11.1: The Dissolution Process
- Describe the basic properties of solutions and how they form.
- Predict whether a given mixture will yield a solution based on molecular properties of its components.
- Explain why some solutions either produce or absorb heat when they form.
An earlier chapter of this text introduced solutions , defined as homogeneous mixtures of two or more substances. Often, one component of a solution is present at a significantly greater concentration, in which case it is called the solvent. The other components of the solution present in relatively lesser concentrations are called solutes. Sugar is a covalent solid composed of sucrose molecules, \(\ce{C12H22O11}\). When this compound dissolves in water, its molecules become uniformly distributed among the molecules of water:
\[\ce{C12H22O11 (s) ⟶ C12H22O11 (aq) } \label{Eq1} \]
The subscript “aq” in the equation signifies that the sucrose molecules are solutes and are therefore individually dispersed throughout the aqueous solution (water is the solvent). Although sucrose molecules are heavier than water molecules, they remain dispersed throughout the solution; gravity does not cause them to “settle out” over time.
Potassium dichromate, \(\ce{K_2Cr_2O_7}\), is an ionic compound composed of colorless potassium ions, \(\mathrm{K^+}\), and orange dichromate ions, \(\ce{Cr_2O_7^{2−}}\). When a small amount of solid potassium dichromate is added to water, the compound dissolves and dissociates to yield potassium ions and dichromate ions uniformly distributed throughout the mixture (Figure \(\PageIndex{1}\)), as indicated in this equation:
\[\ce{K2Cr2O7(s) ⟶ 2K^{+} (aq) + Cr2O7^{2-} (aq)} \label{Eq2} \]
As with the mixture of sugar and water, this mixture is also an aqueous solution. Its solutes, potassium and dichromate ions, remain individually dispersed among the solvent (water) molecules.
Water is used so often as a solvent that the word solution has come to imply an aqueous solution to many people. However, almost any gas, liquid, or solid can act as a solvent. Many alloys are solid solutions of one metal dissolved in another; for example, US five-cent coins contain nickel dissolved in copper. Air is a gaseous solution, a homogeneous mixture of nitrogen, oxygen, and several other gases. Oxygen (a gas), alcohol (a liquid), and sugar (a solid) all dissolve in water (a liquid) to form liquid solutions. Table \(\PageIndex{1}\) gives examples of several different solutions and the phases of the solutes and solvents.
| Solution | Solute | Solvent |
|---|---|---|
| air | O 2 ( g ) | N 2 ( g ) |
| soft drinks | CO 2 ( g ) | H 2 O( l ) |
| hydrogen in palladium | H 2 ( g ) | Pd( s ) |
| rubbing alcohol | H 2 O( l ) | C 3 H 8 O( l ) (2-propanol) |
| saltwater | NaCl( s ) | H 2 O( l ) |
| brass | Zn( s ) | Cu( s ) |
Solutions exhibit these defining traits:
- They are homogeneous; that is, after a solution is mixed, it has the same composition at all points throughout (its composition is uniform).
- The physical state of a solution—solid, liquid, or gas—is typically the same as that of the solvent, as demonstrated by the examples in Table \(\PageIndex{1}\).
- The components of a solution are dispersed on a molecular scale; that is, they consist of a mixture of separated molecules, atoms, and/or ions.
- The dissolved solute in a solution will not settle out or separate from the solvent.
- The composition of a solution, or the concentrations of its components, can be varied continuously, within limits.
The Formation of Solutions
The formation of a solution is an example of a spontaneous process , a process that occurs under specified conditions without the requirement of energy from some external source. Sometimes we stir a mixture to speed up the dissolution process, but this is not necessary; a homogeneous solution would form if we waited long enough. The topic of spontaneity is critically important to the study of chemical thermodynamics and is treated more thoroughly in a later chapter of this text. For purposes of this chapter’s discussion, it will suffice to consider two criteria that favor , but do not guarantee, the spontaneous formation of a solution:
- a decrease in the internal energy of the system (an exothermic change, as discussed in the previous chapter on thermochemistry)
- an increase in the disorder in the system (which indicates an increase in the entropy of the system, as you will learn about in the later chapter on thermodynamics)
In the process of dissolution, an internal energy change often, but not always, occurs as heat is absorbed or evolved. An increase in disorder always results when a solution forms.
When the strengths of the intermolecular forces of attraction between solute and solvent species in a solution are no different than those present in the separated components, the solution is formed with no accompanying energy change. Such a solution is called an ideal solution . A mixture of ideal gases (or gases such as helium and argon, which closely approach ideal behavior) is an example of an ideal solution, since the entities comprising these gases experience no significant intermolecular attractions.
When containers of helium and argon are connected, the gases spontaneously mix due to diffusion and form a solution (Figure \(\PageIndex{2}\)). The formation of this solution clearly involves an increase in disorder, since the helium and argon atoms occupy a volume twice as large as that which each occupied before mixing.
Ideal solutions may also form when structurally similar liquids are mixed. For example, mixtures of the alcohols methanol (CH 3 OH) and ethanol (C 2 H 5 OH) form ideal solutions, as do mixtures of the hydrocarbons pentane, \(\ce{C5H12}\), and hexane, \(\ce{C6H14}\). Placing methanol and ethanol, or pentane and hexane, in the bulbs shown in Figure \(\PageIndex{2}\) will result in the same diffusion and subsequent mixing of these liquids as is observed for the He and Ar gases (although at a much slower rate), yielding solutions with no significant change in energy. Unlike a mixture of gases, however, the components of these liquid-liquid solutions do, indeed, experience intermolecular attractive forces. But since the molecules of the two substances being mixed are structurally very similar, the intermolecular attractive forces between like and unlike molecules are essentially the same, and the dissolution process, therefore, does not entail any appreciable increase or decrease in energy. These examples illustrate how diffusion alone can provide the driving force required to cause the spontaneous formation of a solution. In some cases, however, the relative magnitudes of intermolecular forces of attraction between solute and solvent species may prevent dissolution.
Three types of intermolecular attractive forces are relevant to the dissolution process: solute-solute, solvent-solvent, and solute-solvent. As illustrated in Figure \(\PageIndex{3}\), the formation of a solution may be viewed as a stepwise process in which energy is consumed to overcome solute-solute and solvent-solvent attractions (endothermic processes) and released when solute-solvent attractions are established (an exothermic process referred to as solvation ). The relative magnitudes of the energy changes associated with these stepwise processes determine whether the dissolution process overall will release or absorb energy. In some cases, solutions do not form because the energy required to separate solute and solvent species is so much greater than the energy released by solvation.
For example, cooking oils and water will not mix to any appreciable extent to yield solutions (Figure \(\PageIndex{4}\)). Hydrogen bonding is the dominant intermolecular attractive force present in liquid water; the nonpolar hydrocarbon molecules of cooking oils are not capable of hydrogen bonding, instead being held together by dispersion forces. Forming an oil-water solution would require overcoming the very strong hydrogen bonding in water, as well as the significantly strong dispersion forces between the relatively large oil molecules. And, since the polar water molecules and nonpolar oil molecules would not experience very strong intermolecular attraction, very little energy would be released by solvation.
On the other hand, a mixture of ethanol and water will mix in any proportions to yield a solution. In this case, both substances are capable of hydrogen bonding, and so the solvation process is sufficiently exothermic to compensate for the endothermic separations of solute and solvent molecules.
As noted at the beginning of this module, spontaneous solution formation is favored, but not guaranteed, by exothermic dissolution processes. While many soluble compounds do, indeed, dissolve with the release of heat, some dissolve endothermically. Ammonium nitrate (NH 4 NO 3 ) is one such example and is used to make instant cold packs for treating injuries like the one pictured in Figure \(\PageIndex{5}\). A thin-walled plastic bag of water is sealed inside a larger bag with solid NH 4 NO 3 . When the smaller bag is broken, a solution of NH 4 NO 3 forms, absorbing heat from the surroundings (the injured area to which the pack is applied) and providing a cold compress that decreases swelling. Endothermic dissolutions such as this one require a greater energy input to separate the solute species than is recovered when the solutes are solvated, but they are spontaneous nonetheless due to the increase in disorder that accompanies formation of the solution.
Video \(\PageIndex{1}\) : Watch this brief video illustrating endothermic and exothermic dissolution processes.
Summary
A solution forms when two or more substances combine physically to yield a mixture that is homogeneous at the molecular level. The solvent is the most concentrated component and determines the physical state of the solution. The solutes are the other components typically present at concentrations less than that of the solvent. Solutions may form endothermically or exothermically, depending upon the relative magnitudes of solute and solvent intermolecular attractive forces. Ideal solutions form with no appreciable change in energy.
Footnotes
- If bubbles of gas are observed within the liquid, the mixture is not homogeneous and, thus, not a solution.
Glossary
- alloy
- solid mixture of a metallic element and one or more additional elements
- ideal solution
- solution that forms with no accompanying energy change
- solvation
- exothermic process in which intermolecular attractive forces between the solute and solvent in a solution are established
- spontaneous process
- physical or chemical change that occurs without the addition of energy from an external source


15.1: Achieving Sustainable Development
Economic policies that favour growth are based on the erroneous assumption that natural resources are unlimited. It is thus bound to fail at one point or another.
Efforts to preserve biological diversity are regularly perceived as in conflict with societal progress (Redpath et al., 2013). Perhaps the root of this conflict lies with the fact that most of the development we see today is unsustainable—that is, it risks depleting natural resources to a point where they will no longer be available for use or to provide ecosystem services. Moreover, governments and businesses often measure success in terms of economic growth, which occurs when an economy increasingly produces more goods and services (often measured as GDP). Economic policies that favour economic growth are generally based on an implicit but erroneous assumption that the supply of natural resources is unlimited. A society that aims for economic growth is therefore bound to fail at one point or another.
To overcome these perceptions and conflicts, scientists, policy makers, and conservation managers are increasingly highlighting the need for sustainable development—economic activities that satisfy both present and future needs without compromising the natural world (Figure 15.1). Sustainable development is closely linked to economic development, a multi-dimensional concept that describes economic activities that aim to improve income and health without necessarily increasing consumption of natural resources. We should thus all strive for sustainable development, which emphasises economic development without unsustainable economic growth.
Sustainable development aims to satisfies present and future needs without compromising the natural world.
There are many good examples across Africa that illustrate the progress made towards sustainable development. For instance, many governments are investing in national parks and their infrastructure (such as staff and facilities) to protect biological diversity and provide economic opportunities for local communities. Similarly, stakeholders in large projects are increasingly engaging with one another to mitigate the negative impacts of infrastructure developments. One prime example was the 2015 Pan-African Business and Biodiversity Forum ( http://www.panbbf.org ), where representatives from business, governments, civil society, academia, development organizations, and financial institutions from across Africa came together to discuss how sustainable development can benefit nature, people, and business.
Unfortunately, there are also people and organizations that are taking advantage of this positive energy by misusing the term “sustainable development” to greenwash industrial activities that are harming the environment. For instance, a plan to establish a palm oil plantation that would damage a forested wilderness should not be considered sustainable development simply because the company agrees to protect a small plot of forest adjacent to the damaged area (see biodiversity offsets, Section 10.3.3). Similarly, many environmentally-destructive companies try to mislead customers with “environmentally-friendly” (often green-coloured) imagery on packages which are otherwise no better than the standard manufactured products. It is, therefore, critical for scientists, policy makers, and citizens to carefully study the issues, understand why different groups make arguments, and make thoughtful decisions about which actions or policies will best meet seemingly contradictory demands.


8.5: Imagery and Memory
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Besides the imagery debate, which is concerned with the question how we imagine for example objects, persons, situations and involve our senses in these mental pictures, questions concerning the memory are still untreated. In this part of the chapter about imagery we are dealing with the questions how images are encoded in the brain, and how they are recalled out of our memory. In search of answering these questions three major theories evolved. All of them explain the encoding and recalling processes different, and as usual validating experiments were realised for all these theories.
In search of answering these questions, three major streams evolved. All of them try to explain the encoding and recalling processes differently and, as usual, validating experiments were realised in all streams.
The common-code theory
This view of memory and recall theories that images and words access semantic information in a single conceptual system that is neither word-like nor spatial-like. The model of common-code hypothesis that for example images and words both require analogous processing before accessing semantic information. So the semantic information of all sensational input is encoded in the same way. The consequence is that when you remember for example a situation where you were watching an apple falling down a tree, the visual information about the falling of the apple and the information about the sound, which appeared when the apple hit the ground, both are constructed on – the – fly in the specific brain regions (e.g. visual images in the visual cortex) out of one code stored in the brain. Another difference of this model is that it claims images require less time than words for accessing the common conceptual system. Therefore images need less time to be discriminated, because they share a smaller set of possible alternatives than words. Apart from that words have to be picked out of a much larger set of ambiguous possibilities in the mental dictionary. The heaviest point of criticism on this model is that it does not declare where this common code is stored at the end.
The abstract-propositional theory
This theory rejects any notion of the distinction between verbal and non - verbal modes of representation, but instead describes representations of experience or knowledge in terms of an abstract set of relations and states, in other words propositions . This theory postulates that the recall of images is better if the one who is recalling the image has some connection to the meaning of the image which is recalled. For example if you are looking at an abstract picture on which a bunch of lines is drawn, which you cannot combine in a meaningful way with each other, the recall process of this picture will be very hard (if not impossible). As reason for this it is assumed, that there is no connection to propositions, which can describe some part of the picture, and no connection to a propositional network, which reconstructs parts of the picture. The other case is, that you look at a picture with some lines in it, which you can combine in a meaningful way with each other. The recall process should be successful, because in this case you can scan for a proposition which has at least one attribute with the meaning of the image you recognised. Then this proposition returns the information which is necessary to recall it.
The dual-code theory
Unlike the common-code and abstract-propositional approaches, this model postulates that words and images are represented in functionally distinct verbal and non - verbal memory systems. To establish this model, Roland and Fridberg (1985) had run an experiment, in which the subjects had either to imagine a mnemonic or how they walk the way to their home through their neighbourhoods. While the subjects did one of this tasks, their brain was scanned with the positron emission tomography (PET). Figure 10 is a picture combining the brains of the subjects, which achieved the first and the second task.
Figure 10: Green dots represent regions which showed a higher activity during the walking home task; yellow dots represent regions which showed a higher activity during the mnemonic task.
As we can see on the picture, for the processing of verbal and spatial information different brain areas are involved. The brain areas, which were active during the walking home task, are the same areas which are active during the visual perception and the information processing. And among those areas which showed activity while the mnemonic task was carried out, the Broca-centre is included, where normally language processing is located. This can be considered as an evidence for both representation types to be somehow connected with the modalities, as Paivio’s theory about dual-coding suggests Anderson (1996). Can you imagine other examples, which argue for the dual-code theory? For example, you walk along the beach in the evening, there are some beach bars ahead.  While you drink your drink, you try to remember the name of this person, but you fail stranded, even if you can remember where you have seen the person the last time, and perhaps what you have talked about in that situation. Now imagine another situation.  You are sure that you know that song, but you cannot remember the name of the interpreter, nor the name of the song either where you have heard it. Both examples can be interpreted as indicators for the assumption that in these situations you can recall the information which you perceived in the past, but you fail in remembering the propositions you connected to them.
In this area of research, there are of course other unanswered questions, for example why we cannot imagine smell, how the recall processes are performed or where the storage of images is located. The imagery debate is still going on, and ultimate evidence showing which of the models explains the connection between imagery and memory are missing. For now the dual-code theory seems to be the most promising model.


6.14: Collisions between like Gas Molecules
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When we consider collisions between different gas molecules of the same substance, we can denote the relative velocity and the expected value of the relative velocity as \(v_{11}\) and \(\left\langle v_{11}\right\rangle\), respectively. By the argument we make above, we can find the number of collisions between any one of these molecules and all of the others. Letting this collision frequency be \({\widetilde{\nu }}_{11}\), we find
\[\widetilde{\nu }_{11}=N_1\pi {\sigma }^2_{11}\left\langle v_{11}\right\rangle, \nonumber \]
where \({\sigma }_{11}=2{\sigma }_1\). Since we have
\[\left\langle v_{11}\right\rangle =\sqrt{2}\left\langle v_1\right\rangle, \nonumber \]
while
\[\left\langle v_1\right\rangle =\sqrt{{8kT}/{\pi }m_1}, \nonumber \]
we have \(\left\langle v_{11}\right\rangle =4\sqrt{{kT}/{\pi }m_1}\). The frequency of collisions between molecules of the same substance becomes
\[{\widetilde{\nu }}_{11}=N_1\pi {\sigma }^2_{11}\left\langle v_{11}\right\rangle =4N_1{\sigma }^2_{11}{\left(\frac{\pi kT}{m_1}\right)}^{1/2} \nonumber \]
The mean time between collisions, \({\tau }_{11}\), is
\[{\tau }_{11}={1}/{\widetilde{\nu}_{11}} \nonumber \]
and the mean free path, \({\lambda }_{11}\),
\[{\lambda }_{11}=\left\langle v_1\right\rangle {\tau }_{11}=\frac{1}{\sqrt{2}N}_1\pi {\sigma }^2_{11} \nonumber \]
When we consider the rate of collisions between all of the molecules of type \(1\) in a container, \({\rho }_{11}\), there is a minor complication. If we multiply the collision frequency per molecule, \({\widetilde{\nu }}_{11}\), by the number of molecules available to undergo such collisions, \(N_1\), we count each collision twice, because each such collision involves two type \(1\) molecules. To find the collision rate among like molecules, we must divide this product by 2. That is,
\[{\rho }_{11}=\frac{N_1{\widetilde{\nu }}_{11}}{2}=2N^2_1{\sigma }^2_{11}{\left(\frac{\pi kT}{m_1}\right)}^{1/2} \nonumber \]


4.4: Quantum Field Theory
Field operators
So far, we have been agnostic about the nature of the single-particle states \(\{|\varphi_1\rangle,|\varphi_2\rangle,\dots\}\) used to construct the creation and annihilation operators. Let us now consider the special case where these quantum states are representable by wavefunctions. Let \(|\mathbf{r}\rangle\) denote a position eigenstate for a \(d\) -dimensional space. A single-particle state \(|\varphi_\mu\rangle\) has a wavefunction
\[\varphi_\mu(\mathbf{r}) = \langle\mathbf{r}|\varphi_\mu\rangle.\]
Due to the completeness and orthonormality of the basis, these wavefunctions satisfy
\[\begin{align} \begin{aligned}\int d^dr \; \varphi_\mu^*(\mathbf{r})\, \varphi_\nu(\mathbf{r}) &= \langle\varphi_\mu| \left(\int d^dr\, |\mathbf{r}\rangle\langle\mathbf{r}|\right) |\varphi_\nu\rangle \,= \delta_{\mu\nu}, \\\sum_\mu \varphi_\mu^*(\mathbf{r}) \varphi_\mu(\mathbf{r}') &=\, \langle \mathbf{r}'| \left(\sum_\mu |\varphi_\mu\rangle\langle\varphi_\mu|\right)|\mathbf{r}\rangle \;\;=\, \delta^d(\mathbf{r}-\mathbf{r}').\end{aligned}\end{align}\]
We can use the wavefunctions and the creation/annihilation operators to construct a new and interesting set of operators. For simplicity, suppose the particles are bosons, and let
\[\hat{\psi}(\mathbf{r}) = \sum_\mu \varphi_\mu(\mathbf{r}) \, \hat{a}_\mu, \quad\;\; \hat{\psi}^\dagger(\mathbf{r}) = \sum_\mu \varphi_\mu^*(\mathbf{r}) \, \hat{a}_\mu^\dagger.\]
Using the aforementioned wavefunction properties, we can derive the inverse relations
\[\hat{a}_\mu = \int d^dr \; \varphi_\mu^*(\mathbf{r}) \, \hat{\psi}(\mathbf{r}), \quad\;\; \hat{a}_\mu^\dagger = \int d^dr \; \varphi_\mu(\mathbf{r}) \, \hat{\psi}^\dagger(\mathbf{r}).\]
From the commutation relations for the bosonic \(a_\mu\) and \(a_\mu^\dagger\) operators, we can show that
\[\left[\hat{\psi}(\mathbf{r}), \hat{\psi}(\mathbf{r}')\right] = \left[\hat{\psi}^\dagger(\mathbf{r}), \hat{\psi}^\dagger(\mathbf{r}')\right] = 0, \quad \left[\hat{\psi}(\mathbf{r}), \hat{\psi}^\dagger(\mathbf{r}')\right] = \delta^d(\mathbf{r}-\mathbf{r}').\]
In the original commutation relations, the operators for different single-particle states commute; now, the operators for different positions commute. A straightforward interpretion for the operators \(\hat{\psi}^\dagger(\mathbf{r})\) and \(\hat{\psi}(\mathbf{r})\) is that they respectively create and annihilate one particle at a point \(\mathbf{r}\) (rather than one particle in a given eigenstate).
It is important to note that \(\mathbf{r}\) here does not play the role of an observable. It is an index , in the sense that each \(\mathbf{r}\) is associated with distinct \(\hat{\psi}(\mathbf{r})\) and \(\hat{\psi}^\dagger(\mathbf{r})\) operators. These \(\mathbf{r}\) -dependent operators serve to generalize the classical concept of a field . In a classical field theory, each point \(\mathbf{r}\) is assigned a set of numbers corresponding to physical quantities, such as the electric field components \(E_x(\mathbf{r})\) , \(E_y(\mathbf{r})\) , and \(E_z(\mathbf{r})\) . In the present case, each \(\mathbf{r}\) is assigned a set of quantum operators. This kind of quantum theory is called a quantum field theory .
We can use the \(\hat{\psi}(\mathbf{r})\) and \(\hat{\psi}^\dagger(\mathbf{r})\) operators to write second quantized observables in a way that is independent of the choice of single-particle basis wavefunctions. As discussed in the previous section, given a Hermitian single-particle operator \(\hat{A}_1\) we can define a multi-particle observable \(\hat{A} = \sum_{\mu\nu} \hat{a}^\dagger_\mu A_{\mu\nu} \hat{a}_\nu\) , where \(A_{\mu\nu} = \langle\varphi_\mu|\hat{A}_1|\varphi_\nu\rangle\) . This multi-particle observable can be re-written as
\[\hat{A} = \int d^dr \,d^dr' \;\hat{\psi}^\dagger(\mathbf{r}) \;\langle \mathbf{r}|\hat{A}_1|\mathbf{r}'\rangle \; \hat{\psi}(\mathbf{r}'),\]
which makes no explicit reference to the single-particle basis states.
For example, consider the familiar single-particle Hamiltonian describing a particle in a potential \(V(\mathbf{r})\) :
\[\hat{H}_1 = \hat{T}_1 + \hat{V}_1, \;\;\; \hat{T}_1 = \frac{|\hat{\mathbf{p}}|^2}{2m}, \;\;\; \hat{V}_1 = V(\hat{\mathbf{r}}),\]
where \(\hat{\mathbf{r}}\) and \(\hat{\mathbf{p}}\) are position and momentum operators (single-particle observables). The corresponding second quantized operators for the kinetic energy and potential energy are
\[\begin{align} \begin{aligned}\hat{T} &= \frac{\hbar^2}{2m} \int d^dr\; d^dr'\; \hat{\psi}^\dagger(\mathbf{r}) \left(\int\frac{d^dk}{(2\pi)^d} \; |\mathbf{k}|^2 \; e^{i\mathbf{k}\cdot(\mathbf{r}-\mathbf{r}')}\right) \; \hat{\psi}(\mathbf{r}') \\ &= \frac{\hbar^2}{2m} \int d^d r\; \nabla \hat{\psi}^\dagger (\mathbf{r}) \, \cdot \, \nabla \hat{\psi}(\mathbf{r}) \\ \hat{V} &= \int d^dr \;\hat{\psi}^\dagger(\mathbf{r}) \;V(\mathbf{r}) \; \hat{\psi}(\mathbf{r}).\end{aligned}\end{align}\]
(In going from the first to the second line, we performed integrations by parts.) This result is strongly reminiscent of the expression for the expected kinetic and potential energies in single-particle quantum mechanics:
\[\langle T\rangle = \frac{\hbar^2}{2m} \int d^d r \left|\nabla\psi(\mathbf{r})\right|^2, \;\;\; \langle V\rangle = \int d^dr \; V(\mathbf{r}) \; |\psi(\mathbf{r})|^2,\]
where \(\psi(\mathbf{r})\) is the single-particle wavefunction.
How are the particle creation and annihilation operators related to the classical notion of “the value of a field at point \(\mathbf{r}\) ”, like an electric field \(\mathbf{E}(\mathbf{r})\) or magnetic field \(\mathbf{B}(\mathbf{r})\) ? Field variables are measurable quantities, and should be described by Hermitian operators. As we have just seen, Hermitian operators corresponding to the kinetic and potential energy can be constructed via products of \(\hat{\psi}^\dagger(\mathbf{r})\) with \(\hat{\psi}(\mathbf{r})\) . But there is another type of Hermitian operator that we can construct by taking linear combinations of of \(\hat{\psi}^\dagger(\mathbf{r})\) with \(\hat{\psi}(\mathbf{r})\) . One example is
\[\psi(\mathbf{r}) + \psi(\mathbf{r})^\dagger.\]
Other possible Hermitian operators have the form
\[F(\mathbf{r}) = \int d^dr' \Big(f(\mathbf{r},\mathbf{r}') \hat{\psi}(\mathbf{r}) + f^*(\mathbf{r},\mathbf{r}') \hat{\psi}^\dagger(\mathbf{r}') \Big), \label{fieldop}\]
where \(f(\mathbf{r},\mathbf{r}')\) is some complex function. As we shall see, it is this type of Hermitian operator that corresponds to the classical notion of a field variable like an electric or magnetic field.
In the next two sections, we will try to get a better understanding of the relationship between classical fields and bosonic quantum fields. (For fermionic quantum fields, the situation is more complicated; they cannot be related to classical fields of the sort we are familiar with, for reasons that lie outside the scope of this course.)
Revisiting the harmonic oscillator
Before delving into the links between classical fields and bosonic quantum fields, it is first necessary to revisit the harmonic oscillator, to see how the concept of a mode of oscillation carries over from classical to quantum mechanics.
A classical harmonic oscillator is described by the Hamiltonian
\[H(x,p) = \frac{p^2}{2m} + \frac{1}{2}m\omega^2x^2, \label{Hclass}\]
where \(x\) is the “position” of the oscillator, which we call the oscillator variable ; \(p\) is the corresponding momentum variable; \(m\) is the mass; and \(\omega\) is the natural frequency of oscillation. We know that the classical equation of motion has the general form
\[x(t) = \mathcal{A}\, e^{-i\omega t} + \mathcal{A}^*\, e^{i\omega t}. \label{xt0}\]
This describes an oscillation of frequency \(\omega\) . It is parameterized by the mode amplitude \(\mathcal{A}\) , a complex number that determines the magnitude and phase of the oscillation.
For the quantum harmonic oscillator, \(x\) and \(p\) are replaced by the Hermitian operators \(\hat{x}\) and \(\hat{p}\) . From these, the operators \(\hat{a}\) and \(\hat{a}^\dagger\) can be defined:
\[\begin{align} \left\{ \begin{aligned} \hat{a} &= \sqrt{\frac{m\omega}{2\hbar}} \left(\hat{x} + \frac{i\hat{p}}{m\omega}\right), \\ \hat{a}^\dagger &= \sqrt{\frac{m\omega}{2\hbar}} \left(\hat{x} - \frac{i\hat{p}}{m\omega}\right). \end{aligned}\right. \quad \Leftrightarrow\quad\left\{ \begin{aligned} \hat{x} &= \;\;\;\;\sqrt{\frac{\hbar}{2m\omega}} \, \left(\hat{a}+\hat{a}^\dagger\right) \\ \hat{p} &= -i \sqrt{\frac{m\omega\hbar}{2}} \, \left(\hat{a}-\hat{a}^\dagger\right). \end{aligned}\right. \label{create_annihilate}\end{align}\]
We can then show that
\[\left[\,\hat{a}\,,\,\hat{a}^\dagger\,\right] = 1, \;\;\; \hat{H} = \hbar\omega\left(\hat{a}^\dagger\hat{a}+\frac{1}{2}\right), \label{create_annihilate_qho}\]
and from these the energy spectrum of the quantum harmonic oscillator can be derived. These facts should have been covered in an earlier course.
Here, we are interested in how the creation and annihilation operators relate to the dynamics of the quantum harmonic oscillator. In the Heisenberg picture, with \(t = 0\) as the reference time, we define the time-dependent operator
\[\hat{x}(t) = \hat{U}^\dagger(t) \,\hat{x}\, \hat{U}(t), \;\;\; \hat{U}(t) \equiv \exp\left(-\frac{i}{\hbar}\hat{H}t\right).\]
We will adopt the convention that all operators written with an explicit time dependence are Heisenberg picture operators, while operators without an explicit time dependence are Schrödinger picture operators; hence, \(\hat{x} \equiv \hat{x}(0)\) . The Heisenberg picture creation and annihilation operators, \(\hat{a}^\dagger(t)\) and \(\hat{a}(t)\) , are related to \(\hat{x}(t)\) by
\[\hat{x}(t) = \sqrt{\frac{\hbar}{2m\omega}} \, \Big(\hat{a}(t)+\hat{a}^\dagger(t)\Big). \label{xt1}\]
The Heisenberg equation for the annihilation operator is
\[\begin{align} \begin{aligned} \frac{d\hat{a}(t)}{dt} &= \frac{i}{\hbar} \left[\hat{H},\hat{a}(t)\right] \\ &= \frac{i}{\hbar} \,\hat{U}^\dagger(t) \left[\hat{H},\hat{a}\right] \hat{U}(t) \\ &= \frac{i}{\hbar} \,\hat{U}^\dagger(t) \Big(-\hbar\omega \hat{a}\Big) \hat{U}(t) \\ &= -i\omega \hat{a}(t). \end{aligned} \label{heisenbergeq}\end{align}\]
Hence, the solution for this differential equation is
\[\hat{a}(t) = \hat{a} \, e^{-i\omega t}, \label{at}\]
and Equation \(\eqref{xt1}\) becomes
\[\hat{x}(t) = \sqrt{\frac{\hbar}{2m\omega}} \, \Big(\hat{a} \,e^{-i\omega t} \,+\, \hat{a}^\dagger\, e^{i\omega t}\Big). \label{xtquantum}\]
This has exactly the same form as the classical oscillatory solution \(\eqref{xt0}\)! Comparing the two, we see that \(\hat{a}\) times the scale factor \(\sqrt{\hbar/2m\omega}\) plays the role of the mode amplitude \(\mathcal{A}\) .
Now, suppose we come at things from the opposite end. Let’s say we start with creation and annihilation operators satisfying Equation \(\eqref{create_annihilate_qho}\), from which Equations \(\eqref{heisenbergeq}\)–\(\eqref{at}\) follow. Using the creation and annihilation operators, we would like to construct an observable that corresponds to a classical oscillator variable. A natural Hermitian ansatz is
\[\hat{x}(t) = \mathcal{C} \left(\hat{a} \,e^{-i\omega t} \,+\, \hat{a}^\dagger\, e^{i\omega t}\right), \label{xtc}\]
where \(\mathcal{C}\) is a constant that is conventionally taken to be real.
How might \(\mathcal{C}\) be chosen? A convenient way is to study the behavior of the oscillator variable in the classical limit . The classical limit of a quantum harmonic oscillator is described by a coherent state . The details of how this state is defined need not concern us for now (see Appendix E). The most important things to know are that (i) it can be denoted by \(|\alpha\rangle\) where \(\alpha \in \mathbb{C}\) , (ii) it is an eigenstate of the annihilation operator:
\[\hat{a}|\alpha\rangle = \alpha|\alpha\rangle. \label{coherent}\]
And (iii) its energy expectation value is
\[\langle E\rangle = \ \langle\alpha|\hat{H}|\alpha\rangle = \hbar \omega \left(|\alpha|^2 + \frac{1}{2}\right) \overset{|\alpha|^2\rightarrow\infty}{\longrightarrow} \hbar \omega |\alpha|^2. \label{Havg1}\]
When the system is in a coherent state, we can effectively substitute the \(\hat{a}\) and \(\hat{a}^\dagger\) operators in Equation \(\eqref{xtc}\) with the complex numbers \(\alpha\) and \(\alpha^*\) , which gives a classical trajectory
\[x_{\mathrm{classical}}(t) = \mathcal{C} \left(\alpha \,e^{-i\omega t} \,+\, \alpha^*\, e^{i\omega t}\right). \label{xtc2}\]
This trajectory has amplitude \(2\mathcal{C}|\alpha|\) . At maximum displacement, the classical momentum is zero, so the total energy of the classical oscillator must be
\[E_{\mathrm{classical}} = \frac{1}{2} m\omega^2 \Big(2\mathcal{C}|\alpha|\Big)^2 = 2m\omega^2 \mathcal{C}^2 |\alpha|^2. \label{Eclassical}\]
Equating the classical energy \(\eqref{Eclassical}\) to the coherent state energy \(\eqref{Havg1}\) gives
\[\mathcal{C} = \sqrt{\frac{\hbar}{2m\omega}},\]
which is precisely the scale factor found in Equation \(\eqref{xtquantum}\).
A scalar boson field
We now have the tools available to understand the connection between a very simple classical field and its quantum counterpart. Consider a classical scalar field variable \(f(x,t)\) , defined in one spatial dimension, whose classical equation of motion is the wave equation:
\[\frac{\partial^2\! f(x,t)}{\partial x^2} = \frac{1}{c^2} \frac{\partial^2\! f(x,t)}{\partial t^2}. \label{waveeqn}\]
The constant \(c\) is a wave speed. This sort of classical field arises in many physical contexts, including the propagation of sound through air, in which case \(c\) is the speed of sound.
For simplicity, let us first assume that the field is defined within a finite interval of length \(L\) , with periodic boundary conditions: \(f(x,t) \equiv f(x+L, t)\) . Solutions to the wave equation can be described by the following ansatz:
\[f(x,t) = \sum_n \left( \mathcal{A}_n\, \varphi_n(x) \, e^{-i\omega_n t} + \mathcal{A}_n^*\, \varphi_n^*(x) \, e^{i\omega_n t}\right). \label{fnorm}\]
This ansatz describes a superposition of normal modes . Each normal mode (labelled \(n\) ) varies harmonically in time with a mode frequency \(\omega_n\) , and varies in space according to a complex mode profile \(\varphi_n(x)\) ; its overall magnitude and phase is specified by the mode amplitude \(\mathcal{A}_n\) . The mode profiles are normalized according to some fixed convention, e.g.
\[\int_0^L dx \, |\varphi_n(x)|^2 = 1.\]
Substituting Equation \(\eqref{fnorm}\) into Equation \(\eqref{waveeqn}\), and using the periodic boundary conditions, gives
\[\begin{align} \varphi_n(x) = \frac{1}{\sqrt{L}}\, \exp\left(ik_n x\right), \;\;\; \omega_n = c k_n = \frac{2\pi c n}{L}, \;\;\; n \in \mathbb{Z}. \label{classical_modes}\end{align}\]
These mode profiles are orthonormal:
\[\int_0^L dx \, \varphi_m^*(x) \, \varphi_n(x) = \delta_{mn}.\]
Each normal mode carries energy. By analogy with the classical harmonic oscillator—see Equations \(\eqref{xtc2}\)–\(\eqref{Eclassical}\)—we assume that the energy density (i.e., energy per unit length) is proportional to the square of the field variable. Let it have the form
\[U(x) = 2 \rho \sum_n|\mathcal{A}_n|^2 \, |\varphi_n(x)|^2, \label{Edensity}\]
where \(\rho\) is some parameter that has to be derived from the underlying physical context. For example, for acoustic modes, \(\rho\) is the mass density of the underlying acoustic medium; in the next chapter, we will see a concrete example involving the energy density of an electromagnetic mode. From Equation \(\eqref{Edensity}\), the total energy is
\[E = \int_0^L dx \; U(x) = 2 \rho \sum_n |\mathcal{A}_n|^2. \label{classical_energy}\]
To quantize the classical field, we treat each normal mode as an independent oscillator, with creation and annihilation operators \(\hat{a}_n^\dagger\) and \(\hat{a}_n\) satisfying
\[\big[\,\hat{a}_m\,,\, \hat{a}_n^\dagger\big] = \delta_{mn}, \;\;\; \big[\,\hat{a}_m\,,\, \hat{a}_n\big] = \big[\,\hat{a}_m^\dagger\,,\, \hat{a}_n^\dagger\big] = 0.\]
We then take the Hamiltonian to be that of a set of independent harmonic oscillators:
\[\hat{H} = \sum_n \hbar \omega_n \hat{a}_n^\dagger \hat{a}_n + E_0,\]
where \(E_0\) is the ground-state energy. Just like in the previous section, we can define a Heisenberg-picture annihilation operator, and solving its Heisenberg equation yields
\[\hat{a}_n(t) = \hat{a}_n e^{-i\omega_n t}.\]
We then define a Schrödinger picture Hermitian operator of the form
\[\hat{f}(x) = \sum_n \mathcal{C}_n \Big(\hat{a}_n \varphi_n(x) + \hat{a}_n^\dagger \varphi_n^*(x) \Big),\]
where \(\mathcal{C}_n\) is a real constant (one for each normal mode). The corresponding Heisenberg picture operator is
\[\hat{f}(x,t) = \sum_n \mathcal{C}_n \Big(\hat{a}_n \varphi_n(x) \,e^{-i\omega_nt} + \hat{a}_n^\dagger \varphi_n^*(x) \, e^{i\omega_nt} \Big), \label{fxt}\]
which is the quantum version of the classical solution \(\eqref{fnorm}\).
To determine the \(\mathcal{C}_n\) scale factors, we consider the classical limit. The procedure is a straightforward generalization of the harmonic oscillator case discussed in Section 4.4. We introduce a state \(|\alpha\rangle\) that is a coherent state for all the normal modes; i.e., for any given \(n\) ,
\[\hat{a}_n |\alpha\rangle = \alpha_n |\alpha\rangle\]
for some \(\alpha_n \in \mathbb{C}\) . The energy expectation value is
\[\langle E\rangle = \sum_n \hbar \omega_n |\alpha_n|^2. \label{Ecoherent_field}\]
In the coherent state, the \(\hat{a}_n\) and \(\hat{a}_n^\dagger\) operators in Equation \(\eqref{fxt}\) can be replaced with \(\alpha_n\) and \(\alpha_n^*\) respectively. Hence, we identify \(\mathcal{C}_n \alpha_n\) as the classical mode amplitude \(\mathcal{A}_n\) in Equation \(\eqref{fnorm}\). In order for the classical energy \(\eqref{classical_energy}\) to match the coherent state energy \(\eqref{Ecoherent_field}\), we need
\[2\rho |\mathcal{A}_n|^2 = 2 \rho |\mathcal{C}_n\alpha_n|^2 = \hbar \omega_n |\alpha_n|^2 \quad \Rightarrow \quad \mathcal{C}_n = \sqrt{\frac{\hbar\omega_n}{2\rho}}.\]
Hence, the appropriate field operator is
\[\hat{f}(x,t) = \sum_n \sqrt{\frac{\hbar\omega_n}{2\rho}} \Big(\hat{a}_n \varphi_n(x) \,e^{-i\omega_nt} + \hat{a}_n^\dagger \varphi_n^*(x) \, e^{i\omega_nt} \Big).\]
Returning to the Schrödinger picture, and using the explicit mode profiles from Equation \(\eqref{classical_modes}\), we get
\[\hat{f}(x) = \sum_n \sqrt{\frac{\hbar\omega_n}{2\rho L}} \Big(\hat{a}_n e^{ik_n x} + \hat{a}_n^\dagger e^{-ik_n x} \Big).\]
Finally, if we are interested in the infinite- \(L\) limit, we can convert the sum over \(n\) into an integral. The result is
\[\hat{f}(x) = \int dk\; \sqrt{\frac{\hbar\omega(k)}{4\pi\rho}} \Big(\hat{a}(k)\, e^{ik x} + \hat{a}^\dagger(k)\, e^{-ik x} \Big),\]
where \(\hat{a}(k)\) denotes a rescaled annihilation operator defined by \(\hat{a}_n \rightarrow \sqrt{2\pi/L}\; \hat{a}(k)\) , satisfying
\[\begin{align} \Big[\,\hat{a}(k)\,,\, \hat{a}^\dagger(k')\,\Big] = \delta(k-k').\end{align}\]
Looking ahead
In the next chapter, we will use these ideas to formulate a quantum theory of electromagnetism. This is a bosonic quantum field theory in which the creation and annihilation operators act upon particles called photons —the elementary particles of light. Linear combinations of these photon operators can be used to define Hermitian field operators that correspond to the classical electromagnetic field variables. In the classical limit, the quantum field theory reduces to Maxwell’s theory of the electromagnetic field.
It is hard to overstate the importance of quantum field theories in physics. At a fundamental level, all elementary particles currently known to humanity can be described using a quantum field theory called the Standard Model. These particles are roughly divided into two categories. The first consists of “force-carrying” particles: photons (which carry the electromagnetic force), gluons (which carry the strong nuclear force), and the \(W/Z\) bosons (which carry the weak nuclear force); these particles are excitations of bosonic quantum fields, similar to the one described in the previous section. The second category consists of “particles of matter”, such as electrons, quarks, and neutrinos; these are excitations of fermionic quantum fields, whose creation and annihilation operators obey anticommutation relations.
As Wilczek (1999) has pointed out, the modern picture of fundamental physics bears a striking resemblance to the old idea of “luminiferous ether”: a medium filling all of space and time, whose vibrations are physically-observable light waves. The key difference, as we now understand, is that the ether is not a classical medium, but one obeying the rules of quantum mechanics. (Another difference, which we have not discussed so far, is that modern field theories can be made compatible with relativity.)
It is quite compelling to think of fields, not individual particles, as the fundamental objects in the universe. This point of view “explains”, in a sense, why all particles of the same type have the same properties (e.g., why all electrons in the universe have exactly the same mass). The particles themselves are not fundamental; they are excitations of deeper, more fundamental entities—quantum fields!


1.24: Misc Last updated Save as PDF Page ID 375421 Michael J Blandamer & Joao Carlos R Reis University of Leicester & Faculdade de Ciencias 1.14.1: Adiabatic 1.14.2: Adsorption- Langmuir Adsorption Isotherm- One Adsorbate 1.14.3: Absorption Isotherms - Two Absorbates 1.14.4: Apparent Molar Properties- Solutions- Background 1.14.5: Apparent Molar Properties- Solutions- General 1.14.6: Axioms 1.14.7: Boundary 1.14.8: Calculus 1.14.9: Clausius - Clapeyron Equation 1.14.10: Gibbs - Helmholtz Equation 1.14.11: Guggenheim-Scatchard Equation / Redlich-Kister Equation 1.14.12: Legendre Transformations 1.14.13: Closed System 1.14.14: Cohesive Energy Density 1.14.15: Degree of Dissociation 1.14.16: Energy and First Law of Thermodynamics 1.14.17: Energy and Entropy 1.14.18: Electrochemical Units 1.14.19: Electrical Units 1.14.20: Electric Conductivities of Salt Solutions- Dependence on Composition 1.14.21: Donnan Membrane Equilibria 1.14.22: Descriptions of Systems 1.14.23: Enzyme-Substrate Interaction 1.14.24: Equation of State- General Thermodynamics 1.14.25: Equation of State- Perfect Gas 1.14.26: Equation of State - Real Gases, van der Waals, and Other Equations 1.14.27: Euler's Theorem 1.14.28: First Law of Thermodynamics 1.14.29: Functions of State 1.14.30: Heat, Work, and Energy 1.14.31: Helmholtz Energy 1.14.32: Hildebrand Solubility Parameter 1.14.33: Infinite Dilution 1.14.34: Internal Pressure 1.14.35: Internal Pressure: Liquid Mixtures: Excess Property 1.14.36: Irreversible Thermodynamics 1.14.37: Irreversible Thermodynamics: Onsager Phenomenological Equations 1.14.38: Joule-Thomson Coefficient 1.14.39: Kinetic Salt Effects 1.14.40: Laws of Thermodynamics 1.14.41: Lewisian Variables 1.14.42: L'Hospital's Rule 1.14.43: Master Equation 1.14.44: Maxwell Equations 1.14.45: Moderation 1.14.46: Molality and Mole Fraction 1.14.47: Newton-Laplace Equation 1.14.48: Open System 1.14.49: Osmotic Coefficient 1.14.50: Osmotic Pressure 1.14.51: Partial Molar Properties: General 1.14.52: Partial Molar Properties: Definitions 1.14.53: Phase Rule 1.14.54: Poynting Relation 1.14.55: Process 1.14.56: Properties: Equilibrium and Frozen 1.14.57: Reversible Change 1.14.58: Reversible Chemical Reactions 1.14.59: Salting-In and Salting-Out 1.14.60: Second Law of Thermodynamics 1.14.61: Solubility Products 1.14.62: Solubilities of Gases in Liquids 1.14.63: Solubilities of Solids in Liquids 1.14.64: Solutions: Solute and Solvent 1.14.65: Spontaneous Change: Isothermal and Isobaric 1.14.66: Spontaneous Chemical Reaction 1.14.67: Standard States: Reference States: Processes 1.14.68: Surroundings and System 1.14.69: Temperature of Maximum Density: Aqueous Solutions 1.14.70: Solutions: Neutral Solutes: Inter-Solute Distances 1.14.71: Time and Thermodynamics (Timenote) 1.14.72: Variables: Independent and Dependent 1.14.73: Variables: Gibbsian and Non-Gibbsian 1.14.74: Vaporization 1.14.75: Viscosities: Salt Solutions 1.14.76: Work


3.7: Assignment- Accounting Theory
This assignment is available for download as a Word Document.
Read the following scenario and complete the questions and tasks below.
Marcy recently started her first job after graduating from business school with her accounting degree. She is very excited to be working in her field and although she knows she still has a lot to learn, she already feels like she is getting a good feel for what it takes to be successful. She has been assigned to work under one of the senior accountants, Mrs. Bradlee, and so far, she has been easy to work with and very helpful. For the last several weeks Marcy has spent most of her time entering data into the firm’s computer system in preparation for the issuance of their clients’ annual reports. On Friday, as she is preparing to leave for the day her supervisor calls her into her office.
“Well, Marcy, still enjoying your work here with us?” Mrs. Bradlee asks.
“Oh, yes ma’am, very much. I know I am still new and have a great deal to learn, but I am very happy with my job,” Marcy replies.
“That’s good to hear, Marcy. We are all very happy that you are here and I think you will be a valuable asset to the firm. In fact, I have something that I would like for you to do for me. I understand you are available to work over the weekend, is that correct?” Mrs. Bradlee inquired.
“Yes, I am happy to come in over the weekend if you need me to!” Marcy replied.
“Excellent,” Mrs. Bradlee said as she pulled a folder from her desk drawer. Handing the folder to Marcy she continued, “we identified some errors in the data that was entered for the Dale account and I would consider it a personal favor if you could get these numbers corrected over the weekend. I have already taken the liberty of alerting building security, so you won’t need to worry about getting into the office. I don’t expect anyone else to be here, are you okay with being in the office alone?”
Marcy took the folder from Mrs. Bradlee’s outstretched hand and opened it to see what she had committed to. There were several pages of entries to be corrected, but Marcy breathed a sigh of relief when she realized that the work was not that different from what she had been doing for the last several weeks.
“Oh, I will be just fine. I should be able to get all of these corrections made in a few hours, ma’am.” Marcy said.
“Well, then, that’s that,” said Mrs. Bradlee, smiling at Marcy.
Taking that as her queue to leave, Marcy gathered up the folder and her belongings and headed towards the door. Just as she got to the door Mrs. Bradlee spoke again.
“Oh, and Marcy, I would appreciate it very much if you didn’t discuss this matter with your colleagues. It, ummm, might be an embarrassment to Mark Murphy. He made the original entries and I would hate for him to look bad in front of the other junior partners. You understand, I’m sure.” Mrs. Bradlee said.
“Of course, Mrs. Bradlee, I wouldn’t want Mr. Murphy to be put in a bad position. He has been very helpful and I really like him.” Marcy replied.
“Then it’s settled. If you just get these few minor corrections made over the weekend we will be back on track to get the Dale annual report out on time. And please make sure that you put in overtime for your hours. I will sign off on them personally on Monday morning.” Mrs. Bradlee said as she turned her attention to the stack of phone messages on her desk.
Marcy thanked Mrs. Bradlee for giving her the opportunity to be of assistance as she walked out of her office, pulling the door closed behind her.
On Saturday morning Marcy got dressed and was seated at her desk at 8am. As she waited for her computer to boot up, she began looking through the documents Mrs. Bradlee had given her the day before. There were several pages of entries that had to be made and it slowly dawned on Marcy that all of the entries had something in common – they either increased revenue or decreased an expense. Several of the entries re-classified long-term debt as capital improvements.
“Well, these entries will totally change the financial position of the company,” Marcy said to herself. “How could Mr. Murphy have made this many errors?” she thought. “Something about this isn’t right,” Marcie thought out loud. Slouching back in her chair she suddenly found herself liking her job and Mrs. Bradlee less than she did when she had gotten up that morning.
- Putting yourself in Marcy’s position, what do you do next? Consider the outcome of her various courses of action and the long-term effects of her actions.
- What are the implications of the work Marcy has been asked to perform? Consider all of the stakeholders of accounting information when responding.
- Why would someone in Mrs. Bradlee’s position ask an employee to perform this kind of task?


18.1: Fields of Fractions
Every field is also an integral domain; however, there are many integral domains that are not fields. For example, the integers \({\mathbb Z}\) form an integral domain but not a field. A question that naturally arises is how we might associate an integral domain with a field. There is a natural way to construct the rationals \({\mathbb Q}\) from the integers: the rationals can be represented as formal quotients of two integers. The rational numbers are certainly a field. In fact, it can be shown that the rationals are the smallest field that contains the integers. Given an integral domain \(D\text{,}\) our question now becomes how to construct a smallest field \(F\) containing \(D\text{.}\) We will do this in the same way as we constructed the rationals from the integers.
An element \(p/q \in {\mathbb Q}\) is the quotient of two integers \(p\) and \(q\text{;}\) however, different pairs of integers can represent the same rational number. For instance, \(1/2 = 2/4 = 3/6\text{.}\) We know that
\[ \frac{a}{b} = \frac{c}{d} \nonumber \]
if and only if \(ad = bc\text{.}\) A more formal way of considering this problem is to examine fractions in terms of equivalence relations. We can think of elements in \({\mathbb Q}\) as ordered pairs in \({\mathbb Z} \times {\mathbb Z}\text{.}\) A quotient \(p/q\) can be written as \((p, q)\text{.}\) For instance, \((3, 7)\) would represent the fraction \(3/7\text{.}\) However, there are problems if we consider all possible pairs in \({\mathbb Z} \times {\mathbb Z}\text{.}\) There is no fraction \(5/0\) corresponding to the pair \((5,0)\text{.}\) Also, the pairs \((3,6)\) and \((2,4)\) both represent the fraction \(1/2\text{.}\) The first problem is easily solved if we require the second coordinate to be nonzero. The second problem is solved by considering two pairs \((a, b)\) and \((c, d)\) to be equivalent if \(ad = bc\text{.}\)
If we use the approach of ordered pairs instead of fractions, then we can study integral domains in general. Let \(D\) be any integral domain and let
\[ S = \{ (a, b) : a, b \in D \text{ and } b \neq 0 \}\text{.} \nonumber \]
Define a relation on \(S\) by \((a, b) \sim (c, d)\) if \(ad = bc\text{.}\)
Lemma \(18.1\)
The relation \(\sim\) between elements of \(S\) is an equivalence relation.
- Proof
-
Since \(D\) is commutative, \(ab = ba\text{;}\) hence, \(\sim\) is reflexive on \(D\text{.}\) Now suppose that \((a,b) \sim (c,d)\text{.}\) Then \(ad=bc\) or \(cb = da\text{.}\) Therefore, \((c,d) \sim (a, b)\) and the relation is symmetric. Finally, to show that the relation is transitive, let \((a, b) \sim (c, d)\) and \((c, d) \sim (e,f)\text{.}\) In this case \(ad = bc\) and \(cf = de\text{.}\) Multiplying both sides of \(ad = bc\) by \(f\) yields
\[ a f d = a d f = b c f = b d e = bed\text{.} \nonumber \]
Since \(D\) is an integral domain, we can deduce that \(af = be\) or \((a,b ) \sim (e, f)\text{.}\)
We will denote the set of equivalence classes on \(S\) by \(F_D\text{.}\) We now need to define the operations of addition and multiplication on \(F_D\text{.}\) Recall how fractions are added and multiplied in \({\mathbb Q}\text{:}\)
\begin{align*} \frac{a}{b} + \frac{c}{d} & = \frac{ad + b c}{b d};\\ \frac{a}{b} \cdot \frac{c}{d} & = \frac{ac}{b d}\text{.} \end{align*}
It seems reasonable to define the operations of addition and multiplication on \(F_D\) in a similar manner. If we denote the equivalence class of \((a, b) \in S\) by \([a, b]\text{,}\) then we are led to define the operations of addition and multiplication on \(F_D\) by
\[ [a, b] + [c, d] = [ad + b c,b d] \nonumber \]
and
\[ [a, b] \cdot [c, d] = [ac, b d]\text{,} \nonumber \]
respectively. The next lemma demonstrates that these operations are independent of the choice of representatives from each equivalence class.
Lemma \(18.2\)
The operations of addition and multiplication on \(F_D\) are well-defined.
- Proof
-
We will prove that the operation of addition is well-defined. The proof that multiplication is well-defined is left as an exercise. Let \([a_1, b_1] = [a_2, b_2]\) and \([c_1, d_1] =[ c_2, d_2]\text{.}\) We must show that
\[ [a_1 d_1 + b_1 c_1,b_1 d_1] = [a_2 d_2 + b_2 c_2,b_2 d_2] \nonumber \]
or, equivalently, that
\[ (a_1 d_1 + b_1 c_1)( b_2 d_2) = (b_1 d_1) (a_2 d_2 + b_2 c_2)\text{.} \nonumber \]
Since \([a_1, b_1] = [a_2, b_2]\) and \([c_1, d_1] =[ c_2, d_2]\text{,}\) we know that \(a_1 b_2 = b_1 a_2\) and \(c_1 d_2 = d_1 c_2\text{.}\) Therefore,
\begin{align*} (a_1 d_1 + b_1 c_1)( b_2 d_2) & = a_1 d_1 b_2 d_2 + b_1 c_1 b_2 d_2\\ & = a_1 b_2 d_1 d_2 + b_1 b_2 c_1 d_2\\ & = b_1 a_2 d_1 d_2 + b_1 b_2 d_1 c_2\\ & = (b_1 d_1) (a_2 d_2 + b_2 c_2)\text{.} \end{align*}
Lemma \(18.3\)
The set of equivalence classes of \(S\text{,}\) \(F_D\text{,}\) under the equivalence relation \(\sim\text{,}\) together with the operations of addition and multiplication defined by
\begin{align*} [a, b] + [c, d] & = [ad + b c, b d]\\ [ a, b] \cdot [c, d] & = [ac, b d]\text{,} \end{align*}
is a field.
- Proof
-
The additive and multiplicative identities are \([0,1]\) and \([1,1]\text{,}\) respectively. To show that \([0,1]\) is the additive identity, observe that
\[ [a, b] + [0, 1] = [ a 1 + b 0, b 1] = [a,b]\text{.} \nonumber \]
It is easy to show that \([1, 1]\) is the multiplicative identity. Let \([a, b] \in F_D\) such that \(a \neq 0\text{.}\) Then \([b, a]\) is also in \(F_D\) and \([a,b] \cdot [b, a] = [1,1]\text{;}\) hence, \([b, a]\) is the multiplicative inverse for \([a, b]\text{.}\) Similarly, \([-a,b]\) is the additive inverse of \([a, b]\text{.}\) We leave as exercises the verification of the associative and commutative properties of multiplication in \(F_D\text{.}\) We also leave it to the reader to show that \(F_D\) is an abelian group under addition.
It remains to show that the distributive property holds in \(F_D\text{;}\) however,
\begin{align*} [a, b] [e, f] + [c, d][ e, f ] & = [a e, b f ] + [c e, d f]\\ & = [a e d f + b f c e, b d f^2 ]\\ & = [a e d + b c e, b d f ]\\ & = [a d e + b c e, b d f ]\\ & = ( [a, b] + [c, d] ) [ e, f ] \end{align*}
and the lemma is proved.
The field \(F_D\) in Lemma 18.3 is called the field of fractions or field of quotients of the integral domain \(D\text{.}\)
Theorem \(18.4\)
Let \(D\) be an integral domain. Then \(D\) can be embedded in a field of fractions \(F_D\text{,}\) where any element in \(F_D\) can be expressed as the quotient of two elements in \(D\text{.}\) Furthermore, the field of fractions \(F_D\) is unique in the sense that if \(E\) is any field containing \(D\text{,}\) then there exists a map \(\psi : F_D \rightarrow E\) giving an isomorphism with a subfield of \(E\) such that \(\psi(a) = a\) for all elements \(a \in D\text{,}\) where we identify \(a\) with its image in \(F_D\text{.}\)
- Proof
-
We will first demonstrate that \(D\) can be embedded in the field \(F_D\text{.}\) Define a map \(\phi : D \rightarrow F_D\) by \(\phi(a) = [a, 1]\text{.}\) Then for \(a\) and \(b\) in \(D\text{,}\)
\[ \phi( a + b ) = [a+b, 1] = [a, 1] + [b, 1] = \phi(a ) + \phi(b) \nonumber \]
and
\[ \phi( a b ) = [a b, 1] = [a, 1] [b, 1] = \phi(a ) \phi(b); \nonumber \]
hence, \(\phi\) is a homomorphism. To show that \(\phi\) is one-to-one, suppose that \(\phi(a) = \phi( b)\text{.}\) Then \([a, 1] = [b, 1]\text{,}\) or \(a = a1 = 1b = b\text{.}\) Finally, any element of \(F_D\) can be expressed as the quotient of two elements in \(D\text{,}\) since
\[ \phi(a) [\phi(b)]^{-1} = [a, 1] [b, 1]^{-1} = [a, 1] \cdot [1, b] = [a, b]\text{.} \nonumber \]
Now let \(E\) be a field containing \(D\) and define a map \(\psi :F_D \rightarrow E\) by \(\psi([a, b]) = a b^{-1}\text{.}\) To show that \(\psi\) is well-defined, let \([a_1, b_1] = [a_2, b_2]\text{.}\) Then \(a_1 b_2 = b_1 a_2\text{.}\) Therefore, \(a_1 b_1^{-1} = a_2 b_2^{-1}\) and \(\psi( [a_1, b_1]) = \psi( [a_2, b_2])\text{.}\)
If \([a, b ]\) and \([c, d]\) are in \(F_D\text{,}\) then
\begin{align*} \psi( [a, b] + [c, d] ) & = \psi( [ad + b c, b d ] )\\ & = (ad + b c)(b d)^{-1}\\ & = a b^{-1} + c d^{-1}\\ & = \psi( [a, b] ) + \psi( [c, d] ) \end{align*}
and
\begin{align*} \psi( [a, b] \cdot [c, d] ) & = \psi( [ac, b d ] )\\ & = (ac)(b d)^{-1}\\ & = a b^{-1} c d^{-1}\\ & = \psi( [a, b] ) \psi( [c, d] )\text{.} \end{align*}
Therefore, \(\psi\) is a homomorphism.
To complete the proof of the theorem, we need to show that \(\psi\) is one-to-one. Suppose that \(\psi( [a, b] ) = ab^{-1} = 0\text{.}\) Then \(a = 0b = 0\) and \([a, b] = [0, b]\text{.}\) Therefore, the kernel of \(\psi\) is the zero element \([ 0, b]\) in \(F_D\text{,}\) and \(\psi\) is injective.
Example \(18.5\)
Since \({\mathbb Q}\) is a field, \({\mathbb Q}[x]\) is an integral domain. The field of fractions of \({\mathbb Q}[x]\) is
Solution
the set of all rational expressions \(p(x)/q(x)\text{,}\) where \(p(x)\) and \(q(x)\) are polynomials over the rationals and \(q(x)\) is not the zero polynomial. We will denote this field by \({\mathbb Q}(x)\text{.}\)
We will leave the proofs of the following corollaries of Theorem \(18.4\) as exercises.
Corollary \(18.6\)
Let \(F\) be a field of characteristic zero. Then \(F\) contains a subfield isomorphic to \({\mathbb Q}\text{.}\
Corollary \(18.7\)
Let \(F\) be a field of characteristic \(p\text{.}\) Then \(F\) contains a subfield isomorphic to \({\mathbb Z}_p\text{.}\)


7.2.5: Connections to Other Metabolic Pathways
You have learned about the catabolism of glucose, which provides energy to living cells. But living things consume more than just glucose for food. How does a turkey sandwich, which contains protein, provide energy to your cells? This happens because all of the catabolic pathways for carbohydrates, proteins, and lipids eventually connect into glycolysis and the citric acid cycle pathways (Figure \(\PageIndex{1}\)). Metabolic pathways should be thought of as porous—that is, substances enter from other pathways, and other substances leave for other pathways. These pathways are not closed systems. Many of the products in a particular pathway are reactants in other pathways.
Connections of Other Sugars to Glucose Metabolism
Glycogen, a polymer of glucose, is a short-term energy storage molecule in animals. When there is adequate ATP present, excess glucose is converted into glycogen for storage. Glycogen is made and stored in the liver and muscle. Glycogen will be taken out of storage if blood sugar levels drop. The presence of glycogen in muscle cells as a source of glucose allows ATP to be produced for a longer time during exercise.
Sucrose is a disaccharide made from glucose and fructose bonded together. Sucrose is broken down in the small intestine, and the glucose and fructose are absorbed separately. Fructose is one of the three dietary monosaccharides, along with glucose and galactose (which is part of milk sugar, the disaccharide lactose), that are absorbed directly into the bloodstream during digestion. The catabolism of both fructose and galactose produces the same number of ATP molecules as glucose.
Connections of Proteins to Glucose Metabolism
Proteins are broken down by a variety of enzymes in cells. Most of the time, amino acids are recycled into new proteins. If there are excess amino acids, however, or if the body is in a state of famine, some amino acids will be shunted into pathways of glucose catabolism. Each amino acid must have its amino group removed prior to entry into these pathways. The amino group is converted into ammonia. In mammals, the liver synthesizes urea from two ammonia molecules and a carbon dioxide molecule. Thus, urea is the principal waste product in mammals from the nitrogen originating in amino acids, and it leaves the body in urine.
Connections of Lipids to Glucose Metabolism
The lipids that are connected to the glucose pathways are cholesterol and triglycerides. Cholesterol is a lipid that contributes to cell membrane flexibility and is a precursor of steroid hormones. The synthesis of cholesterol starts with acetyl CoA and proceeds in only one direction. The process cannot be reversed, and ATP is not produced.
Triglycerides are a form of long-term energy storage in animals. Triglycerides store about twice as much energy as carbohydrates. Triglycerides are made of glycerol and three fatty acids. Animals can make most of the fatty acids they need. Triglycerides can be both made and broken down through parts of the glucose catabolism pathways. Glycerol can be phosphorylated and proceeds through glycolysis. Fatty acids are broken into two-carbon units that enter the citric acid cycle.
Figure \(\PageIndex{1}\): Glycogen from the liver and muscles, together with fats, can feed into the catabolic pathways for carbohydrates.EVOLUTION IN ACTION: Pathways of Photosynthesis and Cellular Metabolism
Photosynthesis and cellular metabolism consist of several very complex pathways. It is generally thought that the first cells arose in an aqueous environment—a “soup” of nutrients. If these cells reproduced successfully and their numbers climbed steadily, it follows that the cells would begin to deplete the nutrients from the medium in which they lived, as they shifted the nutrients into their own cells. This hypothetical situation would have resulted in natural selection favoring those organisms that could exist by using the nutrients that remained in their environment and by manipulating these nutrients into materials that they could use to survive. Additionally, selection would favor those organisms that could extract maximal value from the available nutrients.
An early form of photosynthesis developed that harnessed the sun’s energy using compounds other than water as a source of hydrogen atoms, but this pathway did not produce free oxygen. It is thought that glycolysis developed prior to this time and could take advantage of simple sugars being produced, but these reactions were not able to fully extract the energy stored in the carbohydrates. A later form of photosynthesis used water as a source of hydrogen ions and generated free oxygen. Over time, the atmosphere became oxygenated. Living things adapted to exploit this new atmosphere and allowed respiration as we know it to evolve. When the full process of photosynthesis as we know it developed and the atmosphere became oxygenated, cells were finally able to use the oxygen expelled by photosynthesis to extract more energy from the sugar molecules using the citric acid cycle.
Summary
The breakdown and synthesis of carbohydrates, proteins, and lipids connect with the pathways of glucose catabolism. The carbohydrates that can also feed into glucose catabolism include galactose, fructose, and glycogen. These connect with glycolysis. The amino acids from proteins connect with glucose catabolism through pyruvate, acetyl CoA, and components of the citric acid cycle. Cholesterol synthesis starts with acetyl CoA, and the components of triglycerides are picked up by acetyl CoA and enter the citric acid cycle.
Contributors and Attributions
-
Samantha Fowler (Clayton State University), Rebecca Roush (Sandhills Community College), James Wise (Hampton University). Original content by OpenStax (CC BY 4.0; Access for free at https://cnx.org/contents/b3c1e1d2-83...4-e119a8aafbdd ).


12.5: Exercises
1. For the graph in Figure 12.20 , use Kruskal's algorithm (“avoid cycles”) to find a minimum weight spanning tree. Your answer should include a complete list of the edges, indicating which edges you take for your tree and which (if any) you reject in the course of running the algorithm.
2. For the graph in Figure 12.20 , use Prim's algorithm (“build tree”) to find a minimum weight spanning tree. Your answer should list the edges selected by the algorithm in the order they were selected.
3. For the graph in Figure 12.21 , use Kruskal's algorithm (“avoid cycles”) to find a minimum weight spanning tree. Your answer should include a complete list of the edges, indicating which edges you take for your tree and which (if any) you reject in the course of running the algorithm.
4. For the graph in Figure 12.21 , use Prim's algorithm (“build tree”) to find a minimum weight spanning tree. Your answer should list the edges selected by the algorithm in the order they were selected.
5. For the graph in Figure 12.22 , use Kruskal's algorithm (“avoid cycles”) to find a minimum weight spanning tree. Your answer should include a complete list of the edges, indicating which edges you take for your tree and which (if any) you reject in the course of running the algorithm.
6. For the graph in Figure 12.22 , use Prim's algorithm (“build tree”) to find a minimum weight spanning tree. Your answer should list the edges selected by the algorithm in the order they were selected.
7. A new local bank is being created and will establish a headquarters \(h\), two branches \(b_1\) and \(b_2\), and four ATMs \(a_1\), \(a_2\), \(a_3\), and \(a_4\). They need to build a computer network such that the headquarters, branches, and ATMs can all intercommunicate. Furthermore, they will need to be networked with the Federal Reserve Bank of Atlanta, \(f\). The costs of the feasible network connections (in units of $10,000) are listed below:
The bank wishes to minimize the cost of building its network (which must allow for connection, possibly routed through other nodes, from each node to each other node), however due to the need for high-speed communication, they must pay to build the connection from \(h\) to \(f\) as well as the connection from \(b_2\) to \(a_3\). Give a list of the connections the bank should establish in order to minimize their total cost, subject to this constraint. Be sure to explain how you selected the connections and how you know the total cost is minimized.
8. A disconnected weighted graph obviously has no spanning trees. However, it is possible to find a spanning forest of minimum weight in such a graph. Explain how to modify both Kruskal's algorithm and Prim's algorithm to do this.
9. Prove Proposition 12.3 .
10. In the paper where Kruskal's algorithm first appeared, he considered the algorithm a route to a nicer proof that in a connected weighted graph with no two edges having the same weight, there is a unique minimum weight spanning tree. Prove this fact using Kruskal's algorithm.
11. Use Dijkstra's algorithm to find the distance from \(a\) to each other vertex in the digraph shown in Figure 12.23 and a directed path of that length.
12. Figure 12.24 contains the length of the directed edge \((x,y)\) in the intersection of row \(x\) and column \(y\) in a digraph with vertex set \(\{a,b,c,d,e,f\}\). For example, \(w(b,d)=21\). (On the other hand, \(w(d,b)=10\).) Use this data and Dijkstra's algorithm to find the distance from \(a\) to each of the other vertices and a directed path of that length from \(a\).
13. Use Dijkstra's algorithm to find the distance from \(a\) to each other vertex in the digraph shown in Figure 12.25 and a directed path of that length.
14. Figure 12.26 contains the length of the directed edge \((x,y)\) in the intersection of row \(x\) and column \(y\) in a digraph with vertex set \(\{a,b,c,d,e,f\}\). For example, \(w(b,d)=47\). (On the other hand, \(w(d,b)=6\).) Use this data and Dijkstra's algorithm to find the distance from \(a\) to each of the other vertices and a directed path of that length from \(a\).
15. Give an example of a digraph having an undirected path between each pair of vertices, but having a root vertex \(r\) so that Dijkstra's algorithm cannot find a path of finite length from \(r\) to some vertex \(x\).
16. Notice that in our discussion of Dijkstra's algorithm, we required that the edge weights be nonnegative. If the edge weights are lengths and meant to model distance, this makes perfect sense. However, in some cases, it might be reasonable to allow negative edge weights. For example, suppose that a positive weight means there is a cost to travel along the directed edge while a negative edge weight means that you make money for traveling along the directed edge. In this case, a directed path with positive total weight results in paying out to travel it, while one with negative total weight results in a profit.
a. Give an example to show that Dijkstra's algorithm does not always find the path of minimum total weight when negative edge weights are allowed.
b. Bob and Xing are considering this situation, and Bob suggests that a little modification to the algorithm should solve the problem. He says that if there are negative weights, they just have to find the smallest (i.e., most negative weight) and add the absolute value of that weight to every directed edge. For example, if \(w(x,y)≥−10\) for every directed edge \((x,y)\), Bob is suggesting that they add 10 to every edge weight. Xing is skeptical, and for good reason. Give an example to show why Bob's modification won't work.


10.8: Study Points
- If two opposing chemical reactions proceed simultaneously at the same rate, the processes are said to be in equilibrium . The two opposing reactions are shown linked with a double arrow (⇄). An example of opposing chemical reactions and their equilibrium expressions are:
\[\ce{N2(g) + 3 H2(g) -> 2 NH3(g)}\]
and
\[\ce{2 NH3(g) -> N2(g) + 3 H2(g)}\]
The equilibrium would be written as:
\[\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)}\]
- An equilibrium constant ( K ) is a numerical value that relates the concentrations of the products and reactants for a chemical reaction that is at equilibrium. The numeric value of an equilibrium constant is independent of the initial concentrations of reactants , but is dependent on the temperature . Equilibrium constants are generally not written with units (although they may be).
- Because equilibrium constants are written as the concentrations (or partial pressures) of products divided by the concentrations (or partial pressures) of reactants , a large value of K means that there are more products in the equilibrium mixture than there are products. Likewise, a small value of K means that, at equilibrium, there are more reactants than products.
- Equilibrium constants that are based on partial pressures are often written as K P , while equilibrium constants based on molar concentrations are written as K c .
-
An expression for an equilibrium constant can be written from a balanced chemical equation for the reaction.
The Law of Mass Action
states the following regarding equilibrium expressions:
- Partial pressures (or molar concentrations) of products are written in the numerator of the expression and the partial pressures (or concentrations) of the reactants are written in the denominator .
- If there is more that one reactant or more that one product, the partial pressures (or concentrations) are multiplied together.
- The partial pressure (or concentration) of each reactant or product is then raised to the power that numerically equals the stoichiometric coefficient appearing with that term in the balanced chemical equation.
- Reactants or products that are present as solids or liquids do not appear in the equilibrium expression.
- As an example of an equilibrium expression, consider the reaction of nitrogen and hydrogen to form ammonia. The partial pressure of ammonia will be in the numerator, and it will be squared . Because there are two reactants, the partial pressures for nitrogen and hydrogen will be multiplied in the denominator. The partial pressure of nitrogen will be raised to the “first power” (which is not shown) and the partial pressure of hydrogen will be cubed .
\[\ce{N2 (g) + 3 H2 (g) <=> 2 NH3 (g)}\]
\[\frac{(P_{NH_{3}})^{2}}{P_{N_{2}}(P_{H_{2}})^{3}}=K \nonumber \]
- If equilibrium values for a given reaction are known, the equilibrium constant can be calculated simply by substituting those values in the equilibrium expression. Quite often, however, initial and equilibrium values are only given for selected reactants and products. In these cases, initial and equilibrium values are arranged in an ICE Table , and the changes between initial and equilibrium states are calculated based on reaction stoichiometry.
- Because the partial pressure of a gas and the molar concentration of that gas are directly proportional, the ideal gas law can be rearranged as follows, to give an expression relating molarity and partial pressure of a gas.
\[P_{gas}V=nRT \nonumber \]
Dividing by the volume in liters gives the term which is equivalent to molarity , M.
\[P_{gas}= MRT \nonumber \]
- Le Chatelier's Principle states that, if a "stress" is applied to a chemical reaction at equilibrium, the system will readjust in the direction that best reduces the stress imposed on the system. In this context, stress refers to a change in concentration , a change in pressure or a change in temperature , although only concentration is considered here. If pressure or temperature are changed, the numeric value K will change; if only concentration changes are involved, K does not change . In a reaction at equilibrium, the introduction of more products will shift the mass balance towards more reactants, and the introduction of more reactants will lead to the formation of more products, but the ratio of Products/Reactants (as defined by the equilibrium expression) does not change, hence, K is unchanged.
- For a weak acid , dissociating in water to form it's conjugate base and hydronium ion, the equilibrium constant is referred to as K a . Because a weak acid is only "partially dissociated", the concentration of BH in solution is large, thus K a for a weak acid is "small" (in the range of 10 -3 to 10 -6 ). For example, acetic acid (the acidic component of vinegar), has an acid dissociation constant of K a = 1.8 x 10 -5 .
CH 3 COOH (aq) + H 2 O (l) ⇄ CH 3 COO - (aq) + H 3 O + (aq)
- For a solution of a weak acid in water, the concentration of hydronium ion will be very small. If the concentration of the weak acid is fairly large (typically > 0.01 M) the concentration of the undissociated acid will be much larger than [H 3 O + ]. Because of this, the hydronium ion concentration (hence, the pH) can be fairly accurately estimated from the K a of the weak acid and the initial concentration of the acid (C o ), by the equation:
\[[H_{3}O^{+}]=\sqrt{K_{a}\times C_{0}} \nonumber \]
- The equilibrium constant defining the solubility of an ionic compound with low solubility is defined as K sp , where “sp” refers to “solubility product”. Because reactants or products that are present as solids or liquids do not appear in equilibrium expressions, for silver chloride, the expression K sp will be written as:
AgCl (s) ⇄ Ag + (aq) + Cl - (aq)
\[K_{sp}=[Ag^{+}][Cl^{-}] \nonumber \]
- For silver chloride, the solubility at 25 o C is 1.67 × 10 -5 M. That means the concentrations of silver and chloride ions in solution are each 1.67 × 10 -5 M, making the value of K sp under these conditions: \[K_{sp}=[Ag^{+}][Cl^{-}]=(1.67\times 10^{-5})^{2}=2.79\times 10^{-10} \nonumber \]


1.
Of course, it is impossible to force someone to eat. But the real advice here is to avoid turning food into a power struggle so that food doesn’t become a way to gain favor with or express anger toward someone else.
2. Recognize that appetite varies.
Children may eat well at one meal and have no appetite at another. Rather than seeing this as a problem, it may help to realize that appetites do vary. Continue to provide good nutrition at each mealtime (even if children don’t choose to eat the occasional meal).
3. Keep it pleasant.
This tip is designed to help caregivers create a positive atmosphere during mealtime. Mealtimes should not be the time for arguments or expressing tensions. You do not want the child to have painful memories of mealtimes together or have nervous stomachs and problems eating and digesting food due to stress.
4. No short order chefs.
While it is fine to prepare foods that children enjoy, preparing a different meal for each child or family member sets up an unrealistic expectation from others. Children probably do best when they are hungry and a meal is ready. Limiting snacks rather than allowing children to “graze” continuously can help create an appetite for whatever is being served.
5. Limit choices.
If you give your preschool aged child choices, make sure that you give them one or two specific choices rather than asking “What would you like for lunch?” If given an open choice, children may change their minds or choose whatever their sibling does not choose!
6. Serve balanced meals.
Meals prepared at home tend to have better nutritional value than fast food or frozen dinners. Prepared foods tend to be higher in fat and sugar content as these ingredients enhance taste and profit margin because fresh food is often more costly and less profitable. However, preparing fresh food at home is not costly. It does, however, require more activity. Including children in meal preparation can provide a fun and memorable experience.
7. Don’t bribe.
Bribing a child to eat vegetables by promising dessert is not a good idea. First, the child will likely find a way to get the dessert without eating the vegetables (by whining or fidgeting, perhaps, until the caregiver gives in). Secondly, it teaches the child that some foods are better than others. Children tend to naturally enjoy a variety of foods until they are taught that some are considered less desirable than others. A child, for example, may learn the broccoli they have enjoyed is seen as yucky by others unless it’s smothered in cheese sauce! 4
USDA Meal Patterns for Young Children
The United States Department of Agriculture Food and Nutrition Service provides the following guidance for the daytime feeding of children age 3 to 5.
Meal Patterns 6
Table \(\PageIndex{1}\)
Meal
Ages 3-5
Breakfast
3/4 cup milk 1/2 cup vegetables, fruit, or both 1⁄2 ounce equivalent grains
Lunch or Supper
3/4 cup milk 11⁄2 ounces meat or meat alternative 1/4 cup vegetables 1/4 cup fruits 1⁄2 ounce equivalent of grains
Snack
Select two of the following: 1⁄2 cup of milk 1⁄2 ounce meat or meat alternative 1⁄2 cup vegetables 1⁄2 cup fruit 1⁄2 ounce equivalent of grains


7.3.6: Applying Volume and Surface Area
Lesson
Let's explore things that are proportional to volume or surface area.
Exercise \(\PageIndex{1}\): You Decide
For each situation, decide if it requires Noah to calculate surface area or volume. Explain your reasoning.
- Noah is planning to paint the bird house he built. He is unsure if he has enough paint.
- Noah is planning to use a box with a trapezoid base to hold modeling clay. He is unsure if the clay will all fit in the box.
Exercise \(\PageIndex{2}\): Foam Play Structure
At a daycare, Kiran sees children climbing on this foam play structure.
Kiran is thinking about building a structure like this for his younger cousins to play on.
- The entire structure is made out of soft foam so the children don’t hurt themselves. How much foam would Kiran need to build this play structure?
- The entire structure is covered with vinyl so it is easy to wipe clean. How much vinyl would Kiran need to build this play structure?
- The foam costs 0.8¢ per in 3 . Here is a table that lists the costs for different amounts of vinyl. What is the total cost for all the foam and vinyl needed to build this play structure?
| vinyl (in 2 ) | cost ($) |
|---|---|
| \(75\) | \(0.45\) |
| \(125\) | \(0.75\) |
Are you ready for more?
When he examines the play structure more closely, Kiran realizes it is really two separate pieces that are next to each other.
- How does this affect the amount of foam in the play structure?
- How does this affect the amount of vinyl covering the play structure?
Exercise \(\PageIndex{3}\): Filling the Sandbox
The daycare has two sandboxes that are both prisms with regular hexagons as their bases. The smaller sandbox has a base area of 1,146 in 2 and is filled 10 inches deep with sand.
- It took 14 bags of sand to fill the small sandbox to this depth. What volume of sand comes in one bag? (Round to the nearest whole cubic inch.)
- The daycare manager wants to add 3 more inches to the depth of the sand in the small sandbox. How many bags of sand will they need to buy?
- The daycare manager also wants to add 3 more inches to the depth of the sand in the large sandbox. The base of the large sandbox is a scaled copy of the base of the small sandbox, with a scale factor of 1.5. How many bags of sand will they need to buy for the large sandbox?
- A lawn and garden store is selling 6 bags of sand for $19.50. How much will they spend to buy all the new sand for both sandboxes?
Summary
Suppose we wanted to make a concrete bench like the one shown in this picture. If we know that the finished bench has a volume of 10 ft 3 and a surface area of 44 ft 2 we can use this information to solve problems about the bench.
For example,
- How much does the bench weigh?
- How long does it take to wipe the whole bench clean?
- How much will the materials cost to build the bench and to paint it?
To figure out how much the bench weighs, we can use its volume, 10 ft 3 . Concrete weighs about 150 pounds per cubic foot, so this bench weighs about 1,500 pounds, because \(10\cdot 150=1,500\).
To figure out how long it takes to wipe the bench clean, we can use its surface area, 44 ft 2 . If it takes a person about 2 seconds per square foot to wipe a surface clean, then it would take about 88 seconds to clean this bench, because \(44\cdot 2=88\). It may take a little less than 88 seconds, since the surfaces where the bench is touching the ground do not need to be wiped.
Would you use the volume or the surface area of the bench to calculate the cost of the concrete needed to build this bench? And for the cost of the paint?
Glossary Entries
Definition: Base (of a prism or pyramid)
The word base can also refer to a face of a polyhedron.
A prism has two identical bases that are parallel. A pyramid has one base.
A prism or pyramid is named for the shape of its base.
Definition: Cross Section
A cross section is the new face you see when you slice through a three-dimensional figure.
For example, if you slice a rectangular pyramid parallel to the base, you get a smaller rectangle as the cross section.
Definition: Prism
A prism is a type of polyhedron that has two bases that are identical copies of each other. The bases are connected by rectangles or parallelograms.
Here are some drawings of prisms.
Definition: Pyramid
A pyramid is a type of polyhedron that has one base. All the other faces are triangles, and they all meet at a single vertex.
Here are some drawings of pyramids.
Definition: Surface Area
The surface area of a polyhedron is the number of square units that covers all the faces of the polyhedron, without any gaps or overlaps.
For example, if the faces of a cube each have an area of 9 cm 2 , then the surface area of the cube is \(6\cdot 9\), or 54 cm 2 .
Definition: Volume
Volume is the number of cubic units that fill a three-dimensional region, without any gaps or overlaps.
For example, the volume of this rectangular prism is 60 units 3 , because it is composed of 3 layers that are each 20 units 3 .
Practice
Exercise \(\PageIndex{4}\)
A landscape architect is designing a pool that has this top view:
- How much water will be needed to fill this pool 4 feet deep?
- Before filling up the pool, it gets lined with a plastic liner. How much liner is needed for this pool?
- Here are the prices for different amounts of plastic liner. How much will all the plastic liner for the pool cost?
| plastic liner (ft 2 ) | cost ($) |
|---|---|
| \(25\) | \(3.75\) |
| \(50\) | \(7.50\) |
| \(75\) | \(11.25\) |
Exercise \(\PageIndex{5}\)
Shade in a base of the trapezoidal prism. (The base is not the same as the bottom.)
- Find the area of the base you shaded.
- Find the volume of this trapezoidal prism.
(From Unit 7.3.3)
Exercise \(\PageIndex{6}\)
For each diagram, decide if \(y\) is an increase or a decrease of \(x\). Then determine the percentage that \(x\) increased or decreased to result in \(y\).
(From Unit 4.2.4)
Exercise \(\PageIndex{7}\)
Noah is visiting his aunt in Texas. He wants to buy a belt buckle whose price is $25. He knows that the sales tax in Texas is 6.25%.
- How much will the tax be on the belt buckle?
- How much will Noah spend for the belt buckle including the tax?
- Write an equation that represents the total cost, \(c\), of an item whose price is \(p\).
(From Unit 4.3.1)


4.7: Conformations of Monosubstituted Cyclohexanes
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Objectives
After completing this section, you should be able to
- account for the greater stability of the equatorial conformers of monosubstituted cyclohexanes compared to their axial counterparts, using the concept of 1,3‑diaxial interaction.
- compare the gauche interactions in butane with the 1,3‑diaxial interactions in the axial conformer of methylcyclohexane.
- arrange a given list of substituents in increasing or decreasing order of 1,3‑diaxial interactions.
Key Terms
Make certain that you can define, and use in context, the key term below.
- 1,3‑diaxial interaction
Study Notes
1,3-Diaxial interactions are steric interactions between an axial substituent located on carbon atom 1 of a cyclohexane ring and the hydrogen atoms (or other substituents) located on carbon atoms 3 and 5.
Be prepared to draw Newman-type projections for cyclohexane derivatives as the one shown for methylcyclohexane. Note that this is similar to the Newman projections from chapter 3 such as n -butane.
Newman projections of methylcyclohexane and n ‑butane
When a substituent is added to a cyclohexane ring, the two possible chair conformations created during a ring flip are not equally stable. In the example of methylcyclohexane the conformation where the methyl group is in the equatorial position is more stable than the axial conformation by 7.6 kJ/mol at 25 o C. The percentages of the two different conformations at equilibrium can be determined by solving the following equation for K (the equilibrium constant): ΔE = -RTlnK. In this equation ΔE is the energy difference between the two conformations, R is the gas constant (8.314 J/mol•K), T is the temperature in Kelvin, and K is the equilibrium constant for the ring flip conversion. Using this equation, we can calculate a K value of 21 which means about 95% methylcyclohexane molecules have the methyl group in the equatorial position at 25 o C.
The energy difference between the two conformations comes from strain, called 1,3-diaxial interactions, created when the axial methyl group experiences steric crowding with the two axial hydrogens located on the same side of the cyclohexane ring. Because axial bonds are parallel to each other, substituents larger than hydrogen experience greater steric crowding when they are oriented axial rather than equatorial. Consequently, substituted cyclohexanes will preferentially adopt conformations in which the larger substituents are in the equatorial orientation . When the methyl group is in the equatorial position this strain is not present which makes the equatorial conformer more stable and favored in the ring flip equilibrium.
Actually, 1,3-diaxial steric strain is directly related to the steric strain created in the gauche conformer of butane discussed in Section: 3-7. When butane is in the gauche conformation 3.8 kJ/mol of strain was created due the steric crowding of two methyl group with a 60
o
dihedral angle. When looking at the a Newman projection of axial methylcyclohexane the methyl group is at a 60
o
dihedral angle with the ring carbon in the rear. This creates roughly the same amount of steric strain as the gauche conformer of butante. Given that there is actually two such interactions in axial methylcyclohexane, it makes sense that there is 2(3.8 kJ/mol) = 7.6 kJ/mol of steric strain in this conformation. The Newman projection of equatorial methylcyclohexane shows no such interactions and is therefore more stable.
Newman projections of methyl cyclohexane and butane showing similarity of 1,3-diaxial and gauche interactions.
Strain values for other cyclohexane substituents can also be considered. The relative steric hindrance experienced by different substituent groups oriented in an axial versus equatorial location on cyclohexane determined the amount of strain generated. The strain generated can be used to evaluate the relative tendency of substituents to exist in an equatorial or axial location. Looking at the energy values in this table, it is clear that as the size of the substituent increases, the 1,3-diaxial energy tends to increase, also. Note that it is the size and not the molecular weight of the group that is important. Table 4.7.1 summarizes some of these strain values values.
| Substituent | - ΔG ° (kcal/mol) | Substituent | - ΔG ° (kcal/mol) |
| \(\ce{CH_3\bond{-}}\) | 1.7 | \(\ce{O_2N\bond{-}}\) | 1.1 |
| \(\ce{CH_2H_5\bond{-}}\) | 1.8 | \(\ce{N#C\bond{-}}\) | 0.2 |
| \(\ce{(CH_3)_2CH\bond{-}}\) | 2.2 | \(\ce{CH_3O\bond{-}}\) | 0.5 |
| \(\ce{(CH_3)_3C\bond{-}}\) | \(\geq 5.0\) | \(\ce{HO_2C\bond{-}}\) | 0.7 |
| \(\ce{F\bond{-}}\) | 0.3 | \(\ce{H_2C=CH\bond{-}}\) | 1.3 |
| \(\ce{Cl\bond{-}}\) | 0.5 | \(\ce{C_6H_5\bond{-}}\) | 3.0 |
| \(\ce{Br\bond{-}}\) | 0.5 | ||
| \(\ce{I\bond{-}}\) | 0.5 |
Exercises
1) In the molecule, cyclohexyl ethyne there is little steric strain, why?
2) Calculate the energy difference between the axial and equatorial conformations of bromocyclohexane?
3) Using your answer from Question 2) estimate the percentages of axial and equatorial conformations of bromocyclohexane at 25 o C.
4) There very little in 1,3-diaxial strain when going from a methyl substituent (3.8 kJ/mol) to an ethyl substituent (4.0 kJ/mol), why? It may help to use molecular model to answer this question.
Solutions
1) The ethyne group is linear and therefore does not affect the hydrogens in the 1,3 positions to say to the extent as a bulkier or a bent group (e.g. ethene group) would. This leads to less of a strain on the molecule.
2) The equatorial conformation of bromocyclohexane will have two 1,3 diaxial interactions. The table above states that each interaction accounts for 1.2 kJ/mol of strain. The total strain in equatorial bromocyclohexane will be 2(1.2 kJ/mol) = 2.4 kJ/mol.
3) Remembering that the axial conformation is higher in energy, the energy difference between the two conformations is ΔE = (E equatorial - E axial) = (0 - 2.4 kJ/mol) = -2.4 kJ/mol. After converting o C to Kelvin and kJ/mol to J/mol we can use the equation ΔE = -RT lnK to find that -ΔE/RT = lnK or (2.4 x 10 3 J/mol) / (8.313 kJ/mol K • 298 K) = lnK. From this we calculate that K = 2.6. Because the ring flip reaction is an equilibrium we can say that K = [Equatorial] / [Axial]. If assumption is made that [Equatorial] = X then [Axial] must be 1-X. Plugging these values into the equilibrium expression produces K = [X] / [1-X]. After plugging in the calculated value for K, X can be solved algebraically. 2.6 = [X] / [1-X] → 2.6 - 2.6X = X → 2.6 = 3.6X → 2.6/3.6 = X = 0.72. This means that bromocyclohexane is in the equatorial position 72% of the time and in the axial position 28% of the time.
4) The fact that C-C sigma bonds can freely rotate allows the ethyl subsistent to obtain a conformation which places the bulky CH 3 group away from the cyclohexane ring. This forces the ethyl substituent to have only have 1,3- diaxial interactions between hydrogens, which only provides a slight difference to a methyl group.


14.6: Middle Eastern Americans
Little Arabia is an ethnic enclave in Orange County, California, the center for Orange County's Arab Americans, who number more than 24,000 (as of 2000). It is sometimes referred to as "Little Gaza" which was a play on the original designation of this area as the "Garza Island." Little Arabia grew significantly in the 1990s with the arrival of immigrants from the Middle East and is the home to thousands of Arab Americans predominantly hailing from Egypt, Syria, Palestine, and Yemen.
More than 500,000 Iranian Americans live throughout Southern California, including about 20% of the population of Beverly Hills. Iranian American communities also flourish in the San Fernando Valley, Orange County, San Diego, and the San Joaquin Valley. The majority of Iranian Americans immigrated after the Pahlavi dynasty was overthrown in 1979.
California is also home to many Armenian Americans; the highest concentration of Americans of Armenian descent is in the Greater Los Angeles area, where 166,498 people have identified themselves as Armenian to the 2000 Census, comprising over 40% of the 385,488 people who identified Armenian origins in the U.S. at the time. Many of these live in Glendale north of Los Angeles, as well as a large community in Fresno. The size of the Armenian American population is disputed, however. According to a 1988 news article, California had about 500,000 ethnic Armenians with over half of them living in the Greater Los Angeles area.
 In addition, there are more than 250,000 Israeli Americans live in the Los Angeles area, according to the Israeli American Council. There are also significant Israeli American populations in the Bay Area, San Francisco, and San Jose areas of Northern California. The largest Karaite Jewish population outside of Israel exists in the Bay Area, consisting of several hundred descendants of refugees from the Egyptian Karaite community, as well as some recent converts. Their community is centered around the only Karaite synagogue outside the Middle East, Congregation B'nai Israel, located in Daly City. The Moroccan Jewish community in California is one of the largest in North America, approximately 10,000 Moroccan Jews reside in the Greater Los Angeles area, mostly in Pico-Robertson, North Hollywood, and Beverly Hills. Many are the descendants of community members who first emigrated to the United States in the aftermath of World War II. Many others came later in the 20th century from Israel and beginning in the early 21 st century from France due to increasing antisemitism there. The community has their own synagogues as well as a community center.
Over 50,000 Afghan Americans live in California, with large concentrations in the East Bay (primarily in Alameda County and its communities of Fremont and Hayward), Orange County, and Ventura County.
There is also a large population of Assyrian descent living in the Central Valley, with large communities in Modesto, Ceres, and Turlock. San Diego has one of the largest concentrations of Chaldean immigrants in the United States. There is about an estimated 3,000 Moroccan Americans are living in Los Angeles, Long Beach, and Santa Ana. Turkish Americans and Azerbaijani Americans form moderately sized communities in both Los Angeles and San Francisco.