonds, and transfer a proton in their acid-base reactions. Hydrogen bonds are crucial in biochemistry, in part because they help hold proteins in their biologically active folded structures. Hydrogen bonds also connect the two intertwining strands of DNA (deoxyribonucleic acid), the substance that contains the genetic code for all organisms. (For more information on DNA, see .) Because hydrogen bonds are easier to break than the covalent bonds that form the individual DNA strands, the two intertwined strands can be separated to give intact single strands, which is essential for the duplication of genetic information. In addition to the importance of hydrogen bonds in biochemical molecules, the extensive hydrogen-bonding network in water is one of the keys to the existence of life on our planet. Based on its molecular mass, water should be a gas at room temperature (20°C), but the strong intermolecular interactions in liquid water greatly increase its boiling point. Hydrogen bonding also produces the relatively open molecular arrangement found in ice, which causes ice to be less dense than water. Because ice floats on the surface of water, it creates an insulating layer that allows aquatic organisms to survive during cold winter months. These same strong intermolecular hydrogen-bonding interactions are also responsible for the high heat capacity of water and its high heat of fusion. A great deal of energy must be removed from water for it to freeze. Consequently, as noted in , large bodies of water act as “thermal buffers” that have a stabilizing effect on the climate of adjacent land areas. Perhaps the most striking example of this effect is the fact that humans can live comfortably at very high latitudes. For example, palm trees grow in southern England at the same latitude (51°N) as the southern end of frigid Hudson Bay and northern Newfoundland in North America, areas known more for their moose populations than for their tropical vegetation. Warm water from the Gulf Stream current in the Atlantic Ocean flows clockwise from the tropical climate at the equator past the eastern coast of the United States and then turns toward England, where heat stored in the water is released. The temperate climate of Europe is largely attributable to the thermal properties of water. Strong intermolecular hydrogen-bonding interactions are responsible for the high heat capacity of water and its high heat of fusion. The members of group 1 and group 2 that are present in the largest amounts in organisms are sodium, potassium, magnesium, and calcium, all of which form monatomic cations with a charge of +1 (group 1, M ) or +2 (group 2, M ). Biologically, these elements can be classified as . For example, calcium is found in the form of relatively insoluble calcium salts that are used as structural materials in many organisms. Hydroxyapatite [Ca (PO ) OH] is the major component of bones, calcium carbonate (CaCO ) is the major component of the shells of mollusks and the eggs of birds and reptiles, and calcium oxalate (CaO CCO ) is found in many plants. Because calcium and strontium have similar sizes and charge-to-radius ratios, small quantities of strontium are always found in bone and other calcium-containing structural materials. Normally this is not a problem because the Sr ions occupy sites that would otherwise be occupied by Ca ions. When trace amounts of radioactive Sr are released into the atmosphere from nuclear weapons tests or a nuclear accident, however, the radioactive strontium eventually reaches the ground, where it is taken up by plants that are consumed by dairy cattle. The isotope then becomes concentrated in cow’s milk, along with calcium. Because radioactive strontium coprecipitates with calcium in the hydroxyapatite that surrounds the bone marrow (where white blood cells are produced), children, who typically ingest more cow’s milk than adults, are at substantially increased risk for leukemia, a type of cancer characterized by the overproduction of white blood cells. The Na , K , Mg , and Ca ions are important components of intracellular and extracellular fluids. Both Na and Ca are found primarily in cellular fluids, such as blood plasma, whereas K and Mg are found primarily in cellular fluids. Substantial inputs of energy are required to establish and maintain these concentration gradients and prevent the system from reaching equilibrium. Thus energy is needed to transport each ion across the cell membrane toward the side with the higher concentration. The biological machines that are responsible for the selective transport of these metal ions are complex assemblies of proteins called ion pumps . Ion pumps recognize and discriminate between metal ions in the same way that crown ethers and cryptands do, with a high affinity for ions of a certain charge and radius. Defects in the ion pumps or their control mechanisms can result in major health problems. For example, , the most common inherited disease in the United States, is caused by a defect in the transport system (in this case, chloride ions). Similarly, in many cases, , or high blood pressure, is thought to be due to defective Na uptake and/or excretion. If too much Na is absorbed from the diet (or if too little is excreted), water diffuses from tissues into the blood to dilute the solution, thereby decreasing the osmotic pressure in the circulatory system. The increased volume increases the blood pressure, and ruptured arteries called can result, often in the brain. Because high blood pressure causes other medical problems as well, it is one of the most important biomedical disorders in modern society. For patients who suffer from hypertension, low-sodium diets that use NaCl substitutes, such as KCl, are often prescribed. Although KCl and NaCl give similar flavors to foods, the K is not readily taken up by the highly specific Na -uptake system. This approach to controlling hypertension is controversial, however, because direct correlations between dietary Na content and blood pressure are difficult to demonstrate in the general population. More important, recent observations indicate that high blood pressure may correlate more closely with inadequate intake of calcium in the diet than with excessive sodium levels. This finding is important because the typical “low-sodium” diet is also low in good sources of calcium, such as dairy products. Some of the most important biological functions of the group 1 and group 2 metals are due to small changes in the cellular concentrations of the metal ion. The transmission of nerve impulses, for example, is accompanied by an increased flux of Na ions into a nerve cell. Similarly, the binding of various hormones to specific receptors on the surface of a cell leads to a rapid influx of Ca ions; the resulting sudden rise in the intracellular Ca concentration triggers other events, such as muscle contraction, the release of neurotransmitters, enzyme activation, or the secretion of other hormones. Within cells, K and Mg often activate particular enzymes by binding to specific, negatively charged sites in the enzyme structure. , the green pigment used by all plants to absorb light and drive the process of , contains magnesium. During photosynthesis, CO is reduced to form sugars such as glucose. The structure of the central portion of a chlorophyll molecule resembles a crown ether (part (a) in ) with four five-member nitrogen-containing rings linked together to form a large ring that provides a “hole” the proper size to tightly bind Mg . Note the resemblance to the crown ether complexes discussed in . Because the health of cells depends on maintaining the proper levels of cations in intracellular fluids, any change that affects the normal flux of metal ions across cell membranes could well cause an organism to die. Molecules that facilitate the transport of metal ions across membranes are generally called ionophores ( plus from the Greek , meaning “to carry”). Many ionophores are potent antibiotics that can kill or inhibit the growth of bacteria. An example is , a cyclic molecule with a central cavity lined with oxygen atoms (part (a) in ) that is similar to the cavity of a crown ether (part (a) in ). Like a crown ether, valinomycin is highly selective: its affinity for K is about 1000 times greater than that for Na . By increasing the flux of K ions into cells, valinomycin disrupts the normal K gradient across a cell membrane, thereby killing the cell (part (b) in ). A common way to study the function of a metal ion in biology is to replace the naturally occurring metal with one whose reactivity can be traced by spectroscopic methods. The substitute metal ion must bind to the same site as the naturally occurring ion, and it must have a similar (or greater) affinity for that site, as indicated by its charge density. Arrange the following ions in order of increasing effectiveness as a replacement for Ca , which has an ionic radius of 100 pm (numbers in parentheses are ionic radii): Na (102 pm), Eu (117 pm), Sr (118 pm), F (133 pm), Pb (119 pm), and La (103 pm). Explain your reasoning. ions and ionic radii suitability as replacement for Ca Use periodic trends to arrange the ions from least effective to most effective as a replacement for Ca . The most important properties in determining the affinity of a biological molecule for a metal ion are the size and charge-to-radius ratio of the metal ion. Of the possible Ca replacements listed, the F ion has the opposite charge, so it should have no affinity for a Ca -binding site. Na is approximately the right size, but with a +1 charge it will bind much more weakly than Ca . Although Eu , Sr , and Pb are all a little larger than Ca , they are probably similar enough in size and charge to bind. Based on its ionic radius, Eu should bind most tightly of the three. La is nearly the same size as Ca and more highly charged. With a higher charge-to-radius ratio and a similar size, La should bind tightly to a Ca site and be the most effective replacement for Ca . The order is F << Na << Pb ~ Sr ~ Eu < La . Exercise The ionic radius of K is 138 pm. Arrange the following ions in order of increasing affinity for a K -binding site in an enzyme (numbers in parentheses are ionic radii): Na (102 pm), Rb (152 pm), Ba (135 pm), Cl (181 pm), and Tl (150 pm). Cl << Na < Tl ~ Rb < Ba Covalent hydrides in which hydrogen is bonded to oxygen, nitrogen, or sulfur are polar, hydrophilic molecules that form hydrogen bonds and undergo acid–base reactions. Hydrogen-bonding interactions are crucial in stabilizing the structure of proteins and DNA and allow genetic information to be duplicated. The hydrogen-bonding interactions in water and ice also allow life to exist on our planet. The group 1 and group 2 metals present in organisms are , which are important components of intracellular and extracellular fluids. Small changes in the cellular concentration of a metal ion can have a significant impact on biological functions. Metal ions are selectively transported across cell membranes by , which bind ions based on their charge and radius. , many of which are potent antibiotics, facilitate the transport of metal ions across membranes. Explain the thermochemical properties of water in terms of its intermolecular bonding interactions. How does this affect global climate patterns? Of the three classes of hydrides, which is (are) biochemically significant? How do you account for this? Many proteins are degraded and become nonfunctional when heated higher than a certain temperature, even though the individual protein molecules do not undergo a distinct chemical change. Propose an explanation for this observation. Los Angeles has moderate weather throughout the year, with average temperatures between 57°F and 70°F. In contrast, Palm Springs, which is just 100 miles inland, has average temperatures between 55°F and 95°F. Explain the difference in the average temperature ranges between the two cities. Although all group 1 ions have the same charge (+1), Na and K ions are selectively transported across cell membranes. What strategy do organisms employ to discriminate between these two cations? A 0.156 g sample of a chloride salt of an alkaline earth metal is dissolved in enough water to make 20.5 mL of solution. If this solution has an osmotic pressure of 2.68 atm at 25°C, what is the identity of the alkaline earth metal? The thermal buffering capacity of water is one of the reasons the human body is capable of withstanding a wide range of temperatures. How much heat (in kilojoules) is required to raise the temperature of a 70.0 kg human from 37.0°C to 38.0°C? Assume that 70% of the mass of the body is water and that body fluids have the same specific heat as water. During illness, body temperature can increase by more than 2°C. One piece of folklore is that you should “feed a fever.” Using the data in , how many fried chicken drumsticks would a 70.0 kg person need to eat to generate a 2.0°C change in body temperature? Assume the following: there is complete conversion of the caloric content of the chicken to thermal energy, 70% of the mass of the body is solely due to water, and body fluids have the same specific heat as water. Hydrogen bonding is partly responsible for the high enthalpy of vaporization of water (Δ = 43.99 kJ/mol at 25°C), which contributes to cooling the body during exercise. Assume that a 50.0 kg runner produces 20.0 g of perspiration during a race, and all the perspiration is converted to water vapor at 37.0°C. What amount of heat (in joules) is removed from the runner’s skin if the perspiration consists of only water? Ba | 15,187 | 1,101 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_6%3A_Materials/21%3A_Structure_and_Bonding_in_Solids/21.E%3A_Structure_and_Bonding_in_Solids_(Exercises) |
If a great deal of energy is required to form gaseous ions, why do ionic compounds form at all? What are the general physical characteristics of ionic compounds? Ionic compounds consist of crystalline lattices rather than discrete ion pairs. Why? What factors affect the magnitude of the lattice energy of an ionic compound? What is the relationship between ionic size and lattice energy? Which would have the larger lattice energy—an ionic compound consisting of a large cation and a large anion or one consisting of a large anion and a small cation? Explain your answer and any assumptions you made. How would the lattice energy of an ionic compound consisting of a monovalent cation and a divalent anion compare with the lattice energy of an ionic compound containing a monovalent cation and a monovalent anion, if the internuclear distance was the same in both compounds? Explain your answer. Which would have the larger lattice energy—CrCl or CrCl —assuming similar arrangements of ions in the lattice? Explain your answer. Which cation in each pair would be expected to form a chloride salt with the larger lattice energy, assuming similar arrangements of ions in the lattice? Explain your reasoning. Which cation in each pair would be expected to form an oxide with the higher melting point, assuming similar arrangements of ions in the lattice? Explain your reasoning. How can a thermochemical cycle be used to determine lattice energies? Which steps in such a cycle require an input of energy? Although NaOH and CH OH have similar formulas and molecular masses, the compounds have radically different properties. One has a high melting point, and the other is a liquid at room temperature. Which compound is which and why? Arrange SrO, PbS, and PrI in order of decreasing lattice energy. Compare BaO and MgO with respect to each of the following properties. Use a thermochemical cycle and data from Figure 7.13, Table 7.5, Table 8.2, Table 8.3 to calculate the lattice energy ( ) of magnesium chloride (MgCl ). Would you expect the formation of SrO from its component elements to be exothermic or endothermic? Why or why not? How does the valence electron configuration of the component elements help you determine this? Using the information in Problem 4 and Problem 5, predict whether CaO or MgCl will have the higher melting point. Use a thermochemical cycle and data from Table 8.2, Table 8.3, and Chapter 25 to calculate the lattice energy of calcium oxide. The first and second ionization energies of calcium are 589.8 kJ/mol and 1145.4 kJ/mol. Lattice energy is directly proportional to the product of the ionic charges and inversely proportional to the internuclear distance. Therefore, PrI > SrO > PbS. = 2522.2 kJ/mol | 2,759 | 1,103 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/16%3A_Aqueous_AcidBase_Equilibriums/16.03%3A_Molecular_Structure_and_Acid-Base_Strength |
We have seen that the strengths of acids and bases vary over many orders of magnitude. In this section, we explore some of the structural and electronic factors that control the acidity or basicity of a molecule. In general, the stronger the A–H or B–H bond, the less likely the bond is to break to form H ions and thus the less acidic the substance. This effect can be illustrated using the hydrogen halides: The trend in bond energies is due to a steady decrease in overlap between the 1 orbital of hydrogen and the valence orbital of the halogen atom as the size of the halogen increases. The larger the atom to which H is bonded, the weaker the bond. Thus the bond between H and a large atom in a given family, such as I or Te, is weaker than the bond between H and a smaller atom in the same family, such as F or O. As a result, . For example, the order of acidity for the binary hydrides of Group 16 is as follows, with p values in parentheses: H O (14.00 = p ) < H S (7.05) < H Se (3.89) < H Te (2.6). Whether we write an acid–base reaction as \(AH \rightleftharpoons A^−+H^+\) or as \(BH^+ \rightleftharpoons B+H^+\) the conjugate base (A or B) contains one more lone pair of electrons than the parent acid (AH or BH ). Let’s see how this explains the relative acidity of the binary hydrides of the elements in the second row of the periodic table. The observed order of increasing acidity is the following, with p values in parentheses: CH (~50) << NH (~36) < H O (14.00) < HF (3.20). Consider, for example, the compounds at both ends of this series: methane and hydrogen fluoride. The conjugate base of CH is CH , and the conjugate base of HF is F . Because fluorine is much more electronegative than carbon, fluorine can better stabilize the negative charge in the F ion than carbon can stabilize the negative charge in the CH ion. Consequently, HF has a greater tendency to dissociate to form H and F than does methane to form H and CH , making HF a much stronger acid than CH . The same trend is predicted by analyzing the properties of the conjugate acids. For a series of compounds of the general formula HE, as the electronegativity of E increases, the E–H bond becomes more polar, favoring dissociation to form E and H . Due to both the increasing stability of the conjugate base and the increasing polarization of the E–H bond in the conjugate acid, . Acid strengths of binary hydrides as we go a column or across a row of the periodic table. Atoms or groups of atoms in a molecule other than those to which H is bonded can induce a change in the distribution of electrons within the molecule. This is called an , and, much like the coordination of water to a metal ion, it can have a major effect on the acidity or basicity of the molecule. For example, the hypohalous acids (general formula HOX, with X representing a halogen) all have a hydrogen atom bonded to an oxygen atom. In aqueous solution, they all produce the following equilibrium:
\[ HOX_{(aq)} \rightleftharpoons H^+_{(aq)} + OX^−{(aq)} \tag{16.3.1}\] The acidities of these acids vary by about three orders of magnitude, however, due to the difference in electronegativity of the halogen atoms: As the electronegativity of X increases, the distribution of electron density within the molecule changes: the electrons are drawn more strongly toward the halogen atom and, in turn, away from the H in the O–H bond, thus weakening the O–H bond and allowing dissociation of hydrogen as H . The acidity of oxoacids, with the general formula HOXO ( = 0−3), depends strongly on the number of terminal oxygen atoms attached to the central atom X. As shown in , the values of the oxoacids of chlorine increase by a factor of about 10 to 10 with each oxygen as successive oxygen atoms are added. The increase in acid strength with increasing number of terminal oxygen atoms is due to both an inductive effect and increased stabilization of the conjugate base. Any inductive effect that withdraws electron density from an O–H bond increases the acidity of the compound. Because oxygen is the second most electronegative element, adding terminal oxygen atoms causes electrons to be drawn away from the O–H bond, making it weaker and thereby increasing the strength of the acid. The colors in show how the electrostatic potential, a measure of the strength of the interaction of a point charge at any place on the surface of the molecule, changes as the number of terminal oxygen atoms increases. In and , blue corresponds to low electron densities, while red corresponds to high electron densities. The oxygen atom in the O–H unit becomes steadily less red from HClO to HClO (also written as HOClO ), while the H atom becomes steadily bluer, indicating that the electron density on the O–H unit decreases as the number of terminal oxygen atoms increases. The decrease in electron density in the O–H bond weakens it, making it easier to lose hydrogen as H ions, thereby increasing the strength of the acid. Chlorine oxoacids p values from J. R. Bowser, (Pacific Grove, CA: Brooks-Cole,1993). At least as important, however, is the effect of delocalization of the negative charge in the conjugate base. As shown in , the number of resonance structures that can be written for the oxoanions of chlorine increases as the number of terminal oxygen atoms increases, allowing the single negative charge to be delocalized over successively more oxygen atoms. The electrostatic potential plots in demonstrate that the electron density on the terminal oxygen atoms decreases steadily as their number increases. The oxygen atom in ClO is red, indicating that it is electron rich, and the color of oxygen progressively changes to green in ClO , indicating that the oxygen atoms are becoming steadily less electron rich through the series. For example, in the perchlorate ion (ClO ), the single negative charge is delocalized over all four oxygen atoms, whereas in the hypochlorite ion (OCl ), the negative charge is largely localized on a single oxygen atom ( ). As a result, the perchlorate ion has no localized negative charge to which a proton can bind. Consequently, the perchlorate anion has a much lower affinity for a proton than does the hypochlorite ion, and perchloric acid is one of the strongest acids known. Electron delocalization in the conjugate base increases acid strength. Similar inductive effects are also responsible for the trend in the acidities of oxoacids that have the same number of oxygen atoms as we go across a row of the periodic table from left to right. For example, H PO is a weak acid, H SO is a strong acid, and HClO is one of the strongest acids known. The number of terminal oxygen atoms increases steadily across the row, consistent with the observed increase in acidity. In addition, the electronegativity of the central atom increases steadily from P to S to Cl, which causes electrons to be drawn from oxygen to the central atom, weakening the O–H bond and increasing the strength of the oxoacid. Careful inspection of the data in shows two apparent anomalies: carbonic acid and phosphorous acid. If carbonic acid (H CO ) were a discrete molecule with the structure (HO) C=O, it would have a single terminal oxygen atom and should be comparable in acid strength to phosphoric acid (H PO ), for which p = 2.16. Instead, the tabulated value of p for carbonic acid is 6.35, making it about 10,000 times weaker than expected. As we shall see in , however, H CO is only a minor component of the aqueous solutions of CO that are referred to as carbonic acid. Similarly, if phosphorous acid (H PO ) actually had the structure (HO) P, it would have no terminal oxygen atoms attached to phosphorous. It would therefore be expected to be about as strong an acid as HOCl (p = 7.40). In fact, the p for phosphorous acid is 1.30, and the structure of phosphorous acid is (HO) P(=O)H with one H atom directly bonded to P and one P=O bond. Thus the p for phosphorous acid is similar to that of other oxoacids with one terminal oxygen atom, such as H PO . Fortunately, phosphorous acid is the only common oxoacid in which a hydrogen atom is bonded to the central atom rather than oxygen. Inductive effects are also observed in organic molecules that contain electronegative substituents. The magnitude of the electron-withdrawing effect depends on both the nature and the number of halogen substituents, as shown by the p values for several acetic acid derivatives: \[pK_a CH_3CO_2H 4.76< CH_2ClCO_2H 2.87<CHCl_2CO_2H 1.35<CCl_3CO_2H 0.66<CF_3CO_2H 0.52 \notag \] As you might expect, fluorine, which is more electronegative than chlorine, causes a larger effect than chlorine, and the effect of three halogens is greater than the effect of two or one. Notice from these data that inductive effects can be quite large. For instance, replacing the –CH group of acetic acid by a –CF group results in about a 10,000-fold increase in acidity! Arrange the compounds of each series in order of increasing acid or base strength. The structures are shown here. series of compounds relative acid or base strengths Use relative bond strengths, the stability of the conjugate base, and inductive effects to arrange the compounds in order of increasing tendency to ionize in aqueous solution. Trifluoramine is such a weak base that it does not react with aqueous solutions of strong acids. Hence its base ionization constant has not been measured. Exercise Arrange the compounds of each series in order of The acid–base strength of a molecule depends strongly on its structure. The weaker the A–H or B–H bond, the more likely it is to dissociate to form an H ion. In addition, any factor that stabilizes the lone pair on the conjugate base favors the dissociation of H , making the conjugate acid a stronger acid. Atoms or groups of atoms elsewhere in a molecule can also be important in determining acid or base strength through an , which can weaken an O–H bond and allow hydrogen to be more easily lost as H ions. presented several factors that affect the relative strengths of acids and bases. For each pair, identify the most important factor in determining which is the stronger acid or base in aqueous solution. The stability of the conjugate base is an important factor in determining the strength of an acid. Which would you expect to be the stronger acid in aqueous solution—C H NH or NH ? Justify your reasoning. Explain why H Se is a weaker acid than HBr. Arrange the following in order of decreasing acid strength in aqueous solution: H PO , CH PO H , and HClO . Arrange the following in order of increasing base strength in aqueous solution: CH S , OH , and CF S . Arrange the following in order of increasing acid strength in aqueous solution: HClO , HNO , and HNO . Do you expect H SO or H SeO to be the stronger acid? Why? Give a plausible explanation for why CF OH is a stronger acid than CH OH in aqueous solution. Do you expect CHCl CH OH to be a stronger or a weaker acid than CH OH? Why? Do you expect Cl NH or NH to be the stronger base in aqueous solution? Why? CF S < CH S < OH (strongest base) NH ; Cl atoms withdraw electron density from N in Cl NH. | 11,242 | 1,104 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Exercises%3A_Physical_and_Theoretical_Chemistry/Data-Driven_Exercises/Liquid-vapor_Equilibria_-_%CE%94H_and_%CE%94S_for_Vaporization |
: Vapor pressure data is presented for two liquids over a given temperature range and the heat of vaporization and entropy change for vaporization are determined. : An introductory knowledge of thermodynamic relationships that apply to phase equilibrium systems. : This exercise should be carried out within a software environment that is capable of data manipulation and which can generate a best-fit line for an x-y data set. You will also be graphing the data along with the fitted function. One of the simplest equilibrium systems to consider is a pure liquid in contact with its vapor \[liquid \rightleftharpoons vapor\] A liquid-vapor equilibrium can be established by pouring a liquid inside of a small flask and applying a vacuum to degas the liquid and to evacuate the air-space above the liquid, whereupon the flask is sealed. After sealing the vessel, liquid will continue to vaporize and the pressure of the vapor will correspondingly rise. But as the vapor pressure increases, the opposite process can occur. Namely, gas molecules can condense and re-enter the liquid phase. These two competing processes, vaporization and condensation, each occur at their respective rates. And when these rates become equal, the system will exist in a dynamic equilibrium and the vapor pressure will have a constant value. What governs whether the equilibrium lies more toward the liquid phase or the gas phase in a particular system? Temperature is certainly expected to have an impact. As temperature rises, the increased kinetic motion of molecules tends to overcome intermolecular attractions and a greater fraction of the molecules will reside in the vapor phase. Consequently, the vapor pressure of the system is expected to rise with increasing temperature. Thermodynamics provides a framework for describing how vapor pressure depends on temperature. As a starting point, we consider the Clapeyron equation which applies to any type of phase equilibrium \[\dfrac{dp}{dT} = \dfrac{\Delta H_m}{T \Delta V_m} \label{1}\] where \(p\) is the pressure and \(T\) the absolute temperature of a two phase system, and \(ΔH_m\) and \(ΔV_m\) are the molar enthalpy change and molar volume change associated with the phase transition in question. The Clapeyron equation is an exact thermodynamic relationship and its derivation is given in most physical chemistry texts. Written in the form of Equation \ref{1}, the Clapeyron equation tells us that the slope of a vapor pressure versus temperature plot (or more specifically, a tangent line drawn at some point on this plot) is related to \(ΔH_m\) and \(ΔV_m\) for vaporization.
Assuming that the volume occupied by one mole of liquid is negligible in comparison to the volume occupied by one mole of vapor, the volume change that accompanies vaporization can be written as \[\Delta V_m = V_{m,gas} - V_{m,liquid} \approx V_{m,gas} \label{2}\] If we further assume that the volume of one mole of vapor behaves as an ideal gas then \[ V_{m,gas} = \dfrac{RT}{p} \label{3}\] and Equation \ref{1} becomes \[\dfrac{dp}{dT} = \dfrac{p \Delta H_{vap}}{RT^2} \label{4}\] Assuming the molar enthalpy of vaporatization is constant with respect to temperature, Equation \ref{4} can be rearranged and integrated, \[\int \dfrac{dp}{p} = \left( \dfrac{\Delta H_{vap} }{R} \right) \int \dfrac{dT}{T^2} \label{5}\] yielding the following expression for how the vapor pressure depends upon temperature \[ \ln p \left( \dfrac{\Delta H_{vap}}{R} \right) \dfrac{1}{T} + C, \label{6}\] where \(C\) is a constant of integration. According to Equation \ref{6}, a plot of \(\ln p\) versus \(1/T\) should be linear and the slope is related to the heat of vaporization. An expression that is similar to Equation \ref{6} can be obtained by setting the right-hand-side of the expressions \[ \Delta G = -RT \ln K \label{7}\] and \[\Delta G = \Delta H - T \Delta S \label{8}\] equal to one another and rearranging into the form \[ \ln K = - \left( \dfrac{\Delta H}{R} \right) \dfrac{1}{T} + \dfrac{\Delta S}{R} \label{9}\] In a liquid-vapor equilibrium system, the equilibrium constant can be written as \[ K =a_{vap} \approx \dfrac{p}{p^o} \label{10} \] where \(a_{vap}\) is the activity of the vapor, \(p\) is the vapor pressure, and \(p^o\) is standard pressure (760 mmHg). Substitution into Equation \ref{9} gives us \[ \ln \left( \dfrac{p}{p^o} \right) = - \left( \dfrac{\Delta H_{vap}}{R} \right) \dfrac{1}{T} + \dfrac{\Delta S_vap}{R} \label{11}\] Just as we saw earlier, the slope of a logarithmic plot of vapor pressure versus \(1/T\) is related to the heat of vaporization. However, Equation \ref{11} implies that the y-intercept of this plot is related to the entropy change for vaporization. In the exercise below, you will analyze vapor pressure data for two liquids, fluorobenzene and benzenethiol, in order to determine \(ΔH_{vap}\) and \(ΔS_{vap}\) for each. on the following links and save each data set on your computer. Each file consists of two columns of numbers: the values in the first column are temperatures in degrees Celsius and the second column contains the corresponding vapor pressures of the liquid in units of millimeters of mercury (mmHg). | 5,205 | 1,105 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/24%3A_Redox_Couples/IV._Reaction_Mechanism |
Although the radical mechanism shown in offers a reasonable explanation for the reactions pictured in , there is uncertainty in some reactions involving redox couples about whether a free-radical is ever produced. This uncertainty is reflected in the reaction mechanism shown in Scheme 2, which describes two possible pathways for participation of a zinc–copper couple in an addition reaction. One pathway involves radical formation by electron transfer, and the second describes formation of an organozinc intermediate. The stereochemical evidence and solvent effects described in the next two sections offer insight into the nature of the reactive species generated by a typical redox couple. | 713 | 1,106 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Book3A_Bioinorganic_Chemistry_(Bertini_et_al.)/04%3A_Biological_and_Synthetic_Dioxygen_Carriers/4.12%3A_Other_Ligands_for_Biological_Oxygen_Carriers |
As noted above, a variety of other \(\alpha\)-donor or \(\pi\)-acceptor ligands will bind to the active sites of biological oxygen carriers. As documented in Table 4.2, carbon monoxide (CO) generally binds more strongly to hemoglobin than does dioxygen, hence causing carbon-monoxide poisoning. In addition to being readily available from car exhausts and tobacco smoke to convert oxyhemoglobin to carbonmonoxyhemoglobin, CO is produced in the catabolism of heme molecules. Thus under even the most favorable of conditions, about 3 percent of human hemoglobin is in the carbonmonoxy form. When CO binds to a single metal atom in nonbiological systems, it does so through the carbon atom and in a linear manner: \[Fe—C=O \leftarrow Fe^{II} +\;^{-}C=O^{+} \tag{4.40}\] Model systems for carbonmonoxy (also called carbonyl) hemoglobin show a geometry similar to that of the Fe—C\(\equiv\)O group, linear or nearly so and essentially perpendicular to the porphyrin plane. The biochemical literature is littered with reports that this is the geometry adopted by CO in binding to hemoglobins. We will return to this topic later in this chapter, since the physiological consequences are potentially important. Carbon monoxide binds weakly as a \(\sigma\)-donor ligand to four-coordinate cobalt(II) systems. Despite a bout of artifactual excitement, CO has never been observed to bind significantly to five-coordinate Co systems with a nitrogenous axial base to yield octahedral six-coordinate species. The sulfur analogue thiocarbonyl (CS), although not stable as a free entity, binds very strongly to iron-porphyrin species in a linear manner. Nitric oxide (NO) binds to hemes even more strongly than CO (and hence O ), so strongly, in fact, that the Fe—N bond is very weak and easily ruptured. Attachment to the metal is via the nitrogen atom; however, the geometry of attachment is sensitive to the \(\pi\) basicity of the metalloporphyrin, and ranges from linear to strongly bent. In binding to Co the NO ligand is effectively reduced to NO , with concomitant oxidation of Co to Co : \(\tag{4.42}\) In much the same way that cobalt-dioxygen systems are paramagnetic (S = \(\frac{1}{2}\)) and amenable to EPR studies, iron-nitric oxide (also called iron nitrosyl) species are also paramagnetic and isoelectronic with cobalt-dioxygen species. The unpaired spin is localized mostly on the NO group. In contrast to the dioxygen, carbon-monoxide, and nitric-oxide ligands, the isocyanide and nitroso functions bear an organic tail. Moreover, nitroso ligands are isoelectronic with dioxygen. \(\tag{4.42}\) Thus, in principle, not only may the steric bulk of the ligand be varied, in order to probe the dimensions of the dioxygen-binding pocket,* but also the \(\sigma\)-donor/\(\pi\)-acceptor properties of the ligands may be varied by appropriate substituents on the aryl ring. Isocyanide groups may bind to metals in a variety of ways. For 1:1 adducts (Figure 4.19), the isocyanide group is approximately linear, although some flexibility seems to exist in a bis(t-butylisocyanide)iron(Il)tetraphenylporphyrinato complex. For zerovalent metals with much electron density available for donation into ligand \(\pi\)* orbitals, the isocyanide ligand has been observed to bend at the N atom. One prediction exists that an isocyanide ligand binds in this manner to hemoglobin. For 1:1 adducts of nitroso ligands, side-on, O-, and N-ligated modes are possible (Figure 4.19). No O-nitroso complexes have been definitively characterized by diffraction methods. For hemoglobin the N-nitroso mode is likely, since this is the mode found for the nitrosoalkane in Fe(TPP)(amine)(RNO). To date isocyanide ligands have not achieved their potential as probes of the geometry of the ligand-binding pocket in hemoglobin, partly because we lack structural data on the preferred geometry of attachment of these ligands in a sterically uncongested environment. * In fact, this classic experiment of St. George and Pauling established, for the first time and before any crystallographic data were available, that the heme group and the ligand-binding site in hemoglobin reside at least partway inside the protein, rather than on the surface. | 4,241 | 1,107 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/4_f-Block_Elements/The_Actinides/Chemistry_of_Neptunium |
Neptunium (named for the planet Neptune) was the first of the transuranium elements to be synthesized (photo above is of the oxide). The synthesis took place at Berkeley, California, after initial examination of the decay products of U-235 suggested the possibility of a new element. Credit for the discovery goes to McMillan and Abelson in 1940. Although the neptunium on which the characterization work was done was synthesized in a cyclotron, we now know that minute amounts of the element exist in the environment (the longest-lived isotope has a half-life of about 2 million years). All isotopes of the metal are radioactive. | 650 | 1,108 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/02._Fundamental_Concepts_of_Quantum_Mechanics/Tunneling |
Tunneling is a quantum mechanical phenomenon when a particle is able to penetrate through a potential energy barrier that is higher in energy than the particle’s kinetic energy. This amazing property of microscopic particles play important roles in explaining several physical phenomena including radioactive decay. Additionally, the principle of tunneling leads to the development of Scanning Tunneling Microscope (STM) which had a profound impact on chemical, biological and material science research. Consider a ball rolling from one valley to another over a hill (Figure \(\Page {1}\)). If the ball has enough energy (\(E\)) to overcome the potential energy (\(V\)) at the top of the barrier between each valley, then it can roll from one valley to the other. This is the classical picture and is controlled by the simple Law of Conservation of Energy approach taught in beginning physics courses. However, If the ball does not have enough kinetic energy (\(E<V\)), to overcome the barrier it will never roll from one valley to the other. In contrast, when quantum effects are taken into effect, the ball can "tunnel" through the barrier to the other valley, even if its kinetic energy is less than the potential energy of the barrier to the top of one of the hills. The reason for the difference between classical and quantum motion comes from wave-particle nature of matter. One interpretation of this duality involves the , which defines a limit on how precisely the position and the momentum of a particle can be known at the same time. This implies that there are no solutions with a probability of exactly zero (or one), though a solution may approach infinity if, for example, the calculation for its position was taken as a probability of 1, the other, i.e. its speed, would have to be infinity. Hence, the probability of a given particle's existence on the opposite side of an intervening barrier is non-zero, and such particles will appear on the 'other' (a semantically difficult word in this instance) side with a relative frequency proportional to this probability. Microscopic particles such as protons, or electrons would behave differently as a consequence of wave-particle duality. Consider a particle with energy \(E\) that is confined in a box which has a barrier of height \(V\). Classically, the box will prevent these particles from escaping due to the insufficiency in kinetic energy of these particles to get over the barrier. However, if the thickness of the barrier is thin, the particles have some probability of penetrating through the barrier sufficient energy and appear on the other side of the box (Figure \(\Page {2}\)). When it reaches a barrier it cannot overcome, a particle's wave function changes from sinusoidal to exponentially diminishing in form. The solution for the Schrödinger equation in such a medium (Figure \(\Page {2}\); blue region) is: \[ \psi = N e^{-\beta x}\] where For a quantum particle to tunnel through a barrier three conditions must be met (Figure \(\Page {2}\)): If these conditions are met, there would be some probability of finding the particles on the other side of the barrier. Beginning as a sinusoidal wave, a particle begins tunneling through the barrier and goes into exponential decay until it exits the barrier and gets transmitted out the other side as a final sinusoidal wave with a smaller amplitude. The act of tunneling decreases the wave amplitude due the reflection of the incident wave when it comes into the contact with the barrier but does not affect the wave equation. The probability, \(P\), of a particle tunneling through the potential energy barrier is derived from the Schrödinger Equation and is described as, \[ P = \exp\left(\dfrac{-4a\pi}{h} \sqrt{2m(V-E)} \right) \label{prob}\] with \(E<V\) where Therefore, the probability of an object tunneling through a barrier decreases with the object's increasing mass and with the increasing gap between the energy of the object and the energy of the barrier. And although the wave function never quite reaches 0 (as can be determined from the functionality), this explains how tunneling is frequent on nanoscale, but negligible at the macroscopic level. An electron having total kinetic energy \(E\) of 4.50 eV approaches a rectangular energy barrier with \(V= 5.00\, eV\) and \(L= 950\, pm\). Classically, the electron cannot pass through the barrier because \(E<V\). Calculate probability of tunneling of this electron through the barrier. This is a straightforward application of Equation \(\ref{prob}\). \[ P = \exp\left(\dfrac{-4a\pi}{h} \sqrt{2m(V-E)} \right) \nonumber\] The electronvolt (eV) is a unit of energy that is equal to approximately \(1.6 \times 10^{−19}\; J\), which is the conversion used below. The mass of an electron is \(9.10 \times 10^{-31}\; kg\) \[\begin{align*} P &= \exp \left[\left(- \dfrac{ (4) (950 \times 10^{-12}\, m) (\pi)}{ 6.6260 \times 10^{-34} m^2 kg / s} \right) \sqrt{(2)(9.10 \times 10^{-31}\; kg) (5.00 - 4.50 \; \cancel{eV})( 1.60 \times 10^{-19} J/\cancel{eV}) } \right] \\[4pt] &= \exp^{-6.88} = 1.03 \times 10^{-3} \end{align*}\] There is a ~0.1% probability of the electrons tunneling though the barrier. Protons and neutrons in a nucleus have kinetic energy, but it is about 8 MeV less than that needed to get out from attractive nuclear potential (Figure \(\Page {4}\)). Hence, they are bound by an average of 8 MeV per nucleon. The slope of the hill outside the bowl is analogous to the repulsive Coulomb potential for a nucleus, such as for an α particle outside a positive nucleus. In \(\alpha\) decay, two protons and two neutrons spontaneously break away as a He unit. Yet the protons and neutrons do not have enough kinetic energy to classically get over the rim. The \(\alpha\) article tunnels through a region of space it is forbidden to be in, and it comes out of the side of the nucleus. Like an electron making a transition between orbits around an atom, it travels from one point to another without ever having been in between (Figure \(\Page {5}\)). The wave function of a quantum mechanical particle varies smoothly, going from within an atomic nucleus (on one side of a potential energy barrier) to outside the nucleus (on the other side of the potential energy barrier). Inside the barrier, the wave function does not become zero but decreases exponentially, and we do not observe the particle inside the barrier. The probability of finding a particle is related to the square of its wave function, and so there is a small probability of finding the particle outside the barrier, which implies that the particle can tunnel through the barrier. A metal tip usually made out of tungsten is placed between a very small distance above a conducting or semiconducting surface. This distance acts as a potential barrier for tunneling. The space between the tip and the surface normally is vacuum. When electrons tunnel from the metal tip to the surface, a current is created and monitored by a computer (Figure \(\Page {6}\)). The current depends on the distance between the tip and the surface, which is controlled by a piezoelectric cylinder. If there is a strong current, the tip will move away from the surface. The increase of the potential barrier will decrease the probability of tunneling and decrease the current. If the current becomes too weak, the tip moves closer to the surface. The potential barrier will be reduced and the current will increase. The variations in the current as the tip moves over the sample are reconstructed by the computer to produce topological image of the scanned surface. ). | 7,644 | 1,109 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/13%3A_The_Phase_Rule_and_Phase_Diagrams/13.01%3A_The_Gibbs_Phase_Rule_for_Multicomponent_Systems |
\( \newcommand{\tx}[1]{\text{#1}} % text in math mode\)
\( \newcommand{\subs}[1]{_{\text{#1}}} % subscript text\)
\( \newcommand{\sups}[1]{^{\text{#1}}} % superscript text\)
\( \newcommand{\st}{^\circ} % standard state symbol\)
\( \newcommand{\id}{^{\text{id}}} % ideal\)
\( \newcommand{\rf}{^{\text{ref}}} % reference state\)
\( \newcommand{\units}[1]{\mbox{$\thinspace$#1}}\)
\( \newcommand{\K}{\units{K}} % kelvins\)
\( \newcommand{\degC}{^\circ\text{C}} % degrees Celsius\)
\( \newcommand{\br}{\units{bar}} % bar (\bar is already defined)\)
\( \newcommand{\Pa}{\units{Pa}}\)
\( \newcommand{\mol}{\units{mol}} % mole\)
\( \newcommand{\V}{\units{V}} % volts\)
\( \newcommand{\timesten}[1]{\mbox{$\,\times\,10^{#1}$}}\)
\( \newcommand{\per}{^{-1}} % minus one power\)
\( \newcommand{\m}{_{\text{m}}} % subscript m for molar quantity\)
\( \newcommand{\CVm}{C_{V,\text{m}}} % molar heat capacity at const.V\)
\( \newcommand{\Cpm}{C_{p,\text{m}}} % molar heat capacity at const.p\)
\( \newcommand{\kT}{\kappa_T} % isothermal compressibility\)
\( \newcommand{\A}{_{\text{A}}} % subscript A for solvent or state A\)
\( \newcommand{\B}{_{\text{B}}} % subscript B for solute or state B\)
\( \newcommand{\bd}{_{\text{b}}} % subscript b for boundary or boiling point\)
\( \newcommand{\C}{_{\text{C}}} % subscript C\)
\( \newcommand{\f}{_{\text{f}}} % subscript f for freezing point\)
\( \newcommand{\mA}{_{\text{m},\text{A}}} % subscript m,A (m=molar)\)
\( \newcommand{\mB}{_{\text{m},\text{B}}} % subscript m,B (m=molar)\)
\( \newcommand{\mi}{_{\text{m},i}} % subscript m,i (m=molar)\)
\( \newcommand{\fA}{_{\text{f},\text{A}}} % subscript f,A (for fr. pt.)\)
\( \newcommand{\fB}{_{\text{f},\text{B}}} % subscript f,B (for fr. pt.)\)
\( \newcommand{\xbB}{_{x,\text{B}}} % x basis, B\)
\( \newcommand{\xbC}{_{x,\text{C}}} % x basis, C\)
\( \newcommand{\cbB}{_{c,\text{B}}} % c basis, B\)
\( \newcommand{\mbB}{_{m,\text{B}}} % m basis, B\)
\( \newcommand{\kHi}{k_{\text{H},i}} % Henry's law constant, x basis, i\)
\( \newcommand{\kHB}{k_{\text{H,B}}} % Henry's law constant, x basis, B\)
\( \newcommand{\arrow}{\,\rightarrow\,} % right arrow with extra spaces\)
\( \newcommand{\arrows}{\,\rightleftharpoons\,} % double arrows with extra spaces\)
\( \newcommand{\ra}{\rightarrow} % right arrow (can be used in text mode)\)
\( \newcommand{\eq}{\subs{eq}} % equilibrium state\)
\( \newcommand{\onehalf}{\textstyle\frac{1}{2}\D} % small 1/2 for display equation\)
\( \newcommand{\sys}{\subs{sys}} % system property\)
\( \newcommand{\sur}{\sups{sur}} % surroundings\)
\( \renewcommand{\in}{\sups{int}} % internal\)
\( \newcommand{\lab}{\subs{lab}} % lab frame\)
\( \newcommand{\cm}{\subs{cm}} % center of mass\)
\( \newcommand{\rev}{\subs{rev}} % reversible\)
\( \newcommand{\irr}{\subs{irr}} % irreversible\)
\( \newcommand{\fric}{\subs{fric}} % friction\)
\( \newcommand{\diss}{\subs{diss}} % dissipation\)
\( \newcommand{\el}{\subs{el}} % electrical\)
\( \newcommand{\cell}{\subs{cell}} % cell\)
\( \newcommand{\As}{A\subs{s}} % surface area\)
\( \newcommand{\E}{^\mathsf{E}} % excess quantity (superscript)\)
\( \newcommand{\allni}{\{n_i \}} % set of all n_i\)
\( \newcommand{\sol}{\hspace{-.1em}\tx{(sol)}}\)
\( \newcommand{\solmB}{\tx{(sol,$\,$$m\B$)}}\)
\( \newcommand{\dil}{\tx{(dil)}}\)
\( \newcommand{\sln}{\tx{(sln)}}\)
\( \newcommand{\mix}{\tx{(mix)}}\)
\( \newcommand{\rxn}{\tx{(rxn)}}\)
\( \newcommand{\expt}{\tx{(expt)}}\)
\( \newcommand{\solid}{\tx{(s)}}\)
\( \newcommand{\liquid}{\tx{(l)}}\)
\( \newcommand{\gas}{\tx{(g)}}\)
\( \newcommand{\pha}{\alpha} % phase alpha\)
\( \newcommand{\phb}{\beta} % phase beta\)
\( \newcommand{\phg}{\gamma} % phase gamma\)
\( \newcommand{\aph}{^{\alpha}} % alpha phase superscript\)
\( \newcommand{\bph}{^{\beta}} % beta phase superscript\)
\( \newcommand{\gph}{^{\gamma}} % gamma phase superscript\)
\( \newcommand{\aphp}{^{\alpha'}} % alpha prime phase superscript\)
\( \newcommand{\bphp}{^{\beta'}} % beta prime phase superscript\)
\( \newcommand{\gphp}{^{\gamma'}} % gamma prime phase superscript\)
\( \newcommand{\apht}{\small\aph} % alpha phase tiny superscript\)
\( \newcommand{\bpht}{\small\bph} % beta phase tiny superscript\)
\( \newcommand{\gpht}{\small\gph} % gamma phase tiny superscript\) \( \newcommand{\upOmega}{\Omega}\) \( \newcommand{\dif}{\mathop{}\!\mathrm{d}} % roman d in math mode, preceded by space\)
\( \newcommand{\Dif}{\mathop{}\!\mathrm{D}} % roman D in math mode, preceded by space\)
\( \newcommand{\df}{\dif\hspace{0.05em} f} % df\) \(\newcommand{\dBar}{\mathop{}\!\mathrm{d}\hspace-.3em\raise1.05ex{\Rule{.8ex}{.125ex}{0ex}}} % inexact differential \)
\( \newcommand{\dq}{\dBar q} % heat differential\)
\( \newcommand{\dw}{\dBar w} % work differential\)
\( \newcommand{\dQ}{\dBar Q} % infinitesimal charge\)
\( \newcommand{\dx}{\dif\hspace{0.05em} x} % dx\)
\( \newcommand{\dt}{\dif\hspace{0.05em} t} % dt\)
\( \newcommand{\difp}{\dif\hspace{0.05em} p} % dp\)
\( \newcommand{\Del}{\Delta}\)
\( \newcommand{\Delsub}[1]{\Delta_{\text{#1}}}\)
\( \newcommand{\pd}[3]{(\partial #1 / \partial #2 )_{#3}} % \pd{}{}{} - partial derivative, one line\)
\( \newcommand{\Pd}[3]{\left( \dfrac {\partial #1} {\partial #2}\right)_{#3}} % Pd{}{}{} - Partial derivative, built-up\)
\( \newcommand{\bpd}[3]{[ \partial #1 / \partial #2 ]_{#3}}\)
\( \newcommand{\bPd}[3]{\left[ \dfrac {\partial #1} {\partial #2}\right]_{#3}}\)
\( \newcommand{\dotprod}{\small\bullet}\)
\( \newcommand{\fug}{f} % fugacity\)
\( \newcommand{\g}{\gamma} % solute activity coefficient, or gamma in general\)
\( \newcommand{\G}{\varGamma} % activity coefficient of a reference state (pressure factor)\)
\( \newcommand{\ecp}{\widetilde{\mu}} % electrochemical or total potential\)
\( \newcommand{\Eeq}{E\subs{cell, eq}} % equilibrium cell potential\)
\( \newcommand{\Ej}{E\subs{j}} % liquid junction potential\)
\( \newcommand{\mue}{\mu\subs{e}} % electron chemical potential\)
\( \newcommand{\defn}{\,\stackrel{\mathrm{def}}{=}\,} % "equal by definition" symbol\) \( \newcommand{\D}{\displaystyle} % for a line in built-up\)
\( \newcommand{\s}{\smash[b]} % use in equations with conditions of validity\)
\( \newcommand{\cond}[1]{\\[-2.5pt]{}\tag*{#1}}\)
\( \newcommand{\nextcond}[1]{\\[-5pt]{}\tag*{#1}}\)
\( \newcommand{\R}{8.3145\units{J$\,$K$\per\,$mol$\per$}} % gas constant value\)
\( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \)
\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) In Sec. 8.1.7, the Gibbs phase rule for a pure substance was written \(F = 3 - P\). We now consider a system of more than one substance and more than one phase in an equilibrium state. The phase rule assumes the system is at thermal and mechanical equilibrium. We shall assume furthermore that in addition to the temperature and pressure, the only other state functions needed to describe the state are the amounts of the species in each phase; this means for instance that surface effects are ignored. The derivations to follow will show that the phase rule may be written either in the form \begin{equation} F = 2 + C - P \tag{13.1.1} \end{equation} or \begin{equation} F = 2 + s - r - P \tag{13.1.2} \end{equation} where the symbols have the following meanings: \(F=\) the number of degrees of freedom (or variance) \(\hspace2ex =\) the maximum number of intensive variables that can be varied independently while the system remains in an equilibrium state; \(C=\) the number of components \(\hspace2ex =\) the minimum number of substances (or fixed-composition mixtures of substances) \(\hspace 3ex\) that could be used to prepare each phase individually; \(P=\) the number of different phases; \(s=\) the number of different species; \(r=\) the number of independent relations among intensive variables of individual phases other than relations \(\hspace 3ex\) needed for thermal, mechanical, and transfer equilibrium. If we subdivide a phase, that does not change the number of phases \(P\). That is, we treat noncontiguous regions of the system that have identical intensive properties as parts of the same phase. Consider a system in an equilibrium state. In this state, the system has one or more phases; each phase contains one or more species; and intensive properties such as \(T\), \(p\), and the mole fraction of a species in a phase have definite values. Starting with the system in this state, we can make changes that place the system in a new equilibrium state having the same kinds of phases and the same species, but different values of some of the intensive properties. The number of different independent intensive variables that we may change in this way is the or , \(F\), of the system. Clearly, the system remains in equilibrium if we change the of a phase without changing its temperature, pressure, or composition. This, however, is the change of an extensive variable and is not counted as a degree of freedom. The phase rule, in the form to be derived, applies to a system that continues to have complete thermal, mechanical, and transfer equilibrium as intensive variables change. This means different phases are not separated by adiabatic or rigid partitions, or by semipermeable or impermeable membranes. Furthermore, every conceivable reaction among the species is either at reaction equilibrium or else is frozen at a fixed advancement during the time period we observe the system. The number of degrees of freedom is the maximum number of intensive properties of the equilibrium system we may independently vary, or fix at arbitrary values, without causing a change in the number and kinds of phases and species. We cannot, of course, change one of these properties to just any value whatever. We are able to vary the value only within a certain finite (sometimes quite narrow) range before a phase disappears or a new one appears. The number of degrees of freedom is also the number of independent intensive variables needed to specify the equilibrium state in all necessary completeness, aside from the amount of each phase. In other words, when we specify values of \(F\) different independent intensive variables, then the values of all other intensive variables of the equilibrium state have definite values determined by the physical nature of the system. Just as for a one-component system, we can use the terms , , and depending on the value of \(F\) (Sec. 8.1.7). This section derives an expression for the number of degrees of freedom, \(F\), based on . Section 13.1.3 derives an expression based on . Both approaches yield equivalent versions of the phase rule. Recall that a is an entity, uncharged or charged, distinguished from other species by its chemical formula (Sec. 9.1.1). Thus, CO\(_2\) and CO\(_3\)\(^{2-}\) are different species, but CO\(_2\)(aq) and CO\(_2\)(g) is the same species in different phases. Consider an equilibrium system of \(P\) phases, each of which contains the same set of species. Let the number of different species be \(s\). If we could make changes while the system remains in thermal and mechanical equilibrium, but not necessarily in transfer equilibrium, we could independently vary the temperature and pressure of the system as a whole and the amount of each species in each phase; there would then be \(2 + Ps\) independent variables. The equilibrium system is, however, in transfer equilibrium, which requires each species to have the same chemical potential in each phase: \(\mu_i\bph = \mu_i\aph\), \(\mu_i\gph = \mu_i\aph\), and so on. There are \(P - 1\) independent relations like this for each species, and a total of \(s(P - 1)\) independent relations for all species. Each such independent relation introduces a constraint and reduces the number of independent variables by one. Accordingly, taking transfer equilibrium into account, the number of independent variables is \(2 + Ps - s(P - 1) = 2 + s\). We obtain the same result if a species present in one phase is totally excluded from another. For example, solvent molecules of a solution are not found in a pure perfectly-ordered crystal of the solute, undissociated molecules of a volatile strong acid such as HCl can exist in a gas phase but not in aqueous solution, and ions of an electrolyte solute are usually not found in a gas phase. For each such species absent from a phase, there is one fewer amount variable and also one fewer relation for transfer equilibrium; on balance, the number of independent variables is still \(2 + s\). Next, we consider the possibility that further independent relations exist among intensive variables in addition to the relations needed for thermal, mechanical, and transfer equilibrium. (Relations such as \(\sum_i p_i=p\) for a gas phase or \(\sum_i x_i=1\) for a phase in general have already been accounted for in the derivation by the specification of \(p\) and the amount of each species.) If there are \(r\) of these additional relations, the total number of independent variables is reduced to \(2 + s - r\). These relations may come from There is an infinite variety of possible choices of the independent variables (both extensive and intensive) for the equilibrium system, but the total of independent variables is fixed at \(2+s-r\). Keeping intensive properties fixed, we can always vary how much of each phase is present (e.g., its volume, mass, or amount) without destroying the equilibrium. Thus, at least \(P\) of the independent variables, one for each phase, must be extensive. It follows that the maximum number of independent variables is the difference \((2 + s - r) - P\). Since the maximum number of independent intensive variables is the number of degrees of freedom, our expression for \(F\) based on species is \begin{equation} F = 2 + s - r - P \tag{13.1.3} \end{equation} The derivation of the phase rule in this section uses the concept of . The number of components, \(C\), is the minimum number of substances or mixtures of fixed composition from which we could in principle prepare each individual phase of an equilibrium state of the system, using methods that may be hypothetical. These methods include the addition or removal of one or more of the substances or fixed-composition mixtures, and the conversion of some of the substances into others by means of a reaction that is at equilibrium in the actual system. It is not always easy to decide on the number of components of an equilibrium system. The number of components may be less than the number of substances present, on account of the existence of reaction equilibria that produce some substances from others. When we use a reaction to prepare a phase, nothing must remain unused. For instance, consider a system consisting of solid phases of CaCO\(_3\) and CaO and a gas phase of CO\(_2\). Assume the reaction CaCO\(_3\)(s) \(\ra\) CaO(s) + CO\(_2\)(g) is at equilibrium. We could prepare the CaCO\(_3\) phase from CaO and CO\(_2\) by the reverse of this reaction, but we can only prepare the CaO and CO\(_2\) phases from the individual substances. We could not use CaCO\(_3\) to prepare either the CaO phase or the CO\(_2\) phase, because CO\(_2\) or CaO would be left over. Thus this system has three substances but only two components, namely CaO and CO\(_2\). In deriving the phase rule by the components approach, it is convenient to consider only intensive variables. Suppose we have a system of \(P\) phases in which each substance present is a component (i.e., there are no reactions) and each of the \(C\) components is present in each phase. If we make changes to the system while it remains in thermal and mechanical equilibrium, but not necessarily in transfer equilibrium, we can independently vary the temperature and pressure of the whole system, and for each phase we can independently vary the mole fraction of all but one of the substances (the value of the omitted mole fraction comes from the relation \(\sum_i x_i = 1\)). This is a total of \(2 + P(C - 1)\) independent intensive variables. When there also exist transfer and reaction equilibria, not all of these variables are independent. Each substance in the system is either a component, or else can be formed from components by a reaction that is in reaction equilibrium in the system. Transfer equilibria establish \(P - 1\) independent relations for each component (\(\mu_i\bph = \mu_i\aph\), \(\mu_i\gph = \mu_i\aph\), etc.) and a total of \(C(P - 1)\) relations for all components. Since these are relations among chemical potentials, which are intensive properties, each relation reduces the number of independent intensive variables by one. The resulting number of independent intensive variables is \begin{equation} F = [2 + P(C - 1)] - C(P - 1) = 2 + C - P \tag{13.1.4} \end{equation} If the equilibrium system lacks a particular component in one phase, there is one fewer mole fraction variable and one fewer relation for transfer equilibrium. These changes cancel in the calculation of \(F\), which is still equal to \(2 + C - P\). If a phase contains a substance that is formed from components by a reaction, there is an additional mole fraction variable and also the additional relation \(\sum_i\!\nu_i \mu_i = 0\) for the reaction; again the changes cancel. We may need to a component from a phase to achieve the final composition. Note that it is not necessary to consider additional relations for electroneutrality or initial conditions; they are implicit in the definitions of the components. For instance, since each component is a substance of zero electric charge, the electrical neutrality of the phase is assured. We conclude that, regardless of the kind of system, the expression for \(F\) based on components is given by \(F=2+C-P\). By comparing this expression and \(F=2+s-r-P\), we see that the number of components is related to the number of species by \begin{equation} C = s - r \tag{13.1.5} \end{equation} The five examples below illustrate various aspects of using the phase rule. For a single phase of pure water, \(P\) equals \(1\). If we treat the water as the single species H\(_2\)O, \(s\) is 1 and \(r\) is 0. The phase rule then predicts two degrees of freedom: \begin{equation} \begin{split} F & = 2 + s - r - P\cr & = 2 + 1 - 0 - 1 = 2 \end{split} \tag{13.1.6} \end{equation} Since \(F\) is the number of intensive variables that can be varied independently, we could for instance vary \(T\) and \(p\) independently, or \(T\) and \(\rho \), or any other pair of independent intensive variables. Next let us take into account the proton transfer equilibrium \[ \ce{2H2O}\tx{(l)} \arrows \ce{H3O+}\tx{(aq)} + \ce{OH-}\tx{(aq)} \] and consider the system to contain the three species H\(_2\)O, H\(_3\)O\(^+\), and OH\(^-\). Then for the species approach to the phase rule, we have \(s = 3\). We can write two independent relations: If we consider water to contain additional cation species (e.g., \(\ce{H5O2+}\)), each such species would add \(1\) to \(s\) and \(1\) to \(r\), but \(F\) would remain equal to 2. Thus, no matter how complicated are the equilibria that actually exist in liquid water, the number of degrees of freedom remains \(2\). Consider a system containing solid carbon (graphite) and a gaseous mixture of O\(_2\), CO, and CO\(_2\). There are four species and two phases. If reaction equilibrium is absent, as might be the case at low temperature in the absence of a catalyst, we have \(r = 0\) and \(C = s - r = 4\). The four components are the four substances. The phase rule tells us the system has four degrees of freedom. We could, for instance, arbitrarily vary \(T\), \(p\), \(y\subs{O\(_2\)}\), and \(y\subs{CO}\). Now suppose we raise the temperature or introduce an appropriate catalyst to allow the following reaction equilibria to exist: If we wish to calculate \(F\) by the components approach, we must decide on the minimum number of substances we could use to prepare each phase separately. (This does not refer to how we actually prepare the two-phase system, but to a hypothetical preparation of each phase with any of the compositions that can actually exist in the equilibrium system.) Assume equilibria 1 and 2 are present. We prepare the solid phase with carbon, and we can prepare any possible equilibrium composition of the gas phase from carbon and O\(_2\) by using the reactions of both equilibria. Thus, there are two components (C and O\(_2\)) giving the same result of two degrees of freedom. Applying the components approach to this system is straightforward. The solid phase is prepared from PbCl\(_2\) and the aqueous phase could be prepared by dissolving solid PbCl\(_2\) in H\(_2\)O. Thus, there are two components and two phases: \begin{equation} F = 2+C-P=2 \tag{13.1.10} \end{equation} For the species approach, we note that there are four species (PbCl\(_2\), Pb\(^{2+}\), Cl\(^-\), and H\(_2\)O) and two independent relations among intensive variables: If there is no special relation among the total amounts of N\(_2\) and O\(_2\), there are three components and the phase rule gives \begin{equation} F = 2 + C - P = 3 \tag{13.1.12} \end{equation} Since there are three degrees of freedom, we could, for instance, specify arbitrary values of \(T\), \(p\), and \(y\subs{N\(_2\)}\) (arbitrary, that is, within the limits that would allow the two phases to coexist); then the values of other intensive variables such as the mole fractions \(y\subs{H\(_2\)O}\) and \(x\subs{N\(_2\)}\) would have definite values. Now suppose we impose an initial condition by preparing the system with water and dry air of a composition. The mole ratio of N\(_2\) and O\(_2\) in the aqueous solution is not necessarily the same as in the equilibrated gas phase; consequently, the air does not behave like a single substance. The number of components is still three: H\(_2\)O, N\(_2\), and O\(_2\) are all required to prepare each phase individually, just as when there was no initial condition, giving \(F = 3\) as before. The fact that the compositions of both phases depend on the relative amounts of the phases is illustrated in Prob. 9.5. We can reach the same conclusion with the species approach. The initial condition can be expressed by an equation such as \begin{equation} \frac{(n\subs{N\(_2\)}\sups{l} + n\subs{N\(_2\)}\sups{g})} {(n\subs{O\(_2\)}\sups{l} + n\subs{O\(_2\)}\sups{g})} = a \tag{13.1.13} \end{equation} where \(a\) is a constant equal to the mole ratio of N\(_2\) and O\(_2\) in the dry air. This equation cannot be changed to a relation between intensive variables such as \(x\subs{N\(_2\)}\) and \(x\subs{O\(_2\)}\), so that \(r\) is zero and there are still three degrees of freedom. Finally, let us assume that we prepare the system with dry air of fixed composition, as before, but consider the solubilities of N\(_2\) and O\(_2\) in water to be negligible. Then \(n\subs{N\(_2\)}\sups{l} \) and \(n\subs{O\(_2\)}\sups{l} \) are zero and Eq. 13.1.13 becomes \(n\subs{N\(_2\)}\sups{g} / n\subs{O\(_2\)}\sups{g} = a\), or \(y\subs{N\(_2\)} = ay\subs{O\(_2\)}\), which is a relation between intensive variables. In this case, \(r\) is 1 and the phase rule becomes \begin{equation} F = 2 + s - r - P = 2 \tag{13.1.14} \end{equation} The reduction in the value of \(F\) from 3 to 2 is a consequence of our inability to detect any dissolved N\(_2\) or O\(_2\). According to the components approach, we may prepare the liquid phase with H\(_2\)O and the gas phase with H\(_2\)O and air of fixed composition that behaves as a single substance; thus, there are only two components. Consider the following reaction equilibrium: \[ \ce{3CuO}\tx{(s)} + \ce{2NH3}\tx{(g)} \arrows \ce{3Cu}\tx{(s)} + \ce{3H2O}\tx{(g)} + \ce{N2}\tx{(g)} \] According to the species approach, there are five species, one relation (for reaction equilibrium), and three phases. The phase rule gives \begin{equation} F = 2 + s - r - P = 3 \tag{13.1.15} \end{equation} It is more difficult to apply the components approach to this example. As components, we might choose CuO and Cu (from which we could prepare the solid phases) and also NH\(_3\) and H\(_2\)O. Then to obtain the N\(_2\) needed to prepare the gas phase, we could use CuO and NH\(_3\) as reactants in the reaction \(\ce{3CuO} + \ce{2NH3} \arrow \ce{3Cu} + \ce{3H2O} + \ce{N2}\) and remove the products Cu and H\(_2\)O. In the components approach, we are allowed to remove substances from the system provided they are counted as components. | 25,495 | 1,110 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Fundamentals_of_Thermodynamics/Entropy_Changes_in_Reactions |
Entropy is another important aspect of thermodynamics. Enthalpy has something to do with the energetic content of a system or a molecule. Entropy has something to do with how that energy is stored. We sometimes speak of the energy in a system as being "partitioned" or divided into various "states". How this energy is divided up is the concern of entropy. By way of analogy, picture a set of mailboxes. You may have a wall of them in your dormitory or your apartment building. The mailboxes are of several different sizes: maybe there are a few rows of small ones, a couple of rows of medium sized ones, and a row of big mailboxes on the bottom. Instead of putting mail in these boxes, we're going to use them to hold little packages of energy. Later on, you might take the energy packages out of your own mailbox and use them to take a trip to the mall or the gym. But how does the mail get to your mailbox in the first place? The energy packages don't arrive in your molecular dormitory with addresses on them. The packages come in different sizes, because they contain different amounts of energy, but other than that there is no identifying information on them. Some of the packages don't fit into some of the mailboxes, because some of the packages are too big and some of the mailboxes are smaller than the others. The energy packages need to go into mailboxes that they will fit into. Still, there are an awful lot of mailboxes that most of the energy packages could still fit into. There needs to be some system of deciding where to put all of these packages. It turns out that, in the molecular world, there is such a system, and it follows a pretty simple rule. When a whole pile of energy packages arrive, the postmaster does her best to put one package into every mailbox. Then, when every mailbox has one, she starts putting a second one into each box, and so on. It didn't have to be that way. It could have been the case that all the energy was simply put into the first couple of mailboxes and the rest were left empty. In other words, the rule could have been that all the energy must be sorted into the same place, instead of being spread around. But that's not how it is. Entropy is the sorting of energy into different modes or states. When energy is partitioned or sorted into additional states, entropy is said to increase. When energy is bundled into a smaller number of states, entropy is said to decrease. Nature's bias is towards an increase in entropy. This is a fundamental law of the universe; there is no reason that can be used to explain why nature prefers high entropy to low entropy. Instead, increasing entropy is itself the basic reason for a wide range of things that happen in the universe. Entropy is popularly described in terms of "disorder". That can be a useful idea, although it doesn't really describe what is happening energetically. A better picture of entropy can be built by looking at how a goup of molecules might sort some energy that is added to them. In other words, what are some examples of "states" in which energy can be sorted? If you get more energy -- maybe by eating breakfast -- one of the immediate benefits is being able to increase your physical activity. You have more energy to move around, to run, to jump. A similar situation is true with molecules. Molecules have a variety of ways in which they can move, if they are given some energy. They can zip around; this kind of motion is usually called translation. They can tumble and roll; this kind of motion is referred to as rotation. Also, they can wiggle, letting their bonds get longer and shorter by moving individual atoms around a little bit. This type of motion is called vibration. When molecules absorb extra energy, they may be able to sort the energy into rotational, vibrational and translational states. This only works with energy packages of a certain size; other packages would be sorted into other kinds of states. However, these are just a few examples of what we mean by states. Okay, so energy is stored in states, and it is sorted into the maximum possible number of states. But how does entropy change in a reaction? We know that enthalpy may change by breaking or forming certain bonds, but how does the energy get sorted again? The changes in internal entropy during a reaction are often very small. In other words, the energy remaining at the end of the reaction gets sorted more or less the way it was before the reaction. However, there are some very common exceptions. The most common case in which internal entropy changes a lot is when the number of molecules involved changes between the start of the reaction and the end of the reaction. Maybe two molecules react together to form one, new molecule. Maybe one molecule splits apart to make two, new molecules. If one molecule splits apart in the reaction, entropy generally increases. Two molecules can rotate, vibrate and translate (or tumble, wiggle and zip around) independently of each other. That means the number of states available for partitioning energy increases when one molecule splits into two. Apart from a factor like a change in the number of molecules involved, internal entropy changes are often fairly subtle. They are not as easy to predict as enthalpy changes. Nevertheless, there may sometimes be a trade-off between enthalpy and entropy. If a reaction splits a molecule into two, it seems likely that an increase in enthalpy will be involved, so that the bond that held the two pieces together can be broken. That's not favourable. However, when that happens, we've just seen that there will be an increase in entropy, because energy can then be sorted into additional modes in the two, independent molecules. So we have two different factors to balance. There is a tool we often use to decide which factor wins out. It's called free energy, and we will look at it next. , | 5,910 | 1,112 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Effect_of_Pressure_on_Gas-Phase_Equilibria |
states that a system at equilibrium will adjust to relieve stress when there are changes in the concentration of a reactant or product, the partial pressures of components, the volume of the system, and the temperature of reaction. There are three ways to change the pressure of a constant-temperature reaction system involving gaseous components: When a system at equilibrium undergoes a change in pressure, the equilibrium of the system will shift to offset the change and establish a new equilibrium. The system can shift in one of two ways: The effects of changes in pressure can be described as follows ( ): Pressure is inversely related to . Therefore, the effects of changes in pressure are opposite of the effects of changes in volume. Additionally, this does not apply to a change in the pressure in the system due to the addition of an inert gas. | 876 | 1,113 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.10%3A_Standard_Enthalpies_of_Formation/3.10.01%3A_Biology-_Muscle_Energy_from_ATP |
We have looked at sugar as a source of [Weight of Food and Energy Production | energy production] in our bodies in the last few sections, but we also know that the mechanism by which chemical energy is delivered to muscles isn't the same as the mechanism for combustion of a sugar: C H O (s) + 6 O (g) → 6 CO (g) + 6 H O(l) (25 , 1 Atm) Δ = –2808 kJ mol (1) In our bodies, adenosine Triphosphate, ATP, and adsenosine diphosphate, ADP, are central intermediaries in energy delivery. ATP ADP The energy of sugar metabolism is stored in the form of accumulated ATP, and when energy is needed by muscle, it is delivered by the reaction ATP + H O → ADP + HPO + H Δ = ~ –22 kJ mol which is shown in an . But it just doesn't make sense that this reaction should release energy, because it involves breaking a bond to remove HPO , and breaking a bond should require energy. Since so many biologists misinterpret this reaction, we'll try to explain where the heat energy that drives our muscles (and growth) comes from more clearly, in terms of , below. It's clear that there is almost an infinite number of chemical reactions whose heat energy is important to know, and chemists have measured the enthalpy changes for so many reactions that it would take several large volumes to list all the thermochemical equations. Fortunately Hess' law makes it possible to list a single value, Δ , for each compound. The standard enthalpy of formation is the enthalpy change when 1 mol of a pure substance is formed from its elements. Each element must be in the physical and chemical form which is most stable at normal atmospheric pressure and a specified temperature (usually 25°C). For example, if we know that Δ [H O( )] = –285.8 kJ mol , we can immediately write the thermochemical equation H ( ) + ½O ( ) → H O( ) Δ = –285.8 kJ mol (2) The elements H and O appear as diatomic molecules and in gaseous form because these are their most stable chemical and physical states. Note also that 285.8 kJ are given off of H O( ) formed, because stronger bonds are formed (2 H-O bonds), releasing a lot of energy, while weaker bonds (O-O and H-H) are broken, requiring less energy. Equation (1) must specify formation of mol H O( ), and so the coefficient of O must be ½. The heat of formation of Cl atoms makes it clear that bond breaking is endothermic: Cl → 2 Cl Δ = +121.68 kJ mol (3) In some cases, such as that of water, the elements will react directly to form a compound, and measurement of the heat absorbed serves to determine Δ . Quite often, however, elements do not react directly with each other to form the desired compound, and Δ must be calculated by combining the enthalpy changes for other reactions. This is the case for ATP Some Standard Enthalpies of Formation at 25°C. One further point arises from the definition of Δ . . If we form oxygen from its elements, for example, we are talking about the reaction O ( ) → O ( ) Since the oxygen is unchanged, there can be no enthalpy change, and Δ = 0 kJ mol . Standard enthalpies of formation for some common compounds are given in the table above. These values may be used to calculate Δ for any chemical reaction so long as all the compounds involved appear in the tables. To see how and why this may be done, consider the following example. Use standard enthalpies of formation below to estimate Δ for the reaction P O + H O → 2 HPO (1) Note that this is like the decomposition (or "hydrolysis") of ATP to give ADP + HPO , but simpler. It still , even though a bond is ostensibly broken between ATP and HPO , . What isn't seen is very important: The heats of formation of ions include "solvation energies", or in this case, energies released when water molecules bond to the HPO , releasing significant amounts of energy. The energy released by "hydration" is greater than the energy required to break the P-O-P bond in ATP. We can imagine that the reaction takes place in two steps, each of which involves only a standard enthalpy of formation. In the first step the reactants are broken into their elements, and in the second step the elements are recombined to give the products. First, P O ("pyrophosphate ion") is decomposed to its elements: P O ( ) → 2P( ) + 7/2 O ( ) Δ = Δ (2) Since this is the of formation of mol P O from its elements, the enthalpy change is Δ = {–Δ [P O ( )]} = [– (–2286 kJ mol )] = +2286 kJ mol . It's positive (endothermic) because bond breaking takes energy. Next, we break water into its elements: H O (l) → H + 1/2O (3) Again Δ = -(-287) = +287 kJ mol . In the second step the elements are combined to give 2 mol HPO ("inorganic monophosphate ion" or "hydrogen phosphate ion"): 2P( ) + H + 2O ( ) → 2 HPO ( ) Δ = Δ (4) In this case Δ = 2 × Δ [HPO ( )] = 2 × (– 1299 kJ mol ) = – -2598 kJ mol You can easily verify that the sum of Eqs. (2) and (3) is P O + H O → 2 HPO Δ = Δ Therefore Δ = Δ + Δ + Δ = 287 kJ mol +2286 kJ mol – 2598 kJ mol = – 25 kJ mol Note carefully how Example 2 was solved. In step 1 the compound P O (aq) was hypothetically decomposed to its elements. This equation was the reverse of formation of the compound, and so Δ was opposite in sign from Δ . In step 2 we had the hypothetical decomposition of the other reactant, water, again Δ was opposite in sign from Δ . Finally, 2 moles of the HPO ( ) were formed from its elements. Since 2 mol were obtained, the enthalpy change was doubled but its sign remained the same. Any chemical reaction can be approached similarly. To calculate Δ we all the Δ values for the products, multiplying each by the appropriate coefficient, as in step 2 above. Since the signs of Δ for the reactants had to be reversed in step 1, we them, again multiplying by appropriate coefficients. This can he summarized by the equation Δ = ∑ Δ (products) – ∑ Δ (reactants) (4) The symbol Σ means “the sum of.” Since Δ values are given of compound, you must be sure to multiply each Δ by an appropriate coefficient derived from the equation for which Δ is being calculated. Use the table of standard enthalpies of formation at 25°C to calculate Δ for the reaction ATP + H O → ADP + HPO + H Once more, remember that the energy released comes from the fact that bonding between water molecules and the HPO that is released more energy than it to break the P-O-P bond in ATP to give ADP + HPO . Using Eq. (4), we have Δ = ∑ Δ (products) – ∑ Δ (reactants) = [Δ (ADP ) + Δ (HPO )] – [Δ ((H O)+ Δ (ATP )] = (–2000) kJ mol + (-1299) kJ mol – (–2982 kJ mol ) – (–287 kJ mol ) = -30 kJ mol Note that we were careful to use Δ [H O( )] not Δ [H O( )]. Even though water vapor is not the most stable form of water at 25°C, we can still use its Δ value to do an interesting calculation: Find the heat energy required to vaporize 1 mole of water (we know that should be positive. It takes energy to boil water because we're breaking bonds of attraction between water molecules. H O( ) → H O( )] Δ = ? Δ = ∑ Δ (products) – ∑ Δ (reactants) = - 241.8 -(- 285.8) = - 241.8 + 285.8 = + 44 kJ mol | 7,120 | 1,114 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/15%3A_Alcohols_and_Ethers/15.06%3A_Reactions_Involving_the_C-O_Bond_of_Alcohols |
Alkyl halide formation from an alcohol and a hydrogen halide affords an important example of a reaction wherein the \(\ce{C-O}\) bond of the alcohol is broken: The reaction is reversible and the favored direction depends on the water concentration. Primary bromides often are prepared best by passing dry hydrogen bromide into the alcohol heated to just slightly below its boiling point. Halide formation proceeds at a useful rate only in the presence of strong acid, which can be furnished by excess hydrogen bromide or, usually and more economically, by sulfuric acid. The alcohol accepts a proton from the acid to give an alkyloxonium ion, which is more reactive in subsequent displacement with bromide ion than the alcohol (by either \(S_\text{N}2\) or \(S_\text{N}1\) mechanisms) because \(\ce{H_2O}\) is a better leaving group than \(\ce{OH}^\ominus\): or Whether the displacement reaction is an \(S_\text{N}1\) or \(S_\text{N}2\) process depends on the structure of the alcohol. In general, the primary alcohols are considered to react by \(S_\text{N}2\) and the secondary and tertiary alcohols by \(S_\text{N}1\) mechanisms. Hydrogen chloride is less reactive than hydrogen bromide toward primary alcohols, and a catalyst (zinc chloride) may be required. A solution of zinc chloride in concentrated hydrochloric acid (Lucas reagent) is a convenient reagent to differentiate between primary, secondary, and tertiary alcohols with less than eight or so carbons. Tertiary alcohols react very rapidly to give an insoluble layer of alkyl chloride at room temperature. Secondary alcohols react in several minutes, whereas primary alcohols form chlorides only on heating. The order of reactivity is typical of \(S_\text{N}1\) reactions. Zinc chloride probably assists in the breaking of the \(\ce{C-O}\) bond of the alcohol much as silver ion aids ionization of \(\ce{RCl}\) (Section 8-7D): Thionyl chloride, \(\ce{O=SCl_2}\), is useful for the preparation of alkyl chlorides, especially when the use of strongly acidic reagents, such as zinc chloride and hydrochloric acid, is undesirable. Thionyl chloride can be regarded as the acid chloride of sulfurous acid, \(\ce{O=S(OH)_2}\), and like most acid chlorides the halogen is displaced readily by alcohols. Addition of 1 mole of an alcohol to 1 mole of thionyl chloride gives an unstable alkyl chlorosulfite, which generally decomposes on mild heating to yield the alkyl chloride and sulfur dioxide: Chlorides can be prepared in this way from primary and secondary, but not tertiary, alcohols. In practice, an equivalent of a weak base, such as pyridine (azabenzene), is added to neutralize the hydrogen chloride that is formed. If the acid is not removed, undesirable degradation, elimination, and rearrangement reactions may occur. The thionyl chloride reaction apparently can proceed from the alkyl chlorosulfite stage by more than one mechanism: an ionic \(S_\text{N}2\) chain reaction with chloride ion, or an \(S_\text{N}1\)-like ionization and collapse of the resulting \(\ce{R}^\oplus \ce{Cl}^\ominus\) ion pair to give \(\ce{RCl}\): or Obviously, the greater the \(S_\text{N}2\) reactivity associated with the better the \(S_\text{N}2\) reaction will go and, conversely, if \(\ce{R}^\oplus\) is formed easily from the \(S_\text{N}1\) reaction is likely to be favored. Other halides that are useful in converting alcohols to alkyl halides are \(\ce{PCl_5}\), \(\ce{PCl_3}\), \(\ce{PBr_3}\), and \(\ce{PI_3}\), which are acid halides of phosphorus oxyacids. As with thionyl chloride, a weak base often is used to facilitate the reaction. The base acts to neutralize the acid formed, and also to generate bromide ion for \(S_\text{N}\) reactions: It is possible to prefer esters of sulfuric acid by the reaction of an alcohol with the acid: The reaction is closely related to alkyl halide formation under strongly acidic conditions, whereby conversion of the alcohol to an oxonium salt is a first step: Conversion of the oxonium hydrogen sulfate to the ester probably proceeds by an \(S_\text{N}2\) mechanism with primary alcohols and an \(S_\text{N}1\) mechanism with tertiary alcohols: An alternative mechanism, which operates either in \(100\%\), or in fuming sulfuric acid (which contains dissolved \(\ce{SO_3}\)), is addition of sulfur trioxide to the \(\ce{OH}\) group: The sodium salts of alkyl hydrogen sulfate esters have useful properties as detergents if the alkyl group is large, \(\ce{C_{12}}\) or so: The mechanism of detergent action will be considered in more detail in . In principle, dialkyl sulfates could be formed by an \(S_\text{N}2\) reaction between an alkyloxonium salt and an alkyl sulfate ion: Indeed, if methanol is heated with fuming sulfuric acid, dimethyl sulfate, \(\ce{CH_3O(SO_2)OCH_3}\), is obtained; but other alcohols are better converted to dialkyl sulfates by oxidation of the corresponding dialkyl formed by the reaction of 1 mole of thionyl chloride \(\left( \ce{SOCl_2} \right)\) with 2 moles of the alcohol: The reason that dialkyl sulfates seldom are prepared by direct reaction of the alcohol with \(\ce{H_2SO_4}\) is that the mono esters react rapidly on heating to eliminate sulfuric acid and form alkenes, as explained in . , \(\ce{R-SO_2-OH}\) or \(\ce{Ar-SO_2-OH}\), are oxyacids of sulfur that resemble sulfuric acid, \(\ce{HO-SO_2-OH}\), but in which sulfur is in a lower oxidation state. are useful intermediates in displacement reactions (Section 8-7C) and provide a route for the conversion of an alcohol, \(\ce{ROH}\), to \(\ce{RX}\) by the sequence: Sulfonate esters usually are prepared through treatment of the alcohol with the acid chloride (sulfonyl chloride) in the presence of pyridine (azabenzene): In the reaction of an alcohol with hot concentrated sulfuric acid, the alcohol is dehydrated to an alkene: This is the reverse of acid-catalyzed hydration of alkenes discussed previously ( ) and goes to completion if the alkene is allowed to distill out of the reaction mixture as it is formed. One mechanism of dehydration involves proton transfer from sulfuric acid to the alcohol, followed by an \(E2\) reaction of hydrogen sulfate ion or water with the oxonium salt of the alcohol: Alternatively, the alkyl hydrogen sulfate could be formed and eliminate sulfuric acid by an \(E2\) reaction: At lower temperatures the oxonium salt or the alkyl hydrogen sulfate may react by an \(S_\text{N}\) displacement mechanism with excess alcohol in the reaction mixture, thereby forming a dialkyl ether. Although each step in the reaction is reversible, ether formation can be enhanced by distilling away the ether as fast as it forms. Diethyl ether is made commercially by this process: Most alcohols also will dehydrate at fairly high temperatures in the presence of solid catalysts such as silica gel or aluminum oxide to give alkenes or ethers. The behavior of ethanol is reasonably typical of primary alcohols and is summarized in the following equations: Tertiary alcohols react with sulfuric acid at much lower temperatures than do most primary or secondary alcohols. The reactions typically are \(S_\text{N}1\) and \(E1\) by way of a tertiary carbocation, as shown here for -butyl alcohol and sulfuric acid: 2-Methylpropene can be removed from the reaction mixture by distillation and easily is made the principal product by appropriate adjustment of the reaction conditions. If the 2-methylpropene is not removed as it is formed, polymer and oxidation products become important. Sulfuric acid often is an unduly strenuous reagent for dehydration of tertiary alcohols. Potassium hydrogen sulfate, copper sulfate, iodine, phosphoric acid, or phosphorus pentoxide may give better results by causing less polymerization and less oxidative degradation which, with sulfuric acid, results in the formation of sulfur dioxide. The \(S_\text{N}1\)-\(E1\) behavior of tertiary alcohols in strong acids can be used to advantage in the preparation of -butyl ethers. If, for example, a mixture of -butyl alcohol and methanol is heated in the presence of sulfuric acid, the tertiary alcohol reacts rapidly to produce 2-methylpropene by way of the -butyl cation. This cation can be trapped by the methanol to form -butyl methyl ether. High yields of ethers can be obtained in this way: Rearrangement of the alkyl groups of alcohols is very common in dehydration, particularly in the presence of strong acids, which are conducive to carbocation formation. Typical examples showing both methyl and hydrogen migration follow: The key step in each such rearrangement is isomerization of a carbocation, as discussed in . Under kinetic control, the final products always correspond to rearrangement of a less stable carbocation to a more stable carbocation. (Thermodynamic control may lead to quite different results, .) In the dehydration of 3,3-dimethyl-2-butanol, a secondary carbocation is formed initially, which rearranges to a tertiary carbocation when a neighboring methyl group its bonding electron pair migrates to the positive carbon. The charge is thereby transferred to the tertiary carbon: Phosphoric acid \(\left( \ce{H_3PO_4} \right)\) often is used in place of sulfuric acid to dehydrate alcohols. This is because phosphoric acid is less destructive; it is both a weaker acid and a less powerful oxidizing agent than sulfuric acid. Dehydration probably proceeds by mechanisms similar to those described for sulfuric acid ( ) and very likely through intermediate formation of a : The ester can eliminate \(\ce{H_3PO_4}\), as sulfate esters eliminate \(\ce{H_2SO_4}\), to give alkenes: The chemistry of phosphate esters is more complicated than that of sulfate esters because it is possible to have one, two, or three alkyl groups substituted for the acidic hydrogens of phosphoric acid: Also, phosphoric acid forms an extensive series of anhydrides (with \(\ce{P-O-P}\) bonds), which further diversify the number and kind of phosphate esters. The most important phosphate esters are derivatives of mono-, di-, and triphosphoric acid (sometimes classified as ortho-, pyro-, and meta-phosphoric acids, respectively): The equilibrium between the esters of any of these phosphoric acids and water favors hydrolysis: However, phosphate esters are to hydrolyze in water (unless a catalyst is present). The difference in kinetic and thermodynamic stability of phosphate esters toward hydrolysis is used to great effect in biological systems. Of particular importance is the conversion of much of the energy that results from photosynthesis, or from the oxidation of fats, carbohydrates, and proteins in cells into formation of phosphate ester bonds \(\left( \ce{C-O-P} \right)\) or phosphate anhydride bonds \(\left( \ce{P-O-P} \right)\). The energy so stored is used in other reactions, the net result of which is hydrolysis: The substance that is the immediate source of energy for many biological reactions is adenosine triphosphate (ATP). Although this is a rather large and complex molecule, the business end for the purpose of this discussion is the group. Hydrolysis of this group can occur to give adenosine diphosphate (ADP), adenosine monophosphate (AMP), or adenosine itself: (The phosphate groups are repesented here as the major ionized form present at pH \( \cong 7\) in solutions of ATP.) All of these hydrolysis reactions are energetically favorable \(\left( \Delta G^0 < 0 \right)\), but they do not occur directly because ATP reacts slowly with water. However, hydrolysis of ATP is the indirect result of other reactions in which it participates. For example, as we showed in , equilibrium for the direct formation of an ester from a carboxylic acid and an alcohol in the liquid phase is not very favorable (Equation 15-2). However, if esterification can be coupled with ATP hydrolysis (Equation 15-3), the overall reaction (Equation 15-4) becomes much more favorable thermodynamically than is direct esterification. The ATP hydrolysis could be coupled to esterification (or other reactions) in a number of ways. The simplest would be to have the ATP convert one of the participants to a more reactive intermediate. For esterification, the reactive intermediate is an acyl derivative of AMP formed by the displacement of diphosphate from ATP: The acyl AMP is like an acyl chloride, \(\ce{RCOCl}\), in having a leaving group (AMP) that can be displaced with an alcohol: The net result of the sequence in Equations 15-5 and 15-6 is esterification in accord with Equation 15-4. It is not a catalyzed esterification because in the process one molecule of ATP is converted to AMP and diphosphate for each molecule of ester formed. The AMP has to be reconverted to ATP to participate again. These reactions are carried on by cells under the catalytic influence of enzymes. The adenosine part of the molecule is critical for the specificity of action by these enzymes. Just how these enzymes function obviously is of great interest and importance. If the role of phosphate esters, such as ATP, in carrying out reactions such as esterification in aqueous media under the influence of enzymes in cells is not clear to you, think about how you would try to carry out an esterification of ethanol in dilute water solution. Remember that, with water in great excess, the equilibrium will be quite unfavorable for the esterification reaction of Equation 15-2. You might consider adding \(\ce{CH_3COCl}\), for which the equilibrium for ester formation is much more favorable ( ). However, \(\ce{CH_3COCl}\) reacts violently with water to form \(\ce{CH_3CO_2H}\), and this reaction destroys the \(\ce{CH_3COCl}\) before it has much chance to react with ethanol to give the ester. Clearly, what you would need is a reagent that will convert \(\ce{CH_3CO_2H}\) into something that will react with ethanol in water to give the ester with a favorable equilibrium constant and yet not react very fast with water. The phosphate esters provide this function in biochemical systems by being quite unreactive to water but able to react with carboxylic acids under the influence of enzymes to give acyl phosphates. These acyl phosphates then can react with alcohols under the influence of other enzymes to form esters in the presence of water. and (1977) | 14,298 | 1,115 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/02%3A_Properties_of_Gases/2.05%3A_Condensation_of_Gases_and_the_Critical_State |
The most striking feature of real gases is that they cease to remain gases as the temperature is lowered and the pressure is increased. In the last lecture, we saw that for the pair potential \[u(r) = \begin{cases} \infty & r \leq \sigma \\ -\dfrac{C_6}{r^6} & r > \sigma \end{cases} \label{\(\Page {1}\)}\] we could write the second virial coefficient as \[B_2(T) = \dfrac{2}{3} \pi N_0 \sigma^3 \left[ 1 - \dfrac{C_6}{3 k_B T \sigma^6} \right] \label{\(\Page {2}\)}\] Let us introduce to simplifying variables \[\begin{align} b &= \dfrac{2}{3} \pi N_0 \sigma^3 \\ a &= \dfrac{2 \pi N_0^2 C_6}{9 \sigma^3} \end{align} \label{\(\Page {3}\)}\] in terms of which \[B_2(T) = b - \dfrac{a}{RT} \label{\(\Page {4}\)}\] With these definitions, the virial equation of state becomes \[\begin{align} P &= \dfrac{nRT}{V} + \dfrac{n^2}{V^2} RT \left( b - \dfrac{a}{RT} \right) \\ &= \dfrac{nRT}{V} \left(1 + \dfrac{nb}{V} \right) - \dfrac{an^2}{V^2} \end{align} \label{\(\Page {5}\)}\] If we assume \(nb/V\) is small, then we can also write \[1 + \dfrac{nb}{V} \approx \dfrac{1}{1 - \dfrac{nb}{V}} \label{\(\Page {6}\)}\] so that \[P = \dfrac{nRT}{V - nb} - \dfrac{an^2}{V^2} \label{\(\Page {7}\)}\] which is known as the equation of state. The first term in this equation is easy to motivate. In fact, it looks very much like the equation of state for an ideal gas having volume \(V - nb\) rather than \(V\). This part of the van der Waals equation is due entirely to the hard-wall potential \(u_o(r)\). Essentially, this potential energy term describes a system of “billiard balls” of diameter \(\sigma\). The figure below shows two of these billiard ball type particles at the point of contact (also called the distance of closest approach). At this point, they undergo a collision and separate, so that cannot be closer than the distance shown in the figure. At this point, the distance between their centers is also \(\sigma\), as the figure indicates. Because of this distance of closest approach, the total volume available to the particles is not \(V\) but some volume less than \(V\). This reduction in volume can be calculated as follows: Figure \(\Page {1}\) shows a shaded sphere that just contains the pair of billiard ball particles. The volume of this sphere is the volume excluded from any two particles. The radius of the sphere is \(\sigma\) as the figure shows. Hence, the excluded volume for the two particles is \(4 \pi \sigma^3/3\), which is the volume of the shaded sphere. From this, we see that the excluded volume for any particle is just half of this or \(2 \pi \sigma^3/3\). The excluded volume for a mole of such particles is just \(2 \pi \sigma^3 N_0/3\), which is the parameter \(b\): \[b = \dfrac{2}{3} \pi \sigma^3 N_0 \label{\(\Page {8}\)}\] Given \(n\) moles of gas, the total excluded volume is then \(nb\), so that the total volume is simply \(V - nb\). Isotherms of the van der Waals equation are shown in the figure below (left panel). In the figure the volume axis is the molar volume denoted \(\bar{V} = V/n\). At sufficiently high temperature, the isotherms approach those of an ideal gas. However, we also see something strange in some of the isotherms. Specifically, we see a region in which \(P\) and \(V\) increase together, and we know that this cannot actually happen in a real gas. It should be clear that many approximations and assumptions go into the derivation of the van der Waals equation so that some of the important physics is missing from the model. Hence, we should not be surprised if the van der Waals equation has some unphysical behavior buried in it. In fact, we know that at sufficiently low temperatures, any real gas, when compressed, must undergo a transition from gas to liquid. The signature of such a transition is a discontinuous change in the volume, signifying the condensation of the gas into a liquid that occupies a significantly lower volume. Unfortunately, the van der Waals equation does not correctly predict this behavior, and hence, it must be added in . This is done by drawing a horizontal line through the isotherm (Figure \(\Page {2}\)) and in the figure below. The vertical position of the line is chosen so that the area above the line (between the line and the isotherm) and below the line (again between the line and the isotherm) is exactly the same. In this way, we entirely remove the artifact of the unphysical increase of \(P\) with \(V\) when we compute the compressional work on the gas from \(\int P(V) \: dV\), to be discussed in our section on thermodynamics. This horizontal line is called the . As it happens, there is exactly one isotherm along which the van der Waals equation correctly predicts the gas-to-liquid phase transition. Along this isotherm, the volume discontinuity captured by the tie line is shrunken down to a single point (so that there is no possibility of an increase of \(P\) with \(V\)!). This isotherm, in fact, corresponds to the highest possible temperature at which such a transition can occur. As we approach this isotherm from higher temperatures, this isotherm is a kind of dividing line between the system’s remaining a gas at all value of \(P\) and \(V\) and the system’s actually undergoing a gas-to-liquid transition. Hence, this isotherm is called the (Figure \(\Page {4}\)): The temperature of this isotherm is called the critical temperature, denoted \(T_c\). The point at which the curve flattens out, signifying the phase transition, is called the . If we draw a curve through the isotherms joining all points of these isotherms at which the tie lines begin, continue the curve up to the critical isotherm, and down the other side where the tie lines end, this curve reaches a maximum at the critical point. This is illustrated below: The shape of the critical isotherm at the critical point allows us to determine the exact temperature, pressure, and volume at which the phase transition from gas to liquid will occur. At this point, the isotherm is both horizontal and flat. This means that both the first and second derivatives of \(P\) with respect to \(V\) must vanish: \[\dfrac{\partial P}{\partial V} = 0, \: \: \: \dfrac{\partial^2 P}{\partial V^2} = 0 \label{\(\Page {9}\)}\] Substituting the van der Waals equation into these two conditions, we find the following: \[\begin{align} -\dfrac{nRT}{\left( V - nb \right)^2} + \dfrac{2 an^2}{V^3} &= 0 \\ \dfrac{2nRT}{\left( V - nb \right)^3} - \dfrac{6an^2}{V^4} &= 0 \end{align} \label{\(\Page {1}\)0}\] Hence, we have two equations in two unknowns \(V\) and \(T\) for the critical temperature and critical volume. Once these are determined, the van der Waals equation, itself, allows us to determine the critical pressure. To solve the equations, first divide one by the other. This gives us a simple condition for the volume: \[\begin{align} \dfrac{V - nb}{2} &= \dfrac{V}{3} \\ 3V - 3nb &= 2V \\ V &= 3nb \equiv V_c \end{align} \label{\(\Page {1}\)1}\] This is the critical volume. Now use either of the two conditions to obtain the critical temperature \(T_c\). If we use the first one, we find \[\begin{align} \dfrac{nRT_c}{\left( V_c - nb \right)^2} &= \dfrac{2an^2}{V_c^3} \\ \dfrac{nRT_c}{\left( 3nb - nb \right)^2} &= \dfrac{2an^2}{\left( 3nb \right)^3} \\ \dfrac{nRT_c}{4n^2b^2} &= \dfrac{2an^2}{27n^3b^3} \\ RT_c &= \dfrac{8a}{27b} \end{align} \label{\(\Page {1}\)2}\] Finally, plugging the critical temperature and volume into the van der Waals equation, we obtain the critical pressure as \[\begin{align} P &= \dfrac{nRT_c}{V_c - nb} - \dfrac{an^2}{V_c^2} \\ &= \dfrac{8an/27b}{3nb - nb} - \dfrac{an^2}{\left( 3nb \right)^2} \\ &= \dfrac{a}{27b^2} \end{align} \label{\(\Page {1}\)3}\] The plot illustrates this behavior; as the volume is decreased, the lower-temperature isotherms suddenly change into straight lines. Under these conditions, the pressure remains constant as the volume is reduced. This can only mean that the gas is “disappearing" as we squeeze the system down to a smaller volume. In its place, we obtain a new state of matter, the liquid. In the green-shaded region, phases, liquid, and gas, are simultaneously present. Finally, at very small volume all the gas has disappeared and only the liquid phase remains. At this point the isotherms bend strongly upward, reflecting our common experience that a liquid is practically incompressible. To better understand this plot, look at the isotherm labeled . As the gas is compressed from to , the pressure rises in much the same way as Boyle's law predicts. Compression beyond , however, does not cause any rise in the pressure. What happens instead is that some of the gas condenses to a liquid. At , the substance is entirely in its liquid state. The very steep rise to corresponds to our ordinary experience that liquids have very low compressibilities. The range of volumes possible for the liquid diminishes as the critical temperature is approached. Liquid and gas can coexist only within the regions indicated by the green-shaded area in the diagram above. As the temperature and pressure rise, this region becomes more narrow, finally reaching zero width at the . The values of , and at this juncture are known as the , and . The isotherm that passes through the critical point is called the . Beyond this isotherm, the gas and liquids become indistinguishable; there is only a single fluid phase, sometimes referred to as a . At temperatures below 31°C (the ), CO acts somewhat like an ideal gas even at a rather high pressure ( ). Below 31°, an attempt to compress the gas to a smaller volume eventually causes condensation to begin. Thus at 21°C, at a pressure of about 62 atm ( ), the volume can be reduced from 200 cm to about 55 cm without any further rise in the pressure. Instead of the gas being compressed, it is replaced with the far more compact liquid as the gas is essentially being "squeezed" into its liquid phase. After all of the gas has disappeared ( ), the pressure rises very rapidly because now all that remains is an almost incompressible liquid. Above this isotherm ( ), CO exists only as a . What happens if you have some liquid carbon dioxide in a transparent cylinder at just under its of 62 atm, and you then compress it slightly? Nothing very dramatic until you notice that the meniscus has disappeared. By successively reducing and increasing the pressure, you can "turn the meniscus on and off". One intriguing consequence of the very limited bounds of the liquid state is that you could start with a gas at large volume and low temperature, raise the temperature, reduce the volume, and then reduce the temperature so as to arrive at the liquid region at the lower left, without ever passing through the two-phase region, and thus The state of matter, as the fluid above the critical point is often called, possesses the flow properties of a gas and the solvent properties of a liquid. The density of a supercritical fluid can be changed over a wide range by adjusting the pressure; this, in turn, changes its solubility, which can thus be optimized for a particular application. The picture at the right shows a commercial laboratory device used for carrying out chemical reactions under supercritical conditions. is widely used to dissolve the caffeine out of coffee beans and as a dry-cleaning solvent. has recently attracted interest as a medium for chemically decomposing dangerous environmental pollutants such as PCBs. Supercritical fluids are being increasingly employed as as substitutes for organic solvents (so-called "green chemistry") in a range of industrial and laboratory processes. Applications that involve supercritical fluids include extractions, nano particle and nano structured film formation, supercritical drying, carbon capture and storage, as well as enhanced oil recovery studies. ) | 11,874 | 1,117 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Ionization_Constants/Fundamentals_of_Ionization_Constants |
A simple way to understand an ionization constant is to think of it in a clear-cut way: To what degree will a substance produce ions in water? In other words, to what extent will ions be formed? Water has a very low concentration of ions that are detectable. Water undergoes self-ionization, where two water molecules interact to from a hydronium ion and a hydroxide ion. \[H_2O + H_2O \rightleftharpoons H_3O^+ + OH^- \label{1}\] Even though water does not form a lot of ions, the existence of them is evident in pure water by the electrical conductivity measurements. Water undergoes ionization because the powerfully electronegative oxygen atom takes the electron from a hydrogen atom, making the proton dissociate. \[\ce{H-O-H} \rightleftharpoons H^+ + OH^- \label{2}\] with two ions are formed: The hydrogen ions then react with water to form hydronium ions: \[H^+ + H_2O \rightarrow H_3O^+ \label{3}\] Typically, hydrogen atoms are bonded with another water molecule resulting in a hydration that forms a and hydroxide ion. \[H_2O + H_2O \rightarrow H_3O^+ + OH^- \label{4}\] In 1 L of pure water at 25 degrees Celsius, there is 10 moles of hyrodronium ions or hydroxide ions at equilibrium. Let’s come back to the question of to what degree will a substance form ions in water? For water, we have learned that it will occur until 10 moles of either ion will ionize at the previous given conditions. Since this is during equilibrium, a constant can be formed. \[H_2O \rightleftharpoons H^+ + OH^- \label{5}\] \[K_{eq}= \dfrac{[H^+] [OH^-]}{[H_2O]} \label{6}\] This equilibrium constant is commonly referred to as \(K_w\). Now that we have an idea of what an ionization constant is, let’s take a look at how acids and bases play in this scenario. Strong acids and bases are those that completely dissociate into ions once placed in solution. For example: \[KOH + H_2O → K^+ + OH^-\] So, a 2 M solution of KOH would have 2 M concentration of of OH- ions (also 2 M concentration of \(K\)). Weak acids and bases, however do not behave the same way. Their amounts cannot be calculated as easily. This is because the ions do not fully dissociate in the solution. Weak acids have a higher pH than strong acids, because weak acids do not release all of its hydrogens. The acid dissociation constant tell us the extent to which an acid dissociates \[HCN_{(aq)} + H_2O \rightleftharpoons H_3O^+ + CN^- \label{8}\] \[K_a= \dfrac{[H_3O^+] [CN^-]}{[HCN]} \label{9}\] This equation is used fairly often when looking at equilibrium reactions. During equilibrium, the rates of the forward and backward reaction are the same. However, the concentrations tend to be varied. Since concentration is what gives us an idea of how much substance has dissociated, we can relate concentration ratios to give us a constant. K is found by first finding out the molarity of each substance. Then, just as shown in the equation, we divide the products by the reactants, excluding solids and liquids. Also when there is more than one product or reactant, their concentrations must be multiplied together. Even though you will not see a multiplication sign, if there are two molecules associated, remember to multiply them. If there is a coefficient in front of a molecule, the concentration must be raised to that power in the calculations. A weaker acid tends to have a smaller \(K_a\) because the concentration on the bottom of the equation is larger. There is more of the acid, and less of the ions. You can think of Ka as a way of relating concentration in order to find out other calculations, typically the pH of a substance. A pH tells you how basic or acidic something is, and as we have learned that depends on how much ions become dissociated. Calculate the ionization constant of a weak acid. Solve for K given 0.8 M of hydrogen cyanide and 0.0039 M for hyrodronium and cyanide ions: \[HCN_{(aq)} + H_2O → H_3O^+ + CN^-\] We can assume here that [H O ] = [CN ] The equilibrium constant would be \[K_a= \dfrac{0.0039^2}{0.8} = 1.9 \times -^{5}\] We can use the similar method to find the K constant for weak bases. Again, an ionization constant is a measure of the extent a base will dissociate. K relates these molarity quantities. A smaller K corresponds to a weaker base, as a higher one a stronger base. Some common weak bases: NH , NH OH, HS , C H N \[C_5H_5N_{(aq)} + H_2O_{(l)} → C_5H_5N^+_{(aq)} + OH^-_{ (aq)} \label{10}\] We would find K the same way we did K . However, most problems are not as simple and obvious. Let’s do an example that is a little more challenging to help you understand this better. Calculate the concentration of [OH ] in a 3M Pyridine solution with K = 1.5 * 10 . Since we already have our equation, let’s write the expression for the constant as discussed earlier (concentration of products divided by concentration of reactants) \[K_b= \dfrac{[OH^-] [C_5H_5NH^+]}{[C_5H_5N]} = 1.5 \times 10^{-9}\] Because pyridine is a weak base, we can assume that not much of it will disassociate. And since we have 3M, let’s make 3M-X our calculation. Subtracting X is going to be the change in molarity due to dissociation. Since our ions are unknown and they are both one mole, we can solve for them as being X. THUS: (X /3-X) = 1.5 * 10 To make the calculation simple, we can estimate 3-X to be approximately 3 because of the weak base. Now our equation becomes: (X /3) = 1.5 * 10 . When we solve for X our answer is \(4.7 \times 10^{-5}\; M\) This approximation was effective because x is small, which must be less than 5% of the initial concentration for that estimation to be justified. We have learned that an ionization constant quantifies the degree a substance will ionize in a solution (typically water). K , K , and K are constants for acids, bases, and finally water, respectively and are related by \[K_a \times K_b = K_w\] However, these constants are also used to find concentrations as well as pH. | 5,950 | 1,118 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/09%3A_Molecular_Geometry_and_Covalent_Bonding_Models/9.02%3A_VSEPR_-_Molecular_Geometry |
We continue our discussion of structure and bonding by introducing the valence-shell electron-pair repulsion (VSEPR) model (pronounced “vesper”), which is simple to use to predict the shapes of many molecules and polyatomic ions. Keep in mind, however, that the VSEPR model, like any model, is a limited representation of reality; the model provides no information about bond lengths or the presence of multiple bonds. Following sections of this will connect the VSEPR model to molecular orbitals,. The VSEPR model can predict the structure of nearly any molecule or polyatomic ion in which the central atom is a nonmetal, as well as the structures of many molecules and polyatomic ions with a central metal atom. The VSEPR model is a theory; it does not attempt to explain observations. Instead, it is a counting procedure that accurately predicts the three-dimensional structures of a large number of compounds, which cannot be predicted using the Lewis electron-pair approach. Lewis electron structures predict the number and types of bonds, whereas VSEPR can predict the shapes of many molecules and polyatomic ions. We can use the VSEPR model to predict the geometry of most polyatomic molecules and ions by focusing on only the number of electron pairs around the , ignoring all other valence electrons present. According to this model, valence electrons in the Lewis structure form , which may consist of a single bond, a double bond, a triple bond, a lone pair of electrons, or even a single unpaired electron, which in the VSEPR model is counted as a lone pair. Because electrons repel each other electrostatically, the most stable arrangement of electron groups (i.e., the one with the lowest energy) is the one that minimizes repulsions. Groups are positioned around the central atom in a way that produces the molecular structure with the lowest energy, as illustrated in and In the VSEPR model, the molecule or polyatomic ion is given an AX E designation, where A is the central atom, X is a bonded atom, E is a nonbonding valence electron group (usually a lone pair of electrons), and and are integers. Each group around the central atom is designated as a bonding pair (BP) or lone (nonbonding) pair (LP). From the BP and LP interactions we can predict both the relative positions of the atoms and the angles between the bonds, called the bond angles . Using this information, we can describe the molecular geometry , the arrangement of the in a molecule or polyatomic ion. This procedure is summarized as follows: We will illustrate the use of this procedure with several examples, beginning with atoms with two electron groups. In our discussion we will refer to and , which summarize the common molecular geometries and idealized bond angles of molecules and ions with two to six electron groups. Our first example is a molecule with two bonded atoms and no lone pairs of electrons, BeH . 1. The central atom, beryllium, contributes two valence electrons, and each hydrogen atom contributes one. The Lewis electron structure is In this case Be and H each see two electrons which, for those atoms fills the orbitals. This, of course, is an exception to the octet rule, but easy to understand using the aufbau principle 2. There are two electron groups around the central atom. We see from that the arrangement that minimizes repulsions places the groups 180° apart. 3. Both groups around the central atom are bonding pairs (BP). Thus BeH is designated as AX . 4. From we see that with two bonding pairs, the molecular geometry that minimizes repulsions in BeH is . 5. In Section 6.2 we showed that the atomic orbitals of the Be atom were hybridized to for bonds, which gave BeH its linear shape 1. The central atom, carbon, contributes four valence electrons, and each oxygen atom contributes six. The Lewis electron structure is 2. The carbon atom forms two double bonds. Each double bond is a group, so there are two electron groups around the central atom. Like BeH , the arrangement that minimizes repulsions places the groups 180° apart. 3. Once again, both groups around the central atom are bonding pairs (BP), so CO is designated as AX . 4. VSEPR only recognizes groups around the atom. Thus the lone pairs on the oxygen atoms do not influence the molecular geometry. With two bonding pairs on the central atom and no lone pairs, the molecular geometry of CO is linear ( ). The structure of CO is shown in .1 5. If someone asked what the hybridization on the C atom was, we would first draw the Lewis structure. We could then apply VSEPR to the Lewis structure to deduce that the molecular shape of CO was linar and from this we conclude that the hybridization of the C atom is 1. The central atom, boron, contributes three valence electrons, and each chlorine atom contributes seven valence electrons. The Lewis electron structure is Again, we recognize this as an exception to the octet rule, but again, easy to understand based on what we have studied. The three electrons on the boron atom are shared, one each with each chlorine atom. The chlorine atoms each have a complete octet. 2. There are three electron groups around the central atom. To minimize repulsions, the groups are placed 120° apart ( ). 3. All electron groups are bonding pairs (BP), so the structure is designated as AX . 4. From we see that with three bonding pairs around the central atom, the molecular geometry of BCl is , as shown in 5. As discussed in Chapter 6.2, the hybridization on the B atom is and the shape of the orbitals is triangonal planar. 1. The central atom, carbon, has four valence electrons, and each oxygen atom has six valence electrons. As you learned in , the Lewis electron structure of one of three resonance forms is represented as We have already discussed how the electrons in the double bond are smeared over the top and bottom of the ion in a molecular bond and that the bonds linking the carbon and oxygen atoms are identical 2. The structure of CO is a resonance hybrid. It has three identical bonds, each with a bond order of 1 1/3. We minimize repulsions by placing the three groups 120° apart ( ). 3. All electron groups are bonding pairs (BP). With three bonding groups around the central atom, the structure is designated as AX . 4. We see from that the molecular geometry of CO is trigonal planar. 5. Again, the hybridization of the C orbitals is sp but there is also a delocalized & In our next example we encounter the effects of lone pairs and multiple bonds on molecular geometry for the first time. 1. The central atom, sulfur, has 6 valence electrons, as does each oxygen atom. With 18 valence electrons, the Lewis electron structure is shown below. 2. There are three electron groups around the central atom, two double bonds and one lone pair. We initially place the groups in a trigonal planar arrangement to minimize repulsions ( ). 3. There are two bonding pairs and one lone pair, so the structure is designated as AX E. This designation has a total of three electron pairs, two X and one E. Because a lone pair is not shared by two nuclei, it occupies more space near the central atom than a bonding pair ( ). Thus bonding pairs and lone pairs repel each other electrostatically in the order BP–BP < LP–BP < LP–LP. In SO , we have one BP–BP interaction and two LP–BP interactions. 4. The molecular geometry is described only by the positions of the nuclei, by the positions of the lone pairs. Thus with two nuclei and one lone pair the shape is , or , which can be viewed as a trigonal planar arrangement with a missing vertex ( and ). 5, Again the hybridization on the central atom (S) is . Like lone pairs of electrons, multiple bonds occupy more space around the central atom than a single bond, which can cause other bond angles to be somewhat smaller than expected. This is because a multiple bond has a higher electron density than a single bond, so its electrons occupy more space than those of a single bond. For example, in a molecule such as CH O (AX ), whose structure is shown below, the double bond repels the single bonds more strongly than the single bonds repel each other. This causes a deviation from ideal geometry (an H–C–H bond angle of 116.5° rather than 120°). One of the limitations of Lewis structures is that they depict molecules and ions in only two dimensions. With four electron groups, we must learn to show molecules and ions in three dimensions. 1. The central atom, carbon, contributes four valence electrons, and each hydrogen atom has one valence electron, so the full Lewis electron structure is 2. There are four electron groups around the central atom. As shown in repulsions are minimized by placing the groups in the corners of a tetrahedron with bond angles of 109.5°. 3. All electron groups are bonding pairs, so the structure is designated as AX . 4. With four bonding pairs, the molecular geometry of methane is ( ). 5. The hybridization of the C atom orbitals is . 1. In ammonia, the central atom, nitrogen, has five valence electrons and each hydrogen donates one valence electron, producing the Lewis electron structure 2. There are four electron groups around nitrogen, three bonding pairs and one lone pair. Repulsions are minimized by directing each hydrogen atom and the lone pair to the corners of a tetrahedron. 3. With three bonding pairs and one lone pair, the structure is designated as AX E. This designation has a total of four electron pairs, three X and one E. We expect the LP–BP interactions to cause the bonding pair angles to deviate significantly from the angles of a perfect tetrahedron. 4. There are three nuclei and one lone pair, so the molecular geometry is . In essence, this is a tetrahedron with a vertex missing ( ). However, the H–N–H bond angles are less than the ideal angle of 109.5° because of LP–BP repulsions ( and ). 5. The hybridization of the N atom orbitals is . 1. Oxygen has six valence electrons and each hydrogen has one valence electron, producing the Lewis electron structure 2. There are four groups around the central oxygen atom, two bonding pairs and two lone pairs. Repulsions are minimized by directing the bonding pairs and the lone pairs to the corners of a tetrahedron 3. With two bonding pairs and two lone pairs, the structure is designated as AX E with a total of four electron pairs. Due to LP–LP, LP–BP, and BP–BP interactions, we expect a significant deviation from idealized tetrahedral angles. 4. With two hydrogen atoms and two lone pairs of electrons, the structure has significant lone pair interactions. There are two nuclei about the central atom, so the molecular shape is , or , with an H–O–H angle that is even less than the H–N–H angles in NH , as we would expect because of the presence of two lone pairs of electrons on the central atom rather than one.. This molecular shape is essentially a tetrahedron with two missing vertices. 5. The hybridization of the O atom orbitals is . In previous examples it did not matter where we placed the electron groups because all positions were equivalent. In some cases, however, the positions are not equivalent. We encounter this situation for the first time with five electron groups. 1. Phosphorus has five valence electrons and each chlorine has seven valence electrons, so the Lewis electron structure of PCl involves an expanded octet. In the following section we will describe how this expanded octet is formed by the combination of a d orbital with an s and three p orbitals. 2. There are five bonding groups around phosphorus, the central atom. The structure that minimizes repulsions is a , which consists of two trigonal pyramids that share a base ( ): 3. All electron groups are bonding pairs, so the structure is designated as AX . There are no lone pair interactions. 4. The molecular geometry of PCl is , as shown in . The molecule has three atoms in a plane in positions and two atoms above and below the plane in positions. The three equatorial positions are separated by 120° from one another, and the two axial positions are at 90° to the equatorial plane. The axial and equatorial positions are not chemically equivalent, as we will see in our next example. 1. The sulfur atom has six valence electrons and each fluorine has seven valence electrons, so the Lewis electron structure is With an expanded valence, this species is an exception to the octet rule and the hybridization of orbitals on the S atom is .. 2. There are five groups around sulfur, four bonding pairs and one lone pair. With five electron groups, the lowest energy arrangement is a trigonal bipyramid, as shown in 3. We designate SF as AX E; it has a total of five electron pairs. However, because the axial and equatorial positions are not chemically equivalent, where do we place the lone pair? If we place the lone pair in the equatorial position, we have three LP–BP repulsions at 90°. If we place it in the axial position, we have two 90° LP–BP repulsions at 90°. With fewer 90° LP–BP repulsions, we can predict that the structure with the lone pair of electrons in the equatorial position is more stable than the one with the lone pair in the axial position. We also expect a deviation from ideal geometry because a lone pair of electrons occupies more space than a bonding pair. 4. With four nuclei and one lone pair of electrons, the molecular structure is based on a trigonal bipyramid with a missing equatorial vertex; it is described as a . The F –S–F angle is 173° rather than 180° because of the lone pair of electrons in the equatorial plane. 1. The bromine atom has seven valence electrons, and each fluorine has seven valence electrons, so the Lewis electron structure is Once again, we have a compound that is an exception to the octet rule. 2. There are five groups around the central atom, three bonding pairs and two lone pairs and the hybridization of orbitals on the Br atom will be . We again direct the groups toward the vertices of a trigonal bipyramid. 3. With three bonding pairs and two lone pairs, the structural designation is AX E with a total of five electron pairs. Because the axial and equatorial positions are not equivalent, we must decide how to arrange the groups to minimize repulsions. If we place both lone pairs in the axial positions, we have six LP–BP repulsions at 90°. If both are in the equatorial positions, we have four LP–BP repulsions at 90°. If one lone pair is axial and the other equatorial, we have one LP–LP repulsion at 90° and three LP–BP repulsions at 90°: Structure (c) can be eliminated because it has a LP–LP interaction at 90°. Structure (b), with fewer LP–BP repulsions at 90° than (a), is lower in energy. However, we predict a deviation in bond angles because of the presence of the two lone pairs of electrons. 4. The three nuclei in BrF determine its molecular structure, which is described as . This is essentially a trigonal bipyramid that is missing two equatorial vertices. The F –Br–F angle is 172°, less than 180° because of LP–BP repulsions ( ). Because lone pairs occupy more space around the central atom than bonding pairs, electrostatic repulsions are more important for lone pairs than for bonding pairs. 1. Each iodine atom contributes seven electrons and the negative charge one, so the Lewis electron structure is 2. There are five electron groups about the central atom in I , two bonding pairs and three lone pairs. The hybridiztion of the orbitals on the central I atom is To minimize repulsions, the lone pairs are directed to the corners of a trigonal bipyramid. 3. With two bonding pairs and three lone pairs, I has a total of five electron pairs and is designated as AX E . We must now decide how to arrange the lone pairs of electrons in a trigonal bipyramid in a way that minimizes repulsions. Placing them in the axial positions eliminates 90° LP–LP repulsions and minimizes the number of 90° LP–BP repulsions. The three lone pairs of electrons have equivalent interactions with the three iodine atoms, so we do not expect any deviations in bonding angles. 4. With three nuclei and three lone pairs of electrons, the molecular geometry of I is linear. This can be described as a trigonal bipyramid with three equatorial vertices missing. The ion has an I–I–I angle of 180°, as expected. Six electron groups form an , a polyhedron made of identical equilateral triangles and six identical vertices ( ). 1. The central atom, sulfur, contributes six valence electrons, and each fluorine atom has seven valence electrons, so the Lewis electron structure is With an expanded valence, we know from that this species is an exception to the octet rule and we know that the hybridization of atomic orbitals on the S atom is 2. There are six electron groups around the central atom, each a bonding pair. We see from that the geometry that minimizes repulsions is . 3. With only bonding pairs, SF is designated as AX . All positions are chemically equivalent, so all electronic interactions are equivalent. 4. There are six nuclei, so the molecular geometry of SF is octahedral. 1. The central atom, bromine, has seven valence electrons, as does each fluorine, so the Lewis electron structure is With its expanded valence, this species is an exception to the octet rule. 2. There are six electron groups around the Br, five bonding pairs and one lone pair. Placing five F atoms around Br while minimizing BP–BP and LP–BP repulsions gives the following structure: 3. With five bonding pairs and one lone pair, BrF is designated as AX E; it has a total of six electron pairs. The BrF structure has four fluorine atoms in a plane in an equatorial position and one fluorine atom and the lone pair of electrons in the axial positions. We expect all F –Br–F angles to be less than 90° because of the lone pair of electrons, which occupies more space than the bonding electron pairs. 4. With five nuclei surrounding the central atom, the molecular structure is based on an octahedron with a vertex missing. This molecular structure is . The F –B–F angles are 85.1°, less than 90° because of LP–BP repulsions. 1. The central atom, iodine, contributes seven electrons. Each chlorine contributes seven, and there is a single negative charge. The Lewis electron structure is 2. There are six electron groups around the central atom, four bonding pairs and two lone pairs. The structure that minimizes LP–LP, LP–BP, and BP–BP repulsions is 3. ICl is designated as AX E and has a total of six electron pairs. Although there are lone pairs of electrons, with four bonding electron pairs in the equatorial plane and the lone pairs of electrons in the axial positions, all LP–BP repulsions are the same. Therefore, we do not expect any deviation in the Cl–I–Cl bond angles. 4. With five nuclei, the ICl ion forms a molecular structure that is , an octahedron with two opposite vertices missing. If you would like to try building some of these atoms, The relationship between the number of electron groups around a central atom, the number of lone pairs of electrons, and the molecular geometry is summarized in . ( to make full screen) This PheT applet will allow you to create all of the molecular shapes discussed above with and without lone pairs. You can investigate the structure of molecules, or of model shapes using VSEPR. By placing the cursor on any of the non central atoms or lone pairs you can rotate the molecules. You can add atoms and lone pairs to make any of VSEPR shapes Using the VSEPR model, predict the molecular geometry of each molecule or ion. two chemical species molecular geometry Draw the Lewis electron structure of the molecule or polyatomic ion. Determine the electron group arrangement around the central atom that minimizes repulsions. Assign an AX E designation; then identify the LP–LP, LP–BP, or BP–BP interactions and predict deviations in bond angles. Describe the molecular geometry. There are five bonding groups about phosphorus. The structure that minimizes repulsions is a trigonal bipyramid ( ). All electron groups are bonding pairs, so PF is designated as AX . Notice that this gives a total of five electron pairs. With no lone pair repulsions, we do not expect any bond angles to deviate from the ideal. The PF molecule has five nuclei and no lone pairs of electrons, so its molecular geometry is trigonal bipyramidal. There are four electron groups around oxygen, three bonding pairs and one lone pair. Like NH , repulsions are minimized by directing each hydrogen atom and the lone pair to the corners of a tetrahedron. With three bonding pairs and one lone pair, the structure is designated as AX E and has a total of four electron pairs (three X and one E). We expect the LP–BP interactions to cause the bonding pair angles to deviate significantly from the angles of a perfect tetrahedron. There are three nuclei and one lone pair, so the molecular geometry is , in essence a tetrahedron missing a vertex. However, the H–O–H bond angles are less than the ideal angle of 109.5° because of LP–BP repulsions: Exercise Using the VSEPR model, predict the molecular geometry of each molecule or ion. Predict the molecular geometry of each molecule. two chemical compounds molecular geometry Use the strategy given in Example 1. There are five electron groups around the central atom, two bonding pairs and three lone pairs. Repulsions are minimized by placing the groups in the corners of a trigonal bipyramid. From B, XeF is designated as AX E and has a total of five electron pairs (two X and three E). With three lone pairs about the central atom, we can arrange the two F atoms in three possible ways: both F atoms can be axial, one can be axial and one equatorial, or both can be equatorial: The structure with the lowest energy is the one that minimizes LP–LP repulsions. Both (b) and (c) have two 90° LP–LP interactions, whereas structure (a) has none. Thus both F atoms are in the axial positions, like the two iodine atoms around the central iodine in I . All LP–BP interactions are equivalent, so we do not expect a deviation from an ideal 180° in the F–Xe–F bond angle. With two nuclei about the central atom, the molecular geometry of XeF is linear. It is a trigonal bipyramid with three missing equatorial vertices. There are three electron groups around the central atom, two bonding groups and one lone pair of electrons. To minimize repulsions the three groups are initially placed at 120° angles from each other. From B we designate SnCl as AX E. It has a total of three electron pairs, two X and one E. Because the lone pair of electrons occupies more space than the bonding pairs, we expect a decrease in the Cl–Sn–Cl bond angle due to increased LP–BP repulsions. With two nuclei around the central atom and one lone pair of electrons, the molecular geometry of SnCl is bent, like SO , but with a Cl–Sn–Cl bond angle of 95°. The molecular geometry can be described as a trigonal planar arrangement with one vertex missing. Exercise Predict the molecular geometry of each molecule. The VSEPR model can be used to predict the structure of somewhat more complex molecules with no single central atom by treating them as linked AX E fragments. We will demonstrate with methyl isocyanate (CH –N=C=O), a volatile and highly toxic molecule that is used to produce the pesticide Sevin. In 1984, large quantities of Sevin were accidentally released in Bhopal, India, when water leaked into storage tanks. The resulting highly exothermic reaction caused a rapid increase in pressure that ruptured the tanks, releasing large amounts of methyl isocyanate that killed approximately 3800 people and wholly or partially disabled about 50,000 others. In addition, there was significant damage to livestock and crops. We can treat methyl isocyanate as linked AX E fragments beginning with the carbon atom at the left, which is connected to three H atoms and one N atom by single bonds. The four bonds around carbon mean that it must be surrounded by four bonding electron pairs in a configuration similar to AX . We can therefore predict the CH –N portion of the molecule to be roughly tetrahedral, similar to methane: The nitrogen atom is connected to one carbon by a single bond and to the other carbon by a double bond, producing a total of three bonds, C–N=C. For nitrogen to have an octet of electrons, it must also have a lone pair: Because multiple bonds are not shown in the VSEPR model, the nitrogen is effectively surrounded by three electron pairs. Thus according to the VSEPR model, the C–N=C fragment should be bent with an angle less than 120°. The carbon in the –N=C=O fragment is doubly bonded to both nitrogen and oxygen, which in the VSEPR model gives carbon a total of two electron pairs. The N=C=O angle should therefore be 180°, or linear. The three fragments combine to give the following structure: We predict that all four nonhydrogen atoms lie in a single plane, with a C–N–C angle of approximately 120°. The experimentally determined structure of methyl isocyanate confirms our prediction ( ). Certain patterns are seen in the structures of moderately complex molecules. For example, carbon atoms with four bonds (such as the carbon on the left in methyl isocyanate) are generally tetrahedral. Similarly, the carbon atom on the right has two double bonds that are similar to those in CO , so its geometry, like that of CO , is linear. Recognizing similarities to simpler molecules will help you predict the molecular geometries of more complex molecules. Use the VSEPR model to predict the molecular geometry of propyne (H C–C≡CH), a gas with some anesthetic properties. chemical compound molecular geometry Count the number of electron groups around each carbon, recognizing that in the VSEPR model, a multiple bond counts as a single group. Use to determine the molecular geometry around each carbon atom and then deduce the structure of the molecule as a whole. Because the carbon atom on the left is bonded to four other atoms, we know that it is approximately tetrahedral. The next two carbon atoms share a triple bond, and each has an additional single bond. Because a multiple bond is counted as a single bond in the VSEPR model, each carbon atom behaves as if it had two electron groups. This means that both of these carbons are linear, with C–C≡C and C≡C–H angles of 180°. Exercise Predict the geometry of allene (H C=C=CH ), a compound with narcotic properties that is used to make more complex organic molecules. The terminal carbon atoms are trigonal planar, the central carbon is linear, and the C–C–C angle is 180°. The structural formula for water can be drawn as follows, but including the two lone pairs on the oxygen provides more information: Because as we learned from VSEPR, the latter approximates the experimentally determined shape of the water molecule, it is more informative. Similarly, ammonia (NH ) and methane (CH ) are often written as planar molecules but at least for ammonia, it would be more informative to include the lone pair on the nitrogen atom: As shown in , however, we know that the the actual three-dimensional structure of NH looks like a pyramid with a triangular base of three hydrogen atoms. Again using VSEPR, or what we know about sp3 orbitals on the central atom in all three molecules the structure of CH , with four hydrogen atoms arranged around a central carbon atom as shown in , is . That is, the hydrogen atoms are positioned at every other vertex of a cube. Many compounds—carbon compounds, in particular—have four bonded atoms arranged around a central atom to form a tetrahedron. , illustrates different ways to represent the structures of molecules. It should be clear that there is no single “best” way to draw the structure of a molecule; the method you use depends on which aspect of the structure you want to emphasize and how much time and effort you want to spend. shows some of the different ways to portray the structure of a slightly more complex molecule: methanol. These representations differ greatly in their information content. For example, the molecular formula for methanol (part (a) in ) gives only the number of each kind of atom; writing methanol as CH O tells nothing about its structure. In contrast, the structural formula (part (b) in ) indicates how the atoms are connected, but it makes methanol look as if it is planar (which it is not). Both the ball-and-stick model (part (c) in ) and the perspective drawing (part (d) in ) show the three-dimensional structure of the molecule. The latter (also called a representation) is the easiest way to sketch the structure of a molecule in three dimensions. It shows which atoms are above and below the plane of the paper by using wedges and dashes, respectively; the central atom is always assumed to be in the plane of the paper. The space-filling model (part (e) in ) illustrates the approximate relative sizes of the atoms in the molecule, but it does not show the bonds between the atoms. Also, in a space-filling model, atoms at the “front” of the molecule may obscure atoms at the “back.” Although a structural formula, a ball-and-stick model, a perspective drawing, and a space-filling model provide a significant amount of information about the structure of a molecule, each requires time and effort. Consequently, chemists often use a (part (f) in ), which omits the lines representing bonds between atoms and simply lists the atoms bonded to a given atom next to it. Multiple groups attached to the same atom are shown in parentheses, followed by a subscript that indicates the number of such groups. For example, the condensed structural formula for methanol is CH OH, which tells us that the molecule contains a CH unit that looks like a fragment of methane (CH ). Methanol can therefore be viewed either as a methane molecule in which one hydrogen atom has been replaced by an –OH group or as a water molecule in which one hydrogen atom has been replaced by a –CH fragment. Because of their ease of use and information content, we use condensed structural formulas for molecules throughout this text. Ball-and-stick models are used when needed to illustrate the three-dimensional structure of molecules, and space-filling models are used only when it is necessary to visualize the relative sizes of atoms or molecules to understand an important point. In , you learned how to calculate the of simple diatomic molecules. In more complex molecules with polar covalent bonds, the three-dimensional geometry and the compound’s symmetry determine whether there is a net dipole moment. Mathematically, dipole moments are ; they possess both a and a . The dipole moment of a molecule is therefore the of the dipole moments of the individual bonds in the molecule. If the individual bond dipole moments cancel one another, there is no net dipole moment. Such is the case for CO , a linear molecule (part (a) in ). Each C–O bond in CO is polar, yet experiments show that the CO molecule has no dipole moment. Because the two C–O bond dipoles in CO are equal in magnitude and oriented at 180° to each other, they cancel. As a result, the CO molecule has no dipole moment even though it has a substantial separation of charge. In contrast, the H O molecule is not linear (part (b) in ); it is bent in three-dimensional space, so the dipole moments do not cancel each other. Thus a molecule such as H O has a net dipole moment. We expect the concentration of negative charge to be on the oxygen, the more electronegative atom, and positive charge on the two hydrogens. This charge polarization allows H O to hydrogen-bond to other polarized or charged species, including other water molecules. Other examples of molecules with polar bonds are shown in . In molecular geometries that are highly symmetrical (most notably tetrahedral and square planar, trigonal bipyramidal, and octahedral), individual bond dipole moments completely cancel, and there is no net dipole moment. Although a molecule like CHCl is best described as tetrahedral, the atoms bonded to carbon are not identical. Consequently, the bond dipole moments cannot cancel one another, and the molecule has a dipole moment. Due to the arrangement of the bonds in molecules that have V-shaped, trigonal pyramidal, seesaw, T-shaped, and square pyramidal geometries, the bond dipole moments cannot cancel one another. Consequently, molecules with these geometries always have a nonzero dipole moment. Individual bond dipole moments are indicated in red. Due to their different three-dimensional structures, some molecules with polar bonds have a net dipole moment (HCl, CH O, NH , and CHCl ), indicated in blue, whereas others do not because the bond dipole moments cancel (BCl , CCl , PF , and SF ). Molecules with asymmetrical charge distributions have a net dipole moment. Which molecule(s) has a net dipole moment? three chemical compounds net dipole moment For each three-dimensional molecular geometry, predict whether the bond dipoles cancel. If they do not, then the molecule has a net dipole moment. Exercise Which molecule(s) has a net dipole moment? CH Cl; XeO Lewis electron structures give no information about , the arrangement of bonded atoms in a molecule or polyatomic ion, which is crucial to understanding the chemistry of a molecule. The allows us to predict which of the possible structures is actually observed in most cases. It is based on the assumption that pairs of electrons occupy space, and the lowest-energy structure is the one that minimizes electron pair–electron pair repulsions. In the VSEPR model, the molecule or polyatomic ion is given an AX E designation, where A is the central atom, X is a bonded atom, E is a nonbonding valence electron group (usually a lone pair of electrons), and and are integers. Each group around the central atom is designated as a bonding pair (BP) or lone (nonbonding) pair (LP). From the BP and LP interactions we can predict both the relative positions of the atoms and the angles between the bonds, called the . From this we can describe the . A combination of VSEPR and a bonding model, such as Lewis electron structures, however, is necessary to understand the presence of multiple bonds. Molecules with polar covalent bonds can have a , an asymmetrical distribution of charge that results in a tendency for molecules to align themselves in an applied electric field. Any diatomic molecule with a polar covalent bond has a dipole moment, but in polyatomic molecules, the presence or absence of a net dipole moment depends on the structure. For some highly symmetrical structures, the individual bond dipole moments cancel one another, giving a dipole moment of zero. What is the main difference between the VSEPR model and Lewis electron structures? What are the differences between molecular geometry and Lewis electron structures? Can two molecules with the same Lewis electron structures have different molecular geometries? Can two molecules with the same molecular geometry have different Lewis electron structures? In each case, support your answer with an example. How does the VSEPR model deal with the presence of multiple bonds? Three molecules have the following generic formulas: AX , AX E, and AX E . Predict the molecular geometry of each, and arrange them in order of increasing X–A–X angle. Which has the smaller angles around the central atom—H S or SiH ? Why? Do the Lewis electron structures of these molecules predict which has the smaller angle? Discuss in your own words why lone pairs of electrons occupy more space than bonding pairs. How does the presence of lone pairs affect molecular geometry? When using VSEPR to predict molecular geometry, the importance of repulsions between electron pairs decreases in the following order: LP–LP, LP–BP, BP–BP. Explain this order. Draw structures of real molecules that separately show each of these interactions. How do multiple bonds affect molecular geometry? Does a multiple bond take up more or less space around an atom than a single bond? a lone pair? Straight-chain alkanes do not have linear structures but are “kinked.” Using -hexane as an example, explain why this is so. Compare the geometry of 1-hexene to that of -hexane. How is molecular geometry related to the presence or absence of a molecular dipole moment? How are molecular geometry and dipole moments related to physical properties such as melting point and boiling point? What two features of a molecule’s structure and bonding are required for a molecule to be considered polar? Is COF likely to have a significant dipole moment? Explain your answer. When a chemist says that a molecule is , what does this mean? What are the general physical properties of polar molecules? Use the VSPER model and your knowledge of bonding and dipole moments to predict which molecules will be liquids or solids at room temperature and which will be gases. Explain your rationale for each choice. Justify your answers. The idealized molecular geometry of BrF is square pyramidal, with one lone pair. What effect does the lone pair have on the actual molecular geometry of BrF ? If LP–BP repulsions were than BP–BP repulsions, what would be the effect on the molecular geometry of BrF ? Which has the smallest bond angle around the central atom—H S, H Se, or H Te? the largest? Justify your answers. Which of these molecular geometries results in a molecule with a net dipole moment: linear, bent, trigonal planar, tetrahedral, seesaw, trigonal pyramidal, square pyramidal, and octahedral? For the geometries that do not always produce a net dipole moment, what factor(s) will result in a net dipole moment? To a first approximation, the VSEPR model assumes that multiple bonds and single bonds have the same effect on electron pair geometry and molecular geometry; in other words, VSEPR treats multiple bonds like single bonds. Only when considering fine points of molecular structure does VSEPR recognize that multiple bonds occupy more space around the central atom than single bonds. Physical properties like boiling point and melting point depend upon the existence and magnitude of the dipole moment of a molecule. In general, molecules that have substantial dipole moments are likely to exhibit greater intermolecular interactions, resulting in higher melting points and boiling points. The term “polar” is generally used to mean that a molecule has an asymmetrical structure and contains polar bonds. The resulting dipole moment causes the substance to have a higher boiling or melting point than a nonpolar substance. Give the number of electron groups around the central atom and the molecular geometry for each molecule. Classify the electron groups in each species as bonding pairs or lone pairs. Give the number of electron groups around the central atom and the molecular geometry for each species. Classify the electron groups in each species as bonding pairs or lone pairs. Give the number of electron groups around the central atom and the molecular geometry for each molecule. For structures that are not linear, draw three-dimensional representations, clearly showing the positions of the lone pairs of electrons. Give the number of electron groups around the central atom and the molecular geometry for each molecule. For structures that are not linear, draw three-dimensional representations, clearly showing the positions of the lone pairs of electrons. What is the molecular geometry of ClF ? Draw a three-dimensional representation of its structure and explain the effect of any lone pairs on the idealized geometry. Predict the molecular geometry of each of the following. Predict whether each molecule has a net dipole moment. Justify your answers and indicate the direction of any bond dipoles. Predict whether each molecule has a net dipole moment. Justify your answers and indicate the direction of any bond dipoles. Of the molecules Cl C=Cl , IF , and SF , which has a net dipole moment? Explain your reasoning. Of the molecules SO , XeF , and H C=Cl , which has a net dipole moment? Explain your reasoning. The idealized geometry is T shaped, but the two lone pairs of electrons on Cl will distort the structure, making the F–Cl–F angle than 180°. Cl C=CCl : Although the C–Cl bonds are rather polar, the individual bond dipoles cancel one another in this symmetrical structure, and Cl C=CCl does not have a net dipole moment. IF : In this structure, the individual I–F bond dipoles cannot cancel one another, giving IF a net dipole moment. SF : The S–F bonds are quite polar, but the individual bond dipoles cancel one another in an octahedral structure. Thus, SF has no net dipole moment. | 40,880 | 1,119 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Energies_and_Potentials/Free_Energy/What_are_Free_Energies |
Free energy is a composite function that balances the influence of energy vs. entropy. To first define "free" energy, we shall examine the backgrounds of this term, what definitions carry it, and which specific definitions we, as chemists, will choose to refer to: The Gibbs Energy is named after a Josiah William Gibbs, an American physicist in the late 19 century who greatly advanced thermodynamics; his work now serves as a foundation for this branch of science. This energy can be said to be the greatest amount of work (other than expansion work) a system can do on its surroundings, when it operates at a constant pressure and temperature.
First, a modeling of the Gibbs Energy by way of equation: \[G = U + PV - TS\] Where: Of course, we know that \(U + PV\) can also be defined as: \[U + PV = H\] Where: Which leads us to a form of how the Gibbs Energy is related to enthalpy: \[G = H - TS\] All of the members on the right side of this equation are , so G is a state function as well. The change in G is simply: \[\Delta{G} = \Delta{H} - T\Delta{S}\] We will start with an equation for the total entropy change of the universe. Our goal is to whittle it down to a practical form, like a caveman shaping a unwieldy block of stone into a useful hand held tool! \[\Delta{S_{universe}} = \Delta{S_{system}} + \Delta{S_{surroundings}}\] An equation with variables of such scope is difficult to work with. We want to do away with the vagueness, and rewrite a more focused equation. We'll consider the case where temperature and pressure is constant. Here we go: \[(Delta{H}\) \(-q_p = -\Delta{H}_{sys}\), \(\Delta{S}_{surroundings} = \Delta{H_{sys}}/T\). As a quick note, let it be said that the name "free energy", other than being confused with another energy exactly termed, is also somewhat of a misnomer. The multiple meanings of the word "free" can make it seem as if energy can be transferred at no cost; in fact, the word "free" was used to refer to what cost the system was free to pay, in the form of turning energy into work. is useful because it can tell us how a system, when we're given only information on it, will act. \[\Delta{G} < 0\] indicates a spontaneous change to occur. \[\Delta{G} > 0\] indicates an absence of spontaneousness. \[\Delta{G} = 0\] indicates a system at equilibrium. The Gibbs Energy reaches the minimum value when equilibrium is reached. Here, it is represented as a graph, where x represents the extent of how far the reaction has occurred. The minimum of the function has to be smooth, because \(G\) must be differentiable (its first derivative has to exist at the minimum). It was briefly mentioned that is the energy available to be converted to work. The definition is self evident from the equation. Look at \[\Delta{G} = \Delta{H} - T\Delta{S}\]. Recall that is the total energy that can be made into heat. . By a reordering of the Gibbs Energy equation: \[\Delta{H} = \Delta{G} - T\Delta{S}\] Expressed in words: Why constant temperature and pressure? It just so happens that these are regularly occurring factors in the laboratory, making this equation practical to use, and useful as well, for chemists. An example of Gibbs Energy in the real world is the oxidation of glucose; in this case is equal to 2870 kJ, or 686 Calories. For living cells, this is the primary energy reaction. The is deemed as a thermodynamic potential which calculates the “useful” work retrievable from a closed thermodynamic system at a constant temperature and volume. For such a system, the negative of the difference in the Helmholtz energy is equal to the maximum amount of work extractable from a thermodynamic process in which both temperature and volume are kept constant. In these conditions, it is minimized and held constant at equilibrium. The Helmholtz free energy was originally developed by Hermann von Helmotz and is generally denoted by the letter , or the letter . In physics, the letter is mostly used to denote the Helmholtz energy, which is often called the or simple term “free energy." Introduced by German physicist Hermann Helmholtz in 1882, thermodynamic potential found in a system of constant species with constant temperautre and constant volume, given by the formula: \[ΔA = ΔE – TΔS\] Summarily, the Helmholtz free energy is also the measure of an isothermal-isochoric closed system’s ability to do work. If any external field is missing, the Helmholtz free energy formula becomes: \[ΔA = ΔU –TΔS\] The internal energy (U) can be said to be the amount of energy required to create a system in the nonexistant changes of temperature (T) or volume (V). However, if the system is created in an environment of temperature, T, then some of the energy can be captured by spontaneous heat transfer between the environment and system. The amount of this spontaneous energy transfer is TΔS where S is the final entropy of the system. In that case, you don't have to put in as much energy. Note that if a more disordered, resulting in higher entropy, the final state is created, where less work is required to create the system. The Helmholtz free energy becomes a measure of the sum of energy you have to put in to generate a system once the spontaneous energy transfer of the system from the environment is taken into account. Helmholtz Free Energy is generally used in Physics, denoted with the leter , while Chemistry uses, , Gibbs' Free Energy. The Helmholtz Energy is given by the equation: \[A = U - TS\] It is comparable to Gibbs Energy in this way: \[G = A + PV\] The Helmholtz Energy is used when having a constant pressure is not feasible. Along with internal energy and enthalpy, the Helmholtz Energy and Gibbs Energy make up the quad group called the ; these potentials are useful for describing various thermodynamic events. TS represents energy from surroundings, and PV represents work in expansion. If needed, refer to the links above to refresh your memory on enthalpy and internal energy. | 6,001 | 1,121 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Acids_and_Bases_in_Aqueous_Solutions/Water_Autoionization |
Water, even pure water, has an amphiprotic nature. This means that a small amount of ions will form in pure water. Some molecules of H O will act as acids, each donating a proton to a corresponding H O molecule that acts as a base. Thus, the proton-donating molecule becomes a hydroxide ion, OH , while the proton-accepting molecule becomes a hydronium ion, H O . Water molecules are amphiprotic and can function as both acids and bases. One water molecule (acting as a base) can accept a hydrogen ion from a second water molecule (acting as an acid). This will be happening anywhere there is even a trace of water - it does not have to be pure. A and a hydroxide ion are formed. However, the hydroxonium (used unambiguously with hydronium) ion is a very strong acid, and the hydroxide ion is a very strong base; in fact, they are the strongest in water. As fast as they are formed, they react to produce water again. The net effect is that an equilibrium is set up: \[2H_2O\,(l) \rightleftharpoons H_3O^+ \,(aq) + OH^- \, (aq) \tag{1}\] 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. This equilibrium written in a simplified form: \[H_2O\,(l) \rightleftharpoons H^+ \,(aq) + OH^- \, (aq) \tag{2}\] with H (aq) actually referring to a hydronium ion. It is important to remember that water contains VERY low concentration of these ions. In the reversible reaction: \[\underset{\text{base 1}}{H_2O} + \underset{\text{acid 2}}{H_2O} \rightleftharpoons \underset{\text{acid 1}}{H_3O^+} + \underset{\text{base 2}}{OH^-} \tag{4}\] the reaction proceeds by far to the left. Pure water will dissociate to form equal concentrations (here, we are using molarities) of hydronium and hydroxide ions, thus: \[ [H_3O^+] = [OH^-] \tag{5}\] For this equation, we can find K, the equilibrium constant: \[K= [H_3O^+,OH^-] \tag{6}\] At standard temperature and pressure (STP), the equilibrium constant of water, \(K_w\), is equal to: \[K_w= [H_3O^+,OH^-] \tag{7}\] \[K_w=[1.0 \times 10^{-7},1.0 \times 10^{-7}]\tag{8} \] \[K_w=1.0 \times 10^{-14} \tag{9}\] In this equation, [H O ] is the concentration of hydronium ions which, in a chemical equation, is the , \(K_a\). The [OH ] is the concentration of hydroxide ions which, in a chemical equation, is the base constant, \(K_b\). If given a pH, then you can easily calculate the [H O ] by: \[[H_3O^+] = 10^{-pH}. \tag{10}\] The same formula applies to obtaining [OH ] from the pOH: \[ [OH-]=10^{-pOH} \tag{11}\] Adding the pH's gives you the \(pK_w\) \[pK_w= pH + pOH =14.00 \tag{11}\] Since the reaction proceeds so heavily to the left, the concentration of these hydroxide and hydronium ions in pure water is extremely small. When making calculations determining involving acids and bases in solution, you do not need to take into account the effects of water's autoionization unless the acid or base of interest is incredibly dilute. However, it is interesting to note that water's self-ionization is significant in that it makes the substance electrically . In the equation depicting the autoionization of water, \[ H_2O + H_2O \rightleftharpoons H_3O^+ + OH^- \] The reaction proceeds far to the __________. . The concentration of hydroxide and hydronium ions in pure water is very, very small. Although it is rarely something you need to worry about when looking at acids and bases in solution, it does help account for certain properties of water, such as electrical conductivity. If a solution has a pH of 2.1, determine the concentration of hydroxide ions, [OH ]. To solve for this, you must first determine the concentration of the hydronium ion, [H O ]: [H O ] = 10 = 10 = 7.94 x 10 Then, you solve for [OH ] using the Kw constant: Kw = [H O ] [OH ] 1.0 x 10 = [OH-,7.94 x 10 ] [OH-] = (1 x 10 ) / (7.94 x 10 ) = If an aqueous solution has a pOH of 11.2, determine the concentration of hydronium ions. To solve for this, you must first determine the concentration of the hydroxide ion, [OH ]: [OH ] = 10
= 10
= 6.31 x 10 Then, you solve for [H O ] using the K constant: K = [H O ] [OH ]
1.0 x 10 = [H O ,6.31 x 10 ]
[H O ]= (1 x 10 )/ (6.31 x 10 ) = Jim Clark ( ) | 4,348 | 1,122 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/4_f-Block_Elements/The_Lanthanides/Chemistry_of_Cerium |
Discovered by Berzelius and Hisinger in 1803, but not isolated as a metal until 1875, cerium (named for the asteroid Ceres) is the most abundant of the so-called rare-earth metals. It begins the series of lanthanides that runs from elements 58 to 71. In pure form the element is a malleable and ductile metal, similar in coloring to iron. It is much more reactive than iron, however, readily oxidizing in moist air and releasing hydrogen from boiling water. Friction from abrading a sample can cause it to ignite. Although the metal itself is too reactive for most uses, compounds of cerium are used in glass making and photography. It has limited use in some special alloys as well. Most commercial grade cerium is derived from monazite sand which is a mixture of phosphates of many of the rare earth metals along with calcium and thorium. | 860 | 1,123 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/IV._Nucleophilic_Substitution_Reactions/E._Substitution_Reactions_Involving_Cyanide_Ions |
If a halogenoalkane is heated under reflux with a solution of sodium or potassium cyanide in ethanol, the halogen is replaced by a -CN group and a nitrile is produced. Heating under reflux means heating with a condenser placed vertically in the flask to prevent loss of volatile substances from the mixture. The solvent is important. If water is present you tend to get substitution by -OH instead of -CN. For example, using 1-bromopropane as a typical primary halogenoalkane: \[ CH_3CH_2CH_2Br + CN^- \rightarrow CH_3CH_2CH_2CN + Br^-\] The full equation could be written rather than the ionic one, but it obscures what's going on: \[ CH_3CH_2CH_2Br + KCN \rightarrow CH_3CH_2CH_2CN + KBr\] The bromine (or other halogen) in the halogenoalkane is simply replaced by a -CN group - hence a substitution reaction. In this example, butanenitrile is formed. Here is the mechanism for the reaction involving bromoethane: This is an example ofnucleophilic substitution. Because the mechanism involves collision between two species in the slow step (in this case, the only step) of the reaction, it is known as an S 2 reaction. If your examiners want you to show the transition state, draw the mechanism like this: The facts of the reaction are exactly the same as with primary halogenoalkanes. If the halogenoalkane is heated under reflux with a solution of sodium or potassium cyanide in ethanol, the halogen is replaced by -CN, and a nitrile is produced. For example: \[ (CH_3)_3CBr + CN^- \rightarrow (CH_3)_3CCN + Br^-\] Or if you want the full equation rather than the ionic one: \[ (CH_3)_3CBr + KCN \rightarrow (CH_3)_3CCN + KBr\] This mechanism involves an initial ionization of the halogenoalkane: followed by a very rapid attack by the cyanide ion on the carbocation (carbonium ion) formed: This is again an example of nucleophilic substitution. This time the slow step of the reaction only involves one species - the halogenoalkane. It is known as an S 1 reaction. The facts of the reaction are exactly the same as with primary or tertiary halogenoalkanes. The halogenoalkane is heated under reflux with a solution of sodium or potassium cyanide in ethanol. For example: Secondary halogenoalkanes use both S 2 and S 1 mechanisms. For example, the S 2 mechanism is: The two stages of the S 1 mechanism are: Jim Clark ( ) | 2,336 | 1,124 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/20%3A_Entropy_and_The_Second_Law_of_Thermodynamics/20.03%3A_Unlike_heat_Entropy_Is_a_State_Function |
Because entropy is a state function, it integrates to zero over any circular path going back to initial conditions, just like \(U\) and \(H\): \[\oint dS =0 \nonumber \] \[\oint dH =0 \nonumber \] \[\oint dU =0 \nonumber \] As discussed previously, we can use this fact to revisit the isotherm + isochore + adiabat circular path (Figure 20.3.1
). \[q_{rev,B} = 0 \nonumber \] \[\delta w=0 \nonumber \] \[q_{rev,C}=C_V\Delta T \nonumber \] \[q_{rev,A} = nRT \ln \dfrac{V_2}{V_1} \nonumber \] The quantities q , q , and q are the same, which once again underlines that heat is a path function. How about entropy? First, consider the combined paths of B and C: \[q_{rev,B+C} = \int _{T_2}^{T_1} C_v dT \nonumber \] \[\int dS_{B+C} = \int \dfrac{dq_{rev,B+C}}{T} = \int _{T_2}^{T_1} \dfrac{C_v}{T} dT \nonumber \] We had seen this integral before from Section 19-6, albeit from \(T_1\) to \(T_2\): \[\Delta S_{B+C} = nR\ln \dfrac{V_2}{V_1} \label{19.21} \] (Notice sign in Equation \ref{19.21} is positive) Along the isotherm A: \[q_{rev,A} = nRT \ln\frac{V_2}{V_1} \nonumber \] \(T\) is a constant so we can just divide \(q_{rec,A}\) by \(T\) to get \(\Delta S_A\): \[\Delta S_A = nR\ln \dfrac{V_2}{V_1} \nonumber \] We took two different paths to get start and end at the same points. Both paths had the same change in entropy. Clearly entropy is a while \(q_{rev}\) is not. | 1,402 | 1,127 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.10%3A_Standard_Enthalpies_of_Formation/3.10.02%3A_Foods-_Energy_in_a_Marshmallow |
Marshmallow was originally used medicinally to soothe sore throats. It was made from the root of the marsh mallow plant, , sometimes mixed with sugars or other ingredients and whipped to make something like the modern marshmallow. Modern Campfire® marshmallows contain Corn Syrup, Sugar, Modified Food Starch (Corn), Dextrose, Water, Gelatin, Natural and Artificial Flavor, Tetrasodium Pyprophosphate and Blue 1 , and the gelatin protein made from bones and hides makes them off limits to strict vegetarians. We'll consider the marshmallow to be 7.5 g of pure sugar (sucrose) for the calculations below. Let's investigate the fate of a marshmallow when you eat it, and explain in part where the food energy comes from. We'll need the energy supplied by the overall reaction for the metabolism of sucrose, which occurs when plenty of oxygen is available: \[C_{12}H_{22}O_{11} (s) + 12 O_2(g)→12 CO_2(g) + 11 H_2O(l) \,\,\,\, ΔH_{m1} \label{1} \] But that reaction lumps together a lot of interesting processes. The (cleavage by water) of sucrose into the simple sugars glucose and fructose occurs in saliva, but not without the enzyme sucrase that catalyzes the reaction: \[C_{12}H_{22}O_{11} (s) + 1 H_2O(l)→ \underbrace{C_6H_{12}O_6}_{glucose} + \underbrace{C_6H_{12}O_6}_{fructose} \,\,\,\, ΔH_{m2} \label{2} \] in our mouth partially convert the simple sugars to lactic acid (which causes tooth decay), in the overall reaction for glycolysis plus fermentation:
C H O ( ) → 2 C H O Δ (3) This reaction provides energy to sustain the bacteria, but it also occurs in our bodies when sugar is metabolized (with limited oxygen), and the lactic acid is responsible for muscle ache the day after we exert our muscles. Lactobaccilli are used in controlled recipes to create lactic acid that creates the tart or sour taste of yogurt and sauerkraut. If bacteria don't metabolize the glucose, we do, using it to produce ATP in a process called which involves about ten different reactions that end with the production of pyruvic acid (C H O ). If our muscles are well oxygenated, the pyruvic acid is converted to CO and H O and we have overall reaction (1), yielding energy that we'll calculate below. During extended exercise, comes to a halt when it runs out of oxygen to make an essential reactant, NAD . Then takes over, producing the NAD and converting the pyruvic acid to lactic acid (C H O ), which builds up in muscles as a result of reaction (3). This produces a lot less energy than aerobic metabolism, as we'll see below. The process makes just 2 ATP instead of many more that would be produced if the pyruvic acid were metabolized aerobically by overall reaction (1). How do food chemists calculate the energy produced in all these reactions? By now you can imagine that there are innumerable reactions involved in just food metabolism, and it would be virtually impossible to list all the thermochemical equations, along with the corresponding enthalpy changes. Fortunately Hess' law makes it possible to list just Δ , for each compound, and use these Δ values to calculate the Δ for any reaction of interest. Each element must be in the physical and chemical form which is most stable at normal atmospheric pressure and a specified temperature (usually 25°C). For example, since H O( ) appears in Equation (1), we'll need its H to calculate the energy available from a marshmallow. If we know that ΔH [H O(l H ( ) + ½O ( ) → H O( ) Δ = –285.8 kJ mol (4) The elements H and O appear as diatomic molecules and in gaseous form because these are their most stable chemical and physical states. Note also that 285.8 kJ are given off of H O( ) formed. Equation (1) must specify formation of mol H O( ), and so the coefficient of O must be ½. In addition to (4), we'll need two other Δ , values to calculate the energy in a marshmallow. They are the Δ values for the other compounds in Equation (1), CO and C H O . All the Δ values can be found in standard tables like the one at the end of this section, and we can write the equations (5) and (6) knowing the definition of Δ : H ( ) + ½O ( ) → H O( ) Δ = –285.8 kJ mol (4) C( ) + O ( ) → CO ( ) Δ = –393.509 kJ mol (5) 12 C( ) + 11 H ( ) + 11/2 O ( ) → C H O Δ = -2222.1 kJ mol (6) By Hess' Law, we may be able to combine equations 4, 5, and 6 to get Equation (1). First, we notice that (1) has sucrose on the left, but it's on the right in (6); so reversing (6) we get C H O ( ) → 12C( ) + 11 H ( ) + 11/2 O ( ) -Δ = +2222.1 kJ mol (6a) To cancel the 12 C that does not appear in (1), we'll add 12 x Equation (5) (along with 6x its enthalpy change: 12 C( ) + 12 O ( ) → 12 CO ( ) 6 x Δ = 12 x (-393.509) kJ mol (5a) And to add the 11 H O( ) that appears in (1), we'll add 11 x Equation (4): 11 H ( ) + 11/2 O ( ) → 11 H O( ) Δ = 11 x (–285.8) kJ mol (4a) If we combine Equations 6a, 5a, and 4a according to Hess' Law, we notice that 12 H , 12 C, and 11/2 O ( ) appear on both the left and right, and cancel to give Equation (1)! C H O ( ) + 12 O ( ) → 12 CO ( ) + 11 H O( ) Δ (1) We can then combine the enthalpies to get the needed Δ : Δ = 12 Δ + 12 Δ - Δ =12 x (-393.509) + 11 x (–285.8) - (-2222.1)kJ mol = -5643.8 kJ mol Notice that this value appears in the Table at the end of this section. With Hess' Law, we can always calculate an enthalpy of combustion from enthalpies of formation, or vice versa! Reaction (6) corresponding to Δ of sucrose does not occur, but its enthalpy can be calculated from enthalpies of reactions that do occur. Notice that our calculation simplifies to: Δ = ∑ Δ (products) – ∑ Δ (reactants) The symbol Σ means “the sum of.” So we just need to add the Δ values for the products, and subtract the sum of Δ values for the reactants in Equation (1). Since Δ values are given of compound, you must be sure to multiply each Δ by an appropriate coefficient in from Equation (1) (for which Δ is being calculated). Now we can calculate the food energy in the marshmallow: The molar mass of sucrose is 342.3 g/mol, so the energy per gram is -5643.8 kJ/mol / 342.3 g/mol = 16.49 kJ/g. In the 7.5 g marshmallow, remembering that 1 dietary Calorie is 4.184 kJ, we have 7.5 g x 16.49 kJ/g x (1 Cal / 4.184 kJ) = 29.6 Cal. (But who can stop at just one roasted marshmallow?) Note carefully how the problem above was solved. In step 6a the compound C H O ( ) was hypothetically decomposed to its elements. This equation was the reverse of formation of the compound, and so Δ was opposite in sign from Δ . In step 5a we had the hypothetical formation of the CO ( ) from its elements. Since 12 mol were obtained, the enthalpy change was doubled but its sign remained the same. In step 4a we had the hypothetical formation of the H O ( ) from its elements. Since 11 mol were obtained, the enthalpy change was multiplied by 11, but its sign remained the same. Any chemical reaction can be approached similarly. To calculate Δ we all the Δ values for the products, multiplying each by the appropriate coefficient, as in step 2 above. Since the signs of Δ for the reactants had to be reversed in step 1, we them, again multiplying by appropriate coefficients. Again, this can he summarized by the important equation Δ = ∑ Δ (products) – ∑ Δ (reactants) One further point arises from the definition of Δ . . That's why Δ for O doesn't appear in the calculation above; it's value is zero, corresponding to the formation of O from its elements. There's no change in the reaction below, so Δ = 0: O ( ) → O ( ) Δ = 0 Standard enthalpies of formation for some common compounds are given in the table below, and more are given in Table of Some Standard Enthalpies of Formation at 25°C. These values may be used to calculate Δ for any chemical reaction so long as all the compounds involved appear in the tables. To see how and why this may be done, consider the following example. Use standard enthalpies of formation to calculate Δ for the reaction C H O ( ) + 1 H O( ) → C H O ( ) + C H O ( ) Δ = ∑ Δ (products) – ∑ Δ (reactants) From the table below, Δ for glucose, fructose, sucrose and water are -1271, -1266.6 (they may actually be the same, but measured by different methods), -2222.1, and -285.8 kJ mol respectively. Note that we were careful to use Δ [H O( )] not Δ [H O( )] or ( ). Substituting these values in the equation above gives Δ = [1 mol glucose x (-1271 kJ mol ) + 1 mol fructose x (-1266.6 kJ mol ] - [1 mol sucrose x (-2222.1 kJ mol + 1 mol water x -285.8 kJ mol ] = -29.7 kJ mol . The process is actually exothermic, releasing a small amount of heat energy. The measured energy for hydrolysis of maltose to 2 glucose units is only -4.02 kJ . Use the table of standard enthalpies of formation at 25°C to calculate Δ for the reaction below (glycolysis + fermentation), which is associated with the production of 2 mol ATP (as well as NADH) in anaerobic metabolism in your body. The Δ for lactic acid and glucose are -687 and -1271 kJ mol respectively. C H O ( ) → 2 C H O Δ (3) = [2 Δ (C H O ) - [Δ (C H O )] = [2 mol lactic acid x (–687) kJ mol ] – [1 mol glucose x (–1271 kJ mol ) = –1374 + 2222.1 kJ = -103 kJ. This energy is used in part to make 2ATP molecules rather than being released entirely as heat. Note that -5643.8 kJ mol resulted from the aerobic metabolism of sucrose (above) but only 2 (-103)kJ = -206 kJ would result from its anaerobic metabolism (since 1 mol of sucrose yields 2 mol of glucose). 1. Estimate, based on a theoretical calculation The most general references are NIST, this bond energy based calculator and for QSPR calculated values, Int. J. Mol. Sci. 2007, 8, 407-432. From ChemPRIME: 3.9: Standard Enthalpies of Formation | 9,769 | 1,129 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/11%3A_Quantum_Mechanics_and_Atomic_Structure/11.02%3A_Planck's_Quantum_Theory |
By the late 19th century, many physicists thought their discipline was well on the way to explaining most natural phenomena. They could calculate the motions of material objects using Newton’s laws of classical mechanics, and they could describe the properties of radiant energy using mathematical relationships known as , developed in 1873 by James Clerk Maxwell, a Scottish physicist. The universe appeared to be a simple and orderly place, containing matter, which consisted of particles that had mass and whose location and motion could be accurately described, and electromagnetic radiation, which was viewed as having no mass and whose exact position in space could not be fixed. Thus matter and energy were considered distinct and unrelated phenomena. Soon, however, scientists began to look more closely at a few inconvenient phenomena that could not be explained by the theories available at the time. One experimental phenomenon that could not be adequately explained by classical physics was blackbody radiation (Figure 1.2.1
). Attempts to explain or calculate this spectral distribution from classical theory were complete failures. A theory developed by Rayleigh and Jeans predicted that the intensity should go to infinity at short wavelengths. Since the intensity actually drops to zero at short wavelengths, the Rayleigh-Jeans result was called the (Figure 1.2.1
dashed line). There was no agreement between theory and experiment in the ultraviolet region of the blackbody spectrum. In 1900, the German physicist Max Planck (1858–1947) explained the ultraviolet catastrophe by proposing that the energy of electromagnetic waves is rather than continuous. This means that for each temperature, there is a maximum intensity of radiation that is emitted in a blackbody object, corresponding to the peaks in Figure 1.2.1
, so the intensity does not follow a smooth curve as the temperature increases, as predicted by classical physics. Thus energy could be gained or lost only in integral multiples of some smallest unit of energy, a quantum (the smallest possible unit of energy). Energy can be gained or lost only in integral multiples of a quantum. Although quantization may seem to be an unfamiliar concept, we encounter it frequently in quantum mechanics (hence the name). For example, US money is integral multiples of pennies. Similarly, musical instruments like a piano or a trumpet can produce only certain musical notes, such as C or F sharp. Because these instruments cannot produce a continuous range of frequencies, their frequencies are quantized. It is also similar to going up and down a hill using discrete stair steps rather than being able to move up and down a continuous slope. Your potential energy takes on discrete values as you move from step to step. Even electrical charge is quantized: an ion may have a charge of −1 or −2, but −1.33 electron charges. Planck's quantization of energy is described by the his famous equation: \[ E=h \nu \label{Eq1.2.1} \] where the proportionality constant \(h\) is called , one of the most accurately known fundamental constants in science \[h=6.626070040(81) \times 10^{−34}\, J\cdot s \nonumber \] However, for our purposes, its value to four significant figures is sufficient: \[h = 6.626 \times 10^{−34} \,J\cdot s \nonumber \] As the frequency of electromagnetic radiation increases, the magnitude of the associated quantum of radiant energy increases. By assuming that energy can be emitted by an object only in integral multiples of \(hν\), Planck devised an equation that fit the experimental data shown in Figure 1.2.1
. We can understand Planck’s explanation of the ultraviolet catastrophe qualitatively as follows: At low temperatures, radiation with only relatively low frequencies is emitted, corresponding to low-energy quanta. As the temperature of an object increases, there is an increased probability of emitting radiation with higher frequencies, corresponding to higher-energy quanta. At any temperature, however, it is simply more probable for an object to lose energy by emitting a large number of lower-energy quanta than a single very high-energy quantum that corresponds to ultraviolet radiation. The result is a maximum in the plot of intensity of emitted radiation versus wavelength, as shown in Figure 1.2.1
, and a shift in the position of the maximum to lower wavelength (higher frequency) with increasing temperature. At the time he proposed his radical hypothesis, Planck could not explain energies should be quantized. Initially, his hypothesis explained only one set of experimental data—blackbody radiation. If quantization were observed for a large number of different phenomena, then quantization would become a law. In time, a theory might be developed to explain that law. As things turned out, Planck’s hypothesis was the seed from which modern physics grew. Max Planck explain the spectral distribution of blackbody radiation as result from oscillations of electrons. Similarly, oscillations of electrons in an antenna produce radio waves. Max Planck concentrated on modeling the oscillating charges that must exist in the oven walls, radiating heat inwards and—in thermodynamic equilibrium—themselves being driven by the radiation field. He found he could account for the observed curve if he required these oscillators not to radiate energy continuously, as the classical theory would demand, but they could lose or gain energy in chunks, called , of size \(h\nu\), for an oscillator of frequency \(\nu\) (Equation \(\ref{Eq1.2.1} \)). With that assumption, Planck calculated the following formula for the radiation energy density inside the oven: \[ \begin{align} d\rho(\nu,T) &= \rho_\nu (T) d\nu \\[4pt] &= \dfrac {2 h \nu^3}{c^2} \cdot \dfrac {1 }{\exp \left( \dfrac {h\nu}{k_B T}\right)-1} d\nu \label{Eq2a} \end{align} \] with Planck's radiation energy density (Equation \(\ref{Eq2a}\)) can also be expressed in terms of wavelength \(\lambda\). \[\rho (\lambda, T) d \lambda = \dfrac {2 hc^2}{\lambda ^5} \dfrac {1}{ \exp \left(\dfrac {hc}{\lambda k_B T} \right) - 1} d \lambda \label{Eq2b} \] Planck's equation (Equation \(\ref{Eq2b}\)) gave an excellent agreement with the experimental observations for all temperatures (Figure 1.2.2
). Planck made many substantial contributions to theoretical physics, but his fame as a physicist rests primarily on his role as the originator of quantum theory. In addition to being a physicist, Planck was a gifted pianist, who at one time considered music as a career. During the 1930s, Planck felt it was his duty to remain in Germany, despite his open opposition to the policies of the Nazi government. One of his sons was executed in 1944 for his part in an unsuccessful attempt to assassinate Hitler and bombing during the last weeks of World War II destroyed Planck’s home. After WWII, the major German scientific research organization was renamed the Max Planck Society. Use Equation \(\ref{Eq2b}\) to show that the units of \(ρ(λ,T)\,dλ\) are \(J/m^3\) as expected for an energy density. The near perfect agreement of this formula with precise experiments (e.g., Figure 1.2.3
), and the consequent necessity of energy quantization, was the most important advance in physics in the century. His blackbody curve was completely accepted as the correct one: more and more accurate experiments confirmed it time and again, yet the radical nature of the quantum assumption did not sink in. Planck was not too upset—he didn’t believe it either, he saw it as a technical fix that (he hoped) would eventually prove unnecessary. Part of the problem was that Planck’s route to the formula was long, difficult and implausible—he even made contradictory assumptions at different stages, as Einstein pointed out later. However, the result was correct anyway! The mathematics implied that the energy given off by a blackbody was not continuous, but given off at certain specific wavelengths, in regular increments. If Planck assumed that the energy of blackbody radiation was in the form \[E = nh \nu \nonumber \] where \(n\) is an integer, then he could explain what the mathematics represented. This was indeed difficult for Planck to accept, because at the time, there was no reason to presume that the energy should only be radiated at specific frequencies. Nothing in Maxwell’s laws suggested such a thing. It was as if the vibrations of a mass on the end of a spring could only occur at specific energies. Imagine the mass slowly coming to rest due to friction, but not in a continuous manner. Instead, the mass jumps from one fixed quantity of energy to another without passing through the intermediate energies. To use a different analogy, it is as if what we had always imagined as smooth inclined planes were, in fact, a series of closely spaced steps that only presented the illusion of continuity. The agreement between Planck’s theory and the experimental observation provided strong evidence that the energy of electron motion in matter is quantized. In the next two sections, we will see that the energy carried by light also is quantized in units of \(h \bar {\nu}\). These packets of energy are called “photons.” (Beams Professor, , ") | 9,258 | 1,130 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/10%3A_Electrolyte_Solutions/10.06%3A_Mean_Ionic_Activity_Coefficients_from_Osmotic_Coefficients |
\( \newcommand{\tx}[1]{\text{#1}} % text in math mode\)
\( \newcommand{\subs}[1]{_{\text{#1}}} % subscript text\)
\( \newcommand{\sups}[1]{^{\text{#1}}} % superscript text\)
\( \newcommand{\st}{^\circ} % standard state symbol\)
\( \newcommand{\id}{^{\text{id}}} % ideal\)
\( \newcommand{\rf}{^{\text{ref}}} % reference state\)
\( \newcommand{\units}[1]{\mbox{$\thinspace$#1}}\)
\( \newcommand{\K}{\units{K}} % kelvins\)
\( \newcommand{\degC}{^\circ\text{C}} % degrees Celsius\)
\( \newcommand{\br}{\units{bar}} % bar (\bar is already defined)\)
\( \newcommand{\Pa}{\units{Pa}}\)
\( \newcommand{\mol}{\units{mol}} % mole\)
\( \newcommand{\V}{\units{V}} % volts\)
\( \newcommand{\timesten}[1]{\mbox{$\,\times\,10^{#1}$}}\)
\( \newcommand{\per}{^{-1}} % minus one power\)
\( \newcommand{\m}{_{\text{m}}} % subscript m for molar quantity\)
\( \newcommand{\CVm}{C_{V,\text{m}}} % molar heat capacity at const.V\)
\( \newcommand{\Cpm}{C_{p,\text{m}}} % molar heat capacity at const.p\)
\( \newcommand{\kT}{\kappa_T} % isothermal compressibility\)
\( \newcommand{\A}{_{\text{A}}} % subscript A for solvent or state A\)
\( \newcommand{\B}{_{\text{B}}} % subscript B for solute or state B\)
\( \newcommand{\bd}{_{\text{b}}} % subscript b for boundary or boiling point\)
\( \newcommand{\C}{_{\text{C}}} % subscript C\)
\( \newcommand{\f}{_{\text{f}}} % subscript f for freezing point\)
\( \newcommand{\mA}{_{\text{m},\text{A}}} % subscript m,A (m=molar)\)
\( \newcommand{\mB}{_{\text{m},\text{B}}} % subscript m,B (m=molar)\)
\( \newcommand{\mi}{_{\text{m},i}} % subscript m,i (m=molar)\)
\( \newcommand{\fA}{_{\text{f},\text{A}}} % subscript f,A (for fr. pt.)\)
\( \newcommand{\fB}{_{\text{f},\text{B}}} % subscript f,B (for fr. pt.)\)
\( \newcommand{\xbB}{_{x,\text{B}}} % x basis, B\)
\( \newcommand{\xbC}{_{x,\text{C}}} % x basis, C\)
\( \newcommand{\cbB}{_{c,\text{B}}} % c basis, B\)
\( \newcommand{\mbB}{_{m,\text{B}}} % m basis, B\)
\( \newcommand{\kHi}{k_{\text{H},i}} % Henry's law constant, x basis, i\)
\( \newcommand{\kHB}{k_{\text{H,B}}} % Henry's law constant, x basis, B\)
\( \newcommand{\arrow}{\,\rightarrow\,} % right arrow with extra spaces\)
\( \newcommand{\arrows}{\,\rightleftharpoons\,} % double arrows with extra spaces\)
\( \newcommand{\ra}{\rightarrow} % right arrow (can be used in text mode)\)
\( \newcommand{\eq}{\subs{eq}} % equilibrium state\)
\( \newcommand{\onehalf}{\textstyle\frac{1}{2}\D} % small 1/2 for display equation\)
\( \newcommand{\sys}{\subs{sys}} % system property\)
\( \newcommand{\sur}{\sups{sur}} % surroundings\)
\( \renewcommand{\in}{\sups{int}} % internal\)
\( \newcommand{\lab}{\subs{lab}} % lab frame\)
\( \newcommand{\cm}{\subs{cm}} % center of mass\)
\( \newcommand{\rev}{\subs{rev}} % reversible\)
\( \newcommand{\irr}{\subs{irr}} % irreversible\)
\( \newcommand{\fric}{\subs{fric}} % friction\)
\( \newcommand{\diss}{\subs{diss}} % dissipation\)
\( \newcommand{\el}{\subs{el}} % electrical\)
\( \newcommand{\cell}{\subs{cell}} % cell\)
\( \newcommand{\As}{A\subs{s}} % surface area\)
\( \newcommand{\E}{^\mathsf{E}} % excess quantity (superscript)\)
\( \newcommand{\allni}{\{n_i \}} % set of all n_i\)
\( \newcommand{\sol}{\hspace{-.1em}\tx{(sol)}}\)
\( \newcommand{\solmB}{\tx{(sol,$\,$$m\B$)}}\)
\( \newcommand{\dil}{\tx{(dil)}}\)
\( \newcommand{\sln}{\tx{(sln)}}\)
\( \newcommand{\mix}{\tx{(mix)}}\)
\( \newcommand{\rxn}{\tx{(rxn)}}\)
\( \newcommand{\expt}{\tx{(expt)}}\)
\( \newcommand{\solid}{\tx{(s)}}\)
\( \newcommand{\liquid}{\tx{(l)}}\)
\( \newcommand{\gas}{\tx{(g)}}\)
\( \newcommand{\pha}{\alpha} % phase alpha\)
\( \newcommand{\phb}{\beta} % phase beta\)
\( \newcommand{\phg}{\gamma} % phase gamma\)
\( \newcommand{\aph}{^{\alpha}} % alpha phase superscript\)
\( \newcommand{\bph}{^{\beta}} % beta phase superscript\)
\( \newcommand{\gph}{^{\gamma}} % gamma phase superscript\)
\( \newcommand{\aphp}{^{\alpha'}} % alpha prime phase superscript\)
\( \newcommand{\bphp}{^{\beta'}} % beta prime phase superscript\)
\( \newcommand{\gphp}{^{\gamma'}} % gamma prime phase superscript\)
\( \newcommand{\apht}{\small\aph} % alpha phase tiny superscript\)
\( \newcommand{\bpht}{\small\bph} % beta phase tiny superscript\)
\( \newcommand{\gpht}{\small\gph} % gamma phase tiny superscript\) \( \newcommand{\upOmega}{\Omega}\) \( \newcommand{\dif}{\mathop{}\!\mathrm{d}} % roman d in math mode, preceded by space\)
\( \newcommand{\Dif}{\mathop{}\!\mathrm{D}} % roman D in math mode, preceded by space\)
\( \newcommand{\df}{\dif\hspace{0.05em} f} % df\) \(\newcommand{\dBar}{\mathop{}\!\mathrm{d}\hspace-.3em\raise1.05ex{\Rule{.8ex}{.125ex}{0ex}}} % inexact differential \)
\( \newcommand{\dq}{\dBar q} % heat differential\)
\( \newcommand{\dw}{\dBar w} % work differential\)
\( \newcommand{\dQ}{\dBar Q} % infinitesimal charge\)
\( \newcommand{\dx}{\dif\hspace{0.05em} x} % dx\)
\( \newcommand{\dt}{\dif\hspace{0.05em} t} % dt\)
\( \newcommand{\difp}{\dif\hspace{0.05em} p} % dp\)
\( \newcommand{\Del}{\Delta}\)
\( \newcommand{\Delsub}[1]{\Delta_{\text{#1}}}\)
\( \newcommand{\pd}[3]{(\partial #1 / \partial #2 )_{#3}} % \pd{}{}{} - partial derivative, one line\)
\( \newcommand{\Pd}[3]{\left( \dfrac {\partial #1} {\partial #2}\right)_{#3}} % Pd{}{}{} - Partial derivative, built-up\)
\( \newcommand{\bpd}[3]{[ \partial #1 / \partial #2 ]_{#3}}\)
\( \newcommand{\bPd}[3]{\left[ \dfrac {\partial #1} {\partial #2}\right]_{#3}}\)
\( \newcommand{\dotprod}{\small\bullet}\)
\( \newcommand{\fug}{f} % fugacity\)
\( \newcommand{\g}{\gamma} % solute activity coefficient, or gamma in general\)
\( \newcommand{\G}{\varGamma} % activity coefficient of a reference state (pressure factor)\)
\( \newcommand{\ecp}{\widetilde{\mu}} % electrochemical or total potential\)
\( \newcommand{\Eeq}{E\subs{cell, eq}} % equilibrium cell potential\)
\( \newcommand{\Ej}{E\subs{j}} % liquid junction potential\)
\( \newcommand{\mue}{\mu\subs{e}} % electron chemical potential\)
\( \newcommand{\defn}{\,\stackrel{\mathrm{def}}{=}\,} % "equal by definition" symbol\) \( \newcommand{\D}{\displaystyle} % for a line in built-up\)
\( \newcommand{\s}{\smash[b]} % use in equations with conditions of validity\)
\( \newcommand{\cond}[1]{\\[-2.5pt]{}\tag*{#1}}\)
\( \newcommand{\nextcond}[1]{\\[-5pt]{}\tag*{#1}}\)
\( \newcommand{\R}{8.3145\units{J$\,$K$\per\,$mol$\per$}} % gas constant value\)
\( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \)
\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) Recall that \(\g_{\pm}\) is the mean ionic activity coefficient of a strong electrolyte, or the stoichiometric activity coefficient of an electrolyte that does not dissociate completely. The general procedure described in this section for evaluating \(\g_{\pm}\) requires knowledge of the osmotic coefficient \(\phi_m\) as a function of molality. \(\phi_m\) is commonly evaluated by the isopiestic method (Sec. 9.6.4) or from measurements of freezing-point depression (Sec. 12.2). The osmotic coefficient of a binary solution of an electrolyte is defined by \begin{gather} \s{ \phi_m \defn \frac{\mu\A^*-\mu\A}{RTM\A\nu m\B} } \tag{10.6.1} \cond{(binary electrolyte solution)} \end{gather} That is, for an electrolyte the sum \(\sum_{i\neq \tx{A}}m_i\) appearing in the definition of \(\phi_m\) for a nonelectrolyte solution (Eq. 9.6.11) is replaced by \(\nu m\B\), the sum of the ion molalities assuming complete dissociation. It will now be shown that \(\phi_m\) defined this way can be used to evaluate \(\g_{\pm}\). The derivation is like that described in Sec. 9.6.3 for a binary solution of a nonelectrolyte. Solving Eq. 10.6.1 for \(\mu\A\) and taking the differential of \(\mu\A\) at constant \(T\) and \(p\), we obtain \begin{equation} \dif\mu\A = -RTM\A\nu(\phi_m\dif m\B + m\B\dif\phi_m) \tag{10.6.2} \end{equation} From Eq. 10.3.9, we obtain \begin{equation} \dif\mu\B = RT\nu\left(\dif\ln\g_{\pm} + \frac{\dif m\B}{m\B}\right) \tag{10.6.3} \end{equation} Substitution of these expressions in the Gibbs–Duhem equation \(n\A \dif\mu\A + n\B \dif\mu\B = 0\), together with the substitution \(n\A M\A = n\B/m\B\), yields \begin{equation} \dif\ln\g_{\pm} = \dif\phi_m + \frac{\phi_m - 1}{m\B}\dif m\B \tag{10.6.4} \end{equation} Then integration from \(m\B = 0\) to any desired molality \(m'\B\) gives the result \begin{equation} \ln\g_{\pm}(m'\B) = \phi_m (m'\B) - 1 + \int_{0}^{m'\B}\frac{\phi_m - 1}{m\B}\dif m\B \tag{10.6.5} \end{equation} The right side of this equation is the same expression as derived for \(\ln\g\mbB\) for a nonelectrolyte (Eq. 9.6.20). The integrand of the integral on the right side of Eq. 10.6.5 approaches \(-\infty\) as \(m\B\) approaches zero, making it difficult to evaluate the integral by numerical integration starting at \(m\B = 0\). (This difficulty does not exist when the solute is a nonelectrolyte.) Instead, we can split the integral into two parts \begin{equation} \int_{0}^{m'\B}\frac{\phi_m - 1}{m\B}\dif m\B = \int_{0}^{m''\B}\frac{\phi_m - 1}{m\B}\dif m\B + \int_{m''\B}^{m'\B}\frac{\phi_m - 1}{m\B}\dif m\B \tag{10.6.6} \end{equation} where the integration limit \(m''\B\) is a low molality at which the value of \(\phi_m\) is available and at which \(\g_{\pm}\) can either be measured or estimated from the Debye–Hückel equation. We next rewrite Eq. 10.6.5 with \(m\B'\) replaced with \(m\B''\): \begin{equation} \ln\g_{\pm}(m''\B) = \phi_m (m''\B) - 1 + \int_{0}^{m''\B}\frac{\phi_m - 1}{m\B}\dif m\B \tag{10.6.7} \end{equation} By eliminating the integral with an upper limit of \(m\B''\) from Eqs. 10.6.6 and 10.6.7, we obtain \begin{equation} \int_{0}^{m'\B}\frac{\phi_m - 1}{m\B}\dif m\B = \ln\g_{\pm}(m''\B) - \phi_m (m''\B) + 1 + \int_{m''\B}^{m'\B}\frac{\phi_m - 1}{m\B}\dif m\B \tag{10.6.8} \end{equation} Equation 10.6.5 becomes \begin{equation} \ln\g_{\pm}(m'\B) = \phi_m(m'\B) - \phi_m(m''\B) + \ln\g_{\pm}(m''\B) + \int_{m''\B}^{m'\B}\frac{\phi_m - 1}{m\B}\dif m\B \tag{10.6.9} \end{equation} The integral on the right side of this equation can easily be evaluated by numerical integration. | 10,918 | 1,132 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/06%3A_Putting_the_Second_Law_to_Work/6.02%3A_Combining_the_First_and_Second_Laws_-_Maxwell's_Relations |
Modeling the dependence of the Gibbs and Helmholtz functions behave with varying temperature, pressure, and volume is fundamentally useful. But in order to do that, a little bit more development is necessary. To see the power and utility of these functions, it is useful to combine the First and Second Laws into a single mathematical statement. In order to do that, one notes that since \[dS = \dfrac{dq}{T} \nonumber \] for a reversible change, it follows that \[dq= TdS \nonumber \] And since \[dw = TdS - pdV \nonumber \] for a reversible expansion in which only p-V works is done, it also follows that (since \(dU=dq+dw\)): \[dU = TdS - pdV \nonumber \] This is an extraordinarily powerful result. This differential for \(dU\) can be used to simplify the differentials for \(H\), \(A\), and \(G\). But even more useful are the constraints it places on the variables T, S, p, and V due to the mathematics of exact differentials! The above result suggests that the natural variables of internal energy are \(S\) and \(V\) (or the function can be considered as \(U(S, V)\)). So the total differential (\(dU\)) can be expressed: \[dU = \left( \dfrac{\partial U}{\partial S} \right)_V dS + \left( \dfrac{\partial U}{\partial V} \right)_S dV \nonumber \] Also, by inspection (comparing the two expressions for \(dU\)) it is apparent that: \[\left( \dfrac{\partial U}{\partial S} \right)_V = T \label{eq5A} \] and \[\left( \dfrac{\partial U}{\partial V} \right)_S = -p \label{eq5B} \] But the value doesn’t stop there! Since \(dU\) is an exact differential, the Euler relation must hold that \[ \left[ \dfrac{\partial}{\partial V} \left( \dfrac{\partial U}{\partial S} \right)_V \right]_S= \left[ \dfrac{\partial}{\partial S} \left( \dfrac{\partial U}{\partial V} \right)_S \right]_V \nonumber \] By substituting Equations \ref{eq5A} and \ref{eq5B}, we see that \[ \left[ \dfrac{\partial}{\partial V} \left( T \right)_V \right]_S= \left[ \dfrac{\partial}{\partial S} \left( -p \right)_S \right]_V \nonumber \] or \[ \left( \dfrac{\partial T}{\partial V} \right)_S = - \left( \dfrac{\partial p}{\partial S} \right)_V \nonumber \] This is an example of a . These are very powerful relationship that allows one to substitute partial derivatives when one is more convenient (perhaps it can be expressed entirely in terms of \(\alpha\) and/or \(\kappa_T\) for example.) A similar result can be derived based on the definition of \(H\). \[ H \equiv U +pV \nonumber \] Differentiating (and using the chain rule on \(d(pV)\)) yields \[ dH = dU +pdV + Vdp \nonumber \] Making the substitution using the combined first and second laws (\(dU = TdS – pdV\)) for a reversible change involving on expansion (p-V) work \[ dH = TdS – \cancel{pdV} + \cancel{pdV} + Vdp \nonumber \] This expression can be simplified by canceling the \(pdV\) terms. \[ dH = TdS + Vdp \label{eq2A} \] And much as in the case of internal energy, this suggests that the natural variables of \(H\) are \(S\) and \(p\). Or \[dH = \left( \dfrac{\partial H}{\partial S} \right)_p dS + \left( \dfrac{\partial H}{\partial p} \right)_S dV \label{eq2B} \] Comparing Equations \ref{eq2A} and \ref{eq2B} show that \[\left( \dfrac{\partial H}{\partial S} \right)_p= T \label{eq6A} \] and \[\left( \dfrac{\partial H}{\partial p} \right)_S = V \label{eq6B} \] It is worth noting at this point that both (Equation \ref{eq5A}) \[\left( \dfrac{\partial U}{\partial S} \right)_V \nonumber \] and (Equation \ref{eq6A}) \[\left( \dfrac{\partial H}{\partial S} \right)_p \nonumber \] are equation to \(T\). So they are equation to each other \[\left( \dfrac{\partial U}{\partial S} \right)_V = \left( \dfrac{\partial H}{\partial S} \right)_p \nonumber \] Euler \[ \left[ \dfrac{\partial}{\partial p} \left( \dfrac{\partial H}{\partial S} \right)_p \right]_S= \left[ \dfrac{\partial}{\partial S} \left( \dfrac{\partial H}{\partial p} \right)_S \right]_p \nonumber \] so \[ \left( \dfrac{\partial T}{\partial p} \right)_S = \left( \dfrac{\partial V}{\partial S} \right)_p \nonumber \] This is the Maxwell relation on \(H\). Maxwell relations can also be developed based on \(A\) and \(G\). The results of those derivations are summarized in Table 6.2.1.. The Maxwell relations are extraordinarily useful in deriving the dependence of thermodynamic variables on the state variables of p, T, and V. Show that \[ \left( \dfrac{\partial V}{\partial T} \right)_p = T\dfrac{\alpha}{\kappa_T} - p \nonumber \] Start with the combined first and second laws: \[dU = TdS - pdV \nonumber \] Divide both sides by \(dV\) and constraint to constant \(T\): \[\left.\dfrac{dU}{dV}\right|_{T} = \left.\dfrac{TdS}{dV}\right|_{T} - p \left.\dfrac{dV}{dV} \right|_{T} \nonumber \] Noting that \[\left.\dfrac{dU}{dV}\right|_{T} =\left( \dfrac{\partial U}{\partial V} \right)_T \nonumber \] \[ \left.\dfrac{TdS}{dV}\right|_{T} = \left( \dfrac{\partial S}{\partial V} \right)_T \nonumber \] \[\left.\dfrac{dV}{dV} \right|_{T} = 1 \nonumber \] The result is \[ \left( \dfrac{\partial U}{\partial V} \right)_T = T \left( \dfrac{\partial S}{\partial V} \right)_T -p \nonumber \] Now, employ the Maxwell relation on \(A\) (Table 6.2.1) \[ \left( \dfrac{\partial p}{\partial T} \right)_V = \left( \dfrac{\partial S}{\partial V} \right)_T \nonumber \] to get \[ \left( \dfrac{\partial U}{\partial V} \right)_T = T \left( \dfrac{\partial p}{\partial T} \right)_V -p \nonumber \] and since \[\left( \dfrac{\partial p}{\partial T} \right)_V = \dfrac{\alpha}{\kappa_T} \nonumber \] It is apparent that \[ \left( \dfrac{\partial V}{\partial T} \right)_p = T\dfrac{\alpha}{\kappa_T} - p \nonumber \] : How cool is that? This result was given without proof in Chapter 4, but can now be proven analytically using the Maxwell Relations! | 5,758 | 1,136 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Book3A_Bioinorganic_Chemistry_(Bertini_et_al.)/04%3A_Biological_and_Synthetic_Dioxygen_Carriers/4.18%3A_Structural_Basis_of_Ligand_Affinities_of_Oxygen_Carriers |
The interaction of ligands, such as dioxygen, with metal complexes, such as iron-porphyrinato systems, and the means by which this interaction is characterized, have been covered in broad outline in the previous sections. As noted earlier, the affinities of hemoglobins for carbon monoxide and dioxygen span a wide range (see Table 4.2 and Figure 4.24). In this section the active site is examined in much finer detail than before in order to develop relationships between perturbations in structure and affinity (and hence function)—so called structure-function relationships. The reference point is the somewhat hypothetical situation where the dioxygen binder is in the gas phase and independent of interactions with solvent molecules, solute molecules, and itself, and where dioxygen, carbon monoxide, and other small molecules may bind without steric constraints—in other words, a state where intrinsic affinity is measured. In this section attention is focused exclusively on the hemoglobin family and on iron- and cobalt-porphyrinato systems. In recent years structural data on hemoglobin, myoglobin, and their derivatives have become available with a precision that permits meaningful comparison with the more precisely determined model or synthetic systems. In addition, the various hemoglobins and myoglobins, and especially the naturally occurring mutants of hemoglobin A (human Hb), have provided a sort of poor man's site-directed mutagenesis. Now the techniques of molecular biology permit the site of mutation to be selected, the altered gene to be inserted into , and the mutant protein to be expressed in large (mg) quantities. With the conditions for crystallization of hemoglobins now well-established, we can discover quite rapidly what structural perturbations are caused by the substitution of one amino acid for another, and can relate these to the perturbations in properties, such as cooperativity, dioxygen affinity, and kinetics of ligand binding. The principles enunciated here are applicable generally to hemerythrin and hemocyanin; however, we currently lack the thermodynamic and especially structural data we would like to have for these systems. The O2 affinities in biological carriers span five orders of magnitude, which at room temperature corresponds to a difference in the free energy of oxygen binding \[\delta \Delta G = -RTln(K_{max}/K_{min}) = -RTln(P_{1/2min}/P_{1/2max}) \tag{4.47}\] of about 6.0 kcal/mol. This wide range of O and CO affinities has not yet been paralleled in synthetic systems; the values for O affinity do not exceed those for R-state human hemoglobin. A selection of values from model systems is given in Table 4.5. For the flat-open porphyrin system (Figure 4.23) the dioxygen ostensibly binds in an unconstrained manner, but is actually subject to solvent influences. In order to obtain thermodynamic constants on these "unhindered" systems, one must gather data at several low temperatures and then extrapolate to room temperature, or obtain them from kinetic measurements, K = k /k , at room temperature. For the picket-fence porphyrins, dioxygen binds in a protected pocket that is deep enough to accommodate it and to prevent the dimerization that leads to irreversible oxidation, provided that there is a slight excess of base to ensure full saturation of the coordination sites on the unprotected face of the porphyrin. Thus the picket-fence, the capped, and the bis-pocket porphyrins reversibly bind dioxygen at room temperature with little oxidation over many cycles. This stability facilitated isolation of crystals of a synthetic iron-dioxygen species of the picket-fence porphyrin. The capped porphyrin offers a more highly protected site. The low affinity these latter systems have for dioxygen indicates that the binding cavity is so small that repulsive steric interactions between coordinated dioxygen and the cap are unavoidable. The left-hand side of Figure 4.24 depicts on a logarithmic scale the range of O affinities. Each power of 10 corresponds to around 1.2 kcal/mol at 25 °C. The right-hand side of Figure 4.24 illustrates the range of affinities for CO binding. For many synthetic systems the CO affinities are orders of magnitude greater than in the biological systems that have an O affinity similar to the synthetic; for example, see the entries for the picket-fence porphyrin. Comparison of the left- and right-hand sides of Figure 4.24 reveals that the strongest O binder, hemoglobin , is one of the weakest CO binders. The O affinity of the picket-fence porphyrins is very similar to that of myoglobin, but, as will be detailed shortly, one cannot infer from this that the binding sites are strictly comparable. Indeed, similar affinities have been observed with a non-porphyrin iron complex. Moreover, if the CO affinity of myoglobin paralleled that of the picket-fence porphyrins, some 20 percent of myoglobin (and hemoglobin) would be in the carbonmonoxy form (in contrast to the approximately 3 percent that occurs naturally), a level that could render reading this section while chewing gum physically taxing. a) When available P are from thermodynamic measurements, otherwise from k /k , where solubility of O in toluene is 1.02 x 10 M/Torr and of CO in toluene is 1.05 x 10 ; solubilities in benzene are very similar. b) Some k are calculated from K(O ), k (CO), and M. P (O ) Torr \(\Delta\)H kcal/mol \(\Delta\)S eu k \(\mu\)M s k s P (CO) Torr \(\Delta\)H kcal/mol \(\Delta\)S eu k \(\mu\)M s k s There is a convenient index to summarize the extent to which CO (or O ) binding is discriminated against for a given iron-porphyrin system. M is defined as the ratio of O affinity (as P ) to CO affinity for a particular system and experimental conditions: \[M = \frac{P_{1/2}(O_{2})}{P_{1/2}(CO)} \tag{4.48}\] From Figure 4.24 and from Tables 4.2 and 4.5 the M values calculated may be somewhat arbitrarily divided into three classes: those where M > 2 x 10 (good CO binder); those where 2 x 10 < M < 2 x 10 ; and those where M < 2 x 10 (good O binder). An analogous parameter, N, may be defined to summarize the differences in the O affinity between an iron-porphyrin system and its cobalt analogue: \[M = \frac{P_{1/2}(O_{2}—Co)}{P_{1/2}(O_{2}—Fe)} \ldotp \tag{4.49}\] For the picket-fence porphyrins and for vertebrate hemoglobins N is in the range 10 to 250, whereas for the flat-open porphyrins and for some hemoglobins that lack a distal histidine (e.g., hemoglobin and hemoglobin ), N is at least an order of magnitude larger, indicating for these latter species that the cobalt analogue binds O relatively poorly (see Table 4.6). Note that whereas the O binding of the picket-fence porphyrins is similar to that for myoglobin, the kinetics of the process are very different; the synthetic system is more than an order of magnitude faster in k and k (often also referred to as k and k ). On the other hand, O binding to the pocket porphyrin is similar to that for the biological system. The factors by which ligand affinities are modulated, generally to the benefit of the organism, are subtle and varied, and their elucidation requires the structural information that is currently available only from x-ray diffraction experiments. Figure 4.25 shows the structural features of interest that will be elaborated upon in the next subsections. P (Fe—CO) Torr P (Fe—O ) Torr M P (Fe—O )/P (Fe—CO) P (Co—O ) Torr N P (Co—O )/P (Fe—O ) Whale Mb (E7His→Gly) | 7,508 | 1,137 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Nuclear_Chemistry/Nuclear_Energetics_and_Stability/Nuclear_Magic_Numbers |
Nuclear Stability is a concept that helps to identify the stability of an isotope. The two main factors that determine nuclear stability are the neutron/proton ratio and the total number of nucleons in the nucleus. A isotope is an element that has same atomic number but different atomic mass compared to the periodic table. Every element has a proton, neutron, and electron. The number of protons is equal to the atomic number, and the number of electrons is equal the protons, unless it is an ion. To determine the number of neutrons in an element you subtract the atomic number from the atomic mass of the element. Atomic mass is represented as (\(A\)) and atomic number is represented as (\(Z\)) and neutrons are represented as (\(N\)). \[A=N + Z \label{1}\] The principal factor for determining whether a nucleus is stable is the . Elements with (\(Z<20\)) are lighter and these elements' nuclei and have a ratio of 1:1 and prefer to have the same amount of protons and neutrons. Carbon has three isotopes that scientists commonly used: \( \ce{^12C}\), \( \ce{^13C}\), \( \ce{^14C}\). What is the the number of neutron, protons, total nucleons and \(N:Z\) ratio for the \( \ce{^12C}\) nuclide? For this specific isotope, there are 12 total nucleons (\(A\)). From the periodic table, we can see that \(Z\) for carbon (any of the isotopes) is 6, therefore \(N=A-Z\) (from Equation \ref{1}): \[12-6=6 \nonumber\] The N:P ratio therefore is 6:6 or a 1:1. In fact 99% of all carbon in the earth is this isotope. Identify the number of neutron, protons, total nucleons and N:Z ratio in the \( \ce{^12_8O}\) nuclide? Elements that have atomic numbers from 20 to 83 are heavy elements, therefore the ratio is different. The ratio is 1.5:1, the reason for this difference is because of the repulsive force between protons: the the repulsion force, the neutrons are needed to stabilize the nuclei. Neutrons help to separate the protons from each other in a nucleus so that they do not feel as strong a repulsive force from other. The graph of stable elements is commonly referred to as the . The graph consists of a y-axis labeled neutrons, an x-axis labeled protons, and a nuclei. At the higher end (upper right) of the band of stability lies the radionuclides that decay via alpha decay, below is positron emission or electron capture, above is beta emissions and elements beyond the atomic number of 83 are only unstable radioactive elements. Stable nuclei with atomic numbers up to about 20 have an neutron:proton ratio of about 1:1 (solid line). The deviation from the \(N:Z=1\) line on the belt of stability originates from a non-unity \(N:Z\) ratio necessary for total stability of nuclei. That is, more neutrons are required to stabilize the repulsive forces from a fewer number of protons within a nucleus (i.e., \(N>Z\)). The belt of stability makes it is easy to determine where the alpha decay, beta decay, and positron emission or electron capture occurs. As with all decay pathways, if the daughter nuclides are not on the Belt, then subsequent decay pathways will occur until the daughter nuclei are on the Belt. The was formulated from the observation that atoms with eight valence electrons were especially stable (and common). A similar situation applies to nuclei regarding the number of neutron and proton numbers that generate stable (non-radioactive) isotopes. These "magic numbers" are natural occurrences in isotopes that are particularly stable. Table 1 list of numbers of protons and neutrons; isotopes that have these numbers occurring in either the proton or neutron are stable. In some cases there the isotopes can consist of magic numbers for both protons and neutrons; these would be called . The double numbers only occur for isotopes that are heavier, because the repulsion of the forces between the protons. The magic numbers are: Also, there is the concept that isotopes consisting a combination of even-even, even-odd, odd-even, and odd-odd are all stable. There are more nuclides that have a combination of even-even than odd-odd. Just like there exist , many isotopes with no magic numbers of nucleons are stable. Although rare, four stable odd-odd nuclides exist: \(\ce{^2_1H}\), \(\ce{^{6}_3Li}\), \(\ce{^{10}_5B}\), \(\ce{^{14}_7N}\) Here is a simple chart that can help you decide is an element is likely stable. Using the above chart state if this isotope is alpha-emitter, stable, or unstable: If the isotope is located above the band of stability what type of radioactivity is it? what if it was below? Based off the belt of stability: Carbon is stable Carbon is stable Name one of the isotopes that consist of odd-odd combination in the nuclei? Hydrogen-2, Lithium-6, Boron-10, nitrogen-14 | 4,757 | 1,138 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/02%3A_Reaction_Rates/2.06%3A_Reaction_Rates-_A_Microscopic_View |
The Learning Objective of this Module is to determine the individual steps of a simple reaction. One of the major reasons for studying chemical kinetics is to use measurements of the macroscopic properties of a system, such as the rate of change in the concentration of reactants or products with time, to discover the sequence of events that occur at the molecular level during a reaction. This molecular description is the mechanism of the reaction; it describes how individual atoms, ions, or molecules interact to form particular products. The stepwise changes are collectively called the reaction mechanism. In an internal combustion engine, for example, isooctane reacts with oxygen to give carbon dioxide and water: \[2C_8H_{18 \; (l)} + 25O_2(g) \rightarrow 16CO_{2\, (g)} + 18H_2O_{(g)} \tag{14.34} \] For this reaction to occur in a single step, 25 dioxygen molecules and 2 isooctane molecules would have to collide simultaneously and be converted to 34 molecules of product, which is very unlikely. It is more likely that a complex series of reactions takes place in a stepwise fashion. Each individual reaction, which is called an , involves one, two, or (rarely) three atoms, molecules, or ions. The overall sequence of elementary reactions is the mechanism of the reaction. The sum of the individual steps, or elementary reactions, in the mechanism must give the balanced chemical equation for the overall reaction. To demonstrate how the analysis of elementary reactions helps us determine the overall reaction mechanism, we will examine the much simpler reaction of carbon monoxide with nitrogen dioxide. \[NO_{2\, (g)} + CO(g) \rightarrow NO_{(g)} + CO_{2\, (g)} \tag{14.35} \] From the balanced chemical equation, one might expect the reaction to occur via a collision of one molecule of NO with a molecule of CO that results in the transfer of an oxygen atom from nitrogen to carbon. The experimentally determined rate law for the reaction, however, is as follows: \[rate = k[NO_2]^2 \tag{14.36} \] The fact that the reaction is second order in [NO ] and independent of [CO] tells us that it does not occur by the simple collision model outlined previously. If it did, its predicted rate law would be \(rate = k[NO_2,CO]\). The following two-step mechanism is consistent with the rate law if step 1 is much slower than step 2: According to this mechanism, the overall reaction occurs in two steps, or elementary reactions. Summing steps 1 and 2 and canceling on both sides of the equation gives the overall balanced chemical equation for the reaction. The \(NO_3\) molecule is an in the reaction, a species that does not appear in the balanced chemical equation for the overall reaction. It is formed as a product of the first step but is consumed in the second step. The sum of the elementary reactions in a reaction mechanism must give the overall balanced chemical equation of the reaction. The of an elementary reaction is the number of molecules that collide during that step in the mechanism. If there is only a single reactant molecule in an elementary reaction, that step is designated as unimolecular; if there are two reactant molecules, it is bimolecular; and if there are three reactant molecules (a relatively rare situation), it is termolecular. Elementary reactions that involve the simultaneous collision of more than three molecules are highly improbable and have never been observed experimentally. (To understand why, try to make three or more marbles or pool balls collide with one another simultaneously!) Writing the rate law for an elementary reaction is straightforward because we know how many molecules must collide simultaneously for the elementary reaction to occur; hence the order of the elementary reaction is the same as its molecularity ( ). In contrast, the rate law for the reaction cannot be determined from the balanced chemical equation for the overall reaction. The general rate law for a unimolecular elementary reaction (A → products) is rate = [A]. For bimolecular reactions, the reaction rate depends on the number of collisions per unit time, which is proportional to the product of the concentrations of the reactants, as shown in . For a bimolecular elementary reaction of the form A + B → products, the general rate law is rate = [A,B]. The Basis for Writing Rate Laws of Elementary Reactions. Note the important difference between writing rate laws for elementary reactions and the balanced chemical equation of the overall reaction. Because the balanced chemical equation does not necessarily reveal the individual elementary reactions by which the reaction occurs, we cannot obtain the rate law for a reaction from the overall balanced chemical equation alone. In fact, it is the rate law for the slowest overall reaction, which is the same as the rate law for the slowest step in the reaction mechanism, the , that must give the experimentally determined rate law for the overall reaction.This statement is true if one step is substantially slower than all the others, typically by a factor of 10 or more. If two or more slow steps have comparable rates, the experimentally determined rate laws can become complex. Our discussion is limited to reactions in which one step can be identified as being substantially slower than any other. The reason for this is that any process that occurs through a sequence of steps can take place no faster than the slowest step in the sequence. In an automotive assembly line, for example, a component cannot be used faster than it is produced. Similarly, blood pressure is regulated by the flow of blood through the smallest passages, the capillaries. Because movement through capillaries constitutes the rate-determining step in blood flow, blood pressure can be regulated by medications that cause the capillaries to contract or dilate. A chemical reaction that occurs via a series of elementary reactions can take place no faster than the slowest step in the series of reactions. Look at the rate laws for each elementary reaction in our example as well as for the overall reaction. The experimentally determined rate law for the reaction of \(NO_2\) with \(CO\) is the same as the predicted rate law for step 1. This tells us that the first elementary reaction is the rate-determining step, so \(k\) for the overall reaction must equal \(k_1\). That is, NO is formed slowly in step 1, but once it is formed, it reacts very rapidly with CO in step 2. Sometimes chemists are able to propose two or more mechanisms that are consistent with the available data. If a proposed mechanism predicts the wrong experimental rate law, however, the mechanism must be incorrect. Example 12 In an alternative mechanism for the reaction of NO with CO, N O appears as an intermediate. Write the rate law for each elementary reaction. Is this mechanism consistent with the experimentally determined rate law (rate = [NO ] )? elementary reactions rate law for each elementary reaction and overall rate law The rate law for step 1 is rate = [NO ] ; for step 2, it is rate = [N O ,CO]. If step 1 is slow (and therefore the rate-determining step), then the overall rate law for the reaction will be the same: rate = [NO ] . This is the same as the experimentally determined rate law. Hence this mechanism, with N O as an intermediate, and the one described previously, with NO as an intermediate, are kinetically indistinguishable. In this case, further experiments are needed to distinguish between them. For example, the researcher could try to detect the proposed intermediates, NO and N O , directly. Iodine monochloride (ICl) reacts with H as follows: \[2ICl(l) + H_2(g) \rightarrow 2HCl(g) + I_2(s) \nonumber \] The experimentally determined rate law is rate = [ICl,H ]. Write a two-step mechanism for this reaction using only bimolecular elementary reactions and show that it is consistent with the experimental rate law. (Hint: HI is an intermediate.) This mechanism is consistent with the experimental rate law if the first step is the rate-determining step. Is the the reaction between \(NO\) and \(H_2\) occurs via a three-step process: Write the rate law for each elementary reaction, write the balanced chemical equation for the overall reaction, and identify the rate-determining step. Is the rate law for the rate-determining step consistent with the experimentally derived rate law for the overall reaction: rate = \(k[NO]^2[H^2]\)? : Many reaction mechanisms, like those discussed so far, consist of only two or three elementary reactions. Many others consist of long series of elementary reactions. The most common mechanisms are , in which one or more elementary reactions that contain a highly reactive species repeat again and again during the reaction process. Chain reactions occur in fuel combustion, explosions, the formation of many polymers, and the tissue changes associated with aging. They are also important in the chemistry of the atmosphere. Chain reactions are described as having three stages. The first is initiation, a step that produces one or more reactive intermediates. Often these intermediates are , species that have an unpaired valence electron. In the second stage, propagation, reactive intermediates are continuously consumed and regenerated while products are formed. Intermediates are also consumed but not regenerated in the final stage of a chain reaction, termination, usually by forming stable products. Let us look at the reaction of methane with chlorine at elevated temperatures (400°C–450°C), a chain reaction used in industry to manufacture methyl chloride (CH Cl), dichloromethane (CH Cl ), chloroform (CHCl ), and carbon tetrachloride (CCl ): CH + Cl → CH Cl + HCl CH Cl + Cl → CH Cl + HCl CH Cl + Cl → CHCl + HCl CHCl + Cl → CCl + HCl Direct chlorination generally produces a mixture of all four carbon-containing products, which must then be separated by distillation. In our discussion, we will examine only the chain reactions that lead to the preparation of CH Cl. In the initiation stage of this reaction, the relatively weak Cl–Cl bond cleaves at temperatures of about 400°C to produce chlorine atoms (Cl·): Cl → 2Cl· During propagation, a chlorine atom removes a hydrogen atom from a methane molecule to give HCl and CH ·, the methyl radical: Cl· + CH → CH · + HCl The methyl radical then reacts with a chlorine molecule to form methyl chloride and another chlorine atom, Cl·: CH · + Cl → CH Cl + Cl· The sum of the propagation reactions is the same as the overall balanced chemical equation for the reaction: Without a chain-terminating reaction, propagation reactions would continue until either the methane or the chlorine was consumed. Because radical species react rapidly with almost anything, however, including each other, they eventually form neutral compounds, thus terminating the chain reaction in any of three ways: CH · + Cl· → CH Cl CH · + CH · → H CCH Cl· + Cl· → Cl Here is the overall chain reaction, with the desired product (CH Cl) in bold: The chain reactions responsible for explosions generally have an additional feature: the existence of one or more chain branching steps, in which one radical reacts to produce two or more radicals, each of which can then go on to start a new chain reaction. Repetition of the branching step has a cascade effect such that a single initiation step generates large numbers of chain reactions. The result is a very rapid reaction or an explosion. The reaction of H and O , used to propel rockets, is an example of a chain branching reaction: Termination reactions occur when the extraordinarily reactive H· or OH· radicals react with a third species. The complexity of a chain reaction makes it unfeasible to write a rate law for the overall reaction. A reaction mechanism is the microscopic path by which reactants are transformed into products. Each step is an elementary reaction. Species that are formed in one step and consumed in another are intermediates. Each elementary reaction can be described in terms of its molecularity, the number of molecules that collide in that step. The slowest step in a reaction mechanism is the rate-determining step. Chain reactions consist of three kinds of reactions: initiation, propagation, and termination. Intermediates in chain reactions are often radicals, species that have an unpaired valence electron. where M is any molecule, including cyclopropane. Only those cyclopropane molecules with sufficient energy (denoted with an asterisk) can rearrange to propylene. Which step determines the rate constant of the overall reaction? Assume that the rates of the forward and reverse reactions in the first equation are equal. What is the relationship between the relative magnitudes of and if these reactions have the rate law Δ[F]/Δ = [A,B,E]/[C]? How does the magnitude of compare to that of ? Under what conditions would you expect the rate law to be Δ[F]/Δ = ′[A,B]? Assume that the rates of the forward and reverse reactions in the first equation are equal. \(\textrm{rate}=k_2\dfrac{k_1[\mathrm{O_2NNH_2}]}{k_{-1}[\mathrm{H^+}]}=k\dfrac{[\mathrm{O_2NNH_2}]}{[\mathrm{H^+}]}\) | 13,216 | 1,139 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Chemiluminescence/4%3A_Instrumentation/4.11%3A_Chemiluminescence_Detection_in_Capillary_Electrophoresis |
has outstanding resolving power for extremely small samples, but this poses a challenge for detectors. Compared with other candidates, chemiluminescence detection has the advantages of being highly sensitive and requiring inexpensive equipment of simple design. In addition it is not affected by the high voltage used in the separation system, a particular problem for electrochemical detection, which is also highly sensitive. Ultra-violet absorbance detectors also have low cost and are widely used, but narrow capillaries make it difficult to arrange for long enough optical path lengths. Laser induced fluorescence has high sensitivity, but the equipment is costly and pre- or post-column derivatization of nonfluorescent analytes is necessary . As with HPLC, there is an inherent problem of compatibility between the conditions needed for separation and those needed for chemiluminescence. Additionally, there is a potential problem with the stability of chemiluminescence reagents. Both of these are addressed by using the post-column mode rather than pre-column. Post-column interfaces are devices for mixing the eluent with the chemiluminescence reagents and for this purpose designs may make use of merging flow, coaxial flow or reservoir mixing. Interfaces may also be classified as off-, on- or end-column, depending on the site of detection and on whether this is isolated from the high voltage supply used for capillary electrophoresis. The simplest interface, off-column and merging flow, did not find widespread application. Buffer flowed from a reservoir through the separation capillary and merged with the reagent at a four-way connector at the end of the column. The outlet arm carried the mixture to the chemiluminescence reaction coil and flow cell adjacent to a photomultiplier, while the fourth arm connected through a semi-permeable membrane to a second buffer reservoir (containing the ground electrode) immediately downstream of the merging point. This arrangement isolates the high voltage from the detection zone. In contrast, an on-column coaxial flow interface has proved to be effective for a large number of applications . Figure D11.1 shows the detector is located at the capillary outlet tip, so detection is “on-column”. The ground electrode is located in the effluent reservoir so that detection takes place within the high voltage zone. The separation capillary outlet is inserted coaxially into the reaction tube, giving rise to minimum turbulence and reproducible mixing. | 2,529 | 1,140 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Physical_Equilibria/Raoults_Law_and_Ideal_Mixtures_of_Liquids |
This page deals with Raoult's Law and how it applies to mixtures of two volatile liquids. It covers cases where the two liquids are entirely miscible in all proportions to give a single liquid - NOT those where one liquid floats on top of the other (immiscible liquids). The page explains what is meant by an ideal mixture and looks at how the phase diagram for such a mixture is built up and used. An ideal mixture is one which obeys Raoult's Law, but I want to look at the characteristics of an ideal mixture before actually stating Raoult's Law. The page will flow better if I do it this way around. There is actually no such thing as an ideal mixture! However, some liquid mixtures get fairly close to being ideal. These are mixtures of two very closely similar substances. Commonly quoted examples include: In a pure liquid, some of the more energetic molecules have enough energy to overcome the intermolecular attractions and escape from the surface to form a vapor. The smaller the intermolecular forces, the more molecules will be able to escape at any particular temperature. If you have a second liquid, the same thing is true. At any particular temperature a certain proportion of the molecules will have enough energy to leave the surface. In an ideal mixture of these two liquids, the tendency of the two different sorts of molecules to escape is unchanged. You might think that the diagram shows only half as many of each molecule escaping - but the proportion of each escaping is still the same. The diagram is for a 50/50 mixture of the two liquids. That means that there are only half as many of each sort of molecule on the surface as in the pure liquids. If the proportion of each escaping stays the same, obviously only half as many will escape in any given time. If the red molecules still have the same tendency to escape as before, that must mean that the intermolecular forces between two red molecules must be exactly the same as the intermolecular forces between a red and a blue molecule. If the forces were any different, the tendency to escape would change. Exactly the same thing is true of the forces between two blue molecules and the forces between a blue and a red. They must also be the same otherwise the blue ones would have a different tendency to escape than before. If you follow the logic of this through, the intermolecular attractions between two red molecules, two blue molecules or a red and a blue molecule must all be exactly the same if the mixture is to be ideal. This is why mixtures like hexane and heptane get close to ideal behavior. They are similarly sized molecules and so have similarly sized van der Waals attractions between them. However, they obviously are not identical - and so although they get close to being ideal, they are not actually ideal. For the purposes of this topic, getting close to ideal is good enough! When you make any mixture of liquids, you have to break the existing intermolecular attractions (which needs energy), and then remake new ones (which releases energy). If all these attractions are the same, there won't be any heat either evolved or absorbed. That means that an ideal mixture of two liquids will have zero enthalpy change of mixing. If the temperature rises or falls when you mix the two liquids, then the mixture is not ideal. The partial vapor pressure of a component in a mixture is equal to the vapor pressure of the pure component at that temperature multiplied by its mole fraction in the mixture. \[ P_A = \chi_A P^o_A \label{1}\] \[ P_B = \chi_B P^o_B \label{2}\] \[ \chi_A = \dfrac{\text{moles of A}}{\text{total number of moles}} \label{4}\] \[ P_{methanol} = \dfrac{2}{3} \times 81\; kPa\] \[ = 54\; kPa\] \[ P_{ethanol} = \dfrac{1}{3} \times 45\; kPa\] \[ = 15\; kPa\] \[ P_{total} = 54\; kPa + 15 \; kPa = 69 kPa\] Now we'll do the same thing for B - except that we will plot it on the same set of axes. The mole fraction of B falls as A increases so the line will slope down rather than up. As the mole fraction of B falls, its vapor pressure will fall at the same rate. Notice that the vapor pressure of pure B is higher than that of pure A. That means that molecules must break away more easily from the surface of B than of A. B is the more volatile liquid. To get the total vapor pressure of the mixture, you need to add the values for A and B together at each composition. The net effect of that is to give you a straight line as shown in the next diagram. The relationship between boiling point and vapor pressure What do these two aspects imply about the boiling points of the two liquids? There are two ways of looking at the above question: Either: Or: For two liquids at the same temperature, the liquid with the higher vapor pressure is the one with the lower boiling point. To remind you - we've just ended up with this vapor pressure / composition diagram: We're going to convert this into a boiling point / composition diagram. We'll start with the boiling points of pure A and B. Since B has the higher vapor pressure, it will have the lower boiling point. If that is not obvious to you, go back and read the last section again! For mixtures of A and B, you might perhaps have expected that their boiling points would form a straight line joining the two points we've already got. Not so! In fact, it turns out to be a curve. To make this diagram really useful (and finally get to the phase diagram we've been heading towards), we are going to add another line. This second line will show the composition of the vapor over the top of any particular boiling liquid. If you boil a liquid mixture, you would expect to find that the more volatile substance escapes to form a vapor more easily than the less volatile one. That means that in the case we've been talking about, you would expect to find a higher proportion of B (the more volatile component) in the vapor than in the liquid. You can discover this composition by condensing the vapor and analyzing it. That would give you a point on the diagram. The diagram just shows what happens if you boil a particular mixture of A and B. Notice that the vapor over the top of the boiling liquid has a composition which is much richer in B - the more volatile component. If you repeat this exercise with liquid mixtures of lots of different compositions, you can plot a second curve - a vapor composition line. This is now our final phase diagram. The diagram is used in exactly the same way as it was built up. If you boil a liquid mixture, you can find out the temperature it boils at, and the composition of the vapor over the boiling liquid. For example, in the next diagram, if you boil a liquid mixture C , it will boil at a temperature T and the vapor over the top of the boiling liquid will have the composition C . All you have to do is to use the liquid composition curve to find the boiling point of the liquid, and then look at what the vapor composition would be at that temperature. Notice again that the vapor is much richer in the more volatile component B than the original liquid mixture was. Suppose that you collected and condensed the vapor over the top of the boiling liquid and reboiled it. You would now be boiling a new liquid which had a composition C . That would boil at a new temperature T , and the vapor over the top of it would have a composition C . You can see that we now have a vapor which is getting quite close to being pure B. If you keep on doing this (condensing the vapor, and then reboiling the liquid produced) you will eventually get pure B. This is obvious the basis for fractional distillation. However, doing it like this would be incredibly tedious, and unless you could arrange to produce and condense huge amounts of vapor over the top of the boiling liquid, the amount of B which you would get at the end would be very small. Real fractionating columns (whether in the lab or in industry) automate this condensing and reboiling process. How these work will be explored on . Jim Clark ( ) | 8,027 | 1,141 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/18%3A_Partition_Functions_and_Ideal_Gases/18.06%3A_Rotational_Partition_Functions_of_Diatomic_Gases |
The rotational energy levels of a are given by: \[E_\text{rot}(J) = \tilde{B} J (J + 1) \label{Eq0} \] where: \[\tilde{B} = \dfrac{h}{8 π^2 I c} \nonumber \] Here, \(\tilde{B}\) is the rotational constant expressed in cm . The rotational energy levels are given by: \[ E_j = \dfrac{J(J+1) h^2}{8 \pi I} \nonumber \] where \(I\) is the moment of inertia of the molecule given by \(μr^2\) for a diatomic, and \(μ\) is the reduced mass, and \(r\) the bond length (assuming rigid rotor approximation). The energies can be also expressed in terms of the rotational temperature, \(Θ_\text{rot}\), defined as: \[ Θ_\text{rot} = \dfrac{r^2}{8 \pi^2 I k} \label{3.12} \] The interpretation of \(θ_\text{rot}\) is as an estimate of the temperature at which thermal energy (\(\approx kT\)) is comparable to the spacing between rotational energy levels. At about this temperature the population of excited rotational levels becomes important. See Table 1. In the summation for the expression for rotational partition function (\(q_\text{rot}\)), Equation \(\ref{3.13}\), we can do an explicit summation: \[q_\text{rot} = \sum_{j=0} (2J+1) e^{-E_J/ k T} \label{3.13} \] if only a finite number of terms contribute. The factor \((2J+1)\) for each term in the expansion accounts for the degeneracy of a rotational state \(J\). For each allowed energy \(E_J\) from Equation \(\ref{Eq0}\) there are \((2 J + 1)\) eigenstates. The Boltzmann factor must be multiplied by \((2J+ 1)\) to properly account for the degeneracy these states: \[(2J+ 1)e^{ -E_J / k T}\] If the rotational energy levels are lying very close to one another, we can integrate similar to what we did for \(q_{trans}\) previously to get: \[q_\text{rot} = \int _0 ^{\infty} (2J+1) R^{-\tilde{B} J (J+1) / k T} dJ \nonumber \] This integration can easily be done by substituting \(x = J ( J+1)\) and \(dx = (2J + 1) dJ\): \[q_\text{rot} = \dfrac{kT}{\tilde{B}} \label{3.15} \] For a homonuclear diatomic molecule, rotating the molecule by 180° brings the molecule into a configuration which is from the original configuration. This leads to an overcounting of the accessible states. To correct for this, we divide the partition function by \(σ\), which is called the and is equal to the distinct number of ways by which a molecule can be brought into identical configurations by rotations. The rotational partition function becomes: \[q_\text{rot}= \dfrac{kT}{\tilde{B} σ} \label{3.16} \] or commonly expressed in terms of \( Θ_\text{rot}\): \[q_\text{rot}= \dfrac{T}{ Θ_\text{rot} σ} \label{3.17} \] What is the rotational partition function of \(H_2\) at 300 K? The value of \(\tilde{B}\) for \(H_2\) is 60.864 cm . The value of \(k T\) in cm can be obtained by dividing it by \(hc\), i.e., which is \(kT/hc = 209.7\; cm^{-1}\) at 300 K. \(σ = 2\) for a homonuclear molecule. Therefore from Equation \(\ref{3.16}\), \[\begin{align*} q_\text{rot} &= \dfrac{kT}{\tilde{B} σ} \\[4pt] &= \dfrac{209.7 \;cm^{-1} }{(2) (60.864\; cm^{-1})} \\[4pt] &= 1.723 \end{align*} \nonumber \] Since the rotational frequency of \(H_2\) is quite large, only the first few rotational states are accessible at 300 K | 3,169 | 1,142 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Physical_Equilibria/Immiscible_Liquids_and_Steam_Distillation |
This page looks at systems containing two immiscible liquids. Immiscible liquids are those which won't mix to give a single phase. Oil and water are examples of immiscible liquids - one floats on top of the other. It explains the background to steam distillation and looks at a simple way of carrying this out. Obviously if you have two immiscible liquids in a closed flask and keep everything still, the vapor pressure you measure will simply be the vapor pressure of the one which is floating on top. There is no way that the bottom liquid can turn to vapor. The top one is sealing it in. For the purposes of the rest of this topic, we always assume that the mixture is being stirred or agitated in some way so that the two liquids are broken up into drops. At any one time there will be drops of both liquids on the surface. That means that both of them contribute to the overall vapor pressure of the mixture. Assuming that the mixture is being agitated, then both of the liquids will be in equilibrium with their vapors. The total vapor pressure is then simply the sum of the individual vapor pressures: \[\text{Total vapor pressure} = p_A^o + p_B^o\] where \(p^o\) refers to the saturated vapor pressure of the pure liquid. Notice that this is independent of the amount of each sort of liquid present. All you need is enough of each so that both can exist in equilibrium with their vapor. For example, phenylamine and water can be treated as if they were completely immiscible. (That isn't actually true, but they are near enough immiscible to be usable as an example.) At 98°C, the saturated vapor pressures of the two pure liquids are: The total vapor pressure of an agitated mixture would just be the sum of these - in other words, 101.37 kPa Liquids boil when their vapor pressure becomes equal to the external pressure. Normal atmospheric pressure is 101.325 kPa. Compare that with the figure we have just got for the total vapor pressure of a mixture of water and phenylamine at 98°C. Its total vapor pressure is fractionally higher than the normal external pressure. This means that such a mixture would boil at a temperature just a shade less than 98°C - in other words lower than the boiling point of pure water (100°C) and much lower than the phenylamine (184°C). Exactly the same sort of argument could be applied to any other mixture of immiscible liquids. I've chosen the phenylamine-water mixture just because I happen to have some figures for it! Agitated mixtures of immiscible liquids will boil at a temperature lower than the boiling point of either of the pure liquids. Their combined vapor pressures are bound to reach the external pressure before the vapor pressure of either of the individual components get there. Notice that in the presence of water, phenylamine (or any other liquid which is immiscible with water) boils well below its normal boiling point. This has an important advantage in separating molecules like this from mixtures. Normal distillation of these liquids would need quite high temperatures. On the whole these tend to be big molecules we are talking about. Quite a lot of molecules of this sort will be broken up by heating at high temperatures. Distilling them in the presence of water avoids this by keeping the temperature low. That's what steam distillation achieves. We will carry on with the phenylamine example for now. During the preparation of phenylamine it is produced as a part of a mixture containing a solution of all sorts of inorganic compounds. It is removed from this by steam distillation. Steam is blown through the mixture and the water and phenylamine turn to vapor. This vapor can be condensed and collected. The steam can be generated by heating water in another flask (or something similar). As the hot steam passes through the mixture it condenses, releasing heat. This will be enough to boil the mixture of water and phenylamine at 98°C provided the volume of the mixture isn't too great. For large volumes, it is better to heat the flask as well to avoid having to condense too much steam and increase the volume of liquid in the flask too much. The condensed vapor will consist of both water and phenylamine. If these were truly immiscible, they would form two layers which could be separated using a separating funnel. In fact, the phenylamine has a slight solubility in water and various other techniques have to be used in this particular case to get the maximum yield of phenylamine. These aren't relevant to this topic. Steam distillation can be used to extract some natural products - for example, to extract eucalyptus oil from eucalyptus, citrus oils from lemon or orange peel, and to extract oils used in perfumes from various plant materials. Jim Clark ( ) | 4,759 | 1,143 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/02%3A_Properties_of_Gases/2.08%3A_Molecular_Collisions_and_the_Mean_Free_Path |
Collision theory is a theory proposed independently by Max Trautz in 1916 and William Lewis in 1918, that qualitatively explains how chemical reactions occur and why reaction rates differ for different reactions. The collision theory states that when suitable particles of the reactant hit each other, only a certain percentage of the collisions cause any noticeable or significant chemical change; these successful changes are called successful collisions. The successful collisions have enough energy, also known as activation energy, at the moment of impact to break the preexisting bonds and form all new bonds. This results in the products of the reaction. Increasing the concentration of the reactant particles or raising the temperature, thus bringing about more collisions and therefore many more successful collisions, increases the rate of reaction. Consider two particles \(A\) and \(B\) in a system. The kinetic energy of these two particles is \[K_{AB} = \dfrac{\textbf{p}_A^2}{2m_A} + \dfrac{\textbf{p}_B^2}{2m_B} \label{6}\] Let us change to center-of-mass \(\left( \textbf{P} \right)\) and relative \(\left( \textbf{p} \right)\) momenta, which are given by \[\textbf{P} = \textbf{p}_A + \textbf{p}_B, \: \: \: \textbf{p} = \dfrac{m_B \textbf{p}_a - m_A \textbf{p}_B}{M} \label{7}\] where \(M = m_A + m_B\) is the total mass of the two particles. Substituting this into the kinetic energy, we find \[K_{AB} = \dfrac{\textbf{p}_A^2}{2m_A} + \dfrac{\textbf{p}_B^2}{2m_B} = \dfrac{\textbf{P}^2}{2M} + \dfrac{\textbf{p}^2}{2 \mu} \label{8}\] where \[\mu = \dfrac{m_A m_B}{M} \label{9}\] is called the of the two particles. Note that the kinetic energy separates into a sum of a center-of-mass term and a relative term. Now the relative position is \(\textbf{r} = \textbf{r}_A - \textbf{r}_B\) so that the relative velocity is \(\dot{\textbf{r}} = \dot{\textbf{r}}_A - \dot{\textbf{r}}_B\) or \(\textbf{v} = \textbf{v}_A - \textbf{v}_B\). Thus, if the two particles are approaching each other such that \(\textbf{v}_A = - \textbf{v}_B\), then \(\textbf{v} = 2 \textbf{v}_A\). However, by equipartitioning the relative kinetic energy, being mass independent, is \[\left< \dfrac{\textbf{p}^2}{2 \mu} \right> = \dfrac{3}{2} k_B T \label{10}\] which is called the Consider two molecules in a system. The probability that they will collide increases with the effective “size” of each particle. However, the size measure that is relevant is the apparent cross-section area of each particle. For simplicity, suppose the particles are spherical, which is not a bad approximation for small molecules. If we are looking at a sphere, what we perceive as the size of the sphere is the cross section area of a great circle. Recall that each spherical particle has an associated “collision sphere” that just encloses two particles at closest contact, i.e., at the moment of a collision, and that this sphere is a radius \(d\), where \(d\) is the diameter of each spherical particle (see lecture 5). The cross-section of this collision sphere represents an effective cross section for each particle inside which a collision is imminent. The cross-section of the collision sphere is the area of a great circle, which is \(\pi d^2\). We denote this apparent cross section area \(\sigma\). Thus, for spherical particles \(A\) and \(B\) with diameters \(d_A\) and \(d_B\), the individual cross sections are \[\sigma_A = \pi d_A^2, \: \: \: \sigma_B = \pi d_B^2 \label{11}\] The , \(\sigma_{AB}\) is determined by an effective diameter \(d_{AB}\) characteristic of both particles. The collision probability increases of both particles have large diameters and decreases if one of them has a smaller diameter than the other. Hence, a simple measure sensitive to this is the arithmetic average \[d_{AB} = \dfrac{1}{2} \left( d_A + d_B \right) \label{12}\] and the resulting collision cross section becomes \[\begin{align} \sigma_{AB} &= \pi d_{AB}^2 \\ &= \pi \left( \dfrac{d_A + d_B}{2} \right)^2 \\ &= \dfrac{\pi}{4} \left( d_A^2 + 2d_A d_B + d_B^2 \right) \\ &= \dfrac{1}{4} \left( \sigma_A + 2 \sqrt{\sigma_A \sigma_B} + \sigma_B \right) \\ &= \dfrac{1}{2} \left[ \left( \dfrac{\sigma_A + \sigma_B}{2} \right) + \sqrt{\sigma_A \sigma_B} \right] \end{align} \label{13}\] which, interestingly, is an average of the two types of averages of the two individual cross sections, the arithmetic and geometric averages! Consider a system of particles with individual cross sections \(\sigma\). A particle of cross section \(\sigma\) that moves a distance \(l\) in a time \(\Delta t\) will sweep out a cylindrical volume (ignoring the spherical caps) of volume \(\sigma l\) (Figure \(\Page {1}\)). If the system has a number density \(\rho\), then the number of collisions that will occur is \[N_{\text{coll}} = \rho \sigma l \label{14}\] We define the average collision rate as \(N_{\text{coll}}/ \Delta t\), i.e., \[\gamma = \dfrac{N_{\text{coll}}}{\Delta t} = \dfrac{\rho \sigma l}{\Delta t} = \rho \sigma \langle | \textbf{v} | \rangle \label{15}\] where \(\langle | \textbf{v} | \rangle\) is the average relative speed. If all of the particles are of the same type (say, type \(A\)), then performing the average over a Maxwell-Boltzmann speed distribution gives \[\langle | \textbf{v} | \rangle = \sqrt{\dfrac{8 k_B T}{\pi \mu}} \label{16}\] where \(\mu = m_A/2\) is the reduced mass. The average speed of a particle is \[\langle | \textbf{v}_A | \rangle = \sqrt{\dfrac{8 k_B T}{\pi m_A}} \label{17}\] so that \[\langle | \textbf{v} | \rangle = \sqrt{2} \langle | \textbf{v}_A | \rangle \label{18}\] The mean free path is defined as the distance a particle will travel, on average, before experiencing a collision event. This is defined as the product of the speed of a particle and the time between collisions. The former is \(\langle | \textbf{v} | \rangle/ \sqrt{2}\), while the latter is \(1/\gamma\). Hence, we have \[\lambda = \dfrac{\langle | \textbf{v} |\rangle}{\sqrt{2} \rho \sigma \langle | \textbf{v} | \rangle} = \dfrac{1}{\sqrt{2} \rho \sigma} \label{19}\] In any system, a particle undergoing frequent collisions will have the direction of its motion changed with each collision and will trace out a path that appears to be random. In fact, if we treat the process as statistical, then, we are, in fact, treating each collision event as a random event, and the particle will change its direction at random times in random ways! Such a path might appear as shown in Figure \(\Page {2. Such a path is often referred to as a . In order to analyze such paths, let us consider a random walk in one dimension. We’ll assume that the particle move a mean-free path length \(\lambda\) between collisions and that each collision changes the direction of the particles motion, which in one dimension, means that the particle moves either to the right or to the left after each event. This can be mapped onto a metaphoric “coin toss” that can come up heads “H” or tails “T”, with “H” causing motion to the right, and “T” causing motion to the left. \[\mathcal{P}(\textbf{r}) = P(x) \: P(y) \: P(z) = \dfrac{1}{(4 \pi D t) ^{3/2}} \: e^{-\left( x^2 + y^2 + z^2 \right)/4Dt} \label{35}\] \[P(r, t) = \dfrac{4 \pi}{(4 \pi D t)^{3/2}} \: E^{-r^2/4Dt} \label{36}\] Graham’s law is an empirical relationship that states that the ratio of the rates of diffusion or effusion of two gases is the square root of the inverse ratio of their molar masses. The relationship is based on the postulate that all gases at the same temperature have the same average kinetic energy. We can write the expression for the average kinetic energy of two gases with different molar masses: \[KE=\dfrac{1}{2}\dfrac{M_{\rm A}}{N_A}v_{\rm rms,A}^2=\dfrac{1}{2}\dfrac{M_{\rm B}}{N_A}v_{\rm rms,B}^2\label{6.8.2}\] Multiplying both sides by 2 and rearranging give \[\dfrac{v_{\rm rms, B}^2}{v_{\rm rms,A}^2}=\dfrac{M_{\rm A}}{M_{\rm B}}\label{6.8.3}\] Taking the square root of both sides gives \[\dfrac{v_{\rm rms, B}}{v_{\rm rms,A}}=\sqrt{\dfrac{M_{\rm A}}{M_{\rm B}}}\label{6.8.4}\] Thus the rate at which a molecule, or a mole of molecules, diffuses or effuses is directly related to the speed at which it moves. Equation \(\ref{6.8.4}\) shows that Graham’s law is a direct consequence of the fact that gaseous molecules at the same temperature have the same average kinetic energy. The lightest gases have a wider distribution of speeds and the highest average speeds. Molecules with lower masses have a wider distribution of speeds and a higher average speed. Gas molecules do not diffuse nearly as rapidly as their very high speeds might suggest. If molecules actually moved through a room at hundreds of miles per hour, we would detect odors faster than we hear sound. Instead, it can take several minutes for us to detect an aroma because molecules are traveling in a medium with other gas molecules. Because gas molecules collide as often as 10 times per second, changing direction and speed with each collision, they do not diffuse across a room in a straight line. The the gas, the the mean free path. Calculate the rms speed of a sample of -2-butene (C H ) at 20°C. compound and temperature rms speed Calculate the molar mass of cis-2-butene. Be certain that all quantities are expressed in the appropriate units and then use Equation \(\ref{6.8.5}\) to calculate the rms speed of the gas. To use Equation \(\ref{6.8.4}\), we need to calculate the molar mass of -2-butene and make sure that each quantity is expressed in the appropriate units. Butene is C H , so its molar mass is 56.11 g/mol. Thus \[u_{\rm rms}=\sqrt{\dfrac{3RT}{M}}=\rm\sqrt{\dfrac{3\times8.3145\;\dfrac{J}{K\cdot mol}\times(20+273)\;K}{56.11\times10^{-3}\;kg}}=361\;m/s\] Calculate the rms speed of a sample of radon gas at 23°C. 1.82 × 10 m/s (about 410 mi/h) ( ) | 9,836 | 1,144 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/19%3A_The_First_Law_of_Thermodynamics/19.11%3A_Enthalpy_Changes_for_Chemical_Equations_are_Additive |
As enthalpy and energy are state functions we should expect of U and H when we study chemical reactions. This additivity is expressed in . The additivity has important consequences and the law finds wide spread application in the prediction of heats of reaction. By this mechanism it is often possible to calculate the heat of a reaction even if this reaction is hard to carry out. E.g. we could burn both graphite and diamond and measure the heats of combustion for both. The difference would give us the heat of the transformation reaction from graphite to diamond. The enthalpy is for the . That means that if we write: \[2C(s) + O_2(g) \rightarrow 2CO(g) \nonumber \] with Δ H = -221 kJ ( : -110.5 kJ) Because H is a state function the reverse reaction has the same enthalpy but with opposite sign \[ 2CO(g) \rightarrow 2C(s) + O_2(g) \nonumber \] with Δ H = +221 kJ It is quite possible that you cannot really do a certain reaction in practice. For many reactions we can arrive at enthalpy values by doing some bookkeeping. For example, we can calculate the enthalpy for the reaction of PCl with chlorine if we know the two reactions that the elements phosphorous and chlorine can undergo. You do have to make sure you balance your equations properly! | 1,281 | 1,146 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Fundamentals_of_Chemical_Bonding/Bond_Order_and_Lengths |
Bond order is the number of chemical bonds between a pair of atoms and indicates the stability of a bond. For example, in diatomic nitrogen, N≡N, the bond order is 3; in acetylene, H−C≡C−H, the carbon-carbon bond order is also 3, and the C−H bond order is 1. Bond order and bond length indicate the type and strength of between atoms. Bond order and length are inversely proportional to each other: when bond order is increased, bond length is decreased. Chemistry deals with the way in which subatomic particles bond together to form atoms. Chemistry also focuses on the way in which atoms bond together to form molecules. In the atomic structure, electrons surround the atomic nucleus in regions called orbitals. Each orbital shell can hold a certain number of electrons. When the nearest orbital shell is full, new electrons start to gather in the next orbital shell out from the nucleus, and continue until that shell is also full. The collection of electrons continues in ever widening orbital shells as larger atoms have more electrons than smaller atoms. When two atoms bond to form a molecule, their electrons bond them together by mixing into openings in each others' orbital shells. As with the collection of electrons by the atom, the formation of bonds by the molecule starts at the nearest available orbital shell opening and expand outward. Bond order is the number of bonding pairs of electrons between two atoms. In a between two atoms, a single bond has a bond order of one, a double bond has a bond order of two, a triple bond has a bond order of three, and so on. To determine the bond order between two covalently bonded atoms, follow these steps: If the bond order is zero, the molecule cannot form. The higher bond orders indicate greater stability for the new molecule. In molecules that have , the bond order does not need to be an integer. Determine the bond order for cyanide, CN . 1) Draw the Lewis structure. 2) Determine the type of bond between the two atoms. Because there are 3 dashes, the bond is a triple bond. A triple bond corresponds to a bond order of 3. Determine the bond order for hydrogen gas, H . 1) Draw the Lewis structure. 2) Determine the type of bond between the two atoms. There is only one pair of shared electrons (or dash), indicating is a single bond, with a bond order of 1. If there are more than two atoms in the molecule, follow these steps to determine the bond order: Determine the bond order for nitrate, \(NO_3^-\). 1) Draw the Lewis structure. 2) Count the total number of bonds. 4 The total number of bonds is 4. 3) Count the number of bond groups between individual atoms. 3 The number of bond groups between individual atoms is 3. 4) Divide the number of bonds between individual atoms by the total number of bonds. \[\dfrac{4}{3}= 1.33 \] The bond order is 1.33 Determine the bond order for nitronium ion: \(NO_2^+\). 1) Draw the Lewis Structure. 2) Count the total number of bonds. 4 The total number of bonds is 4. 3) Count the number of bond groups between individual atoms. 2 The number of bond groups between atoms is 2. 4) Divide the bond groups between individual atoms by the total number of bonds. \[\frac{4}{2} = 2\] The bond order is 2. A high bond order indicates more attraction between electrons. A higher bond order also means that the atoms are held together more tightly. With a lower bond order, there is less attraction between electrons and this causes the atoms to be held together more loosely. Bond order also indicates the stability of the bond. The higher the bond order, the more electrons holding the atoms together, and therefore the greater the stability. Bond order increases across a period and decreases down a group. Bond length is defined as the distance between the centers of two covalently bonded atoms. The length of the bond is determined by the number of bonded electrons (the bond order). The higher the bond order, the stronger the pull between the two atoms and the shorter the bond length. Generally, the length of the bond between two atoms is approximately the sum of the covalent radii of the two atoms. Bond length is reported in picometers. Therefore, bond length increases in the following order: triple bond < double bond < single bond. To find the bond length, follow these steps: Determine the carbon-to-chlorine bond length in CCl . Using , a C single bond has a length of 75 picometers and that a Cl single bond has a length of 99 picometers. When added together, the bond length of a C-Cl bond is approximately 174 picometers. Determine the carbon-oxygen bond length in CO . Using , we see that a C double bond has a length of 67 picometers and that an O double bond has a length of 57 picometers. When added together, the bond length of a C=O bond is approximately 124 picometers. Because the bond length is proportional to the , the bond length trends in the periodic table follow the same trends as atomic radii: bond length decreases across a period and increases down a group. 1. First, write the Lewis structure for \(O_2\). There is a double bond between the two oxygen atoms; therefore, the bond order of the molecule is 2. 2. The Lewis structure for NO is given below: To find the bond order of this molecule, take the average of the bond orders. N=O has a bond order of two, and both N-O bonds have a bond order of one. Adding these together and dividing by the number of bonds (3) reveals that the bond order of nitrate is 1.33. 3. To find the carbon-nitrogen bond length in HCN, draw the Lewis structure of HCN. The bond between carbon and nitrogen is a triple bond, and a triple bond between carbon and nitrogen has a bond length of approximately 60 + 54 =114 pm. 4. From the Lewis structures for CO and CO, there is a double bond between the carbon and oxygen in CO and a triple bond between the carbon and oxygen in CO. Referring to the table above, a double bond between carbon and oxygen has a bond length of approximately 67 + 57 = 124 pm and a triple bond between carbon and oxygen has a bond length of approximately 60 + 53 =113 pm. Therefore, the bond length is greater in CO . Another method makes use of the fact that the more electron bonds between the atoms, the tighter the electrons are pulling the atoms together. Therefore, the bond length is greater in CO . 5. To find the nitrogen-to-fluorine bond length in NF , draw the Lewis structure. The bond between fluorine and nitrogen is a single bond. From the table above, a single bond between fluorine and nitrogen has a bond length of approximately 64 + 71 =135 pm. | 6,594 | 1,147 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/4_f-Block_Elements/The_Actinides/Chemistry_of_Uranium |
Uranium was named for the planet Uranus and discovered in 1789 by Klaproth. Isolation came in 1841 by Péligot and the radioactivity of the element was noted by Becquerel in 1896. The pure metal is heavy, silver and lustrous. It tarnishes slowly in air \[U(s) + O_2(g) \rightarrow UO_2 (s)\] which at ambient temperatures, \(UO_2\) will gradually convert to \(U_3O_8\). Uranium also reacts with boiling water. Most of the naturally occurring uranium is the isotope U-238. This form of uranium is not fissionable, i.e., it cannot be used in atomic weapons or power plants. A much smaller percentage of naturally occurring uranium is the isotope U-235, which is fissionable. The process of "enriching" uranium to increase the proportion of U-235 in a sample is expensive and tedious but necessary to produce fuel that is usable in power plants and material for weapons. The U-238 is not completely useless, however, as it can be "bred" into Pu-239 by bombardment with slow neutrons. This is how weapons-grade plutonium is produced. The use of Pu-239 in power plants is a controversial subject due to its toxicity and the fear of diversion by terrorist groups. | 1,176 | 1,148 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Physical_Equilibria/Fractional_Distillation_of_Ideal_Mixtures |
This page explains how the (both in the lab and industrially) of an ideal mixture of liquids relates to their phase diagram. On the last page, we looked at how the phase diagram for an ideal mixture of two liquids was built up. I want to start by looking again at material from the last part of that page. The next diagram is new - a modified version of diagrams from the previous page. The secret of getting the more volatile component from a mixture of liquids is obviously to do a succession of boiling-condensing-reboiling operations. It is not quite so obvious how you get a sample of pure A out of this. A typical lab fractional distillation would look like this: The fractionating column is packed with glass beads (or something similar) to give the maximum possible surface area for vapor to condense on. Some fractionating columns have spikes of glass sticking out from the sides which serve the same purpose. If you sketch this, make sure that you do not completely seal the apparatus. There has to be a vent in the system otherwise the pressure build-up when you heat it will blow the apparatus apart. In some cases, where you are collecting a liquid with a very low boiling point, you may need to surround the collecting flask with a beaker of cold water or ice. The mixture is heated at such a rate that the thermometer is at the temperature of the boiling point of the more volatile component. Notice that the thermometer bulb is placed exactly at the outlet from the fractionating column. Suppose you boil a mixture with composition C . The vapor over the top of the boiling liquid will be richer in the more volatile component, and will have the composition C . That vapor now starts to travel up the fractionating column. Eventually it will reach a height in the column where the temperature is low enough that it will condense to give a liquid. The composition of that liquid will, of course, still be C . So what happens to that liquid now? It will start to trickle down the column where it will meet new hot vapor rising. That will cause the already condensed vapor to reboil. Some of the liquid of composition C will boil to give a vapor of composition C . Let's concentrate first on that new vapor and think about the unvaporized part of the liquid afterwards. This new vapor will again move further up the fractionating column until it gets to a temperature where it can condense. Then the whole process repeats itself. Each time the vapor condenses to a liquid, this liquid will start to trickle back down the column where it will be reboiled by up-coming hot vapor. Each time this happens the new vapor will be richer in the more volatile component. The aim is to balance the temperature of the column so that by the time vapor reaches the top after huge numbers of condensing and reboiling operations, it consists only of the more volatile component - in this case, B. Whether or not this is possible depends on the difference between the boiling points of the two liquids. The closer they are together, the longer the column has to be. So what about the liquid left behind at each reboiling? Obviously, if the vapor is richer in the more volatile component, the liquid left behind must be getting richer in the other one. As the condensed liquid trickles down the column constantly being reboiled by up-coming vapor, each reboiling makes it richer and richer in the less volatile component - in this case, A. By the time the liquid drips back into the flask, it will be very rich in A indeed. So, over time, as B passes out of the top of the column into the condenser, the liquid in the flask will become richer in A. If you are very, very careful over temperature control, eventually you will have separated the mixture into B in the collecting flask and A in the original flask. Finally, what is the point of the packing in the column? To make the boiling-condensing-reboiling process as effective as possible, it has to happen over and over again. By having a lot of surface area inside the column, you aim to have the maximum possible contact between the liquid trickling down and the hot vapor rising. If you didn't have the packing, the liquid would all be on the sides of the condenser, while most of the vapor would be going up the middle and never come into contact with it. There is no difference whatsoever in the theory involved. All that is different is what the fractionating column looks like. The diagram shows a simplified cross-section through a small part of a typical column. The column contains a number of trays that the liquid collects on as the vapor condenses. The up-coming hot vapor is forced through the liquid on the trays by passing through a number of bubble caps. This produces the maximum possible contact between the vapor and liquid. This all makes the boiling-condensing-reboiling process as efficient as possible. The overflow pipes are simply a controlled way of letting liquid trickle down the column. If you have a mixture of lots of liquids to separate (such as in petroleum fractionation), it is possible to tap off the liquids from some of the trays rather than just collecting what comes out of the top of the column. That leads to simpler mixtures such as gasoline, kerosene and so on. | 5,272 | 1,150 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Le_Chateliers_Principle/Le_Chatelier's_Principle_Fundamentals |
if a is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change to reestablish an equilibrium. If This page covers changes to the position of equilibrium due to such changes and discusses briefly why catalysts have no effect on the equilibrium position. An action that changes the temperature, pressure, or concentrations of reactants in a system at equilibrium stimulates a response that partially offsets the change while a new equilibrium condition is established (2). Hence, Le principle states that any change to a system at equilibrium will adjust to compensate for that change. In 1884 the French chemist and engineer Henry-Louis Le â proposed one of the central concepts of chemical equilibria, which describes what happens to a system when something briefly removes it from a state of equilibrium. It is important to understand that â 's principle is only a useful guide to identify what happens when the conditions are changed in a reaction in dynamic equilibrium; it does give reasons for the changes at the molecular level (e.g., timescale of change and underlying reaction mechanism). Le â 's principle states that if the system is changed in a way that increases the concentration of one of the reacting species, it must favor the reaction in which that species is consumed. In other words, if there is an increase in products, the reaction quotient, \(Q_c\), is increased, making it greater than the equilibrium constant, \(K_c\). Consider an equilibrium established between four substances, \(A\), \(B\), \(C\), and \(D\): \[ A + 2B \rightleftharpoons C + D\] What happens if conditions are altered by increasing the concentration of A? According to â , the position of equilibrium will move in such a way as to counteract the change. In this case, the equilibrium position will move so that the concentration of A decreases again by reacting it with B to form more C and D. The equilibrium moves to the right (indicated by the green arrow below). In a practical sense, this is a useful way of converting the maximum possible amount of B into C and D; this is advantageous if, for example, B is a relatively expensive material whereas A is cheap and plentiful. In the opposite case in which the concentration of A is decreased, according to â the position of equilibrium will move so that the concentration of A increases again. More C and D will react to replace the A that has been removed. The position of equilibrium moves to the left. This is essentially what happens if one of the products is removed as soon as it is formed. If, for example, C is removed in this way, the position of equilibrium would move to the right to replace it. If it is continually removed, the equilibrium position shifts further and further to the right, effectively creating a one-way, irreversible reaction. This only applies to reactions involving gases, although not necessarily all species in the reaction need to be in the gas phase. A general homogeneous gaseous reaction is given below: \[ A(g) + 2B(g) \rightleftharpoons C(g) + D(g)\] According to â , if the pressure is increased, the position of equilibrium will move so that the pressure is reduced again. Pressure is caused by gas molecules hitting the sides of their container. The more molecules in the container, the higher the pressure will be. The system can reduce the pressure by reacting in such a way as to produce fewer molecules. In this case, there are three moles on the left-hand side of the equation, but only two on the right. By forming more C and D, the system causes the pressure to reduce. Increasing the pressure on a gas reaction shifts the position of equilibrium towards the side with fewer moles of gas molecules. \[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \] If this mixture is transferred from a 1.5 L flask to a 5 L flask, in which direction does a net change occur to return to equilibrium? Because the volume is increased (and therefore the pressure reduced), the shift occurs in the direction that produces more moles of gas. To restore equilibrium the shift needs to occur to the left, in the direction of the reverse reaction. The equilibrium will move in such a way that the pressure increases again. It can do that by producing more gaseous molecules. In this case, the position of equilibrium will move towards the left-hand side of the reaction. In this case, increasing the pressure has no effect on the position of the equilibrium. Because there are equal numbers of molecules on both sides, the equilibrium cannot move in any way that will reduce the pressure again. Again, this is not a rigorous explanation of why the position of equilibrium moves in the ways described. A mathematical treatment of the explanation can be found on this page. Three ways to change the pressure of an equilibrium mixture are: 1. Add or remove a gaseous reactant or product, 2. Add an inert gas to the constant-volume reaction mixture, or 3. Change the volume of the system. To understand how temperature changes affect equilibrium conditions, the sign of the reaction enthalpy must be known. Assume that the forward reaction is exothermic (heat is evolved): In this reaction, 250 kJ is evolved (indicated by the negative sign) when 1 mole of A reacts completely with 2 moles of B. For reversible reactions, the enthalpy value is always given as if the reaction was one-way in the forward direction. The back reaction (the conversion of C and D into A and B) would be endothermic, absorbing the same amount of heat. The main effect of temperature on equilibrium is in changing the value of the equilibrium constant. 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. If the temperature is increased, then the position of equilibrium will move so that the temperature is reduced again. Suppose the system is in equilibrium at 300°C, and the temperature is increased 500°C. To cool down, it needs to absorb the extra heat added. In the case, the back reaction is that in which heat is absorbed. The position of equilibrium therefore moves to the left. The new equilibrium mixture contains more A and B, and less C and D. If the goal is to maximize the amounts of C and D formed, increasing the temperature on a reversible reaction in which the forward reaction is exothermic is a poor approach. The equilibrium will move in such a way that the temperature increases again. Suppose the system is in equilibrium at 500°C and the temperature is reduced to 400°C. The reaction will tend to heat itself up again to return to the original temperature by favoring the exothermic reaction. The position of equilibrium will move to the right with more \(A\) and \(B\) converted into \(C\) and \(D\) at the lower temperature: Consider the formation of water \[O_2 + 2H_2 \rightleftharpoons 2H_2O\;\;\; \Delta{H}= -125.7\, kJ\] Adding a catalyst makes absolutely to the position of equilibrium, and â 's principle does not apply. This is because a catalyst speeds up the forward and back reaction to the same extent and adding a catalyst does not affect the relative rates of the two reactions, it cannot affect the position of equilibrium. However, catalysts have some application to equilibrium systems. For a dynamic equilibrium to be set up, the rates of the forward reaction and the back reaction must be equal. This does not happen instantly and for very slow reactions, it may take years! A catalyst speeds up the rate at which a reaction reaches dynamic equilibrium. You might try imagining how long it would take to establish a dynamic equilibrium if you took the visual model on the introductory page and reduced the chances of the colors changing by a factor of 1000 - from 3 in 6 to 3 in 6000 and from 1 in 6 to 1 in 6000. Starting with blue squares, by the end of the time taken for the examples on that page, you would most probably still have entirely blue squares. Eventually, though, you would end up with the same sort of patterns as before - containing 25% blue and 75% orange squares.
What will happen to the equilibrium when more 2SO (g) is added to the following system? \[2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3 (g) \]
Adding more reactants shifts the equilibrium in the direction of the products; therefore, the equilibrium shifts to the right.
Overall, the concentration of \(2SO_2\) from initial equilibrium to final equilibrium will increase because only a portion of the added amount of \(2SO_2\) will be consumed.
The concentration of \(O_2\) will decrease because as the equilibrium is reestablished, \(O_2\) is consumed with the \(2SO_2\) to create more \(2SO_3\). The concentration of \(2SO_3\) will be greater because none of it is lost and more is being generated.
What will happen to the equilibrium when the volume of the system is decreased? \[2SO_{2(g)} + O_{2 (g)} \rightleftharpoons 2SO_{3 (g)}\]
Decreasing the volume leads to an increase in pressure which will cause the equilibrium to shift towards the side with fewer moles. In this example there are 3 moles on the reactant side and 2 moles on the product side, so the new equilibrium will shift towards the products (to the right).
What will happen to the equilibrium when the temperature of the system is decreased? \[N_{2(g)} + O_{2 (g)} \rightleftharpoons 2NO_{(g)} \;\;\;\; \Delta{H} = 180.5\; kJ\]
Because \(\Delta{H}\) is positive, the reaction is endothermic in the forward direction. Removing heat from the system forces the equilibrium to shift towards the exothermic reaction, so the reverse reaction will occur and more reactants will be produced. Jim Clark ( ) | 10,195 | 1,152 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Nuclear_Chemistry/Radioactivity/Nuclear_Decay_Pathways |
Nuclear reactions that transform atomic nuclei alter their identity and spontaneously emit radiation via processes of radioactive decay. In 1889, Ernest Rutherford recognized and named two modes of radioactive decay, showing the occurrence of both processes in a decaying sample of natural uranium and its daughters. Rutherford named these types of radiation based on their penetrating power: heavier alpha and lighter beta radiation. Gamma rays, a third type of radiation, were discovered by P. Villard in 1900 but weren't recognized as electromagnetic radiation until 1914. Since gamma radiation is only the discharge of a high-energy photon from an over-excited nucleus, it does not change the identity of the atom from which it originates and therefore will not be discussed in depth here. Because nuclear reactions involve the breaking of very powerful intra nuclear bonds, massive amounts of energy can be released. At such high energy levels, the matter can be converted directly to energy according to Einstein's famous Mass-Energy relationship . The sum of mass and energy are conserved in nuclear decay. The of any spontaneous reaction must be negative according to thermodynamics (ΔG < 0), and is essentially equal to the energy change of nuclear reactions because ΔE is so massive. Therefore, a nuclear reaction will occur spontaneously when: \[ΔE = Δmc^2 < 0\] \(ΔE < 0\) or \(Δm < 0\) When the mass of the products of a nuclear reaction weigh less than the reactants, the difference in mass has been converted to energy. There are three types of nuclear reactions that are classified as beta decay processes. Beta decay processes have been observed in 97% of all known unstable nuclides and are thus the most common mechanism for radioactive decay by far. The first type (here referred to as ) is also called Negatron Emission because a negatively charged beta particle is emitted, whereas the second type ( ) emits a positively charged beta particle. In , an orbital electron is captured by the nucleus and absorbed in the reaction. All these modes of decay represent changes of one in the atomic number Z of the parent nucleus but no change in the mass number A. Alpha decay is different because both the atomic and mass number of the parent nucleus decrease. In this article, the term beta decay will refer to the first process described in which a true beta particle is the product of the nuclear reaction. Nuclides can be radioactive and undergo nuclear decay for many reasons. Beta decay can occur in nuclei that are rich in neutrons - that is - the nuclide contains more neutrons than stable isotopes of the same element. These "proton deficient" nuclides can sometimes be identified simply by noticing that their mass number A (the sum of neutrons and protons in the nucleus) is significantly more than twice that of the atomic number Z (number of protons in nucleus). In order to regain some stability, such a nucleus can decay by converting one of its extra neutrons into a proton, emitting an electron and an antineutrino( ). The high energy electron emitted in this reaction is called a and is represented by \( _{-1}^{0}\textrm{e}^{-} \) in nuclear equations. Lighter atoms (Z < 60) are the most likely to undergo beta decay. The decay of a neutron to a proton, a beta particle, and an antineutrino (\(\bar{\nu}\)) is \[ \ce{_{0}^{1}n^0 \rightarrow _{0}^{1}p^+ + _{-1}^{0}e^-}+ \bar{\nu} \] Some examples of are \[ \ce{_{2}^{6}He \rightarrow _{3}^{6}Li + _{-1}^{0}e^-} +\bar{\nu} \] \[ \ce{_{11}^{24}Na \rightarrow _{12}^{24}Mg + _{-1}^{0}e^-} + \bar{\nu} \] In order for beta decay to occur spontaneously according to Δm < 0, the mass of the parent (not atom) must have a mass greater than the sum of the masses of the daughter nucleus and the beta particle: [ Z] > [ (Z+1)] + [ e ]
(Parent nucleus) > (Daughter nucleus) + (electron) The mass of the antineutrino is almost zero and can therefore be neglected. The equation above can be reached easily from any beta decay reaction, however, it is not useful because mass spectrometers measure the mass of rather than just their nuclei. To make the equation useful, we must make these nuclei into neutral atoms by adding the mass of electrons to each side of the equation. The parent nucleus then becomes the neutral atom [ Z] plus the mass of one electron, while the daughter nucleus and the beta particle on the right side of the equation become the neutral atom [ (Z+1)] plus the mass of the beta particle. The extra electron on the left cancels the mass of the beta particle on the right, leaving the inequality [ Z] > [ (Z+1)]
(Parent atom) > (Daughter atom) The change in mass then equals Δm = [ (Z+1)] - [ Z] The energy released in this reaction is carried away as kinetic energy by the beta particle and antineutrino, with an insignificant of energy causing recoil in the daughter nucleus. The beta particle can carry anywhere from all to none of this energy, therefore the maximum kinetic energy of a beta particle in any instance of beta decay is . Nuclides that are imbalanced in their ratio of protons to neutrons undergo decay to correct the imbalance. Nuclei that are rich in protons relative to their number of neutrons can decay by conversion of a proton to a neutron, emitting a (\(^0_1e^+\)) and a neutrino ( . Positrons are the antiparticles of electrons, therefore a positron has the same mass as an electron but with the opposite (positive) charge. In positron emission, the atomic number Z by 1 while the mass number A remains the same. Some examples of are \[\ce{^8_5B \rightarrow ^8_4Be + ^0_1e^{+}} + \nu_e\] \[\ce{^{50}_{25}Mg \rightarrow ^{50}_{24}Cr + ^0_1e^+} + \nu_e\] Positron emission is only one of the two types of decay that tends to happen in "neutron deficient" nuclides, therefore it is very important to establish the correct mass change criterion. Positron emission occurs spontaneously when [ Z] > [ (Z-1)] + [ e ]
(Parent nucleus) > (Daughter nucleus) + (positron) In order to rewrite this inequality in terms of the masses of neutral atoms, we add the mass of electrons to both sides of the equation, giving the mass of a neutral [ Z] atom on the left and the mass of a neutral [ (Z-1)] atom, plus an extra electron, (since only Z-1 electrons are needed to make the neutral atom), and a positron on the right. Because positrons and electrons have equal mass, the inequality can be written as [ Z] > [ (Z-1)] + 2 [ e ]
(Parent atom) > (Daughter atom) + (2 electrons) The change in mass for positron emission decay is Δm = [ (Z)] - [ Z] - 2 [ e ] As with beta decay, the kinetic energy is split between the emitted particles - in this case the positron and neutrino. As mentioned before, there are two ways in which neutron-deficient / proton-rich nuclei can decay. When the mass change \(Δm < 0\) yet is insufficient to cause spontaneous positron emission, a neutron can form by an alternate process known as electron capture. An outside electron is pulled inside the nucleus and combined with a proton to make a neutron, emitting only a neutrino. \[ \ce{^1_1p + ^0_{-1}e^{-} → ^1_0n + \nu }\] Some examples of are \[\ce{^{231}_{92}U + ^0_{-1}e^{-} → ^{231}_{91}Pa + \nu }\] \[\ce{ ^{81}{36}Kr + ^0_{-1}e^- → ^{81}_{35}Br + \nu }\] Electron capture happens most often in the heavier neutron-deficient elements where the mass change is smallest and positron emission isn't always possible. For \(Δm < 0\), the following inequality applies: [ Z] + [ e ] > [ (Z-1)]
(Parent nucleus) + (electron) > (Daughter nucleus) Adding \(Z\) electrons to each side of the inequality changes it to its useful form in which the captured electron on the left cancels out the extra electron on the right [ Z] > [ (Z-1)]
(Parent atom) > (Daughter atom) The change in mass then equals Δm = [ (Z-1)] - [ Z] When the loss of mass in a nuclear reaction is greater than zero, but less than 2 [ e ], the process cannot occur by positron emission and is spontaneous for electron capture. The other three processes of nuclear decay involve the formation of a neutron or a proton inside the nucleus to correct an existing imbalance. In alpha decay, unstable, heavy nuclei (typically \(Z > 83\)) reduce their mass number \(A\) by 4 and their atomic number \(Z\) by 2 with the emission of a helium nuclei (\(\ce{^4_2He^{2+}}\)), known as an . Some examples of are \[ \ce{^{222}_{88}Ra} \rightarrow \ce{^{218}_{86}Rn + ^4_2He^{2+}}\] \[ \ce{^{233}_{92}U} \rightarrow \ce{^{229}_{90}Th + ^4_2He^{2+}}\] As with beta decay and electron capture, Δm must only be less than zero for spontaneous alpha decay to occur. Since the number of total protons on each side of the reaction does not change, equal numbers of electrons are added to each side to make neutral atoms. Therefore, the mass of the parent atom must simply be greater than the sum of the masses of its daughter atom and the helium atom. [ Z] > [ (Z-2)] + [ He ] The change in mass then equals Δm = [ (Z)] - [ (Z-2)] - [ He ] The energy released in an alpha decay reaction is mostly carried away by the lighter helium, with a small amount of energy manifesting itself in the recoil of the much heavier daughter nucleus. Alpha decay is a form of spontaneous fission, a reaction in which a massive nuclei can lower its mass and atomic number by splitting. Other heavy unstable elements undergo fission reactions in which they split into nuclei of about equal size. Proton-deficient or neutron-deficient nuclei undergo nuclear decay reactions that serve to correct unbalanced neutron/proton ratios. Proton-deficient nuclei undergo - emitting a beta particle (electron) and an antineutrino to convert a neutron to a proton - thus raising the elements atomic number Z by one. Neutron-deficient nuclei can undergo or (depending on the mass change), either of which synthesizes a neutron - emitting a positron and a neutrino or absorbing an electron and emitting a neutrino respectively - thus lowering Z by one. Nuclei with Z > 83 which are unstable and too massive will correct by , emitting an alpha particle (helium nucleus) and decreasing both mass and atomic number. Very proton-deficient or neutron-deficient nuclei can also simply eject an excess particle directly from the nucleus. These types of decay are called and . These processes are summarized in the table below. | 10,405 | 1,153 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Atomic_Theory/Dalton's_Atomic_Theory/Proust's_Law_of_Constant_Proportion |
The Law of Constant Composition, discovered by Joseph Proust, is also known as the Law of Definite Proportions. It is different from the Law of Multiple Proportions although both stem from Lavoisier's Law of . The French chemist Joseph Proust stated this law the following way: "A chemical compound always contains the same elements combined together in the same proportion by mass." Joseph Proust was a French Chemist best known for his analytical abilities. He was once recommended for a job as a chemistry professor at Segovia's Royal Artillery School by none other than Antoine Lavoisier! His experiments with inorganic binary compounds - mostly sulfates, sulfides, and metallic oxides - led him to formulate the Law of Constant Composition. The law was first published in a paper on iron oxides in 1794. Proust's law was attacked by the respected French chemist Claude-Louis Berthollet who disagreed that chemical combination was restricted to definite saturation proportions. The confusion was caused by the definition of chemical combination; Berthollet classified solutions as chemical combinations while Proust was careful to distinguish between these and true binary compounds. The conflict lasted until John Dalton, an English chemist, came out with an Atomic Theory that favored Proust's law. Swedish chemist Jons Jacob Berzelius established the relationship between Proust's law and Dalton's theory in 1811. For example, pure water obtained from different sources such as a river, a well, a spring, the sea, etc., always contains hydrogen and oxygen together in the ratio of 1:8 by mass. Similarly, carbon dioxide (CO ) can be obtained by different methods such as, Each sample of CO contains carbon and oxygen in a 3:8 ratio. When 1.375 g of cupric oxide is reduced on heating in a current of hydrogen, the weight of copper remaining 1.098 g. In another experiment, 1.179 g of copper is dissolved in nitric acid and resulting copper nitrate converted into cupric oxide by ignition. The weight of cupric oxide formed is 1.476 g. Show that these results illustrate the law of constant proportion. \[ \text{Percentage of oxygen in CuO} = \dfrac{(0.277)(100\%)}{1.375} = 20.15\% \nonumber\] \[ \text{Percentage of oxygen in CuO} = \dfrac{(0.297)(100\%)}{1.476} = 20.12\% \nonumber\] Percentage of oxygen is approximately (within significant figures) the same in both the above cases. So the law of constant composition is illustrated. | 2,464 | 1,155 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/The_Four_Laws_of_Thermodynamics/0th_Law_of_Thermodynamics |
The Zeroth Law of Thermodynamics states that if two systems are in thermodynamic equilibrium with a third system, the two original systems are in thermal equilibrium with each other. Basically, if system A is in thermal equilibrium with system C and system B is also in thermal equilibrium with system C, system A and system B are in thermal equilibrium with each other. Essentially, two systems that are in thermodynamic equilibrium will not exchange any heat. Systems in thermodynamic equilibrium will have the same temperature. 1 kg of water at 10º C is added to 10 kg of water at 50º C. What is the temperature of the water when it reaches thermal equilibrium? 46.36º C | 691 | 1,156 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Phase_Transitions/Boiling |
Boiling is the process by which a liquid turns into a vapor when it is heated to its boiling point. The change from a liquid phase to a gaseous phase occurs when the vapor pressure of the liquid is equal to the atmospheric pressure exerted on the liquid. Boiling is a physical change and molecules are not chemically altered during the process A liquid will begin to boil when Atmospheric Pressure = Vapor Pressure of Liquid When a liquid boils, what is inside the bubbles? The bubbles in a boiling liquid are made up of molecules of the liquid which have gained enough energy to change to the gaseous phase. Describe the formation of bubbles in a boiling liquid (see video for answer). Based on the heating curve below, when will the temperature of \(H_2O\) exceed 100ºC (in an open system)? The temperature of \(H_2O\) will only exceed 100 ºC once it has entirely changed to the gaseous phase. As long as there is liquid the temperature will remain constant. Based on the atmospheric pressure, predict the boiling point for water at the following locations. Remember that water boils at 100ºC at sea level on earth. Assume constant temperature. Since water boils at 100ºC, water would boil quickly on Mars (actual value us about 10ºC). The boiling point on Mt. Everest would be closer to water (actual value about 70ºC). On Venus water would boil well over 100ºC. The molecules leaving a liquid through evaporation create an upward pressure as they collide with air molecules. This upward push is called the . Different substances have different vapor pressures and therefore different boiling points. This is due to differing between molecules. , NBCT, Ph.D. (Gaithersburg High School) | 1,709 | 1,159 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Energies_and_Potentials/00%3A_Front_Matter/02%3A_InfoPage |
Instructors can adopt existing LibreTexts texts or Remix them to quickly build course-specific resources to meet the needs of their students. Unlike traditional textbooks, LibreTexts’ web based origins allow powerful integration of advanced features and new access texts to improve postsecondary education at all levels of higher learning by developing an Open Access Resource environment. The project currently consists of 14 independently operating and interconnected libraries that are constantly being integrated. and are supported by the Department of Education Open Textbook Pilot 1525057, and 1413739. Unless otherwise noted, LibreTexts content is licensed by . Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not ). and are supported by the Department of Education Open Textbook Pilot Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. . . | 1,070 | 1,160 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/4_f-Block_Elements/The_Actinides/Chemistry_of_Thorium |
Thorium was discovered in 1828 by the Norwegian mineralogist Morten Thrane Esmark and was identified by the Swedish chemist Jons Jakob Berzelius. It was named after Thor, the Norse god of thunder. It is a gray, radioactive metal which is fairly abundant in the earth's crust (more than twice as much as tin) and is the first of the so-called "actinide" series which ends with lawrencium (element 103). The long half-life of the principal isotope, Th-232, (about 10 years) insures that there will be plenty for quite some time to come! The metal is fairly soft and malleable but darkens slowly in air due to oxidation. It reacts slowly with water at room temperature. Applications of thorium include some special magnesium alloys and photosensors. The oxide is used in high-quality lenses. An isotope of thorium can be "bred" into uranium-234 by bombardment with slow neutrons. The U-234 is a fissile form of uranium and can be used in power plants. | 969 | 1,162 |
https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Environmental_Toxicology_(van_Gestel_et_al.)/05%3A_Population_Community_and_Ecosystem_Ecotoxicology/5.02%3A_Population_ecotoxicology_in_laboratory_settings |
: Michiel Kraak : Nico van den Brink and Matthias Liess You should be able to · motivate the importance of studying ecotoxicology at the population level. · name the properties of populations, unique to this level of biological organisation. · explain the implications of age and developmental stage specific sensitivities for population responses to toxicant exposure. Population ecotoxicology, density, age structure, population growth rate The motivation to study ecotoxicological effects at the population level is that generally the targets of environmental protection are indeed populations, communities and ecosystems. Additionally, several phenomena are unique to this level, including age specific sensitivity and interaction between individuals. Studying the population level is distinguished from the individual level and lower by a less direct link between the chemical exposure and the observed effects, due to individual variability and several feedback loops, loosening the dose-response relationships. Research at the population level is thus characterized by an increasing level of uncertainty if these processes are not properly addressed and by increasing time and efforts. Hence, it is not surprising that effects at the population are understudied. This is even more the case for investigations on higher levels like meta-populations, communities and ecosystems (see sections on , and ). It is thus highly important to obtain data and insights into mechanisms leading to effects at the population level, keeping in mind the relevant interactions with lower and higher levels of organisation. Properties of populations are unique to this level of biological organization and include social structure (see section on ), genetic composition (see section on ), density and age structure. This gives room to age and developmental stage specific sensitivities to chemicals. For almost all species, young individuals like neonates or first instars are markedly more sensitive than adults or late instar larvae. This difference may run up to three orders of magnitude and consequently instar specific sensitivities may vary as much as species specific sensitivities (Figure 1). Population developmental stage specific sensitivities have also been reported. Exponentially growing daphnid populations exposed to the insecticide fenvalerate recovered much faster than populations that reached carrying capacity (Pieters and Liess, 2006). Given the age and developmental stage specific sensitivities, the timing of exposure to toxicants in relation to the critical life stage of the organism may seriously affect the extent of the adverse effects, especially in seasonally synchronised populations. A challenging question involved in population ecotoxicology is when a population is considered to be stable or in steady state. In spite of the various types of oscillation all populations depicted in Figure 2 can be considered to be stable. One could even argue that any population that does not go extinct can be considered stable. Hence, a single population could vary considerable in density over time, potentially strongly affecting the impact of exposure to toxicants. When populations suffer from starvation and crowding due to high densities and intraspecific competition, they are markedly more sensitive to toxicants, sometimes even up to a factor of 100 (Liess et al., 2016). This may even lead to unforeseen, indirect effects. Relative population growth rate (individual/individual/day) of high density populations of chironomids actually increased upon exposure to Cd, because Cd induced mortality diminished the food shortage for the surviving larvae (Figure 3). Only at the highest Cd exposure population growth rate decreased again. For populations at low densities, the anticipated decrease in population growth rate with increasing Cd concentrations was observed. Yet, at all Cd exposure levels growth rate of low density populations was markedly higher than that of high density populations. In ecotoxicity studies, preferably cohorts of individuals of the same size and age are selected to minimize variation in the outcome of the test, whereas in ecotoxicology the natural heterogenous population composition is taken into account. This does make it harder though to interpret the obtained experimental data. Especially when studying populations of higher organisms in the wild, the increasing time to complete the research due to the long life span of these organisms imposes practical limitations (see section on ). In the laboratory, this can be circumvented by selecting test species with relatively short life cycles, like algae, bacteria and zooplankton. For algae, a three or four day test can be considered as a multigeneration experiment and during 21 d female daphnids may release up to three clutches of neonates. These population ecotoxicity tests offer the unique possibility to calculate the ultimate population parameter, the population growth rate ( ). This is a demographic population parameter, integrating survival, maturity time and reproduction (see section on ). Yet, such chronic experiments are typically performed with cohorts and not with natural populations, making these experiments rather an extension of chronic toxicity tests than true population ecotoxicity tests. Knillmann, S., Stampfli, N.C., Beketov, M.A., Liess, M. (2012). Intraspecific competition increases toxicant effects in outdoor microcosms. Ecotoxicology 21, 1857-1866. Liess, M., Foit, K., Knillmann, S., Schäfer, R.B., Liess, H.-D. (2016). Predicting the synergy of multiple stress effects. Scientific Reports 6, 32965. Pieters, B.J., Liess, M. (2006). Population developmental stage determines the recovery potential of populations after fenvalerate application. Environmental Science and Technology 40, 6157-6162. Motivate the importance of studying ecotoxicology at the population level and higher. Name the properties of populations that are unique to this level of biological organisation. Why is it important to understand the implications of age and developmental stage specific sensitivities for population responses to toxicant exposure? Explain the results observed in Figure 3. | 6,263 | 1,163 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/25%3A_Solutions_II_-_Nonvolatile_Solutes/25.01%3A_Raoult's_and_Henry's_Laws_Define_Standard_States |
The activity is a relative measure as it measures equilibrium relative to a standard state. The standard state is defined by the International Union of Pure and Applied Chemistry (IUPAC) and followed systematically by chemists around the globe.︎ The standard state for a solution is defined in terms of the infinite-dilution behavior. This is in contrast to the standard state concentration of 1 mol/L. This can be reconciled by considering that the standard state is a hypothetical solution of 1 mol/L in which the solute has infinite-dilution properties, e.g. solute particles do not interact with each other. This means that the activity coefficient describes all non-ideal behavior when the value is not equal to 1. | 737 | 1,164 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/08%3A_Nucleophilic_Substitution_and_Elimination_Reactions/8.06%3A_Stereochemistry_of_(S_N2)_Reactions |
There are two simple ways in which the \(\text{S}_\text{N}2\) reaction of methyl chloride could occur with hydroxide ion. These differ in the direction of approach of the reagents (Figure 8-1). The hydroxide ion could attack chloromethane at the of the carbon where the chlorine is attached or, alternatively, the hydroxide ion could approach the carbon on the side opposite from the chlorine in what is called the approach. In either case, the making of the \(C-O\) bond is essentially with the breaking of the \(C-Cl\) bond. The difference is that for the back-side mechanism the carbon and the attached hydrogens become planar in the transition state. The stereochemical consequences of front- and back-side displacements are different. With cyclic compounds, the two types of displacement lead to products. For example, an \(\text{S}_\text{N}2\) reaction between -3-methylcyclopentyl chloride and hydroxide ion would give the cis alcohol by front-side approach but the trans alcohol by back-side approach. The actual product is the trans alcohol, from which we know that reaction occurs by back-side displacement: displacement to give 2-butanol with the inverted configuration. Similar studies of a wide variety of displacements have established that \(\text{S}_\text{N}2\) reactions invariably proceed with inversion of configuration via back-side attack. This stereochemical course commonly is known as the .\(^5\) An orbital picture of the transition state of an \(\text{S}_\text{N}2\) reaction that leads to inversion of configuration follows: \(^5\)The first documented observation that optically active compounds could react to give products having the opposite configuration was made by P. Walden, in 1895. The implications were not understood, however, until the mechanisms of nucleophilic substitution were elucidated in the 1930's, largely through the work of E. D. Hughes and C. K. Ingold, who established that \(\text{S}_\text{N}2\) substitutions give products of inverted configuration. and (1977) | 2,043 | 1,165 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/11%3A_Quantum_Mechanics_and_Atomic_Structure/11.05%3A_de_Broglie's_Postulate |
The next real advance in understanding the atom came from an unlikely quarter - a student prince in Paris. Prince Louis de Broglie was a member of an illustrious family, prominent in politics and the military since the 1600's. Louis began his university studies with history, but his elder brother Maurice studied x-rays in his own laboratory, and Louis became interested in physics. After World War I, de Broglie focused his attention on Einstein's two major achievements, the theory of special relativity and the quantization of light waves. He wondered if there could be some connection between them. Perhaps the quantum of radiation really should be thought of as a particle. De Broglie suggested that if waves (photons) could behave as particles, as demonstrated by the photoelectric effect, then the converse, namely that particles could behave as waves, should be true. He associated a wavelength \(\lambda\) to a particle with momentum \(p\) using Planck's constant as the constant of proportionality: \[\lambda =\dfrac{h}{p} \label{1.6.1} \] he . The fact that particles can behave as waves but also as particles, depending on which experiment you perform on them, is known as the . From the discussion of the photoelectric effect, we have the first part of the particle-wave duality, namely, that electromagnetic waves can behave like particles. These particles are known as , and they move at the speed of light. Any particle that moves at or near the speed of light has kinetic energy given by Einstein's . In general, a particle of mass \(m\) and momentum \(p\) has an energy \[E=\sqrt{p^2 c^2+m^2 c^4} \label{1.6.2} \] \(p=0\) \(E=mc^2\) \(p\) \[E=pc. \nonumber \] \(E=h\nu\) \[h\nu =\dfrac{hc}{\lambda}=pc \label{1.6.4} \] \(\lambda\) \[\lambda=\dfrac{h}{p}= \dfrac{h}{mv} \nonumber \] where \(v\) is the velocity of the particle. Hence \(\lambda\) \(p\). A Having decided that the photon might well be a particle with a rest mass, even if very small, it dawned on de Broglie that in other respects it might not be too different from other particles, especially the very light electron. In particular, maybe the electron also had an associated wave. The obvious objection was that if the electron was wavelike, why had no diffraction or interference effects been observed? But there was an answer. If de Broglie's relation between momentum and wavelength also held for electrons, the wavelength was sufficiently short that these effects would be easy to miss. As de Broglie himself pointed out, the wave nature of light is not very evident in everyday life. As the next section will demonstrate, the validity of de Broglie’s proposal was confirmed by electron diffraction experiments of G.P. Thomson in 1926 and of C. Davisson and L. H. Germer in 1927. In these experiments it was found that electrons were scattered from atoms in a crystal and that these scattered electrons produced an interference pattern. These diffraction patterns are characteristic of wave-like behavior and are exhibited by both electrons (i.e., matter) and electromagnetic radiation (i.e., light). Calculate the de Broglie wavelength for an electron with a kinetic energy of 1000 eV. To calculate the de Broglie wavelength (Equation \ref{1.6.1}), the momentum of the particle must be established and requires knowledge of both the mass and velocity of the particle. The mass of an electron is \(9.109383 \times 10^{−28}\; g\) and the velocity is obtained from the given kinetic energy of 1000 eV: \[\begin{align*} KE &= \dfrac{mv^2}{2} \\[4pt] &= \dfrac{p^2}{2m} = 1000 \;eV \end{align*} \nonumber \] Solve for momentum \[ p = \sqrt{2 m KE} \nonumber \] convert to SI units \[ p = \sqrt{(1000 \; \cancel{eV}) \left( \dfrac{1.6 \times 10^{-19} \; J}{1\; \cancel{ eV}} \right) (2) (9.109383 \times 10^{-31}\; kg)} \nonumber \] expanding definition of joule into base SI units and cancel \[\begin{align*} p &= \sqrt{(3.1 \times 10^{-16} \;kg \cdot m^2/s^2 ) (9.109383 \times 10^{-31}\; kg)} \\[4pt] &= \sqrt{ 2.9 \times 10^{-40 }\, kg^2 \;m^2/s^2 } \\[4pt] &= 1.7 \times 10^{-23} kg \cdot m/s \end{align*} \nonumber \] Now substitute the momentum into the equation for de Broglie's wavelength (Equation \(\ref{1.6.1}\)) with Planck's constant (\(h = 6.626069 \times 10^{−34}\;J \cdot s\)). After expanding units in Plank's constant \[\begin{align*} \lambda &=\dfrac{h}{p} \\[4pt] &= \dfrac{6.626069 \times 10^{−34}\;kg \cdot m^2/s}{1.7 \times 10^{-23} kg \cdot m/s} \\[4pt] &= 3.87 \times 10^{-11}\; m \\[4pt] &=38.9\; pm \end{align*} \nonumber \] Calculate the de Broglie wavelength for a fast ball thrown at 100 miles per hour and weighing 4 ounces. Comment on whether the wave properties of baseballs could be experimentally observed. Following the unit conversions below, a 4 oz baseball has a mass of 0.11 kg. The velocity of a fast ball thrown at 100 miles per hour in m/s is 44.7 m/s. \[ m = \left(4 \; \cancel{oz}\right)\left(\frac{0.0283 \; kg}{1 \; \cancel{oz}}\right) = 0.11 kg \nonumber \] \[ v = \left(\frac{100 \; \cancel{mi}}{\cancel{hr}}\right) \left(\frac{1609.34 \; m}{\cancel{mi}}\right) \left( \frac{1 \; \cancel{hr}}{3600 \; s}\right) = 44.7 \; m/s \nonumber \] The de Broglie wavelength of this fast ball is: \[ \lambda = \frac{h}{mv} = \frac{6.626069 \times 10^{-34}\;kg \cdot m^2/s}{(0.11 \; kg)(44.7 \;m/s)} = 1.3 \times 10^{-34} m \nonumber \] If an electron and a proton have the same velocity, which would have the longer de Broglie wavelength? Equation \ref{1.6.1} shows that the de Broglie wavelength of a particle's matter wave is inversely proportional to its momentum (mass times velocity). Therefore the smaller mass particle will have a smaller momentum and longer wavelength. The electron is the lightest and will have the longest wavelength. This was the prince's Ph.D. thesis, presented in 1924. His thesis advisor was somewhat taken aback, and was not sure if this was sound work. He asked de Broglie for an extra copy of the thesis, which he sent to Einstein. Einstein wrote shortly afterwards: " and the prince got his Ph.D. (Beams Professor, , ( ) ") | 6,124 | 1,166 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Valence_Bond_Theory/Hybrid_Orbitals_in_Carbon_Compounds |
Diamond crystals such as the one shown here are appreciated by almost everyone, because of their hardness, sparkle, and high value. They are also important in many technical applications. However, in terms of chemistry, diamonds consist of only carbon atoms, except for impurities. Like diamond, the chemistry of carbon is indeed very interesting and valuable. Carbon atoms have the ability to bond to themselves and to other atoms with , , and hybrid orbitals. This link gives you the basics about the hybrid orbitals, and you are introduced to the various bonding of carbon in this document. Compounds containing carbon-hydrogen bonds are called organic compounds. They may also contain \(\ce{C-C}\), \(\ce{C=C}\), \(\ce{C\equiv C}\), \(\ce{C-N}\), \(\ce{C=N}\), \(\ce{C\equiv N}\), \(\ce{C-O}\), and \(\ce{C=O}\) bonds. Such a variety is due to the ability of carbon to make use of , , and hybrid orbitals for the bonding. There are also various inorganic compounds such as carbon monoxide, carbon dioxide, calcium carbonate, sodium bicarbonate, etc. involving carbon. When hybrid orbitals are used for the sigma bond, the two sigma bonds around the carbon are linear. Two other orbitals are available for pi bonding, and a typical compound is the acetylene or ethyne \(\ce{HC\equiv CH}\). The three sigma and two pi bonds of this molecule from University of Florida: General chemistry are shown below. Note that molecules \(\ce{H-C\equiv C-H}\), \(\ce{H-C\equiv N}\), and \(\ce{C\equiv O}\) have the same number of electrons. Bonding in these molecules can be explained by the same theory, and thus their formation is no surprise. The \(\ce{O=C=O}\) molecule is linear, and the carbon atom in this molecule also involves the hybrid orbitals. Two pi bonds are also present in this simple molecule. As an exercise, draw a picture to show the two sigma and two pi bonds for this molecule. When carbon atoms make use of hybrid orbitals for sigma bonding, the three bonds lie on the same plane. One such compound is ethene, in which both carbon atoms make use of hybrid orbitals. One of the remaining orbitals for each carbon overlap to form a pi bond. A pi bond consists of two parts where bonding electrons are supposed to be located. A picture depicting the sigma and pi bonds in ethene from the same source as the previous picture is shown on the right. Carbon atoms make use of hybrid orbitals not only in ethene, but also in many other types of compounds. The following are some of these compounds: Arromatic
compounds Graphite Fullerenes During the lecture on covalent bonding, we can illustrate how atomic orbitals overlap in the formation of bonds. Here, we can only show you the nice picture as a result. In ethane, the carbon atoms use hybrid orbitals for the formation of sigma bonds. The four bonds around each \(\ce{C}\) atom point toward the vertices of a regular tetrahedron, and the ideal bond angles are 109.5°. The simplest compound is methane, \(\ce{CH4}\), which is the first member of the alkane family. The next few members are ethane, \(\ce{CH3CH3}\), propane, \(\ce{CH3CH2CH3}\), butane, \(\ce{CH3CH2CH2CH3}\), . Diamond is a crystal form of elemental carbon, and the structure is particularly interesting. In the crystal, every carbon atom is bonded to four other carbon atoms, and the bonds are arranged in a tetrahedral fashion. The bonding, no doubt, is due to the hybrid orbitals. The bond length of 154 pm is the same as the \(\ce{C-C}\) bond length in ethane, propane and other alkanes. An idealized single crystal of diamond is a gigantic molecule, because all the atoms are inter-bonded. The bonding has given diamond some very unusual properties. It is the hardest stone, much harder than anything else in the material world. It is a poor conductor, because all electrons are localized in the chemical bonds. However, diamond is an excellent heat conductor. A stone made of pure carbon is colorless, but the presence of impurities gives it various colors. The index of refraction is very high, and their glitter (sparkle or splendor) has made them the most precious stones. Some typical bonding features of ethane, ethene, and ethyne are summarized in the table below: H-C C-H As the bond order between carbon atoms increases from 1 to 3 for ethane, ethene, and ethyne, the bond lengths decrease, and the bond energy increases. Note that the bond energies given here are specific for these compounds, and the values may be different from the average values for this type of bonds. Hint: hybrid orbitals
Linear \(\ce{-C -}\) bonds due to hybridized orbitals. This molecule is linear, and it consists of 3 sigma, , bonds, and two pi, , bonds. Compare the bonding of this with \(\ce{C\equiv O}\), \(\ce{H-C\equiv N}\), and \(\ce{CH3-C\equiv N}\). Hint: hybrid orbitals
Planar \(\ce{-C}\textrm{<}\) bonds due to hybridized orbitals. Another orbital is used for the pi, . How many sigma and pi bonds does this molecule have? Do all atoms in this molecule lie on the same plane? Hint: hybrid orbitals
Tetrahedral arrangement around \(\ce{C}\) is due to hybridized orbitals. Hint: hybrid orbitals
Tetrahedral arrangement around \(\ce{C}\) is due to hybridized orbitals. Hint: shortest between triple bonded carbon
The bond length decreases as the bond order increases. Hint: three
Recognize the type of bonding is important. Hint: This is one of the problems for chemists.
The structure is shown below. Can you sketch a bonding structure for caffeine? How many carbon atoms makes use of hybrid orbitals? | 5,649 | 1,167 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/05%3A_Energy_Changes_in_Chemical_Reactions/5.08%3A_Energy_Sources_and_the_Environment |
Our contemporary society requires the constant expenditure of huge amounts of energy to heat our homes, provide telephone and cable service, transport us from one location to another, provide light when it is dark outside, and run the machinery that manufactures material goods. The United States alone consumes almost 10 kJ per person per day, which is about 100 times the normal required energy content of the human diet. This figure is about 30% of the world’s total energy usage, although only about 5% of the total population of the world lives in the United States. In contrast, the average energy consumption elsewhere in the world is about 10 kJ per person per day, although actual values vary widely depending on a country’s level of industrialization. In this section, we describe various sources of fossil fuel energy and their impact on the environment. Driven by environmental concerns about climate change and pollution, the world is undergoing a transformation from fossil fuels to renewable resources such as solar, and wind. The role that hydro and nuclear energy will play is uncertain and especially in the later case a policy rather than a scientific issue. According to the law of conservation of energy, energy can never actually be “consumed”; it can only be changed from one form to another. Fossil fuels, coal, oil and natural gas are the result of anaerobic decay of dead plants and animals laid down hundreds of millions of years ago, most of which took place well before the dinosaurs strode the earth. Fossil fuels slowly formed as further geological layers compressed and heated the dead organic matter. The energy content of fossil fuels results from the transformation of sunlight into vegetation and the chemical transformation brought about by anaerobic cooking at high pressures and temperatures over geological times. Figure \(\Page {1}\) represents a plant for generating electricity using oil or coal where the fuel is burned in a boiler, superheating steam which then powers a turbine for electrical generation. Oil derived fuels are seldom used in large power plants but diesel is used commonly in small electrical generators either in remote locations or as back up for when electrical distribution systems fail. Natural gas fueled power plants burn the fuel directly in the turbine which is similar to a jet engine. Coal power plants can convert ~40% of the energy released from combustion to electricity. In comparison, nuclear power plants can be more than 50% efficient and gas turbines can approach 60% mostly due to higher operating temperatures. Co-generation, using the plant to produce not only electricity but also heat for industrial or other purposes can raise overall efficiency by 10 - 15 % or so. The total expenditure of energy in the world each year is about 3 × 10 kJ. Today, more than 80% of this energy is provided by the combustion of fossil fuels: oil, coal, and natural gas (The sources of the energy consumed in the United States in 2009 are shown in Figure \(\Page {2}\) ) but as Table \(\Page {1}\) shows, energy usage is a complex issue. Petroleum dominates as a source of energy for transportation because gasoline is easy to transport, but is very little used for electrical generation, whereas 91% of coal is used for electrical generation. The other major use of coal is as a reducing agent for metal refining from ores. The former is called thermal coal, the latter metallurgical coal.
Coal was primarily laid down from the large swamp forests of the Carboniferous Period. Coal deposits are found today where those forests were Coal is a complex solid material derived primarily from plants that died and were buried hundreds of millions of years ago and were subsequently subjected to high temperatures and pressures. Because plants contain large amounts of , derived from linked glucose units, the structure of coal is more complex than that of petroleum (Figure \(\Page {3}\) ). In particular, coal contains a large number of oxygen atoms that link parts of the structure together, in addition to the basic framework of carbon–carbon bonds. It is impossible to draw a single structure for coal; however, because of the prevalence of rings of carbon atoms (due to the original high cellulose content), coal is more similar to an aromatic hydrocarbon than an aliphatic one. There are four distinct classes of coal (Table \(\Page {21}\) ); their hydrogen and oxygen contents depend on the length of time the coal has been buried and the pressures and temperatures to which it has been subjected. Lignite, with a hydrogen:carbon ratio of about 1.0 and a high oxygen content, has the lowest Δ . Lignite is extensively mined in Germany and Poland. Anthracite, in contrast, with a hydrogen:carbon ratio of about 0.5 and the lowest oxygen content, has the highest Δ and is the highest grade of coal. Anthracite is the first choice for metallurgical refining. The most abundant form in the Western United States in anthracite while that in the Eastern United States is bituminous coal, which has a high sulfur content because of the presence of small particles of pyrite (FeS ). Combustion of coal releases the sulfur in FeS as SO , which is a major contributor to acid rain. Table \(\Page {3}\) compares the Δ per gram of oil, natural gas, and coal with those of selected organic compounds. Peat, a precursor to coal, is the partially decayed remains of plants that grow in the swampy areas of the Carboniferous Period. It is removed from the ground in the form of soggy bricks of mud that will not burn until they have been dried. Even though peat is a smoky, poor-burning fuel that gives off relatively little heat, humans have burned it since ancient times (Figure \(\Page {4}\) ). If a peat bog were buried under many layers of sediment for a few million years, the peat would eventually be compressed and heated enough to become lignite, the lowest grade of coal; given enough time and heat, lignite would eventually become anthracite, a much better fuel. As a solid, coal is much more difficult to mine and ship than petroleum (a liquid) or natural gas. Consequently, more than 75% of the coal produced each year is simply burned in power plants to produce electricity. Methods to convert coal to gaseous fuels ( ) or liquid fuels ( ) exist, but are not particularly economical unless the prices of oil and natural gas are high. With the development of fracking and the subsequent fall in oil and natural gas prices interest in these processes has fallen however they have played an important role in the past. In the most common approach to coal gasification, coal reacts with steam to produce a mixture of CO and H known as , or : \(\Page {1}\)
\[ C\left ( s \right ) + H_{2}O \left ( g \right ) \rightarrow CO \left ( g \right ) + H_{2} \left ( g \right ) \; \; \; \; \Delta H= 131 \; kJ \] Converting coal to syngas removes any sulfur present and produces a clean-burning mixture of gases. Syngas or town gas was used for cooking until the 1960s when natural gas pipelines were built. Because syngas contains carbon monoxide (CO) it is poisonous, which accounts for scenes in old movies where people were killed by sticking their heads into an oven and allowing the gas to flow. Syngas is can also used as a reactant to produce methane and methanol. A promising approach is to convert coal directly to methane through a series of reactions:
\[ \begin{matrix}
2C\left ( s \right )+2H_{2}O\left ( g \right ) \rightarrow \cancel{2CO\left ( g \right )}+\cancel{2H_{2}\left ( g \right )} & \Delta H_{1}=262 \; kJ \\
\cancel{CO\left ( g \right )}+\cancel{H_{2}O \left ( g \right )} \rightarrow CO_{2}\left ( g \right )+\cancel{H_{2}\left ( g \right )} & \Delta H_{2}=-41 \; kJ \\
\cancel{CO\left ( g \right )}+\cancel{3H_{2}\left ( g \right )} \rightarrow CH_{4}\left ( g \right )+\cancel{H_{2}O\left ( g \right )} & \Delta H_{3}=-206 \; kJ \\
-------------------&--------\\
2C\left ( s \right )+2H_{2}O\left ( g \right ) \rightarrow CH_{4}\left ( g \right ) + CO_{2}\left ( g \right ) & \Delta H_{3}=15 \; kJ
\end{matrix} \] Techniques available for converting coal to liquid fuels are not economically competitive with the production of liquid fuels from petroleum. Current approaches to coal liquefaction use a catalyst to break the complex network structure of coal into more manageable fragments. The products are then treated with hydrogen (from syngas or other sources) under high pressure to produce a liquid more like petroleum. Subsequent distillation, cracking, and reforming can be used to create products similar to those obtained from petroleum. The petroleum that is pumped out of the ground is a complex mixture of several thousand organic compounds including straight-chain alkanes, cycloalkanes, alkenes, and aromatic hydrocarbons with four to several hundred carbon atoms. The identities and relative abundences of the components vary depending on the source. So Texas crude oil is somewhat different from Saudi Arabian crude oil. In fact, the analysis of petroleum from different deposits can produce a “fingerprint” of each, which is useful in tracking down the sources of spilled crude oil. For example, Texas crude oil is “sweet,” meaning that it contains a small amount of sulfur-containing molecules, whereas Saudi Arabian crude oil is “sour,” meaning that it contains a relatively large amount of sulfur-containing molecules. Petroleum is converted to useful products such as gasoline in three steps: distillation, cracking, and reforming. Recall that separates compounds on the basis of their relative volatility, which is usually inversely proportional to their boiling points. Part (a) in \(\Page {5}\) shows a cutaway drawing of a column used in the petroleum industry for separating the components of crude oil. The petroleum is heated to approximately 400°C (750°F), at which temperature it has become a mixture of liquid and vapor. This mixture, called the feedstock, is introduced into the refining tower. The most volatile components (those with the lowest boiling points) condense at the top of the column where it is cooler, while the less volatile components condense nearer the bottom. Some materials are so nonvolatile that they collect at the bottom without evaporating at all. Thus the composition of the liquid condensing at each level is different. These different fractions, each of which usually consists of a mixture of compounds with similar numbers of carbon atoms, are drawn off separately. Part (b) in shows the typical fractions collected at refineries, the number of carbon atoms they contain, their boiling points, and their ultimate uses. These products range from gases used in natural and bottled gas to liquids used in fuels and lubricants to gummy solids used as tar on roads and roofs. The economics of petroleum refining are complex. For example, the market demand for kerosene and lubricants is much lower than the demand for gasoline, yet all three fractions are obtained from the distillation column in comparable amounts. Furthermore, most gasolines and jet fuels are blends with very carefully controlled compositions that cannot vary as their original feedstocks did. To make petroleum refining more profitable, the less volatile, lower-value fractions must be converted to more volatile, higher-value mixtures that have carefully controlled formulas. The first process used to accomplish this transformation is cracking, in which the larger and heavier hydrocarbons in the kerosene and higher-boiling-point fractions are heated to temperatures as high as 900°C. High-temperature reactions cause the carbon–carbon bonds to break, which converts the compounds to lighter molecules similar to those in the gasoline fraction. Thus in cracking, a straight-chain alkane with a number of carbon atoms corresponding to the kerosene fraction is converted to a mixture of hydrocarbons with a number of carbon atoms corresponding to the lighter gasoline fraction. The second process used to increase the amount of valuable products is called reforming; it is the chemical conversion of straight-chain alkanes to either branched-chain alkanes or mixtures of aromatic hydrocarbons. Metal catalysts such as platinum are used to drive the necessary chemical reactions. The mixtures of products obtained from cracking and reforming are separated by fractional distillation. The quality of a fuel is indicated by its octane rating, which is a measure of its ability to burn in a combustion engine without knocking or pinging. Knocking and pinging signal premature combustion ( \(\Page {6}\) ), which can be caused either by an engine malfunction or by a fuel that burns too fast. In either case, the gasoline-air mixture detonates at the wrong point in the engine cycle, which reduces the power output and can damage valves, pistons, bearings, and other engine components. The various gasoline formulations are designed to provide the mix of hydrocarbons least likely to cause knocking or pinging in a given type of engine performing at a particular level. The octane scale was established in 1927 using a standard test engine and two pure compounds: n-heptane and isooctane (2,2,4-trimethylpentane). n-Heptane, which causes a great deal of knocking on combustion, was assigned an octane rating of 0, whereas isooctane, a very smooth-burning fuel, was assigned an octane rating of 100. Chemists assign octane ratings to different blends of gasoline by burning a sample of each in a test engine and comparing the observed knocking with the amount of knocking caused by specific mixtures of n-heptane and isooctane. For example, the octane rating of a blend of 89% isooctane and 11% n-heptane is simply the average of the octane ratings of the components weighted by the relative amounts of each in the blend. Converting percentages to decimals, we obtain the octane rating of the mixture: \[0.89(100)+0.11(0)=89 \] A gasoline that performs at the same level as a blend of 89% isooctane and 11% n-heptane is assigned an octane rating of 89; this represents an intermediate grade of gasoline. Regular gasoline typically has an octane rating of 87; premium has a rating of 93 or higher. As shown in \(\Page {7}\), many compounds that are now available have octane ratings greater than 100, which means they are better fuels than pure isooctane. In addition, antiknock agents, also called octane enhancers, have been developed. One of the most widely used for many years was [(C H ) Pb], which at approximately 3 g/gal gives a 10–15-point increase in octane rating. Since 1975, however, lead compounds have been phased out as gasoline additives because they are highly toxic. Other enhancers, such as methyl t-butyl ether (MTBE), have been developed to take their place that combine a high octane rating with minimal corrosion to engine and fuel system parts. Unfortunately, when gasoline containing MTBE leaks from underground storage tanks, the result has been contamination of the groundwater in some locations, resulting in limitations or outright bans on the use of MTBE in certain areas. As a result, the use of alternative octane enhancers such as ethanol, which can be obtained from renewable resources such as corn, sugar cane, and, eventually, corn stalks and grasses, is increasing. Natural gas is a (mostly) combustible gas found underground. While primarily composed of methane (70-90%) the gas from each well has a different composition and the value of the other components affects the value of the gas. The gas from wells that are rich in methane is called dry and wells that have a considerable amount of higher hydrocarbons produce wet gas. The higher hydrocarbons have value above that of methane so stripping them out is important. Some wells are sour because their gas has hydrogen sulfide which must be removed before the gas can be used for heating or generating electricity. Finally, a few wells in Texas and nearby Oklahoma have a relatively high amount of helium (0.3 - 2.7%). The Helium Act of 1925 established . Political pressure and costs pushed laws to privatize the reserve, but other policy considerations including the need for helium for scientific research has slowed the process. The purification of natural gas is a complex process with many steps as each of the impurities is stripped out Gas turbine power plants to generate electricity are coming increasingly into use as fracking and other advanced drilling technologies have driven the cost of natural gas down and the supply up. While on a continental scale natural gas is transported by pipelines, natural gas can be cooled and compressed to be transported as liquified natural gas. Gas turbine power plants are small and quickly built. They can be rapidly spun up to meet peak demand. More detailed information can be found at the Since 1850 the burning of fossil fuels has increased the concentration of carbon dioxide in the atmosphere from 280 to just over 400 ppmV. A continued increase in the CO burden in the atmosphere will have serious negative effects and this requires shifting our entire energy producing economy from fossil fuels to non-carbon sources such as hydro, solar, wind and nuclear. These include sea level rise that will threaten low lying cities including but not limited to Miami Beach and Norfolk in the US, even a meter or more coupled with storm surge and high tides can cause massive damage as was seen during Hurricane Sandy. Increased carbon dioxide in the atmosphere has already measurably decreased the pH of the oceans. Sea life is adapted to a narrow range of pH. Higher global temperatures of 2 or 3 C may not seem much, but one should keep in mind that the average global temperature during the ice ages was only ~6 C lower than it is today. During the Eemian interglacial the average temperature was only a few degrees higher than the present and the sea level was 6-9 m higher. Finally, humans are mammals who maintain a core temperature within a few degrees of 37.0 C. In hot weather we do so by evaporation of sweat however there are limits to this and by 2100 there is a significant probability even in the US that at least a few days a year will reach this limit by 2100. Given that most people on earth do not have access to air conditioning, parts of the planet may become uninhabitable. Indeed a worst case and a serious problem that merits attention. Given the constraints of this text it is difficult to provide the level of detail needed to understand why this is so. A good source for those interested in learning more is David Archer's and by Andrew Dessler There are a few basic facts that anyone starting to learn about the issue need to know. First, that the Earth gains energy from the Sun, and that it must radiate the energy at the same rate. If more energy is absorbed than radiated the Earth will warm up, if less energy is absorbed than radiated it will cool. Solar radiation follows a 5500 K blackbody distribution while radiation from the surface of the earth is also black body but at ~290K. This is shown schematically in Figure \(\Page {9}\) The atmosphere cools with altitude up to about 15 km where it starts warming again because of absorption of UV radiation by ozone in the stratosphere. The transition between the troposphere and the stratosphere is called the tropopause and is the coldest part of the atmosphere. Figure \(\Page {11}\) shows the IR emission spectrum observed looking down on the earth from a high altitude balloon The dotted lines in Figure \(\Page {11}\) are blackbody curves. The IR window shown schematically in blue in Figure \(\Page {9}\) is the region between the ozone and the carbon dioxide band, where the emission from the hot, 320 K ground follows the blackbody curve with a few sharp water vapor absorption lines. At those wavelengths, in the IR window, the emission comes directly from the surface. The CO band extends down to about a 220 K blackbody curve. What this means is that radiation from CO only escapes to space from the level in the troposphere where that is the temperature. The rate of emission is proportional to T so the rate of emission from higher, therefore colder levels, is slower. Radiation in this area of the spectrum from the surface is blocked and only the greatly reduced emission from the upper troposphere escapes to space. The same is true for the ozone and methane bands as well as the water lines. The net effect is that the surface must warm in order to maintain the balance between incoming solar radiation and the outgoing emission. There is a simple calculation which models the atmosphere as a one dimensional problem and calculates what the temperature of the surface would be if there were no greenhouse gases. The result is 255 K, rather cold. In fact if one attempts a more complex calculation the effective temperature without greenhouse gases would be even colder. What happens if we increase the carbon dioxide in the atmosphere? The altitude at which the atmosphere can emit radiation to space will rise because of increased absorption by the CO . Since in the troposphere the temperature decreases with altitude, the rate of emission from a higher level must decrease. Again, in order to maintain the balance between incoming solar radiation and the outgoing emission the surface will have to warm even more, thus the term global warming. The change is not linear with increasing CO but logarithmic. But, of course it is not so simple, because increasing the surface temperature will increase the water vapor pressure in the atmosphere, which will increase the temperature further. We also have to understand the flow of carbon between the atmosphere, the biosphere, the upper oceans and the deep. Observations to date show that natural emissions of CO from these reservoirs are in balance with absorption, while only about half of fossil fuel emissions remain in the atmosphere, the rest being absorbed by the upper ocean and the biosphere. The three upper reservoirs equilibrate in a decade or less, but flow into the deep ocean requires roughly a thousand years. There is no doubt that atmospheric CO levels are increasing, and the major reason for this increase is the combustion of fossil fuels. An extremely conservative statement of the situation today can be found in the , a consensus between scientists and policymakers. The report starkly states that Cumulative emissions of CO largely determine global mean surface warming by the late 21st century and beyond. and concludes that Continued emission of greenhouse gases will cause further warming and long-lasting changes in all components of the climate system, increasing the likelihood of severe, pervasive and irreversible impacts for people and ecosystems. Limiting climate change would require substantial and sustained reductions in greenhouse gas emissions which, together with adaptation, can limit climate change risks. The situation is serious, but we can work together to limit and even reverse damage while maintaining our standard of living in the developed world while helping the developing world to a better future. However, the issues are complex and we can only touch on some of the basics here. More than 80% of the energy used by modern society (about 3 × 10 kJ/yr) is from the combustion of fossil fuels. Because of their availability, ease of transport, and facile conversion to convenient fuels, natural gas and petroleum are currently the preferred fuels. Coal is primarily used for electricity generation. The combustion of fossil fuels releases large amounts of CO that upset the balance of the and result in a steady increase in atmospheric CO levels. Because CO is a , which absorbs heat before it can be radiated from Earth into space, CO in the atmosphere results in increased surface temperatures (the ). What is meant by the term ? List three greenhouse gases that have been implicated in global warming. Name three factors that determine the rate of planetary CO uptake. The structure of coal is quite different from the structure of gasoline. How do their structural differences affect their enthalpies of combustion? Explain your answer. One of the side reactions that occurs during the burning of fossil fuels is How many kilograms of CO are released during the combustion of 16 gal of gasoline? Assume that gasoline is pure isooctane with a density of 0.6919 g/mL. If this combustion was used to heat 4.5 × 10 L of water from an initial temperature of 11.0°C, what would be the final temperature of the water assuming 42% efficiency in the energy transfer? A 60 W light bulb is burned for 6 hours. If we assume an efficiency of 38% in the conversion of energy from oil to electricity, how much oil must be consumed to supply the electrical energy needed to light the bulb? (1 W = 1 J/s) How many liters of cyclohexane must be burned to release as much energy as burning 10.0 lb of pine logs? The density of cyclohexane is 0.7785 g/mL, and its Δ = −46.6 kJ/g. ( ), Scott Sinex, and Scott Johnson (PGCC) | 25,160 | 1,168 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/06%3A_Structures_and_Energetics_of_Metallic_and_Ionic_solids/6.11%3A_Ionic_Lattices/6.11C%3A_Structure_-_Fluorite_(CaF) |
Calcium Fluoride is a solid and forms a cube like structure that is centralized around the calcium molecules. The crystal lattice structure that Calcium Fluoride is also known as the fluorite structure (Figure \(\Page {1}\)) where the Ca ions are eight-coordinate, being centered in a cube of eight F ions. Each F is coordinated to four Ca in the shape of a tetrahedron. When Calcium Fluoride is in a single molecule it forms a Quasilinear structure. Quasilinear means the molecule resonates between a linear shape and a bent shape.Calcium Fluoride is a polyatomic molecule that contains one calcium molecule and two fluoride molecules. Calcium Fluoride is a quasilinear molecule the bonds are created from the single electrons of calcium and the single electron from fluoride. \(\ce{CaF2}\) has its electrons contained with in the 3d orbitals and are able to move between d and d squared. The molecule in linear when they are in the d orbitals the molecule is also the most stable in this shape. When the electrons are in the d orbitals the molecule becomes bent. The molecule resonates between these two shapes making it quasilinear. Figures two and three show how the d-orbitals cause the molecule to bend. Fig # 1 Fig # 2 Fig # 3 | 1,261 | 1,170 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/12%3A_Chemical_Kinetics_II/12.03%3A_The_Connection_between_Reaction_Mechanisms_and_Reaction_Rate_Laws |
The great value of chemical kinetics is that it can give us insights into the actual reaction pathways (mechanisms) that reactants take to form the products of reactions. Analyzing a reaction mechanism to determine the type of rate law that is consistent (or not consistent) with the specific mechanism can give us significant insight. For example, the reaction \[ A+ B \rightarrow C \nonumber \] might be proposed to follow one of two mechanistic pathways: \[ \underbrace{A + A \xrightarrow{k_1} A_2}_{\text{step 1}} \nonumber \] \[ \underbrace{ A_2 + B \xrightarrow{k_2} C}_{\text{step 2}} \nonumber \] or \[ \underbrace{A \xrightarrow{k_1} A^*}_{\text{step 1}} \nonumber \] \[ \underbrace{ A^* + B \xrightarrow{k_2} C}_{\text{step 2}} \nonumber \] The first rate law will predict that the reaction should be second order in \(A\), whereas the second mechanism predicts that it should be first order in \(A\) (in the limit that the steady state approximation, discussed in the following sections, can be applied to \(A_2\) and \(A^*\)). Based on the observed rate law being first or second order in A, one can rule out one of the rate laws. Unfortunately, this kind of analysis cannot confirm a specific mechanism. Other evidence is needed to draw such conclusions, such as the spectroscopic observation of a particular reaction intermediate that can only be formed by a specific mechanism. In order analyze mechanisms and predict rate laws, we need to build a toolbox of methods and techniques that are useful in certain limits. The next few sections will discuss this kind of analysis, specifically focusing on Each type of approximation is important in certain limits, and they are oftentimes used in conjunction with one another to predict the final forms of rate laws. | 1,792 | 1,171 |
Subsets and Splits