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https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/24%3A_Nuclear_Chemistry/24.03%3A_Nuclear_Reactions |
The two general kinds of nuclear reactions are nuclear decay reactions and nuclear transmutation reactions In a , also called radioactive decay, an unstable nucleus emits radiation and is transformed into the nucleus of one or more other elements. The resulting daughter nuclei have a lower mass and are lower in energy (more stable) than the parent nucleus that decayed. In contrast, in a , a nucleus reacts with a subatomic particle or another nucleus to form a product nucleus that is than the starting material. As we shall see, nuclear decay reactions occur spontaneously under all conditions, but nuclear transmutation reactions occur only under very special conditions, such as the collision of a beam of highly energetic particles with a target nucleus or in the interior of stars. We begin this section by considering the different classes of radioactive nuclei, along with their characteristic nuclear decay reactions and the radiation they emit. Nuclear decay reactions occur spontaneously under all conditions, whereas nuclear transmutation reactions are induced. The three general classes of radioactive nuclei are characterized by a different decay process or set of processes: Nuclear decay reactions always produce daughter nuclei that have a more favorable neutron-to- proton ratio and hence are more stable than the parent nucleus. Just as we use the number and type of atoms present to balance a chemical equation, we can use the number and type of nucleons present to write a balanced nuclear equation for a nuclear decay reaction. This procedure also allows us to predict the identity of either the parent or the daughter nucleus if the identity of only one is known. Regardless of the mode of decay, the total number of nucleons is conserved in all nuclear reactions. To describe nuclear decay reactions, chemists have extended the \(^A _Z \textrm{X}\) notation for nuclides to include radioactive emissions. \(\Page {1}\) lists the name and symbol for each type of emitted radiation. The most notable addition is the , a particle that has the same mass as an electron but a positive charge rather than a negative charge. Like the notation used to indicate isotopes, the upper left superscript in the symbol for a particle gives the mass number, which is the total number of protons and neutrons. For a proton or a neutron, = 1. Because neither an electron nor a positron contains protons or neutrons, its mass number is 0. The numbers should not be taken literally, however, as meaning that these particles have zero mass; ejection of a beta particle (an electron) simply has a negligible effect on the mass of a nucleus. Similarly, the lower left subscript gives the charge of the particle. Because protons carry a positive charge, = +1 for a proton. In contrast, a neutron contains no protons and is electrically neutral, so = 0. In the case of an electron, = −1, and for a positron, = +1. Because γ rays are high-energy photons, both and are 0. In some cases, two different symbols are used for particles that are identical but produced in different ways. For example, the symbol \(^0_{-1}\textrm e\), which is usually simplified to e , represents a free electron or an electron associated with an atom, whereas the symbol \(^0_{-1}\beta\), which is often simplified to β , denotes an electron that originates from within the nucleus, which is a β particle. Similarly, \(^4_{2}\textrm{He}^{2+}\) refers to the nucleus of a helium atom, and \(^4_{2}\alpha\) denotes an identical particle that has been ejected from a heavier nucleus. There are six fundamentally different kinds of nuclear decay reactions, and each releases a different kind of particle or energy. The essential features of each reaction are shown in \(\Page {1}\). The most common are alpha and beta decay and gamma emission, but the others are essential to an understanding of nuclear decay reactions. Many nuclei with mass numbers greater than 200 undergo , which results in the emission of a helium-4 nucleus as an , \(^4_{2}\alpha\). The general reaction is as follows: \[\underset{\textrm{parent}}{^A_Z \textrm X}\rightarrow \underset{\textrm{daughter}}{^{A-4}_{Z-2} \textrm X'}+\underset{\textrm{alpha}\\ \textrm{particle}}{^4_2 \alpha}\label{5.2.1}\] The daughter nuclide contains two fewer protons and two fewer neutrons than the parent. Thus α-particle emission produces a daughter nucleus with a mass number − 4 and a nuclear charge − 2 compared to the parent nucleus. Radium-226, for example, undergoes alpha decay to form radon-222: \[^{226}_{88}\textrm{Ra}\rightarrow ^{222}_{86}\textrm{Rn}+^{4}_{2}\alpha\label{5.2.2}\] Because nucleons are conserved in this and all other nuclear reactions, the sum of the mass numbers of the products, 222 + 4 = 226, equals the mass number of the parent. Similarly, the sum of the atomic numbers of the products, 86 + 2 = 88, equals the atomic number of the parent. Thus the nuclear equation is balanced. Just as the total number of atoms is conserved in a chemical reaction, the total number of nucleons is conserved in a nuclear reaction. Nuclei that contain too many neutrons often undergo , in which a neutron is converted to a proton and a high-energy electron that is ejected from the nucleus as a β particle: \[\underset{\textrm{unstable} \\ \textrm{neutron in} \\ \textrm{nucleus}}{^1_0 \textrm n}\rightarrow \underset{\textrm{proton} \\ \textrm{retained} \\ \textrm{by nucleus}}{^{1}_{1} \textrm p}+\underset{\textrm{beta particle} \\ \textrm{emitted by} \\ \textrm{nucleus}}{^0_{-1} \beta}\label{5.2.3}\] The general reaction for beta decay is therefore Although beta decay does not change the mass number of the nucleus, it does result in an increase of +1 in the atomic number because of the addition of a proton in the daughter nucleus. Thus beta decay decreases the neutron-to-proton ratio, moving the nucleus toward the band of stable nuclei. For example, carbon-14 undergoes beta decay to form nitrogen-14: \[^{14}_{6}\textrm{C}\rightarrow ^{14}_{7}\textrm{N}+\,^{0}_{-1}\beta\label{5.2.5}\] Once again, the number of nucleons is conserved, and the charges are balanced. The parent and the daughter nuclei have the same mass number, 14, and the sum of the atomic numbers of the products is 6, which is the same as the atomic number of the carbon-14 parent. Because a positron has the same mass as an electron but opposite charge, is the opposite of beta decay. Thus positron emission is characteristic of neutron-poor nuclei, which decay by transforming a proton to a neutron and emitting a high-energy positron: \[^{1}_{1}\textrm{p}^+\rightarrow ^{1}_{0}\textrm{n}+\,^{0}_{+1}\beta^+\label{5.2.6}\] The general reaction for positron emission is therefore Like beta decay, positron emission does not change the mass number of the nucleus. In this case, however, the atomic number of the daughter nucleus is lower by 1 than that of the parent. Thus the neutron-to-proton ratio has increased, again moving the nucleus closer to the band of stable nuclei. For example, carbon-11 undergoes positron emission to form boron-11: \[^{11}_{6}\textrm{C}\rightarrow ^{11}_{5}\textrm{B}+\,^{0}_{+1}\beta^+ \label{5.2.8}\] Nucleons are conserved, and the charges balance. The mass number, 11, does not change, and the sum of the atomic numbers of the products is 6, the same as the atomic number of the parent carbon-11 nuclide. A neutron-poor nucleus can decay by either positron emission or , in which an electron in an inner shell reacts with a proton to produce a neutron: \[^{1}_{1}\textrm{p} +\; ^{0}_{-1}\textrm{e}\rightarrow \, ^{1}_{0}\textrm n\label{5.2.9}\] When a second electron moves from an outer shell to take the place of the lower-energy electron that was absorbed by the nucleus, an x-ray is emitted. The overall reaction for electron capture is thus Electron capture does not change the mass number of the nucleus because both the proton that is lost and the neutron that is formed have a mass number of 1. As with positron emission, however, the atomic number of the daughter nucleus is lower by 1 than that of the parent. Once again, the neutron-to-proton ratio has increased, moving the nucleus toward the band of stable nuclei. For example, iron-55 decays by electron capture to form manganese-55, which is often written as follows: \[^{55}_{26}\textrm{Fe}\overset{\textrm{EC}}{\rightarrow}\, ^{55}_{25}\textrm{Mn}+\textrm{x-ray}\label{5.2.11}\] The atomic numbers of the parent and daughter nuclides differ in , although the mass numbers are the same. To write a balanced nuclear equation for this reaction, we must explicitly include the captured electron in the equation: \[^{55}_{26}\textrm{Fe}+\,^{0}_{-1}\textrm{e}\rightarrow \, ^{55}_{25}\textrm{Mn}+\textrm{x-ray}\label{5.2.12}\] Both positron emission and electron capture are usually observed for nuclides with low neutron-to-proton ratios, but the decay rates for the two processes can be very different. Many nuclear decay reactions produce daughter nuclei that are in a nuclear excited state, which is similar to an atom in which an electron has been excited to a higher-energy orbital to give an electronic excited state. Just as an electron in an electronic excited state emits energy in the form of a photon when it returns to the ground state, a nucleus in an excited state releases energy in the form of a photon when it returns to the ground state. These high-energy photons are γ rays. (\(\gamma\)) can occur virtually instantaneously, as it does in the alpha decay of uranium-238 to thorium-234, where the asterisk denotes an excited state: \[^{238}_{92}\textrm{U}\rightarrow \, \underset{\textrm{excited} \\ \textrm{nuclear} \\ \textrm{state}}{^{234}_{90}\textrm{Th*}} + ^{4}_{2}\alpha\xrightarrow {\textrm{relaxation}\,}\,^{234}_{90}\textrm{Th}+^{0}_{0}\gamma\label{5.2.13}\] If we disregard the decay event that created the excited nucleus, then \[^{234}_{88}\textrm{Th*} \rightarrow\, ^{234}_{88}\textrm{Th}+^{0}_{0}\gamma\label{5.2.14}\] or more generally, \[^{A}_{Z}\textrm{X*} \rightarrow\, ^{A}_{Z}\textrm{X}+^{0}_{0}\gamma\label{5.2.15}\] Gamma emission can also occur after a significant delay. For example, technetium-99 has a half-life of about 6 hours before emitting a γ ray to form technetium-99 (the is for metastable). Because γ rays are energy, their emission does not affect either the mass number or the atomic number of the daughter nuclide. Gamma-ray emission is therefore the only kind of radiation that does not necessarily involve the conversion of one element to another, although it is almost always observed in conjunction with some other nuclear decay reaction. Only very massive nuclei with high neutron-to-proton ratios can undergo , in which the nucleus breaks into two pieces that have different atomic numbers and atomic masses. This process is most important for the transactinide elements, with ≥ 104. Spontaneous fission is invariably accompanied by the release of large amounts of energy, and it is usually accompanied by the emission of several neutrons as well. An example is the spontaneous fission of \(^{254}_{98}\textrm{Cf}\), which gives a distribution of fission products; one possible set of products is shown in the following equation: \[^{254}_{98}\textrm{Cf}\rightarrow \,^{118}_{46}\textrm{Pd}+\,^{132}_{52}\textrm{Te}+4^{1}_{0}\textrm{n}\label{5.2.16}\] Once again, the number of nucleons is conserved. Thus the sum of the mass numbers of the products (118 + 132 + 4 = 254) equals the mass number of the reactant. Similarly, the sum of the atomic numbers of the products [46 + 52 + (4 × 0) = 98] is the same as the atomic number of the parent nuclide. Write a balanced nuclear equation to describe each reaction. radioactive nuclide and mode of decay balanced nuclear equation Identify the reactants and the products from the information given. Use the values of and to identify any missing components needed to balance the equation. a. We know the identities of the reactant and one of the products (a β particle). We can therefore begin by writing an equation that shows the reactant and one of the products and indicates the unknown product as \(^{A}_{Z}\textrm{X}\): Because both protons and neutrons must be conserved in a nuclear reaction, the unknown product must have a mass number of = 35 − 0 = 35 and an atomic number of = 16 − (−1) = 17. The element with = 17 is chlorine, so the balanced nuclear equation is as follows: We know the identities of both reactants: \(^{201}_{80}\textrm{Hg}\) and an inner electron, \(^{0}_{-1}\textrm{e}\). The reaction is as follows: \(^{201}_{80}\textrm{Hg}+\,^{0}_{-1}\textrm e\rightarrow\,^{A}_{Z}\textrm{X}\) Both protons and neutrons are conserved, so the mass number of the product must be = 201 + 0 = 201, and the atomic number of the product must be = 80 + (−1) = 79, which corresponds to the element gold. The balanced nuclear equation is thus \(^{201}_{80}\textrm{Hg}+\,^{0}_{-1}\textrm e\rightarrow\,^{201}_{79}\textrm{Au}\) As in part (a), we are given the identities of the reactant and one of the products—in this case, a positron. The unbalanced nuclear equation is therefore \(^{30}_{15}\textrm{P}\rightarrow\,^{A}_{Z}\textrm{X}+\,^{0}_{+1}\beta\) The mass number of the second product is = 30 − 0 = 30, and its atomic number is = 15 − 1 = 14, which corresponds to silicon. The balanced nuclear equation for the reaction is as follows: \(^{30}_{15}\textrm{P}\rightarrow\,^{30}_{14}\textrm{Si}+\,^{0}_{+1}\beta\) Write a balanced nuclear equation to describe each reaction. Predict the kind of nuclear change each unstable nuclide undergoes when it decays. nuclide type of nuclear decay Based on the neutron-to-proton ratio and the value of , predict the type of nuclear decay reaction that will produce a more stable nuclide. Predict the kind of nuclear change each unstable nuclide undergoes when it decays. The nuclei of all elements with atomic numbers greater than 83 are unstable. Thus all isotopes of all elements beyond bismuth in the periodic table are radioactive. Because alpha decay decreases by only 2, and positron emission or electron capture decreases by only 1, it is impossible for any nuclide with > 85 to decay to a stable daughter nuclide in a single step, except via nuclear fission. Consequently, radioactive isotopes with > 85 usually decay to a daughter nucleus that is radiaoctive, which in turn decays to a second radioactive daughter nucleus, and so forth, until a stable nucleus finally results. This series of sequential alpha- and beta-decay reactions is called a . The most common is the uranium-238 decay series, which produces lead-206 in a series of 14 sequential alpha- and beta-decay reactions ( \(\Page {2}\)). Although a radioactive decay series can be written for almost any isotope with > 85, only two others occur naturally: the decay of uranium-235 to lead-207 (in 11 steps) and thorium-232 to lead-208 (in 10 steps). A fourth series, the decay of neptunium-237 to bismuth-209 in 11 steps, is known to have occurred on the primitive Earth. With a half-life of “only” 2.14 million years, all the neptunium-237 present when Earth was formed decayed long ago, and today all the neptunium on Earth is synthetic. Due to these radioactive decay series, small amounts of very unstable isotopes are found in ores that contain uranium or thorium. These rare, unstable isotopes should have decayed long ago to stable nuclei with a lower atomic number, and they would no longer be found on Earth. Because they are generated continuously by the decay of uranium or thorium, however, their amounts have reached a steady state, in which their rate of formation is equal to their rate of decay. In some cases, the abundance of the daughter isotopes can be used to date a material or identify its origin. The discovery of radioactivity in the late 19th century showed that some nuclei spontaneously transform into nuclei with a different number of protons, thereby producing a different element. When scientists realized that these naturally occurring radioactive isotopes decayed by emitting subatomic particles, they realized that—in principle—it should be possible to carry out the reverse reaction, converting a stable nucleus to another more massive nucleus by bombarding it with subatomic particles in a nuclear transmutation reaction. The first successful nuclear transmutation reaction was carried out in 1919 by Ernest Rutherford, who showed that α particles emitted by radium could react with nitrogen nuclei to form oxygen nuclei. As shown in the following equation, a proton is emitted in the process: \[^{4}_{2}\alpha + \, ^{14}_{7}\textrm{N} \rightarrow \,^{17}_{8}\textrm{O}+\,^{1}_{1}\textrm{p}\label{5.2.17}\] Rutherford’s nuclear transmutation experiments led to the discovery of the neutron. He found that bombarding the nucleus of a light target element with an α particle usually converted the target nucleus to a product that had an atomic number higher by 1 and a mass number higher by 3 than the target nucleus. Such behavior is consistent with the emission of a proton after reaction with the α particle. Very light targets such as Li, Be, and B reacted differently, however, emitting a new kind of highly penetrating radiation rather than a proton. Because neither a magnetic field nor an electrical field could deflect these high-energy particles, Rutherford concluded that they were electrically neutral. Other observations suggested that the mass of the neutral particle was similar to the mass of the proton. In 1932, James Chadwick (Nobel Prize in Physics, 1935), who was a student of Rutherford’s at the time, named these neutral particles neutrons and proposed that they were fundamental building blocks of the atom. The reaction that Chadwick initially used to explain the production of neutrons was as follows: \[^{4}_{2}\alpha + \, ^{9}_{4}\textrm{Be} \rightarrow \,^{12}_{6}\textrm{C}+\,^{1}_{0}\textrm{n}\label{5.2.18}\] Because α particles and atomic nuclei are both positively charged, electrostatic forces cause them to repel each other. Only α particles with very high kinetic energy can overcome this repulsion and collide with a nucleus ( \(\Page {3}\)). Neutrons have no electrical charge, however, so they are not repelled by the nucleus. Hence bombardment with neutrons is a much easier way to prepare new isotopes of the lighter elements. In fact, carbon-14 is formed naturally in the atmosphere by bombarding nitrogen-14 with neutrons generated by cosmic rays: \[^{1}_{0}\textrm{n} + \, ^{14}_{7}\textrm{N} \rightarrow \,^{14}_{6}\textrm{C}+\,^{1}_{1}\textrm{p}\label{5.2.19}\] In 1933, Frédéric Joliot and Iréne Joliot-Curie (daughter of Marie and Pierre Curie) prepared the first artificial radioactive isotope by bombarding aluminum-27 with α particles. For each Al that reacted, one neutron was released. Identify the product nuclide and write a balanced nuclear equation for this transmutation reaction. reactants in a nuclear transmutation reaction product nuclide and balanced nuclear equation Based on the reactants and one product, identify the other product of the reaction. Use conservation of mass and charge to determine the values of and of the product nuclide and thus its identity. Write the balanced nuclear equation for the reaction. Bombarding an element with α particles usually produces an element with an atomic number that is 2 greater than the atomic number of the target nucleus. Thus we expect that aluminum ( = 13) will be converted to phosphorus ( = 15). With one neutron released, conservation of mass requires that the mass number of the other product be 3 greater than the mass number of the target. In this case, the mass number of the target is 27, so the mass number of the product will be 30. The second product is therefore phosphorus-30, \(^{30}_{15}\textrm{P}\). The balanced nuclear equation for the reaction is as follows: \[^{27}_{13}\textrm{Al} + \, ^{4}_{2}\alpha \rightarrow \,^{30}_{15}\textrm{P}+\,^{1}_{0}\textrm{n}\] Because all isotopes of technetium are radioactive and have short half-lives, it does not exist in nature. Technetium can, however, be prepared by nuclear transmutation reactions. For example, bombarding a molybdenum-96 target with deuterium nuclei \((^{2}_{1}\textrm{H})\) produces technetium-97. Identify the other product of the reaction and write a balanced nuclear equation for this transmutation reaction. neutron, \(^{1}_{0}\textrm{n}\); \(^{96}_{42}\textrm{Mo} + \, ^{2}_{1}\textrm{H} \rightarrow \,^{97}_{43}\textrm{Tc}+\,^{1}_{0}\textrm{n}\) We noted earlier in this section that very heavy nuclides, corresponding to ≥ 104, tend to decay by spontaneous fission. Nuclides with slightly lower values of , such as the isotopes of uranium ( = 92) and plutonium ( = 94), do not undergo spontaneous fission at any significant rate. Some isotopes of these elements, however, such as \(^{235}_{92}\textrm{U}\) and \(^{239}_{94}\textrm{Pu}\) undergo induced nuclear fission when they are bombarded with relatively low-energy neutrons, as shown in the following equation for uranium-235 and in \(\Page {3}\): \[^{235}_{92}\textrm{U} + \, ^{1}_{0}\textrm{n} \rightarrow \,^{236}_{92}\textrm{U}\rightarrow \,^{141}_{56}\textrm{Ba}+\,^{92}_{36}\textrm{Kr}+3^{1}_{0}\textrm{n}\label{5.2.20}\] Any isotope that can undergo a nuclear fission reaction when bombarded with neutrons is called a . During nuclear fission, the nucleus usually divides asymmetrically rather than into two equal parts, as shown in \(\Page {4}\). Moreover, every fission event of a given nuclide does not give the same products; more than 50 different fission modes have been identified for uranium-235, for example. Consequently, nuclear fission of a fissile nuclide can never be described by a single equation. Instead, as shown in \(\Page {5}\), a distribution of many pairs of fission products with different yields is obtained, but the mass ratio of each pair of fission products produced by a single fission event is always roughly 3:2. Uranium ( = 92) is the heaviest naturally occurring element. Consequently, all the elements with > 92, the , are artificial and have been prepared by bombarding suitable target nuclei with smaller particles. The first of the transuranium elements to be prepared was neptunium ( = 93), which was synthesized in 1940 by bombarding a U target with neutrons. As shown in , this reaction occurs in two steps. Initially, a neutron combines with a U nucleus to form U, which is unstable and undergoes beta decay to produce Np: \(^{238}_{92}\textrm{U} + \, ^{1}_{0}\textrm{n} \rightarrow \,^{239}_{92}\textrm{U}\rightarrow \,^{239}_{93}\textrm{Np}+\,^{0}_{-1}\beta\label{5.2.21}\) Subsequent beta decay of Np produces the second transuranium element, plutonium ( = 94): \(^{239}_{93}\textrm{Np} \rightarrow \,^{239}_{94}\textrm{Pu}+\,^{0}_{-1}\beta\label{5.2.22}\) Bombarding the target with more massive nuclei creates elements that have atomic numbers significantly greater than that of the target nucleus ( \(\Page {2}\)). Such techniques have resulted in the creation of the superheavy elements 114 and 116, both of which lie in or near the “island of stability." A device called a particle accelerator is used to accelerate positively charged particles to the speeds needed to overcome the electrostatic repulsions between them and the target nuclei by using electrical and magnetic fields. Operationally, the simplest particle accelerator is the linear accelerator ( \(\Page {6}\)), in which a beam of particles is injected at one end of a long evacuated tube. Rapid alternation of the polarity of the electrodes along the tube causes the particles to be alternately accelerated toward a region of opposite charge and repelled by a region with the same charge, resulting in a tremendous acceleration as the particle travels down the tube. A modern linear accelerator such as the Stanford Linear Accelerator (SLAC) at Stanford University is about 2 miles long. To achieve the same outcome in less space, a particle accelerator called a cyclotron forces the charged particles to travel in a circular path rather than a linear one. The particles are injected into the center of a ring and accelerated by rapidly alternating the polarity of two large D-shaped electrodes above and below the ring, which accelerates the particles outward along a spiral path toward the target. The length of a linear accelerator and the size of the D-shaped electrodes in a cyclotron severely limit the kinetic energy that particles can attain in these devices. These limitations can be overcome by using a synchrotron, a hybrid of the two designs. A synchrotron contains an evacuated tube similar to that of a linear accelerator, but the tube is circular and can be more than a mile in diameter. Charged particles are accelerated around the circle by a series of magnets whose polarities rapidly alternate. In , the parent nucleus is converted to a more stable daughter nucleus. Nuclei with too many neutrons decay by converting a neutron to a proton, whereas nuclei with too few neutrons decay by converting a proton to a neutron. Very heavy nuclei (with ≥ 200 and > 83) are unstable and tend to decay by emitting an . When an unstable nuclide undergoes radioactive decay, the total number of nucleons is conserved, as is the total positive charge. Six different kinds of nuclear decay reactions are known. results in the emission of an α particle, \(^4 _2 \alpha\), and produces a daughter nucleus with a mass number that is lower by 4 and an atomic number that is lower by 2 than the parent nucleus. converts a neutron to a proton and emits a high-energy electron, producing a daughter nucleus with the same mass number as the parent and an atomic number that is higher by 1. is the opposite of beta decay and converts a proton to a neutron plus a positron. Positron emission does not change the mass number of the nucleus, but the atomic number of the daughter nucleus is lower by 1 than the parent. In , an electron in an inner shell reacts with a proton to produce a neutron, with emission of an x-ray. The mass number does not change, but the atomic number of the daughter is lower by 1 than the parent. In , a daughter nucleus in a nuclear excited state undergoes a transition to a lower-energy state by emitting a γ ray. Very heavy nuclei with high neutron-to-proton ratios can undergo , in which the nucleus breaks into two pieces that can have different atomic numbers and atomic masses with the release of neutrons. Many very heavy nuclei decay via a —a succession of some combination of alpha- and beta-decay reactions. In , a target nucleus is bombarded with energetic subatomic particles to give a product nucleus that is more massive than the original. All —elements with > 92—are artificial and must be prepared by nuclear transmutation reactions. These reactions are carried out in particle accelerators such as linear accelerators, cyclotrons, and synchrotrons. Nuclear decay reactions occur spontaneously under all conditions and produce more stable daughter nuclei, whereas nuclear transmutation reactions are induced and form a product nucleus that is more massive than the starting material. : \(^A_Z \textrm X\rightarrow \, ^{A-4}_{Z-2} \textrm X'+\,^4_2 \alpha\) : \(^A_Z \textrm X\rightarrow \, ^{A}_{Z+1} \textrm X'+\,^0_{-1} \beta\) : \(^A_Z \textrm X\rightarrow \, ^{A}_{Z-1} \textrm X'+\,^0_{+1} \beta\) : \(^A_Z \textrm X+\,^{0}_{-1} \textrm e\rightarrow \, ^{A}_{Z-1} \textrm X'+\textrm{x-ray}\) : \(^A_Z \textrm{X*}\rightarrow \, ^{A}_{Z} \textrm X+\,^0_{0} \gamma\) | 27,726 | 2,062 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/05%3A_Energy_Changes_in_Chemical_Reactions/5.03%3A_Hess's_Law |
Because enthalpy is a state function, the enthalpy change for a reaction depends on two things: (1) the masses of the reacting substances and (2) the physical states of the reactants and products. It does depend on the path by which reactants are converted to products. If you climbed a mountain, for example, the altitude change would not depend on whether you climbed the entire way without stopping or you stopped many times to take a break. If you stopped often, the overall change in altitude would be the sum of the changes in altitude for each short stretch climbed. Similarly, when we add two or more balanced chemical equations to obtain a net chemical equation, H H This principle is called Hess’s law , after the Swiss-born Russian chemist Germain Hess (1802–1850), a pioneer in the study of thermochemistry. Hess’s law allows us to calculate Δ values for reactions that are difficult to carry out directly by adding together the known Δ values for individual steps that give the overall reaction, even though the overall reaction may not actually occur via those steps. We can illustrate Hess’s law using the thermite reaction. The overall reaction shown in Equation 9.3.1 can be viewed as occurring in three distinct steps with known Δ values. As shown in Figure \(\Page {1}\), the first reaction produces 1 mol of solid aluminum oxide (Al O ) and 2 mol of liquid iron at its melting point of 1758°C (part (a) in Equation \(\Page {1}\) ); the enthalpy change for this reaction is −732.5 kJ/mol of Fe O . The second reaction is the conversion of 2 mol of liquid iron at 1758°C to 2 mol of solid iron at 1758°C (part (b) in Equation 9.3.1); the enthalpy change for this reaction is −13.8 kJ/mol of Fe (−27.6 kJ per 2 mol Fe). In the third reaction, 2 mol of solid iron at 1758°C is converted to 2 mol of solid iron at 25°C (part (c) in Equation \(\Page {1}\) ); the enthalpy change for this reaction is −45.5 kJ/mol of Fe (−91.0 kJ per 2 mol Fe). As you can see in Figure \(\Page {1}\), the overall reaction is given by the longest arrow (shown on the left), which is the sum of the three shorter arrows (shown on the right). Adding parts (a), (b), and (c) in Equation \(\Page {1}\) gives the overall reaction, shown in part (d):
\[ \begin{matrix}
2Al\left ( s, \; 25 ^{o}C \right ) + 2Fe_{2}O_{3}\left ( s, \; 25 ^{o}C \right )& \rightarrow & 2Fe\left ( l, \; 1758 ^{o}C \right ) + 2Al_{2}O_{3}\left ( s, \; 1758 ^{o}C \right ) & \Delta H=-732.5 \; kJ& \left ( a \right ) \\
2Fe\left ( l, \; 1758 ^{o}C \right ) & \rightarrow & 2Fe\left ( s, \; 1758 ^{o}C \right ) & \Delta H=-\;\; 27.6 \; kJ & \left ( b \right )\\
2Fe\left ( s, \; 1758 ^{o}C \right ) + 2Al_{2}O_{3}\left ( s, \; 1758 ^{o}C \right ) & \rightarrow & 2Fe\left ( l, \; 25 ^{o}C \right ) + 2Al_{2}O_{3}\left ( s, \; 25 ^{o}C \right ) & \Delta H=-\;\; 91.0 \; kJ & \left ( c \right )\\
2Al\left ( s, \; 25 ^{o}C \right ) + 2Fe_{2}O_{3}\left ( s, \; 25 ^{o}C \right ) & \rightarrow & 2Al_{2}O_{3}\left ( s, \; 25 ^{o}C \right ) + 2Fe_{2}O_{3}\left ( s, \; 25 ^{o}C \right ) & \Delta H=-852.2 \; kJ & \left ( d \right )
\end{matrix} \] The net reaction in part (d) in Equation \(\Page {1}\) is identical to Equation \(\Page {1}\) the sum of parts (a) + (b) +(c). By Hess’s law, the enthalpy change for part (d) is the sum of the enthalpy changes for parts (a), (b), and (c). In essence, Hess’s law enables us to calculate the enthalpy change for the sum of a series of reactions without having to draw a diagram like that in Figure \(\Page {1}\). Comparing parts (a) and (d) in Equation \(\Page {1}\) also illustrates an important point: H When the product is liquid iron at its melting point (part (a) in Equation \(\Page {1}\) ), only 732.5 kJ of heat are released to the surroundings compared with 852 kJ when the product is solid iron at 25°C (part (d) in Equation 9.3.1). The difference, 120 kJ, is the amount of energy that is released when 2 mol of liquid iron solidifies and cools to 25°C. It is important to specify the physical state of all reactants and products when writing a thermochemical equation. When using Hess’s law to calculate the value of Δ for a reaction, follow this procedure: We illustrate how to use this procedure in Example \(\Page {1}\) When carbon is burned with limited amounts of oxygen gas (O ), carbon monoxide (CO) is the main product: \( \left ( 1 \right ) \;2C\left ( s \right ) + O_{2}\left ( g \right ) \rightarrow 2CO\left ( g \right ) \; \;\ \; \Delta H=-221.0 \; kJ \] When carbon is burned in excess O , carbon dioxide (CO ) is produced: \[ \left ( 2 \right ) \;C\left ( s \right ) + O_{2}\left ( g \right ) \rightarrow CO_{2}\left ( g \right ) \; \;\ \; \Delta H=-393.5 \; kJ \] Use this information to calculate the enthalpy change per mole of CO for the reaction of CO with O to give CO . two balanced chemical equations and their Δ values enthalpy change for a third reaction After balancing the chemical equation for the overall reaction, write two equations whose Δ values are known and that, when added together, give the equation for the overall reaction. (Reverse the direction of one or more of the equations as necessary, making sure to also reverse the sign of Δ .) Multiply the equations by appropriate factors to ensure that they give the desired overall chemical equation when added together. To obtain the enthalpy change per mole of CO, write the resulting equations as a sum, along with the enthalpy change for each. We begin by writing the balanced chemical equation for the reaction of interest: \[ \left ( 3 \right ) \;CO\left ( g \right ) + \frac{1}{2}O_{2}\left ( g \right ) \rightarrow CO_{2}\left ( g \right ) \; \;\ \; \Delta H_{rxn}=? \] There are at least two ways to solve this problem using Hess’s law and the data provided. The simplest is to write two equations that can be added together to give the desired equation and for which the enthalpy changes are known. Observing that CO, a reactant in Equation 3, is a product in Equation 1, we can reverse Equation (1) to give \[ 2CO\left ( g \right ) \rightarrow 2C\left ( s \right ) + O_{2}\left ( g \right ) \; \;\ \; \Delta H=+221.0 \; kJ \] Because we have reversed the direction of the reaction, the sign of Δ is changed. We can use Equation 2 as written because its product, CO , is the product we want in Equation 3: \[ C\left ( s \right ) + O_{2}\left ( g \right ) \rightarrow CO_{2}\left ( s \right ) \; \;\ \; \Delta H=-393.5 \; kJ \] Adding these two equations together does not give the desired reaction, however, because the numbers of C(s) on the left and right sides do not cancel. According to our strategy, we can multiply the second equation by 2 to obtain 2 mol of C(s) as the reactant: \[ 2C\left ( s \right ) + 2O_{2}\left ( g \right ) \rightarrow 2CO_{2}\left ( s \right ) \; \;\ \; \Delta H=-787.0 \; kJ \] Writing the resulting equations as a sum, along with the enthalpy change for each, gives \[ \begin{matrix}
2CO\left ( g \right ) & \rightarrow & \cancel{2C\left ( s \right )}+\cancel{O_{2}\left ( g \right )} & \Delta H & = & -\Delta H_{1} & = & +221.0 \; kJ \\
\cancel{2C\left ( s \right )}+\cancel{2}O_{2}\left ( g \right ) & \rightarrow & 2CO_{2} \left ( g \right ) & \Delta H & = & -\Delta 2H_{2} & = & -787.0 \; kJ \\
2CO\left ( g \right ) + O_{2}\left ( g \right ) & \rightarrow & 2CO_{2} \left ( g \right ) & \Delta H & = & & -566.0 \; kJ
\end{matrix} \] Note that the overall chemical equation and the enthalpy change for the reaction are both for the reaction of 2 mol of CO with O , and the problem asks for the amount . Consequently, we must divide both sides of the final equation by 2: \[ \begin{matrix}
CO\left ( g \right ) + \frac{1}{2}O_{2}\left ( g \right ) & \rightarrow & CO_{2} \left ( g \right ) & \Delta H & = & & -283.0 \; kJ
\end{matrix} \] An alternative and equally valid way to solve this problem is to write the two given equations as occurring in steps. Note that we have multiplied the equations by the appropriate factors to allow us to cancel terms: \[ \begin{matrix}
\left ( A \right ) & 2C\left ( s \right ) + O_{2}\left ( g \right ) & \rightarrow & \cancel{2CO\left ( g \right )} & \Delta H_{A} & = & \Delta H_{1} & = & +221.0 \; kJ \\
\left ( B \right ) &\cancel{2CO\left ( g \right )} + O_{2}\left ( g \right ) & \rightarrow & 2CO_{2} \left ( g \right ) & \Delta H_{B} & & & = & ? \\
\left ( C \right ) & 2C\left ( s \right ) + 2O_{2}\left ( g \right ) & \rightarrow & 2CO_{2} \left ( g \right ) & \Delta H & = 2\Delta H_{2} & =2\times \left ( -393.5 \; kJ \right ) & =-787.0 \; kJ
\end{matrix} \] The sum of reactions A and B is reaction C, which corresponds to the combustion of 2 mol of carbon to give CO . From Hess’s law, Δ + Δ = Δ , and we are given Δ for reactions A and C. Substituting the appropriate values gives \[ \begin{matrix}
-221.0 \; kJ + \Delta H_{B} = -787.0 \; kJ \\
\Delta H_{B} = -566.0
\end{matrix} \] This is again the enthalpy change for the conversion of 2 mol of CO to CO . The enthalpy change for the conversion of 1 mol of CO to CO is therefore −566.0 ÷ 2 = −283.0 kJ/mol of CO, which is the same result we obtained earlier. As you can see, . The reaction of acetylene (C H ) with hydrogen (H ) can produce either ethylene (C H ) or ethane (C H ): \[ \begin{matrix}
C_{2}H_{2}\left ( g \right ) + H_{2}\left ( g \right ) \rightarrow C_{2}H_{4}\left ( g \right ) & \Delta H = -175.7 \; kJ/mol \; C_{2}H_{2} \\
C_{2}H_{2}\left ( g \right ) + 2H_{2}\left ( g \right ) \rightarrow C_{2}H_{6}\left ( g \right ) & \Delta H = -312.0 \; kJ/mol \; C_{2}H_{2}
\end{matrix} \] What is Δ for the reaction of C H with H to form C H ? −136.3 kJ/mol of C H Chapter 7 and Chapter 8 presented a wide variety of chemical reactions, and you learned how to write balanced chemical equations that include all the reactants and the products except heat. One way to report the heat absorbed or released would be to compile a massive set of reference tables that list the enthalpy changes for all possible chemical reactions, which would require an incredible amount of effort. Fortunately, Hess’s law allows us to calculate the enthalpy change for virtually any conceivable chemical reaction using a relatively small set of tabulated data, such as the following: The sign convention is the same for all enthalpy changes: negative if heat is released by the system and positive if heat is absorbed by the system. For a chemical reaction, the is the difference in enthalpy between products and reactants; the units of Δ are kilojoules per mole. Reversing a chemical reaction reverses the sign of Δ . The magnitude of Δ also depends on the physical state of the reactants and the products because processes such as melting solids or vaporizing liquids are also accompanied by enthalpy changes: the and the , respectively. The overall enthalpy change for a series of reactions is the sum of the enthalpy changes for the individual reactions, which is . The is the enthalpy change that occurs when a substance is burned in excess oxygen. Based on the following energy diagram, a. write an equation showing how the value of Δ could be determined if the values of Δ and Δ are known. b; identify each step as being exothermic or endothermic. Based on the following energy diagram, a. write an equation showing how the value of Δ could be determined if the values of Δ and Δ are known. b. identify each step as being exothermic or endothermic. Describe how Hess’s law can be used to calculate the enthalpy change of a reaction that cannot be observed directly. When you apply Hess’s law, what enthalpy values do you need to account for each change in physical state? In their elemental form, A and B exist as diatomic molecules. Given the following reactions, each with an associated Δ °, describe how you would calculate ΔH for the compound AB . \[ \begin{matrix}
2AB & \rightarrow & A_{2} + B _{2} & \Delta H_{1}^{o}\\
3AB & \rightarrow & AB_{2} + A _{2}B & \Delta H_{2}^{o} \\
2A_{2}B &\rightarrow & 2A_{2} + B _{2} & \Delta H_{3}^{o}
\end{matrix} \] Methanol is used as a fuel in Indianapolis 500 race cars. Use the following table to determine whether methanol or 2,2,4-trimethylpentane (isooctane) releases more energy per liter during combustion. a. Use the enthalpies of combustion given in the following table to determine which organic compound releases the greatest amount of energy per gram during combustion. b. Calculate the standard enthalpy of formation of 1-ethyl-2-methylbenzene. Given the enthalpies of combustion, which organic compound is the best fuel per gram? 1. 2. a. To one decimal place Octane provides the largest amount of heat per gram upon combustion. b, ( ) | 12,774 | 2,063 |
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Directed motion of molecules in a statistically deterministic manner (i.e., x̅(t) = v̅t) in a thermally fluctuating environment cannot happen spontaneously. It requires a free energy source, which may come from chemical bonds, charge transfer, and electrochemical gradients. From one perspective, displacing a particle requires work, and the force behind this work originates in free energy gradients along the direction of propagation \[ w =-\int_{path}f dx \qquad \qquad f_{rev} = \dfrac{\partial G}{\partial x} \nonumber \] An example of this is steady-state diffusion driven by a spatial difference in chemical potential, for instance the diffusion of ions through a membrane channel driven by a transmembrane potential. This problem is one of passive transport. Although an active input of energy was required to generate the transmembrane potential and the net motion of the ion is directional, the ion itself is a passive participant in this process. Such processes can be modeled as diffusion within a potential. Active transport refers to the direct input of energy into the driving the moving object in a directional manner. At a molecular scale, even with this input of energy, fluctuations and Brownian motion remain very important. Even so, there are multiple ways in which to conceive of directed motion. Step-wise processive motion can also be viewed as a series of states along a free energy or chemical potential gradient. Consider this energy landscape: Under steady state conditions, detailed balance dictates that the ratio of rates for passing forward or reverse over a barrier is dictated by the free energy difference between the initial and final states: \[ \dfrac{k_+}{k_-} = e^{-\Delta G/k_BT} \nonumber \] and thus the active driving force for this downhill process is \[ f \approx -\dfrac{\Delta G}{\Delta x} = \dfrac{k_BT}{\Delta x } \ln{\dfrac{k_+}{k_-} } \nonumber \] This perspective is intimately linked with a biased random walk model when we remember that \[ \dfrac{k_+}{k_-} =\dfrac{P_+}{P_-} \nonumber \] If our free energy is the combination of a chemical process (\(\Delta G_0\)) and an external force, then we can write \[ \dfrac{k_+}{k_-} = \mathrm{exp}[-(\Delta G_0 +f\Delta x)/k_BT] \nonumber \] Feynman used a thought experiment to show you cannot get work from thermal noise. Assume you want to use the thermal kinetic energy from the molecules in a gas, and decide to use the collisions of these molecules with a vane to rotate an axle. The direction or rotation will be based on the velocity of the molecules hitting the vane, so to assure that this rotation proceeds only one way, we use a ratchet with a pawl and spring to catch the ratchet when it advances in one direction. This is the concept of rectified Brownian motion. At a microscopic level, this reasoning does not hold, because the energy used to rotate the ratchet must be enough to lift the pawl against the force of the spring. If we match the thermal energy of gas \(T =\frac{1}{2}m\langle v^2_x \rangle \) to the energy needed to raise the pawl \(U=\frac{1}{2}\kappa x^2\) we find that the pawl will also be undergoing fluctuations in x with similar statistics to the bombardment of the vane \(\kappa = \sqrt{mk_BT/\langle x^2\rangle}\). Therefore, the ratchet will instead thermally diffuse back and forth as a random walk. Further, Feynman showed that if you imbedded the vane and ratchet in reservoirs of temperature T and T , respectively, then the ratchet will advance as desired if T > T , but will move in reverse if T < T . Thus, one cannot extract useful work from thermal fluctuations alone. You need some input of energy—any source of free energy. _________________________________ | 3,722 | 2,064 |
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By now you should be familiar with wherein compounds of the same molecular formula differ because substituents, chain branches, and so on, are not at the same positions in the molecules. 1-Chloropropane and 2-chloropropane are straightforward examples of position isomers. A much more subtle form of isomerism is present when two compounds have the molecular formulas, the substituent and chain-branching positions, and, indeed, even have the names by all of the nomenclature rules we have given you so far. . These are and this type of isomerism, called , is of enormous importance to all areas of organic chemistry and biochemistry. To understand stereoisomerism of carbon compounds, we must understand the ways in which the bonds to carbon atoms are arranged in space. As shown in Section 2-2A, this depends on whether the carbon atoms form single, double, or triple bonds to another atom. Thus, four single bonds to a carbon form a tetrahedral arrangement; two single bonds and one double bond to a carbon give a planar array with bond angles near \(120^\text{o}\), while one single bond and one triple bond (or two double bonds) to a carbon are arranged linearly: Finally, if you have not studied the material already, you may wish to return to the last part of Chapter 3 and become acquainted with the nomenclature of cycloalkanes, alkenes, cycloalkenes, and alkynes (Sections 3-2 to 3-4). and (1977) | 1,440 | 2,065 |
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The difference between a physical reaction and a chemical reaction is . In a chemical reaction, there is a change in the composition of the substances in question; in a physical change there is a difference in the appearance, smell, or simple display of a sample of matter without a change in composition. Although we call them physical "reactions," no reaction is actually occurring. In order for a reaction to take place, there must be a change in the elemental composition of the substance in question. Thus, we shall simply refer to physical "reactions" as physical changes from now on. Physical changes are limited to changes that result in a difference in display without changing the composition. Some common changes (but not limited to) are: Physical properties include many other aspects of a substance. The following are (but not limited to) physical properties. Any change in these physical properties is referred to as a physical change. For further information, please refer to Chemical changes, on the other hand, are quite different. A chemical change occurs when the substance's composition is changed. When bonds are broken and new ones are formed a chemical change occurs. The following are indicators of chemical changes: Note: When two or more reactants are mixed and a change in temperature, color, etc. is noticed, a chemical reaction is probably occurring. These are not definite indicators; a chemical reaction may be occurring. A change in color is not always a chemical change. If one were to change the color of a substance in a non-chemical reaction scenario, such as painting a car, the change is physical and not chemical. This is because the composition of the car has not changed. Proceed with caution. The of a substance can differ with a physical change. For example, if a piece of wood was sanded, waxed, and polished, it would have a very different texture than it initially had as a rough piece of wood. As you can see, the texture of the finished wood is much smoother than the initial grainy wood. The changing of of a substance is not necessarily an indicator of a chemical change. For example, changing the color of a metal does not change its physical properties. However, in a chemical reaction, a color change is usually an indicator that a reaction is occurring. Painting the metal car does changing the composition of the metallic substance. Although we cannot see change, unless if a change of state is occurring, it is a physical change. One cannot see the pan physically changing shape, color, texture, or any of the other physical properties. However, if one were to touch the pan, it would be incredibly hot and could cause a burn. Sitting idle in a cupboard, this pan would be cold. One cannot assess this change only through visual exposure; the use of a thermometer or other instrument is necessary. The of an object can be changed and the object will still remain true to its chemical composition. For example, if one were to fold money, as shown by the figure below, the money is still chemically the same. The is likewise a physical change. In this scenario, one can observe a number of physical properties changing, such as viscosity and shape. As ice turns into water, it does not retain a solid shape and now becomes a viscous fluid. The physical "reaction" for the change of ice into liquid water is: \[H_2O_{(s)} \rightarrow H_2O_{(l)}\] The following are the changes of state: The of an element is defined as the way it reacts to light. Luster is a quality of a metal. Almost all of the metals, transition metals, and metalloids are lustrous. The non-metals and gases are not lustrous. For example, oxygen and bromine are not lustrous. Shown below is are lustrous paper clips. is also a quality of metals. Metals are said to be malleable. This means that the metals can deform under an amount of stress. For example, if you can hit a metal with a mallet and it deforms, it is malleable. Also, a paperclip can be shaped with bare hands. The image shows the malleability of a certain metal as stress is applied to it. In materials science, this property is called . For example, raw copper can be obtained and it can be purified and wrapped into a cord. Once again, this property is characteristic of mainly metals, nonmetals do not possess this quality. The of an object is its mass divided by its volume (d=m/v). A substance will have a higher density if it has more mass in a fixed amount of volume. For example, take a ball of metal, roughly the size of a baseball, compressed from raw metal. Compare this to a baseball made of paper. The baseball made of metal has a much greater weight to it in the same amount of volume. Therefore the baseball made out of metal has a much higher density. The density of an object will also determine whether it will sink or float in a particular chemical. Water for example has a density of 1g/cm . Any substance with a density lower than that will float, while any substance with a density above that will sink. is defined to be the resistance to deformation of a particular chemical substance when a force is applied to it. In the example below, one can see two cubes falling into two different test tubes. The upper substance shows a violent reaction to the dropping of the cube. The lower substance simply engulfs it slowly without much reaction. The upper substance has a lower viscosity relative to the lower substance, which has very high viscosity. One may even think of viscosity in terms of thickness. The substance with more thickness has higher viscosity than a substance that is deemed "thin." Water has a lower viscosity than honey or magma, which have relatively high viscosities. The follow are all indicators of chemical reactions. For further information on chemical reactions, please refer to A is characteristic of a chemical change. During an experiment, one could dip a thermometer into a beaker or Erlenmeyer Flask to verify a temperature change. If temperature increases, as it does in most reactions, a chemical change is likely to be occurring. This is different from the physical temperature change. During a physical temperature change, one substance, such as water is being heated. However, in this case, one compound is mixed in with another, and these reactants produce a product. When the reactants are mixed, the temperature change caused by the reaction is an indicator of a chemical change. As an example of a exothermic reaction, if \(Fe_2O_3\) is mixed with Al and ignighted (often with burning Mg), then the thermite reaciton is initiated \[Fe_2O_3 + 2Al \rightarrow 2Fe + Al_2O_3 + \text{Heat}\] This reaction generates heat as a product and is (very) exothermic. However, physical changes can be exothermic or endothermic. The melting of an ice cube, which is endothermic, is a change in a physical property and not composition. Thus, it is a physical change. A is also another characteristic of a chemical reaction taking place. For example, if one were to observe the rusting of metal over time, one would realized that the metal has changed color and turned orange. This change in color is evidence of a chemical reaction. However, one must be careful; sometimes a change in color is simply the mixing of two colors, but no real change in the composition of the substances in question. The reaction above is that of the rusting of iron. \[4Fe + 3O_2 + 6H_2O \rightarrow 4Fe(OH)_3\] When two or more compounds or elements are mixed and a is present, a chemical reaction has taken place. For example, when an egg begins to smell, (a rotten egg) a chemical reaction has taken place. This is the result of a chemical decomposition. The may be one of the most common signs of a chemical reaction taking place. A precipitate is defined to be a solid that forms inside of a solution or another solid. Precipitates should not be confused with suspensions, which are solutions that are homogeneous fluids with particles floating about in them. For instance, when a soluble carbonate reacts with Barium, a Barium Carbonate precipitate can be observed. Reaction: \[Ba^{2+}_{(aq)} + CO^{2-}_{3\;(aq)} \rightarrow BaCO){3\;(s)}\] For further information, please refer to The , or rather , is another indicator of a chemical reaction taking place. When bubbles form, a temperature change could also be taking place. Temperature change and formation of bubbles often occur together. For example, in the following image, one can see a gas spewing. This is the formation of a gas. However, most reactions are much more subtle. For instance, if the following reaction occurs, one may notice Carbon Dioxide bubbles forming. If there is enough Hydrochloric Acid, bubbles are visible. If there isn't, one can't readily notice the change: \[Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2\] All images are courtesy of , which provides royalty free images that are free to be copied without restrictions. The viscosity image is also free to be duplicated as per permission of author on .com. 1. Which of the following is a chemical reaction? 2. Which of the following is a physical reaction? 3. Which of the following is a chemical reaction? 4. Which of the following is a physical reaction? 5. Write C for Chemical Reaction or P for Physical Reaction. 1. D 2. A 3. B 4. C 5. a) C b) P c) P d) C e) Neither. This is one of the gray areas of chemical change and physical change. Although the salt has dissociated into Sodium and Chloride ions, it is still salt in water. Salt, initially is actually just a conglomerate of sodium and chloride ions and by dissociating them, just the arrangement of the ions has changed. 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Strain in small-ring cycloalkanes has a profound influence on their heats of combustion (Table 12-3). We reasonably expect that other chemical properties also will be affected. Indeed, like alkenes, cyclopropane and cyclobutane undergo \(\ce{C-C}\) bond cleavage reactions that are not observed for cyclopentane and cyclohexane, or for saturated, open-chain hydrocarbons. A summary of these reactions is presented in Table 12-4. It will be seen that the reactions result in cleavage of a \(\ce{C-C}\) bond to give an open-chain compound with normal bond angles. Relief of angle strain is an important contributing factor to the driving force for these reactions. Therefore, ethene is highly reactive, whereas cyclopropane and cyclobutane are somewhat less reactive. The \(\ce{C-C}\) bonds of the larger, relatively strain-free cycloalkanes are inert, so these substances resemble the alkanes in their chemical behavior. Substitution reactions, such as chlorination of cyclopentane and higher cycloalkanes, generally are less complex than those of the corresponding alkanes because there are fewer possible isomeric substitution products. Thus cyclohexane gives only one monochloro product, whereas hexane gives three isomeric monochloro hexanes. Conformation has a major influence on the chemical reactivity of cycloalkanes. To understand its effect in any one reaction, we first need to know what the conformation is of the transition state, and this requires a knowledge of the reaction mechanism. Next, we have to decide what amount of energy is required for the reactants to achieve transition-state conformations. For example, consider the \(E_2\) elimination discussed in Section 8-8D. The preferred transition state requires the leaving groups to be antarafacial and coplanar: For cyclohexane derivatives to react in this way, the transition-state conformation must have both leaving groups : For this reason, compounds such as -4- -butylchlorocyclohexane eliminate \(\ce{HCl}\) much more readily by the \(E_2\) mechanism than do the corresponding trans isomers. To have the antarafacial coplanar mechanism of cycloalkanes operate with the trans isomer, the transition state would have to have the -butyl group in the highly unfavorable axial position. and (1977) | 2,295 | 2,068 |
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More than one stable substance can correspond to a given molecular formula. Examples are butane and 2-methylpropane (isobutane), each of which has the molecular formula \(C_4H_{10}\). Similarly, methoxymethane (dimethyl ether) and ethanol have the same formula, \(C_2H_6O\): Compounds having the same number and kinds of atoms are called .\(^2\) Whereas only one stable substance is known corresponding to the formula \(CH_4\), thirty-five stable isomers have been prepared of the formula \(C_9H_{20}\). From this one may begin to sense the rich variety of organic chemistry, which leads to many problems - in telling one compound from another, in determining structures, and also in finding suitable names for compounds. In the rest of this chapter we will describe one type of isomer - the position isomer - and in later chapters we will discuss another type of isomer - the stereoisomer - and the experimental approaches that are used to establish the purity, identity, structure, and stereochemistry of organic compounds. Compounds having the same number and kind of atoms but having different bonding arrangements between the atoms are called . Butane and 2-methylpropane are examples of position isomers. The atoms are connected differently in the two structures because the carbon chain in butane is a straight or continuous chain, whereas in 2-methylpropane it is branched: Therefore these two molecules are structurally different and, accordingly, do not have the same chemical and physical properties. They cannot be converted one into the other without breaking and remaking \(C-C\) and \(C-H\) bonds. Methoxymethane and ethanol are also position isomers because the oxygen clearly is connected differently in the two molecules: The term means the same as . The designation also is used, but this term is taken by some to include both position isomers and stereoisomers; that is, "structure" can mean both the way in which atoms are connected and their different arrangements in space. The number of position isomers possible for a given formula rapidly increases with the increasing number of carbon atoms, as can be seen from the number of theoretically possible structures of formula \(C_nH_{2n+2}\) up to \(n = 10\) given in Table 2-3. In 1946, it was reported that all of the 75 compounds with values of \(n = 1\) to \(n = 9\) had been prepared in the laboratory. Before we can begin to discuss the chemistry of these compounds it is necessary to know how to name them; without convenient and systematic rules for nomenclature that are adopted universally, catastrophic confusion would result. We shall tackle this problem in the next chapter. \(^2\)The prefix is from the Greek word meaning the same or alike. and (1977) | 2,765 | 2,069 |
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What if the association is influenced by an additional potential for A-B interactions? Following our earlier discussion for diffusion in a potential, the potential U results in an additional contribution to the flux: \[ J_U = -\dfrac{D_AC_A}{k_BT} \dfrac{\partial U_{AB}}{\partial r} \nonumber \] So the total flux of A incident on B from normal diffusion J and the interaction potential J is \[ J_{A \rightarrow B} = -D_A \left[ \dfrac{\partial C_A}{\partial r} + \dfrac{C_A}{k_BT} \dfrac{\partial U_{AB}}{\partial r} \right] \nonumber \] To solve this we make use of a mathematical manipulation commonly used in solving the Smoluchowski equation in which we rewrite the quantity in brackets as \[ J_{A \rightarrow B} = -D_A \left[ e^{U_{AB}/k_BT} \dfrac{d\left[ C_Ae^{U_{AB}/k_BT}\right] }{dr} \right] \] Substitute this into the expression for the rate of collisions of A with B: \[ \begin{aligned} \dfrac{dn_{A \rightarrow B}}{dt} &= A_BJ_{A\rightarrow B} \\ &=4\pi R^2_BJ_{A\rightarrow B} \end{aligned} \] Separate variables and integrate from the surface of the sphere to \(r = \infty \) using the boundary conditions: \(C(R_B)=0, C(\infty )=C_A \): \[ \left( \dfrac{dn_{A \rightarrow B}}{dt} \right) \underbrace{\int^{\infty}_{R_B} e^{U_{AB}/k_BT} \dfrac{dr}{r^2} }_{(R^*)^{-1}} = 4\pi D_A \underbrace{\int^{C_A}_0 d\left[ C_Ae^{U_{AB}/k_BT} \right] }_{C_A} \] Note that integral on the right is just the bulk concentration of A. The integral on the right has units of inverse distance, and we can write this in terms of the variable R*: \[(R^*)^{-1} = \int^{\infty}_{R_B} e^{U_{AB}/k_BT}r^{-2}dr \nonumber \] Note that when no potential is present, then U → 0, and R* = R . Therefore R* is an effective encounter distance which accounts for the added influence of the interaction potential, and we can express it in terms of f, a correction factor the normal encounter radius: R* = f R . For attractive interactions R* > R and f >1, and vice versa. Returning to eq. (16.2.2), we see that the rate of collisions of A with B is \[\dfrac{dn_{A\rightarrow B}}{dt} = 4\pi D_AR_B^*C_A \nonumber \] As before, if we account for the total number of collisions for two diffusing molecules A and B: \[ \begin{aligned} \dfrac{dn_{TOT}}{dt} &= J_{A\rightarrow B}A_{AB}C_B \\ &=k_aC_AC_B \\ k_a&=4\pi (D_A+D_B)R_{AB}^* \\ R_{AB}^* &= R_A^* +R_B^* \end{aligned} \] Let’s calculate the form of the where the interaction is the Coulomb potential. \[U_{AB}(r) = \dfrac{z_Az_Be^2}{4\pi \epsilon r} = k_BT\dfrac{\ell_B}{r} \nonumber \] where the Bjerrum length is \(\ell_B = z_Az_Be^2/(4\pi \epsilon k_BT) \). Then \[ \begin{aligned} (R_{AB}^*)^{-1} &= \int^{\infty}_{R_{AB}} e^{U_{AB}/k_BT} \dfrac{dr}{r^2} \\ &= \ell_B^{-1} \left[ \mathrm{exp}(\ell_B/R_{AB}-1 \right] \end{aligned} \] and \[ R_{AB}^* = \ell_B (e^{\ell_B/R_{AB}-1)^{-1}} \nonumber \] For \( \ell_B \gg R_{AB}, R_{AB}^* \rightarrow R_{AB}\). For \(\ell_B = R_{AB}, R_{AB}^* = 0.58R_{AB} \) if the charges have the same sign (repel), or \(R_{AB}^* = 1.58R_{AB}\) if they are opposite charges (attract). ______________________________________ \[ 4\pi r^2 J_{A\rightarrow B}=\dfrac{4\pi D_A \left[ C_A(\infty )e^{U_{AB}(\infty )/k_BT} -C_A(R_0)e^{U_{AB}(R_0)/k_BT} \right]}{\int^{\infty }_{R_0} r^{-2}e^{U_{AB}(r)/k_BT} dr } \nonumber \] \(C_A(\infty ) \) is the bulk concentration of A. For the perfectly absorbing sphere, the concentration of A at the boundary with B, C (R )=0. For a homogeneous solution we also assume that the interaction potential at long range \( U_{AB}(\infty ) =0\). | 3,600 | 2,070 |
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Constant volume and constant pressure heat capacities are very important in the calculation of many changes. The ratio \(C_p/C_V = \gamma\) appears in many expressions as well (such as the relationship between pressure and volume along an adiabatic expansion.) It would be useful to derive an expression for the difference \(C_p – C_V\) as well. As it turns out, this difference is expressible in terms of measureable physical properties of a substance, such as \(\alpha\), \(\kappa_T\), \(p\), \(V\), and \(T\). In order to derive an expression, let’s start from the definitions. \[C_p= \left( \dfrac{\partial H}{\partial T} \right)_p \nonumber \] and \[ C_V= \left( \dfrac{\partial U}{\partial T} \right)_V \nonumber \] The difference is thus \[ C_p-C_v = \left( \dfrac{\partial H}{\partial T} \right)_p - \left( \dfrac{\partial U}{\partial T} \right)_V \nonumber \] In order to evaluate this difference, consider the definition of enthalpy: \[ H = U + pV \nonumber \] Differentiating this yields \[ dH = dU + pdV + Vdp \nonumber \] Dividing this expression by \(dT\) and constraining to constant \(p\) gives \[\left.\dfrac{dH}{dT} \right\rvert_{p}= \left.\dfrac{dU}{dT} \right\rvert_{p} +p \left.\dfrac{dV}{dT} \right\rvert_{p} + V \left.\dfrac{dp}{dT} \right\rvert_{p} \nonumber \] The last term is kind enough to vanish (since \(dp = 0\) at constant pressure). After converting the remaining terms to partial derivatives: \[ \left( \dfrac{\partial H}{\partial T} \right)_p = \left( \dfrac{\partial U}{\partial T} \right)_p + p \left( \dfrac{\partial V}{\partial T} \right)_p \label{total4} \] This expression is starting to show some of the players. For example, \[ \left( \dfrac{\partial H}{\partial T} \right)_p = C_p \nonumber \] and \[ \left( \dfrac{\partial V}{\partial T} \right)_p = V \alpha \nonumber \] So Equation \ref{total4} becomes \[ C_p = \left( \dfrac{\partial U}{\partial T} \right)_p + pV\alpha \label{eq5} \] In order to evaluate the partial derivative above, first consider \(U(V, T)\). Then the total differential \(du\) can be expressed \[ du = \left( \dfrac{\partial U}{\partial V} \right)_T dV - \left( \dfrac{\partial U}{\partial T} \right)_V dT \nonumber \] Dividing by \(dT\) and constraining to constant \(p\) will generate the partial derivative we wish to evaluate: \[\left.\dfrac{dU}{dT} \right\rvert_{p}= \left( \dfrac{\partial U}{\partial V} \right)_T \left.\dfrac{dV}{dT} \right\rvert_{p} + \left( \dfrac{\partial U}{\partial T} \right)_V \left.\dfrac{dT}{dT} \right\rvert_{p} \nonumber \] \[\left( \dfrac{\partial U}{\partial T} \right)_p= \left( \dfrac{\partial U}{\partial V} \right)_T \left( \dfrac{\partial V}{\partial T} \right)_p+ \left( \dfrac{\partial U}{\partial T} \right)_V \label{eq30} \] \[\left( \dfrac{\partial U}{\partial T} \right)_V = C_V \nonumber \] and \[\left( \dfrac{\partial V}{\partial T} \right)_p = V\alpha \nonumber \] \[\left( \dfrac{\partial U}{\partial T} \right)_p= \left( \dfrac{\partial U}{\partial V} \right)_T V\alpha + C_V \nonumber \] \[\left( \dfrac{\partial U}{\partial V} \right)_T \nonumber \] \[ d = TdS - pdV \nonumber \] \[\left.\dfrac{dU}{dV} \right\rvert_{T}= T \left.\dfrac{dS}{dV} \right\rvert_{T} - p \left.\dfrac{dV}{dV} \right\rvert_{T} \nonumber \] \[\left( \dfrac{\partial U}{\partial V} \right)_T= T \left( \dfrac{\partial S}{\partial V} \right)_T - p \label{eq40} \] \[\left( \dfrac{\partial S}{\partial V} \right)_T = \left( \dfrac{\partial p}{\partial T} \right)_V \nonumber \] \[\left( \dfrac{\partial U}{\partial V} \right)_T= T \left( \dfrac{\partial p}{\partial T} \right)_V - p \nonumber \] \[\left( \dfrac{\partial p}{\partial T} \right)_V = \dfrac{\alpha}{\kappa_T} \nonumber \] then \[\left( \dfrac{\partial U}{\partial V} \right)_T= T \dfrac{\alpha}{\kappa_T} - p \nonumber \] \[ \begin{align*} \left( \dfrac{\partial U}{\partial T} \right)_p &= \left[ T \dfrac{\alpha}{\kappa_T} - p \right] V\alpha + C_V \\[4pt] &= \dfrac{TV \alpha^2}{\kappa_T} - pV\alpha + C_V \end{align*} \] This can now be substituted into the Equation \ref{eq5} \[ C_p = \left[ \dfrac{TV \alpha^2}{\kappa_T} - \cancel{ pV\alpha} + C_V \right] + \cancel{pV\alpha} \nonumber \] he \(pV\alpha\) term ting \(C_V\) fro \[ C_p - C_V = \dfrac{TV \alpha^2}{\kappa_T} \label{final} \] \[ \alpha = \dfrac{1}{T} \nonumber \] \[ \kappa_T=\dfrac{1}{p} \nonumber \] \[\begin{align*} C_p - C_V &= \dfrac{TV \left(\dfrac{1}{T}\right)^2}{\left(\dfrac{1}{p}\right)} \\[4pt] &= \dfrac{pV}{T} \\[4pt] &= R \end{align*} \] as, \(C_p – C_V = R\). Th Derive the expression for the difference between \(C_p\) and \(C_V\) by beginning with the definition of \(H\), differentiating, dividing by \(dV\) (to generate the partial derivative definition of \(C_V\)). In this approach, you will need to find expressions for \[\left( \dfrac{\partial H}{\partial T} \right)_V \nonumber \] and \[\left( \dfrac{\partial U}{\partial p} \right)_T \nonumber \] and also utilize the Maxwell-Relation on \(G\). Begin with the definition of enthalpy. \[ H = U +pV \nonumber \] Differentiate the expression. \[dH = dU + pdV + Vdp \nonumber \] Now, divide by \(dV\) and constrain to constant \(T\) (as described in the instructions) to generate the partial derivative definition of \(C_V\) \[\left.\dfrac{dH}{dT} \right\rvert_{V}= \left.\dfrac{dU}{dT} \right\rvert_{V} + p \left.\dfrac{dV}{dT} \right\rvert_{V} + V \left.\dfrac{dp}{dT} \right\rvert_{V} \nonumber \] \[\left(\dfrac{dH}{dT} \right)_{V}= \left(\dfrac{dU}{dT} \right)_{V} + V \left(\dfrac{dp}{dT} \right)_{V} \label{eq20E} \] Now what is needed is an expression for \[\left(\dfrac{dH}{dT} \right)_{V}. \nonumber \] This can be derived from the total differential for \(H(p,T)\) by dividing by \(dT\) and constraining to constant \(V\). \[dH = \left(\dfrac{dH}{dp} \right)_{T} dp + \left(\dfrac{dH}{dT} \right)_{p} dT \nonumber \] \[\left(\dfrac{dH}{dT} \right)_{V}=\left(\dfrac{dH}{dp} \right)_{T} \left(\dfrac{dp}{dT} \right)_{V} + \left(\dfrac{dH}{dT} \right)_{p} \label{eq30E} \] This again is an example of . To continue, we need an expression for \[\left(\dfrac{dH}{dp} \right)_{T}. \nonumber \] This can be quickly generated by considering the total differential of \(H(p,S)\), its natural variables: \[dH = TdS + Vdp \nonumber \] Dividing by \(dp\) and constraining to constant \(T\) yields \[\left.\dfrac{dH}{dp} \right\rvert_{T}= T \left.\dfrac{dS}{dp} \right\rvert_{T} + V \left.\dfrac{dp}{dp} \right\rvert_{T} \nonumber \] \[\left(\dfrac{dH}{dp} \right)_{T}= T \left(\dfrac{dS}{dp} \right)_{T} + V \label{eq40E} \] Using the Maxwell Relation on \(G\), we can substitute \[- \left(\dfrac{dV}{dT} \right)_{p}= \left(\dfrac{dS}{dp} \right)_{T} \nonumber \] So Equation \ref{eq40E} becomes \[\left(\dfrac{dH}{dp} \right)_{T}= - T \left(\dfrac{dV}{dT} \right)_{p} + V \nonumber \] Now, substitute this back into the expression for (Equation \ref{eq30E}): \[\left(\dfrac{dH}{dT} \right)_{V}= \left[ - T \left(\dfrac{dV}{dT} \right)_{p} + V \right] \left(\dfrac{dp}{dT} \right)_{V} + \left(\dfrac{dH}{dT} \right)_{p} \nonumber \] \[\left(\dfrac{dH}{dT} \right)_{V}= - T \left(\dfrac{dV}{dT} \right)_{p} \left(\dfrac{dp}{dT} \right)_{V} + V \left(\dfrac{dp}{dT} \right)_{V} + \left(\dfrac{dH}{dT} \right)_{p} \nonumber \] This can now substituted for the right-hand side of the initial expression for \(\left(\dfrac{dH}{dT} \right)_{V}\) back into Equation \ref{eq20E}: \[- T \left(\dfrac{dV}{dT} \right)_{p} \left(\dfrac{dp}{dT} \right)_{V} + \cancel{V \left(\dfrac{dp}{dT} \right)_{V}} + \left(\dfrac{dH}{dT} \right)_{p} = \left(\dfrac{dU}{dT} \right)_{V} + \cancel{V \left(\dfrac{dp}{dT} \right)_{V}} \label{eq50E} \] Several terms cancel one another. Equation \ref{eq50E} can then be rearranged to yield \[ \left(\dfrac{dH}{dT} \right)_{p} - \left(\dfrac{dU}{dT} \right)_{V} = T \left(\dfrac{dV}{dT} \right)_{p} \left(\dfrac{dp}{dT} \right)_{V} \nonumber \] or \[ C_p - C_V = \dfrac{TV \alpha^2}{\kappa_T} \nonumber \] which might look familiar (Equation \ref{final})! | 8,012 | 2,071 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/23%3A_Organonitrogen_Compounds_I_-_Amines/23.02%3A_Naturally_Occurring_Amines |
A large and widespread class of naturally occurring amines is known as . These are basic organic nitrogen compounds, mostly of plant origin. The structures of the plant alkaloids are extraordinarily complex, yet they are related to the simple amines in being weak nitrogen bases. In fact, the first investigator to isolate an alkaloid in pure form was F. W. A. Sertürner who, in 1816, described morphine (Figure 23-1) as basic, salt-forming, and ammonia-like. He used the term "organic alkali" from which is derived the name . The structures of some of the better known plant alkaloids are shown in Figure 23-1. You will recognize some of them by name even if you have never seen their structures before. Many of the alkaloids are polycyclic structures and have other functional groups in addition to basic nitrogen. You will see that the nitrogens of alkaloids frequently are tertiary amine functions. All of the alkaloids shown in Figure 23-1 are substances with very pronounced physiological action. Indeed, alkaloids in general have been used and abused for centuries as medicinals, drugs, and poisons. However, only in this century have their structures become known, and we are still a long way from understanding the chemistry that leads to their pronounced physiological effects. It is not even understood what function, if any, these compounds have in the host plant. As you can see from Figure 23-1, alkaloids include compounds that may be classified as antimicrobial (quinine), as analgesics (morphine, codeine), as hallucinogens (mescaline, LSD), as stimulants (cocaine, atropine, caffeine), as topical anaesthetics (cocaine). With the possible exception of caffeine, all may be described as potentially poisonous enough to warrant great care in their use. Although some of these compounds are used as natural medicinals, an entire industry has developed in an effort to produce synthetic analogs with similar, but safer, medicinal properties. Some of the better known of these synthetic drugs are shown in Figure 23-2. They include a group of narcotic substances known as barbiturates, which are used widely as sedatives, anticonvulsants, and sleep-inducing drugs. Several representative nitrogen-containing tranquilizing drugs, synthetic stimulants, and antibiotics also are shown. Basic nitrogen compounds similar to the plant alkaloids also occur in animals, although the description seldom is used. Certain amines and ammonium compounds play key roles in the function of the central nervous system (Figure 23-3) and the balance of amines in the brain is critical for normal brain functioning. Also, many essential vitamins and hormones are basic nitrogen compounds. Nitrogen bases also are vital constituents of nucleic acid polymers (DNA and RNA) and of proteins ( ). and (1977) | 2,819 | 2,072 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Ideal_Systems/Ideal_Gas_Processes |
In this section we will talk about the relationship between ideal gases in relations to thermodynamics. We will see how by using thermodynamics we will get a better understanding of ideal gases. In the realm of Chemistry we often see many relations between the former and its relations to Physics. By utilizing both Chemistry and Physics we can get a better understanding for the both mentioned. We will use what we know about Ideal Gases and Thermodynamics to try to understand specific processes that occur in a system. Before we discuss any further, let’s do a very quick recap on the important aspects of thermodynamics that are important to know for ideal gas processes. Some of this will be a quick review and some will be relatively new unless you have seen it in your Physics class. So first off let’s state the : \(\Delta{U} = Q + W\) The whole point of stating this equation is to remind us that energy within any given system is conserved. What that means is that no energy is EVER created or destroyed, but it is simply converted from one form to another, such as heat to work and vice versa. In case you may be confused with some of these symbols, here is a short explanation in a table. Below are two equations that describe the relationship between the internal energy of the system of a monatomic gas and a diatomic gas. In a monatomic ( : one) gas, since it only has one molecule, the ways for it have energy will be less than a diatomic gas ( : two) since a diatomic gas has more ways to have energy (Hence, diatomic gas has a 5/2 factor while a monatomic gas has a 3/2). Looking at these two equations we have also conclude that the internal energy (ΔU) only has an effect on the kinetic energy of the gas molecules (movement). Nowhere in these two equations do we see that the potential energy being affected. A Monatomic Ideal Gas Equation: \(\Delta{U} = \frac{3}{2}nR\Delta{T}\) In a monatomic gas, it has a total of three translational kinetic energy modes (hence, the /2). A Diatomic Ideal Gas Equation: \(\Delta{U} = \frac{5}{2}nR\Delta{T}\) In a diatomic gas, it has a total of three translational kinetic energy modes and two rotational energy modes (hence, the /2). In relations to the first law of thermodynamics, we can see that by adding heat (Q) or work (W) the internal energy of the gaseous system can be increased. Also, that during compression of the system, the volume of the gas will decrease and response its temperature will increase and thus the internal energy of the system will also increase since temperature is related to energy. And this is true except in an isothermal system (which we’ll talk more about later). That is why when a gas is compressed the work is positive and when it is being compressed, it is negative. It may also be good to know that the area under the curve is work. If you have taken Calculus, you may remember the integral as it is used to find the area under the curve (or graph) as shown below. \[W = -\displaystyle \int P dV\] In this case, you can literally take the area of the triangle or work with integrals. Work = Area = (1/2)base x height
or
Work = ∫F(x) dx When certain state functions (P, V, T) are held constant, the specific heat of the gas is affected. Below is the universal formula for a gas molecule when its pressure is held constant: \( c_p = c_v + R\) When this formula is rearranged we get the heat capcity of the gas when its volume is held constant: \( c_v = R - c_p\) There are four types of thermodynamics processes. What this basically means is that in a system, one or more variable is held constant. To keep things simple, below are examples as to how keeping a certain variable in a system constant can lead to. •This is a process where the pressure of the system is kept constant.
•P = 0
•An example of this would be when water is boiling in a pot over a burner. In this case, heat is being exchanged between the burner and pot but the pressure stays constant. To derive this process we start off by using what we know, and that is the first law of thermodynamics: \(\Delta{U} = Q + W\) Rearranging this equation a bit we get: \( Q = \Delta{U} + W\) Next, since pressure is equal to W ΔV, it can be denoted as: \(Q = \Delta{U} + p\Delta{V}\) Now, the ideal gas law can be applied (PV=nRΔT) and since pressure is constant: \(Q = ΔU + nR\Delta{T}\) For the next step, we will assume that this number of moles of gas stays constant throughout this process: \(Q = n\c_V\ \Delta{T} + nR\Delta{T}\) Simplifying the equation some more by taking out the nΔT from both equations we get: \(Q = n\(c_V + R)\ \Delta{T}\) Knowing that c = c + R we can substitute for cp: \(Q = n\c_P\ \Delta{T} \) Now we got the equation for an isobaric process! •This is a process where the volume of the system is kept constant.
•ΔV = 0
•An example of this would be when you have Helium gas sealed up in a container and there is an object (like a piston) pushing down the container (exerting pressure). But, gas molecules is neither entering nor exiting out of the system.
Let’s find the equation for this process, as before let’s start off with the first law of thermodynamics: \(\Delta{U} = Q + W\) Rearranging this equation a bit we get: \( Q = \Delta{U} + W\) In this case, since volume is constant, ΔV = 0: \(Q = \Delta{U}\) Since the internal energy of the system equals to the amount of heat transferred we can replace ΔU with the ideal gas equation for heat: \(Q = nC_V\Delta{T}\) Above is the ideal gas equation for an isochoric process!
•This is a process where the temperature of the system is kept constant.
ΔU = 0, ΔT = 0
•When volume increases, the pressure will decrease, and vice versa.
ΔT = 0 then: and and (inverse relationship)
•As an example, gas molecules are sealed up in a container but an object on top of the container (such as a piston) pushes down on the container in a very slow fashion that there is not enough to change its temperature. To derive the equation for an isothermal process we must first write out the first law of thermodynamics: \(\Delta{U} = Q + W\) Rearranging this equation a bit we get: \( Q = \Delta{U} + W\) Since ΔT = 0. Therefore we are only left with work: \( Q = W \) As such we get: \( W = -p\Delta{V} \) Making this equation into an ideal gas equation we get: \( W = \frac{nRt}{V} \) In order to get to the next step we need to use some calculus: \(Q = nRT ln \frac{V_f}{V_i} \) And there you go! The equation for an isothermal process. •This is a process where no heat is being added or removed from the system.
•Or can be simply stated as: no heat transfer (or heat flow) happening in a system.
•In freshman chemistry, only the basic idea of this process is needed and that is when there is no heat transfer, Q = 0. The volume of a gas in a container expanded from 1L to 3L upon releasing the piston upward. From following graph, find the amount work associated with the expansion of the gas. \(f(x) = x + 3\) from [1, 3] Calvin is observing an unknown monatomic gas molecule (sealed inside a container) in his Freshman Chemistry Lab. He has been told by his lab instructor that there are four moles of this unidentified gas in the container. The laboratory room’s temperature was initially set at room temperature when he started the lab, but by the time he was almost finished with the lab the temperature has gone up 10°C. What is the total internal energy of this unknown gaseous substance by the time Calvin’s lab session ended? In an isochoric system, three moles of hydrogen gas is trapped inside an enclosed container with a piston on top of it. The total amount of internal energy of the gaseous system is 65 Joules, and the temperature of the system decreased from 25°C to 19°C. What is the specific heat of the gas molecules? While looking over some of her lab data, a chemistry student notices she forgot to record the numbers of moles of the gas molecule she looked at. The experiment she did that day had kept the pressure constant and the temperature went down two degrees in the process of the experiment. Assuming all variables are also ideal, how many moles was she dealing with? A chemistry student is looking at 5.00 grams of the monatomic Helium gas that is put into a container that expands from 10L to 13L. The container is confined at a constant temperature of 30°C in an enclosed system. (a) What is the total energy of the system? (b) Is the pressure of the system increasing, decreasing, or not affected? As stated above, we know that the amount of work is the same as saying the area under the curve. In this case we can look for shapes we can easily find the area of. Here, we have a shape that is similar to a rectangle. From the problem, we know that the unknown gaseous substance is monatomic, so we would have to use the equation for a monatomic gas: \(\Delta{U} = \frac{3}{2}nR\Delta{T}\) Also from the problem, we know that the substance is at room temperature with is 25°C, but that was just extra information given and we didn’t really need to know that to solve the problem. Since we know that the temperature change from start to finish was +10°. \(\Delta{T} = +10K\) The problem also given us the number of moles that was in the container: \(n = 4 mol\) For the R-value, we can choose any R constant, but to make this problem a little easier we will choose a R-constant that will cancel out any other units except for joules. \(R = 8.3145 \frac{J}{mol K}\) Plugging in all these values into the equation we have: \(\Delta{U} = \frac{3}{2}(4 mol)(8.3145 \frac{J}{mol K})(10K)\) \(= 498 J\) Since the system is an isochoric system, there will be zero change in volume, therefore: \(\Delta{V} = 0\) And the equation for an isochoric system will be: \(Q = nC_V\Delta{T}\) From the equation we know the following: \( n=3mol\) \(Q = \Delta{T}=(25-19)K=6K\) \( Q=65J\) Before plugging the values in, let’s rearrange the isochoric equation and set it to what we are trying to find: \( C_V = \frac{Q}{n\Delta{T}}\) Now, we can plug the values in: \( C_V = \frac{65J}{(3mol)(6K)}\) \( = 3.6\frac{J}{mol k}\) From the problem we are given the following: \( \Delta{T} = +2K\) \( \Delta{U} = Q = 110 J\) Since are given that the pressure of the gas was kept constant: \( C_p = 14\frac{J}{K mol}\) We have enough evidence to conclude that this is an isobaric process: \(Q = n\c_P\ \Delta{T} \) Rearranging the equation to fit what we are trying to find we get: \(n = \frac{Q}{\c_P\ \Delta{T}} \) Now we can plug in our values: \(n = \frac{110 J}{14\frac{J}{K mol}2K} \) \(= 4 moles of gas\) A) First, let’s list what is given in the problem: \( V_i = 10 L\) \( V_f = 13 L\) Converting temperature from Celsius to Kelvin we have: \(T = (30+273.15)K = 303.15K\) And since temperature is constant, this is an isothermal process: \(Q = nRT ln \frac{V_f}{V_i} \) In order to use this equation we will need to convert Helium gas to moles: \(n = \frac{grams}{\frac{grams}{mole}}\) \( = \frac{5.00 g He}{\frac{4.00 g He}{mol He}}\) \( = 1.25 mol He\) We will choose a universal R-constant that will easily cancel out all the units that will help us get the answer without any other conversions: \( = 8.31451\frac{J}{K mol}\) Plugging in all these values into the isothermal equation we get: \(Q = (1.25 mol)(8.31451 \frac{J}{K mol})(303.15 K) ln \frac{13 L}{10 L} \) \( = 827 J\) B) It is a rule that in an isothermal process that volume and pressure have an inverse relationship. Meaning in one goes up, the other must go down. In this case, volume went up so the . | 11,561 | 2,073 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/13%3A_Solutions/13.01%3A_Factors_Affecting_Solution_Formation |
In all solutions, whether gaseous, liquid, or solid, the substance present in the greatest amount is the solvent, and the substance or substances present in lesser amounts are the solute(s). The solute does not have to be in the same physical state as the solvent, but the physical state of the solvent usually determines the state of the solution. As long as the solute and solvent combine to give a homogeneous solution, the solute is said to be soluble in the solvent. Table 13.1 lists some common examples of gaseous, liquid, and solid solutions and identifies the physical states of the solute and solvent in each. Types of Solutions The formation of a solution from a solute and a solvent is a physical process, not a chemical one. That is, both solute and solvent can be recovered in chemically unchanged forms using appropriate separation methods. For example, solid zinc nitrate dissolves in water to form an aqueous solution of zinc nitrate: \[Zn(NO_3)_{2(s)} + H_2O(l) \rightarrow Zn^{2+}_{(aq)}+2NO^-_{3(aq)} \tag{13.1}\] Because \(Zn(NO_3)_2\) can be recovered easily by evaporating the water, this is a physical process. In contrast, metallic zinc appears to dissolve in aqueous hydrochloric acid. In fact, the two substances undergo a chemical reaction to form an aqueous solution of zinc chloride with evolution of hydrogen gas: \[ Zn_{(s)} + 2H^+_{(aq)} + 2Cl^-_{(aq)} \rightarrow Zn^{2+}_{(aq)} + 2Cl^-_{(aq)} + H_{2(g)} \tag{13.2}\] When the solution evaporates, we do not recover metallic zinc, so we cannot say that metallic zinc is soluble in aqueous hydrochloric acid because it is chemically transformed when it dissolves. The dissolution of a solute in a solvent to form a solution does not involve a chemical transformation. Substances that form a single homogeneous phase in all proportions are said to be completely miscible in one another. Ethanol and water are miscible, just as mixtures of gases are miscible. If two substances are essentially insoluble in each other, such as oil and water, they are immiscible. Examples of gaseous solutions that we have already discussed include Earth’s atmosphere. Energy is required to overcome the intermolecular interactions in a solute, which can be supplied only by the new interactions that occur in the solution, when each solute particle is surrounded by particles of the solvent in a process called solvation, or hydration when the solvent is water. Thus all of the solute–solute interactions and many of the solvent–solvent interactions must be disrupted for a solution to form. In this section, we describe the role of enthalpy in this process. Because enthalpy is a , we can use a thermochemical cycle to analyze the energetics of solution formation. The process occurs in three discrete steps, indicated by ΔH , ΔH , and ΔH in Figure 13.1. The overall enthalpy change in the formation of the solution (\( \Delta H_{soln}\)) is the sum of the enthalpy changes in the three steps: \[ \Delta H_{soln} = \Delta H_1 + \Delta H_2 + \Delta H_3 \tag{13.3}\] When a solvent is added to a solution, steps 1 and 2 are both endothermic because energy is required to overcome the intermolecular interactions in the solvent (\(\Delta H_1\)) and the solute (\(\Delta H_2\)). Because \(ΔH\) is positive for both steps 1 and 2, the solute–solvent interactions (\(\Delta H_3\)) must be stronger than the solute–solute and solvent–solvent interactions they replace in order for the dissolution process to be exothermic (\(\Delta H_{soln} < 0\)). When the solute is an ionic solid, ΔH2 corresponds to the lattice energy that must be overcome to form a solution. The higher the charge of the ions in an ionic solid, the higher the lattice energy. Consequently, solids that have very high lattice energies, such as \(MgO\) (−3791 kJ/mol), are generally insoluble in all solvents. A positive value for \(ΔH_{soln}\) does not mean that a solution will not form. Whether a given process, including formation of a solution, occurs spontaneously depends on whether the total energy of the system is lowered as a result. Enthalpy is only one of the contributing factors. A high \(ΔH_{soln}\) is usually an indication that the substance is not very soluble. Instant cold packs used to treat athletic injuries, for example, take advantage of the large positive \(ΔH_{soln}\) of ammonium nitrate during dissolution (+25.7 kJ/mol), which produces temperatures less than 0°C (Figure 13.2). The enthalpy change that accompanies a process is important because processes that release substantial amounts of energy tend to occur spontaneously. A second property of any system, its entropy, is also important in helping us determine whether a given process occurs spontaneously. We will discuss entropy in more detailelsehwere, but for now we can state that entropy (S) is a thermodynamic property of all substances that is proportional to their degree of disorder. A perfect crystal at 0 K, whose atoms are regularly arranged in a perfect lattice and are motionless, is arbitrarily assigned an entropy of zero. In contrast, gases have large positive entropies because their molecules are highly disordered and in constant motion at high speeds. The formation of a solution disperses molecules, atoms, or ions of one kind throughout a second substance, which generally increases the disorder and results in an increase in the entropy of the system. Thus entropic factors almost always favor formation of a solution. In contrast, a change in enthalpy may or may not favor solution formation. The London dispersion forces that hold cyclohexane and n-hexane together in pure liquids, for example, are similar in nature and strength. Consequently, ΔHsoln should be approximately zero, as is observed experimentally. Mixing equal amounts of the two liquids, however, produces a solution in which the n-hexane and cyclohexane molecules are uniformly distributed over approximately twice the initial volume. In this case, the driving force for solution formation is not a negative \(ΔH_{soln}\) but rather the increase in entropy due to the increased disorder in the mixture. All spontaneous processes with \(ΔH \ge 0\) are characterized by an increase in entropy. In other cases, such as mixing oil with water, salt with gasoline, or sugar with hexane, the enthalpy of solution is large and positive, and the increase in entropy resulting from solution formation is not enough to overcome it. Thus in these cases a solution does not form. Table 13.2 summarizes how enthalpic factors affect solution formation for four general cases. The column on the far right uses the relative magnitudes of the enthalpic contributions to predict whether a solution will form from each of the four. Keep in mind that in each case entropy favors solution formation. In two of the cases the enthalpy of solution is expected to be relatively small and can be either positive or negative. Thus the entropic contribution dominates, and we expect a solution to form readily. In the other two cases the enthalpy of solution is expected to be large and positive. The entropic contribution, though favorable, is usually too small to overcome the unfavorable enthalpy term. Hence we expect that a solution will not form readily. Relative Changes in Enthalpies for Different Solute–Solvent Combinations* In contrast to liquid solutions, the intermolecular interactions in gases are weak (they are considered to be nonexistent in ideal gases). Hence mixing gases is usually a thermally neutral process (ΔH ≈ 0), and the entropic factor due to the increase in disorder is dominant (Figure 13.3). Consequently, all gases dissolve readily in one another in all proportions to form solutions. We will return to a discussion of enthalpy and entropy in Chapter 18, where we treat their relationship quantitatively. Considering LiCl, benzoic acid (C H CO H), and naphthalene, which will be most soluble and which will be least soluble in water? : three compounds relative solubilities in water : Assess the relative magnitude of the enthalpy change for each step in the process shown in Figure 13.1. Then use Table 13.2 to predict the solubility of each compound in water and arrange them in order of decreasing solubility. : The first substance, LiCl, is an ionic compound, so a great deal of energy is required to separate its anions and cations and overcome the lattice energy (ΔH is far greater than zero in Equation 13.3). Because water is a polar substance, the interactions between both Li and Cl ions and water should be favorable and strong. Thus we expect ΔH to be far less than zero, making LiCl soluble in water. In contrast, naphthalene is a nonpolar compound, with only London dispersion forces holding the molecules together in the solid state. We therefore expect ΔH to be small and positive. We also expect the interaction between polar water molecules and nonpolar naphthalene molecules to be weak ΔH ≈ 0. Hence we do not expect naphthalene to be very soluble in water, if at all. Benzoic acid has a polar carboxylic acid group and a nonpolar aromatic ring. We therefore expect that the energy required to separate solute molecules (ΔH ) will be greater than for naphthalene and less than for LiCl. The strength of the interaction of benzoic acid with water should also be intermediate between those of LiCl and naphthalene. Hence benzoic acid is expected to be more soluble in water than naphthalene but less soluble than LiCl. We thus predict LiCl to be the most soluble in water and naphthalene to be the least soluble. Considering ammonium chloride, cyclohexane, and ethylene glycol (HOCH CH OH), which will be most soluble and which will be least soluble in benzene? The most soluble is cyclohexane; the least soluble is ammonium chloride. Solutions are homogeneous mixtures of two or more substances whose components are uniformly distributed on a microscopic scale. The component present in the greatest amount is the solvent, and the components present in lesser amounts are the solute(s). The formation of a solution from a solute and a solvent is a physical process, not a chemical one. Substances that are miscible, such as gases, form a single phase in all proportions when mixed. Substances that form separate phases are immiscible. Solvation is the process in which solute particles are surrounded by solvent molecules. When the solvent is water, the process is called hydration. The overall enthalpy change that accompanies the formation of a solution, ΔH , is the sum of the enthalpy change for breaking the intermolecular interactions in both the solvent and the solute and the enthalpy change for the formation of new solute–solvent interactions. Exothermic (ΔH < 0) processes favor solution formation. In addition, the change in entropy, the degree of disorder of the system, must be considered when predicting whether a solution will form. An increase in entropy (a decrease in order) favors dissolution. The magnitude of the changes in both enthalpy and entropy must be considered when predicting whether a given solute–solvent combination will spontaneously form a solution. 1. Classify each of the following as a heterogeneous mixture or homogeneous mixture. Explain your rationale in each case. a. aqueous ammonia b. liquid decongestant c. vinegar d. seawater e. gasoline f. fog 2. Solutions and heterogeneous mixtures are at the extreme ends of the solubility scale. Name one type of mixture that is intermediate on this scale. How are the properties of the mixture you have chosen different from those of a solution or a heterogeneous mixture? 3. Classify each process as simple dissolution or a chemical reaction. a. a naphthalene mothball dissolving in benzene b. a sample of a common drain cleaner that has a mixture of NaOH crystals and Al chunks dissolving in water to give H2 gas and an aqueous solution of Na , OH , and Al ions c. an iron ship anchor slowly dissolving in seawater d. sodium metal dissolving in liquid ammonia 4. Classify each process as simple dissolution or a chemical reaction. a. a sugar cube dissolving in a cup of hot tea b. SO gas dissolving in water to produce sulfuric acid c. calcium oxide dissolving in water to produce a basic solution d. metallic gold dissolving in a small quantity of liquid mercury 5. You notice that a gas is evolved as you are dissolving a solid in a liquid. Will you be able to recover your original solid by evaporation? Why or why not? 6. Why is heat evolved when sodium hydroxide pellets are dissolved in water? Does this process correspond to simple dissolution or a chemical reaction? Justify your answer. 7. Which process(es) is the simple formation of a solution, and which process(es) involves a chemical reaction? a. mixing an aqueous solution of NaOH with an aqueous solution of HCl b. bubbling HCl gas through water c. adding iodine crystals to CCl d. adding sodium metal to ethanol to produce sodium ethoxide (C H O Na ) and hydrogen gas 8. Using thermochemical arguments, explain why some substances that do not form a solution at room temperature will form a solution when heated. Explain why a solution can form even when ΔH is positive. 9. If you wanted to formulate a new compound that could be used in an instant cold pack, would you select a compound with a positive or negative value of ΔH in water? Justify your answer. 10. Why is entropy the dominant factor in the formation of solutions of two or more gases? Is it possible for two gases to be immiscible? Why or why not? | 13,528 | 2,074 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/Electrophilic_Substitution_Reactions/Electrophilic_Substitution |
Electrophilic substitution happens in many of the reactions of compounds containing benzene rings - the arenes. For simplicity, we'll only look for now at benzene itself. This is what you need to understand for the purposes of the electrophilic substitution mechanisms: Because of the delocalized electrons exposed above and below the plane of the rest of the molecule, benzene is obviously going to be highly attractive to electrophiles - species which seek after electron rich areas in other molecules. The electrophile will either be a positive ion, or the slightly positive end of a polar molecule. The delocalized electrons above and below the plane of the benzene molecule are open to attack in the same way as those above and below the plane of an ethene molecule. However, the end result will be different. If benzene underwent addition reactions in the same way as ethene, it would need to use some of the delocalized electrons to form bonds with the new atoms or groups. This would break the delocalization - and this costs energy. Instead, it can maintain the delocalization if it replaces a hydrogen atom by something else - a substitution reaction. The hydrogen atoms aren't involved in any way with the delocalized electrons. In most of benzene's reactions, the electrophile is a positive ion, and these reactions all follow a general pattern. Suppose the electrophile is a positive ion \(X^+\). Two of the electrons in the delocalized system are attracted towards the \(X^+\) and form a bond with it. This has the effect of breaking the delocalization, although not completely. The ion formed in this step isn't the final product. It immediately goes on to react with something else. It is just an intermediate. There is still delocalization in the intermediate formed, but it only covers part of the ion. When you write one of these mechanisms, draw the partial delocalization to take in all the carbon atoms apart from the one that the \(X\) has become attached to. The intermediate ion carries a positive charge because you are joining together a neutral molecule and a positive ion. This positive charge is spread over the delocalized part of the ring. Simply draw the "+" in the middle of the ring. The hydrogen at the top isn't new - it's the hydrogen that was already attached to that carbon. We need to show that it is there for the next stage. Here we've introduced a new ion, \(Y^-\). W here did this come from? You have to remember that it is impossible to get a positive ion on its own in a chemical system - so \(Y^-\) is simply the negative ion that was originally associated with \(X^+\). Don't w orry about this at the moment - it's much easier to see when you've got a real example in front of you. A lone pair of electrons on Y forms a bond with the hydrogen atom at the top of the ring. That means that the pair of electrons joining the hydrogen onto the ring aren't needed any more. These then move down to plug the gap in the delocalized electrons, so restoring the delocalized ring of electrons which originally gave the benzene its special stability. The complete delocalization is temporarily broken as \(X\) replaces \(H\) on the ring, and this costs energy. However, that energy is recovered when the delocalization is re-established. This initial input of energy is simply the activation energy for the reaction. In this case, it is going to be high (something around 150 kJ mol ), and this means that benzene's reactions tend to be slow. In these reactions, the electrophiles are polar molecules rather than fully positive ions. Because these mechanisms are different from what's gone before (and from each other), there isn't any point in dealing with them in a general way. Jim Clark ( ) | 3,746 | 2,079 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/Electrophilic_Substitution_Reactions/The_Sulfonation_of_Benzene |
This page gives you the facts and a simple, uncluttered mechanism for the electrophilic substitution reaction between benzene and sulfuric acid (or sulfur trioxide). There are two equivalent ways of sulfonating benzene: \[C_6H_6 + H_2SO_4 \rightarrow C_6H_5SO_3H + H_2O\] Or: The product is benzenesulfonic acid. The electrophile is actually sulfur trioxide, SO , and you may find the equation for the sulfonation reaction written: The sulfur trioxide electrophile arises in one of two ways depending on which sort of acid you are using. Concentrated sulfuric acid contains traces of SO due to slight dissociation of the acid. \[ H_2SO_4 \rightleftharpoons H_2O + SO_3\] , \(H_2S_2O_7\), can be thought of as a solution of \(SO_3\) i n sulfuric acid - and so is a much richer source of the SO . Sulfur trioxide is an electrophile because it is a highly polar molecule with a fair amount of positive charge on the sulfur atom. It is this which is attracted to the ring electrons. The second stage of the reaction involves a transfer of the hydrogen from the ring to the negative oxygen. Jim Clark ( ) | 1,112 | 2,081 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/05%3A_Solutions/5.03%3A_The_Thermodynamics_of_Mixing |
When solids, liquids or gases are combined, the thermodynamic quantities of the system experience a change as a result of the mixing. This module will discuss the effect that mixing has on a solution’s , , and , with a specific focus on the mixing of two gases. A solution is created when two or more components mix homogeneously to form a single phase. Studying solutions is important because most chemical and biological life processes occur in systems with multiple components. Understanding the thermodynamic behavior of mixtures is integral to the study of any system involving either ideal or non-ideal solutions because it provides valuable information on the molecular properties of the system. Most real gases behave like ideal gases at standard temperature and pressure. This allows us to combine our knowledge of ideal systems and solutions with standard state thermodynamics in order to derive a set of equations that quantitatively describe the effect that mixing has on a given gas-phase solution’s thermodynamic quantities. Unlike the extensive properties of a one-component system, which rely only on the amount of the system present, the extensive properties of a solution depend on its temperature, pressure and composition. This means that a mixture must be described in terms of the partial molar quantities of its components. The total Gibbs free energy of a two-component solution is given by the expression \[ G=n_1\overline{G}_1+n_2\overline{G} _2 \label{1}\] where The molar Gibbs energy of an ideal gas can be found using the equation \[\overline{G}=\overline{G}^\circ+RT\ln \frac{P}{1 bar} \label{2}\] where \( \overline{G}^\circ\) is the standard molar Gibbs energy of the gas at 1 bar, and P is the pressure of the system. In a mixture of ideal gases, we find that the system’s partial molar Gibbs energy is equivalent to its chemical potential, or that \[ \overline{G}_i=\mu_i \label{3} \] This means that for a solution of ideal gases, Equation \(\ref{2}\) can become \[\overline{G}_i=\mu_i=\mu^\circ_i+RT \ln \frac{P_i}{1 bar} \label{4}\] where Now pretend we have two gases at the same temperature and pressure, gas 1 and gas 2. The Gibbs energy of the system before the gases are mixed is given by Equation \(\ref{1}\), which can be combined with Equation \(\ref{4}\) to give the expression \[G_{initial}=n_1(\mu^\circ_1+RT \ln P)+n_2(\mu^\circ_2+RT \ln P) \label{5}\] If gas 1 and gas 2 are then mixed together, they will each exert a partial pressure on the total system, \(P_1\) and \(P_2\), so that \(P_1+ P_2= P\). This means that the final Gibbs energy of the final solution can be found using the equation \[G_{final}=n_1(\mu^\circ_1+RT \ln P_1)+n_2(\mu^\circ_2+RT \ln P_2) \label{6}\] The Gibbs energy of mixing, \(Δ_{mix}G\), can then be found by subtracting \(G_{initial}\) from \(G_{final}\). \[ \begin{align} Δ_{mix}G &= G_{final} - G_{initial}\\[4pt] &=n_1RT \ln \frac{P_1}{P}+n_2RT \ln \frac{P_2}{P} \\[4pt] &=n_1 RT \ln \chi_1+n_2 RT \ln \chi_2 \label{7} \end{align} \] where \[P_i = \chi_iP\] and \(\chi_i\) is the mole fraction of gas \(i\). This equation can be simplified further by knowing that the mole fraction of a component is equal to the number of moles of that component over the total moles of the system, or \[\chi_i = \dfrac{n_i}{n}.\] Equation \ref{7} then becomes \[\Delta_{mix} G=nRT(\chi_1 \ln \chi_1 + \chi_2 \ln \chi_2) \label{8}\] This expression gives us the effect that mixing has on the Gibbs free energy of a solution. Since \(\chi_1\) and \(\chi_2\) are mole fractions that range from 0 to 1, we can conclude that \(Δ_{mix}G \) will be a negative number. This is consistent with the idea that gases mix spontaneously at constant pressure and temperature. Figure \(\Page {1}\) shows that when two gases mix, it can really be seen as two gases expanding into twice their original volume. This greatly increases the number of available microstates, and so we would therefore expect the entropy of the system to increase as well. Thermodynamic studies of an ideal gas’s dependence of Gibbs free energy of temperature have shown that \[ \left( \dfrac {d G} {d T} \right )_P=-S \label{9}\] This means that differentiating Equation \(\ref{8}\) at constant pressure with respect to temperature will give an expression for the effect that mixing has on the entropy of a solution. We see that \[ \begin{align} \left( \dfrac {d G_{mix}} {d T} \right)_P &=nR(x_1 \ln x_1+x_2 \ln x_2) \\[4pt] &=-\Delta_{mix} S \end{align}\] \[\Delta_{mix} S=-nR(x_1 \ln x_1+x_2 \ln x_2) \label{10}\] Since the mole fractions again lead to negative values for ln x and ln x , the negative sign in front of the equation makes Δ S positive, as expected. This agrees with the idea that mixing is a spontaneous process. We know that in an ideal system \(\Delta G= \Delta H-T \Delta S\), but this equation can also be applied to the thermodynamics of mixing and solved for the enthalpy of mixing so that it reads \[\Delta_{mix} H=\Delta_{mix} G+T\Delta_{mix} S \label{11}\] Plugging in our expressions for \(Δ_{mix}G\) (Equation \(\ref{8}\)) and \(Δ_{mix}S\) (Equation \(\ref{10}\)) , we get \[\Delta_{mix} H=nRT(x_1 \ln x_1+x_2 \ln x_2)+T \left[-nR(x_1 \ln x_1+x_2 \ln x_2) \right] = 0\] This result makes sense when considering the system. The molecules of ideal gas are spread out enough that they do not interact with one another when mixed, which implies that no heat is absorbed or produced and results in a \(Δ_{mix}H\) of zero. Figure \(\Page {2}\) illustrates how \(TΔ_{mix}S\) and \(Δ_{mix}G\) change as a function of the mole fraction so that \(Δ_{mix}H\) of a solution will always be equal to zero (this is for the mixing of two ideal gasses). | 5,718 | 2,082 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkanes/Properties_of_Alkanes/Cycloalkanes/Rings%3A_cis_trans_and_axial_equatorial_relationships |
The purpose of this page is to help organic chem students show how substituent groups are located on ring structures. We focus here on six-membered rings (6-rings); these are among the most common rings in organic chem (and biochem), and they suffice to raise the main issues. We will look at how to show cis and trans relationships in simple hexagon structural formulas, and we will look at structures showing the common "chair" conformation, focusing on axial vs equatorial orientations. We will also discuss the relationship between cis/trans and axial/equatorial. The basic approach here is to look at a series of compounds, of generally increasing complexity. They are chosen to illustrate one new feature at a time of how to draw the structures and how to see cis/trans and axial/equatorial features. This basic 6-ring cycloalkane, without any special substituents (i.e., other than hydrogen), is easily shown by either of the formulas in Fig 1. Both structures clearly imply six C in a ring, with two H at each position. Fig 1A is a simple structural formula, condensed. We say it is "condensed" because not all features are shown explicitly. In fact, in this case, not much at all is shown explicitly: the C atoms are understood (at each vertex), the H atoms are understood (enough at each C to give C its normal four bonds), and the C-H bonds are understood. About all that is explicit is the basic carbon skeleton (C-C bonds), showing that there are six C atoms in a ring. Fig 1B is very much like Fig 1A, but it adds one new feature: Fig 1B attempts to show the conformation of the molecule. In this case, we understand that cyclohexane and its simple derivatives tend to spend most of their time near this "chair" conformation. Figure 2: If we want to show the hydrogen atoms explicitly, we can do so with structural formulas such as in Fig 2. There is little room for confusion here, since all six C atoms are equivalent, as are all 12 H atoms. For notes on how to draw chairs, see the section . So what is new here? Not much, with the hexagon formula, Fig 3A. That type of formula shows the basic "connectivity" of the atoms -- who is connected to whom. This chemical has one Cl on the ring, and it does not matter where we show it. There is now only one H on that C, but since we are not showing H explicitly here, that is not an issue in drawing the structure. (It is an issue when you look at it and want to count H.) With the chair formula (Fig 3B), which shows information not only about connectivity but also about conformation, there is important new information here. In a chair, there are two "types" of substituents: those pointing up or down, and called axial, and those pointing "outward", and called equatorial. I have shown the chlorine atom in an equatorial position. Why? Two reasons: it is what we would predict, and it is what is found. Why do we predict that the Cl is equatorial? Because it is bigger than H, and there is more room in the equatorial positions. The next level of complexity is a di-substituted cycloalkane, "dichlorocyclohexane". The first question we must ask is which C the two chlorine substituents are on. For now, I want to discuss 1,3- dichlorocyclohexane. This introduces another issue: are the two Cl on the same side of the ring, or on opposite sides? We call these "cis" (same side) and "trans" (opposite sides). We focus on one of these, cis-1,3-dichlorocyclohexane. And for now, we will just look at hexagon structural formulas, leaving the question of conformation for later. Let's go through this one step at a time. The structure in Fig 4 is indeed a dichlorocyclohexane. It is even a 1,3-dichlorocyclohexane. However, this structure provides no information about the orientation of the two Cl atoms relative to the plane of the ring. To show a specific isomer -- cis or trans -- we must somehow show how the two Cl atoms are oriented relative to the plane of the ring. The basic idea in both of these is that we can imagine the ring to be planar, and then show the groups above or below the plane of the ring. How we show this is different in the two parts of Fig 5. In Fig 5A, we look at the ring "edge-on". The thick line for the bottom bond is intended to convey the edge-on view (or side view); this is sometimes omitted, especially with hand-drawn structures, but be careful then that the meaning is clear. Once we understand that we are now looking at the ring edge-on, it is clear that the two Cl atoms are both above the ring, hence cis. In Fig 5B, we view the ring "face-on" (or top view), and use special bond symbols -- "stereo bonds" -- to convey up and down: the heavy wedge -- an "up wedge" -- points upward, toward you, and the dashed bond -- a "down bond" -- points downward, away from you. Again, both Cl are "up", hence cis. Notes... For some notes on how to draw the stereo bonds, see the section . In discussing Fig 5, I started by saying that we imagine the ring to be planar. Emphasize that cycloalkane rings are not really planar (except for cyclopropane rings). As so often, the structural formula represents the general layout of the atoms, but not the actual molecular geometry. Those with the Ouellette book can see examples of these two ways of showing up/down on p 81 (top) and p 80 (middle). Most organic chemistry books will show you this. The structure shows -1,3-dichlorocyclohexane: a 6-ring; 2 Cl atoms, at positions 1 and 3; and cis, with both Cl on the same side of -- above -- the H that is on the same C. I showed the 2 Cl atoms at corner positions, and I showed the H at the key positions explicitly. These points follow from some of the discussed earlier. It is not required that you do these things, but they can make things easier for you -- and for anyone reading your structures. Now, what is new here? The conformation. We start with the notion that the conformation of cyclohexane derivatives is based on the "chair". At each position, one substituent is axial (loosely, perpendicular to the ring), and one is equatorial (loosely, in the plane of the ring). There is more room in the equatorial positions (not easily seen with these simple drawings, but ordinary ball and stick models do help with this point). Thus we try to put the larger substituents in the equatorial positions. In this case, we put the Cl equatorial and the H axial at each position 1 and position 3. We are now done with this compound, cis-1,3-dichlorocyclohexane. However, we have missed one very important concern -- because it is not an issue in this case. So let's look at another compound. Figures 7 and 8 above introduce no new ideas or complications. These two figures should be straightforward. So, what is the preferred conformation of cis-1,2-dichlorocyclohexane? This requires careful consideration; an important lesson from this exercise is to realize that we cannot propose a good conformation based simply on what we have learned so far. The two guidelines we have so far for conformation of 6-rings are: Let's explore the difficulty here by looking at some things people might naively draw. Why is Fig 9 wrong? It is the wrong chemical. The structure shown in Fig 9 is trans, not cis. Look carefully at the 1 and 2 positions. At one of them, the H is above the Cl; at the other, the Cl is above the H. Trans. Wrong chemical. The structure shown in Fig 9 is not the requested chemical. Why is Fig 10 wrong? After all, it seems to address the criteria presented. It contains both of the larger atoms (Cl) equatorial, and they are cis as desired. However, in Fig 10, the two axial groups on carbons # 1 and 2 (the two H that are shown) are both pointing up. This is impossible. In a valid chair, the axial groups alternate up/down as one goes around the ring; see . This follows from the tetrahedral bonding of C. , as relevant here, ; that condition is violated here. That is, Fig 10 is not a valid chair. Those who find the above point new or surprising should check their textbook. If possible, look at models of cyclohexane and simple derivatives such as the one here. , is correct. In my experience, many students have not yet noticed this feature of chair conformations. If you think you have an alternative that is better (or even satisfactory), please show it to me. I suspect it will turn out to be equivalent to Fig 9 or Fig 10, above. But if you think it is good, let's discuss it. So now what? We have a contradiction. And that really is the most important point here. It is important to realize that we cannot draw a conformation for cis-1,2-dichlorocyclohexane which easily fits the criteria we have used so far: a chair, with large groups equatorial. So what do we do? Clearly, the conformation of this compound must, in some way, involve more issues than what we have considered so far. Instructors and books will vary in how much further explanation they want to give on this matter. Therefore, how you proceed from here must take into account the preferences in your course. Here is one way to proceed. One simple way to proceed is to re-examine the two criteria we have been using, and then state a generality about what to do in the event of a conflict. That generality is: use the chair, and then fit the groups as best you can. That is, try to put as many of the larger groups equatorial as you can, but realize that you may not get them all equatorial. What have we accomplished here? First, this is the correct compound. Convince yourself that this really is cis-1,2-dichlorocyclohexane. In particular, it is cis because at each substituted position the Cl is "above" the H; that is, both Cl are on the same side of the ring. Second, this is a proper chair. Adjacent axial groups point in opposite directions. Only two of them are shown here: the axial Cl on the upper right C is "up", and the axial H on the lower right C is "down". Thus, these aspects satisfy the proper way to orient things, in general, on a cyclohexane chair. How good a conformation is the one shown in Fig 11? Actually, it is quite good, in this case. The actual conformation has been measured, and it follows the basic ideas shown here. Is this the end of the story? No, but it is about enough for now. The main purpose here was to show how one must carefully look at conformation, within the constraints of the specific isomer one is trying to draw. Some compounds cannot be easily drawn within the common "rules". cis-1,2-dichlorocyclohexane is one such example. In this case, we kept the basic chair conformation, but put one larger group in the less favored axial orientation. Measurements on many such chemicals have shown that the energetic penalty of moving a cyclohexane ring much away from the basic chair conformation is quite large -- certainly larger than the energetic penalty of putting one "somewhat large" group in an axial position. Of course, with larger groups or more groups, this might not hold. There are various ways to show these orientations. The solid (dark) "up wedge" I used is certainly common. Some people use an analogous "down wedge", which is light, to indicate a down bond; unfortunately, there is no agreement as to which way the wedge should point, and you are left relying on the lightness of the wedge to know it is "down". The "down bond" avoids this wedge ambiguity, and just uses some kind of light line. The down bond I used (e.g., in ) is a dashed line; IUPAC encourages a series of parallel lines, something like . What I did is a variation of what is recommended by IUPAC: . In , the "up wedge" and "down bond" that I used, along with other variations, are available from a tool button that may be labeled with any of them, depending on most recent use. It is located directly below the tool button for ordinary C-C bonds. In , the "up wedge" and "down bond", along with other variations, are available from a tool button that may be labeled with any of them, depending on most recent use. It is located directly below the "Chain" tool button. provides up and down wedges, but not the simple up and down bonds discussed above. The wedges are available from the second toolbar across the top. For an expanded discussion of using these wedges, see the section of my ChemSketch Guide on . As always, the information provided on these pages in intended to help you get started. Each program has more options for drawing bonds than discussed here. When you feel the need, look around! Most of the structures shown on this page were drawn with the free program . I have posted a guide to help you get started with . ISIS/Draw provides a simple cyclohexane (6-ring) hexagon template on the toolbar across the top. It provides templates for various 6-ring chair structures from the Templates menu; choose Rings. There are templates for simple chairs, without substituents (e.g., ), and for chairs showing all the substituents (e.g., ). In either case, you can add, delete, or change things as you wish. Various kinds of stereo bonds (wedges and bars) are available by clicking the left-side tool button that is just below the regular C-C single bond button. It may have a wedge shown on it, but this will vary depending on how it has been used. To choose a type of stereo bond, click on the button and hold the mouse click; a new menu will appear to the right of the button. The free drawing program , the successor to ISIS/Draw, provides similar templates and tools. A basic chair structure is provided on the default template bar that is shown. More options are available by choosing the Rings template. See my page for a general guide for getting started with this program. The free drawing program provides similar templates and tools. To find the special templates for chairs, go to the menu, choose , and then choose "Rings" from the drop-down menu near upper left. See my page for a general guide for getting started with this program. If you want to draw chair structures by hand (and if you are going on in organic chemistry, you should)... Be careful. The precise zigs and zags, and the angles of substituents are all important. Your textbook may offer you some hints for how to draw chairs. A short item in the Journal of Chemical Education offers a nice trick, showing how the chair can be thought of as consisting of an M and a W. The article is V Dragojlovic, A method for drawing the cyclohexane ring and its substituents. J Chem Educ 78:923, 7/01. (I thank M Farooq Wahab, Chemistry, Univ Karachi, for suggesting that this article be noted here.) Aside from drawing the basic chair, the key points in adding substituents are: The main issues raised here for showing cis and trans hold for other ring sizes. Specifically, holds for other ring sizes. The only difference would be the geometric figure used to show the ring. Discussion of conformation is more complex, and must be considered for each ring size. The "chair" is a likely conformation for 6-rings, but not for other sizes. Discussing conformations for other ring sizes is beyond the scope of this page. 6-rings are among the most common, in both organic and biochemistry. >Robert Bruner ( ) | 15,152 | 2,083 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/14%3A_Chemical_Kinetics/14.03%3A_Reaction_Rates_and_Rate_Laws |
The factors discussed in affect the reaction rate of a chemical reaction, which may determine whether a desired product is formed. In this section, we will show you how to quantitatively determine the reaction rate. Reaction rates are usually expressed as the concentration of reactant consumed or the concentration of product formed per unit time. The units are thus moles per liter per unit time, written as M/s, M/min, or M/h. To measure reaction rates, chemists initiate the reaction, measure the concentration of the reactant or product at different times as the reaction progresses, perhaps plot the concentration as a function of time on a graph, and then calculate the in the concentration per unit time. The progress of a simple reaction (A → B) is shown in where the beakers are snapshots of the composition of the solution at 10 s intervals. The number of molecules of reactant (A) and product (B) are plotted as a function of time in the graph. Each point in the graph corresponds to one beaker in . The reaction rate is the change in the concentration of either the reactant or the product over a period of time. The concentration of A decreases with time, while the concentration of B increases with time.
\( rate=\dfrac{\Delta \left [ B \right ]}{\Delta t}=-\dfrac{\Delta \left [ A \right ]}{\Delta t} \tag{14.2.1} \) Square brackets indicate molar concentrations, and the capital Greek delta (Δ) means “change in.” Because chemists follow the convention of expressing all reaction rates as positive numbers, however, a negative sign is inserted in front of Δ[A]/Δ to convert that expression to a positive number. The reaction rate we would calculate for the reaction A → B using would be different for each interval. (This is not true for every reaction, as you will see later.) A much greater change occurs in [A] and [B] during the first 10 s interval, for example, than during the last, which means that the reaction rate is fastest at first. This is consistent with the concentration effects described in because the concentration of A is greatest at the beginning of the reaction. Reaction rates generally decrease with time as reactant concentrations decrease. We can use to determine the reaction rate of hydrolysis of aspirin, probably the most commonly used drug in the world. (More than 25,000,000 kg are produced annually worldwide.) Aspirin (acetylsalicylic acid) reacts with water (such as water in body fluids) to give salicylic acid and acetic acid. Because salicylic acid is the actual substance that relieves pain and reduces fever and inflammation, a great deal of research has focused on understanding this reaction and the factors that affect its rate. Data for the hydrolysis of a sample of aspirin are in and are shown in the graph in . These data were obtained by removing samples of the reaction mixture at the indicated times and analyzing them for the concentrations of the reactant (aspirin) and one of the products (salicylic acid). We can calculate the average reaction rate \( rate_{\left ( t=0-2.0\;h \right )}=\dfrac{\left [ salicylic\;acid \right ]_{2}-\left [ salicylic\;acid \right ]_{0}}{2.0\;h-0.0\;h} \) \( =\dfrac{0.040\times 10^{-3}\;M-0.000\;M}{2.0\;h-0.0\;h}=2\times 10^{-5}\;M/h \) We can also calculate the reaction rate from the concentrations of aspirin at the beginning and the end of the same interval, remembering to insert a negative sign, because its concentration decreases: \( rate_{\left ( t=0-2.0\;h \right )}=\dfrac{\left [ aspirin \right ]_{2}-\left [ aspirin \right ]_{0}}{2.0\;h-0.0\;h} \) \( rate_{\left ( t=0-2.0\;h \right )}=\dfrac{5.51\times 10^{-3}\;M-5.55\times 10^{-3}\;M}{2.0\;h-0.0\;h}=2.0\times 10^{-5}\;M/h \)\;h} \) If we now calculate the reaction rate during the last interval given in (the interval between 200 h and 300 h after the start of the reaction), we find that the reaction rate is significantly slower than it was during the first interval ( = 0–2.0 h): \( rate_{\left ( t=200-300\;h \right )}=\dfrac{\left [ salicylic\;acid \right ]_{300}-\left [ salicylic\;acid \right ]_{200}}{300\;h-200\;h} \) \( =\dfrac{3.73\times 10^{-3}\;M-2.91\times 10^{-3}\;M}{100;h}=8.2\times 10^{-6}\;M/h \) (You should verify from the data in that you get the same rate using the concentrations of aspirin measured at 200 h and 300 h.) In the preceding example, the stoichiometric coefficients in the balanced chemical equation are the same for all reactants and products; that is, the reactants and products all have the coefficient 1. Let us look at a reaction in which the coefficients are all the same: the fermentation of sucrose to ethanol and carbon dioxide, which we encountered in
\( \underset{sucrose}{C_{12}H_{22}O_{11}\left ( aq \right )}+H_{2}O\left ( l \right )\rightarrow 4C_{2}H_{5}OH \left ( aq \right ) + CO_{2}\left ( g \right ) \tag{14.2.2}\) The coefficients show us that the reaction produces four molecules of ethanol and four molecules of carbon dioxide for every one molecule of sucrose consumed. As before, we can find the reaction rate by looking at the change in the concentration of any reactant or product. In this particular case, however, a chemist would probably use the concentration of either sucrose or ethanol because gases are usually measured as volumes and, as you learned in , the volume of CO gas formed will depend on the total volume of the solution being studied and the solubility of the gas in the solution, not just the concentration of sucrose. The coefficients in the balanced chemical equation tell us that the reaction rate at which ethanol is formed is always four times faster than the reaction rate at which sucrose is consumed:
\( rate=\dfrac{\Delta \left [ C_{2}H_{5}OH \right ]}{\Delta t}=-\dfrac{4 \Delta \left [ sucrose ] \right )}{\Delta t} \tag{14.2.3} \) The concentration of the reactant—in this case sucrose— with increasing time, so the value of Δ[sucrose] is negative. Consequently, a minus sign is inserted in front of Δ[sucrose] in .3 so that the rate of change of the sucrose concentration is expressed as a positive value. Conversely, the ethanol concentration with increasing time, so its rate of change is automatically expressed as a positive value. Often the reaction rate is expressed in terms of the reactant or product that has the smallest coefficient in the balanced chemical equation. The smallest coefficient in the sucrose fermentation reaction ( ) corresponds to sucrose, so the reaction rate is generally defined as follows:
\( rate=-\dfrac{\Delta \left [ sucrose \right ]}{\Delta t}=\dfrac{1}{4}\left (\dfrac{\Delta \left [ C_{2}H_{5}OH \right ]}{\Delta t} \right ) \tag{14.2.4} \) Consider the thermal decomposition of gaseous N O to NO and O via the following equation: \( 2N_{2}O_{5}\left ( g \right )\overset{\Delta }{\rightarrow} 4NO_{2}\left ( g \right )+O_{2}\left ( g \right ) \) Write expressions for the reaction rate in terms of the rates of change in the concentrations of the reactant and each product with time. balanced chemical equation reaction rate expressions Choose the species in the equation that has the smallest coefficient. Then write an expression for the rate of change of that species with time. For the remaining species in the equation, use molar ratios to obtain equivalent expressions for the reaction rate. Because O has the smallest coefficient in the balanced chemical equation for the reaction, we define the reaction rate as the rate of change in the concentration of O and write that expression. We know from the balanced chemical equation that 2 mol of N O must decompose for each 1 mol of O produced and that 4 mol of NO are produced for every 1 mol of O produced. The molar ratios of O to N O and to NO are thus 1:2 and 1:4, respectively. This means that we divide the rate of change of [N O ] and [NO ] by its stoichiometric coefficient to obtain equivalent expressions for the reaction rate. For example, because NO is produced at four times the rate of O , we must divide the rate of production of NO by 4. The reaction rate expressions are as follows: \( rate=\dfrac{\Delta \left [ O_{2} \right ]}{\Delta t} = \dfrac{\Delta \left [ NO_{2} \right ]}{4\Delta t}= -\dfrac{\Delta \left [ N_{2}O_{5} \right ]}{2\Delta t} \) Exercise The key step in the industrial production of sulfuric acid is the reaction of SO with O to produce SO . \( 2SO_{2}\left ( g \right )+ O_{2}\left ( g \right ) \rightarrow 2SO_{3}\left ( g \right ) \) Write expressions for the reaction rate in terms of the rate of change of the concentration of each species. \( rate=-\dfrac{\Delta \left [ O_{2} \right ]}{\Delta t} = -\dfrac{\Delta \left [ SO_{2} \right ]}{\Delta t}=\dfrac{\Delta \left [ SO_{3} \right ]}{2\Delta t} \) Using the reaction shown in Example 1, calculate the reaction rate from the following data taken at 56°C: balanced chemical equation and concentrations at specific times reaction rate Using the equations in Example 1, subtract the initial concentration of a species from its final concentration and substitute that value into the equation for that species. Substitute the value for the time interval into the equation. Make sure your units are consistent. We are asked to calculate the reaction rate in the interval between = 240 s and = 600 s. From Example 1, we see that we can evaluate the reaction rate using any of three expressions: \( rate=\dfrac{\Delta \left [ O_{2} \right ]}{\Delta t} = \dfrac{\Delta \left [ NO_{2}\right ]}{4\Delta t}= -\dfrac{\Delta \left [ N_{2}O_{5} \right ]}{2\Delta t} \) Subtracting the initial concentration from the final concentration of N O and inserting the corresponding time interval into the rate expression for N O , \( rate= -\dfrac{\Delta \left [ N_{2}O_{5} \right ]}{2\Delta t} =-\dfrac{\left [ N_{2}O_{5} \right ]_{600}-\left [ N_{2}O_{5} \right ]_{240}}{2\left ( 600\;s-240\;s \right )} \) Substituting actual values into the expression, \( rate=-\dfrac{0.197\;M-0.0388\;M}{2\left (360 \;s \right )} \) Similarly, we can use NO to calculate the reaction rate: \( rate= -\dfrac{\Delta \left [ NO_{2} \right] }{4\Delta t} =-\dfrac{\left [ NO_{2} \right ]_{600}-\left [ NO_{2} \right ]_{240}}{4\left ( 600\;s-240\;s \right )} =\dfrac{0.0699\;M-0.0314\;M}{4\left ( 360 \right )}=2.67\times 10^{-5}\;M/s \) If we allow for experimental error, this is the same rate we obtained using the data for N O , as it should be because the reaction rate should be the same no matter which concentration is used. We can also use the data for O : \( rate= -\dfrac{\Delta \left [ O_{2} \right] }{\Delta t} =-\dfrac{\left [ O_{2} \right ]_{600}-\left [ O_{2} \right ]_{240}}{\left ( 600\;s-240\;s \right )} =\dfrac{0.0175\;M-0.00792\;M}{360\;s}=2.66\times 10^{-5}\;M/s \) Again, this is the same value we obtained from the N O and NO data. Thus the reaction rate does not depend on which reactant or product is used to measure it. Exercise Using the data in the following table, calculate the reaction rate of SO (g) with O (g) to give SO (g). 9.0 × 10 M/s So far, we have determined average reaction rates over particular intervals of time. We can also determine the instantaneous rate of a reaction, which is the reaction rate at any given point in time. As the period of time used to calculate an average rate of a reaction becomes shorter and shorter, the average rate approaches the instantaneous rate. Think of the distinction between the instantaneous and average rates of a reaction as being similar to the distinction between the actual speed of a car at any given time on a trip and the average speed of the car for the entire trip. Although you may travel for a long time at 65 mph on an interstate highway during a long trip, there may be times when you travel only 25 mph in construction zones or 0 mph if you stop for meals or gas. Thus your average speed on the trip may be only 50 mph, whereas your instantaneous speed on the interstate at a given moment may be 65 mph. Whether you are able to stop the car in time to avoid an accident depends on your instantaneous speed, not your average speed. There are important differences between the speed of a car during a trip and the speed of a chemical reaction, however. The speed of a car may vary unpredictably over the length of a trip, and the initial part of a trip is often one of the slowest. In a chemical reaction, the initial interval normally has the fastest rate (though this is not always the case), and the reaction rate generally changes smoothly over time. In chemical kinetics, we generally focus on one particular instantaneous rate, which is the initial reaction rate, = 0. Initial rates are determined by measuring the reaction rate at various times and then extrapolating a plot of rate versus time to = 0. In , you learned that reaction rates generally decrease with time because reactant concentrations decrease as reactants are converted to products. You also learned that reaction rates generally increase when reactant concentrations are increased. We now examine the mathematical expressions called rate laws , which describe the relationships between reactant rates and reactant concentrations. Rate laws are laws as defined in ; that is, they are mathematical descriptions of experimentally verifiable data. Rate laws may be written from either of two different but related perspectives. A differential rate law expresses the reaction rate in terms of in the concentration of one or more reactants (Δ[R]) over a specific time interval (Δ ). In contrast, an integrated rate law describes the reaction rate in terms of the concentration ([R] ) and the concentration of one or more reactants ([R]) after a given amount of time ( ); we will discuss integrated rate laws in . The integrated rate law can be found by using calculus to integrate the differential rate law, although the method of doing so is beyond the scope of this text. For a reaction with the general equation
\( aA+bB\rightarrow cC+dD \tag{14.2.5} \) the experimentally determined rate law usually has the following form:
\( rate=k\left [ A \right ]^{m}\left [ B \right ]^{n} \tag{14.2.6} \) The ( ) is called the rate constant , and its value is characteristic of the reaction and the reaction conditions. A given reaction has a particular value of the rate constant under a given set of conditions, such as temperature, pressure, and solvent; varying the temperature or the solvent usually changes the value of the rate constant. The numerical value of , however, does change as the reaction progresses under a given set of conditions. Thus the reaction rate depends on the rate constant for the given set of reaction conditions and the concentration of A and B raised to the powers and , respectively. The values of and are derived from experimental measurements of the changes in reactant concentrations over time and indicate the reaction order , the degree to which the reaction rate depends on the concentration of each reactant; and need not be integers. For example, tells us that is th order in reactant A and th order in reactant B. It is important to remember that and are not related to the stoichiometric coefficients and in the balanced chemical equation and must be determined experimentally. The is the sum of all the exponents in the rate law: + . Under a given set of conditions, the value of the rate constant does not change as the reaction progresses. Although differential rate laws are generally used to describe what is occurring on a molecular level during a reaction, integrated rate laws are used to determine the reaction order and the value of the rate constant from experimental measurements. (We present general forms for integrated rate laws in .) To illustrate how chemists interpret a differential rate law, we turn to the experimentally derived rate law for the hydrolysis of -butyl bromide in 70% aqueous acetone. This reaction produces -butanol according to the following equation:
\( \left ( CH_{3} \right )_{3}CBr \left ( soln \right )+H_{2}O\left ( soln \right )\rightarrow \left ( CH_{3} \right )_{3}COH \left( soln \right )+HBr\left ( soln \right ) \tag{14.7.7} \) Combining the rate expression in and gives us a general expression for the differential rate law:
\( rate=-\dfrac{\Delta \left [A \right ]}{\Delta t}= k\left [ A \right ]^{m}\left [ B \right ]^{n} \tag{14.2.8} \) Inserting the identities of the reactants into gives the following expression for the differential rate law for the reaction:
\( rate=-\dfrac{ \Delta \left [\left (CH_{3} \right )_{3}CBr \right ]}{\Delta t}= k\left [ \left (CH_{3} \right )_{3}CBr \right ]^{m}\left [ H_{2}O \right ]^{n} \tag{14.2.9} \) Experiments done to determine the rate law for the hydrolysis of -butyl bromide show that the reaction rate is directly proportional to the concentration of (CH ) CBr but is independent of the concentration of water. Thus and in are 1 and 0, respectively, and
\( rate= k\left [ \left (CH_{3} \right )_{3}CBr \right ]^{1}\left [ H_{2}O \right ]^{0}=k\left [ \left (CH_{3} \right )_{3}CBr \right ] \tag{14.2.10} \) Because the exponent for the reactant is 1, the reaction is in (CH ) CBr. It is in water because the exponent for [H O] is 0. (Recall that anything raised to the zeroth power equals 1.) Thus the overall reaction order is 1 + 0 = 1. What the reaction orders tell us in practical terms is that doubling the concentration of (CH ) CBr doubles the reaction rate of the hydrolysis reaction, halving the concentration of (CH ) CBr halves the reaction rate, and so on. Conversely, increasing or decreasing the concentration of water has on the reaction rate. (Again, when you work with rate laws, there is no simple correlation between the stoichiometry of the reaction and the rate law. The values of , , and in the rate law be determined experimentally.) Experimental data show that has the value 5.15 × 10 s at 25°C. The rate constant has units of reciprocal seconds (s ) because the reaction rate is defined in units of concentration per unit time (M/s). The units of a rate constant depend on the rate law for a particular reaction. Under conditions identical to those for the -butyl bromide reaction, the experimentally derived differential rate law for the hydrolysis of methyl bromide (CH Br) is as follows:
\( rate=-\dfrac{ \Delta \left [CH_{3} Br \right ]}{\Delta t}= k'\left [ CH_{3} Br \right ] \tag{14.2.11} \) This reaction also has an overall reaction order of 1, but the rate constant in .11 is approximately 10 times smaller than that for -butyl bromide. Thus methyl bromide hydrolyzes about 1 million times more slowly than -butyl bromide, and this information tells chemists how the reactions differ on a molecular level. Frequently, changes in reaction conditions also produce changes in a rate law. In fact, chemists often change reaction conditions to obtain clues about what is occurring during a reaction. For example, when -butyl bromide is hydrolyzed in an aqueous acetone solution containing OH ions rather than in aqueous acetone alone, the differential rate law for the hydrolysis reaction does not change. For methyl bromide, in contrast, the differential rate law becomes rate = ″[CH Br,OH ], with an overall reaction order of 2. Although the two reactions proceed similarly in neutral solution, they proceed very differently in the presence of a base, which again provides clues as to how the reactions differ on a molecular level. Differential rate laws are generally used to describe what is occurring on a molecular level during a reaction, whereas integrated rate laws are used for determining the reaction order and the value of the rate constant from experimental measurements. We present three reactions and their experimentally determined differential rate laws. For each reaction, give the of the rate constant, give the reaction with respect to each reactant, give the , and predict what happens to the reaction rate when the concentration of the first species in each chemical equation is doubled. balanced chemical equations and differential rate laws units of rate constant, reaction orders, and effect of doubling reactant concentration Express the reaction rate as moles per liter per second [mol/(L·s), or M/s]. Then determine the units of each chemical species in the rate law. Divide the units for the reaction rate by the units for all species in the rate law to obtain the units for the rate constant. Identify the exponent of each species in the rate law to determine the reaction order with respect to that species. Sum all exponents to obtain the overall reaction order. Use the mathematical relationships as expressed in the rate law to determine the effect of doubling the concentration of a single species on the reaction rate. The exponent in the rate law is 2, so the reaction is second order in HI. Because HI is the only reactant and the only species that appears in the rate law, the reaction is also second order overall. If the concentration of HI is doubled, the reaction rate will increase from [HI] to (2[HI]) = 4 [HI] . The reaction rate will therefore quadruple. The rate law tells us that the reaction rate is constant and independent of the N O concentration. That is, the reaction is zeroth order in N O and zeroth order overall. Because the reaction rate is independent of the N O concentration, doubling the concentration will have no effect on the reaction rate. The only concentration in the rate law is that of cyclopropane, and its exponent is 1. This means that the reaction is first order in cyclopropane. Cyclopropane is the only species that appears in the rate law, so the reaction is also first order overall. Doubling the initial cyclopropane concentration will increase the reaction rate from [cyclopropane] to 2 [cyclopropane] . This doubles the reaction rate. Exercise Given the following two reactions and their experimentally determined differential rate laws: determine the of the rate constant if time is in seconds, determine the reaction with respect to each reactant, give the , and predict what will happen to the reaction rate when the concentration of the first species in each equation is doubled. Reaction rates are reported either as the over a period of time or as the at a single time. The for a reaction is a mathematical relationship between the reaction rate and the concentrations of species in solution. Rate laws can be expressed either as a , describing the change in reactant or product concentrations as a function of time, or as an , describing the actual concentrations of reactants or products as a function of time. The ( ) of a rate law is a constant of proportionality between the reaction rate and the reactant concentration. The power to which a concentration is raised in a rate law indicates the , the degree to which the reaction rate depends on the concentration of a particular reactant. : \( rate=\dfrac{\Delta \left [ B \right ]}{\Delta t}=-\dfrac{\Delta \left [ A \right ]}{\Delta t} \) : \( rate=k\left [ A \right ]^{m}\left [ B \right ]^{n} \tag{14.2.6} \) Explain why the reaction rate is generally fastest at early time intervals. For the second-order A + B → C, what would the plot of the concentration of C versus time look like during the course of the reaction? Explain the differences between a differential rate law and an integrated rate law. What two components do they have in common? Which form is preferred for obtaining a reaction order and a rate constant? Why? Diffusion-controlled reactions have rates that are determined only by the reaction rate at which two reactant molecules can diffuse together. These reactions are rapid, with second-order rate constants typically on the order of 10 L/(mol·s). Would you expect the reactions to be faster or slower in solvents that have a low viscosity? Why? Consider the reactions H O + OH → 2H O and H O + N(CH ) → H O + HN(CH ) in aqueous solution. Which would have the higher rate constant? Why? What information can you get from the reaction order? What correlation does the reaction order have with the stoichiometry of the overall equation? During the hydrolysis reaction A + H O → B + C, the concentration of A decreases much more rapidly in a polar solvent than in a nonpolar solvent. How do you expect this effect to be reflected in the overall reaction order? Reactant concentrations are highest at the beginning of a reaction. The plot of [C] versus is a curve with a slope that becomes steadily less positive. Faster in a less viscous solvent because the rate of diffusion is higher; the H O /OH reaction is faster due to the decreased relative size of reactants and the higher electrostatic attraction between the reactants. The reaction rate of a particular reaction in which A and B react to make C is as follows: Write a reaction equation that is consistent with this rate law. What is the rate expression with respect to time if 2A are converted to 3C? While commuting to work, a person drove for 12 min at 35 mph, then stopped at an intersection for 2 min, continued the commute at 50 mph for 28 min, drove slowly through traffic at 38 mph for 18 min, and then spent 1 min pulling into a parking space at 3 mph. What was the average rate of the commute? What was the instantaneous rate at 13 min? at 28 min? Why do most studies of chemical reactions use the initial rates of reaction to generate a rate law? How is this initial rate determined? Given the following data, what is the reaction order? Estimate. Predict how the reaction rate will be affected by doubling the concentration of the first species in each equation. Cleavage of C H to produce two CH · radicals is a gas-phase reaction that occurs at 700°C. This reaction is first order, with = 5.46 × 10 s . How long will it take for the reaction to go to 15% completion? to 50% completion? Three chemical processes occur at an altitude of approximately 100 km in Earth’s atmosphere. \( N_{2}^{+}+O_{2}\overset{k_{1}}{\rightarrow} N_{2}+O_{2}^{+}\) \( O_{2}^{+}+O\overset{k_{2}}{\rightarrow} O_{2}+O^{+}\) \( O^{+}+N_{2}\overset{k_{3}}{\rightarrow} N+NO^{+} \) Write a rate law for each elementary reaction. If the rate law for the overall reaction were found to be rate = [N ,O ], which one of the steps is rate limiting? The oxidation of aqueous iodide by arsenic acid to give I and arsenous acid proceeds via the following reaction: Write an expression for the initial rate of decrease of [I ], Δ[I ]/Δ . When the reaction rate of the forward reaction is equal to that of the reverse reaction: / = [H AsO ,I ]/[H AsO ,I ] [H ] . Based on this information, what can you say about the nature of the rate-determining steps for the reverse and the forward reactions? 298 s; 1270 s | 26,913 | 2,084 |
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The common-ion effect is used to describe the effect on an equilibrium involving a substance that adds an ion that is a part of the equilibrium. The solubility products 's are equilibrium constants in hetergeneous equilibria (i.e., between two different phases). If several salts are present in a system, they all ionize in the solution. If the salts contain a common cation or anion, these salts contribute to the concentration of the common ion. Contributions from all salts must be included in the calculation of concentration of the common ion. For example, a solution containing sodium chloride and potassium chloride will have the following relationship: \[\mathrm{[Na^+] + [K^+] = [Cl^-]} \label{1}\] Consideration of or or both leads to the same conclusion. The solubility product expression tells us that the equilibrium concentrations of the cation and the anion are inversely related. That is, as the concentration of the anion increases, the maximum concentration of the cation needed for precipitation to occur decreases—and vice versa—so that is constant. Consequently, the solubility of an ionic compound depends on the concentrations of other salts that contain the same ions. Adding a common cation or anion shifts a solubility equilibrium in the direction predicted by Le Chatelier’s principle. As a result, the solubility of any sparingly soluble salt is almost always decreased by the presence of a soluble salt that contains a common ion. The exceptions generally involve the formation of complex ions, which is discussed later. When \(\ce{NaCl}\) and \(\ce{KCl}\) are dissolved in the same solution, the \(\mathrm{ {\color{Green} Cl^-}}\) ions are to both salts. In a system containing \(\ce{NaCl}\) and \(\ce{KCl}\), the \(\mathrm{ {\color{Green} Cl^-}}\) ions are common ions. \(\mathrm{NaCl \rightleftharpoons Na^+ + {\color{Green} Cl^-}}\)
\(\mathrm{KCl \rightleftharpoons K^+ + {\color{Green} Cl^-}}\)
\(\mathrm{CaCl_2 \rightleftharpoons Ca^{2+} + {\color{Green} 2 Cl^-}}\)
\(\mathrm{AlCl_3 \rightleftharpoons Al^{3+} + {\color{Green} 3 Cl^-}}\)
\(\mathrm{AgCl \rightleftharpoons Ag^+ + {\color{Green} Cl^-}}\) For example, when \(\ce{AgCl}\) is dissolved into a solution already containing \(\ce{NaCl}\) (actually \(\ce{Na+}\) and \(\ce{Cl-}\) ions), the \(\ce{Cl-}\) ions come from the ionization of both \(\ce{AgCl}\) and \(\ce{NaCl}\). Thus, \(\ce{[Cl- ]}\) differs from \(\ce{[Ag+]}\). The following examples show how the concentration of the common ion is calculated. What are \(\ce{[Na+]}\), \(\ce{[Cl- ]}\), \(\ce{[Ca^2+]}\), and \(\ce{[H+]}\) in a solution containing 0.10 M each of \(\ce{NaCl}\), \(\ce{CaCl2}\), and \(\ce{HCl}\)? Due to the conservation of ions, we have \[\mathrm{[Na^+] = [Ca^{2+}] = [H^+] = 0.10\: \ce M}\nonumber.\] but \[\begin{alignat}{3}
&\ce{[Cl- ]} &&= && && \:\textrm{0.10 (due to NaCl)}\nonumber \\
& && && + &&\mathrm{\:0.20\: (due\: to\: CaCl_2)}\nonumber\\
& && && + &&\mathrm{\:0.10\: (due\: to\: HCl)}\nonumber\\
& &&= && &&\mathrm{\:0.40\: M}\nonumber
\end{alignat}\] John poured 10.0 mL of 0.10 M \(\ce{NaCl}\), 10.0 mL of 0.10 M \(\ce{KOH}\), and 5.0 mL of 0.20 M \(\ce{HCl}\) solutions together and then he made the total volume to be 100.0 mL. What is \(\ce{[Cl- ]}\) in the final solution? \[\mathrm{[Cl^-] = \dfrac{0.1\: M\times 10\: mL+0.2\: M\times 5.0\: mL}{100.0\: mL} = 0.020\: M}\nonumber\] states that if an equilibrium becomes unbalanced, the reaction will shift to restore the balance. If a common ion is added to a weak acid or weak base equilibrium, then the equilibrium will shift towards the reactants, in this case the weak acid or base. Consider the lead(II) ion concentration in this solution of PbCl . The balanced reaction is \[ PbCl_{2 (s)} \rightleftharpoons Pb^{2+} _{(aq)} + 2Cl^-_{(aq)}\nonumber\] Defining \(s\) as the concentration of dissolved lead(II) chloride, then: \[[Pb^{2+}] = s\nonumber \] \[[Cl^- ] = 2s\nonumber\] These values can be substituted into the solubility product expression, which can be solved for \(s\): \[\begin{align*} K_{sp} &= [Pb^{2+}] [Cl^-]^2 \\[4pt] &= s \times (2s)^2 \\[4pt] 1.7 \times 10^{-5} &= 4s^3 \\[4pt] s^3 &= \frac{1.7 \times 10^{-5}}{4} \\[4pt] &= 4.25 \times 10^{-6} \\[4pt] s &= \sqrt[3]{4.25 \times 10^{-6}} \\[4pt] &= 1.62 \times 10^{-2}\, mol\ dm^{-3} \end{align*}\] The concentration of lead(II) ions in the solution is 1.62 x 10 M. Consider what happens if sodium chloride is added to this saturated solution. Sodium chloride shares an ion with lead(II) chloride. The chloride ion is to both of them; this is the origin of the term "common ion effect". Look at the original equilibrium expression again: \[ PbCl_2 \; (s) \rightleftharpoons Pb^{2+} \; (aq) + 2Cl^- \; (aq)\nonumber \] What happens to that equilibrium if extra chloride ions are added? According to the position of equilibrium will shift to counter the change, in this case, by removing the chloride ions by making extra solid lead(II) chloride. Of course, the concentration of lead(II) ions in the solution is so small that only a tiny proportion of the extra chloride ions can be converted into solid lead(II) chloride. The lead(II) chloride becomes even , and the concentration of lead(II) ions in the solution . This type of response occurs with any sparingly soluble substance: it is less soluble in a solution which contains any ion which it has in common. This is the common ion effect. If an attempt is made to dissolve some lead(II) chloride in some 0.100 M sodium chloride solution instead of in water, what is the equilibrium concentration of the lead(II) ions this time? As before, define s to be the concentration of the lead(II) ions. \[[Pb^{2+}] = s \label{2}\] The calculations are different from before. This time the concentration of the chloride ions is governed by the concentration of the sodium chloride solution. The number of ions coming from the lead(II) chloride is going to be tiny compared with the 0.100 M coming from the sodium chloride solution. In calculations like this, it can be assumed that the concentration of the common ion is entirely due to the other solution. This simplifies the calculation. So we assume: \[[Cl^- ] = 0.100\; M \label{3}\] The rest of the mathematics looks like this: \begin{equation} \begin{split} K_{sp}& = [Pb^{2+},Cl^-]^2 \\ & = s \times (0.100)^2 \\ 1.7 \times 10^{-5} & = s \times 0.00100 \end{split} \end{equation} therefore: \begin{equation} \begin{split} s & = \dfrac{1.7 \times 10^{-5}}{0.0100} \\ & = 1.7 \times 10^{-3} \, \text{M} \end{split} \label{4} \end{equation} Finally, compare that value with the simple saturated solution: Original solution: \[[Pb^{2+}] = 0.0162 \, M \label{5}\] Solution in 0.100 M NaCl solution: \[ [Pb^{2+}] = 0.0017 \, M \label{6}\] The concentration of the lead(II) ions has decreased by a factor of about 10. If more concentrated solutions of sodium chloride are used, the solubility decreases further. Adding a common ion to a system at equilibrium affects the equilibrium composition, but the ionization constant. Adding a common ion prevents the weak acid or weak base from ionizing as much as it would without the added common ion. The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium. The common ion effect of H O on the ionization of acetic acid The common ion effect suppresses the ionization of a weak base by adding more of an ion that is a product of this equilibrium. Consider the common ion effect of OH on the ionization of ammonia Adding the common ion of hydroxide shifts the reaction towards the left to decrease the stress (in accordance with ), forming more reactants. This decreases the reaction quotient, because the reaction is being pushed towards the left to reach equilibrium. The equilibrium constant, \(K_b=1.8 \times 10^{-5}\), does not change. The reaction is put out of balance, or equilibrium. \[Q_a = \dfrac{[NH_4^+,OH^-]}{[NH_3]}\nonumber \] At first, when more hydroxide is added, the quotient is greater than the equilibrium constant. The reaction then shifts right, causing the denominator to increase, decreasing the reaction quotient and pulling towards equilibrium and causing \(Q\) to decrease towards \(K\). Consider, for example, the effect of adding a soluble salt, such as CaCl , to a saturated solution of calcium phosphate [Ca (PO ) ]. \[\ce{Ca3(PO4)2(s) <=> 3Ca^{2+}(aq) + 2PO^{3−}4(aq)} \label{Eq1}\] We have seen that the solubility of Ca (PO ) in water at 25°C is 1.14 × 10 M ( = 2.07 × 10 ). Thus a saturated solution of Ca (PO ) in water contains \[3 × (1.14 × 10^{−7}\, M) = 3.42 × 10^{−7}\, M\, \ce{Ca^{2+}} \] and \[2 × (1.14 × 10^{−7}\, M) = 2.28 × 10^{−7}\, M\, \ce{PO4^{3−}}\] according to the stoichiometry shown in (neglecting hydrolysis to form HPO ). If CaCl is added to a saturated solution of Ca (PO ) , the Ca ion concentration will increase such that [Ca ] > 3.42 × 10 M, making > . The only way the system can return to equilibrium is for the reaction in to proceed to the left, resulting in precipitation of \(\ce{Ca3(PO4)2}\). This will decrease the concentration of both Ca and PO until = . Adding a common ion decreases solubility, as the reaction shifts toward the left to relieve the stress of the excess product. Adding a common ion to a dissociation reaction causes the equilibrium to shift left, toward the reactants, causing precipitation. Consider the reaction: \[ PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq)\nonumber \] What happens to the solubility of PbCl (s) when 0.1 M NaCl is added? \[K_{sp}=1.7 \times 10^{-5}\nonumber\] \[Q_{sp}= 1.8 \times 10^{-5}\nonumber\] Identify the common ion: Cl Notice: Q > K The addition of NaCl has caused the reaction to shift out of equilibrium because there are more dissociated ions. Typically, solving for the molarities requires the assumption that the solubility of PbCl is equivalent to the concentration of Pb produced because they are in a 1:1 ratio. Because for the reaction is 1.7×10 , the overall reaction would be (s)(2s) = 1.7×10 . Solving the equation for s gives s= 1.62×10 M. The coefficient on Cl is 2, so it is assumed that twice as much Cl is produced as Pb , hence the '2s.' The solubility equilibrium constant can be used to solve for the molarities of the ions at equilibrium. The molarity of Cl added would be 0.1 M because Na and Cl are in a 1:1 ration in the ionic salt, NaCl. Therefore, the overall molarity of Cl would be 2s + 0.1, with 2s referring to the contribution of the chloride ion from the dissociation of lead chloride. \[\begin{eqnarray} Q_{sp} &=& [Pb^{2+},Cl^-]^2\nonumber \\ 1.8 \times 10^{-5} &=& (s)(2s + 0.1)^2 \\ s &=& [Pb^{2+}]\nonumber \\ &=& 1.8 \times 10^{-3} M\nonumber\\ 2s &=& [Cl^-]\nonumber\\ &\approx & 0.1 M \end{eqnarray} \] Notice that the molarity of Pb is lower when NaCl is added. The equilibrium constant remains the same because of the increased concentration of the chloride ion. To simplify the reaction, it can be assumed that [Cl ] is approximately 0.1M since the formation of the chloride ion from the dissociation of lead chloride is so small. The reaction quotient for PbCl is greater than the equilibrium constant because of the added Cl . This therefore shift the reaction left towards equilibrium, causing precipitation and lowering the current solubility of the reaction. Overall, the solubility of the reaction decreases with the added sodium chloride. The common ion effect usually decreases the solubility of a sparingly soluble salt. Calculate the solubility of calcium phosphate [Ca (PO ) ] in 0.20 M CaCl . concentration of CaCl solution solubility of Ca (PO ) in CaCl solution The balanced equilibrium equation is given in the following table. If we let equal the solubility of Ca (PO ) in moles per liter, then the change in [Ca ] is once again +3 , and the change in [PO ] is +2 . We can insert these values into the ICE table. \[Ca_3(PO_4)_{2(s)} \rightleftharpoons 3Ca^{2+}_{(aq)} + 2PO^{3−}_{4(aq)}\] The expression is as follows: Because Ca (PO ) is a sparingly soluble salt, we can reasonably expect that << 0.20. Thus (0.20 + 3 ) M is approximately 0.20 M, which simplifies the expression as follows: \[\begin{align*}K_{\textrm{sp}}=(0.20)^3(2x)^2&=2.07\times10^{-33}
\\[4pt] x^2&=6.5\times10^{-32}
\\[4pt] x&=2.5\times10^{-16}\textrm{ M}\end{align*}\] This value is the solubility of Ca (PO ) in 0.20 M CaCl at 25°C. It is approximately nine orders of magnitude less than its solubility in pure water, as we would expect based on Le Chatelier’s principle. With one exception, this example is identical to Example \(\Page {2}\)—here the initial [Ca ] was 0.20 M rather than 0. Calculate the solubility of silver carbonate in a 0.25 M solution of sodium carbonate. The solubility of silver carbonate in pure water is 8.45 × 10 at 25°C. 2.9 × 10 M (versus 1.3 × 10 M in pure water) The Common Ion Effect in Solubility Products: Adding a common cation or common anion to a solution of a sparingly soluble salt shifts the solubility equilibrium in the direction predicted by Le Chatelier’s principle. The solubility of the salt is almost always by the presence of a common ion. Jim Clark ( ) | 13,302 | 2,085 |
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If we combine the splitting schemes for the 2s and 2p orbitals, we can predict bond order in all of the diatomic molecules and ions composed of elements in the first complete row of the periodic table. Remember that only the valence orbitals of the atoms need be considered; as we saw in the cases of lithium hydride and dilithium, the inner orbitals remain tightly bound and retain their localized atomic character. We now describe examples of systems involving period 2 homonuclear diatomic molecules, such as N , O , and F . We illustrate how to use these points by constructing a molecular orbital energy-level diagram for F . We use the diagram in part (a) in Figure \(\Page {1}\); the = 1 orbitals (σ and σ ) are located well below those of the = 2 level and are not shown. As illustrated in the diagram, the σ and σ molecular orbitals are much lower in energy than the molecular orbitals derived from the 2 atomic orbitals because of the large difference in energy between the 2 and 2 atomic orbitals of fluorine. The lowest-energy molecular orbital derived from the three 2 orbitals on each F is \( \sigma _{2p_{z}} \) and the next most stable are the two degenerate orbitals, \( \pi _{2p_{x}} \) and \( \pi _{2p_{y}} \) For each bonding orbital in the diagram, there is an antibonding orbital, and the antibonding orbital is destabilized by about as much as the corresponding bonding orbital is stabilized. As a result, the \( \sigma ^{\star }_{2p_{z}} \) orbital is higher in energy than either of the degenerate \( \pi _{2p_{x}}^{\star } \) and \( \pi _{2p_{y}}^{\star } \) orbitals. We can now fill the orbitals, beginning with the one that is lowest in energy. Each fluorine has 7 valence electrons, so there are a total of 14 valence electrons in the F molecule. Starting at the lowest energy level, the electrons are placed in the orbitals according to the Pauli principle and Hund’s rule. Two electrons each fill the σ and σ orbitals, 2 fill the \( \sigma _{2p_{z}} \) orbital, 4 fill the two degenerate π orbitals, and 4 fill the two degenerate π orbitals, for a total of 14 electrons. To determine what type of bonding the molecular orbital approach predicts F to have, we must calculate the bond order. According to our diagram, there are 8 bonding electrons and 6 antibonding electrons, giving a bond order of (8 − 6) ÷ 2 = 1. Thus F is predicted to have a stable F–F single bond, in agreement with experimental data. We now turn to a molecular orbital description of the bonding in O . It so happens that the molecular orbital description of this molecule provided an explanation for a long-standing puzzle that could not be explained using other bonding models. To obtain the molecular orbital energy-level diagram for O , we need to place 12 valence electrons (6 from each O atom) in the energy-level diagram shown in part (b) in Figure \(\Page {1}\). We again fill the orbitals according to Hund’s rule and the Pauli principle, beginning with the orbital that is lowest in energy. Two electrons each are needed to fill the σ and σ orbitals, 2 more to fill the \( \sigma _{2p_{z}} \) orbital, and 4 to fill the degenerate \( \pi _{2p_{x}}^{\star } \) and \( \pi _{2p_{y}}^{\star } \) orbitals. According to Hund’s rule, the last 2 electrons must be placed in separate π orbitals with their spins parallel, giving two unpaired electrons. This leads to a predicted bond order of (8 − 4) ÷ 2 = 2, which corresponds to a double bond, in agreement with experimental data (Table 4.5): the O–O bond length is 120.7 pm, and the bond energy is 498.4 kJ/mol at 298 K. None of the other bonding models can predict the presence of two unpaired electrons in O . Chemists had long wondered why, unlike most other substances, liquid O is attracted into a magnetic field. As shown in Figure \(\Page {2}\), it actually remains suspended between the poles of a magnet until the liquid boils away. The only way to explain this behavior was for O to have unpaired electrons, making it paramagnetic, exactly as predicted by molecular orbital theory. This result was one of the earliest triumphs of molecular orbital theory over the other bonding approaches we have discussed. The magnetic properties of O are not just a laboratory curiosity; they are absolutely crucial to the existence of life. Because Earth’s atmosphere contains 20% oxygen, all organic compounds, including those that compose our body tissues, should react rapidly with air to form H O, CO , and N in an exothermic reaction. Fortunately for us, however, this reaction is very, very slow. The reason for the unexpected stability of organic compounds in an oxygen atmosphere is that virtually all organic compounds, as well as H O, CO , and N , have only paired electrons, whereas oxygen has two unpaired electrons. Thus the reaction of O with organic compounds to give H O, CO , and N would require that at least one of the electrons on O change its spin during the reaction. This would require a large input of energy, an obstacle that chemists call a . Consequently, reactions of this type are usually exceedingly slow. If they were not so slow, all organic substances, including this book and you, would disappear in a puff of smoke! For period 2 diatomic molecules to the left of N in the periodic table, a slightly different molecular orbital energy-level diagram is needed because the \( \sigma _{2p_{z}} \) molecular orbital is slightly in energy than the degenerate \( \pi ^{\star }_{np_{x}} \) and \( \pi ^{\star }_{np_{y}} \) orbitals. The difference in energy between the 2 and 2 atomic orbitals increases from Li to F due to increasing nuclear charge and poor screening of the 2 electrons by electrons in the 2 subshell. The bonding interaction between the 2 orbital on one atom and the 2 orbital on the other is most important when the two orbitals have similar energies. This interaction decreases the energy of the σ orbital and increases the energy of the \( \sigma _{2p_{z}} \) orbital. Thus for Li , Be , B , C , and N , the \( \sigma _{2p_{z}} \) orbital is higher in energy than the \( \sigma _{3p_{z}} \) orbitals, as shown in Figure \(\Page {3}\) Experimentally, it is found that the energy gap between the and atomic orbitals as the nuclear charge increases (Figure \(\Page {3}\) ). Thus for example, the \( \sigma _{2p_{z}} \) molecular orbital is at a lower energy than the \( \pi _{2p_{x,y}} \) pair. Completing the diagram for N in the same manner as demonstrated previously, we find that the 10 valence electrons result in 8 bonding electrons and 2 antibonding electrons, for a predicted bond order of 3, a triple bond. Experimental data show that the N–N bond is significantly shorter than the F–F bond (109.8 pm in N versus 141.2 pm in F ), and the bond energy is much greater for N than for F (945.3 kJ/mol versus 158.8 kJ/mol, respectively). Thus the N bond is much shorter and stronger than the F bond, consistent with what we would expect when comparing a triple bond with a single bond. Use a qualitative molecular orbital energy-level diagram to predict the electron configuration, the bond order, and the number of unpaired electrons in S , a bright blue gas at high temperatures. chemical species molecular orbital energy-level diagram, bond order, and number of unpaired electrons Sulfur has a [Ne]3 3 valence electron configuration. To create a molecular orbital energy-level diagram similar to those in Figure \(\Page {1}\) and Figure \(\Page {3}\), we need to know how close in energy the 3 and 3 atomic orbitals are because their energy separation will determine whether the \( \pi _{3p_{x,y}} \) or the \( \sigma _{3p_{z}} \) molecular orbital is higher in energy. Because the – energy gap as the nuclear charge increases (Figure \(\Page {3}\)), the \( \sigma _{3p_{z}} \) molecular orbital will be lower in energy than the \( \pi _{3p_{x,y}} \) pair. The molecular orbital energy-level diagram is as follows: Each sulfur atom contributes 6 valence electrons, for a total of 12 valence electrons. Ten valence electrons are used to fill the orbitals through \( \pi _{3p_{x}} \) and \( \pi _{3p_{y}} \), leaving 2 electrons to occupy the degenerate \( \pi ^{\star }_{3p_{x}} \) and \( \pi ^{\star }_{3p_{y}} \) pair. From Hund’s rule, the remaining 2 electrons must occupy these orbitals separately with their spins aligned. With the numbers of electrons written as superscripts, the electron configuration of S is \( \left ( \sigma _{3s} \right )^{2}\left ( \sigma ^{\star }_{3s} \right )^{2}\left ( \sigma _{3p_{z}} \right )^{2}\left ( \pi _{3p_{x,y}} \right )^{4}\left ( \pi _{3p ^{\star }_{x,y}} \right )^{2} \) with 2 unpaired electrons. The bond order is (8 − 4) ÷ 2 = 2, so we predict an S=S double bond. Use a qualitative molecular orbital energy-level diagram to predict the electron configuration, the bond order, and the number of unpaired electrons in the peroxide ion (O ). \( \left ( \sigma _{2s} \right )^{2}\left ( \sigma ^{\star }_{2s} \right )^{2}\left ( \sigma _{2p_{z}} \right )^{2}\left ( \pi _{2p_{x,y}} \right )^{4}\left ( \pi _{2p ^{\star }_{x,y}} \right )^{4} \) bond order of 1; no unpaired electrons Diatomic molecules with two different atoms are called heteronuclear diatomic molecules. When two nonidentical atoms interact to form a chemical bond, the interacting atomic orbitals do not have the same energy. If, for example, element B is more electronegative than element A (χ > χ ), the net result is a “skewed” molecular orbital energy-level diagram, such as the one shown for a hypothetical A–B molecule in Figure \(\Page {4}\). The atomic orbitals of element B are uniformly lower in energy than the corresponding atomic orbitals of element A because of the enhanced stability of the electrons in element B. The molecular orbitals are no longer symmetrical, and the energies of the bonding molecular orbitals are more similar to those of the atomic orbitals of B. Hence the electron density of bonding electrons is likely to be closer to the more electronegative atom. In this way, molecular orbital theory can describe a polar covalent bond. Nitric oxide (NO) is an example of a heteronuclear diatomic molecule. The reaction of O with N at high temperatures in internal combustion engines forms nitric oxide, which undergoes a complex reaction with O to produce NO , which in turn is responsible for the brown color we associate with air pollution. Recently, however, nitric oxide has also been recognized to be a vital biological messenger involved in regulating blood pressure and long-term memory in mammals. Because NO has an odd number of valence electrons (5 from nitrogen and 6 from oxygen, for a total of 11), its bonding and properties cannot be successfully explained by either the Lewis electron-pair approach or valence bond theory. The molecular orbital energy-level diagram for NO (Figure \(\Page {13}\)) shows that the general pattern is similar to that for the O molecule (Figure \(\Page {11}\)). Because 10 electrons are sufficient to fill all the bonding molecular orbitals derived from 2 atomic orbitals, the 11th electron must occupy one of the degenerate π orbitals. The predicted bond order for NO is therefore (8-3) ÷ 2 = 2 1/2 . Experimental data, showing an N–O bond length of 115 pm and N–O bond energy of 631 kJ/mol, are consistent with this description. These values lie between those of the N and O molecules, which have triple and double bonds, respectively. As we stated earlier, molecular orbital theory can therefore explain the bonding in molecules with an odd number of electrons, such as NO, whereas Lewis electron structures cannot. Note that electronic structure studies show the ground state configuration of \(\ce{NO}\) to be \( \left ( \sigma _{2s} \right)^{2}\left ( \sigma ^{\star }_{2s} \right)^{2}\left ( \pi _{2p_{x,y}} \right)^{4} \left ( \sigma _{2p_{z}} \right)^{2} \left ( \pi _{2p ^{\star }_{x,y}} \right)^{1} \) in order of increasing energy. Hence, the \( \pi _{2p_{x,y}}\) orbitals are lower in energy than the \(\sigma _{2p_{z}}\) orbital. This is because the \(\ce{NO}\) molecule is near the transition of flipping energies levels observed in homonuclear diatomics where the sigma bond drops below the pi bond ( Molecular orbital theory can also tell us something about the of \(NO\). As indicated in the energy-level diagram in Figure \(\Page {13}\), NO has a single electron in a relatively high-energy molecular orbital. We might therefore expect it to have similar reactivity as alkali metals such as Li and Na with their single valence electrons. In fact, \(NO\) is easily oxidized to the \(NO^+\) cation, which is isoelectronic with \(N_2\) and has a bond order of 3, corresponding to an N≡O triple bond. Molecular orbital theory is also able to explain the presence of lone pairs of electrons. Consider, for example, the HCl molecule, whose Lewis electron structure has three lone pairs of electrons on the chlorine atom. Using the molecular orbital approach to describe the bonding in HCl, we can see from Figure \(\Page {6}\) that the 1 orbital of atomic hydrogen is closest in energy to the 3 orbitals of chlorine. Consequently, the filled Cl 3 atomic orbital is not involved in bonding to any appreciable extent, and the only important interactions are those between the H 1 and Cl 3 orbitals. Of the three orbitals, only one, designated as 3 , can interact with the H 1 orbital. The 3 and 3 atomic orbitals have no net overlap with the 1 orbital on hydrogen, so they are not involved in bonding. Because the energies of the Cl 3 , 3 , and 3 orbitals do not change when HCl forms, they are called . A nonbonding molecular orbital occupied by a pair of electrons is the molecular orbital equivalent of a lone pair of electrons. By definition, electrons in nonbonding orbitals have no effect on bond order, so they are not counted in the calculation of bond order. Thus the predicted bond order of HCl is (2 − 0) ÷ 2 = 1. Because the σ bonding molecular orbital is closer in energy to the Cl 3 than to the H 1 atomic orbital, the electrons in the σ orbital are concentrated closer to the chlorine atom than to hydrogen. A molecular orbital approach to bonding can therefore be used to describe the polarization of the H–Cl bond to give \( H^{\delta +} -- Cl^{\delta -} \). Use a “skewed” molecular orbital energy-level diagram like the one in Figure \(\Page {4}\) to describe the bonding in the cyanide ion (CN ). What is the bond order? chemical species “skewed” molecular orbital energy-level diagram, bonding description, and bond order The CN ion has a total of 10 valence electrons: 4 from C, 5 from N, and 1 for the −1 charge. Placing these electrons in an energy-level diagram like Figure \(\Page {4}\) fills the five lowest-energy orbitals, as shown here: Because \(\chi_N > \chi_C\), the atomic orbitals of N (on the right) are lower in energy than those of C. The resulting valence electron configuration gives a predicted bond order of (8 − 2) ÷ 2 = 3, indicating that the CN ion has a triple bond, analogous to that in N . Use a qualitative molecular orbital energy-level diagram to describe the bonding in the hypochlorite ion (OCl ). What is the bond order? All molecular orbitals except the highest-energy σ* are filled, giving a bond order of 1. Although the molecular orbital approach reveals a great deal about the bonding in a given molecule, the procedure quickly becomes computationally intensive for molecules of even moderate complexity. Furthermore, because the computed molecular orbitals extend over the entire molecule, they are often difficult to represent in a way that is easy to visualize. Therefore we do not use a pure molecular orbital approach to describe the bonding in molecules or ions with more than two atoms. Instead, we use a valence bond approach and a molecular orbital approach to explain, among other things, the concept of resonance, which cannot adequately be explained using other methods. Molecular orbital energy-level diagrams for diatomic molecules can be created if the electron configuration of the parent atoms is known, following a few simple rules. Most important, the number of molecular orbitals in a molecule is the same as the number of atomic orbitals that interact. The difference between bonding and antibonding molecular orbital combinations is proportional to the overlap of the parent orbitals and decreases as the energy difference between the parent atomic orbitals increases. With such an approach, the electronic structures of virtually all commonly encountered , molecules with two identical atoms, can be understood. The molecular orbital approach correctly predicts that the O molecule has two unpaired electrons and hence is attracted into a magnetic field. In contrast, most substances have only paired electrons. A similar procedure can be applied to molecules with two dissimilar atoms, called , using a molecular orbital energy-level diagram that is skewed or tilted toward the more electronegative element. Molecular orbital theory is able to describe the bonding in a molecule with an odd number of electrons such as NO and even to predict something about its chemistry. ( ) | 17,290 | 2,086 |
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It is important to remember that an equilibrium constant is always tied to a specific chemical equation, and if we write the equation in reverse or multiply its coefficients by a common factor, the value of \(K\) will change. Fortunately, the rules are very simple: Here are some of the possibilities for the reaction involving the equilibrium between gaseous water and its elements: \(K_p = \dfrac{P_{H_2O}^2}{P_{H_2}^2P_{O_2}}\) \(K_p = \dfrac{P_{H_2O}^{10}}{P_{H_2}^{10}P_{O_2}^5}\) \(= \left(\dfrac{P_{H_2O}^2}{P_{H_2}^2P_{O_2}}\right)^{5}\) \(K_p = \dfrac{P_{H_2O}}{P_{H_2}P_{O_2}^{1/2}}\) \(= \left(\dfrac{P_{H_2O}^2}{P_{H_2}^2P_{O_2}}\right)^{1/2}\) \(K_p = \dfrac{P_{H_2}P_{O_2}^{1/2}}{P_{H_2O}}\) \(= \left(\dfrac{P_{H_2O}^2}{P_{H_2}^2P_{O_2}}\right)^{-1/2}\) Many chemical changes can be regarded as the sum or difference of two or more other reactions. If we know the equilibrium constants of the individual processes, we can easily calculate that for the overall reaction according to the following rule: T Calculate the value of \(K\) for the reaction \[CaCO_{3(s)} + H^+_{(aq)} \rightarrow Ca^{2+}_{(aq)} + HCO^–_{3(aq)}\] given the following equilibrium constants: \(CaCO_{3(s)} \rightleftharpoons Ca^{2+}_{(aq)} + CO^{2–}_{3(aq)}\) \(K_1 = 10^{–6.3}\) \(HCO^–_{3(aq)} \rightleftharpoons H^+_{(aq)} + CO^{2–}_{3(aq)}\) \(K_2 = 10^{–10.3}\) The net reaction is the sum of reaction 1 and the reverse of reaction 2: \(CaCO_{3(s)} \rightleftharpoons Ca^{2+}_{(aq)} + CO^{2–}_{3(aq)}\) \(K_1 = 10^{–6.3}\) \( H^+_{(aq)} + CO^{2–}_{3(aq)} \rightleftharpoons HCO^–_{3(aq)} \) \(CaCO_{3(s)} + H^+_{(aq)} \rightarrow Ca^{2+}_{(aq)} + HCO^–_{3(aq)}\) t \(K_p = 4.5 \times 10^{15}\) a Relationships Involving Equilibrium Constants: For reactions that involve species in solution, the concentrations used in equilibrium calculations are usually expressed in moles/liter. For gases, however, the concentrations are usually expressed in terms of partial pressures rather than molarity, where the standard state is 1 atm of pressure. The symbol \(K_p\) is used to denote equilibrium constants calculated from partial pressures. For the general reaction \(aA+bB \rightleftharpoons cC+dD\), in which all the components are gases, the equilibrium constant expression can be written as the ratio of the partial pressures of the products and reactants (each raised to its coefficient in the chemical equation): \[K_p=\dfrac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b} \label{15.3.1}\] Thus \(K_p\) for the decomposition of \(N_2O_4\) is as follows: \[K_p=\dfrac{(P_{NO_2})^2}{P_{N_2O_4}} \label{15.3.2}\] Like K, \(K_p\) is a unitless quantity because the quantity that is actually used to calculate it is an “effective pressure,” the ratio of the measured pressure to a standard state of 1 bar (approximately 1 atm), which produces a unitless quantity.The “effective pressure” is called the , just as activity is the effective concentration. Because partial pressures are usually expressed in atmospheres or mmHg, the molar concentration of a gas and its partial pressure do not have the same numerical value. Consequently, the numerical values of \(K\) and \(K_p\) are usually different. They are, however, related by the ideal gas constant (\(R\)) and the absolute temperature (\(T\)): \[ \color{red} {K_p = K_c(RT)^{Δn} \label{15.3.3}}\] where \(K\) is the equilibrium constant expressed in units of concentration and \(Δn\) is the difference between the numbers of moles of gaseous products and gaseous reactants (\(n_p − n_r\)). The temperature is expressed as the absolute temperature in Kelvin. According to Equation 15.3.3, \(K_p = K\) only if the moles of gaseous products and gaseous reactants are the same (i.e., \(Δn = 0\)). For the decomposition of \(N_2O_4\), there are 2 mol of gaseous product and 1 mol of gaseous reactant, so \(Δn = 1\). Thus, for this reaction, \[K_p = K(RT)^1 = K_cRT \label{15.3.4}\] The equilibrium constant for the reaction of nitrogen and hydrogen to give ammonia is 0.118 at 745 K. The balanced equilibrium equation is as follows: \[N_{2(g)}+3H_{2(g)} \rightleftharpoons 2NH_{3(g)}\] What is \(K_p\) for this reaction at the same temperature? : equilibrium equation, equilibrium constant, and temperature : \(K_p\) : Use the coefficients in the balanced chemical equation to calculate \(Δn\). Then use Equation 15.3.3 to calculate \(K\) from \(K_p\). : This reaction has 2 mol of gaseous product and 4 mol of gaseous reactants, so \(\Delta{n} = (2 − 4) = −2\). We know \(K\), and \(T = 745\; K\). Thus, from Equation 15.2.15, we have the following: \[K_p=K(RT)^{−2}=\dfrac{K}{(RT)^2}=\dfrac{0.118}{\{ [0.08206(L \cdot atm)/(mol \cdot K),745\; K]\}^2}=3.16 \times 10^{−5}\] Because \(K_p\) is a unitless quantity, the answer is \( K_p = 3.16 \times 10^{−5}\). Calculate \(K_p\) for the reaction \[2SO_{2(g)}+O_{2(g)} \rightleftharpoons 2SO_{3(g)}\] at 527°C, if \(K = 7.9 \times 10^4\) at this temperature. : \(K_p = 1.2 \times 10^3\) Converting Kc to Kp: Chemists frequently need to know the equilibrium constant for a reaction that has not been previously studied. In such cases, the desired reaction can often be written as the sum of other reactions for which the equilibrium constants are known. The equilibrium constant for the unknown reaction can then be calculated from the tabulated values for the other reactions. To illustrate this procedure, let’s consider the reaction of \(N_2\) with \(O_2\) to give \(NO_2\). This reaction is an important source of the \(NO_2\) that gives urban smog its typical brown color. The reaction normally occurs in two distinct steps. In the first reaction (1), \(N_2\) reacts with \(O_2\) at the high temperatures inside an internal combustion engine to give \(NO\). The released \(NO\) then reacts with additional \(O_2\) to give \(NO_2\) (2). The equilibrium constant for each reaction at 100°C is also given. Summing reactions (1) and (2) gives the overall reaction of \(N_2\) with \(O_2\): The equilibrium constant expressions for the reactions are as follows: \[K_1=\dfrac{[NO]^2}{[N_2,O_2]}\;\;\; K_2=\dfrac{[NO_2]^2}{[NO]^2[O_2]}\;\;\; K_3=\dfrac{[NO_2]^2}{[N_2,O_2]^2} \label{15.3.5}\] What is the relationship between \(K_1\), \(K_2\), and \(K_3\), all at 100°C? The expression for \(K_1\) has \([NO]^2\) in the numerator, the expression for \(K_2\) has \([NO]^2\) in the denominator, and \([NO]^2\) does not appear in the expression for \(K_3\). Multiplying \(K_1\) by \(K_2\) and canceling the \([NO]^2\) terms, \[ K_1K_2=\dfrac{\cancel{[NO]^2}}{[N_2,O_2]} \times \dfrac{[NO_2]^2}{\cancel{[NO]^2}[O_2]}=\dfrac{[NO_2]^2}{[N_2,O_2]^2}=K_3 \label{15.3.6}\] Thus the product of the equilibrium constant expressions for \(K_1\) and \(K_2\) is the same as the equilibrium constant expression for \(K_3\): \[K_3 = K_1K_2 = (2.0 \times 10^{−25})(6.4 \times 10^9) = 1.3 \times 10^{−15} \label{15.3.7}\] The equilibrium constant for a reaction that is the sum of two or more reactions is equal to the product of the equilibrium constants for the individual reactions. In contrast, recall that according to , \(ΔH\) for the sum of two or more reactions is the sum of the ΔH values for the individual reactions. To determine \(K\) for a reaction that is the of two or more reactions, add the reactions but the equilibrium constants. The following reactions occur at 1200°C: Calculate the equilibrium constant for the following reaction at the same temperature. : two balanced equilibrium equations, values of \(K\), and an equilibrium equation for the overall reaction : equilibrium constant for the overall reaction : Arrange the equations so that their sum produces the overall equation. If an equation had to be reversed, invert the value of \(K\) for that equation. Calculate \(K\) for the overall equation by multiplying the equilibrium constants for the individual equations. : The key to solving this problem is to recognize that reaction 3 is the sum of reactions 1 and 2: \[CO_{(g)}+ \cancel{3H_{2(g)}} \rightleftharpoons \cancel{CH_{4(g)}} + H_2O_{(g)}\] \[\cancel{CH_{4(g)}} +2H_2S_{(g)} \rightleftharpoons CS_{2(g)} + \cancel{3H_{2(g)}} + H_{2(g)}\] \[ CO_{(g)} + 3H_{2(g)} \rightleftharpoons CS_{2(g)}+H_2O_{(g)}+H_{2(g)}\] The values for \(K_1\) and \(K_2\) are given, so it is straightforward to calculate \(K_3\): \[K_3 = K_1K_2 = (9.17 \times 10^{−2})(3.3 \times 10^4) = 3.03 \times 10^3\] In the first of two steps in the industrial synthesis of sulfuric acid, elemental sulfur reacts with oxygen to produce sulfur dioxide. In the second step, sulfur dioxide reacts with additional oxygen to form sulfur trioxide. The reaction for each step is shown, as is the value of the corresponding equilibrium constant at 25°C. Calculate the equilibrium constant for the overall reaction at this same temperature. : \(K_3 = 1.1 \times 10^{66}\) When the products and reactants of an equilibrium reaction form a single phase, whether gas or liquid, the system is a homogeneous equilibrium. In such situations, the concentrations of the reactants and products can vary over a wide range. In contrast, a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibrium, such as the reaction of a gas with a solid or liquid. Because the molar concentrations of pure liquids and solids normally do not vary greatly with temperature, their concentrations are treated as constants, which allows us to simplify equilibrium constant expressions that involve pure solids or liquids.The reference states for pure solids and liquids are those forms stable at 1 bar (approximately 1 atm), which are assigned an . (Recall that the density of water, and thus its volume, changes by only a few percentage points between 0°C and 100°C.) Consider the following reaction, which is used in the final firing of some types of pottery to produce brilliant metallic glazes: \[CO_{2(g)}+C_{(s)} \rightleftharpoons 2CO_{(g)} \label{15.3.8}\] The glaze is created when metal oxides are reduced to metals by the product, carbon monoxide. The equilibrium constant expression for this reaction is as follows: \[K=\dfrac{[CO]^2}{[CO_2,C]} \label{15.3.9}\] Because graphite is a solid, however, its molar concentration, determined from its density and molar mass, is essentially constant and has the following value: \[ [C] =\dfrac{2.26 \cancel{g}/{\cancel{cm^3}}}{12.01\; \cancel{g}/mol} \times 1000 \; \cancel{cm^3}/L = 188 \; mol/L = 188\;M \label{15.3.10}\] We can rearrange Equation 15.3.8 so that the constant terms are on one side: \[ K[C]=K(188)=\dfrac{[CO]^2}{[CO_2]} \label{15.3.11}\] Incorporating the constant value of \([C]\) into the equilibrium equation for the reaction in Equation 15.3.7, \[K'=\dfrac{[CO]^2}{[CO_2]} \label{15.3.12}\] The equilibrium constant for this reaction can also be written in terms of the partial pressures of the gases: \[K_p=\dfrac{(P_{CO})^2}{P_{CO_2}} \label{15.3.13}\] Incorporating all the constant values into \(K′\) or \(K_p\) allows us to focus on the substances whose concentrations change during the reaction. Although the concentrations of pure liquids or solids are not written explicitly in the equilibrium constant expression, these substances must be present in the reaction mixture for chemical equilibrium to occur. Whatever the concentrations of \(CO\) and \(CO_2\), the system described in Equation 15.3.7 will reach chemical equilibrium only if a stoichiometric amount of solid carbon or excess solid carbon has been added so that some is still present once the system has reached equilibrium. As shown in Figure \(\Page {1}\), it does not matter whether 1 g or 100 g of solid carbon is present; in either case, the composition of the gaseous components of the system will be the same at equilibrium. Write each expression for \(K\), incorporating all constants, and \(K_p\) for the following equilibrium reactions. : balanced equilibrium equations : expressions for \(K\) and \(K_p\) : Find \(K\) by writing each equilibrium constant expression as the ratio of the concentrations of the products and reactants, each raised to its coefficient in the chemical equation. Then express \(K_p\) as the ratio of the partial pressures of the products and reactants, each also raised to its coefficient in the chemical equation. This reaction contains a pure solid (\(PCl_5\)) and a pure liquid (\(PCl_3\)). Their concentrations do not appear in the equilibrium constant expression because they do not change significantly. So \[K=\dfrac{1}{[Cl_2]}\] and \[K_p=\dfrac{1}{P_{Cl_2}}\] This reaction contains two pure solids (\(Fe_3O_4\) and \(Fe\)), which do not appear in the equilibrium constant expressions. The two gases do, however, appear in the expressions: \[K=\dfrac{[H_2O]^4}{[H_2]^4}\] and \[K_p=\dfrac{(P_{H_2O})^4}{(P_{H_2})^4}\] Write the expressions for \(K\) and \(K_p\) for the following reactions. : For reactions carried out in solution, the concentration of the solvent is omitted from the equilibrium constant expression even when the solvent appears in the balanced chemical equation for the reaction. The concentration of the solvent is also typically much greater than the concentration of the reactants or products (recall that pure water is about 55.5 M, and pure ethanol is about 17 M). Consequently, the solvent concentration is essentially constant during chemical reactions, and the solvent is therefore treated as a pure liquid. The equilibrium constant expression for a reaction contains only those species whose concentrations could change significantly during the reaction. The concentrations of pure solids, pure liquids, and solvents are omitted from equilibrium constant expressions because they do not change significantly during reactions when enough is present to reach equilibrium. An equilibrated system that contains products and reactants in a single phase is a homogeneous equilibrium; a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibrium. For gases, the equilibrium constant expression can be written as the ratio of the partial pressures of the products to the partial pressures of the reactants, each raised to a power matching its coefficient in the chemical equation. An equilibrium constant calculated from partial pressures (\(K_p\)) is related to \(K\) by the ideal gas constant (\(R\)), the temperature (\(T\)), and the change in the number of moles of gas during the reaction. An equilibrium system that contains products and reactants in a single phase is a homogeneous equilibrium; a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibrium. When a reaction can be expressed as the sum of two or more reactions, its equilibrium constant is equal to the product of the equilibrium constants for the individual reactions. | 14,823 | 2,087 |
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If we plot the partial pressure of one component, \(P_1\), above a mixture with a mole fraction \(x_1\), we should get a straight line with a slope of \(P^*_1\) (Raoult's law). Above non-ideal solutions the graph will no longer be a straight line but a curve. However towards \(x_1=1\) the curve typically approaches the Raoult line. On the other extreme, there often is a more or less linear region as well, (Figure 24.5.1
). This means that we can identify two laws: \[P_1 = K_H x_1 \nonumber \] \[P_1 = P^*_1 x_1 \nonumber \] This implies that the straight line that indicates the Henry expression will intersect the y-axis at \(x=1\) (pure compound) at a point than \(P^*\). For \(x \rightarrow 0\) (low concentrations) we can speak of component 1 being the (the minority component). At the other end \(x \rightarrow 1\) it plays the role of the (majority component). Another thing to note is that \(P^*\) is a property of , the value of \(K_H\) by contrast is a property of the , so it needs to be measured for each solute-solvent combination. As you can see we have a description for both the high and the low end, but not in the middle. In general, the more modest the deviations from ideality the larger the range of validity of the two limiting laws. The way to determine \(K_H\) would be to actually determine vapor pressures. How about the other component? Do we need to measure them too? Fortunately we can use thermodynamics to answer this question with no. There is a handy expression that saves us the trouble. | 1,555 | 2,088 |
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The modern periodic table has evolved through a long history of attempts by chemists to arrange the elements according to their properties as an aid in predicting chemical behavior. One of the first to suggest such an arrangement was the German chemist Johannes Dobereiner (1780–1849), who noticed that many of the known elements could be grouped in triads —for example, chlorine, bromine, and iodine; or copper, silver, and gold. Dobereiner proposed that all elements could be grouped in such triads, but subsequent attempts to expand his concept were unsuccessful. We now know that portions of the periodic table—the block in particular—contain triads of elements with substantial similarities. The middle three members of most of the other columns, such as sulfur, selenium, and tellurium in group 16 or aluminum, gallium, and indium in Group 13, also have remarkably similar chemistry. By the mid-19th century, the atomic masses of many of the elements had been determined. The English chemist John Newlands (1838–1898), hypothesizing that the chemistry of the elements might be related to their masses, arranged the known elements in order of increasing atomic mass and discovered that every seventh element had similar properties (Figure \(\Page {1}\)). Newlands therefore suggested that the elements could be classified into octaves Unfortunately, Newlands’s “law of octaves” did not seem to work for elements heavier than calcium, and his idea was publicly ridiculed. At one scientific meeting, Newlands was asked why he didn’t arrange the elements in alphabetical order instead of by atomic mass, since that would make just as much sense! Actually, Newlands was on the right track—with only a few exceptions, atomic mass does increase with atomic number, and similar properties occur every time a set of subshells is filled. Despite the fact that Newlands’s table had no logical place for the -block elements, he was honored for his idea by the Royal Society of London in 1887. John Alexander Reina Newlands was an English chemist who worked on the development of the periodic table. He noticed that elemental properties repeated every seventh (or multiple of seven) element, as musical notes repeat every eighth note. The periodic table achieved its modern form through the work of the German chemist Julius Lothar Meyer (1830–1895) and the Russian chemist Dimitri Mendeleev (1834–1907), both of whom focused on the relationships between atomic mass and various physical and chemical properties. In 1869, they independently proposed essentially identical arrangements of the elements. Meyer aligned the elements in his table according to periodic variations in simple atomic properties, such as “atomic volume” (Figure \(\Page {2}\)), which he obtained by dividing the atomic mass (molar mass) in grams per mole by the density (\(\rho\)) of the element in grams per cubic centimeter. This property is equivalent to what is today defined as molar volume, t (measured in cubic centimeters per mole): \[ \dfrac{molar\; mass\left ( \cancel{g}/mol \right )}{density\left ( \cancel{g}/cm^{3} \right )}=molar\; volume\left ( cm^{3}/mol \right ) \label{7.1.1} \] As shown in Figure \(\Page {2}\), the alkali metals have the highest molar volumes of the solid elements. In Meyer’s plot of atomic volume versus atomic mass, the nonmetals occur on the rising portion of the graph, and metals occur at the peaks, in the valleys, and on the downslopes. When his family’s glass factory was destroyed by fire, Mendeleev moved to St. Petersburg, Russia, to study science. He became ill and was not expected to recover, but he finished his PhD with the help of his professors and fellow students. In addition to the periodic table, another of Mendeleev’s contributions to science was an outstanding textbook, , which was used for many years. | 3,860 | 2,089 |
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The molecules of hydrogen can exist in two forms depending on the spins on the two hydrogen nuclei. If both the nuclear spins are parallel, the molecule is called ortho and if the spins are antiparallel, it is referred to as para (in disubstituted benzene, refers to the two groups at two opposite ends, while in , they are adjacent or “parallel” to each other). The spin on the hydrogen nucleus has a magnitude of \(½ ħ\). The presence of nuclear spins leads to very interesting consequences for the populations of the rotational states and on a macroscopic scale, has consequences on measured entropies and heat capacities as well. The total partition function of \(\ce{H_2}\) can be written as: \[q_\text{tot} = q_\text{elec} q_\text{vib} q_\text{rot} q_\text{trans} q_\text{nuc} \nonumber \] where, the subscripts refer to the respective motions. After “half” a rotation, the nuclei are superimposed on each other. Since a proton is a spin half nucleus, the total wave function must be antisymmetic with respect to the exchange of the particles. i.e.: \[ \psi(1,2) = - \psi (2,1) \nonumber \] The translational motion refers to the motion of the molecular center of mass and has no influence on the symmetry of the nuclear wave function. Vibrational motion depends on the magnitude of the internuclear distance and has no effect on the particle exchange. The electronic motion also has no effect on the symmetry properties of the nuclear wave function. Therefore, the product of the nuclear spin and rotational wave functions must be antisymmetric with respect to the particle exchange. For the nuclear spin functions, there are four combinations. One combination is a singlet: \[ | \psi_{\nu,s} \rangle = \alpha(1) \beta(2) - \alpha (2) \beta (1) \nonumber \] And the other three combinations are the three states of a triplet: \[ \begin{align*} | \psi_{\nu,s} \rangle &= \alpha(1) \alpha(2) \\[4pt] | \psi_{\nu,s} \rangle &= \alpha(1) \beta(2) + \alpha (2) \beta (1) \\[4pt] | \psi_{\nu,s} \rangle &= \beta(1) \beta(2) \end{align*} \] The rotational wavefunctions (\(ψ_r\)) are given in terms of the associated Legendre polynomials \(P^{|m|}_l (x)\) where \(x = \cos θ\): \[ | ψ_r \rangle = e^{im\phi} P^{|m|}_l (\cos \theta) \label{3.91} \] with: \[\begin{align} P_{l}^{(m)}(x) &=\left(1-x^{2}\right)^{| m / 2} \frac{d^{|m|} P_{l}(x)}{d x^{m |}} \\[4pt] \quad P_{l}(x) &=\frac{1}{2^{l} l !} \frac{d^{l}\left(x^{2}-1\right)^{l}}{d x^{l}} \label{3.92} \end{align} \] When the nuclei are interchanged, \(θ\) becomes \(π - θ\) and \(φ\) is changed to \(φ + π\). The polynomials change as: \[P_{l}(-x)=(-1)^{\prime} P_{l}(x) ; \quad P_{l}^{|m|}(-x)=(-1)^{l-|m|||} P_{l}^{|m|}(x) \label{3.93} \] The exponential function changes as: \[e^{i m(\phi+\pi)}=e^{i m \pi} e^{i m \varphi}=(-1)^{|m|} e^{i m \varphi} \label{3.94} \] Therefore, the rotational wave function changes as: \[ \begin{align} P_l(-x) &= (-1)^l P_l(x) \\[4pt] P^{|m|}_l(-x) &= (-1)^{l-|m|} P^{|m|}_l(x) \label{3.95} \end{align} \] Combining the nuclear spin and the rotational parts, we see that, the product \(\psi_{r}, \psi_{m}\) must be antisymmetrical (with respect to the exchange of nuclei) for half integral nuclear spins and symmetrical for integral spins. To accomplish this, the singlet nuclear states (para) must be combined with the even rotational functions and the triplet nuclear states must be combined with the odd rotational states. The rotational partition functions for ortho and para hydrogens are, thus: \[q_{\text{ortho}}=q_{\nu, t} q_{r, \text{odd}}=3 \sum_{J=1,3,5, \ldots}(2 J+1) e^{-J(J+1) \Theta_{R} / T} \label{3.97} \] and: \[q_{\text{para}}=q_{\text {\nu,s }} q_{r, \text{even}}=1 \sum_{J=0,2,4, \ldots}(2 J+1) e^{-J(J+1) \Theta_{R} / T} \label{3.98} \] where \(Θ_R\) is the rotational temperature defined previously. The total partition function consisting both ortho and para hydrogens is given by: \[q_{\text{rot}, nu}=1 \sum_{j=0.2 .4}(2 J+1) e^{-J(J+1) \Theta_{k} / T}+3 \sum_{j=1,3,5, n}(2 J+1) e^{-J(J+1) \Theta_{R} / T} \label{3.99} \] The ratio of ortho to para hydrogens at thermal equilibrium is given by: \[\frac{N_{o}}{N_{p}}=\frac{3 \sum_{j=1,2,3 \ldots}(2 J+1) e^{-J(J+1) \Theta_{R} / T}}{\sum_{j=0} 2(2 J+1) e^{-J(J+1) \Theta_{R} / T}} \label{3.100} \] At high temperature, the two summations become equal and therefore, the high temperature limit of \(N_{o} / N_{p}\) is 3. At low temperature, the ratio becomes: \[\frac{N_{o}}{N_{p}}=\frac{3\left(3 e^{-2 \Theta_{k} / T}+\ldots \ldots\right)}{1\left(1+5 e^{-5 \Theta_{n} / T} + \ldots \right)} \rightarrow 0, \text { as } T \rightarrow 0 \label{3.101} \] A good experimental verification of the above analysis is a comparison between the calculated rotational heat capacities at constant volume (\(C_V\)) rot, nu (calculated from: \[C_V= \dfrac{\partial \langle E \rangle }{\partial T} \nonumber \] where: \[\langle E \rangle = \dfrac{\partial \ln q_{\text{rot},\nu} }{\partial \beta} \nonumber \] The heat capacities are shown as a function of temperature in Figure 18.10.1
. The purple curve marked 3:1 gives the experimental data, the curve eq represents the data for an equilibrated mixture of o- and p- at a given temperature. The curves o- and p- represent the heat capacities of pure o- and pure p- hydrogens calculated from the o- and p- partition functions given by Equations \(\ref{3.97}\) and \(\ref{3.98}\) respectively. Initially it was a puzzle as to why the experimental data differs from the calculated values. In fact, the experimental data seemed to agree very well with the following equation: \[ (C_V)_{\text{rot},\nu} = \dfrac{3}{4} (C_V)_{\text{rot},\nu} (\text{ortho}) + \dfrac{1}{4} (C_V)_{\text{rot},\nu} (\text{para}) \label{3.102} \] The reason for this is that, when \(\ce H_2\) is cooled down from a higher temperature, the ortho:para ratio continues to remain 0.75 / 0.25 (the high temperature value) because the \(\text{ortho} \rightarrow \text{para}\) interconversion rate is very very small and we do not reach the equilibrium composition unless a catalyst such as activated charcoal is added to the gas mixture. Equation \(\ref{3.102}\) corresponds to a “frozen high temperature mixture” of ortho:para hydrogens. In the presence of the catalyst, the experiments also give the curve labeled as eq in the graph. This in indeed a very nice case where the experiments support not only the methods of statistical thermodynamics but also of the antisymmetry principle for bosons and fermions. If we consider the case of \(\ce{^{16}O2}\), where the nuclear spins are zero, the rotational wave function has to be symmetric as only symmetric wave functions are permitted for bosons. Thus, only even rotational states contribute to the partition function: \[ q_{\text{rot}, V} (^{16}\ce{O_2}) = \sum_{J=0,2,4,\ldots} (2J+1) e^{-J(J+1) \theta_R/T} \nonumber \] | 6,869 | 2,092 |
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An element is the simplest form of matter that cannot be split into simpler substances or built from simpler substances by any ordinary chemical or physical method. There are 118 elements known to us, out of which 92 are naturally occurring, while the rest have been prepared artificially. Elements are further classified into metals, non-metals, and metalloids based on their properties, which are correlated with their placement in the periodic table. With the exception of hydrogen, all elements that form positive ions by losing electrons during chemical reactions are called metals. Thus metals are electropositive elements with relatively low ionization energies. They are characterized by bright luster, hardness, ability to resonate sound and are excellent conductors of heat and electricity. Metals are solids under normal conditions except for Mercury. Metals are lustrous, malleable, ductile, good conductors of heat and electricity. Other properties include: Metals are electropositive elements that generally form or oxides with oxygen. Other chemical properties include: \[\ce{Na^0 \rightarrow Na^+ + e^{-}} \label{1.1} \] \[\ce{Mg^0 \rightarrow Mg^{2+} + 2e^{-}} \label{1.2} \] \[\ce{Al^0 \rightarrow Al^{3+} + 3e^{-}} \label{1.3} \] Compounds of metals with non-metals tend to be in nature. Most metal oxides are basic oxides and dissolve in water to form : \[\ce{Na2O(s) + H2O(l) \rightarrow 2NaOH(aq)}\label{1.4} \] \[\ce{CaO(s) + H2O(l) \rightarrow Ca(OH)2(aq)} \label{1.5} \] Metal oxides exhibit their chemical nature by reacting with to form metal and water: \[\ce{MgO(s) + HCl(aq) \rightarrow MgCl2(aq) + H2O(l)} \label{1.6} \] \[\ce{NiO(s) + H2SO4(aq) \rightarrow NiSO4(aq) + H2O(l)} \label{1.7} \] What is the chemical formula for aluminum oxide? Al has a 3+ charge, the oxide ion is \(O^{2-}\), thus \(Al_2O_3\). Would you expect it to be solid, liquid or gas at room temperature? Oxides of metals are characteristically solid at room temperature Write the balanced chemical equation for the reaction of aluminum oxide with nitric acid: \[\ce{Al2O3(s) + 6HNO3(aq) \rightarrow 2Al(NO3)3(aq) + 3H2O(l)} \nonumber \] Elements that tend to gain electrons to form anions during chemical reactions are called non-metals. These are electronegative elements with high ionization energies. They are non-lustrous, brittle and poor conductors of heat and electricity (except graphite). Non-metals can be gases, liquids or solids. Non-metals have a tendency to gain or share electrons with other atoms. They are electronegative in character. Nonmetals, when reacting with metals, tend to gain electrons (typically and become \[\ce{3Br2(l) + 2Al(s) \rightarrow 2AlBr3(s)} \nonumber \] Compounds composed entirely of nonmetals are covalent substances. They generally form acidic or neutral oxides with oxygen that that dissolve in water to form acids: \[\ce{CO2(g) + H2O(l)} \rightarrow \underset{\text{carbonic acid}}{\ce{H2CO3(aq)}} \nonumber \] As you may know, carbonated water is slightly acidic (carbonic acid). Nonmetal oxides can combine with bases to form salts. \[\ce{CO2(g) + 2NaOH(aq) \rightarrow Na2CO3(aq) + H2O(l)} \nonumber \] Metalloids have properties intermediate between the metals and nonmetals. Metalloids are useful in the semiconductor industry. Metalloids are all solid at room temperature. They can form alloys with other metals. Some metalloids, such as silicon and germanium, can act as electrical conductors under the right conditions, thus they are called semiconductors. Silicon for example appears lustrous, but is malleable nor ductile (it is - a characteristic of some nonmetals). It is a much poorer conductor of heat and electricity than the metals. The physical properties of metalloids tend to be metallic, but their chemical properties tend to be non-metallic. The oxidation number of an element in this group can range from +5 to -2, depending on the group in which it is located. Metallic character is strongest for the elements in the leftmost part of the periodic table, and tends to decrease as we move to the right in any period (nonmetallic character increases with increasing electronegativity and ionization energy values). Within any group of elements (columns), the metallic character increases from top to bottom (the electronegativity and ionization energy values generally decrease as we move down a group). This general trend is not necessarily observed with the transition metals. ( ) | 4,491 | 2,094 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Nucleophilic_Substitution_at_Tetrahedral_Carbon/NS12._Elimination_Reactions |
Sometimes, elimination reactions occur instead of aliphatic nucleophilic substitutions. In an elimination reaction, instead of connecting to the electrophilic carbon, the nucleophile takes a proton from the next carbon away from it. The halide or other leaving group is still displaced. A double bond forms between the two carbons. Thus, there are actually more than two competing mechanisms occurring at once here. In addition to unimolecular and bimolecular substitution, a reaction involving deprotonation is also possible. Draw a mechanism for the elimination reaction above. Assume the reaction is bimolecular. The mechanism of an elimination reaction is almost exactly the same as an aliphatic nucleophilic substitution, except that the nucelophile misses its mark. It hits a proton instead of a carbon and acts as a base instead of a nucleophile. This process can happen at the same time as the leaving group's departure or it can happen afterwards. These mechanisms are called E1 and E2. Draw another mechanism for the elimination reaction above, but this time, suppose the reaction is unimolecular. Given the mechanism in NS12.2., other products would also be expected. Why might a reaction undergo elimination rather than substitution? The most important reason concerns the nature of the nucleophile. The more basic the nucleophile, the more likely it will induce elimination. Very strong bases include carbon and nitrogen anions and semi-anions. Examples include butyllithium and sodium amide. Very strong bases are highly likely to engage in elimination, rather than substitution. Strong bases include non-stabilized oxygen anions. Examples include sodium hydroxide as well as alkoxides such as potassium tert-butoxide or sodium ethoxide. Strong bases favour elimination, too. Nevertheless, they can sometimes undergo either elimination or substitution, depending on other factors (see below). Weak bases include cyanide, stabilized oxygen anions such as carboxylates and aryloxides, fluoride ion and neutral amines. Weak bases are much more likely to undergo substitution than elimination. Very weak bases include heavy halides such as chloride, bromide or iodide, as well as phosphorus and sulfur nucleophiles. Very weak bases undergo elimination only rarely. Typically, strong bases and very strong bases are more likely to react via the E2 mechanism; they react so quickly that the deprotonation step triggers C-LGp ionization, rather than the other way around. However, E1 mechanisms also occur with these bases, especially at low concentrations. Explain why. Why is it that an anion such as cyanide is a weak base, whereas CH Li is a strong base? Give two reasons. Another factor is sterics. The more crowded the electyrophile, the more likely the nucleophile will encounter a proton on its way to the electrophilic carbon. As a nucleophile approaches -butyl bromide, coming from the side opposite the bromine in order to undergo nucleophilic substitution, it is pretty likely to collide with a proton on its way to the electrophilic carbon. The same thing has a good chance of happening with propyl bromide. However, it is much less likely to happen with bromoethane. Finally, bromomethane doesn't even have a beta-hydrogen, so the chance of elimination in that case is zero. Note that the crowding could involve either the structure of the base or the structure of the electrophile. A large, bulky base may be more likely to deprotonate than find its way in to the electrophilic carbon atom. Given the following pairs of nucleophiles, which one is more likely to undergo elimination? Although acetylides (such as sodium acetylide, Na CCH) are actually more basic than alkoxides (such as sodium isopropoxide, Na OCH(CH ) ), acetylides frequently undergo substitution rather than elimination. Propose a reason for this difference. A third factor is temperature. An elimination reaction involves the cleavage of two bonds, whereas a substitution reaction requires only one bond to break. Thus, an elimination reaction is more energy-intensive, and it is more likely to occur at higher temperatures, when more energy is available. An additional factor in the energy dependence of eliminations and substitutions is entropy. , | 4,256 | 2,095 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumentation_and_Analysis/Capillary_Electrophoresis/Ionic_Mobility_and_Electrophoresis |
When an electric field is applied to ions in a medium, a phenomenon causes the ions to move with the electric field. The motion of the ion itself can be characterized by charge, shape, and size. The ion moves through the medium and migrates with a characteristic velocity, which is in turn dependent on the characteristics of the ion itself and the properties of the medium. This phenomenon is called . Many different biochemical techniques use the principles of electrophoresis to identify compounds of interest in a sample and understand interactions at the molecular and ionic level. The basic concept of the techniques uses an electric field that attracts or repels certain ions, and measurements of the mobility of the ions through the medium (Figure \(\Page {1}\)). The cations move toward the cathode, and the anions move toward the anode. Heavier or bulkier ions move more slowly through the medium. Visual patterns formed on the various media (through the use of dyes and other techniques, such as Coomassie Brilliant Blue G 250 dye) can be analyzed for useful information. The velocity of the ion is given by the following equation: \[ v = \dfrac{E \times q}{f} \label{1}\] where: The (\(f\)) in Equation \(\ref{1}\) is introduced due to the fact that when the ion moves through the medium, the medium exerts a frictional force on the ion that inhibits the movement of the ion. Usually the voltage applied is held constant, so the (\(η\)) of the ion can be measured; it is defined as follows: \[ \eta = \dfrac{v}{E} \label{2} \] where: The higher the current, the faster the migration of the ions (due to increased Coulombic attractions). Because current is affected by voltage, as the difference in potential between the electrodes increases, the rate of migration increases. The resistance is dependent on the properties of the medium; a denser or more saturated medium will retard migration, as will a longer or narrower medium. Mediums with different densities can be made from the same compound, and are quite useful for separating and identifying molecules (through molecular sieving, for example). The most common mediums are agar and polyacrylamide gels. Of key importance is the buffer used. Different compounds have different stability conditions, and the buffer must be chosen carefully so as not to “harm” the molecules’ native states. The main factors to consider when choosing a buffer are its concentration and its pH. Concentration in terms of electrophoresis buffer refers to the ionic strength of the buffer; a buffer with a higher ionic strength will conduct the current more than the sample, leading to slower migration of the molecules. For compounds that have different ionization forms, pH is a major factor in choosing a buffer. The buffer must have a pH that matches the specific ionization forms’ pH range. | 2,863 | 2,097 |
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Exquisite mechanisms have evolved that control the flux of metabolites through metabolic pathways to insure that the output of the pathways meets biological demand and that energy in the form of ATP is not wasted by having opposing pathways run concomitantly in the same cell. Enzymes can be regulated by changing the activity of a preexisting enzyme or changing the amount of an enzyme. A. Changing the activity of a pre-existing enzyme: The quickest way to modulate the activity of an enzyme is to alter the activity of an enzyme that already exists in the cell. The list below, illustrated in the following figure, gives common ways to regulate enzyme activity Jmol: Erk2 - B. Changing the amount of an enzyme: Another and less immediate but longer duration method to modulate the activity of an enzyme is to alter the activity of an enzyme that already exists in the cell. The list below, illustrated in the following figure, shows way in which enzyme concentration is regulated. , | 1,006 | 2,098 |
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The cycloalkyl halides, except for cyclopropyl halides, have physical and chemical properties that are similar to those of the open-chain secondary halides and can be prepared by the same types of reactions (Table 14-5). All the cycloalkyl halides undergo \(S_\text{N}2\) reactions rather slowly and, with nucleophiles that are reasonably basic (\(^\ominus \ce{OH}\), \(^\ominus \ce{OC_2H_5}\), \(^\ominus \ce{C \equiv N}\), etc.), \(E2\) reactions can be expected to predominate (Table 14-6). The rate of carbocation formation leading to \(S_\text{N}1\) and \(E1\) reactions is sensitive to ring size but, except for the small-ring halides, the carbocation reactions are normal in most other respects. The cyclopropyl halides are exceptional in that their behavior is much more like alkenyl halides than like secondary alkyl halides. Thus cyclopropyl chloride undergoes \(S_\text{N}1\) and \(S_\text{N}2\) reactions much less rapidly than isopropyl or cyclohexyl chlorides. A relationship between the reactivity of cyclopropyl chloride and chloroethene is not surprising in view of the general similarity between cyclopropane rings and double bonds ( ). This similarity extends to cyclopropylmethyl derivatives as well. Cyclopropylmethyl chloride is reactive in both \(S_\text{N}1\) and \(S_\text{N}2\) reactions in much the same way as 3-chloropropene: and (1977) | 1,384 | 2,099 |
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A phase is a certain form of matter that includes a specific set of physical properties. That is, the atoms, the molecules, or the ions that make up the phase do so in a consistent manner throughout the phase. Science recognizes three stable phases: the , in which individual particles can be thought of as in contact and held in place; the , in which individual particles are in contact but moving with respect to each other; and the , in which individual particles are separated from each other by relatively large distances (see Figure 8.1.1). Not all substances will readily exhibit all phases. For example, carbon dioxide does not exhibit a liquid phase unless the pressure is greater than about six times normal atmospheric pressure. Other substances, especially complex organic molecules, may decompose at higher temperatures, rather than becoming a liquid or a gas. For many substances, there are different arrangements the particles can take in the solid phase, depending on temperature and pressure. Which phase a substance adopts depends on the pressure and the temperature it experiences. Of these two conditions, temperature variations are more obviously related to the phase of a substance. When it is very cold, H O exists in the solid form as ice. When it is warmer, the liquid phase of H O is present. At even higher temperatures, H O boils and becomes steam. Pressure changes can also affect the presence of a particular phase (as we indicated for carbon dioxide), but its effects are less obvious most of the time. We will mostly focus on the temperature effects on phases, mentioning pressure effects only when they are important. Most chemical substances follow the same pattern of phases when going from a low temperature to a high temperature: the solid phase, then the liquid phase, and then the gas phase. However, the temperatures at which these phases are present differ for all substances and can be rather extreme. Table \(\Page {1}\) shows the temperature ranges for solid, liquid, and gas phases for three substances. As you can see, there is extreme variability in the temperature ranges. What accounts for this variability? Why do some substances become liquids at very low temperatures, while others require very high temperatures before they become liquids? It all depends on the strength of the between the of substances and the (KE) of its molecules. (Although ionic compounds are not composed of discrete molecules, we will still use the term to include interactions between the ions in such compounds.) Substances that experience require Substances that experience do not need much energy (as measured by temperature) to become liquids and gases and will exhibit these phases at Substances with the highest melting and boiling points have bonding. This type of intermolecular interaction is actually a covalent bond. In these substances, all the atoms in a sample are covalently bonded to one another; in effect, the entire sample is essentially . Many of these substances are solid over a large temperature range because it takes a lot of energy to disrupt all the covalent bonds at once. One example of a substance that shows covalent network bonding is diamond (Figure \(\Page {2}\)). Diamond is composed entirely of carbon atoms, each bonded to four other carbon atoms in a tetrahedral geometry. Melting a covalent network solid is not accomplished by overcoming the relatively weak intermolecular forces. Rather, all of the covalent bonds must be broken, a process that requires extremely high temperatures. Diamond, in fact, does not melt at all. Instead, it vaporizes to a gas at temperatures above 3,500°C. Diamond is extremely hard and is one of the few materials that can cut glass. The strongest force between any two particles is , in which two ions of opposing charge are attracted to each other. Thus, ionic interactions between particles are another type of intermolecular interaction. Substances that contain ionic interactions are relatively strongly held together, so these substances typically have high melting and boiling points. Sodium chloride (Figure \(\Page {3}\)) is an example of a substance whose particles experience ionic interactions (Table \(\Page {1}\)). These attractive forces are sometimes referred to as ion-ion interactions. There are two different covalent structures: molecular and network. Covalent network compounds like SiO (quartz) have structures of atoms in a network like diamond described earlier. In this section, we are dealing with the molecular type that contains individual molecules. The bonding between atoms in the individual molecule is covalent but the attractive forces between the molecules are called . In contrast to forces (see Figure 8.1.4), such as the covalent bonds that hold atoms together in molecules and polyatomic ions, intermolecular forces hold molecules together in a . Intermolecular forces are generally much weaker than covalent bonds. For example, it requires 927 kJ to overcome the intramolecular forces and break both O–H bonds in 1 mol of water, but it takes only about 41 kJ to overcome the intermolecular attractions and convert 1 mol of liquid water to water vapor at 100°C. (Despite this seemingly low value, the intermolecular forces in liquid water are among the strongest such forces known!) Given the large difference in the strengths of intra- and intermolecular forces, changes between the solid, liquid, and gaseous states almost invariably occur for molecular substances In this section, we will discuss the three types of IMF in molecular compounds: dipole-dipole, hydrogen bonding and London dispersion forces. As discussed in Section 4.4, covalent bond that has an unequal sharing of electrons is called a bond. (A covalent bond that has an equal sharing of electrons, as in a covalent bond with the same atom on each side, is called a bond.) A molecule with a net unequal distribution of electrons in its covalent bonds is a polar molecule. is an example of a polar molecule (see Figure 8.1.5). The charge separation in a is not as extreme as is found in ionic compounds, but there is a related result: oppositely charged ends of different molecules will attract each other. This type of intermolecular interaction is called a . Many molecules with polar covalent bonds experience dipole-dipole interactions. The covalent bonds in some molecules are oriented in space in such a way that the bonds in the molecules cancel each other out. The individual bonds are polar, but due to molecular symmetry, the overall molecule is not polar; rather, the molecule is Such molecules experience little or no dipole-dipole interactions. Carbon dioxide (CO ) and carbon tetrachloride (CCl ) are examples of such molecules (Figure \(\Page {6}\)). Recall from the Sections 4.4 and 4.5, on chemical bonding and molecular geometry that molecules have a partial positive charge on one side and a partial negative charge on the other side of the molecule—a separation of charge called a Consider a polar molecule such as hydrogen chloride, HCl. In the HCl molecule, the more electronegative Cl atom bears the partial negative charge, whereas the less electronegative H atom bears the partial positive charge. An attractive force between HCl molecules results from the attraction between the positive end of one HCl molecule and the negative end of another. This attractive force is called a —the electrostatic force between the partially positive end of one polar molecule and the partially negative end of another, as illustrated in Figure \(\Page {7}\). The effect of a dipole-dipole attraction is apparent when we compare the properties of HCl molecules to nonpolar F molecules. Both HCl and F consist of the same number of atoms and have approximately the same molecular mass. At a temperature of 150 K, molecules of both substances would have the same average kinetic energy. However, the dipole-dipole attractions between HCl molecules are sufficient to cause them to “stick together” to form a liquid, whereas the relatively weaker dispersion forces between nonpolar F molecules are not, and so this substance is gaseous at this temperature. The higher normal boiling point of HCl (188 K) compared to F (85 K) is a reflection of the greater strength of dipole-dipole attractions between HCl molecules, compared to the attractions between nonpolar F molecules. We will often use values such as boiling or freezing points as indicators of the relative strengths of IMFs of attraction present within different substances. Predict which will have the higher boiling point: N or CO. Explain your reasoning. CO and N are both diatomic molecules with masses of about 28 amu, so they experience similar London dispersion forces. Because CO is a polar molecule, it experiences dipole-dipole attractions. Because N is nonpolar, its molecules cannot exhibit dipole-dipole attractions. The dipole-dipole attractions between CO molecules are comparably stronger than the dispersion forces between nonpolar N molecules, so CO is expected to have the higher boiling point. Predict which will have the higher boiling point: \(\ce{ICl}\) or \(\ce{Br2}\). Explain your reasoning. ICl. ICl and Br have similar masses (~160 amu) and therefore experience similar London dispersion forces. ICl is polar and thus also exhibits dipole-dipole attractions; Br is nonpolar and does not. The relatively stronger dipole-dipole attractions require more energy to overcome, so ICl will have the higher boiling point. Molecules with bonded to electronegative atoms such as tend to exhibit unusually strong intermolecular interactions due to a particularly strong type of dipole-dipole attraction called . The very large difference in electronegativity between the H atom (2.1) and the atom to which it is bonded (4.0 for an F atom, 3.5 for an O atom, or 3.0 for a N atom), combined with the very small size of a H atom and the relatively small sizes of F, O, or N atoms, leads to with these atoms. Because the hydrogen atom does not have any electrons other than the ones in the covalent bond, its positively charged nucleus is almost completely exposed, allowing strong attractions to other nearby lone pairs of electrons. Examples of hydrogen bonds include HF⋯HF, H O⋯HOH, and H N⋯HNH , in which the are denoted by . Figure \(\Page {8}\) illustrates hydrogen bonding between water molecules. The physical properties of water, which has two O–H bonds, are strongly affected by the presence of hydrogen bonding between water molecules. Most molecular compounds that have a mass similar to water are gases at room temperature. However, because of the strong hydrogen bonds, water molecules are able to stay condensed in the liquid state. A is an intermolecular attractive force in which , that is covalently bonded to a small, highly electronegative atom, is attracted to a on an atom in a neighboring molecule Figure \(\Page {9}\) shows how methanol (CH OH) molecules experience hydrogen bonding. Methanol contains both a hydrogen atom attached to O; methanol can thus form hydrogen bonds by acting as either a hydrogen bond donor or a hydrogen bond acceptor (lone pair). The hydrogen-bonded structure of methanol is as follows: Despite use of the word “bond,” keep in mind that hydrogen bonds are attractive forces, not attractive forces (covalent bonds). Hydrogen bonds are much weaker than covalent bonds, but are generally much stronger than other dipole-dipole attractions and dispersion forces. Consider the compounds dimethylether (CH OCH ), ethanol (CH CH OH), and propane (CH CH CH ). Their boiling points, not necessarily in order, are −42.1 °C, −24.8 °C, and 78.4 °C. Match each compound with its boiling point. Explain your reasoning. The shapes of CH OCH , CH CH OH, and CH CH CH are similar, as are their molar masses (46 g/mol, 46 g/mol, and 44 g/mol, respectively), so they will exhibit similar dispersion forces. Since CH CH CH is nonpolar, it may exhibit dispersion forces. Because CH OCH is polar, it will also experience dipole-dipole attractions. Finally, CH CH OH has an −OH group, and so it will experience the uniquely strong dipole-dipole attraction known as hydrogen bonding. So the ordering in terms of strength of IMFs, and thus boiling points, is CH CH CH < CH OCH < CH CH OH. The boiling point of propane is −42.1 °C, the boiling point of dimethylether is −24.8 °C, and the boiling point of ethanol is 78.5 °C. Ethane (CH CH ) has a melting point of −183 °C and a boiling point of −89 °C. Predict the melting and boiling points for methylamine (CH NH ). Explain your reasoning. The melting point and boiling point for methylamine are predicted to be significantly greater than those of ethane. CH CH and CH NH are similar in size and mass, but methylamine possesses an −NH group and therefore may exhibit hydrogen bonding. This greatly increases its IMFs, and therefore its melting and boiling points. It is difficult to predict values, but the known values are a melting point of −93 °C and a boiling point of −6 °C. Deoxyribonucleic acid (DNA) is found in every living organism and contains the genetic information that determines the organism’s characteristics, provides the blueprint for making the proteins necessary for life, and serves as a template to pass this information on to the organism’s offspring. A DNA molecule consists of two (anti-)parallel chains of repeating nucleotides, which form its well-known double helical structure, as shown in Figure \(\Page {10}\). Each nucleotide contains a (deoxyribose) sugar bound to a phosphate group on one side, and one of four nitrogenous bases on the other. Two of the bases, cytosine (C) and thymine (T), are single-ringed structures known as pyrimidines. The other two, adenine (A) and guanine (G), are double-ringed structures called purines. These bases form complementary base pairs consisting of one purine and one pyrimidine, with adenine pairing with thymine, and cytosine with guanine. Each base pair is held together by hydrogen bonding. A and T share two hydrogen bonds, C and G share three, and both pairings have a similar shape and structure Figure \(\Page {11}\) The cumulative effect of millions of hydrogen bonds effectively holds the two strands of DNA together. Importantly, the two strands of DNA can relatively easily “unzip” down the middle since hydrogen bonds are relatively weak compared to the covalent bonds that hold the atoms of the individual DNA molecules together. This allows both strands to function as a template for replication. Finally, there are forces between all molecules that are caused by electrons being in different places in a molecule at any one time, which sets up a temporary separation of charge that disappears almost as soon as it appears. These are very weak intermolecular interactions and are called . (An alternate name is London dispersion forces.) Molecules that experience no other type of intermolecular interaction will at least experience dispersion forces. Substances that experience only dispersion forces are typically soft in the solid phase and have relatively low melting points. Examples include , which are long hydrocarbon chains that are solids at room temperature because the molecules have so many electrons. The resulting dispersion forces between these molecules make them assume the solid phase at normal temperatures. Dispersion forces that develop between atoms in different molecules can attract the two molecules to each other. The forces are relatively weak, however, and become significant only when the molecules are very close. exhibit than do smaller and lighter atoms and molecules. F and Cl are gases at room temperature (reflecting weaker attractive forces); Br is a liquid, and I is a solid (reflecting stronger attractive forces). Trends in observed melting and boiling points for the halogens clearly demonstrate this effect, as seen in Table 8.1.2. The increase in melting and boiling points with increasing atomic/molecular size may be rationalized by considering how the strength of dispersion forces is affected by the electronic structure of the atoms or molecules in the substance. In a larger atom, the valence electrons are, on average, farther from the nuclei than in a smaller atom. Thus, they are less tightly held and can more easily form the temporary dipoles that produce the attraction. The measure of how easy or difficult it is for another electrostatic charge (for example, a nearby ion or polar molecule) to distort a molecule’s charge distribution (its electron cloud) is known as . A molecule that has a charge cloud that is easily distorted is said to be very polarizable and will have large dispersion forces; one with a charge cloud that is difficult to distort is not very polarizable and will have small dispersion forces. Order the following compounds of a group 14 element and hydrogen from lowest to highest boiling point: CH , SiH , GeH , and SnH . Explain your reasoning. Applying the skills acquired in the chapter on chemical bonding and molecular geometry, all of these compounds are predicted to be nonpolar, so they may experience only dispersion forces: the smaller the molecule, the less polarizable and the weaker the dispersion forces; the larger the molecule, the larger the dispersion forces. The molar masses of CH , SiH , GeH , and SnH are approximately 16 g/mol, 32 g/mol, 77 g/mol, and 123 g/mol, respectively. Therefore, CH is expected to have the lowest boiling point and SnH the highest boiling point. The ordering from lowest to highest boiling point is expected to be CH < SiH < GeH < SnH A graph of the actual boiling points of these compounds versus the period of the elementsshows this prediction to be correct: Order the following hydrocarbons from lowest to highest boiling point: C H , C H , and C H . All of these compounds are nonpolar and only have London dispersion forces: the larger the molecule, the larger the dispersion forces and the higher the boiling point. The ordering from lowest to highest boiling point is therefore C H < C H < C H . Geckos have an amazing ability to adhere to most surfaces. They can quickly run up smooth walls and across ceilings that have no toe-holds, and they do this without having suction cups or a sticky substance on their toes. And while a gecko can lift its feet easily as it walks along a surface, if you attempt to pick it up, it sticks to the surface. How are geckos (as well as spiders and some other insects) able to do this? Although this phenomenon has been investigated for hundreds of years, scientists only recently uncovered the details of the process that allows geckos’ feet to behave this way. Geckos’ toes are covered with hundreds of thousands of tiny hairs known as , with each seta, in turn, branching into hundreds of tiny, flat, triangular tips called . The huge numbers of spatulae on its setae provide a gecko, shown in Figure 8.1.12, with a large total surface area for sticking to a surface. In 2000, Kellar Autumn, who leads a multi-institutional gecko research team, found that geckos adhered equally well to both polar silicon dioxide and nonpolar gallium arsenide. This proved that geckos stick to surfaces because of dispersion forces—weak intermolecular attractions arising from temporary, synchronized charge distributions between adjacent molecules. Although dispersion forces are very weak, the total attraction over millions of spatulae is large enough to support many times the gecko’s weight. In 2014, two scientists developed a model to explain how geckos can rapidly transition from “sticky” to “non-sticky.” Alex Greaney and Congcong Hu at Oregon State University described how geckos can achieve this by changing the angle between their spatulae and the surface. Geckos’ feet, which are normally nonsticky, become sticky when a small shear force is applied. By curling and uncurling their toes, geckos can alternate between sticking and unsticking from a surface, and thus easily move across it. Further investigations may eventually lead to the development of better adhesives and other applications. In order for a substance to enter the gas phase, its particles must completely overcome the intermolecular forces holding them together. Therefore, a comparison of boiling points is essentially equivalent to comparing the strengths of the attractive intermolecular forces exhibited by the individual molecules. For small molecular compounds, London dispersion forces are the weakest intermolecular forces. Dipole-dipole forces are somewhat stronger, and hydrogen bonding is a particularly strong form of dipole-dipole interaction. However, when the mass of a nonpolar molecule is sufficiently large, its dispersion forces can be stronger than the dipole-dipole forces in a lighter polar molecule. Thus, nonpolar Cl has a higher boiling point than polar HCl. What intermolecular forces besides dispersion forces, if any, exist in each substance? Are any of these substances solids at room temperature? What intermolecular forces besides dispersion forces, if any, exist in each substance? Are any of these substances solids at room temperature? a. dipole-dipole, hydrogen bonding b. ionic forces (solid at room temperature) c. dipole-dipole | 21,370 | 2,100 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Periodic_Trends_of_Elemental_Properties/The_Periodic_Law |
The periodic law was developed independently by Dmitri Mendeleev and Lothar Meyer in 1869. Mendeleev created the first periodic table and was shortly followed by Meyer. They both arranged the elements by their mass and proposed that certain properties periodically reoccur. Meyer formed his periodic law based on the atomic volume or molar volume, which is the atomic mass divided by the density in solid form. Mendeleev's table is noteworthy because it exhibits mostly accurate values for atomic mass and it also contains blank spaces for unknown elements. In 1804 physicist John Dalton advanced helping scientists determine the mass of the known elements. Around the same time, two chemists Sir Humphry Davy and Michael Faraday developed electrochemistry which aided in the discovery of new elements. By 1829, chemist Johann Wolfgang Doberiner observed that certain elements with similar properties occur in group of three such as; chlorine, bromine, iodine; calcium, strontium, and barium; sulfur, selenium, tellurium; iron, cobalt, manganese. However, at the time of this discovery too few elements had been discovered and there was confusion between molecular weight and atomic weights; therefore, chemists never really understood the significance of Doberiner's triad. In 1859 two physicists Robert Willhem Bunsen and Gustav Robert Kirchoff discovered spectroscopy which allowed for discovery of many new elements. This gave scientists the tools to reveal the relationships between elements. Thus in 1864, chemist John A. R Newland arranged the elements in increasing of atomic weights. Explaining that a given set of properties reoccurs every eight place, he named it the law of Octaves. In 1869, and individually came up with their own periodic law "when the elements are arranged in order of increasing atomic mass, certain sets of properties recur periodically." Meyer based his laws on the atomic volume (the atomic mass of an element divided by the density of its solid form), this property is called . \[\text{Atomic (molar) volume (cm}^3\text{/mol)} = \dfrac{\text{ molar mass (g/ mol)}}{\rho \text{ (cm}^3\text{/g)}}\] Mendeleev's periodic table is an arrangement of the elements that group similar elements together. He left blank spaces for the undiscovered elements (atomic masses, element: 44, scandium; 68, gallium; 72, germanium; & 100, technetium) so that certain elements can be grouped together. However, Mendeleev had not predicted the noble gases, so no spots were left for them. Li (174 C) > Na (97.8 C) > K (63.7 C) > Rb (38.9 C) > Cs (28.5 C) Assuming there were errors in atomic masses, Mendeleev placed certain elements not in order of increasing atomic mass so that they could fit into the proper groups (similar elements have similar properties) of his periodic table. An example of this was with argon (atomic mass 39.9), which was put in front of potassium (atomic mass 39.1). Elements were placed into groups that expressed similar chemical behavior. In 1913 Henry G.J. Moseley did researched the X-Ray spectra of the elements and suggested that the energies of electron orbitals depend on the nuclear charge and the nuclear charges of atoms in the target, which is also known as anode, dictate the frequencies of emitted X-Rays. Moseley was able to tie the X-Ray frequencies to numbers equal to the nuclear charges, therefore showing the placement of the elements in Mendeleev's periodic table. The equation he used: \[\nu = A(Z-b)^2\] with With Moseley's contribution the Periodic Law can be restated: Similar properties recur periodically when elements are arranged according to increasing atomic number." Atomic numbers, not weights, determine the factor of chemical properties. As mentioned before, argon weights more than potassium (39.9 vs. 39.1, respectively), yet argon is in front of potassium. Thus, we can see that elements are arranged based on their atomic number. The periodic law is found to help determine many patterns of many different properties of elements; melting and boiling points, densities, electrical conductivity, reactivity, acidic, basic, valance, polarity, and solubility. The table below shows that elements increase from left to right accordingly to their atomic number. The vertical columns have similar properties within their group for example Lithium is similar to sodium, beryllium is similar to magnesium, and so on. Elements in Group 1 (periodic table) have similar chemical properties and are called . Elements in Group 2 have similar chemical properties, they are called the . The short form periodic table is a table where elements are arranged in 7 rows, periods, with increasing atomic numbers from left to right. There are 18 vertical columns known as groups. This table is based on Mendeleev's periodic table and the periodic law. In the long form, each period correlates to the building up of electronic shell; the first two groups (1-2) (s-block) and the last 6 groups (13-18) (p-block) make up the main-group elements and the groups (3-12) in between the s and p blocks are called the transition metals. Group 18 elements are called noble gases, and group 17 are called halogens. The f-block elements, called inner transition metals, which are at the bottom of the periodic table (periods 8 and 9); the 15 elements after barium (atomic number 56) are called and the 14 elements after radium (atomic number 88) are called . ) | 5,443 | 2,102 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Main_Group_Reactions/Compounds/Aluminosilicates/Aluminas |
Aside from silicates, aluminas are the most abundant mineral of the earth crust. Thus, it is important for chemical engineers to know some chemistry about aluminas, because they are found and used in so many different places and technologies. Furthermore, aluminum ions often replace silicon ions in silicates forming aluminosilicates, which is discussed in the next page. From the discussion on this page, you will be introduced to various forms of alumina, their structures, and properties so that when you encounter them, you can associate their properties with their chemical identities (compositions) and structures. The most common ore is , which is aluminum oxide, Al O , mixed with oxides of silicon, iron, and other elements and varying small percentages of clay and other silicates. Physically, bauxite can be as hard as rock or as soft as mud, and its color may be red, white, buff, pink, yellow or any combination of these. The picture shows the mining of bauxite at Gove in Australia Bauxite is the product of extreme chemical weathering of aluminum-rich rocks. Australia produces the largest amount of alumina because she has a large body of bauxite. Jamica, South Africa and some other countries also have a good reserve. Aluminum oxides often coexist with silicates. Natural aluminum oxide minerals include Bauxites are mainly used for producing pure alumina, which is the feed stock for aluminum metal production, raw material for ceramics, and other applications. There are a few forms of aluminum oxide, and corundum being the most common. The structure of corundum can be viewed as a hexagonal close packed array of oxygen atoms with / of the octahedral sites occupied by Al ions. Thus, the Al ions are bonded to 6 oxygen in a distorted octahedron. Each such octahedron share a face with one on the upper and one on the lower layers. The distortion is caused by repulsion between Al ions in octahedra sharing the faces. Corundum is a dense (specific gravity of 3.97), hard (9 on the Mohs' scale, next only to diamond), high melting (melting point 2288 K), and insoluble in water. Crystals of corundum are usually prismatic or barrel-shaped bounded by steep pyramids. A massive grey granular corundum powder is called . Colored corundum are called (deep red due to presence of Cr ions) and (blue, pink, yellow or green due to various degrees of Fe , and Ti ). The color may be modified by heating or irradiation. Some ruby crystals are shown here exposed in a piece of bauxite ore. Grey corundum or emery are used as abrasive, for example, emery paper (sand paper) and ruby and sapphire are for gemstones. They do have technical applications, for example, the first LASER was produced using a ruby crystal. Pure aluminas are used for pottery, ceramics, refractories, catalyst supports, and for the production of aluminum by the Hall process. Thus, bauxite and other aluminum containing minerals such as kaolinite (\(Al_4Si_4O_{10}(OH)_8\)) must be mined and processed to produce pure alumina. Australia produces about $3 billion worth of alumina a year from six Australian refineries. These refineries use the to extract aluminum hydroxide from the bauxite using hot caustic liquor. Aluminum oxide, Al O , is a typical , which dissolves in a strong acid and a strong base. \[Al_2O_3 + 6H^+ \rightarrow 2 Al^{3+} + 3 H_2O \tag{1}\] \[Al_2O_3 + 6 OH^- + 3 H_2O \rightarrow 2 Al(OH)_6^{3-} \tag{2}\] After separation of the solids residue, the clear liquor is cooled. Depending on the pH of the solution, the aluminate ion Al(OH) bears various amounts of charge due to these reactions: \[Al(OH)_6^{3-} + H6+ \rightarrow Al(OH)_5^{2-}(H_2O) \tag{3}\] \[Al(OH)_5^{2} \cdot (H_2O) + H^+ = Al(OH)_4 \cdot (H_2O)_2 \tag{4}\] \[Al(OH)_4 \cdot (H_2O)_2 + H^+ \rightarrow \underset{\text{a precipitate}}{Al(OH)_3(H_2O)_3} \tag{5} \] In a neutral solution, the compound, Al(OH) (H O) or Al(OH) if water is ignored, forms a gelatinous precipitate. Under controlled manner, the liquor crystallizes to give particles of gibbsite of the desired chemical purity and physical characteristics. A lot of research and development has gone into this crystallization process alone. The hydroxide ions of gibbsite form two layers similar to layers of closest packed spheres with Al ions filling in some of the octahedral sites. The crystal structure of gibbsite consists of stacked double layers. It is expected that the hydroxide ions form extensive intra- and inter-layer . Further dehydration converts Al(OH) into and , both of which have the stoichiometry AlO(OH). Gibbsite is is converted to , Al O by calcination. Alumina is marketed as the feed stock to smelters for the production of aluminium metal, ceramics, catalyst supports and other applications. Mineralogists consider a mineral a homogeneous solid body, formed by natural process that has a regular crystal structure with a limit range of atomic compositions. Engineers are mainly interested in properties and their applications. Scientists are interested in correlate the relationship of structures and properties. Engineers deal with natural and synthetic materials alike. Aluminum oxide is a basic material for the ceramic industry. For more details regarding the properties of alumina, consult the data sheet for alumina ceramics. Aluminas are basic materials for ceramics, and they are useful for lining containers and mass transferring pipes, especially if heat resistance is required. Intricate tools such as the 95 % alumina ceramic rotor for 20 cm rotary valve have been made of these materials. Ceramics are related to many technologies. A related to ceramics gives many companies, whose main products are made of ceramic materials. For example, aluminas are used for paint, ink, coating and filling paper, adhesives, rubber, pharmaceuticals, tiles, bricks, cooking utilities, table wares, electronic components, porcelain, pottery, dental restoration, and plastics. Technological changes demand materials with new specific properties. Since changes take place all the time, new materials are also developed all the time. Additions of specific amounts of other oxides to aluminas produce composite materials whose properties differ from both parent materials. This type of blending is a new frontier of material engineering. Aluminum is a very reactive metal if it is not protected by its aluminum oxide film. It is much more reactive than zinc and iron, but far less reactive than magnesium. Their oxidation reduction potentials are given below for you to compare. Fe + 3 e = Fe(s), = -0.037 V.
Fe + 2 e = Fe(s), = -0.447 V.
Zn + 2 e = Zn(s), = -0.76 V.
Al + 3 e = Al(s), = -1.67 V.
Mg + 2 e = Mg(s), = -2.70 V. If the oxide film is cracked under aerated water, Al is formed instantly along with OH ions. Thus, an oxide film is formed immediately, sealing it from further corrosion at the anodic site. Another important fact is that Al ions will not form, and the aluminum oxide is an inert substance. Calculate the molar volume of aluminum and aluminum oxide.
The question requires the densities of aluminum and aluminum oxides. The CRC Handbook give their densities as 2.702 and 3.97 g/cc respectively. Thus, molar volumes of Al and aluminum oxides are of Al = 26.98 / 2.702 = 9.96 cc per mol of Al
of Al O = (26.98 + 24.0) / 3.97 = 12.84 cc
By applying the , aluminum oxide formed has a larger molar volume per Al than the metal itself. Thus, the Pilling and Bedworth model also apply to aluminum and its oxide. Furthermore, aluminum oxide and some hydrates have different densities, Al O H O, 3.014 g/cc.
Al O 3 H O, 2.42 g/cc. The formation of Al(OH) forms a protective layer. This formula suggest the formation of gibbsite, density 2.44 g/cc leading to a molar volume of 32.0 cc.
To be able to identify the ore bauxite from the description given in this page.
Explain the relationship of various materials, even if they appear very different.
Explain the properties of corundum and relate them to its applications.
Describe the structure of corundum, an important mineral related to ruby and sapphire.
Explain why a concentrated basic solution is used in the Bayer process.
Describe the chemistry of the Bayer process for the extraction of alumina.
Explain the chemical property base on electromotive potentials.
The thin layer does not cover the luster of aluminum metal. | 8,476 | 2,105 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/03%3A_Functional_Groups_and_Nomenclature |
After reading this chapter and completing ALL the exercises, a student can be able to A Pearl of Wisdom: Most common names were derived from older systems of nomenclature that some may argue were "not systematic at all". However, it is helpful to note that the older systems of nomenclature were often based on shared structural features and/or chemical reactivity. Learning carefully selected common names can offer insights into chemical reactivity and structural patterns. Additionally, there are some common names that are so prevalent, they need to be memorized. Please note: The nomenclature for organic compounds with sulfur and phosphorus are introduced so that students can interpret a given name and draw the correct structure. Derivation of names can be required by the professor and requires additional instruction. | 850 | 2,106 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Electrophilic_Addition_Reactions/Reactions_of_Alkenes_with_Bromine |
This page gives you the facts and a simple, uncluttered mechanism for the electrophilic addition reactions between bromine (and the other halogens) and alkenes like ethene and cyclohexene Alkenes react in the cold with pure liquid bromine, or with a solution of bromine in an organic solvent like tetrachloromethane. The double bond breaks, and a bromine atom becomes attached to each carbon. The bromine loses its original red-brown color to give a colorless liquid. In the case of the reaction with ethene, 1,2-dibromoethane is formed. This decoloration of bromine is often used as a test for a carbon-carbon double bond. If an aqueous solution of bromine is used ("bromine water"), you get a mixture of products. The other halogens, apart from fluorine, behave similarly. (Fluorine reacts explosively with all hydrocarbons - including alkenes - to give carbon and hydrogen fluoride.) If you are interested in the reaction with, say, chlorine, all you have to do is to replace Br by Cl. The reaction is an example of .The bromine is a very "polarizable" molecule and the approaching pi bond in the ethene induces a dipole in the bromine molecule. If you draw this mechanism in an exam, write the words "induced dipole" next to the bromine molecule - to show that you understand what's going on. In the first stage of the reaction, one of the bromine atoms becomes attached to both carbon atoms, with the positive charge being found on the bromine atom. A bromonium ion is formed. The bromonium ion is then attacked from the back by a bromide ion formed in a nearby reaction. Cyclohexene reacts with bromine in the same way and under the same conditions as any other alkene. 1,2-dibromocyclohexane is formed. The reaction is an example of electrophilic addition. Again, the bromine is polarized by the approaching \(\pi\) bond in the cyclohexene. Do not forget to write the words "induced dipole" next to the bromine molecule. In the first stage of the reaction, one of the bromine atoms becomes attached to both carbon atoms, with the positive charge being found on the bromine atom. A bromonium ion is formed. The bromonium ion is then attacked from the back by a bromide ion formed in a nearby reaction. Jim Clark ( ) | 2,234 | 2,107 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Chemiluminescence/4%3A_Instrumentation/4.08%3A_Photo-induced_chemiluminescence |
Photo-induced chemiluminescence (PICL) involves irradiating an analyte with ultra-violet light in order to convert it to a photoproduct of different chemiluminescence behaviour, usually substantially increased emission. Such reactions form the basis of highly sensitive and selective analytical techniques. Irradiating a molecule can break it into fragments of smaller molecular weight (photolysis) or can induce reactions such as oxidation, reduction, cyclization or isomerization. Direct photolysis involves absorption of photons by the target molecule; in indirect photolysis, the target molecule absorbs energy from another molecule that has previously absorbed photons . Photochemistry is concerned with excited electronic states induced by the absorption of photons. Photoexcitation is more selective than thermal excitation and leads to a different energy distribution within the molecule. The excited molecule can undergo photochemical processes, the products of which are sometimes involved in side processes. In analytically useful photochemical reactions the light is strongly absorbed by the analyte but not by the photoproducts; the photochemical yield is high; the photoproducts are stable for as long as is needed to complete the analysis and are structurally rigid enough for the emission to have an adequate quantum yield. Successful analytical application also depends on appropriately designed photoreactors. When those conditions are fulfilled, using light has several advantages over the use of chemical derivatization. Lamps are inexpensive and their stable light output allows reproducible results. They differ in their spectral characteristics, which gives scope for increasing selectivity. The use of light has minimal environmental impact and can be effected in ambient conditions. Analysis times are shorter because photochemical reactions are fast and can be shortened further by optimizing reactor configuration or increasing lamp power. PICL has a linear relationship with analyte concentration over a wide concentration range and extends the range of analytes that can be detected by chemiluminescence. It is not necessary to identify or separate the photoproducts. In PICL-based methods, the sample is irradiated on-line and subsequently merged with the chemiluminescence reagents prior to reaching the flow cell in front of the detector. Flow methods allow the irradiation time to be easily controlled and provide better reproducibility than stationary methods, coping better with the very fast rate of chemiluminescence reactions. Sample throughput, ease of automation and reagent consumption are also improved using flow methods. PICL has the same instrumentation as other chemiluminescence but, in addition, a photoreactor is required and this has two essential elements – a light source and a container for the sample. Lamps are selected on the basis of power and spectrum (continuous or discrete). Continuous spectra span a wide zone, whereas discrete spectra are series of individual lines in a narrow wavelength range. The mercury-xenon lamp provides a continuous spectrum and is used when its high power is necessary, though it needs cooling. The low-pressure mercury lamp generates little heat and is a typical discrete-spectrum lamp, emitting over the range 200-320 nm, maximally at 254 nm; most substances absorb in this zone. The absorption zone of the selected lamp must be the most useful for excitation and bond-breaking. In flow systems, the sample is usually contained in PTFE tubing, which admits little light but maximises its effect by repeated reflections from the inner tube surfaces; the tubing can be coiled around a low-power lamp. Batch methods instead use quartz cells, which are transparent to ultra-violet. Quartz is more inert than PTFE, but also more fragile. | 3,841 | 2,109 |
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A buffer is a solution that can resist pH change upon the addition of an acidic or basic components. It is able to neutralize small amounts of added acid or base, thus maintaining the pH of the solution relatively stable. This is important for processes and/or reactions which require specific and stable pH ranges. Buffer solutions have a working pH range and capacity which dictate how much acid/base can be neutralized before pH changes, and the amount by which it will change. To effectively maintain a pH range, a buffer must consist of a weak conjugate acid-base pair, meaning either a. a weak acid and its conjugate base, or b. a weak base and its conjugate acid. The use of one or the other will simply depend upon the desired pH when preparing the buffer. For example, the following could function as buffers when together in solution: A buffer is able to resist pH change because the two components (conjugate acid and conjugate base) are both present in appreciable amounts at equilibrium and are able to neutralize small amounts of other acids and bases (in the form of H O and OH ) when the are added to the solution. To clarify this effect, we can consider the simple example of a Hydrofluoric Acid (HF) and Sodium Fluoride (NaF) buffer. Hydrofluoric acid is a weak acid due to the strong attraction between the relatively small F ion and solvated protons (H O ), which does not allow it to dissociate completely in water. Therefore, if we obtain HF in an aqueous solution, we establish the following equilibrium with only slight dissociation (K (HF) = 6.6x10 , strongly favors reactants): \[HF_{(aq)} + H_2O_{(l)} \rightleftharpoons F^-_{(aq)} + H_3O^+_{(aq)} \nonumber \] We can then add and dissolve sodium fluoride into the solution and mix the two until we reach the desired volume and pH at which we want to buffer. When Sodium Fluoride dissolves in water, the reaction goes to completion, thus we obtain: \[NaF_{(aq)} + H_2O_{(l)} \rightarrow Na^+_{(aq)} + F^-_{(aq)} \nonumber \] Since Na is the conjugate of a strong base, it will have no effect on the pH or reactivity of the buffer. The addition of \(NaF\) to the solution will, however, increase the concentration of F in the buffer solution, and, consequently, by , lead to slightly less dissociation of the HF in the previous equilibrium, as well. The presence of significant amounts of both the conjugate acid, \(HF\), and the conjugate base, F , allows the solution to function as a buffer. This buffering action can be seen in the titration curve of a buffer solution. As we can see, over the working range of the buffer. pH changes very little with the addition of acid or base. Once the buffering capacity is exceeded the rate of pH change quickly jumps. This occurs because the conjugate acid or base has been depleted through . This principle implies that a larger amount of conjugate acid or base will have a greater buffering capacity. \[F^-_{(aq)} + H_3O^+_{(aq)} \rightleftharpoons HF_{(aq)} + H_2O_{(l)} \nonumber \] In this reaction, the conjugate base, F , will neutralize the added acid, H O , and this reaction goes to completion, because the reaction of F with H O has an equilibrium constant much greater than one. (In fact, the equilibrium constant the reaction as written is just the inverse of the K for HF: 1/K (HF) = 1/(6.6x10 ) = 1.5x10 .) So long as there is more F than H O , almost all of the H O will be consumed and the equilibrium will shift to the right, slightly increasing the concentration of HF and slightly decreasing the concentration of F , but resulting in hardly any change in the amount of H O present once equilibrium is re-established. \[HF_{(aq)} + OH^-_{(aq)} \rightleftharpoons F^-_{(aq)} + H_2O_{(l)} \nonumber \] In this reaction, the conjugate acid, HF, will neutralize added amounts of base, OH , and the equilibrium will again shift to the right, slightly increasing the concentration of F in the solution and decreasing the amount of HF slightly. Again, since most of the OH is neutralized, little pH change will occur. These two reactions can continue to alternate back and forth with little pH change. Buffers function best when the pK of the conjugate weak acid used is close to the desired working range of the buffer. This turns out to be the case when the concentrations of the conjugate acid and conjugate base are approximately equal (within about a factor of 10). For example, we know the K for hydroflouric acid is 6.6 x 10 so its pK = -log(6.6 x 10 ) = 3.18. So, a hydrofluoric acid buffer would work best in a buffer range of around pH = 3.18. For the weak base ammonia (NH ), the value of K is 1.8x10 , implying that the K for the dissociation of its conjugate acid, NH , is K /K =10 /1.8x10 = 5.6x10 . Thus, the pK for NH = 9.25, so buffers using NH /NH will work best around a pH of 9.25. (It's always the pK of the conjugate acid that determines the approximate pH for a buffer system, though this is dependent on the pK of the conjugate base, obviously.) When the desired pH of a buffer solution is near the pK of the conjugate acid being used (i.e., when the amounts of conjugate acid and conjugate base in solution are within about a factor of 10 of each other), the can be applied as a simple approximation of the solution pH, as we will see in the next section. In this example we will continue to use the hydrofluoric acid buffer. We will discuss the process for preparing a buffer of HF at a pH of 3.0. We can use the to calculate the necessary ratio of F and HF. \[pH = pKa + \log\dfrac{[Base]}{[Acid]} \nonumber \] \[3.0 = 3.18 + \log\dfrac{[Base]}{[Acid]} \nonumber \] \[\log\dfrac{[Base]}{[Acid]} = -0.18 \nonumber \] \[\dfrac{[Base]}{[Acid]} = 10^{-0.18} \nonumber \] \[\dfrac{[Base]}{[Acid]} = 0.66 \nonumber \] This is simply the ratio of the concentrations of conjugate base and conjugate acid we will need in our solution. However, what if we have 100 ml of 1 M HF and we want to prepare a buffer using NaF? How much Sodium Fluoride would we need to add in order to create a buffer at said pH (3.0)? We know from our Henderson-Hasselbalch calculation that the ratio of our base/acid should be equal to 0.66. From a table of molar masses, such as a periodic table, we can calculate the molar mass of NaF to be equal to 41.99 g/mol. HF is a weak acid with a K = 6.6 x 10 and the concentration of HF is given above as 1 M. Using this information, we can calculate the amount of F we need to add. The dissociation reaction is: \[HF_{(aq)} + H_2O_{(l)} \rightleftharpoons F^-_{(aq)} + H_3O^+_{(aq)} \nonumber \] We could use to calculate the concentration of F from HF dissociation, but, since K is so small, we can approximate that virtually all of the HF will remain undissociated, so the amount of F in the solution from HF dissociation will be negligible. Thus, the [HF] is about 1 M and the [F ] is close to 0. This will be especially true once we have added more F , the addition of which will even further suppress the dissociation of HF. We want the ratio of Base/Acid to be 0.66, so we will need [Base]/1M = 0.66. Thus, [F ] should be about 0.66 M. For 100 mL of solution, then, we will want to add 0.066 moles (0.1 L x 0.66 M) of F . Since we are adding NaF as our source of F , and since NaF completely dissociates in water, we need 0.066 moles of NaF. Thus, 0.066 moles x 41.99 g/mol = 2.767 g. Note that, since the conjugate acid and the conjugate base are both mixed into the same volume of solution in the buffer, the ratio of "Base/Acid" is the same whether we use a ratio of the "concentration of base over concentration of acid," OR a ratio of "moles of base over moles of acid." The pH of the solution does not, it turns out, depend on the volume! (This is only true so long as the solution does not get so dilute that the autoionization of water becomes an important source of H or OH . Such dilute solutions are rarely used as buffers, however.) Now that we have this nice F /HF buffer, let's see what happens when we add strong acid or base to it. Recall that the amount of F in the solution is 0.66M x 0.1 L = 0.066 moles and the amount of HF is 1.0 M x 0.1L = 0.10 moles. Let's double check the pH using the , but using moles instead of concentrations: pH = pK + log(Base/Acid) = 3.18 + log(0.066 moles F /0.10 moles HF) = 3.00 Good. Now let's see what happens when we add a small amount of strong acid, such as HCl. When we put HCl into water, it completely dissociates into H O and Cl . The Cl is the conjugate base of a strong acid so is inert and doesn't affect pH, and we can just ignore it. However, the H O can affect pH and it can also react with our buffer components. In fact, we already discussed what happens. The equation is: \[F^-_{(aq)} + H_3O^+_{(aq)} \rightleftharpoons HF_{(aq)} + H_2O_{(l)} \nonumber \] For every mole of H O added, an equivalent amount of the conjugate base (in this case, F ) will also react, and the equilibrium constant for the reaction is large, so the reaction will continue until one or the other is essentially used up. If the F is used up before reacting away all of the H O , then the remaining H O will affect the pH directly. In this case, the capacity of the buffer will have been exceeded - a situation one tries to avoid. However, for our example, let's say that the amount of added H O is smaller than the amount of F present, so our buffer capacity is NOT exceeded. For the purposes of this example, we'll let the added H O be equal to 0.01 moles (from 0.01 moles of HCl). Now, if we add 0.01 moles of HCl to 100 mL of pure water, we would expect the pH of the resulting solution to be 1.00 (0.01 moles/0.10 L = 0.1 M; pH = -log(0.1) = 1.0). However, we are adding the H O to a solution that has F in it, so the H O will all be consumed by reaction with F . In the process, the 0.066 moles of F is reduced: 0.066 initial moles F - 0.010 moles reacted with H O = 0.056 moles F remaining Also during this process, more HF is formed by the reaction: 0.10 initial moles HF + 0.010 moles from reaction of F with H O = 0.11 moles HF after reaction Plugging these new values into Henderson-Hasselbalch gives: pH = pK + log (base/acid) = 3.18 + log (0.056 moles F /0.11 moles HF) = 2.89 Thus, our buffer did what it should - it resisted the change in pH, dropping only from 3.00 to 2.89 with the addition of 0.01 moles of strong acid. | 10,440 | 2,110 |
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Protons and other electrophiles are not the only reactive species that initiate addition reactions to carbon-carbon double bonds. Curiously, this first became evident as a result of conflicting reports concerning the regioselectivity of HBr additions. As noted earlier, the acid-induced addition of HBr to 1-butene gave predominantly 2-bromobutane, the product. However, in some early experiments in which peroxide contaminated reactants were used, 1-bromobutane was the chief product. Further study showed that an alternative radical chain-reaction, initiated by peroxides, was responsible for the anti-Markovnikov product. This is shown by the following equations. The weak O–O bond of a peroxide initiator is broken homolytically by thermal or hight energy. The resulting alkoxy radical then abstracts a hydrogen atom from HBr in a strongly exothermic reaction. Once a bromine atom is formed it adds to the π-bond of the alkene in the first step of a chain reaction. This addition is regioselective, giving the more stable as an intermediate. The second step is carbon radical abstraction of another hydrogen from HBr, generating the anti-Markovnikov alkyl bromide and a new bromine atom. Each of the steps in this chain reaction is exothermic, so once started the process continues until radicals are lost to termination events. This free radical chain addition competes very favorably with the slower ionic addition of HBr described earlier, especially in non-polar solvents. It is important to note, however, that HBr is unique in this respect. The radical addition process is unfavorable for HCl and HI because one of the chain steps becomes endothermic (the second for HCl & the first for HI). Other radical addition reactions to alkenes have been observed, one example being the peroxide induced addition of carbon tetrachloride shown in the following equation The best known and most important use of free radical addition to alkenes is probably . Since the addition of carbon radicals to double bonds is energetically favorable, concentrated solutions of alkenes are prone to radical-initiated polymerization, as illustrated for propene by the following equation. The blue colored R-group represents an initiating radical species or a growing polymer chain; the propene monomers are colored maroon. The addition always occurs so that the more stable radical intermediate is formed. CH (CH )CH + CH (CH )CH- + CH (CH )CH | 2,457 | 2,111 |
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is a polar functional group that is made up a carbon and oxygen double bonded together. There are two simple classes of the carbonyl group: . Aldehydes have the carbon atom of the carbonyl group is bound to a hydrogen and ketones have the carbon atom of the carbonyl group is bound to two other carbons. Since the carbonyl group is extremely polar across the carbon-oxygen double bond, this makes it susceptible to addition reactions like the ones that occur in the pi bond of , especially by nucleophilic and electrophilic attack. There are three regions of reactivity for both aldehydes and ketones: the electron donating oxygen, the electron withdrawing carbon, and the carbon adjacent to the carbonyl group (labeled "alpha"). This module will only address the oxygen and electron withdrawing carbon areas of reactivity. Resonance structure is defined as any of two or more possible structures of the same compound that have identical geometry but different arrangements of their paired electrons; none of the structures have physical reality or adequately account for the properties of the compound. Examination of the resonance structures of the carbonyl group clearly shows its polar nature, and highlights the areas for either electrophilic or nucleophilic attack in the addition reaction. As a result of the dipole shown in the resonance structures, polar reagents such as LiAlH and NaBH (hydride reagents) or R'MgX (Grignard reagent) will reduce the carbonyl groups, and ultimately convert unsaturated aldehydes and ketones into unsaturated alcohols. Since these reagents are extremely basic, their addition reactions are irreversible. There are, however, addition reactions with less basic nucleophiles such as water, , and that are capable of establishing equilibria or reversible reactions. These less basic reagents can react with the carbonyl group via two pathways: nucleophilic addition-protonation and electrophilic addition-protonation. Under neutral or basic conditions, nucleophilic attack of the electrophilic carbon takes place. As the nucleophile approaches the electrophilic carbon, two valence electrons from the nucleophile form a covalent bond to the carbon. As this occurs, the electron pair from the pie bond transfers completely over to the oxygen which produces the intermediate . This , with a negative charge on oxygen is susceptible to protonation from a protic solvent like water or alcohol, giving the final addition reaction. Under acidic conditions, electrophilic attack of the carbonyl oxygen takes place. Initially, protonation of the carbonyl group at the oxygen takes place because of excess \(H^+\) all around. Once protonation has occurred, nucleophilic attack by the nucleophile finishes the addition reaction. It should be noted that electrophilic attack is extremely unlikely, however, a few carbonyl groups do become protonated initially to initiate addition through electrophilic attack. This type of reaction works best when the reagent being used is a very mildly basic nucleophile. 1. What is the best reagent to do the following conversion? 2. What is the Product of this reaction? 3. What is the Product of this reaction? 4. What is the Product of this reaction? 5. What is the product of this reaction? 1. D 2. 3. 4. 5. | 3,303 | 2,112 |
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As ionic solutions are very common in chemistry, having to measure all activity coefficients (\(γ_{\pm}\)) for all possible solute-solvent combinations is a pretty daunting task, even though in times past extensive tabulation has taken place. We should be grateful for the rich legacy that our predecessors have left us in this respect (it would be hard to get any funding to do such tedious work today). Of course, it would be very desirable to be able to \(γ_{\pm}\) values from first principles or if that fails by semi-empirical means. Fortunately, considerable progress has been made on this front as well. We can only scratch the surface of that topic in this course and will briefly discuss the simplest approach due to Debye and Hückel Debye and Hückel came up with a theoretical expression that makes is possible to predict mean ionic activity coefficients as dilute concentrations. The theory considers the vicinity of each ion as an atmosphere-like cloud of charges of opposite sign that cancels out the charge of the central ion (Figure \(\Page {1}\)). From a distance the cloud looks neutral. The quantity \(1/κ\) is a measure for the size of this cloud and \(\kappa\) is the . Its size depends on the concentration of all other ions. To take the effect from all other ions into account, it is useful to define the (\(I\)) as: \[I =\dfrac{1}{2} \sum m_iz_i^2 \nonumber \] where \(m_i\) is the molality of ion \(i\) and \(z_i\) is its charge coefficient. Note that highly charged ions (e.g. \(z=3+\)) contribute strongly (nine times more than +1 ions), but the formula is in the molality. Using the ionic strength the Debye-length becomes: \[κ^2 = constant \, I \nonumber \] The constant contains \(kT\) and \(ε_rε_o\) in the denominator and the number of Avogadro \(N_A\) and the square of the charge of the electron \(e\) in the numerator: \[constant= 2000 \dfrac{e^2N_A}{ε_rε_okT} \nonumber \] The Debye length and the logarithmic mean ionic activity coefficient are proportional: \[\ln γ_{\pm} \propto κ \nonumber \] Again there are a number of factors in the proportionality constant: \[\ln γ_± = -|q_+q_-| \dfrac{κ}{8πε_rε_okT} \nonumber \] The factors \(ε_r\) and \(ε_o\) are the relative permittivity of the medium and the permittivity of vacuum, respectively. Note that the factor \(8πε_rε_o\) is to the SI system of units. In cgs units the expression would look different, because the permittivities are defined differently in that system If there is only one salt being dissolved, the ionic strength depends linearly on its concentration, the Debye length κ and \(\ln γ_±\), therefore, go as the (usually molality): \[\ln \gamma_{\pm} \propto \sqrt{m} \nonumber \] If there are other ions present the ionic strength involves of them. This fact is sometimes used to keep ionic strength constant while changing the concentration of one particular ion. Say we wish to lower the concentration of Cu in a redox reaction but we want to keep activity coefficients the same as much as possible. We could then it by an ion of the same charge say Zn that does not partake in the reaction. A good way to do that is to dilute the copper solution with a zinc solution of the same concentration instead of with just solvent. The mean activity coefficient is given by the logarithm of this quantity as follows: \[\log _{10}\gamma _{\pm }=-Az_{j}^{2}{\frac {\sqrt {I}}{1+Ba_{0}{\sqrt {I}}}} \label{DH}\] with: \[ A={\frac {e^{2}B}{2.303\times 8\pi \epsilon _{0}\epsilon _{r}k_{\rm {B}}T}}\] \[B=\left({\frac {2e^{2}N}{\epsilon _{0}\epsilon _{r}k_{\rm {B}}T}}\right)^{1/2}\] where \(I\) is the ionic strength and \(a_0\) is a parameter that represents the distance of closest approach of ions. For aqueous solutions at 25 °C \(A = 0.51\, mol^{−1}/2dm^{3/2}\) and \(B = 3.29 nm^{−1}mol^{−1}/2dm^{3/2}\). Unfortunately this theory only works at very low concentrations and is therefore also known as the (Figure \(\Page {2}\)). There are a number of refinements that aim at extending the range of validity of the theory to be able to work at somewhat higher concentrations. These are discussed in the next section. The most significant aspect of Equation \ref{DH} is the prediction that the mean activity coefficient is a function of rather than the electrolyte concentration. For very low values of the ionic strength the value of the denominator in the expression above becomes nearly equal to one. In this situation the mean activity coefficient is proportional to the square root of the ionic strength. When a solid is formed by a reaction from solution it is sometimes possible that it remains dispersed as very small particles in the solvent. The sizes typically range in the nanometers This is why it has become fashionable to call them , although they had been known as since the mid nineteenth century. They are smaller than the wavelength of the visible reason. This causes liquids that contain them to remain , although they can at times be beautifully colored. A good example is the reduction of AuCl with citrate to metallic gold. This produces clear wine red solutions, even at tiny gold concentrations. \[\ce{2n AuCl4(aq)^{-} + n \,citrate^{3-}(aq) + 2n\, H_2O(l) \rightarrow} \\[4pt] \ce{2n\, Au(colloid) + 3n\, CH2O(aq) + 3n\, CO2(g) + 8n \,Cl^{-}(aq) + 3n\, H^{+}(aq)} \nonumber \] The reason the gold does not precipitate completely is typically that the nanoparticle (AuNP) formed during the reaction are by the attachment of some of the ionic species in solution to its surface. This results in an charged particle with an atmosphere with a certain Debye length around it (Figure \(\Page {3}\)). This charged cloud prevents the particle form coalescing with other particles by electrostatic repulsion. Such a system is called a Of course these systems are . Often they have a pretty small threshold to crashing to a real precipitate under influence of the strong van der Waals interactions that the particles experience once they manage to get in close contact. Under the right conditions colloids can survive for a long time. Some gold colloids prepared by Faraday in the 1850's are still stable today. It will be clear from the above that addition of a salt -particularly containing highly charged ions like 3+ or 3-- may destabilize the colloid because the ionic strength will changed drastically and this will affect the Debye length. | 6,414 | 2,113 |
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A comparison of the properties and reactivity of aldehydes and ketones with those of the alkenes is warranted, since both have a double bond functional group. Because of the greater electronegativity of oxygen, the carbonyl group is polar, and aldehydes and ketones have larger molecular (D) than do alkenes. The resonance structures in Figure 1 illustrate this polarity, and the relative dipole moments of formaldehyde, other aldehydes and ketones confirm the stabilizing influence that alkyl substituents have on carbocations (the larger the dipole moment the greater the polar character of the carbonyl group). We expect, therefore, that aldehydes and ketones will have higher boiling points than similar sized alkenes. Furthermore, the presence of oxygen with its non-bonding electron pairs makes aldehydes and ketones hydrogen-bond acceptors, and should increase their water solubility relative to hydrocarbons. Specific examples of these relationships are provided in the following table. : Resonance structures (CH ) C=CH 56 -7.0 ºC 0.04 g/100 (CH ) C=O 58 56.5 ºC infinite CH CH CH CH=CH 70 30.0 ºC 0.03 g/100 CH CH CH CH=O 72 76.0 ºC 7 g/100
96 103.0 ºC insoluble 98 155.6 ºC 5 g/100 The polarity of the carbonyl group also has a profound effect on its chemical reactivity, compared with the non-polar double bonds of alkenes. Thus, reversible addition of water to the carbonyl function is fast, whereas water addition to alkenes is immeasurably slow in the absence of a strong acid catalyst. Curiously, relative bond energies influence the thermodynamics of such addition reactions in the opposite sense. The C=C of alkenes has an average bond energy of 146 kcal/mole. Since a C–C σ-bond has a bond energy of 83 kcal/mole, the π-bond energy may be estimated at 63 kcal/mole (i.e. less than the energy of the sigma bond). The C=O bond energy of a carbonyl group, on the other hand, varies with its location, as follows: The C–O σ-bond is found to have an average bond energy of 86 kcal/mole. Consequently, with the exception of formaldehyde, the carbonyl function of aldehydes and ketones has a π-bond energy greater than that of the sigma-bond, in contrast to the pi-sigma relationship in C=C. This suggests that addition reactions to carbonyl groups should be thermodynamically disfavored, as is the case for the addition of water. All of this is summarized in the following diagram (ΔHº values are for the addition reaction). Although the addition of water to an alkene is exothermic and gives a stable product (an alcohol), the uncatalyzed reaction is extremely slow due to a high activation energy. The reverse reaction (dehydration of an alcohol) is even slower, and because of the kinetic barrier, both reactions are practical only in the presence of a strong acid. In contrast, both the endothermic addition of water to a carbonyl function, and the exothermic elimination of water from the resulting -diol are fast. The inherent polarity of the carbonyl group, together with its increased basicity (compared with alkenes), lowers the transition state energy for both reactions, with a resulting increase in rate. Acids and bases catalyze both the addition and elimination of water. Proof that rapid and reversible addition of water to carbonyl compounds occurs is provided by experiments using isotopically labeled water. If a carbonyl reactant composed of O (colored blue above) is treated with water incorporating the O isotope (colored red above), a rapid exchange of the oxygen isotope occurs. This can only be explained by the addition-elimination mechanism shown here. ), | 3,634 | 2,114 |
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The boron family contains elements in group 13 of the periodic talbe and include the semi-metal (B) and the metals (Al), (Ga), (In), and (Tl). Aluminum, gallium, indium, and thallium have three electrons in their outermost shell (a full s orbital and one electron in the p orbital) with the valence electron configuration ns np . The elments of the boron family adopts oxidation states +3 or +1. The +3 oxidation states are favorable except for the heavier elements, such as Tl, which prefer the +1 oxidation state due to its stability; this is known as the . The elements generally follow except for certain Tl deviations:
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https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumentation_and_Analysis/Mass_Spectrometry/Organic_Compounds_Containing_Halogen_Atoms |
This page explains how the M+2 peak in a mass spectrum arises from the presence of chlorine or bromine atoms in an organic compound. It also deals briefly with the origin of the M+4 peak in compounds containing two chlorine atoms. The molecular ion peaks (M+ and M+2) each contain one chlorine atom - but the chlorine can be either of the two chlorine isotopes, Cl and Cl. The molecular ion containing the Cl isotope has a relative formula mass of 78. The one containing Cl has a relative formula mass of 80 - hence the two lines at m/z = 78 and m/z = 80. Notice that the peak heights are in the ratio of 3 : 1. That reflects the fact that chlorine contains 3 times as much of the Cl isotope as the Cl one. That means that there will be 3 times more molecules containing the lighter isotope than the heavier one. So . . . if you look at the molecular ion region, and find two peaks separated by 2 m/z units and with a ratio of 3 : 1 in the peak heights, that tells you that the molecule contains 1 chlorine atom. You might also have noticed the same pattern at m/z = 63 and m/z = 65 in the mass spectrum above. That pattern is due to fragment ions also containing one chlorine atom - which could either be Cl or Cl. The fragmentation that produced those ions was: The lines in the molecular ion region (at m/z values of 98, 100 ands 102) arise because of the various combinations of chlorine isotopes that are possible. The carbons and hydrogens add up to 28 - so the various possible molecular ions could be: 28 + 35 + 35 = 98 28 + 35 + 37 = 100 28 + 37 + 37 = 102 If you have the necessary math, you could show that the chances of these arrangements occurring are in the ratio of 9:6:1 - and this is the ratio of the peak heights. If you don't know the right bit of math, just learn this ratio! So . . . if you have 3 lines in the molecular ion region (M+, M+2 and M+4) with gaps of 2 m/z units between them, and with peak heights in the ratio of 9:6:1, the compound contains 2 chlorine atoms. Bromine has two isotopes, Br and Br in an approximately 1:1 ratio (50.5 : 49.5 if you want to be fussy!). That means that a compound containing 1 bromine atom will have two peaks in the molecular ion region, depending on which bromine isotope the molecular ion contains. Unlike compounds containing chlorine, though, the two peaks will be very similar in height. The carbons and hydrogens add up to 29. The M+ and M+2 peaks are therefore at m/z values given by: 29 + 79 = 108 29 + 81 = 110 Hence, if two lines in the molecular ion region are observed with a gap of 2 m/z units between them and with almost equal heights, this suggests the presence of a bromine atom in the molecule. Jim Clark ( ) | 2,714 | 2,116 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Electrophilic_Addition_Reactions/What_is_Electrophilic_Addition%3F |
Electrophilic addition happens in many of the reactions of compounds containing carbon-carbon double bonds - the alkenes. We are going to start by looking at ethene, because it is the simplest molecule containing a carbon-carbon double bond. What is true of C=C in ethene will be equally true of C=C in more complicated alkenes. Ethene, C H , is often modeled as shown on the right. The double bond between the carbon atoms is, of course, two pairs of shared electrons. What the diagram doesn't show is that the two pairs aren't the same as each other. One of the pairs of electrons is held on the line between the two carbon nuclei as you would expect, but the other is held in a molecular orbital above and below the plane of the molecule. A molecular orbital is a region of space within the molecule where there is a high probability of finding a particular pair of electrons. In this diagram, the line between the two carbon atoms represents a normal bond - the pair of shared electrons lies in a molecular orbital on the line between the two nuclei where you would expect them to be. This sort of bond is called a sigma bond. The other pair of electrons is found somewhere in the shaded part above and below the plane of the molecule. This bond is called a pi bond. The electrons in the pi bond are free to move around anywhere in this shaded region and can move freely from one half to the other. The pi electrons are not as fully under the control of the carbon nuclei as the electrons in the sigma bond and, because they lie exposed above and below the rest of the molecule, they are relatively open to attack by other things. An electrophile is something which is attracted to electron-rich regions in other molecules or ions. Because it is attracted to a negative region, an electrophile must be something which carries either a full positive charge, or has a slight positive charge on it somewhere. Ethene and the other alkenes are attacked by electrophiles. The electrophile is normally the slightly positive ( +) end of a molecule like hydrogen bromide, HBr. Electrophiles are strongly attracted to the exposed electrons in the pi bond and reactions happen because of that initial attraction - as you will see shortly. You might wonder why fully positive ions like sodium, Na , don't react with ethene. Although these ions may well be attracted to the pi bond, there is no possibility of the process going any further to form bonds between sodium and carbon, because sodium forms ionic bonds, whereas carbon normally forms covalent ones. In a sense, the pi bond is an unnecessary bond. The structure would hold together perfectly well with a single bond rather than a double bond. The pi bond often breaks and the electrons in it are used to join other atoms (or groups of atoms) onto the ethene molecule. In other words, ethene undergoes addition reactions. For example, using a general molecule X-Y . . . An addition reaction is a reaction in which two molecules join together to make a bigger one. Nothing is lost in the process. All the atoms in the original molecules are found in the bigger one. An electrophilic addition reaction is an addition reaction which happens because what we think of as the "important" molecule is attacked by an electrophile. The "important" molecule has a region of high electron density which is attacked by something carrying some degree of positive charge. The mechanism for the reaction between ethene and a molecule X-Y. The slightly positive X atom is an electrophile and is attracted to the exposed pi bond in the ethene. Now imagine what happens as they approach each other. You are now much more likely to find the electrons in the half of the pi bond nearest the XY. As the process continues, the two electrons in the pi bond move even further towards the X until a covalent bond is made. The electrons in the X-Y bond are pushed entirely onto the Y to give a negative Y ion. An ion in which the positive charge is carried on a carbon atom is called a carbocation or a carbonium ion (an older term). In the final stage of the reaction the electrons in the lone pair on the Y ion are strongly attracted towards the positive carbon atom. They move towards it and form a co-ordinate (dative covalent) bond between the Y and the carbon. The movements of the various electron pairs are shown using curly arrows. Don't leave this page until you are sure that you understand how this relates to the electron pair movements drawn in the previous diagrams. Jim Clark ( ) | 4,535 | 2,117 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/10%3A_Electrochemistry/10.03%3A_Half_Cells_and_Standard_Reduction_Potentials |
Much like \(G\) itself, \(E\) can only be measured as a , so a convention is used to set a zero to the scale. Toward this end, convention sets the reduction potential of the standard hydrogen electrode (SHE) to 0.00 V. \[\ce{Zn \rightarrow Zn^{2+} + 2e^{-}} \nonumber \] wtih \(E_{ox}^o = 0.763\, V\) \[\ce{2 H^{+} + 2 e^{-} \rightarrow H2 } \nonumber \] with \(E_{red}^o = 0.000 \,V\) The standard hydrogen electrode is constructed so that H gas flows over an inert electrode made of platinum, and can interact with an acid solution which provides H for the half reaction \[\ce{2 H^+(aq) + 2 e^{-} -> H_2(g)} \nonumber \] Both H and H need to have unit activity (or fugacity), which if the solution and gas behave ideally means a concentration of 1 M and a pressure of 1 bar. Standard reduction potentials can be measured relative to the convention of setting the reduction potential of the (SHE) to zero. A number of values are shown in Table P1. Which pair of reactants will produce a spontaneous reaction if everything is present in its standard state at 25 °C? The species with the standard reduction potential ( ) will force the other to oxidize. From the table, \[\ce{Cu^{2+} + 2 e^{-} \rightarrow Cu} \nonumber \] with \(0.337\, V\) \[ \ce{Fe^{2+} + 2 e^{-} \rightarrow Fe} \nonumber \] with \(-0.440\, V\) So the iron half-reaction will flip (so that iron is oxidizing) and the spontaneous reaction under standard conditions will be \[Cu^{2+} + Fe \rightarrow Cu + Fe^{2+} \nonumber \] with \(E^o = 0.777\, V\) Using values measured relative to the SHE, it is fairly easy to calculate the standard cell potential of a given reaction. For example, consider the reaction \[\ce{ 2 Ag^{+}(aq) + Cu(s) \rightarrow 2 Ag(s) + Cu^{2+}(aq)} \nonumber \] Before calculating the cell potential, we should review a few definitions. The anode half reaction, which is defined by the half-reaction in which oxidation °Ccurs, is \[\ce{Cu(s) \rightarrow Cu^{2+}(aq) + 2 e^{-}} \nonumber \] And the cathode half-reaction, defined as the half-reaction in which reduction takes place, is \[\ce{Ag^+(aq) + e- \rightarrow Ag(s)}\nonumber \] Using , the conditions (such as the concentrations of the ions in solution) can be represented. In the standard cell notation, the anode is on the left-hand side, and the cathode on the right. The two are typically separated by a , which is designated by a double vertical line. A single vertical line indicates a phase boundary. Hence for the reaction above, if the silver ions are at a concentration of 0.500 M, and the copper (II) ions are at a concentration of 0.100 M, the standard cell notation would be Calculate the cell potential at 25 °C for the cell indicated by \[\ce{Cu(s) | Cu^{2+}(aq, \,0.100\, M) || Ag^+ (aq,\, 0.500\, M) | Ag(s)} \nonumber \] In order to calculate the cell potential (\(E\)), the standard cell potential must first be obtained. The standard cell potential at 25 °C is given by \[\begin{align*} E_{cell} = E^o_{cathode} -E^o_{anode} \\[4pt] &= 0.799 \,V - 0.337\,V \\[4pt] &=0.462\,V \end{align*} \] And for a cell at non-standard conditions, such as those indicated above, the Nernst equation can be used to calculate the cell potential. At 25 °C, The cell potential is given by \[ \begin{align*} E_{cell} &= E^o_{cell} - \dfrac{RT}{nF} \ln \left( \dfrac{[Cu^{2+}]}{[Ag^+]} \right) \\[4pt] &= 0.462\,V - \dfrac{(8.314 \,J/(mol\,K) (298\,K) }{2(96484\,C)} \ln \left( \dfrac{0.100\,M}{0.500\,M} \right) \end{align*} \] Noting that \(1\, J/C = 1\, V\), \[E = 0.483\,V \nonumber \] Calculate the cell potential at 25 °C for the cell defined by \[Ni(s) | Ni^{2+}\, (aq, \,0.500\, M) || Cu(s) | Cu^{2+}(aq, \,0.100\, M) \nonumber \] We will use the Nernst equation. First, we need to determine \(E^o\). Using Table P1, it is apparent that \[ \ce{Cu^{2 }+ 2 e^{-} \rightarrow Cu } \nonumber \] \(E^o = 0.337 \,V\) \[\ce{ Ni^{2+} + 2 e^{-} \rightarrow Ni} \nonumber \] with \(E^o = -0.250\, V\) So copper, having the larger reduction potential will be the cathode half-reaction while forcing nickel to oxidize, making it the anode. So E for the cell will be given by \[ \begin{align*} E_{cell} &= E^o_{cathode} -E^o_{anode} \\[4pt] &= 0.337 \,V -(-0.250\,V) \\[4pt] = 0.587\,V \end{align*} \] And the cell potential is then given by the Nernst Equation \[ \begin{align*} E_{cell} &= E^o_{cell} - \dfrac{RT}{nF} \ln Q \\[4pt] &= 0.587 - \dfrac{(8.314 \,J/(mol\,K) (298\,K) }{2(96484\,C)} \ln \left( \dfrac{0.500\,M}{0.100\,M} \right) \\[4pt] &= 0.566\,V \end{align*} \] A typical galvanic electrochemical cell can be constructed similar to what is shown in the diagram above. The electrons flow from the anode (the electron source) to the cathode (the electron sink.) The salt bridge allows for the flow of ions to complete the circuit while minimizing the introduction of a junction potential. | 4,872 | 2,118 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/01%3A_Introduction_to_Organic_Chemistry/1.01%3A_Prelude_to_Organic_Chemistry |
You now are starting the study of organic chemistry, which is the chemistry of compounds of carbon. In this introductory chapter, we will tell you something of the background and history of organic chemistry, something of the problems and the rewards involved, and something of our philosophy of what is important for you to learn so that you will have a reasonable working knowledge of the subject, whether you are just interested in chemistry or plan for a career as a chemist, an engineer, a physician, a biologist, and so on. The subject is very large; more than two million organic compounds have been isolated or prepared and characterized, yet the number of guiding principles is relatively small. You certainly will not learn everything about organic chemistry from this book, but with a good knowledge of the guiding principles, you will be able later to find out what you need to know either from the chemical literature, or directly by experiment in the laboratory. Unfortunately, learning about and learning how to use organic chemistry is not a straightforward process, wherein one step leads to another in a simple, logical way like Euclidean geometry. A more realistic analogy would be to consider yourself thrust into and required to deal successfully with a sizable group of strangers speaking a new and complex language. In such a situation, one has to make many decisions- how much of the language to learn at the outset? Which people are the best to interact with first? Which will be the most important to know in the long run? How well does one have to know each person? How much does one have to know about the history of the group to understand their interactions? These are difficult questions, and a period of confusion, if not anxiety, is expected in any attempt to complete a task of this kind in a set, brief period of time. Clearly, it would be difficult to learn all at once the language, the people, and the interactions between them. Nonetheless, this is pretty much what is expected of you in learning organic chemistry. A number of approaches have been devised to help you become familiar with and use organic chemistry. In terms of our analogy, one way is to learn the language, then the relationships between the people, and finally, well prepared, to proceed to interact with the people singly and then in groups. Such an approach may be logical in concept, but is not to everyone's taste as a way to learn. Many of us do better with an interactive approach, where language, relationships, and people are worked out more or less in concert, with attendant misunderstandings and ambiguities. What we will try to do is to introduce some of the important basic concepts and the elements of the language of organic chemistry, then show how these are used in connection with various classes of compounds. The initial round will be a fairly extensive one and you should not expect to be able to master everything at once. This will take practice and we will provide opportunity for practice. One of the appealing yet bothersome features of modern organic chemistry is its extraordinary vitality. Unlike Euclidean geometry or classical mechanics, it is evolving rapidly and many of the concepts introduced in this book are either new or have been drastically modified in the past ten years. Every issue of the current chemical journals has material of such basic interest that one would like to include it in an introductory course. Truly, those who write organic textbooks write on water, with no hope of producing definitive book. Things just change too fast. Despite this, one of the great ideas of modern civilization, namely that organic compounds can be described in terms of more or less simple three-dimensional molecular structures with atoms held together by chemical bonds, has persisted for more than one hundred years and seems unlikely to be superseded, no matter how much it is refined and modified. and (1977) | 3,977 | 2,119 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/03%3A_First_Law_of_Thermodynamics/3.05%3A_Temperature_Dependence_of_Enthalpy |
It is often required to know thermodynamic functions (such as enthalpy) at temperatures other than those available from tabulated data. Fortunately, the conversion to other temperatures is not difficult. At constant pressure \[ dH = C_p \,dT \nonumber \] And so for a temperature change from \(T_1\) to \(T_2\) \[ \Delta H = \int_{T_2}^{T_2} C_p\, dT \label{EQ1} \] Equation \ref{EQ1} is often referred to as . If \(C_p\) is independent of temperature, then \[\Delta H = C_p \,\Delta T \label{intH} \] If the temperature dependence of the heat capacity is known, it can be incorporated into the integral in Equation \ref{EQ1}. A common model used to fit heat capacities over broad temperature ranges is \[C_p(T) = a+ bT + \dfrac{c}{T^2} \label{EQ15} \] After combining Equations \ref{EQ15} and \ref{EQ1}, the enthalpy change for the temperature change can be found obtained by a simple integration \[ \Delta H = \int_{T_1}^{T_2} \left(a+ bT + \dfrac{c}{T^2} \right) dT \label{EQ2} \] Solving the definite integral yields \[ \begin{align} \Delta H &= \left[ aT + \dfrac{b}{2} T^2 - \dfrac{c}{T} \right]_{T_1}^{T_2} \\ &= a(T_2-T_1) + \dfrac{b}{2}(T_2^2-T_1^2) - c \left( \dfrac{1}{T_2} - \dfrac{1}{T_1} \right) \label{ineq} \end{align} \] This expression can then be used with experimentally determined values of \(a\), \(b\), and \(c\), some of which are shown in the following table. What is the molar enthalpy change for a temperature increase from 273 K to 353 K for Pb(s)? The enthalpy change is given by Equation \ref{EQ1} with a temperature dependence \(C_p\) given by Equation \ref{EQ1} using the parameters in Table \(\Page {1}\). This results in the integral form (Equation \ref{ineq}): \[ \Delta H = a(T_2-T_1) + \dfrac{b}{2}(T_2^2-T_1^2) - c \left( \dfrac{1}{T_2} - \dfrac{1}{T_1} \right) \nonumber \] when substituted with the relevant parameters of Pb(s) from Table \(\Page {1}\). \[ \begin{align*} \Delta H = \,& (22.14\, \dfrac{J}{mol\,K} ( 353\,K - 273\,K) \\ & + \dfrac{1.172 \times 10^{-2} \frac{J}{mol\,K^2}}{2} \left( (353\,K)^2 - (273\,K)^2 \right) \\ &- 9.6 \times 10^4 \dfrac{J\,K}{mol} \left( \dfrac{1}{(353\,K)} - \dfrac{1}{(273\,K)} \right) \\ \Delta H = \, & 1770.4 \, \dfrac{J}{mol}+ 295.5\, \dfrac{J}{mol}+ 470.5 \, \dfrac{J}{mol} \\ = & 2534.4 \,\dfrac{J}{mol} \end {align*} \] For chemical reactions, the reaction enthalpy at differing temperatures can be calculated from \[\Delta H_{rxn}(T_2) = \Delta H_{rxn}(T_1) + \int_{T_1}^{T_2} \Delta C_p \Delta T \nonumber \] The enthalpy of formation of NH (g) is -46.11 kJ/mol at 25 C. Calculate the enthalpy of formation at 100 C. \[\ce{N2(g) + 3 H2(g) \rightleftharpoons 2 NH3(g)} \nonumber \] with \(\Delta H \,(298\, K) = -46.11\, kJ/mol\) \[ \begin{align*} \Delta H (373\,K) & = \Delta H (298\,K) + \Delta C_p\Delta T \\ & = -46110 +\dfrac{J}{mol} \left[ 2 \left(35.06 \dfrac{J}{mol\,K}\right) - \left(29.12\, \dfrac{J}{mol\,K}\right) - 3\left(28.82\, \dfrac{J}{mol\,K}\right) \right] (373\,K -298\,K) \\ & = -49.5\, \dfrac{kJ}{mol} \end{align*} \] | 3,053 | 2,120 |
https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Environmental_Toxicology_(van_Gestel_et_al.)/05%3A_Population_Community_and_Ecosystem_Ecotoxicology/5.04%3A_Trait-based_approaches |
: Paul J. Van den Brink : Nico van den Brink, Michiel Kraak, Alexa Alexander-Trusiak You should be able to Sensitivity, levels of biological organisation, species traits, recovery, indirect effects It is impossible to assess the sensitivity of all species to all chemicals. Risk assessments therefore, need methods to extrapolate the sensitivity of a limited number of species to all species present in the environment is desired. Statistical approaches, like the species sensitivity distribution concept, perform this extrapolation by fitting a statistical distribution (e.g. log-normal distribution) to a selected set of sensitivity data (e.g. 96h-EC50 data) in order to obtain a distribution of the sensitivity of all species. From this distribution a threshold value associated with the lower end of the distribution can be chosen and used as a protective threshold value (Figure 1). The disadvantage of this approach is that it does not include mechanistic knowledge on what determines species' sensitivity and uses species taxonomy rather than their characteristics. To overcome these and other problems associated with a taxonomy based approach (see Van den Brink et al., 2011 for a review) traits-based bioassessment approaches have been developed for assessing the effects of chemicals on aquatic ecosystem. In traits-based bioassessment approaches, species are not represented by their taxonomy but by their traits. A trait is a phenotypic or ecological characteristic of an organism, usually measured at the individual level but often applied as the average state/condition of a species. Examples of traits are body size, feeding habits, food preference, mode of respiration and lipid content. Traits describe the physical characteristics, ecological niche, and functional role of a species within the ecosystem. The recognized strengths of traits-based bioassessment approaches include: (1) traits add mechanistic and diagnostic knowledge, (2) traits are transferrable across geographies, (3) traits require no new sampling methodology as data that are currently collected can be used, (4) the use of traits has a long-standing tradition in ecology and can supplement taxonomic analysis. When traits are used to study effects of chemical stressors on ecosystem structure (community composition) and function (e.g. nutrient cycling) it is important to make a distinction between response and effects traits (Figure 2). Response traits are traits that enable a response of the species to the exposure to a chemical. An example of a response trait may be size related surface area of an organism. Smaller organisms have relatively large surface areas, because their surface to volume ratio is higher than for larger animals Herewith, the uptake rate of the chemical stressor is generally higher in smaller animals compared to larger ones (Rubach et al., 2012). Effects traits of organisms influence the surrounding environment by the organisms, by altering the structure and functioning of the ecosystem. An example of an effect trait is the food preference of an organism. For instance, if the small (response trait) and herewith sensitive organisms happen to be herbivorous (effect trait) an increase in algal biomass may be expected when the organisms are affected (Van den Brink, 2008). So, to be able to predict ecosystem level responses it is important to know the (cor)relations between response and effect traits as traits are not independent from each other but can be linked phylogenetically or mechanistically and thus form trait syndromes (Van den Brink et al., 2011). One of the holy grails of ecotoxicology is to find out which species traits make one species more sensitive to a chemical stressor than another one. In the past, two approaches have been used to assess the (cor)relationships between species traits and their sensitivity, one based on empirical correlations between species' traits and their sensitivity as represented by EC50's (Rico and Van den Brink, 2015) and one based on a more mechanistic approach using toxicokinetic/toxicodynamic experiments and models (Rubach et al., 2012). Toxicokinetic-toxicodynamic models (TKTD models) simulate the time-course of processes leading to toxic effects on organisms (Jager et al., 2011). Toxicokinetics describe what an individual does with the chemical and, in their simplest form, include the processes of uptake and elimination, thereby translating an external concentration of a toxicant to an internal body concentration over time (see Section on ). Toxicodynamics describes what the chemical does with the organism, herewith linking the internal concentration to the effect at the level of the individual organism over time (e.g., mortality) (Jager et al., 2011) (see Sections on and on ). Rubach et al. (2012) showed that almost 90% of the variation in uptake rates and 80% of the variation in elimination rates of an insecticide in a range of 15 freshwater arthropod species could be explained by 4 species traits. These traits were: i) surface area (without gills), ii) detritivorous feeding, iii) using atmospheric oxygen and iv) phylogeny in case of uptake, and i) thickness of exoskeleton, ii) complete sclerotization, iii) using dissolved oxygen and iv) % lipid of dry weight in case of elimination. For most of these traits, a mechanistic hypothesis between the traits and their influence on the uptake and elimination can be made (Rubach et al., 2012). For instance, a higher surface area to volume ratio increases the uptake of the chemical, so uptake is expected to be higher in small animals compared to larger animals. This shows that it is possible to construct mechanistic models that are able to predict the toxicokinetics of chemicals in species and herewith the sensitivity of species to chemicals based on their traits. Traits determining the way organisms within an ecosystem react to chemical stress are related to the intrinsic sensitivity of the organisms on the one hand (response traits) and their recovery potential and food web relations (effect traits) on the other hand (Van den Brink, 2008). Recovery of aquatic invertebrates is, for instance, determined by traits like number of life cycles per year, the presence of insensitive life stages like resting eggs, dispersal ability and having an aerial life stage (Gergs et al., 2011) (Figure 3). Besides recovery, effect traits will also determine how individual level effects will propagate to higher levels of biological organisation like the community or ecosystem level. For instance, when are affected by a chemical, their trait related to food preference (algae) will ensure that, under nutrient-rich conditions, the algae will not be subjected to top-down control and will increase in abundance. The latter effects are called indirect effects, which are not a direct result of the exposure to the toxicant but an indirect one through competition, food-web relationships, etc.. Gergs, A., Classen, S., Hommen, U. (2011). Identification of realistic worst case aquatic macroinvertebrate species for prospective risk assessment using the trait concept. Environmental Science and Pollution Research 18, 1316-1323. Jager, T., Albert, C., Preuss, T.G., Ashauer, R. (2011). General unified threshold model of survival-A toxicokinetic-toxicodynamic framework for ecotoxicology. Environmental Science and Technology 45, 2529-2540 Rico, A., Van den Brink, P.J. (2015). Evaluating aquatic invertebrate vulnerability to insecticides based on intrinsic sensitivity, biological traits and toxic mode-of-action. Environmental Toxicology and Chemistry 34, 1907-1917. Rubach, M.N., D.J. Baird, M-C. Boerwinkel, S.J. Maund, I. Roessink, Van den Brink P.J. (2012). Species traits as predictors for intrinsic sensitivity of aquatic invertebrates to the insecticide chlorpyrifos. Ecotoxicology 21, 2088-2101. Van den Brink, P.J. (2008). Ecological risk assessment: from book-keeping to chemical stress ecology. Environmental Science and Technology 42, 8999-9004. Van den Brink P.J., Alexander, A., Desrosiers, M., Goedkoop, W., Goethals, P., Liess, M., Dyer, S. (2011). Traits-based approaches in bioassessment and ecological risk assessment: strengths, weaknesses, opportunities and threats. Integrated Environmental Assessment and Management 7, 198-208. Which two traits may determine the sensitivity of invertebrates to insecticides? Will two daphnids which are equal except in their size be equally sensitive to a chemical? What is the difference between a response and an effect trait? Which traits are of importance for the recovery of impacted populations? How will the insecticide-induced death of water fleas propagate to the ecosystem level under eutrophic circumstances, and what are these types of effects called? | 8,797 | 2,121 |
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Many of the physical properties of solutions differ significantly from those of the pure substances discussed in earlier chapters, and these differences have important consequences. For example, the limited temperature range of liquid water (0°C–100°C) severely limits its use. Aqueous solutions have both a lower freezing point and a higher boiling point than pure water. Probably one of the most familiar applications of this phenomenon is the addition of ethylene glycol (“antifreeze”) to the water in an automobile radiator. This solute lowers the freezing point of the water, preventing the engine from cracking in very cold weather from the expansion of pure water on freezing. Antifreeze also enables the cooling system to operate at temperatures greater than 100°C without generating enough pressure to explode. Changes in the freezing point and boiling point of a solution depend primarily on the number of solute particles present rather than the kind of particles. Such properties of solutions are called colligative properties (from the Latin colligatus, meaning “bound together” as in a quantity). As we will see, the vapor pressure and osmotic pressure of solutions are also colligative properties. When we determine the number of particles in a solution, it is important to remember that not all solutions with the same molarity contain the same concentration of solute particles. Consider, for example, 0.01 M aqueous solutions of sucrose, \(\ce{NaCl}\), and \(\ce{CaCl_2}\). Because sucrose dissolves to give a solution of neutral molecules, the concentration of solute particles in a 0.01 M sucrose solution is 0.01 M. In contrast, both \(\ce{NaCl}\) and \(\ce{CaCl_2}\) are ionic compounds that dissociate in water to yield solvated ions. As a result, a 0.01 M aqueous solution of \(\ce{NaCl}\) contains 0.01 M \(\ce{Na^{+}}\) ions and 0.01 M \(\ce{Cl^{−}}\) ions, for a total particle concentration of 0.02 M. Similarly, the \(\ce{CaCl_2}\) solution contains 0.01 M \(\ce{Ca^{2+}}\) ions and 0.02 M \(\ce{Cl^{−}}\) ions, for a total particle concentration of 0.03 M. These values are correct for dilute solutions, where the dissociation of the compounds to form separately solvated ions is complete. At (typically >1 M), especially with salts of small, highly charged ions (such as \(\ce{Mg^{2+}}\) or \(\ce{Al^{3+}}\)), or in solutions with less polar solvents, dissociation to give separate ions is often incomplete. The sum of the concentrations of the dissolved solute particles dictates the physical properties of a solution. In the following discussion, we must therefore keep the chemical nature of the solute firmly in mind. Adding a nonvolatile solute, one whose vapor pressure is too low to measure readily, to a volatile solvent decreases the vapor pressure of the solvent. We can understand this phenomenon qualitatively by examining Figure \(\Page {1}\), which is a schematic diagram of the surface of a solution of glucose in water. In an aqueous solution of glucose, a portion of the surface area is occupied by nonvolatile glucose molecules rather than by volatile water molecules. As a result, fewer water molecules can enter the vapor phase per unit time, even though the surface water molecules have the same kinetic energy distribution as they would in pure water. At the same time, the rate at which water molecules in the vapor phase collide with the surface and reenter the solution is unaffected. The net effect is to shift the dynamic equilibrium between water in the vapor and the liquid phases, decreasing the vapor pressure of the solution compared with the vapor pressure of the pure solvent. Figure \(\Page {2}\) shows two beakers, one containing pure water and one containing an aqueous glucose solution, in a sealed chamber. We can view the system as having two competing equilibria: water vapor will condense in both beakers at the same rate, but water molecules will evaporate more slowly from the glucose solution because fewer water molecules are at the surface. Eventually all of the water will evaporate from the beaker containing the liquid with the higher vapor pressure (pure water) and condense in the beaker containing the liquid with the lower vapor pressure (the glucose solution). If the system consisted of only a beaker of water inside a sealed container, equilibrium between the liquid and vapor would be achieved rather rapidly, and the amount of liquid water in the beaker would remain constant. If the particles of a solute are essentially the same size as those of the solvent and both solute and solvent have roughly equal probabilities of being at the surface of the solution, then the effect of a solute on the vapor pressure of the solvent is proportional to the number of sites occupied by solute particles at the surface of the solution. Doubling the concentration of a given solute causes twice as many surface sites to be occupied by solute molecules, resulting in twice the decrease in vapor pressure. The relationship between solution composition and vapor pressure is therefore \[P_A=\chi_AP^0_A \label{13.5.1} \] where \(P_A\) is the vapor pressure of component A of the solution (in this case the solvent), \(\chi_A\) is the mole fraction of \(A\) in solution, and \(P^0_A\) is the vapor pressure of pure \(A\). Equation \(\ref{13.5.1}\) is known as , after the French chemist who developed it. If the solution contains only a single nonvolatile solute (B), then \(\chi_A + \chi_B = 1\), and we can substitute \(\chi_A = 1 − \chi_B\) to obtain \[\begin{align} P_A &=(1−\chi_B)P^0_A \\[4pt] &=P^0_A−\chi_BP^0_A \label{13.5.2} \end{align} \] Rearranging and defining \(ΔP_A=P^0_A−P_A\), we obtain a relationship between the decrease in vapor pressure and the mole fraction of nonvolatile solute: \[ \begin{align} P^0_A−P_A &=ΔP_A \\[4pt] &=\chi_BP^0_A \label{13.5.3} \end{align} \] We can solve vapor pressure problems in either of two ways: by using Equation \(\ref{13.5.1}\) to calculate the actual vapor pressure above a solution of a nonvolatile solute, or by using Equation \(\ref{13.5.3}\) to calculate the decrease in vapor pressure caused by a specified amount of a nonvolatile solute. Ethylene glycol (\(\ce{HOCH_2CH_2OH}\)), the major ingredient in commercial automotive antifreeze, increases the boiling point of radiator fluid by lowering its vapor pressure. At 100°C, the vapor pressure of pure water is 760 mmHg. Calculate the vapor pressure of an aqueous solution containing 30.2% ethylene glycol by mass, a concentration commonly used in climates that do not get extremely cold in winter. : identity of solute, percentage by mass, and vapor pressure of pure solvent : vapor pressure of solution : : A 30.2% solution of ethylene glycol contains 302 g of ethylene glycol per kilogram of solution; the remainder (698 g) is water. To use Raoult’s law to calculate the vapor pressure of the solution, we must know the mole fraction of water. Thus we must first calculate the number of moles of both ethylene glycol (EG) and water present: \[moles \;EG=(302 \;\cancel{g}) \left( \dfrac{1\; mol}{62.07\; \cancel{g}} \right)=4.87\; mol\; EG \nonumber \] \[moles \; H_2O=(698 \;\cancel{g}) \left( \dfrac{1\; mol}{18.02\; \cancel{g}} \right)=38.7\; mol\; H_2O \nonumber \] The mole fraction of water is thus \[\chi_{H_2O}=\dfrac{38.7\; \cancel{mol} \; H_2O}{38.7\; \cancel{mol}\; H_2O +4.87 \cancel{mol}\; EG} =0.888 \nonumber \] From Raoult’s law (Equation \(\ref{13.5.1}\)), the vapor pressure of the solution is \[ \begin{align*} P_{H_2O} &=(\chi_{H2_O})(P^0_{H_2O}) \\[4pt] &=(0.888)(760\; mmHg) =675 \;mmHg \end{align*} \nonumber \] Alternatively, we could solve this problem by calculating the mole fraction of ethylene glycol and then using Equation \(\ref{13.5.3}\) to calculate the resulting decrease in vapor pressure: \[\chi_{EG}=\dfrac{4.87\; mol\; EG}{4.87\; mol\; EG+38.7\; mol\; H_2O}=0.112 \nonumber \] \[ΔP_{H2_O}=(\chi_{EG})(P^0_{H_2O})=(0.112)(760\; mmHg)=85.1\; mmHg \nonumber \] \[P_{H_2O}=P^0_{H_2O}−ΔP_{H_2O}=760\; mmHg−85.1\; mmHg=675\; mmHg \nonumber \] The same result is obtained using either method. Seawater is an approximately 3.0% aqueous solution of \(\ce{NaCl}\) by mass with about 0.5% of other salts by mass. Calculate the decrease in the vapor pressure of water at 25°C caused by this concentration of \(\ce{NaCl}\), remembering that 1 mol of \(\ce{NaCl}\) produces 2 mol of solute particles. The vapor pressure of pure water at 25°C is 23.8 mmHg. 0.45 mmHg. This may seem like a small amount, but it constitutes about a 2% decrease in the vapor pressure of water and accounts in part for the higher humidity in the north-central United States near the Great Lakes, which are freshwater lakes. The decrease therefore has important implications for climate modeling. Even when a solute is volatile, meaning that it has a measurable vapor pressure, we can still use Raoult’s law. In this case, we calculate the vapor pressure of each component separately. The total vapor pressure of the solution (\(P_{tot}\)) is the sum of the vapor pressures of the components: \[P_{tot}=P_A+P_B=\chi_AP^0_A+\chi_BP^0_B \label{13.5.4} \] Because \(\chi_B = 1 − \chi_A\) for a two-component system, \[P_{tot}=\chi_AP^0_A+(1−\chi_A)P^0_B \label{13.5.5} \] Thus we need to specify the mole fraction of only one of the components in a two-component system. Consider, for example, the vapor pressure of solutions of benzene and toluene of various compositions. At 20°C, the vapor pressures of pure benzene and toluene are 74.7 and 22.3 mmHg, respectively. The vapor pressure of benzene in a benzene–toluene solution is \[P_{C_6H_6}=\chi_{C_6H_6}P^0_{C_6H_6} \label{13.5.6} \] and the vapor pressure of toluene in the solution is \[P{C_6H_5CH_3}=\chi_{C_6H_5CH_3}P^0_{C_6H_5CH3} \label{13.5.7} \] Equations \(\ref{13.5.6}\) and \(\ref{13.5.7}\) are both in the form of the equation for a straight line: \(y = mx + b\), where \(b = 0\). Plots of the vapor pressures of both components versus the mole fractions are therefore straight lines that pass through the origin, as shown in Figure \(\Page {3}\). Furthermore, a plot of the total vapor pressure of the solution versus the mole fraction is a straight line that represents the sum of the vapor pressures of the pure components. Thus the vapor pressure of the solution is always greater than the vapor pressure of either component. A solution of two volatile components that behaves like the solution in Figure \(\Page {3}\), which is defined as a solution that obeys Raoult’s law. Like an ideal gas, an ideal solution is a hypothetical system whose properties can be described in terms of a simple model. Mixtures of benzene and toluene approximate an ideal solution because the intermolecular forces in the two pure liquids are almost identical in both kind and magnitude. Consequently, the change in enthalpy on solution formation is essentially zero (\(ΔH_{soln} ≈ 0\)), which is one of the defining characteristics of an ideal solution. Ideal solutions and ideal gases are both simple models that ignore intermolecular interactions. Most real solutions, however, do not obey Raoult’s law precisely, just as most real gases do not obey the ideal gas law exactly. Real solutions generally deviate from Raoult’s law because the intermolecular interactions between the two components A and B differ. We can distinguish between two general kinds of behavior, depending on whether the intermolecular interactions between molecules A and B are stronger or weaker than the A–A and B–B interactions in the pure components. If the A–B interactions are stronger than the A–A and B–B interactions, each component of the solution exhibits a lower vapor pressure than expected for an ideal solution, as does the solution as a whole. The favorable A–B interactions effectively stabilize the solution compared with the vapor. This kind of behavior is called a negative deviation from Raoult’s law. Systems stabilized by hydrogen bonding between two molecules, such as acetone and ethanol, exhibit from Raoult’s law. Conversely, if the A–B interactions are weaker than the A–A and B–B interactions yet the entropy increase is enough to allow the solution to form, both A and B have an increased tendency to escape from the solution into the vapor phase. The result is a higher vapor pressure than expected for an ideal solution, producing a positive deviation from Raoult’s law. In a solution of \(\ce{CCl_4}\) and methanol, for example, the nonpolar \(\ce{CCl_4}\) molecules interrupt the extensive hydrogen bonding network in methanol, and the lighter methanol molecules have weaker London dispersion forces than the heavier \(\ce{CCl_4}\) molecules. Consequently, solutions of \(\ce{CCl_4}\) and methanol exhibit from Raoult’s law. For each system, compare the intermolecular interactions in the pure liquids and in the solution to decide whether the vapor pressure will be greater than that predicted by Raoult’s law (positive deviation), approximately equal to that predicted by Raoult’s law (an ideal solution), or less than the pressure predicted by Raoult’s law (negative deviation). : identity of pure liquids : predicted deviation from Raoult’s law (Equation \ref{13.5.1}) : Identify whether each liquid is polar or nonpolar, and then predict the type of intermolecular interactions that occur in solution. : For each system, compare the intermolecular interactions in the pure liquids with those in the solution to decide whether the vapor pressure will be greater than that predicted by Raoult’s law (positive deviation), approximately equal to that predicted by Raoult’s law (an ideal solution), or less than the pressure predicted by Raoult’s law (negative deviation): approximately equal positive deviation (vapor pressure greater than predicted) negative deviation (vapor pressure less than predicted) A Discussing Roult's Law. Link: A Discussing How to find the Vapor Pressure of a Solution. Link: Recall that the normal boiling point of a substance is the temperature at which the vapor pressure equals 1 atm. If a nonvolatile solute lowers the vapor pressure of a solvent, it must also affect the boiling point. Because the vapor pressure of the solution at a given temperature is less than the vapor pressure of the pure solvent, achieving a vapor pressure of 1 atm for the solution requires a higher temperature than the normal boiling point of the solvent. Thus the boiling point of a solution is always greater than that of the pure solvent. We can see why this must be true by comparing the phase diagram for an aqueous solution with the phase diagram for pure water (Figure \(\Page {4}\)). The vapor pressure of the solution is less than that of pure water at all temperatures. Consequently, the liquid–vapor curve for the solution crosses the horizontal line corresponding to P = 1 atm at a higher temperature than does the curve for pure water. Figure \(\Page {4}\): Phase Diagrams of Pure Water and an Aqueous Solution of a Nonvolatile Solute. The vaporization curve for the solution lies below the curve for pure water at all temperatures, which results in an increase in the boiling point and a decrease in the freezing point of the solution. The boiling point of a solution with a nonvolatile solute is greater than the boiling point of the pure solvent. The magnitude of the increase in the boiling point is related to the magnitude of the decrease in the vapor pressure. As we have just discussed, the decrease in the vapor pressure is proportional to the concentration of the solute in the solution. Hence the magnitude of the increase in the boiling point must also be proportional to the concentration of the solute (Figure \(\Page {5}\)). We can define the boiling point elevation (\(ΔT_b\)) as the difference between the boiling points of the solution and the pure solvent: \[ΔT_b=T_b−T^0_b \label{13.5.8} \] where \(T_b\) is the boiling point of the solution and \(T^0_b\) is the boiling point of the pure solvent. We can express the relationship between \(ΔT_b\) and concentration as follows \[ΔT_b = mK_b \label{13.5.9} \] where m is the concentration of the solute expressed in molality, and \(K_b\) is the of the solvent, which has units of °C/m. Table \(\Page {1}\) lists characteristic \(K_b\) values for several commonly used solvents. For relatively dilute solutions, the magnitude of both properties is proportional to the solute concentration. The concentration of the solute is typically expressed as molality rather than mole fraction or molarity for two reasons. First, because the density of a solution changes with temperature, the value of molarity also varies with temperature. If the boiling point depends on the solute concentration, then by definition the system is not maintained at a constant temperature. Second, molality and mole fraction are proportional for relatively dilute solutions, but molality has a larger numerical value (a mole fraction can be only between zero and one). Using molality allows us to eliminate nonsignificant zeros. According to Table \(\Page {1}\), the molal boiling point elevation constant for water is 0.51°C/m. Thus a 1.00 m aqueous solution of a nonvolatile molecular solute such as glucose or sucrose will have an increase in boiling point of 0.51°C, to give a boiling point of 100.51°C at 1.00 atm. The increase in the boiling point of a 1.00 m aqueous \(\ce{NaCl}\) solution will be approximately twice as large as that of the glucose or sucrose solution because 1 mol of \(\ce{NaCl}\) produces 2 mol of dissolved ions. Hence a 1.00 m \(\ce{NaCl}\) solution will have a boiling point of about 101.02°C. In Example \(\Page {1}\), we calculated that the vapor pressure of a 30.2% aqueous solution of ethylene glycol at 100°C is 85.1 mmHg less than the vapor pressure of pure water. We stated (without offering proof) that this should result in a higher boiling point for the solution compared with pure water. Now that we have seen why this assertion is correct, calculate the boiling point of the aqueous ethylene glycol solution. : composition of solution : boiling point : Calculate the molality of ethylene glycol in the 30.2% solution. Then use Equation \(\ref{13.5.9}\) to calculate the increase in boiling point. : From Example \(\Page {1}\), we know that a 30.2% solution of ethylene glycol in water contains 302 g of ethylene glycol (4.87 mol) per 698 g of water. The molality of the solution is thus \[\text{molality of ethylene glycol}= \left(\dfrac{4.87 \;mol}{698 \; \cancel{g} \;H_2O} \right) \left(\dfrac{1000\; \cancel{g}}{1 \;kg} \right)=6.98\, m \nonumber \] From Equation \(\ref{13.5.9}\), the increase in boiling point is therefore \[ΔT_b=m K_b=(6.98 \cancel{m})(0.51°C/\cancel{m})=3.6°C \nonumber \] The boiling point of the solution is thus predicted to be 104°C. With a solute concentration of almost 7 m, however, the assumption of a dilute solution used to obtain Equation \(\ref{13.5.9}\) may not be valid. Assume that a tablespoon (5.00 g) of \(\ce{NaCl}\) is added to 2.00 L of water at 20.0°C, which is then brought to a boil to cook spaghetti. At what temperature will the water boil? 100.04°C, or 100°C to three significant figures. (Recall that 1 mol of \(\ce{NaCl}\) produces 2 mol of dissolved particles. The small increase in temperature means that adding salt to the water used to cook pasta has essentially no effect on the cooking time.) A Discussing Boiling Point Elevation and Freezing Point Depression. Link: The phase diagram in Figure \(\Page {4}\) shows that dissolving a nonvolatile solute in water not only raises the boiling point of the water but also lowers its freezing point. The solid–liquid curve for the solution crosses the line corresponding to \(P = 1\,atm\) at a lower temperature than the curve for pure water. We can understand this result by imagining that we have a sample of water at the normal freezing point temperature, where there is a dynamic equilibrium between solid and liquid. Water molecules are continuously colliding with the ice surface and entering the solid phase at the same rate that water molecules are leaving the surface of the ice and entering the liquid phase. If we dissolve a nonvolatile solute such as glucose in the liquid, the dissolved glucose molecules will reduce the number of collisions per unit time between water molecules and the ice surface because some of the molecules colliding with the ice will be glucose. Glucose, though, has a very different structure than water, and it cannot fit into the ice lattice. Consequently, the presence of glucose molecules in the solution can only decrease the rate at which water molecules in the liquid collide with the ice surface and solidify. Meanwhile, the rate at which the water molecules leave the surface of the ice and enter the liquid phase is unchanged. The net effect is to cause the ice to melt. The only way to reestablish a dynamic equilibrium between solid and liquid water is to lower the temperature of the system, which decreases the rate at which water molecules leave the surface of the ice crystals until it equals the rate at which water molecules in the solution collide with the ice. By analogy to our treatment of boiling point elevation,the freezing point depression (\(ΔT_f\)) is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution: \[ ΔT_f=T^0_f−T_f \label{13.5.10} \] where \(T^0_f\) is the freezing point of the pure solvent and \(T_f\) is the freezing point of the solution. The order of the terms is reversed compared with Equation \(\ref{13.5.8}\) to express the freezing point depression as a positive number. The relationship between \(ΔT_f\) and the solute concentration is given by an equation analogous to Equation \(\ref{13.5.9}\): \[ΔT_f = mK_f \label{13.5.11} \] where \(m\) is the molality of the solution and \(K_f\) is the molal freezing point depression constant for the solvent (in units of °C/m). Like \(K_b\), each solvent has a characteristic value of \(K_f\) (see Table \(\Page {1}\)). Freezing point depression depends on the total number of dissolved nonvolatile solute particles, just as with boiling point elevation. Thus an aqueous \(\ce{NaCl}\) solution has twice as large a freezing point depression as a glucose solution of the same molality. People who live in cold climates use freezing point depression to their advantage in many ways. For example, salt is used to melt ice and snow on roads and sidewalks, ethylene glycol is added to engine coolant water to prevent an automobile engine from being destroyed, and methanol is added to windshield washer fluid to prevent the fluid from freezing. The decrease in vapor pressure, increase in boiling point, and decrease in freezing point of a solution versus a pure liquid all depend on the total number of dissolved nonvolatile solute particles. In colder regions of the United States, \(\ce{NaCl}\) or \(\ce{CaCl_2}\) is often sprinkled on icy roads in winter to melt the ice and make driving safer. Use the data in the Figure below to estimate the concentrations of two saturated solutions at 0°C, one of \(\ce{NaCl}\) and one of \(\ce{CaCl_2}\), and calculate the freezing points of both solutions to see which salt is likely to be more effective at melting ice. : solubilities of two compounds : concentrations and freezing points : : A From Figure above, we can estimate the solubilities of \(\ce{NaCl}\) and \(\ce{CaCl_2}\) to be about 36 g and 60 g, respectively, per 100 g of water at 0°C. The corresponding concentrations in molality are \[m_{NaCl}=\left(\dfrac{36 \; \cancel{g \;NaCl}}{100 \;\cancel{g} \;H_2O}\right)\left(\dfrac{1\; mol\; NaCl}{58.44\; \cancel{ g\; NaCl}}\right)\left(\dfrac{1000\; \cancel{g}}{1\; kg}\right)=6.2\; m \nonumber \] \[m_{CaCl_2}=\left(\dfrac{60\; \cancel{g\; CaCl_2}}{100\;\cancel{g}\; H_2O}\right)\left(\dfrac{1\; mol\; CaCl_2}{110.98\; \cancel{g\; CaCl_2}}\right)\left(\dfrac{1000 \;\cancel{g}}{1 kg}\right)=5.4\; m \nonumber \] The lower formula mass of \(\ce{NaCl}\) more than compensates for its lower solubility, resulting in a saturated solution that has a slightly higher concentration than \(CaCl_2\). B Because these salts are ionic compounds that dissociate in water to yield two and three ions per formula unit of \(\ce{NaCl}\) and \(CaCl_2\), respectively, the actual concentrations of the dissolved species in the two saturated solutions are 2 × 6.2 m = 12 m for \(\ce{NaCl}\) and 3 × 5.4 m = 16 m for \(CaCl_2\). The resulting freezing point depressions can be calculated using Equation \(\ref{13.5.11}\): \[\ce{NaCl}: ΔT_f=mK_f=(12\; \cancel{m})(1.86°C/\cancel{m})=22°C \nonumber \] \[\ce{CaCl2}: ΔT_f=mK_f=(16\;\cancel{m})(1.86°C/\cancel{m})=30°C \nonumber \] Because the freezing point of pure water is 0°C, the actual freezing points of the solutions are −22°C and −30°C, respectively. Note that \(CaCl_2\) is substantially more effective at lowering the freezing point of water because its solutions contain three ions per formula unit. In fact, \(CaCl_2\) is the salt usually sold for home use, and it is also often used on highways. Because the solubilities of both salts decrease with decreasing temperature, the freezing point can be depressed by only a certain amount, regardless of how much salt is spread on an icy road. If the temperature is significantly below the minimum temperature at which one of these salts will cause ice to melt (say −35°C), there is no point in using salt until it gets warmer Calculate the freezing point of the 30.2% solution of ethylene glycol in water whose vapor pressure and boiling point we calculated in Examples \(\Page {5}\) and \(\Page {5}\). −13.0°C Arrange these aqueous solutions in order of decreasing freezing points: 0.1 m \(KCl\), 0.1 m glucose, 0.1 m SrCl2, 0.1 m ethylene glycol, 0.1 m benzoic acid, and 0.1 m HCl. : molalities of six solutions relative freezing points : : Because the molal concentrations of all six solutions are the same, we must focus on which of the substances are strong electrolytes, which are weak electrolytes, and which are nonelectrolytes to determine the actual numbers of particles in solution. \(\ce{KCl}\), \(\ce{SrCl_2}\), and \(\ce{HCl}\) are , producing two, three, and two ions per formula unit, respectively. Benzoic acid is a weak electrolyte (approximately one particle per molecule), and glucose and ethylene glycol are both nonelectrolytes (one particle per molecule). The molalities of the solutions in terms of the total particles of solute are: \(\ce{KCl}\) and \(\ce{HCl}\), 0.2 m; \(SrCl_2\), 0.3 m; glucose and ethylene glycol, 0.1 m; and benzoic acid, 0.1–0.2 m. Because the magnitude of the decrease in freezing point is proportional to the concentration of dissolved particles, the order of freezing points of the solutions is: glucose and ethylene glycol (highest freezing point, smallest freezing point depression) > benzoic acid > \(\ce{HCl}\) = \(\ce{KCl}\) > \(\ce{SrCl_2}\). Arrange these aqueous solutions in order of increasing freezing points: 0.2 m \(\ce{NaCl}\), 0.3 m acetic acid, 0.1 m \(\ce{CaCl_2}\), and 0.2 m sucrose. 0.2 m \(\ce{NaCl}\) (lowest freezing point) < 0.3 m acetic acid ≈ 0.1 m \(\ce{CaCl_2}\) < 0.2 m sucrose (highest freezing point) Colligative properties can also be used to determine the molar mass of an unknown compound. One method that can be carried out in the laboratory with minimal equipment is to measure the freezing point of a solution with a known mass of solute. This method is accurate for dilute solutions (≤1% by mass) because changes in the freezing point are usually large enough to measure accurately and precisely. By comparing \(K_b\) and \(K_f\) values in Table \(\Page {1}\), we see that changes in the boiling point are smaller than changes in the freezing point for a given solvent. Boiling point elevations are thus more difficult to measure precisely. For this reason, freezing point depression is more commonly used to determine molar mass than is boiling point elevation. Because of its very large value of \(K_f\) (37.8°C/m), d-(+)-camphor (Table \(\Page {1}\)) is often used to determine the molar mass of organic compounds by this method. A 7.08 g sample of elemental sulfur is dissolved in 75.0 g of \(CS_2\) to create a solution whose freezing point is −113.5°C. Use these data to calculate the molar mass of elemental sulfur and thus the formula of the dissolved \(\ce{S_n}\) molecules (i.e., what is the value of \(n\)?). : masses of solute and solvent and freezing point : molar mass and number of \(\ce{S}\) atoms per molecule : : A The first step is to calculate the freezing point depression using Equation \(\ref{13.5.10}\): \[ΔT_f=T^0_f−T_f=−112.1°C−(−113.5°C)=1.4°C \nonumber \] Then Equation \(\ref{13.5.11}\) gives B The total number of moles of solute present in the solution is \[\text{moles solute}=\left(\dfrac{0.37 mol}{\cancel{kg}}\right) (75.0\; g) \left(\dfrac{1 kg}{1000\; g}\right)=0.028 \;mol \nonumber \] C We now know that 0.708 g of elemental sulfur corresponds to 0.028 mol of solute. The molar mass of dissolved sulfur is thus \[\text{molar mass}=\dfrac{7.08\; g}{0.028\; mol}=260\; g/mol \nonumber \] The molar mass of atomic sulfur is 32 g/mol, so there must be 260/32 = 8.1 sulfur atoms per mole, corresponding to a formula of \(\ce{S_8}\). One of the byproducts formed during the synthesis of \(C_{60}\) is a deep red solid containing only carbon. A solution of 205 mg of this compound in 10.0 g of \(CCl_4\) has a freezing point of −23.38°C. What are the molar mass and most probable formula of the substance? 847 g/mol; \(\ce{C_{70}}\) A Discussing how to find the Molecular Weight of an Unknown using Colligative Properties. Link: Osmotic pressure is a colligative property of solutions that is observed using a semipermeable membrane, a barrier with pores small enough to allow solvent molecules to pass through but not solute molecules or ions. The net flow of solvent through a semipermeable membrane is called osmosis (from the Greek osmós, meaning “push”). The direction of net solvent flow is always from the side with the lower concentration of solute to the side with the higher concentration. Osmosis can be demonstrated using a U-tube like the one shown in Figure \(\Page {6}\), which contains pure water in the left arm and a dilute aqueous solution of glucose in the right arm. A net flow of water through the membrane occurs until the levels in the arms eventually stop changing, which indicates that equilibrium has been reached. The osmotic pressure (\(\Pi\)) of the glucose solution is the difference in the pressure between the two sides, in this case the heights of the two columns. Although the semipermeable membrane allows water molecules to flow through in either direction, the rate of flow is not the same in both directions because the concentration of water is not the same in the two arms. The net flow of water through the membrane can be prevented by applying a pressure to the right arm that is equal to the osmotic pressure of the glucose solution. Just as with any other colligative property, the osmotic pressure of a solution depends on the concentration of dissolved solute particles. Osmotic pressure obeys a law that resembles the ideal gas equation: \[\Pi=\dfrac{nRT}{V}=MRT \label{13.5.12} \] where \(M\) is the number of moles of solute per unit volume of solution (i.e., the molarity of the solution), \(R\) is the ideal gas constant, and \(T\) is the absolute temperature. As shown in Example \(\Page {7}\), osmotic pressures tend to be quite high, even for rather dilute solutions. When placed in a concentrated salt solution, certain yeasts are able to produce high internal concentrations of glycerol to counteract the osmotic pressure of the surrounding medium. Suppose that the yeast cells are placed in an aqueous solution containing 4.0% \(\ce{NaCl}\) by mass; the solution density is 1.02 g/mL at 25°C. : concentration, density, and temperature of \(\ce{NaCl}\) solution; internal osmotic pressure of cell : osmotic pressure of \(\ce{NaCl}\) solution and concentration of glycerol needed : : A The solution contains 4.0 g of \(\ce{NaCl}\) per 100 g of solution. Using the formula mass of \(\ce{NaCl}\) (58.44 g/mol) and the density of the solution (1.02 g/mL), we can calculate the molarity: \[ \begin{align*} M_{NaCl} &=\dfrac{moles\; NaCl}{\text{liter solution}} \\[4pt] &=\left(\dfrac{4.0 \; \cancel{g} \;NaCl}{58.44\; \cancel{g}/mol\; NaCl}\right)\left(\dfrac{1}{100\; \cancel{g \;solution}}\right)\left(\dfrac{1.02\; \cancel{g\; solution}}{1.00\; \cancel{mL}\; solution}\right)\left(\dfrac{1000\; \cancel{mL}}{1\; L}\right) \\[4pt] &= 0.70\; M\; \ce{NaCl} \end{align*} \nonumber \] Because 1 mol of \(\ce{NaCl}\) produces 2 mol of particles in solution, the total concentration of dissolved particles in the solution is (2)(0.70 M) = 1.4 M. B Now we can use Equation \ref{13.5.12} to calculate the osmotic pressure of the solution: \[ \begin{align*} \Pi &=MRT \\[4pt] &=(1.4 \;mol/L)\left[ 0.0821\; (L⋅atm)/(K⋅mol) \right ] (298\; K)\\[4pt] &=34 \;atm\end{align*} \nonumber \] C If the yeast cells are to exactly balance the external osmotic pressure, they must produce enough glycerol to give an additional internal pressure of (34 atm − 7.3 atm) = 27 atm. Glycerol is a nonelectrolyte, so we can solve Equation \ref{13.5.12} for the molarity corresponding to this osmotic pressure: \[ \begin{align*} M&=\dfrac{\Pi}{RT}\\[4pt] &=\dfrac{27\; \cancel{atm}}{[0.0821(L⋅\cancel{atm})/(\cancel{K}⋅mol)] (298 \;\cancel{K})}\\[4pt] &=1.1 \;M \;\text{glycerol} \end{align*} \nonumber \] In solving this problem, we could also have recognized that the only way the osmotic pressures can be the same inside the cells and in the solution is if the concentrations of dissolved particles are the same. We are given that the normal concentration of dissolved particles in the cells is 0.3 M, and we have calculated that the \(\ce{NaCl}\) solution is effectively 1.4 M in dissolved particles. The yeast cells must therefore synthesize enough glycerol to increase the internal concentration of dissolved particles from 0.3 M to 1.4 M—that is, an additional 1.1 M concentration of glycerol. Assume that the fluids inside a sausage are approximately 0.80 M in dissolved particles due to the salt and sodium nitrite used to prepare them. Calculate the osmotic pressure inside the sausage at 100°C to learn why experienced cooks pierce the semipermeable skin of sausages before boiling them. 24 atm Because of the large magnitude of osmotic pressures, osmosis is extraordinarily important in biochemistry, biology, and medicine. Virtually every barrier that separates an organism or cell from its environment acts like a semipermeable membrane, permitting the flow of water but not solutes. The same is true of the compartments inside an organism or cell. Some specialized barriers, such as those in your kidneys, are slightly more permeable and use a related process called dialysis, which permits both water and small molecules to pass through but not large molecules such as proteins. The same principle has long been used to preserve fruits and their essential vitamins over the long winter. High concentrations of sugar are used in jams and jellies not for sweetness alone but because they greatly increase the osmotic pressure. Thus any bacteria not killed in the cooking process are dehydrated, which keeps them from multiplying in an otherwise rich medium for bacterial growth. A similar process using salt prevents bacteria from growing in ham, bacon, salt pork, salt cod, and other preserved meats. The effect of osmotic pressure is dramatically illustrated in Figure \(\Page {7}\), which shows what happens when red blood cells are placed in a solution whose osmotic pressure is much lower or much higher than the internal pressure of the cells. In addition to capillary action, trees use osmotic pressure to transport water and other nutrients from the roots to the upper branches. Evaporation of water from the leaves results in a local increase in the salt concentration, which generates an osmotic pressure that pulls water up the trunk of the tree to the leaves. Finally, a process called reverse osmosis can be used to produce pure water from seawater. As shown schematically in Figure \(\Page {8}\), applying high pressure to seawater forces water molecules to flow through a semipermeable membrane that separates pure water from the solution, leaving the dissolved salt behind. Large-scale desalinization plants that can produce hundreds of thousands of gallons of freshwater per day are common in the desert lands of the Middle East, where they supply a large proportion of the freshwater needed by the population. Similar facilities are now being used to supply freshwater in southern California. Small, hand-operated reverse osmosis units can produce approximately 5 L of freshwater per hour, enough to keep 25 people alive, and are now standard equipment on Navy lifeboats. A Discussing Osmotic Pressure. Link: Thus far we have assumed that we could simply multiply the molar concentration of a solute by the number of ions per formula unit to obtain the actual concentration of dissolved particles in an electrolyte solution. We have used this simple model to predict such properties as freezing points, melting points, vapor pressure, and osmotic pressure. If this model were perfectly correct, we would expect the freezing point depression of a 0.10 m solution of sodium chloride, with 2 mol of ions per mole of \(\ce{NaCl}\) in solution, to be exactly twice that of a 0.10 m solution of glucose, with only 1 mol of molecules per mole of glucose in solution. In reality, this is not always the case. Instead, the observed change in freezing points for 0.10 m aqueous solutions of \(\ce{NaCl}\) and \(\ce{KCl}\) are significantly less than expected (−0.348°C and −0.344°C, respectively, rather than −0.372°C), which suggests that fewer particles than we expected are present in solution. The relationship between the actual number of moles of solute added to form a solution and the apparent number as determined by colligative properties is called the van’t Hoff factor (\(i\)) and is defined as follows: \[i=\dfrac{\text{apparent number of particles in solution}}{\text{ number of moles of solute dissolved}} \label{13.5.13} \] Named for Jacobus Hendricus van’t Hoff (1852–1911), a Dutch chemistry professor at the University of Amsterdam who won the first Nobel Prize in Chemistry (1901) for his work on thermodynamics and solutions. As the solute concentration increases, the van’t Hoff factor decreases. The van’t Hoff factor is therefore . The lower the van’t Hoff factor, the greater the deviation. As the data in Table \(\Page {2}\) show, the van’t Hoff factors for ionic compounds are somewhat lower than expected; that is, their solutions apparently contain fewer particles than predicted by the number of ions per formula unit. As the concentration of the solute increases, the van’t Hoff factor decreases because ionic compounds generally do not totally dissociate in aqueous solution. Instead, some of the ions exist as ion pairs, a cation and an anion that for a brief time are associated with each other without an intervening shell of water molecules (Figure \(\Page {9}\)). Each of these temporary units behaves like a single dissolved particle until it dissociates. Highly charged ions such as \(Mg^{2+}\), \(Al^{3+}\), \(SO_4^{2−}\), and \(PO_4^{3−}\) have a greater tendency to form ion pairs because of their strong electrostatic interactions. The actual number of solvated ions present in a solution can be determined by measuring a colligative property at several solute concentrations. A 0.0500 M aqueous solution of \(FeCl_3\) has an osmotic pressure of 4.15 atm at 25°C. Calculate the van’t Hoff factor \(i\) for the solution. : solute concentration, osmotic pressure, and temperature : van’t Hoff factor : : A If \(\ce{FeCl_3}\) dissociated completely in aqueous solution, it would produce four ions per formula unit [Fe3+(aq) plus 3Cl−(aq)] for an effective concentration of dissolved particles of 4 × 0.0500 M = 0.200 M. The osmotic pressure would be \[\begin{align*} \Pi &=MRT \\[4pt] &=(0.200 \;mol/L) \left[0.0821\;(L⋅atm)/(K⋅mol) \right] (298\; K)=4.89\; atm \end{align*} \nonumber \] B The observed osmotic pressure is only 4.15 atm, presumably due to ion pair formation. The ratio of the observed osmotic pressure to the calculated value is 4.15 atm/4.89 atm = 0.849, which indicates that the solution contains (0.849)(4) = 3.40 particles per mole of \(\ce{FeCl_3}\) dissolved. Alternatively, we can calculate the observed particle concentration from the osmotic pressure of 4.15 atm: \[4.15\; atm=M \left( \dfrac{0.0821 \;(L⋅atm)}{(K⋅mol)}\right) (298 \;K) \nonumber \] or after rearranging \[M = 0.170 mol \nonumber \] The ratio of this value to the expected value of 0.200 M is 0.170 M/0.200 M = 0.850, which again gives us (0.850)(4) = 3.40 particles per mole of \(\ce{FeCl_3}\) dissolved. From Equation \(\ref{13.5.13}\), the van’t Hoff factor for the solution is \[i=\dfrac{\text{3.40 particles observed}}{\text{1 formula unit}\; \ce{FeCl_3}}=3.40 \nonumber \] Calculate the van’t Hoff factor for a 0.050 m aqueous solution of \(MgCl_2\) that has a measured freezing point of −0.25°C. 2.7 (versus an ideal value of 3). A Discussing the Colligative Properties in Solutions. Link: The colligative properties of a solution depend on only the total number of dissolved particles in solution, not on their chemical identity. Colligative properties include vapor pressure, boiling point, freezing point, and osmotic pressure. The addition of a nonvolatile solute (one without a measurable vapor pressure) decreases the vapor pressure of the solvent. The vapor pressure of the solution is proportional to the mole fraction of solvent in the solution, a relationship known as . Solutions that obey Raoult’s law are called ideal solutions. Most real solutions exhibit positive or negative deviations from Raoult’s law. The boiling point elevation (\(ΔT_b\)) and freezing point depression (\(ΔT_f\)) of a solution are defined as the differences between the boiling and freezing points, respectively, of the solution and the pure solvent. Both are proportional to the molality of the solute. When a solution and a pure solvent are separated by a semipermeable membrane, a barrier that allows solvent molecules but not solute molecules to pass through, the flow of solvent in opposing directions is unequal and produces an osmotic pressure, which is the difference in pressure between the two sides of the membrane. Osmosis is the net flow of solvent through such a membrane due to different solute concentrations. Dialysis uses a semipermeable membrane with pores that allow only small solute molecules and solvent molecules to pass through. In more concentrated solutions, or in solutions of salts with highly charged ions, the cations and anions can associate to form ion pairs, which decreases their effect on the colligative properties of the solution. The extent of ion pair formation is given by the van’t Hoff factor (i), the ratio of the apparent number of particles in solution to the number predicted by the stoichiometry of the salt. | 43,430 | 2,122 |
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Two criteria are usually applied to any theory. First, does it agree with facts which are already known? Second, does it predict new relationships and stimulate additional observation and experimentation? Dalton’s atomic theory was able to do both of these things.It was especially useful in dealing with data regarding the masses of different elements which were involved in chemical compounds or chemical reactions. To test a theory, we first use it to make a prediction about the macroscopic world. If the prediction agrees with existing data, the theory passes the test. If it does not, the theory must be discarded or modified. If data are not available, then more research must be done. Eventually the results of new experiments can be compared with the predictions of the theory. One hallmark of a theory is that it suggests tests that may falsify the theory. Non-scientific theories do not. For example, Intelligent Design (ID) theory proposes that the world was created by an intelligent designer, but proposes no tests of that claim. It's a theory, but not a theory. Several examples of this process of testing a theory against the facts are afforded by Dalton’s work. For example, postulate 3 in states that atoms are not created, destroyed, or changed in a chemical reaction. Postulate 2 says that atoms of a given element have a characteristic mass: By logical deduction, then, equal numbers of each type of atom must appear on left and right sides of chemical equations such as and the total mass of reactants must equal the total mass of products. Dalton’s atomic theory predicts Lavoisier’s experimental law of conservation of mass. A second prediction of the atomic theory is a bit more complex. A compound is made up of molecules, each of which contains a certain number of each type of atom. No matter how, when, or where a compound is made, its molecules will always be the same. Thus water molecules always have the formula H O. No matter how much we have or where the compound came from, there will always be twice as many hydrogen atoms as oxygen atoms. Since each type of atom has a characteristic mass, the mass of one element which combines with a fixed mass of the other should always be the same. In water, for example, if each oxygen atom is 15.873 times as heavy as a hydrogen atom, the ratio of masses would be \[\dfrac{\text{mass of 1 O atom}}{\text{mass of 2 H atoms}}= \dfrac{\text{15.873 x mass of 1 H atom}}{\text{2 x mass of 1 H atom}} \nonumber \] and cancelling "mass of 1 H atom" in the numerator and denominator we get \[\dfrac{\text {15.873}}{2} = \dfrac{7.937}{1} \nonumber \] No matter how many water molecules we have, each has the same proportion of oxygen, and so sample of water must have 7.937 times as much oxygen as hydrogen. We have just derived the , sometimes called the . When elements combine to form a compound, they always do so in exactly the same ratio of masses. This law had been postulated in 1799 by the French chemist Proust (1754 to 1826) four years before Dalton proposed the atomic theory, and its logical derivation from the theory contributed to the latter’s acceptance. The law of constant composition makes the important point that the composition and other properties of a compound are independent of who prepared it or where it came from. The carbon dioxide found on Mars, for instance, can be expected to have the same composition as that on Earth, while the natural vitamin C extracted and purified from rose hips has exactly the same composition as the synthetic vitamin C prepared by a drug company. Absolute purity is, however, an ideal limit which we can only approach, and the properties of many substances may be affected by the presence of very small quantities of impurities. A third law of chemical composition may be deduced from the atomic theory. It involves the situation where two elements can combine in more than one way, forming more than one compound. For example, hydrogen and oxygen form another compound, hydrogen peroxide, modeled here: This is a "Jmol" model. If you place the mouse pointer on the molecule, hold down the left mouse key, and move the mouse, you can rotate the model to get a 3D perspective. The oxygen atoms are red, hydrogen gray. Hydrogen peroxide is a pale blue liquid that freezes just 0.4 C below 0 , boils at 150.2 C, and is slightly more viscous than water. It is used in 3% water solution as an antiseptic (it's used to disinfect biological safety cabinets) and bleaching agent. Pure hydrogen peroxide is a dangerous oxidizer, used in rocket engines, which burns skin even in 10% water solutions. Surprisingly, is naturally produced in organisms as a byproduct of oxygen metabolism. Nearly all living things possess enzymes known as peroxidases, which very rapidly catalytically decompose low concentrations of hydrogen peroxide to water and oxygen. From the molecular model of hydrogen peroxide, you can readily see that its chemical formula is H O . (Since there are two atoms of each kind in the molecule, it would be incorrect to write the formula as HO). Hydrogen peroxide cannot be synthesized directly by the reaction of hydrogen and oxygen, and it decomposes to water and oxygen: From the formulas H O and H O we can see that water has only 1 oxygen atom for every 2 hydrogens, while hydrogen peroxide has 2 oxygen atoms for every 2 hydrogens. Thus, for a given number of bromine atoms, hydrogen peroxide will always have twice as many oxygen atoms as water. Again using postulate 2 from , the atoms have characteristic masses, and so a given number of hydrogen atoms corresponds to a fixed mass of hydrogen. Twice as many oxygen atoms correspond to twice the mass of oxygen. Therefore we can say that for a given mass of hydrogen, hydrogen peroxide will contain twice the mass of oxygen that water will. Given that the mass of an oxygen atom is 7.937 times the mass of a hydrogen atom, calculate the mass ratio of oxygen to hydrogen in hydrogen peroxide. The formula H O tells us that there are 2 oxygen atoms and 2 hydrogen atoms in each molecule. Thus the mass ratio is \[\dfrac{\text{mass of 2 O atoms}}{\text{mass of 2 H atoms}}= \dfrac{\text{2 x 15.873 x mass of 1 H atom}}{\text{2 x mass of 1 H atom}} \nonumber \] and again cancelling "mass of 1 H atom" in the numerator and denominator we get \[\dfrac{\text {2 x 15.873}}{2} = \dfrac{15.873}{1} \nonumber \] Note that the mass of oxygen per unit mass of hydrogen is double that calculated earlier for water. The reasoning and calculations above illustrate the law of multiple proportions. When two elements form several compounds, the mass ratio in one compound will be a small whole-number multiple of the mass ratio in another. In the case of water and hydrogen peroxide, the mass ratios of mercury to bromine are 7.937 and 15.873, respectively. The second value is a small whole-number multiple of (2 times) the first. Until the atomic theory was proposed, no one had expected relationship to exist between mass ratios in two or more compounds containing the same elements. Because the theory predicted such relationships, Dalton and other chemists began to look for them. Before long, a great deal of experimental evidence was accumulated to show that the law of multiple proportions was valid. Thus the atomic theory was able to account for previously known facts and laws, and it also . In the process of verifying that prediction, Dalton and his contemporaries did many additional quantitative experiments. These led onward to more facts, more laws, and, eventually, new or modified theories. This characteristic of stimulating more research and thought put Dalton’s postulates in the distinguished company of other good scientific theories. From ChemPRIME: | 7,787 | 2,123 |
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Water, sand, and oil baths are related heat sources as they envelop a flask in a warm material (liquid or sand). A thermometer is often used to monitor the temperature of the bath, and is used to the internal temperature of liquid in a flask (the bath is often slightly hotter than the liquid in the flask). , heated on a hotplate, are most commonly used to heat solutions to \(100^\text{o} \text{C}\) (boiling baths, Figures 1.53 + 1.54a). They may also be used to heat to lower temperatures, although it can be difficult to maintain a constant temperature. Water baths can be covered with aluminum foil to prevent excessive evaporation, or to prevent excess moisture from entering open vessels. Cold water baths can also be used to cool apparatuses in a quick manner (Figure 1.54b). , also heated on a hotplate, can be used to heat solutions to a wide variety of temperatures, including very high temperatures (> \(250^\text{o} \text{C}\)). Sand can be placed inside Pyrex crystallizing dishes (as is used with the water bath in Figure 1.54b), although Pyrex may crack if heated at too fast a rate. Sand can also be heated to moderate temperatures inside thin aluminum foil pie dishes, but should not be used at high temperatures as aluminum will then oxidize, causing the thin foil to disintegrate. Thick metal pie tins (Figure 1.54c) are indestructible alternatives when high temperatures are required (but may interfere with the stirring mechanism if used). A vessel should be buried in a sand bath as much as possible as the surface is often much cooler than the sand below. A scoopula or metal spatula can be used to pile the sand up to at least the height of liquid inside the flask. Sand takes a long time to heat up, and a long time to cool down. To save time, a sand bath may be preheated while an apparatus is assembled as long as it is preheated a distance away from volatile organic liquids. If the sand overheats and causes a liquid to boil uncontrollably, the flask can be partially lifted out of the sand, or the sand moved with a metal spatula away from contact with the liquid. Sand will remain warm even after turning off the hotplate, and therefore flasks have to be lifted out of the sand bath in order to cool (Figure 1.54d). Hot sand baths can be cooled atop a ceramic tile. are much like water baths, but use silicone or mineral oils in order to enable temperatures hotter than the boiling point of water (> \(100^\text{o} \text{C}\)). Silicone oil baths can be heated to \(250^\text{o} \text{C}\), while mineral oil baths can be heated to \(300^\text{o} \text{C}\).\(^7\) Mineral oil is composed of mixtures of long-chain alkanes, and so is combustible. Direct contact with open flames should therefore be avoided. Oil baths can be heated in a Pyrex crystallizing dish atop a hotplate. It is also quite common for the oil to be electrically heated, through immersion of a coiled wire connected to a " " (light blue piece of equipment in Figure 1.55a). A Variac connects to the outlet and can deliver variable voltage through the wire. A Variac set to "100" would be equivalent to plugging the system directly into the wall \(\left( 100\% \right)\), while a setting of "50" means the delivered voltage is halved \(\left( 50\% \right)\). By controlling the delivered voltage, Variacs are used to regulate the temperature of the oil. If you have previous experience using a Variac with heating mantles, the settings will not translate to the oil bath as oil bath wires heat more rapidly than a heating mantle's wires. A paper clip can also be used in an oil bath (Figure 1.55c) and stirred with a stirring plate in order to dissipate heat. This allows for the temperature of an oil bath to quickly respond to adjustments up or down. \(^7\)J. A. Dean, , 15\(^\text{th}\) ed., McGraw-Hill, , Sect. 11.3. Additionally, the boiling point of mineral oil is \(310^\text{o} \text{C}\). | 3,918 | 2,124 |
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Suction filtration (vacuum filtration) is the standard technique used for separating a solid-liquid mixture when the goal is to retain the solid (for example in crystallization). Similar to gravity filtration, a solid-liquid mixture is poured onto a filter paper, with the main difference being that the process is aided by suction beneath the funnel (Figures 1.70 + 1.71). The process has advantages and disadvantages in comparison to gravity filtration. Advantages: 1) Suction filtration is much faster than gravity filtration, often taking less than one minute with good seals and a good vacuum source. 2) Suction filtration is more efficient at removing residual liquid, leading to a purer solid. This is especially important in crystallization, as the liquid may contain soluble impurities which could adsorb back onto the solid surface when the solvent evaporates. Disadvantages: The force of suction may draw fine crystals through the filter paper pores, leading to a quantity of material that cannot be recovered from the filter paper, and possibly an additional quantity that is lost in the filtrate. This method therefore works best with large crystals. On small scales, the loss of material to the filter paper and filtrate is significant, and so other methods are recommended for microscale work. As the goal of suction filtration is to fully separate a solid from its surrounding liquid, rinsing the solid is necessary if the liquid cannot easily evaporate. In the case of crystallization, the liquid may contain impurities that can reincorporate into the solid if not removed. To rinse a suction-filtered solid, the vacuum is removed and a small portion of cold solvent is poured over the solid (the " "). In the case of crystallization, the same solvent from the crystallization is used. The solid is then delicately slushed around in the solvent with a glass rod, and the vacuum is reapplied to remove the rinse solvent. To demonstrate the importance of a rinse, Figure 1.72 shows the recovery of a white solid from a yellow liquid using suction filtration. The yellow liquid seemed to be somewhat retained by the solid, as the first crystals collected had a yellow tint (Figure 1.72b). However, rinsing with a few portions of cold solvent were effective at removing the yellow liquid (Figure 1.72d), which could have been reincorporated into the solid without the rinse. A vacuum source is necessary for suction filtration (and vacuum distillation). Although many science buildings come equipped with a house vacuum system (Figure 1.73a), solvents evaporating from a suction filter flask over time can degrade the oil pumps used in a house vacuum. Therefore, it is recommended to instead connect a suction flask to a water aspirator. A water aspirator is an inexpensive attachment to a water spigot, and the nub on the aspirator connects with tubing to the vessel to be evacuated (Figure 1.73b). As water flows through the faucet and the aspirator, suction is created in the flask. A water aspirator creates suction through the (technically, the , for liquids). Water coming from the faucet is constricted inside the aspirator (Figure 1.73c). As the water flow must be the same going into the aspirator as it is going out, the water speed must increase in the constricted area in the direction of flow. A similar phenomenon can be seen in creeks and rivers where the water flows the fastest at the narrowest portions of streams. When the water increases its velocity in the direction of the water flow, conservation of energy dictates that its velocity in perpendicular directions must decrease. The result is a lowered pressure adjacent to the fast-moving liquid. In other words, the gain in velocity of the constricted liquid is balanced by a reduction in pressure on the surrounding material (the gas). For this reason, the speed at which the water flows through the faucet is correlated with the amount of suction experienced in the connected flask. A strong flow of water will have the fastest speeds through the aspirator and the greatest reduction in pressure. Clamp a side-armed Erlenmeyer flask. Connect thick-walled hosing from the side arm to a vacuum trap and the water aspirator. Place a vacuum sleeve on the Buchner (or Hirsch) funnel, then filter paper on the funnel so it arches downward. Turn on the aspirator. Add a few \(\text{mL}\) of the same solvent used in the flask to wet the filter. The solvent should drain with suction. Swirl the mixture to be filtered to dislodge the solid from the sides of the flask. With a quick motion, pour the slurry into the funnel in portions. In some applications, (e.g. crystallization), rinse with solvent: Apply suction again for a few minutes (repeat the rinse step if necessary). Dry the solid on a watch glass along with the filter paper, overnight if possible. The solid will flake off the paper when dried. | 4,901 | 2,125 |
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The Atomic Theory in Foods Nutrition books are often confusing when they discuss human nutrient requirements. For example, one, in discussing essential minerals in sports nutrition says "Minerals are elements". But only one common essential mineral, iron (symbolized Fe for the Latin "Ferrum") is commonly ingested as an element. Iron is an because it is made up of a , designated by the symbol Fe. Other symbols are given in the Table below. Iron metal particles are actually found in breakfast cereals , and the easiest way to prove it is to float a cereal flake on water and pull it around with a magnet. You can also see it extracted in several YouTube videos like . Sometimes iron is added as a , like "ferrous sulfate", FeSO . A compound has atoms of several elements bonded together is specific ratios. The ratios are given by the subscripts in the formula; in this case, ferrous sulfate is made from 1 atom of iron, 1 of the element sulfur, and 4 atoms of the element oxygen. Another essential mineral that is commony misidentified is "iodine". The nutrient we actually ingest is not the element iodine, but a compound of iodine like potassium iodide, KI (K is the symbol for potassium, after the German Kallium). While the mineral KI looks and tastes like table salt, the element iodine is a dark purple (almost black) solid, sometimes dissolved in alcohol to give a brown solution used to disinfect minor wounds. It's a toxic element with the formula I , showing that an element may be made up of which contain two or more atoms bonded together, as long as the atoms are the same. When iodine crystals are heated, they "sublime" (turn directly to a gas) to give purple vapors composed of diatomic molecules. The liquid phase, which also contains diatomic molecules, only forms under higher than atmospheric pressure. All the phases are the element iodine, because they contain only I atoms. ( ) Other essential minerals are referred to as "elements", but are actually either toxic or useless to our body as elements. Foods must contain nutrients in chemical compounds that are easily digested and absorbed in our bodies in order for them to be nutritious. For example, the phosphorus may have several forms. One, P , is very toxic and actually burns when exposed to air, making a glow that's responsible for its name. The nutrient form of phosporus is illustrated by mineral phosphates, like Ca (PO ) , which contains 3 calcium atoms (we will see later, they're in the form of each with a 2+ charge, Ca ) and 2 "phosphate ions", each made of 1 phosphorus atom bonded to 4 oxygen atoms, and having an overall 3- charge. Phosphorus is also supplied in found in meats and vegetables. An "Elemental diet" is a solution of nutrients that can be administered intravenously (or with a gastric feeding tube) for people with digestive disorders, like or colitis. There are no in an elemental diet, but the simplest chemical compounds that can provide complete (or near complete) nutrition. A typical elemental diet shows us what is required to maintain our bodies. We require ounces to pounds of macronutrients daily, but typically less than 5 g (.2 oz) of micronutrients. The ratio of chemical elements in our bodies is sometimes presented as a "formula for a human being", which does not represent any actual chemical compound, but just tells us the relative numbers of atoms: \[\ce{H_{375,000,000}O_{132,000,000}C_{85,700,000}N_{6,430,000}Ca_{1,500,000}P_{1,020,000}S_{206,000}Na_{183,000}K_{177,000}Cl_{127,000}Mg_{40,000}Si_{38, 600}Fe_{2,680}Zn_{2,110}Cu_{76}I_{14}Mn_{13}F_{13}Cr_{7}Se_{4}Mo_{3}Co_{1}} \nonumber \] So for just 1 cobalt (Co) atom, we need 2,680 iron atoms, 1,500,000 calcium (Ca) atoms, etc. Clearly, there's a lot that chemistry can tell us about nutrition. But how do we know what is an "element" and what is a "compound", and what compounds provide the elements we need to maintain our bodies? The early greek philosophers (Empedocles, Lucretius and Democritus) proposed that everything was made of atoms, but had no way of providing evidence for the claim. Evidence that each chemical element is composed of one kind of atom (that is unchanged during chemical reactions) finally developed between 1750 and 1850. Looking at the pictures of potassium iodide (KI) and iodine (I ) above, will help to imagine how implausible it must have seemed that they share a common iodine atom. Evidence for unchanging atoms as the components of elements came from weight measurements. Joseph Priestley and Antoine Lavoisier finally correctly interpreted the loss of weight when some substances burn (wood or sugar) and the gain in weight of others, by recognizing the fact that the mass of atoms in gases had been neglected. Lavoisier showed that mercury (Hg) gains weight because it combines with oxygen molecules from air to make solid mercuric oxide: \[\ce{2 Hg + O2 → 2 HgO.} \nonumber \] But wood loses weight because it is converted to gaseous molecules of carbon dioxide and water of equal mass. The combustion reaction is similar to the combusion of the sugar, glucose, which the basis for the metabolism of carbohydrates in our bodies to provide energy: \[\ce{C6H12O6 + 6 O2 → 6 CO2 + 6 H2O} \nonumber \] Earlier, van Helmont had failed to recognize that incorporation of carbon dioxide gas contributed the most to the weight gain of trees; he thought it was the water, because the soil showed virtually no change in mass. Nonetheless, he showed committment to the idea of , that no mass should be lost or gained during a chemical change. The equation for photosynthesis is just the reverse of the combustion equation above, \[\begin{align*} \ce{6 CO2} + \ce{6 H2O} &\rightarrow \ce{C6H12O6} + \ce{6 O2} \\[4pt]
\text{carbon dioxide} + \text{water} + \text{light energy} &\rightarrow \text{carbohydrate} + \text{oxygen} \end{align*} \nonumber \] so we could say that trees are mostly carbohydrate, and that when they burn, they release the energy that they had absorbed from the sun while growing: As Lavoisier continued his experiments with oxygen, he noticed something else. Although oxygen combined with many other substances, it never behaved as though it were itself a combination of other substances. Lavoisier was able to decompose the red calx into mercury and oxygen, but he could find no way to break down oxygen into two or more new substances. Because of this he suggested that oxygen must be an —an ultimately simple substance which could not be decomposed by chemical changes. This was the fundamental discovery that allows us to identify tricalcium phosphate as a good source of "phosphorus" or potassium iodide as a good source of "iodine". John Dalton (1766 to 1844) was a generation younger than Lavoisier and different from him in almost every respect. Dalton came from a working class family and only attended elementary school. Apart from this, he was entirely self-taught. Even after he became famous, he never aspired beyond a modest bachelor’s existence in which he supported himself by teaching mathematics to private pupils. Dalton made many contributions to science, and he seems not to have realized that his atomic theory was the most important of them. In his “New System of Chemical Philosophy” published in 1808, only the last seven pages out of a total of 168 are devoted to it! | 7,370 | 2,127 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Surface_Science_(Nix)/03%3A_The_Langmuir_Isotherm/3.03%3A_Langmuir_Isotherm_from_a_Kinetics_Consideration |
The equilibrium that may exist between gas adsorbed on a surface and molecules in the gas phase is a dynamic state, i.e. . It should therefore be possible to derive an isotherm for the adsorption process simply by considering and equating the rates for these two processes. Expressions for the rate of adsorption and rate of desorption have been derived in Sections 2.3 & 2.6 respectively: specifically, \[ R_{ads}=\dfrac{f(\theta)P}{\sqrt{2\pi mkT}} \exp (-E_a^{ads}/RT)\] \[R_{des} = v\; f'(\theta) \exp (-E_a^{ads}/RT)\] Equating these two rates yields an equation of the form: \[ \dfrac{P \; f(\theta)}{f'(\theta)} = C(T) \label{33Eq1}\] where \(\theta\) is the fraction of sites occupied at equilibrium and the terms \(f(\theta)\) and \(f '(\theta)\) contain the pre-exponential surface coverage dependence of the rates of adsorption and desorption respectively and all other factors have been taken over to the right hand side to give a temperature-dependent "constant" characteristic of this particular adsorption process, \(C(T)\). We now need to make certain simplifying assumptions. The first is one of the key assumptions of the Langmuir isotherm. Adsorption takes place only at specific localized sites on the surface and the saturation coverage corresponds to complete occupancy of these sites. Let us initially further restrict our consideration to a simple case of , i.e. \[S_- + \ce{M_{(g)}} \rightleftharpoons \ce{S-M} \label{33Eq2}\] where Under these circumstances it is reasonable to assume coverage dependencies for rates of the two processes of the form: These coverage dependencies in Equations \(\ref{Eqabsorb}\) and \(\ref{Eqdesorb}\) are exactly what would be predicted by noting that the forward and reverse processes are elementary reaction steps, in which case it follows from standard chemical kinetic theory that Substitution of Equations \(\ref{Eqabsorb}\) and \(\ref{Eqdesorb}\) into Equation \(\ref{33Eq1}\) yields: \[\dfrac{P(1-\theta)}{\theta}=B(T)\] where \[B(T) = \left(\dfrac{c'}{c}\right) C(T).\] After rearrangement this gives the expression for the surface coverage \[ \theta = \dfrac{bP}{1+bP}\] where \(b = 1/B(T)\) is a function of temperature and contains an exponential term of the form \[b \propto \exp [ ( E_a^{des} - E_a^{ads} ) / R T ] = \exp [ - ΔH_{ads} / R T ]\] with \[ΔH_{ads} = E_a^{des} - E_a^{ads}.\] \(b\) can be regarded as a with respect to coverage only if the enthalpy of adsorption is itself of coverage; this is the second major assumption of the Langmuir Isotherm. | 2,549 | 2,128 |
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Because an equilibrium state is achieved when the forward reaction rate equals the reverse reaction rate, under a given set of conditions there must be a relationship between the composition of the system at equilibrium and the kinetics of a reaction (represented by rate constants). We can show this relationship using the system described in , the decomposition of N O to NO . Both the forward and reverse reactions for this system consist of a single elementary reaction, so the reaction rates are as follows:
\[\text{forward rate} = k_f[N_2O_4] \tag{15.2.1}\] and \[\text{reverse rate} = k_r[NO_2]^2 \tag{15.2.2}\] At equilibrium, the forward rate equals the reverse rate:
\[ k_f[N_2O_4] = k_r[NO_2]^2 \tag{15.2.3}\] so \[\dfrac{k_f}{k_r}=\dfrac{[NO_2]^2}{[N_2O_4]} \tag{15.2.4}\] The ratio of the rate constants gives us a new constant, the equilibrium constant ( ) which is defined as follows:
\[K=\dfrac{k_f}{k_r} \tag{15.2.5}\] Hence there is a fundamental relationship between chemical kinetics and chemical equilibrium: The equilibrium constant is equal to the rate constant for the reaction divided by the rate constant for the reaction. lists the initial and equilibrium concentrations from five different experiments using the reaction system described by . At equilibrium the magnitude of the quantity [NO ] /[N O ] is essentially the same for all five experiments. In fact, no matter what the initial concentrations of NO and N O are, at equilibrium the quantity [NO ] /[N O ] will be 6.53 ± 0.03 × 10 at 25°C, which corresponds to the ratio of the rate constants for the forward and reverse reactions. That is, at a given temperature, the equilibrium constant for a reaction always has the same value, even though the specific concentrations of the reactants and products vary depending on their initial concentrations. In 1864, the Norwegian chemists Cato Guldberg (1836–1902) and Peter Waage (1833–1900) carefully measured the compositions of many reaction systems at equilibrium. They discovered that for reversible reaction of the general form
\[aA+bB \rightleftharpoons cC+dD \tag{15.2.6}\] where A and B are reactants, C and D are products, and , , , and are the stoichiometric coefficients in the balanced chemical equation for the reaction, the ratio of the product of the equilibrium concentrations of the products (raised to their coefficients in the balanced chemical equation) to the product of the equilibrium concentrations of the reactants (raised to their coefficients in the balanced chemical equation) is always a constant under a given set of conditions. This relationship is known as the law of mass action and can be stated as follows:
\[K=\dfrac{[C]^c[D]^d}{[A]^a[B]^b} \tag{15.2.7}\] where is the equilibrium constant for the reaction. is called the equilibrium equation aA+bB = cC+ dD , and the right side of is called the equilibrium constant expression. The relationship shown in is true for pair of opposing reactions regardless of the mechanism of the reaction or the number of steps in the mechanism. The equilibrium constant can vary over a wide range of values. The values of shown in , for example, vary by 60 orders of magnitude. Because products are in the numerator of the equilibrium constant expression and reactants are in the denominator, values of greater than 10 indicate a strong tendency for reactants to form products. In this case, chemists say that equilibrium lies to the right as written, favoring the formation of products. An example is the reaction between H and Cl to produce HCl, which has an equilibrium constant of 1.6 × 10 at 300 K. Because H is a good reductant and Cl is a good oxidant, the reaction proceeds essentially to completion. In contrast, values of less than 10 indicate that the ratio of products to reactants at equilibrium is very small. That is, reactants do not tend to form products readily, and the equilibrium lies to the left as written, favoring the formation of reactants. You will also notice in that equilibrium constants have no units, even though suggests that the units of concentration might not always cancel because the exponents may vary. In fact, equilibrium constants are calculated using “effective concentrations,” or , of reactants and products, which are the ratios of the measured concentrations to a standard state of 1 M. As shown in , the units of concentration cancel, which makes unitless as well:
\[ \dfrac{[A]_{measured}}{[A]_{standard\; state}}=\dfrac{\cancel{M}}{\cancel{M}} = \dfrac{\cancel{\frac{mol}{L}}}{\cancel{\frac{mol}{L}}} \tag{15.2.8}\] Many reactions have equilibrium constants between 1000 and 0.001 (10 ≥ ≥ 10 ), neither very large nor very small. At equilibrium, these systems tend to contain significant amounts of both products and reactants, indicating that there is not a strong tendency to form either products from reactants or reactants from products. An example of this type of system is the reaction of gaseous hydrogen and deuterium, a component of high-stability fiber-optic light sources used in ocean studies, to form HD:
\[H_{2(g)}+D_{2(g)} \rightleftharpoons 2HD_{(g)} \tag{15.10}\] The equilibrium constant expression for this reaction is \[K= \dfrac{[HD]^2}{[H_2,D_2]}\] varying between 1.9 and 4 over a wide temperature range (100–1000 K). Thus an equilibrium mixture of H , D , and HD contains significant concentrations of both product and reactants. summarizes the relationship between the magnitude of and the relative concentrations of reactants and products at equilibrium for a general reaction, written as Because there is a direct relationship between the kinetics of a reaction and the equilibrium concentrations of products and reactants ( and ), when >> , is a large number, and the concentration of products at equilibrium predominate. This corresponds to an essentially irreversible reaction. Conversely, when << , is a very small number, and the reaction produces almost no products as written. Systems for which ≈ have significant concentrations of both reactants and products at equilibrium. A large value of the equilibrium constant means that products predominate at equilibrium; a small value means that reactants predominate at equilibrium. Write the equilibrium constant expression for each reaction. balanced chemical equations equilibrium constant expressions Refer to . Place the arithmetic product of the concentrations of the products (raised to their stoichiometric coefficients) in the numerator and the product of the concentrations of the reactants (raised to their stoichiometric coefficients) in the denominator. Exercise Write the equilibrium constant expression for each reaction. Predict which systems at equilibrium will (a) contain essentially only products, (b) contain essentially only reactants, and (c) contain appreciable amounts of both products and reactants. systems and values of composition of systems at equilibrium Use the value of the equilibrium constant to determine whether the equilibrium mixture will contain essentially only products, essentially only reactants, or significant amounts of both. Exercise Hydrogen and nitrogen react to form ammonia according to the following balanced chemical equation: \[3H_{2(g)}+N_{2(g)} \rightleftharpoons 2NH_{3(g)} \notag \] Values of the equilibrium constant at various temperatures were reported as = 3.3 × 10 , = 2.6 × 10 , and = 4.1. Because equilibrium can be approached from either direction in a chemical reaction, the equilibrium constant expression and thus the magnitude of the equilibrium constant depend on the form in which the chemical reaction is written. For example, if we write the reaction described in in reverse, we obtain the following:
\[cC+dD \rightleftharpoons aA+bB \tag{15.2.10}\] The corresponding equilibrium constant \(K′\) is as follows: \[K'=\dfrac{[A]^a[B]^b}{[C]^c[D]^d} \tag{15.2.11}\] This expression is the inverse of the expression for the original equilibrium constant, so ′ = 1/ . That is, when we write a reaction in the reverse direction, the equilibrium constant expression is inverted. For instance, the equilibrium constant for the reaction is as follows:
\[K=\dfrac{[NO_2]^2}{[N_2O_4]} \tag{15.2.12}\] but for the opposite reaction, 2NO ⇌ the equilibrium constant ′ is given by the inverse expression:
\[K'=\dfrac{[N_2O_4]}{[NO_2]^2} \tag{15.2.13}\] Consider another example, the formation of water: 2H (g) + O (g) 2H O(g) Because H is a good reductant and O is a good oxidant, this reaction has a very large equilibrium constant ( = 2.4 × 10 at 500 K). Consequently, the equilibrium constant for the reverse reaction, the decomposition of water to form O and H , is very small: ′ = 1/ = 1/(2.4 × 10 ) = 4.2 × 10 . As suggested by the very small equilibrium constant, and fortunately for life as we know it, a substantial amount of energy is indeed needed to dissociate water into H and O . The equilibrium constant for a reaction written in reverse is the of the equilibrium constant for the reaction as written originally. Writing an equation in different but chemically equivalent forms also causes both the equilibrium constant expression and the magnitude of the equilibrium constant to be different. For example, we could write the equation for the reaction \[2NO_2 \rightleftharpoons N_2O_4\] as \[NO_2 \rightleftharpoons \frac{1}{2}N_2O_4\] with the equilibrium constant ″ is as follows:
\[ K′′=\dfrac{[N_2O_4]^{1/2}}{[NO_2]} \tag{15.2.14}\] The values for ′ ( ) and ″ are related as follows:
\[ K′′=(K')^{1/2}=\sqrt{K'} \tag{15.2.15}\] In general, if all the coefficients in a balanced chemical equation are subsequently multiplied by , then the new equilibrium constant is the original equilibrium constant raised to the th power. At 745 K, is 0.118 for the following reaction: \[N_{2(g)}+3H_{2(g)} \rightleftharpoons 2NH_{3(g)} \notag \] What is the equilibrium constant for each related reaction at 745 K? balanced equilibrium equation, at a given temperature, and equations of related reactions values of for related reactions Write the equilibrium constant expression for the given reaction and for each related reaction. From these expressions, calculate for each reaction. The equilibrium constant expression for the given reaction of N (g) with H (g) to produce NH (g) at 745 K is as follows: \[K=\dfrac{[NH_3]^2}{[N_2,H_2]^3}=0.118 \notag \] Exercise At 527°C, the equilibrium constant for the reaction \[2SO_{2(g)}+O_{2(g)} \rightleftharpoons 2SO_{3(g)} \notag \] is 7.9 × 10 . Calculate the equilibrium constant for the following reaction at the same temperature: \[SO_{3(g)} \rightleftharpoons SO_{2(g)}+\frac{1}{2}O_{2(g) \notag \] 3.6 × 10 For reactions that involve species in solution, the concentrations used in equilibrium calculations are usually expressed in moles/liter. For gases, however, the concentrations are usually expressed in terms of partial pressures rather than molarity, where the standard state is 1 atm of pressure. The symbol is used to denote equilibrium constants calculated from partial pressures. For the general reaction aA + bB in which all the components are gases, we can write the equilibrium constant expression as the ratio of the partial pressures of the products and reactants (each raised to its coefficient in the chemical equation):
\[K_p=\dfrac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b} \tag{15.2.16}\] Thus for the decomposition of N O ( ) is as follows:
\[K_p=\dfrac{(P_{NO_2})^2}{P_{N_2O_4}} \tag{15.2.17}\] Like , is a unitless quantity because the quantity that is actually used to calculate it is an “effective pressure,” the ratio of the measured pressure to a standard state of 1 bar (approximately 1 atm), which produces a unitless quantity. Because partial pressures are usually expressed in atmospheres or mmHg, the molar concentration of a gas and its partial pressure do not have the same numerical value. Consequently, the numerical values of and are usually different. They are, however, related by the ideal gas constant ( ) and the temperature ( ):
\[K_p = K(RT)^{Δn} \tag{15.2.18}\] where is the equilibrium constant expressed in units of concentration and Δ is the difference between the numbers of moles of gaseous products and gaseous reactants ( − ). The temperature is expressed as the absolute temperature in kelvins. According to , = only if the moles of gaseous products and gaseous reactants are the same (i.e., Δ = 0). For the decomposition of N O , there are 2 mol of gaseous product and 1 mol of gaseous reactant, so Δ = 1. Thus, for this reaction, = ( ) = . The equilibrium constant for the reaction of nitrogen and hydrogen to give ammonia is 0.118 at 745 K. The balanced equilibrium equation is as follows: \[N_{2(g)}+3H_{2(g)} \rightleftharpoons 2NH_{3(g)} \notag \] What is for this reaction at the same temperature? equilibrium equation, equilibrium constant, and temperature Use the coefficients in the balanced chemical equation to calculate Δ . Then use to calculate from . This reaction has 2 mol of gaseous product and 4 mol of gaseous reactants, so Δ = (2 − 4) = −2. We know , and = 745 K. Thus, from , we have the following: \[K_p=K(RT)^{−2}=\dfrac{K}{(RT)^2}=\dfrac{0.118}{\{ [0.08206(L \cdot atm)/(mol \cdot K),745\; K]\}^2}=3.16 \times 10^{−5} \notag \] Because is a unitless quantity, the answer is = 3.16 × 10 . Exercise Calculate for the reaction \[2SO_{2(g)}+O_{2(g)} \rightleftharpoons 2SO_{3(g)} \notag \] at 527°C, if = 7.9 × 10 at this temperature. = 1.2 × 10 When the products and reactants of an equilibrium reaction form a single phase, whether gas or liquid, the system is a homogeneous equilibrium . In such situations, the concentrations of the reactants and products can vary over a wide range. In contrast, a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibrium , such as the reaction of a gas with a solid or liquid. Because the molar concentrations of pure liquids and solids normally do not vary greatly with temperature, their concentrations are treated as constants, which allows us to simplify equilibrium constant expressions that involve pure solids or liquids. (Recall from , for example, that the density of water, and thus its volume, changes by only a few percentage points between 0°C and 100°C.) Consider the following reaction, which is used in the final firing of some types of pottery to produce brilliant metallic glazes:
\[CO_{2(g)}+C_{(s)} \rightleftharpoons 2CO_{(g)} \tag{15.2.19}\] The glaze is created when metal oxides are reduced to metals by the product, carbon monoxide. The equilibrium constant expression for this reaction is as follows:
\[K=\dfrac{[CO]^2}{[CO_2,C]} \tag{15.2.20}\] Because graphite is a solid, however, its molar concentration, determined from its density and molar mass, is essentially constant and has the following value:
\[ [C] =\dfrac{2.26 \cancel{g}/{\cancel{cm^3}}}{12.01\; \cancel{g}/mol} \times 1000 \; \cancel{cm^3}/L = 188 \; mol/L = 188\;M \tag{15.2.21}\] We can rearrange so that the constant terms are on one side:
\[ K[C]=K(188)=\dfrac{[CO]^2}{[CO_2]} \tag{15.2.22}\] Incorporating the constant value of [C] into the equilibrium equation for the reaction in ,
\[K'=\dfrac{[CO]^2}{[CO_2]} \tag{15.2.23}\] The equilibrium constant for this reaction can also be written in terms of the partial pressures of the gases:
\[K_p=\dfrac{(P_{CO})^2}{P_{CO_2}} \tag{15.2.24}\] Incorporating all the constant values into ′ or allows us to focus on the substances whose concentrations change during the reaction. Although the concentrations of pure liquids or solids are not written explicitly in the equilibrium constant expression, these substances must be present in the reaction mixture for chemical equilibrium to occur. Whatever the concentrations of CO and CO , the system described in will reach chemical equilibrium only if a stoichiometric amount of solid carbon or excess solid carbon has been added so that some is still present once the system has reached equilibrium. As shown in , it does not matter whether 1 g or 100 g of solid carbon is present; in either case, the composition of the gaseous components of the system will be the same at equilibrium. Write each expression for , incorporating all constants, and for the following equilibrium reactions. balanced equilibrium equations expressions for and Find by writing each equilibrium constant expression as the ratio of the concentrations of the products and reactants, each raised to its coefficient in the chemical equation. Then express as the ratio of the partial pressures of the products and reactants, each also raised to its coefficient in the chemical equation. and \[K_p=\dfrac{1}{P_{Cl_2}} \notag \] Exercise Write the expressions for and for the following reactions. For reactions carried out in solution, the concentration of the solvent is omitted from the equilibrium constant expression even when the solvent appears in the balanced chemical equation for the reaction. The concentration of the solvent is also typically much greater than the concentration of the reactants or products (recall that pure water is about 55.5 M, and pure ethanol is about 17 M). Consequently, the solvent concentration is essentially constant during chemical reactions, and the solvent is therefore treated as a pure liquid. The equilibrium constant expression for a reaction contains only those species whose concentrations could change significantly during the reaction. The concentrations of pure solids, pure liquids, and solvents are omitted from equilibrium constant expressions because they do not change significantly during reactions when enough is present to reach equilibrium. Chemists frequently need to know the equilibrium constant for a reaction that has not been previously studied. In such cases, the desired reaction can often be written as the sum of other reactions for which the equilibrium constants are known. The equilibrium constant for the unknown reaction can then be calculated from the tabulated values for the other reactions. To illustrate this procedure, let’s consider the reaction of N with O to give NO . As we stated in , this reaction is an important source of the NO that gives urban smog its typical brown color. The reaction normally occurs in two distinct steps. In the first reaction (1), N reacts with O at the high temperatures inside an internal combustion engine to give NO. The released NO then reacts with additional O to give NO (2). The equilibrium constant for each reaction at 100°C is also given. The equilibrium constant expressions for the reactions are as follows: \[K_1=\dfrac{[NO]^2}{[N_2,O_2]}\;\;\; K_2=\dfrac{[NO_2]^2}{[NO]^2[O_2]}\;\;\; K_3=\dfrac{[NO_2]^2}{[N_2,O_2]^2} \notag \]v What is the relationship between , , and all at 100°C? The expression for has [NO] in the numerator, the expression for has [NO] in the denominator, and [NO] does not appear in the expression for . Multiplying by and canceling the [NO] terms, \[ K_1K_2=\dfrac{\cancel{[NO]^2}}{[N_2,O_2]} \times \dfrac{[NO_2]^2}{\cancel{[NO]^2}[O_2]}=\dfrac{[NO_2]^2}{[N_2,O_2]^2}=K_3 \notag \] Thus the product of the equilibrium constant expressions for and is the same as the equilibrium constant expression for : \[K_3 = K_1K_2 = (2.0 \times 10^{−25})(6.4 \times 10^9) = 1.3 \times 10^{−15} \notag \] In contrast, recall that according to Hess’s Law, Δ for the sum of two or more reactions is the of the Δ values for the individual reactions. To determine for a reaction that is the sum of two or more reactions, add the reactions but multiply the equilibrium constants. The following reactions occur at 1200°C: two balanced equilibrium equations, values of , and an equilibrium equation for the overall reaction equilibrium constant for the overall reaction Arrange the equations so that their sum produces the overall equation. If an equation had to be reversed, invert the value of for that equation. Calculate for the overall equation by multiplying the equilibrium constants for the individual equations. The key to solving this problem is to recognize that reaction 3 is the sum of reactions 1 and 2: \[CO_{(g)}+ \cancel{3H_{2(g)}} \rightleftharpoons \cancel{CH_{4(g)}} + H_2O_{(g)} \notag \] \[\cancel{CH_{4(g)}} +2H_2S_{(g)} \rightleftharpoons CS_{2(g)} + \cancel{3H_{2(g)}} + H_{2(g)} \notag \] \[ CO_{(g)} + 3H_{2(g)} \rightleftharpoons CS_{2(g)}+H_2O_{(g)}+H_{2(g)} \notag \] The values for and are given, so it is straightforward to calculate : \[K_3 = K_1K_2 = (9.17 \times 10^{−2})(3.3 \times 10^4) = 3.03 \times 10^3 \notag \] Exercise In the first of two steps in the industrial synthesis of sulfuric acid, elemental sulfur reacts with oxygen to produce sulfur dioxide. In the second step, sulfur dioxide reacts with additional oxygen to form sulfur trioxide. The reaction for each step is shown, as is the value of the corresponding equilibrium constant at 25°C. Calculate the equilibrium constant for the overall reaction at this same temperature. = 1.1 × 10 The ratio of the rate constants for the forward and reverse reactions at equilibrium is the , a unitless quantity. The composition of the equilibrium mixture is therefore determined by the magnitudes of the forward and reverse rate constants at equilibrium. Under a given set of conditions, a reaction will always have the same . For a system at equilibrium, the relates to the ratio of the equilibrium concentrations of the products to the concentrations of the reactants raised to their respective powers to match the coefficients in the . The ratio is called the . When a reaction is written in the reverse direction, and the equilibrium constant expression are inverted. For gases, the equilibrium constant expression can be written as the ratio of the partial pressures of the products to the partial pressures of the reactants, each raised to a power matching its coefficient in the chemical equation. An equilibrium constant calculated from partial pressures ( ) is related to by the ideal gas constant ( ), the temperature ( ), and the change in the number of moles of gas during the reaction. An equilibrium system that contains products and reactants in a single phase is a ; a system whose reactants, products, or both are in more than one phase is a . When a reaction can be expressed as the sum of two or more reactions, its equilibrium constant is equal to the product of the equilibrium constants for the individual reactions. : \( K=\dfrac{k_f}{k_r} \) : \( K=\dfrac{\left [ C \right ]^c \left [ D \right ]^d}{\left [ A \right ]^a \left [ B \right ]^b} \) : \( K_p=\dfrac{P_C^c P_D^d}{P_A^aP_B^b} \) : \( K_p=K\left ( RT \right )^{\Delta n} \) For an equilibrium reaction, what effect does reversing the reactants and products have on the value of the equilibrium constant? Which of the following equilibriums are homogeneous and which are heterogeneous? Classify each equilibrium system as either homogeneous or heterogeneous. If an equilibrium reaction is endothermic, what happens to the equilibrium constant if the temperature of the reaction is increased? if the temperature is decreased? Industrial production of NO by the reaction is carried out at elevated temperatures to drive the reaction toward the formation of product. After sufficient product has formed, the reaction mixture is quickly cooled. Why? How would you differentiate between a system that has reached chemical equilibrium and one that is reacting so slowly that changes in concentration are difficult to observe? What is the relationship between the equilibrium constant, the concentration of each component of the system, and the rate constants for the forward and reverse reactions? Write the equilibrium constant expressions for and for each reaction. Write the equilibrium constant expressions for and as appropriate for each reaction. Why is it incorrect to state that pure liquids, pure solids, and solvents are not part of an equilibrium constant expression? Write the equilibrium constant expressions for and for each equilibrium reaction. Write the equilibrium constant expressions for and for each equilibrium reaction. At room temperature, the equilibrium constant for the reaction is 1. What does this indicate about the concentrations of A and B at equilibrium? Would you expect and to vary significantly from each other? If so, how would their difference be affected by temperature? For a certain series of reactions, if [OH ,HCO ]/[CO ] = and [OH ,H CO ]/[HCO ] = , what is the equilibrium constant expression for the overall reaction? Write the overall equilibrium equation. In the equation for an enzymatic reaction, ES represents the complex formed between the substrate S and the enzyme protein E. In the final step of the following oxidation reaction, the product P dissociates from the ESO complex, which regenerates the active enzyme: Give the overall reaction equation and show that = × × . The equilibrium constant for the reaction written in reverse: ′ = 1/ . Each system is heterogeneous. Rapid cooling “quenches” the reaction mixture and prevents the system from reverting to the low-temperature equilibrium composition that favors the reactants. At equilibrium, \([A]=\sqrt{B}\) and \(Δn =−1\) so \(K_p=K(RT)^{Δn}=\dfrac{K}{RT}\) the difference increases as increases. Explain what each of the following values for tells you about the relative concentrations of the reactants versus the products in a given equilibrium reaction: = 0.892; = 3.25 × 10 ; = 5.26 × 10 . Are products or reactants favored at equilibrium? Write the equilibrium constant expression for each reaction. Are these equilibrium constant expressions equivalent? Explain. Write the equilibrium constant expression for each reaction. How are these two expressions mathematically related to the equilibrium constant expression for \[N_{2(g)}+3H_{2(g)} \rightleftharpoons 2NH_{3(g)} ? \notag \] Write an equilibrium constant expression for each reaction. Give an equilibrium constant expression for each reaction. Calculate and for each reaction. Calculate and for each reaction. Determine and (where applicable) for each reaction. Determine and for each reaction. The equilibrium constant expression for a reaction is [CO ] /[SO ] [O ]. What is the balanced chemical equation for the overall reaction if one of the reactants is Na CO (s)? The equilibrium constant expression for a reaction is [NO,H O] /[NH ,O ] . What is the balanced chemical equation for the overall reaction? Given = / , what happens to the magnitude of the equilibrium constant if the reaction rate of the forward reaction is doubled? What happens if the reaction rate of the reverse reaction for the overall reaction is decreased by a factor of 3? The value of the equilibrium constant for \[ 2H_{2(g)}+S_{2(g)}⇌2H_2S_{(g)} \notag \] is 1.08 × 10 at 700°C. What is the value of the equilibrium constant for the following related reactions? = 0.892: the concentrations of the products and the reactants are approximately equal at equilibrium so neither is favored; = 3.25 × 10 : the ratio of the concentration of the products to the reactants at equilibrium is very large so the formation of products is favored; = 5.26 × 10 : the ratio of the concentration of the products to the reactants at equilibrium is very small so the formation of products is not favored. \(NH_3 + \frac{5}{4}O_2⇌NO +\frac{3}{2}H_2O\), which can also be written as follows: \[4NH_{3(g)} + 5O_{2(g)}⇌4NO_{(g)} + 6H_2O_{(g)}\] | 27,839 | 2,129 |
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The interactions between ions (ion - ion interactions) are the easiest to understand: like charges repel each other and opposite charges attract. These Coulombic forces operate over relatively long distances in the gas phase. The force depends on the product of the charges (\(Z_1\), \(Z_2\)) divided by the square of the distance of separation (\(d^2\)): \[F \propto \dfrac{- Z_1Z_2}{d^2} \tag{1}\] Two oppositely-charged particles flying about in a vacuum will be attracted toward each other, and the force becomes stronger and stronger as they approach until eventually they will stick together and a considerable amount of energy will be required to separate them.
They form an , a new particle which has a positively-charged area and a negatively-charged area. There are fairly strong interactions between these ion pairs and free ions, so that these the clusters tend to grow, and they will eventually fall out of the gas phase as a liquid or solid (depending on the temperature). , Professor of Chemistry, University of Missouri-Rolla | 1,061 | 2,130 |
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In some contexts, the choice of what heat source to use is critical while in other contexts several could work equally well. The choice of which heat source to use depends on several factors: As safety is an important factor in making laboratory choices, it's important to consider the flammability of the liquid to be heated. Almost all organic liquids are considered "flammable," meaning they are capable of catching on fire and sustaining combustion (an important exception is that halogenated solvents tend to be non-flammable). However, this doesn't mean that all organic liquids will immediately if placed near a heat source. Many liquids require an ignition source (a spark, match, or flame) in order for their vapors to catch on fire, a property often described by the liquid's . The flash point is the temperature where the vapors can be ignited with an ignition source. For example, the flash point of \(70\%\) ethanol is \(16.6^\text{o} \text{C}\),\(^2\) meaning it can catch on fire at room temperature using a match (Figure 1.38). A Bunsen burner is an excellent ignition source (and can reach temperatures of approximately \(1500^\text{o} \text{C}\)),\(^3\) making burners a serious fire hazard with organic liquids, and a heat source that should often be avoided. Another important property in discussing flammability is a liquid's : the temperature where the substance spontaneously ignites under normal pressure and the presence of an ignition source. This property is particularly insightful because it does not require a flame (which is often avoided in the organic lab), but only a hot area. A hotplate surface turned up to "high" can reach temperatures up to \(350^\text{o} \text{C}\).\(^3\) as diethyl ether, pentane, hexane, and low-boiling petroleum ether have autoignition temperatures below this value (Table 1.7), it would be dangerous to boil these solvents on a hotplate as vapors could spill out of the container and ignite upon contact with the surface of the hotplate. In general, caution should be used when using a hotplate for heating volatile, flammable liquid in an open vessel as it's possible that vapors can overrun the hotplate's ceramic covering and contact the heating element beneath, which may be hotter than \(350^\text{o} \text{C}\). It is for this reason that hotplates are not the optimal choice when heating open vessels of volatile organic liquids, although in some cases they may be used cautiously when set to "low" and used in a well-ventilated fume hood. As combustion is a reaction in the vapor phase, liquids with low boiling points (< \(40^\text{o} \text{C}\)) tend to have low flash points and autoignition temperatures as they have significant vapor pressures (Table 1.7). All low boiling liquids should be treated more cautiously than liquids with moderate boiling points (> \(60^\text{o} \text{C}\)). \(^2\)From the SDS (Safety Data Sheets) of \(70\%\) denatured ethanol. \(^3\) As reported in the Fischer Scientific catalog. \(^4\) Data from , 84\(^\text{th}\) ed., CRC Press, , 16-16 to 16-31. Petroleum ether autoignition temperature is from the SDS. | 3,138 | 2,131 |
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The polar oxidative addition mechanism is very similar to an aliphatic nucleophilic substitution (S 1 or S 2) reaction. Figure OA3.1. An example of a polar oxidative addition. In an oxidative addition, the metal can act as a nucleophile in the first step in an S 2 process. In the second step, the liberated halide binds to the metal. That doesn't happen in a normal nucleophilic substitution. In this case, the metal has donated its electrons and is able to accept another pair from the halide. Figure OA3.2. Mechanistic steps in a polar oxidative addition. Polar oxidative addition has some requirements similar to a regular S 1 or S 2 reaction: The platinum compound shown below is capable of reductively eliminating a molecule of iodobenzene. a) Show the products of this reaction. The starting platinum compound is completely stable in benzene; no reaction occurs in that solvent. However, reductive elimination occurs quickly when the compound is dissolved in methanol instead. b) Explain why the solvents may play a role in how easily this compound reacts. The reaction in methanol is inhibited by added iodide salts, such as sodium iodide. c) Provide a mechanism for the reductive elimination of iodobenzene from the platinum complex, taking into account the solvent dependence and the inhibition by iodide ion. For the following reaction, Reaction of the following deuterium-labeled alkyl chloride with tetrakis(triphenylphosphine) palladium produces an enantiomerically pure product (equation a). Draw the expected product. However, reaction of a very similar alkyl halide produces a compound that is only 90% enantiomerically pure. Draw the major product and explain the reason that there is some racemization. Frequently, oxidative additions and reductive eliminations are preceded or followed other reactions. Draw a mechanism for the following transformation. , | 1,888 | 2,132 |
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In all solutions, whether gaseous, liquid, or solid, the substance present in the greatest amount is the solvent, and the substance or substances present in lesser amounts are the solute(s). The solute does not have to be in the same physical state as the solvent, but the physical state of the solvent usually determines the state of the solution. As long as the solute and solvent combine to give a homogeneous solution, the solute is said to be soluble in the solvent. Table \(\Page {1}\) lists some common examples of gaseous, liquid, and solid solutions and identifies the physical states of the solute and solvent in each. The formation of a solution from a solute and a solvent is a physical process, not a chemical one. That is, both solute and solvent can be recovered in chemically unchanged forms using appropriate separation methods. For example, solid zinc nitrate dissolves in water to form an aqueous solution of zinc nitrate: \[\ce{Zn(NO3)2 (s) + H2O(l) \rightarrow Zn^{2+} (aq) +2NO^{-}3 (aq)} \nonumber\] Because \(\ce{Zn(NO3)2}\) can be recovered easily by evaporating the water, this is a physical process. In contrast, metallic zinc appears to dissolve in aqueous hydrochloric acid. In fact, the two substances undergo a chemical reaction to form an aqueous solution of zinc chloride with evolution of hydrogen gas: \[\ce{ Zn (s) + 2H^{+}(aq) + 2Cl^{-} (aq) \rightarrow Zn^{2+} (aq) + 2Cl^{-} (aq) + H_2 (g) } \nonumber\] When the solution evaporates, we do not recover metallic zinc, so we cannot say that metallic zinc is soluble in aqueous hydrochloric acid because it is chemically transformed when it dissolves. The dissolution of a solute in a solvent to form a solution does not involve a chemical transformation. An amalgam is an alloy of mercury with another metal. It may be a liquid, a soft paste or a solid, depending upon the proportion of mercury. These alloys are formed through metallic bonding, with the electrostatic attractive force of the conduction electrons working to bind all the positively charged metal ions together into a crystal lattice structure. Almost all metals can form amalgams with mercury including aluminum ( \(\Page {1}\)), the notable exceptions being iron, platinum, tungsten, and tantalum. Silver-mercury amalgams are important in dentistry, and gold-mercury amalgam is used in the extraction of gold from ore. Substances that form a single homogeneous phase in all proportions are said to be completely in one another. Ethanol and water are miscible, just as mixtures of gases are miscible. If two substances are essentially insoluble in each other, such as oil and water, they are immiscible. Examples of gaseous solutions that we have already discussed include Earth’s atmosphere. | 2,766 | 2,133 |
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There are two basic steps in ligand substitution: association and dissociation. Association, in this case, refers to the binding of a ligand to the metal. The ligand donates an electron pair to the metal and the two molecules come together to form a new bond. Dissociation, in this case, refers to the release of a ligand from a metal. The metal-ligand bond breaks and the ligand leaves with its electron pair. Two mechanistic possibilities seem pretty obvious. Either the new ligand binds first and then the old one leaves, or the old ligand leaves first and then the new one binds. Knowing the mechanism is important because the mechanism has an impact on what factors affect the reaction. For example, if the reaction is associative, adding lots more new ligand may speed up the reaction, because then it becomes more likely that the new ligand will find the metal complex and bind with it. However, if the old ligand is supposed to leave before the new ligand arrives, then it doesn’t matter how much new ligand is around. It has to wait for the old ligand to leave before it can bind, anyway, so adding a lot more new ligand won’t speed things up. Draw a mechanism, with arrows, for the following substitution. Assume an associative mechanism. Draw a mechanism, with arrows, for the following substitution. Assume a dissociative mechanism. Draw a mechanism, with arrows, for the following isomerization. The ability to substitute for a ligand depends partly on its ability to leave. Rank the following ligands, from the easiest to replace to the hardest to replace: CO Cl PPh NH NO H O Draw curved arrows for the following steps. Classify each step as associative or dissociative. , | 1,703 | 2,134 |
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In all solutions, whether gaseous, liquid, or solid, the substance present in the greatest amount is the solvent, and the substance or substances present in lesser amounts are the solute(s). The solute does not have to be in the same physical state as the solvent, but the physical state of the solvent usually determines the state of the solution. As long as the solute and solvent combine to give a homogeneous solution, the solute is said to be soluble in the solvent. Table \(\Page {1}\) lists some common examples of gaseous, liquid, and solid solutions and identifies the physical states of the solute and solvent in each. The formation of a solution from a solute and a solvent is a physical process, not a chemical one. That is, both solute and solvent can be recovered in chemically unchanged forms using appropriate separation methods. For example, solid zinc nitrate dissolves in water to form an aqueous solution of zinc nitrate: \[\ce{Zn(NO3)2(s) + H2O(l) \rightarrow Zn^{2+}(aq) + 2NO^{-}3(aq)} \label{13.1.1} \] Because \(Zn(NO_3)_2\) can be recovered easily by evaporating the water, this is a physical process. In contrast, metallic zinc appears to dissolve in aqueous hydrochloric acid. In fact, the two substances undergo a chemical reaction to form an aqueous solution of zinc chloride with evolution of hydrogen gas: \[\ce{ Zn(s) + 2H^{+}(aq) + 2Cl^{-}(aq) \rightarrow Zn^{2+}(aq) + 2Cl^{-}(aq) + H2(g)} \label{13.1.2} \] When the solution evaporates, we do not recover metallic zinc, so we cannot say that metallic zinc is soluble in aqueous hydrochloric acid because it is chemically transformed when it dissolves. The dissolution of a solute in a solvent to form a solution does not involve a chemical transformation (that it is a ). Dissolution of a solute in a solvent to form a solution does not involve a chemical transformation. Substances that form a single homogeneous phase in all proportions are said to be completely in one another. Ethanol and water are miscible, just as mixtures of gases are miscible. If two substances are essentially insoluble in each other, such as oil and water, they are . Examples of gaseous solutions that we have already discussed include Earth’s atmosphere. Energy is required to overcome the intermolecular interactions in a solute, which can be supplied only by the new interactions that occur in the solution, when each solute particle is surrounded by particles of the solvent in a process called (or hydration when the solvent is water). Thus all of the solute–solute interactions and many of the solvent–solvent interactions must be disrupted for a solution to form. In this section, we describe the role of enthalpy in this process. Because enthalpy is a , we can use a thermochemical cycle to analyze the energetics of solution formation. The process occurs in three discrete steps, indicated by \(ΔH_1\), \(ΔH_2\), and \(ΔH_3\) in Figure \(\Page {2}\). The overall enthalpy change in the formation of the solution (\( \Delta H_{soln}\)) is the sum of the enthalpy changes in the three steps: \[ \Delta H_{soln} = \Delta H_1 + \Delta H_2 + \Delta H_3 \label{13.1.3} \] When a solvent is added to a solution, steps 1 and 2 are both endothermic because energy is required to overcome the intermolecular interactions in the solvent (\(\Delta H_1\)) and the solute (\(\Delta H_2\)). Because \(ΔH\) is positive for both steps 1 and 2, the solute–solvent interactions (\(\Delta H_3\)) must be stronger than the solute–solute and solvent–solvent interactions they replace in order for the dissolution process to be exothermic (\(\Delta H_{soln} < 0\)). When the solute is an ionic solid, \(ΔH_2\) corresponds to the lattice energy that must be overcome to form a solution. The higher the charge of the ions in an ionic solid, the higher the lattice energy. Consequently, solids that have very high lattice energies, such as \(MgO\) (−3791 kJ/mol), are generally insoluble in all solvents. A positive value for \(ΔH_{soln}\) does not mean that a solution will not form. Whether a given process, including formation of a solution, occurs spontaneously depends on whether the total energy of the system is lowered as a result. Enthalpy is only one of the contributing factors. A high \(ΔH_{soln}\) is usually an indication that the substance is not very soluble. Instant cold packs used to treat athletic injuries, for example, take advantage of the large positive \(ΔH_{soln}\) of ammonium nitrate during dissolution (+25.7 kJ/mol), which produces temperatures less than 0°C (Figure \(\Page {3}\)). The enthalpy change that accompanies a process is important because processes that release substantial amounts of energy tend to occur spontaneously. A second property of any system, its entropy, is also important in helping us determine whether a given process occurs spontaneously. We will discuss entropy in more detail elsewhere, but for now we can state that entropy (\(S\)) is a thermodynamic property of all substances that is proportional to their degree of disorder. A perfect crystal at 0 K, whose atoms are regularly arranged in a perfect lattice and are motionless, has an entropy of zero. In contrast, gases have large positive entropies because their molecules are highly disordered and in constant motion at high speeds. The formation of a solution disperses molecules, atoms, or ions of one kind throughout a second substance, which generally increases the disorder and results in an increase in the entropy of the system. Thus entropic factors almost always favor formation of a solution. In contrast, a change in enthalpy may or may not favor solution formation. The London dispersion forces that hold cyclohexane and n-hexane together in pure liquids, for example, are similar in nature and strength. Consequently, \(ΔH_{soln}\) should be approximately zero, as is observed experimentally. Mixing equal amounts of the two liquids, however, produces a solution in which the n-hexane and cyclohexane molecules are uniformly distributed over approximately twice the initial volume. In this case, the driving force for solution formation is not a negative \(ΔH_{soln}\) but rather the increase in entropy due to the increased disorder in the mixture. All spontaneous processes with \(ΔH \ge 0\) are characterized by an in entropy. In other cases, such as mixing oil with water, salt with gasoline, or sugar with hexane, the enthalpy of solution is large and positive, and the increase in entropy resulting from solution formation is not enough to overcome it. Thus in these cases a solution does not form. All spontaneous processes with ΔH ≥ 0 are characterized by an increase in entropy. Table \(\Page {2}\) summarizes how enthalpic factors affect solution formation for four general cases. The column on the far right uses the relative magnitudes of the enthalpic contributions to predict whether a solution will form from each of the four. Keep in mind that in each case entropy favors solution formation. In two of the cases the enthalpy of solution is expected to be relatively small and can be either positive or negative. Thus the entropic contribution dominates, and we expect a solution to form readily. In the other two cases the enthalpy of solution is expected to be large and positive. The entropic contribution, though favorable, is usually too small to overcome the unfavorable enthalpy term. Hence we expect that a solution will not form readily. In contrast to liquid solutions, the intermolecular interactions in gases are weak (they are considered to be nonexistent in ideal gases). Hence mixing gases is usually a thermally neutral process (\(ΔH_{soln} \approx 0\)), and the entropic factor due to the increase in disorder is dominant (Figure \(\Page {4}\)). Consequently, all gases dissolve readily in one another in all proportions to form solutions. Considering \(\ce{LiCl}\), benzoic acid (\(\ce{C6H5CO2H}\)), and naphthalene, which will be most soluble and which will be least soluble in water? : three compounds relative solubilities in water : Assess the relative magnitude of the enthalpy change for each step in the process shown in Figure \(\Page {2}\). Then use Table \(\Page {2}\) to predict the solubility of each compound in water and arrange them in order of decreasing solubility. The first substance, \(\ce{LiCl}\), is an ionic compound, so a great deal of energy is required to separate its anions and cations and overcome the lattice energy (ΔH is far greater than zero in Equation \(\ref{13.1.1}\)). Because water is a polar substance, the interactions between both Li and Cl ions and water should be favorable and strong. Thus we expect \(ΔH_3\) to be far less than zero, making LiCl soluble in water. In contrast, naphthalene is a nonpolar compound, with only London dispersion forces holding the molecules together in the solid state. We therefore expect \(ΔH_2\) to be small and positive. We also expect the interaction between polar water molecules and nonpolar naphthalene molecules to be weak \(ΔH_3 \approx 0\). Hence we do not expect naphthalene to be very soluble in water, if at all. Benzoic acid has a polar carboxylic acid group and a nonpolar aromatic ring. We therefore expect that the energy required to separate solute molecules (ΔH ) will be greater than for naphthalene and less than for LiCl. The strength of the interaction of benzoic acid with water should also be intermediate between those of LiCl and naphthalene. Hence benzoic acid is expected to be more soluble in water than naphthalene but less soluble than \(\ce{LiCl}\). We thus predict \(\ce{LiCl}\) to be the most soluble in water and naphthalene to be the least soluble. Considering ammonium chloride, cyclohexane, and ethylene glycol (\(HOCH_2CH_2OH\)), which will be most soluble and which will be least soluble in benzene? The most soluble is cyclohexane; the least soluble is ammonium chloride. Solutions are homogeneous mixtures of two or more substances whose components are uniformly distributed on a microscopic scale. The component present in the greatest amount is the solvent, and the components present in lesser amounts are the solute(s). The formation of a solution from a solute and a solvent is a physical process, not a chemical one. Substances that are miscible, such as gases, form a single phase in all proportions when mixed. Substances that form separate phases are immiscible. Solvation is the process in which solute particles are surrounded by solvent molecules. When the solvent is water, the process is called hydration. The overall enthalpy change that accompanies the formation of a solution, \(ΔH_{soln}\), is the sum of the enthalpy change for breaking the intermolecular interactions in both the solvent and the solute and the enthalpy change for the formation of new solute–solvent interactions. Exothermic (\(ΔH_{soln} < 0\)) processes favor solution formation. In addition, the change in entropy, the degree of disorder of the system, must be considered when predicting whether a solution will form. An increase in entropy (a decrease in order) favors dissolution. | 11,128 | 2,135 |
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Let us assume that you are faced with a specific problem. Then we can see how scientific thinking might help solve it. Suppose that you live near a large plant which manufactures cement. Smoke from the plant settles on your car and house, causing small pits in the paint. You would like to stop this air-pollution problem—but how? As an individual you will probably have little influence, and even as part of a group of concerned citizens you may be ineffective, unless you can prove that there is a problem. Scientists have had a hand in writing most air-pollution regulations, and so you will have to employ some scientific techniques (or a scientist) to help solve your problem. It will probably be necessary to determine how much air pollution the plant is producing. This might be done by comparing the smoke with a scale which ranges from white to gray to black, the assumption being that the darker the smoke, the more there is. For white cement dust, however, this is much less satisfactory than for black coal smoke. A better way to determine how much pollution there is would be to measure the mass of smoke particles which could be collected near your house or car. This could be done by using a pump (such as a vacuum cleaner) to suck the polluted gas through a filter. Weighing the filter before and after such an experiment would determine what mass of smoke had been collected. Because is the force of gravity on an object, which varies from place to place by about 0.5% as shown in the Table, we must use for accurate measurements of how much matter we have. We still call the process of obtaining an accurate mass "weighing". The of an object is actually the force of gravity, and is calculated as \[F=\text{“W”}=m \times g \label{1} \] Weight is measured in newtons (kg m s ) or pounds (lb), where 1 lb = 4.44822162 newtons. The base unit for mass is kilograms (kg), but the pound may also be used as a unit of mass (1 pound = 0.45359237 kilograms). The average value of g is usually taken to be 9.80665 m s , so the weight of a 1 kg mass would be \[F = \text{ “W” } = m \times g = 1 \text{ kg} \times 9.80665 \text{ m s}^{-2} = 9.80665 \text{ N or } 2.2046 \text{ lb} \nonumber \] If the weight of the 1 kg mass were measured on an arctic exploration camp with a balance (see below) still calibrated for the average g, its weight would be: \[ F = \text{ “W” } = m \times g = 10 \text{ kg} \times 9.83208 \text{ m s}^{-2} = 9.83208 \text{ N or } 2.2103 \text{ lb} \nonumber \] Accurate weighing is usually done with a single pan balance. The empty pan is balanced by a counterweight. When an object is placed on the pan, gravitational attraction forces the pan downward. To restore balance, a series of weights (objects of known mass) are removed from holders above the pan. The force of gravitational attraction is proportional to mass, and when the pan is balanced, the force on it must always be the same. Therefore the mass of the object being weighed must equal that of the weights that were removed. A gives the same mass readout regardless of the force of gravity. Modern laboratory "balances" are based on that convert the force exerted by an object on the balance pan to an electrical signal. The load cell generally is coupled with a dedicated converter and digital display. Because the force exerted by the object depends on gravity, these are really that measure weight, and must be calibrated frequently (against standard masses) to read a . If you kept a notebook or other record of your measurements, you would probably write down something like 0.0342 g to represent how much smoke had been collected. Such a result, which describes (in this case the magnitude of the parameter, mass), is called a . Notice that it consists of two parts, a number and a unit. It would be ambiguous to write just 0.0342—you might not remember later whether that was measured in units of grams, ounces, pounds, or something else. A quantity always behaves as though the number and the units are multiplied together. For example, we could write the quantity already obtained as 0.0342 × g. Using this simple rule of number × units, we can apply arithmetic and algebra to any quantity: \[\begin{align} & 5 \text{g} + 2 \text{g} = (5 + 2) \text{g} = 7 \text{g} \\ & 5 \text{g} \div 2 \text{g} = \dfrac{5 \text{g}}{2 \text{g}} = 2.5 \text{ (the units cancel, and so we get a pure number)} \\ & 5 \text{ in} \times 2 \text{ in} = 10 \text{ in}^{2} \text{ (10 square inches)}\end{align} \nonumber \] This works perfectly well as long as we do not write equations with different (i.e., those having units which measure different properties, like mass and length, temperature and energy, or volume and amount) on opposite sides of the equal sign. For example, applying algebra to the equation \[ 5 \text{g} = 2 \text{ in}^{2} \nonumber \] can lead to trouble in much the same way that dividing by zero does and should be avoided, because grams (g) is a unit of the , and the inch (in) is a unit of the . Notice also that whether a quantity is large or small depends on the size of the units as well as the size of the number. For example, the mass of smoke has been measured as 0.0342 g, but the balance might have been set to read in milligrams (or grains in the English system). Then the reading would be 34.2 mg (or 0.527 grains). The results (0.0342 g, 34.2 mg, or 0.527 gr) are the quantity, the mass of smoke. One involves a smaller number and larger unit (0.0342 g), while the others have a larger number and smaller unit. So long as we are talking about the same quantity, it is a simple matter to adjust the number to go with any units we want. We can convert among the different ways of expressing the mass with as follows: Since 1 mg and 0.001 g are the same parameter (mass), we can write the equation \[ 1 \text{ mg} = 0.001 \text{ g} \nonumber \] Dividing both sides by 1 mg, we have \[1 = \dfrac{1\text{ mg}}{1\text{ mg}} = \dfrac{0.001\text{ g}}{1\text{ mg}} \nonumber \] Since the last term of this equation equals one, it is called a . It can be multiplied by any quantity, leaving the quantity unchanged. We can generate another unity factor by dividing both sides by 0.001 g: \[1 = \dfrac{1\text{ mg}}{0.001\text{ g}} = \dfrac{0.001\text{ g}}{0.001\text{ g}} \nonumber \] What is the mass in grams of a 5.0 grain (5 gr) aspirin tablet, given that 1 gram = 15.4323584 grains? \[ 5.0 \text{ gr} = 5.0 \text{ gr} \times 1 = 5.0 \text{ gr}\times \dfrac{1.0\text{ g}}{\text{15.4323 gr}} \nonumber \] The units cancel, yielding the result \[ 5.0 \text{ gr} = 5.0 \div 15.4323 \text{ g} = 0.324 \text{ g} \nonumber \] The length may be measured in inches (in) in the English system, but scientific measurements (all measurements in the world exclusive of the U.S.) are reported in the metric units of meters (m) or some more convenient derived unit like centimeters (cm). Express the length 8.89 cm in inches, given that 1 cm = 0.3937 in. Since 1 cm and 0.3937 in are the same quantity, we can write the equation \[1 \text{ cm} = 0.3937 \text{ in} \nonumber \] Dividing both sides by 1 cm, we have \[1 = \dfrac{0.3937\text{ in}}{1\text{ cm}} \nonumber \] Since the right side of this equation equals one, it is called a . It can be multiplied by any quantity, leaving the quantity unchanged. \[ 8.89 \text{ cm} = 8.89 \text{ cm} \times 1 = 8.89 \text{ cm} \times \dfrac{0.3937\text{ in}}{\text{1 cm}} \nonumber \] The units cancel, yielding the result 8.89 cm = 8.89 × 0.3937 in = 3.50 in This agrees with the direct observation made in the figure. Let us look at . It has probably already occurred to you that simply measuring the mass of smoke collected is not enough. Some other variables may affect your experiment and should also be measured if the results are to be reproducible. For example, wind direction and speed would almost certainly be important. The time of day and date when a measurement was made should be noted too. In addition you should probably specify what kind of filter you are using. Some are not fine enough to catch all the smoke particles. Another variable which is almost always recorded is the temperature. A thermometer is easy to use, and temperature can vary a good deal outdoors, where your experiments would have to be done. In scientific work, temperatures are usually reported in degrees Celsius (°C), a scale in which the freezing point of pure water is 0°C and the normal boiling point 100°C. In the United States, however, you would be more likely to have available a thermometer calibrated in degrees Fahrenheit (°F). The relationship between these two scales of temperature is \[\dfrac{\text{T}_{(^{o}\text{F)}} - 32}{\text{T}_{(^{o}\text{C)}}}=\dfrac{9}{5} \nonumber \] Note that the temperature scales cannot be interconverted with simple , because they do not have a common zero point (0°C = 32°F). Rather, the mathematical above must be used. The equation above is written in terms of the temperature ( ) with the or subscripted in parentheses. More important than any of the above variables is the fact that the more air you pump through the filter, the more smoke you will collect. Since air is a gas, it is easier to measure how much you collect in terms of volume than in terms of mass, and so you might decide to do it that way. Running your pump until it had filled a plastic weather balloon would provide a crude, inexpensive volume measurement. Assuming the balloon to be approximately spherical, you could measure its diameter and calculate its volume from the formula \[V=\dfrac{4}{3}\pi r^{3} \nonumber \] Calculate the volume of gas in a sphere whose diameter is 106 in. Express your result in cubic centimeters (cm ). Since the radius of a sphere is half its diameter, \[ r = \dfrac{1}{2} \times 106 \text{ in} = 53 \text{ in} \nonumber \] We can use the same equality of quantities as in Example 1 to convert the radius to centimeters. When we cube the number and units, our result will be in cubic centimeters. \[ 1 \text{ cm} = 0.3937 \text{ in} \nonumber \] \[\dfrac{\text{1 cm}}{\text{0}\text{.3937 in}} = 1 \nonumber \] \[ R = 53 \text{ in} \times \dfrac{\text{1 cm}}{\text{0}\text{.3937 in}} = \dfrac{\text{53}}{\text{0}\text{.3937}} \text{ cm} \nonumber \] Using the formula \[\begin{align} & V =\dfrac{\text{4}}{\text{3}}\pi r^{\text{3}}=\dfrac{\text{4}}{\text{3}}\times \text{3}\text{.14159}\times ( \dfrac{\text{53}}{\text{0}\text{.3937}}\text{cm} )^{3} \\ & \text{ }=\text{10219264 cm}^{\text{3}} \\ & \end{align} \nonumber \] You can see from Examples 1 and 2 that two unity factors may be obtained from the equality \[ 1 \text{ cm} = 0.3937 \text{ in} \nonumber \] We can use one of them to convert inches to centimeters and the other to convert centimeters to inches The result in Example 2 also shows that cubic centimeters are rather small units for expressing the volume of the balloon. If we used larger units, as shown in the following example, we would not need more than 10 million of them to report our answer. Express the result of Example 3 in cubic meters, given that 1 m = 100 cm. Again we wish to use a unity factor, and since we are trying to get rid of cubic centimeters, centimeters must be in the denominator: \[\text{1 = }\dfrac{\text{1 m}}{\text{100 cm}} \nonumber \] But this will not allow cancellation of cubic centimeters. However, note that 1 = 1 That is, we can raise a unity factor to any power, and it remains unity. Thus \[\begin{align} & \text{1 =}\left( \text{ }\dfrac{\text{1 m}}{\text{100 cm}} \right)^{\text{3}}\text{ = }\dfrac{\text{1 m}^{\text{3}}}{\text{100}^{\text{3}}\text{ cm}^{\text{3}}}\text{ = }\dfrac{\text{1 m}^{\text{3}}}{\text{1 000 000 cm}^{\text{3}}} \\ \text{ and } \\ & \text{10 219 264 cm}^{\text{3}}\text{ = 10 219 264 cm}^{\text{3}}\text{ }\times \text{ }\left( \dfrac{\text{1 m}}{\text{100 cm}} \right)^{\text{3}} \\ & \text{ = 10 219 264 cm}^{\text{3}}\text{ }\times \text{ }\dfrac{\text{1 m}^{\text{3}}}{\text{100 000 cm}^{\text{3}}} \\ & \text{ = 10.219 264 m}^{\text{3}} \\ & \end{align} \nonumber \] | 12,117 | 2,136 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Nucleophilic_Substitution_at_Tetrahedral_Carbon/NS8._Nucleophilicity_in_NS |
The nucleophile can sometimes play a pronounced role in nucleophilic substitutions. The following relative rates have been observed when these nucleophiles reacted with methyl bromide in methanol: note: Ph = phenyl, C H ; Ac = acetyl, CH C=O; Et = ethyl, CH CH . Presumably, some of the species react much more quickly with methyl bromide because they are better nucleophiles than others. Sometimes we can draw general conclusions about kinetic factors by looking at sub-groups among the data. Determine how the following factors influence nucleophilicity (the ability of a species to act as a nucleophile). Support your ideas with groups of examples from the data (preferably more than just a pair of entries). Nucleophilicity plays a strong role in the rate of one type of substitution mechanism, but not the other. A trend very similar to the data above is found in substitution reactions of py PtCl (py = pyridine) in methanol. Draw a mechanism for this substitution and explain why nucleophilicity plays an important role. Very fast nucleophiles are sometimes more likely to undergo S 2 reactions than S 1 reactions. Explain why. , | 1,150 | 2,137 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/01%3A_Introduction_-_The_Ambit_of_Chemistry/1.06%3A_Measurements_Quantities_and_Unity_Factors |
Let us assume that you are faced with a specific problem. Then we can see how scientific thinking might help solve it. Suppose that you live near a large plant which manufactures cement. Smoke from the plant settles on your car and house, causing small pits in the paint. You would like to stop this air-pollution problem—but how? As an individual you will probably have little influence, and even as part of a group of concerned citizens you may be ineffective, unless you can prove that there is a problem. Scientists have had a hand in writing most air-pollution regulations, and so you will have to employ some scientific techniques (or a scientist) to help solve your problem. It will probably be necessary to determine how much air pollution the plant is producing. This might be done by comparing the smoke with a scale which ranges from white to gray to black, the assumption being that the darker the smoke, the more there is. For white cement dust, however, this is much less satisfactory than for black coal smoke. A better way to determine how much pollution there is would be to measure the mass of smoke particles which could be collected near your house or car. This could be done by using a pump (such as a vacuum cleaner) to suck the polluted gas through a filter. Weighing the filter before and after such an experiment would determine what mass of smoke had been collected. Because is the force of gravity on an object, which varies from place to place by about 0.5% as shown in the Table, we must use for accurate measurements of how much matter we have. We still call the process of obtaining an accurate mass "weighing". The of an object is actually the force of gravity, and is calculated as \[F=\text{“W”}=m \times g \label{1} \] Weight is measured in newtons (kg m s ) or pounds (lb), where 1 lb = 4.44822162 newtons. The base unit for mass is kilograms (kg), but the pound may also be used as a unit of mass (1 pound = 0.45359237 kilograms). The average value of g is usually taken to be 9.80665 m s , so the weight of a 1 kg mass would be \[F = \text{ “W” } = m \times g = 1 \text{ kg} \times 9.80665 \text{ m s}^{-2} = 9.80665 \text{ N or } 2.2046 \text{ lb} \nonumber \] If the weight of the 1 kg mass were measured on an arctic exploration camp with a balance (see below) still calibrated for the average g, its weight would be: \[ F = \text{ “W” } = m \times g = 10 \text{ kg} \times 9.83208 \text{ m s}^{-2} = 9.83208 \text{ N or } 2.2103 \text{ lb} \nonumber \] Accurate weighing is usually done with a single pan balance. The empty pan is balanced by a counterweight. When an object is placed on the pan, gravitational attraction forces the pan downward. To restore balance, a series of weights (objects of known mass) are removed from holders above the pan. The force of gravitational attraction is proportional to mass, and when the pan is balanced, the force on it must always be the same. Therefore the mass of the object being weighed must equal that of the weights that were removed. A gives the same mass readout regardless of the force of gravity. Modern laboratory "balances" are based on that convert the force exerted by an object on the balance pan to an electrical signal. The load cell generally is coupled with a dedicated converter and digital display. Because the force exerted by the object depends on gravity, these are really that measure weight, and must be calibrated frequently (against standard masses) to read a . If you kept a notebook or other record of your measurements, you would probably write down something like 0.0342 g to represent how much smoke had been collected. Such a result, which describes (in this case the magnitude of the parameter, mass), is called a . Notice that it consists of two parts, a number and a unit. It would be ambiguous to write just 0.0342—you might not remember later whether that was measured in units of grams, ounces, pounds, or something else. A quantity always behaves as though the number and the units are multiplied together. For example, we could write the quantity already obtained as 0.0342 × g. Using this simple rule of number × units, we can apply arithmetic and algebra to any quantity: \[\begin{align} & 5 \text{g} + 2 \text{g} = (5 + 2) \text{g} = 7 \text{g} \\ & 5 \text{g} \div 2 \text{g} = \dfrac{5 \text{g}}{2 \text{g}} = 2.5 \text{ (the units cancel, and so we get a pure number)} \\ & 5 \text{ in} \times 2 \text{ in} = 10 \text{ in}^{2} \text{ (10 square inches)}\end{align} \nonumber \] This works perfectly well as long as we do not write equations with different (i.e., those having units which measure different properties, like mass and length, temperature and energy, or volume and amount) on opposite sides of the equal sign. For example, applying algebra to the equation \[ 5 \text{g} = 2 \text{ in}^{2} \nonumber \] can lead to trouble in much the same way that dividing by zero does and should be avoided, because grams (g) is a unit of the , and the inch (in) is a unit of the . Notice also that whether a quantity is large or small depends on the size of the units as well as the size of the number. For example, the mass of smoke has been measured as 0.0342 g, but the balance might have been set to read in milligrams (or grains in the English system). Then the reading would be 34.2 mg (or 0.527 grains). The results (0.0342 g, 34.2 mg, or 0.527 gr) are the quantity, the mass of smoke. One involves a smaller number and larger unit (0.0342 g), while the others have a larger number and smaller unit. So long as we are talking about the same quantity, it is a simple matter to adjust the number to go with any units we want. We can convert among the different ways of expressing the mass with as follows: Since 1 mg and 0.001 g are the same parameter (mass), we can write the equation \[ 1 \text{ mg} = 0.001 \text{ g} \nonumber \] Dividing both sides by 1 mg, we have \[1 = \dfrac{1\text{ mg}}{1\text{ mg}} = \dfrac{0.001\text{ g}}{1\text{ mg}} \nonumber \] Since the last term of this equation equals one, it is called a . It can be multiplied by any quantity, leaving the quantity unchanged. We can generate another unity factor by dividing both sides by 0.001 g: \[1 = \dfrac{1\text{ mg}}{0.001\text{ g}} = \dfrac{0.001\text{ g}}{0.001\text{ g}} \nonumber \] What is the mass in grams of a 5.0 grain (5 gr) aspirin tablet, given that 1 gram = 15.4323584 grains? \[ 5.0 \text{ gr} = 5.0 \text{ gr} \times 1 = 5.0 \text{ gr}\times \dfrac{1.0\text{ g}}{\text{15.4323 gr}} \nonumber \] The units cancel, yielding the result \[ 5.0 \text{ gr} = 5.0 \div 15.4323 \text{ g} = 0.324 \text{ g} \nonumber \] The length may be measured in inches (in) in the English system, but scientific measurements (all measurements in the world exclusive of the U.S.) are reported in the metric units of meters (m) or some more convenient derived unit like centimeters (cm). Express the length 8.89 cm in inches, given that 1 cm = 0.3937 in. Since 1 cm and 0.3937 in are the same quantity, we can write the equation \[1 \text{ cm} = 0.3937 \text{ in} \nonumber \] Dividing both sides by 1 cm, we have \[1 = \dfrac{0.3937\text{ in}}{1\text{ cm}} \nonumber \] Since the right side of this equation equals one, it is called a . It can be multiplied by any quantity, leaving the quantity unchanged. \[ 8.89 \text{ cm} = 8.89 \text{ cm} \times 1 = 8.89 \text{ cm} \times \dfrac{0.3937\text{ in}}{\text{1 cm}} \nonumber \] The units cancel, yielding the result 8.89 cm = 8.89 × 0.3937 in = 3.50 in This agrees with the direct observation made in the figure. Let us look at . It has probably already occurred to you that simply measuring the mass of smoke collected is not enough. Some other variables may affect your experiment and should also be measured if the results are to be reproducible. For example, wind direction and speed would almost certainly be important. The time of day and date when a measurement was made should be noted too. In addition you should probably specify what kind of filter you are using. Some are not fine enough to catch all the smoke particles. Another variable which is almost always recorded is the temperature. A thermometer is easy to use, and temperature can vary a good deal outdoors, where your experiments would have to be done. In scientific work, temperatures are usually reported in degrees Celsius (°C), a scale in which the freezing point of pure water is 0°C and the normal boiling point 100°C. In the United States, however, you would be more likely to have available a thermometer calibrated in degrees Fahrenheit (°F). The relationship between these two scales of temperature is \[\dfrac{\text{T}_{(^{o}\text{F)}} - 32}{\text{T}_{(^{o}\text{C)}}}=\dfrac{9}{5} \nonumber \] Note that the temperature scales cannot be interconverted with simple , because they do not have a common zero point (0°C = 32°F). Rather, the mathematical above must be used. The equation above is written in terms of the temperature ( ) with the or subscripted in parentheses. More important than any of the above variables is the fact that the more air you pump through the filter, the more smoke you will collect. Since air is a gas, it is easier to measure how much you collect in terms of volume than in terms of mass, and so you might decide to do it that way. Running your pump until it had filled a plastic weather balloon would provide a crude, inexpensive volume measurement. Assuming the balloon to be approximately spherical, you could measure its diameter and calculate its volume from the formula \[V=\dfrac{4}{3}\pi r^{3} \nonumber \] Calculate the volume of gas in a sphere whose diameter is 106 in. Express your result in cubic centimeters (cm ). Since the radius of a sphere is half its diameter, \[ r = \dfrac{1}{2} \times 106 \text{ in} = 53 \text{ in} \nonumber \] We can use the same equality of quantities as in Example 1 to convert the radius to centimeters. When we cube the number and units, our result will be in cubic centimeters. \[ 1 \text{ cm} = 0.3937 \text{ in} \nonumber \] \[\dfrac{\text{1 cm}}{\text{0}\text{.3937 in}} = 1 \nonumber \] \[ R = 53 \text{ in} \times \dfrac{\text{1 cm}}{\text{0}\text{.3937 in}} = \dfrac{\text{53}}{\text{0}\text{.3937}} \text{ cm} \nonumber \] Using the formula \[\begin{align} & V =\dfrac{\text{4}}{\text{3}}\pi r^{\text{3}}=\dfrac{\text{4}}{\text{3}}\times \text{3}\text{.14159}\times ( \dfrac{\text{53}}{\text{0}\text{.3937}}\text{cm} )^{3} \\ & \text{ }=\text{10219264 cm}^{\text{3}} \\ & \end{align} \nonumber \] You can see from Examples 1 and 2 that two unity factors may be obtained from the equality \[ 1 \text{ cm} = 0.3937 \text{ in} \nonumber \] We can use one of them to convert inches to centimeters and the other to convert centimeters to inches The result in Example 2 also shows that cubic centimeters are rather small units for expressing the volume of the balloon. If we used larger units, as shown in the following example, we would not need more than 10 million of them to report our answer. Express the result of Example 3 in cubic meters, given that 1 m = 100 cm. Again we wish to use a unity factor, and since we are trying to get rid of cubic centimeters, centimeters must be in the denominator: \[\text{1 = }\dfrac{\text{1 m}}{\text{100 cm}} \nonumber \] But this will not allow cancellation of cubic centimeters. However, note that 1 = 1 That is, we can raise a unity factor to any power, and it remains unity. Thus \[\begin{align} & \text{1 =}\left( \text{ }\dfrac{\text{1 m}}{\text{100 cm}} \right)^{\text{3}}\text{ = }\dfrac{\text{1 m}^{\text{3}}}{\text{100}^{\text{3}}\text{ cm}^{\text{3}}}\text{ = }\dfrac{\text{1 m}^{\text{3}}}{\text{1 000 000 cm}^{\text{3}}} \\ \text{ and } \\ & \text{10 219 264 cm}^{\text{3}}\text{ = 10 219 264 cm}^{\text{3}}\text{ }\times \text{ }\left( \dfrac{\text{1 m}}{\text{100 cm}} \right)^{\text{3}} \\ & \text{ = 10 219 264 cm}^{\text{3}}\text{ }\times \text{ }\dfrac{\text{1 m}^{\text{3}}}{\text{100 000 cm}^{\text{3}}} \\ & \text{ = 10.219 264 m}^{\text{3}} \\ & \end{align} \nonumber \] | 12,117 | 2,138 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/19%3A_The_First_Law_of_Thermodynamics/19.04%3A_Energy_is_a_State_Function |
The change in internal energy of a system is the sum of \(w\) and \(q\), which is a state function. The realization that work and heat are both forms of energy transfer undergoes quite an extension by saying that internal energy is a state function. It means that although heat and work can be produced and destroyed (and transformed into each other), energy is conserved. This allows us to do some serious bookkeeping! We can write the law as: \[\Delta U = w + q \nonumber \] But the (important!) bit about the state function is better represented if we talk about small changes of the energy: \[dU = \delta w +\delta q \nonumber \] We write a straight Latin \(d\) for \(U\) to indicate when the change in a state function, where as the changes in work and heat are path-dependent. This is indicated by the 'crooked' \(\delta\). We can represent changes as integrals, but only for \(U\) can we say that regardless of path we get \(\Delta U = U_2-U_1\) if we go from state one to state two. (I.e. it only depends on the end points, not the path). Notice that when we write \(dU\) or \(\delta q\), we always mean infinitesimally small changes, i.e. we are implicitly taking a limit for the change approaching zero. To arrive at a macroscopic difference like \(\Delta U\) or a macroscopic (finite) amount of heat \(q\) or work \(w\) . We will now invoke the : These are all ways of saying that internal energy is a state function. | 1,448 | 2,139 |
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Oil has been a staple of our every day lives for thousands of years. In ancient times it was used as a fuel source, but more widely as a medicine. In the late 19th century oil was found in the Texas area, and from then on the world would never be the same. Oil was used for cars, planes, fuel, and helped to spur the industrial revolution. With the profitability of oil rising, many oil companies and drilling areas evolved around the world. The production and extraction of oil became one of the most profitable business in the world. However, a major problem with the oil industry was that it was all around the world. In order to get oil to the largest consumers the oil had to be shipped. Some of the largest oil reserves in the world are in the middle east, Alaska, and in the deep ocean. Each of these places has great access to shipping lanes, and therefore have been very profitable. However, a major problem is that along with the shipping lanes comes huge bodies of water, and oil and water definitely don't mix. In 1989 an Exxon Valdez tanker was taking oil from Alaska to California. It ran aground and spilled 32 million gallons of oil all over the Alaskan coastline. The spill took place in Prince William Sound, which is in Southern Alaska by the Kenai Peninsula. The oil spread to over 1,300 miles of coastline and 11,000 square miles in the ocean. 250,000 animals were killed in the spill, including endangered species like the Bald Eagle. The reason that the oil is so deadly is that it sits on top of the water, because the two are not miscible, or do not mix. The sea life or birds come to the surface and get caked in oil. When covered in oil the birds wings do not work, and gills of fish are covered and they can't breathe. No animal is safe from the oil. The Exxon Valdez spill was extremely destructive, and was the United States' largest oil spill until the 2010 sinking of the oil rig Deepwater Horizon off the coast of Louisiana. The Louisiana oil spill released over 90 million gallons of oil into the ocean. However, these two oil spills are minute compared to some of the biggest in the world. These include the Kuwait oil spill (336 million gallons) and the Ixtoc, Mexico oil well collapse (140 million gallons). All oil spills no matter how big have devastating effects on the environments for many years because of the difficulty of cleaning up oil. It could be seen that oil would be easy to clean up because it is not miscible with water. It seems easy to simply take off the top layer of oil and that would be it. However, the problem is that there is no great solvent to dissolve the oil. Many different techniques are used to clean up oil. One of these techniques is called using a dispersant. The dispersant takes large clumps of oil and breaks them down to very small particles. Since there is so much water available, and the particles are so small, the oil starts to be dissolved into the water column. The problem though is that with excessive amounts of oil, the water column can become toxic because of the large amounts of oil being dissolved. Another technique is to use skimmers. The skimmers take the oil off the water and get rid of it. The skimmers are prone to clogging though, especially with thick oil. One option that many people ask about is burning the oil. This is done to some extent, but burning oil creates many bad after effects for the atmosphere and people to breathe in. There really is no silver bullet in treating an oil spill, and because of this all of the above are done to clean up an oil spill. In basic chemistry classes everyone learns that in solvent chemistry like dissolves like. In a more in depth sense, polar solutions dissolve polar solutes, and non-polar solvents dissolve non-polar solutes. In the case of oil and water, one is polar (water), and the other is non-polar (oil). Water is polar because of the bonds between oxygen and hydrogen are polar. The electronegativity difference between these two elements is about 1.24, meaning it is very polar. The water molecule is in the bent shape, which is important because the two polar bonds don't cancel each other out. For oil on the other hand, none of the bonds are polar. Oil is made of alkanes, or long carbon chains. Gas fuel for cars is made up of carbon chains between 5-8 carbons long and diesel and jet fuel is made up of carbon chains of 9-16 carbons long. These are the kinds of oil that the Exxon Valdez was carrying, and are components of every oil spill. Alkanes are simply made up of carbons and hydrogens, and the electronegativity difference between these elements is only .35. This means that the bonds are not polar. Since alkanes are not polar and water is polar, the two do not mix (not miscible), which creates the many problems with oil spills. From ChemPRIME: 1. 2. 3. iml.jou.ufl.edu/projects/Fall02/Susi/exxon.htm 4. | 4,893 | 2,141 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/06%3A_Electronic_Structure_of_Atoms/6.S%3A_Electronic_Structure_of_Atoms_(Summary) |
Understanding the electronic structure of atoms requires an understanding of the properties of waves and electromagnetic radiation. A basic knowledge of the electronic structure of atoms requires an understanding of the properties of waves and electromagnetic radiation. A wave is a periodic oscillation by which energy is transmitted through space. All waves are periodic, repeating regularly in both space and time. Waves are characterized by several interrelated properties. \[\nu λ = c \nonumber \] where \(\nu\) = frequency, \(λ\)= wavelength, and \(c\) = speed of light Wavelength is expressed in units of length. Frequency is expressed in Hertz (Hz), also denoted by s or /s Quantum Effects and Photons \[E = hv \nonumber \] where \(E\) = energy, \(h\) = Planck’s constant, \(\nu\) = frequency According to Planck’s theory, energy is always emitted or absorbed in whole-number multiples of , for example, , , , and so forth. We say that the allowed energies are (that is, their values are restricted to certain quantities). Blackbody radiation is the radiation emitted by hot objects and could not be explained with classical physics. Max Planck postulated that energy was quantized and may be emitted or absorbed only in integral multiples of a small unit of energy, known as a quantum. The energy of a quantum is proportional to the frequency of the radiation; the proportionality constant h is a fundamental constant (Planck’s constant). Albert Einstein used the quantization of energy to explain the photoelectric effect There is an intimate connection between the atomic structure of an atom and its spectral characteristics. Most light is polychromatic and contains light of many wavelengths. Light that has only a single wavelength is monochromatic and is produced by devices called lasers, which use transitions between two atomic energy levels to produce light in a very narrow range of wavelengths. Atoms can also absorb light of certain energies, resulting in a transition from the ground state or a lower-energy e Radiation composed of a single wavelength is said to be . Electrons in a permitted orbit have a specific energy and are said to be in an "allowed" energy state. An electron in an allowed energy state will not radiate energy and therefore will not spiral into the nucleus. \[E_n = -R_H \dfrac{1}{n^2} \nonumber \] \[E_∞ = (-2.18 \times 10^{-18} J) \left(\dfrac{1}{∞^2}\right) = 0 \nonumber \] Electrons can change from one energy state to another by absorbing or emitting radiant energy. Radiant energy must be absorbed for an electron to move to a higher energy state, but is emitted when the electron moves to a lower energy state. \[\Delta E = E_f – E_i \nonumber \] An electron possesses both particle and wave properties. Louis de Broglie showed that the wavelength of a particle is equal to Planck’s constant divided by the mass times the velocity of the particle. The electron in Bohr’s circular orbits could thus be described as a standing wave, one that does not move through space. Werner Heisenberg’s uncertainty principle states that it is impossible to precisely describe both the location and the speed of particles that exhibit wavelike behavior. \[λ = h / mv \nonumber \] The Uncertainty Principle There is a relationship between the motions of electrons in atoms and molecules and their energies that is described by quantum mechanics. Because of wave–particle duality, scientists must deal with the probability of an electron being at a particular point in space. To do so required the development of quantum mechanics, which uses wavefunctions to describe the mathematical relationship between the motion of electrons in atoms and molecules and their energies. Orbitals and Quantum Numbers - : allowed energy state of an electron in the quantum-mechanical model of the atom; also used to describe the spatial distribution of an electron. Defined by the value of 3 quantum numbers; . : collection of orbitals with the same value of : one or more orbitals with the same set of and values Orbitals with l = 0 are s orbitals and are spherically symmetrical, with the greatest probability of finding the electron occurring at the nucleus. Orbitals with values of n > 1 and l = 0 contain one or more nodes. Orbitals with l = 1 are p orbitals and contain a nodal plane that includes the nucleus, giving rise to a dumbbell shape. Orbitals with l = 2 are d orbitals and have more complex shapes with at least two nodal surfaces. l = 3 orbitals are f orbitals, which are still more complex. 1s orbital: most stable, spherically symmetric, figure indicates that the probability decreases as we move away from the nucleus. . Electron density is concentrated on two sides of the nucleus, separated by a node at the nucleus. The orbitals of a given subshell have the same size and shape but differ from each other in orientation. The axis along which the orbital is oriented is not related to . In addition to the three quantum numbers (n, l, ml) dictated by quantum mechanics, a fourth quantum number is required to explain certain properties of atoms. This is the electron spin quantum number (ms), which can have values of +½ or −½ for any electron, corresponding to the two possible orientations of an electron in a magnetic field. This is important for chemistry because the Pauli exclusion principle implies that no orbital can contain more than two electrons (with opposite spin). Although the shapes of the orbitals for many-electron atoms are the same as those for hydrogen, the presence of more than one electron greatly changes the energies of the orbitals. In hydrogen, the energy of an orbital depends only on its principal quantum number, however, in many-electron atoms, electron-electron repulsions cause different subshells to be at different energies \[Z_{eff} = Z – S \nonumber \] The extent to which an electron will be screened by the other electrons depends on its electron distribution as we move outward from the nucleus. : orbitals that have the same energy Electron Spin and the Pauli Exclusion Principle Based on the Pauli principle and a knowledge of orbital energies obtained using hydrogen-like orbitals, it is possible to construct the periodic table by filling up the available orbitals beginning with the lowest-energy orbitals (the aufbau principle), which gives rise to a particular arrangement of electrons for each element (its electron configuration). Hund’s rule says that the lowest-energy arrangement of electrons is the one that places them in degenerate orbitals with parallel spins. Writing Electron Configurations The arrangement of atoms in the periodic table results in blocks corresponding to filling of the ns, np, nd, and nf orbitals to produce the distinctive chemical properties of the elements in the s block, p block, d block, and f block, respectively. | 6,890 | 2,142 |
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Have you ever estimated a distance by “stepping it off”— that is, by counting the number of steps required to take you a certain distance? Or perhaps you have used the width of your hand, or the distance from your elbow to a fingertip to compare two dimensions. If so, you have engaged in what is probably the first kind of measurement ever undertaken by primitive mankind. The results of a measurement are always expressed on some kind of a that is defined in terms of a particular kind of . The first scales of distance were likely related to the human body, either directly (the length of a limb) or indirectly (the distance a man could walk in a day). As civilization developed, a wide variety of measuring scales came into existence, many for the same quantity (such as length), but adapted to particular activities or trades. Eventually, it became apparent that in order for trade and commerce to be possible, these scales had to be defined in terms of standards that would allow measures to be verified, and, when expressed in different units (bushels and pecks, for example), to be correlated or converted. Over the centuries, hundreds of measurement units and scales have developed in the many civilizations that achieved some literate means of recording them. Some, such as those used by the Aztecs, fell out of use and were largely forgotten as these civilizations died out. Other units, such as the various systems of measurement that developed in England, achieved prominence through extension of the Empire and widespread trade; many of these were confined to specific trades or industries. The examples shown here are only some of those that have been used to measure length or distance. The history of measuring units provides a fascinating reflection on the history of industrial development. The most influential event in the history of measurement was undoubtedly the French Revolution and the Age of Rationality that followed. This led directly to the metric system that attempted to do away with the confusing multiplicity of measurement scales by reducing them to a few fundamental ones that could be combined in order to express any kind of quantity. The metric system spread rapidly over much of the world, and eventually even to England and the rest of the U.K. when that country established closer economic ties with Europe in the latter part of the 20th Century. The United States is presently the only major country in which “metrication” has made little progress within its own society, probably because of its relative geographical isolation and its vibrant internal economy. Science, being a truly international endeavor, adopted metric measurement very early on; engineering and related technologies have been slower to make this change, but are gradually doing so. Even the within the metric system, however, a variety of units were employed to measure the same fundamental quantity; for example, energy could be expressed within the metric system in units of ergs, electron-volts, joules, and two kinds of calories. This led, in the mid-1960s, to the adoption of a more basic set of units, the ( ) units that are now recognized as the standard for science and, increasingly, for technology of all kinds. In principle, any physical quantity can be expressed in terms of only seven base units (Table \(\Page {1}\)), with each base unit defined by a standard described in the . A few special points about some of these units are worth noting: Owing to the wide range of values that quantities can have, it has long been the practice to employ prefixes such as milli and mega to indicate decimal fractions and multiples of metric units. As part of the SI standard, this system has been extended and formalized (Table \(\Page {2}\)). There is a category of units that are “honorary” members of the SI in the sense that it is acceptable to use them along with the base units defined above. These include such mundane units as the hour, minute, and degree (of angle), etc., but the three shown here are of particular interest to chemistry, and you will need to know them. Most of the physical quantities we actually deal with in science and also in our daily lives, have units of their own: volume, pressure, energy and electrical resistance are only a few of hundreds of possible examples. It is important to understand, however, that all of these can be expressed in terms of the SI base units; they are consequently known as . In fact, most physical quantities can be expressed in terms of one or more of the following five fundamental units: Consider, for example, the unit of , which we denote as V. To measure the volume of a rectangular box, we need to multiply the lengths as measured along the three coordinates: \[V = x · y · z\] We say, therefore, that volume has the dimensions of length-cubed: \[dim\{V\} = L^3\] Thus the units of volume will be m (in the SI) or cm , ft (English), etc. Moreover, any formula that calculates a volume must contain within it the L dimension; thus the volume of a sphere is \(4/3 πr^3\). The of a unit are the powers which M, L, t, Q and must be given in order to express the unit. Thus, \[dim\{V\} = M^0L^3T^0Q^0 Θ^0\] as given above. There are several reasons why it is worthwhile to consider the dimensions of a unit. The formal, detailed study of dimensions is known as and is a topic in any basic physics course. Find the dimensions of energy. When mechanical work is performed on a body, its energy increases by the amount of work done, so the two quantities are equivalent and we can concentrate on work. The latter is the product of the force applied to the object and the distance it is displaced. From Newton’s law, force is the product of mass and acceleration, and the latter is the rate of change of velocity, typically expressed in meters per second per second. Combining these quantities and their dimensions yields the result shown in Table \(\Page {1}\). Dimensional analysis is widely employed when it is necessary to convert one kind of unit into another, and chemistry students often use it in "chemical arithmetic" calculations, in which context it is also known as the "Factor-Label" method. In this section, we will look at some of the quantities that are widely encountered in Chemistry, and at the units in which they are commonly expressed. In doing so, we will also consider the actual range of values these quantities can assume, both in nature in general, and also within the subset of nature that chemistry normally addresses. In looking over the various units of measure, it is interesting to note that their unit values are set close to those encountered in everyday human experience These two quantities are widely confused. Although they are often used synonymously in informal speech and writing, they have different dimensions: is the exerted on a mass by the local gravational field: \[f = m a = m g \label{Eq1}\] where is the acceleration of gravity. While the nominal value of the latter quantity is 9.80 m s at the Earth’s surface, its exact value varies locally. Because it is a force, the SI unit of weight is properly the , but it is common practice (except in physics classes!) to use the terms "weight" and "mass" interchangeably, so the units and are acceptable in almost all ordinary laboratory contexts. The range of masses spans 90 orders of magnitude, more than any other unit. The range that chemistry ordinarily deals with has greatly expanded since the days when a microgram was an almost inconceivably small amount of material to handle in the laboratory; this lower limit has now fallen to the atomic level with the development of tools for directly manipulating these particles. The upper level reflects the largest masses that are handled in industrial operations, but in the recently developed fields of geochemistry and enivonmental chemistry, the range can be extended indefinitely. Flows of elements between the various regions of the environment (atmosphere to oceans, for example) are often quoted in teragrams. Chemists tend to work mostly in the moderately-small part of the distance range. Those who live in the lilliputian world of crystal- and molecular structures and atomic radii find the a convenient currency, but one still sees the older non-SI unit called the used in this context; 1Å = 10 m = 100pm. Nanotechnology, the rage of the present era, also resides in this realm. The largest polymeric molecules and colloids define the top end of the particulate range; beyond that, in the normal world of doing things in the lab, the and occasionally the commonly rule. For humans, time moves by the heartbeat; beyond that, it is the motions of our planet that count out the hours, days, and years that eventually define our lifetimes. Beyond the few thousands of years of history behind us, those years-to-the-powers-of-tens that are the fare for such fields as evolutionary biology, geology, and cosmology, cease to convey any real meaning for us. Perhaps this is why so many people are not very inclined to accept their validity. Most of what actually takes place in the chemist’s test tube operates on a far shorter time scale, although there is no limit to how slow a reaction can be; the upper limits of those we can directly study in the lab are in part determined by how long a graduate student can wait around before moving on to gainful employment. Looking at the microscopic world of atoms and molecules themselves, the time scale again shifts us into an unreal world where numbers tend to lose their meaning. You can gain some appreciation of the duration of a nanosecond by noting that this is about how long it takes a beam of light to travel between your two outstretched hands. In a sense, the material foundations of chemistry itself are defined by time: neither a new element nor a molecule can be recognized as such unless it lasts long enough to have its “picture” taken through measurement of its distinguishing properties. Temperature, the measure of thermal intensity, spans the narrowest range of any of the base units of the chemist’s measurement toolbox. The reason for this is tied into temperature’s meaning as a measure of the intensity of thermal kinetic energy. Chemical change occurs when atoms are jostled into new arrangements, and the weakness of these motions brings most chemistry to a halt as absolute zero is approached. At the upper end of the scale, thermal motions become sufficiently vigorous to shake molecules into atoms, and eventually, as in stars, strip off the electrons, leaving an essentially reaction-less gaseous fluid, or plasma, of bare nuclei (ions) and electrons. The degree is really an of temperature, a fixed fraction of the distance between two defined reference points on a . Although rough means of estimating and comparing temperatures have been around since AD 170, the first mercury thermometer and temperature scale were introduced in Holland in 1714 by Gabriel Daniel Fahrenheit. Fahrenheit established three fixed points on his thermometer. Zero degrees was the temperature of an ice, water, and salt mixture, which was about the coldest temperature that could be reproduced in a laboratory of the time. When he omitted salt from the slurry, he reached his second fixed point when the water-ice combination stabilized at "the thirty-second degree." His third fixed point was "found at the ninety-sixth degree, and the spirit expands to this degree when the thermometer is held in the mouth or under the armpit of a living man in good health." After Fahrenheit died in 1736, his thermometer was recalibrated using 212 degrees, the temperature at which water boils, as the upper fixed point. Normal human body temperature registered 98.6 rather than 96. In 1743, the Swedish astronomer Anders Celsius devised the aptly-named scale that places exactly 100 degrees between the two reference points defined by the freezing and boiling points of water. When we say that the temperature is so many degrees, we must specify the particular scale on which we are expressing that temperature. A temperature scale has two defining characteristics, both of which can be chosen arbitrarily: To express a temperature given on one scale in terms of another, it is necessary to take both of these factors into account. The key to temperature conversions is easy if you bear in mind that between the so-called ice- and steam-points of water there are 180 Fahrenheit degrees, but only 100 Celsius degrees, making the F° 100/180 = 5/9 the magnitude of the C°. Note the distinction between “°C” (a ) and “C°” (a temperature ). Because the ice point is at 32°F, the two scales are offset by this amount. If you remember this, there is no need to memorize a conversion formula; you can work it out whenever you need it. Near the end of the 19th Century when the physical significance of temperature began to be understood, the need was felt for a temperature scale whose zero really means zero— that is, the complete absence of thermal motion. This gave rise to the whose zero point is –273.15 °C, but which retains the same degree magnitude as the Celsius scale. This eventually got renamed after Lord Kelvin (William Thompson); thus the Celsius degree became the . Thus we can now express an increment such as five C° as “five kelvins” In 1859 the Scottish engineer and physicist William J. M. Rankine proposed an absolute temperature scale based on the Fahrenheit degree. Absolute zero (0° Ra) corresponds to –459.67°F. The Rankine scale has been used extensively by those same American and English engineers who delight in expressing heat capacities in units of BTUs per pound per F°. The importance of absolute temperature scales is that absolute temperatures can be entered directly in all the fundamental formulas of physics and chemistry in which temperature is a variable. is the measure of the exerted on a unit area of surface. Its SI units are therefore newtons per square meter, but we make such frequent use of pressure that a derived SI unit, the , is commonly used: \[1\; Pa = 1\; N \;m^{–2}\] The concept of pressure first developed in connection with studies relating to the atmosphere and vacuum that were carried out in the 17th century. Atmospheric pressure is caused by the weight of the column of air molecules in the atmosphere above an object, such as the tanker car below. At sea level, this pressure is roughly the same as that exerted by a full-grown African elephant standing on a doormat, or a typical bowling ball resting on your thumbnail. These may seem like huge amounts, and they are, but life on earth has evolved under such atmospheric pressure. If you actually perch a bowling ball on your thumbnail, the pressure experienced is twice the usual pressure, and the sensation is unpleasant. The molecules of a gas are in a state of constant thermal motion, moving in straight lines until experiencing a collision that exchanges momentum between pairs of molecules and sends them bouncing off in other directions. This leads to a completely random distribution of the molecular velocities both in speed and direction— or it would in the absence of the Earth’s gravitational field which exerts a tiny downward force on each molecule, giving motions in that direction a very slight advantage. In an ordinary container this effect is too small to be noticeable, but in a very tall column of air the effect adds up: the molecules in each vertical layer experience more downward-directed hits from those above it. The resulting force is quickly randomized, resulting in an increased pressure in that layer which is then propagated downward into the layers below. At sea level, the total mass of the sea of air pressing down on each 1-cm of surface is about 1034 g, or 10340 kg m . The force (weight) that the Earth’s gravitional acceleration g exerts on this mass is \[f = ma = mg = (10340 \;kg)(9.81\; m\; s^{–2}) = 1.013 \times 10^5 \;kg \;m \;s^{–2} = 1.013 \times 10^5\; N\] resulting in a pressure of 1.013 × 10 n m = 1.013 × 10 Pa. The actual pressure at sea level varies with atmospheric conditions, so it is customary to define standard atmospheric pressure as 1 atm = 1.01325 x 10 Pa or 101.325 kPa. Although the standard atmosphere is not an SI unit, it is still widely employed. In meteorology, the , exactly 1.000 × 10 = 0.967 atm, is often used. In the early 17th century, the Italian physicist and mathematician Evangalisto Torricelli invented a device to measure atmospheric pressure. The Torricellian consists of a vertical glass tube closed at the top and open at the bottom. It is filled with a liquid, traditionally mercury, and is then inverted, with its open end immersed in the container of the same liquid. The liquid level in the tube will fall under its own weight until the downward force is balanced by the vertical force transmitted hydrostatically to the column by the downward force of the atmosphere acting on the liquid surface in the open container. Torricelli was also the first to recognize that the space above the mercury constituted a vacuum, and is credited with being the first to create a vacuum. One will support a column of mercury that is 760 mm high, so the “millimeter of mercury”, now more commonly known as the , has long been a common pressure unit in the sciences: . ) ). | 17,428 | 2,143 |
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This use of Avogadro’s number raises the question of how we know its value. There are numerous ways to measure Avogadro’s number. One such method is to divide the charge of one mole of electrons by the charge of a single electron. We can obtain the charge of a mole of electrons from electrolysis experiments. The charge of one electron can be determined in a famous experiment devised by Robert Millikan, the “Millikan oil-drop experiment”. The charge on a mole of electrons is called the . Experimentally, it has the value \(96,485\ \mathrm{C\ }{\mathrm{mol}}^{\mathrm{-1}}\)\({}^{\ }\)(coulombs per mole). As determined by Millikan’s experiment, the charge on one electron is \(1.6022\times {10}^{-19}\ \mathrm{C}\). Then \[ \left(\frac{96,485\ C}{\mathrm{mole\ electrons}}\right)\left(\frac{1\ \mathrm{electron}}{1.6022\times {10}^{-19\ }\ C}\right) =6.022\times {10}^{23\ \ }\frac{\mathrm{electrons}}{\mathrm{mole\ electrons}}\] | 946 | 2,144 |
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Group 13 is the first group to span the dividing line between metals and nonmetals, so its chemistry is more diverse than that of groups 1 and 2, which include only metallic elements. Except for the lightest element (boron), the group 13 elements are all relatively electropositive; that is, they tend to lose electrons in chemical reactions rather than gain them. Although group 13 includes aluminum, the most abundant metal on Earth, none of these elements was known until the early 19th century because they are never found in nature in their free state. Elemental boron and aluminum, which were first prepared by reducing B O and AlCl , respectively, with potassium, could not be prepared until potassium had been isolated and shown to be a potent reductant. Indium (In) and thallium (Tl) were discovered in the 1860s by using spectroscopic techniques, long before methods were available for isolating them. Indium, named for its indigo (deep blue-violet) emission line, was first observed in the spectrum of zinc ores, while thallium (from the Greek thallos, meaning “a young, green shoot of a plant”) was named for its brilliant green emission line. Gallium (Ga; Mendeleev’s eka-aluminum) was discovered in 1875 by the French chemist Paul Émile Lecoq de Boisbaudran during a systematic search for Mendeleev’s “missing” element in group 13. Group 13 elements are never found in nature in their free state. As reductants, the group 13 elements are less powerful than the alkali metals and alkaline earth metals. Nevertheless, their compounds with oxygen are thermodynamically stable, and large amounts of energy are needed to isolate even the two most accessible elements—boron and aluminum—from their oxide ores. Although boron is relatively rare (it is about 10,000 times less abundant than aluminum), concentrated deposits of borax [Na B O (OH) ·8H O] are found in ancient lake beds (Figure \(\Page {1}\)) and were used in ancient times for making glass and glazing pottery. Boron is produced on a large scale by reacting borax with acid to produce boric acid [B(OH) ], which is then dehydrated to the oxide (B O ). Reduction of the oxide with magnesium or sodium gives amorphous boron that is only about 95% pure: \[\mathrm{Na_2B_4O_5(OH)_4\cdot8H_2O(s)}\xrightarrow{\textrm{acid}}\mathrm{B(OH)_3(s)}\xrightarrow{\Delta}\mathrm{B_2O_3(s)} \label{Eq1}\] \[\mathrm{B_2O_3(s)}+\mathrm{3Mg(s)}\xrightarrow{\Delta}\mathrm{2B(s)}+\mathrm{3MgO(s)} \label{Eq2}\] Pure, crystalline boron, however, is extremely difficult to obtain because of its high melting point (2300°C) and the highly corrosive nature of liquid boron. It is usually prepared by reducing pure BCl with hydrogen gas at high temperatures or by the thermal decomposition of boron hydrides such as diborane (B H ): \[\mathrm{BCl_3(g)}+\frac{3}{2}\mathrm{H_2(g)}\rightarrow\mathrm{B(s)}+\mathrm{3HCl(g)} \label{Eq3}\] \[B_2H_{6(g)} \rightarrow 2B_{(s)} + 3H_{2(g)} \label{Eq4}\] The reaction shown in Equation \(\ref{Eq3}\) is used to prepare boron fibers, which are stiff and light. Hence they are used as structural reinforcing materials in objects as diverse as the US space shuttle and the frames of lightweight bicycles that are used in races such as the Tour de France. Boron is also an important component of many ceramics and heat-resistant borosilicate glasses, such as Pyrex, which is used for ovenware and laboratory glassware. In contrast to boron, deposits of aluminum ores such as bauxite, a hydrated form of Al O , are abundant. With an electrical conductivity about twice that of copper on a weight for weight basis, aluminum is used in more than 90% of the overhead electric power lines in the United States. However, because aluminum–oxygen compounds are stable, obtaining aluminum metal from bauxite is an expensive process. Aluminum is extracted from oxide ores by treatment with a strong base, which produces the soluble hydroxide complex [Al(OH) ] . Neutralization of the resulting solution with gaseous CO results in the precipitation of Al(OH) : \[2[Al(OH)_4]^−_{(aq)} + CO_{2(g)} \rightarrow 2Al(OH)_{3(s)} + CO^{2−}_{3(aq)} + H_2O_{(l)} \label{Eq5}\] Thermal dehydration of Al(OH) produces Al O , and metallic aluminum is obtained by the electrolytic reduction of Al O using the . Of the group 13 elements, only aluminum is used on a large scale: for example, each Boeing 777 airplane is about 50% aluminum by mass. The other members of group 13 are rather rare: gallium is approximately 5000 times less abundant than aluminum, and indium and thallium are even scarcer. Consequently, these metals are usually obtained as by-products in the processing of other metals. The extremely low melting point of gallium (29.6°C), however, makes it easy to separate from aluminum. Due to its low melting point and high boiling point, gallium is used as a liquid in thermometers that have a temperature range of almost 2200°C. Indium and thallium, the heavier group 13 elements, are found as trace impurities in sulfide ores of zinc and lead. Indium is used as a crushable seal for high-vacuum cryogenic devices, and its alloys are used as low-melting solders in electronic circuit boards. Thallium, on the other hand, is so toxic that the metal and its compounds have few uses. Both indium and thallium oxides are released in flue dust when sulfide ores are converted to metal oxides and SO . Until relatively recently, these and other toxic elements were allowed to disperse in the air, creating large “dead zones” downwind of a smelter. The flue dusts are now trapped and serve as a relatively rich source of elements such as In and Tl (as well as Ge, Cd, Te, and As). Table \(\Page {1}\) summarizes some important properties of the group 13 elements. Notice the large differences between boron and aluminum in size, ionization energy, electronegativity, and standard reduction potential, which is consistent with the observation that boron behaves chemically like a nonmetal and aluminum like a metal. All group 13 elements have ns np valence electron configurations, and all tend to lose their three valence electrons to form compounds in the +3 oxidation state. The heavier elements in the group can also form compounds in the +1 oxidation state formed by the formal loss of the single np valence electron. Because the group 13 elements generally contain only six valence electrons in their neutral compounds, these compounds are all moderately strong Lewis acids. Neutral compounds of the group 13 elements are electron deficient, so they are generally moderately strong Lewis acids. In contrast to groups 1 and 2, the group 13 elements show no consistent trends in ionization energies, electron affinities, and reduction potentials, whereas electronegativities actually increase from aluminum to thallium. Some of these anomalies, especially for the series Ga, In, Tl, can be explained by the increase in the effective nuclear charge (Z ) that results from poor shielding of the nuclear charge by the filled (n − 1)d and (n − 2)f subshells. Consequently, although the actual nuclear charge increases by 32 as we go from indium to thallium, screening by the filled 5d and 4f subshells is so poor that Z increases significantly from indium to thallium. Thus the first ionization energy of thallium is actually greater than that of indium. Anomalies in periodic trends among Ga, In, and Tl can be explained by the increase in the effective nuclear charge due to poor shielding. Elemental boron is a semimetal that is remarkably unreactive; in contrast, the other group 13 elements all exhibit metallic properties and reactivity. We therefore consider the reactions and compounds of boron separately from those of other elements in the group. All group 13 elements have fewer valence electrons than valence orbitals, which generally results in delocalized, metallic bonding. With its high ionization energy, low electron affinity, low electronegativity, and small size, however, boron does not form a metallic lattice with delocalized valence electrons. Instead, boron forms unique and intricate structures that contain multicenter bonds, in which a pair of electrons holds together three or more atoms. Elemental boron forms multicenter bonds, whereas the other group 13 elements exhibit metallic bonding. The basic building block of elemental boron is not the individual boron atom, as would be the case in a metal, but rather the B icosahedron. Because these icosahedra do not pack together very well, the structure of solid boron contains voids, resulting in its low density (Figure \(\Page {3}\)). Elemental boron can be induced to react with many nonmetallic elements to give binary compounds that have a variety of applications. For example, plates of boron carbide (B C) can stop a 30-caliber, armor-piercing bullet, yet they weigh 10%–30% less than conventional armor. Other important compounds of boron with nonmetals include boron nitride (BN), which is produced by heating boron with excess nitrogen (Equation \(\ref{Eq22.6}\)); boron oxide (B O ), which is formed when boron is heated with excess oxygen (Equation \(\ref{Eq22.7}\)); and the boron trihalides (BX ), which are formed by heating boron with excess halogen (Equation \(\ref{Eq22.8}\)). \[\mathrm{2B(s)}+\mathrm{N_2(g)}\xrightarrow{\Delta}\mathrm{2BN(s)} \label{Eq22.6}\] \[\mathrm{4B(s)} + \mathrm{3O_2(g)}\xrightarrow{\Delta}\mathrm{2B_2O_3(s)}\label{Eq22.7}\] \[\mathrm{2B(s)} +\mathrm{3X_2(g)}\xrightarrow{\Delta}\mathrm{2BX_3(g)}\label{Eq22.8}\] Boron nitride is similar in many ways to elemental carbon. With eight electrons, the B–N unit is isoelectronic with the C–C unit, and B and N have the same average size and electronegativity as C. The most stable form of BN is similar to graphite, containing six-membered B N rings arranged in layers. At high temperature and pressure, hexagonal BN converts to a cubic structure similar to diamond, which is one of the hardest substances known. Boron oxide (B O ) contains layers of trigonal planar BO groups (analogous to BX ) in which the oxygen atoms bridge two boron atoms. It dissolves many metal and nonmetal oxides, including SiO , to give a wide range of commercially important borosilicate glasses. A small amount of CoO gives the deep blue color characteristic of “cobalt blue” glass. At high temperatures, boron also reacts with virtually all metals to give metal borides that contain regular three-dimensional networks, or clusters, of boron atoms. The structures of two metal borides—ScB and CaB —are shown in Figure \(\Page {4}\). Because metal-rich borides such as ZrB and TiB are hard and corrosion resistant even at high temperatures, they are used in applications such as turbine blades and rocket nozzles. Boron hydrides were not discovered until the early 20th century, when the German chemist Alfred Stock undertook a systematic investigation of the binary compounds of boron and hydrogen, although binary hydrides of carbon, nitrogen, oxygen, and fluorine have been known since the 18th century. Between 1912 and 1936, Stock oversaw the preparation of a series of boron–hydrogen compounds with unprecedented structures that could not be explained with simple bonding theories. All these compounds contain multicenter bonds. The simplest example is diborane (B H ), which contains two bridging hydrogen atoms (part (a) in Figure \(\Page {5}\). An extraordinary variety of polyhedral boron–hydrogen clusters is now known; one example is the B H ion, which has a polyhedral structure similar to the icosahedral B unit of elemental boron, with a single hydrogen atom bonded to each boron atom. A related class of polyhedral clusters, the carboranes, contain both CH and BH units; an example is shown here. Replacing the hydrogen atoms bonded to carbon with organic groups produces substances with novel properties, some of which are currently being investigated for their use as liquid crystals and in cancer chemotherapy. The enthalpy of combustion of diborane (B H ) is −2165 kJ/mol, one of the highest values known: \[B_2H_{6(g)} + 3O_{2(g)} \rightarrow B_2O_{3(s)} + 3H_2O(l)\;\;\; ΔH_{comb} = −2165\; kJ/mol \label{Eq 22.9}\] Consequently, the US military explored using boron hydrides as rocket fuels in the 1950s and 1960s. This effort was eventually abandoned because boron hydrides are unstable, costly, and toxic, and, most important, B O proved to be highly abrasive to rocket nozzles. Reactions carried out during this investigation, however, showed that boron hydrides exhibit unusual reactivity. Because boron and hydrogen have almost identical electronegativities, the reactions of boron hydrides are dictated by minor differences in the distribution of electron density in a given compound. In general, two distinct types of reaction are observed: electron-rich species such as the BH ion are reductants, whereas electron-deficient species such as B H act as oxidants. For each reaction, explain why the given products form. balanced chemical equations why the given products form Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the reaction products form. Predict the products of the reactions and write a balanced chemical equation for each reaction. All four of the heavier group 13 elements (Al, Ga, In, and Tl) react readily with the halogens to form compounds with a 1:3 stoichiometry: \[ 2M_{(s)} + 3X_{2(s,l,g)} \rightarrow 2MX_{3(s)} \text{ or } M_2X_6 \label{Eq10}\] The reaction of Tl with iodine is an exception: although the product has the stoichiometry TlI , it is not thallium(III) iodide, but rather a thallium(I) compound, the Tl salt of the triiodide ion (I ). This compound forms because iodine is not a powerful enough oxidant to oxidize thallium to the +3 oxidation state. Of the halides, only the fluorides exhibit behavior typical of an ionic compound: they have high melting points (>950°C) and low solubility in nonpolar solvents. In contrast, the trichorides, tribromides, and triiodides of aluminum, gallium, and indium, as well as TlCl and TlBr , are more covalent in character and form halogen-bridged dimers (part (b) in Figure \(\Page {4}\)). Although the structure of these dimers is similar to that of diborane (B H ), the bonding can be described in terms of electron-pair bonds rather than the delocalized electron-deficient bonding found in diborane. Bridging halides are poor electron-pair donors, so the group 13 trihalides are potent Lewis acids that react readily with Lewis bases, such as amines, to form a Lewis acid–base adduct: \[Al_2Cl_{6(soln)} + 2(CH_3)_3N_{(soln)} \rightarrow 2(CH_3)_3N:AlCl_{3(soln)} \label{Eq11}\] In water, the halides of the group 13 metals hydrolyze to produce the metal hydroxide (\[M(OH)_3\)): \[MX_{3(s)} + 3H_2O_{(l)} \rightarrow M(OH)_{3(s)} + 3HX_{(aq)} \label{Eq12}\] In a related reaction, Al (SO ) is used to clarify drinking water by the precipitation of hydrated Al(OH) , which traps particulates. The halides of the heavier metals (In and Tl) are less reactive with water because of their lower charge-to-radius ratio. Instead of forming hydroxides, they dissolve to form the hydrated metal complex ions: [M(H O) ] . Of the group 13 halides, only the fluorides behave as typical ionic compounds. Like boron (Equation \(\ref{Eq22.7}\)), all the heavier group 13 elements react with excess oxygen at elevated temperatures to give the trivalent oxide (M O ), although Tl O is unstable: \[\mathrm{4M(s)}+\mathrm{3O_2(g)}\xrightarrow{\Delta}\mathrm{2M_2O_3(s)} \label{Eq13}\] Aluminum oxide (Al O ), also known as alumina, is a hard, high-melting-point, chemically inert insulator used as a ceramic and as an abrasive in sandpaper and toothpaste. Replacing a small number of Al ions in crystalline alumina with Cr ions forms the gemstone ruby, whereas replacing Al with a mixture of Fe , Fe , and Ti produces blue sapphires. The gallium oxide compound MgGa O gives the brilliant green light familiar to anyone who has ever operated a xerographic copy machine. All the oxides dissolve in dilute acid, but Al O and Ga O are amphoteric, which is consistent with their location along the diagonal line of the periodic table, also dissolving in concentrated aqueous base to form solutions that contain M(OH) ions. Group 13 trihalides are potent Lewis acids that react with Lewis bases to form a Lewis acid–base adduct. Aluminum, gallium, and indium also react with the other group 16 elements (chalcogens) to form chalcogenides with the stoichiometry M Y . However, because Tl(III) is too strong an oxidant to form a stable compound with electron-rich anions such as S , Se , and Te , thallium forms only the thallium(I) chalcogenides with the stoichiometry Tl Y. Only aluminum, like boron, reacts directly with N (at very high temperatures) to give AlN, which is used in transistors and microwave devices as a nontoxic heat sink because of its thermal stability; GaN and InN can be prepared using other methods. All the metals, again except Tl, also react with the heavier group 15 elements (pnicogens) to form the so-called III–V compounds, such as GaAs. These are semiconductors, whose electronic properties, such as their band gaps, differ from those that can be achieved using either pure or doped group 14 elements. For example, nitrogen- and phosphorus-doped gallium arsenide (GaAs P N ) is used in the displays of calculators and digital watches. All group 13 oxides dissolve in dilute acid, but Al O and Ga O are amphoteric. Unlike boron, the heavier group 13 elements do not react directly with hydrogen. Only the aluminum and gallium hydrides are known, but they must be prepared indirectly; AlH is an insoluble, polymeric solid that is rapidly decomposed by water, whereas GaH is unstable at room temperature. Boron has a relatively limited tendency to form complexes, but aluminum, gallium, indium, and, to some extent, thallium form many complexes. Some of the simplest are the hydrated metal ions [M(H O) ], which are relatively strong Brønsted–Lowry acids that can lose a proton to form the M(H O) (OH) ion: \[[M(H_2O)_6]^{3+}_{(aq)} \rightarrow M(H_2O)_5(OH)^{2+}_{(aq)} + H^+_{(aq)} \label{Eq14}\] Group 13 metal ions also form stable complexes with species that contain two or more negatively charged groups, such as the oxalate ion. The stability of such complexes increases as the number of coordinating groups provided by the ligand increases. For each reaction, explain why the given products form. balanced chemical equations why the given products form Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the reaction products form. Predict the products of the reactions and write a balanced chemical equation for each reaction. Compounds of the group 13 elements with oxygen are thermodynamically stable. Many of the anomalous properties of the group 13 elements can be explained by the increase in Z moving down the group. Isolation of the group 13 elements requires a large amount of energy because compounds of the group 13 elements with oxygen are thermodynamically stable. Boron behaves chemically like a nonmetal, whereas its heavier congeners exhibit metallic behavior. Many of the inconsistencies observed in the properties of the group 13 elements can be explained by the increase in Z that arises from poor shielding of the nuclear charge by the filled (n − 1)d and (n − 2)f subshells. Instead of forming a metallic lattice with delocalized valence electrons, boron forms unique aggregates that contain multicenter bonds, including metal borides, in which boron is bonded to other boron atoms to form three-dimensional networks or clusters with regular geometric structures. All neutral compounds of the group 13 elements are electron deficient and behave like Lewis acids. The trivalent halides of the heavier elements form halogen-bridged dimers that contain electron-pair bonds, rather than the delocalized electron-deficient bonds characteristic of diborane. Their oxides dissolve in dilute acid, although the oxides of aluminum and gallium are amphoteric. None of the group 13 elements reacts directly with hydrogen, and the stability of the hydrides prepared by other routes decreases as we go down the group. In contrast to boron, the heavier group 13 elements form a large number of complexes in the +3 oxidation state. | 20,473 | 2,145 |
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The ability to write an organic reaction mechanism properly is key to success in organic chemistry classes. Organic chemists use a technique called arrow pushing to depict the flow or movement of electrons during chemical reactions. Arrow pushing helps chemists keep track of the way in which electrons and their associated atoms redistribute as bonds are made and broken. The first essential rule to keep in mind is the following: : Arrows are used to indicate movement of electrons A regular arrow (double-sided arrowhead) is used to indicate the movement of two electrons, while a line with a single-sided arrowhead (sometimes called a “fish hook arrow”) is used for single electron movement involved with radical reactions that are first described in Chapter 8. The great majority of reactions that will be discussed in this book involve movement of pairs of electrons, so they are represented by double-sided arrowheads. Arrow pushing was first introduced in Section 1.8A in the discussion of resonance contributing structures. Recall that when comparing two or more contributing structures, an arrow was used to show how two electrons (lines representing bonds or pairs of dots representing lone pairs) could be redistributed within a single chemical structure to create an alternative Lewis line structure representation of the bonding. By convention, arrows are used to keep track of all pairs of electrons that are in different locations in the two different contributing Lewis line structures, shown here for the acetate anion and benzene molecule. Keep in mind that in the case of resonance, 1) the atoms do not move between contributing structures, and 2) the electrons are not actually moving. The true chemical structure should be thought of as a hybrid of the contributing Lewis line structures. It is worth pointing out that when used with contributing structures, arrows generally indicate only the interconversion of p bonds and lone pairs (acetate ions) or just p bonds (benzene), not the formation or breaking of s bonds. In chemical reactions, both electrons and atoms change positions as both p and s bonds are formed and broken. Arrow pushing is used to keep track of the movement of all electrons involved with each step of the overall transformation. Because electrons are located in orbitals surrounding atoms, when bonds are formed or broken, the movement of electrons between orbitals is necessarily accompanied by the movement of the associated atoms, which leads to the second rule of arrow pushing when depicting chemical reaction mechanisms: Arrows are never used to indicate the movement of atoms directly. The arrows only show atom movement indirectly as a consequence of electron movement when covalent bonds are made and broken. We have already used arrow pushing to show proton transfer several times in Chapter 4. The example below shows the transfer of a proton from the relatively acidic acetic acid molecule to the relatively basic hydroxide anion. We show this process with one arrow (labeled “a” in the diagram) that starts at a lone pair of electrons on the basic oxygen atom of the hydroxide anion, then points to the acidic H atom of acetic acid to indicate formation of the new bond being made. A second arrow originates at the line representing the breaking O-H bond and points to the O atom to denote creation of a lone pair (arrow “b”). In this reaction, the proton is being transferred between molecules, and the arrows indicate movement of the electrons involved. A common mistake beginning students make is that they will erroneously write an arrow pointing from the H of the acetic acid to the O atom of the hydroxide anion. This is wrong, because such an arrow would be indicating the H atom movement directly, not electron movement! Other common mistakes in arrow pushing are given at the end. Combined with the arrows shown for the contributing structures shown previously, we have now seen all three of the situations illustrated by arrows with double-sided arrowheads, namely the redistribution of p bonds and/or lone pairs, formation of a new s bond (generally from a lone pair or sometimes a new p bond), and breaking of a s bond (generally to form a new lone pair or sometimes a new p bond). Often, as in the case of the acetate-hydroxide ion reaction, more than one arrow is used in a given mechanism step. Now that you have seen all of the important types of arrows, we can point out the most important common feature between them: : Arrows always start at an electron source and end at an electron sink. An electronsource is a bond or a lone pair of electrons. It is either a p bond or a lone pair on an atom of relatively high electron density in a molecule or ion, or a bond that must break during a reaction. An electron sink is an atom on a molecule or ion that can accept a new bond or lone pair of electrons. Learning to identify the characteristic sources and sinks in different functional groups is the key to learning organic chemistry reaction mechanisms. For example, for arrows that depict the formation of new s bonds, the electron source is often readily identified as being a lone pair on the most electron rich atom of a molecule or ion, and the electron sink is readily identified as the most electron poor atom of a molecule or ion. Thus, the prediction of many of the most important electron sources and sinks comes down to lessons concerning the differences in electronegativity between atoms that were presented in Section 1.2, which allow you to identify partial and formal negative and positive charges in molecules. As an aid to your analysis, the red and blue colors of the various electrostatic surface maps given throughout this book indicate the negative and positive regions of molecules. We will have more to say about this reactivity pattern a little bit later. This leads us to another commonly encountered type of process that deserves mention. As you will see in this and many later chapters, making a new bond to an electron sink often requires the simultaneous breaking of one of the bonds present at the sink atom to avoid overfilling its valence orbitals, a situation referred to as . : Breaking a bond will occur to avoid overfilling valence (hypervalence) on an atom serving as an electron sink. In these cases, the electron source for the arrow is the bond being broken, and the sink is an atom able to accommodate the electrons as a lone pair, generally an electronegative atom such as an O atom or a halogen. If an ion is created, that ion is often stabilized by resonance delocalization or other stabilizing interactions. Returning to the proton transfer reaction between acetic acid and hydroxide, we can now summarize our analysis of this simple one-step mechanism. Viewed in the context of the third rule, when considering the arrow used to make a new s bond (arrow a), the hydroxide O atom is the electron source (most negatively charged atom) and the acetic acid H atom is the electron sink (atom with highest partial positive charge). This is illustrated using the electrostatic molecular surfaces shown below the reaction equation. The O atom of hydroxide ion has the greatest localized negative charge as indicated by the most intense red color and the acetic acid proton being transferred has the most intense positive charge character indicated by the most intense blue color. In order to avoid overfilling the valence of the H atom during the reaction (fourth rule), the O-H bond of acetic acid must be broken (arrow “b”). In so doing, the acetate ion is formed. Note that the acetate ion is stabilized by resonance delocalization. Based on our analysis of the reaction between acetic acid and the hydroxide anion, you should now appreciate that the transfer of a proton (a so-called Brønsted acid-base reaction) is really just a special case of the common pattern of reactivity between an electron source (the base) and the proton as an electron sink, combined with breaking a bond to satisfy valence and create a relatively stable ion. The addition or removal of protons during chemical reactions is so common that proton transfer steps are referred to by name directly, and we will use phrases such as “add a proton” or “take a proton away” when referring to them. However, proton transfer reactions are not the only case in which we use special names to describe a particular type of common reaction that involves arrows between electron sources and electron sinks. As briefly mentioned in Section 4.7, a broader terminology is applied to the very common case of reactions in which new s bonds form between electron rich and electron poor regions of molecules. Nucleophiles (meaning nucleus seeking) are molecules that have relatively electron rich p bonds or lone pairs that act as electron sources for arrows making new bonds. Electrophiles (meaning electron seeking) are molecules with relatively electron poor atoms that serve as sinks for these arrows. Analogously, a molecule, or region of a molecule, that is a source for such an arrow is called nucleophilic, while a molecule or region of a molecule that is a sink for these arrows is referred to as being electrophilic. Based on this description, it should be clear that nucleophiles are analogous to Lewis bases and electrophiles are analogous to Lewis acids. Chemists use these terms interchangeably, although nucleophile and electrophile are more commonly used in kinetics discussions while Lewis acid and Lewis base are more commonly used in discussions about reaction thermodynamics. We will use all of these terms throughout the rest of the book. It is helpful to summarize the appropriate use of key terms associated with arrow pushing and reaction mechanisms. The terms “source” and “sink” are used to identify the start and end of each reaction mechanism arrow, which is indicating the change in location of electron pairs. The terms “nucleophile” and “electrophile” (as well as “Lewis base” and “Lewis acid”) are used to describe molecules based on their chemical reactivity and propensity to either donate or receive electrons when they interact. Protons can be thought of as a specific type of electrophile, and for reactions in which a proton is transferred, the nucleophile is called a base. The following two sets of reactions (A and B) show possibilities for arrow pushing in individual reaction steps. Identify which is wrong and explain why. Next, using arrow pushing correctly, label which molecule is the nucleophile and which is the electrophile. In each case the first arrow pushing scenario is wrong. The arrows shown below with stars over them do not start at a source of electrons, but rather they start at positions of relative positive charge, which is incorrect. In the correct arrow pushing, the arrow labeled “a” depicts the interaction of a region of relative high negative charge (a p-bond or lone pair) with an atom of relatively high partial positive charge on the other reactant. Therefore, the molecule acting as the source for arrow the s bond-forming arrow “a” is the nucleophile while the molecule containing the sink atom is the electrophile. The arrow labeled “b” is needed to satisfy valence, and is not considered when defining the nucleophile and electrophile. In the sections and chapters that follow, many different reaction mechanisms will be described in a stepwise fashion. Each arrow can be classified according to one of the three overall situations we have already encountered (redistribution of \(\pi\) bonds and/or lone pairs, formation of a new s bond from a lone pair or \(\pi\) bond, breaking a s bond to give a new lone pair or \(\pi\) bond). When learning new mechanisms, first focus on the overall transformation that takes place. It might be a reaction in which atoms or groups are added (an addition reaction), a reaction in which atoms or groups are removed (an elimination reaction), a reaction in which atoms or groups replace an atom or group (a substitution reaction), or other processes we will encounter. Often, the overall process is composed of multiple steps. Once you have the overall process in mind, it is time to think about the individual steps that convert starting material(s) into product(s). Predicting complete multi-step mechanisms, then, comes down to learning how to predict the individual steps. Understanding, as opposed to memorizing, mechanisms is critical to mastering organic chemistry. Although the mechanisms you encounter throughout the course may seem entirely different, they are actually related in fundamental ways. In fact, almost all of the organic reaction mechanisms you will learn are composed of only a few different individual elements (steps) that are put together in various combinations. Your job is to learn these individual mechanism elements, and then understand how to assemble them into the steps of the correct mechanism for the overall reaction. Fortunately, there are a surprisingly small number of different types of characteristic mechanism elements (patterns of arrows) to be considered when trying to predict individual steps of even complex chemical reactions. For this reason, you should view the prediction of each step in an organic mechanism as essentially a multiple choice situation in which your most common choices are the following: The situation is even simpler than you might expect because 1. and 2. are the functional reverse of each other, as are 3. and 4. in many cases. Many times, more than one of the four choices occurs simultaneously in the same mechanism step and there are some special situations in which unique or different processes such as electrophilic addition or 1,2 shifts occur. These different processes are described in detail as they are encountered. In the following sections and chapters of the book, you will learn important properties of the different functional groups that allow you to deduce the appropriate choices for the individual steps in reaction mechanisms. To help you accomplish this, as new mechanisms are introduced throughout the rest of the book, we will label each mechanistic step as one of the four mentioned here when appropriate, emphasizing the common features between even complex mechanisms. When you are able to predict which of the above choices is(are) the most appropriate for a given step in a mechanism, you will then be able to push electrons correctly without relying on memorization. At that point, you will have taken a major step toward mastering organic chemistry! Throughout this book arrow pushing is used to indicate the flow of electrons in the various organic reaction mechanisms that are discussed. A few simple rules for properly performing arrow pushing were introduced in Section 6.2. In this Appendix we examine some of the most common mistakes that students make when first learning arrow-pushing methods and tell you how to avoid them. The mistakes given below are the ones seen most often by the authors during their cumulative dozens of year of experience in teaching Introductory Organic Chemistry. Reversing the direction of one or more arrows during a chemical step is the most common mistake made by students when writing organic reaction mechanisms. Backwards arrow pushing usually derives from a student thinking about the movement of atoms, not the movement of electrons. Hence, to avoid this mistake it is important to remember that arrows depict how electrons move, not where atoms move, within or between chemical structures. Further, one can avoid this mistake by remembering that every arrow must start at an electron source (a bond or lone pair) and terminate at an electron sink (an atom that can accept a new bond or lone pair). A second common mistake in writing arrow-pushing schemes is to not use enough arrows. This usually results from not keeping track of all lone pairs, bonds made, or bonds broken in a mechanism step. In other words, if you analyze exactly the new position of electrons resulting from each arrow, missing arrows will become evident. In the following example we compare two arrow-pushing scenarios, one of which is missing an arrow. In the incorrect scheme there is no arrow that indicates breaking of the C-H bond of the reactant and formation of the p-bond in the alkene product. Note that when an arrow is missing, the result is commonly too many bonds and/or lone pairs on one atom (see the next section on hypervalency) and not enough bonds or lone pairs on another. Another frequent mistake when writing arrow-pushing schemes is to expand the valency of an atom to more electrons than an atom can accommodate, a situation referred to as hypervalency. An overarching principle of organic chemistry is that carbon has eight electrons in its valence shell when present in stable organic molecules (the Octet Rule, Section 1.2). Analogously, many of the other most common elements in organic molecules, such as nitrogen, oxygen, and chlorine, also obey the Octet Rule. There are three common ways in which students incorrectly draw hypervalent atoms: 1) Too many bonds to an atom, 2) Forgetting the presence of hydrogens, and 3) Forgetting the presence of lone pairs. In the following case an arrow is used to depict a potential resonance structure of nitromethane. However, the result is a nitrogen atoms with 10 electrons in its valence shell because there are too many bonds to N. Such mistakes can be avoided by remembering to draw all bonds and lone pairs on an atom so that the total number of electrons in each atoms valence shell is apparent. Another common way students mistakenly end up with a hypervalent atom is to forget the presence of hydrogens that are not explicitly written. When using stick diagrams to write organic chemical structures not all the hydrogens are drawn, and hence it is common to forget them during an arrow pushing exercise. The following example shows two proposed resonance contributing structures of an amide anion. The arrow drawn on the molecule to the left is incorrect because it depicts the formation of a new bond to a carbon that already has four bonds. When both bonds to hydrogen are drawn explicitly as on the structure farthest to the right, it is clear there are now five bonds around the indicated carbon atom. Notice also that the negative charge was lost upon drawing the contributing structures on the right, providing another clear signal that something was wrong because overall charge is always conserved when arrows are drawn correctly. Another common way to make a hypervalency mistake is by forgetting to count all lone pairs of electrons. The following example shows a negatively charged nucleophile incorrectly adding to the formal positive charge on an alkylated ketone. This may look correct because atoms with positive and negative charges are being directly combined, but when counting bonds and lone pairs of electrons, it is found that the oxygen ends up with 10 electrons overall. Hence, this is a mistake. Acids and bases are catalysts, reactants, products, and intermediates in many organic chemistry transformations. When writing mechanisms for reactions involving acids and bases, there are three general rules that will help guide you in depicting the correct mechanism. In strongly acidic media, all the intermediates and products will be either neutral or positively charged, while in strongly basic media, all the products and intermediates will be neutral or negatively charged. The reason for these rules is that significant extents of strong acids and bases cannot co-exist simultaneously in the same medium because they would rapidly undergo a proton transfer reaction before anything else would happen in the solution. An example of a mixed media error is given below. The first example shows a strong base being created although the reaction is performed under acidic conditions (see conditions over the first equilibrium arrows). Not shown are the three steps that lead to the intermediate drawn. A mistake is made in the arrow pushing because a strong base (methoxide) is generated as the leaving group even though the reaction is run in strong acid. In the correct mechanism, the next step would be protonation of the ether oxygen atom followed by loss of methanol in the last step (not shown) to give a carboxylic acid product. Overall charge must be conserved in all mechanism steps. Failure to conserve overall charge could be caused by some of the preceding errors (hypervalency, failure to draw arrows, mixed media errors), but we mention it by itself because it is always helpful to check that your arrow pushing is consistent by confirming that overall charge conservation is obeyed. In the example shown below, an arrow is missing leading to a neutral intermediate even thought the overall charge on the left side of the equation was minus one. Notice there are five bonds to carbon on the intermediate (hypervalency), providing another obvious indication that something was incorrect in the mechanism step as drawn. | 21,065 | 2,146 |
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There are often several factors that can influence the course of a reaction. Probably the most important is the structure of the alkyl halide, but the solvent can also play a role. The crucial difference between S 1 and S 2 reacions is the ionization step in the S 1 pathway. Factors that stabilize ions, and assist in ionization, promote this pathway. In general, more polar solvents are often helpful in nucleophilic substitutions; the nucleophile may be an ionic compound itself, and a more polar solvent will help it to dissolve. However, especially polar solvents may provide additional stability to ions. The most polar solvents tend to be those that are capable of hydrogen bonding, such as water and alcohols. These are sometimes called "polar, protic solvents." The "protic" part refers to hydrogen bonding; in hydrogen bonding, the hydrogen atom attached to a very electronegative oxygen or nitrogen develops a significant positive charge, like a proton. Polar, protic solvents can stabilize ions through very strong intermolecular attractions. The protic hydrogen can strongly interact with anions, whereas the lone pair on the oxygen atom can stabilize cations. These stabilizing interactions can strongly stabilize the intermediates in S 1 reactions. In the same way, the transition state leading into the intermediate is also significantly stabilized. The barrier for this reaction is lowered and the reaction can occur more quickly. In addition, polar, protic solvents may play an additional role in stabilizing the nucleophile. If the nucleophile is stabilized, it is less likely to react until a sufficiently strong electrophile becomes available. As a result, polar, protic solvents may also depress the rate of S 2 reactions. Once the alkyl halide ionizes and a more attractive electrophile becomes available, the nucleophile can spring into action. The following are a baker's dozen of potential solvents. Describe the S 1 reaction as slow, medium or fast in the following cases. Describe the S 2 reaction as slow, medium or fast in the following cases. , | 2,088 | 2,147 |
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Oxygen is a very common element in all kinds of compounds, whether they are biological molecules, minerals from the earth or petrochemicals. Exploiting oxygen's electronegativity and giving it a little help to become a leaving group is a common way to make connections and build new molecules in nature, the laboratory or the production facility. Sometimes oxygen does not need much help to become a leaving group. , or oxiranes, are three-membered ring ethers. They are good electrophiles, and a C-O bond breaks easily when a nucleophile donates electrons to the carbon. Explain why the C-O bond in an epoxide breaks easily. Use a potential energy diagram to show why epoxides are susceptible to react with nucleophiles, whereas other ethers are not. Epoxides are very useful in the synthesis of important molecules. The Nu-C-C-O motif that is formed in nucleophilic addition to an epoxide is very valuable. Whereas other nucleophilic additions simply replace a halide or leaving group with a nucleophile, exchanging one reactive site with another, addition to an epoxide makes a product that has gone from having one reatcive site to two reactive sites. That can open the door to lots of useful strategies when trying to make a valuable commodity. Show how you could carry out the following transformation. More than one step is involved. One of the most widespread uses of epoxides is in making polymers. The polyethylene glycol produced in polymerization of an epoxide is frequently used in biomedical applications. Provide a mechanism with arrows for the following polymerization of ethylene oxide, in the presence of: Tetrahydrofuran can also be polymerized, forming polytetramethylene glycol. , | 1,715 | 2,148 |
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The goal of an analytical separation is to remove either the analyte or the interferent from the sample’s matrix. To achieve this separation we must identify at least one significant difference between the analyte’s and the interferent’s chemical or physical properties. A significant difference in properties, however, is not sufficient to effect a separation if the conditions that favor the extraction of interferent from the sample also removes a small amount of analyte. Two factors limit a separation’s efficiency: failing to recover all the analyte and failing to remove all the interferent. We define the analyte’s , , as \[R_{A}=\frac{C_{A}}{\left(C_{A}\right)_{\mathrm{o}}} \label{7.1}\] where is the concentration of analyte that remains after the separation, and ( ) is the analyte’s initial concentration. A recovery of 1.00 means that no analyte is lost during the separation. The interferent’s recovery, , is defined in the same manner \[R_{I}=\frac{C_{I}}{\left(C_{I}\right)_{o}} \label{7.2}\] where is the concentration of interferent that remains after the separation, and ( ) is the interferent’s initial concentration. We define the extent of the separation using a , [(a) Sandell, E. B. , Interscience Publishers: New York, 1950, pp. 19–20; (b) Sandell, E. B. , , 834–835]. \[S_{I, A}=\frac{R_{I}}{R_{A}} \label{7.3}\] In general, an of approximately 10 is needed for the quantitative analysis of a trace analyte in the presence of a macro interferent, and 10 when the analyte and interferent are present in approximately equal amounts. The meaning of trace and macro, as well as other terms for describing the concentrations of analytes and interferents, is presented in . An analytical method for determining Cu in an industrial plating bath gives poor results in the presence of Zn. To evaluate a method for separating the analyte from the interferent, samples with known concentrations of Cu or Zn were prepared and analyzed. When a sample of 128.6 ppm Cu was taken through the separation, the concentration of Cu that remained was 127.2 ppm. Taking a 134.9 ppm solution of Zn through the separation left behind a concentration of 4.3 ppm Zn. Calculate the recoveries for Cu and Zn, and the separation factor. Using Equation \ref{7.1} and Equation \ref{7.2}, the recoveries for the analyte and interferent are \[R_{\mathrm{Cu}}=\frac{127.2 \ \mathrm{ppm}}{128.6 \ \mathrm{ppm}}=0.9891 \text { or } 98.91 \% \nonumber\] \[R_{\mathrm{zn}}=\frac{4.3 \ \mathrm{ppm}}{134.9 \ \mathrm{ppm}}=0.032 \text { or } 3.2 \% \nonumber\] and the separation factor is \[S_{\mathrm{Zn}, \mathrm{Cu}}=\frac{R_{\mathrm{Zn}}}{R_{\mathrm{Cu}}}=\frac{0.032}{0.9891}=0.032 \nonumber\] Recoveries and separation factors are useful tools for evaluating a separation’s potential effectiveness; they do not, however, give a direct indication of the error that results from failing to remove all the interferent or from failing to completely recover the analyte. The relative error due to the separation, , is \[E=\frac{S_{s a m p}-S_{s a m p}^*}{S_{samp}} \label{7.4}\] where \(S_{samp}^*\) is the sample’s signal for an ideal separation in which we completely recover the analyte. \[S_{samp}^{*}=k_{A}\left(C_{A}\right)_{\mathrm{o}} \label{7.5}\] Substituting and Equation \ref{7.5} into Equation \ref{7.4}, and rearranging \[E=\frac{k_{A}\left(C_{A}+K_{A, l} \times C_{I}\right)-k_{A}\left(C_{A}\right)_{o}}{k_{A}\left(C_{A}\right)_{o}} \nonumber\] \[E=\frac{C_{A}+K_{A, I} \times C_{I}-\left(C_{A}\right)_{\circ}}{\left(C_{A}\right)_{\circ}} \nonumber\] \[E=\frac{C_{A}}{\left(C_{A}\right)_{\mathrm{o}}}-\frac{\left(C_{A}\right)_{o}}{\left(C_{A}\right)_{o}}+\frac{K_{A, I} \times C_{I}}{\left(C_{A}\right)_{o}} \nonumber\] leaves us with \[E=\left(R_{A}-1\right)+\frac{K_{A, I} \times C_{I}}{\left(C_{A}\right)_{o}} \label{7.6}\] A more useful equation is obtained by solving Equation \ref{7.2} for and substituting into Equation \ref{7.6}. \[E=\left(R_{A}-1\right)+\frac{K_{A, I} \times\left(C_{I}\right)_{o}}{\left(C_{A}\right)_{o}} \times R_{I} \label{7.7}\] The first term of Equation \ref{7.7} accounts for the analyte’s incomplete recovery and the second term accounts for a failure to remove all the interferent. Following the separation outlined in , an analysis is carried out to determine the concentration of Cu in an industrial plating bath. Analysis of standard solutions that contain either Cu or Zn give the following linear calibrations. \[S_{\mathrm{Cu}}=1250 \ \mathrm{ppm}^{-1} \times C_{\mathrm{Cu}} \text { and } S_{\mathrm{Zn}}=2310 \ \mathrm{ppm}^{-1} \times C_{\mathrm{Zn}} \nonumber\] (a) What is the relative error if we analyze a sample without removing the Zn? Assume the initial concentration ratio, Cu:Zn, is 7:1. (b) What is the relative error if we first complete the separation with the recoveries determined in Example 7.5.1
? (c) What is the maximum acceptable recovery for Zn if the recovery for Cu is 1.00 and if the error due to the separation must be no greater than 0.10%? (a) If we complete the analysis without separating Cu and Zn, then and are exactly 1 and Equation \ref{7.7} simplifies to \[E=\frac{K_{\mathrm{Cu}, \mathrm{Zn}} \times\left(C_{\mathrm{Zn}}\right)_{\mathrm{o}}}{\left(C_{\mathrm{Cu}}\right)_{\mathrm{o}}} \nonumber\] Using , we find that the selectivity coefficient is \[K_{\mathrm{Cu}, \mathrm{Zn}}=\frac{k_{\mathrm{Zn}}}{k_{\mathrm{Cu}}}=\frac{2310 \ \mathrm{ppm}^{-1}}{1250 \ \mathrm{ppm}^{-1}}=1.85 \nonumber\] Given the initial concentration ratio of 7:1 for Cu and Zn, the relative error without the separation is \[E=\frac{1.85 \times 1}{7}=0.264 \text { or } 26.4 \% \nonumber\] (b) To calculate the relative error we substitute the recoveries from Example 7.5.1
into Equation \ref{7.7}, obtaining \[E=(0.9891-1)+\frac{1.85 \times 1}{7} \times 0.032= -0.0109+0.085=-0.0024 \nonumber\] or –0.24%. Note that the negative determinate error from failing to recover all the analyte is offset partially by the positive determinate error from failing to remove all the interferent. (c) To determine the maximum recovery for Zn, we make appropriate substitutions into Equation \ref{7.7} \[E=0.0010=(1-1)+\frac{1.85 \times 1}{7} \times R_{\mathrm{Zn}} \nonumber\] and solve for , obtaining a recovery of 0.0038, or 0.38%. Thus, we must remove at least \[100.00 \%-0.38 \%=99.62 \% \nonumber\] of the Zn to obtain an error of 0.10% when is exactly 1. | 6,454 | 2,149 |
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When a manufacturer lists a chemical as ACS Reagent Grade, they must demonstrate that it conforms to specifications set by the American Chemical Society (ACS). For example, the ACS specifications for commercial NaBr require that the concentration of iron is less than 5 ppm. To verify that a production lot meets this standard, the manufacturer collects and analyzes several samples, reporting the average result on the product’s label (Figure 7.1.1
). If the individual samples do not represent accurately the population from which they are drawn—a population that we call the —then even a careful analysis will yield an inaccurate result. Extrapolating a result from a sample to its target population always introduces a determinate sampling error. To minimize this determinate sampling error, we must collect the right sample. Even if we collect the right sample, indeterminate sampling errors may limit the usefulness of our analysis. Equation \ref{7.1} shows that a confidence interval about the mean, \(\overline{X}\) , is proportional to the standard deviation, , of the analysis \[\mu=\overline{X} \pm \frac{t s}{\sqrt{n}} \label{7.1}\] where is the number of samples and is a statistical factor that accounts for the probability that the confidence interval contains the true value, \(\mu\). Equation \ref{7.1} should be familiar to you. See to review confidence intervals and see for values of . Each step of an analysis contributes random error that affects the overall standard deviation. For convenience, let’s divide an analysis into two steps—collecting the samples and analyzing the samples—each of which is characterized by a variance. Using a propagation of uncertainty, the relationship between the overall variance, , and the variances due to sampling, \(s_{samp}^2\), and the variance due to the analytical method, \(s_{meth}^2\), is \[s^{2}=s_{samp}^{2}+s_{meth}^{2} \label{7.2}\] Although Equation \ref{7.1} is written in terms of a standard deviation, , a propagation of uncertainty is written in terms of variances, . In this section, and those that follow, we will use both standard deviations and variances to discuss sampling uncertainty. For a review of the propagation of uncertainty, see and . Equation \ref{7.2} shows that the overall variance for an analysis is limited by either the analytical method or sampling, or by both. Unfortunately, analysts often try to minimize the overall variance by improving only the method’s precision. This is a futile effort, however, if the standard deviation for sampling is more than three times greater than that for the method [Youden, Y. J. , , 1007–1013]. Figure 7.1.2
shows how the ratio / affects the method’s contribution to the overall variance. As shown by the dashed line, if the sample’s standard deviation is \(3 \times\) the method’s standard deviation, then indeterminate method errors explain only 10% of the overall variance. If indeterminate sampling errors are significant, decreasing provides only limited improvement in the overall precision. A quantitative analysis gives a mean concentration of 12.6 ppm for an analyte. The method’s standard deviation is 1.1 ppm and the standard deviation for sampling is 2.1 ppm. (a) What is the overall variance for the analysis? (b) By how much does the overall variance change if we improve by 10% to 0.99 ppm? (c) By how much does the overall variance change if we improve by 10% to 1.9 ppm? (a) The overall variance is \[s^{2}=s_{samp}^{2}+s_{meth}^{2}=(2.1 \ \mathrm{ppm})^{2}+(1.1 \ \mathrm{ppm})^{2}=5.6 \ \mathrm{ppm}^{2} \nonumber\] (b) Improving the method’s standard deviation changes the overall variance to \[s^{2}=(2.1 \ \mathrm{ppm})^{2}+(0.99 \ \mathrm{ppm})^{2}=5.4 \ \mathrm{ppm}^{2} \nonumber\] Improving the method’s standard deviation by 10% improves the overall variance by approximately 4%. (c) Changing the standard deviation for sampling \[s^{2}=(1.9 \ \mathrm{ppm})^{2}+(1.1 \ \mathrm{ppm})^{2}=4.8 \ \mathrm{ppm}^{2} \nonumber\] improves the overall variance by almost 15%. As expected, because is larger than , we achieve a bigger improvement in the overall variance when we focus our attention on sampling problems. Suppose you wish to reduce the overall variance in Example 7.1.1
to 5.0 ppm . If you focus on the method, by what percentage do you need to reduce ? If you focus on the sampling, by what percentage do you need to reduce ? To reduce the overall variance by improving the method’s standard deviation requires that \[s^{2}=5.00 \ \mathrm{ppm}^{2} = s_{samp}^{2}+s_{m e t h}^{2} = (2.1 \mathrm{ppm})^{2}+s_{m e t h}^{2} \nonumber\] Solving for gives its value as 0.768 ppm. Relative to its original value of 1.1 ppm, this is a reduction of \(3.0 \times 10^1\)%. To reduce the overall variance by improving the standard deviation for sampling requires that \[s^{2}=5.00 \ \mathrm{ppm}^{2} = s_{samp}^{2}+s_{meth}^{2} = s_{samp}^{2}+(1.1 \ \mathrm{ppm})^{2} \nonumber\] Solving for gives its value as 1.95 ppm. Relative to its original value of 2.1 ppm, this is reduction of 7.1%. To determine which step has the greatest effect on the overall variance, we need to measure both and . The analysis of replicate samples provides an estimate of the overall variance. To determine the method’s variance we must analyze samples under conditions where we can assume that the sampling variance is negligible; the sampling variance is then determined by difference. There are several ways to minimize the standard deviation for sampling. Here are two examples. One approach is to use a standard reference material (SRM) that has been carefully prepared to minimize indeterminate sampling errors. When the sample is homogeneous—as is the case, for example, with an aqueous sample—then another useful approach is to conduct replicate analyses on a single sample. The following data were collected as part of a study to determine the effect of sampling variance on the analysis of drug-animal feed formulations [Fricke, G. H.; Mischler, P. G.; Staffieri, F. P.; Houmyer, C. L. , , 1213– 1217]. The data on the left were obtained under conditions where both and contribute to the overall variance. The data on the right were obtained under conditions where is insignificant. Determine the overall variance, and the standard deviations due to sampling and the analytical method. To which source of indeterminate error—sampling or the method—should we turn our attention if we want to improve the precision of the analysis? Using the data on the left, the overall variance, , is \(4.71 \times 10^{-7}\). To find the method’s contribution to the overall variance, \(s_{meth}^2\), we use the data on the right, obtaining a value of \(7.00 \times 10^{-8}\). The variance due to sampling, \(s_{samp}^2\), is \[s_{samp}^{2}=s^{2}-s_{meth}^{2} = 4.71 \times 10^{-7}-7.00 \times 10^{-8}=4.01 \times 10^{-7} \nonumber\] Converting variances to standard deviations gives as \(6.33 \times 10^{-4}\) and as \(2.65 \times 10^{-4}\). Because is more than twice as large as , improving the precision of the sampling process will have the greatest impact on the overall precision. A polymer’s density provides a measure of its crystallinity. The standard deviation for the determination of density using a single sample of a polymer is \(1.96 \times 10^{-3}\) g/cm . The standard deviation when using different samples of the polymer is \(3.65 \times 10^{-2}\) g/cm . Determine the standard deviations due to sampling and to the analytical method. The analytical method’s standard deviation is \(1.96 \times 10^{-3}\) g/cm as this is the standard deviation for the analysis of a single sample of the polymer. The sampling variance is \[s_{sa m p}^{2}=s^{2}-s_{meth}^{2}= \left(3.65 \times 10^{-2}\right)^{2}-\left(1.96 \times 10^{-3}\right)^{2}=1.33 \times 10^{-3} \nonumber\] Converting the variance to a standard deviation gives as \(3.64 \times 10^{-2}\) g/cm . | 7,958 | 2,150 |
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These are homework exercises to accompany of McQuarrie and Simon's "Physical Chemistry: A Molecular Approach" Textmap. One liter of N (g) at 2.1 bar and two liters of Ar(g) at 3.4 bar are mixed in a 4.0-L flask to form an ideal-gas mixture. Calculate the value of the final pressure of the mixture if the initial and final temperature of the gases are the same. Then, repeat this calculation if the initial temperatures of the N (g) and Ar(g) are 304 K and 402 K, respectively, and the final temperature of the mixture is 377 K. (Assume ideal-gas behavior.) Using the ideal gas law, we can find the number of moles of each gas. \( n_{N_2} = \dfrac{P_{N_2} V_{N_2}}{RT} = \dfrac{(2.1x10^{5} Pa)*(1x10^{-3} m^3)}{RT} = \dfrac{210 Pa * m^3}{RT} \) \( n_{Ar} = \dfrac{P_{Ar} V_{Ar}}{RT} = \dfrac{(3.4x10^{5} Pa)*(2x10^{-3} m^3)}{RT} = \dfrac{680 Pa * m^3}{RT} \) The total moles of gas in the final mixture is the sum of the moles of each gas in the mixture, which is \( \dfrac{210 Pa * m^3}{RT} + \dfrac{680 Pa * m^3}{RT} = \dfrac{890 Pa * m^3}{RT} \) Therefore, \( P = \dfrac{nRT}{V} = \dfrac{890 Pa * m^3}{0.0040 m^3} = 2.2 x 10^5 Pa = 2.2 bar \) Now, considering the initial temperatures of the gases are different from each other and from the final temperature of the mixture, we calculate the total number of moles to be \( n_{total} = n_{N_2} + n_{Ar} = \dfrac{210 Pa * m^3}{R*(304 K)} + \dfrac{680 Pa * m^3}{R*(402 K)} \) Substituting this into the ideal gas law, we get the final pressure to be \( P= [ \dfrac{210 Pa * m^3}{R*(304 K)} + \dfrac{680 Pa * m^3}{R*(402 K)} ] * \dfrac{R*(377 K)}{0.0040 m^3} = 2.2 x 10^5 Pa = 2.2 bar \) What is the molar Gas Constant in units of cm torr K mol . We know \(R=0.082058 L\cdot atm\cdot mol^{-1}K^{-1}\) using dimensional analysis: \[R=0.082058 L\cdot atm\cdot mol^{-1}K^{-1}\times \frac{760torr}{1 atm} \times \frac{1 dm^3}{1L} \times \frac{1 dm^3}{1000cm^3}\] \[R=0.0623639 cm^3\cdot atm\cdot mol^{-1}K^{-1}\] Use the van der Waals equation to plot the compressibility factor, Z, against P for methane for T = 180 K, 189 K, 190 K, 200 K, and 250 K. First, calculate Z as a function of \(\bar{V}\) and \(P\) as a function of \(\bar{V}\) and plot \(Z\) versus \(P\) For methane, a = 2.3026 \(dm^{6}\cdot bar\cdot mol^{-2} \) and b = 0.043067 \(dm^{3}\cdot mol^{-1} \) \[ Z = \dfrac{P\bar{V}}{RT} \] and the van der Waals equation of state is: \[ P = \dfrac{RT}{\bar{V} - b} - \dfrac{a}{\bar{V}^2}\] Then create a parametric plot of Z versus P for the suggested temperatures as shown below. Define the equation that could be solved by using the Netwon-Raphson method to find the molar volume of O2 at 300K and 200atm. Use the van der Waals anf Redlich-Kwong equations. Constants: Solving van der Waals equation for volume produces: $$\bar{V}^{3} - (b + \dfrac{RT}{P})\bar{V}^{2} + \dfrac{a}{P}\bar{V} - \dfrac{ab}{P}=0\] Plugging in constants gives: $$\bar{V}^{3} - 0.15486\bar{V}^{2} + 0.00691\bar{V} - 0.00022015=0\] which can be solved numerically to find molar volume. The Redlich-Kwong equation can be written as a cubic in volume: $$\bar{V}^{3} - \dfrac{RT}{P}\bar{V}^{2} - \left(B^{2} + \dfrac{BRT}{P} - \dfrac{A}{T^{1/2}P}\right)\bar{V} - \dfrac{AB}{T^{1/2}P} = 0\] Plugging in constants gives: $$\bar{V}^{3} - 0.123\bar{V}^{2} + 0.001822\bar{V} - 0.00011099 = 0\] which can be solved numerically to find molar volume. Compare the Redlich-Kwong and the van der Waals equations for ethane. Take the molar volume to be 0.09416 . van der Waals: $$P = \dfrac{RT}{\bar{V}-b}-\dfrac{a}{(\bar{V})^2}\] Reference Table 16.3 for a, b $$P = \dfrac{0.083145*400}{.09416-0.065144}-\dfrac{5.5818}{(0.09416)^2}\] equation of state improved the accuracy of the van der Waals equation by adding a temperature dependence for the attractive term: $$P = \dfrac{RT}{\bar{V}-B} - \dfrac{A}{T^(\dfrac{1}{2})\bar{V}(\bar{V}+B)}\] Reference Table 16.4 for A, B $$P = \dfrac{0.083145*400}{0.09416-0.045153} - \dfrac{98.831}{400^(\dfrac{1}{2})0.09416*(0.09416+0.62723)}\] \[p=\dfrac{RT}{\dfrac{V}{n}-b}-\dfrac{a}{\sqrt{T}\dfrac{V}{n}\big( \dfrac{V}{n}+b\big) }\] \[ \big (p+\dfrac{n^{2}a}{V^{2}} \big) (V-nb)=nRT\]Use the equations above to calculate the temperature of a system with \(1.5 \ mol\) of methane at \(300 \ K\) confined to a volume of \(100 \ cm^{3}\). Compare the two calculated pressures. \(R = 82.057338 \dfrac{cm^{2} atm }{K \ mol} \) Redlich-Kwong constants: \(a = 31.59\times 10^{6} \ (atm)(K^{\frac{1}{2}})\big( \dfrac{cm^{3}}{mol} \big) ^{2} \) and \(b = 29.6 \ \dfrac{cm^{3}}{mol}\) Van der Waals constants: \(a = 2.25 \times 10^{6} \ (atm) \big( \dfrac{cm^{3}}{mol} \big) ^{2} \) and \(b = 42.8 \ \dfrac{cm^{3}}{mol}\) Redlich-Kwong equation: \(p=\dfrac{82.057338 \dfrac{cm^{2} atm }{K \ mol} \ 300\ K}{\dfrac{100 \ cm^{3}}{1.5 \ mol}-29.6 \ \dfrac{cm^{3}}{mol}}-\dfrac{31.59 \times 10^{6} \ (atm)(K^{\frac{1}{2}})\big( \dfrac{cm^{3}}{mol} \big) ^{2}}{\sqrt{300 \ K}\dfrac{100 \ cm^{3}}{1.5 \ mol}\big( \dfrac{100 \ cm^{3}}{1.5 \ mol}+29.6 \ \dfrac{cm^{3}}{mol}\big) } = 379.95\ atm\) Van der Waals equation: \( p =\dfrac{(1.5 \ mol)(82.057338 \dfrac{cm^{2} atm }{K \ mol})(300\ K)}{(100 \ cm^{3})-(1.5\ mol)(42.8 \ \dfrac{cm^{3}}{mol})} -\dfrac{(1.5\ mol)^{2}(2.25 \times 10^{6} \ (atm) \big( \dfrac{cm^{3}}{mol} \big) ^{2} )}{(100 \ cm^{3})^{2}} = 525.2\ atm\) There is a \(145.25\ atm\) difference in the calculated pressures. The following equation can be used to relate the pressure of propane to its density at temperatures below 400 K: \[ P = 33.258\rho - 7.5884\rho^{2} + 1.0306\rho^{3} - 0.058757\rho^{4} - 0.0033566\rho^{5} + 0.00060696\rho^{6} \] where the pressure (P) is in bar and the density ( \(\rho \)) is in \(\frac{mol}{L} \) and ranges from \( 0 \frac{mol}{L} \leq \rho \leq 12.3 \frac{mol}{L} \). Calculate and plot the pressure for the given density range using: Compare your results from the above parts. a.) Straight forward to calculate. b.) The following constants were used in the equation below: \(a = 9.3919 \dfrac{L^{2} bar}{mol^{2}} \) and \(b = 0.09049 \dfrac{L}{mol} \) \[P_{VDW} = \dfrac{RT}{\overline{V}-b} - \dfrac{a}{\overline{V}^{2}} \] c.) The following constants were used in the equation below: \(a = 183.01 \dfrac{L^{2} bar}{mol^{2}} \) and \(b = 0.06272 \dfrac{L}{mol} \) \[P_{RK} = \dfrac{RT}{\overline{V}-b} - \dfrac{a}{\sqrt{T}\overline{V} (\overline{V}+b)} \] As can be seen by the plot, the Van der Waals equation of state deviates greatly from the empirical fit at higher densities. The Redlich-Kwong equation provides a reasonable approximation to the empirical data. (All calculations were done in MATLAB, and the above plot was produced by me) Use the data below to evaluate the van der waals and Redlich Kwong constants for benzene. *n to indicate molar volume \(a = \dfrac{27(RT_c)^2}{64P_c} \,and \,b = \dfrac{RT_c}{8P_c}\) \(A = 0.42748\dfrac{R^2T_c^{\dfrac{5}{2}}}{P_c} \,and \,B = 0.08664\dfrac{RT_c}{P_c}\) a = \(\dfrac{27*(0.083145 dm^3 \cdot bar \cdot mol^{-1}K^{-1})^2*(561.75 K)^2}{64*48.758 bar} = {\bf 18.8754 \, dm^6 \cdot atm \cdot mol^{-2}}\) b = \(\dfrac{(0.083145 dm^3 \cdot bar \cdot mol^{-1}K^{-1})^2*(561.75 K)^2}{8*48.758 bar} = {\bf 5.5927 \, dm^3 \cdot mol^{-1}}\) \(A = 0.42748*\dfrac{(0.083145 dm^3 \cdot bar \cdot mol^{-1}K^{-1})^2*(561.75 K)^{\dfrac{5}{2}}}{48.758 bar} = {\bf 453.315 \, dm^6 \cdot atm \cdot mol^{-2}\cdot K^{\dfrac{1}{2}}} \) \(B = 0.42748*\dfrac{(0.083145 dm^3 \cdot bar \cdot mol^{-1}K^{-1})^2*(561.75 K)^2}{48.758 bar} = {\bf 19.126\, dm^3 \cdot mol^{-1}}\) Show that the van der Waals equation for argon at T = 142.69 K and P = 35.00 atm can be written as \[\overline {V}^3 - 0.3664\overline {V}^2+0.001210=0 \] where, for convenience, we have suppressed the units in the coefficients. Use the Newton-Raphson method (MathChapter G) to find the three roots to this equation, and calculate the values of the density of liquid and vapor in equilibrium with each other under these conditions. Using Table 16.3, we can get the values of a and b for argon to use in Van der Waals equation of state. By doing this, you can isolate \( \overline {V} \) in van der Waals equation of state. Next, by applying the Newton-Raphson method to the function given by van der Waals equation (which is only dependent on \( \overline {V} \) ), and the derivative of the function, we find the roots to the equation. On the case here, the smallest root represents the molar volume of liquid argon, and the largest root represents the molar volume of vapor argon. Calculate the Volume occupied by \(50\ kg\) of propane at \(50^{\circ}C\) and \(35\ bar\), using the Redlich-Kwong equation of state and the Peng-Robinson equation of state. \[P = \frac{RT}{v-b} - \frac{a}{v(v+b)}\] \(v\) is volume of gas per mole. we have \(T = 50^{\circ} C = 323\ K\) , \(P = 35\ bar = 35 \times (10^5) Pa\) For propane, we have \(P_{c} =42.4924\ bar = 42.4924 \times 10^5 Pa\) and \(T_{c} = 369.522\ K\) so we have \(a = \frac{0.42748 R^2 T_{c}^{2.5}}{P_{c}T^{0.5}}\) ; \(b = \frac{0.08664 * R*T_{c}}{P_{c}}\) so, \(a = \frac{0.42748 (8.314^2) 369.522^{2.5}}{42.4924 \times 10^5(323^{0.5})} = 1.015603\) \(b = \frac{(0.08664) (8.314)(369.522)}{42.4924 \times 10^5} = 6.2640829 \times 10^{-5}\) so, \(P = \frac{RT}{v-b} - \frac{a}{v(v+b)}\) \((35*(10^5)) = \frac{(8.314)(323)}{ v-(6.2640829 \times 10^{-5})} - \frac{1.015603}{v(v+6.2640829 \times 10^{-5})}\) The only real solution is \(v = 0.00010933 \frac{m^3}{mol}\) so In \(50\ Kg\) propane, number of moles = \(\frac{50*1000}{44.10 } = 1133.78684\) total volume occupied by \(50\ Kg\) propane at the specified conditions = \((1133.78684)(0.00010933) = 0.1239569 m^3 = 123.9569\) Liters is the answer. \(P = \frac{RT} {v-b} -\frac{ a}{v(v+b) + b(v-b)}\) \(a = 0.45724R^2 * T_{c}^2 * \frac{\alpha}{P_c}\) \(b = 0.0778 R \frac{T_c}{P_c}\) \(\alpha = 1 + S(1- \sqrt{T_r})\) \(S = 0.37464 + 1.54226w -0.26992w^2\) \(T_r = \frac{T}{T_c}\) we have \(P = 35 \times 10^5 Pa\) and \(T = 323\ K\) \(P_c =42.4924\ bar = 42.4924 \times 10^5 Pa\) and \(T_c = 369.522\ K\) we have accentric factor = \(w = 0.152\). \(T_r = \frac{T}{T_c} = \frac{323}{369.522} = 0.87410\) \(S = 0.37464 + (1.54226)(0.152) -(0.26992)(0.152^2) =0.602827\) \(\alpha = 1 + 0.602827*(1- \sqrt{0.87410}) = 1.0392240523\) \(a = (0.45724)(8.314^2 )( 369.522^2 )\frac{ 1.0392240523}{42.4924 \times 10^5} = 1.055462444\) \(b = (0.0778 ) (8.314)\frac{369.522} {42.4924 \times 10^5} = 0.0000562494\) so, by the equation of state: \((35 \times 10^5) = \frac{(8.314)(323)}{v-0.0000562494} - \frac{1.055462444}{v(v+ 0.0000562494+ 0.0000562494(v- 0.0000562494)}\) \(v = 0.0000988504 \frac{m^3}{mol}\) so In \(50\ Kg\) propane, number of moles = \(\frac{(50)(1000)}{44.10} = 1133.78684\) total volume occupied by \(50\ Kg\) propane at the specified conditions = \((1133.78684)(0.0000988504) = 0.1120752 m^3 = 112.0752\) Liters is the answer. A way to obtain the expressions for the van der Waals constants is to make \((\frac{\partial P}{\partial\bar{V}})_T\) and \((\frac{\partial^2 P}{\partial\bar{V}^{2}})_T\) zero at the critical point, but why do they equal to zero at critical point? Obtain the equation \(\bar{V_c}\) = \(3b\)from this procedure. \((\frac{\partial P}{\partial\bar{V}})_T\) = 0 \((\frac{\partial^2 P}{\partial\bar{V}^{2}})_T\) = 0 \((\frac{\partial P}{\partial\bar{V}})_T\) = \(\frac{\partial}{\partial\bar{V}}\) \((\frac{RT}{\bar{V}-b} - \frac{a}{\bar{T}^{2}})_T\) = \(RT\) \((-\frac{1}{(\bar{V}-b)^{2}})\) - \(a\) \((\frac{-2}{\bar{V}^{3}})\) = \(-\frac{RT}{(\bar{V}-b)^{2}}\) +\(\frac{2a}{\bar{V}^{3}}\) At critical point, \(\frac{RT_c}{(\bar{V_c}-b)^2}\) = \(\frac{2a}{\bar{V_c}^{3}}\) by equating \((\frac{\partial P}{\partial\bar{V}})_T\) = 0. By differentiating the equation \(RT\) \((-\frac{1}{(\bar{V}-b)^{2}})\) - \(a\) \((\frac{-2}{\bar{V}^{3}})\) with respect to \(\bar{V}\), \((\frac{\partial^{2} P}{\partial\bar{V}^{2}})_T\) = \(\frac{\partial}{\partial \bar{V}}\) \((-\frac{RT}{(\bar{V}-b)^2}\) +\(\frac{2a}{\bar{V}^{3}})_T\) = \(RT\) \((-\frac{2}{(\bar{V}-b)^{3}})\) + \(2a\) \((\frac{-3}{\bar{V}^{4}})\) = \(\frac{2RT}{(\bar{V}-b)^{3}}\) - \(\frac{6a}{\bar{V}^{4}}\) At critical point, \(\frac{2RT_c}{(\bar{V_c}-b)^{3}}\) = \(\frac{6a}{\bar{V_c}^{4}}\) by equating \((\frac{\partial^2 P}{\partial\bar{V}^{2}})_T\) = 0. Now, \(\frac{\frac{2a}{\bar{V_c}^{3}}}{\frac{6a}{\bar{V_c}^{4}}}\) = \(\frac{\frac{RT_c}{(\bar{V_c}-b)^2}}{\frac{2RT_c}{(\bar{V_c}-b)^{3}}}\) \(\frac{\bar{V_c}}{3}\) = \(\frac{(\bar{V_c}-b)}{2}\) \(\bar{V_c}\) = \(3b\) Show that the Redlich-Kwong equation can be rewritten in the following form: \[\bar(V)^3 - \frac{RT}{P}\bar{V}^2 - (B^2 + \frac{BRT}{P} - \frac{A}{PT^{1/2}})\bar{V} - \frac{AB}{PT^{1/2}} = 0 \] Starting with the Redlich-Kwong equation and manipulating it we get the following: \[P = \frac{RT}{\bar{V}-B} - \frac{A}{T^{1/2}\bar{V}(\bar{V} + B)}\] \[ P(\bar{V}-B)T^{1/2}\bar{V}(\bar{V} + B) = RT^{3/2}\bar{V}(\bar{V} + B) - A(\bar{V}-B)\] \[ PT^{1/2}\bar{V}(\bar{V}^2 - B^2) = RT^{3/2}\bar{V}^2 + RT^{3/2}\bar{V}B - A\bar{V} + AB\] \[\bar(V)^3 - \frac{RT}{P}\bar{V}^2 - (B^2 + \frac{BRT}{P} - \frac{A}{PT^{1/2}})\bar{V} - \frac{AB}{PT^{1/2}} = 0 \] Use the results of the previous question, Q16.26, to derive Equations 16.14. \((a) \bar{V}_{c} = 3.8473B \), \((b) P_{c} = .029894\dfrac{A^{2/3}R^{1/3}}{B^{5/3}}\), and \((c) T_{c} = .34504\left(\frac{A}{BR}\right) ^{2/3}\) The following are solved from Q16.26 and are necessary to solve for Equation 16.14: \[1) B = .25992\bar{V}_{c} \] \[2) A = 0.42748\dfrac{R^{2}T^{5/2}_{c}}{P_{c}}\] \[3) \dfrac{P_{c}\bar{V_{C}}}{RT_{c}} = \frac{1}{3} \] From equation (1) we see that solving for \(\bar{V}_{c}\) yields the first equation \[\bar{V}_{c} = 3.8473B \]. From equation (2) we can rewrite the expression with some manipulation as \[ A = .42748RT^{3/2}_{c}\left(\frac{RT_{c}}{P_{c}\bar{V}_{c}}\right)\bar{V}_{c} \] Substituting equation (3) and 16.14a into the previous equation gives the expression \[A = 3(.42748)RT^{3/2}_{c}(3.8473B) \] Lastly, solving for \(T_{c}\) gives the equation for 16.14c \[T_{c} = .34504\left(\frac{A}{BR}\right) ^{2/3}\] Finally, solving for equation 16.14b we must substitute the previously solved 16.14c into equation (2),solving for \(P_{c}\) \[P_{c} = 0.42748\dfrac{R^{2}T^{5/2}_{c}}{A} = \left(\frac{.42748}{A}\right)R^{2}\left[.34504\left(\frac{A}{BR}\right)^{2/3}\right]^{5/2} = .029894\dfrac{A^{2/3}R^{1/3}}{B^{5/3}}\] which is equation 16.14b. Derive the Van der Waals cubic equation of state from \[\big( P+\frac{a}{\bar{V}^2}\big)\big(\bar{V}-b\big)=RT\] The Van der Waal equation can be rewritten in terms of \(Z\): \[Z=\frac{P\bar{V}}{RT}=\frac{\bar{V}}{\bar{V}-b}-\frac{a}{RT\bar{V}}\] solving for \(\bar{V}\) \[\bar{V}^3-\big(b+\frac{RT}{P}\big)\bar{V}^2 + \frac{a}{P}\bar{V}-\frac{ab}{P}=0\] | 14,949 | 2,151 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/07%3A_Obtaining_and_Preparing_Samples_for_Analysis/7.07%3A_Liquid-Liquid_Extractions |
A liquid–liquid extraction is an important separation technique for environmental, clinical, and industrial laboratories. A standard environmental analytical method illustrates the importance of liquid–liquid extractions. Municipal water departments routinely monitor public water supplies for trihalomethanes (CHCl , CHBrCl , CHBr Cl, and CHBr ) because they are known or suspected carcinogens. Before their analysis by gas chromatography, trihalomethanes are separated from their aqueous matrix using a liquid–liquid extraction with pentane [“The Analysis of Trihalomethanes in Drinking Water by Liquid Extraction,”EPAMethod501.2 (EPA 500-Series, November 1979)]. The Environmental Protection Agency (EPA) also publishes two additional methods for trihalomethanes. Method 501.1 and Method 501.3 use a purge-and-trap to collect the trihalomethanes prior to a gas chromatographic analysis with a halide-specific detector (Method 501.1) or a mass spectrometer as the detector (Method 501.3). You will find more details about gas chromatography, including detectors, in . In a simple liquid–liquid extraction the solute partitions itself between two immiscible phases. One phase usually is an aqueous solvent and the other phase is an organic solvent, such as the pentane used to extract trihalomethanes from water. Because the phases are immiscible they form two layers, with the denser phase on the bottom. The solute initially is present in one of the two phases; after the extraction it is present in both phases. —that is, the percentage of solute that moves from one phase to the other—is determined by the equilibrium constant for the solute’s partitioning between the phases and any other side reactions that involve the solute. Examples of other reactions that affect extraction efficiency include acid–base reactions and complexation reactions. As we learned earlier in this chapter, a solute’s partitioning between two phases is described by a partition coefficient, . If we extract a solute from an aqueous phase into an organic phase \[S_{a q} \rightleftharpoons S_{o r g} \nonumber\] then the partition coefficient is \[K_{\mathrm{D}}=\frac{\left[S_{org}\right]}{\left[S_{a q}\right]} \nonumber\] A large value for indicates that extraction of solute into the organic phase is favorable. To evaluate an extraction’s efficiency we must consider the solute’s total concentration in each phase, which we define as a , . \[D=\frac{\left[S_{o r g}\right]_{\text { total }}}{\left[S_{a q}\right]_{\text { total }}} \nonumber\] The partition coefficient and the distribution ratio are identical if the solute has only one chemical form in each phase; however, if the solute exists in more than one chemical form in either phase, then and usually have different values. For example, if the solute exists in two forms in the aqueous phase, and , only one of which, , partitions between the two phases, then \[D=\frac{\left[S_{o r g}\right]_{A}}{\left[S_{a q}\right]_{A}+\left[S_{a q}\right]_{B}} \leq K_{\mathrm{D}}=\frac{\left[S_{o r g}\right]_{A}}{\left[S_{a q}\right]_{A}} \nonumber\] This distinction between and is important. The partition coefficient is a thermodynamic equilibrium constant and has a fixed value for the solute’s partitioning between the two phases. The distribution ratio’s value, however, changes with solution conditions if the relative amounts of and change. If we know the solute’s equilibrium reactions within each phase and between the two phases, we can derive an algebraic relationship between and . In a simple liquid–liquid extraction, the only reaction that affects the extraction efficiency is the solute’s partitioning between the two phases (Figure 7.7.1
). In this case the distribution ratio and the partition coefficient are equal. \[D=\frac{\left[S_{o r g}\right]_{\text { total }}}{\left[S_{aq}\right]_{\text { total }}} = K_\text{D} = \frac {[S_{org}]} {[S_{aq}]} \label{7.1}\] Let’s assume the solute initially is present in the aqueous phase and that we wish to extract it into the organic phase. A conservation of mass requires that the moles of solute initially present in the aqueous phase equal the combined moles of solute in the aqueous phase and the organic phase after the extraction. \[\left(\operatorname{mol} \ S_{a q}\right)_{0}=\left(\operatorname{mol} \ S_{a q}\right)_{1}+\left(\operatorname{mol} \ S_{org}\right)_{1} \label{7.2}\] where the subscripts indicate the extraction number with 0 representing the system before the extraction and 1 the system following the first extraction. After the extraction, the solute’s concentration in the aqueous phase is \[\left[S_{a q}\right]_{1}=\frac{\left(\operatorname{mol} \ S_{a q}\right)_{1}}{V_{a q}} \label{7.3}\] and its concentration in the organic phase is \[\left[S_{o r g}\right]_{1}=\frac{\left(\operatorname{mol} \ S_{o r g}\right)_{1}}{V_{o r g}} \label{7.4}\] where and are the volumes of the aqueous phase and the organic phase. Solving Equation \ref{7.2} for (mol ) and substituting into Equation \ref{7.4} leave us with \[\left[S_{o r g}\right]_{1} = \frac{\left(\operatorname{mol} \ S_{a q}\right)_{0}-\left(\operatorname{mol} \ S_{a q}\right)_{1}}{V_{o r g}} \label{7.5}\] Substituting Equation \ref{7.3} and Equation \ref{7.5} into Equation \ref{7.1} gives \[D = \frac {\frac {(\text{mol }S_{aq})_0-(\text{mol }S_{aq})_1} {V_{org}}} {\frac {(\text{mol }S_{aq})_1} {V_{aq}}} = \frac{\left(\operatorname{mol} \ S_{a q}\right)_{0} \times V_{a q}-\left(\operatorname{mol} \ S_{a q}\right)_{1} \times V_{a q}}{\left(\operatorname{mol} \ S_{a q}\right)_{1} \times V_{o r g}} \nonumber\] Rearranging and solving for the fraction of solute that remains in the aqueous phase after one extraction, ( ) , gives \[\left(q_{aq}\right)_{1} = \frac{\left(\operatorname{mol} \ S_{aq}\right)_{1}}{\left(\operatorname{mol} \ S_{a q}\right)_{0}} = \frac{V_{aq}}{D V_{o r g}+V_{a q}} \label{7.6}\] The fraction present in the organic phase after one extraction, ( ) , is \[\left(q_{o r g}\right)_{1}=\frac{\left(\operatorname{mol} S_{o r g}\right)_{1}}{\left(\operatorname{mol} S_{a q}\right)_{0}}=1-\left(q_{a q}\right)_{1}=\frac{D V_{o r g}}{D V_{o r g}+V_{a q}} \nonumber\] Example 7.7.1
shows how we can use Equation \ref{7.6} to calculate the efficiency of a simple liquid-liquid extraction. A solute has a between water and chloroform of 5.00. Suppose we extract a 50.00-mL sample of a 0.050 M aqueous solution of the solute using 15.00 mL of chloroform. (a) What is the separation’s extraction efficiency? (b) What volume of chloroform do we need if we wish to extract 99.9% of the solute? For a simple liquid–liquid extraction the distribution ratio, , and the partition coefficient, , are identical. (a) The fraction of solute that remains in the aqueous phase after the extraction is given by Equation \ref{7.6}. \[\left(q_{aq}\right)_{1}=\frac{V_{a q}}{D V_{org}+V_{a q}}=\frac{50.00 \ \mathrm{mL}}{(5.00)(15.00 \ \mathrm{mL})+50.00 \ \mathrm{mL}}=0.400 \nonumber\] The fraction of solute in the organic phase is 1–0.400, or 0.600. Extraction efficiency is the percentage of solute that moves into the extracting phase; thus, the extraction efficiency is 60.0%. (b) To extract 99.9% of the solute ( ) must be 0.001. Solving Equation \ref{7.6} for , and making appropriate substitutions for ( ) and gives \[V_{o r g}=\frac{V_{a q}-\left(q_{a q}\right)_{1} V_{a q}}{\left(q_{a q}\right)_{1} D}=\frac{50.00 \ \mathrm{mL}-(0.001)(50.00 \ \mathrm{mL})}{(0.001)(5.00 \ \mathrm{mL})}=999 \ \mathrm{mL} \nonumber\] This is large volume of chloroform. Clearly, a single extraction is not reasonable under these conditions. In Example 7.7.1
, a single extraction provides an extraction efficiency of only 60%. If we carry out a second extraction, the fraction of solute remaining in the aqueous phase, ( ) , is \[\left(q_{a q}\right)_{2}=\frac{\left(\operatorname{mol} \ S_{a q}\right)_{2}}{\left(\operatorname{mol} \ S_{a q}\right)_{1}}=\frac{V_{a q}}{D V_{org}+V_{a q}} \nonumber\] If and are the same for both extractions, then the cumulative fraction of solute that remains in the aqueous layer after two extractions, ( ) , is the product of ( ) and ( ) , or \[\left(Q_{aq}\right)_{2}=\frac{\left(\operatorname{mol} \ S_{aq}\right)_{2}}{\left(\operatorname{mol} \ S_{aq}\right)_{0}}=\left(q_{a q}\right)_{1} \times\left(q_{a q}\right)_{2}=\left(\frac{V_{a q}}{D V_{o r g}+V_{a q}}\right)^{2} \nonumber\] In general, for a series of identical extractions, the fraction of analyte that remains in the aqueous phase after the last extraction is \[\left(Q_{a q}\right)_{n}=\left(\frac{V_{a q}}{D V_{o r g}+V_{a q}}\right)^{n} \label{7.7}\] For the extraction described in Example 7.7.1
, determine (a) the extraction efficiency for two identical extractions and for three identical extractions; and (b) the number of extractions required to ensure that we extract 99.9% of the solute. (a) The fraction of solute remaining in the aqueous phase after two extractions and three extractions is \[\left(Q_{aq}\right)_{2}=\left(\frac{50.00 \ \mathrm{mL}}{(5.00)(15.00 \ \mathrm{mL})+50.00 \ \mathrm{mL}}\right)^{2}=0.160 \nonumber\] \[\left(Q_{a q}\right)_{3}=\left(\frac{50.0 \ \mathrm{mL}}{(5.00)(15.00 \ \mathrm{mL})+50.00 \ \mathrm{mL}}\right)^{3}=0.0640 \nonumber\] The extraction efficiencies are 84.0% for two extractions and 93.6% for three extractions. (b) To determine the minimum number of extractions for an efficiency of 99.9%, we set ( ) to 0.001 and solve for using Equation \ref{7.7}. \[0.001=\left(\frac{50.00 \ \mathrm{mL}}{(5.00)(15.00 \ \mathrm{mL})+50.00 \ \mathrm{mL}}\right)^{n}=(0.400)^{n} \nonumber\] Taking the log of both sides and solving for \[\begin{aligned} \log (0.001) &=n \log (0.400) \\ n &=7.54 \end{aligned} \nonumber\] we find that a minimum of eight extractions is necessary. The last two examples provide us with an important observation—for any extraction efficiency, we need less solvent if we complete several extractions using smaller portions of solvent instead of one extraction using a larger volume of solvent. For the conditions in and , an extraction efficiency of 99.9% requires one extraction with 9990 mL of chloroform, or 120 mL when using eight 15-mL portions of chloroform. Although extraction efficiency increases dramatically with the first few multiple, the effect diminishes quickly as we increase the number of extractions (Figure 7.7.2
). In most cases there is little improvement in extraction efficiency after five or six extractions. For the conditions in Example 7.7.2
, we reach an extraction efficiency of 99% after five extractions and need three additional extractions to obtain the extra 0.9% increase in extraction efficiency. To plan a liquid–liquid extraction we need to know the solute’s distribution ratio between the two phases. One approach is to carry out the extraction on a solution that contains a known amount of solute. After the extraction, we isolate the organic phase and allow it to evaporate, leaving behind the solute. In one such experiment, 1.235 g of a solute with a molar mass of 117.3 g/mol is dissolved in 10.00 mL of water. After extracting with 5.00 mL of toluene, 0.889 g of the solute is recovered in the organic phase. (a) What is the solute’s distribution ratio between water and toluene? (b) If we extract 20.00 mL of an aqueous solution that contains the solute using 10.00 mL of toluene, what is the extraction efficiency? (c) How many extractions will we need to recover 99.9% of the solute? (a) The solute’s distribution ratio between water and toluene is \[D=\frac{\left[S_{o r g}\right]}{\left[S_{a q}\right]}=\frac{0.889 \ \mathrm{g} \times \frac{1 \ \mathrm{mol}}{117.3 \ \mathrm{g}} \times \frac{1}{0.00500 \ \mathrm{L}}}{(1.235 \ \mathrm{g}-0.889 \ \mathrm{g}) \times \frac{1 \ \mathrm{mol}}{117.3 \ \mathrm{g}} \times \frac{1}{0.01000 \ \mathrm{L}}}=5.14 \nonumber\] (b) The fraction of solute remaining in the aqueous phase after one extraction is \[\left(q_{a q}\right)_{1}=\frac{V_{a q}}{D V_{org}+V_{a q}}=\frac{20.00 \ \mathrm{mL}}{(5.14)(10.00 \ \mathrm{mL})+20.00 \ \mathrm{mL}}=0.280 \nonumber\] The extraction efficiency, therefore, is 72.0%. (c) To extract 99.9% of the solute requires \[\left(Q_{aq}\right)_{n}=0.001=\left(\frac{20.00 \ \mathrm{mL}}{(5.14)(10.00 \ \mathrm{mL})+20.00 \ \mathrm{mL}}\right)^{n}=(0.280)^{n} \nonumber\] \[\begin{aligned} \log (0.001) &=n \log (0.280) \\ n &=5.4 \end{aligned} \nonumber\] a minimum of six extractions. As we see in Equation \ref{7.1}, in a simple liquid–liquid extraction the distribution ratio and the partition coefficient are identical. As a result, the distribution ratio does not depend on the composition of the aqueous phase or the organic phase. A change in the pH of the aqueous phase, for example, will not affect the solute’s extraction efficiency when and have the same value. If the solute participates in one or more additional equilibrium reactions within a phase, then the distribution ratio and the partition coefficient may not be the same. For example, Figure 7.7.3
shows the equilibrium reactions that affect the extraction of the weak acid, HA, by an organic phase in which ionic species are not soluble. In this case the partition coefficient and the distribution ratio are \[K_{\mathrm{D}}=\frac{\left[\mathrm{HA}_{org}\right]}{\left[\mathrm{HA}_{a q}\right]} \label{7.8}\] \[D=\frac{\left[\mathrm{HA}_{org}\right]_{\text { total }}}{\left[\mathrm{HA}_{a q}\right]_{\text { total }}} =\frac{\left[\mathrm{HA}_{org}\right]}{\left[\mathrm{HA}_{a q}\right]+\left[\mathrm{A}_{a q}^{-}\right]} \label{7.9}\] Because the position of an acid–base equilibrium depends on pH, the distribution ratio, , is pH-dependent. To derive an equation for that shows this dependence, we begin with the acid dissociation constant for HA. \[K_{\mathrm{a}}=\frac{\left[\mathrm{H}_{3} \mathrm{O}_{\mathrm{aq}}^{+}\right]\left[\mathrm{A}_{\mathrm{aq}}^{-}\right]}{\left[\mathrm{HA}_{\mathrm{aq}}\right]} \label{7.10}\] Solving Equation \ref{7.10} for the concentration of A in the aqueous phase \[\left[\mathrm{A}_{a q}^{-}\right]=\frac{K_{\mathrm{a}} \times\left[\mathrm{HA}_{a q}\right]}{\left[\mathrm{H}_{3} \mathrm{O}_{a q}^{+}\right]} \nonumber\] and substituting into Equation \ref{7.9} gives \[D = \frac {[\text{HA}_{org}]} {[\text{HA}_{aq}] + \frac {K_a \times [\text{HA}_{aq}]}{[\text{H}_3\text{O}_{aq}^+]}} \nonumber\] Factoring [HA ] from the denominator, replacing [HA ]/[HA ] with (Equation \ref{7.8}), and simplifying leaves us with the following relationship between the distribution ratio, , and the pH of the aqueous solution. \[D=\frac{K_{\mathrm{D}}\left[\mathrm{H}_{3} \mathrm{O}_{aq}^{+}\right]}{\left[\mathrm{H}_{3} \mathrm{O}_{aq}^{+}\right]+K_{a}} \label{7.11}\] An acidic solute, HA, has a of \(1.00 \times 10^{-5}\) and a between water and hexane of 3.00. Calculate the extraction efficiency if we extract a 50.00 mL sample of a 0.025 M aqueous solution of HA, buffered to a pH of 3.00, with 50.00 mL of hexane. Repeat for pH levels of 5.00 and 7.00. When the pH is 3.00, [\(\text{H}_3\text{O}_{aq}^+\)] is \(1.0 \times 10^{-3}\) and the distribution ratio is \[D=\frac{(3.00)\left(1.0 \times 10^{-3}\right)}{1.0 \times 10^{-3}+1.00 \times 10^{-5}}=2.97 \nonumber\] The fraction of solute that remains in the aqueous phase is \[\left(Q_{aq}\right)_{1}=\frac{50.00 \ \mathrm{mL}}{(2.97)(50.00 \ \mathrm{mL})+50.00 \ \mathrm{mL}}=0.252 \nonumber\] The extraction efficiency, therefore, is almost 75%. The same calculation at a pH of 5.00 gives the extraction efficiency as 60%. At a pH of 7.00 the extraction efficiency is just 3% . The extraction efficiency in Example 7.7.3
is greater at more acidic pH levels because HA is the solute’s predominate form in the aqueous phase. At a more basic pH, where A is the solute’s predominate form, the extraction efficiency is smaller. A graph of extraction efficiency versus pH is shown in Figure 7.7.4
. Note that extraction efficiency essentially is independent of pH for pH levels more acidic than the HA’s p , and that it is essentially zero for pH levels more basic than HA’s p . The greatest change in extraction efficiency occurs at pH levels where both HA and A are predominate species. The ladder diagram for HA along the graph’s -axis helps illustrate this effect. The liquid–liquid extraction of the weak base B is governed by the following equilibrium reactions: \[\begin{array}{c}{\mathrm{B}(a q) \rightleftharpoons \mathrm{B}(org) \quad K_{D}=5.00} \\ {\mathrm{B}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \mathrm{OH}^{-}(a q)+\mathrm{HB}^{+}(a q) \quad K_{b}=1.0 \times 10^{-4}}\end{array} \nonumber\] Derive an equation for the distribution ratio, , and calculate the extraction efficiency if 25.0 mL of a 0.025 M solution of B, buffered to a pH of 9.00, is extracted with 50.0 mL of the organic solvent. Because the weak base exists in two forms, only one of which extracts into the organic phase, the partition coefficient, , and the distribution ratio, , are not identical. \[K_{\mathrm{D}}=\frac{\left[\mathrm{B}_{org}\right]}{\left[\mathrm{B}_{aq}\right]} \nonumber\] \[D = \frac {[\text{B}_{org}]} {[\text{B}_{aq}]} = \frac {[\text{B}_{org}]} {[\text{B}_{aq}] + [\text{HB}_{aq}^+]} \nonumber\] Using the expression for the weak base \[K_{\mathrm{b}}=\frac{\left[\mathrm{OH}_{a q}^{-}\right]\left[\mathrm{HB}_{a q}^{+}\right]}{\left[\mathrm{B}_{a q}\right]} \nonumber\] we solve for the concentration of HB and substitute back into the equation for , obtaining \[D = \frac {[\text{B}_{org}]} {[\text{B}_{aq}] + \frac {K_b \times [\text{B}_{aq}]} {[\text{OH}_{aq}^-]}} = \frac {[\text{B}_{org}]} {[\text{B}_{aq}]\left(1+\frac {K_b} {[\text{OH}_{aq}^+]} \right)} =\frac{K_{D}\left[\mathrm{OH}_{a q}^{-}\right]}{\left[\mathrm{OH}_{a q}^{-}\right]+K_{\mathrm{b}}} \nonumber\] At a pH of 9.0, the [OH ] is \(1 \times 10^{-5}\) M and the distribution ratio has a value of \[D=\frac{K_{D}\left[\mathrm{OH}_{a q}^{-}\right]}{\left[\mathrm{OH}_{aq}^{-}\right]+K_{\mathrm{b}}}=\frac{(5.00)\left(1.0 \times 10^{-5}\right)}{1.0 \times 10^{-5}+1.0 \times 10^{-4}}=0.455 \nonumber\] After one extraction, the fraction of B remaining in the aqueous phase is \[\left(q_{aq}\right)_{1}=\frac{25.00 \ \mathrm{mL}}{(0.455)(50.00 \ \mathrm{mL})+25.00 \ \mathrm{mL}}=0.524 \nonumber\] The extraction efficiency, therefore, is 47.6%. At a pH of 9, most of the weak base is present as HB , which explains why the overall extraction efficiency is so poor. One important application of a liquid–liquid extraction is the selective extraction of metal ions using an organic ligand. Unfortunately, many organic ligands are not very soluble in water or undergo hydrolysis or oxidation reactions in aqueous solutions. For these reasons the ligand is added to the organic solvent instead of the aqueous phase. Figure 7.7.5
shows the relevant equilibrium reactions (and equilibrium constants) for the extraction of M by the ligand HL, including the ligand’s extraction into the aqueous phase ( ), the ligand’s acid dissociation reaction ( ), the formation of the metal–ligand complex (\(\beta_n\)), and the complex’s extraction into the organic phase ( ). If the ligand’s concentration is much greater than the metal ion’s concentration, then the distribution ratio is \[D=\frac{\beta_{n} K_{\mathrm{D}, c}\left(K_{a}\right)^{n}\left(C_{\mathrm{HL}}\right)^{n}}{\left(K_{\mathrm{D}, \mathrm{HL}}\right)^{n}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]^{n}+\beta_{n}\left(K_{\mathrm{a}}\right)^{n}\left(C_{\mathrm{HL}}\right)^{n}} \label{7.12}\] where is the ligand’s initial concentration in the organic phase. As shown in Example 7.7.4
, the extraction efficiency for metal ions shows a marked pH dependency. A liquid–liquid extraction of the divalent metal ion, M , uses the scheme outlined in Figure 7.7.5
. The partition coefficients for the ligand, , and for the metal–ligand complex, , are \(1.0 \times 10^4\) and \(7.0 \times 10^4\), respectively. The ligand’s acid dissociation constant, , is \(5.0 \times 10^{-5}\), and the formation constant for the metal–ligand complex, \(\beta_2\), is \(2.5 \times 10^{16}\). What is the extraction efficiency if we extract 100.0 mL of a \(1.0 \times 10^{-6}\) M aqueous solution of M , buffered to a pH of 1.00, with 10.00 mL of an organic solvent that is 0.1 mM in the chelating agent? Repeat the calculation at a pH of 3.00. When the pH is 1.00 the distribution ratio is \[D=\frac{\left(2.5 \times 10^{16}\right)\left(7.0 \times 10^{4}\right)\left(5.0 \times 10^{-5}\right)^{2}\left(1.0 \times 10^{-4}\right)^{2}}{\left(1.0 \times 10^{4}\right)^{2}(0.10)^{2}+\left(2.5 \times 10^{16}\right)\left(5.0 \times 10^{-5}\right)^{2}\left(1.0 \times 10^{-4}\right)^{2}} \nonumber\] or a of 0.0438. The fraction of metal ion that remains in the aqueous phase is \[\left(Q_{aq}\right)_{1}=\frac{100.0 \ \mathrm{mL}}{(0.0438)(10.00 \ \mathrm{mL})+100.0 \ \mathrm{mL}}=0.996 \nonumber\] At a pH of 1.00, we extract only 0.40% of the metal into the organic phase. Changing the pH to 3.00, however, increases the extraction efficiency to 97.8%. Figure 7.7.6
shows how the pH of the aqueous phase affects the extraction efficiency for M . One advantage of using a ligand to extract a metal ion is the high degree of selectivity that it brings to a liquid–liquid extraction. As seen in Figure 7.7.6
, a divalent metal ion’s extraction efficiency increases from approximately 0% to 100% over a range of 2 pH units. Because a ligand’s ability to form a metal–ligand complex varies substantially from metal ion to metal ion, significant selectivity is possible if we carefully control the pH. Table 7.7.1
shows the minimum pH for extracting 99% of a metal ion from an aqueous solution using an equal volume of 4 mM dithizone in CCl . Using Table 7.7.1
, explain how we can separate the metal ions in an aqueous mixture of Cu , Cd , and Ni by extracting with an equal volume of dithizone in CCl . From Table 7.7.1
, a quantitative separation of Cu from Cd and from Ni is possible if we acidify the aqueous phase to a pH of less than 1. This pH is greater than the minimum pH for extracting Cu and significantly less than the minimum pH for extracting either Cd or Ni . After the extraction of Cu is complete, we shift the pH of the aqueous phase to 4.0, which allows us to extract Cd while leaving Ni in the aqueous phase. | 22,485 | 2,153 |
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A simple distillation is incapable of significant purification if the boiling points of the components are too close. When the difference in boiling points is less than \(100^\text{o} \text{C}\), a modification is necessary, namely insertion of a fractionating column between the distilling flask and three-way adapter (Figure 5.37a). The distillate of a simple distillation is always enriched in the lower boiling compound. As previously discussed, the simple distillation of a \(75 \: \text{mol}\%\) ethylbenzene/\(25 \: \text{mol}\%\) -cymene mixture resulted in a distillate that was \(90 \: \text{mol}\%\) ethylbenzene (ethylbenzene had the lower boiling point). Imagine if a subsequent distillation were performed on the \(90 \: \text{mol}\%\) solution: the distillate of process would contain an even higher percentage of ethylbenzene. A fractionating column essentially allows for many successive distillations to take place at once, without dismantling the apparatus. A fractionating column contains indentations (a Vigreux column, Figure 5.37) or a packing material with lots of surface area. The vapors temporarily condense on these surfaces (see Figure 5.37b) and the heat of the distillation allows those pools of liquid to vaporize again. Every vaporization-condensation event (called a "theoretical plate") is similar to a simple distillation, and each event enriches the distillate in the lower boiling component. The concepts of a fractional distillation can be shown through a distillation curve. In the distillation of an equimolar mixture of compounds A + B (which is described by the distillation curve in Figure 5.38), one vaporization-condensation event is represented in Figure 5.38 by following points to to . This process represents one theoretical plate, and would produce a distillate that is \(81\%\) A and \(19\%\) B. A second vaporization-condensation event is represented in Figure 5.38 by following points to to , and would produce a distillate that is \(96\%\) A and \(4\%\) B. A \(50\%\)/\(50\%\) mixture of two components whose boiling points differ by only \(20\)-\(30^\text{o} \text{C}\) would require at least three theoretical plates to obtain a distillate with \(> 95\%\) purity. In practice, fractional distillations still often produce mixtures. The best chance of obtaining purity via fractional distillation is when there is very little impurity to begin with. | 2,432 | 2,154 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Valence_Bond_Theory/Resonance |
Resonance is a mental exercise within the of bonding that describes the delocalization of electrons within molecules. It involves constructing multiple Lewis structures that, when combined, represent the full electronic structure of the molecule. are used when a single Lewis structure cannot fully describe the bonding; the combination of possible resonance structures is defined as a , which represents the overall delocalization of electrons within the molecule. In general, molecules with multiple resonance structures will be more stable than one with fewer and some resonance structures contribute more to the stability of the molecule than others - formal charges aid in determining this. Resonance is a way of describing delocalized electrons within certain molecules or polyatomic ions where the bonding cannot be expressed by a single Lewis formula. A molecule or ion with such delocalized electrons is represented by several resonance structures. The nuclear skeleton of the Lewis Structure of these resonance structures remains the same, only the electron locations differ. Such is the case for (\(\ce{O3}\)), an allotrope of oxygen with a V-shaped structure and an O–O–O angle of 117.5°. Let's motivate the discussion by building the Lewis structure for ozone. 1. We know that ozone has a V-shaped structure, so one O atom is central: 2. Each O atom has 6 valence electrons, for a total of 18 valence electrons. 3. Assigning one bonding pair of electrons to each oxygen–oxygen bond gives with 14 electrons left over. 4. If we place three lone pairs of electrons on each terminal oxygen, we obtain and have 2 electrons left over. 5. At this point, both terminal oxygen atoms have octets of electrons. We therefore place the last 2 electrons on the central atom: 6. The central oxygen has only 6 electrons. We must convert one lone pair on a terminal oxygen atom to a bonding pair of electrons—but which one? Depending on which one we choose, we obtain either Which is correct? In fact, neither is correct. Both predict one O–O single bond and one O=O double bond. As you will learn, if the bonds were of different types (one single and one double, for example), they would have different lengths. It turns out, however, that both O–O bond distances are identical, 127.2 pm, which is shorter than a typical O–O single bond (148 pm) and longer than the O=O double bond in O (120.7 pm). Equivalent Lewis dot structures, such as those of ozone, are called . The position of the is the same in the various resonance structures of a compound, but the position of the is different. Double-headed arrows link the different resonance structures of a compound: The double-headed arrow indicates that the actual electronic structure is an of those shown, not that the molecule oscillates between the two structures. When it is possible to write more than one equivalent resonance structure for a molecule or ion, the actual structure is the average of the resonance structures. The electrons appear to "shift" between different resonance structures and while not strictly correct as each resonance structure is just a limitation of using the Lewis structure perspective to describe these molecules. A more accurate description of the electron structure of the molecule requires considering multiple resonance structures simultaneously. Most arrows in chemistry cannot be used interchangeably and care must be given to selecting the correct arrow for the job. \(\ce{CO3^{2-}}\). 1. Because carbon is the least electronegative element, we place it in the central position: 2. Carbon has 4 valence electrons, each oxygen has 6 valence electrons, and there are 2 more for the −2 charge. This gives 4 + (3 × 6) + 2 = 24 valence electrons. 3. Six electrons are used to form three bonding pairs between the oxygen atoms and the carbon: 4. We divide the remaining 18 electrons equally among the three oxygen atoms by placing three lone pairs on each and indicating the −2 charge: 5. No electrons are left for the central atom. 6. At this point, the carbon atom has only 6 valence electrons, so we must take one lone pair from an oxygen and use it to form a carbon–oxygen double bond. In this case, however, there are possible choices: As with ozone, none of these structures describes the bonding exactly. Each predicts one carbon–oxygen double bond and two carbon–oxygen single bonds, but experimentally all C–O bond lengths are identical. We can write resonance structures (in this case, three of them) for the carbonate ion: The actual structure is an average of these three resonance structures. Like ozone, the electronic structure of the carbonate ion cannot be described by a single Lewis electron structure. Unlike O , though, the actual structure of CO is an average of resonance structures. While each resonance structure contributes to the total electronic structure of the molecule, they may not contribute equally. Assigning to atoms in the molecules is one mechanism to identify the viability of a resonance structure and determine its relative magnitude among other structures. The formal charge on an atom in a covalent species is the net charge the atom would bear if the electrons in all the bonds to the atom were equally shared. Alternatively the formal charge on an atom in a covalent species is the net charge the atom would bear if all bonds to the atom were nonpolar covalent bonds. To determine the formal charge on a given atom in a covalent species, use the following formula: \[\text{Formal Charge} = (\text{number of valence electrons in free orbital}) - (\text{number of lone-pair electrons}) - \frac{1}{2} (\text{ number bond pair electrons}) \label{FC}\] (\(CNS^-\)) ion. 1. Find the Lewis Structure of the molecule. (Remember the Lewis Structure rules.) 2. Resonance: All elements want an octet, and we can do that in multiple ways by moving the terminal atom's electrons around (bonds too). 3. Assign Formal Charges via Equation \ref{FC}. Formal Charge = (number of valence electrons in free orbital) - (number of lone-pair electrons) - ( \( \frac{1}{2} \) number bond pair electrons) Remember to determine the number of valence electron each atom has before assigning Formal Charges C = 4 valence e , N = 5 valence e , S = 6 valence e , also add an extra electron for the (-1) charge. The total of valence electrons is 16. 4. Find the most ideal resonance structure. (Note: It is the one with the least formal charges that adds up to zero or to the molecule's overall charge.) 5. Now we have to look at electronegativity for the "Correct" Lewis structure. The most atom usually has the negative formal charge, while the least electronegative atom usually has the positive formal charges. It is useful to combine the resonance structures into a single structure called the that describes the bonding of the molecule. The general approach is described below: Benzene is a common organic solvent that was previously used in gasoline; it is no longer used for this purpose, however, because it is now known to be a carcinogen. The benzene molecule (\(\ce{C6H6}\)) consists of a regular hexagon of carbon atoms, each of which is also bonded to a hydrogen atom. Use resonance structures to describe the bonding in benzene. molecular formula and molecular geometry resonance structures Each hydrogen atom contributes 1 valence electron, and each carbon atom contributes 4 valence electrons, for a total of (6 × 1) + (6 × 4) = 30 valence electrons. If we place a single bonding electron pair between each pair of carbon atoms and between each carbon and a hydrogen atom, we obtain the following: Each carbon atom in this structure has only 6 electrons and has a formal charge of +1, but we have used only 24 of the 30 valence electrons. If the 6 remaining electrons are uniformly distributed pairwise on alternate carbon atoms, we obtain the following: Three carbon atoms now have an octet configuration and a formal charge of −1, while three carbon atoms have only 6 electrons and a formal charge of +1. We can convert each lone pair to a bonding electron pair, which gives each atom an octet of electrons and a formal charge of 0, by making three C=C double bonds. There are, however, two ways to do this: Each structure has alternating double and single bonds, but experimentation shows that each carbon–carbon bond in benzene is identical, with bond lengths (139.9 pm) intermediate between those typically found for a C–C single bond (154 pm) and a C=C double bond (134 pm). We can describe the bonding in benzene using the two resonance structures, but the actual electronic structure is an average of the two. The existence of multiple resonance structures for aromatic hydrocarbons like benzene is often indicated by drawing either a circle or dashed lines inside the hexagon: Draw the possible resonance structures for the Nitrate ion \(\ce{NO_3^{-}}\). 1. Count up the valence electrons: (1*5) + (3*6) + 1(ion) = electrons 2. Draw the bond connectivities: 3. Add octet electrons to the atoms bonded to the center atom: 4. Place any leftover electrons (24-24 = ) on the center atom: 5. Does the central atom have an octet? 6. Does the central atom have an octet? Note: We would expect that the bond lengths in the \(\ce{NO_3^{-}}\) ion to be somewhat shorter than a single bond. 2. Below are the all Lewis dot structure with formal charges (in red) for Sulfate ( ). There isn't a most favorable resonance of the Sulfate ion because they are all identical in charge and there is no change in Electronegativity between the Oxygen atoms. 3. Below is the resonance for , formal charges are displayed in red. The Lewis Structure with the most formal charges is not desirable, because we want the Lewis Structure with the least formal charge. 4. The resonance for , and the formal charges (in red). 5. The resonance for , and the formal charges (in red). 6. The resonance hybrid for , hybrid bonds are in red. 7. The resonance hybrid for , hybrid bonds are in red. | 10,041 | 2,155 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Atomic_Theory/The_Mole_and_Avogadro's_Constant |
The number of moles in a system can be determined using the atomic mass of an element, which can be found on the periodic table. This mass is usually an of the abundant forms of that element found on earth. An element's mass is listed as the average of all its isotopes on earth. s \(6.02214179 \times 10^{23}\) s \(6.02214179 \times 10^{23}\) er \(6.02214179 \times 10^{23}\) is cal (\(N_A\)) , Each carbon-12 atom weighs about \(1.99265 \times 10^{-23}\; g\); therefore, \[(1.99265 \times 10^{-23}\; g) \times (6.02214179 \times 10^{23}\; atoms) = 12\; g\; \text{ of carbon-12} \nonumber \] The mass of a mole of substance is called the of that substance. The molar mass is used to convert grams of a substance to moles and is used often in chemistry. The molar mass of an element is found on the periodic table, and it is the element's atomic weight in grams/mole (g/mol). If the mass of a substance is known, the number of moles in the substance can be calculated. Converting the mass, in grams, of a substance to moles requires a conversion factor of (one mole of substance/molar mass of substance). The mole concept is also applicable to the composition of chemical compounds. For instance, consider methane, CH . This molecule and its molecular formula indicate that per mole of methane there is 1 mole of carbon and 4 moles of hydrogen. In this case, the mole is used as a common unit that can be applied to a ratio as shown below: \[2 \text{ mol H } + 1 \text{ mol O }= 1 \text{ mol } \ce{H2O} \nonumber\] In this this chemical reactions, the moles of H and O describe the number of atoms of each element that react to form 1 mol of \(\ce{H_2O}\). To think about what a mole means, one should relate it to quantities such as dozen or pair. Just as a pair can mean two shoes, two books, two pencils, two people, or two of anything else, a mole means 6.02214179×10 of anything. Using the following relation: \[\text{1 mole} = 6.02214179 \times 10^{23}\] is analogous to saying: \[\text{1 Dozen} = \text{12 eggs}\] It is quite difficult to visualize a mole of something because Avogadro's constant is extremely large. For instance, consider the size of one single grain of wheat. If all the people who have existed in Earth's history did nothing but count individual wheat grains for their entire lives, the total number of wheat grains counted would still be much less than Avogadro's constant; the number of wheat grains produced throughout history does not even approach Avogadro's Number. In this example, multiply the mass of by the conversion factor (inverse molar mass of potassium): \[\dfrac{1\; mol\; K}{39.10\; grams \;K} \nonumber \] 39.10 grams is the molar mass of one mole of ; cancel out grams, leaving the moles of : \[3.04\; \cancel{g\; K} \left(\dfrac{1\; mol\; K}{39.10\; \cancel{g\; K}}\right) = 0.0778\; mol\; K \nonumber \] Similarly, if the moles of a substance are known, the number grams in the substance can be determined. Converting moles of a substance to grams requires a conversion factor of . One simply needs to follow the same method but in the opposite direction. How many grams are 10.78 moles of Calcium ( )? Multiply moles of Ca by the conversion factor (molar mass of calcium) which then allows the cancelation of moles, leaving grams of Ca. \[10.78 \cancel{\;mol\; Ca} \left(\dfrac{40.08\; g\; Ca}{1\; \cancel{mol\; Ca}}\right) = 432.1\; g\; Ca \nonumber \] The total number of atoms in a substance can also be determined by using the relationship between grams, moles, and atoms. If given the mass of a substance and asked to find the number of atoms in the substance, one must first convert the mass of the substance, in grams, to moles, as in Example \(\Page {1}\). Then the number of moles of the substance must be converted to atoms. Converting moles of a substance to atoms requires a conversion factor of . Verifying that the units cancel properly is a good way to make sure the correct method is used. \[3.5\; \cancel{g\; Na} \left(\dfrac{1\; mol\; Na}{22.98\; \cancel{g\; Na}}\right) = 0.152\; mol\; Na \nonumber \] \[0.152\; \cancel{mol\; Na} \left(\dfrac{6.02214179\times 10^{23}\; atoms\; Na}{1\;\cancel{ mol\; Na}}\right) = 9.15 \times 10^{22}\; atoms\; of\; Na \nonumber \] In this example, multiply the grams of Na by the conversion factor 1 mol Na/ 22.98 g Na, with 22.98g being the molar mass of one mole of Na, which then allows cancelation of grams, leaving moles of Na. Then, multiply the number of moles of Na by the conversion factor 6.02214179×10 atoms Na/ 1 mol Na, with 6.02214179×10 atoms being the number of atoms in one mole of Na (Avogadro's constant), which then allows the cancelation of moles, leaving the number of atoms of Na. Using Avogadro's \(\Page {1}\)) Avogadro's mol \[3.00 \; \cancel{g\; K} \left(\dfrac{1\; mol\; K}{39.10\; \cancel{g\; K}}\right) = 0.0767\; mol\; K \nonumber \] In this example, multiply the mass of K by the conversion factor: \[\dfrac{1\; mol\; K}{39.10\; grams\; K} \nonumber \] 39.10 grams is the molar mass of one mole of K. Grams can be canceled, leaving the moles of K. This is the calculation in Example \(\Page {2}\) performed in reverse. Multiply moles of Ca by the conversion factor with 40.08 g being the molar mass of one mole of Ca. The moles cancel, leaving grams of Ca: \[10.00\; \cancel{mol\; Ca} \left(\dfrac{40.08\; g\; Ca}{1\;\cancel{ mol\; Ca}}\right) = 400.8\; grams \;of \;Ca \nonumber \] The number of atoms can also be calculated using . Convert grams to moles \[3.0\; \cancel{g\; Na} \left(\dfrac{1\; mol\; Na}{22.98\; \cancel{g\; Na}}\right) = 0.130\; mol\; Na \nonumber \] Convert moles to atoms \[0.130548\; \cancel{ mol\; Na} \left(\dfrac{6.02214179 \times 10^{23}\; atoms \;Na}{1\; \cancel{ mol\; Na}}\right) = 7.8 \times 10^{22} \; atoms\; of\; \; Na \nonumber \] The mole, abbreviated mol, is an SI unit which measures the number of particles in a specific substance. One mole is equal to \(6.02214179 \times 10^{23}\) atoms, or other elementary units such as molecules. 5.06g O (1mol/16.00g)= 0.316 mols (6.022x10 atoms/ 1mol) = 2.14g K (1mol/39.10g) 0.055 mols (6.022x10 atoms/ 1mol) 0.134kg Li (1000g/1kg)= 134g Li (1mol/6.941g)= 19.3 (6.022x10 atoms/ 1mol) = | 6,249 | 2,157 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Chemiluminescence/4%3A_Instrumentation/4.01%3A_Detection_of_chemiluminescence |
The detector of choice for chemiluminescence is the photomultiplier tube, a development of the vacuum phototube that permits considerable amplification of the signal. Figure D1.1 shows how the photomultiplier works. The surface of the cathode supports a photoemissive layer that ejects electrons in direct proportion to the intensity of the incident light; several electrons are emitted for each photon and are attracted towards a positively-charged dynode. When the electron beam meets the dynode several electrons (E in figure D1.1) are ejected for each incident electron and these are attracted to a second dynode at a higher positive potential. This process is repeated along a series of dynodes, the intensity of the electron beam increasing continually until when it reaches the anode (at the greatest positive potential) there are over a million electrons for each photon incident at the cathode. The resulting current can be amplified electronically. In the absence of light, the photomultiplier generates a dark current, chiefly due to thermal emission. Thermal dark currents can be eliminated by cooling to ─30°C.
Diodes are also used for chemiluminescence detection , especially in low-cost applications. Photographic detection was also used in very early work . | 1,295 | 2,158 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Ideal_Systems/Entropy_of_Mixing |
A gas will always flow into a newly available volume and does so because its molecules are rapidly bouncing off one another and hitting the walls of their container, readily moving into a new allowable space. It follows from the second law of thermodynamics that a process will occur in the direction towards a more probable state. In terms of entropy, this can be expressed as a system going from a state of lesser probability (less microstates) towards a state of higher probability (more microstates). This corresponds to increasing the \(W\) in the equation \( S=k_B\ln W \). For our example, we shall again consider a simple system of two ideal gases, A and B, with a number of moles \( n_A \) and \( n_B \), at a certain constant temperature and pressure in volumes of \( V_A \) and \( V_B \), as shown in Figure \(\Page {1A}\). These two gases are separated by a partition so they are each sequestered in their respective volumes. If we now remove the partition (like opening a window in the example above), we expect the two gases to randomly diffuse and form a homogenous mixture as we see in Figure \(\Page {1B}\). To calculate the entropy change, let us treat this mixing as two separate gas expansions, one for gas A and another for B. From the statistical definition of entropy, we know that \[ \Delta S=nR\ln \dfrac{V_2}{V_1} \;. \nonumber \] Now, for each gas, the volume \( V_1 \) is the initial volume of the gas, and \( V_2 \) is the final volume, which is both the gases combined, \( V_A+V_B \). So for the two separate gas expansions, \[ \Delta S_A=n_A R\ln \dfrac{V_A+V_B}{V_A} \nonumber \] \[ \Delta S_B=n_B R\ln \dfrac{V_A+V_B}{V_B} \nonumber \] So to find the total entropy change for both these processes, because they are happening at the same time, we simply add the two changes in entropy together. \[ \Delta_{mix}S = \Delta S_{A}+\Delta S_{B}=n_{A}R\ln \dfrac{V_{A}+V_{B}}{V_{A}}+n_{B}R\ln \dfrac{V_{A}+V_{B}}{V_{B}} \nonumber \] Recalling the ideal gas law, PV=nRT, we see that the volume is directly proportional to the number of moles ( ), and since we know the number of moles we can substitute this for the volume: \[ \Delta_{mix}S=n_{A}R\ln \dfrac{n_{A}+n_{B}}{n_{A}}+n_{B}R\ln \dfrac{n_{A}+n_{B}}{n_{B}} \nonumber \] Now we recognize that the inverse of the term \( \frac{n_{A}+n_{B}}{n_{A}} \) is the mole fraction \( \chi_{A}=\frac{n_{A}}{n_{A}+n_{B}} \), and taking the inverse of these two terms in the above equation, we have: \[ \Delta_{mix}S=-n_{A}R\ln \dfrac{n_A}{n_A+n_B}-n_BR\ln \dfrac{n_A}{n_A+n_B}\chi_{B} = -n_A R\ln \chi_A -n_B R\ln \chi_B \nonumber \] since \(\ln x^{-1}=-\ln x\) from the rules for logarithms. If we now factor out R from each term: \[ \Delta_{mix}S=-R(n_{A}\ln \chi_{A}+n_{B}\ln \chi_{B}) \nonumber \] represents the equation for the entropy change of mixing. This equation is also commonly written with the total number of moles: \[ \Delta_{mix}S=-nR(\chi_A \ln \chi_A+\chi_B\ln \chi_B) \label{Final} \] where the total number of moles is \( n=n_A+n_B \) Notice that when the two gases will be mixed, their mole fraction will be less than one, making the term inside the parentheses negative, and thus the entropy of mixing will always be positive. This observation makes sense, because as you add a component to another for a two-component solution, the mole fraction of the other component will decrease, and the log of a number less than 1 is negative. Multiplied by the negative in the front of the equation gives a positive quantity. Equation \(\ref{Final}\) applies to both ideal solutions and ideal gases. | 3,607 | 2,160 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Mass_Action_Law |
Heat is energy flowing from a high temperature object to a low temperature object. When the two objects are at the same temperature, there is no net flow of energy or heat. That is why a covered cup of coffee will not be colder than or warmer than the room temperature after it has been in there for a few hours. This phenomenon is known as . In this example, we deal with the flow of energy. Equilibria happen in phase transitions. For example, if the temperature in a system containing a mixture of ice and water is uniformly 273.15 K, the net amount of ice formed and melted will be zero. The amount of liquid water will also remain constant, if no vapor escapes from the system. In this case, three phases, ice (solid) water (liquid), and vapor (gas) are in equilibrium with one another. Similarly, equilibrium can also be established between the vapor phase and the liquid at a particular temperature. Equilibrium conditions also exist between solid phase and vapor phases. These are phase equilibria. Chemical reactions may not be as complete as we have assumed in Stoichiometry calculations. For example, the following reactions are far short of completion. \[\ce{2 NO2 \rightleftharpoons N2O4}\] \[\ce{3 H2 + N2 \rightleftharpoons 2 NH3}\] \[\ce{H2O + CO \rightleftharpoons H2 + CO2}\] Let us consider only the first reaction in this case. At room temperature, it is impossible to have pure \(\ce{NO2}\) or \(\ce{N2O4}\). However, in a sealed tube (closed system), the ratio \[\ce{\dfrac{[N2O4]}{[NO2]^2}}\] is a constant. This phenomenon is known as . Such a law of nature is called the or . Of course, when conditions, such as pressure and temperature, change, a period of time is required for the system to establish an equilibrium. Before we introduce the mass action law, it is important for us to identify a or a in our discussion. The law provides an expression for a constant for all reversible reactions. For systems that are not at equilibrium yet, the ratio calculated from the mass action law is called a reaction quotient . The values of a closed system have a tendency to reach a limiting value called over time. A system has a tendency to reach an . To discuss equilibrium, we must define a system, which may be a cup of water, a balloon, a laboratory, a planet or a universe. Thus, for discussion purposes, we define an isolated portion of the universe under consideration as a , and anything outside of the system is called . When the system under consideration is isolated from its environment in such a way that there is no energy or mass transferred into or out of the system, the system is said to be an . In a isolated system, changes continue, but eventually there is no NET change over time; such a state is called an . For example, a glass containing water is an open system. Evaporation lets water molecules escape into the air by absorbing energy from the environment until the glass is empty. When covered and insulated it is a closed system. Water vapor in the space above water eventually reaches an . In fact, measuring of temperature itself requires the thermometer to be at the same state as the system it measures. We read the temperature of the thermometer when heat transfer between the thermometer and the system stops (at equilibrium). Equilibrium states are reached for physical as well as chemical reactions. Equilibrium is dynamic in the sense that changes continue, but the net change is zero. Heat transfer, vaporization, melting, and other phase changes are physical changes. These changes are reversible and you have already experienced them. Many chemical reactions are also reversible. For example \[ \underset{colorless}{\ce{N2O4}} \rightleftharpoons \underset{brown}{\ce{2 NO2}}\] \[\ce{N2 + 3 H2 \rightleftharpoons 2 NH3}\] are reversible chemical reactions. The law of mass action is universal, applicable under any circumstance. However, for reactions that are complete, the result may not be very useful. We introduce the mass action law by using a general chemical reaction equation in which reactants \(\ce{A}\) and \(\ce{B}\) react to give products \(\ce{C}\) and \(\ce{D}\). \[a\, \ce A + b\, \ce B \rightarrow c\, \ce C + d\, \ce D\] where are the coefficients for a balanced chemical equation. The states that if the system is at equilibrium at a given temperature, then the following ratio is a constant: \[\dfrac{[\ce C]^c [\ce D]^d}{[\ce A]^a [\ce B]^b} = K_{\ce{eq}}\] The square brackets "[ ]" around the chemical species represent their concentrations. This is the or . If the system is NOT at equilibrium, the ratio is different from the equilibrium constant. In such cases, the ratio is called a which is designated as . \[\dfrac{[\ce C]^c [\ce D]^d}{[\ce A]^a [\ce B]^b} = Q\] A system not at equilibrium tends to become at equilibrium, and the changes will cause changes in so that its value approaches the equilibrium constant, : \[Q \rightarrow K_{\ce{eq}}\] The mass action law gives us a general method to write the expression for the equilibrium constant of any reaction. At this stage, you should be able to write the equilibrium expression for any reaction equation. If you are not sure from the above general theory, here are some examples. It is more important for you to understand WHY the equilibrium constants are expressed this way than what the equilibrium expression is. Write the the equilibrium constant expression for the reaction equation: \[\ce{NH3 + HOAc \rightleftharpoons NH4+ + OAc-} \nonumber\] Hint \[\ce{\dfrac{[NH4+] [OAc- ]}{[NH3] [HOAc]}} = K\: (\textrm{unitless constant}) \nonumber\] For the ionization of an acid, \[\ce{H2SO4 \rightleftharpoons 2 H+ + SO4^2-} \nonumber\] what is the equilibrium constant expression? Hint The equilibrium constant is \[\ce{\dfrac{[H+]^2 [SO4^2- ]}{[H2SO4]}} = K \ce M^2 \nonumber\] where M = mol/L. Note the unit for . For the reaction equation: \(\ce{Cu^2+ + 6 NH3 \rightleftharpoons Cu(NH3)6^2+} \nonumber\) what is the equilibrium constant expression? Hint The expression is \(\ce{\dfrac{[Cu(NH3)6^2+]}{[Cu^2+] [NH3]^6}} = K \ce M^{-6} \nonumber\) The equilibrium constant depends on the written equation. Discuss the ionization of oxalic acid, \(\ce{H2C2O4}\), in two stages. Experimentally, it has been shown that \(\ce{H2C2O4 \rightleftharpoons H+ + HC2O4- }\tag{1} \) \(\ce{\dfrac{[HC2O4- ] [H+]}{[H2C2O4]}} = K_1 = \mathrm{0.059\: M} \nonumber\) The second ionization constant is much smaller: \(\ce{HC2O4- \rightleftharpoons H+ + C2O4^2- }\tag{2}\) \[\ce{\dfrac{[H+] [C2O4^2- ]}{[HC2O4- ]}} = K_2 = \mathrm{0.000064\: M} \nonumber\] The overall ionization can be obtained by adding (1) and (2) to give (3). \[\ce{H2C2O4 \rightleftharpoons 2 H+ + C2O4^2- }\tag{3} \nonumber\] and the equilibrium constant is \[\ce{\dfrac{[H+]^2 [C2O4^2- ]}{[H2C2O4]}} = K_{3}\: \ce M^2 \nonumber\] It is obvious that \[\begin{align*}
K_{3} &= K_1 \times K_2\\
&= 3.8\times10^{-6}\: \ce M^2
\end{align*}\] Please confirm the above obvious relationship to satisfy yourself. At some temperature, the equilibrium constant is 4.0 for the reaction \[\ce{CO_{\large{(g)}} + 3 H_{2\large{(g)}} \rightleftharpoons CH_{4\large{(g)}} + H2O_{\large{(g)}}} \nonumber\] What is the equilibrium constant expression and value for the reaction, \[\ce{CH_{4\large{(g)}} + H2O_{\large{(g)}} \rightleftharpoons CO_{\large{(g)}} + 3 H_{2\large{(g)}}}? \nonumber\] Hint \(\ce{\dfrac{[CO] [H2]^3}{[CH4] [H2O]}} = \dfrac{1}{4.0} = 0.25\) At some temperature, the equilibrium constant is 4.0 for the reaction \(\ce{CO_{\large{(g)}} + 3 H_{2\large{(g)}} \rightleftharpoons CH_{4\large{(g)}} + H2O_{\large{(g)}}}\). What is the equilibrium constant expression and value for the reaction, \(\ce{\dfrac{1}{3} CO_{\large{(g)}} + H_{2\large{(g)}} \rightleftharpoons \dfrac{1}{3} CH_{4\large{(g)}} + \dfrac{1}{3} H2O_{\large{(g)}}}\)? Hint \(\mathrm{\dfrac{[CH_4]^{1/3} [H_2O]^{1/3}}{[CO]^{1/3} [H_2]} = 4.0^{1/3} = 1.59}\) The application of the mass action law leads to the method to write the expression for the equilibrium constant. The law is given in a general form, and these examples should help you grasp the method. \(\ce{N2O4 \rightleftharpoons 2 NO2}\) \(\ce{N2O4 \rightleftharpoons 2 NO2}\) \(\ce{HCN + HCOO- \rightleftharpoons HCOOH + CN-}\)? \(K_{reverse} = \dfrac{1}{K_{forward}}\). \(C = \dfrac{n}{V}\) \(n R T = P V\) | 8,398 | 2,162 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Catalysis/Catalyst_Examples/Olefin_Polymerization_with_Ziegler-Natta_Catalyst |
The Ziegler-Natta (ZN) catalyst, named after two chemists: Karl Ziegler and Giulio Natta, is a powerful tool to polymerize α-olefins with high linearity and stereoselectivity (Figure 1). A typical ZN catalyst system usually contains two parts: a transition metal (Group IV metals, like Ti, Zr, Hf) compound and an organoaluminum compound (co-catalyst). The common examples of ZN catalyst systems include TiCl + Et Al and TiCl + AlEt Cl. In 1953, German chemist Karl Ziegler discovered a catalytic system able to polymerize ethylene into linear, high molecular weight polyethylene which conventional polymerization techniques could not make. The system contained a transition metal halide with a main group element alkyl compound (Figure 2). Following the catalytic design, Italian chemist Giulio Natta found that polymerization of α-olefins resulted in stereoregular structures, either syndiotactic or isotactic, depending on the catalyst used (Figure 3). Because of these important discoveries, Karl Ziegler and Giulio Natta shared the Nobel Prize in Chemistry in 1963. Traditionally, polymerization of α-olefins was done by radical polymerization (Figure 4). Problem with this technique was that the formation of undesired allylic radicals leaded to branched polymers. For example, radical polymerization of propene gived branched polymers with large molecular weight distribution. Also, radical polymerization had no control over stereochemistry. Linear unbranched polyethylene and stereoregulated polypropylene could not be fabricated by free radical polymerization. This technique largely limited the potential applications of these polymeric materials. The invention of ZN catalyst successfully addressed these two problems. The catalyst can give linear α-olefin polymers with high and controllable molecular weights. Moreover, it makes the fabrication of polymers with specific tacticity possible. By controlling the stereochemistry of products, either syndiotactic or isotactic polymers can be achieved. It is necessary to understand the catalyst’s structure before understanding how this catalyst system works. Herein, TiCl +AlEt catalyst system is taken as an example. The titanium chloride compound has a crystal structure in which each Ti atom is coordinated to 6 chlorine atoms. On the crystal surface, a Ti atom is surrounded by 5 chlorine atoms with one empty orbital to be filled. When Et Al comes in, it donates an ethyl group to Ti atom and the Al atom is coordinated to one of the chlorine atoms. Meanwhile, one chlorine atom from titanium is kicked out during this process. Thus, the catalyst system still has an empty orbital (Figure 5). The catalyst is activated by the coordination of AlEt to Ti atom. The polymerization reaction is initiated by forming alkene-metal complex. When a vinyl monomer like propylene comes to the active metal center, it can be coordinated to Ti atom by overlapping their orbitals. As shown in Figure 6, there is an empty dxy orbital and a filled dx -y orbital in Ti’s outermost shell (the other four orbitals are not shown here). The carbon-carbon double bond of alkene has a pi bond, which consists of a filled pi-bonding orbital and an empty pi-antibonding orbital. So, the alkene's pi-bonding orbital and the Ti's dxy orbital come together and share a pair of electrons. Once they're together, that Ti’s dx -y orbital comes mighty close to the pi-antibonding orbital, sharing another pair of electrons. The formed alkene-metal complex (1) then goes through electron shuffling, with several pairs of electrons shifting their positions: The pair of electrons from the carbon-carbon pi-bond shifts to form Ti-carbon bond, while the pair of electrons from the bond between Ti and AlEt3’ ethyl group shifts to form a bond between the ethyl group and the methyl-substituted carbon of propylene (Figure 7). After electron shuffling, Ti is back with an empty orbital again, needing electrons to fill it (2). When other propylene molecules come in, this process starts over and over, giving linear polypropylene (Figure 8). Termination is the final step of a chain-growth polymerization, forming “dead” polymers (desired products). Figure 9 illustrates several termination approaches developed with the aid of co-catalyst AlEt . Unlike the mechanism discussed above, there is also a competing mechanism proposed by Ivin and coworkers. They proposed that a 1,2-hydrogen shift occurs prior to monomer association, giving a carbene intermediate (Figure 10). To determine the actual mechanism, Grubbs conducted kinetic isotope effect (KIE) experiments. Due to their different weights, carbon-deuterium bond reacts slower than carbon-hydrogen bond (Figure 11). If such bonds are involved in the rate-determining step, the isotopic species should proceed in lower rate. In Grubbs’s experiment, deuterated ethylene and normal ethylene were catalyzed by Ziegler-Natta catalyst with 1:1 ratio (Figure 12). The result showed that H/D ratio in the resulting polymers was still 1:1.6 This indicated that no carbon-hydrogen bond cleavage or formation is involved in the rate determining step. Therefore, the mechanism proposed by Ivin was excluded. For propene polymerization, most ZN catalysts are highly regioselective, favoring 1,2-primary insertion [M-R + CH2=CH(Me)=M-CH2-CH(Me)(R)] (Figure 13), due to electronic and steric effects. Stereochemistry of polymers made from ZN-catalyst can be well regulated by rational design of ligands. By using different ligand system, either syndiotactic or isotactic polymers can be obtained (Figure 14). The relative stereochemistry of adjacent chiral centers within a macromolecule is defined as tacticity. Three kinds of stereochemistry are possible: isotactic, syndiotactic and atactic. In isotactic polymers, substituents are located on the same side of the polymer backbone, while substituents on syodiotactic polymers have alternative positions. In atactic polymers, substituents are placed randomly along the chain. Choice of ZN catalyst regulates the stereochemistry. We use propylene polymerization as an example here. Recall the mechanism section, a monomer approaches the metal center and forms a four-membered ring intermediate. The binding of a monomer to the reactive metal-carbon bond should occur in from the least hindered site. As shown in Figure 15, the trans-complex in which methyl on the growing chain is trans to the methyl group on the incoming monomer, should be energetically favored than the cis-complex, as the trans-complex experience less steric effect. Following the trans- complex, the methyl group on the newly added monomer is trans to that on the previous monomer. The step repeats so that a syndiotactic polypropylene is obtained (Figure 16a). However, if the metal center is coordinated with bulky ligands (e.g. –iBu, -Et2 groups), as denoted by Y in Figure 16b, the incoming monomer will adopt a cis-conformation to avoid serious steric effect with the bulky ligand. Thus, at the presence of bulky ligand, the propylene monomer is cis to the growing chain. This results in a isotactic product. ZN catalysts have provided a worldwide profitable industry with production of more than 160 billion pounds and creation of numerous positions. These products have been extensively applied in different areas, largely improving the quality of people’s life. They can catalyze α-olefins to produce various commercial polymers, like polyethylene, polypropylene and Polybutene-1. Polyethylene and polypropylene is reported to be the top two widely used synthetic plastic in the word. Among the polyethylene fabricated by ZN catalysts, there are three major classes: high density polyethylene (HDPE), linear low density polyethylene (LLDPE), and ultra-high molecular weight polyethylene (UHMWPE). HDPE, a linear homopolymer, is widely applied in garbage containers, detergent bottles and water pipes because of its high tensile strength. Compared with HDPE whose branching degree is quite low, LLDPE has many short branches. Its better toughness, flexibility and stress-cracking resistance makes it suitable for materials like cable coverings, bubble wrap and so on. UHMWPE is polyethylene with a molecular weight between 3.5 and 7.5 million. This material is extremely tough and chemically resistant. Therefore, it is often used to make gears and artificial joints. Compared to polyethylene, polypropylene has enhanced mechanical properties and thermal resistance because of the additional methyl group. Moreover, isotactic polypropylene is stiffer and more resistant to creep than atactic polypropylene. Polypropylene has a wide range of applications in clothing, medical plastics, food packing, and building construction. ZN catalysts are effective for polymerization of α-Olefins (ethylene, propylene) and some dienes (butadiene, isoprene). However, they don’t work for some other monomers, such as 1.2 disubstituted double bonds. Vinyl chloride cannot be polymerized by ZN catalyst either, because free radical vinyl polymerization is initiated during the reaction. Another situation that ZN catalysts don’t work is when the substrate is acrylate. The reason is that ZN catalysts often initiate anionic vinyl polymerization in those monomers. | 9,280 | 2,163 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/20%3A_Carbohydrates/20.02%3A_Classification_and_Occurrence_of_Carbohydrates |
The simple sugars, or , are the building blocks of carbohydrate chemistry. They are polyhydroxy aldehydes or ketones with five, six, seven, or eight carbon atoms that are classified appropriately as , , , or , respectively. They can be designated by more specific names, such as or , to denote the kind of carbonyl compound they represent. For example, and aldopentose is a five-carbon sugar with an aldehyde carbonyl; a ketohexose is a six-carbon sugar with a ketone carbonyl: However, it is important to keep in mind that the carbonyl groups of sugars usually are combined with one of the hydroxyl groups in the same molecule to form a cyclic hemiacetal or hemiketal. These structures once were written as follows, and considerable stretch of the imagination is needed to recognize that they represent oxacycloalkane ring systems: The saccharides have long and awkward names by the IUPAC system, consequently a highly specialized nomenclature system has been developed for carbohydrates. Because this system (and those like it for other natural products) is unlikely to be replaced by more systematic names, you will find it necessary to memorize some names and structures. It will help you to remember the meaning of names such as aldopentose and ketohexose, and to learn the names and details of the structures of glucose, fructose, and ribose. For the rest of the carbohydrates, the nonspecialist needs only to remember the kind of compounds that they are. The most abundant five-carbon sugars are \(L\)-arabinose, \(D\)-ribose, 2-deoxy-\(D\)-ribose,\(^1\) and \(D\)-xylose, which all are . Both the open-chain and cyclic structures of the \(D\)-aldoses up to \(\ce{C_6}\) are shown in Figure 20-1. The common six-carbon sugars (hexoses) are \(D\)-glucose, \(D\)-fructose, \(D\)-galactose, and \(D\)-mannose. They all are , except \(D\)-fructose, which is a . The structures of the ketoses up to \(\ce{C_6}\) are shown for reference in Figure 20-2. The occurrence and uses of the more important ketoses and aldoses are summarized in Table 20-1. \(^1\) means lacking a hydroxyl group, and 2-deoxyribose is simply ribose without an \(\ce{OH}\) group at the 2-carbon. and (1977) | 2,218 | 2,164 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/24%3A_Nuclear_Chemistry/24.07%3A_The_Origin_of_the_Elements |
The relative abundances of the elements in the known universe vary by more than 12 orders of magnitude. For the most part, these differences in abundance cannot be explained by differences in nuclear stability. Although the Fe nucleus is the most stable nucleus known, the most abundant element in the known universe is not iron, but hydrogen ( H), which accounts for about 90% of all atoms. In fact, H is the raw material from which all other elements are formed. In this section, we explain why H and He together account for at least 99% of all the atoms in the known universe. We also describe the nuclear reactions that take place in stars, which transform one nucleus into another and create all the naturally occurring elements. The relative abundances of the elements in the known universe and on Earth relative to silicon are shown in . The data are estimates based on the characteristic emission spectra of the elements in stars, the absorption spectra of matter in clouds of interstellar dust, and the approximate composition of Earth as measured by geologists. The data in illustrate two important points. First, except for hydrogen, the most abundant elements have atomic numbers. Not only is this consistent with the trends in nuclear stability discussed in , but it also suggests that heavier elements are formed by combining helium nuclei ( = 2). Second, the relative abundances of the elements in the known universe and on Earth are often very different, as indicated by the data in for some common elements. Some of these differences are easily explained. For example, nonmetals such as H, He, C, N, O, Ne, and Kr are much less abundant relative to silicon on Earth than they are in the rest of the universe. These elements are either noble gases (He, Ne, and Kr) or elements that form volatile hydrides, such as NH , CH , and H O. Because Earth’s gravity is not strong enough to hold such light substances in the atmosphere, these elements have been slowly diffusing into outer space ever since our planet was formed. Argon is an exception; it is relatively abundant on Earth compared with the other noble gases because it is continuously produced in rocks by the radioactive decay of isotopes such as K. In contrast, many metals, such as Al, Na, Fe, Ca, Mg, K, and Ti, are relatively abundant on Earth because they form nonvolatile compounds, such as oxides, that cannot escape into space. Other metals, however, are much abundant on Earth than in the universe; some examples are Ru and Ir. You may recall from that the anomalously high iridium content of a 66-million-year-old rock layer was a key finding in the development of the asteroid-impact theory for the extinction of the dinosaurs. This section explains some of the reasons for the great differences in abundances of the metallic elements. All the elements originally present on Earth (and on other planets) were synthesized from hydrogen and helium nuclei in the interiors of stars that have long since exploded and disappeared. Six of the most abundant elements in the universe (C, O, Ne, Mg, Si, and Fe) have nuclei that are integral multiples of the helium-4 nucleus, which suggests that helium-4 is the primary building block for heavier nuclei. Elements are synthesized in discrete stages during the lifetime of a star, and some steps occur only in the most massive stars known ( ). Initially, all stars are formed by the aggregation of interstellar “dust,” which is mostly hydrogen. As the cloud of dust slowly contracts due to gravitational attraction, its density eventually reaches about 100 g/cm , and the temperature increases to about 1.5 × 10 K, forming a dense plasma of ionized hydrogen nuclei. At this point, self-sustaining nuclear reactions begin, and the star “ignites,” creating a like our sun. In the first stage of its life, the star is powered by a series of nuclear fusion reactions that convert hydrogen to helium: \( _{1}^{1}H + \;_{1}^{1}H \rightarrow \;_{1}^{2}H + \;_{+1}^{\;\;0}\beta \tag{24.6.1} \)
\( _{1}^{2}H + _{1}^{1}H \rightarrow _{2}^{3}He + _{0}^{0}\gamma \) \( _{2}^{3}He + _{2}^{3}He \rightarrow _{2}^{4}He + 2 _{1}^{1}H \) The overall reaction is the conversion of four hydrogen nuclei to a helium-4 nucleus, which is accompanied by the release of two positrons, two γ rays, and a great deal of energy:
\( 4_{1}^{1}H \rightarrow _{2}^{4}He + 2_{+1}^{\;\;0}\beta + 2_{0}^{0}\gamma \tag{24.6.2} \) These reactions are responsible for most of the enormous amount of energy that is released as sunlight and solar heat. It takes several billion years, depending on the size of the star, to convert about 10% of the hydrogen to helium. Once large amounts of helium-4 have been formed, they become concentrated in the core of the star, which slowly becomes denser and hotter. At a temperature of about 2 × 10 K, the helium-4 nuclei begin to fuse, producing beryllium-8:
\( 2_{2}^{4}He \rightarrow _{4}^{8}Be \tag{24.6.3} \) Although beryllium-8 has both an even mass number and an even atomic number, its also has a low neutron-to-proton ratio (and other factors beyond the scope of this text) that makes it unstable; it decomposes in only about 10 s. Nonetheless, this is long enough for it to react with a third helium-4 nucleus to form carbon-12, which is very stable. Sequential reactions of carbon-12 with helium-4 produce the elements with even numbers of protons and neutrons up to magnesium-24:
\( _{4}^{8}Be \overset{_{2}^{4}He}{\rightarrow} \;_{6}^{12}C \overset{_{2}^{4}He}{\rightarrow} \;_{8}^{16}O \overset{_{2}^{4}He}{\rightarrow} \;_{10}^{20}Ne \overset{_{2}^{4}He}{\rightarrow} \;_{12}^{24}Mg \tag{24.6.4} \) So much energy is released by these reactions that it causes the surrounding mass of hydrogen to expand, producing a that is about 100 times larger than the original yellow star. As the star expands, heavier nuclei accumulate in its core, which contracts further to a density of about 50,000 g/cm , so the core becomes even hotter. At a temperature of about 7 × 10 K, carbon and oxygen nuclei undergo nuclear fusion reactions to produce sodium and silicon nuclei:
\( _{6}^{12}C + \;_{6}^{12}C \rightarrow \;_{11}^{23}Na + \;_{1}^{1}H \tag{24.6.5} \)
\( _{6}^{12}C + \;_{8}^{16}O \rightarrow \;_{14}^{28}Si + \;_{0}^{0}\gamma \tag{24.6.6} \) At these temperatures, carbon-12 reacts with helium-4 to initiate a series of reactions that produce more oxygen-16, neon-20, magnesium-24, and silicon-28, as well as heavier nuclides such as sulfur-32, argon-36, and calcium-40:
\( _{6}^{12}C \overset{_{2}^{4}He}{\rightarrow} \;_{8}^{16}O \overset{_{2}^{4}He}{\rightarrow} \;_{10}^{20}Ne \overset{_{2}^{4}He}{\rightarrow} \;_{12}^{24}Mg \overset{_{2}^{4}He}{\rightarrow} \;_{14}^{28}Si \overset{_{2}^{4}He}{\rightarrow} \;_{16}^{32}S \overset{_{2}^{4}He}{\rightarrow} \;_{18}^{36}Ar \overset{_{2}^{4}He}{\rightarrow} \;_{20}^{40}Ca \tag{24.6.7} \) The energy released by these reactions causes a further expansion of the star to form a , and the core temperature increases steadily. At a temperature of about 3 × 10 K, the nuclei that have been formed exchange protons and neutrons freely. This equilibration process forms heavier elements up to iron-56 and nickel-58, which have the most stable nuclei known. None of the processes described so far produces nuclei with > 28. All naturally occurring elements heavier than nickel are formed in the rare but spectacular cataclysmic explosions called ( ). When the fuel in the core of a very massive star has been consumed, its gravity causes it to collapse in about 1 s. As the core is compressed, the iron and nickel nuclei within it disintegrate to protons and neutrons, and many of the protons capture electrons to form neutrons. The resulting is a dark object that is so dense that atoms no longer exist. Simultaneously, the energy released by the collapse of the core causes the supernova to explode in what is arguably the single most violent event in the universe. The force of the explosion blows most of the star’s matter into space, creating a gigantic and rapidly expanding dust cloud, or ( ). During the extraordinarily short duration of this event, the concentration of neutrons is so great that multiple neutron-capture events occur, leading to the production of the heaviest elements and many of the less stable nuclides. Under these conditions, for example, an iron-56 nucleus can absorb as many as 64 neutrons, briefly forming an extraordinarily unstable iron isotope that can then undergo multiple rapid β-decay processes to produce tin-120:
\( _{26}^{56}Fe + \; 64_{0}^{1}n \rightarrow \;_{26}^{120}Fe \rightarrow \; _{50}^{120 }Sn + 24_{-1}^{\;\;0}\beta \tag{24.6.8} \) A view of the remains of Supernova 1987A, located in the Large Magellanic Cloud, showing the circular halo of expanding debris produced by the explosion. Multiple neutron-capture events occur during a supernova explosion, forming both the heaviest elements and many of the less stable nuclides. Although a supernova occurs only every few hundred years in a galaxy such as the Milky Way, these rare explosions provide the only conditions under which elements heavier than nickel can be formed. The force of the explosions distributes these elements throughout the galaxy surrounding the supernova, and eventually they are captured in the dust that condenses to form new stars. Based on its elemental composition, our sun is thought to be a second- or third-generation star. It contains a considerable amount of cosmic debris from the explosion of supernovas in the remote past. The reaction of two carbon-12 nuclei in a carbon-burning star can produce elements other than sodium. Write a balanced nuclear equation for the formation of reactant and product nuclides balanced nuclear equation Use conservation of mass and charge to determine the type of nuclear reaction that will convert the reactant to the indicated product. Write the balanced nuclear equation for the reaction. Exercise How many neutrons must an iron-56 nucleus absorb during a supernova explosion to produce an arsenic-75 nucleus? Write a balanced nuclear equation for the reaction. 19 neutrons; \( _{26}^{56}Fe + \; 19_{0}^{1}n \rightarrow \;_{26}^{75}Fe \rightarrow \; _{33}^{75 }As + 7_{-1}^{\;\;0}\beta \) By far the most abundant element in the universe is hydrogen. The fusion of hydrogen nuclei to form helium nuclei is the major process that fuels young stars such as the sun. Elements heavier than helium are formed from hydrogen and helium in the interiors of stars. Successive fusion reactions of helium nuclei at higher temperatures create elements with even numbers of protons and neutrons up to magnesium and then up to calcium. Eventually, the elements up to iron-56 and nickel-58 are formed by exchange processes at even higher temperatures. Heavier elements can only be made by a process that involves multiple neutron-capture events, which can occur only during the explosion of a supernova. Why do scientists believe that hydrogen is the building block of all other elements? Why do scientists believe that helium-4 is the building block of the heavier elements? How does a star produce such enormous amounts of heat and light? How are elements heavier than Ni formed? Propose an explanation for the observation that elements with even atomic numbers are more abundant than elements with odd atomic numbers. During the lifetime of a star, different reactions that form different elements are used to power the fusion furnace that keeps a star “lit.” Explain the different reactions that dominate in the different stages of a star’s life cycle and their effect on the temperature of the star. A line in a popular song from the 1960s by Joni Mitchell stated, “We are stardust….” Does this statement have any merit or is it just poetic? Justify your answer. If the laws of physics were different and the primary element in the universe were boron-11 ( = 5), what would be the next four most abundant elements? Propose nuclear reactions for their formation. The raw material for all elements with > 2 is helium ( = 2), and fusion of helium nuclei will always produce nuclei with an even number of protons. Write a balanced nuclear reaction for the formation of each isotope. At the end of a star’s life cycle, it can collapse, resulting in a supernova explosion that leads to the formation of heavy elements by multiple neutron-capture events. Write a balanced nuclear reaction for the formation of each isotope during such an explosion. When a star reaches middle age, helium-4 is converted to short-lived beryllium-8 (mass = 8.00530510 amu), which reacts with another helium-4 to produce carbon-12. How much energy is released in each reaction (in megaelectronvolts)? How many atoms of helium must be “burned” in this process to produce the same amount of energy obtained from the fusion of 1 mol of hydrogen atoms to give deuterium? | 12,960 | 2,165 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Lattices/Thermodynamics_of_Lattices/00%3A_Front_Matter/02%3A_InfoPage |
Instructors can adopt existing LibreTexts texts or Remix them to quickly build course-specific resources to meet the needs of their students. Unlike traditional textbooks, LibreTexts’ web based origins allow powerful integration of advanced features and new access texts to improve postsecondary education at all levels of higher learning by developing an Open Access Resource environment. The project currently consists of 14 independently operating and interconnected libraries that are constantly being integrated. and are supported by the Department of Education Open Textbook Pilot 1525057, and 1413739. Unless otherwise noted, LibreTexts content is licensed by . Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not ). and are supported by the Department of Education Open Textbook Pilot Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. . . | 1,070 | 2,166 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Surface_Science_(Nix)/03%3A_The_Langmuir_Isotherm/3.05%3A_Applications_-_Kinetics_of_Catalytic_Reactions |
It is possible to predict how the kinetics of certain heterogeneously-catalyzed reactions might vary with the partial pressures of the reactant gases above the catalyst surface by using the Langmuir isotherm expression for equilibrium surface coverages. Consider the surface decomposition of a molecule A, i.e. the process \[A_{(g)} \rightleftharpoons A_{(ads)} → Products\] Let us assume that: Under these circumstances, the molecules of A adsorbed on the surface are in equilibrium with those in the gas phase and we may predict the surface concentration of A from the Langmuir isotherm, i.e. \[θ = \dfrac{bP}{1 + bP}\] The rate of the surface decomposition (and hence of the reaction) is given by an expression of the form \[rate = k θ\] This is assuming that the decomposition of A(ads) occurs in a simple unimolecular elementary reaction step and that the kinetics are first order with respect to the surface concentration of this adsorbed intermediate). Substituting for the coverage, θ, gives us the required expression for the rate in terms of the pressure of gas above the surface \[rate = \dfrac{k b P}{1 + b P} \label{rate}\] It is useful to consider two extremes: This is the low pressure (or weak binding. i.e., small \(b\)) limit: under these conditions the steady state surface coverage, \(θ\), of the reactant molecule is very small. \[b P \ll 1\] then \[1 + bP \approx 1\] and Equation \(\ref{rate}\) can be simplified to \[rate \approx kbP\] Under this limiting case, the kinetics follow a first order reaction (with respect to the partial pressure of \(A\)) with an apparent first order rate constant \(k' = kb\). This is the high pressure (or strong binding, i.e., large \(b\)) limit: under these conditions the steady state surface coverage, \(θ\), of the reactant molecule is almost unity and \[bP \gg 1\] then \[1 + bP \approx bP\] and Equation \(\ref{rate}\) can be simplified to \[rate \approx k\] under this limiting case, the kinetics follow a zero order reaction (with respect to the partial pressure of \(A\)). The rate shows the same pressure variation as does the surface coverage, but this hardly surprising since it is directly proportional to θ. These two limiting cases can be identified in the general kinetics from Equation \(\ref{rate}\) in Figure 3.5.1. Consider a Langmuir-Hinshelwood reaction of the following type: \[A_{(g)} \rightleftharpoons A_{(ads)} \label{Eq2.1}\] \[B_{(g)} \rightleftharpoons B_{(ads)} \label{Eq2.2} \] \[A_{(ads)} + B_{(ads)} \overset{slow}{\longrightarrow} AB_{(ads)} \overset{fast}{\longrightarrow} AB_{(g)} \label{Eq2.3} \] We will further assume, as noted in the above scheme, that the surface reaction between the two adsorbed species (left side of Equation \(\ref{Eq2.3}\) is the rate determining step. If the two adsorbed molecules are mobile on the surface and freely intermix then the rate of the reaction will be given by the following rate expression for the surface combination step \[Rate = k θ_A θ_B\] For a single molecular adsorbate the surface coverage (as given by the standard Langmuir isotherm) is \[ θ = \dfrac{bP}{1 + bP}\] Where two molecules (\(A\) & \(B\) ) are competing for the same adsorption sites then the relevant expressions are (see derivation): \[\theta_A = \dfrac{b_AP_A}{1+b_AP_A + b_BP_B}\] and \[\theta_B = \dfrac{b_BP_B}{1+b_AP_A + b_BP_B}\] Substituting these into the rate expression gives: \[Rate = k \theta_A \theta_B = \dfrac{k b_AP_A b_BP_B }{( 1+b_AP_A + b_BP_B )^2}\] Once again, it is interesting to look at several extreme limits \[b_A P_A \ll 1\] and \[b_B P_B \ll 1\] In this limit that \(θ_A\) & \(θ_B\) are both very low, and \[rate → k b_A P_A b_B P_B = k' P_A P_B\] i.e. in both reactants \[b_A P_A \ll 1 \ll b_B P_B\] In this limit \(θ_A → 0\), \(θ_B → 1\), and \[Rate → \dfrac{k b_A P_A }{b_B P_B } = \dfrac{k' P_A}{P_B}\] i.e. in \(A\), but in \(B\) Clearly, depending upon the partial pressure and binding strength of the reactants, a given model for the reaction scheme can give rise to a variety of apparent kinetics: this highlights the dangers inherent in the reverse process - namely trying to use kinetic data to obtain information about the reaction mechanism. On precious metal surfaces (e.g. Pt), the \(CO\) oxidation reaction is generally believed to by a of the following type: \[CO_{(g)} \rightleftharpoons CO_{(ads)}\] \[O_{2 (g)} \rightleftharpoons 2 O_{(ads)}\] \[CO_{(ads)} + O_{(ads)} \overset{slow}{\longrightarrow} CO_{2 (ads)} \overset{fast}{\longrightarrow} CO_{2 (g)}\] As CO is comparatively weakly-bound to the surface, the desorption of this product molecule is relatively fast and in many circumstances it is the surface reaction between the two adsorbed species that is the rate determining step. If the two adsorbed molecules are assumed to be mobile on the surface and freely intermix then the rate of the reaction will be given by the following rate expression for the bimolecular surface combination step \[Rate = k \,θ_{CO}\, θ_O\] Where two such species (one of which is molecularly adsorbed, and the other dissociatively adsorbed) are competing for the same adsorption sites then the relevant expressions are (see derivation): \[\theta_{CO} = \dfrac{b_{CO}P_{CO}}{1+ \sqrt{b_OP_{O_2}} + b_{CO}P_{CO}}\] and \[\theta_{O} = \dfrac{ \sqrt{b_OP_{O_2}} }{1+ \sqrt{b_OP_{O_2}} + b_{CO}P_{CO}}\] Substituting these into the rate expression gives: \[rate = k \theta_{CO} \theta_O = \dfrac{ k b_{CO}P_{CO} \sqrt{b_OP_{O_2}} }{(1+ \sqrt{b_OP_{O_2}} + b_{CO}P_{CO})^2} \label{Ex1.1}\] Once again, it is interesting to look at certain limits. If the \(CO\) is much more strongly bound to the surface such that \[b_{CO}P_{CO} \gg 1 + \sqrt{b_OP_{O_2}}\] then \[1 + \sqrt{b_OP_{O_2}} + b_{CO}P_{CO} \approx b_{CO}P_{CO}\] and the Equation \(\ref{Ex1.1}\) simplifies to give \[rate \approx \dfrac{k \sqrt{b_OP_{O_2}} } {b_{CO}P_{CO}} = k' \dfrac{P^{1/2}_{O_2}}{P_{CO}}\] In this limit the kinetics are half-order with respect to the gas phase pressure of molecular oxygen, but negative order with respect to the \(CO\) partial pressure, i.e. \(CO\) acts as a poison (despite being a reactant) and increasing its pressure slows down the reaction. This is because the CO is so strongly bound to the surface that it blocks oxygen adsorbing, and without sufficient oxygen atoms on the surface the rate of reaction is reduced. | 6,388 | 2,167 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Acid_Base_Reactions/Hydrolysis |
A salt is formed between the reaction of an acid and a base. Usually, a neutral salt is formed when a strong acid and a strong base are neutralized in the reaction: \[\ce{H+ + OH- \rightleftharpoons H2O} \label{1}\] The bystander ions in an acid-base reaction form a salt solution. Most neutral salts consist of cations and anions listed in the table below. These ions have little tendency to react with water. Thus, salts consisting of these ions are neutral salts. For example: \(\ce{NaCl}\), \(\ce{KNO3}\), \(\ce{CaBr2}\), \(\ce{CsClO4}\) are neutral salts. When weak acids and bases react, the relative strength of the conjugated acid-base pair in the salt determines the pH of its solutions. The salt, or its solution, so formed can be acidic, neutral or basic. A salt formed between a strong acid and a weak base is an acid salt, for example \(\ce{NH4Cl}\). A salt formed between a weak acid and a strong base is a basic salt, for example \(\ce{NaCH3COO}\). These salts are acidic or basic due to their acidic or basic ions as shown in the Table \(\Page {1}\). A salt formed between a strong acid and a weak base is an acid salt. Ammonia is a weak base, and its salt with any strong acid gives a solution with a pH lower than 7. For example, let us consider the reaction: \[\ce{HCl + NH4OH \rightleftharpoons NH4+ + Cl- + H2O} \label{2}\] In the solution, the \(\ce{NH4+}\) ion reacts with water (called hydrolysis) according to the equation: \[\ce{NH4+ + H2O \rightleftharpoons NH3 + H3O+}. \label{3}\] The acidity constant can be derived from \(K_w\) and \(K_b\). \[\begin{align}
K_{\large\textrm a} &= \dfrac{\ce{[H3O+] [NH3]}}{\ce{[NH4+]}} \dfrac{\ce{[OH- ]}}{\ce{[OH- ]}}\\
&= \dfrac{K_{\large\textrm w}}{K_{\large\textrm b}}\\
&= \dfrac{1.00 \times 10^{-14}}{1.75 \times 10^{-5}} = 5.7 \times 10^{-10}
\end{align}\] What is the concentration of \(\ce{NH4+}\), \(\ce{NH3}\), and \(\ce{H+}\) in a 0.100 M \(\ce{NH4NO3}\) solution? Assume that \(\ce{[NH3]} = x\), then \(\ce{[H3O+]} = x\), and you write the concentration below the formula in the reaction: \(\begin{array}{ccccccc}
\ce{NH4+ &+ &H2O &\rightleftharpoons &NH3 &+ &H3O+}\\
0.100-x &&&&x &&x
\end{array}\) \(\begin{align}
K_{\large\textrm a} &= \textrm{5.7E-10}\\
&= \dfrac{x^2}{0.100-x}
\end{align}\) Since the concentration has a value much greater than , you may use \(\begin{align}
x &= (0.100\times\textrm{5.7E(-10)})^{1/2}\\
&= \textrm{7.5E-6}
\end{align}\) \(\begin{align}
\ce{[NH3]} &= \ce{[H+]} = x = \textrm{7.5E-6 M}\\
\ce{pH} &= -\log\textrm{7.5e-6} = 5.12
\end{align}\) \(\ce{[NH4+]} = \textrm{0.100 M}\) Since pH = 5.12, the contribution of \(\ce{[H+]}\) due to self ionization of water may therefore be neglected. A basic salt is formed between a weak acid and a strong base. The basicity is due to the hydrolysis of the conjugate base of the (weak) acid used in the neutralization reaction. For example, sodium acetate formed between the weak acetic acid and the strong base \(\ce{NaOH}\) is a basic salt. When the salt is dissolved, ionization takes place: \[\ce{NaAc \rightleftharpoons Na+ + Ac-} \label{4}\] In the presence of water, \(\ce{Ac-}\) undergoes hydrolysis: \[\ce{H2O + Ac- \rightleftharpoons HAc + OH-} \label{5}\] And the equilibrium constant for this reaction is of the conjugate base \(\ce{Ac-}\) of the acid \(\ce{HAc}\). Note the following equilibrium constants: Acetic acid (\(K_a=1.75 \times 10^{-5}\)) and Ammonia (\( K_b=1.75 \times 10^{-5}\))
\(\begin{align}
K_{\large\textrm b} &= \ce{\dfrac{[HAc] [OH- ]}{[Ac- ]}}\\
K_{\large\textrm b} &= \ce{\dfrac{[HAc] [OH- ]}{[Ac- ]} \dfrac{[H+]}{[H+]}}\\
K_{\large\textrm b} &= \ce{\dfrac{[HAc]}{[Ac- ,H+]} [OH- ,H+]}\\
&= \dfrac{K_{\large\textrm w}}{K_{\large\textrm a}}\\
&= \dfrac{\textrm{1.00e-14}}{\textrm{1.75e-5}} = \textrm{5.7e-10}
\end{align}\) Thus, \(K_{\large\ce a} K_{\large\ce b} = K_{\large\ce w}\) or \(\mathrm{p\mathit K_{\large a} + p\mathit K_{\large b} = 14}\) for a conjugate acid-base pair. Let us look at a numerical problem of this type. Calculate the \(\ce{[Na+]}\), \(\ce{[Ac- ]}\), \(\ce{[H+]}\) and \(\ce{[OH- ]}\) of a solution of 0.100 M \(\ce{NaAc}\) (at 298 K). ( = 1.8E-5) Let x represent \(\ce{[H+]}\), then \(\begin{array}{ccccccc}
\ce{H2O &+ &Ac- &\rightleftharpoons &HAc &+ &OH-}\\
&&0.100-x &&x &&x
\end{array}\) \(\dfrac{x^2}{0.100-x} = \dfrac{\textrm{1E-14}}{\textrm{1.8E-5}} = \textrm{5.6E-10}\) Solving for x results in \(\begin{align}
x &= \sqrt{0.100\times\textrm{5.6E-10}}\\
&= \textrm{7.5E-6}
\end{align}\) \(\ce{[OH- ]} = \ce{[HAc]} = \textrm{7.5E-6}\) \(\ce{[Na+]} = \textrm{0.100 F}\) This corresponds to a pH of 8.9 or \(\ce{[H+]} = \textrm{1.3E-9}\). Note that \(\dfrac{K_{\large\ce w}}{K_{\large\ce a}} = K_{\large\ce b}\) of \(\ce{Ac-}\), so that rather than may be given as data in this question. A salt formed between a weak acid and a weak base can be neutral, acidic, or basic depending on the relative strengths of the acid and base. Arrange the three salts according to their acidity. \(\ce{NH4CH3COO}\) (ammonium acetate), \(\ce{NH4CN}\) (ammonium cyanide), and \(\ce{NH4HC2O4}\) (ammonium oxalate). Solution ammonium oxalate -- acidic, \(K_{\large\ce a}(\ce o) > K_{\large\ce b}(\ce{NH3})\)
ammonium acetate -- neutral, \(K_{\large\ce a} = K_{\large\ce b}\)
ammonium cyanide -- basic, \(K_{\large\ce a}(\ce c) < K_{\large\ce b}(\ce{NH3})\) \(\ce{HS- \rightleftharpoons H+ + S^2-}\) \(\begin{array}{ccccccccccc} \ce{KCN &\rightarrow &K+ &+ &CN- &&&&&&}\\ \ce{&&& &CN- &+ &H2O &\rightleftharpoons &HCN &+ &OH-}\\ &&& &(0.100-x) &&&&x &&x \end{array}\) \(\begin{align}
x &= (0.100\times\textrm{1.5E-5})^{1/2}\\
&= \textrm{1.2E-3}\\
\ce{pOH} &= 2.9\\
\ce{pH} &= 11.1
\end{align}\) | 5,775 | 2,168 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Concepts_in_Biophysical_Chemistry_(Tokmakoff)/04%3A_Transport/15%3A_Passive_Transport/15.03%3A_Search_Times_in_Facilitated_Diffusion |
Consider a series of repetitive 1D and 3D diffusion cycles. The search time for a protein to find its target is \[ t_s = \sum^k_{i=1} \left( \tau_{1D,i}+\tau_{3D,i} \right) \nonumber \] where k is the number of cycles. If the genome has a length of M bases and the average number of bases scanned per cycle is n, the average number of cycles k̅ = M/n̅, and the average search time can be written as \[ \overline{t}_s = \dfrac{M}{\overline{n}} \left( \overline{\tau}_{1D,i}+\overline{\tau}_{3D,i} \right) \] \(\overline{\tau}\) is the mean search time during one cycle. If we assume that sliding occurs through normal 1D diffusion, then we expect that \(\overline{n} \propto \sqrt{D_{1D}\overline{\tau}_{1D}}\), where the diffusion constant is expressed in units of bp /s. More accurately, it is found that if you executed a random walk with an exponentially weighted distribution of search times: \[ \begin{aligned} P(\tau_{1D}) &= \overline{\tau}_{1D}^{\: -1}\exp (-\tau_{1D}/\overline{\tau}_{1D}) \\ \overline{n} &= \sqrt{4D_{1D}\overline{\tau}_{1D}} \end{aligned} \] \[ \overline{t}_s = \dfrac{M}{\sqrt{4D_{1D}\overline{\tau}_{1D}}}(\overline{\tau}_{1D} +\overline{\tau}_{3D} \nonumber \] Let’s calculate the optimal search time, t . In the limits that \(\overline{\tau}_1\) or \(\overline{\tau}_3\rightarrow 0\), you just have pure 1D or 3D diffusion, but this leads to suboptimal search times because a decrease in \(\overline{\tau}_{1D}\) or \(\overline{\tau}_{3D}\) leads to an increase in the other. To find the minimum search time we solve: \[ \dfrac{\partial \overline{t}_s}{\partial \tau_{1D}} = 0 \nonumber \] and find that t corresponds to the condition \[ \overline{\tau}_{1D} = \overline{\tau}_{3D} \nonumber \] Using this in eq. (15.3.1) we have \[ \begin{aligned} t_{opt} &= \dfrac{2M}{\overline{n}} \overline{\tau}_{3D} = M \sqrt{\dfrac{\overline{\tau}_{3D}}{D_{1D}}} \\ \overline{n}_{opt} &= \sqrt{4D_{1D}\overline{\tau}_{3D}} \end{aligned} \] Now let’s find out how much this 1D + 3D search process speeds up over the pure 1D or 3D search. \[ \overline{t}_{3D} = M\overline{\tau}_{3D} \nonumber \] Facilitated diffusion speeds up the search relative to pure 3D diffusion by a factor proportional to the average number of bases searched during the 1D sliding. \[ \dfrac{\overline{t}_{3D}}{(\overline{t}_s)_{opt}}=\dfrac{\overline{n}}{2} \nonumber \] \[ \begin{aligned} \overline{t}_1D &\approx \dfrac{M^2}{4D_{1D}} \\ \dfrac{\overline{\tau}_{1D}}{(\overline{t}_s)_{opt}} &= \dfrac{M}{4} \sqrt{\dfrac{1}{D_{1D}\tau_{1D}}} = \dfrac{M}{\overline{n}} \end{aligned} \] Facilitated diffusion speeds up the search over pure 1D diffusion by a factor or M/n̅. \[ \begin{aligned} &M \approx 5\mathrm{x}10^6 \: \text{bp} \\ &\overline{n} \approx 200- 500 \: \text{bp} \\ \text{Optimal facilitated diffusion is} \sim &10^2 \text{faster than 3D} \\ \sim &10^4 \text{faster than 1D} \end{aligned} \] What determines the diffusion coefficient for sliding and \(\overline{\tau}_1 \)? We need the non-specific protein interaction to be strong enough that it doesn’t dissociate too rapidly, but also weak enough that it can slide rapidly. To analyze this, we use a model in which the protein is diffusing on a modulated energy landscape looking for a low energy binding site. Model Calculating the mean first passage time to reach a target site at a distance of L base pairs from the original position yields \[ \overline{\tau}_{1D} = L^2\Delta \tau \left( 1+\dfrac{1}{2} \left( \dfrac{\sigma}{k_BT} \right)^2 \right)^{-1/2} e^{-7\sigma^2/4(k_BT)^2} \nonumber \] Which follows a diffusive form with a diffusion constant \[ D_{1D} = \dfrac{L^2}{2\overline{\tau}_{1D}}=\dfrac{1}{2\Delta \tau} \left( 1+\dfrac{1}{2} \left( \dfrac{\sigma}{k_BT} \right)^2 \right)^{1/2} e^{-7\sigma^2/4(k_BT)^2} \] Using this to find conditions for the fastest search time: \[ t_{opt} = \dfrac{M}{2} \sqrt{\dfrac{\pi \overline{\tau}_{3D}}{4D_{1D}}} \qquad \qquad \overline{n}_{opt} = \sqrt{\dfrac{16}{\pi}D_{1D}\overline{\tau}_{3D}} \qquad \qquad \overline{\tau}_{1D} = \overline{\tau}_{3D} \nonumber \] Speed: Fast speed \(\rightarrow \) fast search in 1D. From eq. (15.3.2), we see that \[ D_{1D} \propto \exp \left[ - \left( \dfrac{\sigma^2}{k_BT} \right) \right] \] With this strong dependence on σ, effective sliding with proper \(\overline{n}\) requires \[ \sigma < 2k_BT \nonumber \] Stability: On the other hand, we need to remain stably bound for proper recognition and activity. To estimate we argue that we want the equilibrium probability of having the protein bound at the target site be \(P_{eq} \approx 0.25\). If E is minimum energy of the binding site, and the probability of occupying the binding site is the following. First we can estimate that \[ E_0 \approx - \sigma \sqrt{2 \log M} \nonumber \] which suggests that for adequate binding: \[ \sigma > 5 k_BT \nonumber \] To account for these differences, a model has been proposed: \[ > \dfrac{\overline{n}}{\overline{\tau}_{1D}} \sim 10^4 s^{-1} \nonumber \] \[ \begin{aligned} \langle E_R \rangle &< \langle E_S \rangle \\ \sigma_R&>\sigma_S \end{aligned} \] The observation in eq. (15.3.3), relating the roughness of an energy landscape to an effective diffusion rate is quite general. If we are diffusing over a distance long enough that the corrugation of the energy landscape looks like Gaussian random noise with a standard deviation σ, we expect the effective diffusion coefficient to scale as \[ D_{eff} = D_0 \exp \left[ - \left( \dfrac{\sigma^2}{k_BT} \right) \right] \] where D is the diffusion constant in the absence of the energy roughness. To now there still is no definitive evidence for coupled 1D + 3D transport, although there is a lot of data now showing 1D sliding. These studies used flow to stretch DNA and followed the position of fluorescently labelled proteins as they diffused along the DNA. Austin: Lac Repression follow up \(\rightarrow \) observed D varies by many orders of magnitude. \[ \begin{aligned} D_{1D} &:10^2-10^5 \: \mathrm{nm}^2/\mathrm{s} \\ \overline{n} &\approx 500 \: \mathrm{nm} \end{aligned} \] Blainey and Xie: hOGG1 DNA repair protein: \[ \begin{aligned} \Delta G^{\dagger}_{slide} &\approx 0.5 \text{kcal/mol} \approx k_BT \\ D_{1D} &\sim 5\mathrm{x}10^6 \: \mathrm{bp}^2/\mathrm{s} \\ \overline{n} &\approx 440 \: \mathrm{bp} \end{aligned} \] \[ D_{1D} \qquad 10^6 - 10^7 \: \mathrm{bp}^2/\mathrm{s} \approx 10^{-1}-10^0 \: \mu \mathrm{m}^2/\mathrm{s} \nonumber \] ______________________________________ | 6,611 | 2,170 |
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All ionic compounds have numerous properties in common. Consequently, the ability to recognize an ionic compound from its formula will allow you to predict many of its properties. This is often possible in the case of a (one which contains only two elements), because formation of a binary ionic compound places quite severe restrictions on the elements involved. One element must be a metal and must have a very low ionization energy (see Figure for the IE of various metals). The other element must be a nonmetal and must have a very high electron affinity. In the following paragraphs, many references will be made to groups on the periodic table. Use the figure below for reference to keep track of the groups referred to in the paragraph. Even though metals in general have low ionization energies, not all of them are low enough to form binary ionic compounds with a large fraction of the nonmetals. Although it is impossible to draw an exact line of demarcation, a good working rule is that essentially all binary compounds involving metals from periodic group 1, group 2, group 3 (Sc, Y, Lu), and the lanthanoids will be ionic. (Hydrogen is not a metal and is, therefore, an exception to the rule for group 1. Beryllium, whose ionization energy of 899 kJ mol is quite high for a metal, also forms many binary compounds which are not ionic. Beryllium is the only exception to the rule from group 2.) The transition metals to the right of group 3 in the periodic table form numerous binary compounds which involve covalent bonding, so they cannot be included in our rule. The same is true of the metals in periodic groups 13, 14, and 15. The number of nonmetals with which a group 1, 2, 3, or lanthanoid metal can combine to form a binary ionic compound is even more limited than the number of appropriate metals. Such nonmetals are found mainly in periodic groups 16 and 17. The only other elements which form monatomic anions under normal circumstances are hydrogen (which forms H ions) and nitrogen (which forms N ions). In addition to combining with metals to form ionic compounds, all of the nonmetals can combine with other nonmetals to form covalent compounds as well. Therefore, presence of a particular nonmetal does not guarantee that a binary compound is ionic. It is necessary, however, for a group 16 or 17 nonmetal, nitrogen, or hydrogen to be present if a binary compound is to be classified as definitely ionic. Which of the following compounds can be identified as definitely ionic? Which are definitely not ionic? CuO HgBr H S CaO BaBr InF MgH B H BrCl According to the guidelines in the previous two paragraphs, only compounds containing metals from groups IA, IIA, and IIIB, or the lanthanoids are definitely ionic, as long as the metal is combined with an appropriate nonmetal. CaO, MgH and BaBr fall into this category. Compounds which do not contain a metallic element, such as B H , H S, and BrCl, cannot possibly be ionic. This leaves CuO, HgBr , and InF in the category of possiblity, but not definitely, ionic. | 3,073 | 2,171 |
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The approximation is one of the simplest approximations one can make to analyze a proposed mechanism to deduce the rate law it predicts. Simply stated, the rate determining step approximation says that a mechanism can proceed no faster than its slowest step. So, for example, if the reaction \[ A + B \rightarrow C \nonumber \] is proposed to follow the mechanism \[ \underbrace{A +A \xrightarrow{k_1} A_2}_{\text{slow}} \nonumber \] \[ \underbrace{ A_2 \xrightarrow{k_2} C + A}_{\text{fast}} \nonumber \] the rate determining step approximation suggests that the rate (expressed in terms of the appearance of product \(C\)) should be determined by the slow initial step, and so the rate law will be \[\dfrac{[C]}{dt} = k_1[A]^2 \nonumber \] matching the order of the rate law to the molecularity of the slow step. Conversely, if the reaction mechanism is proposed as \[ \underbrace{A \xrightarrow{k_1} A^*}_{\text{slow}} \nonumber \] \[ \underbrace{ A^* + B \xrightarrow{k_2} C}_{\text{fast}} \nonumber \] the rate determining step approximation suggests that the rate of the reaction should be \[\dfrac{[C]}{dt} = k_1[A] \nonumber \] again, with the order of the rate law matching the molecularity of the rate determining step. | 1,244 | 2,172 |
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We can also use an energy level diagram to show the relative content of energy. The energy content of \(\mathrm{H_{2\large{(g)}} + 0.5\, O_2}\) is 285.8 kJ higher than a mole of water, \(\ce{H2O}\). Oil, gas, and food are often called energy by the news media, but more precisely they are sources of (chemical) energy -- energy stored in chemicals with a potential to be released in a chemical reaction. The released energy performs work or causes physical and chemical changes. It is obvious that the amount of energy released in a chemical reaction is related to the amount of reactants. For example, when the amount is doubled, so is the amount of energy released. \(\mathrm{2 H_{2\large{(g)}} + O_2 \rightarrow 2 H_2O_{\large{(l)}}}, \hspace{20px} dH = \mathrm{-571.6\: kJ/mol}\) Example 1 shows the calculation when the amount of reactants is only a fraction of a mole. | 895 | 2,173 |
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The structures of most binary compounds can be described using the packing schemes we have just discussed for metals. To do so, we generally focus on the arrangement in space of the largest species present. In ionic solids, this generally means the anions, which are usually arranged in a simple cubic, bcc, fcc, or hcp lattice. Often, however, the anion lattices are not truly “close packed”; because the cations are large enough to prop them apart somewhat, the anions are not actually in contact with one another. In ionic compounds, the cations usually occupy the “holes” between the anions, thus balancing the negative charge. The ratio of cations to anions within a unit cell is required to achieve electrical neutrality and corresponds to the bulk stoichiometry of the compound. As shown in part (a) in Figure 12.8, a simple cubic lattice of anions contains only one kind of hole, located in the center of the unit cell. Because this hole is equidistant from all eight atoms at the corners of the unit cell, it is called a cubic hole. An atom or ion in a cubic hole therefore has a coordination number of 8. Many ionic compounds with relatively large cations and a 1:1 cation:anion ratio have this structure, which is called the cesium chloride structure (Figure 12.9) because CsCl is a common example.Solid-state chemists tend to describe the structures of new compounds in terms of the structure of a well-known reference compound. Hence you will often read statements such as “Compound X possesses the cesium chloride (or sodium chloride, etc.) structure” to describe the structure of compound X. Notice in Figure 12.9 that the z = 0 and the z = 1.0 planes are always the same. This is because the z = 1.0 plane of one unit cell becomes the z = 0 plane of the succeeding one. The unit cell in CsCl contains a single Cs+ ion as well as 8×18Cl−=1Cl− ion, for an overall stoichiometry of CsCl. The cesium chloride structure is most common for ionic substances with relatively large cations, in which the ratio of the radius of the cation to the radius of the anion is in the range shown in Table 12.2. Relationship between the Cation:Anion Radius Ratio and the Site Occupied by the Cations In contrast, a face-centered cubic (fcc) array of atoms or anions contains two types of holes: octahedral holes, one in the center of the unit cell plus a shared one in the middle of each edge (part (b) in Figure 12.8), and tetrahedral holes, located between an atom at a corner and the three atoms at the centers of the adjacent faces (part (c) in Figure 12.8). As shown in Table 12.2, the ratio of the radius of the cation to the radius of the anion is the most important determinant of whether cations occupy the cubic holes in a cubic anion lattice or the octahedral or tetrahedral holes in an fcc lattice of anions. Very large cations occupy cubic holes in a cubic anion lattice, cations of intermediate size tend to occupy the octahedral holes in an fcc anion lattice, and relatively small cations tend to occupy the tetrahedral holes in an fcc anion lattice. In general, larger cations have higher coordination numbers than small cations. The most common structure based on a fcc lattice is the sodium chloride structure (Figure 12.10), which contains an fcc array of Cl− ions with Na+ ions in all the octahedral holes. We can understand the sodium chloride structure by recognizing that filling all the octahedral holes in an fcc lattice of Cl− ions with Na+ ions gives a total of 4 Cl− ions (one on each face gives 6× \({1 \over 2} \)=3 plus one on each corner gives 8× \({1 \over 8} \)=1, for a total of 4) and 4 Na+ ions (one on each edge gives 12×\({1 \over 4}\)=3 plus one in the middle, for a total of 4). The result is an electrically neutral unit cell and a stoichiometry of NaCl. As shown in Figure 12.10, the Na+ ions in the sodium chloride structure also form an fcc lattice. The sodium chloride structure is favored for substances with two atoms or ions in a 1:1 ratio and in which the ratio of the radius of the cation to the radius of the anion is between 0.414 and 0.732. It is observed in many compounds, including MgO and TiC. The structure shown in Figure 12.11 is called the zinc blende structure, from the common name of the mineral ZnS. It results when the cation in a substance with a 1:1 cation:anion ratio is much smaller than the anion (if the cation:anion radius ratio is less than about 0.414). For example, ZnS contains an fcc lattice of S ions, and the cation:anion radius ratio is only about 0.40, so we predict that the cation would occupy either a tetrahedral hole or an octahedral hole. In fact, the relatively small Zn cations occupy the tetrahedral holes in the lattice. If all 8 tetrahedral holes in the unit cell were occupied by Zn ions, however, the unit cell would contain 4 S and 8 Zn ions, giving a formula of Zn S and a net charge of +4 per unit cell. Consequently, the Zn ions occupy every other tetrahedral hole, as shown in Figure 12.11, giving a total of 4 Zn and 4 S ions per unit cell and a formula of ZnS. The zinc blende structure results in a coordination number of 4 for each Zn ion and a tetrahedral arrangement of the four S ions around each Zn ion. a. If all the tetrahedral holes in an fcc lattice of anions are occupied by cations, what is the stoichiometry of the resulting compound? b. Use the ionic radii given in Figure 7.9 to identify a plausible oxygen-containing compound with this stoichiometry and structure. : lattice, occupancy of tetrahedral holes, and ionic radii : stoichiometry and identity : : a Figure 12.8 shows that the tetrahedral holes in an fcc unit cell of anions are located entirely within the unit cell, for a total of eight (one near each corner). Because the tetrahedral holes are located entirely within the unit cell, there are eight cations per unit cell. We calculated previously that an fcc unit cell of anions contains a total of four anions per unit cell. The stoichiometry of the compound is therefore M Y or, reduced to the smallest whole numbers, M Y. b. The M Y stoichiometry is consistent with a lattice composed of M ions and Y ions. If the anion is O (ionic radius 140 pm), we need a monocation with a radius no larger than about 140 × 0.414 = 58 pm to fit into the tetrahedral holes. According to Figure 7.9, none of the monocations has such a small radius; therefore, the most likely possibility is Li at 76 pm. Thus we expect Li O to have a structure that is an fcc array of O anions with Li cations in all the tetrahedral holes. If only half the octahedral holes in an fcc lattice of anions are filled by cations, what is the stoichiometry of the resulting compound? : MX ; an example of such a compound is cadmium chloride (CdCl ), in which the empty cation sites form planes running through the crystal. We examine only one other structure of the many that are known, the perovskite structure. Perovskite is the generic name for oxides with two different kinds of metal and have the general formula MM′O , such as CaTiO . The structure is a body-centered cubic (bcc) array of two metal ions, with one M (Ca in this case) located at the corners of the cube, and the other M′ (in this case Ti) in the centers of the cube. The oxides are in the centers of the square faces (part (a) in Figure 12.12). The stoichiometry predicted from the unit cell shown in part (a) in Figure 12.12 agrees with the general formula; each unit cell contains 8× \({1 \over 8}\) =1 Ca, 1 Ti, and 6× \({1 \over 2} \) =3 O atoms. The Ti and Ca atoms have coordination numbers of 6 and 12, respectively. We will return to the perovskite structure when we discuss high-temperature superconductors in Section 12.7. As you learned in Chapter 6, the wavelengths of x-rays are approximately the same magnitude as the distances between atoms in molecules or ions. Consequently, x-rays are a useful tool for obtaining information about the structures of crystalline substances. In a technique called x-ray diffraction, a beam of x-rays is aimed at a sample of a crystalline material, and the x-rays are diffracted by layers of atoms in the crystalline lattice (part (a) in Figure 12.13). When the beam strikes photographic film, it produces an x-ray diffraction pattern, which consists of dark spots on a light background (part (b) in Figure 12.13). In 1912, the German physicist Max von Laue (1879–1960; Nobel Prize in Physics, 1914) predicted that x-rays should be diffracted by crystals, and his prediction was rapidly confirmed. Within a year, two British physicists, William Henry Bragg (1862–1942) and his son, William Lawrence Bragg (1890–1972), had worked out the mathematics that allows x-ray diffraction to be used to measure interatomic distances in crystals. The Braggs shared the Nobel Prize in Physics in 1915, when the son was only 25 years old. Virtually everything we know today about the detailed structures of solids and molecules in solids is due to the x-ray diffraction technique. Recall from Chapter 6 that two waves that are in phase interfere constructively, thus reinforcing each other and generating a wave with a greater amplitude. In contrast, two waves that are out of phase interfere destructively, effectively canceling each other. When x-rays interact with the components of a crystalline lattice, they are scattered by the electron clouds associated with each atom. As shown in Figure 12.5, Figure 12.7, and Figure 12.8, the atoms in crystalline solids are typically arranged in planes. 12.14 illustrates how two adjacent planes of atoms can scatter x-rays in a way that results in constructive interference. If two x-rays that are initially in phase are diffracted by two planes of atoms separated by a distance d, the lower beam travels the extra distance indicated by the lines BC and CD. The angle of incidence, designated as θ, is the angle between the x-ray beam and the planes in the crystal. Because BC = CD = d sin θ, the extra distance that the lower beam in Figure 12.14 must travel compared with the upper beam is 2d sin θ. For these two x-rays to arrive at a detector in phase, the extra distance traveled must be an integral multiple n of the wavelength λ: \[2d \sin \theta = n\lambda \tag{12.1}\] Equation 12.1 is the Bragg equation. The structures of crystalline substances with both small molecules and ions or very large biological molecules, with molecular masses in excess of 100,000 amu, can now be determined accurately and routinely using x-ray diffraction and the Bragg equation. Example 4 illustrates how to use the Bragg equation to calculate the distance between planes of atoms in crystals. X-rays from a copper x-ray tube (λ = 1.54062 Å or 154.062 pm)In x-ray diffraction, the angstrom (Å) is generally used as the unit of wavelength. are diffracted at an angle of 10.89° from a sample of crystalline gold. Assuming that n = 1, what is the distance between the planes that gives rise to this reflection? Give your answer in angstroms and picometers to four significant figures. : wavelength, diffraction angle, and number of wavelengths : distance between planes : Substitute the given values into the Bragg equation and solve to obtain the distance between planes. : We are given n, θ, and λ and asked to solve for d, so this is a straightforward application of the Bragg equation. For an answer in angstroms, we do not even have to convert units. Solving the Bragg equation for d gives \[ d = {n \lambda \over 2 \sin \theta } \] \[ d = {(1) (1.54062 \, \overset {0}{A} ) \over 2 \, sin \, 10.89^0 } = 4.077 \, \overset {0}{A} = 407.7 \, pm \] X-rays from a molybdenum x-ray tube (λ = 0.709300 Å) are diffracted at an angle of 7.11° from a sample of metallic iron. Assuming that n = 1, what is the distance between the planes that gives rise to this reflection? Give your answer in angstroms and picometers to three significant figures. : 2.87 Å or 287 pm (corresponding to the edge length of the bcc unit cell of elemental iron) The structures of most binary compounds are dictated by the packing arrangement of the largest species present (the anions), with the smaller species (the cations) occupying appropriately sized holes in the anion lattice. A simple cubic lattice of anions contains a single cubic hole in the center of the unit cell. Placing a cation in the cubic hole results in the cesium chloride structure, with a 1:1 cation:anion ratio and a coordination number of 8 for both the cation and the anion. An fcc array of atoms or ions contains both octahedral holes and tetrahedral holes. If the octahedral holes in an fcc lattice of anions are filled with cations, the result is a sodium chloride structure. It also has a 1:1 cation:anion ratio, and each ion has a coordination number of 6. Occupation of half the tetrahedral holes by cations results in the zinc blende structure, with a 1:1 cation:anion ratio and a coordination number of 4 for the cations. More complex structures are possible if there are more than two kinds of atoms in a solid. One example is the perovskite structure, in which the two metal ions form an alternating bcc array with the anions in the centers of the square faces. Because the wavelength of x-ray radiation is comparable to the interatomic distances in most solids, x-ray diffraction can be used to provide information about the structures of crystalline solids. X-rays diffracted from different planes of atoms in a solid reinforce one another if they are in phase, which occurs only if the extra distance they travel corresponds to an integral number of wavelengths. This relationship is described by the Bragg equation: 2d sin θ = nλ. \[2d \sin \theta = n\lambda \tag{12.1}\] 1. Using circles or spheres, sketch a unit cell containing an octahedral hole. Which of the basic structural types possess octahedral holes? If an ion were placed in an octahedral hole, what would its coordination number be? 2. Using circles or spheres, sketch a unit cell containing a tetrahedral hole. Which of the basic structural types possess tetrahedral holes? If an ion were placed in a tetrahedral hole, what would its coordination number be? 3. How many octahedral holes are there in each unit cell of the sodium chloride structure? Potassium fluoride contains an fcc lattice of F ions that is identical to the arrangement of Cl ions in the sodium chloride structure. Do you expect K ions to occupy the tetrahedral or octahedral holes in the fcc lattice of F ions? 4. The unit cell of cesium chloride consists of a cubic array of chloride ions with a cesium ion in the center. Why then is cesium chloride described as having a simple cubic structure rather than a bcc structure? The unit cell of iron also consists of a cubic array of iron atoms with an iron atom in the center of the cube. Is this a bcc or a simple cubic unit cell? Explain your answer. 5. Why are x-rays used to determine the structure of crystalline materials? Could gamma rays also be used to determine crystalline structures? Why or why not? 6. X-rays are higher in energy than most other forms of electromagnetic radiation, including visible light. Why can’t you use visible light to determine the structure of a crystalline material? 7. When x-rays interact with the atoms in a crystal lattice, what relationship between the distances between planes of atoms in the crystal structure and the wavelength of the x-rays results in the scattered x-rays being exactly in phase with one another? What difference in structure between amorphous materials and crystalline materials makes it difficult to determine the structures of amorphous materials by x-ray diffraction? 8. It is possible to use different x-ray sources to generate x-rays with different wavelengths. Use the Bragg equation to predict how the diffraction angle would change if a molybdenum x-ray source (x-ray wavelength = 70.93 pm) were used instead of a copper source (x-ray wavelength = 154.1 pm). 9. Based on the Bragg equation, if crystal A has larger spacing in its diffraction pattern than crystal B, what conclusion can you draw about the spacing between layers of atoms in A compared with B? 1. Thallium bromide crystallizes in the cesium chloride structure. This bcc structure contains a Tl ion in the center of the cube with Br ions at the corners. Sketch an alternative unit cell for this compound. 2. Potassium fluoride has a lattice identical to that of sodium chloride. The potassium ions occupy octahedral holes in an fcc lattice of fluoride ions. Propose an alternative unit cell that can also represent the structure of KF. 3. Calcium fluoride is used to fluoridate drinking water to promote dental health. Crystalline CaF (d = 3.1805 g/cm ) has a structure in which calcium ions are located at each corner and the middle of each edge of the unit cell, which contains eight fluoride ions per unit cell. The length of the edge of this unit cell is 5.463 Å. Use this information to determine Avogadro’s number. 4. Zinc and oxygen form a compound that is used as both a semiconductor and a paint pigment. This compound has the following structure: What is the empirical formula of this compound? 5. Here are two representations of the perovskite structure: Are they identical? What is the empirical formula corresponding to each representation? 6. The salt MX has a cubic close-packed (ccp) structure in which all the tetrahedral holes are filled by anions. What is the coordination number of M? of X? 7. A compound has a structure based on simple cubic packing of the anions, and the cations occupy half of the cubic holes. What is the empirical formula of this compound? What is the coordination number of the cation? 8. Barium and fluoride form a compound that crystallizes in the fluorite structure, in which the fluoride ions occupy all the tetrahedral holes in a ccp array of barium ions. This particular compound is used in embalming fluid. What is its empirical formula? 9. Cadmium chloride is used in paints as a yellow pigment. Is the following structure consistent with an empirical formula of CdCl ? If not, what is the empirical formula of the structure shown? 10. Use the information in the following table to decide whether the cation will occupy a tetrahedral hole, an octahedral hole, or a cubic hole in each case. 11. Calculate the angle of diffraction when x-rays from a copper tube (λ = 154 pm) are diffracted by planes of atoms parallel to the faces of the cubic unit cell for Mg (260 pm), Zn (247 pm), and Ni (216 pm). The length on one edge of the unit cell is given in parentheses; assume first-order diffraction (n = 1). 12. If x-rays from a copper target (λ = 154 pm) are scattered at an angle of 17.23° by a sample of Mg, what is the distance (in picometers) between the planes responsible for this diffraction? How does this distance compare with that in a sample of Ni for which θ = 20.88°? 3. d = 3.1805 g/cm ; Avogadro’s number = 6.023 × 10 mol 5. Both have same stoichiometry, CaTiO 7. Stoichiometry is MX ; coordination number of cations is 8 9. No, the structure shown has an empirical formula of Cd Cl . 11. Mg: 17.2°, Zn: 18.2°, Ni: 20.9° | 19,147 | 2,174 |
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The heat capacity (\(C\)) of a substance is a measure of how much heat is required to raise the temperature of that substance by one degree Kelvin. For a simple molecular gas, the molecules can simultaneously store kinetic energy in the translational, vibrational, and rotational motions associated with the individual molecules. In this case, the heat capacity of the substance can be broken down into translational, vibrational, and rotational contributions; \[ C = C_{trans} + C_{vib} + C_{rot}\] Monoatomic crystalline solids represent a much simpler case. Einstein proposed a simple model for such substances whereby the atoms only have vibrational energy (each atom can vibrate in three perpendicular directions around its lattice position). Specifically, the ‘Einstein Solid Model’ assumes that the atoms act like three-dimensional harmonic oscillators (with the vibrational motion of each atom in each perpendicular dimension entirely independent). Statistical mechanics provides a relatively simple expression for the constant volume molar heat capacity (\(C_{V,m}\)) of a one-dimensional harmonic oscillator \[ C_{V,m}^{1-D} = R \left( \dfrac{\Theta_v }{T} \right)^2 \left( \dfrac{e^{-\Theta_v/2T} }{1- e^{-\Theta_v/T}} \right) ^2 \label{1}\] where \(R\) is the universal gas constant, \(T\) is absolute temperature, and \(Θ_v\) is called the ‘characteristic vibrational temperature’ of the oscillator and depends on the vibrational frequency (\(ν\)) according to \[ \Theta_v = \dfrac{hv}{k} \label{2}\] with \(h\) representing Plank’s constant and \(k\) representing Boltzmann’s constant. Since the vibrations in each dimension are assumed to be independent, the expression for the constant volume molar heat capacity of a 'three-dimensional' Einstein Solid is obtained by simply multiplying Equation \ref{1} by three; \[ C_{V,m}^{3-D} = 3R \left( \dfrac{\Theta_v }{T} \right)^2 \left( \dfrac{e^{-\Theta_v/2T} }{1- e^{-\Theta_v/T}} \right) ^2 \label{3}\] The temperature variation of the heat capacity of most metallic solids is well described by Equation \ref{3}. Furthermore, plots of Equation \ref{3} as a function of temperature for metals with widely varying vibrational frequencies reveal that the heat capacity always approaches the same asymptotic limit of \(3R\) at high temperatures. Stated another way, at high temperatures \[ \lim_{T \to \infty} \left[ \left( \dfrac{\Theta_v }{T} \right)^2 \left( \dfrac{e^{-\Theta_v/2T} }{1- e^{-\Theta_v/T}} \right) ^2 \right] = 1 \label{4}\] and Equation \ref{3} reduces to \[ \lim_{T \to \infty} \left[ C_{V,m}^{3D} \right] = 3R \label{5}\] (You will be asked to verify this result in the exercise below). According to Equation \ref{5}, the molar heat capacities of metallic solids should approach 24.9 J/(K mol) at high temperatures, regardless of the identity of the metal. The vibrational frequencies of most metallic solids are usually small enough so that \(Θ_v\) lies considerably below room temperature (\(Θ_v \ll 298\, K\)). For these substances, the limits implied by Equations \ref{4} and \ref{5} are well approximated even at room temperature, leading to the result that \(C_{v,m} = 24.9\, J/(K·mol)\) for most metals at room temperature. In the early 1800s, two French scientists by the names of Pierre Louis Dulong and Alexis Therese Petit empirically discovered the same remarkable result. The Dulong-Petit Law is normally expressed in terms of the specific heat capacity (\(C_s\)) and the molar mass (\(M\)) of the metal \[C_s M = C_{V,m} \approx 25 (J\, K^{-1} \, mol^{-1}) \label{6}\] where \(C_s\) represents how much heat is required to raise the temperature of 'one gram' of that substance by one degree Kelvin. Dulong and Petit, as well as other scientists of their time, used this famous relationship as a means of establishing more accurate values for the atomic weight of metallic elements (by instead measuring the specific heat capacity of the element and using the Dulong-Petit relationship, which is a relatively simple method of establishing weights in comparison to the more disputable gravimetric methods that were being used at the time to establish the equivalent weights of elements). In the exercise below, you will look up the specific heat capacities of a number of elements that exist as simple monoatomic solids at room temperature and assess the accuracy of the Dulong-Petit law. Consult the CRC Handbook of Chemistry and Physics (CRC Press: Boca Raton, FL) and compile a table of specific heat capacities for a large number of elements that are known to exist as monoatomic solids at room temperature. Also look up and record the molar mass of these elements. The elements that you consider should be restricted to those appearing in groups 1-14 of the periodic table. Make sure you generate a fairly large list which includes a number of elements that are normally considered as metallic in character (such as copper, iron, sodium, lithium, gold, platinum, barium, and aluminum), but also some non-metallic elements that are nonetheless monoatomic isotropic solids (such as carbon-diamond, beryllium, boron, and silicon). Heat capacities that are usually reported in the literature are not actual constant volume heat capacities (\(C_v\)), but are instead constant pressure heat capacities (\(C_p\)). Fortunately, \(C_p\) and \(C_v\) are essentially equal for simple solids (within the level of precision that we consider in this exercise), and you can assume that the values from the CRC Handbook represent \(C_s\). | 5,531 | 2,175 |
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For a mole of ideal gas we can use the gas law to integrate volume over pressure and we get \[ΔG_{molar} = RT \ln \left(\dfrac{P_2}{P_1}\right) \nonumber \] It is customary to identify one of the pressures (P } with the of and use the plimsoll to indicate the fact that we are referring to a standard state by writing: \[G_{molar}(P) = G^o_{molar} + RT \ln \left(\dfrac{P}{1}\right)=G^o_{molar} + RT \ln[P] \nonumber \] The fact that we are making the function intensive (per mole) is usually indicated by putting a bar over the \(G\) symbol, although this is often omitted for G For solids the volume does not change very much with pressure (the \(κ\) is small), so can assume it more or less constant: \[G(P_{final})=G(P_{initial})+ \int VdP \text{(from init to final)} ≈ G(P_{initial})+ V \int dP \text{(from init to final)}=G(P_{initial})+ VΔP \nonumber \] \[ \frac { G } { T } = \frac { H } { T } - S \nonumber \] Take the derivative under constant pressure of each side to get \[ \left( \frac { \partial G / T } { \partial T } \right) _ { P } = - \frac { H } { T ^ { 2 } } + \frac { 1 } { T } \left( \frac { \partial H } { \partial T } \right) _ { P } - \left( \frac { \partial S } { \partial T } \right) _ { P } \nonumber \] We make use of the relationship between \(C_p\) and \(H\) and \(C_p\) and \(S\) \[\begin{align} \left( \frac { \partial G / T } { \partial T } \right) _ { P } &= - \frac { H } { T ^ { 2 } } + \cancel{ \frac { C _ { P } } { T }} - \cancel{\frac { C _ { P } } { T }} \\ &= - \frac { H } { T ^ { 2 } } \end{align} \nonumber \] We said before that \(S\) is a first order derivative of \(G\). As you can see from this derivation the enthalpy \(H\) is also a first order derivative, albeit not of \(G\) itself, but of \(G/T\). \[\left( \frac { \partial \Delta G / T } { \partial T } \right) _ { P } = - \frac { \triangle H } { T ^ { 2 } } \nonumber \] The last step in the derivation simply takes the step before twice -say for the \(G\) and \(H\) at the begin and end of a - and subtracts the two identical equations leading to a \(Δ\) symbol. In this differential form the Gibbs-Helmholtz equation can be applied to any process. If heat capacities are know from 0 K we could determine both enthalpy and entropy by integration: \[S(T) = S(0) + \int_0^T \dfrac{C_p}{T} dT \nonumber \] \[H(T) = H(0) + \int_0^T C_p\; dT \nonumber \] As we have seen we must be careful at phase transitions such as melting or vaporization. At these points the curves are and the derivative \(C_p\) is undefined. \[H ( T ) = H ( 0 ) + \int_0^{T_{fus}} C_p(T)_{solid} dT + \Delta H_{fus} + \int_{T_{fus}}^{T_{boil}} C_p(T)_{liquid} dT + \Delta H_{vap} + etc. \label{Hcurve} \] \[\begin{align} S ( T ) &= S ( 0 ) + \int_0^{T_{fus}} \dfrac{C_p(T)_{solid}}{T} dT + \Delta S_{fus} + \int_{T_{fus}}^{T_{boil}} \dfrac{C_p(T)_{liquid}}{T} dT + \Delta S_{vap} + etc. \\[4pt] &= S(0) + \int_0^{T_{fus}} \dfrac{C_p(T)_{solid}}{T} dT + \dfrac{\Delta H_{fus}}{T_{fus}} + \int_{T_{fus}}^{T_{boil}} \dfrac{C_p(T)_{liquid}}{T} dT + \dfrac{\Delta H_{vap}}{T_{boil}} + etc. \label{Scurve} \end{align} \] with \(H(T=0)= \text{undefined}\) and \(S(T=0)=0\) from the third law of thermodynamics. We also discussed the fact that the third law allows us to define \(S(0)\) as zero in most cases. For the enthalpy we cannot do that so that our curve is with respect to an undefined zero point. We really should plot \(H(T) - H(0)\) and leave \(H(0)\) undefined. Because the Gibbs free energy \(G= H-TS\) we can also construct a curve for \(G\) as a function of temperature, simply by combining the \(H\) and the \(S\) curves (Equations \ref{Hcurve} and \ref{Scurve}): \[ G(T) = H(T) - TS(T) \nonumber \] Interestingly, if we do so, the at the phase transition points will for \(G\) because at these points \(Δ_{trs}H = T_{trs}Δ_{trs}S\). Therefore, \(G\) is continuous. The \(H(0)\) problem does not disappear so that once again our curve is subject to an arbitrary offset in the y-direction. The best thing we can do is plot the quantity \(G(T) - H(0)\) and leave the offset \(H(0)\) undefined. We have seen above that the derivative of \(G\) with temperature is \(-S\). As entropy is always positive, this means that the G-curve is always with temperature. It also means that although the curve is even at the phase transitions, the slope of the G curve is not, because the derivative \(-S\) makes a jump there. Fig. 22.7 in the book shows an example of such a curve for benzene. Note the kinks in the curve at the mp and the boiling point. | 4,573 | 2,177 |
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When a solute dissolves, its individual atoms, molecules, or ions interact with the solvent, become solvated, and are able to diffuse independently throughout the solution (part (a) in Figure 13.4). This is not, however, a unidirectional process. If the molecule or ion happens to collide with the surface of a particle of the undissolved solute, it may adhere to the particle in a process called crystallization. Dissolution and crystallization continue as long as excess solid is present, resulting in a dynamic equilibrium analogous to the equilibrium that maintains the vapor pressure of a liquid. We can represent these opposing processes as follows: \[ \text{solute} + \text{solvent} \ce{<=>[\ce{crystallization},\ce{dissolution}]} \text{solution} \tag{13.4} \] Although the terms precipitation and crystallization are both used to describe the separation of solid solute from a solution, crystallization refers to the formation of a solid with a well-defined crystalline structure, whereas precipitation refers to the formation of any solid phase, often one with very small particles. | 1,103 | 2,178 |
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How does pressure affect enthalpy \(H\)? As we showed above we have the following relations of first and second order for \(G\) \[\left( \dfrac{\partial G}{\partial T} \right)_P = -S \nonumber \] \[ \left( \dfrac{\partial G}{\partial P} \right)_T = -V \nonumber \] \[ -\left (\dfrac{\partial S}{\partial P }\right)_T = \left (\dfrac{\partial V}{\partial T} \right)_P \nonumber \] We also know that by definition: \[G = H - TS \label{def} \] Consider an isothermal change in pressure, so taking the partial derivative of each side of Equation \(\ref{def}\), we get: \[ \left( \dfrac{\partial G}{\partial P}\right)_T = \left( \dfrac{\partial H}{ \partial P}\right)_T -T \left( \dfrac{\partial S}{\partial P}\right)_T \nonumber \] \[ \left( \dfrac{\partial H}{\partial P}\right)_T = V -T \left( \dfrac{\partial V}{\partial T}\right)_P \label{Eq12} \] For an ideal gas \[\dfrac{\partial V}{\partial T} = \dfrac{nR}{P} \nonumber \] so Equation \(\ref{Eq12}\) becomes \[ \left( \dfrac{\partial H}{\partial P}\right)_T = V - T \left( \dfrac{nR}{P}\right) = 0 \nonumber \] As we can see for an ideal gas, there is no dependence of \(H\) on \(P\). | 1,170 | 2,179 |
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A very important property of a substance is how compressible it is. Gases are very compressible, so when subjected to high pressures, their volumes decrease significantly (think Boyle’s Law!) Solids and liquids however are not as compressible. However, they are not entirely incompressible! High pressure will lead to a decrease in volume, even if it is only slight. And, of course, different substances are more compressible than others. To quantify just how compressible substances are, it is necessary to define the property. The is defined by the fractional differential change in volume due to a change in pressure. \[ \kappa_T \equiv - \dfrac{1}{V} \left( \dfrac{\partial V}{\partial p} \right)_T \label{compress} \] The negative sign is important in order to keep the value of \(\kappa_T\) positive, since an increase in pressure will lead to a decrease in volume. The \(1/V\) term is needed to make the property intensive so that it can be tabulated in a useful manner. Another very important property of a substance is how its volume will respond to changes in temperature. Again, gases respond profoundly to changes in temperature (think Charles’ Law!) whereas solids and liquid will have more modest (but not negligible) responses to changes in temperature. (For example, If mercury or alcohol didn’t expand with increasing temperature, we wouldn’t be able to use those substances in thermometers.) The definition of the (or sometimes called the expansion coefficient) is \[ \alpha \equiv \dfrac{1}{V} \left( \dfrac{\partial V}{\partial T} \right)_p \label{expand} \] As was the case with the compressibility factor, the \(1/V\) term is needed to make the property intensive, and thus able to be tabulated in a useful fashion. In the case of expansion, volume tends to increase with increasing temperature, so the partial derivative is positive. Consider a system that is described by three variables, and for which one can write a mathematical constraint on the variables \[F(x, y, z) = 0 \nonumber \] Under these circumstances, one can specify the state of the system varying only two parameters independently because the third parameter will have a fixed value. As such one could define two functions: \(z(x, y)\) and \(y(x,z)\). This allows one to write the total differentials for \(dz\) and \(dy\) as follows \[dz = \left( \dfrac{\partial z}{\partial x} \right)_y dx + \left( \dfrac{\partial z}{\partial y} \right)_x dy \label{eq5} \] and \[dy= \left( \dfrac{\partial y}{\partial x} \right)_z dx + \left( \dfrac{\partial y}{\partial z} \right)_x dz \label{eq6} \] Substituting the Equation \ref{eq6} expression into Equation \ref{eq5}: \[ \begin{align} dz &= \left( \dfrac{\partial z}{\partial x} \right)_y dx + \left( \dfrac{\partial z}{\partial y} \right)_x \left[ \left( \dfrac{\partial y}{\partial x} \right)_z dx + \left( \dfrac{\partial y}{\partial z} \right)_x dz \right] \\[4pt] &= \left( \dfrac{\partial z}{\partial x} \right)_y dx + \left( \dfrac{\partial z}{\partial y} \right)_x \left( \dfrac{\partial y}{\partial x} \right)_z dx + \left( \dfrac{\partial z}{\partial y} \right)_x \left( \dfrac{\partial y}{\partial z} \right)_x dz \label{eq7} \end{align} \] If the system undergoes a change following a pathway where \(x\) is held constant (\(dx = 0\)), this expression simplifies to \[dz = \left( \dfrac{\partial z}{\partial y} \right)_x \left( \dfrac{\partial y}{\partial z} \right)_x dz \nonumber \] And so for changes for which \(dz \neq 0\), \[\left( \dfrac{\partial z}{\partial y} \right)_x = \dfrac{1}{\left( \dfrac{\partial y}{\partial z} \right)_x } \nonumber \] This is very convenient in the manipulation of partial derivatives. But it can also be derived in a straight-forward, albeit less rigorous, manner. Begin by writing the total differential for \(z(x,y)\) (Equation \ref{eq5}): \[dz = \left( \dfrac{\partial z}{\partial x} \right)_y dx + \left( \dfrac{\partial z}{\partial y} \right)_x dy \nonumber \] Now, divide both sides by \(dz\) and constrain to constant \(x\). \[\left.\dfrac{dz}{dz} \right\rvert_{x}= \left( \dfrac{\partial z}{\partial x} \right)_y \left.\dfrac{dx}{dz} \right\rvert_{x} + \left( \dfrac{\partial z}{\partial y} \right)_x \left.\dfrac{dy}{dz} \right\rvert_{x} \label{eq10} \] Noting that \[\left.\dfrac{dz}{dz} \right\rvert_{x} =1 \nonumber \] \[ \left.\dfrac{dx}{dz} \right\rvert_{x} = 0 \nonumber \] and \[\left.\dfrac{dy}{dz} \right\rvert_{x} = \left( \dfrac{\partial y}{\partial z} \right)_{x} \nonumber \] Equation \ref{eq10} becomes \[ 1= \left( \dfrac{\partial z}{\partial y} \right)_z \left( \dfrac{\partial y}{\partial z} \right)_x \nonumber \] or \[ \left( \dfrac{\partial z}{\partial y} \right)_z = \dfrac{1}{\left( \dfrac{\partial y}{\partial z} \right)_x} \nonumber \] This “formal” method of partial derivative manipulation is convenient and useful, although it is not mathematically rigorous. However, it does work for the kind of partial derivatives encountered in thermodynamics because the variables are and the differentials are . This alternative derivation follow the initial steps in the derivation above to Equation \ref{eq7}: \[dz = \left( \dfrac{\partial z}{\partial x} \right)_y dx + \left( \dfrac{\partial z}{\partial y} \right)_x \left( \dfrac{\partial y}{\partial x} \right)_z dx + \left( \dfrac{\partial z}{\partial y} \right)_x \left( \dfrac{\partial y}{\partial z} \right)_x dz \nonumber \] If the system undergoes a change following a pathway where \(z\) is held constant (\(dz = 0\)), this expression simplifies to \[0 = \left( \dfrac{\partial z}{\partial x} \right)_y dy + \left( \dfrac{\partial z}{\partial y} \right)_x \left( \dfrac{\partial y}{\partial x} \right)_z dx \nonumber \] And so for and changes in which \(dx \neq 0\) \[\left( \dfrac{\partial z}{\partial x} \right)_y = - \left( \dfrac{\partial z}{\partial y} \right)_x \left( \dfrac{\partial y}{\partial x} \right)_z \nonumber \] This is very convenient in the manipulation of partial derivatives. But it can also be derived in a straight-forward, albeit less rigorous, manner. As with the derivation above, we wegin by writing the total differential of \(z(x,y)\) \[dz = \left( \dfrac{\partial z}{\partial x} \right)_y dx + \left( \dfrac{\partial z}{\partial y} \right)_x dy \nonumber \] Now, divide both sides by \(dx\) and constrain to constant \(z\). \[\left.\dfrac{dz}{dx} \right\rvert_{z}= \left( \dfrac{\partial z}{\partial x} \right)_y \left.\dfrac{dx}{dx} \right\rvert_{z} + \left( \dfrac{\partial z}{\partial y} \right)_x \left.\dfrac{dy}{dx} \right\rvert_{z} \label{eq21} \] Note that \[\left.\dfrac{dz}{dx} \right\rvert_{z} =0 \nonumber \] \[ \left.\dfrac{dx}{dx} \right\rvert_{z} =1 \nonumber \] and \[\left.\dfrac{dy}{dx} \right\rvert_{z} = \left( \dfrac{\partial y}{\partial x} \right)_{z} \nonumber \] Equation \ref{eq21} becomes \[ 0 = \left( \dfrac{\partial z}{\partial x} \right)_y + \left( \dfrac{\partial z}{\partial y} \right)_x \left( \dfrac{\partial y}{\partial x} \right)_{z} \nonumber \] which is easily rearranged to \[ \left( \dfrac{\partial z}{\partial x} \right)_y = - \left( \dfrac{\partial z}{\partial y} \right)_x \left( \dfrac{\partial y}{\partial x} \right)_{z} \nonumber \] This type of transformation is very convenient, and will be used often in the manipulation of partial derivatives in thermodynamics. Derive an expression for \[\dfrac{\alpha}{\kappa_T}. \label{e1} \] in terms of derivatives of thermodynamic functions using the definitions in Equations \ref{compress} and \ref{expand}. Substituting Equations \ref{compress} and \ref{expand} into the Equation \ref{e1} \[\dfrac{\alpha}{\kappa_T}= \dfrac{\dfrac{1}{V} \left( \dfrac{\partial V}{\partial T} \right)_p}{- \dfrac{1}{V} \left( \dfrac{\partial V}{\partial p} \right)_T} \nonumber \] Simplifying (canceling the \(1/V\) terms and using transformation Type I to invert the partial derivative in the denominator) yields \[\dfrac{\alpha}{\kappa_T} = - \left( \dfrac{\partial V}{\partial T} \right)_p \left( \dfrac{\partial p}{\partial V} \right)_T \nonumber \] Applying Transformation Type II give the final result: \[ \dfrac{\alpha}{\kappa_T} = \left( \dfrac{\partial p}{\partial T} \right)_V \nonumber \] | 8,216 | 2,180 |
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Two families of metalloproteins are excellent catalysts for the disproportionation of superoxide (Reaction 5.95). \[2O_{2}^{-} + 2 H^{+} \xrightarrow{SOD} O_{2} + H_{2}O_{2} \tag{5.95}\] These are (1) the copper-zinc superoxide dismutases, CuZnSOD, found in almost all eukaryotic cells and a very few prokaryotes, and (2) the manganese and iron superoxide dismutases, MnSOD and FeSOD, the former found in the mitochondria of eukaryotic cells, and both found in many prokaryotes. Recent studies of bacterial and yeast mutants that were engineered to contain no superoxide dismutases demonstrated that the cells were unusually sensitive to dioxygen and that the sensitivity to dioxygen was relieved when an SOD gene was reintroduced into the cells. These results indicate that the superoxide dismutase enzymes playa critical role in dioxygen metabolism, but they do not define the chemical agent responsible for dioxygen toxicity (see Section III). Several transition-metal complexes have been observed to catalyze superoxide disproportionation; in fact, aqueous copper ion, Cu , is an excellent SOD catalyst, comparable in activity to CuZnSOD itself! Free aqueous Cu would not itself be suitable for use as an SOD , however, because it is too toxic (see Section III) and because it binds too strongly to a large variety of cellular components and thus would not be present as the free ion. (Most forms of complexed cupric ion show much less superoxide dismutase activity than the free ion.) Aside from aqueous copper ion, few other complexes are as effective as the SOD enzymes. Two mechanisms (Reactions 5.96 to 5.99) have been proposed for catalysis of superoxide disproportionation by metal complexes and metalloenzymes. Mechanism I: $$M^{n+} + O_{2}^{-} \rightarrow M^{(n-1)+} + O_{2} \tag{5.96}\] \[M^{(n-1)+} + O_{2}^{-} \rightarrow M^{n+}(O_{2}^{2-}) \xrightarrow{2H^{+}} M^{n+} + H_{2}O_{2} \tag{5.97}\] Mechanism II: $$M^{n+} + O_{2}^{-} \rightarrow M^{n+} (O_{2}^{-}) \tag{5.98}\] \[M^{n+} (O_{2}^{-}) + O_{2}^{-} \rightarrow M^{n+}(O_{2}^{2-}) \xrightarrow{2 H^{+}} M^{n} + H_{2}O_{2} \tag{5.99}\] \[+O_{2}\] In Mechanism I, which is favored for the SOD enzymes and most redox-active metal complexes with SOD activity, superoxide reduces the metal ion in the first step, and then the reduced metal ion is reoxidized by another superoxide, presumably via a metal-peroxo complex intermediate. In Mechanism II, which is proposed for nonredox metal complexes but may be operating in other situations as well, the metal ion is never reduced, but instead forms a superoxo complex, which is reduced to a peroxo complex by a second superoxide ion. In both mechanisms, the peroxo ligands are protonated and dissociate to give hydrogen peroxide. Analogues for each of the separate steps of Reactions (5.96) to (5.99) have been observed in reactions of superoxide with transition-metal complexes, thereby establishing the feasibility of both mechanisms. For example, superoxide was shown to reduce Cu phen) to give Cu (phen) (phen = 1,10-phenanthroline), a reaction analogous to Reaction (5.96). On the other hand, superoxide reacts with Cu (tet b) to form a superoxo complex (a reaction analogous to Reaction 5.98), presumably because Cu (tet b) is not easily reduced to the cuprous state, because the ligand cannot adjust to the tetrahedral geometry that Cu prefers. \(\tag{5.100}\) Reaction of superoxide with a reduced metal-ion complex to give oxidation of the complex and release of hydrogen peroxide (analogous to Reaction 5.97) has been observed in the reaction of Fe EDTA with superoxide. Reduction of a Co superoxo complex by free superoxide to give a peroxo complex (analogous to Reaction 5.99) has also been observed. If a metal complex can be reduced by superoxide and if its reduced form can be oxidized by superoxide, both at rates competitive with superoxide disproportionation, the complex can probably act as an SOD by Mechanism I. Mechanism II has been proposed to account for the apparent catalysis of superoxide disproportionation by Lewis acidic nonredox-active metal ions under certain conditions. However, this mechanism should probably be considered possible for redox metal ions and the SOD enzymes as well. It is difficult to distinguish the two mechanisms for redox-active metal ions and the SOD enzymes unless the reduced form of the catalyst is observed directly as an intermediate in the reaction. So far it has not been possible to observe this intermediate in the SOD enzymes or the metal complexes. The x-ray crystal structure of the oxidized form of CuZnSOD from bovine erythrocytes shows a protein consisting of two identical subunits held together almost entirely by hydrophobic interactions. Each subunit consists of a flattened cylindrical barrel of \(\beta\)-pleated sheet from which three external loops of irregular structure extend (Figure 5.15). The metal-binding region of the protein binds Cu and Zn in close proximity to each other, bridged by the imidazolate ring of a histidyI side chain. Figure 5.16 represents the metal-binding region. The Cu ion is coordinated to four histidyl imidazoles and a water in a highly distorted square-pyramidal geometry with water at the apical position. The Zn ion is coordinated to three histidyl imidazoles (including the one shared with copper) and an aspartyl carboxylate group, forming a distorted tetrahedral geometry around the metal ion. One of the most unusual aspects of the structure of this enzyme is the occurrence of the bridging imidazolate ligand, which holds the copper and zinc ions 6 Å apart. Such a configuration is not unusual for imidazole complexes of metal ions, which sometimes form long polymeric imidazolate-bridged structures. \(\tag{5.101}\) However, no other imidazolate-bridged bi- or polymetallic metalloprotein has yet been identified. The role of the zinc ion in CuZnSOD appears to be primarily structural. There is no evidence that water, anions, or other potential ligands can bind to the zinc, so it is highly unlikely that superoxide could interact with that site. Moreover, removal of zinc under conditions where the copper ion remains bound to the copper site does not significantly diminish the SOD activity of the enzyme. However, such removal does result in a diminished thermal stability, i.e., the zinc-depleted protein denatures at a lower temperature than the native protein, supporting the hypothesis that the role of the zinc is primarily structural in nature. The copper site is clearly the site of primary interaction of superoxide with the protein. The x-ray structure shows that the copper ion lies at the bottom of a narrow channel that is large enough to admit only water, small anions, and similarly small ligands (Figure 5.17). In the lining of the channel is the positively charged side chain of an arginine residue, 5 Å away from the copper ion and situated in such a position that it could interact with superoxide and other anions when they bind to copper. Near the mouth of the channel, at the surface of the protein, are two positively charged lysine residues, which are believed to play a role in attracting anions and guiding them into the channel. Chemical modification of these lysine or arginine residues substantially diminishes the SOD activity, supporting their role in the mechanism of reaction with superoxide. The x-ray structural results described above apply only to the oxidized form of the protein, i.e., the form containing Cu . The reduced form of the enzyme containing Cu is also stable and fully active as an SOD. If, as is likely, the mechanism of CuZnSOD-catalyzed superoxide disproportionation is Mechanism I (Reactions 5.96-5.97), the structure of the reduced form is of critical importance in understanding the enzymatic mechanism. Unfortunately, that structure is not yet available. The mechanism of superoxide disproportionation catalyzed by CuZnSOD is generally believed to go by Mechanism I (Reactions 5.96-5.97), i.e., reduction of Cu to Cu by superoxide with the release of dioxygen, followed by reoxidation of Cu to Cu by a second superoxide with the release of HO or H O . The protonation of peroxide dianion, O , prior to its release from the enzyme is required, because peroxide dianion is highly basic and thus too unstable to be released in its unprotonated form. The source of the proton that protonates peroxide in the enzymatic mechanism is the subject of some interest. Reduction of the oxidized protein has been shown to be accompanied by the uptake of one proton per subunit. That proton is believed to protonate the bridging imidazolate in association with the breaking of the bridge upon reduction of the copper. Derivatives with Co substituted for Zn at the native zinc site have been used to follow the process of reduction of the oxidized Cu form to the reduced Cu form. The Co in the zinc site does not change oxidation state, but acts instead as a spectroscopic probe of changes occurring at the native zinc-binding site. Upon reduction (Reaction 5.102), the visible absorption band due to Co shifts in a manner consistent with a change occurring in the ligand environment of Co . The resulting spectrum of the derivative containing Cu in the copper site and Co in the zinc site is very similar to the spectrum of the derivative in which the copper site is empty and the zinc site contains Co . This result suggests strongly that the imidazolate bridge is cleaved and protonated and that the resulting imidazole ligand is retained in the coordination sphere of Co (Reaction 5.102). \(\tag{5.102}\) The same proton is thus an attractive possibility for protonation of peroxide as it is formed in the enzymatic mechanism (Reactions 5.103 and 5.104). \(\tag{5.103}\) \(\tag{5.104}\) Attractive as this picture appears, there are several uncertainties about it. For example, the turnover of the enzyme may be too fast for protonation and deprotonation of the bridging histidine to occur. Moreover, the mechanism proposed would require the presence of a metal ion at the zinc site to hold the imidazole in place and to regulate the pK of the proton being transferred. The observation that removal of zinc gives a derivative with almost full SOD activity is thus surprising and may also cast some doubt on this mechanism. Other criticisms of this mechanism have been recently summarized. Studies of CuZnSOD derivatives prepared by site-directed mutagenesis are also providing interesting results concerning the SOD mechanism. For example, it has been shown that mutagenized derivatives of human CuZnSOD with major differences in copper-site geometry relative to the wild-type enzyme may nonetheless remain fully active. Studies of these and similar derivatives should provide considerable insight into the mechanism of reaction of CuZnSOD with superoxide. Studies of the interaction of CuZnSOD and its metal-substituted derivatives with anions have been useful in predicting the behavior of the protein in its reactions with its substrate, the superoxide anion, O . Cyanide, azide, cyanate, and thiocyanate bind to the copper ion, causing dissociation of a histidyl ligand and the water ligand from the copper. Phosphate also binds to the enzyme at a position close to the Cu center, but it apparently does not bind directly to it as a ligand. Chemical modification of Arg-141 with phenylglyoxal blocks the interaction of phosphate with the enzyme, suggesting that this positively charged residue is the site of interaction with phosphate. Electrostatic calculations of the charges on the CuZnSOD protein suggest that superoxide and other anions entering into the vicinity of the protein will be drawn toward and into the channel leading down to the copper site by the distribution of positive charges on the surface of the protein, the positively charged lysines at the mouth of the active-site cavity, and the positively charged arginine and copper ion within the active-site region. Some of the anions studied, e.g., CN , F , N ,and phosphate, have been shown to inhibit the SOD activity of the enzyme. The source of the inhibition is generally assumed to be competition with superoxide for binding to the copper, but it may sometimes result from a shift in the redox potential of copper, which is known to occur sometimes when an anion binds to copper. In the example described above, studies of a metal-substituted derivative helped in the evaluation of mechanistic possibilities for the enzymatic reaction. In addition, studies of such derivatives have provided useful information about the environment of the metal-ion binding sites. For example, metal-ion-substituted derivatives of CuZnSOD have been prepared with Cu , Cu , Zn , Ag , Ni , or Co bound to the native copper site, and with Zn , Cu , Cu , Co , Hg , Cd , Ni , or Ag bound to the native zinc site. The SOD activities of these derivatives are interesting; only those derivatives with copper in the copper site have a high degree of SOD activity, whereas the nature of the metal ion in the zinc site or even its absence has little or no effect. Derivatives of CuZnSOD are known with Cu ion bound either to the native copper site or to the native zinc site. The electronic absorption spectra of these derivatives indicate that the ligand environments of the two sites are very different. Copper(II) is a d transition-metal ion, and its d-d transitions are usually found in the visible and near-IR regions of the spectrum. Copper(II) complexes with coordinated nitrogen ligands are generally found to have an absorption band between 500 and 700 nm, with an extinction coefficient below 100 M cm . Bands in the absorption spectra of complexes with geometries that are distorted away from square planar tend to be red-shifted because of a smaller d-d splitting, and to have higher extinction coefficients because of the loss of centrosymmetry. Thus the optical spectrum of CuZnSOD with an absorption band with a maximum at 680 nm (14,700 cm ; see Figure 5.18A) and an extinction coefficient of 155 M cm per Cu is consistent with the crystal structural results that indicate that copper(II) is bound to four imidazole nitrogens and a water molecule in a distorted square-pyramidal geometry. Metal-substituted derivatives with Cu at the native copper site but with Co , Cd , Hg , or Ni substituted for Zn at the native zinc site all have a band at 680 nm, suggesting that the substitution of another metal ion for zinc perturbs the copper site very little, despite the proximity of the two metal sites. The absorption spectra of native CuZnSOD and these CuMSOD derivatives also have a shoulder at 417 nm (24,000 cm ; see Figure 5.18A), which is at lower energy than normal imidazole-to-Cu charge-transfer transitions, and has been assigned to an imidazolate-to-Cu charge transfer, indicating that the imidazolate bridge between Cu and the metal ion in the native zinc site is present, as observed in the crystal structure of CuZnSOD. Derivatives with the zinc site empty, which therefore cannot have an imidazolate bridge, are lacking this 417 nm shoulder. Small but significant changes in the absorption spectrum are seen when the metal ion is removed from the zinc site, e.g., in copper-only SOD (Figure 5.18B). The visible absorption band shifts to 700 nm (14,300 cm ), presumably due to a change in ligand field strength upon protonation of the bridging imidazolate. In addition, the shoulder at 417 nm has disappeared, again due to the absence of the imidazolate ligand. The spectroscopic properties due to copper in the native zinc site are best observed in the derivative Ag CuSOD, which has Ag in the copper site and Cu in the zinc site (see Figure 5.18C), since the d Ag ion is spectroscopically silent. In this derivative, the d-d transition is markedly red-shifted from the visible region of the spectrum into the near-IR, indicating that the ligand environment of Cu in that site is either tetrahedral or five coordinate. The EPR properties of Cu in this derivative are particularly interesting (as discussed below). The derivative with Cu bound at both sites, CuCuSOD, has a visible-near IR spectrum that is nearly a superposition of the spectra of CuZnSOD and Ag CuSOD (see Figure 5.19), indicating that the geometry of Cu in each of these sites is little affected by the nature of the metal ion in the other site. EPR spectroscopy has also proven to be particularly valuable in characterizing the metal environments in CuZnSOD and derivatives. The EPR spectrum of native CuZnSOD is shown in Figure 5.20A. The g resonance is split by the hyperfine coupling between the unpaired electron on Cu and the I = \(\frac{3}{2}\) nuclear spin of copper. The A value, 130 G, is intermediate between the larger A typical of square-planar Cu complexes with four nitrogen donor ligands and the lower A observed in blue copper proteins (see Chapter 6). The large linewidth seen in the g region indicates that the copper ion is in a rhombic (i.e., distorted) environment. Thus, the EPR spectrum is entirely consistent with the distorted square-pyramidal geometry observed in the x-ray structure. Removal of zinc from the native protein to give copper-only SOD results in a perturbed EPR spectrum, with a narrower g resonance and a larger A value (142 G) more nearly typical of Cu in an axial N environment (Figure 5.20B). Apparently the removal of zinc relaxes some constraints imposed on the geometry of the active-site ligands, allowing the copper to adopt to a geometry closer to its preferred tetragonal arrangement. The EPR spectrum due to Cu in the native Zn site in the Ag CuSOD derivative indicates that Cu is in a very different environment than when it is in the native copper site (Figure 5.20C). The spectrum is strongly rhombic, with a low value of A (97 G), supporting the conclusion based on the visible spectrum that copper is bound in a tetrahedral or five-coordinate environment. This type of site is unusual either for copper coordination complexes or for copper proteins in general, but does resemble the Cu EPR signal seen when either laccase or cytochrome c oxidase is partially reduced (see Figure 5.21). Partial reduction disrupts the magnetic coupling between these Cu centers that makes them EPR-silent in the fully oxidized protein. The EPR spectrum of CuCuSOD is very different from that of any of the other copper-containing derivatives (Figure 5.22) because the unpaired spins on the two copper centers interact and magnetically couple across the imidazolate bridge, resulting in a triplet EPR spectrum. This spectrum is virtually identical with that of model imidazolate-bridged binuclear copper complexes. Electronic absorption and EPR studies of derivatives of CuZnSOD containing Cu have provided useful information concerning the nature of the metal binding sites of those derivatives. H NMR spectra of those derivatives are generally not useful, however, because the relatively slowly relaxing paramagnetic Cu center causes the nearby proton resonances to be extremely broad. This difficulty has been overcome in two derivatives, CuCoSOD and CuNiSOD, in which the fast-relaxing paramagnetic Co and Ni centers at the zinc site interact across the imidazolate bridge and increase the relaxation rate of the Cu center, such that well-resolved paramagnetically shifted H NMR spectra of the region of the proteins near the two paramagnetic metal centers in the protein can be obtained and the resonances assigned. The use of H NMR to study CuCoSOD derivatives of CuZnSOD in combination with electronic absorption and EPR spectroscopies has enabled investigators to compare active-site structures of a variety of wild-type and mutant CuZnSOD proteins in order to find out if large changes in active-site structure have resulted from replacement of nearby amino-acid residues. | 19,934 | 2,181 |
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The factors that affect the reaction rate of a chemical reaction, which may determine whether a desired product is formed. In this section, we will show you how to quantitatively determine the reaction rate. Typically, reaction rates decrease with time because reactant concentrations decrease as reactants are converted to products. Reaction rates generally increase when reactant concentrations are increased. This section examines mathematical expressions called , which describe the relationships between reactant rates and reactant concentrations. Rate laws are mathematical descriptions of experimentally verifiable data. Rate laws may be written from either of two different but related perspectives. A expresses the reaction rate in terms of changes in the concentration of one or more reactants (Δ[R]) over a specific time interval (Δt). In contrast, an describes the reaction rate in terms of the initial concentration ([R] ) and the measured concentration of one or more reactants ([R]) after a given amount of time (t); integrated rate laws are discussed in more detail later The integrated rate law is derived by using calculus to integrate the differential rate law. Whether using a differential rate law or integrated rate law, always make sure that the rate law gives the proper units for the reaction rate, usually moles per liter per second (M/s). For a reaction with the general equation: \[aA + bB \rightarrow cC + dD \label{14.3.1} \] the experimentally determined rate law usually has the following form: \[\text{rate} = k[A]^m[B]^n \label{14.3.2} \] The proportionality constant ( ) is called the , and its value is characteristic of the reaction and the reaction conditions. A given reaction has a particular rate constant value under a given set of conditions, such as temperature, pressure, and solvent; varying the temperature or the solvent usually changes the value of the rate constant. The numerical value of , however, does not change as the reaction progresses under a given set of conditions. Under a given set of conditions, the value of the rate constant does change as the reaction progresses. The reaction rate thus depends on the rate constant for the given set of reaction conditions and the concentration of A and B raised to the powers and , respectively. The values of and are derived from experimental measurements of the changes in reactant concentrations over time and indicate the , the degree to which the reaction rate depends on the concentration of each reactant; and need not be integers. For example, tells us that is order in reactant A and order in reactant B. It is important to remember that and are not related to the stoichiometric coefficients and in the balanced chemical equation and must be determined . The overall reaction order is the sum of all the exponents in the rate law: + . and are related to the stoichiometric coefficients in the balanced chemical (e.g., and To illustrate how chemists interpret a differential rate law, consider the experimentally derived rate law for the of -butyl bromide in 70% aqueous acetone. This reaction produces -butanol according to the following equation: \[(CH_3)_3CBr_{(soln)} + H_2O_{(soln)} \rightarrow (CH_3)_3COH_{(soln)} + HBr_{(soln)} \label{14.3.3} \] Combining the rate expression in Equation \(\ref{14.3.2}\) with the definition of average reaction rate \[\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t} \nonumber \] gives a general expression for the differential rate law: \[\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm A]^m[\textrm B]^n \label{14.3.4} \] Inserting the identities of the reactants into gives the following expression for the differential rate law for the reaction: \[\textrm{rate}=-\dfrac{\Delta[\mathrm{(CH_3)_3CBr}]}{\Delta t}=k[\mathrm{(CH_3)_3CBr}]^m[\mathrm{H_2O}]^n \label{14.3.5} \] Experiments to determine the rate law for the hydrolysis of -butyl bromide show that the reaction rate is directly proportional to the concentration of (CH ) CBr but is independent of the concentration of water. Therefore, m and n in Equation \(\ref{14.3.4}\) are 1 and 0, respectively, and, \[\text{rate} = k[(CH_3)_3CBr]^1[H_2O]^0 = k[(CH_3)_3CBr] \label{14.3.6} \] Because the exponent for the reactant is 1, the reaction is first order in (CH ) CBr. It is zeroth order in water because the exponent for [H O] is 0. (Recall that anything raised to the zeroth power equals 1.) Thus, the overall reaction order is 1 + 0 = 1. The reaction orders state in practical terms that doubling the concentration of (CH ) CBr doubles the reaction rate of the hydrolysis reaction, halving the concentration of (CH ) CBr halves the reaction rate, and so on. Conversely, increasing or decreasing the concentration of water has no effect on the reaction rate. (Again, when working with rate laws, there is no simple correlation between the stoichiometry of the reaction and the rate law. The values of , , and in the rate law must be determined experimentally.) Experimental data show that has the value 5.15 × 10 s at 25°C. The rate constant has units of reciprocal seconds (s ) because the reaction rate is defined in units of concentration per unit time (M/s). The units of a rate constant depend on the rate law for a particular reaction. Under conditions identical to those for the -butyl bromide reaction, the experimentally derived differential rate law for the hydrolysis of methyl bromide (CH Br) is as follows: \[\textrm{rate}=-\dfrac{\Delta[\mathrm{CH_3Br}]}{\Delta t}=k'[\mathrm{CH_3Br}] \label{14.3.7} \] This reaction also has an overall reaction order of 1, but the rate constant in is approximately 10 times smaller than that for -butyl bromide. Thus, methyl bromide hydrolyzes about 1 million times more slowly than -butyl bromide, and this information tells chemists how the reactions differ on a molecular level. Frequently, changes in reaction conditions also produce changes in a rate law. In fact, chemists often alter reaction conditions to study the mechanics of a reaction. For example, when -butyl bromide is hydrolyzed in an aqueous acetone solution containing OH ions rather than in aqueous acetone alone, the differential rate law for the hydrolysis reaction does not change. In contrast, for methyl bromide, the differential rate law becomes \[\text{rate} =k″[CH_3Br,OH^−] \nonumber \] with an overall reaction order of 2. Although the two reactions proceed similarly in neutral solution, they proceed very differently in the presence of a base, providing clues as to how the reactions differ on a molecular level. An experiment shows that the reaction of nitrogen dioxide with carbon monoxide: is second order in NO and zero order in at 100 °C. What is the rate law for the reaction? The reaction will have the form: \[\ce{rate}=k[\ce{NO2}]^m[\ce{CO}]^n \nonumber \] The reaction is second order in NO ; thus = 2. The reaction is zero order in CO; thus = 0. The rate law is: \[\ce{rate}=k[\ce{NO2}]^2[\ce{CO}]^0=k[\ce{NO2}]^2 \nonumber \] Remember that a number raised to the zero power is equal to 1, thus [CO] = 1, which is why we can simply drop the concentration of CO from the rate equation: the rate of reaction is solely dependent on the concentration of NO . When we consider rate mechanisms later in this chapter, we will explain how a reactant’s concentration can have no effect on a reaction despite being involved in the reaction. The rate law for the reaction: \[\ce{H2}(g)+\ce{2NO}(g)⟶\ce{N2O}(g)+\ce{H2O}(g) \nonumber \] has been experimentally determined to be \(rate = k[NO]^2[H_2]\). What are the orders with respect to each reactant, and what is the overall order of the reaction? In a transesterification reaction, a triglyceride reacts with an alcohol to form an ester and glycerol. Many students learn about the reaction between methanol (CH OH) and ethyl acetate (CH CH OCOCH ) as a sample reaction before studying the chemical reactions that produce biodiesel: \[\ce{CH3OH + CH3CH2OCOCH3 ⟶ CH3OCOCH3 + CH3CH2OH} \nonumber \] The rate law for the reaction between methanol and ethyl acetate is, under certain conditions, experimentally determined to be: \[\ce{rate}=k[\ce{CH3OH}] \nonumber \] What is the order of reaction with respect to methanol and ethyl acetate, and what is the overall order of reaction? Below are three reactions and their experimentally determined differential rate laws. For each reaction, give the units of the rate constant, give the reaction order with respect to each reactant, give the overall reaction order, and predict what happens to the reaction rate when the concentration of the first species in each chemical equation is doubled. balanced chemical equations and differential rate laws units of rate constant, reaction orders, and effect of doubling reactant concentration \[k\textrm M^2=\dfrac{\textrm M}{\textrm s}k=\dfrac{\textrm{M/s}}{\textrm M^2}=\dfrac{1}{\mathrm{M\cdot s}}=\mathrm{M^{-1}\cdot s^{-1}} \nonumber \] The exponent in the rate law is 2, so the reaction is second order in HI. Because HI is the only reactant and the only species that appears in the rate law, the reaction is also second order overall. If the concentration of HI is doubled, the reaction rate will increase from [HI] to (2[HI]) = 4 [HI] . The reaction rate will therefore quadruple. The rate law tells us that the reaction rate is constant and independent of the N O concentration. That is, the reaction is zeroth order in N O and zeroth order overall. Because the reaction rate is independent of the N O concentration, doubling the concentration will have no effect on the reaction rate. The only concentration in the rate law is that of cyclopropane, and its exponent is 1. This means that the reaction is first order in cyclopropane. Cyclopropane is the only species that appears in the rate law, so the reaction is also first order overall. Doubling the initial cyclopropane concentration will increase the reaction rate from [cyclopropane] to 2 [cyclopropane] . This doubles the reaction rate. Given the following two reactions and their experimentally determined differential rate laws: determine the units of the rate constant if time is in seconds, determine the reaction order with respect to each reactant, give the overall reaction order, and predict what will happen to the reaction rate when the concentration of the first species in each equation is doubled. s ; first order in CH N=NCH ; first order overall; doubling [CH N=NCH ] will double the reaction rate. M ·s ; first order in NO , first order in F ; second order overall; doubling [NO ] will double the reaction rate. The number of fundamentally different mechanisms (sets of steps in a reaction) is actually rather small compared to the large number of chemical reactions that can occur. Thus understanding can simplify what might seem to be a confusing variety of chemical reactions. The first step in discovering the reaction mechanism is to determine the reaction’s rate law. This can be done by designing experiments that measure the concentration(s) of one or more reactants or products as a function of time. For the reaction \(A + B \rightarrow products\), for example, we need to determine and the exponents and in the following equation: \[\text{rate} = k[A]^m[B]^n \label{14.4.11} \] To do this, we might keep the initial concentration of B constant while varying the initial concentration of A and calculating the initial reaction rate. This information would permit us to deduce the reaction order with respect to A. Similarly, we could determine the reaction order with respect to B by studying the initial reaction rate when the initial concentration of A is kept constant while the initial concentration of B is varied. In earlier examples, we determined the reaction order with respect to a given reactant by comparing the different rates obtained when only the concentration of the reactant in question was changed. An alternative way of determining reaction orders is to set up a proportion using the rate laws for two different experiments. Rate data for a hypothetical reaction of the type \(A + B \rightarrow products\) are given in . The general rate law for the reaction is given in . We can obtain or directly by using a proportion of the rate laws for two experiments in which the concentration of one reactant is the same, such as Experiments 1 and 3 in . \[\dfrac{\mathrm{rate_1}}{\mathrm{rate_3}}=\dfrac{k[\textrm A_1]^m[\textrm B_1]^n}{k[\textrm A_3]^m[\textrm B_3]^n} \nonumber \] Inserting the appropriate values from , \[\dfrac{8.5\times10^{-3}\textrm{ M/min}}{34\times10^{-3}\textrm{ M/min}}=\dfrac{k[\textrm{0.50 M}]^m[\textrm{0.50 M}]^n}{k[\textrm{1.00 M}]^m[\textrm{0.50 M}]^n} \nonumber \] Because 1.00 to any power is 1, [1.00 M] = 1.00 M. We can cancel like terms to give 0.25 = [0.50] , which can also be written as 1/4 = [1/2] . Thus we can conclude that = 2 and that the reaction is second order in A. By selecting two experiments in which the concentration of B is the same, we were able to solve for . Conversely, by selecting two experiments in which the concentration of A is the same (e.g., Experiments 5 and 1), we can solve for . \(\dfrac{\mathrm{rate_1}}{\mathrm{rate_5}}=\dfrac{k[\mathrm{A_1}]^m[\mathrm{B_1}]^n}{k[\mathrm{A_5}]^m[\mathrm{B_5}]^n}\) Substituting the appropriate values from , \[\dfrac{8.5\times10^{-3}\textrm{ M/min}}{8.5\times10^{-3}\textrm{ M/min}}=\dfrac{k[\textrm{0.50 M}]^m[\textrm{0.50 M}]^n}{k[\textrm{0.50 M}]^m[\textrm{1.00 M}]^n} \nonumber \] Canceling leaves 1.0 = [0.50] , which gives \(n = 0\); that is, the reaction is zeroth order in \(B\). The experimentally determined rate law is therefore rate = [A] [B] = [A] We can now calculate the rate constant by inserting the data from any row of into the experimentally determined rate law and solving for \(k\). Using Experiment 2, we obtain 19 × 10 M/min = (0.75 M) 3.4 × 10 M ·min = k You should verify that using data from any other row of Table \(\Page {1}\) gives the same rate constant. This must be true as long as the experimental conditions, such as temperature and solvent, are the same. Nitric oxide is produced in the body by several different enzymes and acts as a signal that controls blood pressure, long-term memory, and other critical functions. The major route for removing NO from biological fluids is via reaction with \(O_2\) to give \(NO_2\), which then reacts rapidly with water to give nitrous acid and nitric acid: These reactions are important in maintaining steady levels of NO. The following table lists kinetics data for the reaction of NO with O at 25°C: \[2NO(g) + O_2(g) \rightarrow 2NO_2(g) \nonumber \] Determine the rate law for the reaction and calculate the rate constant. balanced chemical equation, initial concentrations, and initial rates rate law and rate constant Comparing Experiments 1 and 2 shows that as [O ] is doubled at a constant value of [NO ], the reaction rate approximately doubles. Thus the reaction rate is proportional to [O ] , so the reaction is first order in O . Comparing Experiments 1 and 3 shows that the reaction rate essentially quadruples when [NO] is doubled and [O ] is held constant. That is, the reaction rate is proportional to [NO] , which indicates that the reaction is second order in NO. Using these relationships, we can write the rate law for the reaction: rate = [NO] [O ] The data in any row can be used to calculate the rate constant. Using Experiment 1, for example, gives \[k=\dfrac{\textrm{rate}}{[\mathrm{NO}]^2[\mathrm{O_2}]}=\dfrac{7.98\times10^{-3}\textrm{ M/s}}{(0.0235\textrm{ M})^2(0.0125\textrm{ M})}=1.16\times10^3\;\mathrm{ M^{-2}\cdot s^{-1}} \nonumber \] Alternatively, using Experiment 2 gives \[k=\dfrac{\textrm{rate}}{[\mathrm{NO}]^2[\mathrm{O_2}]}=\dfrac{15.9\times10^{-3}\textrm{ M/s}}{(0.0235\textrm{ M})^2(0.0250\textrm{ M})}=1.15\times10^3\;\mathrm{ M^{-2}\cdot s^{-1}} \nonumber \] The difference is minor and associated with significant digits and likely experimental error in making the table. The overall reaction order \((m + n) = 3\), so this is a third-order reaction whose rate is determined by three reactants. The units of the rate constant become more complex as the overall reaction order increases. The peroxydisulfate ion (S O ) is a potent oxidizing agent that reacts rapidly with iodide ion in water: \[S_2O^{2−}_{8(aq)} + 3I^−_{(aq)} \rightarrow 2SO^{2−}_{4(aq)} + I^−_{3(aq)} \nonumber \] The following table lists kinetics data for this reaction at 25°C. Determine the rate law and calculate the rate constant. A Discussing Initial Rates and Rate Law Expressions. Link: The rate law for a reaction is a mathematical relationship between the reaction rate and the concentrations of species in solution. Rate laws can be expressed either as a differential rate law, describing the change in reactant or product concentrations as a function of time, or as an integrated rate law, describing the actual concentrations of reactants or products as a function of time. The rate constant ( ) of a rate law is a constant of proportionality between the reaction rate and the reactant concentration. The exponent to which a concentration is raised in a rate law indicates the reaction order, the degree to which the reaction rate depends on the concentration of a particular reactant. | 17,457 | 2,182 |
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Scientists of the 18th and early 19th centuries studied compounds obtained from plants and animals and labeled them because they were isolated from “organized” (living) systems. Compounds isolated from nonliving systems, such as rocks and ores, the atmosphere, and the oceans, were labeled . For many years, scientists thought organic compounds could be made by only living organisms because they possessed a vital force found only in living systems. The vital force theory began to decline in 1828, when the German chemist Friedrich Wöhler synthesized urea from inorganic starting materials. He reacted silver cyanate (AgOCN) and ammonium chloride (NH Cl), expecting to get ammonium cyanate (NH OCN). What he expected is described by the following equation. \[AgOCN + NH_4Cl \rightarrow AgCl + NH_4OCN \label{Eq1}\] Instead, he found the product to be urea (NH CONH ), a well-known organic material readily isolated from urine. This result led to a series of experiments in which a wide variety of organic compounds were made from inorganic starting materials. The vital force theory gradually went away as chemists learned that they could make many organic compounds in the laboratory. Today organic chemistry is the study of the chemistry of the carbon compounds, and inorganic chemistry is the study of the chemistry of all other elements. It may seem strange that we divide chemistry into two branches—one that considers compounds of only one element and one that covers the 100-plus remaining elements. However, this division seems more reasonable when we consider that of tens of millions of compounds that have been characterized, the overwhelming majority are carbon compounds. The word has different meanings. Organic fertilizer, such as cow manure, is organic in the original sense; it is derived from living organisms. Organic foods generally are foods grown without synthetic pesticides or fertilizers. Organic chemistry is the chemistry of compounds of carbon. Carbon is unique among the other elements in that its atoms can form stable covalent bonds with each other and with atoms of other elements in a multitude of variations. The resulting molecules can contain from one to millions of carbon atoms. We previously surveyed organic chemistry by dividing its compounds into families based on functional groups. We begin with the simplest members of a family and then move on to molecules that are organic in the original sense—that is, they are made by and found in living organisms. These complex molecules (all containing carbon) determine the forms and functions of living systems and are the subject of biochemistry. Organic compounds, like inorganic compounds, obey all the natural laws. Often there is no clear distinction in the chemical or physical properties among organic and inorganic molecules. Nevertheless, it is useful to compare typical members of each class, as in Table \(\Page {1}\). Keep in mind, however, that there are exceptions to every category in this table. To further illustrate typical differences among organic and inorganic compounds, Table \(\Page {1}\) also lists properties of the inorganic compound sodium chloride (common table salt, NaCl) and the organic compound hexane (C H ), a solvent that is used to extract soybean oil from soybeans (among other uses). Many compounds can be classified as organic or inorganic by the presence or absence of certain typical properties, as illustrated in Table \(\Page {1}\). We begin our study of organic chemistry with the hydrocarbons, the simplest organic compounds, which are composed of carbon and hydrogen atoms only. As we noted, there are several different kinds of hydrocarbons. They are distinguished by the types of bonding between carbon atoms and the properties that result from that bonding. Hydrocarbons with only carbon-to-carbon single bonds (C–C) and existing as a continuous chain of carbon atoms also bonded to hydrogen atoms are called alkanes (or saturated hydrocarbons). , in this case, means that each carbon atom is bonded to four other atoms (hydrogen or carbon)—the most possible; there are no double or triple bonds in the molecules. The word has the same meaning for hydrocarbons as it does for the dietary fats and oils: the molecule has no carbon-to-carbon double bonds (C=C). We previously introduced the three simplest alkanes—methane (CH ), ethane (C H ), and propane (C H ) and they are shown again in Figure \(\Page {1}\). The flat representations shown do not accurately portray bond angles or molecular geometry. Methane has a tetrahedral shape that chemists often portray with wedges indicating bonds coming out toward you and dashed lines indicating bonds that go back away from you. An ordinary solid line indicates a bond in the plane of the page. Recall that the VSEPR theory correctly predicts a tetrahedral shape for the methane molecule (Figure \(\Page {2}\)). Methane (CH ), ethane (C H ), and propane (C H ) are the beginning of a series of compounds in which any two members in a sequence differ by one carbon atom and two hydrogen atoms—namely, a CH unit. The first 10 members of this series are given in Table \(\Page {2}\). Consider the series in Figure \(\Page {3}\). The sequence starts with C H , and a CH unit is added in each step moving up the series. Any family of compounds in which adjacent members differ from each other by a definite factor (here a CH group) is called a homologous series. The members of such a series, called , have properties that vary in a regular and predictable manner. The principle of gives organization to organic chemistry in much the same way that the periodic table gives organization to inorganic chemistry. Instead of a bewildering array of individual carbon compounds, we can study a few members of a homologous series and from them deduce some of the properties of other compounds in the series. The principle of homology allows us to write a general formula for alkanes: C H . Using this formula, we can write a molecular formula for any alkane with a given number of carbon atoms. For example, an alkane with eight carbon atoms has the molecular formula C H = C H . We use several kinds of formulas to describe organic compounds. A shows only the kinds and numbers of atoms in a molecule. For example, the molecular formula C H tells us there are 4 carbon atoms and 10 hydrogen atoms in a molecule, but it doesn’t distinguish between butane and isobutane. A structural formula shows all the carbon and hydrogen atoms and the bonds attaching them. Thus, structural formulas identify the specific isomers by showing the order of attachment of the various atoms. Unfortunately, structural formulas are difficult to type/write and take up a lot of space. Chemists often use condensed structural formulas to alleviate these problems. The condensed formulas show hydrogen atoms right next to the carbon atoms to which they are attached, as illustrated for butane: The ultimate condensed formula is a line-angle formula, in which carbon atoms are implied at the corners and ends of lines, and each carbon atom is understood to be attached to enough hydrogen atoms to give each carbon atom four bonds. For example, we can represent pentane (CH CH CH CH CH ) and isopentane [(CH ) CHCH CH ] as follows: Parentheses in condensed structural formulas indicate that the enclosed grouping of atoms is attached to the adjacent carbon atom. As noted in Table \(\Page {2}\):, the number of isomers increases rapidly as the number of carbon atoms increases. There are 3 pentanes, 5 hexanes, 9 heptanes, and 18 octanes. It would be difficult to assign unique individual names that we could remember. A systematic way of naming hydrocarbons and other organic compounds has been devised by the International Union of Pure and Applied Chemistry (IUPAC). These rules, used worldwide, are known as the IUPAC System of Nomenclature. (Some of the names we used earlier, such as isobutane, isopentane, and neopentane, do not follow these rules and are called .) A stem name (Table \(\Page {3}\)) indicates the number of carbon atoms in the longest continuous chain (LCC). Atoms or groups attached to this carbon chain, called , are then named, with their positions indicated by numbers. For now, we will consider only those substituents called alkyl groups. An alkyl group is a group of atoms that results when one hydrogen atom is removed from an alkane. The group is named by replacing the suffix of the parent hydrocarbon with . For example, the CH group derived from methane (CH ) results from subtracting one hydrogen atom and is called a . The alkyl groups we will use most frequently are listed in Table \(\Page {4}\). Alkyl groups are not independent molecules; they are parts of molecules that we consider as a unit to name compounds systematically. Simplified IUPAC rules for naming alkanes are as follows (demonstrated in Example \(\Page {1}\)). When these rules are followed, every unique compound receives its own exclusive name. The rules enable us to not only name a compound from a given structure but also draw a structure from a given name. The best way to learn how to use the IUPAC system is to put it to work, not just memorize the rules. It’s easier than it looks. Name each compound. Name each compound. Draw the structure for each compound. In drawing structures, always start with the parent chain. Then add the groups at their proper positions. You can number the parent chain from either direction as long as you are consistent; just don’t change directions before the structure is done. The name indicates two methyl (CH ) groups, one on the second carbon atom and one on the third. Finally, fill in all the hydrogen atoms, keeping in mind that each carbon atom must have four bonds. Adding the groups at their proper positions gives Filling in all the hydrogen atoms gives the following condensed structural formulas: Note that the bonds (dashes) can be shown or not; sometimes they are needed for spacing. Draw the structure for each compound. are atoms or small groups of atoms (two to four) that exhibit a characteristic reactivity. A particular functional group will almost always display its characteristic chemical behavior when it is present in a compound. Because of their importance in understanding organic chemistry, functional groups have characteristic names that often carry over in the naming of individual compounds incorporating specific groups. In our study of organic chemistry, it will become extremely important to be able to quickly recognize the most common functional groups, because they are the key structural elements that define how organic molecules react. For now, we will only worry about drawing and recognizing each functional group, as depicted by Lewis and line structures. Much of the remainder of your study of organic chemistry will be taken up with learning about how the different functional groups tend to behave in organic reactions. Previously, we considered several kinds of hydrocarbons. Now we examine some of the many organic compounds that contain functional groups. We first introduced the idea of the functional group, a specific structural arrangement of atoms or bonds that imparts a characteristic chemical reactivity to the molecule. If you understand the behavior of a particular functional group, you will know a great deal about the general properties of that class of compounds. In this chapter, we make a brief yet systematic study of some of organic compound families. Each family is based on a common, simple functional group that contains an oxygen atom or a nitrogen atom. Some common functional groups are listed in Table \(\Page {5}\) and a more comprehensive list if found here. by (University of Minnesota, Morris) | 11,830 | 2,184 |
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