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https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Principles_of_Chemical_Equilibria/Basic_Concepts |
Consider the following gas-phase chemical reaction: \(H_2 + I_2 \rightleftharpoons 2HI\) The apparatus shown belows contains 0.02 mol/L I gas in the left syringe and 0.01 mol/L H gas in the right syringe. When the two syringes are depressed, the two gases are mixed and the above reaction occurs. The graph at the right shows the variations of the H , I , and HI concentrations as a function of time. Run the experiment and examine the concentration-time plots. As the reaction progresses, the concentrations of iodine and hydrogen decrease as they are consumed while the concentration of hydrogen iodide increases as it is formed. Eventually, however, the concentrations of all three species reach constant values. This behavior is a result of the reverse in which hydrogen iodide reacts to form iodine and hydrogen. Initially there is no hydrogen iodide, so the reverse reaction cannot occur. As hydrogen iodide accumulates, the reverse reaction can progress and becomes significant. Eventually the rate at which iodine is being consumed by the forward reaction is perfectly balanced by the rate at which it is being produced by the back reaction. The same is true for hydrogen and hydrogen iodide. When the rates of forward and reverse reactions are perfectly balanced in this way, the reaction is said to be at equilibrium. | 1,347 | 918 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Fundamentals_of_Thermodynamics/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 | 919 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Electronic_Structure_of_Atoms_and_Molecules/Hydrogen's_Atomic_Emission_Spectrum |
A hydrogen discharge tube is a slim tube containing hydrogen gas at low pressure with an electrode at each end. If a high voltage (5000 volts) is applied, the tube lights up with a bright pink glow. If the light is passed through a prism or diffraction grating, it is split into its various colors. This is a small part of the hydrogen emission spectrum. Most of the spectrum is invisible to the eye because it is either in the infrared or the ultraviolet region of the electromagnetic spectrum. The photograph shows part of a hydrogen discharge tube on the left, and the three most apparent lines in the visible part of the spectrum on the right. (Ignore the "smearing," particularly to the left of the red line. This is caused by flaws in the way the photograph was taken. See note below.) The hydrogen spectrum is complex, comprising more than the three lines visible to the naked eye. It is possible to detect patterns of lines in both the ultraviolet and infrared regions of the spectrum as well. These fall into a number of "series" of lines named after the person who discovered them. The diagram below shows three of these series, but there are others in the infrared to the left of the Paschen series shown in the diagram. The diagram is quite complicated. Consider first at the Lyman series on the right of the diagram; this is the broadest series, and the easiest to decipher. The Lyman series is a series of lines in the ultraviolet region. The lines grow closer and closer together as the frequency increases. Eventually, they are so close together that it becomes impossible to see them as anything other than a continuous spectrum. This is suggested by the shaded part on the right end of the series. At one particular point, known as the series limit, the series ends. In Balmer series or the Paschen series, the pattern is the same, but the series are more compact. In the Balmer series, notice the position of the three visible lines from the photograph further up the page. The hydrogen spectrum is often drawn using wavelengths of light rather than frequencies. Unfortunately, because of the mathematical relationship between the frequency of light and its wavelength, two completely different views of the spectrum are obtained when it is plotted against frequency or against wavelength. The mathematical relationship between frequency and wavelength is the following: Rearranging this gives equations for either wavelength or frequency: \[\lambda =\dfrac{c}{\nu}\] or \[\nu=\dfrac{c}{\lambda }\] There is an inverse relationship between the two variables—a high frequency means a low wavelength and vice versa. This is what the spectrum looks like plotted in terms of wavelength instead of frequency: Compare this to the same spectrum in terms of frequency: In an amazing demonstration of mathematical insight, in 1885 Balmer came up with a simple formula for predicting the wavelength of any of the lines in what we now know as the Balmer series. Three years later, Rydberg generalized this so that it was possible to determine the wavelengths of any of the lines in the hydrogen emission spectrum. Rydberg's equation is as follows: \[ \dfrac{1}{\lambda}=R_H \left( \dfrac{1}{n^2_1}-\dfrac{1}{n^2_2}\right)\] where The various combinations of numbers that can be substituted into this formula allow the calculation the wavelength of any of the lines in the hydrogen emission spectrum; there is close agreement between the wavelengths generated by this formula and those observed in a real spectrum. A modified version of the Rydberg equation can be used to calculate the frequency of each of the lines: The lines in the hydrogen emission spectrum form regular patterns and can be represented by a (relatively) simple equation. Each line can be calculated from a combination of simple whole numbers. Why does hydrogen emit light when excited by a high voltage and what is the significance of those whole numbers? When unexcited, hydrogen's electron is in the first energy level—the level closest to the nucleus. But if energy is supplied to the atom, the electron is excited into a higher energy level, or even removed from the atom altogether. The high voltage in a discharge tube provides that energy. Hydrogen molecules are first broken up into hydrogen atoms (hence the atomic hydrogen emission spectrum) and electrons are then promoted into higher energy levels. Suppose a particular electron is excited into the third energy level. It would tend to lose energy again by falling back down to a lower level. It can do this in two different ways. It could fall all the way back down to the first level again, or it could fall back to the second level and then, in a second jump, down to the first level. If an electron falls from the 3-level to the 2-level, it must lose an amount of energy exactly equal to the energy difference between those two levels. That energy which the electron loses is emitted as light (which "light" includes UV and IR as well as visible radiation). Each frequency of light is associated with a particular energy by the equation: The higher the frequency, the higher the energy of the light. If an electron falls from the 3-level to the 2-level, red light is seen. This is the origin of the red line in the hydrogen spectrum. From the frequency of the red light, its energy can be calculated. That energy must be exactly the same as the energy gap between the 3-level and the 2-level in the hydrogen atom. The last equation can therefore be rewritten as a measure of the energy gap between two electron levels: The greatest possible fall in energy will therefore produce the highest frequency line in the spectrum. The greatest fall will be from the infinity level to the 1-level. (The significance of the infinity level will be made clear later.) The next few diagrams are in two parts, with the energy levels at the top and the spectrum at the bottom. If an electron falls from the 6-level, the difference is slightly less than before, and so the frequency is slightly lower (because of the scale of the diagram, it is impossible to depict the levels beyond 7). All other possible jumps to the first level make up the whole Lyman series. The spacings between the lines in the spectrum reflect the changes in spacings between the energy levels. If the same is done for the 2-level, the Balmer series is shown. These energy gaps are all much smaller than in the Lyman series, and so the frequencies produced are also much lower. The Paschen series is made up of the transitions to the 3-level, but they are omitted to avoid cluttering the diagram. In the Rydberg equation, n and n n is the level being jumped from. In the case before, in which a red line is produced by electrons falling from the 3-level to the 2-level, n is equal to 3. The infinity level represents the highest possible energy an electron can have as a part of a hydrogen atom. If the electron exceeds that energy, it is no longer a part of the atom. The infinity level represents the point at which ionization of the atom occurs to form a positively charged ion. When there is no additional energy supplied to it, hydrogen atom's electron is found at the 1-level. This is known as its ground state. If enough energy is supplied to move the electron up to the infinity level, the atom is ionized. The ionization energy per electron is therefore a measure of the difference in energy between the 1-level and the infinity level. In above diagrams, that particular energy jump produces the series limit of the Lyman series. A problem with this approach is that the frequency of a series limit is quite difficult to find accurately from a spectrum because the lines are so close together in that region that the spectrum looks continuous. The following is a list of the frequencies of the seven most widely spaced lines in the Lyman series, together with the increase in frequency between successive lines. As the lines become closer together, the increase in frequency is lessened. At the series limit, the gap between the lines is zero. Consequently, if the increase in frequency is plotted against the actual frequency, the curve can be extrapolated to the point at which the increase becomes zero, the frequency of the series limit. In fact, two graphs can be plotted from the data in the table above. The frequency difference is related to two frequencies. For example, the figure of 0.457 is found by subtracting 2.467 from 2.924. Which of the two values should be plotted against 0.457 does not matter, as long as consistency is maintained—the difference must always be plotted a As illustrated in the graph below, plotting both of the possible curves on the same graph makes it easier to decide exactly how to extrapolate the curves. Because these are curves, they are much more difficult to extrapolate than straight lines. Both lines indicate a series limit at about 3.28 x 10 Hz. With this information, it is possible calculate the energy needed to remove a single electron from a hydrogen atom. Recall the equation above: The energy gap between the ground state and the point at which the electron leaves the atom can be determined by substituting the frequency and looking up the value of Planck's constant from a data book. \[ \begin{eqnarray} \Delta E &=& h\nu \\ &=& (6.626 \times 10^{-34})(3.28 \times 10^{15}) \\ &=& 2.173 \times 10^{-18}\ J \end{eqnarray} \] This is the ionization energy for a single atom. To find the normally quoted ionization energy, this value is multiplied by the number of atoms in a mole of hydrogen atoms (the Avogadro constant) and then dividing by 1000 to convert joules to kilojoules. \[\begin{eqnarray} Ionization\ energy &=& (2.173 \times 10^{-18})( 6.022 \times 10^{23})( \frac{1}{1000}) \\ &=& 1310\ kJ\ mol^{-1} \end{eqnarray} \] This compares well with the normally quoted value for hydrogen's ionization energy of 1312 kJ mol . | 9,917 | 920 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/4_f-Block_Elements/The_Lanthanides/Chemistry_of_Praseodymium |
Praseodymium, which is named from the Greek prasios + didymos (green twin), was isolated and identified by von Welsbach in 1885 from what was known at the time as didymium. von Welsbach's work revealed that this "substance" actually contained two new elements, one of which was praseodymium (neodymium was the other). Pure praseodymium is silvery-white and fairly soft. It oxidizes slowly in air and reacts vigorously with water to release hydrogen gas. It is used as an alloying agent along with magnesium for parts in aircraft engines. Misch metal is 5% praseodymium and is used for alloying steel and in flints used to create sparks in lighters. The glass in welder's goggles contains a mixture of praseodymium and neodymium. | 748 | 921 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Phase_Transitions/Vapor_Pressure |
This page looks at how the equilibrium between a liquid (or a solid) and its vapor leads to the idea of a saturated vapor pressure. It also looks at how saturated vapor pressure varies with temperature, and the relationship between saturated vapor pressure and boiling point. The average energy of the particles in a liquid is governed by the temperature. The higher the temperature, the higher the average energy. But within that average, some particles have energies higher than the average, and others have energies lower than the average. Some of the more energetic particles on the surface of the liquid can be moving fast enough to escape from the attractive forces holding the liquid together. They evaporate. The diagram shows a small region of a liquid near its surface. Notice that evaporation only takes place on the surface of the liquid. That's quite different from boiling which happens when there is enough energy to disrupt the attractive forces throughout the liquid. That's why, if you look at boiling water, you see bubbles of gas being formed all the way through the liquid. If you look at water which is just evaporating in the sun, you don't see any bubbles. Water molecules are simply breaking away from the surface layer. Eventually, the water will all evaporate in this way. The energy which is lost as the particles evaporate is replaced from the surroundings. As the molecules in the water jostle with each other, new molecules will gain enough energy to escape from the surface. Now imagine what happens if the liquid is in a closed container. Common sense tells you that water in a sealed bottle does not seem to evaporate - or at least, it does not disappear over time. But there is constant evaporation from the surface. Particles continue to break away from the surface of the liquid - but this time they are trapped in the space above the liquid. As the gaseous particles bounce around, some of them will hit the surface of the liquid again, and be trapped there. There will rapidly be an equilibrium set up in which the number of particles leaving the surface is exactly balanced by the number rejoining it. In this equilibrium, there will be a fixed number of the gaseous particles in the space above the liquid. When these particles hit the walls of the container, they exert a pressure. This pressure is called the saturated vapor pressure (also known as ) of the liquid. It is not difficult to show the existence of this saturated vapor pressure (and to measure it) using a simple piece of apparatus. If you have a mercury barometer tube in a trough of mercury, at 1 atmosphere pressure the column will be 760 mm tall. 1 atmosphere is sometimes quoted as 760 mmHg ("millimetres of mercury"). If you squirt a few drops of liquid into the tube, it will rise to form a thin layer floating on top of the mercury. Some of the liquid will evaporate and you will get the equilibrium we've just been talking about - provided there is still some liquid on top of the mercury. It is only an equilibrium if both liquid and vapor are present. The saturated vapor pressure of the liquid will force the mercury level down a bit. You can measure the drop - and this gives a value for the saturated vapor pressure of the liquid at this temperature. In this case, the mercury has been forced down by a distance of 760 - 630 mm. The saturated vapor pressure of this liquid at the temperature of the experiment is 130 mmHg. You could convert this into proper SI units (pascals) if you wanted to. 760 mmHg is equivalent to 101,325 Pa. A value of 130 mmHg is quite a high vapor pressure if we are talking about room temperature. Water's saturated vapor pressure is about 20 mmHg at this temperature. A high vapor pressure means that the liquid must be volatile - molecules escape from its surface relatively easily, and aren't very good at sticking back on again either. That will result in larger numbers of them in the gas state once equilibrium is reached. The liquid in the example must have significantly weaker intermolecular forces than water. You can look at this in two ways. (1) There is a common sense way. If you increase the temperature, you are increasing the average energy of the particles present. That means that more of them are likely to have enough energy to escape from the surface of the liquid. That will tend to increase the saturated vapor pressure. (2) Or you can look at it in terms of Le Chatelier's Principle - which works just as well in this kind of physical situation as it does in the more familiar chemical examples. When the space above the liquid is saturated with vapor particles, you have this equilibrium occurring on the surface of the liquid: The forward change (liquid to vapor) is endothermic. It needs heat to convert the liquid into the vapor. According to Le Chatelier, increasing the temperature of a system in a dynamic equilibrium favors the endothermic change. That means that increasing the temperature increases the amount of vapor present, and so increases the saturated vapor pressure. The graph shows how the saturated vapor pressure (svp) of water varies from 0°C to 100 °C. The pressure scale (the vertical one) is measured in kilopascals (kPa). 1 atmosphere pressure is 101.325 kPa. A liquid boils when its saturated vapor pressure becomes equal to the external pressure on the liquid. When that happens, it enables bubbles of vapor to form throughout the liquid - those are the bubbles you see when a liquid boils. If the external pressure is higher than the saturated vapor pressure, these bubbles are prevented from forming, and you just get evaporation at the surface of the liquid. If the liquid is in an open container and exposed to normal atmospheric pressure, the liquid boils when its saturated vapor pressure becomes equal to 1 atmosphere (or 101325 Pa or 101.325 kPa or 760 mmHg). This happens with water when the temperature reaches 100°C. But at different pressures, water will boil at different temperatures. For example, at the top of Mount Everest the pressure is so low that water will boil at about 70°C. Whenever we just talk about "the boiling point" of a liquid, we always assume that it is being measured at exactly 1 atmosphere pressure. In practice, of course, that is rarely exactly true. Solids can also lose particles from their surface to form a vapor, except that in this case we call the effect sublimation rather than evaporation. Sublimation is the direct change from solid to vapor (or vice versa) without going through the liquid stage. In most cases, at ordinary temperatures, the saturated vapor pressures of solids range from low to very, very, very low. The forces of attraction in many solids are too high to allow much loss of particles from the surface. However, there are some which do easily form vapors. For example, naphthalene (used in old-fashioned "moth balls" to deter clothes moths) has quite a strong smell. Molecules must be breaking away from the surface as a vapor, because otherwise you would not be able to smell it. Another fairly common example (discussed in detail on another page) is solid carbon dioxide - "dry ice". This never forms a liquid at atmospheric pressure and always converts directly from solid to vapor. That's why it is known as dry ice. Jim Clark ( ) | 7,316 | 922 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Rotational_Spectroscopy/Microwave_Rotational_Spectroscopy |
Microwave rotational spectroscopy uses microwave radiation to measure the energies of rotational transitions for molecules in the gas phase. It accomplishes this through the interaction of the of the molecules with the electromagnetic field of the exciting microwave photon. To probe the pure rotational transitions for molecules, scientists use microwave rotational spectroscopy. This spectroscopy utilizes photons in the microwave range to cause transitions between the quantum rotational energy levels of a gas molecule. The reason why the sample must be in the gas phase is due to intermolecular interactions hindering rotations in the liquid and solid phases of the molecule. For microwave spectroscopy, molecules can be broken down into 5 categories based on their shape and the inertia around their 3 orthogonal rotational axes. These 5 categories include diatomic molecules, linear molecules, spherical tops, symmetric tops and asymmetric tops. The to the rigid rotor is \[H = T\] since, \[H = T + V\] Where \(T\) is kinetic energy and \(V\) is potential energy. Potential energy, \(V\), is 0 because there is no resistance to the rotation (similar to a model). Since \(H = T\), we can also say that: \[{T = }\dfrac{1}{2}\sum{m_{i}v_{i}^2}\] However, we have to determine \(v_i\) in terms of rotation since we are dealing with rotation. Since, \[\omega = \dfrac{v}{r}\] where \(\omega\) = angular velocity, we can say that: \[v_{i} = \omega{X}r_{i}\] Thus we can rewrite the T equation as: \[T = \dfrac{1}{2}\sum{m_{i}v_{i}\left(\omega{X}r_{i}\right)}\] Since \(\omega\) is a scalar constant, we can rewrite the T equation as: \[T = \dfrac{\omega}{2}\sum{m_{i}\left(v_{i}{X}r_{i}\right)} = \dfrac{\omega}{2}\sum{l_{i}} = \omega\dfrac{L}{2}\] where \(l_i\) is the of the particle, and is the angular momentum of the . Also, we know from physics that, \[L = I\omega\] where is the moment of inertia of the rigid body relative to the axis of rotation. W \[T = \omega\dfrac{{I}\omega}{2} = \dfrac{1}{2}{I}\omega^2\] \[E_n = \dfrac{J(J+1)h^2}{8\pi^2I} \] and \(\hbar\) However, if we let: \[B = \dfrac {h}{8 \pi^2I} \] where \(B\) is a rotational constant, then we can substitute it into the \(E_n\) equation and get: \[E_{n} = J(J+1)Bh\] Considering the transition energy between two energy levels, the difference is a multiple of 2. That is, from \(J = 0\) to \(J = 1\), the \(\Delta{E_{0 \rightarrow 1}}\) is 2Bh and from J = 1 to J = 2, the \(\Delta{E}_{1 \rightarrow 2}\) is 4Bh. When a gas molecule is irradiated with microwave radiation, a photon can be absorbed through the interaction of the photon’s electronic field with the electrons in the molecules. For the microwave region this energy absorption is in the range needed to cause transitions between rotational states of the molecule. However, only molecules with a permanent dipole that changes upon rotation can be investigated using microwave spectroscopy. This is due to the fact that their must be a charge difference across the molecule for the oscillating electric field of the photon to impart a torque upon the molecule around an axis that is perpendicular to this dipole and that passes through the molecules center of mass. This interaction can be expressed by the transition dipole moment for the transition between two rotational states \[ \text{Probability of Transition}=\int \psi_{rot}(F)\hat\mu \psi_{rot}(I)d\tau\] Where Ψ is the complex conjugate of the wave function for the final rotational state, Ψ is the wave function of the initial rotational state , and μ is the dipole moment operator with Cartesian coordinates of μ , μ , μ . For this integral to be nonzero the integrand must be an even function. This is due to the fact that any odd function integrated from negative infinity to positive infinity, or any other symmetric limits, is always zero. In addition to the constraints imposed by the transition moment integral, transitions between rotational states are also limited by the nature of the photon itself. A photon contains one unit of angular momentum, so when it interacts with a molecule it can only impart one unit of angular momentum to the molecule. This leads to the selection rule that a transition can only occur between rotational energy levels that are only one quantum rotation level (J) away from another . \[ \Delta\textrm{J}=\pm 1 \] The transition moment integral and the selection rule for rotational transitions tell if a transition from one rotational state to another is allowed. However, what these do not take into account is whether or not the state being transitioned from is actually populated, meaning that the molecule is in that energy state. This leads to the concept of the Boltzmann distribution of states. The is a statistical distribution of energy states for an ensemble of molecules based on the temperature of the sample . \[ \dfrac{n_J}{n_0} = \dfrac{e^{(-E_{rot}(J)/RT)}}{\sum_{J=1}^{J=n} e^{(-E_{rot}(J)/RT)}}\] where E is the molar energy of the J rotational energy state of the molecule, This distribution of energy states is the main contributing factor for the observed absorption intensity distributions seen in the microwave spectrum. This distribution makes it so that the absorption peaks that correspond to the transition from the energy state with the largest population based on the Boltzmann equation will have the largest absorption peak, with the peaks on either side steadily decreasing. A molecule can have three types of degrees of freedom and a total of 3N degrees of freedom, where N equals the number of atoms in the molecule. These degrees of freedom can be broken down into 3 categories . Each of these degrees of freedom is able to store energy. However, In the case of rotational and vibrational degrees of freedom, energy can only be stored in discrete amounts. This is due to the quantized break down of energy levels in a molecule described by quantum mechanics. In the case of rotations the energy stored is dependent on the rotational inertia of the gas along with the corresponding quantum number describing the energy level. To analyze molecules for rotational spectroscopy, we can break molecules down into 5 categories based on their shapes and their moments of inertia around their 3 orthogonal rotational axes: The rotations of a diatomic molecule can be modeled as a rigid rotor. This rigid rotor model has two masses attached to each other with a fixed distance between the two masses. It has an inertia (I) that is equal to the square of the fixed distance between the two masses multiplied by the reduced mass of the rigid rotor. \[ \large I_e= \mu r_e^2\] \[ \large \mu=\dfrac{m_1 m_2} {m_1+m_2} \] Using quantum mechanical calculations it can be shown that the energy levels of the rigid rotator depend on the inertia of the rigid rotator and the quantum rotational number J . \[ E(J) = B_e J(J+1) \] \[ B_e = \dfrac{h}{8 \pi^2 cI_e}\] However, this rigid rotor model fails to take into account that bonds do not act like a rod with a fixed distance, but like a spring. This means that as the angular velocity of the molecule increases so does the distance between the atoms. This leads us to the in which a centrifugal distortion term (\(D_e\)) is added to the energy equation to account for this stretching during rotation. \[ E(J)(cm^{-1}) = B_e J(J+1) – D_e J^2(J+1)^2\] This means that for a diatomic molecule the transitional energy between two rotational states equals \[ E=B_e[J'(J'+1)-J''(J''+1)]-D_e[J'^2(J'+1)^2-J''^2(J'+1)^2]\label{8} \] Where J’ is the quantum number of the final rotational energy state and J’’ is the quantum number of the initial rotational energy state. Using the selection rule of \(\Delta{J}= \pm 1\) the spacing between peaks in the microwave absorption spectrum of a diatomic molecule will equal \[ E_R =(2B_e-4D_e)+(2B_e-12D_e){J}''-4D_e J''^3\] Linear molecules behave in the same way as diatomic molecules when it comes to rotations. For this reason they can be modeled as a non-rigid rotor just like diatomic molecules. This means that linear molecule have the same equation for their rotational energy levels. The only difference is there are now more masses along the rotor. This means that the inertia is now the sum of the distance between each mass and the center of mass of the rotor multiplied by the square of the distance between them . \[ \large I_e=\sum_{j=1}^{n} m_j r_{ej}^2 \] Where m is the mass of the jth mass on the rotor and r is the equilibrium distance between the j mass and the center of mass of the rotor. Spherical tops are molecules in which all three orthogonal rotations have equal inertia and they are highly symmetrical. This means that the molecule has no dipole and for this reason spherical tops do not give a microwave rotational spectrum. Examples: Symmetrical tops are molecules with two rotational axes that have the same inertia and one unique rotational axis with a different inertia. Symmetrical tops can be divided into two categories based on the relationship between the inertia of the unique axis and the inertia of the two axes with equivalent inertia. If the unique rotational axis has a greater inertia than the degenerate axes the molecule is called an oblate symmetrical top. If the unique rotational axis has a lower inertia than the degenerate axes the molecule is called a prolate symmetrical top. For simplification think of these two categories as either frisbees for oblate tops or footballs for prolate tops. In the case of linear molecules there is one degenerate rotational axis which in turn has a single rotational constant. With symmetrical tops now there is one unique axis and two degenerate axes. This means an additional rotational constant is needed to describe the energy levels of a symmetrical top. In addition to the rotational constant an additional quantum number must be introduced to describe the rotational energy levels of the symmetric top. These two additions give us the following rotational energy levels of a prolate and oblate symmetric top \[E_{(J,K)}(cm^{-1})=B_e*J(J+1)+(A_e-B_e)K^2 \] Where B is the rotational constant of the unique axis, A is the rotational constant of the degenerate axes, \(J\) is the total rotational angular momentum quantum number and K is the quantum number that represents the portion of the total angular momentum that lies along the unique rotational axis. This leads to the property that \(K\) is always equal to or less than \(J\). Thus we get the two selection rules for symmetric tops \[\Delta J = 0, \pm1 \] \[\Delta K=0\] when \( K\neq 0 \) \[\Delta J = \pm1 \] \[\Delta K=0 \] when \( K=0 \) However, like the rigid rotor approximation for linear molecules, we must also take into account the elasticity of the bonds in symmetric tops. Therefore, in a similar manner to the rigid rotor we add a centrifugal coupling term, but this time we have one for each quantum number and one for the coupling between the two. \[E_{(J,K)}(cm^{-1})=B_e J(J+1)-D_{eJ} J^2(J+1)^2+(A_e-B_e)*K^2 \label{13}\] \[-D_{ek} K^4-D_{ejk} J(J +1)K^2 \label{14}\] Asymmetrical tops have three orthogonal rotational axes that all have different moments of inertia and most molecules fall into this category. Unlike linear molecules and symmetric tops these types of molecules do not have a simplified energy equation to determine the energy levels of the rotations. These types of molecules do not follow a specific pattern and usually have very complex microwave spectra. In addition to microwave spectroscopy, IR spectroscopy can also be used to probe rotational transitions in a molecule. However, in the case of IR spectroscopy the rotational transitions are coupled to the vibrational transitions of the molecule. One other spectroscopy that can probe the rotational transitions in a molecule is Raman spectroscopy, which uses UV-visible light scattering to determine energy levels in a molecule. However, a very high sensitivity detector must be used to analyze rotational energy levels of a molecule. | 12,069 | 923 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/06._One_Dimensional_Harmonic_Oscillator/Anharmonic_Oscillator |
Anharmonic oscillation is defined as the deviation of a system from harmonic oscillation, or an oscillator not oscillating in simple harmonic motion. A obeys Hooke's Law and is an idealized expression that assumes that a system displaced from equilibrium responds with a restoring force whose magnitude is proportional to the displacement. In nature, idealized situations break down and fails to describe linear equations of motion. Anharmonic oscillation is described as the restoring force is no longer proportional to the displacement. Two forms of nonlinearity are used to describe real-world situations: Anharmonic oscillators can be approximated to a harmonic oscillator and the anharmonicity can be calculated using perturbation theory. Figure \(\Page {1}\) shows the ground state potential well and is calculated using the energy levels of a harmonic oscillator with the first anharmonic correction. \(D_o\) is the dissociation energy, which is different from the well depth \(D_e\). The vibrational energy levels of this plot are calculated using the harmonic oscillator model: \[ E_v = \left(v + \dfrac{1}{2}\right) v_e - \left(v + \dfrac{1}{2}\right)^2 v_e x_e + \left(v + \dfrac{1}{2}\right)^3 v_e y_e + higher \; terms\] where \( v \) is the vibrational quantum number and \( x_e\) and \( y_e\) are the first and second anharmonicity constants, respectively. The v = 0 level is the vibrational ground state. The lines in the first figure represent overtones correspond to the transitions of the quantum number \(v\) which terminate at the top line = \( v_{max}\). Because this line is less confining than a parabola, the energy levels become less widely spaced at high excitation. These overtones are present because the selection rule is derived from the properties of harmonic oscillator wavefunctions, which are only approximately valid in the presence of anharmonicity. | 1,906 | 924 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/11%3A_Quantum_Mechanics_and_Atomic_Structure/11.07%3A_The_Schrodinger_Wave_Equation |
Beginning in the early 20th century, physicists began to acknowledge that matter--much like electromagnetic radiation--possessed wave-like behaviors. While electromagnetic radiation were well understood to obey Maxwell's Equations, matter obeyed no known equations. In 1926, the Austrian physicist Erwin Schrödinger formulated what came to be known as the (time-dependent) Schrödinger Equation: \[ i\hbar\dfrac{\partial}{\partial t}\Psi(x,t)=\dfrac{-\hbar}{2m}\nabla^2\Psi(x,t) +V(x)\Psi(x,t) \label{1.1}\] Equation \(\ref{1.1}\) effectively describes matter as a wave that fluctuates with both displacement and time. Since the imaginary portion of the equation dictates its time dependence, it is sufficed to say that for most purposes it can be treated as time-independent. The result is seen in Equation \(\ref{1.3}\): \[\dfrac{-\hbar^2}{2m}\dfrac{d^2\psi(x)}{dx^2}+V(x)\psi(x) = E\psi(x) \label{1.3}\] Although this time-independent Schrödinger Equation can be useful to describe a matter wave in free space, we are most interested in waves when confined to a small region, such as an electron confined in a small region around the nucleus of an atom. Several different models have been developed that provide a means by which to study a matter-wave when confined to a small region: the particle in a box (infinite well), finite well, and the Hydrogen atom. We will discuss each of these in order to develop a greater understanding for how a wave behaves when it is in a bound state. There are four general aspects that are applicable to an acceptable wavefunction Since wavefunctions can in general be complex functions, the physical significance cannot be found from the function itself because the \(\sqrt {-1}\) is not a property of the physical world. Rather, the physical significance is found in the product of the wavefunction and its complex conjugate, i.e. the absolute square of the wavefunction, which also is called the square of the modulus. \[ \psi ^* (r , t ) \psi (r , t ) = {|\psi (r , t ) |}^2 \label{3-34}\] where \(r\) is a vector (x, y, z) specifying a point in three-dimensional space. The square is used, rather than the modulus itself, just like the intensity of a light wave depends on the square of the electric field. At one time it was thought that for an electron described by the wavefunction \(\psi(r)\), the quantity \(e\psi^*(r_i)\psi(r_i)d\tau\) was the amount of charge to be found in the volume \(d\tau\) located at \(r_i\). However, Max Born found this interpretation to be inconsistent with the results of experiments. The Born interpretation, which generally is accepted today, is that \(\psi^*(r_i)\psi(r_i)\, d\tau\) is the probability that the electron is in the volume dτ located at \(r_i\). The Born interpretation therefore calls the wavefunction the probability amplitude, the absolute square of the wavefunction is called the probability density, and the probability density times a volume element in three-dimensional space (\(d\tau\)) is the probability. The idea that we can understand the world of atoms and molecules only in terms of probabilities is disturbing to some, who are seeking more satisfying descriptions through ongoing research. A probability is a real number between 0 and 1. An outcome of a measurement which has a probability 0 is an impossible outcome, whereas an outcome which has a probability 1 is a certain outcome. The probability of a measurement of \(x\) yielding a result between \(-\infty\) and \(+\infty\) is \[P_{x \in -\infty:\infty}(t) = \int_{-\infty}^{\infty}\vert\psi(x,t)\vert^{ 2} dx. \label{3.6.2}\] However, a measurement of \(x\) yield a value between \(-\infty\) and \(+\infty\), since the particle has to be located somewhere. It follows that \(P_{x \in -\infty:\infty}(t) =1\), or \[\int_{-\infty}^{\infty}\vert\psi(x,t)\vert^{ 2} dx = 1, \label{3.6.3}\] which is generally known as the for the wavefunction. Normalize the wavefunction of a Gaussian wave packet, centered on \(x=x_o\), and of characteristic width \(\sigma\): i.e., \[\psi(x) = \psi_0 {\rm e}^{-(x-x_0)^{ 2}/(4 \sigma^2)}. \label{3.6.4}\] To determine the normalization constant \(\psi_0\), we simply substitute Equation \(\ref{3.6.4}\) into Equation \(\ref{3.6.3}\), to obtain \[\vert\psi_0\vert^{ 2}\int_{-\infty}^{\infty}{\rm e}^{-(x-x_0)^{ 2}/(2 \sigma^2)} dx = 1. \label{3.6.5}\] Changing the variable of integration to \(y=(x-x_0)/(\sqrt{2} \sigma)\), we get \[\vert\psi_0\vert^{ 2}\sqrt{2} \sigma \int_{-\infty}^{\infty}{\rm e}^{-y^2} dy=1. \label{3.6.6}\] However, \[\int_{-\infty}^{\infty}{\rm e}^{-y^2} dy = \sqrt{\pi}, \label{3.6.7}\] which implies that \[\vert\psi_0\vert^{ 2} = \dfrac{1}{(2\pi \sigma^2)^{1/2}}. \label{3.6.8}\] Hence, a general Gaussian wavefunction takes the form \[\psi(x) = \dfrac{e^{\rm{i} \phi}}{(2\pi \sigma^2)^{1/4}}e^{-(x-x_0)^2/(4 \sigma^2)} \label{3.6.9}\] where \(\phi\) is an arbitrary real phase-angle. ") | 4,938 | 925 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Rotational_Spectroscopy/Microwave_Rotational_Spectroscopy |
Microwave rotational spectroscopy uses microwave radiation to measure the energies of rotational transitions for molecules in the gas phase. It accomplishes this through the interaction of the of the molecules with the electromagnetic field of the exciting microwave photon. To probe the pure rotational transitions for molecules, scientists use microwave rotational spectroscopy. This spectroscopy utilizes photons in the microwave range to cause transitions between the quantum rotational energy levels of a gas molecule. The reason why the sample must be in the gas phase is due to intermolecular interactions hindering rotations in the liquid and solid phases of the molecule. For microwave spectroscopy, molecules can be broken down into 5 categories based on their shape and the inertia around their 3 orthogonal rotational axes. These 5 categories include diatomic molecules, linear molecules, spherical tops, symmetric tops and asymmetric tops. The to the rigid rotor is \[H = T\] since, \[H = T + V\] Where \(T\) is kinetic energy and \(V\) is potential energy. Potential energy, \(V\), is 0 because there is no resistance to the rotation (similar to a model). Since \(H = T\), we can also say that: \[{T = }\dfrac{1}{2}\sum{m_{i}v_{i}^2}\] However, we have to determine \(v_i\) in terms of rotation since we are dealing with rotation. Since, \[\omega = \dfrac{v}{r}\] where \(\omega\) = angular velocity, we can say that: \[v_{i} = \omega{X}r_{i}\] Thus we can rewrite the T equation as: \[T = \dfrac{1}{2}\sum{m_{i}v_{i}\left(\omega{X}r_{i}\right)}\] Since \(\omega\) is a scalar constant, we can rewrite the T equation as: \[T = \dfrac{\omega}{2}\sum{m_{i}\left(v_{i}{X}r_{i}\right)} = \dfrac{\omega}{2}\sum{l_{i}} = \omega\dfrac{L}{2}\] where \(l_i\) is the of the particle, and is the angular momentum of the . Also, we know from physics that, \[L = I\omega\] where is the moment of inertia of the rigid body relative to the axis of rotation. W \[T = \omega\dfrac{{I}\omega}{2} = \dfrac{1}{2}{I}\omega^2\] \[E_n = \dfrac{J(J+1)h^2}{8\pi^2I} \] and \(\hbar\) However, if we let: \[B = \dfrac {h}{8 \pi^2I} \] where \(B\) is a rotational constant, then we can substitute it into the \(E_n\) equation and get: \[E_{n} = J(J+1)Bh\] Considering the transition energy between two energy levels, the difference is a multiple of 2. That is, from \(J = 0\) to \(J = 1\), the \(\Delta{E_{0 \rightarrow 1}}\) is 2Bh and from J = 1 to J = 2, the \(\Delta{E}_{1 \rightarrow 2}\) is 4Bh. When a gas molecule is irradiated with microwave radiation, a photon can be absorbed through the interaction of the photon’s electronic field with the electrons in the molecules. For the microwave region this energy absorption is in the range needed to cause transitions between rotational states of the molecule. However, only molecules with a permanent dipole that changes upon rotation can be investigated using microwave spectroscopy. This is due to the fact that their must be a charge difference across the molecule for the oscillating electric field of the photon to impart a torque upon the molecule around an axis that is perpendicular to this dipole and that passes through the molecules center of mass. This interaction can be expressed by the transition dipole moment for the transition between two rotational states \[ \text{Probability of Transition}=\int \psi_{rot}(F)\hat\mu \psi_{rot}(I)d\tau\] Where Ψ is the complex conjugate of the wave function for the final rotational state, Ψ is the wave function of the initial rotational state , and μ is the dipole moment operator with Cartesian coordinates of μ , μ , μ . For this integral to be nonzero the integrand must be an even function. This is due to the fact that any odd function integrated from negative infinity to positive infinity, or any other symmetric limits, is always zero. In addition to the constraints imposed by the transition moment integral, transitions between rotational states are also limited by the nature of the photon itself. A photon contains one unit of angular momentum, so when it interacts with a molecule it can only impart one unit of angular momentum to the molecule. This leads to the selection rule that a transition can only occur between rotational energy levels that are only one quantum rotation level (J) away from another . \[ \Delta\textrm{J}=\pm 1 \] The transition moment integral and the selection rule for rotational transitions tell if a transition from one rotational state to another is allowed. However, what these do not take into account is whether or not the state being transitioned from is actually populated, meaning that the molecule is in that energy state. This leads to the concept of the Boltzmann distribution of states. The is a statistical distribution of energy states for an ensemble of molecules based on the temperature of the sample . \[ \dfrac{n_J}{n_0} = \dfrac{e^{(-E_{rot}(J)/RT)}}{\sum_{J=1}^{J=n} e^{(-E_{rot}(J)/RT)}}\] where E is the molar energy of the J rotational energy state of the molecule, This distribution of energy states is the main contributing factor for the observed absorption intensity distributions seen in the microwave spectrum. This distribution makes it so that the absorption peaks that correspond to the transition from the energy state with the largest population based on the Boltzmann equation will have the largest absorption peak, with the peaks on either side steadily decreasing. A molecule can have three types of degrees of freedom and a total of 3N degrees of freedom, where N equals the number of atoms in the molecule. These degrees of freedom can be broken down into 3 categories . Each of these degrees of freedom is able to store energy. However, In the case of rotational and vibrational degrees of freedom, energy can only be stored in discrete amounts. This is due to the quantized break down of energy levels in a molecule described by quantum mechanics. In the case of rotations the energy stored is dependent on the rotational inertia of the gas along with the corresponding quantum number describing the energy level. To analyze molecules for rotational spectroscopy, we can break molecules down into 5 categories based on their shapes and their moments of inertia around their 3 orthogonal rotational axes: The rotations of a diatomic molecule can be modeled as a rigid rotor. This rigid rotor model has two masses attached to each other with a fixed distance between the two masses. It has an inertia (I) that is equal to the square of the fixed distance between the two masses multiplied by the reduced mass of the rigid rotor. \[ \large I_e= \mu r_e^2\] \[ \large \mu=\dfrac{m_1 m_2} {m_1+m_2} \] Using quantum mechanical calculations it can be shown that the energy levels of the rigid rotator depend on the inertia of the rigid rotator and the quantum rotational number J . \[ E(J) = B_e J(J+1) \] \[ B_e = \dfrac{h}{8 \pi^2 cI_e}\] However, this rigid rotor model fails to take into account that bonds do not act like a rod with a fixed distance, but like a spring. This means that as the angular velocity of the molecule increases so does the distance between the atoms. This leads us to the in which a centrifugal distortion term (\(D_e\)) is added to the energy equation to account for this stretching during rotation. \[ E(J)(cm^{-1}) = B_e J(J+1) – D_e J^2(J+1)^2\] This means that for a diatomic molecule the transitional energy between two rotational states equals \[ E=B_e[J'(J'+1)-J''(J''+1)]-D_e[J'^2(J'+1)^2-J''^2(J'+1)^2]\label{8} \] Where J’ is the quantum number of the final rotational energy state and J’’ is the quantum number of the initial rotational energy state. Using the selection rule of \(\Delta{J}= \pm 1\) the spacing between peaks in the microwave absorption spectrum of a diatomic molecule will equal \[ E_R =(2B_e-4D_e)+(2B_e-12D_e){J}''-4D_e J''^3\] Linear molecules behave in the same way as diatomic molecules when it comes to rotations. For this reason they can be modeled as a non-rigid rotor just like diatomic molecules. This means that linear molecule have the same equation for their rotational energy levels. The only difference is there are now more masses along the rotor. This means that the inertia is now the sum of the distance between each mass and the center of mass of the rotor multiplied by the square of the distance between them . \[ \large I_e=\sum_{j=1}^{n} m_j r_{ej}^2 \] Where m is the mass of the jth mass on the rotor and r is the equilibrium distance between the j mass and the center of mass of the rotor. Spherical tops are molecules in which all three orthogonal rotations have equal inertia and they are highly symmetrical. This means that the molecule has no dipole and for this reason spherical tops do not give a microwave rotational spectrum. Examples: Symmetrical tops are molecules with two rotational axes that have the same inertia and one unique rotational axis with a different inertia. Symmetrical tops can be divided into two categories based on the relationship between the inertia of the unique axis and the inertia of the two axes with equivalent inertia. If the unique rotational axis has a greater inertia than the degenerate axes the molecule is called an oblate symmetrical top. If the unique rotational axis has a lower inertia than the degenerate axes the molecule is called a prolate symmetrical top. For simplification think of these two categories as either frisbees for oblate tops or footballs for prolate tops. In the case of linear molecules there is one degenerate rotational axis which in turn has a single rotational constant. With symmetrical tops now there is one unique axis and two degenerate axes. This means an additional rotational constant is needed to describe the energy levels of a symmetrical top. In addition to the rotational constant an additional quantum number must be introduced to describe the rotational energy levels of the symmetric top. These two additions give us the following rotational energy levels of a prolate and oblate symmetric top \[E_{(J,K)}(cm^{-1})=B_e*J(J+1)+(A_e-B_e)K^2 \] Where B is the rotational constant of the unique axis, A is the rotational constant of the degenerate axes, \(J\) is the total rotational angular momentum quantum number and K is the quantum number that represents the portion of the total angular momentum that lies along the unique rotational axis. This leads to the property that \(K\) is always equal to or less than \(J\). Thus we get the two selection rules for symmetric tops \[\Delta J = 0, \pm1 \] \[\Delta K=0\] when \( K\neq 0 \) \[\Delta J = \pm1 \] \[\Delta K=0 \] when \( K=0 \) However, like the rigid rotor approximation for linear molecules, we must also take into account the elasticity of the bonds in symmetric tops. Therefore, in a similar manner to the rigid rotor we add a centrifugal coupling term, but this time we have one for each quantum number and one for the coupling between the two. \[E_{(J,K)}(cm^{-1})=B_e J(J+1)-D_{eJ} J^2(J+1)^2+(A_e-B_e)*K^2 \label{13}\] \[-D_{ek} K^4-D_{ejk} J(J +1)K^2 \label{14}\] Asymmetrical tops have three orthogonal rotational axes that all have different moments of inertia and most molecules fall into this category. Unlike linear molecules and symmetric tops these types of molecules do not have a simplified energy equation to determine the energy levels of the rotations. These types of molecules do not follow a specific pattern and usually have very complex microwave spectra. In addition to microwave spectroscopy, IR spectroscopy can also be used to probe rotational transitions in a molecule. However, in the case of IR spectroscopy the rotational transitions are coupled to the vibrational transitions of the molecule. One other spectroscopy that can probe the rotational transitions in a molecule is Raman spectroscopy, which uses UV-visible light scattering to determine energy levels in a molecule. However, a very high sensitivity detector must be used to analyze rotational energy levels of a molecule. | 12,069 | 926 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Rotational_Spectroscopy/Microwave_Rotational_Spectroscopy |
Microwave rotational spectroscopy uses microwave radiation to measure the energies of rotational transitions for molecules in the gas phase. It accomplishes this through the interaction of the of the molecules with the electromagnetic field of the exciting microwave photon. To probe the pure rotational transitions for molecules, scientists use microwave rotational spectroscopy. This spectroscopy utilizes photons in the microwave range to cause transitions between the quantum rotational energy levels of a gas molecule. The reason why the sample must be in the gas phase is due to intermolecular interactions hindering rotations in the liquid and solid phases of the molecule. For microwave spectroscopy, molecules can be broken down into 5 categories based on their shape and the inertia around their 3 orthogonal rotational axes. These 5 categories include diatomic molecules, linear molecules, spherical tops, symmetric tops and asymmetric tops. The to the rigid rotor is \[H = T\] since, \[H = T + V\] Where \(T\) is kinetic energy and \(V\) is potential energy. Potential energy, \(V\), is 0 because there is no resistance to the rotation (similar to a model). Since \(H = T\), we can also say that: \[{T = }\dfrac{1}{2}\sum{m_{i}v_{i}^2}\] However, we have to determine \(v_i\) in terms of rotation since we are dealing with rotation. Since, \[\omega = \dfrac{v}{r}\] where \(\omega\) = angular velocity, we can say that: \[v_{i} = \omega{X}r_{i}\] Thus we can rewrite the T equation as: \[T = \dfrac{1}{2}\sum{m_{i}v_{i}\left(\omega{X}r_{i}\right)}\] Since \(\omega\) is a scalar constant, we can rewrite the T equation as: \[T = \dfrac{\omega}{2}\sum{m_{i}\left(v_{i}{X}r_{i}\right)} = \dfrac{\omega}{2}\sum{l_{i}} = \omega\dfrac{L}{2}\] where \(l_i\) is the of the particle, and is the angular momentum of the . Also, we know from physics that, \[L = I\omega\] where is the moment of inertia of the rigid body relative to the axis of rotation. W \[T = \omega\dfrac{{I}\omega}{2} = \dfrac{1}{2}{I}\omega^2\] \[E_n = \dfrac{J(J+1)h^2}{8\pi^2I} \] and \(\hbar\) However, if we let: \[B = \dfrac {h}{8 \pi^2I} \] where \(B\) is a rotational constant, then we can substitute it into the \(E_n\) equation and get: \[E_{n} = J(J+1)Bh\] Considering the transition energy between two energy levels, the difference is a multiple of 2. That is, from \(J = 0\) to \(J = 1\), the \(\Delta{E_{0 \rightarrow 1}}\) is 2Bh and from J = 1 to J = 2, the \(\Delta{E}_{1 \rightarrow 2}\) is 4Bh. When a gas molecule is irradiated with microwave radiation, a photon can be absorbed through the interaction of the photon’s electronic field with the electrons in the molecules. For the microwave region this energy absorption is in the range needed to cause transitions between rotational states of the molecule. However, only molecules with a permanent dipole that changes upon rotation can be investigated using microwave spectroscopy. This is due to the fact that their must be a charge difference across the molecule for the oscillating electric field of the photon to impart a torque upon the molecule around an axis that is perpendicular to this dipole and that passes through the molecules center of mass. This interaction can be expressed by the transition dipole moment for the transition between two rotational states \[ \text{Probability of Transition}=\int \psi_{rot}(F)\hat\mu \psi_{rot}(I)d\tau\] Where Ψ is the complex conjugate of the wave function for the final rotational state, Ψ is the wave function of the initial rotational state , and μ is the dipole moment operator with Cartesian coordinates of μ , μ , μ . For this integral to be nonzero the integrand must be an even function. This is due to the fact that any odd function integrated from negative infinity to positive infinity, or any other symmetric limits, is always zero. In addition to the constraints imposed by the transition moment integral, transitions between rotational states are also limited by the nature of the photon itself. A photon contains one unit of angular momentum, so when it interacts with a molecule it can only impart one unit of angular momentum to the molecule. This leads to the selection rule that a transition can only occur between rotational energy levels that are only one quantum rotation level (J) away from another . \[ \Delta\textrm{J}=\pm 1 \] The transition moment integral and the selection rule for rotational transitions tell if a transition from one rotational state to another is allowed. However, what these do not take into account is whether or not the state being transitioned from is actually populated, meaning that the molecule is in that energy state. This leads to the concept of the Boltzmann distribution of states. The is a statistical distribution of energy states for an ensemble of molecules based on the temperature of the sample . \[ \dfrac{n_J}{n_0} = \dfrac{e^{(-E_{rot}(J)/RT)}}{\sum_{J=1}^{J=n} e^{(-E_{rot}(J)/RT)}}\] where E is the molar energy of the J rotational energy state of the molecule, This distribution of energy states is the main contributing factor for the observed absorption intensity distributions seen in the microwave spectrum. This distribution makes it so that the absorption peaks that correspond to the transition from the energy state with the largest population based on the Boltzmann equation will have the largest absorption peak, with the peaks on either side steadily decreasing. A molecule can have three types of degrees of freedom and a total of 3N degrees of freedom, where N equals the number of atoms in the molecule. These degrees of freedom can be broken down into 3 categories . Each of these degrees of freedom is able to store energy. However, In the case of rotational and vibrational degrees of freedom, energy can only be stored in discrete amounts. This is due to the quantized break down of energy levels in a molecule described by quantum mechanics. In the case of rotations the energy stored is dependent on the rotational inertia of the gas along with the corresponding quantum number describing the energy level. To analyze molecules for rotational spectroscopy, we can break molecules down into 5 categories based on their shapes and their moments of inertia around their 3 orthogonal rotational axes: The rotations of a diatomic molecule can be modeled as a rigid rotor. This rigid rotor model has two masses attached to each other with a fixed distance between the two masses. It has an inertia (I) that is equal to the square of the fixed distance between the two masses multiplied by the reduced mass of the rigid rotor. \[ \large I_e= \mu r_e^2\] \[ \large \mu=\dfrac{m_1 m_2} {m_1+m_2} \] Using quantum mechanical calculations it can be shown that the energy levels of the rigid rotator depend on the inertia of the rigid rotator and the quantum rotational number J . \[ E(J) = B_e J(J+1) \] \[ B_e = \dfrac{h}{8 \pi^2 cI_e}\] However, this rigid rotor model fails to take into account that bonds do not act like a rod with a fixed distance, but like a spring. This means that as the angular velocity of the molecule increases so does the distance between the atoms. This leads us to the in which a centrifugal distortion term (\(D_e\)) is added to the energy equation to account for this stretching during rotation. \[ E(J)(cm^{-1}) = B_e J(J+1) – D_e J^2(J+1)^2\] This means that for a diatomic molecule the transitional energy between two rotational states equals \[ E=B_e[J'(J'+1)-J''(J''+1)]-D_e[J'^2(J'+1)^2-J''^2(J'+1)^2]\label{8} \] Where J’ is the quantum number of the final rotational energy state and J’’ is the quantum number of the initial rotational energy state. Using the selection rule of \(\Delta{J}= \pm 1\) the spacing between peaks in the microwave absorption spectrum of a diatomic molecule will equal \[ E_R =(2B_e-4D_e)+(2B_e-12D_e){J}''-4D_e J''^3\] Linear molecules behave in the same way as diatomic molecules when it comes to rotations. For this reason they can be modeled as a non-rigid rotor just like diatomic molecules. This means that linear molecule have the same equation for their rotational energy levels. The only difference is there are now more masses along the rotor. This means that the inertia is now the sum of the distance between each mass and the center of mass of the rotor multiplied by the square of the distance between them . \[ \large I_e=\sum_{j=1}^{n} m_j r_{ej}^2 \] Where m is the mass of the jth mass on the rotor and r is the equilibrium distance between the j mass and the center of mass of the rotor. Spherical tops are molecules in which all three orthogonal rotations have equal inertia and they are highly symmetrical. This means that the molecule has no dipole and for this reason spherical tops do not give a microwave rotational spectrum. Examples: Symmetrical tops are molecules with two rotational axes that have the same inertia and one unique rotational axis with a different inertia. Symmetrical tops can be divided into two categories based on the relationship between the inertia of the unique axis and the inertia of the two axes with equivalent inertia. If the unique rotational axis has a greater inertia than the degenerate axes the molecule is called an oblate symmetrical top. If the unique rotational axis has a lower inertia than the degenerate axes the molecule is called a prolate symmetrical top. For simplification think of these two categories as either frisbees for oblate tops or footballs for prolate tops. In the case of linear molecules there is one degenerate rotational axis which in turn has a single rotational constant. With symmetrical tops now there is one unique axis and two degenerate axes. This means an additional rotational constant is needed to describe the energy levels of a symmetrical top. In addition to the rotational constant an additional quantum number must be introduced to describe the rotational energy levels of the symmetric top. These two additions give us the following rotational energy levels of a prolate and oblate symmetric top \[E_{(J,K)}(cm^{-1})=B_e*J(J+1)+(A_e-B_e)K^2 \] Where B is the rotational constant of the unique axis, A is the rotational constant of the degenerate axes, \(J\) is the total rotational angular momentum quantum number and K is the quantum number that represents the portion of the total angular momentum that lies along the unique rotational axis. This leads to the property that \(K\) is always equal to or less than \(J\). Thus we get the two selection rules for symmetric tops \[\Delta J = 0, \pm1 \] \[\Delta K=0\] when \( K\neq 0 \) \[\Delta J = \pm1 \] \[\Delta K=0 \] when \( K=0 \) However, like the rigid rotor approximation for linear molecules, we must also take into account the elasticity of the bonds in symmetric tops. Therefore, in a similar manner to the rigid rotor we add a centrifugal coupling term, but this time we have one for each quantum number and one for the coupling between the two. \[E_{(J,K)}(cm^{-1})=B_e J(J+1)-D_{eJ} J^2(J+1)^2+(A_e-B_e)*K^2 \label{13}\] \[-D_{ek} K^4-D_{ejk} J(J +1)K^2 \label{14}\] Asymmetrical tops have three orthogonal rotational axes that all have different moments of inertia and most molecules fall into this category. Unlike linear molecules and symmetric tops these types of molecules do not have a simplified energy equation to determine the energy levels of the rotations. These types of molecules do not follow a specific pattern and usually have very complex microwave spectra. In addition to microwave spectroscopy, IR spectroscopy can also be used to probe rotational transitions in a molecule. However, in the case of IR spectroscopy the rotational transitions are coupled to the vibrational transitions of the molecule. One other spectroscopy that can probe the rotational transitions in a molecule is Raman spectroscopy, which uses UV-visible light scattering to determine energy levels in a molecule. However, a very high sensitivity detector must be used to analyze rotational energy levels of a molecule. | 12,069 | 927 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/09%3A_Separation_Purification_and_Identification_of_Organic_Compounds/9.07%3A_Microwave_(Rotational)_Spectra |
Rotational energy levels normally are very closely spaced so low-energy radiation, such as is produced by radio transmitters operating in the microwave region, suffices to change molecular rotational energies. Because electronic and vibrational energy levels are spaced much more widely, and because changes between them, are induced only by higher-energy radiation, microwave absorptions by gaseous substances can be characterized as essentially pure “rotational spectra.” It is possible to obtain rotational moments of inertia from microwave spectra, and from these moments to obtain bond angles and bond distances for simple molecules. An example of the use of microwave spectra is provided by Figure 9-8, which shows separate rotational absorptions observed for trans and gauche conformations of propyl iodide (cf. ). Although microwave spectroscopy, being confined to gases, is not a routine method in the organic laboratory, it is important to us here in setting the stage for the consideration of more complex absorptions that occur with infrared radiation. | 1,079 | 928 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Rotational_Spectroscopy/Microwave_Rotational_Spectroscopy |
Microwave rotational spectroscopy uses microwave radiation to measure the energies of rotational transitions for molecules in the gas phase. It accomplishes this through the interaction of the of the molecules with the electromagnetic field of the exciting microwave photon. To probe the pure rotational transitions for molecules, scientists use microwave rotational spectroscopy. This spectroscopy utilizes photons in the microwave range to cause transitions between the quantum rotational energy levels of a gas molecule. The reason why the sample must be in the gas phase is due to intermolecular interactions hindering rotations in the liquid and solid phases of the molecule. For microwave spectroscopy, molecules can be broken down into 5 categories based on their shape and the inertia around their 3 orthogonal rotational axes. These 5 categories include diatomic molecules, linear molecules, spherical tops, symmetric tops and asymmetric tops. The to the rigid rotor is \[H = T\] since, \[H = T + V\] Where \(T\) is kinetic energy and \(V\) is potential energy. Potential energy, \(V\), is 0 because there is no resistance to the rotation (similar to a model). Since \(H = T\), we can also say that: \[{T = }\dfrac{1}{2}\sum{m_{i}v_{i}^2}\] However, we have to determine \(v_i\) in terms of rotation since we are dealing with rotation. Since, \[\omega = \dfrac{v}{r}\] where \(\omega\) = angular velocity, we can say that: \[v_{i} = \omega{X}r_{i}\] Thus we can rewrite the T equation as: \[T = \dfrac{1}{2}\sum{m_{i}v_{i}\left(\omega{X}r_{i}\right)}\] Since \(\omega\) is a scalar constant, we can rewrite the T equation as: \[T = \dfrac{\omega}{2}\sum{m_{i}\left(v_{i}{X}r_{i}\right)} = \dfrac{\omega}{2}\sum{l_{i}} = \omega\dfrac{L}{2}\] where \(l_i\) is the of the particle, and is the angular momentum of the . Also, we know from physics that, \[L = I\omega\] where is the moment of inertia of the rigid body relative to the axis of rotation. W \[T = \omega\dfrac{{I}\omega}{2} = \dfrac{1}{2}{I}\omega^2\] \[E_n = \dfrac{J(J+1)h^2}{8\pi^2I} \] and \(\hbar\) However, if we let: \[B = \dfrac {h}{8 \pi^2I} \] where \(B\) is a rotational constant, then we can substitute it into the \(E_n\) equation and get: \[E_{n} = J(J+1)Bh\] Considering the transition energy between two energy levels, the difference is a multiple of 2. That is, from \(J = 0\) to \(J = 1\), the \(\Delta{E_{0 \rightarrow 1}}\) is 2Bh and from J = 1 to J = 2, the \(\Delta{E}_{1 \rightarrow 2}\) is 4Bh. When a gas molecule is irradiated with microwave radiation, a photon can be absorbed through the interaction of the photon’s electronic field with the electrons in the molecules. For the microwave region this energy absorption is in the range needed to cause transitions between rotational states of the molecule. However, only molecules with a permanent dipole that changes upon rotation can be investigated using microwave spectroscopy. This is due to the fact that their must be a charge difference across the molecule for the oscillating electric field of the photon to impart a torque upon the molecule around an axis that is perpendicular to this dipole and that passes through the molecules center of mass. This interaction can be expressed by the transition dipole moment for the transition between two rotational states \[ \text{Probability of Transition}=\int \psi_{rot}(F)\hat\mu \psi_{rot}(I)d\tau\] Where Ψ is the complex conjugate of the wave function for the final rotational state, Ψ is the wave function of the initial rotational state , and μ is the dipole moment operator with Cartesian coordinates of μ , μ , μ . For this integral to be nonzero the integrand must be an even function. This is due to the fact that any odd function integrated from negative infinity to positive infinity, or any other symmetric limits, is always zero. In addition to the constraints imposed by the transition moment integral, transitions between rotational states are also limited by the nature of the photon itself. A photon contains one unit of angular momentum, so when it interacts with a molecule it can only impart one unit of angular momentum to the molecule. This leads to the selection rule that a transition can only occur between rotational energy levels that are only one quantum rotation level (J) away from another . \[ \Delta\textrm{J}=\pm 1 \] The transition moment integral and the selection rule for rotational transitions tell if a transition from one rotational state to another is allowed. However, what these do not take into account is whether or not the state being transitioned from is actually populated, meaning that the molecule is in that energy state. This leads to the concept of the Boltzmann distribution of states. The is a statistical distribution of energy states for an ensemble of molecules based on the temperature of the sample . \[ \dfrac{n_J}{n_0} = \dfrac{e^{(-E_{rot}(J)/RT)}}{\sum_{J=1}^{J=n} e^{(-E_{rot}(J)/RT)}}\] where E is the molar energy of the J rotational energy state of the molecule, This distribution of energy states is the main contributing factor for the observed absorption intensity distributions seen in the microwave spectrum. This distribution makes it so that the absorption peaks that correspond to the transition from the energy state with the largest population based on the Boltzmann equation will have the largest absorption peak, with the peaks on either side steadily decreasing. A molecule can have three types of degrees of freedom and a total of 3N degrees of freedom, where N equals the number of atoms in the molecule. These degrees of freedom can be broken down into 3 categories . Each of these degrees of freedom is able to store energy. However, In the case of rotational and vibrational degrees of freedom, energy can only be stored in discrete amounts. This is due to the quantized break down of energy levels in a molecule described by quantum mechanics. In the case of rotations the energy stored is dependent on the rotational inertia of the gas along with the corresponding quantum number describing the energy level. To analyze molecules for rotational spectroscopy, we can break molecules down into 5 categories based on their shapes and their moments of inertia around their 3 orthogonal rotational axes: The rotations of a diatomic molecule can be modeled as a rigid rotor. This rigid rotor model has two masses attached to each other with a fixed distance between the two masses. It has an inertia (I) that is equal to the square of the fixed distance between the two masses multiplied by the reduced mass of the rigid rotor. \[ \large I_e= \mu r_e^2\] \[ \large \mu=\dfrac{m_1 m_2} {m_1+m_2} \] Using quantum mechanical calculations it can be shown that the energy levels of the rigid rotator depend on the inertia of the rigid rotator and the quantum rotational number J . \[ E(J) = B_e J(J+1) \] \[ B_e = \dfrac{h}{8 \pi^2 cI_e}\] However, this rigid rotor model fails to take into account that bonds do not act like a rod with a fixed distance, but like a spring. This means that as the angular velocity of the molecule increases so does the distance between the atoms. This leads us to the in which a centrifugal distortion term (\(D_e\)) is added to the energy equation to account for this stretching during rotation. \[ E(J)(cm^{-1}) = B_e J(J+1) – D_e J^2(J+1)^2\] This means that for a diatomic molecule the transitional energy between two rotational states equals \[ E=B_e[J'(J'+1)-J''(J''+1)]-D_e[J'^2(J'+1)^2-J''^2(J'+1)^2]\label{8} \] Where J’ is the quantum number of the final rotational energy state and J’’ is the quantum number of the initial rotational energy state. Using the selection rule of \(\Delta{J}= \pm 1\) the spacing between peaks in the microwave absorption spectrum of a diatomic molecule will equal \[ E_R =(2B_e-4D_e)+(2B_e-12D_e){J}''-4D_e J''^3\] Linear molecules behave in the same way as diatomic molecules when it comes to rotations. For this reason they can be modeled as a non-rigid rotor just like diatomic molecules. This means that linear molecule have the same equation for their rotational energy levels. The only difference is there are now more masses along the rotor. This means that the inertia is now the sum of the distance between each mass and the center of mass of the rotor multiplied by the square of the distance between them . \[ \large I_e=\sum_{j=1}^{n} m_j r_{ej}^2 \] Where m is the mass of the jth mass on the rotor and r is the equilibrium distance between the j mass and the center of mass of the rotor. Spherical tops are molecules in which all three orthogonal rotations have equal inertia and they are highly symmetrical. This means that the molecule has no dipole and for this reason spherical tops do not give a microwave rotational spectrum. Examples: Symmetrical tops are molecules with two rotational axes that have the same inertia and one unique rotational axis with a different inertia. Symmetrical tops can be divided into two categories based on the relationship between the inertia of the unique axis and the inertia of the two axes with equivalent inertia. If the unique rotational axis has a greater inertia than the degenerate axes the molecule is called an oblate symmetrical top. If the unique rotational axis has a lower inertia than the degenerate axes the molecule is called a prolate symmetrical top. For simplification think of these two categories as either frisbees for oblate tops or footballs for prolate tops. In the case of linear molecules there is one degenerate rotational axis which in turn has a single rotational constant. With symmetrical tops now there is one unique axis and two degenerate axes. This means an additional rotational constant is needed to describe the energy levels of a symmetrical top. In addition to the rotational constant an additional quantum number must be introduced to describe the rotational energy levels of the symmetric top. These two additions give us the following rotational energy levels of a prolate and oblate symmetric top \[E_{(J,K)}(cm^{-1})=B_e*J(J+1)+(A_e-B_e)K^2 \] Where B is the rotational constant of the unique axis, A is the rotational constant of the degenerate axes, \(J\) is the total rotational angular momentum quantum number and K is the quantum number that represents the portion of the total angular momentum that lies along the unique rotational axis. This leads to the property that \(K\) is always equal to or less than \(J\). Thus we get the two selection rules for symmetric tops \[\Delta J = 0, \pm1 \] \[\Delta K=0\] when \( K\neq 0 \) \[\Delta J = \pm1 \] \[\Delta K=0 \] when \( K=0 \) However, like the rigid rotor approximation for linear molecules, we must also take into account the elasticity of the bonds in symmetric tops. Therefore, in a similar manner to the rigid rotor we add a centrifugal coupling term, but this time we have one for each quantum number and one for the coupling between the two. \[E_{(J,K)}(cm^{-1})=B_e J(J+1)-D_{eJ} J^2(J+1)^2+(A_e-B_e)*K^2 \label{13}\] \[-D_{ek} K^4-D_{ejk} J(J +1)K^2 \label{14}\] Asymmetrical tops have three orthogonal rotational axes that all have different moments of inertia and most molecules fall into this category. Unlike linear molecules and symmetric tops these types of molecules do not have a simplified energy equation to determine the energy levels of the rotations. These types of molecules do not follow a specific pattern and usually have very complex microwave spectra. In addition to microwave spectroscopy, IR spectroscopy can also be used to probe rotational transitions in a molecule. However, in the case of IR spectroscopy the rotational transitions are coupled to the vibrational transitions of the molecule. One other spectroscopy that can probe the rotational transitions in a molecule is Raman spectroscopy, which uses UV-visible light scattering to determine energy levels in a molecule. However, a very high sensitivity detector must be used to analyze rotational energy levels of a molecule. | 12,069 | 929 |
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Molecular term symbols specify molecular electronic energy levels. Term symbols for diatomic molecules are based on irreducible representations in linear symmetry groups, derived from spectroscopic notations. They usually consist of four parts: spin multiplicity, azimuthal angular momentum, total angular momentum and symmetry. All molecular term symbols discussed here are based on . Molecular term symbols mark different electronic energy levels of a diatomic molecule. These symbols are similar to , since both follow the Russell-Saunders coupling scheme. Molecular term symbols employ symmetry labels from group theory. The possibility of an electronic transition can be deducted from molecular term symbols following selection rules. For multi-atomic molecules, symmetry labels play most of term symbols' roles. For homonuclear diatomics, the term symbol has the following form: \[ ^{2S+1}\Lambda_{\Omega,(g/u)}^{(+/-)} \] whereas Λ is the projection of the orbital angular momentum along the internuclear axis: Ω is the projection of the total angular momentum along the internuclear axis; / is the parity; and +/− is the reflection symmetry along an arbitrary plane containing the internuclear axis. Λ may be one of the greek letters in the sequence: Σ Π Δ Φ... when Λ = 0, 1, 2, 3..., respectively. For heteronuclear diatomics, the term symbol does not include the / part, for there is not inversion center in the molecule. Let's start with CO again. As we have seen before, the molecule has a close-shell configuration. Its ground state is a totally symmetric singlet, Σ , since the only possible values of ( , Λ) are (0, 0). If one of the HOMO electrons on the 5σ orbital has jumped to the LUMO, this molecule will be in an excited state as follows. Suppose a CO molecule is in the excited state shown above. In order to know the term symbol of this state, a direct product of the labels is required for the two MO's with unpaired electrons. The multiplication is such as \(\Pi \times \Sigma^+ = \Pi\). According to Pauli's exclusion rule, these two unpaired electrons can never share the same set of quantum numbers, therefore the spin degeneracy can reach its maximum 3. The resulting term symbols are Π and Π. Now if we look at O , it does not have a close-shell configuration at its ground state. There are two unpaired electrons each occupying one of the two degenerate 2 orbitals, which can be seen in the diagram below. The term symbol for oxygen molecule at its ground state is therefore derived such as Π x Π = Σ + Σ + [Δ], as the symbol in brackets does not allow the oxygen atoms to commute. We'll focus on selection rules. Like atomic electronic states, different selection rules apply when differently incurred transitions occur. Usually for electric dipole field induced transitions, the selection rules are the same as for atoms. | 2,888 | 930 |
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We can solve Boyle’s law and Charles’ law for the volume. Equating the two, we have \[\dfrac{n\alpha (T)}{P}=n\beta \left(P\right)T\] The number of moles, \(n\), cancels. Rearranging gives \[\dfrac{\alpha (T)}{T}=P\beta (P)\] In this equation, the left side is a function only of temperature, the right side only of pressure. Since pressure and temperature are independent of one another, this can be true only if each side is in fact constant. If we let this constant be \(R\), we have \[\alpha \left(T\right)=RT\] and \[\beta \left(P\right)={R}/{P}\] Since the values of \(\alpha (T)\) and \(\beta (P)\) are independent of the gas being studied, the value of \(R\) is also the same for any gas. \(R\) is called the , the , or the . Substituting the appropriate relationship into either Boyle’s law or Charles’ law gives the \[PV=nRT\] The product of pressure and volume has the units of work or energy, so the gas constant has units of energy per mole per degree. (Remember that we simplified the form of Charles’s law by defining the Kelvin temperature scale; temperature in the ideal gas equation is in degrees Kelvin.) | 1,139 | 931 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Structure_and_Nomenclature_of_Coordination_Compounds/Ligands/The_Irving-Williams_Series |
The general stability sequence of high spin octahedral metal complexes for the replacement of water by other ligands is: \[\ce{Mn(II) < Fe(II) < Co(II) < Ni(II) < Cu(II) > Zn(II)}\] This trend is essentially independent of the ligand. In the case of 1,2-diaminoethane (en), the first step-wise stability constants (logK ) for M(II) ions are shown below. The Irving-Williams sequence is generally quoted ONLY for Mn(II) to Zn(II) since there is little data available for the other first row transition metal ions because their M(II) oxidation states are not very stable. The position of Cu(II) is considered out-of-line with predictions based on Crystal Field Theory and is probably a consequence of the fact that Cu(II) often forms distorted octahedral complexes. Crystal Field Theory is based on the idea that a purely electrostatic interaction exists between the central metal ion and the ligands. This suggests that the stability of the complexes should be related to the ionic potential; that is, the . In the Irving-Williams series, the trend is based on high-spin M(II) ions, so what needs to be considered is how the ionic radii vary across the d-block. For free metal ions in the gaseous phase it might be expected that the ionic radius of each ion on progressing across the d-block should show a gradual decrease in size. This would come about due to the incomplete screening of the additional positive charge by the additional electron, as is observed in the . For high-spin octahedral complexes it is essential to consider the effect of the removal of the degeneracy of the d-orbitals by the crystal field. Here the d-electrons will initially add to the lower t orbitals before filling the e orbitals since for octahedral complexes, the t subset are directed in between the incoming ligands whilst the e subset are directed towards the incoming ligands and cause maximum repulsion. For d -d (and d -d ) the addition of the electrons to the t orbitals will mean that the screening of the increasing attractive nuclear charge is not very effective and the radius should be smaller than for the free ion. The position of d and d on the plot is difficult to ascertain with certainty since six-coordinate complexes are expected to be distorted due to the . Cr(II) is not very stable so few measurements are available. For Cu(II) however, most complexes are found to have 4 short bonds and 2 long bonds although 2 short and 4 long bonds is feasible. The radii are expected to show an increase over the d and d situation since electrons are being added to the e subset. The reported values have been found to lie on both sides of the predicted value. For d , d and d the screening expected is essentially that of a spherical arrangement equivalent to the absence of a crystal field. The plot above shows that these points return to the line drawn showing a gradual decrease of the radius on moving across the d-block. Once the decrease in radius with Z pattern is understood, it is a small step to move to a pattern for q/r since this only involves taking the reciprocal of the radius and holding the charge constant. The radius essentially decreases with increasing \(Z\), therefore 1/r must increase with increasing \(Z\). For the sequence Mn(II) to Zn(II), the crystal field (q/r) trend expected would be \[\ce{Mn(II) < Fe(II) < Co(II) < Ni(II) > Cu(II) > Zn(II)}\] Apart from the position of Cu(II), this corresponds to the Irving-Williams series (Eqution 1). The discrepancy is once again accounted for by the fact that copper(II) complexes are often distorted or not octahedral at all. When this is taken into consideration, it is seen that the Irving-Williams series can be explained quite well using . | 3,743 | 932 |
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Sulfur is a chemical element that is represented with the chemical symbol "S" and the atomic number 16 on the periodic table. Because it is 0.0384% of the Earth's crust, sulfur is the seventeenth most abundant element following strontium. Sulfur also takes on many forms, which include elemental sulfur, organo-sulfur compounds in oil and coal, H S(g) in natural gas, and mineral sulfides and sulfates. This element is extracted by using the Frasch process (discussed below), a method where superheated water and compressed air is used to draw liquid sulfur to the surface. Offshore sites, Texas, and Louisiana are the primary sites that yield extensive amounts of elemental sulfur. However, elemental sulfur can also be produced by reducing H S, commonly found in oil and natural gas. For the most part though, sulfur is used to produce SO (g) and H SO . Known from ancient times (mentioned in the Hebrew scriptures as brimstone) sulfur was classified as an element in 1777 by Lavoisier. Pure sulfur is tasteless and odorless with a light yellow color. Samples of sulfur often encountered in the lab have a noticeable odor. Sulfur is the tenth most abundant element in the known universe. Sulfur has an atomic weight of 32.066 grams per mole and is part of group 16, the oxygen family. It is a nonmetal and has a specific heat of 0.706 J g C . The electron affinity if 200 kJ mol and the electronegativity is 2.58 (unitless). Sulfur is typically found as a light-yellow, opaque, and brittle solid in large amounts of small orthorhombic crystals. Not only does sulfur have twice the density of water, it is also insoluble in water. On the other hand, sulfur is highly soluble in carbon disulfide and slightly soluble in many common solvents. Sulfur can also vary in color and blackens upon boiling due to carbonaceous impurities. Even as little as 0.05% of carbonaceous matter darkens sulfur significantly. Most sulfur is recovered directly as the element from underground deposits by injecting super-heated water and piping out molten sulfur (sulfur melts at 112 C). Compared to other elements, sulfur has the most allotropes. While the S ring is the most common allotrope, there are 6 other structures with up to 20 sulfur atoms per ring. While has fewer allotropes than sulfur, including \(\ce{O}\), \(\ce{O_2}\), \(\ce{O_3}\), \(\ce{O_4}\), \(\ce{O_8}\), metallic \(\ce{O}\) (and four other solid phases), many of these actually have a corresponding sulfur variant. However, sulfur has more tendency to (t ). Here are the values of the single and double bond enthalpies: \[\begin{array}{c|r} \ce {O-O} & \ce{142\ kJ/mol} \\ \ce {S–S} & \ce{268\ kJ/mol} \\ \ce {O=O} & \ce{499\ kJ/mol} \\ \ce {S=S} & \ce{352\ kJ/mol} \\ \end{array}\] This means that \(\ce{O=O}\) is stronger than \(\ce{S=S}\), while \(\ce{O–O}\) is weaker than \(\ce{S–S}\). So, in sulfur, single bonds are favored and catenation is easier than in oxygen compounds. It seems that the reason for the weaker \(\ce{S=S}\) double bonds has its roots in the size of the atom: it's harder for the two atoms to come at a small enough distance, so that the \(p\) orbitals overlap is small and the \(\pi\) bond is weak. This is attested by looking down the periodic table: \(\ce{Se=Se}\) has an even weaker bond enthalpy of \(\ce{272 kJ/mol}\). What happens when the solid sulfur melts? The \(\ce{S8}\) molecules break up. When suddenly cooled, long chain molecules are formed in the plastic sulfur which behave as rubber. Plastic sulfur transform into rhombic sulfur over time. Reading the following reactions, figure out and notice the change of the oxidation state of \(\ce{S}\) in the reactants and products. Common oxidation states of sulfur are -2, 0, +4, and +6. Sulfur (brimstone, stone that burns) reacts with \(\ce{O2}\) giving a blue flame (Figure \(\Page {1}\)): \[\ce{S + O_2 \rightarrow SO_2}\] \(\ce{SO2}\) is produced whenever metal sulfide is oxidized. It is recovered and oxidized further to give \(\mathrm{SO_3}\), for production of \(\mathrm{H_2SO_4}\). \(\mathrm{SO_2}\) reacts with \(\mathrm{H_2S}\) to form \(\mathrm{H_2O}\) and \(\ce{S}\). \[\mathrm{2 SO_2 + O_2 \rightleftharpoons 2 SO_3}\] \[\mathrm{SO_3 + H_2O \rightleftharpoons H_2SO_4} \;\;(\text{a valuable commodity})\] \[\mathrm{SO_3 + H_2SO_4 \rightleftharpoons H_2S_2O_7} \;\;\; (\text{pyrosulfuric acid})\] Sulfur reacts with sulfite ions in solution to form thiosulfate, \[\ce{S + SO_3^{2-} -> S_2O_3^{2-}}\] but the reaction is reversed in an acidic solution. There are many different stable sulfur oxides, but the two that are commonly found are sulfur dioxide and sulfur trioxide. Sulfur dioxide is a commonly found oxide of sulfur. It is a colorless, pungent, and nonflammable gas. It has a density of 2.8 kg/m and a melting point of -72.5 C. Because organic materials are more soluble in \(SO_2\) than in water, the liquid form is a good solvent. \(SO_2\) is primarily used to produce \(SO_3\). The direct combustion of sulfur and the roasting of metal sulfides yield \(SO_2\) via : \[\underbrace{S(s) + O_2(g) \rightarrow SO_2(g)}_{\text{Direct combustion}}\] \[\underbrace{2 ZnS(s) + 3 O_2(g) \rightarrow 2 ZnO(s) + 2 SO_2(g)}_{\text{Roasting of metal sulfides}}\] Sulfur trioxide is another one of the commonly found oxides of sulfur. It is a colorless liquid with a melting point of 16.9 C and a density of kg/m . \(SO_3\) is used to produce sulfuric acid. \(SO_2\) is used in the synthesis of \(SO_3\): \[\underbrace{2 SO_2 (g) + O_2(g) \rightleftharpoons 2 SO_3(g)}_{\text{Exothermic, reversible reaction}}\] This reaction needs a catalyst to be completed in a reasonable amount of time with \(V_2O_5\) is the catalyst most commonly used. Perhaps the most significant compound of sulfur used in modern industrialized societies is sulfuric acid (\(H_2SO_4\)). Sulfur dioxide (\(SO_2\)) finds practical applications in bleaching and refrigeration but it is also a nuisance gas resulting from the burning of sulfurous coals. Sulfur dioxide gas then reacts with the water vapor in the air to produce a weak acid, sulfurous acid (\(H_2SO_3\)) which contributes to the acid rain problem. \[C_{12}H_{22}O_{11}(s) \rightarrow 12 C(s) + 11 H_2O(l)\] (Concentrated sulfuric acid used in forward reaction to remove H and O atoms.) O (g) + 2 SO (aq) \rightarrow 2 SO (aq) (Reducing agent) 2 H S(g) + 2 H (aq) + SO (aq) \rightarrow 3 H O(l) + 3 S(s) (Oxidizing agent) H SO is a diprotic acid that acts as a weak acid in both steps and H SO is also a diprotic acid but acts as a strong acid in the first step and a weak acid in the second step. Acids like NaHSO and NaHSO are called acid salts because they are the product of the first step of these diprotic acids. Boiling elemental sulfur in a solution of sodium sulfite yields . Not only are thiosulfates important in photographic processing, but they are also common analytical reagents used with iodine (like in the following two reactions). \[2 Cu^{2+}_{(aq)} + 5 I^-_{(aq)} \rightarrow 2 CuI_{(s)} + I^-_{3(aq)} \] \[I^-_{3(aq)} + 2 S_2O^{2-}_{3(aq)} \rightarrow 3 I^-_{(aq)} + S_4O^{2-}_{6(aq)}\] with excess triiodide ion titrated with Na S O (aq). Other than sulfuric acid, perhaps the most familiar compound of sulfur in the chemistry lab is the foul-smelling hydrogen sulfide gas, \(H_2S\), which smells like rotten eggs. \[SCl_2 + 2CH_2CH_2 \rightarrow S(CH_2CH_2Cl)_2 \] Sulfur can be mined by the Frasch process. This process has made sulfur a high purity (up to 99.9 percent pure) chemical commodity in large quantities. Most sulfur-containing minerals are metal sulfides, and the best known is perhaps pyrite (\(\mathrm{FeS_2}\), known as fools gold because of its golden color). The most common sulfate containing mineral is gypsum, \(\mathrm{CaSO_4 \cdot 2H_2O}\), also known as plaster of Paris. The Frasch process forces (99.5% pure) sulfur out by using hot water and air. In this process, superheated water is forced down the outermost of three concentric pipes. Compressed air is pumped down the center tube, and a mixture of elemental sulfur, hot water, and air comes up the middle pipe. Sulfur is melted with superheated water (at 170 °C under high pressure) and forced to the surface of the earth as a slurry. Sulfur is mostly used for the production of sulfuric acid, \(\ce{H2SO4}\). Most sulfur mined by Frasch process is used in industry for the manufacture of sulfuric acid. Sulfuric acid, the most abundantly produced chemical in the United States, is manufactured by the . Most (about 70%) of the sulfuric acid produced in the world is used in the fertilizer industry. Sulfuric acid can act as a strong acid, a dehydrating agent, and an oxidizing agent. Its applications use these properties. Sulfur is an essential element of life in sulfur-containing proteins. Sulfur has many practical applications. As a fungicide, sulfur is used to counteract apple scab in organically farmed apple production. Other crops that utilize sulfur fungicides include grapes, strawberries, and many vegetables. In general, sulfur is effective against mildew diseases and black spot. Sulfur can also be used as an organic insecticide. Sulfites are frequently used to bleach paper and preserve dried fruit. The vulcanization of rubber includes the use of sulfur as well. Cellophane and rayon are produced with carbon disulfide, a product of sulfur and methane. Sulfur compounds can also be found in detergents, acne treatments, and agrichemicals. Magnesium sulfate (epsom salt) has many uses, ranging from bath additives to exfoliants. Sulfur is being increasingly used as a fertilizer as well. Because standard sulfur is hydrophobic, it is covered with a surfactant by bacteria before oxidation can occur. Sulfur is therefore a slow-release fertilizer. Lastly, sulfur functions as a light-generating medium in sulfur lamps. Concentrated sulfuric acid was once one of the most produced chemicals in the United States, the majority of the H SO that is now produced is used in fertilizer. It is also used in oil refining, production of titanium dioxide, in emergency power supplies and car batteries. The mineral gypsum is calcium sulfate dihydrate is used in making plaster of Paris. Over one million tons of aluminum sulfate is produced each year in the United States by reacting H SO and Al O . This compound is important in water purification. Copper sulfate is used in electroplating. Sulfites are used in the paper making industry because they produce a substance that coats the cellulose in the word and frees the fibers of the wood for treatment. Particles, SO (g), and H SO mist are the components of . Because power plants burn coal or high-sulfur fuel oils, SO (g) is released into the air. When catalyzed on the surfaces of airborne particles, SO can be oxidized to SO . A reaction with NO works as well as shown in the following reaction: \[ SO_{2(g)} + NO_{2(g)} \rightarrow SO_{3(g)} + NO_{(g)}\] H SO mist is then produced after SO reacts with water vapor in the air. If H SO reacts with airborne NH , (NH ) SO is produced. When SO (g) and H SO reach levels that exceed 0.10 ppm, they are potentially harmful. By removing sulfur from fuels and controlling emissions, and industrial smog can be kept under control. Processes like the have been presented to remove SO from smokestack gases. | 11,387 | 933 |
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A large number of reactions proceed through a series of steps that can collectively be classified as a . The reactions contain steps that can be classified as These types of reactions are very common when the intermediates involved are radicals. An example, is the reaction \[H_2 + Br_2 \rightarrow 2HBr \nonumber \] The observed rate law for this reaction is \[ \text{rate} = \dfrac{k [H_2,Br_2]^{3/2}}{[Br_2] + k'[HBr]} \label{exp} \] A proposed mechanism is \[Br_2 \ce{<=>[k_1,k_{-1}]} 2Br^\cdot \label{step1} \] \[ 2Br^\cdot + H_2 \ce{<=>[k_2,k_{-2}]} HBr + H^\cdot \label{step2} \] \[ H^\cdot + Br_2 \xrightarrow{k_3} HBr + Br^\cdot \label{step3} \] Based on this mechanism, the rate of change of concentrations for the intermediates (\(H^\cdot\) and \(Br^\cdot\)) can be written, and the steady state approximation applied. \[\dfrac{d[H^\cdot]}{dt} = k_2[Br^\cdot,H_2] - k_{-2}[HBr,H^\cdot] - k_3[H^\cdot,Br_2] =0 \nonumber \] \[\dfrac{d[Br^\cdot]}{dt} = 2k_1[Br_2] - 2k_{-1}[Br^\cdot]^2 - k_2[Br^\cdot,H_2] + k_{-2}[HBr,H^\cdot] + k_3[H^\cdot,Br_2] =0 \nonumber \] Adding these two expressions cancels the terms involving \(k_2\), \(k_{-2}\), and \(k_3\). The result is \[ 2 k_1 [Br_2] - 2k_{-1} [Br^\cdot]^2 = 0 \nonumber \] Solving for \(Br^\cdot\) \[ Br^\cdot = \sqrt{\dfrac{k_1[Br_2]}{k_{-1}}} \nonumber \] This can be substituted into an expression for the \(H^\cdot\) that is generated by solving the steady state expression for \(d[H^\cdot]/dt\). \[[H^\cdot] = \dfrac{k_2 [Br^\cdot] [H_2]}{k_{-2}[HBr] + k_3[Br_2]} \nonumber \] so \[[H^\cdot] = \dfrac{k_2 \sqrt{\dfrac{k_1[Br_2]}{k_{-1}}} [H_2]}{k_{-2}[HBr] + k_3[Br_2]} \nonumber \] Now, armed with expressions for \(H^\cdot\) and \(Br^\cdot\), we can substitute them into an expression for the rate of production of the product \(HBr\): \[ \dfrac{[HBr]}{dt} = k_2[Br^\cdot] [H_2] + k_3 [H^\cdot] [Br_2] - k_{-2}[H^\cdot] [HBr] \nonumber \] After substitution and simplification, the result is \[ \dfrac{[HBr]}{dt} = \dfrac{2 k_2 \left( \dfrac{k_1}{k_{-1}}\right)^{1/2} [H_2,Br_2]^{1/2}}{1+ \dfrac{k_{-1}}{k_3} \dfrac{[HBr]}{[Br_2]} } \nonumber \] Multiplying the top and bottom expressions on the right by \([Br_2]\) produces \[ \dfrac{[HBr]}{dt} = \dfrac{2 k_2 \left( \dfrac{k_1}{k_{-1}}\right)^{1/2} [H_2,Br_2]^{3/2}}{[Br_2] + \dfrac{k_{-1}}{k_3} [HBr] } \nonumber \] which matches the form of the rate law found experimentally (Equation \ref{exp})! In this case, \[ k = 2k_2 \sqrt{ \dfrac{k_1}{k_{-1}}} \nonumber \] and \[ k'= \dfrac{k_{-2}}{k_3} \nonumber \] | 2,557 | 935 |
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As seen in the chapter on the stoichiometry of chemical reactions, titrations can be used to quantitatively analyze solutions for their acid or base concentrations. In this section, we will explore the changes in the concentrations of the acidic and basic species present in a solution during the process of a titration. Previously, when we studied acid-base reactions in solution, we focused only on the point at which the acid and base were stoichiometrically equivalent. No consideration was given to the pH of the solution before, during, or after the neutralization. A titration is carried out for 25.00 mL of 0.100 HCl (strong acid) with 0.100 of a strong base NaOH the titration curve is shown in Figure \(\Page {1}\). Calculate the pH at these volumes of added base solution: Since HCl is a strong acid, we can assume that all of it dissociates. The initial concentration of H O is \(\ce{[H3O+]_0}=0.100\:M\). When the base solution is added, it also dissociates completely, providing OH ions. The H O and OH ions neutralize each other, so only those of the two that were in excess remain, and their concentration determines the pH. Thus, the solution is initially acidic (pH < 7), but eventually all the hydronium ions present from the original acid are neutralized, and the solution becomes neutral. As more base is added, the solution turns basic. The total initial amount of the hydronium ions is: \[\mathrm{n(H^+)_0=[H_3O^+]_0×0.02500\: L=0.002500\: mol} \nonumber \] Once X mL of the 0.100- base solution is added, the number of moles of the OH ions introduced is: The total volume becomes: \[V=\mathrm{(25.00\: mL+X\: mL)\left(\dfrac{1\: L}{1000\: mL}\right)} \nonumber \] The number of moles of H O becomes: \[\mathrm{n(H^+)=n(H^+)_0-n(OH^-)_0=0.002500\: mol-0.100\:\mathit{M}×X\: mL×\left(\dfrac{1\: L}{1000\: mL}\right)} \nonumber \] The concentration of H O is: \[\mathrm{[H_3O^+]=\dfrac{n(H^+)}{V}=\dfrac{0.002500\: mol-0.100\:\mathit{M}×X\: mL×\left(\dfrac{1\: L}{1000\: mL}\right)}{(25.00\: mL+X\: mL)\left(\dfrac{1\: L}{1000\: mL}\right)}} \nonumber \] \[\mathrm{=\dfrac{0.002500\: mol×\left(\dfrac{1000\: mL}{1\: L}\right)-0.100\:\mathit{M}×X\: mL}{25.00\: mL+X\: mL}} \nonumber \] with the definition of \(\mathrm{pH}\): \[\mathrm{pH=−\log([H_3O^+])} \label{phdef} \] The preceding calculations work if \(\mathrm{n(H^+)_0-n(OH^-)_0>0}\) and so n(H ) > 0. When \(\mathrm{n(H^+)_0=n(OH^-)_0}\), the H O ions from the acid and the OH ions from the base mutually neutralize. At this point, the only hydronium ions left are those from the autoionization of water, and there are no OH particles to neutralize them. Therefore, in this case: \[\ce{[H3O+]}=\ce{[OH- ]},\:\ce{[H3O+]}=K_\ce{w}=1.0\times 10^{-14};\:\ce{[H3O+]}=1.0\times 10^{-7} \nonumber \] \[\mathrm{pH=-log(1.0\times 10^{-7})=7.00} \nonumber \] Finally, when \(\mathrm{n(OH^-)_0>n(H^+)_0}\), there are not enough H O ions to neutralize all the OH ions, and instead of \(\mathrm{n(H^+)=n(H^+)_0-n(OH^-)_0}\), we calculate: \(\mathrm{n(OH^-)=n(OH^-)_0-n(H^+)_0}\) In this case: \[\mathrm{[OH^-]=\dfrac{n(OH^-)}{\mathit{V}}=\dfrac{0.100\:\mathit{M}×X\: mL×\left(\dfrac{1\: L}{1000\: mL}\right)-0.002500\: mol}{(25.00\: mL+X\: mL)\left(\dfrac{1\: L}{1000\: mL}\right)}} \nonumber \] \[\mathrm{=\dfrac{0.100\:\mathit{M}×X\: mL-0.002500\: mol×\left(\dfrac{1000\: mL}{1\: L}\right)}{25.00\: mL+X\: mL}} \nonumber \] then using the definition of \(pOH\) and its relationship to \(pH\) in room temperature aqueous solutios (Equation \ref{phdef}): \[\begin{align} pH &=14-pOH \nonumber \\&=14+\log([OH^-]) \nonumber\end{align} \nonumber \] Let us now consider the four specific cases presented in this problem: (a) X = 0 mL \[\mathrm{[H_3O^+]=\dfrac{n(H^+)}{\mathit{V}}=\dfrac{0.002500\: mol×\left(\dfrac{1000\: mL}{1\: L}\right)}{25.00\: mL}=0.1\:\mathit{M}} \nonumber \] then using the definition of \(pH\) (Equation \ref{phdef}): \[\begin{align} pH &= −\log(0.100) \nonumber \\ &= 1.000 \nonumber\end{align} \nonumber \] (b) X = 12.50 mL \[\mathrm{[H_3O^+]=\dfrac{n(H^+)}{\mathit{V}}=\dfrac{0.002500\: mol×\left(\dfrac{1000\: mL}{1\: L}\right)-0.100\:\mathit{M}×12.50\: mL}{25.00\: mL+12.50\: mL}=0.0333\:\mathit{M}} \nonumber \] then using the definition of \(pH\) (Equation \ref{phdef}): \[ \begin{align} pH &= −\log(0.0333) \nonumber \\ &= 1.477 \nonumber\end{align} \nonumber \] (c) X = 25.00 mL Since the volumes and concentrations of the acid and base solutions are the same: \[\mathrm{n(H^+)_0=n(OH^-)_0} \nonumber \] and \[pH = 7.000 \nonumber \] as described earlier. (d) X = 37.50 mL In this case: \[\mathrm{n(OH^-)_0>n(H^+)_0} \nonumber \] \[\mathrm{[OH^-]=\dfrac{n(OH^-)}{\mathit{V}}=\dfrac{0.100\:\mathit{M}×35.70\: mL-0.002500\: mol×\left(\dfrac{1000\: mL}{1\: L}\right)}{25.00\: mL+37.50\: mL}=0.0200\:\mathit{M}} \nonumber \] then using the definition of \(pH\) (Equation \ref{phdef}): \[ \begin{align}[pH = 14 − pOH \nonumber\\ &= 14 + \log([OH^{−}]) \nonumber \\ &= 14 + \log(0.0200) \nonumber \\ &= 12.30 \nonumber \end{align} \nonumber \] Calculate the pH for the strong acid/strong base titration between 50.0 mL of 0.100 HNO ( ) and 0.200 NaOH (titrant) at the listed volumes of added base: 0.00: 1.000 15.0: 1.5111 25.0: 7e. Do not delete this text first. 40.0: 12.523 In Example \(\Page {1}\), we calculated pH at four points during a titration. Table \(\Page {1}\) shows a detailed sequence of changes in the pH of a strong acid and a weak acid in a titration with NaOH. The simplest acid-base reactions are those of a strong acid with a strong base. Table \(\Page {1}\) shows data for the titration of a 25.0-mL sample of 0.100 hydrochloric acid with 0.100 sodium hydroxide. The values of the pH measured after successive additions of small amounts of NaOH are listed in the first column of this table, and are graphed in Figure \(\Page {1}\), in a form that is called a . The pH increases slowly at first, increases rapidly in the middle portion of the curve, and then increases slowly again. The point of inflection (located at the midpoint of the vertical part of the curve) is the equivalence point for the titration. It indicates when equivalent quantities of acid and base are present. For the titration of a strong acid with a strong base, the equivalence point occurs at a pH of 7.00 and the points on the titration curve can be calculated using solution stoichiometry (Table \(\Page {1}\) and Figure \(\Page {1}\)). The titration of a weak acid with a strong base (or of a weak base with a strong acid) is somewhat more complicated than that just discussed, but it follows the same general principles. Let us consider the titration of 25.0 mL of 0.100 acetic acid (a weak acid) with 0.100 sodium hydroxide and compare the titration curve with that of the strong acid. Table \(\Page {1}\) gives the pH values during the titration, Figure \(\Page {1b}\) shows the titration curve. Although the initial volume and molarity of the acids are the same, there are important differences between the two titration curves. The titration curve for the weak acid begins at a higher value (less acidic) and maintains higher pH values up to the equivalence point. This is because acetic acid is a weak acid, which is only partially ionized. The pH at the equivalence point is also higher (8.72 rather than 7.00) due to the hydrolysis of acetate, a weak base that raises the pH: \[\ce{CH3CO2-}(aq)+\ce{H2O}(l)⇌\ce{CH3CO2H}(l)+\ce{OH-}(aq) \nonumber \] After the equivalence point, the two curves are identical because the pH is dependent on the excess of hydroxide ion in both cases. The titration curve shown in Figure \(\Page {1b}\) is for the titration of 25.00 mL of 0.100 CH CO H with 0.100 NaOH. The reaction can be represented as: \[\ce{CH3CO2H + OH- ⟶ CH3CO2- + H2O} \nonumber \] (a) Assuming that the dissociated amount is small compared to 0.100 , we find that: \[K_\ce{a}=\ce{\dfrac{[H3O+,CH3CO2- ]}{[CH3CO2H]}}≈\ce{\dfrac{[H3O+]^2}{[CH3CO2H]_0}} \nonumber \] and \[\ce{[H3O+]}=\sqrt{K_\ce{a}×\ce{[CH3CO2H]}}=\sqrt{1.8\times 10^{-5}×0.100}=1.3\times 10^{-3} \nonumber \] \[\mathrm{pH=-\log(1.3\times 10^{-3})=2.87} \nonumber \] (b) After 25.00 mL of NaOH are added, the number of moles of NaOH and CH CO H are equal because the amounts of the solutions and their concentrations are the same. All of the CH CO H has been converted to \(\ce{CH3CO2-}\). The concentration of the \(\ce{CH3CO2-}\) ion is: \[\mathrm{\dfrac{0.00250\: mol}{0.0500\: L}=0.0500\: \ce{MCH3CO2-}} \nonumber \] The equilibrium that must be focused on now is the basicity equilibrium for \(\ce{CH3CO2-}\): \[\ce{CH3CO2-}(aq)+\ce{H2O}(l)⇌\ce{CH3CO2H}(aq)+\ce{OH-}(aq) \nonumber \] so we must determine for the base by using the ion product constant for water: \[K_\ce{b}=\ce{\dfrac{[CH3CO2H,OH- ]}{[CH3CO2- ]}} \nonumber \] \[K_\ce{a}=\ce{\dfrac{[CH3CO2- ,H+]}{[CH3CO2H]}},\textrm{ so }\ce{\dfrac{[CH3CO2H]}{[CH3CO2- ]}}=\dfrac{\ce{[H+]}}{K_\ce{a}}. \nonumber \] Since = [H ,OH ]: \[\begin{align} K_\ce{b} &=\dfrac{\ce{[H+,OH- ]}}{K_\ce{a}} \\ &=\dfrac{K_\ce{w}}{K_\ce{a}} \\ &=\dfrac{1.0\times 10^{-14}}{1.8\times 10^{-5}} \\ &=5.6\times 10^{-10} \end{align} \nonumber \] Let us denote the concentration of each of the products of this reaction, CH CO H and OH , as . Using the assumption that is small compared to 0.0500 , \(K_\ce{b}=\dfrac{x^2}{0.0500\:M}\), and then: \[\ce{pOH}=-\log(5.3\times 10^{-6})=5.28 \nonumber \] \[\ce{pH}=14.00−5.28=8.72 \nonumber \] Note that the pH at the equivalence point of this titration is significantly greater than 7. (c) In (a), 25.00 mL of the NaOH solution was added, and so practically all the CH CO H was converted into \(\ce{CH3CO2-}\). In this case, only 12.50 mL of the base solution has been introduced, and so only half of all the CH CO H is converted into \(\ce{CH3CO2-}\). The total initial number of moles of CH CO H is 0.02500L × 0.100 = 0.00250 mol, and so after adding the NaOH, the numbers of moles of CH CO H and \(\ce{CH3CO2-}\) are both approximately equal to \(\mathrm{\dfrac{0.00250\: mol}{2}=0.00125\: mol}\), and their concentrations are the same. Since the amount of the added base is smaller than the original amount of the acid, the equivalence point has not been reached, the solution remains a buffer, and we can use the Henderson-Hasselbalch equation: (as the concentrations of \(\ce{CH3CO2-}\) and CH CO H are the same) Thus: \[\ce{pH}=−\log(1.8\times 10^{−5})=4.74 \nonumber \] (the pH = the p at the halfway point in a titration of a weak acid) (d) After 37.50 mL of NaOH is added, the amount of NaOH is 0.03750 L × 0.100 = 0.003750 mol NaOH. Since this is past the equivalence point, the excess hydroxide ions will make the solution basic, and we can again use stoichiometric calculations to determine the pH: So: Note that this result is the same as for the strong acid-strong base titration example provided, since the amount of the strong base added moves the solution past the equivalence point. Calculate the pH for the weak acid/strong base titration between 50.0 mL of 0.100 HCOOH( ) (formic acid) and 0.200 NaOH (titrant) at the listed volumes of added base: 0.00 mL: 2.37 15.0 mL: 3.92 25.00 mL: 8.29 30.0 mL: 12.097 Certain organic substances change color in dilute solution when the hydronium ion concentration reaches a particular value. For example, phenolphthalein is a colorless substance in any aqueous solution with a hydronium ion concentration greater than 5.0 × 10 (pH < 8.3). In more basic solutions where the hydronium ion concentration is less than 5.0 × 10 (pH > 8.3), it is red or pink. Substances such as phenolphthalein, which can be used to determine the pH of a solution, are called . Acid-base indicators are either weak organic acids or weak organic bases. The equilibrium in a solution of the acid-base indicator methyl orange, a weak acid, can be represented by an equation in which we use \(\ce{HIn}\) as a simple representation for the complex methyl orange molecule: \[\underbrace{\ce{HIn (aq)}}_{\ce{red}}+\ce{H2O (l)}⇌\ce{H3O^{+} (aq)}+\underbrace{\ce{In^{-} (aq)}}_{\ce{yellow}} \nonumber \] \[K_\ce{a}=\ce{\dfrac{[H3O+,In- ]}{[HIn]}}=4.0\times 10^{−4} \nonumber \] The anion of methyl orange, \(\ce{In^{-}}\), is yellow, and the nonionized form, \(\ce{HIn}\), is red. When we add acid to a solution of methyl orange, the increased hydronium ion concentration shifts the equilibrium toward the nonionized red form, in accordance with Le Chatelier’s principle. If we add base, we shift the equilibrium towards the yellow form. This behavior is completely analogous to the action of buffers. An indicator’s color is the visible result of the ratio of the concentrations of the two species In and \(\ce{HIn}\). If most of the indicator (typically about 60−90% or more) is present as \(\ce{In^{-}}\), then we see the color of the \(\ce{In^{-}}\) ion, which would be yellow for methyl orange. If most is present as \(\ce{HIn}\), then we see the color of the \(\ce{HIn}\) molecule: red for methyl orange. For methyl orange, we can rearrange the equation for and write: \[\mathrm{\dfrac{[In^-]}{[HIn]}=\dfrac{[substance\: with\: yellow\: color]}{[substance\: with\: red\: color]}=\dfrac{\mathit{K}_a}{[H_3O^+]}} \label{ABeq2} \] Equation \ref{ABeq2} shows us how the ratio of \(\ce{\dfrac{[In- ]}{[HIn]}}\) varies with the concentration of hydronium ion. The above expression describing the indicator equilibrium can be rearranged: \[ \begin{align} \dfrac{[H_3O^+]}{\mathit{K}_a} &=\dfrac{[HIn]}{[In^- ]} \\[8pt] \log\left(\dfrac{[H_3O^+]}{\mathit{K}_a}\right) &= \log\left(\dfrac{[HIn]}{[In^- ]}\right) \\[8pt] \log([H_3O^+])-\log(\mathit{K}_a) &=-\log\left(\dfrac{[In^-]}{[HIn]}\right) \\[8pt] -pH+p\mathit{K}_a & =-\log\left(\dfrac{[In^-]}{[HIn]}\right) \\[8pt] pH &=p\mathit{K}_a+\log\left(\dfrac{[In^-]}{[HIn]}\right) \end {align} \nonumber \] or in general terms \[pH=p\mathit{K}_a+\log\left(\dfrac{[base]}{[acid]}\right) \label{HHeq} \] Equation \ref{HHeq} is the same as the , which can be used to describe the equilibrium of indicators. When [H O ] has the same numerical value as , the ratio of [In ] to [HIn] is equal to 1, meaning that 50% of the indicator is present in the red form (HIn) and 50% is in the yellow ionic form (In ), and the solution appears orange in color. When the hydronium ion concentration increases to 8 × 10 (a pH of 3.1), the solution turns red. No change in color is visible for any further increase in the hydronium ion concentration (decrease in pH). At a hydronium ion concentration of 4 × 10 (a pH of 4.4), most of the indicator is in the yellow ionic form, and a further decrease in the hydronium ion concentration (increase in pH) does not produce a visible color change. The pH range between 3.1 (red) and 4.4 (yellow) is the of methyl orange; the pronounced color change takes place between these pH values. There are many different acid-base indicators that cover a wide range of pH values and can be used to determine the approximate pH of an unknown solution by a process of elimination. Universal indicators and pH paper contain a mixture of indicators and exhibit different colors at different pHs. Figure \(\Page {2}\) presents several indicators, their colors, and their color-change intervals. Titration curves help us pick an indicator that will provide a sharp color change at the equivalence point. The best selection would be an indicator that has a color change interval that brackets the pH at the equivalence point of the titration. The color change intervals of three indicators are shown in Figure \(\Page {3}\). The equivalence points of both the titration of the strong acid and of the weak acid are located in the color-change interval of phenolphthalein. We can use it for titrations of either strong acid with strong base or weak acid with strong base. Litmus is a suitable indicator for the HCl titration because its color change brackets the equivalence point. However, we should not use litmus for the CH CO H titration because the pH is within the color-change interval of litmus when only about 12 mL of NaOH has been added, and it does not leave the range until 25 mL has been added. The color change would be very gradual, taking place during the addition of 13 mL of NaOH, making litmus useless as an indicator of the equivalence point. We could use methyl orange for the HCl titration, but it would not give very accurate results: (1) It completes its color change slightly before the equivalence point is reached (but very close to it, so this is not too serious); (2) it changes color, as Figure \(\Page {2}\) demonstrates, during the addition of nearly 0.5 mL of NaOH, which is not so sharp a color change as that of litmus or phenolphthalein; and (3) it goes from yellow to orange to red, making detection of a precise endpoint much more challenging than the colorless to pink change of phenolphthalein. Figure \(\Page {2}\) shows us that methyl orange would be completely useless as an indicator for the CH CO H titration. Its color change begins after about 1 mL of NaOH has been added and ends when about 8 mL has been added. The color change is completed long before the equivalence point (which occurs when 25.0 mL of NaOH has been added) is reached and hence provides no indication of the equivalence point. We base our choice of indicator on a calculated pH, the pH at the equivalence point. At the equivalence point, equimolar amounts of acid and base have been mixed, and the calculation becomes that of the pH of a solution of the salt resulting from the titration. A titration curve is a graph that relates the change in pH of an acidic or basic solution to the volume of added titrant. The characteristics of the titration curve are dependent on the specific solutions being titrated. The pH of the solution at the equivalence point may be greater than, equal to, or less than 7.00. The choice of an indicator for a given titration depends on the expected pH at the equivalence point of the titration, and the range of the color change of the indicator. | 18,118 | 936 |
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\[\ce{2Mg(s) + O2(g) → 2MgO(s)} \qquad 1 \text{atm}, \quad 298 \text{K} \label{1} \] Since this reaction occurs at the standard pressure and at 298 K, we can find Δ from the : \[\begin{align} \Delta S &= \Delta S_{m}^{º}(298\text{K})=2S_{m}^{º}(\text{Mg}) - S_{m}^{º}\text{O}_{2} \\ &=(2 \times 26.8 - 2 \times 32.6 - 205.0) \frac { \text{J}}{\text{mol K}} \\&=-216.6 \frac{\text{J}}{\text{mol K}} \end{align} \nonumber \] This result would suggest that the reaction is not spontaneous, but in fact it is. Once ignited, a ribbon of magnesium metal burns freely in air to form solid magnesium oxide in the form of a white powder. The reaction is plainly spontaneous even though Δ is negative. Why is this not a contradiction of the second law? The answer is that we have failed to realize that the entropy change which the magnesium and oxygen atoms undergo as a result of the reaction is not the only entropy change which occurs. The oxidation of magnesium is a highly exothermic reaction, and the heat which is evolved flows into the surroundings, . There are thus entropy changes which we must take into account in deciding whether a reaction will be spontaneous or not: (1) the change in entropy of the actually undergoing the chemical change, which we will indicate with the symbol Δ ; and (2) the change in entropy of the , Δ , which occurs as the surroundings absorb the heat energy liberated by an exothermic reaction or supply the heat energy absorbed by an endothermic reaction. Of these two changes, the first is readily obtained from tables of entropy values. Thus, for the oxidation of magnesium, according to Eq. \(\ref{1}\), Δ has the value already found, namely, –216.6 J K mol . The second entropy change, Δ , can also be derived from tables, as we shall now show. When a chemical reaction occurs at atmospheric pressure and its surroundings are maintained at a constant temperature , then the surroundings will absorb a quantity of heat, q equal to the heat energy given off by the reaction. \[q_{surr}=- \Delta H \label{4} \] (The negative sign before Δ is needed because q is positive if the surroundings absorb heat energy, but Δ is negative if the system gives off heat energy for them to absorb.) If we now feed Eq. \(\ref{4}\) into , we obtain an expression for the entropy change of the surroundings in terms of Δ : \[\Delta S_{\text{surr}}=\frac{q_{\text{surr}}}{T}=\frac{-\Delta H}{T} \label{5} \] Using this equation it is now possible to find the value of Δ from . In the case of the oxidation of magnesium, for example, we easily find that the enthalpy change for \[\ce{2Mg(s) + O2(g) -> 2MgO(s)} \qquad 1 \text{atm, 298 K } \nonumber \] is given by \[\Delta H = \Delta H_{m}^{º} (MgO)=2 \times -601.8 \frac{\text{J}}{\text{mol K}}=-1204 \frac{\text{J}}{\text{mol K}} \nonumber \] Substituting this result into Eq. \(\ref{5}\), we then find \[\Delta S_{\text{surr}}=\frac{-\Delta H}{T}=\frac{\text{1204 }\times \text{ 10}^{\text{3}}\text{ J mol}^{-\text{1}}}{\text{298 K}}=\text{4040 J K}^{-\text{1}}\text{ mol}^{-\text{1}} \nonumber \] If a reaction is spontaneous, then it is the total entropy change Δ , given by the Δ and Δ , which must be positive in order to conform to the second law. In the oxidation of magnesium, for example, we find that the total entropy change is given by \[\Delta S_{tot} = \Delta S_{surr} + \Delta S_{sys} = (4040 - 216.6) \frac{\text{J}}{\text{ mol K}}=3823 \frac{\text{J}}{\text{mol K}} \nonumber \] This is a positive quantity because the entropy increase in the surroundings is more than enough to offset the decrease in the system itself, and the second law is satisfied. In the general case the total entropy change is given by \[\Delta S_{tot}= \Delta S_{surr} + \Delta S_{sys} = -\frac{\Delta H}{T} + \Delta S_{sys} \nonumber \] The second law requires that this sum must be positive; i.e., \[\frac{-\Delta H}{T} + \Delta S_{sys}>0\qquad \label{11} \] This simple inequality gives us what we have been looking for: a simple criterion for determining whether a reaction is spontaneous or not. Since both Δ and Δ can be obtained from tables, and is presumably known, we are now able to predict in advance whether a reaction will be uphill or downhill. Using the and the find Δ °(298 K) and Δ °(298 K) for the reaction \[ \ce{N2 (g) + 3H2 (g) -> 2NH3 (g)} \qquad 1 \text{atm} \nonumber \] Predict whether the reaction will be spontaneous or not at a temperature of (a) 298 K, and (b) 1000 K. We find from the Table of Some Standard Enthalpies of Formation that \[ \begin{align} \Delta H_{m}^{º} (298 \text{K}) & = \sum \Delta H_{f}^{º} \text{(products)} - \sum \Delta H_{f}^{º} \text{(reactants)} \\ & = 2 \Delta H_{f}^{º} \text{(NH}_3 ) - \Delta H_{f}^{º} ( \text{N}_2 ) - 3 \Delta H_{f}^{º} ( \text{H}_2 ) \\ & = (-2 \times 46.1 - 0.0 - 0.0 ) \text{kJ mol}^{-1} = -92.2 \text{kJ mol}^{-1} \end{align} \nonumber \] and from the Table of Standard Molar Entropies: \[ \begin{align}\Delta S_m^{º} (298 \text{K}) & = \sum \Delta S_m^{º} \text{(products)} - \sum \Delta S_m^{º} \text{(reactants)} \\ & = 2 \Delta S_m^º (\text{NH}_3 ) - \Delta S_m^º ( \text{N}_2 ) - 3 \Delta S_m^º ( \text{H}_2 ) \\ & = (2 \times 192.2 – 191.5 – 3 \times 130.6) \text{ J K}^{-1} \text{ mol}^{-1} = –198.9 \text{ J K}^{-1} \text{ mol}^{-1} \end{align} \nonumber \] At 298 K the total entropy change per mol N is given by \[ \begin{align} \Delta S_{tot} & = \frac{-\Delta H_{m}^{ o}}{T} + \Delta S_{sys}^{º} = \frac{\text{92}\text{.2 }\times \text{ 10}^{\text{3}}\text{ J mol}^{-\text{1}}}{\text{298 K}} – 198.9 \text{ J K}^{-1} \text{ mol}^{-1} \\ & = (309.4 – 198.9) \text{ J K}^{-1} \text{ mol}^{-1} = 110.5 \text{ J K}^{-1} \text{ mol}^{-1} \end{align} \nonumber \] Since the total entropy change is positive, the reaction is spontaneous: Since tables are available only for 298 K, we must make the approximate assumption that neither Δ nor Δ varies greatly with temperature. Accordingly we assume \[ \begin{align} & \Delta H_m^º (1000 \text{ K} ) = \Delta H_m^º (298 \text{ K} ) = -92.2 \text{kJ mol}^{-1} \\ \text{and} ~~~~~ & \Delta S_m^º (1000 \text{ K} = \Delta S_m^º (298 \text{ K} ) = -198.9 \text{ J K}^{-1} \text{ mol}^{-1}\end{align} \nonumber \] Δ °(1000 K) = Δ °(298 K) = –92.2 kJ mol and Δ °(1000 K) = Δ °(298 K) = – 198.9 J K mol Thus for 1 mol N reacted, \[\begin{align} \Delta S_{tot} & = \frac{-\Delta H_{m}^{ o}}{T} + \Delta S_{sys}^ º = \frac{\text{92}\text{.2 }\times \text{ 10}^{\text{3}}\text{ J mol}^{-\text{1}}}{\text{1000 K}} – 198.9 \text{ J K}^{-1} \text{ mol}^{-1} \\ & = (92.2 – 198.9) \text{ J K}^{-1} \text{ mol}^{-1} = –106.7 \text{ J K}^{-1} \text{ mol}^{-1} \end{align} \nonumber \] Since the total entropy change is negative at this high temperature, we conclude that N and H will not react to form NH , but that rather NH will decompose into its elements. Apart from enabling us to predict the direction of a chemical reaction from tables of thermodynamic data, the inequality [Eq. \(\ref{11}\)] shows that three factors determine whether a reaction is spontaneous or not: the enthalpy change Δ , the entropy change Δ , and the temperature . Let us examine each of these to see what effect they have and why. At a very low temperature, therefore, whether the reaction is spontaneous or not will depend on the sign of Δ , i.e., on whether the reaction is exothermic or endothermic. By contrast, as we raise the temperature to very high values, the “boy in the bedroom” effect takes over and the reaction affects the entropy of its surroundings to an increasingly smaller extent until finally it is only the value of Δ which determines the behavior of the reaction. In short, Δ Δ . Since Δ can be positive or negative and so can Δ , there are four possible combinations of these two factors, each of which exhibits a different behavior at high and low temperatures. All four cases are listed and described in the next table, and they are also illustrated by simple examples in Figure \(\Page {1}\). In this figure the surroundings are indicated by a shaded border around the reaction system. If the reaction is exothermic, the border changes from gray to pink, indicating that the surroundings have absorbed heat energy and thus increased in entropy. When the border changes from pink to gray, this indicates an endothermic reaction and a decrease in the entropy of the surroundings. In each case the change in entropy of the system should be obvious from an increase or decrease in the freedom of movement of the molecules and/or atoms. The example illustrated in Figure \(\Page {1}\) I is the decomposition of ozone to oxygen. \[\ce{2O3(g)→3O2(g)}\qquad 1 \text{ atm} \nonumber \] for which Δ °(298 K) = –285 kJ mol and Δ °(298 K) = +137 J K mol . A reaction of this type is always spontaneous because the entropy of both the surroundings and the system are increased by its occurrence. The example illustrated in Figure \(\Page {1}\) (II) is the reaction of magnesium metal with hydrogen gas to form magnesium hydride: \[\ce{Mg(s) + H2(g) → MgH2(g)} \qquad 1 \text{ atm} \nonumber \] for which Δ °(298 K) = –76.1 kJ mol and Δ °(298 K) = –132.1 J K mol . Reactions of this type can be either spontaneous or nonspontaneous depending on the temperature. At low temperatures when the effect on the surroundings is most important, the exothermic nature of the reaction makes it spontaneous. At high temperatures the effect of Δ predominates. Since Δ is negative (free H molecules becoming fixed H ions), the reaction must become nonspontaneous at high temperatures. Experimentally, solid MgH will not form from its elements above 560 K, and any formed at a lower temperature will decompose. The example illustrated in Figure \(\Page {2}\) (III) is the vaporization of liquid bromine: \[\ce{Br2(l) → Br2(g)}\qquad 1 \text{ atm} \nonumber \] for which Δ °(298 K) = +31.0 kJ mol and Δ °(298 K) = 93.1 J K mol . This example is usually classified as a physical rather than as a chemical change, but such distinctions are not important in thermodynamics. As in the previous case this reaction can be spontaneous or nonspontaneous at different temperatures. At low temperatures, bromine will not boil because the entropy increase occurring in the bromine as it turns to vapor is not enough to offset the decrease in entropy, which the surroundings experience in supplying the heat energy which is needed for the change in state. At higher temperatures the entropy effect on the surroundings becomes less pronounced, and the positive value of Δ makes the reaction spontaneous. At 101.3 kPa (1atm) pressure, bromine will not boil below 331 K (58°C), but above this temperature it will. \[\ce{2Ag(s) + 3N2(g) → 2AgN3(s)} \nonumber \] for which Δ °(298 K) = +620.6 kJ mol and Δ °(298 K) = –461.5 J K mol . Reactions of this type can never be spontaneous. If this reaction were to occur, it would reduce the entropy of both the system and the surroundings in contradiction of the second law. Since the forward reaction is nonspontaneous, we expect the reverse reaction to be spontaneous. This prediction is borne out experimentally. When silver azide is struck by a hammer, it decomposes explosively into its elements! Classify the following reactions as one of the four possible types (cases) just described. Hence suggest whether the reaction will be spontaneous at (i) a very low temperature, and (ii) at a very high temperature. Δ °(298 K)/ kJ mol Δ °(298 K)/ J K mol This reaction is exothermic, and Δ is negative. It belongs to type 2 and will be spontaneous at low temperatures but nonspontaneous at high temperatures. Since this reaction is endothermic and Δ is negative, it belongs to type 4. It cannot be spontaneous at any temperature. Since this reaction is endothermic but Δ is positive, it belongs to type 3. It will be spontaneous at high temperatures and nonspontaneous at ow temperatures. (All dissociation reactions belong to this class.) This reaction belongs to type 1 and is spontaneous at all temperatures. Above is a table of standard enthalpies of formation at 25°C. This is the enthalpy change which occurs when a compound is formed from its constituent elements. For instance, the enthalpy of formation value of Δ = –285.8 kJ mol is the enthalpy change for the reaction: H ( ) + ½O ( ) → H O( ). It is important to note that the enthalpy of formation for an element in its most stable state is 0. Using these enthalpies of formation and the enthalpy change of any reaction involving the compounds with a known enthalpy of formation can be determined. This table is found on CoreChem:Standard Enthalpies of Formation | 12,713 | 937 |
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The Electronic Structure of Atoms in Biology Since had discovered that the nucleus occupies only 10 % (a millionth of a millionth of a percent) of an atom's volume, it became clear that the nucleus was essentially a tiny point, and that the shape and size of atoms is entirely due to a very sparse "electron cloud" of some kind. This realization was critical to biology, because the shape and size of molecules is critical to understanding their biological function, and the "electron cloud" of constituent atoms determines their shape and size. This is sometimes very subtle. For example, there are two different forms of the carvone molecule, C H O (like right and left hands): The two forms of carvone The carvone on the left (R-(-)-Carvone) smells like spearmint, and is used for aromatherapy, alternative medicine, and room fresheners. The carvone on the right, (S-(+)-carvone) smells like caraway. The difference in smell is an indication of the difference in biological activity that comes from very small differences in shape, a principal that is encountered again and again in the drug industry. But it would also be impossible to understand photosynthesis and vision, and indeed, the color of leaves and flowers without understanding the electronic structure of atoms, because and their ability to absorb or release energy. Naturally, all chemists, biologists, and physicists became interested in the electronic structure of atoms—the chemists because they wanted to explain , chemical properties and bonding, the physicists because they wanted to explain the color and spectra of atoms (the light emitted when gaseous atoms were raised to a high temperature or bombarded by electrons) and biologists for the reasons above. The chief contributors to fundamental understanding of electronic structure, which developed mainly during the 15 years between 1910 and 1925, were the U.S. chemist Gilbert Newton Lewis (1875 to 1946), and the Danish physicist Niels Bohr (1885 to 1962). Ideas about the electronic structures of atoms progressed during the first half of the twentieth century. The periodic repetition of chemical properties discovered by Mendeleev led G. N. Lewis to the conclusion that atoms must have a . This was confirmed by , which developed very beautiful wave patterns which replaced the circular or elliptical shells of earlier models. The energy of each electron in an atom depends on how strongly the electron is attracted by the positive charge on the nucleus and on how much it is repelled by other electrons. Although each electron cannot be assigned a precise trajectory or orbit in an atom, its wave pattern allows us to . From this the energy of each electron and the order of filling can be obtained. Thus we can determine the electron configuration for an atom of any element. Such correlate with . Because electrons in inner orbitals screen outer electrons from nuclear charge, the fourth and higher shells begin to fill before (and sometimes ) subshells in previous shells are occupied. This overlap in energies of shells explains why Lewis’ ideas are less useful for elements in the fourth and subsequent rows of the periodic table, like the cobalt in vitamin B12 or iron in hemoglobin. It also accounts for the steady variation in properties of transition metals, from titanium to zinc, across the middle of the table, and for the nearly identical characteristics of inner transition elements as opposed to the large differences from one group of representative elements to the next. Although some added complication arises from the wave-mechanical picture, it does confirm Lewis’ basic postulate that valence electrons determine chemical properties and influence the bonding of one atom to another. In other pages you will see how rearrangement of valence electrons can hold atoms together, and how different kinds of bonds result in different macroscopic properties. | 3,934 | 938 |
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are used to relate energy changes to the chemical reactions that produce them. For example, we've already seen in Metabolism of Dietary Sugar that sugar is metabolized according to the equation : It is important to notice that Δ is the energy change for the equation as written. This is necessary because the quantity of heat released or absorbed by a reaction is proportional to the amount of each substance consumed or produced by the reaction. Thus Eq. (1) tells us that 2805 kJ of heat energy is given off for every 6 mole of H for Equation (1) also tells us that 2808 kJ of heat is released when 6 mol of carbon dioxide is produced, or 6 mol of oxygen is consumed. Seen in this way, ΔH is a conversion factor enabling us to calculate the heat absorbed when a given amount of substance is consumed or produced. If q n How much heat energy is obtained if we assume that the eagle's diet of 250-550 g includes 350 g of glucose, C H O , which is burned in oxygen according to the equation: C H O (s) + 6 O (g) → 6 CO (g) + 6 H O(l) The mass of C H O is easily converted to the amount of C H O from which the heat energy is easily calculated by means of Eq. (2). The value of Δ is –2805 kJ per mole of C H O , \(m_{\text{C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}\text{ }\xrightarrow{M}\text{ }n_{\text{C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}\text{ }\xrightarrow{\Delta H_{m}}\text{ }q\) so that It is important to realize that the value of Δ given in thermochemical equations like (1) or (3) depends on the physical state of both the reactants and the products. Thus, if water were obtained as a gas instead of a liquid in the reaction in Eq. (1), the value of Δ would be different from -2808 kJ. It is also necessary to specify both the temperature and pressure since the value of Δ depends very slightly on these variables. If these are not specified [as in Eq. (3)] they usually refer to 25°C and to normal atmospheric pressure. Two more characteristics of thermochemical equations arise from the law of conservation of energy. The first is that For example, Ed Vitz (Kutztown University), (University of | 2,175 | 939 |
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When a stepwise ionic addition reaction involves attack at carbon as a first step, it is described as a . Reactions of this type often are catalyzed by bases, which generate the required nucleophile. For example, consider the addition of some weakly acidic reagent \(\ce{HX}\) to an alkene. In the presence of a strong base \(\left( ^\ominus \ce{OH} \right)\), \(\ce{HX}\) could give up its proton to form the conjugate base \(\ce{X}^\ominus\), which is expected to be a much better nucleophile than \(\ce{HX}\): \[\ce{H}:\ce{X} + ^\ominus \ce{OH} \rightleftharpoons \ce{H_2O} + :\ce{X}^\ominus\] What can follow with an alkene is an with the following two propagating steps. First, the nucleophile attacks at carbon to form a carbon anion (carbanion) intermediate (Equation 10-8). Second, electrophilic transfer of a proton from \(\ce{HX}\) to the carbanion forms the adduct and regenerates the nucleophile (Equation 10-9). The overall reaction is the addition of \(\ce{HX}\) to the double bond: Net reaction: The \(\ce{HX}\) reagent can be water, an alcohol \(\left( \ce{ROH} \right)\), a thiol \(\left( \ce{RSH} \right)\), an amine \(\left( \ce{RNH_2} \right)\), or hydrogen cyanide \(\left( \ce{HCN} \right)\) or other carbon acids (i.e., compounds with acidic \(\ce{C-H}\) bonds). However, nucleophilic addition of these reagents to simple alkenes is encountered. To have nucleophilic addition the double bond must be substituted with strongly electron-withdrawing groups such as carbonyl-containing groups, \(\ce{NO_2}\), \(\ce{C \equiv N}\), or positively charged ammonium or sulfonium groups. However, alkynes generally are more reactive towards nucleophiles than they are toward electrophiles. For example, with a base catalyst, 2-hexen-4-yne adds methanol across the triple bond, leaving the double bond untouched: (Nonetheless, the double bond seems to be necessary because a corresponding addition is not observed for 2-butyne, \(\ce{CH_3C \equiv CCH_3}\).) Many nucleophilic addition reactions have considerable synthetic value, particularly those involving addition of carbon acids, such as \(\ce{HCN}\), because they provide ways of forming carbon-carbon bonds. More of their utility will be discussed in Chapters 14, 17, and 18. and (1977) | 2,283 | 940 |
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This page describes and explains the trends in atomic and physical properties of the Period 3 elements from sodium to argon. It covers ionization energy, atomic radius, electronegativity, electrical conductivity, melting point and boiling point. Across Period 3 of the Periodic Table, the 3s and 3p orbitals fill with electrons. Below are the abbreviated electronic configurations for the eight Period 3 elements: In each case, [Ne] represents the complete electronic configuration of a neon atom. The first ionization energy is the energy required to remove the most loosely held electron from one mole of gaseous atoms to produce 1 mole of gaseous ions each with a charge of +1. \[ X (g) \rightarrow X^+ (g) + e^-\] The molar first ionization energy is the energy required to carry out this change per mole of \(X\). There is a general upward trend across the period, but this trend is broken by decreases between magnesium and aluminum, and between phosphorus and sulfur. First ionization energy is dependent on four factors: In the whole of period 3, the outer electrons are in 3-level orbitals. These electrons are at approximately the same distance from the nucleus, and are screened by corresponding electrons in orbitals with principal atomic numbers n=1 and n=2. The determining factor in the increase in energy is the increasing number of protons in the nucleus from sodium across to argon. This creates greater attraction between the nucleus and the electrons and thus increases the ionization energies. The increasing nuclear charge also pulls the outer electrons toward the nucleus, further increasing ionization energies across the period. : The value for aluminum might be expected to be greater than that of magnesium due to the extra proton. However, this effect is offset by the fact that the outer electron of aluminum occupies a 3p orbital rather than a 3s orbital. The 3p electron is slightly farther from the nucleus than the 3s electron, and partially screened by the 3s electrons as well as the inner electrons. Both of these factors offset the effect of the extra proton. In this case something other than the transition from a 3s orbital to a 3p orbital must offset the effect of an extra proton. The screening 3s 3p The diagram below shows how atomic radius changes across Period 3. The figures used to construct this diagram are based on: It is appropriate to compare metallic and covalent radii because they are both being measured in tightly bonded circumstances. These radii cannot be compared with a van der Waals radius, however, making the diagram deceptive. A metallic or covalent radius is a measure of the distance from the nucleus to the bonding pair of electrons. From sodium to chlorine, the bonding electrons are all in the 3-level, screened by the electrons in the first and second levels. The increasing number of protons in the nucleus across the period attracts the bonding electrons more strongly. The amount of screening is constant across Period 3. is a measure of the tendency of an atom to attract a bonding pair of electrons. The Pauling scale is most commonly used. Fluorine (the most electronegative element) is assigned a value of 4.0, and values decrease toward cesium and francium which are the least electronegative at 0.7. The trend across Period 3 looks like this: Argon is not included; because it does not form covalent bonds, its electronegativity cannot be assigned. The explanation is the same as that for the trend in atomic radii. Across the period, the valence electrons for each atom are in the 3-level. They are screened by the same inner electrons. The only difference is the number of protons in the nucleus. From sodium to chlorine, the number of protons steadily increases and so attracts the bonding pair more closely. This section discusses electrical conductivity and the melting and boiling points of the Period 3 elements. An understanding of the structure of each element is necessary for this discussion. The structures of the elements vary across the period. The first three are metallic, silicon is network covalent, and the rest are simple molecules. Sodium, magnesium and aluminum all have metallic structures. In sodium, only one electron per atom is involved in the metallic bond, the single 3s electron. In magnesium, both of its outer electrons are involved, and in aluminum all three are involved. One key difference to be aware of is the way the atoms are packed in the metal crystal. Sodium is 8-coordinated with each sodium atom interacting with only 8 other atoms. Magnesium and aluminum are each 12-coordinated, and therefore packed more efficiently, creating less empty space in the metal structures and stronger bonding in the metal. Silicon has a network covalent structure like that of diamond. A representative section of this structure is shown: The structure is held together by strong covalent bonds in all three dimensions. The structures of phosphorus and sulfur vary depending on the type of phosphorus or sulfur in question. In this case, white phosphorus and one of the crystalline forms of sulfur—rhombic or monoclinic—are considered. These structures are shown below: Aside from argon, the atoms in each of these molecules are held together by covalent bonds. In the liquid or solid state, the molecules are held in close proximity by . The three metals conduct electricity because the delocalized electrons (as in the "sea of electrons" model) are free to move throughout the solid or the liquid metal. Semiconductor chemistry for substances such as silicon is beyond the scope of most introductory level chemistry courses. The other elements do not conduct electricity because they are simple molecular substances. without free, delocalized electrons.. The chart shows how the melting and boiling points of the elements change as you go across the period. The figures are plotted in kelvin rather than °C to avoid showing negative temperatures. Melting and boiling points increase across the three metals because of the increasing strength of their metallic bonds. The number of electrons which each atom can contribute to the delocalized "sea of electrons" increases. The atoms also get smaller and have more protons as you go from sodium to magnesium to aluminum. The attractions and therefore the melting and boiling points increase because: Silicon has high melting and boiling points due to its network covalent structure. Melting or boiling silicon requires the breaking of strong covalent bonds. Because of the two different types of bonding in silicon and aluminum, it makes little sense to directly compare the two melting and boiling points. Phosphorus, sulfur, chlorine and argon are simple molecular substances with only van der Waals attractions between the molecules. Their melting or boiling points are lower than those of the first four members of the period which have complex structures. The magnitudes of the melting and boiling points are governed entirely by the sizes of the molecules, which are shown again for reference: Jim Clark ( ) | 7,091 | 941 |
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Molecules with hydrogen atoms bonded to electronegative atoms such as O, N, and F (and to a much lesser extent Cl and S) tend to exhibit unusually strong intermolecular interactions. These result in much higher boiling points than are observed for substances in which dominate, as illustrated for the covalent hydrides of elements of groups 14–17 in Figure \(\Page {1}\). Methane and its heavier congeners in group 14 form a series whose boiling points increase smoothly with increasing molar mass. This is the expected trend in nonpolar molecules, for which London dispersion forces are the exclusive intermolecular forces. In contrast, the hydrides of the lightest members of groups 15–17 have boiling points that are more than 100°C greater than predicted on the basis of their molar masses. The effect is most dramatic for water: if we extend the straight line connecting the points for H Te and H Se to the line for period 2, we obtain an estimated boiling point of −130°C for water! Imagine the implications for life on Earth if water boiled at −130°C rather than 100°C. Why do strong intermolecular forces produce such anomalously high boiling points and other unusual properties, such as high enthalpies of vaporization and high melting points? The answer lies in the highly polar nature of the bonds between hydrogen and very electronegative elements such as O, N, and F. The large difference in electronegativity results in a large partial positive charge on hydrogen and a correspondingly large partial negative charge on the O, N, or F atom. Consequently, H–O, H–N, and H–F bonds have very large bond dipoles that can interact strongly with one another. Because a hydrogen atom is so small, these dipoles can also approach one another more closely than most other dipoles. The combination of large bond dipoles and short dipole–dipole distances results in very strong dipole–dipole interactions called , as shown for ice in Figure \(\Page {2}\). A hydrogen bond is usually indicated by a dotted line between the hydrogen atom attached to O, N, or F (the ) and the atom that has the lone pair of electrons (the ). Because each water molecule contains two hydrogen atoms and two lone pairs, a tetrahedral arrangement maximizes the number of hydrogen bonds that can be formed. In the structure of ice, each oxygen atom is surrounded by a distorted tetrahedron of hydrogen atoms that form bridges to the oxygen atoms of adjacent water molecules. The bridging hydrogen atoms are equidistant from the two oxygen atoms they connect, however. Instead, each hydrogen atom is 101 pm from one oxygen and 174 pm from the other. In contrast, each oxygen atom is bonded to two H atoms at the shorter distance and two at the longer distance, corresponding to two O–H covalent bonds and two O⋅⋅⋅H hydrogen bonds from adjacent water molecules, respectively. The resulting open, cagelike structure of ice means that the solid is actually slightly less dense than the liquid, which explains why ice floats on water rather than sinks. Each water molecule accepts two hydrogen bonds from two other water molecules and donates two hydrogen atoms to form hydrogen bonds with two more water molecules, producing an open, cagelike structure. The structure of liquid water is very similar, but in the liquid, the hydrogen bonds are continually broken and formed because of rapid molecular motion. Hydrogen bond formation requires a hydrogen bond donor a hydrogen bond acceptor. Because ice is less dense than liquid water, rivers, lakes, and oceans freeze from the top down. In fact, the ice forms a protective surface layer that insulates the rest of the water, allowing fish and other organisms to survive in the lower levels of a frozen lake or sea. If ice were denser than the liquid, the ice formed at the surface in cold weather would sink as fast as it formed. Bodies of water would freeze from the bottom up, which would be lethal for most aquatic creatures. The expansion of water when freezing also explains why automobile or boat engines must be protected by “antifreeze” and why unprotected pipes in houses break if they are allowed to freeze. Considering CH OH, C H , Xe, and (CH ) N, which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures. compounds formation of hydrogen bonds and structure Of the species listed, xenon (Xe), ethane (C H ), and trimethylamine [(CH ) N] do not contain a hydrogen atom attached to O, N, or F; hence they cannot act as hydrogen bond donors. The one compound that can act as a hydrogen bond donor, methanol (CH OH), contains both a hydrogen atom attached to O (making it a hydrogen bond donor) and two lone pairs of electrons on O (making it a hydrogen bond acceptor); methanol can thus form hydrogen bonds by acting as either a hydrogen bond donor or a hydrogen bond acceptor. The hydrogen-bonded structure of methanol is as follows: Considering \(\ce{CH3CO2H}\), \(\ce{(CH3)3N}\), \(\ce{NH3}\), and \(\ce{CH3F}\), which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures. \(\ce{CH3CO2H}\) and \(\ce{NH3}\); Although hydrogen bonds are significantly weaker than covalent bonds, with typical dissociation energies of only 15–25 kJ/mol, they have a significant influence on the physical properties of a compound. Compounds such as \(\ce{HF}\) can form only two hydrogen bonds at a time as can, on average, pure liquid NH . Consequently, even though their molecular masses are similar to that of water, their boiling points are significantly lower than the boiling point of water, which forms hydrogen bonds at a time. Arrange C (buckminsterfullerene, which has a cage structure), NaCl, He, Ar, and N O in order of increasing boiling points. compounds order of increasing boiling points Identify the intermolecular forces in each compound and then arrange the compounds according to the strength of those forces. The substance with the weakest forces will have the lowest boiling point. Electrostatic interactions are strongest for an ionic compound, so we expect NaCl to have the highest boiling point. To predict the relative boiling points of the other compounds, we must consider their polarity (for dipole–dipole interactions), their ability to form hydrogen bonds, and their molar mass (for London dispersion forces). Helium is nonpolar and by far the lightest, so it should have the lowest boiling point. Argon and N O have very similar molar masses (40 and 44 g/mol, respectively), but N O is polar while Ar is not. Consequently, N O should have a higher boiling point. A C molecule is nonpolar, but its molar mass is 720 g/mol, much greater than that of Ar or N O. Because the boiling points of nonpolar substances increase rapidly with molecular mass, C should boil at a higher temperature than the other nonionic substances. The predicted order is thus as follows, with actual boiling points in parentheses: He (−269°C) < Ar (−185.7°C) < N O (−88.5°C) < C (>280°C) < NaCl (1465°C). Arrange 2,4-dimethylheptane, Ne, CS , Cl , and KBr in order of decreasing boiling points. KBr (1435°C) > 2,4-dimethylheptane (132.9°C) > CS (46.6°C) > Cl (−34.6°C) > Ne (−246°C) Besides mercury, water has the highest for all liquids. Water's high surface tension is due to the hydrogen bonding in water molecules. Water also has an exceptionally high . Vaporization occurs when a liquid changes to a gas, which makes it an endothermic reaction. Water's heat of vaporization is 41 kJ/mol. Vapor pressure is inversely related to intermolecular forces, so those with stronger intermolecular forces have a lower vapor pressure. Water has very strong intermolecular forces, hence the low vapor pressure, but it's even lower compared to larger molecules with low vapor pressures. Because of water's polarity, it is able to dissolve or dissociate many particles. Oxygen has a slightly negative charge, while the two hydrogens have a slightly positive charge. The slightly negative particles of a compound will be attracted to water's hydrogen atoms, while the slightly positive particles will be attracted to water's oxygen molecule; this causes the compound to dissociate. Besides the explanations above, we can look to some attributes of a water molecule to provide some more reasons of water's uniqueness: The properties of water make it suitable for organisms to survive in during differing weather conditions. Water expands as it freezes, which explains why ice is able to float on liquid water. During the winter when lakes begin to freeze, the surface of the water freezes and then moves down toward deeper water; this explains why people can ice skate on or fall through a frozen lake. If ice was not able to float, the lake would freeze from the bottom up killing all ecosystems living in the lake. However ice floats, so the fish are able to survive under the surface of the ice during the winter. The surface of ice above a lake also shields lakes from the cold temperature outside and insulates the water beneath it, allowing the lake under the frozen ice to stay liquid and maintain a temperature adequate for the ecosystems living in the lake to survive. Intermolecular forces are electrostatic in nature and include van der Waals forces and hydrogen bonds. Molecules in liquids are held to other molecules by intermolecular interactions, which are weaker than the intramolecular interactions that hold the atoms together within molecules and polyatomic ions. Transitions between the solid and liquid or the liquid and gas phases are due to changes in intermolecular interactions but do not affect intramolecular interactions. The three major types of intermolecular interactions are dipole–dipole interactions, London dispersion forces (these two are often referred to collectively as ), and hydrogen bonds. arise from the electrostatic interactions of the positive and negative ends of molecules with permanent dipole moments; their strength is proportional to the magnitude of the dipole moment and to 1/ , where is the distance between dipoles. are due to the formation of in polar or nonpolar molecules as a result of short-lived fluctuations of electron charge distribution, which in turn cause the temporary formation of an in adjacent molecules. Like dipole–dipole interactions, their energy falls off as 1/ . Larger atoms tend to be more than smaller ones because their outer electrons are less tightly bound and are therefore more easily perturbed. are especially strong dipole–dipole interactions between molecules that have hydrogen bonded to a highly electronegative atom, such as O, N, or F. The resulting partially positively charged H atom on one molecule (the ) can interact strongly with a lone pair of electrons of a partially negatively charged O, N, or F atom on adjacent molecules (the ). Because of strong O⋅⋅⋅H hydrogen bonding between water molecules, water has an unusually high boiling point, and ice has an open, cagelike structure that is less dense than liquid water. | 11,039 | 942 |
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For a polyatomic molecule containing \(N\) atoms, the total number of degrees of freedom is \(3N\). Out of these, three degrees of freedom are taken up for the translational motion of the molecule as a whole. For nonlinear molecules, there are three rotational degrees of freedom and the \(3N – 6\) vibrational degrees. For linear molecules, the rotational motion along the molecular axis is quantum mechanically not meaningful as the rotated configuration is indistinguishable from the original configuration. Therefore, linear molecules have two rotational degrees of freedom and \(3N – 5\) vibrational degrees of freedom. To investigate the rotational motion, we need to fix the center of mass of the molecule and calculate the three principal moments of inertia \(I_A\), \(I_B\), and \(I_C\) of the ellipsoid of inertia. The center of mass is defined as the point for which the following identities hold: \[ \sum_i m_i x_i = \sum_i m_i y_i = \sum_i m_i z_i \nonumber \] The inertia products are defined by: \[ I_{xx} = \sum_i m_i \left(y_i^2+ z^2_i\right) \nonumber \] \[ I_{xy} = \sum_i m_i \left(x_i y_i\right) \nonumber \] The other components \(I_{yy}\), \(I_{xz}\),.. are defined analogously. To find the direction cosines \(\alpha_i, \beta_o, \gamma_i\) of the three principle moments of inertia, we need to solve the following matrix equations: \[ \alpha (I_{xx} - \eta) - \beta I_{xy} - \gamma I_{xz} =0\nonumber \] \[ \alpha I_{xy} - \beta (I_{yy} - \eta) - \gamma I_{yz} = 0\nonumber \] \[ -\alpha I_{xz} - \beta I_{yz} + \gamma (I_{zz} - \eta ) =0\nonumber \] If the off diagonal terms \(I_{xy}\) are zero in the above equations, then the \(x\), \(y\), \(z\) axis will be the principal axis. The energy of a rotor with the three moments of inertia \(I_A\), \(I_B\). and \(I_C\) is given by: \[\begin{align*} \epsilon &= \dfrac{1}{2} I_A \omega_A^2 + \dfrac{1}{2} I_B \omega_B^2 + \dfrac{1}{2} I_C \omega_C^2 \\[4pt] &= \dfrac{L_A^2}{2I_A} + \dfrac{L_B^2}{2I_B} + \dfrac{L_C^2}{2I_C} \end{align*} \nonumber \] Each of the rotational degrees of freedom will have a characteristic rotational temperature in terms of the moment of inertia: \[ \Theta_{rot,i} = \frac{hbar^2}{2I_ik} \qquad i=A,B,C \] There are many different shapes of molecules and these shapes affect the rotational behavior of the molecules. Molecules are therefore classified according to symmetry into different groups called tops. The three different tops are: \[ \text{Spherical top} \qquad \Theta_{rot,A} = \Theta_{rot,B} = \Theta_{rot,C} \] \[ \text{Symmetric top} \qquad \Theta_{rot,A} = \Theta_{rot,B} \neq \Theta_{rot,C} \nonumber\] \[ \text{Asymmetric top} \qquad \Theta_{rot,A} \neq \Theta_{rot,B} \neq \Theta_{rot,C} \nonumber\] The spherical top can be solved exactly to give: \[ E_J = \frac{J(J+1)\hbar^2}{2I} \] \[ g_J = (2J+1^2) \qquad J=0,1,2,\ldots \] The rotational partition function is: \[\begin{split} q_\text{rot} &= \sum_{J=0}^\infty (2J+1)^2 e^{\hbar^2J(J+1)/2IkT} \\ &= \sum_{J=0}^\infty (2J+1)^2 e^{\Theta_\text{rot}J(J+1)/T} \end{split}\] For almost all spherical top molecules: \[ \Theta_\text{rot} \gg T \nonumber\] Therefore, we can convert the sum to an integral: \[ q_\text{rot} = \frac{1}{\sigma}\int_{0}^\infty (2J+1)^2 e^{\Theta_\text{rot}J(J+1)/T} \] where we have now included the symmetry term, \(\sigma\). Solving for this integral, we get: \[ q_\text{rot} = \frac{\pi^{1/2}}{\sigma}\left(\frac{T}{q_\text{rot}}\right)^{3/2} \] Here, \(\omega_A\), \(\omega_B\), and \(\omega_C\) are the three angular speeds and \(L_A\), \(L_B\), and \(L_C\) are the three angular momenta. For a symmetric top molecule such as ammonia, or chloromethane, two components of the moments of inertia are equal, i.e., \(I_B = I_C\). The rotational energy levels of such a molecule are specified by two quantum numbers \(J\) and \(K\). The total angular momentum is determined by \(J\) and the component of this angular momentum along the unique molecular axis is determined by \(K\). The energy levels are given by: \[ \epsilon_{J,K} = \tilde{B} J(J+1) + (\tilde{A} - \tilde{B}) K^2\nonumber \] with rotational constants in units of wavenumbers: \[ \tilde{B} = \dfrac{h}{8\pi^2 c I_B}\nonumber \] and: \[ \tilde{A} = \dfrac{h}{8\pi^2 c I_A}\nonumber \] where The rotational partition function is given by: \[ q_{rot} = \dfrac{1}{\sigma} \sum_{J=0}^{\infty} (2J+1) e^{-\tilde{B} J(J+1)/kT} \sum_{K=-J}^{J} (2J+1) e^{\tilde{A} - \tilde{B} )K^2/kT} \nonumber \] This can be converted to an integral and the result is: \[ q_{rot} = \dfrac{ \sqrt{\pi}}{\sigma} \left( \dfrac{8 \pi I_B kT}{h^2} \right) \left( \dfrac{8 \pi I_A kT}{h^2} \right)^{1/2} \nonumber \] For asymmetric tops, the expressions for rotational energies are more complex and the conversions to integrations are not easy. One can actually calculate the sum of terms using a computer. An intuitive answer can be obtained by integrating over the angular momenta \(L_A\), \(L_B\) and \(L_C\) as: \[ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{H(p,q)/kT} \,dL_A \,dL_B \,dL_C = \sqrt{ 2\pi I_A kT} \sqrt{ 2\pi I_B kT} \sqrt{ 2\pi I_C kT} \nonumber \] And then multiplying by a factor of \(8 π^2 /σ h^3\), we get the rotational partition function. The factor of \(8π^2\) accounts for the angular integration. For any axis chosen in a molecule, a complete rotation contributes a factor of \(2π\). Integration over all possible orientations of this axis contribute another factor of \(4π\). The factor of \(h^3\) is for the conversion from the classical phase space to the quantum mechanical phase space. As discussed , the \(σ\) corrects for overcounting of rotational configurations. If the molecule has no center of symmetry (e.g. \(\ce{HCN}\)) \(\sigma = 1\) whereas if the molecule has a center of symmetry (e.g. \(\ce{CO2}\)) then \(\sigma = 2\). The final result is: \[ q_{rot} = \dfrac{\pi^2}{\sigma} \sqrt{\dfrac{8 \pi I_A kT}{h^2}} \sqrt{\dfrac{8 \pi I_B kT}{h^2}} \sqrt{\dfrac{8 \pi I_C kT}{h^2}} \label{big0} \] We can explicitly obtain the classical rotational partition function of an asymmetric top by writing the classical expression for the rotational energy in terms of the Euler angles. The orientation of a rigid rotor can be specified by there Euler angles \(θ\), \(φ\), and \(ψ\) with the ranges of angles \(0\) to \(π\), \(0\) to \(2π\) and \(0\) to \(2π\) respectively. The rotational Hamiltonian for the kinetic energy can be written in terms of the angles and their conjugate momenta ( \(p_{\theta}\), \( p_{\phi}\), \(p_{\psi}\)) \[ H = \dfrac{\sin^2 \psi}{2I_A} \left( p_{\theta} - \dfrac{\cos \psi}{\sin \theta \sin \psi} ( p_{\phi} - \cos \theta p_{\psi} ) \right)^2 \nonumber \] \[+ \dfrac{\cos^2 \psi}{2I_B} \left( p_{\theta} - \dfrac{\sin \psi}{\cos \theta \cos \psi} ( p_{\phi} - \cos \theta p_{\psi} ) \right)^2 + \dfrac{1}{2I_C} p_{\psi}^2\nonumber \] The classical rotational partition function is given by \[ q_{rot} = \int_{-\infty}^ {\infty} \int_{-\infty}^ {\infty} \int_{-\infty}^ {\infty} \int_{0}^ {\pi} \int_{0}^ {2\pi} \int_{0}^ {2\pi} \dfrac{1}{h^3} e ^{-H(p,q)/kT} dp_{\theta}\, dp_{\phi} \, dp_{\psi} \,d \theta \,d \phi \,d \psi \nonumber \] The integrations can be simplified by rewriting \(H(p,q) / kT\) as \[ \dfrac{H}{kT} = \dfrac{1}{2I_AkT} \left( \dfrac{\sin^2 \psi}{I_A} + \dfrac{\cos^2 \psi}{I_B} \right) \left( p_{\theta} + \left( \dfrac{1}{I_B} - \dfrac{1}{I_A} \right) \dfrac{\sin \psi \cos \psi}{\sin \theta \left(\dfrac{\sin^ \psi}{I_A} + \dfrac{\cos^2 \psi}{I_B} \right) } (p_{\phi} - \cos \theta p_{\psi}) \right)^2\nonumber \] \[+ \dfrac{1}{2 kT I_AI_B \sin^2 \theta} \left( \dfrac{1}{\sin \theta \left(\dfrac{\sin^ \psi}{I_A} + \dfrac{\cos^2 \psi}{I_B} \right) } (p_{\phi} - \cos \theta p_{\psi}) \right)^2 + \dfrac{1}{2KT I_c} p^2_{\psi}\nonumber \] Using the following integral, \[ \int_{-\infty}^ {\infty} e^{-a(x+b)^2} dx = \int_{-\infty}^{\infty} e ^{-ax^2} dx = \sqrt{\dfrac{\pi}{a}}\nonumber \] Integration over \(p_θ\) gives using the above expression \[ \sqrt{ 2 \pi kT} \left( \dfrac{\sin^2 \psi}{I_A} + \dfrac{\cos^2 \psi}{I_B} \right)^{-1/2} \label{SA1} \] Integration over \(p_φ\) gives the factor, \[ \sqrt{ 2 \pi kT I_AI_B} \sin \theta \left( \dfrac{\sin^2 \psi}{I_A} + \dfrac{\cos^2 \psi}{I_B} \right)^{1/2}\nonumber \] This cancels partly the second square root in Equation \(\ref{SA1}\). Integration over \(p_ψ\) gives the factor \[\sqrt{2 \pi k T I_c}\nonumber \] Integration over \(θ\), \(φ\) and \(ψ\) gives a factor of \( 8 π^ 2\). \[ \int _0^{\pi} \sin \theta \,d\theta =2\nonumber \] \[ \int_0^{2\pi} d\phi = 2 \pi\nonumber \] \[ \int_0^{2\pi} d \psi = 2\pi\nonumber \] Combining all the integrals, we finally get Equation \(\ref{big0}\) after reintroduced the symmetry number \(σ\) as before with diatomic molecular rotation. We can simplify calculations by defining for each axis of rotational: \[ \Theta _A = \dfrac{h^2}{8 \pi^2 I_A k} \nonumber \] \[ \Theta _B = \dfrac{h^2}{8 \pi^2 I_B k} \nonumber \] \[ \Theta_C = \dfrac{h^2}{8 \pi^2 I_C k} \nonumber \] The polyatomic rotational partition function expressed in Equation \(\ref{big0}\) can then be re-expressed as: \[ q_{rot} = \dfrac{\sqrt{\pi}}{\sigma}\sqrt{\dfrac{T^3}{ \Theta_A \Theta_B \Theta_C }} \label{RotQ1} \] or alternatively: \[ \ln q_{rot} = \dfrac{1}{2} \ln \dfrac{\pi T}{ \Theta_A \Theta_B \Theta_C \sigma^2}\nonumber \] The three characteristic rotational temperatures for \(\ce{NO_2}\) are \(111.5\, K\), \(0.624\, K\) and \(0.590\, K\). Calculate the rotational partition function at 300 K. The rotational temperature is given by Equation \(\ref{RotQ1}\): \[ q_{rot} = \dfrac{\sqrt{\pi}}{\sigma} \sqrt{\dfrac{T^3}{ \Theta_A \Theta_B \Theta_C }} \nonumber \] The rotational partition function becomes, \[ q_{rot} = \dfrac{1.772}{2} \sqrt{\dfrac{300\; K}{ (11.5\; K)(0.624\,K) ( 22.55\;K)}} = 2242.4 \nonumber \] The molar thermodynamic functions can be readily calculated including average rotation energy and molar heat capacity: \[ E_{rot} = \dfrac{3}{2} RT\nonumber \] and: \[\bar{C_V}= \dfrac{3}{2} R\nonumber \] Improvements over the classical approximation for the rotational partition function derived above have been obtained. One of the improved versions (with no derivation) is: \[ q_{rot} = q_{rot}^0 \left[ 1+ \dfrac{h^2}{96 \pi^2 kT} \left( \dfrac{2}{I_A} + \dfrac{2}{I_C} + \dfrac{2}{I_C} + \dfrac{I_C}{I_BI_B} - \dfrac{I_A}{I_BI_C} - \dfrac{I_B}{I_AI_C} \right) \right]\nonumber \] where \( q_{rot}^0 \) is the classical approximation in Equation \(\ref{big0}\). Comparing this result with the vibrational partition function calculation before (\(q_{vib} = 1.0035\)), give the implication that while multiple rotational states are accessible at room temperature, very few vibrational states (other than the ground vibrational state) are accessible. | 10,835 | 943 |
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The magnetic properties of a compound can be determined from its electron configuration and the size of its atoms. Because magnetism is generated by electronic spin, the number of unpaired electrons in a specific compound indicates how magnetic the compound is. In this section, the magnetism of the d-block elements (or transition metals) are evaluated. These compounds tend to have a large number of unpaired electrons. An interesting characteristic of transition metals is their ability to form magnets. Metal complexes that have unpaired electrons are magnetic. Since the last electrons reside in the d orbitals, this magnetism must be due to having unpaired d electrons. The spin of a single electron is denoted by the quantum number \(m_s\) as +(1/2) or –(1/2). This spin is negated when the electron is paired with another, but creates a weak magnetic field when the electron is unpaired. More unpaired electrons increase the paramagnetic effects. The electron configuration of a transition metal (d-block) changes in a coordination compound; this is due to the repulsive forces between electrons in the ligands and electrons in the compound. Depending on the strength of the ligand, the compound may be paramagnetic or diamagnetic. Ferromagnetism is the basic mechanism by which certain materials (such as iron) form permanent magnets. This means the compound shows permanent magnetic properties rather than exhibiting them only in the presence of a magnetic field (Figure \(\Page {1}\)). In a ferromagnetic element, electrons of atoms are grouped into domains in which each domain has the same charge. In the presence of a magnetic field, these domains line up so that charges are parallel throughout the entire compound. Whether a compound can be ferromagnetic or not depends on its number of unpaired electrons and on its atomic size. Ferromagnetism, the permanent magnetism associated with nickel, cobalt, and iron, is a common occurrence in everyday life. Examples of the knowledge and application of ferromagnetism include Aristotle's discussion in 625 BC, the use of the compass in 1187, and the modern-day refrigerator. Einstein declared that electricity and magnetism are inextricably linked in his theory of special relativity. Experimental evidence of magnetic measurements supports the theory of high- and low-spin complexes. Remember that molecules such as O that contain unpaired electrons are paramagnetic. Paramagnetic substances are attracted to magnetic fields. Many transition metal complexes have unpaired electrons and hence are paramagnetic. Molecules such as N and ions such as Na and [Fe(CN) ] that contain no unpaired electrons are diamagnetic. Diamagnetic substances have a slight tendency to be repelled by magnetic fields. When an electron in an atom or ion is unpaired, the magnetic moment due to its spin makes the entire atom or ion paramagnetic. The size of the magnetic moment of a system containing unpaired electrons is related directly to the number of such electrons: the greater the number of unpaired electrons, the larger the magnetic moment. Therefore, the observed magnetic moment is used to determine the number of unpaired electrons present. The measured magnetic moment of low-spin [Fe(CN) ] confirms that iron is diamagnetic, whereas high-spin [Fe(H O) ] has four unpaired electrons with a magnetic moment that confirms this arrangement. states that electrons fill all available orbitals with single electrons before pairing up, while maintaining parallel spins (paired electrons have opposing spins). For a set of five degenerate d-orbitals in an uncomplexed metal atom, electrons fill all orbitals before pairing to conserve pairing energy. With the addition of ligands, the situation is more complicated. The splitting energy between the d-orbitals increases the energy required to place single electrons into the higher-energy orbitals. Once the lower-energy orbitals have been half-filled (one electron per orbital), an electron can either be placed in a higher-energy orbital or paired with an electron in a lower-energy orbital. The strength of the ligands determine which option is chosen. If the splitting energy is greater than the pairing energy, the electrons will pair up; if the pairing energy is greater, unpaired electrons will occupy higher energy orbitals. In other words, with a strong-field ligand, low-spin complexes are usually formed; with a weak-field ligand, a high-spin complex is formed. Low-spin complexes contain more paired electrons because the splitting energy is larger than the pairing energy. These complexes, such as [Fe(CN) ] , are more often diamagnetic or weakly paramagnetic. Likewise, high-spin complexes usually contain more unpaired electrons because the pairing energy is larger than the splitting energy. With more unpaired electrons, high-spin complexes are often paramagnetic. The unpaired electrons in paramagnetic compounds create tiny magnetic fields, similar to the domains in ferromagnetic materials. The strength of the paramagnetism of a coordination complex increases with the number of unpaired electrons; a higher-spin complex is more paramagnetic. The occurrence and relative strength of paramagnetism can be predicted by determining whether the compound is coordinated to a weak field ligand or a strong field ligand. Which ligand generates a stronger magnetic complex ion when bound to \(Fe^{+2}\): EDTA or \(CN^-\)? Since \([Fe(EDTA)_3^{-2}]\) has more unpaired electrons than \([FeCN_6^{-3}]\), it is more paramagnetic. \(ZnI_4\) have eight valence electrons. If it is found to be diamagnetic, then does it occupy a tetrahedral or square plan geometry? The splitting pattern for the two geometries differ and hence the electron configuration from adding the eight electrons also differ. The tetrahedral geometry has two unpaired electrons and the square planer geometry has zero. Since \(ZnI_4\) is diamagnetic, it must have a square planar geometry. For each of the following coordination complexes, identify if it is paramagnetic or diamagnetic? The Gouy balance is used to measure paramagnetism by suspending the complex in question against an equivalent weight with access to a magnetic field. In principle, a magnetic measurement can be done very simply. All it takes is a balance and a magnetic field. We all know how a balance works. In a simple model from an earlier time, we place the sample in one pan. The balance tips over. We find the appropriate weight - a little piece of carefully prepared metal, certified by some bureau of standards. The balance balances. The masses in the two pans must be equal. With a Gouy balance, the same idea applies, but we throw in a magnetic field, too. Even when it should be balanced, the balance tips, because of an attraction to the magnetic field. We need to find another weight that will get the balance even again. The weight needed to balance the scale is proportional to the attraction of the material to the magnetic field. If the compound is paramagnetic, it will be pulled visibly towards the electromagnet, which is the distance proportional to the magnitude of the compound's paramagnetism. If the compound, however, is diamagnetic, it will not be pulled towards the electromagnet, instead, it might even slightly be repelled by it. This will be proven by the decreased weight or the no change in weight. The change in weight directly corresponds to the amount of unpaired electrons in the compound. , ). | 7,515 | 944 |
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A novel ultrasonic flow injection chemiluminescence (FI-CL) manifold for determining hydrogen peroxide (H O ) has been designed . Chemiluminescence obtained from the luminol-H -cobalt (II) reaction was enhanced by applying 120 W of ultrasound for a period of 4 s to the reaction coil in the FI-CL system and this enhancement was verified by comparison with an identical manifold without ultrasound. The method was applied to the determination of trace amounts of H O in purified water and natural water samples without any special pre-treatments. It is well-known that alkaline solutions of luminol emit light when subject to ultrasound of sufficient intensity to produce acoustic cavitation. Light emission is believed to occur through a process of oxidative chemiluminescence involving sonochemically generated HO . The cyclic pressure variations associated with the propagation of ultrasound waves in aqueous solution are known to result in the growth and periodic collapse of microscopic cavitation bubbles filled with gas and/or vapour [40]. Furthermore, it has been shown that extremely high local temperatures and pressures may be generated during the collapse or implosion of such bubbles. Consequently, it is generally accepted that it is within the cavitation bubble, or the layer of solution immediately contacting the cavitation bubble, that the sonochemical effects take place. Luminol chemiluminescence has been described in section B1 (ADD LINK). Light emission from the reaction between luminol and hydrogen peroxide can be induced by the presence of cobalt(II) at concentrations low enough to be regarded as catalytic. The effect of ultrasound on hydrogen peroxide is to produce hydroxyl radicals by homolytic fission of the O―O bond: \[H_2O_2 \rightarrow 2HO^•\] Hydroxyl radicals in aqueous solution are short-lived. The consumption of these radicals by recombination is very rapid and attenuates the ultrasound enhancement: \[ 2HO^• \rightarrow H_2O_2 \] Because of this, the concentration of hydrogen peroxide soon greatly exceeds that of hydroxyl radical, even if the radicals are initially produced in high yield. There is then a greater probability that radicals will instead react with hydroen peroxide molecules, forming superoxide: \[ HO^• + H_2O_2 \rightarrow O_2^{•-} + H_3O^+\] As a result, the effect of sonication is the production in the sample of superoxide rather than hydroxyl radicals. The hydroxyl radicals initially formed would have reacted with luminol to initiate the light-emitting pathway but the primary oxidation of luminol by superoxide is negligible. Instead, when the sample merges with luminol/buffer/cobalt, the effect of this enhanced superoxide concentration is to increase the concentration of the hydroperoxide intermediate, enhancing the light emitting pathway where it has already been initiated by cobalt/hydrogen peroxide; this leads to a fivefold improvement in the detection limit. The practical implementation of this ultrasound enhancement proved to be exacting. Small changes in the FIA manifold were found to have a considerable effect on the chemiluminescence intensity. In spite of this it was found possible to optimize a range of relevant variables. Some variables were concerned with the arrangements for administering a dose of ultrasound energy to the sample as it flowed through a coil immersed in the sonication bath. To achieve this, the coil had to be long enough to contain the sample all the time that sonication was occurring, but not so long that the enhancement of the chemiluminescence signal would be abolished either by dispersion of the sample into the carrier or by decay of the short-lived radicals generated by sonication. The optimum distances between the water surface and the probe tip and between the probe tip and the upper edge of the sonication coil correspond closely to the conditions for the establishment of standing waves in the sonic bath. Cavitation when present is the predominant mechanism of acoustic energy absorption as well as providing the collapsing bubbles that are the sites of the sonochemical reactions. Absorption by bubbles is so effective that they provide a shielding effect and so could explain the difficulty in predicting the effect of small changes in the position of the coil within the sonication bath. It was necessary to vary the sonication arrangements in order to optimise them, but operational analytical applications of ultrasound enhancement would be more easily carried out using fixed sonication arrangements in a permanent and purpose-designed apparatus. | 4,607 | 945 |
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The quantity , or "power of hydrogen," is a numerical representation of the acidity or basicity of a solution. It can be used to calculate the concentration of hydrogen ions [H ] or ions [H O ] in an aqueous solution. Solutions with low pH are the most acidic, and solutions with high pH are most basic. Although pH is formally defined in terms of , it is often estimated using free proton or hydronium concentration: \[ pH \approx -\log[H_3O^+] \label{eq1}\] or \[ pH \approx -\log[H^+] \label{eq2}\] \(K_a\), the , is the equilibrium constant for chemical reactions involving weak acids in aqueous solution. The numerical value of \(K_a\) is used to predict the extent of acid dissociation. A large \(K_a\) value indicates a stronger acid (more of the acid dissociates) and small \(K_a\) value indicates a weaker acid (less of the acid dissociates). For a chemical equation of the form \[ HA + H_2O \leftrightharpoons H_3O^+ + A^- \] \(K_a\) is express as \[ K_a = \dfrac{[H_3O^+,A^-]}{[HA]} \label{eq3} \] where Since \(H_2O\) is a pure liquid, it has an activity equal to one and is ignored in the equilibrium constant expression in (Equation \ref{eq3}) like in other equilibrium constants. When given the pH value of a solution, solving for \(K_a\) requires the following steps: Calculate the \(K_a\) value of a 0.2 M aqueous solution of propionic acid (\(\ce{CH3CH2CO2H}\)) with a pH of 4.88. \[ \ce{CH_3CH_2CO_2H + H_2O \leftrightharpoons H_3O^+ + CH_3CH_2CO_2^- } \nonumber\] ICE TABLE According to the definition of pH (Equation \ref{eq1}) \[\begin{align*} -pH = \log[H_3O^+] &= -4.88 \\[4pt] [H_3O^+] &= 10^{-4.88} \\[4pt] &= 1.32 \times 10^{-5} \\[4pt] &= x \end{align*}\] According to the definition of \(K_a\) (Equation \ref{eq3} \[\begin{align*} K_a &= \dfrac{[H_3O^+,CH_3CH_2CO_2^-]}{[CH_3CH_2CO_2H]} \\[4pt] &= \dfrac{x^2}{0.2 - x} \\[4pt] &= \dfrac{(1.32 \times 10^{-5})^2}{0.2 - 1.32 \times 10^{-5}} \\[4pt] &= 8.69 \times 10^{-10} \end{align*}\] | 1,987 | 946 |
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Viscosity is another type of bulk property defined as a liquid’s resistance to flow. When the intermolecular forces of attraction are strong within a liquid, there is a larger viscosity. An example of this phenomenon is imagining a race between two liquids down a windshield. Which would you expect to roll down the windshield faster honey or water? Obviously from experience one would expect water to easily speed right past the honey, a fact that reveals honey has a much higher viscocity than water. Viscosity can be not only a fluid’s resistance to flow but also a gas’ resistance to flow, change shape or movement. The opposite of viscosity is fluidity which measures the ease of flow while liquids such as motor oil or honey which are “sluggish” and high in viscosity are known as . One may ask the question of what is actually going on in the liquids to make one type flow faster and the other more resistant to flow such as the comparison between honey and water earlier. Because part of a fluid moves, it forces other adjacent parts of the liquid to move along with it causing an internal friction between the molecules which ultimately leads to a reduced rate of flow. It is also important to note that the viscosity of liquids and gases are but in opposite ways meaning that upon heating, the viscosity of a liquid decreases rapidly, whereas gases flow more sluggishly. Why is this the case? As temperature increases, the average speed of molecules in a liquid also increases and as a result, they spend less time with their "neighbors." Therefore, as temperature increases, the average decrease and the molecules are able to interact without being "weighed down" by one another. The viscosity of a gas, however, increases as temperature increases because there is an increase in frequency of intermolecular collisions at higher temperatures. Since the molecules are flying around in the void most of the time, any increase in the contact they have with one another will increase the intermolecular force which will ultimately lead to a disability for the whole substance to move. There are numerous ways to measure viscosity. One of the most elementary ways is to allow a sphere, such as a metal ball, to drop through a fluid and time the fall of the metal ball: the slower the sphere falls, the greater the viscosity that is measured. Another more advanced design of measuring viscosity known as the that is much more accurate than dropping a metal ball. An consists of two reservoir bulbs and a capillary tube. The viscometer is filled with liquid until the liquid reaches the mark A with the aid of a pipette to accurately measure out the volume of needed liquid. The viscometer is then put into a water bath which equilibrates the temperature of the test liquid. As noted before, the equilibration is important to maintain a constant temperature as to not affect the viscosity otherwise. The liquid is then drawn through the side 2 of the U-tube by use of suction and lastly, the flow is time between marks C and B. The viscosity is calculated with Equation \(\ref{1}\) \[ \eta =K t \label{1}\] where \(K\) is the value of a liquid with known viscosity and density such as water. Once the value of is known, the viscosity can be determined by measuring the amount of time the test liquid flows between the two graduated marks. 1 Pascal-second (Pa•s) = 1 kg•m?1•s?1 1 Poise = 1 g•cm?1•s?1 10 P = 1 kg•m?1•s?1 = 1 Pa•s 1 cP = 0.001 Pa•s = 1 mPa•s When measuring viscosity with any type of viscometer, accurate temperature is so important that viscosity can double with a change of only 5 Celsius. | 3,638 | 947 |
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The equation for \(\log \gamma _{\pm }\) predicted from is: \[\log _{10}\gamma _{\pm }=-Az_{j}^{2}{\frac {\sqrt {I}}{1+Ba_{0}{\sqrt {I}}}} \label{DH}\] It gives satisfactory agreement with experimental measurements for low electrolyte concentrations, typically less than \(10^{−3} mol/L\). Deviations from the theory occur at higher concentrations and with electrolytes that produce ions of higher charges, particularly asymmetrical electrolytes. These deviations occur because the model is oversimplified, so there is little to be gained by making small adjustments to the model. Instead, we must challenge the individual assumptions of the model: Most extensions to the Debye–Hückel theory are empirical in nature. They usually allow the Debye–Hückel equation to be followed at low concentration and add further terms in some power of the ionic strength to fit experimental observations. Several approaches have been proposed to extend the validity of the Debye–Hückel theory. One such approach is the Extended Debye-Hückel Equation: \[- \log(\gamma) = \dfrac{A|z_+z_-|\sqrt{I}}{1 + Ba\sqrt{I}}\] where \(\gamma\) is the activity coefficient, \(z\) is the integer charge of the ion \(\mu\) is the ionic strength of the aqueous solution, and \(a\), is the size or effective diameter of the ion in angstrom. The effective hydrated radius of the ion, \(a\) is the radius of the ion and its closely bound water molecules. Large ions and less highly charged ions bind water less tightly and have smaller hydrated radii than smaller, more highly charged ions. Typical values are 3 Å for ions such as H+,Cl-,CN-, and HCOO-. The effective diameter for the hydronium ion is 9 Å. \ (A\) and \(B\) are constants with values of respectively 0.5085 and 0.3281 at 25°C in water. Other approaches include the Davies equation, Pitzer equations and specific ion interaction theory. The Davies equation is an empirical extension of Debye–Hückel theory which can be used to calculate activity coefficients of electrolyte solutions at relatively high concentrations at 25 °C. The equation, originally published in 1938, was refined by fitting to experimental data. The final form of the equation gives the mean molal activity coefficient f± of an electrolyte that dissociates into ions having charges z1 and z2 as a function of ionic strength I: \[ -\log f_{\pm }=0.5z_{1}z_{2}\left({\frac {\sqrt {I}}{1+{\sqrt {I}}}}-0.30I\right).\] The second term, 0.30 \(I\), goes to zero as the ionic strength goes to zero, so the equation reduces to the Debye–Hückel equation at low concentration. However, as concentration increases, the second term becomes increasingly important, so the Davies equation can be used for solutions too concentrated to allow the use of the Debye–Hückel equation. For 1:1 electrolytes the difference between measured values and those calculated with this equation is about 2% of the value for 0.1 M solutions. The calculations become less precise for electrolytes that dissociate into ions with higher charges. Further discrepancies will arise if there is association between the ions, with the formation of ion pairs, such as \(\ce{Mg^{2+}SO^{2-}4}\). Plot of activity coefficients calculated using the Davies equation. Pitzer equations are important for the understanding of the behaviour of ions dissolved in natural waters such as rivers, lakes and sea-water. They were first described by physical chemist Kenneth Pitzer. The parameters of the Pitzer equations are linear combinations of parameters, of a virial expansion of the excess Gibbs free energy, which characterize interactions amongst ions and solvent. The derivation is thermodynamically rigorous at a given level of expansion. The parameters may be derived from various experimental data such as the osmotic coefficient, mixed ion activity coefficients, and salt solubility. They can be used to calculate mixed ion activity coefficients and water activities in solutions of high ionic strength for which the Debye–Hückel theory is no longer adequate. An expression is obtained for the mean activity coefficient. \[\ln \gamma _{\pm }= \dfrac {p\ln \gamma _{M}+q\ln \gamma _{X}}{p+q}\]
\[\ln \gamma _{\pm }=|z^{+}z^{-}|f^{\gamma }+m\left({\frac {2pq}{p+q}}\right)B_{MX}^{\gamma }+m^{2}\left[2{\frac {(pq)^{3/2}}{p+q}}\right]C_{MX}^{\gamma }\] These equations were applied to an extensive range of experimental data at 25 °C with excellent agreement to about 6 mol kg−1 for various types of electrolyte. The treatment can be extended to mixed electrolytes and to include association equilibria. Values for the parameters β(0), β(1) and C for inorganic and organic acids, bases and salts have been tabulated. Temperature and pressure variation is also discussed. Specific ion Interaction Theory (SIT theory) is a theory used to estimate single-ion activity coefficients in electrolyte solutions at relatively high concentrations. It does so by taking into consideration interaction coefficients between the various ions present in solution. Interaction coefficients are determined from equilibrium constant values obtained with solutions at various ionic strengths. The determination of SIT interaction coefficients also yields the value of the equilibrium constant at infinite dilution. The activity coefficient of the jth ion in solution is written as \(γ_j\) when concentrations are on the molal concentration scale and as yj when concentrations are on the molar concentration scale. (The molality scale is preferred in thermodynamics because molal concentrations are independent of temperature). The basic idea of SIT theory is that the activity coefficient can be expressed as \[\log \gamma _{j}=-z_{j}^{2}{\frac {0.51{\sqrt {I}}}{1+1.5{\sqrt {I}}}}+\sum _{k}\epsilon _{jk}m_{k}\] where z is the electrical charge on the ion, \(I\) is the ionic strength, \(ε\) and \(b\) are interaction coefficients and \(m\) are concentrations. The summation extends over the other ions present in solution, which includes the ions produced by the background electrolyte. The first term in these expressions comes from Debye-Hückel theory. The second term shows how the contributions from "interaction" are dependent on concentration. Thus, the interaction coefficients are used as corrections to Debye-Hückel theory when concentrations are higher than the region of validity of that theory. | 6,363 | 948 |
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Both the MO and VB theories are used to help determine the structure of a molecule. Unlike the VB theory, which is largely based off of valence electrons, the MO theory describes structure more in depth by taking into consideration, for example, the overlap and energies of the bonding and antibonding electrons residing in a particular molecular orbital. While MO theory is more involved and difficult, it results in a more complete picture of the structure of a chosen molecule. Despite various shortcomings, complete disregard of one theory and not the other would hinder our ability to describe the bonding in molecules. | 642 | 951 |
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Electrostatic potential maps, also known as electrostatic potential energy maps, or molecular electrical potential surfaces, illustrate the charge distributions of molecules three dimensionally. These maps allow us to visualize variably charged regions of a molecule. Knowledge of the charge distributions can be used to determine how molecules interact with one another. Electrostatic potential maps are very useful three dimensional diagrams of molecules. They enable us to visualize the charge distributions of molecules and charge related properties of molecules. They also allow us to visualize the size and shape of molecules. In organic chemistry, electrostatic potential maps are invaluable in predicting the behavior of complex molecules. The first step involved in creating an electrostatic potential map is collecting a very specific type of data: electrostatic potential energy. An advanced computer program calculates the electrostatic potential energy at a set distance from the nuclei of the molecule. Electrostatic potential energy is fundamentally a measure of the strength of the nearby charges, nuclei and electrons, at a particular position. To accurately analyze the charge distribution of a molecule, a very large quantity of electrostatic potential energy values must be calculated. The best way to convey this data is to visually represent it, as in an electrostatic potential map. A computer program then imposes the calculated data onto an electron density model of the molecule derived from the Schrödinger equation. To make the electrostatic potential energy data easy to interpret, a color spectrum, with red as the lowest electrostatic potential energy value and blue as the highest, is employed to convey the varying intensities of the electrostatic potential energy values. Electrostatic potential maps involve a number of basic concepts. The actual process of mapping the electrostatic potentials of a molecule, however, involves factors that complicate these fundamental concepts. An analogous system will be employed to introduce these basic concepts. Imagine that there is a special type of mine. This mine is simply an explosive with some charged components on top of it. The circles with positive and negative charges in them are the charged components. If the electric field of the electric components are significantly disturbed, the mine triggers and explodes. The disarming device is positively charged. To disarm the mine, the disarming device must take the path of least electric resistance and touch the first charged mine component on this path. Deviating from this minimal energy path will cause a significant disturbance and the mine will explode. The specific charged components within the mine are known. Q. How do you disarm the following mine? The mine with positive charge and negative charge. A. Touch the bottom most portion of negatively charged component, red, with the disarming device. Coulomb's Law Formula \[F=k \dfrac{q_aq_b}{r^2}\] with Here is a simplified visual representation of the relationship between charge distribution and electrostatic potential. Keep in mind the equation used to find the electrostatic potential. \[ \text{Total Electrostatic Potential Energy= \sum_i \text{Electrostatic Potential Energy} \] where \[ \text{Potential Energy}=K \dfrac{q_1q_2}{r}\] and \(K= \text{Coulomb's Constant}\). | 3,395 | 952 |
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Lattice structures are not perfect; in fact most of the time they experience defects. Lattice structures (or crystals) are prone to defects especially when their temperature is greater than 0 K [1]. One of these defects is known as the Schottky defect, which occurs when oppositely charged ions vacant their sites [1]. Like the human body, lattice structures (most commonly known as crystals) are far from perfection. Our body works hard to keep things proportional but occasionally our right foot is bigger than our left; similarly, crystals may try to arrange it's ions under a strict layout, but occasionally an ion slips to another spot or simply goes missing. Realistically speaking, it should be expected that crystals will depart itself from order (not surprising considering defects occurs at temperature greater than 0 K). There are many ways a crystal can depart itself from order (thus experiences defects); these defects can be grouped in different categories such as Point Defects, Line Defects, Planar Defects, or Volume or Bulk Defects [2]. We will focus on Point Defects, specifically the defect that occurs in ionic crystal structures (i.e. NaCl) called the Schottky Defect. Lattice structures (or crystals) undergoing point defects experience one of two types: By the simplest definition, the Schottky defect is defined by type one, while type two defects are known as the Frenkel defect. The Schottky defect is often visually demonstrated using the following layout of anions and cations: In addition, this layout is applicable only for ionic crystal compounds of the formula MX--layout for ionic crystals with formula MX and M X will be discussed later--where M is metal and X is nonmetal. Notice the figure has exactly one cation and one anion vacating their sites; that is what defines a (one) Schottky Defect for a crystal of MX formula--for every cation that vacant its site, the same number of anion will follow suit; essentially the vacant sites come in pairs. This also means the crystal will neither be too positive or too negative because the crystal will always be in equilibrium in respect to the number of anions and cations. It is possible to approximate the number of Schottky defects (n ) in a MX ionic crystal compound by using the equation: \[N= \exp^{-\dfrac{\Delta H}{2RT}} \label{3}\] where N can be calculated by: \[N = \dfrac{\text{density of the ionic crystal compound} \times N_A}{\text{molar mass of the ionic crystal compound}} \label{4}\] From Equation \(\ref{3}\), it is also possible to calculate the fraction of vacant sites by using the equation: \[\dfrac{n_s}{N} = \exp^{-\dfrac{\Delta H}{2RT}} \label{5}\] As mentioned earlier, a Schottky defect will always result a crystal structure in equilibrium--where no crystal is going to be too positive or too negative; thus in the case of: | 2,860 | 953 |
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Like the alkali metals, the alkaline earth metals are so reactive that they are never found in elemental form in nature. Because they form +2 ions that have very negative reduction potentials, large amounts of energy are needed to isolate them from their ores. Four of the six group 2 elements—magnesium (Mg), calcium (Ca), strontium (Sr), and barium (Ba)—were first isolated in the early 19th century by Sir Humphry Davy, using a technique similar to the one he used to obtain the first alkali metals. In contrast to the alkali metals, however, compounds of the alkaline earth metals had been recognized as unique for many centuries. In fact, the name alkali comes from the Arabic al-qili, meaning “ashes,” which were known to neutralize acids. Medieval alchemists found that a portion of the ashes would melt on heating, and these substances were later identified as the carbonates of sodium and potassium (\(M_2CO_3\)). The ashes that did not melt (but did dissolve in acid), originally called alkaline earths, were subsequently identified as the alkaline earth oxides (MO). In 1808, Davy was able to obtain pure samples of Mg, Ca, Sr, and Ba by electrolysis of their chlorides or oxides. Beryllium (Be), the lightest alkaline earth metal, was first obtained in 1828 by Friedrich Wöhler in Germany and simultaneously by Antoine Bussy in France. The method used by both men was reduction of the chloride by the potent “new” reductant, potassium: Radium was discovered in 1898 by Pierre and Marie Curie, who processed tons of residue from uranium mines to obtain about 120 mg of almost pure \(RaCl_2\). Marie Curie was awarded the Nobel Prize in Chemistry in 1911 for its discovery. Because of its low abundance and high radioactivity however, radium has few uses. The alkaline earth metals are produced for industrial use by electrolytic reduction of their molten chlorides, as indicated in this equation for calcium: \[CaCl_{2\;(l)} \rightarrow Ca_{(l)} + Cl_{2\;(g)} \label{Eq2}\] The group 2 metal chlorides are obtained from a variety of sources. For example, \(BeCl_2\) is produced by reacting \(HCl\) with beryllia (\(BeO\)), which is obtained from the semiprecious stone beryl \([Be_3Al_2(SiO_3)_6]\). Chemical reductants can also be used to obtain the group 2 elements. For example, magnesium is produced on a large scale by heating a form of limestone called dolomite (CaCO ·MgCO ) with an inexpensive iron/silicon alloy at 1150°C. Initially \(CO_2\) is released, leaving behind a mixture of \(CaO\) and MgO; Mg is then reduced: \[2CaO·MgO_{(s)} + Fe/Si_{(s)} \rightarrow 2Mg(l) + Ca_2SiO_{4\;(s)} + Fe(s) \label{Eq3}\] An early source of magnesium was an ore called magnesite (\(MgCO_3\)) from the district of northern Greece called Magnesia. Strontium was obtained from strontianite (\(SrCO_3\)) found in a lead mine in the town of Strontian in Scotland. The alkaline earth metals are somewhat easier to isolate from their ores, as compared to the alkali metals, because their carbonate and some sulfate and hydroxide salts are insoluble. Several important properties of the alkaline earth metals are summarized in Table \(\Page {1}\). Although many of these properties are similar to those of the alkali metals (Table \(\Page {1}\)), certain key differences are attributable to the differences in the valence electron configurations of the two groups (ns for the alkaline earth metals versus ns for the alkali metals). As with the alkali metals, the atomic and ionic radii of the alkaline earth metals increase smoothly from Be to Ba, and the ionization energies decrease. As we would expect, the first ionization energy of an alkaline earth metal, with an ns valence electron configuration, is always significantly greater than that of the alkali metal immediately preceding it. The group 2 elements do exhibit some anomalies, however. For example, the density of Ca is less than that of Be and Mg, the two lightest members of the group, and Mg has the lowest melting and boiling points. In contrast to the alkali metals, the heaviest alkaline earth metal (Ba) is the strongest reductant, and the lightest (Be) is the weakest. The standard electrode potentials of Ca and Sr are not very different from that of Ba, indicating that the opposing trends in ionization energies and hydration energies are of roughly equal importance. One major difference between the group 1 and group 2 elements is their electron affinities. With their half-filled ns orbitals, the alkali metals have a significant affinity for an additional electron. In contrast, the alkaline earth metals generally have little or no tendency to accept an additional electron because their ns valence orbitals are already full; an added electron would have to occupy one of the vacant np orbitals, which are much higher in energy. With their low first and second ionization energies, the group 2 elements almost exclusively form ionic compounds that contain M ions. As expected, however, the lightest element (Be), with its higher ionization energy and small size, forms compounds that are largely covalent. Some compounds of Mg also have significant covalent character. Hence organometallic compounds like those discussed for Li in group 1 are also important for Be and Mg in group 2. The group 2 elements almost exclusively form ionic compounds containing M ions. All alkaline earth metals react vigorously with the halogens (group 17) to form the corresponding halides (MX ). Except for the beryllium halides, these compounds are all primarily ionic in nature, containing the M cation and two X anions. The beryllium halides, with properties more typical of covalent compounds, have a polymeric halide-bridged structure in the solid state, as shown for BeCl . These compounds are volatile, producing vapors that contain the linear X–Be–X molecules predicted by the valence-shell electron-pair repulsion (VSEPR) model. As expected for compounds with only four valence electrons around the central atom, the beryllium halides are potent Lewis acids. They react readily with Lewis bases, such as ethers, to form tetrahedral adducts in which the central beryllium is surrounded by an octet of electrons: \[ BeCl_{2(s)} + 2(CH_3CH_2)_2O_{(l)} \rightarrow BeCl_2[O(CH_2CH_3)_2]_{2(soln)} \label{Eq4}\] Because of their higher ionization energy and small size, both Be and Mg form organometallic compounds. The reactions of the alkaline earth metals with oxygen are less complex than those of the alkali metals. All group 2 elements except barium react directly with oxygen to form the simple oxide MO. Barium forms barium peroxide (BaO ) because the larger O ion is better able to separate the large Ba ions in the crystal lattice. In practice, only BeO is prepared by direct reaction with oxygen, and this reaction requires finely divided Be and high temperatures because Be is relatively inert. The other alkaline earth oxides are usually prepared by the thermal decomposition of carbonate salts: \[\mathrm{MCO_3(s)}\xrightarrow\Delta\mathrm{MO(s)}+\mathrm{CO_2(g)} \label{Eq5}\] The reactions of the alkaline earth metals with the heavier chalcogens (Y) are similar to those of the alkali metals. When the reactants are present in a 1:1 ratio, the binary chalcogenides (MY) are formed; at lower M:Y ratios, salts containing polychalcogenide ions (Y ) are formed. In the reverse of Equation \(\ref{Eq5}\), the oxides of Ca, Sr, and Ba react with CO to regenerate the carbonate. Except for BeO, which has significant covalent character and is therefore amphoteric, all the alkaline earth oxides are basic. Thus they react with water to form the hydroxides—M(OH) : \[MO_{(s)} + H_2O_{(l)} \rightarrow M^{2+}_{(aq)} + 2OH^−_{(aq)} \label{Eq6}\] and they dissolve in aqueous acid. Hydroxides of the lighter alkaline earth metals are insoluble in water, but their solubility increases as the atomic number of the metal increases. Because BeO and MgO are much more inert than the other group 2 oxides, they are used as refractory materials in applications involving high temperatures and mechanical stress. For example, MgO (melting point = 2825°C) is used to coat the heating elements in electric ranges. The carbonates of the alkaline earth metals also react with aqueous acid to give CO and H O: \[MCO_{3(s)} + 2H^+_{(aq)} \rightarrow M^{2+}_{(aq)} + CO_{2(g)} + H_2O_{(l)} \label{Eq7}\] The reaction in Equation \(\ref{Eq7}\) is the basis of antacids that contain MCO , which is used to neutralize excess stomach acid. The trend in the reactivities of the alkaline earth metals with nitrogen is the opposite of that observed for the alkali metals. Only the lightest element (Be) does not react readily with N to form the nitride (M N ), although finely divided Be will react at high temperatures. The higher lattice energy due to the highly charged M and N ions is apparently sufficient to overcome the chemical inertness of the N molecule, with its N≡N bond. Similarly, all the alkaline earth metals react with the heavier group 15 elements to form binary compounds such as phosphides and arsenides with the general formula M Z . Higher lattice energies cause the alkaline earth metals to be more reactive than the alkali metals toward group 15 elements. When heated, all alkaline earth metals, except for beryllium, react directly with carbon to form ionic carbides with the general formula MC . The most important alkaline earth carbide is calcium carbide (CaC ), which reacts readily with water to produce acetylene. For many years, this reaction was the primary source of acetylene for welding and lamps on miners’ helmets. In contrast, beryllium reacts with elemental carbon to form Be C, which formally contains the C ion (although the compound is covalent). Consistent with this formulation, reaction of Be C with water or aqueous acid produces methane: \[Be_2C_{(s)} + 4H_2O_{(l)} \rightarrow 2Be(OH)_{2(s)} + CH_{4(g)} \label{Eq8}\] Beryllium does not react with hydrogen except at high temperatures (1500°C), although BeH can be prepared at lower temperatures by an indirect route. All the heavier alkaline earth metals (Mg through Ba) react directly with hydrogen to produce the binary hydrides (MH ). The hydrides of the heavier alkaline earth metals are ionic, but both BeH and MgH have polymeric structures that reflect significant covalent character. All alkaline earth hydrides are good reducing agents that react rapidly with water or aqueous acid to produce hydrogen gas: \[CaH_{2(s)} + 2H_2O_{(l)} \rightarrow Ca(OH)_{2(s)} + 2H_{2(g)} \label{Eq9} \] Like the alkali metals, the heavier alkaline earth metals are sufficiently electropositive to dissolve in liquid ammonia. In this case, however, two solvated electrons are formed per metal atom, and no equilibriums involving metal dimers or metal anions are known. Also, like the alkali metals, the alkaline earth metals form a wide variety of simple ionic salts with oxoanions, such as carbonate, sulfate, and nitrate. The nitrate salts tend to be soluble, but the carbonates and sulfates of the heavier alkaline earth metals are quite insoluble because of the higher lattice energy due to the doubly charged cation and anion. The solubility of the carbonates and the sulfates decreases rapidly down the group because hydration energies decrease with increasing cation size. The solubility of alkaline earth carbonate and sulfates decrease down the group because the hydration energies decrease. Because of their higher positive charge (+2) and smaller ionic radii, the alkaline earth metals have a much greater tendency to form complexes with Lewis bases than do the alkali metals. This tendency is most important for the lightest cation (Be ) and decreases rapidly with the increasing radius of the metal ion. The alkaline earth metals have a substantially greater tendency to form complexes with Lewis bases than do the alkali metals. The chemistry of Be is dominated by its behavior as a Lewis acid, forming complexes with Lewis bases that produce an octet of electrons around beryllium. For example, Be salts dissolve in water to form acidic solutions that contain the tetrahedral [Be(H O) ] ion. Because of its high charge-to-radius ratio, the Be ion polarizes coordinated water molecules, thereby increasing their acidity: \[ [Be(H_2O)_4]^{2+}_{(aq)} \rightarrow [Be(H_2O)_3(OH)]^+_{(aq)} + H^+_{(aq)} \label{Eq10}\] Similarly, in the presence of a strong base, beryllium and its salts form the tetrahedral hydroxo complex: [Be(OH) ] . Hence beryllium oxide is amphoteric. Beryllium also forms a very stable tetrahedral fluoride complex: [BeF ] . Recall that beryllium halides behave like Lewis acids by forming adducts with Lewis bases (Equation \(\ref{Eq4}\)). The heavier alkaline earth metals also form complexes, but usually with a coordination number of 6 or higher. Complex formation is most important for the smaller cations (Mg and Ca ). Thus aqueous solutions of Mg contain the octahedral [Mg(H O) ] ion. Like the alkali metals, the alkaline earth metals form complexes with neutral cyclic ligands like the crown ethers and cryptands discussed in Section 21.3. Like the alkali metals, the lightest alkaline earth metals (Be and Mg) form the most covalent-like bonds with carbon, and they form the most stable organometallic compounds. Organometallic compounds of magnesium with the formula RMgX, where R is an alkyl or aryl group and X is a halogen, are universally called , after Victor Grignard (1871–1935), the French chemist who discovered them. Grignard reagents can be used to synthesize various organic compounds, such as alcohols, aldehydes, ketones, carboxylic acids, esters, thiols, and amines. Elemental magnesium is the only alkaline earth metal that is produced on a large scale (about 5 × 10 tn per year). Its low density (1.74 g/cm compared with 7.87 g/cm for iron and 2.70 g/cm for aluminum) makes it an important component of the lightweight metal alloys used in aircraft frames and aircraft and automobile engine parts (Figure \(\Page {1}\)). Most commercial aluminum actually contains about 5% magnesium to improve its corrosion resistance and mechanical properties. Elemental magnesium also serves as an inexpensive and powerful reductant for the production of a number of metals, including titanium, zirconium, uranium, and even beryllium, as shown in the following equation: \[TiCl_{4\;(l)} + 2Mg(s) \rightarrow Ti_{(s)} + 2MgCl_{2\;(s)} \label{11}\] The only other alkaline earth that is widely used as the metal is beryllium, which is extremely toxic. Ingestion of beryllium or exposure to beryllium-containing dust causes a syndrome called berylliosis, characterized by severe inflammation of the respiratory tract or other tissues. A small percentage of beryllium dramatically increases the strength of copper or nickel alloys, which are used in nonmagnetic, nonsparking tools (such as wrenches and screwdrivers), camera springs, and electrical contacts. The low atomic number of beryllium gives it a very low tendency to absorb x-rays and makes it uniquely suited for applications involving radioactivity. Both elemental Be and BeO, which is a high-temperature ceramic, are used in nuclear reactors, and the windows on all x-ray tubes and sources are made of beryllium foil. Millions of tons of calcium compounds are used every year. As discussed in earlier chapters, CaCl is used as “road salt” to lower the freezing point of water on roads in cold temperatures. In addition, CaCO is a major component of cement and an ingredient in many commercial antacids. “Quicklime” (CaO), produced by heating CaCO (Equation \(\ref{Eq5}\)), is used in the steel industry to remove oxide impurities, make many kinds of glass, and neutralize acidic soil. Other applications of group 2 compounds described in earlier chapters include the medical use of BaSO in “barium milkshakes” for identifying digestive problems by x-rays and the use of various alkaline earth compounds to produce the brilliant colors seen in fireworks. For each application, choose the most appropriate substance based on the properties and reactivities of the alkaline earth metals and their compounds. Explain your choice in each case. Use any tables you need in making your decision, such as K values (Table 17.1), lattice energies (Table 8.1), and band-gap energies. application and selected alkaline earth metals most appropriate substance for each application Based on the discussion in this section and any relevant information elsewhere in this book, determine which substance is most appropriate for the indicated use. Which of the indicated alkaline earth metals or their compounds is most appropriate for each application? Predict the products of each reaction and then balance each chemical equation. reactants products and balanced chemical equation Follow the procedure given in Example 3 to predict the products of each reaction and then balance each chemical equation. The balanced chemical equation is \[CaO_{(s)} + 2HCl_{(g)} → CaCl_{2(aq)} + H_2O_{(l)}\] We conclude that no reaction occurs. The balanced chemical equation is \(\mathrm{CaH_2(s)}+\mathrm{TiO_2(s)}\xrightarrow\Delta\mathrm{Ti(s)}+\mathrm{CaO(s)}+\mathrm{H_2O(l)}\) We could also write the products as Ti(s) + Ca(OH) (s). Predict the products of each reaction and then balance each chemical equation. Group 2 elements almost exclusively form ionic compounds containing the M ion, they are more reactive toward group 15 elements, and they have a greater tendency to form complexes with Lewis bases than do the alkali metals. Pure samples of most of the alkaline earth metals can be obtained by electrolysis of the chlorides or oxides. Beryllium was first obtained by the reduction of its chloride; radium chloride, which is radioactive, was obtained through a series of reactions and separations. In contrast to the alkali metals, the alkaline earth metals generally have little or no affinity for an added electron. All alkaline earth metals react with the halogens to produce the corresponding halides, with oxygen to form the oxide (except for barium, which forms the peroxide), and with the heavier chalcogens to form chalcogenides or polychalcogenide ions. All oxides except BeO react with CO to form carbonates, which in turn react with acid to produce CO and H O. Except for Be, all the alkaline earth metals react with N to form nitrides, and all react with carbon and hydrogen to form carbides and hydrides. Alkaline earth metals dissolve in liquid ammonia to give solutions that contain two solvated electrons per metal atom. The alkaline earth metals have a greater tendency than the alkali metals to form complexes with crown ethers, cryptands, and other Lewis bases. The most important alkaline earth organometallic compounds are Grignard reagents (RMgX), which are used to synthesize organic compounds. | 18,910 | 954 |
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The value of \(K_p\) is independent of pressure, although the composition of a system at equilibrium may be very much dependent on pressure. Temperature dependence is another matter. Because the value of \(\Delta G_{rxm}^o\) is dependent on temperature, the value of \(K_p\) is as well. The form of the temperature dependence can be taken from the definition of the Gibbs function. At constant temperature and pressure \[ \dfrac{\Delta G^o_{T_2}}{T_2} - \dfrac{\Delta G^o_{T_1}}{T_1} = \Delta H^o \left(\dfrac{1}{T_2} - \dfrac{1}{T_1} \right) \nonumber \] Substituting \[ \Delta G^o = -RT \ln K \nonumber \] For the two values of \(\Delta G_{}^o\) and using the appropriate temperatures, yields \[ \dfrac{-R{T_2} \ln K_2}{T_2} - \dfrac{-R{T_1} \ln K_1}{T_1} = \Delta H^o \left(\dfrac{1}{T_2} - \dfrac{1}{T_1} \right) \nonumber \] And simplifying the expression so that only terms involving \(K\) are on the left and all other terms are on the right results in the , which describes the temperature dependence of the equilibrium constant. \[ \ln \left(\dfrac{\ K_2}{\ K_1}\right) = - \dfrac{\Delta H^o}{R} \left(\dfrac{1}{T_2} - \dfrac{1}{T_1} \right) \label{vH} \] Because of the assumptions made in the derivation of the Gibbs-Helmholtz equation, this relationship only holds if \(\Delta H^o\) is independent of temperature over the range being considered. This expression also suggests that a plot of \(\ln(K)\) as a function of \(1/T\) should produce a straight line with a slope equal to \(–\Delta H^o/R\). Such a plot is known as a , and can be used to determine the reaction enthalpy. A certain reaction has a value of \(K_p = 0.0260\) at 25 °C and \(\Delta H_{rxm}^o = 32.4 \,kJ/mol\). Calculate the value of \(K_p\) at 37 °C. This is a job for the van ’t Hoff equation! So Equation \ref{vH} becomes \[ \begin{align*} \ln \left( \dfrac{\ K_2}{0.0260} \right) &= - \dfrac{32400 \,J/mol}{8.314 \,K/(mol \,K)} \left(\dfrac{1}{310\, K} - \dfrac{1}{298 \,K} \right) \\[4pt] K_2 &= 0.0431 \end{align*} \] Note: the value of \(K_2\) with temperature, which is what is expected for an reaction. An increase in temperature should result in an increase of product formation in the equilibrium mixture. But unlike a change in pressure, a change in temperature actually leads to a change in the value of the equilibrium constant! Given the following average bond enthalpies for \(\ce{P-Cl}\) and \(\ce{Cl-Cl}\) bonds, predict whether or not an increase in temperature will lead to a larger or smaller degree of dissociation for the reaction \[\ce{PCl_5 \rightleftharpoons PCl_3 + Cl_2} \nonumber \] The estimated reaction enthalpy is given by the total energy expended breaking bonds minus the energy recovered by the formation of bonds. Since this reaction involves breaking two P-Cl bonds (costing 652 kJ/mol) and the formation of one Cl-Cl bond (recovering 240 kJ/mol), it is clear that the reaction is endothermic (by approximately 412 kJ/mol). As such, an increase in temperature should increase the value of the equilibrium constant, causing the degree of dissociation to be increased at the higher temperature. | 3,132 | 955 |
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Tunneling is a quantum mechanical phenomenon when a particle is able to penetrate through a potential energy barrier that is higher in energy than the particle’s kinetic energy. This amazing property of microscopic particles play important roles in explaining several physical phenomena including radioactive decay. Additionally, the principle of tunneling leads to the development of Scanning Tunneling Microscope (STM) which had a profound impact on chemical, biological and material science research. Consider a ball rolling from one valley to another over a hill (Figure \(\Page {1}\)). If the ball has enough energy (\(E\)) to overcome the potential energy (\(V\)) at the top of the barrier between each valley, then it can roll from one valley to the other. This is the classical picture and is controlled by the simple Law of Conservation of Energy approach taught in beginning physics courses. However, If the ball does not have enough kinetic energy (\(E<V\)), to overcome the barrier it will never roll from one valley to the other. In contrast, when quantum effects are taken into effect, the ball can "tunnel" through the barrier to the other valley, even if its kinetic energy is less than the potential energy of the barrier to the top of one of the hills. The reason for the difference between classical and quantum motion comes from wave-particle nature of matter. One interpretation of this duality involves the , which defines a limit on how precisely the position and the momentum of a particle can be known at the same time. This implies that there are no solutions with a probability of exactly zero (or one), though a solution may approach infinity if, for example, the calculation for its position was taken as a probability of 1, the other, i.e. its speed, would have to be infinity. Hence, the probability of a given particle's existence on the opposite side of an intervening barrier is non-zero, and such particles will appear on the 'other' (a semantically difficult word in this instance) side with a relative frequency proportional to this probability. Microscopic particles such as protons, or electrons would behave differently as a consequence of wave-particle duality. Consider a particle with energy \(E\) that is confined in a box which has a barrier of height \(V\). Classically, the box will prevent these particles from escaping due to the insufficiency in kinetic energy of these particles to get over the barrier. However, if the thickness of the barrier is thin, the particles have some probability of penetrating through the barrier sufficient energy and appear on the other side of the box (Figure \(\Page {2}\)). When it reaches a barrier it cannot overcome, a particle's wave function changes from sinusoidal to exponentially diminishing in form. The solution for the Schrödinger equation in such a medium (Figure \(\Page {2}\); blue region) is: \[ \psi = N e^{-\beta x}\] where For a quantum particle to tunnel through a barrier three conditions must be met (Figure \(\Page {2}\)): If these conditions are met, there would be some probability of finding the particles on the other side of the barrier. Beginning as a sinusoidal wave, a particle begins tunneling through the barrier and goes into exponential decay until it exits the barrier and gets transmitted out the other side as a final sinusoidal wave with a smaller amplitude. The act of tunneling decreases the wave amplitude due the reflection of the incident wave when it comes into the contact with the barrier but does not affect the wave equation. The probability, \(P\), of a particle tunneling through the potential energy barrier is derived from the Schrödinger Equation and is described as, \[ P = \exp\left(\dfrac{-4a\pi}{h} \sqrt{2m(V-E)} \right) \label{prob}\] with \(E<V\) where Therefore, the probability of an object tunneling through a barrier decreases with the object's increasing mass and with the increasing gap between the energy of the object and the energy of the barrier. And although the wave function never quite reaches 0 (as can be determined from the functionality), this explains how tunneling is frequent on nanoscale, but negligible at the macroscopic level. An electron having total kinetic energy \(E\) of 4.50 eV approaches a rectangular energy barrier with \(V= 5.00\, eV\) and \(L= 950\, pm\). Classically, the electron cannot pass through the barrier because \(E<V\). Calculate probability of tunneling of this electron through the barrier. This is a straightforward application of Equation \(\ref{prob}\). \[ P = \exp\left(\dfrac{-4a\pi}{h} \sqrt{2m(V-E)} \right) \nonumber\] The electronvolt (eV) is a unit of energy that is equal to approximately \(1.6 \times 10^{−19}\; J\), which is the conversion used below. The mass of an electron is \(9.10 \times 10^{-31}\; kg\) \[\begin{align*} P &= \exp \left[\left(- \dfrac{ (4) (950 \times 10^{-12}\, m) (\pi)}{ 6.6260 \times 10^{-34} m^2 kg / s} \right) \sqrt{(2)(9.10 \times 10^{-31}\; kg) (5.00 - 4.50 \; \cancel{eV})( 1.60 \times 10^{-19} J/\cancel{eV}) } \right] \\[4pt] &= \exp^{-6.88} = 1.03 \times 10^{-3} \end{align*}\] There is a ~0.1% probability of the electrons tunneling though the barrier. Protons and neutrons in a nucleus have kinetic energy, but it is about 8 MeV less than that needed to get out from attractive nuclear potential (Figure \(\Page {4}\)). Hence, they are bound by an average of 8 MeV per nucleon. The slope of the hill outside the bowl is analogous to the repulsive Coulomb potential for a nucleus, such as for an α particle outside a positive nucleus. In \(\alpha\) decay, two protons and two neutrons spontaneously break away as a He unit. Yet the protons and neutrons do not have enough kinetic energy to classically get over the rim. The \(\alpha\) article tunnels through a region of space it is forbidden to be in, and it comes out of the side of the nucleus. Like an electron making a transition between orbits around an atom, it travels from one point to another without ever having been in between (Figure \(\Page {5}\)). The wave function of a quantum mechanical particle varies smoothly, going from within an atomic nucleus (on one side of a potential energy barrier) to outside the nucleus (on the other side of the potential energy barrier). Inside the barrier, the wave function does not become zero but decreases exponentially, and we do not observe the particle inside the barrier. The probability of finding a particle is related to the square of its wave function, and so there is a small probability of finding the particle outside the barrier, which implies that the particle can tunnel through the barrier. A metal tip usually made out of tungsten is placed between a very small distance above a conducting or semiconducting surface. This distance acts as a potential barrier for tunneling. The space between the tip and the surface normally is vacuum. When electrons tunnel from the metal tip to the surface, a current is created and monitored by a computer (Figure \(\Page {6}\)). The current depends on the distance between the tip and the surface, which is controlled by a piezoelectric cylinder. If there is a strong current, the tip will move away from the surface. The increase of the potential barrier will decrease the probability of tunneling and decrease the current. If the current becomes too weak, the tip moves closer to the surface. The potential barrier will be reduced and the current will increase. The variations in the current as the tip moves over the sample are reconstructed by the computer to produce topological image of the scanned surface. ). | 7,644 | 957 |
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Equilibrium between two phases, e.g. water ice (\(ice\)) and liquid water (\(water\)), is at constant \(T\) and \(P\). Therefore: \[dG = \cancel{-SdT + VdP} +μ_{ice}dn_{ice}+μ_{water}dn_{water} \nonumber \] There is a relationship between the amount of ice and water: \[dn_{ice} = -dn_{water} \nonumber \] From this, we get: \[0 = [μ_{ice}-μ_{water}]dn_{water} = \Delta \mu\; dn_{water} \nonumber \] As \(dn_{water}\) is not zero, this means that \(\Delta \mu\) must be zero! This must hold true for any set of points where ice and water are in equilibrium. The statement is not just for liquid and solid water, but for any two phases in equilibrium. That is, any two phases in equilbrium will always have the same chemical potential. That is the almost vertical line in the diagram. Its points are not at the same \(P\) and \(T\), but we can find out where they should be by considering the thermodynamic potential \(\mu\) as a of \(T\) and \(P\): \[dμ = \left(\dfrac{∂μ}{∂P}\right)_TdT + \left(\dfrac{∂μ}{∂T}\right)_PdP \nonumber \] Because \(\mu= \left(\frac{\partial G}{\partial n}\right)_{T,P} = \bar{G}\), it is not hard to identify the partial derivatives: \[\left(\dfrac{∂μ}{∂P}\right) = \left(\dfrac{∂\bar{G}}{∂P}\right) = \bar{V} \nonumber \] \[\left(\dfrac{∂μ}{∂T}\right) = \left(\dfrac{∂\bar{G}}{∂T}\right) = -\bar{S} \nonumber \] This is true for both water and ice, or any two phases in equilibrium. As the \(Δμ=0\), we can equate the \(dμ\) expressions for both water and ice: \[\left(\dfrac{∂μ_{ice}}{∂P}\right)_TdT + \left(\dfrac{∂μ_{ice}}{∂T}\right)_P dP=\left(\dfrac{∂μ_{water}}{∂P}\right)_TdT + \left(\dfrac{∂μ_{water}}{∂T}\right)_PdP \nonumber \] Rearranging and identifying the partials gives: \[\bar{V}_{ice}dP -\bar{S}_{ice}dT= \bar{V}_{water}dP -\bar{S}_{water}dT \nonumber \] Solving for \(dP/dT\) we get: \[\dfrac{dP}{dT} = \dfrac{Δ\bar{S}}{Δ\bar{V}} \nonumber \] As \(\Delta \bar{G}= \Delta \bar{H}-T\Delta \bar{S} = 0\), we have: \[Δ\bar{S} = \dfrac{Δ\bar{H}}{T_m} \nonumber \] So: \[\dfrac{dP}{dT} = \dfrac{Δ\bar{H}}{TΔ\bar{V}} \nonumber \] This expression should be valid for all points along a phase boundary, such as the melt line. In fact, it tells use that the phase boundary is by \(\Delta\bar{H}/T\Delta\bar{V}\). For water and ice, we immediately see why the melt line runs a little to the left: exceptionally \(\Delta\bar{V}\) is negative for ice, because water is actually a little denser that ice. The above expression(s) are named after The values of \(\Delta\bar{H}\) and \(\Delta\bar{V}\) do not change much with pressure and can often be considered for the melting line. When gases are involved that is not really true. | 2,688 | 958 |
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In the , we discussed the principles of chemical kinetics, which deal with the , or how quickly a given chemical reaction occurs. We now turn our attention to the to which a reaction occurs and how reaction conditions affect the final concentrations of reactants and products. For most of the reactions that we have discussed so far, you may have assumed that once reactants are converted to products, they are likely to remain that way. In fact, however, virtually all chemical reactions are to some extent. That is, an opposing reaction occurs in which the products react, to a greater or lesser degree, to re-form the reactants. Eventually, the forward and reverse reaction rates become the same, and the system reaches the point at which the composition of the system no longer changes with time. Chemical equilibrium is a dynamic process that consists of a forward reaction, in which reactants are converted to products, and a reverse reaction, in which products are converted to reactants. At equilibrium, the forward and reverse reactions proceed at equal rates. Consider, for example, a simple system that contains only one reactant and one product, the reversible dissociation of dinitrogen tetroxide (\(\ce{N_2O_4}\)) to nitrogen dioxide (\(\ce{NO_2}\)). You may recall that \(\ce{NO_2}\) is responsible for the brown color we associate with smog. When a sealed tube containing solid \(\ce{N_2O_4}\) (mp = −9.3°C; bp = 21.2°C) is heated from −78.4°C to 25°C, the red-brown color of \(\ce{NO_2}\) appears (Figure \(\Page {1}\)). The reaction can be followed visually because the product (\(\ce{NO_2}\)) is colored, whereas the reactant (\(\ce{N_2O_4}\)) is colorless: \[\underset{colorless }{\ce{N2O4 (g)}} \ce{ <=>[k_f,k_r] } \underset{red-brown }{\ce{2NO2(g)}}\label{Eq1} \] The double arrow indicates that both the forward reaction \[\ce{N2O4 (g) ->[k_f] 2NO2(g)} \label{eq1B} \] and reverse reaction \[\ce{2NO2(g) ->[k_r] N2O4 (g) } \label{eq1C} \] occurring simultaneously (i.e, the reaction is reversible). However, this does not necessarily mean the system is equilibrium as the following chapter demonstrates. Figure \(\Page {2}\) shows how the composition of this system would vary as a function of time at a constant temperature. If the initial concentration of \(\ce{NO_2}\) were zero, then it increases as the concentration of \(\ce{N_2O_4}\) decreases. Eventually the composition of the system stops changing with time, and chemical equilibrium is achieved. Conversely, if we start with a sample that contains no \(\ce{N_2O_4}\) but an initial \(\ce{NO_2}\) concentration twice the initial concentration of \(\ce{N_2O_4}\) (Figure \(\Page {2a}\)), in accordance with the stoichiometry of the reaction, we reach exactly the same equilibrium composition (Figure \(\Page {2b}\)). Thus equilibrium can be approached from either direction in a chemical reaction. Figure \(\Page {3}\) shows the forward and reverse reaction rates for a sample that initially contains pure \(\ce{NO_2}\). Because the initial concentration of \(\ce{N_2O_4}\) is zero, the forward reaction rate (dissociation of \(\ce{N_2O_4}\)) is initially zero as well. In contrast, the reverse reaction rate (dimerization of \(\ce{NO_2}\)) is initially very high (\(2.0 \times 10^6\, M/s\)), but it decreases rapidly as the concentration of \(\ce{NO_2}\) decreases. As the concentration of \(\ce{N_2O_4}\) increases, the rate of dissociation of \(\ce{N_2O_4}\) increases—but more slowly than the dimerization of \(\ce{NO_2}\)—because the reaction is only first order in \(\ce{N_2O_4}\) (rate = \(k_f[N_2O_4]\), where \(k_f\) is the rate constant for the forward reaction in Equations \(\ref{Eq1}\) and \(\ref{eq1B}\)). Eventually, the forward and reverse reaction rates become identical, \(k_f = k_r\), and the system has reached chemical equilibrium. If the forward and reverse reactions occur at different rates, then the system is not at equilibrium. The rate of dimerization of \(\ce{NO_2}\) (reverse reaction) decreases rapidly with time, as expected for a second-order reaction. Because the initial concentration of \(\ce{N_2O_4}\) is zero, the rate of the dissociation reaction (forward reaction) at \(t = 0\) is also zero. As the dimerization reaction proceeds, the \(\ce{N_2O_4}\) concentration increases, and its rate of dissociation also increases. Eventually the rates of the two reactions are equal: chemical equilibrium has been reached, and the concentrations of \(\ce{N_2O_4}\) and \(\ce{NO_2}\) no longer change. At equilibrium, the forward reaction rate is equal to the reverse reaction rate. The three reaction systems (1, 2, and 3) depicted in the accompanying illustration can all be described by the equation: \[2A \rightleftharpoons B \nonumber \] where the blue circles are \(A\) and the purple ovals are \(B\). Each set of panels shows the changing composition of one of the three reaction mixtures as a function of time. Which system took the longest to reach chemical equilibrium? : three reaction systems relative time to reach chemical equilibrium : Compare the concentrations of A and B at different times. The system whose composition takes the longest to stabilize took the longest to reach chemical equilibrium. : In systems 1 and 3, the concentration of A decreases from \(t_0\) through \(t_2\) but is the same at both \(t_2\) and \(t_3\). Thus systems 1 and 3 are at equilibrium by \(t_3\). In system 2, the concentrations of A and B are still changing between \(t_2\) and \(t_3\), so system 2 may not yet have reached equilibrium by \(t_3\). Thus system 2 took the longest to reach chemical equilibrium. In the following illustration, A is represented by blue circles, B by purple squares, and C by orange ovals; the equation for the reaction is A + B ⇌ C. The sets of panels represent the compositions of three reaction mixtures as a function of time. Which, if any, of the systems shown has reached equilibrium? system 2 A Introduction to Dynamic Equilibrium: At equilibrium, the forward and reverse reactions of a system proceed at equal rates. Chemical equilibrium is a dynamic process consisting of forward and reverse reactions that proceed at equal rates. At equilibrium, the composition of the system no longer changes with time. The composition of an equilibrium mixture is independent of the direction from which equilibrium is approached. | 6,401 | 959 |
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Discovered in 1913 by Fajans and Göhring, and then isolated in 1934 by Grosse, protactinium is named from the Greek proto + actinium (parent of actinium). The silvery-white metal is extremely rare, very radioactive and highly poisonous. Originally called brevium by its discoverers because of the short life-time of the transition between Th-234 and U-234, a longer-lived isotope was eventually isolated and called protoactinium by Grosse. The name was shortened to its present form in 1949. About 60 tons of pitchblende ore (which contains uranium, radium, and a host of other radioactive elements) yields about 125 grams of protactinium. | 659 | 960 |
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This page deals with the basic ideas about energy changes during chemical reactions, including simple energy diagrams and the terms exothermic and endothermic. Obviously, lots of chemical reactions give out energy as heat. Getting heat by burning a fuel is a simple example, but you will probably have come across lots of others in the lab. Other reactions need a continuous supply of heat to make them work. Splitting calcium carbonate into calcium oxide and carbon dioxide is a simple example of this. Any chemical reaction will involve breaking some bonds and making new ones. Energy is needed to break bonds, and is given out when the new bonds are formed. It is very unlikely that these two processes will involve exactly the same amount of energy - and so some energy will either be absorbed or released during a reaction. You can show this on simple energy diagrams. For an exothermic change: Notice that in an exothermic change, the products have a lower energy than the reactants. The energy that the system loses is given out as heat. The surroundings warm up. For an endothermic change: This time the products have a higher energy than the reactants. The system absorbs this extra energy as heat from the surroundings. Here is an exothermic reaction, showing the amount of heat evolved: \[ C + O_2 \rightarrow CO_2 \;\;\; \Delta H = -394 \text{kJ} \, mol^{-1}\] This shows that 394 kJ of heat energy are evolved when equation quantities of carbon and oxygen combine to give carbon dioxide. The mol (per mole) refers to the whole equation in mole quantities. How do you know that heat is evolved? That is shown by the negative sign. You always think of the energy change during a reaction from the point of view of the reactants. The reactants (carbon and oxygen) have lost energy during the reaction. When you burn carbon in oxygen, that is the energy which is causing the surroundings to get hotter. And here is an endothermic change: \[ CaCO_3 \rightarrow CaO + CO_2 \;\;\; \Delta H = +178 \text{kJ} \, mol^{-1}\] In this case, 178 kJ of heat are absorbed when 1 mole of calcium carbonate reacts to give 1 mole of calcium oxide and 1 mole of carbon dioxide. You can tell that energy is being absorbed because of the plus sign. A simple energy diagram for the reaction looks like this: The products have a higher energy than the reactants. Energy has been gained by the system - hence the plus sign. Whenever you write values for any energy change, you must always write a plus or a minus sign in front of it. Chemists often express statements that something is energetically more stable than something else. For example, that oxygen, O , is more energetically stable than ozone, O . What does this mean? If you plot the positions of oxygen and ozone on an energy diagram, it looks like this: The lower down the energy diagram something is, the more energetically stable it is. If ozone converted into ordinary oxygen, heat energy would be released, and the oxygen would be in a more energetically stable form than it was before. So why does not ozone immediately convert into the more energetically stable oxygen? Similarly, if you mix gasoline and air at ordinary temperatures (when you are filling up a car, for example), why does not it immediately convert into carbon dioxide and water? It would be much more energetically stable if it turned into carbon dioxide and water - you can tell that, because lots of heat is given out when gasoline burns in air. But there is no reaction when you mix the two. For any reaction to happen, bonds have to be broken, and new ones made. Breaking bonds takes energy. There is a minimum amount of energy needed before a reaction can start - activation energy. If the molecules don't, for example, hit each other with enough energy, then nothing happens. We say that the mixture is kinetically stable, even though it may be energetically unstable with respect to its possible products. So a gasoline/air mixture at ordinary temperatures does not react, even though a lot of energy would be released if the reaction took place. gasoline and air are energetically unstable with respect to carbon dioxide and water - they are much higher up the energy diagram. But a gasoline and air mixture is kinetically stable at ordinary temperatures, because the activation energy barrier is too high.If you expose the mixture to a flame or a spark, then you get a major fire or explosion. The initial flame supplies activation energy. The heat given out by the molecules that react first is more than enough to supply the activation energy for the next molecules to react - and so on. The moral of all this is that you should be very careful using the word "stable" in chemistry! Jim Clark ( ) | 4,745 | 961 |
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We can put together the first and the second law for a reversible process with no other work than volume (\(PV\)) work and obtain: \[dU= δq_{rev} + δw_{rev} \nonumber \] Entropy is the dispersal of energy and is related to heat: \[δq_{rev}= TdS \nonumber \] Work is related to the change in volume: \[δw_{rev}= -PdV \nonumber \] Plugging these into our expression for \(dU\) for reversible changes: \[dU= TdS -PdV \nonumber \] We no longer have any path functions in the expression, as \(U\), \(S\) and \(V\) are . This means this expression must be an exact differential. We can generalize the expression to hold for irreversible processes, but then the expression becomes an inequality: \[dU≤ TdS - PdV \nonumber \] This equality expresses \(U\) as a function of two variables, entropy and volume: \(U(S,V)\). \(S\) and \(V\) are the of \(U\). At constant volume, \(dU\) becomes: \[dU=TdS \nonumber \] Recall that internal energy is related to constant volume heat capacity, \(C_V\): \[C_V=\left(\frac{dU}{dT}\right)_V \nonumber \] Combining these two expressions, we obtain: \[dS=\frac{C_V}{T}dT \nonumber \] Integrating: \[\Delta S=\int_{T_1}^{T_2}{\frac{C_V(T)}{T}dT} \nonumber \] If we know how \(C_V\) changes with temperature, we can calculate the change in entropy, \(\Delta S\). Since heat capacity is always a positive value, entropy must increase as the temperature increases. There is nothing to stop us from expressing \(U\) in variables, e.g. \(T\) and \(V\). In fact, we can derive some interesting relationships if we do. We can also derive an expression for the change in entropy as a function of constant pressure heat capacity, \(C_P\). To start, we need to change from internal energy, \(U\), to enthalpy, \(H\): \[\begin{align*} H &= U + PV \\[4pt] dH &= dU +d(PV) \\[4pt] &= dU + PdV + VdP \end{align*} \nonumber \] For reversible processes: \[\begin{align*} dH &= dU + PdV + VdP \\[4pt] &= TdS -PdV + PdV + VP \\[4pt] &= TdS + VdP\end{align*} \nonumber \] The natural variables of the enthalpy are \(S\) and \(P\) (not: \(V\)). A similar derivation as above shows that the temperature change of entropy is related to the constant pressure heat capacity: \[dH=TdS+VdP \nonumber \] At constant pressure: \[dH=TdS+VdP \nonumber \] Recall that: \[C_P=\frac{dH}{dT} \nonumber \] Combining, we obtain: \[dS=\frac{C_P}{T}dT \nonumber \] Integrating: \[\Delta S=\int_{T_1}^{T_2}{\frac{C_P(T)}{T}dT} \nonumber \] This means that if we know the heat capacities as a function of temperature we can calculate how the entropy changes with temperature. Usually it is easier to obtain data under constant \(P\) conditions than for constant \(V\), so that the route with \(C_p\) is the more common one. | 2,734 | 962 |
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Ionic or polar reactions of alkyl halides rarely are observed in the vapor phase because the energy required to dissociate a carbon-halogen bond heterolytically is almost prohibitively high. For example, while the heat of dissociation of chloromethane to a methyl radical and a chlorine atom is \(84 \: \text{kcal mol}^{-1}\) (Table 4-6), dissociation to a methyl cation and a chloride ion requires about \(227 \: \text{kcal mol}^{-1}\): However, the heat of ionic dissociation of methyl chloride in aqueous solution is estimated to be \(63 \: \text{kcal}\), and while this reaction is still substantially endothermic, it requires about \(227 - 63 = 164 \: \text{kcal}\) energy than in the gas phase: The reason is that ions are much more stable in water than in the gas phase; for example, the transfer of a chloride ion from the gas to water is exothermic by \(-85 \: \text{kcal}\). The \(\Delta H^\text{0}\) value for the corresponding transfer of a methyl cation, \(CH_3^\oplus\), is not known with certainty, but is about \(-80 \: \text{kcal}\). These ionic are clearly large. In contrast, the \(\Delta H^\text{0}\) for solution of methyl chloride in water is small (about \(1 \: \text{kcal}\)). We can use these data to calculate the heat of ionic dissociation of chloromethane in water: Thermochemical data for the solvation of ions as used in the preceding calculations are difficult to measure and even to estimate. Therefore this kind of calculation of \(\Delta H^\text{0}\) for ionic reactions involving organic molecules in solution usually cannot be made. As a result, we have considerably fewer possibilities to assess the thermodynamic feasibility of the individual steps of polar reactions in solution than we do of vapor-phase radical processes. Bond energies are not of much use in predicting or explaining reactivity in ionic reactions unless we have some information that can be used to translate gas-phase \(\Delta H^\text{0}\) values to solution \(\Delta H^\text{0}\) values. \(^2\)Calculated from the following data: and (1977) | 2,073 | 963 |
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Oxygen is an element that is widely known by the general public because of the large role it plays in sustaining life. Without oxygen, animals would be unable to breathe and would consequently die. Oxygen is not only important to supporting life, but plays an important role in many other chemical reactions. Oxygen is the most common element in the earth's crust and makes up about 20% of the air we breathe. Historically the discovery of oxygen as an element essential for combustion stands at the heart of the phlogiston controversy (see below). Oxygen is found in the group 16 elements and is considered as a chalcogen. Named from the Greek oxys + genes, "acid-former", oxygen was discovered in 1772 by Scheele and independently by Priestly in 1774. Oxygen was given its name by the French scientist, Antoine Lavoisier. Scheele descovered oxygen, through an experiment, which involved burning manganese oxide. Scheele came to find that the hot manganese oxide produced a gas which he called "fire air". He also came to find that when this gas was able to come into contact with charcoal, it produced beautiful bright sparks. All of the other elements produced the same gas. Although Scheele discovered oxygen, he did not publish his work until three years after another chemist, Joseph Priestly, discovered oxygen. Joseph Priestly, an English chemist, repeated Scheele's experiment in 1774 using a slightly different set up. Priestly used a 12in burning glass and aimed the sunlight directly towards the compound that he was testing, mercuric oxide. As a result, he was able to "discover better air" that was shown to expand a mouse's lifetime to four times as long and caused a flame to burn with higher intensity. Despite these experiments, both chemists were not able to pinpoint exactly what this element was. It was not until 1775 that Antoine Lavoisier, a French chemist, was able to recognize this unknown gas as an element. Our atmosphere currently contains about 21% of free oxygen. Oxygen is produced in various ways. The process of photochemical dissociation in which water molecules are broken up by ultraviolet rays produces about 1-2% of our oxygen. Another process that produces oxygen is which is performed by plants and photosynthetic bacteria. Photosynthesis occurs through the following general reaction: \[\ce{CO2 + H2O + h\nu \rightarrow} \text{organic compounds} \ce{+ O2}\] Phlogiston theory is the outdated belief that a fire-like element called phlogiston is contained within combustible bodies and released during combustion. The name comes from the Ancient Greek φλογιστόν phlogistón (burning up), from φλόξ phlóx (flame). It was first stated in 1667 by Johann Joachim Becher, and then put together more formally by Georg Ernst Stahl. The theory attempted to explain burning processes such as combustion and rusting, which are now collectively known as oxidation. As shown, there are two unpaired electrons which causes O to be paramagnetic. There are also eight valence electrons in the bonding orbitals and four in antibonding orbitals which makes the bond order 2. This accounts for the double covalent bond that is present in O . As shown in \(\Page {1}\), since molecular oxygen (\(O_2\) has unpaired electrons, it is paramagnetic and is attracted to the magnet. In contrast, molecular nitrogen (\(N_2\)) has no unpaired electrons and it not attracted to the magnet. Oxygen does not react with itself, nitrogen, or water under normal conditions. Oxygen does, however, dissolve in water at 20 degrees Celsius and 1 atmosphere. Oxygen also does not normally react with bases or acids. Group 1 metals (alkaline metals) are very reactive with oxygen and must be stored away from oxygen in order to prevent them from becoming oxidized. The metals at the bottom of the group are more reactive than those at the top. The reactions of a few of these metals are explored in more detail below. Reacts with oxygen to form white lithium oxide in the reaction below. \[\ce{4Li + O_2 \rightarrow 2Li_2O} \label{1}\] Reacts with oxygen to form a white mixture of sodium oxide and sodium peroxide. The reactions are shown below. Reacts with oxygen to form a mixture of potassium peroxide and potassium superoxide. The reactions are shown below. Both metals react to produce superoxides through the same process as that of the potassium superoxide reaction. The oxides of these metals form metal hydroxides when reacting with water. These metal hydroxides make the solution basic or alkaline hence the name alkaline metals. Group 2 metals (alkaline earth metals) react with oxygen through the process of burning to form metal oxides but there are a few exceptions. Beryllium is very difficult to burn because it has a layer of beryllium oxide on its surface which prevents further interaction with oxygen. Strontium and barium react with oxygen to form peroxides. The reaction of barium and oxygen is shown below and the reaction with strontium would be the same. \[\ce{Ba(s) + O2 (g) \rightarrow BaO2 (s) }\label{6}\] Group 13 reacts with oxygen in order to form oxides and hydroxides that are of the form \(X_2O_3\) and \(X(OH)_3\). The variable X represents the various group 13 elements. As you go down the group, the oxides and hydroxides get increasingly basic. elements react with oxygen to form oxides. The oxides formed at the top of the group are more acidic than those at the bottom of the group. Oxygen reacts with silicon and carbon to form silicon dioxide and carbon dioxide. Carbon is also able to react with oxygen to form carbon monoxide which is slightly acidic. Germanium, tin, and lead react with oxygen to form monoxides and dioxides that are amphoteric which means that they react both with acids and bases. Group 15 elements react with oxygen to form oxides. The most important are listed below. Group 16 elements react with oxygen to form various oxides. Some of the oxides are listed below. Group 17 elements (halogens) fluorine, chlorine, bromine, and iodine react with oxygen to form oxides. Fluorine forms two oxides with oxygen which are F O and F O . Both fluorine oxides are called oxygen fluorides because fluorine is the more electronegative element. One of the fluorine reactions is shown below. \[\ce{O2 (g) + F2 (g) \rightarrow F2O2 (g)} \label{7}\] Group 18 Some would assume that the Noble Gases would not react with oxygen. However, xenon does react with oxygen to form \(\ce{XeO_3}\) and \(\ce{XeO_4}\). The ionization energy of xenon is low enough for the electronegative oxygen atom to "steal away" electrons. Unfortunately, \(\ce{XeO_3}\) is HIGHLY unstable, and it has been known to spontaneously detonated in a clean, dry environment. Transition metals react with oxygen to form metal oxides. However, gold, silver, and platinum do not react with oxygen. A few reactions involving transition metals are shown below. \[2Sn_{(s)} + O_{2(g)} \rightarrow 2SnO_{(s)} \label{8} \] \[4Fe_{(s)} + 3O_{2(g)} \rightarrow 2Fe_2O_{3(s)} \label{9A}\] \[4Al_{(s)} + 3O_{2(g)} \rightarrow 2Al_2O_{3(s)} \label{9B}\] We will be discussing metal oxides of the form \(X_2O\). The variable \(X\) represents any metal that is able to bond to oxygen to form an oxide. \[X_2O + H_2O \rightarrow 2XOH\] \[X_2O + 2HCl \rightarrow 2XCl + H_2O\] The peroxides we will be discussing are of the form \(X_2O_2\). The variable \(X\) represents any metal that can form a peroxide with oxygen. If the temperature of the reaction is kept constant despite the fact that the reaction is exothermic then the reaction proceeds as follows. \[X_2O_2+ 2H_2O \rightarrow 2XOH + H_2O_2\] If the reaction is not done at a constant temperature, then the reaction of the peroxide and water will result in the hydrogen peroxide that is produced to decompose into water and oxygen. This reaction is more exothermic than that with water. The heat produced causes the hydrogen peroxide to decompose to water and oxygen. The reaction is shown below. \[X_2O_2 + 2HCl \rightarrow 2XCl + H_2O_2\] The superoxides we will be talking about are of the form \(XO_2\). with \(X\) representing any metal that forms a superoxide when reacting with oxygen. The superoxide and water react in a very exothermic reaction that is shown below. The heat that is produced in forming the hydrogen peroxide will cause the hydrogen peroxide to decompose to water and oxygen. \[2XO_2 + 2H_2O \rightarrow 2XOH + H_2O_2 + O_2\] The superoxide and dilute acid react in a very exothermic reaction that is shown below. The heat produced will cause the hydrogen peroxide to decompose to water and oxygen. \[2XO_2 + 2HCl \rightarrow 2XCl + H_2O_2 + O_2\] There are two allotropes of oxygen; dioxygen (O ) and trioxygen (O ) which is called . The reaction of converting dioxygen into ozone is very endothermic causing it to occur rarely and only in the low atmosphere. The reaction is shown below: \[3O_{2 (g)} \rightarrow 2O_{3 (g)} \;\;\; ΔH^o= +285 \;kJ\] Ozone is unstable and quickly decomposes back to oxygen but is a great oxidizing agent. The most common reactions that involve occur with oxygen. Alkanes are able to burn and it is the process of oxidizing the hydrocarbons that makes them important as fuels. An example of an alkane reaction is the reaction of octane with oxygen as showed below. C H (l) + 25/2 O (g) → 8CO (g) + 9H O(l) ∆H = -5.48 X 10 kJ Oxygen is able to react with ammonia to produce dinitrogen (N ) and water (H O) through the reaction shown below. \[4NH_3 + 3O_2 \rightarrow 2N_2 + 6H_2O\] Oxygen is able to react with nitrogen oxide in order to produce nitrogen dioxide through the reaction shown below. \[NO + O_2 \rightarrow NO_2\] | 9,687 | 964 |
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To investigate reactions that are complete in less than a millisecond, one can start with a pre-mixed sample in which one of active reactants is generated . Alternatively, a rapid change in pressure or temperature can alter the composition of a reaction that has already achieved equilibrium. Many reactions are known which do not take place without light of wavelength sufficiently short to supply the activation energy needed to break a bond, often leading to the creation of a highly reactive radical. A good example is the combination of gaseous Cl with H , which proceeds explosively when the system is illuminated with visible light. In , a short pulse of light is used to initiate a reaction whose progress can be observed by optical or other means. refers to the use of light to decompose a molecule into simpler units, often ions or free radicals. In contrast to (decomposition induced by high temperature), photolysis is able to inject energy into a molecule almost instantaneously and can be much "cleaner," meaning that there are fewer side reactions that often lead to complex mixtures of products. Photolysis can also be highly ; the wavelength of the light that triggers the reaction can often be adjusted to activate one particular kind of molecule without affecting others that might be present. In 1945, Ronald Norrish of Cambridge University and his graduate student George Porter conceived the idea of using a short-duration flash lamp to generate gas-phase CH radicals, and then following the progress of the reaction of these radicals with other species by means of absorption spectroscopy. In a flash photolysis experiment, recording of the absorbance of the sample cell contents is timed to follow the flash by an interval that can be varied in order to capture the effects produced by the product or intermediate as it is formed or decays. Norrish and Porter shared the 1967 Nobel Prize in Chemistry for this work. Flash durations of around 1 millisecond permitted one to follow processes having lifetimes in the microsecond range, but the advent of fast lasers gradually extended this to picoseconds and femtoseconds. Flash photolysis revolutionized the study of organic photochemistry, especially that relating to the chemistry of free radicals and other reactive species that cannot be isolated or stored, but which can easily be produced by photolysis of a suitable precursor. It has proven invaluable for understanding the complicated kinetics relating to atmospheric chemistry and smog formation. More recently, flash photolysis has become an important tool in biochemistry and cellular physiology. Many reactions, especially those that take place in solution, occur too rapidly to follow by flow techniques, and can therefore only be observed when they are already at equilibrium. The classical examples of such reactions are two of the fastest ones ever observed, the dissociation of water \[ 2 H_2O \rightarrow H_3O^+ + OH^- \nonumber \] and the formation of the triiodide ion in aqueous solution \[ I^– + I_2 \rightarrow I_3^– \nonumber \] Reactions of these kinds could not be studied until the mid-1950s when techniques were developed to shift the equilibrium by imposing an abrupt physical change on the system. For example, if the reaction A B is endothermic, then according to the Le Chatelier principle, subjecting the system to a rapid jump in temperature will shift the equilibrium state to one in which the product B has a higher concentration. The composition of the system will than begin to shift toward the new equilibrium composition at a rate determined by the kinetics of the process. For the general case illustrated here, the quantity " " being plotted is a measurable quantity such as light absorption or electrical conductivity that varies linearly with the composition of the system. In a first-order process, will vary with time according to \[ x_t = x_o e^{-kt} \nonumber \] After the abrupt perturbation at time , the relaxation time is defined as the half-time for the return to equilibrium — that is, as the time required for xo to decrease by Δ /e = Δ /2.718. The derivation of and the relations highlighted in yellow can be found in most standard kinetics textbooks. are likely most commonly used. This is the method that Manfred Eigen (Germany, 1927-) pioneered when, in the early 1960's, he measured the rate constant of what was then the fastest reaction ever observed: \[ H^+ + OH^– \rightarrow H_2O \;\;\;\; k = 1.3 \times 10^{11} \; M^{–1}\; sec^{–1} \nonumber \] According to the Le Chatelier principle, a change in the applied pressure will shift the equilibrium state of any reaction which involves a change in the volume of a system. Aside from the obvious examples associated with changes in the number of moles of gases, there are many more subtle cases involving formation of complexes, hydration shells and surface adsorption, and phase changes. One area of considerable interest is the study of protein folding, which has implications in diseases such as Parkinson's and Alzheimer's. The pressure-jump is applied to the cell through a flexible membrane that is activated by a high-pressure gas supply, or through an electrically-actuated piezoelectric crystal. The latter method is employed in the device shown here, which can produce P-jumps of around 1 GPa over sub-millisecond time intervals. When a change in pressure propagates through a gas at a rate greater than the ordinary compressions and rarefactions associated with the travel of sound, a moving front (a ) of very high pressure forms. This in turn generates an almost instantaneous rise in the temperature that can approach several thousand degrees in magnitude. A shock tube is an apparatus in which shock waves can be generated and used to study the kinetics of gas-phase reactions that are otherwise inaccessible to kinetic measurements. Since all molecules tend to dissociate at high temperatures, shock tubes are widely used to study dissociation processes and the chemistry of the resulting fragments. For example, the shock-induced decomposition of carbon suboxide provides an efficient means of investigating carbon atom reactions: \[ C_3O_2 \rightarrow C + CO \nonumber \] Shock tube techniques are also useful for studying combustion reactions, including those that proceed explosively. The shock tube itself consists of two sections separated by a breakable diaphragm of metal or plastic. One section is filled with a "driver" gas at a very high pressure, commonly helium, but often mixed with other inert gases to adjust the properties of the shock. The other, longer section of the tube contains the "driven" gas—the reactants—at a low pressure, usually less than 1 atmosphere. The reaction is initiated by causing the diaphragm to rupture, either by means of a mechanical plunger or by raising the pressure beyond its bursting point. The kinetics of the reaction are monitored by means of an absorption or other optical monitoring device that is positioned at a location along the reaction tube that is appropriate to the time course of the reaction. ) | 7,144 | 965 |
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Catalase and peroxidase are heme enzymes that catalyze reactions of hydrogen peroxide. In catalase, the enzymatic reaction is the disproportionation of hydrogen peroxide (Reaction 5.82) and the function of the enzyme appears to be prevention of any buildup of that potentially dangerous oxidant (see the discussion of dioxygen toxicity in Section III). \[2H_{2}O_{2} \xrightarrow{catalase} 2H_{2}O + O_{2} \tag{5.82}\] Peroxidase reacts by mechanisms similar to catalase, but the reaction catalyzed is the oxidation of a wide variety of organic and inorganic substrates by hydrogen peroxide (Reaction 5.83). \[H_{2}O_{2} + AH_{2} \xrightarrow{peroxidase} 2H_{2}O + A \tag{5.83}\] (The catalase reaction can be seen to be a special case of Reaction 5.83 in which the substrate, AH , is hydrogen peroxide.) Some examples of peroxidases that have been characterized are horseradish peroxidase, cytochrome c peroxidase, glutathione peroxidase, and myeloperoxidase. X-ray crystal structures have been determined for beef-liver catalase and for horseradish peroxidase in the resting, high-spin ferric state. In both, there is a single heme b group at the active site. In catalase, the axial ligands are a phenolate from a tyrosyl residue, bound to the heme on the side away from the active-site cavity, and water, bound to heme within the cavity and presumably replaced by hydrogen peroxide in the catalytic reaction. In horseradish peroxidase, the axial ligand is an imidazole from a histidyl residue. Also within the active-site cavity are histidine and aspartate or asparagine side chains that appear to be ideally situated to interact with hydrogen peroxide when it is bound to the iron. These residues are believed to play an important part in the mechanism by facilitating O—O bond cleavage (see Section VI.B below). Three other forms of catalase and peroxidase can be generated, which are referred to as compounds I, II, and III. Compound I is generated by reaction of the ferric state of the enzymes with hydrogen peroxide. Compound I is green and has spectral characteristics very similar to the Fe (P )(O) complex prepared at low temperatures by reaction of ferric porphyrins with single-oxygenatom donors (see Section V.C.1.a). Titrations with reducing agents indicate that it is oxidized by two equivalents above the ferric form. It has been proposed (see 5.84) that the anionic nature of the tyrosinate axial ligand in catalase may serve to stabilize the highly oxidized iron center in compound I of that enzyme, and furthermore that the histidyl imidazole ligand in peroxidase may deprotonate, forming imidazolate, or may be strongly hydrogen bonded, thus serving a similar stabilizing function for compound I in that enzyme. \(\tag{5.84}\) Reduction of compound I by one electron produces compound II, which has the characteristics of a normal ferryl-porphyrin complex, analogous to , i.e., (L)Fe (P)(O). Reaction of compound II with hydrogen peroxide produces compound III, which can also be prepared by reaction of the ferrous enzyme with dioxygen. It is an oxy form, analogous to oxymyoglobin, and does not appear to have a physiological function. The reactions producing these three forms and their proposed formulations are summarized in Reactions (5.85) to (5.88). \[Fe^{III}(P)^{+} + H_{2}O_{2} \rightarrow Fe^{IV}(P^{-})(O)^{+} + H_{2}O \tag{5.85}\] \[ferric\; form \quad \qquad \qquad Compound\; I \qquad \qquad \] \[Fe^{IV}(P^{\cdotp -})(O)^{+} + e^{-} \rightarrow Fe^{IV}(P)(O) \tag{5.86}\] \[Compound\; I \qquad \qquad Compound\; II \quad \] \[Fe^{IV}(P)(O) + H_{2}O_{2} \rightarrow Fe(P)O_{2} + H_{2}O \tag{5.87}\] \[Compound\; II \qquad \qquad Compound\; III \qquad \qquad \] \[Fe{II}(P) + O_{2} \rightarrow Fe(P)O_{2} \tag{5.88}\] \[ferrous\; form \qquad Compound\; III\] The accepted mechanisms for catalase and peroxidase are described in Reactions (5.89) to (5.94). \[Fe^{III}(P)^{+} + H_{2}O_{2} \rightarrow Fe^{III}(P)(H_{2}O_{2})^{+} \rightarrow Fe^{IV}(P^{\cdotp -})(O)^{+} + H_{2}O \tag{5.89}\] \[\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad Compound\; I\] catalase: \[ Fe^{IV}(P^{\cdotp -})(O)^{+} + H_{2}O_{2} \rightarrow Fe^{III}(P)^{+} + H_{2}O + O_{2} \tag{5.90}\] \[Compound\; I \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \] peroxidase: \[ Fe^{IV}(P^{\cdotp -})(O)^{+} + AH_{2} \rightarrow Fe^{IV}(P)(O) +HA^{\cdotp} + H^{+} \tag{5.91}\] \[Compound\; I \qquad \qquad \qquad \qquad Compound\; II\] \[Fe^{IV}(P)(O) + AH_{2} \rightarrow Fe^{III}(P)^{+} +HA^{\cdotp} + OH^{-} \tag{5.92}\] \[Compound\; II \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \] \[2HA^{\cdotp} \rightarrow A + AH_{2} \tag{5.93}\] or \[2HA^{\cdotp} \rightarrow HA - AH \tag{5.94}\] In the catalase reaction, it has been established by use of H O that the dioxygen formed is derived from hydrogen peroxide, i.e., that O—O bond cleavage does not occur in Reaction (5.90), which is therefore a two-electron reduction of compound I by hydrogen peroxide, with the oxo ligand of the former being released as water. For the peroxidase reaction under physiological conditions, it is believed that the oxidation proceeds in one-electron steps (Reactions 5.91 and 5.92), with the final formation of product occurring by disproportionation (Reaction 5.93) or coupling (Reaction 5.94) of the one-electron oxidized intermediate. The proposal that these three enzymes all go through a similar high-valent oxo intermediate, i.e., or compound I, raises two interesting questions. The first of these is why the same high-valent metal-oxo intermediate gives two very different types of reactions, i.e., oxygen-atom transfer with cytochrome P-450 and electron transfer with catalase and peroxidase. The answer is that, although the high-valent metal-oxo heme cores of these intermediates are in fact very similar, the substrate-binding cavities seem to differ substantially in how much access the substrate has to the iron center. With cytochrome P-450, the substrate is jammed right up against the location where the oxo ligand must reside in the high-valent oxo intermediate. But the same location in the peroxidase enzymes is blocked by the protein structure so that substrates can interact only with the heme edge. Thus oxidation of the substrate by electron transfer is possible for catalase and peroxidase, but the substrate is too far away from the oxo ligand for oxygen-atom transfer. The second question is about how the the high-valent oxo intermediate forms in both enzymes. For catalase and peroxidase, the evidence indicates that hydrogen peroxide binds to the ferric center and then undergoes heterolysis at the O—O bond. Heterolytic cleavage requires a significant separation of positive and negative charge in the transition state. In catalase and peroxidase, analysis of the crystal structure indicates strongly that amino-acid side chains are situated to aid in the cleavage by stabilizing a charge-separated transition state (Figure 5.14). In cytochrome P-450, as mentioned in Section V.C.1, no such groups are found in the hydrophobic substrate-binding cavity. It is possible that the cysteinyl axial ligand in cytochrome P-450 plays an important role in O—O bond cleavage, and that the interactions found in catalase and peroxidase that appear to facilitate such cleavage are therefore not necessary. | 7,402 | 966 |
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Changes in reaction conditions can have a tremendous effect on the course of a redox reaction. For example, under standard conditions, the reaction of \(\ce{Co(s)}\) with \(\ce{Ni^{2+}(aq)}\) to form \(\ce{Ni(s)}\) and \(\ce{Co^{2+}(aq)}\) occurs spontaneously, but if we reduce the concentration of \(\ce{Ni^{2+}}\) by a factor of 100, so that \(\ce{[Ni^{2+}]}\) is 0.01 M, then the reverse reaction occurs spontaneously instead. The relationship between voltage and concentration is one of the factors that must be understood to predict whether a reaction will be spontaneous. Electrochemical cells convert chemical energy to electrical energy and vice versa. The total amount of energy produced by an electrochemical cell, and thus the amount of energy available to do electrical work, depends on both the cell potential and the total number of electrons that are transferred from the reductant to the oxidant during the course of a reaction. The resulting electric current is measured in , an unit that measures the number of electrons passing a given point in 1 s. A coulomb relates energy (in joules) to electrical potential (in volts). Electric current is measured in ; 1 A is defined as the flow of 1 C/s past a given point (1 C = 1 A·s): \[\dfrac{\textrm{1 J}}{\textrm{1 V}}=\textrm{1 C}=\mathrm{A\cdot s} \label{20.5.1} \] In chemical reactions, however, we need to relate the coulomb to the charge on a mole of electrons. Multiplying the charge on the electron by Avogadro’s number gives us the charge on 1 mol of electrons, which is called the , named after the English physicist and chemist Michael Faraday (1791–1867): \[\begin{align}F &=(1.60218\times10^{-19}\textrm{ C})\left(\dfrac{6.02214 \times 10^{23}\, J}{\textrm{1 mol e}^-}\right)
\\[4pt] &=9.64833212\times10^4\textrm{ C/mol e}^- \\[4pt] &\simeq 96,485\, J/(\mathrm{V\cdot mol\;e^-})\end{align} \label{20.5.2} \] The total charge transferred from the reductant to the oxidant is therefore \(nF\), where \(n\) is the number of moles of electrons. Faraday was a British physicist and chemist who was arguably one of the greatest experimental scientists in history. The son of a blacksmith, Faraday was self-educated and became an apprentice bookbinder at age 14 before turning to science. His experiments in electricity and magnetism made electricity a routine tool in science and led to both the electric motor and the electric generator. He discovered the phenomenon of electrolysis and laid the foundations of electrochemistry. In fact, most of the specialized terms introduced in this chapter (electrode, anode, cathode, and so forth) are due to Faraday. In addition, he discovered benzene and invented the system of oxidation state numbers that we use today. Faraday is probably best known for “The Chemical History of a Candle,” a series of public lectures on the chemistry and physics of flames. The maximum amount of work that can be produced by an electrochemical cell (\(w_{max}\)) is equal to the product of the cell potential (\(E^°_{cell}\)) and the total charge transferred during the reaction (\(nF\)): \[ w_{max} = nFE_{cell} \label{20.5.3} \] Work is expressed as a negative number because work is being done by a system (an electrochemical cell with a positive potential) on its surroundings. The change in free energy (\(\Delta{G}\)) is also a measure of the maximum amount of work that can be performed during a chemical process (\(ΔG = w_{max}\)). Consequently, there must be a relationship between the potential of an electrochemical cell and \(\Delta{G}\); this relationship is as follows: \[\Delta{G} = −nFE_{cell} \label{20.5.4} \] A spontaneous redox reaction is therefore characterized by a negative value of \(\Delta{G}\) and a positive value of \(E^°_{cell}\), consistent with our earlier discussions. When both reactants and products are in their standard states, the relationship between ΔG° and \(E^°_{cell}\) is as follows: \[\Delta{G^°} = −nFE^°_{cell} \label{20.5.5} \] A spontaneous redox reaction is characterized by a negative value of ΔG°, which corresponds to a positive value of E° . Suppose you want to prepare elemental bromine from bromide using the dichromate ion as an oxidant. Using the data in , calculate the free-energy change (ΔG°) for this redox reaction under standard conditions. Is the reaction spontaneous? redox reaction \(ΔG^o\) for the reaction and spontaneity As always, the first step is to write the relevant half-reactions and use them to obtain the overall reaction and the magnitude of \(E^o\). From , we can find the reduction and oxidation half-reactions and corresponding \(E^o\) values: \[\begin{align*}
& \textrm{cathode:} &\quad & \mathrm{Cr_2O_7^{2-}(aq)} + \mathrm{14H^+(aq)}+\mathrm{6e^-}\rightarrow \mathrm{2Cr^{3+}(aq)} +\mathrm{7H_2O(l)}
&\quad & E^\circ_{\textrm{cathode}} =\textrm{1.36 V} \\
& \textrm{anode:} &\quad & \mathrm{2Br^{-}(aq)} \rightarrow \mathrm{Br_2(aq)} +\mathrm{2e^-}
&\quad & E^\circ_{\textrm{anode}} =\textrm{1.09 V}
\end{align*} \nonumber \] To obtain the overall balanced chemical equation, we must multiply both sides of the oxidation half-reaction by 3 to obtain the same number of electrons as in the reduction half-reaction, remembering that the magnitude of \(E^o\) is not affected: \[\begin{align*}
& \textrm{cathode:} &\quad & \mathrm{Cr_2O_7^{2-}(aq)} + \mathrm{14H^+(aq)}+\mathrm{6e^-}\rightarrow \mathrm{2Cr^{3+}(aq)} +\mathrm{7H_2O(l)}
&\quad & E^\circ_{\textrm{cathode}} =\textrm{1.36 V} \\[4pt]
& \textrm{anode:} &\quad & \mathrm{6Br^{-}(aq)} \rightarrow \mathrm{3Br_2(aq)} +\mathrm{6e^-}
&\quad & E^\circ_{\textrm{anode}} =\textrm{1.09 V} \\[4pt] \hline
& \textrm{overall:} &\quad & \mathrm{Cr_2O_7^{2-}(aq)} + \mathrm{6Br^{-}(aq)} + \mathrm{14H^+(aq)} \rightarrow \mathrm{2Cr^{3+}(aq)} + \mathrm{3Br_2(aq)} +\mathrm{7H_2O(l)}
&\quad & E^\circ_{\textrm{cell}} =\textrm{0.27 V}
\end{align*} \nonumber \] We can now calculate ΔG° using Equation \(\ref{20.5.5}\). Because six electrons are transferred in the overall reaction, the value of \(n\) is 6: \[\begin{align*}\Delta G^\circ &=-(n)(F)(E^\circ_{\textrm{cell}}) \\[4pt] & =-(\textrm{6 mole})[96,485\;\mathrm{J/(V\cdot mol})(\textrm{0.27 V})] \\& =-15.6 \times 10^4\textrm{ J} \\ & =-156\;\mathrm{kJ/mol\;Cr_2O_7^{2-}} \end{align*} \nonumber \] Thus \(ΔG^o\) is −168 kJ/mol for the reaction as written, and the reaction is spontaneous. Use the data in to calculate \(ΔG^o\) for the reduction of ferric ion by iodide: \[\ce{2Fe^{3+}(aq) + 2I^{−}(aq) → 2Fe^{2+}(aq) + I2(s)}\nonumber \] Is the reaction spontaneous? −44 kJ/mol I ; yes Although list several half-reactions, many more are known. When the standard potential for a half-reaction is not available, we can use relationships between standard potentials and free energy to obtain the potential of any other half-reaction that can be written as the sum of two or more half-reactions whose standard potentials are available. For example, the potential for the reduction of \(\ce{Fe^{3+}(aq)}\) to \(\ce{Fe(s)}\) is not listed in the table, but two related reductions are given: \[\ce{Fe^{3+}(aq) + e^{−} -> Fe^{2+}(aq)} \;\;\;E^° = +0.77 V \label{20.5.6} \] \[\ce{Fe^{2+}(aq) + 2e^{−} -> Fe(s)} \;\;\;E^° = −0.45 V \label{20.5.7} \] Although the sum of these two half-reactions gives the desired half-reaction, we cannot simply add the potentials of two reductive half-reactions to obtain the potential of a third reductive half-reaction because \(E^o\) is not a state function. However, because \(ΔG^o\) is a state function, the sum of the \(ΔG^o\) values for the individual reactions gives us \(ΔG^o\) for the overall reaction, which is proportional to both the potential and the number of electrons (\(n\)) transferred. To obtain the value of \(E^o\) for the overall half-reaction, we first must add the values of \(ΔG^o (= −nFE^o)\) for each individual half-reaction to obtain \(ΔG^o\) for the overall half-reaction: \[\begin{align*}\ce{Fe^{3+}(aq)} + \ce{e^-} &\rightarrow \ce\mathrm{Fe^{2+}(aq)} &\quad \Delta G^\circ &=-(1)(F)(\textrm{0.77 V})\\[4pt]
\ce{Fe^{2+}(aq)}+\ce{2e^-} &\rightarrow\ce{Fe(s)} &\quad\Delta G^\circ &=-(2)(F)(-\textrm{0.45 V})\\[4pt]
\ce{Fe^{3+}(aq)}+\ce{3e^-} &\rightarrow \ce{Fe(s)} &\quad\Delta G^\circ & =[-(1)(F)(\textrm{0.77 V})]+[-(2)(F)(-\textrm{0.45 V})] \end{align*} \nonumber \] Solving the last expression for ΔG° for the overall half-reaction, \[\Delta{G^°} = F[(−0.77 V) + (−2)(−0.45 V)] = F(0.13 V) \label{20.5.9} \] Three electrons (\(n = 3\)) are transferred in the overall reaction, so substituting into Equation \(\ref{20.5.5}\) and solving for \(E^o\) gives the following: \[\begin{align*}\Delta G^\circ & =-nFE^\circ_{\textrm{cell}} \\[4pt]
F(\textrm{0.13 V}) & =-(3)(F)(E^\circ_{\textrm{cell}}) \\[4pt]
E^\circ & =-\dfrac{0.13\textrm{ V}}{3}=-0.043\textrm{ V}\end{align*} \nonumber \] This value of \(E^o\) is very different from the value that is obtained by simply adding the potentials for the two half-reactions (0.32 V) and even has the opposite sign. Values of \(E^o\) for half-reactions cannot be added to give \(E^o\) for the sum of the half-reactions; only values of \(ΔG^o = −nFE^°_{cell}\) for half-reactions can be added. We can use the relationship between \(\Delta{G^°}\) and the equilibrium constant \(K\), to obtain a relationship between \(E^°_{cell}\) and \(K\). Recall that for a general reaction of the type \(aA + bB \rightarrow cC + dD\), the standard free-energy change and the equilibrium constant are related by the following equation: \[\Delta{G°} = −RT \ln K \label{20.5.10} \] Given the relationship between the standard free-energy change and the standard cell potential (Equation \(\ref{20.5.5}\)), we can write \[−nFE^°_{cell} = −RT \ln K \label{20.5.12} \] Rearranging this equation, \[E^\circ_{\textrm{cell}}= \left( \dfrac{RT}{nF} \right) \ln K \label{20.5.12B} \] For \(T = 298\, K\), Equation \(\ref{20.5.12B}\) can be simplified as follows: \[ \begin{align} E^\circ_{\textrm{cell}} &=\left(\dfrac{RT}{nF}\right)\ln K \\[4pt] &=\left[ \dfrac{[8.314\;\mathrm{J/(mol\cdot K})(\textrm{298 K})]}{n[96,485\;\mathrm{J/(V\cdot mol)}]}\right]2.303 \log K \\[4pt] &=\left(\dfrac{\textrm{0.0592 V}}{n}\right)\log K \label{20.5.13} \end{align} \] Thus \(E^°_{cell}\) is directly proportional to the logarithm of the equilibrium constant. This means that large equilibrium constants correspond to large positive values of \(E^°_{cell}\) and vice versa. Use the data in to calculate the equilibrium constant for the reaction of metallic lead with PbO in the presence of sulfate ions to give PbSO under standard conditions. (This reaction occurs when a car battery is discharged.) Report your answer to two significant figures. redox reaction \(K\) The relevant half-reactions and potentials from are as follows: \[\begin{align*} & \textrm {cathode:}
& & \mathrm{PbO_2(s)}+\mathrm{SO_4^{2-}(aq)}+\mathrm{4H^+(aq)}+\mathrm{2e^-}\rightarrow\mathrm{PbSO_4(s)}+\mathrm{2H_2O(l)}
& & E^\circ_\textrm{cathode}=\textrm{1.69 V} \\[4pt]
& \textrm{anode:}
& & \mathrm{Pb(s)}+\mathrm{SO_4^{2-}(aq)}\rightarrow\mathrm{PbSO_4(s)}+\mathrm{2e^-}
& & E^\circ_\textrm{anode}=-\textrm{0.36 V} \\[4pt] \hline
& \textrm {overall:}
& & \mathrm{Pb(s)}+\mathrm{PbO_2(s)}+\mathrm{2SO_4^{2-}(aq)}+\mathrm{4H^+(aq)}\rightarrow\mathrm{2PbSO_4(s)}+\mathrm{2H_2O(l)}
& & E^\circ_\textrm{cell}=\textrm{2.05 V} \end{align*} \nonumber \] Two electrons are transferred in the overall reaction, so \(n = 2\). Solving Equation \(\ref{20.5.13}\) for log K and inserting the values of \(n\) and \(E^o\), \[\begin{align*}\log K & =\dfrac{nE^\circ}{\textrm{0.0591 V}}=\dfrac{2(\textrm{2.05 V})}{\textrm{0.0591 V}}=69.37 \\[4pt]
K & =2.3\times10^{69}\end{align*} \nonumber \] Thus the equilibrium lies far to the right, favoring a discharged battery (as anyone who has ever tried unsuccessfully to start a car after letting it sit for a long time will know). Use the data in to calculate the equilibrium constant for the reaction of \(\ce{Sn^{2+}(aq)}\) with oxygen to produce \(\ce{Sn^{4+}(aq)}\) and water under standard conditions. Report your answer to two significant figures. The reaction is as follows: \[\ce{2Sn^{2+}(aq) + O2(g) + 4H^{+}(aq) <=> 2Sn^{4+}(aq) + 2H2O(l)} \nonumber \] \(5.7 \times 10^{72}\) Figure \(\Page {1}\) summarizes the relationships that we have developed based on properties of the system—that is, based on the equilibrium constant, standard free-energy change, and standard cell potential—and the criteria for spontaneity (ΔG° < 0). Unfortunately, these criteria apply only to systems in which all reactants and products are present in their standard states, a situation that is seldom encountered in the real world. A more generally useful relationship between cell potential and reactant and product concentrations, as we are about to see, uses the relationship between \(\Delta{G}\) and the reaction quotient \(Q\). A coulomb (C) relates electrical potential, expressed in volts, and energy, expressed in joules. The current generated from a redox reaction is measured in amperes (A), where 1 A is defined as the flow of 1 C/s past a given point. The faraday (F) is Avogadro’s number multiplied by the charge on an electron and corresponds to the charge on 1 mol of electrons. The product of the cell potential and the total charge is the maximum amount of energy available to do work, which is related to the change in free energy that occurs during the chemical process. Adding together the ΔG values for the half-reactions gives ΔG for the overall reaction, which is proportional to both the potential and the number of electrons (n) transferred. Spontaneous redox reactions have a negative ΔG and therefore a positive E . Because the equilibrium constant K is related to ΔG, E° and K are also related. Large equilibrium constants correspond to large positive values of E°. | 13,917 | 967 |
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In order for dioxygen transport to be more efficient than simple diffusion through cell membranes and fluids, it is not sufficient that a metalloprotein merely binds dioxygen. Not only is there an optimal affinity of the carrier for dioxygen, but also, and more importantly, the carrier must bind and release dioxygen at a rapid rate. These thermodynamic and kinetic aspects are illustrated in Figure 4.3, a general diagram of energy vs. reaction coordinate for the process \[M + O_{2} \xrightleftharpoons[k_{-1}]{k_{1}} MO_{2}\tag{4.2}\] where M is an oxygen carrier, for example hemocyanin or a simple nonbiological metal complex. Thermodynamic or equilibrium aspects are summarized by \(\Delta\)G in Figure 4.3. As illustrated there, \(\Delta\)G is negative, and thus the forward reaction, dioxygen binding, is spontaneous. The equilibrium constant (K) is given by \[K = \frac{a(MO_{2})}{a(M)a(O_{2})} , \tag{4.3}\] where \(a\) is the (i.e., effective concentration) of the component. The equilibrium constant is related to the change in free energy by \[\Delta G^{o} = -RT\ln K \ldotp \tag{4.4}\] The rate of the forward reaction (\(k_1\)) is related to \(\Delta G_1^*\); the rate of the reverse reaction (\(k_{-1}\)) is related to \(\Delta G^*_{-1}\). Provided that oxygen binding is effectively a single-step process, then \[K = \frac{k_{1}}{k_{-1}} \ldotp \tag{4.5}\] Usually the rates of the forward and reverse reactions are related by the empirical Arrhenius expression to quantities termed the activation energies (E and E ) of the reactions, where \[k_{1} = A_{1}^{\frac{-E_{1}}{RT}}\; and\; k_{-1} = A_{-1}^{\frac{-E_{-1}}{RT}} \ldotp \tag{4.6}\] These quantities are experimentally accessible through the change in rate as a function of temperature. It is of little benefit to the organism if its dioxygen carrier, such as hemoglobin, binds and releases O at such slow rates that O is not delivered faster than it would be by simple diffusive processes. Thus, a binding rate within a couple of orders of magnitude of the rate of diffusion, together with the high carrying capacity of O that high concentrations of oxygen carrier enable (noted ), and a pumping system ensure adequate O supplies under all but the most physiologically stressful conditions. Whereas measurements of equilibrium give little or no molecular information, rather more molecular information may be inferred from kinetic data. The processes of binding and release can be examined by a variety of techniques, with timescales down to the picosecond range. The temperature behavior of the rates gives information on the heights of energy barriers that are encountered as dioxygen molecules arrive at or depart from the binding site. The quantitative interpretation of kinetic data generally requires a molecular model of some sort. It is because of this multibarrier pathway that the equilibrium constant measured as k /k (Equation 4.5) may differ substantially from the thermodynamically measured value (Equation 4.3). The simple Adair scheme outlined above is readily adapted to cater to kinetic data. Most biological conversions involving dioxygen require enzymatic catalysis. It is reasonable then that metals found in the proteins involved in the transport and storage of O also frequently appear, with minor modification of ligands, in enzymes that incorporate oxygen from dioxygen into some substrate. Dioxygen, in this case, is not only coordinated, but also activated and made available to the substrate. In the family of proteins with heme groups, hemoglobin is a dioxygen carrier and cytochrome P-450 is an oxygenase. A similar differentiation in function is also found for hemocyanin and tyrosinase from the family of proteins with a dinuclear copper complex at the active site. Note that not all enzymes that mediate the incorporation of oxygen from O into some substrate coordinate and activate dioxygen. For example, lipoxygenase probably catalyzes the conversion of a 1,4-diene to a 1,3-diene-4-hydroxyperoxy species by activation of the organic substrate. The active site does not resemble that of any known oxygen-carrier protein. This topic is discussed more fully in . | 4,201 | 968 |
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A system, such as a container of gas, can consist of a large number of subsystems. How is the partition function of the system built up from those of the subsystems? This depends on whether the subsystems are distinguishable or indistinguishable. Let's start with energy. Energy is additive so that: \[E_\text{tot}(N,V) = \epsilon_1(V) + \epsilon_2(V) + \epsilon_3(V) + \cdots \nonumber \] Each of the molecules can have their energy distributed over their respective energy states (e.g., vibrations, rotations, etc.). This means that each \(\epsilon_i\) is already a summation over the states of the molecule. Let us assume that we can somehow distinguish all the molecules as: \(a\), \(b\), \(c\), \(d\)… and denote the energy state they are in by \(i\), \(j\), \(k\): \[E_l(N,V) = \epsilon_i^a (V) + \epsilon_j^b (V) + \epsilon_k^c (V) + \cdots \nonumber \] A good example would be the molecules in a molecular crystal. They only move around a fixed site and so we can distinguish by how far molecule '\(a\)' is from a given corner of the crystal. The systems partition function becomes: \[Q(N,V,\beta) = \sum_i e^{-\beta E_l} = \sum_{i,j,k,…} e^{-\beta (\epsilon_i^a + \epsilon_j^b + \epsilon_k^c)} \nonumber \] So far, we have done little effort to distinguish between the partition function of a molecular system \(q\) and the whole ensemble \(Q\) (e.g. the gas). If the entities that we called systems are and independent, the whole ensemble partition function is the product of the molecular system partition functions. We get: \[Q(N,V,\beta) = \prod_i{q_i} \nonumber \] for \(N\) distinguishable systems. We can split up the summation into a product of molecular partition functions: \[Q(N,V,β) = \prod_i^N q_i(V,\beta)= q_a(V,\beta) q_b(V,\beta) q_c(V,\beta) \cdots \nonumber \] This results in each molecular system partition function being summed over independently: \[ Q(N,V,\beta) = \sum_{i} e^{-\beta \epsilon_i^a} \sum_{j} e^{-\beta \epsilon_i^b} \sum_{k} e^{-\beta \epsilon_i^c} \cdots \nonumber \] If the energy states of all the particles are the same, then the equation simplifies to: \[ Q(N,V,\beta) = [q(V,\beta)]^N \nonumber \] We can do this if, for example, the particles are embedded in a crystal where we can distinguish them by their location. We will see in the next chapter that for indistinguishable particles, such as those in a gas, we get a different result. | 2,412 | 969 |
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The great value of chemical kinetics is that it can give us insights into the actual reaction pathways (mechanisms) that reactants take to form the products of reactions. Analyzing a reaction mechanism to determine the type of rate law that is consistent (or not consistent) with the specific mechanism can give us significant insight. For example, the reaction \[ A+ B \rightarrow C \nonumber \] might be proposed to follow one of two mechanistic pathways: \[ \underbrace{A + A \xrightarrow{k_1} A_2}_{\text{step 1}} \nonumber \] \[ \underbrace{ A_2 + B \xrightarrow{k_2} C}_{\text{step 2}} \nonumber \] or \[ \underbrace{A \xrightarrow{k_1} A^*}_{\text{step 1}} \nonumber \] \[ \underbrace{ A^* + B \xrightarrow{k_2} C}_{\text{step 2}} \nonumber \] The first rate law will predict that the reaction should be second order in \(A\), whereas the second mechanism predicts that it should be first order in \(A\) (in the limit that the steady state approximation, discussed in the following sections, can be applied to \(A_2\) and \(A^*\)). Based on the observed rate law being first or second order in A, one can rule out one of the rate laws. Unfortunately, this kind of analysis cannot confirm a specific mechanism. Other evidence is needed to draw such conclusions, such as the spectroscopic observation of a particular reaction intermediate that can only be formed by a specific mechanism. In order analyze mechanisms and predict rate laws, we need to build a toolbox of methods and techniques that are useful in certain limits. The next few sections will discuss this kind of analysis, specifically focusing on Each type of approximation is important in certain limits, and they are oftentimes used in conjunction with one another to predict the final forms of rate laws. | 1,798 | 970 |
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The important chemistry of alkyl halides, \(\ce{RX}\), includes the nucleophilic \(\left( S_\text{N} \right)\) displacement and elimination \(\left( E \right)\) reactions discussed in Chapter 8. Recall that alkyl halides normally are reactive in ionization \(\left( S_\text{N}1 \right)\) reactions, whereas halides, and to a lesser extent halides, are reactive in \(S_\text{N}2\) reactions, which occur by a concerted mechanism with inversion of configuration ( to 8-7). Elimination competes with substitution in many \(S_\text{N}\) reactions and can become the major pathway at high temperatures or in the presence of strong base. Elimination \(\left( E_2 \right)\), unlike displacement \(\left( S_\text{N}2 \right)\), is insensitive to steric hindrance in the alkyl halide. In fact, the \(E_2\) reactivity of alkyl halides is \(\ce{RX} >\) \(\ce{RX} >\) \(\ce{RX}\), which is opposite to their \(S_\text{N}2\) reactivity. Several useful reactions for the synthesis of alkyl halides that we already have encountered are summarized below with references to the sections that supply more detail: A summary of these and some other reactions for the synthesis of alkyl halides or organohalogen compounds is given in Table 14-5. Halogen compounds in which the carbon-halogen bond is adjacent to a double bon, as in are known as . The simplest example is 3-chloropropene, \(\ce{CH_2=CHCH_2Cl}\), which is made on a large scale by the radical chlorination of propene at \(400^\text{o}\): Most of the 3-chloropropene prepared in this manner is converted to other important compounds. For example, addition of hypochlorous acid gives a mixture of dichloropropanols, which on treatment with base gives a substance known commercially as "epichlorohydrin": The ring closure reaction with \(\ce{Ca(OH)_2}\) is an internal \(S_\text{N}2\) reaction. Hydroxide ion converts the alcohol to an alkoxide ion that acts as a nucleophile in displacing the neighboring chlorine: The epichlorohydrice so produced is used primarily to make epoxy resins (see ), although some of it is hydrolyzed to glycerol: A general method for preparing allylic halides is by addition of hydrogen halides to conjugated dienes. This reaction usually produces a mixture of 1,2- and 1,4-addition products (see ): A second general method involves the bromination of alkene with -bromosuccinimide (the reaction). A radical-chain reaction takes place between -bromosuccinimide (NBS) and alkenes, which commonly is initiated by light, peroxides, or other catalysts, and yields allylic bromides: This reaction, like the chlorination of propene, is highly selective in that the so-called allylic \(\ce{C-H}\) is attacked preferentially. From bond energies (Table 4-6) we know that the weakest \(\ce{C-H}\) bonds of propene are to the allylic hydrogens, \(\ce{H_2C=CHCH_2-H}\). Therefore, in the first step of radical-chain chlorination of propene, an allylic hydrogen is removed by a chlorine atom (Equation 14-1). The allylic \(\ce{C-H}\) bonds are weaker than the alkenic \(\ce{C-H}\) bonds because of the extra stabilization of the radical obtained on hydrogen abstraction (Equation 14-1). Two equivalent valence-bond structures (\(1a\) and \(1b\)) can be written for the 2-propenyl radical; the electron delocalization enhances the stability of the radical (see ): In the second step of the chain reaction (Equation 14-2) the propenyl radical can form a carbon-halogen bond at either end by abstracting a halogen atom from the halogenating agent: The \(\ce{Cl} \cdot\) atom produced now can participate in Reaction 14-1, thereby continuing the chain. With propene the intermediate radical gives the same product, 2-propenyl chloride, irrespective of whether a chlorine atom is transferred to the 1- or 3-carbon. However, the radical formed by removal of an allylic hydrogen from 2-butene gives a mixture of products: The carbon-halogen bonds of allylic halides are especially reactive in both \(S_\text{N}1\) and \(S_\text{N}2\) reactions (Table 14-6). The reasons for the enhanced \(S_\text{N}1\) reactivity have been discussed previously (Section 8-7B). For example, the ease with which 1-chloro-2-butene ionizes compared to 1-chlorobutane is attributed to the stability of the 2-butenyl cation, which is distributed between \(\ce{C_1}\) and \(\ce{C_3}\), and the nucleophile (water) attacks at both positions to give mixtures of products. The same results are obtained if one starts with 3-chloro-1-butene because the same cation is formed: Reactivities comparable to allylic halides are found in the nucleophile displacement reactions of benzylic halides by \(S_\text{N}1\) and \(S_\text{N}2\) mechanisms (Table 14-6). The ability of the benzylic halides to undergo \(S_\text{N}1\) reactions clearly is related to the stability of the resulting benzylic cations, the electrons of which are extensively delocalized. Thus, for phenylmethyl chloride, When the halogen substituent is located two or more carbons from the aryl group as in 2-phenylethyl bromide, \(\ce{C_6H_5CH_2CH_2Br}\), the pronounced activating effect evident in benzylic halides disappears, and the reactivity of the halides is essentially that of a primary alkyl halide (e.g., \(\ce{CH_3CH_2CH_2Br}\)). Benzylic halides can be prepared by the same radical-halogenating agents that give allylic halides from alkenes. These include \(\ce{Cl_2}\), \(\ce{Br_2}\), -bromosuccinimide ( ), \(\ce{SO_2Cl_2}\), and -butyl hypochlorite: The benzylic \(\ce{C-H}\) bond is weaker and more restrictive then primary alkane \(\ce{C-H}\) bonds because of the stabilization of benzylic radicals (see Table 4-6). and (1977) | 5,677 | 971 |
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Matter can be identified by its characteristic inertial and gravitational mass and the space that it occupies. Matter is typically commonly found in three different states: solid, liquid, and gas. A is a sample of matter whose physical and chemical properties are the same throughout the sample because the matter has a constant composition. It is common to see substances changing from one state of matter to another. To differentiate the at least at a particle level, we look at the behavior of the particles within the substance. . For example, we all have probably observed that water can exist in three forms with different characteristic ways of behaving: the solid state (ice), liquid state (water), and gaseous state (water vapor and steam). Due to water's prevalence, we use it to exemplify and describe the three different states of matter. As ice is heated and the particles of matter that make up water gain energy, eventually the ice melts in to water that eventually boils and turns into steam. Before we examine the states of matter, we will consider some ways samples of matter have been classified by those who have studied how matter behaves. Evidence suggests that substances are made up of smaller particles that are ordinarily moving around. Some of those particles of matter can be split into smaller units using fairly strong heat or electricity into smaller rather uniform bits of matter called atoms. Atoms are the building blocks of elements. Elements are all those substances that have not ever been decomposed or separated into any other substances through chemical reactions, by the application of heat, or by attempting to force an direct electric current through the sample. Atoms in turn have been found to be made up of yet smaller units of matter called electrons, protons, and neutrons. Elements can be arranged into what is called the periodic table of elements based on observed similarities in chemical and physical properties among the different elements. When atoms of two or more elements come together and bond, a compound is formed. The compound formed can later be broken down into the pure substances that originally reacted to form it. Compounds such as water are composed of smaller units of bonded atoms called molecules. Molecules of a compound are composed of the same proportion of elements as the compound as a whole since they are the smallest units of that compound. For example, every portion of a sample of water is composed of water molecules. Each water molecule contains two hydrogen atoms and one oxygen atom, and so water as a whole has, in a combined state, twice as many hydrogen atoms as oxygen atoms.. Water can still consist of the same molecules, but its physical properties may change. For instance, water at a temperature below 0° Celsius is ice, whereas water above the temperature of 100° C is a gas, water vapor. When matter changes from one state to another, temperature and pressure may be involved in the process and the density and other physical properties change. The temperature and pressure exerted on a sample of matter determines the resulting form of that the matter takes, whether solid, liquid, or gas. Since the properties of compounds and elements are uniform, they are classified as . When two or more substances are mixed together, the result is called a mixture. Mixtures can be classified into two main categories: homogeneous and heterogeneous. A mixture is one in which the composition of its constituents are uniformly mixed throughout. A homogeneous mixture in which on substance, the solute, dissolves completely in another substance, the solvent, may also be called a Usually the solvent is a liquid, however the solute can be either a liquid, solid, or a gas. In a homogeneous solution, the particles of solute are spread evenly among the solvent particles and the extremely small particles of solute cannot be separated from the solvent by filtration through filter paper because the spaces between paper fibers are much greater than the size of the solute and solvent particles. Other examples of homogeneous mixtures include sugar water, which is the mixture of sucrose and water, and gasoline, which is a mixture of dozens of compounds. A mixture is a nonuniform mixture in which the components separate and the composition varies. Unlike the homogeneous mixture, heterogeneous mixtures can be separated through physical processes. An example of a physical process used is filtration, which can easily separate the sand from the water in a sand-water mixture by using a filter paper. Some more examples of heterogeneous mixtures include salad dressing, rocks, and oil and water mixtures. Heterogeneous mixtures involving at least one fluid are also called mixtures and separate if they are left standing long enough. Consider the idea of mixing oil and water together. Regardless of the amount of time spent shaking the two together, eventually oil and water mixtures will separate with the oil rising to the top of the mixture due to its lower density. Mixtures that fall between a solution and a heterogeneous mixture are called colloidal suspensions (or just colloids). A mixture is considered colloidal if it typically does not spontaneously separate or settle out as time passes and cannot be completely separated by filtering through a typical filter paper. It turns out that a mixture is colloidal in its behavior if one or more of its dimensions of length, width, or thickness is in the range of 1-1000 nm. A colloidal mixture can also be recognized by shining a beam of light through the mixture. If the mixture is colloidal, the beam of light will be partially scattered by the suspended nanometer sized particles and can be observed by the viewer. This is known as the Tyndall effect. In the case of the Tyndall effect, some of the light is scattered since the wavelengths of light in the visible range, about 400 nm to 700 nm, are encountering suspended colloidal sized particles of about the same size. In contrast, if the beam of light were passed through a solution, the observer standing at right angles to the direction of the beam would see no light being reflected from either the solute or solvent formula units that make up the solution because the particles of solute and solvent are so much smaller than the wavelength of the visible light being directed through the solution. Most substances are naturally found as mixtures, therefore it is up to the chemist to separate them into their natural components. One way to remove a substance is through the physical property of magnetism. For example, separating a mixture of iron and sulfur could be achieved because pieces of iron would be attracted to a magnet placed into the mixture, removing the iron from the remaining sulfur. Filtration is another way to separate mixtures. Through this process, a solid is separated from a liquid by passing through a fine pored barrier such as filter paper. Sand and water can be separated through this process, in which the sand would be trapped behind the filter paper and the water would strain through. Another example of filtration would be separating coffee grounds from the liquid coffee through filter paper. Distillation is another technique to separate mixtures. By boiling a solution of a non-volatile solid disolved in a liquid in a flask, vapor from the lower boiling point solvent can be driven off from the solution by heat, be condensed back into the liquid phase as it comes in contact with cooler surfaces, and be collected in another container. Thus a solution such as this may be separated into its original components, with the solvent collected in a separate flask and the solute left behind in the original distillation flask. An example of a solution being separated through distillation would be the distillation of a solution of copper(II) sulfate in water, in which the water would be boiled away and collected and the copper(II) sulfate would remain behind in the distillation flask. Everything that is familiar to us in our daily lives - from the land we walk on, to the water we drink and the air we breathe - is based upon the called gases, liquids, and solids. When the temperature of a liquid is lowered to the freezing point of the substance (for water the freezing point is 0 C), the movement of the particles slows with the spacing between the particles changing until the attractions between the particles lock the particles into a solid form. At the freezing point, the particles are closely packed together and tend to block the motions of each other. The attractions between the particles hold the particles tightly together so that the entire ensemble of particles takes on a fixed shape. The volume of the solid is constant and the shape of a solid is constant unless deformed by a sufficiently strong external force. (Solids are thus unlike liquids whose particles are slightly less attracted to one another because the particles of a liquid are a bit further apart than those in the corresponding solid form of the same substance.) In a solid the particles remain in a relatively fixed positions but continue to vibrate. The vibrating particles in a solid do not completely stop moving and can slowly move into any voids that exist within the solid. When the temperature of a sample increases above the melting point of a solid, that sample can be found in the liquid state of matter. The particles in the liquid state are much closer together than those in the gaseous state, and still have a quite an attraction for each other as is apparent when droplets of liquid form. In this state, the weak attractive forces within the liquid are unable to hold the particles into a mass with a definite shape. Thus a liquid's shape takes on the shape of any particular container that holds it. A liquid has a definite volume but not a definite shape. Compared to to the gaseous state there is less freedom of particle movement in the liquid state since the moving particles frequently are colliding with one another, and slip and slide over one another as a result of the attractive forces that still exist between the particles, and hold the particles of the liquid loosely together. At a given temperature the volume of the liquid is constant and its volume typically only varies slightly with changes in temperature. In the gas phase, matter does not have a fixed volume or shape. This occurs because the molecules are widely separated with the spaces between the particles typically around ten times further apart in all three spatial directions, making the gas around 1000 times less dense than the corresponding liquid phase at the same temperature. (A phase is a uniform portion of mater.) As the temperature of a gas is increased, the particles to separate further from each other and move at faster speeds. The particles in a gas move in a rather random and independent fashion, bouncing off each other and the walls of the container. Being so far apart from one another, the particles of a real gas only weakly attract each other such that the gas has no ability to have a shape of its own. The extremely weak forces acting between the particles in a gas and the greater amount of space for the particles to move in results in almost independent motion of the moving, colliding particles. The particles freely range within any container in which they are put, filling its entire volume with the net result that the sides of the container determine the shape and volume of gas. If the container has an opening, the particles heading in the direction of the opening will escape with the result that the gas as a whole slowly flows out of the container. Besides of the three classical states of matter, there are many other states of matter that share characteristics of one more of the classical states of matter. Most of these states of matter can be put into three categories according to the degrees in varying temperature. At room temperature, the states of matters include liquid crystal, amorphous solid, and magnetically ordered states. At low temperatures the states of matter include superconductors, superfluids, and Bose-Einstein condensate state of matter. At high temperatures the states of matter include, plasma and Quark-gluon plasma. These other states of matter are not typically studied in general chemistry. | 12,422 | 972 |
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A unit cell is the most basic and least volume consuming repeating structure of any solid. It is used to visually simplify the crystalline patterns solids arrange themselves in. When the unit cell repeats itself, the network is called a . The work of Auguste Bravais in the early 19th century revealed that there are only fourteen different lattice structures (often referred to as ). These fourteen different structures are derived from seven , which indicate the different shapes a unit cell take and four , which tells how the atoms are arranged within the unit. The kind of cell one would pick for any solid would be dependent upon how the latices are arranged on top of one another. A method called is used to determine how the crystal is arranged. X-ray Diffraction consists of a X-ray beam being fired at a solid, and from the diffraction of the beams calculated by the configuration can be determined. The unit cell has a number of shapes, depending on the angles between the cell edges and the relative lengths of the edges. It is the basic building block of a crystal with a special arrangement of atoms. The unit cell of a crystal can be completely specified by three vectors, a, b, c that form the edges of a parallelepiped. A crystal structure and symmetry is also considered very important because it takes a role in finding a cleavage, an electronic band structure, and an optical property. There are seven crystal systems that atoms can pack together to produce 3D space lattice. The unit cell is generally chosen so as to contain only complete complement of the asymmetric units. In a geometrical arrangement, the crystal systems are made of three set of (\(a\), \(b\), \(c\)), which are coincident with unit cell edges and lengths. The lengths of a, b, c are called the unit cell dimensions, and their directions define the major crystallographic axes. A unit cell can be defined by specifying a, b, c, or alternately by specifying the lengths |a|, |b|, |c| and the angles between the vectors, \(\alpha\), \(\beta\), and \(\gamma\) as shown in Fig. 1.1. Unit cell have lower or higher symmetry than the aggregate of asymmetric units. There are seven crystal systems and particular kind of unit cell chosen will be determined by the type of symmetry elements relating the asymmetric units in the cell. The unit cell is chosen to contain only one complete complement of the asymmetric units, which is called primitive (P). Unit cells that contain an asymmetric unit greater than one set are called centered or nonprimitive unit cells. The additional asymmetric unit sets are related to the first simple fractions of unit cells edges. For example, (1/2, 1/2, 1/2) for the body centered cell \(I\) and (1/2,1/2, 0) for the single-face-centered cell \(C\). The units can be completely specified by three vectors (a, b, c) and the lengths of vectors in angstroms are called the unit cell dimensions. Vectors directions are defined the major crystallographic axes. Unit cell can also be defied by specifying the lengths (|a|, |b|, |c|) and the angles between the vectors (\(\alpha\), \(\beta\), and \(\gamma\)) as shown in Fig.1.1. This table describes the fourteen different kind of unit cells available. As you can see not every crystal systems can have all the different types of lattices. Calculating the volume for a unit cell is the same as calculating the volume for any prism - base area multiplied by height. The equations for each different crystal system are as follow: Most calculations involving unit cells can be solved with the formula: density = Mass/Volume. Then in addition to the obvious three the number of particles per cell can also be calculated by the density/molar mass. It can easily be seen that not all the particles are complete in the unit cell form. For the fractional particles in unit cell, its will always sum to one whole particle, its (for face centered lattices) will sum to three whole particles, and for the will sum to one. | 4,007 | 974 |
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Rules are algorithms, by which we generate possible quantum numbers. The lowest value of is 1 (NOT zero). For = 1, the only possible value for quantum number is 0, and = 0. Each set of quantum numbers is called a . Thus, for = 1, there is only one state (1,0,0). The states are represented by symbols, and special symbols have been used to represent the quantum number as follows: Using symbols, the valid quantum states can be listed in the following manner: 1s
2s 2p
3s 3p 3d
4s 4p 4d 4f
5s 5p 5d 5f 5g
6s 6p 6d 6f 6g 7h
7s 7p 7d 7f 7g 7h 8i For hydrogen-like atoms, that is atoms or ions with one electron, the energy level is solely determined by the principle quantum number n, and the energy levels of the subshells np and nd etc. are the same as the ns. For these species, the energy levels have the reverse order as the list given earlier: Hint: There are nine (9) possible orbitals.
3s, 3 3p, 5 3d; total = 1 + 3 + 5 = ? Hint: There are 9 such orbitals.
For 5g, = 4; = -4, -3, -2, -1, 0, 1, 2, 3, 4. A total of 9. Hint: A total of 14 electrons.
Each of the 7 4 orbitals accommodates a pair of electrons. There are 14 elements in the lanthanides, the 4 block elements. Hint: There are 5 3d orbitals.
The magnetic q.n. = -2, 1, 0, 1, 2. The number of orbitals associated with a given value of is (2 + 1). Hint: The symbol is 3d.
For symbols, consider the following: | 1,440 | 976 |
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are also known as " There are three types of : single, double, and triple. The name "Non-singular covalent bonds" speaks for itself. Non-singular covalent bonds are covalent bonds that need to share more then one electron pair, so they create double and triple bonds. The main motive an atom has to bond with other atoms is to fulfill it's need to have eight valence shell electrons (with some , i.e. Hydrogen). This defines the octet rule. Two different orbital overlaps occur with multiple bonds. The difference between single bonds and multiple bonds is that multiple bonds have one or two pi bonds (one for a double and two for a triple) in addition to the sigma bonds that a single bond creates. The sigma bond with the pi bond is what makes double and triple bonds so strong compared to single bonds. The more bonds there are means there is more overlap between the orbitals. Bond length is also effected by the overlap of the two orbitals, the more overlap the shorter the bond length. A Sigma bond "σ" is the strongest chemical covalent bond. It is created by the "end-to-end" overlap of atomic orbitals. Going more in depth, it is in which the region of electron sharing is along the imaginary line which connects the bonded atoms. They can be formed from two s-orbitals, two p-orbitals, one s in the region of space within the sigma bond. The sigma bond is symmetric and can freely rotate around the bond axis. Pi Bonds " " are created by the "side-to-side" overlapping of two parallel p-orbitals (pictured below). A pi bond is a weaker chemical covalent bond than a sigma bond , thus double and triple bonds are stronger then single bonds.The pi bond looks like two macaroni's sandwiching the sigma bond. Above and below the Pi bonds molecular plane has high electron charge densities. Electrons in a pi bond are sometimes referred to as "Pi electrons." The pi bond is a region of space where you can find the two pi electrons that create the bond. Pi bonds create since they prevent free rotation around the bond. Example of Pi bond formation with ethylene, These are where four lobes of one involved electron orbital overlap four lobes of the other involved electron orbital. Of the orbital's node planes, two (and no more), go through both atoms. The Greek letter in their name refers to d orbitals, since the orbital symmetry of the delta bond is the same as that of the usual (4-lobed) type of d orbital when seen down the bond axis. In sufficiently-large atoms, occupied d-orbitals are low enough in energy to participate in bonding. Delta bonds are usually observed in organometallic species. Some ruthenium and molybdenum compounds contain a quadruple bond, which can only be explained by invoking the delta bond. It is possible to excite electrons in acetylene from lower-energy nonbonding orbitals to form a delta bond between the two carbon triple bonds. This is because the orbital symmetry of the pi antibonding orbital is the same as that of the delta bond. Theoretical chemists have conjectured that higher-order bonds ( and , corresponding to overlap of f and g orbitals) are possible, with even more overlapping lobes of their component atomic orbitals, but no experimental evidence for these has yet been observed. 1) 2 electrons 2) 5 sigma bonds and 1 pi bond (-) sigma bonds (=) one sigma and one pi bond 3) A compound with one pi bond would have a longer bond length since it is a double bond. | 3,453 | 977 |
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Gases that deviate from ideality are known as Real Gases, which originate from two factors: (1) First, the theory assumes that as pressure increases, the volume of a gas becomes very small and approaches zero. While it does approach a small number, it will not be zero because molecules do occupy space (i.e. have volume) and cannot be compressed. (2) Intermolecular forces do exist in gases. These become increasingly important in low temperatures, when translational (definition of translational, please) molecular motion slows down, almost to a halt. However, at high temperatures, or even normal, every day temperatures, the intermolecular forces are very small and tend to be considered negligible. Jim Clark ( ) | 729 | 978 |
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The harmonic oscillator is a model which has several important applications in both classical and quantum mechanics. It serves as a prototype in the mathematical treatment of such diverse phenomena as elasticity, acoustics, AC circuits, molecular and crystal vibrations, electromagnetic fields and optical properties of matter. A simple realization of the harmonic oscillator in classical mechanics is a particle which is acted upon by a restoring force proportional to its displacement from its equilibrium position. Considering motion in one dimension, this means \[ F = −kx \label{1}\] Such a force might originate from a spring which obeys Hooke’s law, as shown in Figure \(\Page {1}\). According to Hooke’s law, which applies to real springs for sufficiently small displacements, the restoring force is proportional to the displacement—either stretching or compression—from the equilibrium position. The \(k\) is a measure of the stiffness of the spring. The variable \(x\) is chosen equal to zero at the equilibrium position, positive for stretching, negative for compression. The negative sign in Equation \(\ref{1}\) reflects the fact that \(F\) is a force, always in the opposite sense to the displacement \(x\). Applying Newton’s second law to the force from Equation \(\ref{1}\), we find \(x\) \[ F = m \dfrac{d^2 x}{dx^2} = -kx \label{2} \] where \(m\) is the mass of the body attached to the spring, which is itself assumed massless. This leads to a differential equation of familiar form, although with different variables: \[ \ddot{x}(t)+ \omega^2x(t)= 0 \label{3}\] with \[\omega^2 \equiv \dfrac{k}{m}\] The dot notation (introduced by Newton himself) is used in place of primes when the independent variable is time. The general solution to Equation \(\ref{3}\) is \[ x(t) = A\sin ωt + B\cos ωt \label{4}\] which represents periodic motion with a sinusoidal time dependence. This is known as and the corresponding system is known as a . The oscillation occurs with a constant angular frequency \[ \omega = \sqrt{\dfrac{k}{m}}\; \text{radians per second} \label{5} \] This is called the of the oscillator. The corresponding in Hertz (cycles per second) is \[ \nu = \dfrac{\omega}{2\pi } = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{m}}\; \text{Hz} \label{6}\] The general relation between force and potential energy in a conservative system in one dimension is \[ F =\dfrac{−dV}{dx} \label{7}\] Thus the potential energy of a harmonic oscillator is given by \[ V(x) = \dfrac{1}{2}kx^2 \label{8}\] which has the shape of a parabola, as drawn in Figure \(\Page {2}\). A simple computation shows that the oscillator moves between positive and negative turning points \(\pm x_{max}\) where the total energy \(E\) equals the potential energy \(\dfrac{1}{2} k x_{max}^{2}\) while the kinetic energy is momentarily zero. In contrast, when the oscillator moves past \(x = 0\), the kinetic energy reaches its maximum value while the potential energy equals zero. Given the potential energy in Equation \(\ref{8}\), we can write down the Schrödinger equation for the one-dimensional harmonic oscillator: \[ -\dfrac{\hbar^{2}}{2m} \psi''(x) + \dfrac{1}{2}kx^2 \psi(x) = E \psi(x) \label{9}\] For the first time we encounter a differential equation with coefficients, which is a much greater challenge to solve. We can combine the constants in Equation \(\ref{9}\) to two parameters \[ \alpha^2 = \dfrac{mk}{\hbar^2}\] and \[\lambda = \dfrac{2mE}{\hbar^2\alpha} \label{10}\] and redefine the independent variable as \[ \xi = \alpha^{1/2}x \label{11} \] This reduces the Schrödinger equation to \[ \psi''(\xi) + (\lambda-\xi^2)\psi(\xi) = 0\label{12} \] The range of the variable \(x\) (also \(\xi\)) must be taken from \(−\infty\) to \(+\infty\), there being no finite cutoff as in the case of the particle in a box. A useful first step is to determine the asymptotic solution to Equation \(\ref{11}\), that is, the form of \(\psi(\xi)\) as \(\xi\rightarrow\pm\infty\). For sufficiently large values of \(\lvert\xi\rvert\), \(\xi^{2} \gg \lambda\) and the differential equation is approximated by \[ \psi''(\xi) - \xi^2\psi(\xi) \approx 0 \label{13}\] This suggests the following manipulation: \[ \left(\dfrac{d^2}{d\xi^2} - \xi^2 \right) \psi(\xi) \approx \left( \dfrac{d}{d\xi}-\xi \right) \left( \dfrac{d}{d\xi}+\xi \right) \psi(\xi) \approx 0 \label{14}\] The first-order differential equation \[ \psi'(\xi) + \xi\psi(\xi)=0 \label{15}\] can be solved exactly to give \[ \psi(\xi) = \text{const.}\, e^{-\xi^2/2} \label{16}\] Remarkably, this turns out to be an solution of the Schrödinger equation (Equation \(\ref{12}\)) with \(\lambda=1\). Using Equation \(\ref{10}\), this corresponds to an energy \[ E=\dfrac{\lambda\hbar^2\alpha}{2m} = \dfrac{1}{2}\hbar\sqrt{\dfrac{k}{m}} = \dfrac{1}{2} \hbar\omega \label{17} \] where \(\omega\) is the natural frequency of the oscillator according to classical mechanics. The function in Equation \(\ref{16}\) has the form of a Gaussian, the bell-shaped curve so beloved in the social sciences. The function has no nodes, which leads us to conclude that this represents the ground state of the system.The ground state is usually designated with the quantum number \(n = 0\) (the particle in a box is a exception, with \(n = 1\) labeling the ground state). Reverting to the original variable \(x\), we write \[ \psi_{0}(x) = \text{const} e^{-\alpha x^2/2}\] with \[\alpha=(mk/\hbar^2)^{1/2} \label{18}\] With help of the well-known definite integral (Laplace 1778) \[\int^{\infty}_{-\infty} e^{- \alpha x^{2}} dx= \sqrt{\dfrac{\pi}{\alpha}} \label{19}\] we find the normalized eigenfunction \[\psi_{0}(x)=(\dfrac{\alpha}{\pi})^{1/4} e^{-\alpha x^{2}/2} \label{20}\] with the corresponding eigenvalue \[E_{0}=\dfrac{1}{2}\hbar\omega \label{21}\] Drawing from our experience with the particle in a box, we might surmise that the first excited state of the harmonic oscillator would be a function similar to Equation \(\ref{20}\), but with a node at \(x=0\), say, \[\psi_{1}(x)=const x e^{-\alpha x^{2}/2} \label{22}\] This is orthogonal to \(\psi_0(x)\) by symmetry and is indeed an eigenfunction with the eigenvalue \[E_{1}=\dfrac{3}{2}\hbar\omega \label{23}\] Continuing the process, we try a function with two nodes \[\psi_{2}= const (x^{2}-a) e^{-\alpha x^{2}/2} \label{24}\] Using the integrals tabulated in the Supplement 5, on Gaussian Integrals, we determine that with \(a=\dfrac{1}{2}\) makes \(\psi_{2}(x)\) orthogonal to \(\psi_{0}(x)\) and \(\psi_{1}(x)\). We verify that this is another eigenfunction, corresponding to \[E_{2}=\dfrac{5}{2}\hbar\omega \label{25}\] The general result, which follows from a more advanced mathematical analysis, gives the following formula for the normalized eigenfunctions: \[\psi_{n}(x)=(\dfrac{\sqrt{\alpha}}{2^{n}n!\sqrt{\pi}})^{1/2} H_{n}(\sqrt{\alpha}x) e^{-\alpha x^{2}/2} \label{26}\] where \(H_{n}(\xi)\) represents the Hermite polynomial of degree \(n\). The first few Hermite polynomials are \[H_{0}(\xi)=1\] \[H_{1}(\xi)=2\xi\] \[H_{2}(\xi)=4\xi^{2}-2\] \[H_{3}(\xi)=8\xi^{3}-12\xi \label{27}\] The four lowest harmonic-oscillator eigenfunctions are plotted in Figure \(\Page {3}\). Note the topological resemblance to the corresponding particle-in-a-box eigenfunctions. The eigenvalues are given by the simple formula \[E_{n}=\left(n+\dfrac{1}{2}\right)\hbar\omega \label{28}\] These are drawn in Figure \(\Page {2}\), on the same scale as the potential energy. The ground-state energy \(E_{0}=\dfrac{1}{2}\hbar\omega\) is greater than the classical value of zero, again a consequence of the uncertainty principle. This means that the oscillator is always oscillating. It is remarkable that the difference between successive energy eigenvalues has a constant value \[\Delta E=E_{n+1}-E_{n}=\hbar\omega=h\nu \label{29}\] This is reminiscent of Planck’s formula for the energy of a photon. It comes as no surprise then that the quantum theory of radiation has the structure of an assembly of oscillators, with each oscillator representing a mode of electromagnetic waves of a specified frequency. (Professor Emeritus of Chemistry and Physics at the , ) | 8,179 | 979 |
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The existence of promethium (for the Greek god, Promethius) was predicted in 1912 when Henry Moseley developed an x-ray method for determining integer atomic numbers of elements. An element was clearly missing between neodymium and samarium. It's existence was not confirmed until 1947 by Marinsky, Glendenin and Coryell. Historical claims for the discovery of element 61 create an interesting trail from around 1925 in Florence (suggested name: florentium) to America in 1926 (suggested name: illinium). None of the claims, however, could be substantiated and today we know they were not simply a result of fleetingly small samples but rather poor work. While spectral lines of promethium are evident in the light from some stars, it now seems apparent that no promethium is found in accessible areas of the earth--hence the difficulty in finding any! Initial attempts at synthesis of element 61 in a cyclotron at Ohio State University in 1941 led to the suggested name cyclonium. But the recognized synthesis and identification finally came at Oak Ridge in 1947. The longest-lived isotope of promethium is Pm-145 with a half-life of 17.7 years. There are no significant commercial uses of the metal and so very little has been produced except for theoretical studies. | 1,289 | 981 |
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Amino acids are the building blocks of the polyamide structures of peptides and proteins. Each amino acid is linked to another by an amide (or peptide) bond formed between the \(\ce{NH_2}\) group of one and the \(\ce{CO_2H}\) group of the other: In this manner a polymeric structure of repeating amide links is built into a chain or ring. The amide groups are planar and configuration about the \(\ce{C-N}\) bond is usually, but not always, trans ( ). The pattern of bonds in a peptide or protein is called its : The distinction between a protein and a peptide is not completely clear. One arbitrary choice is to call proteins only those substances with molecular weights greater than 5000. The distinction might also be made in terms of differences in physical properties, particularly hydration and conformation. Thus proteins, in contrast to peptides, have very long chains that are coiled and folded in particular ways, with water molecules filling the voids in the coils and folds. Hydrogen bonding between the amide groups plays a decisive role in holding the chains in juxtaposition to one another, in what is sometimes called the and .\(^5\) Under the influence of heat, organic solvents, salts, and so on, protein molecules undergo changes, often irreversibly, called . The conformations of the chains and the degree of hydration are thereby altered, with the result that solubility and ability to crystallize decreases. Most importantly, the physiological properties of the protein usually are destroyed permanently on denaturation. Therefore, if a synthesis of a protein is planned, it would be necessary to duplicate not only the amino-acid sequences but also the exact conformations of the chains and the manner of hydration characteristic of the native protein. With peptides, the chemical and physiological properties of natural and synthetic materials usually are identical, provided the synthesis duplicates all of the structural and configurational elements. What this means is that a peptide automatically assumes the secondary and tertiary structure characteristic of the native peptide on crystallization or dissolution in solvents. Representation of peptide structures of any length with conventional structural formulas is cumbersome. As a result, abbreviations are universally used that employ three-letter symbols for the component amino acids. It is important that you know the conventions for these abbreviations. The two possible dipeptides made up of one glycien and one alanine are Notice that in the conventions used for names and abbreviated formulas the amino acid with the free amino group ( ) always is written on the . The amino acid with the free carboxyl group ( ) always is written on the . The dash between the three-letter abbreviations for the acids designates that they are linked together by an bond. The general procedure for determining the primary structure of a peptide or protein consists of three main steps. First, the number and kind of amino-acid units in the primary structure must be determined. Second, the amino acids at the ends of the chains are identified, and third, the sequence of the component amino acids in the chains is determined. The amino-acid composition usually is obtained by complete acid hydrolysis of the peptide into its component amino acids and analysis of the mixture by ion-exchange chromatography ( ). This procedure is complicated by the fact that tryptophan is destroyed under acidic conditions. Also, asparagine and glutamine are converted to aspartic and glutamic acids, respectively. Determination of the \(\ce{N}\)-terminal acid in the peptide can be made by treatment of the peptide with 2,4-dinitrofluorobenzene, a substance very reactive in nucleophilic displacements with amines but not amides (see ). The product is an \(\ce{N}\)-2,4-dinitrophenyl derivative of the peptide which, after hydrolysis of the amide linkages, produces an \(\ce{N}\)-2,4-dinitrophenylamino acid: These amino-acid derivatives can be separated from the ordinary amino acids resulting from hydrolysis of the peptide because the low basicity of the 2,4-dinitrophenyl-substituted nitrogen ( ) greatly reduces the solubility of the compound in acid solution and alters its chromatographic behavior. The main disadvantage to the method is that the entire peptide must be destroyed in order to identify the one \(\ce{N}\)-terminal acid. A related and more sensitive method makes a sulfonamide of the terminal \(\ce{NH_2}\) group with a reagent called "dansyl chloride". As with 2,4-dinitrofluorobenzene, the peptide must be destroyed by hydrolysis to release the \(\ce{N}\)-sulfonated amino acid, which can be identified spectroscopically in microgram amounts: A powerful method of sequencing a peptide from the \(\ce{N}\)-terminal end is the in which phenyl isothiocyanate, \(\ce{C_6H_5N=C=S}\), reacts selectively with the terminal amino acid under mildly basic conditions. If the reaction mixture is then acidified, the terminal amino acid is cleaved from the peptide as a cyclic , \(8\): There are simple reagents that react selectively with the carboxyl terminus of a peptide, but they have not proved as generally useful for analysis of the \(\ce{C}\)-terminal amino acids as has the enzyme . This enzyme catalyzes the hydrolysis of the peptide bond connecting the amino acid with the terminal carboxyl groups to the rest of the peptide. Thus the amino acids at the carboxyl end will be removed one by one through the action of the enzyme. Provided that appropriate corrections are made for different rates of hydrolysis of peptide bonds for different amino acids at the carboxyl end of the peptide, the sequence of up to five or six amino acids in the peptide can be deduced from the order of their release by carboxypeptidase. Thus a sequence such as peptide-Ser-Leu-Tyr could be established by observing that carboxypeptidase releases amino acids from the peptide in the order Tyr, Leu, Ser: Determining the amino-acid sequences of large peptides and proteins is very difficult. Although the Edman degradation and even carboxypeptidase can be used to completely sequence small peptides, they cannot be applied successfully to peptide chains with several hundred amino acid units. Success has been obtained with long peptide chains by employing reagents, often enzymes, to selectively cleave certain peptide bonds. In this way the chain can be broken down into several smaller peptides that can be separated and sequenced. The problem then is to determine the sequence of these small peptides in the original structure. To do this, alternative procedures for selective cleavages are carried out that produce different sets of smaller peptides. It is not usually necessary to sequence completely all of the peptide sets. The overall amino-acid composition and the respective end groups of each peptide may suffice to show overlapping sequences from which the complete amino-acid sequence logically can be deduced. The best way to show you how the overlap method of peptide sequencing works is by a specific example. In this example, we will illustrate the use of the two most commonly used enzymes for selective peptide cleavage. One is , a proteolytic enzyme of the pancreas (MW 24,000) that selectively catalyzes the hydrolysis of the peptide bonds of amino acids, lysine and arginine. Cleavage occurs on the of lysine or arginine: is a proteolytic enzyme of the pancreas (MW 24,500) that catalyzes the hydrolysis of peptide bonds to the aromatic amino acids, tyrosine, tryptophan, and phenylalanine, more rapidly than to other amino acids. Cleavage occurs on the of the aromatic amino acid: Our example is the sequencing of a peptide (P) derived from partial hydrolysis of a protein which, on complete acid hydrolysis, gave Ala, 3 Gly, Glu, His, 3 Lys, Phe, Tyr, 2 Val, and one molar equivalent of ammonia. 1. Treatment of the peptide (P) with carboxypeptidase released alanine, and with 2,4-dinitrofluorobenzene followed by hydrolysis gave the 2,4-dinitrophenyl derivative of valine. These results establish the \(\ce{N}\)-terminus as valine and the \(\ce{C}\)-terminus as alanine. The known structural elements now are
2. Partial hydrolysis of the peptide (P) with trypsin gave a hexapeptide, a tetrapeptide, a dipeptide, and one molar equivalent of lysine. The peptides, which we will designated respectively as M, N, and O, were sequenced by Edman degradation and found to have structures: \[\begin{array}{ll} \text{Gly}-\text{Ala} & \text{O} \\ \text{Val}-\text{Tyr}-\text{Glu}-\text{Lys} & \text{N} \\ \text{Val}-\text{Gly}-\text{Phe}-\text{Gly}-\text{His}-\text{Lys} & \text{M} \end{array}\] With this information, four possible structures can be written for the original peptide P that are consistent with the known end groups and the fact that trypsin cleaves the peptide P on the side of the lysine unit. Thus \[\begin{array}{cc} \text{N}-\text{M}-\text{Lys}-\text{O} & \text{M}-\text{N}-\text{Lys}-\text{O} \\ \text{N}-\text{Lys}-\text{M}-\text{O} & \text{M}-\text{Lys}-\text{N}-\text{O} \end{array}\] 3. Partial hydrolysis of the peptide P using chymotrypsin as catalyst gave three peptides, X, Y, and Z. These were not sequenced, but their amino-acid composition was determined: \[\begin{array}{ll} \text{Gly, Phe, Val} & \text{X} \\ \text{Gly, His, Lys, Tyr, Val} & \text{Y} \\ \text{Ala, Glu, Gly, 2 Lys} & \text{Z} \end{array}\] This information can be used to decide which of the alternative structures deduced above is correct. Chymotrypsin cleaves the peptide on the carboxyl side of the phenylalanine and tyrosine units. Only peptide M contains Phe, and if we compare M with the compositions of X, Y, and Z, we see that only X and Y overlap with M. Peptide Z contains the only Ala unit and must be the \(\ce{C}\)-terminus. If we put together these pieces to get a peptide, P' (which differs from P by not having the nitrogen corresponding to the ammonia formed on complete hydrolysis) then P' must have the structure X-Y-Z: This may not be completely clear, and it will be well to consider the logic in some detail. Peptides M and N both have \(\ce{N}\)-terminal valines, and one of them must be the \(\ce{N}\)-terminal unit. Peptide M overlaps with X and Y, and because X and Y are produced by a cleavage on the side of Phe, the X and Y units have to be connected in the order X-Y. Because the other Val is in Y, the \(\ce{N}\)-terminus must be M. This narrows the possibilities to \[\begin{array}{c} \text{M}-\text{N}-\text{Lys}-\text{O} \\ \text{M}-\text{Lys}-\text{N}-\text{O} \end{array}\] There are two Lys units in Z, and this means that only the sequence M-N-Lys-O is consistent with the sequence X-Y-Z, as shown: The final piece of the puzzle is the placement of the mole of ammonia released from the original peptide on acid hydrolysis. The ammonia comes from a primary amide function: \[\ce{R-CONH_2} \overset{\ce{H_3O}^\oplus}{\longrightarrow} \ce{RCO_2H} + \ce{NH_4^-}\] The amide group cannot be at the \(\ce{C}\)-terminus because the peptide would then be inert to carboxypeptidase. The only other possible place is on the side-chain carboxyl of glutamic acid. The complete structure may be written as Using procedures such as those outlined in this section more than 100 proteins have been sequenced. This is an impressive accomplishment considering the complexity and size of many of these molecules (see, for example, Table 25-3). It has been little more than two decades since the first amino acid sequence of a protein was reported by F. Sanger, who determined the primary structure of insulin (1953). This work remains a landmark in the history of chemistry because it established for the first time that proteins have definite primary structures in the same way that other organic molecules do. Up until that time, the concept of definite primary structures for proteins was openly questioned. Sanger developed the method of analysis for \(\ce{N}\)-terminal amino acids using 2,4-dinitrofluorobenzene and received a Nobel Prize in 1958 for his success in determining the amino-acid sequence of insulin. The problems involved in peptide syntheses are of much practical importance and have received considerable attention. The major difficulty in putting together a chain of say 100 amino acids in a particular order is one of overall yield. At least 100 separate synthetic steps would be required, and, if the yield in each step were equal to \(n \times 100\%\), the overall yield would be \(\left( n^{100} \times 100\% \right)\). If the yield in each step were \(90\%\), the overall yield would be only \(0.003\%\). Obviously, a practical laboratory synthesis of a peptide chain must be a highly efficient process. The extraordinary ability of living cells to achieve syntheses of this nature, not of just one but of a wide variety of such substances, is truly impressive. Several methods for the formation of amide bonds have been discussed in and . The most general reaction is shown below, in which X is some reactive leaving group (see Table 24-1): When applied to coupling two different amino acids, difficulty is to be expected because these same reactions can link two amino acids in a total of four different ways. Thus if we started with a mixture of glycine and alanine, we could generate for dipeptides, Gly-Ala, Ala-Gly, Gly-Gly, and Ala-Ala. To avoid unwanted coupling reactions a protecting group is substituted on the amino function of the acid that is to act as the acylating agent. Furthermore, all of the amino, hydroxyl, and thiol functions that may be acylated to give undesired products usually must be protected. For instance, to synthesize Gly-Ala free of other possible dipeptides, we would have to protect the amino group of glycine and the carboxyl group of alanine: Some methods of protecting amine and hydroxyl functions were discussed previously in and , respectively. A summary of some commonly used protecting groups for \(\ce{NH_2}\), \(\ce{OH}\), \(\ce{SH}\), and \(\ce{CO_2H}\) functions is in Table 25-2, together with the conditions by which the protecting groups may be removed. The best protecting groups for \(\ce{NH_2}\) functions are phenylmethoxycarbonyl (benzyloxycarbonyl) and -butoxycarbonyl. Both groups can be removed by treatment with acid, although the -butoxycarbonyl group is more reactive. The phenylmethoxycarbonyl group can be removed by reduction with either hydrogen over a metal catalyst or with sodium in liquid ammonia. This method is most useful when, in the removal step, it is necessary to avoid treatment with acid: In most cases, formation of the ethyl ester provides a satisfactory protecting group for the carboxyl function. Conversion of the carboxyl group to a more reactive group and coupling are key steps in peptide synthesis. The coupling reaction must occur readily and quantitatively, and with a minimum of racemization of the chiral centers in the molecule. This last criterion is the Achilles' heel of many possible coupling sequences. The importance of nonracemization can best be appreciated by an example. Consider synthesis of a tripeptide from three protected \(L\)-amino acids, A, B, and C, in two sequential coupling steps, \(\text{C} \overset{\text{B}}{\rightarrow} \text{B}-\text{C} \overset{\text{A}}{\rightarrow} \text{A}-\text{B}-\text{C}\). Suppose that the coupling yield is quantitative, but there is \(20\%\) formation of the \(D\) isomer in the component in each coupling step. Then the tripeptide will consist of a mixture of four diastereomers, only \(64\%\) of which will be the desired \(L\),\(L\),\(L\) diastereomer (Equation 25-7): \(\tag{25-7}\) This is clearly unacceptable, especially for longer-chain peptides. Nine coupling steps with \(20\%\) of the wrong isomer formed in each would give only \(13\%\) of the decapeptide with the correct stereochemistry. The most frequently used carboxyl derivatives in amide coupling are azides, \(\ce{RCO-N_3}\), mixed anhydrides, \(\ce{RCO-O-COR'}\), and esters of moderately acidic phenols, \(\ce{RCO-OAr}\) (see Table 24-1). It also is possible to couple free acid with an amine group using a diimide, \(\ce{R-N=C=N-R}\), most frequently \(\ce{N}\),\(\ce{N'}\)-dicyclohexylcarbodiimide. The diimide reagent may be thought of as a dehydrating agent. The "elements of water" eliminated in the coupling are consumed by the diimide to form a substituted urea. The overall reaction is This reaction takes place because diimides, \(\ce{-N=C=N}-\), have reactive cumulated double-bond systems like those of ketenes, \(\ce{-C=C=O}\); isocyanates, \(\ce{-N=C=O}\); and isothiocyanates, \(\ce{-N=C=S}\); and are susceptible to nucleophilic attack at the central carbon. In the first step of the diimide-coupling reaction, the carboxyl function adds to the imide to give an acyl intermediate, \(9\). This intermediate is an activated carboxyl derivative \(\ce{RCO-X}\) and is much more reactive toward an amino function than is the parent acid. The second step therefore is the aminolysis of \(9\) to give the coupled product and \(\ce{N}\),\(\ce{N'}\)-dicyclohexylurea: After completion of a coupling reaction, and before another amino acid can be added to the \(\ce{N}\)-terminus, it is necessary to remove the protecting group. This must be done by selective reactions that do not destroy the peptide bonds or side-chain protecting groups. This part of peptide synthesis is discussed in , and some reactions useful for removal of the \(\ce{N}\)-terminal protecting groups are summarized in Table 25-2. In spite of the large number of independent steps involved in the synthesis of even small peptides, each with its attendant problems of yield, racemization, and selectivity, remarkable success has been achieved in the synthesis of large peptides and certain of the smaller peptides. The synthesis of insulin (Figure 25-8) with its 51 amino acid units and 3-disulfide bridges has been achieved by several investigators. Several important hormonal peptides, namely glutathione, oxytocin, vasopressin, and thyrotropic hormone (see Figure 25-9) have been synthesized. A major accomplishment has been the synthesis of an enzyme with ribonuclease activity reported independently by two groups of investigators, led by R. Hirschman (Merck) and R. B. Merrifield (Rockefeller University). This enzyme is one of the simpler proteins, having a linear stricture of 124 amino-acid residues. It is like a peptide, not a protein, in that it assumes the appropriate secondary and tertiary structure without biochemical intervention ( ). As a specific example of the strategy involved in peptide synthesis, the stepwise synthesis of oxytocin is outlined in Figure 25-10, using the abbreviated notation in common usage. not The overall yield in a multistep synthesis of a peptide of even modest size is very poor unless each step can be carried out very efficiently. An elegant modification of classical peptide synthesis has been developed by R. B. Merrifield, which offers improved yields by minimizing manipulative losses that normally attend each step of a multistage synthesis. The key innovation is to anchor the \(\ce{C}\)-terminal amino acid to an insoluble support, and then add amino-acid units by the methods used for solution syntheses. After the desired sequence of amino acids has been achieved, the peptide can be cleaved from the support and recovered from solution. All the reactions involved in the synthesis must, of course, be brought to essentially \(100\%\) completion so that a homogeneous product can be obtained. The advantage of having the peptide anchored to a solid support is that laborious purification steps are virtually eliminated; solid material is purified simply by washing and filtering without transferring the material from one container to another. The method has become known as . More of the details of the solid-phase synthesis follow. The nature of the polymer support is of great importance for a successful peptide synthesis. One that is widely used is a cross-linked polystyrene resin of the type employed in ion-exchange chromatography ( ). It is necessary that the resin be insoluble but have a loose enough structure to absorb organic solvents. Otherwise, the reagents will not be able to penetrate into the spaces between the chains. This is undesirable because the reactions occur on the surface of the resin particles and poor penetration greatly reduces the number of equivalents of reactive sites that can be obtained per gram of resin. Finally, to anchor a peptide chain to the resin, a reactive functional group (usually a chloromethyl group) must be introduced into the resin. This can be done by a Friedel-Crafts chloromethylation reaction, which substitutes the \(\ce{ClCH_2}-\) group in the 4-position of the phenyl groups in the resin: At the start of the peptide synthesis, the \(\ce{C}\)-terminal amino acid is bonded through its carboxyl group to the resin by a nucleophilic attack of the carboxylate ion on the chloromethyl groups. The \(\alpha\)-amino group must be suitably protected, as with -butoxycarbonyl, before carrying out this step: Next, the amine protecting group must be removed without cleaving the ester bond to the resin. The coupling step to a second \(\ce{N}\)-protected amino acid follows, with \(\ce{N}\),\(\ce{N'}\)-dicyclohexylcarbodiimide as the coupling reagent of choice: The peptide-bond-forming steps are repeated as many times as needed to build up the desired sequence. Ultimately, the peptide chain is removed from the resin, usually with \(\ce{HBr}\) in anhydrous trifluoroethanoic acid, \(\ce{CF_3CO_2H}\), or with anhydrous \(\ce{HF}\). This treatment also removes the other acid-sensitive protecting groups. The method lends itself beautifully to automatic control, and machines suitably programmed to add reagents and wash the product at appropriate times have been developed. At present, the chain can be extended by six or so amino acid units a day. It is necessary to check the homogeneity of the growing peptide chain at intervals because if any step does not proceed properly, the final product can be seriously contaminated with peptides with the wrong sequence. In the synthesis of the enzyme ribonuclease by the Merrifield method, the 124 amino acids were arranged in the ribonuclease sequence through 369 reactions and some 12,000 individual operations of the automated peptide-synthesis machine without isolation of any intermediates. In many problems of peptide sequencing and peptide synthesis it is necessary to be able to separate mixtures of peptides and proteins. The principal methods used for this purpose depend on acid-base properties or on molecular sizes and shapes. is widely used for the purification, separation, and molecular-weight determination of proteins. A centrifugal field, up to 500,000 times that of gravity, is applied to the solution, and molecules move downward in the field according to their mass and size. Large molecules also can be separated by (or gel chromatography), wherein small molecules are separated from large ones by passing a solution over a gel that has pores of a size that the small molecules can penetrate into and be trapped. Molecules larger than the pore size are carried on with the solvent. This form of chromatographic separation is based on "sieving" rather than on chemical affinity. A wide range of gels with different pore sizes is available, and it is possible to fractionate molecules with molecular weights ranging from 700 to 200,000. The molecular weight of a protein can be estimated by the sizes of the pores that it will, or will not, penetrate. The acid-base properties, and hence ionic character, of peptides and proteins also can be used to achieve separations. , similar to that described for amino acids ( ), is an important separation method. Another method based on acid-base character and molecular size depends on differential rates of migration of the ionized forms of a protein in an electric field ( ). Proteins, like amino acids, have isoelectric points, which are the pH values at which the molecules have no net charge. At all other pH values there will be some degree of net ionic charge. Because different proteins have different ionic properties, they frequently can be separated by electrophoresis in buffered solutions. Another method, which is used for the separation and purification of enzymes, is , which was described briefly in . \(^5\) The distinction between secondary and tertiary structure is not sharp. Secondary structure involves consideration of the interactions and spatial relationships of the amino acids in the peptide chains that are together in the primary structure, whereas tertiary structure is concerned with those that are in the primary structure. and (1977) | 24,889 | 982 |
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Electronic Spectroscopy relies on the quantized nature of energy states. Given enough energy, an electron can be excited from its initial ground state or initial excited state (hot band) and briefly exist in a higher energy excited state. Electronic transitions involve exciting an electron from one principle quantum state to another. Without incentive, an electron will not transition to a higher level. Only by absorbing energy, can an electron be excited. Once it is in the excited state, it will relax back to it's original more energetically stable state, and in the process, release energy as photons. Often, during electronic spectroscopy, the electron is excited first from an initial low energy state to a higher state by absorbing photon energy from the spectrophotometer. If the wavelength of the incident beam has enough energy to promote an electron to a higher level, then we can detect this in the absorbance spectrum. Once in the excited state, the electron has higher potential energy and will relax back to a lower state by emitting photon energy. This is called fluorescence and can be detected in the spectrum as well. Embedded into the electronic states (n=1,2,3...) are vibrational levels (v=1,2,3...) and within these are rotational energy levels (j=1,2,3...). Often, during electronic transitions, the initial state may have the electron in a level that is excited for both vibration and rotation. In other words, n=0, v does not = 0 and r does not =0. This can be true for the ground state and the excited state. In addition, due to the Frank Condon Factor, which describes the overlap between vibrational states of two electronic states, there may be visible vibrational bands within the absorption bands. Therefore, vibrational fine structure that can be seen in the absorption spectrum gives some indication of the degree of Frank Condon overlap between electronic states. When interpreting the absorbance and fluorescence spectra of a given molecule, compound, material, or an elemental material, understanding the possible electronic transitions is crucial. Assigning the peaks in the absorption spectrum can become easier when considering which transitions are allowed by symmetry, the Laporte Rules, electron spin, or vibronic coupling. Knowing the degree of allowedness, one can estimate the intensity of the transition, and the extinction coefficient associated with that transition. These guidelines are a few examples of the selection rules employed for interpreting the origin of spectral bands. Only a complete model of molecular energy diagrams for the species under investigation can make clear the possible electronic transitions. Every different compound will have unique energy spacing between electronic levels, and depending on the type of compound, one can categorize these spacings and find some commonality. For example, aromatic compounds pi to pi* and n to pi* transitions where as inorganic compounds can have similar transitions with Metal to Ligand Charge Transfer (MLCT) and Ligand to Metal Charge Transfer (LMCT) in addition to d-d transitions, which lead to the bright colors of transition metal complexes. Although surprises in science often lead to discovery, it is more fortuitous for the interpreter to predict the spectra rather than being baffled by the observation. The following section will discuss the interpretation of electronic absorption spectra given the nature of the chemical species being studied. This includes an understanding of the molecular or elemental electronic state symmetries, Russell-Sanders states, spin multiplicities, and forbidden and allowed transitions of a given species. As the light passes through the monochrometer of the spectrophotometer, it hits the sample with some wavelength and corresponding energy. The ratio of the initial intensity of this light and the final intensity after passing through the sample is measured and recorded as absorbance (Abs). When absorbance is measured at different wavelengths, an absorbance spectrum of Abs vs wavelength can be obtained. This spectra reveals the wavelengths of light that are absorbed by the chemical specie, and is specific for each different chemical. Many electronic transitions can be visible in the spectrum if the energy of the incident light matches or surpasses the quantum of energy separating the ground state and that particular excited state. An example of an absorbance spectrum is given below. Here we can see the effect of temperature and also the effect of solvents on the clarity of the spectrum. We can see from these anonymous compounds that decreasing the temperature allows the vibrational fine structure to emerge. These vibrational bands embedded within the electronic bands represent the transitions from v=n to v'=n. Generally, the v=0 to v'=0 transition is the one with the lowest frequency. From there, increasing energy, the transitions can be from v=0 to v'=n, where n=1,2,3... With a higher temperature, the vibrational transitions become averaged in the spectrum due to the presence of vibrational hot bands and Fermi Resonance, and with this, the vibrational fine structure is lost at higher temperatures. The effect that the solvent plays on the absorption spectrum is also very important. It is clear that polar solvents give rise to broad bands, non-polar solvents show more resolution, though, completely removing the solvent gives the best resolution. This is due to solvent-solute interaction. The solvent can interact with the solute in its ground state or excited state through intermolecular bonding. For example, a polar solvent like water has the ability of hydrogen bonding with the solute if the solute has a hydrogen bonding component, or simply through induced dipole-dipole interactions. The non-polar solvents can interact though polarizability via London interactions also causing a blurring of the vibronic manifold. This is due to the solvent's tendency to align its dipole moment with the dipole moment of the solute. Depending on the interaction, this can cause the ground state and the excited state of the solute to increase or decrease, thus changing the frequency of the absorbed photon. Due to this, there are many different transition energies that become average together in the spectra. This causes peak-broadening. The effects of peak broadening are most severe for polar solvent, less so for non-polar solvents, and absent when the solute is in vapor phase. When estimating the intensities of the absorption peaks, we use the molar absorptivity constant (epsilon). If the transition is "allowed" then the molar absorptivity constant from the Beer's Law Plot will be high. This means that the probability of transition is large. If the transition is not allowed, then there will be no intensity and no peak on the spectrum. Transitions can be "partially allowed" as well, and these bands appear with a lower intensity than the full allowed transitions. One way to decide whether a transition will be allowed or not is to use symmetry arguments with Group Theory. If the symmetries of the ground and final state of a transition are correct, then the transition is symmetry allowed. We express this by modifying the transition moment integral from an integral of eigenstates to an orthogonally expressed direct product of the symmetries of the states. \[\int \psi_{2} \mu \psi_{1} d \mathcal{T} \longrightarrow \Gamma_{2} \otimes \Gamma \mu_{x y z} \otimes \Gamma_{1} \label{1}\] \[\int \psi_{v_{2}} \psi_{e l_{2}} \mu \psi_{v_{1}} \psi_{e l_{1}} d \mathcal{T} \longrightarrow \Gamma_{\mathrm{v}_{2}} \otimes \Gamma_{\mathrm{el}_{2}} \otimes \Gamma \mu_{x y z} \otimes \Gamma_{\mathrm{v}_{1}} \otimes \Gamma_{\mathrm{el}_{1}} \label{2}\] The conversions of integration to direct products of symmetry as shown gives spectroscopists a short cut into deciding whether the transition will be allowed or forbidden. A transition will be forbidden if the direct products of the symmetries of the electronic states with the coupling operator is odd. More specifically, if the direct product does not contain the totally symmetric representation, then the transition is forbidden by symmetry arguments. If the product does contain the totally symmetric representation (A, A , A ...etc) then the transition is symmetry allowed. Some transitions are forbidden by the Equation \ref{1} and one would not expect to be able to see the band that corresponds to the transition; however, a weak absorbance band is quite clear on the spectrum of many compounds. The transition may be forbidden via pure electronic symmetries; however, for an octahedral complex for example since it has a center of inversion, the transition is weakly allowed because of vibronic coupling. When the octahedra of a transition metal complex is completely symmetric (without vibrations), the transition cannot occur. However, when vibrations exist, they temporarily perturb the symmetry of the complex and allow the transition by equation (2). If the product of all of these representations contains the totally symmetric representation, then the transition will be allowed via vibronic coupling even if it forbidden electronically. Some transitions are forbidden by symmetry and do not appear in the absorption spectrum. If the symmetries are correct, then another state besides the ground state can be used to make the otherwise forbidden transition possible. This is accomplished by hot bands, meaning the electrons in the ground state are heated to a higher energy level that has a different symmetry. When the transition moment integral is solved with the new hot ground state, then the direct product of the symmetries may contain the totally symmetric representation. If we employ the old saying, "You can't get there from here!" then we would be referring to the transition from the ground state to the excited state. However, if we thermally excite the molecules from out of the ground state, then, "we can get there from here!" Knowing whether a transition will be allowed by symmetry is an essential component to interpreting the spectrum. If the transition is allowed, then it should be visible with a large extinction coefficient. If it is forbidden, then it should only appear as a weak band if it is allowed by vibronic coupling. In addition to this, a transition can also be spin forbidden. The examples below of excited state symmetries, give an indication of what spin forbidden means: These states are derived from the electron configuration of benzene. Once we have the molecular orbital energy diagram for benzene, we can assign symmetries to each orbital arrangement of the ground state. From here, we can excite an electron from the Highest Occupied Molecular Orbital (HOMO) to the Lowest Unoccupied Molecular Orbital (LUMO). This is the lowest energy transition. Other transitions include moving the electron above the LUMO to higher energy molecular orbitals. To solve for the identity of the symmetry of the excited state, one can take the direct product of the HOMO symmetry and the excited MO symmetry. This give a letter (A, B, E..) an the subscript (1u, 2u, 1g...). The superscript is the spin multiplicity, and from single electron transitions, the spin multiplicity is 2S+1 = M, where S = 1 with two unpaired electrons having the same spin and S=0 when the excited electron flips its spin so that the two electrons have opposite spin. This gives M=1 and M=3 for benzene above. From the results above, we have three transitions that are spin allowed and three that are spin forbidden. Once we take the direct product of the symmetries and the coupling operator for each of these states given above, we find that only the A to E transition is allowed by symmetry. Therefore, we have information regarding spin and symmetry allowedness and we have an idea of what the spectra will look like: When interpreting the spectrum, it is clear that some transitions are more probable than others. According to the symmetry of excited states, we can now order them from low energy to high energy based on the position of the peaks (E1u is the highest, then B1u, and B2u is lowest). The A to E transition is fully allowed and therefore the most intense peak. The A to B and A to B transitions are symmetry forbidden and thus have a lower probability which is evident from the lowered intensity of their bands. The singlet A to triplet B transition is both symmetry forbidden and spin forbidden and therefore has the lowest intensity. This transition is forbidden by spin arguments; however, a phenomenon known as spin-orbit coupling can allow this transition to be weakly allowed as well. If spin-orbit coupling exists, then the singlet state has the same total angular momentum as the triplet state so the two states can interact. A small amount of singlet character in the triplet state leads to a transition moment integral that is non-zero, so the transition is allowed. From the example of benzene, we have investigated the characteristic pi to pi* transitions for aromatic compounds. Now we can move to other organic molecules, which involves n to pi* as well as pi to pi*. Two examples are given below: The highest energy transition for both of these molecules has an intensity around 10,000 cm and the second band has an intensity of approximately 100 cm . In the case of formaldehyde, the \(n \rightarrow \pi^*\) transition is forbidden by symmetry where as the pi to pi* is allowed. The opposite is true for As(Ph) and the difference in molar absorptivity is evidence of this. These transitions involve moving an electron from a nonbonding electron pair to a antibonding \(\pi^*\) orbital. They tend to have molar absorbtivities less than 2000 and undergo a blue shift with solvent interactions (a shift to higher energy and shorter wavelengths). This is because the lone pair interacts with the solvent, especially a polar one, such that the solvent aligns itself with the ground state. When the excited state emerges, the solvent molecules do not have time to rearrange in order to stabilize the excited state. This causes a lowering of energy of the ground state and not the excited state. Because of this, the energy of the transition increases, hence the "blue shift". These transitions involve moving an electron from a bonding \(\pi\) orbital to an antibonding \(\pi^*\) orbital. They tend to have molar absorptivities on the order of 10,000 and undergo a red shift with solvent interactions (a shift to lower energy and longer wavelengths). This could either be due to a raising of the ground state energy or lowering of the excited state energy. If the excited state is polar, then it will be solvent stabilized, thus lowering its energy and the energy of the transition. Speaking of transition probabilities in organic molecules is a good introduction to interpreting the spectra of inorganic molecules. Three types of transitions are important to consider are Metal to Ligand Charge Transfer (MLCT), Ligand to Metal Charge Transfer (LMCT), and d-d transitions. To understand the differences of these transitions we must investigate where these transitions originate. To do this, we must define the difference between pi accepting and pi donating ligands: From these two molecular orbital energy diagrams for transition metals, we see that the pi donor ligands lie lower in energy than the pi acceptor ligands. According to the spectral chemical series, one can determine whether a ligand will behave as a pi accepting or pi donating. When the ligand is more pi donating, its own orbitals are lower in energy than the t2g metal orbitals forcing the frontier orbitals to involve an antibonding pi* (for t2g) and an antibonding sigma* (for eg). This is in contrast to the pi accepting ligands which involve a bonding pi (t2g) and an antibonding sigma* (eg). Because of this, the d-d transition (denoted above by delta) for the pi acceptor ligand complex is larger than the pi donor ligand. In the spectra, we would see the d-d transitions of pi acceptor ligands to be of a higher frequency than the pi donor ligands. In general though, these transitions appear as weakly intense on the spectrum because they are Laporte forbidden. Due to vibronic coupling; however, they are weakly allowed and because of their relatively low energy of transition, they can emit visible light upon relaxation which is why many transition metal complexes are brightly colored. The molar extinction coefficients for these transition hover around 100. At an even higher energy are the LMCT which involve pi donor ligands around the metal. These transitions arise because of the low-lying energy of the ligand orbitals. Therefore, we can consider this as a transition from orbitals that are ligand in character to orbitals that are more metal in character, hence the name, Ligand to Metal Charge Transfer. The electron travels from a bonding pi or non-bonding pi orbital into a sigma* orbital. These transitions are very strong and appear very intensely in the absorbance spectrum. The molar extinction coefficients for these transitions are around 10 . Examples of pi donor ligands are as follows: F , Cl , Br-, I , H O, OH , RS , S , NCS , NCO ,... The somewhat less common MLCT has the same intensity and energy of the LMCT as they involve the transition of an electron from the t2g (pi) and the eg (sigma*) to the t1u (pi*/sigma*). These transitions arise from pi acceptor ligands and metals that are willing to donate electrons into the orbitals of Ligand character. This is the reason that they are less frequent since metals commonly accept electrons rather than donate them. All the same, both types of Charge Transfer bands are more intense than d-d bands since they are not Laporte Rule forbidden. Examples of pi accepting ligands are as follows: CO, NO, CN , N , bipy, phen, RNC, C H , C=C double bonds, C C triple bonds,... From this spectra of an octahedral Chromium complex, we see that the d-d transitions are far weaker than the LMCT. Since Chlorine is a pi donor ligand in this example, we can label the CT band as LMCT since we know the electron is transitioning from a MO of ligand character to a MO of metal character. The Laporte forbidden (symmetry forbidden) d-d transitions are shown as less intense since they are only allowed via vibronic coupling. In addition, the d-d transitions are lower in energy than the CT band because of the smaller energy gap between the t2g and eg in octahedral complexes (or eg to t2g in tetrahedral complexes) than the energy gap between the ground and excited states of the charge transfer band. These transitions abide by the same selection rules that organic molecules follow: spin selection and symmetry arguments. The Tanabe and Sugano diagrams for transition metal complexes can be a guide for determining which transitions are seen in the spectrum. We will use the [CrCl(NH3)5]2+ ion as an example for determining the types of transitions that are spin allowed. To do this we look up the Tanabe and Sugano diagrams for Octahedral fields. Since Cr in the complex has three electrons, it is a d3 and so we find the diagram that corresponds to d3 metals: Based on the TS diagram on the left, and the information we have already learned, can you predict which transition will be spin allowed and which ones will be forbidden? From the diagram we see that the ground state is a A . This is because of the three unpaired electrons which make M=2S+1= 4. The A comes from the fact that there is only one combination of electrons possible. With a spin multiplicity of 4, by the spin selection rules, we can only expect intense transitions between the ground state A and T , T , and the other T excited state. The other transitions are spin forbidden. Therefore, we would expect to see three d-d transitions on the absorption spectra. For us to visualize this, we can draw these transitions in order of increasing energy and then plot the spectrum as we would expect it for only the d-d transitions in a d octahedral complex: From three spin allowed transitions, we would expect to see three d-d bands appear on the spectrum. In addition to these of course, the LMCT band will appear as well. Now that we have discussed the nature of absorption involving an electron absorbing photon energy to be excited to a higher energy level, now we can discuss what happens to that excited electron. Due to its higher potential energy, the electron will relax back to its initial ground state, and in the process, emit electromagnetic radiation. The energy gap between the excited state and the state to which the electron falls determines the wavelength of light that will be emitted. This process is called fluorescence. Generally, the wavelengths of fluorescence are longer than absorbance, can you explain why? Given the following diagram, one can see that vibrational relaxation occurs in the excited electronic state such that the electronic relaxation occurs from the ground vibrational state of the excited electronic state. This causes lower energy electronic relaxations than the previous energy of absorption. Here we see that the absorption transitions by default involve a greater energy change than the emission transitions. Due to vibrational relaxation in the excited state, the electron tends to relax only from the v'=0 ground state vibrational level. This gives emission transitions of lower energy and consequently, longer wavelength than absorption. When obtaining fluorescence, we have to block out the transmitted light and only focus on the light being emitted from the sample, so the detector is usually 90 degrees from the incident light. Because of this emission spectra are generally obtained separately from the absorption spectra; however, they can be plotted on the same graph as shown. Generally separated by ~10 nm, the fluorescence peak follows the absorption peak according to the spectrum. With that, we conclude our discussion of electronic spectroscopy interpretation. Refer to outside links and references for additional information. | 22,138 | 983 |
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Silicon dioxide, \(\ce{SiO2}\), also known as silica is a linear molecule that is formed by one silicon atom and two oxygen atoms with two sets of doubles bonds and 4 single bonds. Because of its main component: glass, silicon dioxide is a very common and important molecule in the construction industry. One of the forms of silicon dioxide is quartz, which is found in sand. SiO is a 3 dimensional structure and comes from the tetrahedral structure, SiO . Each of the Silicon atoms are connected to each other with an oxygen atom, which creates a "diamond type network". All forms of SiO possess a 3 dimensional shape and has a diamond structure. The bonding angle of Si-O-Si, which is the building block of the SiO molecule, is 144 degrees. These are called polymorph and in order to be stable, 3 of these polymorph are suppose to exist. This stable unit creates a a temperature for each of the different forms of SiO . Forms that have (alpha) are at low temperature, while forms with (beta) are at high temperature. The structure of the different forms of SiO is important because it gives each of the different forms of SiO different characteristics and functions. Commercially, SiO is very important in steel, electronic, and semiconductor industries because of its structure, SiO is able to undergo rapid temperature changes and still maintain its shape and structure. There are many different forms of SiO , which mainly derived quartz glass. One form that is derived from quartz is beta-cristobalite, which is found in high temperature. Some other forms are beta-quartz, alpha-quartz, beta-tridymite, alpha-tridymite, alpha=cristobalite, and many more. The alpha and beta stands for the temperature range. Alpha molecules have low temperature while beta molecules have high temperature. | 1,819 | 984 |
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Let’s try to construct an antisymmetric function that describes the two electrons in the ground state of helium. Blindly following the first statement of the Pauli Exclusion Principle, then each electron in a multi-electron atom be described by a different spin-orbital. For the ground-state helium atom, this gives a \(1s^22s^02p^0\) configuration (Figure 8.6.1
). We try constructing a simple product wavefunction for helium using two different spin-orbitals. Both have the 1s spatial component, but one has spin function \(\alpha\) and the other has spin function \(\beta\) so the product wavefunction matches the form of the ground state electron configuration for He, \(1s^2\). \[ | \psi (\mathbf{r}_1, \mathbf{r}_2 ) \rangle = \varphi _{1s}\alpha (\mathbf{r}_1) \varphi _{1s}\beta ( \mathbf{r}_2) \label {8.6.1} \] After permutation of the electrons, this becomes \[ | \psi ( \mathbf{r}_2,\mathbf{r}_1 ) \rangle = \varphi _{1s}\alpha ( \mathbf{r}_2) \varphi _{1s}\beta (\mathbf{r}_1) \label {8.6.2} \] which is different from the starting function since \(\varphi _{1s\alpha}\) and \(\varphi _{1s\beta}\) are different spin-orbital functions. Hence, the simple product wavefunction in Equation \ref{8.6.1} does not satisfy the indistinguishability requirement since an antisymmetric function must produce the same function multiplied by (–1) after permutation of two electrons, and that is not the case here. We must try something else. To avoid getting a totally different function when we permute the electrons, we can make . A very simple way of taking a linear combination involves making a new function by simply adding or subtracting functions. The function that is created by subtracting the right-hand side of Equation \(\ref{8.6.2}\) from the right-hand side of Equation \(\ref{8.6.1}\) has the desired antisymmetric behavior. The constant on the right-hand side accounts for the fact that the total wavefunction must be normalized. \[ | \psi (\mathbf{r}_1, \mathbf{r}_2) \rangle = \dfrac {1}{\sqrt {2}} [ \varphi _{1s}\alpha (\mathbf{r}_1) \varphi _{1s}\beta ( \mathbf{r}_2) - \varphi _{1s} \alpha( \mathbf{r}_2) \varphi _{1s} \beta (\mathbf{r}_1)] \label{8.6.3} \] In this orbital approximation, a single electron is held in a single s with an orbital component (e.g., the \(1s\) orbital) determined by the \(n\), \(l\), \(m_l\) quantum numbers and a spin component determined by the \(m_s\) quantum number. The wavefunction in Equation \ref{8.6.3} can be decomposed into spatial and spin components: \[ | \psi (\mathbf{r}_1, \mathbf{r}_2) \rangle = \dfrac {1}{\sqrt {2}} \underbrace{[ \varphi _{1s}(1) \varphi _{1s}(2)]}_{\text{spatial component}} \underbrace{[ \alpha(1) \beta( 2) - \alpha( 2) \beta(1)]}_{\text{spin component}} \label{8.6.3B} \] Show that the linear combination of spin-orbitals in Equation \(\ref{8.6.3}\) is antisymmetric with respect to permutation of the two electrons. Replace the minus sign with a plus sign (i.e. take the positive linear combination of the same two functions) and show that the resultant linear combination is symmetric. First a reminder of permutation symmetries: We start with the original wavefunction \[ | \psi (\mathbf{r}_1, \mathbf{r}_2) \rangle = \dfrac {1}{\sqrt {2}} [ \varphi _{1s\alpha}(\mathbf{r}_1) \varphi _{1s\beta}( \mathbf{r}_2) - \varphi _{1s\alpha}( \mathbf{r}_2) \varphi _{1s\beta}(\mathbf{r}_1)] \nonumber \] and flip the position of electron 1 with electron 2 and vice versa \[ | \psi (\mathbf{r}_2, \mathbf{r}_1) \rangle = \dfrac {1}{\sqrt {2}} [ \varphi _{1s\alpha}(\mathbf{r}_2) \varphi _{1s\beta}( \mathbf{r}_1) - \varphi _{1s\alpha}( \mathbf{r}_1) \varphi _{1s\beta}(\mathbf{r}_2)] \label{permute1} \] We then we ask if we can rearrange the left side of Equation \ref{permute1} to either become \( + | \psi(\mathbf{r}_1, \mathbf{r}_2)\rangle\) (symmetric to permutation) or \( - | \psi(\mathbf{r}_1, \mathbf{r}_2)\rangle \) (antisymmetric to permutation). \[ | \psi (\mathbf{r}_2, \mathbf{r}_1) \rangle = \dfrac {1}{\sqrt {2}} [ - \varphi _{1s\alpha}( \mathbf{r}_1) \varphi _{1s\beta}(\mathbf{r}_2) + \varphi _{1s\alpha}(\mathbf{r}_2) \varphi _{1s\beta}( \mathbf{r}_1) ] \nonumber \] or \[ | \psi (\mathbf{r}_2, \mathbf{r}_1) \rangle = - \dfrac {1}{\sqrt {2}} [ \varphi _{1s\alpha}( \mathbf{r}_1) \varphi _{1s\beta}(\mathbf{r}_2) - \varphi _{1s\alpha}(\mathbf{r}_2) \varphi _{1s\beta}( \mathbf{r}_1) ] \nonumber \] This is just the negative of the original wavefunction, therefore \[| \psi (\mathbf{r}_2, \mathbf{r}_1) \rangle = - | \psi (\mathbf{r}_1, \mathbf{r}_2) \rangle \nonumber \] The wavefunction is antisymemtric. Is this linear combination of spin-orbitals \[ | \psi (\mathbf{r}_1, \mathbf{r}_2) \rangle = \dfrac {1}{\sqrt {2}} [ \varphi _{1s\alpha}(\mathbf{r}_1) \varphi _{1s\beta}( \mathbf{r}_2) + \varphi _{1s\alpha}( \mathbf{r}_2) \varphi _{1s\beta}(\mathbf{r}_1)] \nonumber \] symmetric or antisymmetric with respect to permutation of the two electrons? Symmetric The electronic configuration of the first excited state of He is \(1s^12s^12p^0\) and we can envision four microstates for this configuration (Figure 8.6.2
). As spected, the wavefunctions associated for of these microstate must satisfy indistinguishability requirement just like the ground state. These electron configurations are used to construct four possible excited-state two-electron wavefunctions (but not necessarily in a one-to-one correspondence): \[ \begin{align} | \psi_1 (\mathbf{r}_1, \mathbf{r}_2) \rangle &= \dfrac {1}2 \underbrace{[ \varphi _{1s}(1) \varphi _{2s}(2)+\varphi _{1s}(2) \varphi _{2s}(1)]}_{\text{spatial component}} \underbrace{[ \alpha(1) \beta( 2) - \alpha( 2) \beta(1)]}_{\text{spin component}} \label{8.6.3C1} \\[4pt] | \psi_2 (\mathbf{r}_1, \mathbf{r}_2) \rangle &= \dfrac {1}{\sqrt {2}} \underbrace{[ \varphi _{1s}(1) \varphi _{2s}(2) - \varphi _{1s}(2) \varphi _{2s}(1)]}_{\text{spatial component}} \underbrace{[ \alpha(1) \alpha( 2)]}_{\text{spin component}} \label{8.6.3C2} \\[4pt] | \psi_3 (\mathbf{r}_1, \mathbf{r}_2) \rangle &= \dfrac {1}{\sqrt {2}} \underbrace{[ \varphi _{1s}(1) \varphi _{2s}(2) - \varphi _{1s}(2) \varphi _{2s}(1)]}_{\text{spatial component}} \underbrace{[ \alpha(1) \beta( 2) + \alpha( 2) \beta(1)]}_{\text{spin component}} \label{8.6.3C3} \\[4pt] | \psi_4 (\mathbf{r}_1, \mathbf{r}_2) \rangle &= \dfrac {1}2 \underbrace{[ \varphi _{1s}(1) \varphi _{2s}(2) - \varphi _{1s}(2) \varphi _{2s}(1)]}_{\text{spatial component}} \underbrace{[ \beta(1) \beta( 2)]}_{\text{spin component}} \label{8.6.3C4} \end{align} \] All four wavefunctions are antisymmetric as required for fermionic wavefunctions (which is left to an exercise). Wavefunctions \(| \psi_2 \rangle \) and \(| \psi_4 \rangle\) correspond to the two electrons both having spin up or both having spin down (Configurations 2 and 3 in Figure 8.6.2
, respectively). Wavefunctions \(| \psi_1 \rangle \) and \(| \psi_3 \rangle \) are more complicated and are antisymmetric (Configuration 1 - Configuration 4) and symmetric combinations (Configuration 1 + 4). That is, a single electron configuration does not describe the wavefunction. For many electrons, this ad hoc construction procedure would obviously become unwieldy. However, there is an elegant way to construct an antisymmetric wavefunction for a system of \(N\) identical particles. A linear combination that describes an appropriately antisymmetrized multi-electron wavefunction for any desired orbital configuration is easy to construct for a two-electron system. However, interesting chemical systems usually contain more than two electrons. For these multi-electron systems a relatively simple scheme for constructing an antisymmetric wavefunction from a product of one-electron functions is to write the wavefunction in the form of a determinant. John Slater introduced this idea so the determinant is called a . John C. Slater introduced the determinants in 1929 as a means of ensuring the antisymmetry of a wavefunction, however the determinantal wavefunction first appeared three years earlier independently in Heisenberg's and Dirac's papers. The Slater determinant for the two-electron ground-state wavefunction of helium is \[ | \psi (\mathbf{r}_1, \mathbf{r}_2) \rangle = \dfrac {1}{\sqrt {2}} \begin {vmatrix} \varphi _{1s} (1) \alpha (1) & \varphi _{1s} (1) \beta (1) \\ \varphi _{1s} (2) \alpha (2) & \varphi _{1s} (2) \beta (2) \end {vmatrix} \label {8.6.4} \] A shorthand notation for the determinant in Equation \(\ref{8.6.4}\) is then \[ | \psi (\mathbf{r}_1 , \mathbf{r}_2) \rangle = 2^{-\frac {1}{2}} Det | \varphi _{1s\alpha} (\mathbf{r}_1) \varphi _{1s\beta} ( \mathbf{r}_2) | \label {8.6.5} \] The determinant is written so the electron coordinate changes in going from one row to the next, and the spin orbital changes in going from one column to the next. The advantage of having this recipe is clear if you try to construct an antisymmetric wavefunction that describes the orbital configuration for uranium! Note that the normalization constant is \((N!)^{-\frac {1}{2}}\) for \(N\) electrons. The generalized Slater determinant for a multi-electrom atom with \(N\) electrons is then \[ \psi(\mathbf{r}_1, \mathbf{r}_2, \ldots, \mathbf{r}_N)=\dfrac{1}{\sqrt{N!}} \left| \begin{matrix} \varphi_1(\mathbf{r}_1) & \varphi_2(\mathbf{r}_1) & \cdots & \varphi_N(\mathbf{r}_1) \\ \varphi_1(\mathbf{r}_2) & \varphi_2(\mathbf{r}_2) & \cdots & \varphi_N(\mathbf{r}_2) \\ \vdots & \vdots & \ddots & \vdots \\ \varphi_1(\mathbf{r}_N) & \varphi_2(\mathbf{r}_N) & \cdots & \varphi_N(\mathbf{r}_N) \end{matrix} \right| \label{5.6.96} \] Expand the Slater determinant in Equation \(\ref{8.6.4}\) for the \(\ce{He}\) atom. To expand the Slater determinant of the Helium atom, the wavefunction in the form of a two-electron system: \[ | \psi (\mathbf{r}_1, \mathbf{r}_2) \rangle = \dfrac {1}{\sqrt {2}} \begin {vmatrix} \varphi _{1s} (1) \alpha (1) & \varphi _{1s} (1) \beta (1) \\ \varphi _{1s} (2) \alpha (2) & \varphi _{1s} (2) \beta (2) \end {vmatrix} \nonumber \] This is a simple expansion exercise of a \(2 \times 2\) determinant \[ | \psi (\mathbf{r}_1, \mathbf{r}_2) \rangle = \dfrac {1}{\sqrt {2}} \left[ \varphi _{1s} (1) \alpha (1) \varphi _{1s} (2) \beta (2) - \varphi _{1s} (2) \alpha (2) \varphi _{1s} (1) \beta (1) \right] \nonumber \] It is not unexpected that the determinant wavefunction in Equation \ref{8.6.4} is the same as the form for the helium wavefunction that is given in Equation \ref{8.6.3}. Write and expand the Slater determinant for the ground-state \(\ce{Li}\) atom. Slater determinant for \(\ce{Li}\) atom: \[\psi(1,2,3)=\frac{1}{\sqrt{6}} \operatorname{det}\left(\begin{array}{ccc} {\varphi _{1s} \alpha(1)} & {\varphi _{1s} \beta(1)} & {\varphi _{2s} \alpha(1)} \\ \varphi _{1s} \alpha(2) & {\varphi _{1s} \beta(2)} & {\varphi _{2s} \alpha(2)} \\ {\varphi _{1s} \alpha(3)} & {\varphi _{1s} \beta(3)} & {\varphi _{2s} \alpha(3)} \end{array}\right)\nonumber \] Expansion of Slater determinant: \[\psi(1,2,3)=\frac{1}{\sqrt{6}}[\varphi _{1s} \alpha(1) \varphi _{1s} \beta(2) \varphi _{2s} \alpha(3)-\varphi _{1s} \alpha(1) \varphi _{1s} \beta(3) \varphi _{2s} \alpha(2)+ \varphi _{1s} \alpha(3) \varphi _{1s} \beta(1) \varphi _{2s} \alpha(2) - \varphi _{1s} \alpha(3) \varphi _{1s} \beta(2) \varphi _{1s} \alpha(1)+ \varphi _{1s} \alpha(2) \varphi _{1s} \beta(3) \varphi _{2s} \alpha(3) ] \nonumber \] Note that this is also a valid ground state wavefunction \[\psi(1,2,3)=\frac{1}{\sqrt{6}} \operatorname{det}\left(\begin{array}{ccc} {\varphi _{1s} \alpha(1)} & {\varphi _{1s} \beta(1)} & {\varphi _{2s} \beta(1)} \\ \varphi _{1s} \alpha(2) & {\varphi _{1s} \beta(2)} & {\varphi _{2s} \beta(2)} \\ {\varphi _{1s} \alpha(3)} & {\varphi _{1s} \beta(3)} & {\varphi _{2s} \beta(3)} \end{array}\right)\nonumber \] What is the difference between these two wavefunctions? Now that we have seen how acceptable multi-electron wavefunctions can be constructed, it is time to revisit the “guide” statement of conceptual understanding with which we began our deeper consideration of electron indistinguishability and the Pauli Exclusion Principle. What does a multi-electron wavefunction constructed by taking specific linear combinations of product wavefunctions mean for our physical picture of the electrons in multi-electron atoms? Overall, the antisymmetrized product function describes the configuration (the orbitals, regions of electron density) for the multi-electron atom. Because of the requirement that electrons be indistinguishable, we cannot visualize specific electrons assigned to specific spin-orbitals. Instead, we construct functions that allow each electron’s probability distribution to be dispersed across each spin-orbital. The total charge density described by any one spin-orbital cannot exceed one electron’s worth of charge, and each electron in the system is contributing a portion of that charge density. The four configurations in Figure 8.6.2
for first-excited state of the helium atom can be expressed as the following Slater Determinants \[ | \phi_a (\mathbf{r}_1, \mathbf{r}_2) \rangle = \dfrac {1}{\sqrt {2}} \begin {vmatrix} \varphi _{1s} (1) \alpha (1) & \varphi _{2s} (1) \beta(1) \\ \varphi _{1s} (2) \alpha (2) & \varphi _{2s} (2) \beta (2) \end {vmatrix} \label {8.6.10A} \] \[ | \phi_b (\mathbf{r}_1, \mathbf{r}_2) \rangle = \dfrac {1}{\sqrt {2}} \begin {vmatrix} \varphi _{1s} (1) \alpha (1) & \varphi _{2s} (1) \alpha (1) \\ \varphi _{1s} (2) \alpha (2) & \varphi _{2s} (2) \alpha(2) \end {vmatrix} \label {8.6.10B} \] \[ | \phi_c (\mathbf{r}_1, \mathbf{r}_2) \rangle = \dfrac {1}{\sqrt {2}} \begin {vmatrix} \varphi _{1s} (1) \beta(1) & \varphi _{2s} (1) \alpha(1) \\ \varphi _{1s} (2) \beta(2) & \varphi _{2s} (2) \alpha(2) \end {vmatrix} \label {8.6.10D} \] \[ | \phi_d (\mathbf{r}_1, \mathbf{r}_2) \rangle = \dfrac {1}{\sqrt {2}} \begin {vmatrix} \varphi _{1s} (1) \beta(1) & \varphi _{2s} (1) \beta (1) \\ \varphi _{1s} (2) \beta(2) & \varphi _{2s} (2) \beta (2) \end {vmatrix} \label {8.6.10C} \] Slater determinants are constructed by arranging spinorbitals in columns and electron labels in rows and are normalized by dividing by \(\sqrt{N!}\), where \(N\) is the number of occupied spinorbitals. As you can imagine, the algebra required to compute integrals involving Slater determinants is extremely difficult. It is therefore most important that you realize several things about these states so that you can avoid unnecessary algebra: The wavefunctions in \ref{8.6.3C1}-\ref{8.6.3C4} can be expressed in term of the four determinants in Equations \ref{8.6.10A}-\ref{8.6.10C}. \[ \begin{align*} | \psi_2 \rangle &= |\phi_b \rangle \\[4pt] &= \dfrac {1}{\sqrt {2}} \begin {vmatrix} \varphi _{1s} (1) \alpha (1) & \varphi _{2s} (1) \alpha (1) \\ \varphi _{1s} (2) \alpha (2) & \varphi _{2s} (2) \alpha(2) \end {vmatrix} \end{align*} \nonumber \] \[ \begin{align*} | \psi_4 \rangle &= |\phi_d \rangle \\[4pt] &= \dfrac {1}{\sqrt {2}} \begin {vmatrix} \varphi _{1s} (1) \beta(1) & \varphi _{2s} (1) \beta (1) \\ \varphi _{1s} (2) \beta(2) & \varphi _{2s} (2) \beta (2) \end {vmatrix} \end{align*} \nonumber \] but the wavefunctions that represent combinations of spinorbitals and hence combinations of electron configurations (e.g., igure 8.6.2
) are combinations of Slater determinants (Equation \ref{8.6.10A}-\ref{8.6.10D}) \[ \begin{align*} | \psi_1 \rangle & = |\phi_a \rangle - |\phi_c \rangle \\[4pt] &= \dfrac {1}{2} \left( \begin {vmatrix} \varphi _{1s} (1) \alpha (1) & \varphi _{2s} (1) \beta(1) \\ \varphi _{1s} (2) \alpha (2) & \varphi _{2s} (2) \beta (2) \end {vmatrix} - \begin {vmatrix} \varphi _{1s} (1) \beta(1) & \varphi _{2s} (1) \alpha(1) \\ \varphi _{1s} (2) \beta(2) & \varphi _{2s} (2) \alpha(2) \end {vmatrix} \right) \end{align*} \nonumber \] \[\begin{align*} | \psi_3 \rangle &= |\phi_a \rangle + |\phi_c \rangle \\[4pt] &= \dfrac {1}{2} \left( \begin {vmatrix} \varphi _{1s} (1) \alpha (1) & \varphi _{2s} (1) \beta(1) \\ \varphi _{1s} (2) \alpha (2) & \varphi _{2s} (2) \beta (2) \end {vmatrix} + \begin {vmatrix} \varphi _{1s} (1) \beta(1) & \varphi _{2s} (1) \alpha(1) \\ \varphi _{1s} (2) \beta(2) & \varphi _{2s} (2) \alpha(2) \end {vmatrix} \right) \end{align*} \nonumber \] Note the expected change in the normalization constants. Write the Slater determinant for the ground-state carbon atom. If you expanded this determinant, how many terms would be in the linear combination of functions? Carbon has 6 electrons which occupy the 1s 2s and 2p orbitals. Each row in the determinant represents a different electron and each column a unique spin-obital where the electron could be found. There are 6 rows, 1 for each electron, and 6 columns, with the two possible p orbitals both alpha (spin up), in the determinate. There are two columns for each s orbital to account for the alpha and beta spin possibilities. There are two different p orbitals because the electrons in their ground state will be in the different p orbitals and both spin up. N=6 so the normalization constant out front is 1 divided by the square-root of 6! \begin{align*}\psi(1,2,3,4,5,6)=\frac{1}{6!^{1/2}}\begin{vmatrix}\varphi _{1s} (1) \alpha (1) & \varphi _{1s} (1) \beta (1) & \varphi _{2s} (1) \alpha (1) & \varphi _{2s} (1) \beta (1) & \varphi _{2px} (1) \alpha (1) & \varphi _{2py} (1) \alpha (1) \\ \varphi _{1s} (2) \alpha (2) & \varphi _{1s} (2) \beta (2) & \varphi _{2s} (2) \alpha (2) & \varphi _{2s} (2) \beta (2) & \varphi _{2px} (2) \alpha (2) & \varphi _{2py} (2) \alpha (2) \\ \varphi _{1s} (3) \alpha (3) & \varphi _{1s} (3) \beta (3) & \varphi _{2s} (3) \alpha (3) & \varphi _{2s} (3) \beta (3) & \varphi _{2px} (3) \alpha (3) & \varphi _{2py} (3) \alpha (3) \\ \varphi _{1s} (4) \alpha (4) & \varphi _{1s} (4) \beta (4) & \varphi _{2s} (4) \alpha (4) & \varphi _{2s} (4) \beta (4) & \varphi _{2px} (4) \alpha (4) & \varphi _{2py} (4) \alpha (4)\\ \varphi _{1s} (5) \alpha (5) & \varphi _{1s} (5) \beta (5) & \varphi _{2s} (5) \alpha (5) & \varphi _{2s} (5) \beta (5) & \varphi _{2px} (5) \alpha (5) & \varphi _{2py} (5) \alpha (5)\\ \varphi _{1s} (6) \alpha (6) & \varphi _{1s} (6) \beta (6) & \varphi _{2s} (6) \alpha (6) & \varphi _{2s} (6) \beta (6) & \varphi _{2px} (6) \alpha (6) & \varphi _{2py} (6) \alpha (6)\end{vmatrix} \end{align*} Expanding this determinant would result in a linear combination of functions containing 720 terms. An expanded determinant will contain N! factorial terms, where N is the dimension of the matrix. Write the Slater determinant for the \(1s^12s^1\) excited state orbital configuration of the helium atom. Since there are 2 electrons in question, the Slater determinant should have 2 rows and 2 columns exactly. Additionally, this means the normalization constant is \(1/\sqrt{2}\). Each element of the determinant is a different combination of the spatial component and the spin component of the \(1 s^{1} 2 s^{1}\) atomic orbitals \[
\frac{1}{\sqrt{2}}\left[\begin{array}{cc}
{\varphi _{1_s}(1) \alpha(1)} & {\varphi {2_s}(1) \beta(1)} \\
{\varphi {1_s}(2) \alpha(2)} & {\varphi {2_s}(2) \beta(2)}
\end{array}\right] \nonumber
\nonumber \] Critique the energy level diagram and shorthand electron configuration notation from the perspective of the indistinguishability criterion. Can you imagine a way to represent the wavefunction expressed as a Slater determinant in a schematic or shorthand notation that more accurately represents the electrons? (This is not a solved problem!) | 19,542 | 986 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/06%3A_Structures_and_Energetics_of_Metallic_and_Ionic_solids/6.17%3A_Defects_in_Solid_State_Lattices/6.17B%3A_Frenkel_Defect |
The (also known as the Frenkel pair/disorder) is a defect in the lattice crystal where an atom or ion occupies a normally vacant site other than its own. As a result the atom or ion leaves its own lattice site vacant. The Frenkel Defect explains a defect in the molecule where an atom or ion (normally the cation) leaves its own lattice site vacant and instead occupies a normally vacant site (Figure \(\Page {1}\)). The cation leaves its own lattice site open and places itself between the area of all the other cations and anions. This defect is only possible if the cations are smaller in size when compared to the anions. The number of Frenkel Defects can be calculated using the equation: Frenkel The crystal lattices are relatively open and the coordination number is low. | 798 | 987 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/11%3A_Reactions_and_Other_Chemical_Processes/11.02%3A_The_Advancement_and_Molar_Reaction_Quantities |
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\( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \)
\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) Many of the processes of interest to chemists can be described by balanced reaction equations, or chemical equations, for the conversion of reactants into products. Thus, for the vaporization of water we write \[ \ce{H2O}\tx{(l)} \arrow \ce{H2O}\tx{(g)} \] For the dissolution of sodium chloride in water, we write \[ \ce{NaCl}\tx{(s)} \arrow \ce{Na+}\tx{(aq)} + \ce{Cl-}\tx{(aq)} \] For the Haber synthesis of ammonia, the reaction equation can be written \begin{equation*} \ce{N2}\tx{(g)} + \ce{3H2}\tx{(g)} \arrow \ce{2NH3}\tx{(g)} \end{equation*} The essential feature of a reaction equation is that equal amounts of each element and equal net charges appear on both sides; the equation is said to be . Thus, matter and charge are conserved during the process, and the process can take place in a closed system. The species to the left of a single arrow are called , the species to the right are called , and the arrow indicates the direction of the process. A reaction equation is sometimes written with right and left arrows \[ \ce{N2}\tx{(g)} + \ce{3H2}\tx{(g)} \arrows \ce{2NH3}\tx{(g)} \] to indicate that the process is at reaction equilibrium. It can also be written as a with an equal sign: \[ \ce{N2}\tx{(g)} + \ce{3H2}\tx{(g)} = \ce{2NH3}\tx{(g)} \] A reaction equation shows stoichiometric relations among the reactants and products. It is important to keep in mind that it specifies neither the initial and final states of a chemical process, nor the change in the amount of a reactant or product during the process. For example, the reaction equation N\(_2\) + 3 H\(_2\) \(\ra\) 2 NH\(_3\) does not imply that the system initially contains only N\(_2\) and H\(_2\), or that only NH\(_3\) is present in the final state; and it does not mean that the process consists of the conversion of exactly one mole of N\(_2\) and three moles of H\(_2\) to two moles of NH\(_3\) (although this is a possibility). Instead, the reaction equation tells us that a change in the amount of N\(_2\) is accompanied by three times this change in the amount of H\(_2\) and by twice this change, with the opposite sign, in the amount of NH\(_3\). It is convenient to indicate the progress of a chemical process with a variable called the . The reaction equation
N\(_2\) + 3 H\(_2\) \(\ra\) 2 NH\(_3\) for the synthesis of ammonia synthesis will serve to illustrate this concept. Let the system be a gaseous mixture of N\(_2\), H\(_2\), and NH\(_3\). If the system is and the intensive properties remain uniform throughout the gas mixture, there are five independent variables. We can choose them to be \(T\), \(p\), and the amounts of the three substances. We can write the total differential of the enthalpy, for instance, as \begin{equation} \begin{split} \dif H & = \Pd{H}{T}{p,\allni}\dif T + \Pd{H}{p}{T,\allni}\difp \cr & \quad + H\subs{N\(_2\)}\dif n\subs{N\(_2\)} + H\subs{H\(_2\)}\dif n\subs{H\(_2\)} + H\subs{NH\(_3\)}\dif n\subs{NH\(_3\)} \end{split} \tag{11.2.1} \end{equation} The notation \(\allni\) stands for the set of amounts of all substances in the mixture, and the quantities \(H\subs{N\(_2\)}\), \(H\subs{H\(_2\)}\), and \(H\subs{NH\(_3\)}\) are partial molar enthalpies. For example, \(H\subs{N\(_2\)}\) is defined by \begin{equation} H\subs{N\(_2\)} = \Pd{H}{n\subs{N\(_2\)}}{T, p, n\subs{H\(_2\)}, n\subs{NH\(_3\)}} \tag{11.2.2} \end{equation} If the system is , the amounts of the three substances can still change because of the reaction N\(_2\) + 3 H\(_2\) \(\ra\) 2 NH\(_3\), and the number of independent variables is reduced from five to three. We can choose them to be \(T\), \(p\), and a variable called advancement. The (or extent of reaction), \(\xi\), is the amount by which the reaction defined by the reaction equation has advanced in the forward direction from specified initial conditions. The quantity \(\xi\) has dimensions of amount of substance, the usual unit being the mole. Let the initial amounts be \(n_{\tx{N}_2,0}\), \(n_{\tx{H}_2,0}\), and \(n_{\tx{NH}_3,0}\). Then at any stage of the reaction process in the closed system, the amounts are given by \begin{equation} n\subs{N\(_2\)} = n_{\tx{N}_2,0} - \xi \qquad n\subs{H\(_2\)} = n_{\tx{H}_2,0} - 3\xi \qquad n\subs{NH\(_3\)} = n_{\tx{NH}_3,0} + 2\xi \tag{11.2.3} \end{equation} These relations come from the stoichiometry of the reaction as expressed by the stoichiometric coefficients in the reaction equation. The second relation, for example, expresses the fact that when one mole of reaction has occurred (\(\xi=1\mol\)), the amount of H\(_2\) in the closed system has decreased by three moles. Taking the differentials of Eqs. 11.2.3, we find that infinitesimal changes in the amounts are related to the change of \(\xi\) as follows: \begin{equation} \dif n\subs{N\(_2\)} = - \dif\xi \qquad \dif n\subs{H\(_2\)} = - 3\dif\xi \qquad \dif n\subs{NH\(_3\)} = 2\dif\xi \tag{11.2.4} \end{equation} These relations show that in a closed system, the changes in the various amounts are not independent. Substitution in Eq. 11.2.1 of the expressions for \(\dif n\subs{N\(_2\)}\), \(\dif n\subs{H\(_2\)}\), and \(\dif n\subs{NH\(_3\)}\) gives \begin{gather} \s{ \begin{split} \dif H & = \Pd{H}{T}{p, \xi}\dif T + \Pd{H}{p}{T, \xi}\difp \cr & \quad + \left( -H\subs{N\(_2\)} - 3H\subs{H\(_2\)} + 2H\subs{NH\(_3\)} \right)\dif\xi \end{split} } \tag{11.2.5} \cond{(closed system)} \end{gather} (The subscript \(\allni\) on the partial derivatives has been replaced by \(\xi\) to indicate the same thing: that the derivative is taken with the amount of each species held constant.) Equation 11.2.5 gives an expression for the total differential of the enthalpy with \(T\), \(p\), and \(\xi\) as the independent variables. The coefficient of \(\dif \xi\) in this equation is called the , or molar enthalpy of reaction, \(\Delsub{r}H\): \begin{equation} \Delsub{r}H = -H\subs{N\(_2\)} - 3H\subs{H\(_2\)} + 2H\subs{NH\(_3\)} \tag{11.2.6} \end{equation} We identify this coefficient as the partial derivative \begin{equation} \Delsub{r}H = \Pd{H}{\xi}{T, p} \tag{11.2.7} \end{equation} That is, the molar reaction enthalpy is the rate at which the enthalpy changes with the advancement as the reaction proceeds in the forward direction at constant \(T\) and \(p\). The partial molar enthalpy of a species is the enthalpy change per amount of the species added to an system. To see why the particular combination of partial molar enthalpies on the right side of Eq. 11.2.6 is the rate at which enthalpy changes with advancement in the system, we can imagine the following process at constant \(T\) and \(p\): An infinitesimal amount \(\dif n\) of N\(_2\) is removed from an open system, three times this amount of H\(_2\) is removed from the same system, and twice this amount of NH\(_3\) is added to the system. The total enthalpy change in the open system is \(\dif H = (-H\subs{N\(_2\)} - 3H\subs{H\(_2\)} + 2H\subs{NH\(_3\)})\dif n\). The net change in the state of the system is equivalent to an advancement \(\dif\xi = \dif n\) in a closed system, so \(\dif H/\dif\xi\) in the closed system is equal to \((-H\subs{N\(_2\)} - 3H\subs{H\(_2\)} + 2H\subs{NH\(_3\)})\) in agreement with Eqs. 11.2.6 and 11.2.7. Note that because the advancement is defined by how we write the reaction equation, the value of \(\Delsub{r}H\) also depends on the reaction equation. For instance, if we change the reaction equation for ammonia synthesis from N\(_2\) + 3 H\(_2\) \(\ra\) 2 NH\(_3\) to \[ \ce{1/2N2} + \ce{3/2H2} \arrow \ce{NH3} \] then the value of \(\Delsub{r}H\) is halved. Now let us generalize the relations of the preceding section for any chemical process in a closed system. Suppose the stoichiometric equation has the form \begin{equation} a\tx{A} + b\tx{B} = d\tx{D} + e\tx{E} \tag{11.2.8} \end{equation} where A and B are reactant species, D and E are product species, and \(a\), \(b\), \(d\), and \(e\) are the corresponding stoichiometric coefficients. We can rearrange this equation to \begin{equation} 0 = - a\tx{A} - b\tx{B} + d\tx{D} + e\tx{E} \tag{11.2.9} \end{equation} In general, the stoichiometric relation for any chemical process is \begin{equation} 0 = \sum_i\nu_i \tx{A}_i \tag{11.2.10} \end{equation} where \(\nu_i\) is the of species A\(_i\), a dimensionless quantity taken as negative for a reactant and positive for a product. In the ammonia synthesis example of the previous section, the stoichiometric relation is \(0 = -\tx{N}_2 - 3\tx{H}_2 + 2\tx{NH}_3\) and the stoichiometric numbers are \(\nu\subs{N\(_2\)} = -1\), \(\nu\subs{H\(_2\)} = -3\), and \(\nu\subs{NH\(_3\)} = +2\). In other words, each stoichiometric number is the same as the stoichiometric coefficient in the reaction equation, except that the sign is negative for a reactant. The amount of reactant or product species \(i\) present in the closed system at any instant depends on the advancement at that instant, and is given by \begin{gather} \s{ n_i = n_{i,0} + \nu_i\xi } \tag{11.2.11} \cond{(closed system)} \end{gather} The infinitesimal change in the amount due to an infinitesimal change in the advancement is \begin{gather} \s{ \dif n_i = \nu_i\dif\xi } \tag{11.2.12} \cond{(closed system)} \end{gather} In an open system, the total differential of extensive property \(X\) is \begin{equation} \dif X = \Pd{X}{T}{p, \allni}\dif T + \Pd{X}{p}{T, \allni}\difp + \sum_i X_i\dif n_i \tag{11.2.13} \end{equation} where \(X_i\) is a partial molar quantity. We restrict the system to a closed one with \(T\), \(p\), and \(\xi\) as the independent variables. Then, with the substitution \(\dif n_i = \nu_i\dif\xi\) from Eq. 11.2.12, the total differential of \(X\) becomes \begin{gather} \s{ \dif X = \Pd{X}{T}{p, \xi}\dif T + \Pd{X}{p}{T, \xi}\difp + \Delsub{r}X\dif\xi } \tag{11.2.14} \cond{(closed system)} \end{gather} where the coefficient \(\Delsub{r}X\) is the defined by \begin{equation} \Delsub{r}X \defn \sum_i\nu_i X_i \tag{11.2.15} \end{equation} Equation 11.2.14 allows us to identify the molar reaction quantity as a partial derivative: \begin{gather} \s{ \Delsub{r}X = \Pd{X}{\xi}{T, p} } \tag{11.2.16} \cond{(closed system)} \end{gather} It is important to observe the distinction between the notations \(\Del X\), the finite change of \(X\) during a process, and \(\Delsub{r}X\), a differential quantity that is a property of the system in a given state. The fact that both notations use the symbol \(\Del\) can be confusing. Equation 11.2.16 shows that we can think of \(\Delsub{r}\) as an . In dealing with the change of an extensive property \(X\) as \(\xi\) changes, we must distinguish between molar integral and molar differential reaction quantities. The notation for a molar differential reaction quantity such as \(\Delsub{r}H\) includes a subscript following the \(\Del\) symbol to indicate the kind of chemical process. The subscript “r” denotes a reaction or process in general. The meanings of “vap,” “sub,” “fus,” and “trs” were described in Sec. 8.3.1. Subscripts for specific kinds of reactions and processes are listed in Sec. D.2 of Appendix D and are illustrated in sections to follow. For certain kinds of processes, it may happen that a partial molar quantity \(X_i\) remains constant for each species \(i\) as the process advances at constant \(T\) and \(p\). If \(X_i\) remains constant for each \(i\), then according to Eq. 11.2.15 the value of \(\Delsub{r}X\) must also remain constant as the process advances. Since \(\Delsub{r}X\) is the rate at which \(X\) changes with \(\xi\), in such a situation \(X\) is a linear function of \(\xi\). This means that the molar integral reaction quantity \(\Del X\m\rxn\) defined by \(\Del X/\Del\xi\) is equal, for any finite change of \(\xi\), to \(\Delsub{r}X\). Enthalpy and entropy as functions of advancement at constant \(T\) and \(p\). The curves are for a reaction A\(\rightarrow\)2B with positive \(\Delsub{r}H\) taking place in an ideal gas mixture with initial amounts \(n_{\tx{A},0}=1\mol\) and \(n_{\tx{B},0}=0\). An example is the partial molar enthalpy \(H_i\) of a constituent of an ideal gas mixture, an ideal condensed-phase mixture, or an ideal-dilute solution. In these ideal mixtures, \(H_i\) is independent of composition at constant \(T\) and \(p\) (Secs. 9.3.3, 9.4.3, and 9.4.7). When a reaction takes place at constant \(T\) and \(p\) in one of these mixtures, the molar differential reaction enthalpy \(\Delsub{r}H\) is constant during the process, \(H\) is a linear function of \(\xi\), and \(\Delsub{r}H\) and \(\Del H\m\rxn\) are equal. Figure 11.6(a) illustrates this linear dependence for a reaction in an ideal gas mixture. In contrast, Fig. 11.6(b) shows the nonlinearity of the entropy as a function of \(\xi\) during the same reaction. The nonlinearity is a consequence of the dependence of the partial molar entropy \(S_i\) on the mixture composition (Eq. 11.1.24). In the figure, the slope of the curve at each value of \(\xi\) equals \(\Delsub{r}S\) at that point; its value changes as the reaction advances and the composition of the reaction mixture changes. Consequently, the molar integral reaction entropy \(\Del S\m\rxn=\Del S\rxn/\Del\xi\) approaches the value of \(\Delsub{r}S\) only in the limit as \(\Del\xi\) approaches zero. If a chemical process takes place at constant temperature while each reactant and product remains in its standard state of unit activity, the molar reaction quantity \(\Delsub{r}X\) is called the and is denoted by \(\Delsub{r}X\st\). For instance, \(\Delsub{vap}H\st\) is a standard molar enthalpy of vaporization (already discussed in Sec. 8.3.3), and \(\Delsub{r}G\st\) is the standard molar Gibbs energy of a reaction. From Eq. 11.2.15, the relation between a standard molar reaction quantity and the standard molar quantities of the reactants and products at the same temperature is \begin{equation} \Delsub{r}X\st \defn \sum_i\nu_i X_i\st \tag{11.2.18} \end{equation} Two comments are in order. These general concepts will now be applied to some specific chemical processes. | 21,385 | 989 |
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To some degree the reactions of amino acids are typical of isolated carboxylic acid and amine functions. Thus the carboxyl function can be esterified with an excess of an alcohol under acidic conditions, and the amine function can be acylated with acid chlorides or anhydrides under basic conditions: The products, however, are not indefinitely stable because the functional groups can, and eventually will, react with each other. For example, in the acylation of glycine with ethanoic anhydride, the first-formed product may cyclize to the "azlactone" if the reaction is prolonged or excess anhydride is used: Esters of amino acids also cyclize, but they do so intermolecularly to give "diketopiperazines". These compounds are cyclic amides: The amine function of \(\alpha\)-amino acids and esters reacts with nitrous acid in a manner similar to that described for primary amines ( ). The diazonium ion intermediate loses molecular nitrogen in the case of the acid, but the diazonium ester loses a proton and forms a relatively stable diazo compound known as ethyl diazoethanoate: This diazo ester is formed because loss of \(\ce{N_2}\) from the diazonium ion results in formation of a quite unfavorable carbocation. \(\alpha\)-Amino acids react with aldehydes to form decarboxylation and/or deamination products. The reaction sequence is shown in Figure 25-5 and closely resembles the ninhydrin reaction ( ). In the first step the amine condenses with the aldehyde to give an imine or Schiff base, \(2\). What happens next depends on the relative rates of proton shift and decarboxylation of \(2\). Proton shift produces a rearranged imine, \(3\), which can hydrolyze to the keto acid \(4\). The keto acid is a deamination product. Alternatively, decarboxylation can occur (see ) and the resulting imine, \(5\), can either hydrolyze or rearrange by a proton shift to a new imine, \(6\). Hydrolysis of \(5\) or \(6\) gives an aldehyde and an amine. There is an important biochemical counterpart of the deamination reaction that utilizes pyridoxal phosphate, \(7\), as the aldehyde. Each step in the sequence is catalyzed by a specific enzyme. The \(\alpha\)-amino group of the amino acid combines with \(7\) and is converted to a keto acid. The resulting pyridoxamine then reacts to form an imine with a different \(\alpha\)-keto acid, resulting in formation of a new \(\alpha\)-amino acid and regenerating \(7\). The overall process is shown in Equation 25-6 and is called . It is a key part of the process whereby amino acids are metabolized. \(\tag{25-6}\) The biochemical process occurs with complete preservation of the \(L\) configuration at the \(\alpha\) carbon. The same reactions can be carried out nonenzymatically using pyridoxal phosphate, but they are nonstereospecific, require metal ions as a catalyst, and give mixtures of products. and (1977) | 2,882 | 990 |
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A good map will take you to your destination with ease, provided you know how to read it. A map is an example of a , a pictorial representation of a body of knowledge. In science they play a considerable role. Next to plots and tables are an important means of making information and/or theoretical knowledge accessible. Constructing them takes quite a bit of thought. You want to represent as much of what you know and give as accurate a picture of it without conveying anything incorrect. If the drawing can be made to scale that makes it quite a bit more powerful, but this is not strictly necessary. A remark like or does need to be given if applicable. A good caption or description is essential. There are different kinds of equilibrium, besides the equilibrium that represents an absolute minimum in the \(G\) function. Of course \(G\) is potentially a function of a great number of variables, but let us look at a in which \(G\) is shown as a function of only one unspecified variable. You could think of the density, the mole fraction of one of the components of a mixture or an applied electrical field or whatever, but the argument is general. Figure 23.1.1
is helpful to point out that besides a stable equilibrium (\(A\)) there can also be an equilibrium (B) or an equilibrium (\(C\)). The local derivatives of \(G\) (versus variables of which we only show one) are zero in all three cases, which means that changes in the variables are . For a equilibrium (\(D\)) the opposite is true. Any small deviation will make the system role down hill. (Note that the second derivative has the opposite sign compared to cas A) and B)) A labile equilibrium is seldom or never observed except in a circus where artists delight in balancing objects on their heads (because you pay for it).. This usually requires continuous small corrections to maintain the precarious balance. All other points in our diagram represent state of because locally \(dG\) is not zero and a spontaneous process can take place. The fact that \(dG=0\) in the equilibrium points does not mean small deviations from the minimum cannot happen at times. We have seen e.g. that the Boltzmann distribution was simply distribution. The most likely one is the one that has the highest number of realizations W. Another way of saying that is that it is the one with the highest entropy S. A slightly less likely distribution may occur from time to time by chance. It will have a little less entropy, but the same \(\langle E \rangle\). That means it will have a slightly higher \(G\) (\(G=H-TS\)). From time to time therefore \(G\) will fluctuate a bit. Such fluctuations are very small for large systems, but they are of greater relative importance for small systems (like a nanoparticle). (Statistical averaging works best on large ensembles.) The fluctuations in \(G\) mean that small fluctuations in its variables like density etc. can also occur. They are usually kept in check, because \(dG\) is no longer zero when moving away from the equilibrium state. This drives the system back to the minimum spontaneously. You could picture the system wobbling a bit around in its G-well. This holds for stable and metastable equilibria alike. In the indifferent case (\(C\)) however the derivative is zero (or very close to zero) for a of neighboring values of some variable. In contrast to \(A\) and \(B\) also the is zero. This means that there is little penalty to much larger deviations in the variable. If this variable is the density the system becomes milky and shows a strong scattering of light because the refractive index depends on the strongly fluctuating density. This is observed near critical points and is called . A unary phase diagram summarizes the equilibrium states of a single pure substance. We will see that we can also look at mixtures of two components (binary diagrams) or more (ternary, quaternary, quinary, senary etc.). Usually a phase diagram only maps out stable equilibria, but occasionally metastable ones may be given too (e.g., with a dashed line). We have seen that the Gibbs function \(G\) depends strongly (logarithmically) on pressure for a gas, but only slightly (and linearly) for a liquid. The two curves intersect in a point representing the of the liquid. At lower pressures the vapor is more stable, at higher ones the liquid. (For a solid the same holds as for the liquid). This means that except at the intersection point we only will observe one phase. It is important to stress that this holds , e.g., when we put only water into an evacuated cylinder (Figure 23.1.2
). We may get three cases: At room temperature \(P_{eq}\) for water is only about 15 Torr. If we apply 1 bar -or let the atmosphere do the job- we will only have liquid water. If other gases are present, e.g., air, we must distinguish between the total pressure (e.g., 1 bar) and the equilibrium vapor pressure which will now be the pressure. In a cylinder with water and one bar of air just enough water will evaporate to establish equilibrium. The evaporation will be limited to the gas-liquid interface unless the partial pressure equals the total pressure. Then the liquid will . (Do allow the volume to expand, though., why? If the volume is constant the pressure builds up and boiling will stop. If we consider the set of equilibrium pressures as a function of temperature and plot that in a P vs. T diagram we have one component of our phase diagram. For solids the situation is similar as the \(G(P)\) curve is once again an almost flat straight line. The intersection with the logarithmic curve for the gas will define an equilibrium pressure for gas-solid co-existence. Generally vapor pressures above solids are quite small, but not negligible. As for liquids we can construct a line representing the equilibrium pressures for sublimation as function of temperature and add it to the phase diagram. The \(\text{solid} \rightleftharpoons \text{liquid}\) equilibrium is known as or is not very dependent on pressure. Usually melting points increase a little bit with pressure, although water is a peculiar exception. It expands upon freezing and the melting point goes down (a bit) with pressure. In our diagram this will represent an almost vertical line leaning a little forwards for most substances, but backwards for water and a few others. The three lines come together in the , the only point where all three phases are at equilibrium with each other. For water, its temperature is only 0.01 K different from the normal melting point (273.16 K) and its pressure is only 4.58 Torr. The intersection points with a line representing atmospheric pressure give the melting and boiling points at that pressure. If the triple point lies above the line that represents atmospheric pressure this implies that a liquid is never observed. On earth CO is such a substance. The intersection of the solid-vapor equilibrium line with the 1 bar line represents a state where the solid will (evaporate from inside out). This is known as the . The melting points at P=1 bar are known as the , the only slightly different one at 760 Torr = 1 atm is called . The same goes for boiling and sublimation points. There is about \(P=1 \,bar\). It just happens to be the pressure of our home planet. On a planet with higher atmospheric pressures \(CO_2\) may well be a liquid and on such a planet, all boiling points will be quite different (higher than on earth). The melting points will also differ, but only slightly so. On Mars where atmospheric pressure is much lower water can not occur in liquid form, much like carbon dioxide on earth - it sublimes. We should also realize that in a (glass ampoule, hermetically sealed DSC pan), we can observe melting points at only very slightly different temperature values, but we will see a boiling effect. Why? To see a boiling point the container must be open to the ( ) 1 bar pressure of the earths atmosphere that it and causes the boiling phenomenon. If the ampoule is sealed it will generate its (autogenous) pressure, depending on what you put in, how much of it in relation to the volume, how volatile it is and the temperature. The autogenous pressure does not interfere with the melting point much (the melting line is almost vertical), but as \(P\) changes with temperature you may never reach boiling conditions. In DSC experiments it is possible to observe boiling points only if the pan has been carefully with a hole of known size. It must be big enough that the pressure inside the pan does not build up above atmospheric, but small enough that it does not cause premature loss of mass during the run. The latter spoils the calculation of the intensive value (per mole, per gram) of the heat of vaporization. The liquid evaporation line ends in a point that we have encountered before: the critical point \(T_C\). As temperature increases the liquid and vapor phases in equilibrium with each other start to resemble each other more and more and at \(T_c\) they coalesce. At this point the liquid-gas equilibrium becomes indifferent with respect to density and large fluctuations occur leading to critical opalescence. Notice that there is a relationship of dimensionality between the objects in the diagram and the number of phases present: As you see the sum is always three. So far we have typically considered one substance at the time, but for chemists it is imperative to deal with more than one because we are typically changing one into the other in our reactions. This means that the number of moles \(n\), that we often simply set equal to one now becomes an important variable in its own right. Besides we will actually have two (or more) of them: the number of moles of component one and the one for the other component. This makes n a much less trivial variable. This is already the case at a simple melting point, say when ice melts, because we are dealing with changing quantities of ice and water: \[n_{ice} + n_{water} = n_{total} \nonumber \] If all we do is turn water into ice or vice versa, we have \(dn_{total}=0\), so that: \[dn_{ice} =- dn_{water} \nonumber \] To deal with changing n's, we need to expand our mathematical notation a bit. So far we have simply divided our thermodynamic functions if they were extensive by the number of moles and arrived at intensive molar values: \[G_{molar} = \dfrac{G}{ n} \nonumber \] \[V_{molar} = \dfrac{V}{n} \nonumber \] We have written such intensive molar values by writing a bar over the symbol G or V. We should note that scaling the function this way departs from the assumption that the function \(G\) depends on the variable n as a straight line that passes through the origin. If we have the same pure compound in two phases, like ice and water we can still apply this principle and write: \[G_{molar}^{ice} = \dfrac{G^{ice}}{n^{ice}} \nonumber \] \[V_{molar}^{ice} = \dfrac{V^{ice}}{n^{ice}} \nonumber \] \[G_{molar}^{water} = \dfrac{G^{water}}{n^{water}} \nonumber \] \[V_{molar}^{water} = \dfrac{V^{water}}{n^{water}} \nonumber \] If we have a of two substances present as \(n_1\) and \(n_2\) moles the dependency need not be linear on either if the two substances interact with each other. This is also true for function like the volume of a liquid mixture. In the presence of interactions volumes do have to be linearly additive. We can define a partial molar value of e.g. for the volume: \[V_{partial molar,1} = \dfrac{\partial V}{\partial n_1} \nonumber \] at \(n_2\) = constant \[V_{partial molar,2} = \dfrac{\partial V}{\partial n_2} \nonumber \] at \(n_1\) = constant The notation of putting a bar over the \(V\) symbol is used for these partial quantities as well. Partial molar volumes have been measured for many binary systems. They are functions of the composition (mole fraction) as well as the temperature and to a lesser extent the pressure. The partial molar Gibbs free energy (\(\left(\dfrac{∂G}{∂n_i}\right)_{P,T}\), all other n's) is denoted with \(μ\) and is called the thermodynamic potential. When numbers of moles can change we can write the corresponding change in \(G\) as: \[dG = -SdT + VdP + \sum_i^N \left( \dfrac{\partial G}{\partial n_1} \right)_{P,T,n_{j\neq i}}dn_i \nonumber \] \[dG = -SdT + VdP + \sum_i^N μ_idn_i \nonumber \] over \(N\) phases in the system. As you can see we are adding a set of conjugate variables \(μ_in_i\) for each phase \(i\). If we are considering a component (but in different modifications, like ice and steam), we can still write: \[ μ_i = \left( \dfrac{\partial G}{\partial n_1} \right)_{P,T,n_{j\neq i}} = \dfrac{G_i}{n_i} \nonumber \] As soon as we are dealing with mixtures we really do have derivatives. | 12,766 | 991 |
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The search for life elsewhere in the universe has centered on finding because they are the stuff of all life on earth. Organic compounds are easily synthesized by abiotic (non-living) processes, however, so the search for extraterrestrial life has occasionally centered on other "biomarkers" (chemical structures that are typically found in living things). A meteorite from Mars collided with the earth and was collected in 1984 from the Allan Hills region of Antarctica and thus designated ALH84001. An asteroid probably collided with Mars, breaking off the chunk which circled the sun for eons before colliding with the earth as a meteorite. The meteorite contained PAHs (Polycyclic Aromatic Hydrocarbons) containing rings of six carbon atoms, fused together in arrays like those shown below. Benzo-α-pyrene is formed when meat is barbecued (it may be a carcinogen) and anthracene is found in coal as a relic of it's biological origin. Note that these organic compounds have even numbers of carbon atoms (a sign of biotic origin). They are considered because they contains carbon chains and hydrogen atoms (H atoms are not shown in the figure by convention; organic chemists know that there must be hydrogen atoms where necessary to make a total of four bonds to each carbon). Carbon-carbon bonds are quite strong, allowing formation of long chains to which side branches and a variety of functional groups may be attached. Hence the number of molecular structures which can be adopted by organic compounds is extremely large. PAHs are because they contain only carbon and hydrogen, like the simpler hydrocarbon ethane (shown below). Hydrocarbons are further categorized as alkanes, cycloalkanes, aromatic compounds, alkenes and alkynes, Organic compounds often also contain oxygen, nitrogen, and small proportions of other elements. Addition elements other than carbon and hydrogen creates different classes of organic compounds. For example, alcohols contain the -OH group, as shown below in ethanol (drinking alcohol). The meteorite ALH84001 also contained carbonate minerals (like CaCO ), which are not organic, but are the stuff of stalactites formed when water dissolves carbon dioxide and passes through cracks in rocks. Since water may be a prerequisite of life, carbonates are considered a biomarker, especially in the shapes found on ALH84001, which are reminiscent of cell fossils. Carbonates are not considered organic, because they don't contain carbon atoms bonded to each other as described above. Carbonates are typical inorganic, ionic compounds. Several other types of evidence, including the structures shown in the figure, convinced David McKay and his NASA collaborators to claim "Although there are alternative explanations for each of these phenomena taken individually, when they are considered collectively, particularly in view of heir spatial association, we conclude that they are evidence for primitive life on early Mars." Many skeptics, notably paleonologist J. William Schopf, pointed out that "organic compounds" come from sources other than organisms (life), and are often created by abiotic processes, as are carbonate rocks. Furthermore, all meteorites, when carefully inspected, show signs of life, because they are almost immediately colonized by Earth's biota. In the end, the meteorite held no convincing evidence for life. There are several telltale markers that distinguish abiotic "organic" compounds, like those that are synthesized in the laboratory or under the extreme conditions found on most planets where life as we know it cannot survive: , Unique groupings of atoms like the -OH group in alcohols mentioned above, are called . Functional groups determine the chemical and physical properties of organic molecules. Living things are characterized by involvement of molecules containing the carbonyl (-C=O) functional group which appears in aldehydes and ketones as well as carboxylic acids and esters. Organic Nitrogen Compounds like amino acids and nucleotides are also nearly universally found in terran living things. Amino acids have been detected in comets . It is possible that life elsewhere ("nonterran" life is NASA's term ) may be entirely different, and a NASA conference on "Weird Life" explored other possibilities for what "organic" might mean elsewhere in the universe. Extraterrestrial life is studied by "Astrobiologists" or "Exobiologists" (working with astronomers at NASA ). Astronomers interested in nonterran life need to understand the types of organic molecules found on Earth, their properties that make them useful in living things, and how these properties differ from other classes of substances. Macroscopic physical properties such as melting and boiling points depend on the strengths of the forces which hold microscopic particles together. In the case of molecules whose atoms are connected by covalent bonds, such intermolecular forces may be of three types. All molecules are attracted together by weak London forces. These depend on instantaneous polarization and increase in strength with the size of the molecular electron cloud. When a molecule contains atoms whose electronegativities differ significantly and the resulting bond dipoles do not cancel each other’s effects, dipole forces occur. This results in higher melting and boiling points than for nonpolar substances. The third type of intermolecular force, the hydrogen bond, occurs when one molecule contains a hydrogen atom connected to a highly electronegative partner. The other molecule must contain an electronegative atom, like fluorine, oxygen, or nitrogen, which has a lone pair. Although each hydrogen bond is weak compared with a covalent bond, large numbers of hydrogen bonds can have very significant effects. One example of this is in the properties of water. This highly unusual liquid plays a major role in making living systems and the earth’s environment behave as they do. | 5,968 | 992 |
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Nitrogen has a wide range of oxidation states in organic compounds. We can arrive at an arbitrary scale for the oxidation of nitrogen in much the same way as we did for carbon ( ). We simply define elementary nitrogen as the zero oxidation state, and every atom bonded to nitrogen contributes -1 to the oxidation state if it is more electropositive than nitrogen (e.g., \(\ce{H}\), \(\ce{C}\), \(\ce{Li}\), \(\ce{B}\), \(\ce{Mg}\)) and +1 if it is more electronegative (e.g., \(\ce{O}\), \(\ce{F}\), \(\ce{Cl}\)). Doubly bonded atoms are counted twice, and a formal positive charge associated with nitrogen counts as +1. To illustrate, the oxidation states of several representative compounds are as follows: Several types of nitrogen compounds are listed in Table 23-5 to illustrate the range of oxidation states that are possible. For the oxidation of a tertiary amine by reagents such as hydrogen peroxide, \(\ce{H_2O_2}\), or peroxycarboxylic acids, \(\ce{RCOOOH}\), which can supply an oxygen atom with six electrons, the expected product is an azane oxide (amine oxide). Thus \(\ce{N}\),\(\ce{N}\)-diethylethanamine (triethylamine) can be oxidized to triethylazane oxide (triethylamine oxide): Amine oxides are interesting for two reasons. First, amine oxides decompose when strongly heated, and this reaction provides a useful preparation of alkenes. With triethylazane oxide (triethylamine oxide), ethene is formed: The second interesting point about amine oxides is that, unlike amines, they do not undergo rapid inversion at the nitrogen atom, and the oxides from amines with three different \(\ce{R}\) groups are resolvable into optically active forms. This has been achieved for several amine oxides, including the one from \(\ce{N}\)-ethyl-\(\ce{N}\)-methyl-2-propanamine. Addition of an oxygen atom from hydrogen peroxide or a peroxyacid to a primary or secondary amine might be expected to yield an amine oxide-type intermediate, which then could rearrange to an azanol (hydroxylamine): However, these oxidations usually take a more complicated course, because the azanols themselves are oxidized easily, and in the case of primary amines, oxidation occurs all the way to nitro compounds, in fair-to-good yields: We shall use benzenamine to illustrate some typical oxidation reactions of arenamines. The course of oxidation depends on the nature of the oxidizing agent and on the arenamine. With hydrogen peroxide or peroxycarboxylic acids, each of which functions , oxidation to the azanol, the nitroso, or the nitro compound may occur, depending on the temperature, the pH, and the amount of oxidizing agent: Oxidizing agents that lead to more complex reactions, which often result in highly colored products. One of the best black dyes for fabric (Aniline Black) is produced by impregnating cloth with phenylammonium chloride solution and then oxidizing, first with sodium chlorate \(\left( \ce{NaClO_3} \right)\) and finally with sodium dichromate \(\left( \ce{Na_2Cr_2O_7} \right)\). Aniline Black probably is not a single substance, and its exact structure(s) is not known; but its formation certainly involves addition reactions in which carbon-nitrogen bonds are made. A possible structure is shown in which there are seven aniline units: Oxidation of benzenamine with sodium dichromate in aqueous sulfuric acid solution produces 1,4-cyclohexadienedione ( -benzoquinone), which is the simplest member of an interesting class of conjugated cyclic diketones that will be discussed in more detail in Chapter 26: and (1977) | 3,563 | 993 |
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Acids and bases have been known for a long time. When Robert characterized them in 1680, he noted that acids dissolve many substances, change the color of certain natural dyes (for example, they change litmus from blue to red), and lose these characteristic properties after coming into contact with alkalis (bases). In the eighteenth century, it was recognized that acids have a sour taste, react with limestone to liberate a gaseous substance (now known to be CO ), and interact with alkalis to form neutral substances. In 1815, Humphry contributed greatly to the development of the modern acid-base concept by demonstrating that hydrogen is the essential constituent of acids. Around that same time, Joseph Louis Gay-Lussac concluded that acids are substances that can neutralize bases and that these two classes of substances can be defined only in terms of each other. The significance of hydrogen was reemphasized in 1884 when Carl Axel defined an acid as a compound that dissolves in water to yield hydrogen cations (now recognized to be hydronium ions) and a base as a compound that dissolves in water to yield hydroxide anions. Acids and bases are common solutions that exist everywhere. Almost every liquid that we encounter in our daily lives consists of acidic and basic properties, with the exception of water. They have completely different properties and are able to neutralize to form H O, which will be discussed later in a subsection. Acids and bases can be defined by their physical and chemical observations (Table \(\Page {1}\)). Common Examples: Soap, toothpaste, bleach, cleaning agents, limewater, ammonia water, sodium hydroxide. Acids and bases in aqueous solutions will conduct electricity because they contain dissolved ions. Therefore, acids and bases are . Strong acids and bases will be strong . Weak acids and bases will be weak electrolytes. This affects the amount of conductivity. In chemistry, acids and bases have been defined differently by : One is the Arrhenius definition defined above, which revolves around the idea that acids are substances that ionize (break off) in an aqueous solution to produce hydrogen (H ) ions while bases produce hydroxide (OH ) ions in solution. The other two definitions are discussed in detail include the the defines acids as substances that donate protons (H ) whereas bases are substances that accept protons and the of acids and bases states that acids are electron pair acceptors while bases are electron pair donors. In 1884, the Swedish chemist Svante Arrhenius proposed two specific classifications of compounds, termed acids and bases. When dissolved in an aqueous solution, certain ions were released into the solution. The Arrhenius definition of acid-base reactions is a development of the "hydrogen theory of acids". It was used to provide a modern definition of acids and bases, and followed from Arrhenius's work with Friedrich Wilhelm Ostwald in establishing the presence of ions in aqueous solution in 1884. This led to Arrhenius receiving the Nobel Prize in Chemistry in 1903. An is a compound that increases the concentration of \(\ce{H^{+}}\) ions that are present when added to water. \[ \ce{HCl(g) \rightarrow H^{+}(aq) + Cl^{-}(aq)} \label{eq1}\] In this reaction, hydrochloric acid (\(\ce{HCl}\)) gas dissociates into hydrogen (\(\ce{H^{+}}\)) and chloride (\(\ce{Cl^{-}}\)) ions when dissolved in water, thereby releasing \(\ce{H^{+}}\) ions into solution. These \(\ce{H^{+}}\) ions form the hydronium ion (\(\ce{H_3O^{+}}\)) when they combine with water molecules. Both processes can be represented in a chemical equation by adding \(\ce{H2O}\) to the reactants side of Equation \ref{eq1} and switching hydronium ions for free protons. \[ \ce{ HCl(g) + H2O(l) \rightarrow H_3O^{+}(aq) + Cl^{-}(aq)} \label{eq2} \] An is a compound that dissociates to yield hydroxide ions (\(\ce{OH^{−}}\)) in aqueous solution. Common examples of Arrhenus bases include Sodium hydroxide (\(\ce{NaOH}\)), Potassium hydroxide (\(\ce{KOH}\)), Magnesium hydroxide \(\ce{Mg(OH)2}\), and Calcium hydroxide (\(\ce{Ca(OH)2}\). All of these bases are solids at room temperature and when dissolved in water, will generate a metal cation and the hydroxide ion (\(\ce(OH^{-}}\)), for example, Sodium hydroxide \[\ce{NaOH(s) -> Na^{+}(aq) + OH^{-}(aq)} \label{eq3}\] All Arrhenius acids have easily detachable hydrogen that leave to form hydronium ions in solution and all Arrhenius bases have easily detachable OH groups that form hydroxide ions in solution. The Arrhenius definitions of acidity and alkalinity are restricted to aqueous solutions and refer to the concentration of the solvated ions. Under this definition, pure \(\ce{H2SO4}\) or \(\ce{HCl}\) dissolved in toluene are not acidic, despite the fact that both of these acids will donate a proton to toluene. In addition, under the Arrhenius definition, a solution of sodium amide (\(\ce{NaNH2}\)) in liquid ammonia is not alkaline, despite the fact that the amide ion (\(\ce{NH2^{-}}\)) will readily deprotonate ammonia. Thus, the Arrhenius definition can only describe acids and bases in an aqueous environment. The Arrhenius definition can describe acids and bases in protic solvents and environments (e.g., water, alcohols, within proteins etc.). The Arrhenius definitions identified an acid as a compound that dissolves in water to yield hydronium ions (Equation \ref{eq2}\) and a base as a compound that dissolves in water to yield hydroxide ions (Equation \ref{eq3}). As mentioned above, this definition limited. We extended the definition of an acid or a base using the more general definition proposed in 1923 by the Danish chemist Johannes Brønsted and the English chemist Thomas Lowry. Their definition centers on the proton, \(\ce{H^+}\). A proton is what remains when a normal hydrogen atom, \(\ce{^1_1H}\), loses an electron. A compound that donates a proton to another compound is called a , and a compound that accepts a proton is called a . An acid-base reaction is the transfer of a proton from a proton donor (acid) to a proton acceptor (base). Acids may be compounds such as \(\ce{HCl}\) or \(\ce{H2SO4}\), organic acids like acetic acid (\(\ce{CH_3COOH}\)) or ascorbic acid (vitamin C), or \(\ce{H2O}\). Anions (such as \(\ce{HSO_4^-}\), \(\ce{H_2PO_4^-}\), \(\ce{HS^-}\), and \(\ce{HCO_3^-}\)) and cations (such as \(\ce{H_3O^+}\), \(\ce{NH_4^+}\), and \(\ce{[Al(H_2O)_6]^{3+}}\)) may also act as acids. Bases fall into the same three categories and may be neutral molecules (such as \(\ce{H_2O}\), \(\ce{NH_3}\), and \(\ce{CH_3NH_2}\)), anions (such as \(\ce{OH^-}\), \(\ce{HS^-}\), \(\ce{HCO_3^-}\), \(\ce{CO_3^{2−}}\), \(\ce{F^-}\), and \(\ce{PO_4^{3−}}\)), or cations (such as \(\ce{[Al(H_2O)_5OH]^{2+}}\)). The most familiar bases are ionic compounds such as \(\ce{NaOH}\) and \(\ce{Ca(OH)_2}\), which contain the hydroxide ion, \(\ce{OH^-}\). The hydroxide ion in these compounds accepts a proton from acids to form water: \[\ce{H^+ + OH^- \rightarrow H_2O} \label{15.1.1}\] We call the product that remains after an acid donates a proton the of the acid. This species is a base because it can accept a proton (to re-form the acid): \[\begin{align*} \text{acid} &\rightleftharpoons \text{proton} + \text{conjugate base} \\[4pt] \ce{HF} &\rightleftharpoons \ce{H^+ + F^-} \\[4pt] \ce{H_2SO_4} &\rightleftharpoons \ce{H^+ + HSO_4^{−}} \\[4pt] \ce{H_2O} &\rightleftharpoons \ce{H^+ + OH^-} \\[4pt] \ce{HSO_4^-} &\rightleftharpoons \ce{H^+ + SO_4^{2−}} \\[4pt] \ce{NH_4^+} &\rightleftharpoons \ce{H^+ + NH_3} \end{align*}\] We call the product that results when a base accepts a proton the base’s . This species is an acid because it can give up a proton (and thus re-form the base): \[\begin{align*} \text{base} + \text{proton} &\rightleftharpoons \text{conjugate acid} \\[4pt] \ce{OH^- + H^+} &\rightleftharpoons \ce{H2O} \\[4pt] \ce{H_2O + H^+} &\rightleftharpoons \ce{H3O+} \\[4pt] \ce{NH_3 + H^+} &\rightleftharpoons \ce{NH4+} \\[4pt] \ce{S^{2-} + H^+} &\rightleftharpoons \ce{HS-} \\[4pt] \ce{CO_3^{2-} + H^+} &\rightleftharpoons \ce{HCO3-} \\[4pt] \ce{F^- + H^+} &\rightleftharpoons \ce{HF} \end{align*}\] In these two sets of equations, the behaviors of acids as proton donors and bases as proton acceptors are represented in isolation. In reality, all acid-base reactions involve the transfer of protons between acids and bases. For example, consider the acid-base reaction that takes place when ammonia is dissolved in water. A water molecule (functioning as an acid) transfers a proton to an ammonia molecule (functioning as a base), yielding the conjugate base of water, \(\ce{OH^-}\), and the conjugate acid of ammonia, \(\ce{NH4+}\): The reaction between a Brønsted-Lowry acid and water is called . For example, when hydrogen fluoride dissolves in water and ionizes, protons are transferred from hydrogen fluoride molecules to water molecules, yielding hydronium ions and fluoride ions: When we add a base to water, a reaction occurs in which protons are transferred from water molecules to base molecules. For example, adding ammonia to water yields hydroxide ions and ammonium ions: Notice that both these ionization reactions are represented as equilibrium processes. The relative extent to which these acid and base ionization reactions proceed is an important topic treated in a later section of this chapter. In the preceding paragraphs we saw that water can function as either an acid or a base, depending on the nature of the solute dissolved in it. In fact, in pure water or in any aqueous solution, water acts both as an acid and a base. A very small fraction of water molecules donate protons to other water molecules to form hydronium ions and hydroxide ions: This type of reaction, in which a substance ionizes when one molecule of the substance reacts with another molecule of the same substance, is referred to as . Pure water undergoes autoionization to a very slight extent. Only about two out of every \(10^9\) molecules in a sample of pure water are ionized at 25 °C. The equilibrium constant for the ionization of water is called the ion-product constant for water (\(K_w\)): \[\ce{H_2O(l) + H2O(l) <=> H_3O^{+}(aq) + OH^{-}(aq)} \] with \[K_\ce{w}=\ce{[H_3O^+,OH^- ]} \label{15.1.4}\] The slight ionization of pure water is reflected in the small value of the equilibrium constant; at 25 °C, \(K_w\) has a value of \(1.0 \times 10^{−14}\). The process is endothermic, and so the extent of ionization and the resulting concentrations of hydronium ion and hydroxide ion increase with temperature. For example, at 100 °C, the value for \(K_\ce{w}\) is approximately \(5.1 \times 10^{−13}\), roughly 100-times larger than the value at 25 °C. What are the hydronium ion concentration and the hydroxide ion concentration in pure water at 25 °C? The autoionization of water yields the same number of hydronium and hydroxide ions. Therefore, in pure water, \(\ce{[H_3O^+]} = \ce{[OH^- ]}\). At 25 °C: \[\begin{align*} K_\ce{w} &=\ce{[H_3O^+,OH^- ]} \\[4pt] &= [\ce{H3O^{+}}]^2 \\[4pt] &= [\ce{OH^{-}} ]^2 \\[4pt] &=1.0 \times 10^{−14} \end{align*}\] So \[\ce{[H_3O^+]}=\ce{[OH^- ]}=\sqrt{1.0 \times 10^{−14}} =1.0 \times 10^{−7}\; M \nonumber\] The hydronium ion concentration and the hydroxide ion concentration are the same, and we find that both equal \(1.0 \times 10^{−7}\; M\). The ion product of water at 80 °C is \(2.4 \times 10^{−13}\). What are the concentrations of hydronium and hydroxide ions in pure water at 80 °C? \[\ce{[H_3O^+]} = \ce{[OH^- ]} = 4.9 \times 10^{−7} \; M\nonumber\] It is important to realize that the autoionization equilibrium for water is established in all aqueous solutions. Adding an acid or base to water will not change the position of the equilibrium. Example \( \Page {2}\) demonstrates the quantitative aspects of this relation between hydronium and hydroxide ion concentrations. The Inverse Proportionality of [H O ] and [OH ] A solution of carbon dioxide in water has a hydronium ion concentration of \(2.0 \times 10^{−6}\; M\). What is the concentration of hydroxide ion at 25 °C? We know the value of the ion-product constant for water at 25 °C: \[\ce{2 H_2O}_{(l)} \rightleftharpoons \ce{H_3O^+}_{(aq)} + \ce{OH^-}_{(aq)} \nonumber\] \[K_\ce{w}=\ce{[H3O+,OH^- ]}=1.0 \times 10^{−14} \nonumber\] Thus, we can calculate the missing equilibrium concentration. Rearrangement of the expression yields that \([\ce{OH^- }]\) is directly proportional to the inverse of [H O ]: \[[\ce{OH^- }]=\dfrac{K_{\ce w}}{[\ce{H_3O^+}]}=\dfrac{1.0 \times 10^{−14}}{2.0 \times 10^{−6}}=5.0 \times 10^{−9} \nonumber\] The hydroxide ion concentration in water is reduced to \(5.0 \times 10^{−9}\: M\) as the hydrogen ion concentration increases to \(2.0 \times 10^{−6}\; M\). This is expected from Le Chatelier’s principle; the autoionization reaction shifts to the left to reduce the stress of the increased hydronium ion concentration and the \(\ce{[OH^- ]}\) is reduced relative to that in pure water. A check of these concentrations confirms that our arithmetic is correct: \[K_\ce{w}=\ce{[H_3O^+,OH^- ]}=(2.0 \times 10^{−6})(5.0 \times 10^{−9})=1.0 \times 10^{−14} \nonumber\] What is the hydronium ion concentration in an aqueous solution with a hydroxide ion concentration of 0.001 M at 25 °C? \[\ce{[H3O+]} = 1 \times 10^{−11} M \nonumber\] Like water, many molecules and ions may either gain or lose a proton under the appropriate conditions. Such species are said to be . Another term used to describe such species is , which is a more general term for a species that may act either as an acid or a base by any definition (not just the Brønsted-Lowry one). Consider for example the bicarbonate ion, which may either donate or accept a proton as shown here: \[\ce{HCO^-3(aq) + H2O(l) <=> CO^{2-}3(aq) + H3O^+(aq)} \nonumber\] \[ \ce{HCO^{-}3(aq) + H2O(l) <=> H2CO3(aq) + OH^{-}(aq)} \nonumber\] Write separate equations representing the reaction of \(\ce{HSO3-}\) Write separate equations representing the reaction of \(\ce{H2PO4-}\) \(\ce{H2PO4-}(aq)+\ce{HBr}(aq)\rightleftharpoons \ce{H3PO4}(aq)+\ce{Br-}(aq)\) \(\ce{H2PO4-}(aq)+\ce{OH^-} (aq)\rightleftharpoons \ce{HPO4^2-}(aq)+ \ce{H_2O}_{(l)} \) The Brønsted–Lowry concept of acids and bases defines a base as any species that can accept a proton, and an acid as any substance that can donate a proton. Lewis proposed an alternative definition that focuses on instead. According to Lewis: In modern chemistry, electron donors are often referred to as nucleophiles, while acceptors are electrophiles. Just as any Arrhenius acid is also a Brønsted acid, any Brønsted acid is also a Lewis acid, so the various acid-base concepts are all "upward compatible". Although we do not really need to think about electron-pair transfers when we deal with ordinary aqueous-solution acid-base reactions, it is important to understand that it is the opportunity for electron-pair sharing that enables proton transfer to take place. This equation for a simple acid-base neutralization shows how the Brønsted and Lewis definitions are really just different views of the same process. Take special note of the following points: The point about the electron-pair remaining on the donor species is especially important to bear in mind. For one thing, it distinguishes a from an , in which a physical transfer of one or more electrons from donor to acceptor does occur. The product of a Lewis acid-base reaction is known formally as an " " or "complex", although we do not ordinarily use these terms for simple proton-transfer reactions such as the one in the above example. Here, the proton combines with the hydroxide ion to form the "adduct" \(\ce{H2O}\). The following examples illustrate these points for some other proton-transfer reactions that you should already be familiar with. Another example, showing the autoprotolysis of water. Note that the conjugate acid is also the adduct. Ammonia is both a Brønsted and a Lewis base, owing to the unshared electron pair on the nitrogen. The reverse of this reaction represents the of the ammonium ion. Because \(\ce{HF}\) is a weak acid, fluoride salts behave as bases in aqueous solution. As a Lewis base, \(\ce{F^{–}}\) accepts a proton from water, which is transformed into a hydroxide ion. The bisulfite ion is and can act as an electron donor or acceptor. All Brønsted–Lowry bases (proton acceptors), such as \(\ce{OH^{−}}\), \(\ce{H2O}\), and \(\ce{NH3}\), are also electron-pair donors. Thus the Lewis definition of acids and bases does not contradict the Brønsted–Lowry definition. Rather, it expands the definition of acids to include substances other than the \(\ce{H^{+}}\) ion. Electron-deficient molecules, such as BCl , contain less than an octet of electrons around one atom and have a strong tendency to gain an additional pair of electrons by reacting with substances that possess a lone pair of electrons. Lewis’s definition, which is less restrictive than either the Brønsted–Lowry or the Arrhenius definition, grew out of his observation of this tendency. A general Brønsted–Lowry acid–base reaction can be depicted in Lewis electron symbols as follows: The proton (H ), which has no valence electrons, is a Lewis acid because it accepts a lone pair of electrons on the base to form a bond. The proton, however, is just one of many electron-deficient species that are known to react with bases. For example, neutral compounds of boron, aluminum, and the other Group 13 elements, which possess only six valence electrons, have a very strong tendency to gain an additional electron pair. Such compounds are therefore potent Lewis acids that react with an electron-pair donor such as ammonia to form an acid–base adduct, a new covalent bond, as shown here for boron trifluoride (BF ): The bond formed between a Lewis acid and a Lewis base is a because both electrons are provided by only one of the atoms (N, in the case of F B:NH ). After it is formed, however, a coordinate covalent bond behaves like any other covalent single bond. Species that are very weak Brønsted–Lowry bases can be relatively strong Lewis bases. For example, many of the group 13 trihalides are highly soluble in ethers (R–O–R′) because the oxygen atom in the ether contains two lone pairs of electrons, just as in H O. Hence the predominant species in solutions of electron-deficient trihalides in ether solvents is a Lewis acid–base adduct. A reaction of this type is shown in Figure \(\Page {1}\) for boron trichloride and diethyl ether: Many molecules with multiple bonds can act as Lewis acids. In these cases, the Lewis base typically donates a pair of electrons to form a bond to the central atom of the molecule, while a pair of electrons displaced from the multiple bond becomes a lone pair on a terminal atom. Identify the acid and the base in each Lewis acid–base reaction. reactants and products identity of Lewis acid and Lewis base In each equation, identify the reactant that is electron deficient and the reactant that is an electron-pair donor. The electron-deficient compound is the Lewis acid, whereas the other is the Lewis base. Identify the acid and the base in each Lewis acid–base reaction. Lewis base: (CH ) O; Lewis acid: BF Lewis base: H O; Lewis acid: SO Here are several more examples of Lewis acid-base reactions that be accommodated within the Brønsted or Arrhenius models. Identify the Lewis acid and Lewis base in each reaction. A compound that can donate a proton (a hydrogen ion) to another compound is called a Brønsted-Lowry acid. The compound that accepts the proton is called a Brønsted-Lowry base. The species remaining after a Brønsted-Lowry acid has lost a proton is the conjugate base of the acid. The species formed when a Brønsted-Lowry base gains a proton is the conjugate acid of the base. Thus, an acid-base reaction occurs when a proton is transferred from an acid to a base, with formation of the conjugate base of the reactant acid and formation of the conjugate acid of the reactant base. Amphiprotic species can act as both proton donors and proton acceptors. Water is the most important amphiprotic species. It can form both the hydronium ion, H O , and the hydroxide ion, \(\ce{OH^-}\) when it undergoes autoionization: \[\ce{2 H_2O}_{(l)} \rightleftharpoons \ce{H_3O^+}(aq)+\ce{OH^-} (aq) \nonumber\] The ion product of water, is the equilibrium constant for the autoionization reaction: \[K_\ce{w}=\mathrm{[H_3O^+,OH^- ]=1.0 \times 10^{−14} \; at\; 25°C} \nonumber\] ) ). | 20,594 | 994 |
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Another approach to describing reaction rates is based on the time required for the concentration of a reactant to decrease to one-half its initial value. This period of time is called the of the reaction, written as . Thus the half-life of a reaction is the time required for the reactant concentration to decrease from [A] to [A] . If two reactions have the same order, the faster reaction will have a shorter half-life, and the slower reaction will have a longer half-life. The half-life of a first-order reaction under a given set of reaction conditions is a constant. This is not true for zeroth- and second-order reactions. The half-life of a first-order reaction is independent of the concentration of the reactants. This becomes evident when we rearrange the integrated rate law for a first-order reaction to produce the following equation: Substituting [A] for [A] and for (to indicate a half-life) into gives Substituting \(\ln{2} \approx 0.693\) into the equation results in the expression for the half-life of a first-order reaction: \[t_{1/2}=\dfrac{0.693}{k} \label{21.4.2}\] Thus, for a first-order reaction, each successive half-life is the same length of time, as shown in , and is of [A]. If we know the rate constant for a first-order reaction, then we can use half-lives to predict how much time is needed for the reaction to reach a certain percent completion. As you can see from this table, the amount of reactant left after half-lives of a first-order reaction is (1/2) times the initial concentration. For a first-order reaction, the concentration of the reactant decreases by a constant with each half-life and is independent of [A]. The anticancer drug cis-platin hydrolyzes in water with a rate constant of 1.5 × 10 min at pH 7.0 and 25°C. Calculate the half-life for the hydrolysis reaction under these conditions. If a freshly prepared solution of cis-platin has a concentration of 0.053 M, what will be the concentration of cis-platin after 5 half-lives? after 10 half-lives? What is the percent completion of the reaction after 5 half-lives? after 10 half-lives? rate constant, initial concentration, and number of half-lives half-life, final concentrations, and percent completion We can calculate the half-life of the reaction using : \(t_{1/2}=\dfrac{0.693}{k}=\dfrac{0.693}{1.5\times10^{-3}\textrm{ min}^{-1}}=4.6\times10^2\textrm{ min}\) Thus it takes almost 8 h for half of the cis-platin to hydrolyze. After 5 half-lives (about 38 h), the remaining concentration of cis-platin will be as follows: After 10 half-lives (77 h), the remaining concentration of cis-platin will be as follows: The percent completion after 5 half-lives will be as follows: The percent completion after 10 half-lives will be as follows: Thus a first-order chemical reaction is 97% complete after 5 half-lives and 100% complete after 10 half-lives. Ethyl chloride decomposes to ethylene and HCl in a first-order reaction that has a rate constant of 1.6 × 10 s at 650°C. 4.3 × 10 s = 120 h = 5.0 days; 4.8 × 10 M Radioactivity, or radioactive decay, is the emission of a particle or a photon that results from the spontaneous decomposition of the unstable nucleus of an atom. The rate of radioactive decay is an intrinsic property of each radioactive isotope that is independent of the chemical and physical form of the radioactive isotope. The rate is also independent of temperature. In this section, we will describe radioactive decay rates and how half-lives can be used to monitor radioactive decay processes. In any sample of a given radioactive substance, the number of atoms of the radioactive isotope must decrease with time as their nuclei decay to nuclei of a more stable isotope. Using to represent the number of atoms of the radioactive isotope, we can define the of the sample, which is also called its as the decrease in the number of the radioisotope’s nuclei per unit time: Activity is usually measured in disintegrations per second (dps) or disintegrations per minute (dpm). The activity of a sample is directly proportional to the number of atoms of the radioactive isotope in the sample: \[A = kN \label{21.4.4}\] Here, the symbol is the radioactive decay constant, which has units of inverse time (e.g., s , yr ) and a characteristic value for each radioactive isotope. If we combine and , we obtain the relationship between the number of decays per unit time and the number of atoms of the isotope in a sample: is the same as the equation for the reaction rate of a first-order reaction, except that it uses numbers of atoms instead of concentrations. In fact, radioactive decay is a first-order process and can be described in terms of either the differential rate law ( ) or the integrated rate law: \[N = N_0e^{−kt} \] \[\ln \dfrac{N}{N_0}=-kt \label{21.4.6}\] Because radioactive decay is a first-order process, the time required for half of the nuclei in any sample of a radioactive isotope to decay is a constant, called the half-life of the isotope. The half-life tells us how radioactive an isotope is (the number of decays per unit time); thus it is the most commonly cited property of any radioisotope. For a given number of atoms, isotopes with shorter half-lives decay more rapidly, undergoing a greater number of radioactive decays per unit time than do isotopes with longer half-lives. The half-lives of several isotopes are listed in , along with some of their applications. Radioactive decay is a first-order process. In our earlier discussion, we used the half-life of a first-order reaction to calculate how long the reaction had been occurring. Because nuclear decay reactions follow first-order kinetics and have a rate constant that is independent of temperature and the chemical or physical environment, we can perform similar calculations using the half-lives of isotopes to estimate the ages of geological and archaeological artifacts. The techniques that have been developed for this application are known as radioisotope dating techniques. The most common method for measuring the age of ancient objects is carbon-14 dating. The carbon-14 isotope, created continuously in the upper regions of Earth’s atmosphere, reacts with atmospheric oxygen or ozone to form CO . As a result, the CO that plants use as a carbon source for synthesizing organic compounds always includes a certain proportion of CO molecules as well as nonradioactive CO and CO . Any animal that eats a plant ingests a mixture of organic compounds that contains approximately the same proportions of carbon isotopes as those in the atmosphere. When the animal or plant dies, the carbon-14 nuclei in its tissues decay to nitrogen-14 nuclei by a radioactive process known as beta decay, which releases low-energy electrons (β particles) that can be detected and measured: \[ \ce{^{14}C \rightarrow ^{14}N + \beta^{−}} \label{21.4.7}\] The half-life for this reaction is 5700 ± 30 yr. The C/ C ratio in living organisms is 1.3 × 10 , with a decay rate of 15 dpm/g of carbon ( ). Comparing the disintegrations per minute per gram of carbon from an archaeological sample with those from a recently living sample enables scientists to estimate the age of the artifact, as illustrated in Example 11.Using this method implicitly assumes that the CO / CO ratio in the atmosphere is constant, which is not strictly correct. Other methods, such as tree-ring dating, have been used to calibrate the dates obtained by radiocarbon dating, and all radiocarbon dates reported are now corrected for minor changes in the CO / CO ratio over time. In 1990, the remains of an apparently prehistoric man were found in a melting glacier in the Italian Alps. Analysis of the C content of samples of wood from his tools gave a decay rate of 8.0 dpm/g carbon. How long ago did the man die? isotope and final activity elapsed time Use to calculate / . Then substitute the value for the half-life of C into to find the rate constant for the reaction. Using the values obtained for / and the rate constant, solve to obtain the elapsed time. We know the initial activity from the isotope’s identity (15 dpm/g), the final activity (8.0 dpm/g), and the half-life, so we can use the integrated rate law for a first-order nuclear reaction ( ) to calculate the elapsed time (the amount of time elapsed since the wood for the tools was cut and began to decay). From , we know that = . We can therefore use the initial and final activities ( = 15 dpm and = 8.0 dpm) to calculate / : Now we need only calculate the rate constant for the reaction from its half-life (5730 yr) using : This equation can be rearranged as follows: Substituting into the equation for , From our calculations, the man died 5200 yr ago. It is believed that humans first arrived in the Western Hemisphere during the last Ice Age, presumably by traveling over an exposed land bridge between Siberia and Alaska. Archaeologists have estimated that this occurred about 11,000 yr ago, but some argue that recent discoveries in several sites in North and South America suggest a much earlier arrival. Analysis of a sample of charcoal from a fire in one such site gave a C decay rate of 0.4 dpm/g of carbon. What is the approximate age of the sample? 30,000 yr The half-life of a reaction is the time required for the reactant concentration to decrease to one-half its initial value. The half-life of a first-order reaction is a constant that is related to the rate constant for the reaction: t = 0.693/ . Radioactive decay reactions are first-order reactions. The rate of decay, or activity, of a sample of a radioactive substance is the decrease in the number of radioactive nuclei per unit time. | 9,755 | 995 |
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Need to define a new variable. The thermodynamic activity, \(a\), is the effective concentration of a species in a mixture. It is a dimensionless quantity that are calculated with respect to standard states. For a gas, this would be related to the fugacity and for a solution, to the concentration. The activity for a real gas: \[a_i=\frac{f_i}{P^{\circ}}=\frac{\phi_iP_i}{P^{\circ}}=\frac{\phi_i(y_iP)}{P^{\circ}}\] For systems where we treat the gases as ideal: \[ \phi_i=1 \] \[ a_i=\frac{P_i}{P^{\circ}}=y_i\frac{P}{P^{\circ}} \] The activity for a solution: \[ a_i=\gamma i\sf\frac{[A]}{1\;\underline{M}} \] General chemistry and organic chemistry use ideal reactants where \(\gamma_i=1\): \[ a_i=\sf\frac{[A]}{1\;\underline{M}} \] The activity for a solid or liquid: \[ a_i=1 \] | 802 | 997 |
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This page explains the relationship between the physical properties of the oxides of Period 3 elements and their structures (including sodium to chlorine; argon is omitted because it does not form an oxide). : The oxides in the top row are the highest known oxides of the various elements, in which the Period 3 elements are in their highest oxidation states. In these oxides, all the outer electrons in the Period 3 elements are involved in bonding. The attractive forces between these molecules include van der Waals dispersion and . These vary in size depending on the size, shape and polarity of the various molecules, but will always be much weaker than the ionic or covalent bonds in a giant structure. These oxides tend to be gases, liquids, or low melting point solids. The electronegativity of the elements increases across the period; silicon and oxygen do not differ enough in electronegativity to form an ionic bond. There are three different crystal forms of silicon dioxide; the most convenient structure to visualize is similar to that of diamond. In silicon dioxide, oxygen atoms fill the empty spaces along the silicon-silicon bond axes, as shown in a representative structure below: This structure is repeated many times throughout the silicon dioxide substance. Silicon dioxide has no mobile electrons or ions, and hence does not conduct electricity either as a solid or a liquid. Phosphorus, sulfur and chlorine form molecular oxides. Some of these molecules are fairly simple—others are polymeric. Here the simple structures are considered. Melting and boiling points of these oxides are much lower than those of the metal oxides or silicon dioxide. The intermolecular forces binding one molecule to its neighbors are van der Waals dispersion forces or dipole-dipole interactions. The strength of these vary depending on the size of the molecules. None of these oxides conduct electricity either as solids or as liquids, because none of them contain ions or free electrons. Phosphorus has two common oxides, phosphorus(III) oxide, P O , and phosphorus(V) oxide, P O . : Phosphorus(III) oxide is a white solid, melting at 24°C and boiling at 173°C. To understand its structure, consider a tetrahedral P molecule: The structure is expanded to display the bonds: The phosphorus-phosphorus bonds are interrupted with oxygen atoms, in a bent shape similar to water, as shown below: Only 3 of the valence electrons of phosphorus (the 3 unpaired p electrons) are involved in the phosphorus-oxygen bonds. Phosphorus(V) oxide is also a white solid, which sublimes at 300°C. In this case, the phosphorus uses all five of its outer electrons in the bonding. Solid phosphorus(V) oxide exists in several different forms - some of them polymeric. We are going to concentrate on a simple molecular form, and this is also present in the vapor. This is most easily drawn starting from P O . The other four oxygens are attached to the four phosphorus atoms via double bonds. Sulfur has two common oxides, sulfur dioxide (sulfur(IV) oxide), SO , and sulfur trioxide (sulfur(VI) oxide), SO . : S ulfur dioxide is a colorless gas at room temperature with an easily recognized rotten-egg smell. It consists of simple SO molecules as shown: The sulfur uses four of its six valence electrons to form the double bonds with oxygen, leaving the other two as a lone pair on the sulfur. The bent shape of SO is due to this lone pair. : Pure sulfur trioxide is a white solid with a low melting and boiling point. It reacts very rapidly with water vapour in the air to form sulfuric acid. Under laboratory conditions, it forms a white sludge which fumes dramatically in moist air (forming a fog of sulfuric acid droplets). Gaseous sulfur trioxide consists of simple SO molecules in which all six of the sulfur's outer electrons are involved in the bonding. There are various forms of solid sulfur trioxide. The simplest one is a trimer, S O , in which three SO molecules are joined in a ring. There are also other polymeric forms in which the SO molecules join together in long chains. For example: The tendency of sulfur trioxide to form polymers determines to its solid nature. Chlorine forms several oxides. Two are considered here: chlorine(I) oxide, Cl O, and chlorine(VII) oxide, Cl O . : Chlorine(I) oxide is a yellowish-red gas at room temperature. It consists of simple, small molecules. The physical properties of chlorine (I) oxide are consistent with those predicted for a molecule of its size. In chlorine(VII) oxide, the chlorine involves all of its seven valence electrons in bonds with oxygen. This produces a molecule much larger than chlorine (I) oxide, suggesting higher melting and boiling points. Chlorine(VII) oxide is a colorless oily liquid at room temperature. The diagram below shows a simple structural formula, neglecting three-dimensional structure; the geometry is tetrahedral around both chlorine atoms, and V-shaped around the central oxygen. Jim Clark ( ) | 5,002 | 998 |
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With the development of more precise ideas on elements, compounds and mixtures, scientists began to investigate how and why substances react. French chemist A. Lavoisier laid the foundation to the scientific investigation of matter by describing that substances react by following certain laws. These laws are called the laws of chemical combination. These eventually formed the basis of Dalton's Atomic Theory of Matter. According to this law, during any physical or chemical change, the total mass of the products remains equal to the total mass of the reactants. The law of conservation of mass is also known as the "law of indestructibility of matter." If heating 10 grams of \(\ce{CaCO3}\) produces 4.4 g of \(\ce{CO2}\) and 5.6 g of \(\ce{CaO}\), show that these observations are in agreement with the law of conservation of mass. Because the mass of the reactants is equal to the mass of the products, the observations are in agreement with the law of conservation of mass. | 998 | 999 |
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The Molecular Orbital Theory, initially developed by Robert S. Mullikan, incorporates the wave like characteristics of electrons in describing bonding behavior. In Molecular Orbital Theory, the bonding between atoms is described as a combination of their atomic orbitals. While the Valence Bond Theory and Lewis Structures sufficiently explain simple models, the Molecular Orbital Theory provides answers to more complex questions. In the Molecular Orbital Theory, the electrons are delocalized. Electrons are considered delocalized when they are not assigned to a particular atom or bond (as in the case with Lewis Structures). Instead, the electrons are “smeared out” across the molecule. The Molecular Orbital Theory allows one to predict the distribution of electrons in a molecule which in turn can help predict molecular properties such as shape, magnetism, and Bond Order. Atoms form bonds by sharing electrons. Atoms can share two, four, or six electrons, forming single, double, and triple bonds respectively. Although it is impossible to determine the exact position of an electron, it is possible to calculate the probability that one will find the electron at any point around the nucleus using the Schrödinger Equation. This equation can help predict and determine the energy and spatial distribution of the electron, as well as the shape of each orbital. The figure below shows the first five solutions to the equation in a three dimensional space for a one electron atom. The colors show the phase of the function. In this diagram, blue stands for negative and red stands for positive. Note, however, that the 2s orbital has 2 phases, one of which is not visible because it is inside the other. In molecules, atomic orbitals combine to form molecular orbitals which surround the molecule. Similar to atomic orbitals, molecular orbitals are wave functions giving the probability of finding an electron in certain regions of a molecule. Each molecular orbital can only have 2 electrons, each with an opposite spin. Each molecular orbital can only have 2 electrons, each with an opposite spin. Once you have the molecular orbitals and their energy ordering the ground state configuration is found by applying the Pauli principle, the principle and Hund's rule just as with atoms. The principles to apply when forming pictorial molecular orbitals from atomic orbitals are summarized in the table below: Total number of molecular orbitals is equal to the total number of atomic orbitals used to make them. The molecule H is composed of two H atoms. Both H atoms have a 1s orbital, so when bonded together, there are therefore two molecular orbitals. molecular orbitals energy than the atomic orbitals from which they were formed. molecular orbitals energy than the atomic orbitals from which they were formed. Electrons in molecular orbitals help stabilize a system of atoms since less energy is associated with bonded atoms as opposed to a system of unbound atoms. Bonding orbitals are formed by in-phase combinations of atomic orbitals and increase the electron density between the atoms (see figure 2 below) Electrons in molecular orbitals cause a system to be destabilized since more energy is associated with bonded atoms than that of a system of unbound atoms. Antibonding orbitals are formed by out-of-phase combinations of atomic orbitals and decrease the electron density between atoms (see figure 2 below). Following both the Pauli exclusion principle and Hund's rule, electrons fill in orbitals of increasing energy. Molecular orbitals are best formed when composed of Atomic orbitals of like energies. When Li forms the two lowest energy orbitals are the pair of bonding and antibonding orbitals formed from the two possible combinations of the 1s on each atom. The 2s orbitals combine primarily with each other to form another pair of bonding and antibonding orbitals at a higher energy. Figure 2 (below) shows how bonding and antibonding σ orbitals can be formed by combining s orbitals in-phase (bonding, bottom) and out-of-phase (antibonding, top). If the atomic orbitals are combined with the same phase they interfere constructively and a bonding orbital is formed. Bonding molecular orbitals have lower energy than the atomic orbitals from which they were formed. The lowering of the energy is attributed to the increase in shielding of the nuclear repulsion because of the increase in electron density between the nuclei. If the atomic orbitals are combined with different phases, they interfere destructively and an antibonding molecular orbital is formed. Antibonding molecular orbitals have a higher energy than the atomic orbitals from which they were formed. The higher energy is attributed to the reduced shielding of the nuclear repulsion because of the lower electron probability density between the nuclei. Molecular orbitals that are symmetrical about the axis of the bond are called sigma molecular orbitals, often abbreviated by the Greek letter \(\sigma\). Figure 2 shows the 1s orbitals of 2 Hydrogen atoms forming sigma orbitals. There are two types of sigma orbitals formed, antibonding sigma orbitals (abbreviated \(\sigma^*\)), and bonding sigma orbitals (abbreviated \(\sigma\)). In sigma bonding orbitals, the in phase atomic orbitals overlap causing an increase in electron density along the bond axis. Where the atomic orbitals overlap, there is an increase in electron density and therefore an increase in the intensity of the negative charge. This increase in negative charge causes the nuclei to be drawn closer together. In sigma antibonding orbitals (\(\sigma^*\)), the out of phase 1s orbitals interfere destructively which results in a low electron density between the nuclei as seen on the top of the diagram. The diagram below (figure 3) is a representation of the energy levels of the bonding and antibonding orbitals formed in the hydrogen molecule. Two molecular orbitals were formed: one antibonding (\(\sigma^*\)) and one bonding (\(\sigma\)).The two electrons in the hydrogen molecule have antiparallel spins. Notice that the \(\sigma^*\) orbital is empty and has a higher energy than the \(\sigma\) orbital. Sigma bonding orbitals and antibonding orbitals can also be formed between p orbitals (figure 4). Notice that the orbitals have to be in phase in order to form bonding orbitals. Sigma molecular orbitals formed by p orbitals are often differentiated from other types of sigma orbitals by adding the subscript p below it. So the antibonding orbital shown in the diagram below would be σ* . The \(\pi\) bonding is a side to side overlap of orbitals, which then causes there to be no electron density along the axis, but there is density above and below the axis. The diagram below (figure 5) shows a \(\pi\) molecular orbital and a \(\pi\) molecular orbital. \(\pi\) The two 2p orbitals overlap to create another pair of pi 2p and pi *2p molecular orbitals. The 2p -2p overlap is similar to the 2p -2p overlap because it is just the orbitals of the 2pz rotated 90 degrees about the axis. The new molecular orbitals have the same potential energies as those from the 2p -2p overlap. In summary the three pairs of p orbitals can combine to form one set of \sigma\ orbitals and two sets of \pi\ orbitals. Note that the σ and π orbitals formed from the p's are not always in the order π energy less than σ energy. For first row diatomics the ordering shown above is valid for Z ≤ 7. Thus for oxygen and fluorine the σ is below the π orbitals. Bond Order indicates the strength of the bond with the greater the bond order, the stronger the bond. \[\text{Bond Order}= \dfrac{1}{2} \left(a-b\right)\] where If the bond order is zero, then no bonds are produced and the molecule is not stable (for example \(He_2\)). If the Bond Order is 1, then it is a single covalent bond. The higher the Bond Order, the more stable the molecule is. An advantage of Molecular Orbital Theory when it comes to Bond Order is that it can more accurately describe partial bonds (for example in H , where the Bond Order=1/2), than Lewis Structures. 2. The molecular orbital diagram for a diatomic helium molecule, He , shows the following. 3. The molecular orbital diagram for a diatomic oxygen molecule, O , is 4.The molecular orbital diagram for a diatomic Neon molecule, Ne is 5. The molecular orbital diagram for the diatomic fluorine molecule, F is | 8,433 | 1,000 |
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The caloric value for fats is about 9 Cal/g, while for carbohydrates (sugars or starches) and proteins, it's about 4 Cal/g , so a teaspoon of sugar is only about 20 Calories, but a teaspoon of oil is about 45 calories. Our body stores fats as a long term, high energy per gram energy source, while sugars can be metabolized quickly, but don't give as much energy per gram. The energy is released when each is metabolized, giving the same amount of energy as combustion in air. Vegetable fats and oils are all triglycerides, which contain a glycerol ( ) three carbon "backbone" with 3 long chain attached through ester linkages, as in the figure below. The actual shape is shown in the Jmol model, which can be rotated with the mouse. Triglycerides are called "fats" when they're solids or semisolids, and "oils" when they're liquids. A triglyceride, overall unsaturated, with the glycerol "backbone" on the left, and saturated palmitic acid, monounsaturated oleic acid, and polyunsaturated alpha-linolenic acid. Carbon atoms are at each bend in the structure, and hydrogen atoms are omitted. The long chain fatty acids may be with hydrogen atoms, in which case they have all single bonds like the top fatty acid in the Figure (which is palmitic acid). If they have fewer hydrogen atoms, they are and have double bonds like the middle fatty acid in the Figure (which is oleic acid). The bottom fatty acid is , with multiple double bonds (it is linolenic acid). Various cooking oils have |known concentrations of saturated and unsaturated fatty acids. Carbohydrates are made up of simple sugar units. The sucrose (ordinary table sugar) molecule shown below is made of a glucose and fructose "monosaccharides". Sucrose,C H O (C atoms are at each bend, H atoms are not shown) A key to why fats have more than twice the caloric value of sugars comes from the combustion reactions: A typical sugar combustion reaction is C H O + 6 O → 6 CO + 6 H O It requires 6 mol O for every 6 mol of C, a 1:1 ratio. Since the molar mass of sucrose is 180 g/mol, about 0.033 mol O is required per gram. While a typical fat combustion might be C H O + 80 O → 56 CO + 54 H O It requires 80 mol O for every 56 mol of C, a 1.42:1 ratio. Or, since the molar mass of the fat shown is 878 g/mol, about 0.091 mol O is required per gram. We see why by looking at the C:O mole ratios in fats and sugars: In the fat above, \(\text{S}\left( \frac{\text{C}}{\text{O}} \right)=\frac{\text{56 mol C}}{\text{6 mol O}}\) (a 9.3:1 ratio). Or for the sugar,
\(\text{S}\left( \frac{\text{C}}{\text{O}} \right)=\frac{\text{6 mol C}}{\text{6 mol O}}\) (a 1:1 ratio). Since the sugar is already more oxygenated, it produces less energy when it is burned. Looking at it another way, the carbon, which reacts with oxygen and releases energy in combustion, is a bigger part of the fat: In fat: \(S\left( \frac{\text{C}}{\text{C}_{56}\text{H}_{108}\text{O}_{6}} \right)=\frac{\text{56 mol C}}{\text{1 mol C}_{56}\text{H}_{108}\text{O}_{6}}\) or 56 mol C/877.5 g C H O = 0.064 mol C/g fat. In Sugar: \(S\left( \frac{\text{C}}{\text{C}_{6}\text{H}_{12}\text{O}_{6}} \right)=\frac{\text{6 mol C}}{\text{1 mol C}_{6}\text{H}_{12}\text{O}_{6}}\) or 6 mol C/180 g C H O = 0.033 mol C/g fat. We'll see below that these ratios actually allow us to determine the chemical formula for a fat, sugar, or any other compound. Stoichiometric ratios derived from formulas instead of equations are involved in the most common procedure for determining the empirical formulas of compounds which contain only C, H, and O. A weighed quantity of the substance to be analyzed is placed in a combustion train and heated in a stream of dry O . All the H in the compound is converted to H O(g) which is trapped selectively in a previously weighed absorption tube. All the C is converted to CO (g) and this is absorbed selectively in a second tube. The increase of mass of each tube tells, respectively, how much H O and CO were produced by combustion of the sample A 1.000 g sample of a fat was burned in a combustion train, producing 2.784 g of CO and 1.140 g of H O. Determine the empirical formula of the fat. We need to know the amount of C, the amount of H, and the amount of O in the sample. The ratio of these gives the subscripts in the formula. The first two may be obtained from the masses of CO and H O using the molar masses and the stoichiometric ratios \(\text{S}\left( \frac{\text{C}}{\text{CO}_{\text{2}}} \right)=\frac{\text{1 mol C}}{\text{1 mol CO}_{\text{2}}}\) \(\text{S}\left( \frac{\text{H}}{\text{H}_{\text{2}}\text{O}} \right)=\frac{\text{2 mol H}}{\text{1 mol H}_{\text{2}}\text{O}}\) Thus \(n_{\text{C}}=\text{2.784 g CO}_{\text{2}}\times \frac{\text{1 mol CO}_{\text{2}}}{\text{44}\text{.01 g CO}_{\text{2}}}\times \frac{\text{1 mol C}}{\text{1 mol CO}_{\text{2}}}=\text{0.06326 mol C}\) \(n_{\text{H}}=\text{ 1.140 g H}_{\text{2}}\text{O}\times \frac{\text{1 mol H}_{\text{2}}\text{O}}{\text{18}\text{.02 g H}_{\text{2}}\text{O}}\times \frac{\text{2 mol H}}{\text{1 mol H}_{\text{2}}\text{O}}=\text{0.1265 mol H}\) The compound may also have contained oxygen. To see if it does, calculate the masses of C and H and subtract from the total mass of sample \(\begin{align}
& m_{\text{C}}=\text{0.06326 mol C}\times \frac{\text{12}\text{.01 g C}}{\text{1 mol C}}=\text{0.7598 g C} \\
& \\
& m_{\text{H}}=\text{0.1265 mol H}\times \frac{\text{1}\text{.008 g H}}{\text{1 mol H}}=\text{0.1275g H} \\
\end{align}\) Thus we have 1.000 g sample – 0.7598 g C – 0.1275 g H = 0.1128 g O and \(n_{\text{O}}=\text{0.1128 g O}\times \frac{\text{1 mol O}}{\text{16}\text{.00 g O}}=\text{0.007050 mol O}\) The ratios of the amounts of the elements in ascorbic acid are therefore \(\frac{n_{\text{H}}}{n_{\text{O}}}=\frac{\text{0.1265 mol H}}{\text{0.00705 mol C}}=\frac{\text{17.94 mol H}}{\text{1 mol O}}\) \(\frac{n_{\text{C}}}{n_{\text{O}}}=\frac{\text{20.06326 mol O}}{\text{0.00705 mol O}}=\frac{\text{8.97 mol C}}{\text{1 mol C}}\) Since : : is 9 mol C : 18 mol H : 1 mol O, the empirical formula is C H O . Because most fats have 3 fatty acids bonded to glycerol with 2 oyxgen atoms in each "ester" bond, the molecular formula is probably C H O . This molecule would contain the 3 carbon glycerol backbone, the 3 fatty acid chains would share the remaining 51 carbon atoms, and would be of average length 51/3 = 17 carbon atoms. A 6.49-mg sample of ascorbic acid (vitamin C) was burned in a combustion train. 9.74 mg CO and 2.64 mg H O were formed. Determine the empirical formula of ascorbic acid. We need to know the amount of C, the amount of H, and the amount of O in the sample. The ratio of these gives the subscripts in the formula. The first two may be obtained from the masses of CO and H O using the molar masses and the stoichiometric ratios \(\text{S}\left( \frac{\text{C}}{\text{CO}_{\text{2}}} \right)=\frac{\text{1 mol C}}{\text{1 mol CO}_{\text{2}}}\) \(\text{S}\left( \frac{\text{H}}{\text{H}_{\text{2}}\text{O}} \right)=\frac{\text{2 mol H}}{\text{1 mol H}_{\text{2}}\text{O}}\) Thus \(n_{\text{C}}=\text{9}\text{.74}\times \text{10}^{\text{-3}}\text{g CO}_{\text{2}}\times \frac{\text{1 mol CO}_{\text{2}}}{\text{44}\text{.01 g CO}_{\text{2}}}\times \frac{\text{1 mol C}}{\text{1 mol CO}_{\text{2}}}=\text{2}\text{.21}\times \text{10}^{\text{-4}}\text{ mol C}\) \(n_{\text{H}}=\text{2}\text{.64}\times \text{10}^{\text{-3}}\text{g H}_{\text{2}}\text{O}\times \frac{\text{1 mol H}_{\text{2}}\text{O}}{\text{18}\text{.02 g H}_{\text{2}}\text{O}}\times \frac{\text{2 mol H}}{\text{1 mol H}_{\text{2}}\text{O}}=\text{2}\text{.93}\times \text{10}^{\text{-4}}\text{ mol H}\) The compound may also have contained oxygen. To see if it does, calculate the masses of C and H and subtract from the total mass of sample \(\begin{align}
& m_{\text{C}}=\text{2}\text{.21}\times \text{10}^{\text{-4}}\text{ mol C}\times \frac{\text{12}\text{.01 g C}}{\text{1 mol C}}=\text{2}\text{.65}\times \text{10}^{\text{-3}}\text{g C}=\text{2}\text{.65 mg C} \\
& \\
& m_{\text{C}}=\text{2}\text{.93}\times \text{10}^{\text{-4}}\text{ mol H}\times \frac{\text{1}\text{.008 g H}}{\text{1 mol H}}=\text{2}\text{.95}\times \text{10}^{\text{-4}}\text{g H}=\text{0}\text{.295 mg H} \\
\end{align}\) Thus we have 6.49 mg sample – 2.65 mg C – 0.295 mg H = 3.54 mg O and \(n_{\text{O}}=\text{3}\text{.54}\times \text{10}^{\text{-3}}\text{ g O}\times \frac{\text{1 mol O}}{\text{16}\text{.00 g O}}=\text{2}\text{.21}\times \text{10}^{\text{-4}}\text{ mol O}\) The ratios of the amounts of the elements in ascorbic acid are therefore \(\frac{n_{\text{H}}}{n_{\text{C}}}=\frac{\text{2}\text{.93}\times \text{10}^{\text{-4}}\text{ mol H}}{\text{2}\text{.21}\times \text{10}^{\text{-4}}\text{ mol C}}=\frac{\text{1}\text{.33 mol H}}{\text{1 mol C}}=\frac{\text{1}\tfrac{1}{3}\text{mol H}}{\text{1 mol C}}=\frac{\text{4 mol H}}{\text{3 mol C}}\) \(\frac{n_{\text{O}}}{n_{\text{C}}}=\frac{\text{2}\text{.21}\times \text{10}^{\text{-4}}\text{ mol O}}{\text{2}\text{.21}\times \text{10}^{\text{-4}}\text{ mol C}}=\frac{\text{1 mol O}}{\text{1 mol C}}=\frac{\text{3 mol O}}{\text{3 mol C}}\) Since : : is 3 mol C:4 mol H:3 mol O, the empirical formula is C H O . A drawing of a molecule of ascorbic acid is shown here. You can determine by counting the atoms that the molecular formula is C H O —exactly double the empirical formula. It is also evident that there is more to know about a molecule than just how many atoms of each kind are present. In ascorbic acid, as in other molecules, the way the atoms are connected together and their arrangement in three-dimensional space are quite important. A picture showing which atoms are connected to which is called a . Empirical formulas may be obtained from percent composition or combustion-train experiments, and, if the molecular weight is known, molecular formulas may be determined from the same data. More complicated experiments are required to find structural formulas. In Example 2 we obtained the mass of O by subtracting the masses of C and H from the total mass of sample. This assumed that only C, H, and O were present. Sometimes such an assumption may be incorrect. When penicillin was first isolated and analyzed, the fact that it contained sulfur was missed. This mistake was not discovered for some time because the atomic weight of sulfur is almost exactly twice that of oxygen. Two atoms of oxygen were substituted in place of one sulfur atom in the formula. A 3D representation of L-Ascorbic Acid <chemeddl-jmol2>ascorbic|size=300</chemeddl-jmol2> | 10,606 | 1,002 |
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confined to a cubic box of length \(L\). The particle inside the box has translational energy levels given by: \[E_\text{trans}= \dfrac{h^2 \left(n_x^2+ n_y^2+ n_z^2 \right)}{8 mL^2} \nonumber \] where \(n_x\), \(n_y\) and \(n_z\) are the quantum numbers in the three directions. The translational partition function is given by: \[q_\text{trans} = \sum_i e^{−\epsilon_i/kT} \nonumber \] which is the product of translational partition functions in the three dimensions. We can write the translation partition function as the product of the translation partition function for each direction: \[\begin{align} q_\text{trans} &= q_{x} q_{y} q_{z} \label{times1} \\[4pt] &= \sum_{n_x=1}^{\infty} e^{−\epsilon_x/kT} \sum_{n_y=1}^{\infty} e^{−\epsilon_y/kT} \sum_{n_z=1}^{\infty} e^{−\epsilon_z/kT} \label{sum1} \end{align} \] Since the levels are very closely spaced (continuous), we can replace each sum in Equation \(\ref{sum1}\) with an integral. For example: \[\begin{align} q_{x} &= \sum_{n_x=1}^{\infty} e^{−\epsilon_x/kT} \\[4pt] &\approx \int_{n_x=1}^{\infty} e^{−\epsilon_x/kT} \label{int1} \end{align} \] and after substituting the energy for the relevant dimension: \[ \epsilon_x = \dfrac{h^2 n_x^2}{8mL^2} \nonumber \] we can extend the lower limit of integration in the approximation of Equation \(\ref{int1}\): \[ q_x= \int_{1}^{\infty} e^{− \frac{h^2 n_x^2}{8mL^2 kT}} \approx \int_{0}^{\infty} e^{− \frac{h^2 n_x^2}{8mL^2 kT}} \nonumber \] We then use the following solved Gaussian integral: \[ \int_o^{\infty} e^{-an^2} dn = \sqrt{\dfrac{\pi}{4a}} \nonumber \] with the following substitution: \[a = \dfrac{h^2}{8mL^2 kT} \nonumber \] we get: \[q_x= \dfrac{1}{2} \sqrt{\dfrac{π}{a}} = \dfrac{1}{2} \sqrt{\dfrac{π 8m kT }{ h^2 }} L \nonumber \] or more commonly presented as: \[q_x = \dfrac{L}{\Lambda} \nonumber \] where \(\Lambda\) is the and is given by \[ \Lambda = \dfrac{h}{\sqrt{2 π 8m kT}} \nonumber \] Multiplying the expressions for \(q_x\), \(q_y\) and \(q_z\) (Equation \(\ref{times1}\)) and using \(V\) as the volume of the box \(L^3\), we arrive at: \[q_\text{trans} = \left( \dfrac{\sqrt{2 π 8m kT}} {h} \right)^{3/2} V = \dfrac{ V}{\Lambda^3} \label{parttransation} \] This is usually a very large number (10 ) for volumes of 1 cm for small molecular masses. This means that such a large number of translational states are accessible and available for occupation by the molecules of a gas. This result is very similar to the result of the classical kinetic gas theory that said that the observed energy of an ideal gas is: \[U=\dfrac{3}{2} nRT \nonumber \] We postulate therefore that the observed energy of a macroscopic system should equal the statistical average over the partition function as shown above. In other words: if you know the particles your system is composed of and their energy states you can use statistics to calculate what you should observe on the whole ensemble. Calculate the translational partition function of an \(I_2\) molecule at 300K. Assume V to be 1 liter. Mass of \(I_2\) is \(2 \times 127 \times 1.6606 \times 10^{-27} kg\) \[\begin{align*} 2πmkT &= 2 \times 3.1415 \times (2 \times 127 \times 1.6606 \times 10^{-27}\, kg) \times 1.3807 \times 10^{-23} \, J/K \times 300 K \\[4pt] &= 1.0969 \times 10^{-44}\; J\, kg \end{align*} \nonumber \] \[\begin{align*} Λ &= \dfrac{h}{\sqrt{2 π m kT}} \\[4pt] &= \dfrac{6.6262 \times 10^{-34}\;J\, s}{ \sqrt{1.0969 \times 10^{-44}\, J \, kg}} = 6.326 \times 10^{-12}\;m \end{align*} \nonumber \] The via Equation \ref{parttransation} \[q_\text{trans}= \dfrac{V}{Λ^3}= \dfrac{1000 \times 10^{-6} m^3}{(6.326 \times 10^{-12} \; m)^3}= 3.95 \times 10^{30} \nonumber \] This means that \(3.95 \times 10^{30}\) quantum states are thermally accessible to the molecular system | 3,796 | 1,003 |
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The mean free path is the average distance traveled by a moving molecule between collisions. Imagine gas leaking out of a pipe. It would take a while for the gas to diffuse and spread into the environment. This is because gas molecules collide with each other, causing them to change in speed and direction. Therefore, they can never move in a straight path without interruptions. Between every two consecutive collisions, a gas molecule travels a straight path. The average distance of all the paths of a molecule is the mean free path. Imagine a ball traveling in a box ; the ball represents a moving molecule. Every time the ball hits a wall, a collision occurs and the direction of the ball changes (Figure 1). The ball hits the wall five times, causing five collisions. Between every two consecutive collisions, the ball travels an individual path. It travels a total of four paths between the five collisions; each path has a specific distance, d. The mean free path, \lambda, of this ball is the average length of all four paths. Each path traveled by the ball has a distance, denoted d : \[\lambda = \dfrac{d_1 + d_2 + d_3 + d_4}{4} \nonumber \] In reality, the mean free path cannot be calculated by taking the average of all the paths because it is impossible to know the distance of each path traveled by a molecule. However, we can calculate it from the average speed (\(\langle c \rangle\)) of the molecule divided by the (\(Z\)). The formula for this is: \[\lambda = \dfrac{\langle c \rangle}{Z} \nonumber \] Because \(Z\) is equal to \(1/ t\), where \(t\) is the average time between collisions, the formula can also be: \[\begin{eqnarray} \lambda &=& \dfrac{\langle c \rangle}{\dfrac{1}{t}} \\ &=& \langle c \rangle \times t \end{eqnarray} \nonumber \] In addition, because \[\lambda = \sqrt{2} \pi d^2 \langle c \rangle (N/V) \nonumber \] where The formula can be further modified to: \[ \begin{align*} \lambda &= \dfrac{\langle c \rangle}{\sqrt{2} \pi d^2 \langle c \rangle \dfrac{N}{V}} \\[4pt] &= \dfrac{1}{\sqrt{2} \pi d^2 \dfrac{N}{V}} \end{align*} \] A gas has an average speed of 10 m/s and a collision frequency of 10 s . What is its mean free path? l = <c> / = 10 m/s / 10 s^(-1) = 1 m A gas has an average speed of 10 m/s and an average time of 0.1 s between collisions. What is its mean free path? l = <c> X (average time between collisions) = 10 m/s X 0.1 s = 1 m l = 1 / [ (pi) d^2 (N/v)] = 1 / [ (pi)(0.1 m)^2 (10 m^(-3))] = 2.25 m A gas in a 1 m container has a molecular diameter of 0.1 m. There are 10 molecules. What is its mean free path? l = 1 / [ (pi) d^2 (N/v)] = 1 / [ (pi)(0.1 m)^2 (10 / 1 m^3)] = 2.25 m A gas has a molecular diameter of 0.1 m. It also has a mean free path of 2.25 m. What is its density? l = 1 / [ (pi) d^2 (N/v)]
2.25 m = 1 / [ (pi)(0.1 m)^2 (N/v)]
N/v = 10 m^(-3) | 2,862 | 1,004 |
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Organic substances have been used throughout this text to illustrate the differences between ionic and covalent bonding and to demonstrate the intimate connection between the structures of compounds and their chemical reactivity. You learned, for example, that even though NaOH and alcohols ( ) both have in their formula, NaOH is an ionic compound that dissociates completely in water to produce a basic solution containing Na and OH ions, whereas alcohols are covalent compounds that do not dissociate in water and instead form neutral aqueous solutions. You also learned that an amine (RNH ), with its lone pairs of electrons, is a base, whereas a carboxylic acid (RCO H), with its dissociable proton, is an acid. 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 ). What he expected is described by the following equation. \[AgOCN + NH_4Cl \rightarrow AgCl + NH_4OCN \label{25.1.1} \] 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 is the study of the chemistry of the carbon compounds, and 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 \(\Page {1}\). | 4,176 | 1,005 |
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This page looks at the phase diagram for mixtures of salt and water - how the diagram is built up, and how to interpret it. It includes a brief discussion of solubility curves. A solubility curve shows how the solubility of a salt like sodium chloride or potassium nitrate varies with temperature. The solubility is often (although not always) measured as the mass of salt which would saturate 100 grams of water at a particular temperature. A solution is saturated if it will not dissolve any more of the salt at that particular temperature - in the presence of crystals of the salt. For most, but not all, substances, . For some (like potassium nitrate), the increase is quite fast. For others (like sodium chloride), there is only a small change in solubility with temperature. Obviously, a solubility curve shows you the solubility of a substance at a particular temperature. For phase diagram purposes, one important way of looking at this is to examine what happens if you decrease the temperature of a solution with some given concentration. For example, suppose you have a near-boiling solution of potassium nitrate in water. We'll take a solution containing 100 g of potassium nitrate and 100 g of water. Now let the solution cool. At all temperatures above that marked on the graph (about 57°C), 100 g of water will dissolve more than 100 g of potassium nitrate. All the potassium nitrate will stay in solution. At 57°C, you hit the solubility curve. This is the temperature at which 100 g of water will dissolve 100 g of potassium nitrate to give a saturated solution. If the temperature falls even the tiniest bit below 57°C, the water will no longer dissolve as much potassium nitrate - and so some crystallises out. The lower the temperature falls, the more potassium nitrate crystallizes, because the water will dissolve less and less of it. You can think of this as a simple phase diagram. If you have a mixture of 100 g of potassium nitrate and 100 g of water and the temperature is above 57°C, you have a single phase - a solution of potassium nitrate. If the temperature is below 57°C for this mixture, you will have a mixture of two phases - the solution and some solid potassium nitrate. The solubility curve represents the boundary between these two different situations. We'll look now at the phase diagram for sodium chloride solution in some detail. We'll talk through what everything means in detail, but first of all notice two things compared with the similar I hope you have already read about. First we are only looking at a very restricted temperature range. The top of this particular diagram is only about 50°C - although it could go higher than that. Secondly, all the action takes place in the left-hand side of the diagram. That's why a lot of the horizontal scale is missing (represented by the broken zigzag in the scale). These areas all show what you would see if you had a particular mixture of salt and water at a given temperature. For example, if the temperature was below -21.1°C, you would always see a mixture of solid salt and ice. There would never be any liquid whatever proportions of salt and water you had. At temperatures above this, what you would see would depend on where your particular set of conditions fell in the diagram. You just need to notice what area the conditions fall into. Let's take the lines one at a time. The first one to look at is the one in bold green in the next diagram. This line represents the effect of increasing amounts of salt on the freezing point of water. Up to the point where there is 23.3% of salt in the mixture, the more salt the lower the freezing point of the water. The second line is actually a solubility curve for salt in water - although it doesn't look quite like the one we described earlier on this page. The reason that it looks different is because the axes have been reversed. On a normal solubility curve, temperature is the horizontal axis and the vertical axis shows the solubility in grams per 100 g of water - a measure of concentration. This time the measure of concentration is the horizontal axis and temperature the vertical one. If you have already read about the , you might think it strange to include a solubility curve in the phase diagram. In the tin-lead system, both of the sloping lines showed the effect of one of the components on the freezing point of the other. But if you think what this means, it isn't so odd. The freezing point is the temperature at which crystals start to appear when you cool a liquid mixture. In the case of a salt solution with concentrations of salt greater than 23.3%, the solubility curve shows the temperature at which crystals of salt will appear when you cool a solution of a given concentration. If you aren't clear about that, go back and re-read the beginning of this page. There is also a hidden difference between this line and the corresponding line on the tin-lead diagram. In that case, both of the sloping lines eventually reached the the left-hand or the right-hand axis. It was possible to talk about the case where you had 100% lead or 100% tin. In this case it is impossible to get to 100% salt. This line eventually comes to an end. At 1 atmosphere pressure it won't get very far above 100°C because the water will boil and you won't have a solution any more. Even if you raise the pressure, the maximum temperature you could achieve would be 374°C - the critical temperature of the water. Water doesn't exist as a liquid above this temperature. To get to the right-hand side of the graph where you have 100% salt, you would have to get the temperature up to 801°C - the melting point of the salt. After all this hassle, the final line is easy! This line is simply drawn across at the lowest temperature at which a mixture of salt and water can contain any liquid. This is known as the eutectic temperature. The particular mixture of salt and water (containing 23.3% salt) which freezes at this temperature is known as a eutectic mixture. The phase diagram is used to find out what happens if you cool salt solution of a particular concentration. We need to look at three separate cases. This is the easy one! Nothing happens until you get down to the eutectic temperature. At that point, you will start to get both ice crystals and salt crystals forming. If you keep cooling it, you will obviously end up with a solid mixture of ice and salt. You are moving straight from the "salt solution" area of the phase diagram into the "solid salt + ice" area. When the temperature drops enough so that it reaches the boundary between the two areas of the phase diagram, ice crystals start to form - in other words, the solution starts to freeze. As the solution continues to cool, it moves down into the "ice + salt solution" region. Obviously, the composition of the solution has changed because it contains less water - some of it has frozen to give ice. But the composition of the system as a whole is still the same, and so we continue down the same line. To find out what is actually in the mixture, you draw a horizontal tie line through the point showing the temperature you have got to, and look at what it hits at either end. As the mixture continues to cool, it will eventually reach the eutectic temperature, and at this point all of the rest of the solution will turn to solid - a mixture of ice and salt crystals. The phase diagram is interpreted in just the same way - except that this time, salt crystals are formed at first rather than ice crystals. In the next diagram, the first salt crystals will form when the temperature hits the boundary line. As the temperature continues to drop into the "solid salt + salt solution" area, more and more salt will crystallize out. To find out exactly what is present at any temperature, you can again draw a tie line and look at what is at either end. The diagram shows the tie line when the temperature has dropped to some random place in the "solid salt + salt solution" area. Again, by looking at the ends of this tie line, you can see that the mixture contains solid salt (the 100% salt at the right-hand end of the line) and a solution whose concentration can be found by looking at the left-hand end of the line. As the temperature continues to fall, you will eventually reach the eutectic temperature, when everything left - salt and water - will turn to solid giving you a mixture of solid salt and ice. Jim Clark ( ) | 8,464 | 1,006 |
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Dipole moments occur when there is a separation of charge. They can occur between two ions in an ionic bond or between atoms in a covalent bond; dipole moments arise from differences in electronegativity. The larger the difference in electronegativity, the larger the dipole moment. The distance between the charge separation is also a deciding factor in the size of the dipole moment. The dipole moment is a measure of the polarity of the molecule. When atoms in a molecule share electrons unequally, they create what is called a dipole moment. This occurs when one atom is more electronegative than another, resulting in that atom pulling more tightly on the shared pair of electrons, or when one atom has a lone pair of electrons and the difference of electronegativity vector points in the same way. One of the most common examples is the water molecule, made up of one oxygen atom and two hydrogen atoms. The differences in electronegativity and lone electrons give oxygen a partial negative charge and each hydrogen a partial positive charge. (\(\mu\)). Dip uals \(3.34 \times 10^{-30}\; C\, m\)). The dipole moment of a molecule can be calculated by Equation \(\ref{1}\): \[ \vec{\mu} = \sum_i q_i \, \vec{r}_i \label{1}\] where The dipole moment acts in the direction of the vector quantity. s \(\ce{H_2O}\). Bec \(\ce{H_2O}\) The vector points from positive to negative, on both the molecular (net) dipole moment and the individual bond dipoles. shows the electronegativity of some of the common elements. The larger the difference in electronegativity between the two atoms, the more electronegative that bond is. To be considered a polar bond, the difference in electronegativity must be large. The dipole moment points in the direction of the vector quantity of each of the bond electronegativities added together. It is relatively easy to measure dipole moments: just place a substance between charged plates (Figure \(\Page {2}\)); polar molecules increase the charge stored on the plates, and the dipole moment can be obtained (i.e., via the capacitance of the system). Nonpolar \(\ce{CCl_4}\) is not deflected; moderately polar acetone deflects slightly; highly polar water deflects strongly. In general, polar molecules will align themselves: (1) in an electric field, (2) with respect to one another, or (3) with respect to ions (Figure \(\Page {2}\)). Equation \(\ref{1}\) can be simplified for a simple separated two-charge system like diatomic molecules or when considering a bond dipole within a molecule \[ \mu_{diatomic} = Q \times r \label{1a}\] This bond dipole is interpreted as the dipole from a charge separation over a distance \(r\) between the partial charges \(Q^+\) and \(Q^-\) (or the more commonly used terms \(δ^+\) - \(δ^-\)); the orientation of the dipole is along the axis of the bond. Consider a simple system of a single electron and proton separated by a fixed distance. When the proton and electron are close together, the dipole moment (degree of polarity) decreases. However, as the proton and electron get farther apart, the dipole moment increases. In this case, the dipole moment is calculated as (via Equation \(\ref{1a}\)): \[\begin{align*} \mu &= Qr \nonumber \\[4pt] &= (1.60 \times 10^{-19}\, C)(1.00 \times 10^{-10} \,m) \nonumber \\[4pt] &= 1.60 \times 10^{-29} \,C \cdot m \label{2} \end{align*}\] The Debye characterizes the size of the dipole moment. When a proton and electron are 100 pm apart, the dipole moment is \(4.80\; D\): \[\begin{align*} \mu &= (1.60 \times 10^{-29}\, C \cdot m) \left(\dfrac{1 \;D}{3.336 \times 10^{-30} \, C \cdot m} \right) \nonumber \\[4pt] &= 4.80\; D \label{3} \end{align*}\] \(4.80\; D\) is a key reference value and represents a pure charge of +1 and -1 separated by 100 pm. If the charge separation is increased then the dipole moment increases (linearly): The water molecule in Figure \(\Page {1}\) can be used to determine the direction and magnitude of the dipole moment. From the electronegativities of oxygen and hydrogen, the difference in electronegativity is 1.2e for each of the hydrogen-oxygen bonds. Next, because the oxygen is the more electronegative atom, it exerts a greater pull on the shared electrons; it also has two lone pairs of electrons. From this, it can be concluded that the dipole moment points from between the two hydrogen atoms toward the oxygen atom. Using the equation above, the dipole moment is calculated to be 1.85 D by multiplying the distance between the oxygen and hydrogen atoms by the charge difference between them and then finding the components of each that point in the direction of the net dipole moment (the angle of the molecule is 104.5˚). The bond moment of the O-H bond =1.5 D, so the net dipole moment is \[\mu=2(1.5) \cos \left(\dfrac{104.5˚}{2}\right)=1.84\; D \nonumber\] The shape of a molecule and the polarity of its bonds determine the OVERALL POLARITY of that molecule. A molecule that contains polar bonds might not have any overall polarity, depending upon its shape. The simple definition of whether a complex molecule is polar or not depends upon whether its overall centers of positive and negative charges overlap. If these centers lie at the same point in space, then the molecule has no overall polarity (and is non polar). If a molecule is completely symmetric, then the dipole moment vectors on each molecule will cancel each other out, making the molecule nonpolar. A molecule can only be polar if the structure of that molecule is not symmetric. Although a polar bond is a prerequisite for a molecule to have a dipole, not all molecules with polar bonds exhibit dipoles \(AB_n\) molecul Although the C–Cl bonds are rather polar, the individual bond dipoles cancel one another in this symmetrical structure, and \(\ce{Cl_2C=CCl_2}\) does not have a net dipole moment. C-Cl, the key polar bond, is 178 pm. Measurement reveals 1.87 D. From this data, % ionic character can be computed. If this bond were 100% ionic (based on proton & electron), \[\begin{align*} \mu &= \dfrac{178}{100}(4.80\; D) \nonumber \\[4pt] &= 8.54\; D \nonumber \end{align*} \] Although the bond length is , the dipole is as you move down the halogen group. The electronegativity decreases as we move down the group. Thus, the greater influence is the electronegativity of the two atoms (which influences the at the ends of the dipole). | 6,403 | 1,007 |
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These methods are applicable to reactions that are not excessively fast, typically requiring a few minutes or hours to run to completion. They were about the only methods available before 1950. Most first-year laboratory courses students will include at least one experiment based on one of these methods. is perhaps the most widely-used technique. If either a reactant or a product is colored, the reaction is easily followed by recording the change in transmission of an appropriate wavelength after the reaction is started. When a beam of light passes through a solution containing a colored substance, the fraction that is absorbed is directly proportional to the concentration of that substance and to the length of the light's path through the solution. The latter can be controlled by employing a cell or cuvette having a fixed path length. If is the intensity of the light incident on the cell and is the intensity that emerges on the other side, then the is just 100 × / . Because a limited 1-100 scale of light absorption is often inadequate to express the many orders of magnitude frequently encountered, a logarithmic term optical density is often employed. The relation between this, the cell path length, concentration, and innate absorption ability of the colored substance is expressed by Beer's Law. The simplest absorbance measuring device is a in which a beam of white light from an incandescent lamp is passed through a cell (often just a test tube) and onto a photodetector whose electrical output is directly proportional to the light intensity. Before beginning the experiment, a sets the meter to zero when the light path is blocked off, and a sets it to 100 when a cell containing an uncolored solution (a "blank") is inserted. The sensitivity and selectivity of such an arrangement is greatly enhanced by adjusting the wavelength of the light to match the absorption spectrum of the substance being measured. Thus if the substance has a yellow color, it is because blue light is being absorbed, so a blue color filter is placed in the light path. More sophisticated employ two cells, one for the sample and another for the blank. They also allow one to select the particular wavelength range that is most strongly absorbed by the substance under investigation, which can often extend into the near-ultraviolet region. measurements (made with a ) can be useful for reactions that lead to the formation a fine precipitate. A very simple student laboratory experiment of this kind can be carried out by placing a conical flask containing the reaction mixture on top of a marked piece of paper. The effects of changing the temperature or reactant concentrations can be made by observing how long it takes for precipitate formation to obscure the mark on the paper. such a fluorescence and polarimetry (measurement of the degree to which a solution rotates the plane of polarized light) are also employed when applicable. | 2,976 | 1,008 |
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The traditional experimental methods described above all assume the possibility of following the reaction after its components have combined into a homogeneous mixture of known concentrations. But what can be done if the time required to complete the mixing process is comparable to or greater than the time needed for the reaction to run to completion? For reactions that take place in milliseconds, the standard approach since the 1950s has been to employ a flow technique of some kind. An early example was used to study fast gas-phase reactions in which one of the reactants is a free radical such as OH that can be produced by an intense microwave discharge acting on a suitable source gas mixture. This gas, along with the other reactant being investigated, is made to flow through a narrow tube at a known velocity. If the distance between the point at which the reaction is initiated and the product detector is known, then the time interval can be found from the flow rate. By varying this distance, the time required to obtain the maximum yield can then be determined. Although this method is very simple in principle, it can be complicated in practice, as the illustration shows. Owing to the rather large volumes required, his method is more practical for the study of gas-phase reactions than for solutions, for which the stopped-flow method described below is generally preferred. These are by far the most common means of studying fast solution-phase reactions over time intervals of down to a fraction of a millisecond. The use of reasonably simple devices is now practical even in student laboratory experiments. These techniques make it possible to follow not only changes in the concentrations of reactants and products, but also the buildup and decay of reaction intermediates. The basic stopped-flow apparatus consists of two or more coupled syringes that rapidly inject the reactants into a small mixing chamber and then through an observation cell that can be coupled to instruments that measure absorption, fluorescence, light scattering, or other optical or electrical properties of the solution. As the solution flows through the cell, it empties into a stopping syringe that, when filled, strikes a backstop that abruptly stops the flow. The volume that the stopping syringe can accept is adjusted so that the mixture in the cell has just become uniform and has reached a steady state; at this point, recording of the cell measurement begins and its change is followed. Of course, there are many reactions that cannot be followed by changes in light absorption or other physical properties that are conveniently monitored. In such cases, it is often practical to (stop) the reaction after a desired interval by adding an appropriate quenching agent. For example, an enzyme-catalyzed reaction can be stopped by adding an acid, base, or salt solution that denatures (destroys the activity of) the protein enzyme. Once the reaction has been stopped, the mixture is withdrawn and analyzed in an appropriate manner. works something like the stopped-flow method described above, with a slightly altered plumbing arrangement. The reactants A and B are mixed and fed directly through the diverter valve to the measuring cell, which is not shown in this diagram. After a set interval that can vary from a few milliseconds to 200 sec or more, the controller activates the quenching syringe and diverter valve, flooding the cell with the quenching solution. ) | 3,493 | 1,009 |
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The electronic angular wavefunction is one spatial component of the electronic Schrödinger wave equation, which describes the motion of an electron. It depends on angular variables, \(\theta\) and \(\phi\), and describes the direction of the orbital that the electron may occupy. Some of its solutions are equal in energy and are therefore called degenerate. Electrons can be described as a particle or a wave. Because they exhibit wave behavior, there is a wavefunction that is a solution to the Schrödinger wave equation: \[\hat{H}\Psi(r,\phi,\theta,t)=E\Psi(r,\phi,\theta,t)\] This equation has eigenvalues, \(E\), which are energy values that correspond to the different wavefunctions. To solve the Shrödinger equation, spherical coordinates are used. Spherical coordinates are in terms of a radius \(r\), as well as angles \(\phi\), which is measured from the positive x axis in the xy plane and may be between 0 and \(2\pi\), and \(\theta\), which is measured from the positive z axis towards the xy plane and may be between 0 and \(\pi\). \(x=rsin(\theta)cos(\phi)\) \(y=rsin(\theta)sin(\phi)\) \(z=rcos(\theta)\) The electronic wavefunction, \(\Psi(r,\phi ,\theta ,t)\), describes the wave behavior of an electron. Its value is purely mathematical and has no corresponding measurable physical quantity. However, the square modulus of the wavefunction, \(\mid \Psi(r,\phi ,\theta ,t)\mid ^2\) gives the probability of locating the electron at a given set of values. To use separation of variables, the wavefunction can be expressed as \(\Psi(r,\phi, \theta ,t)=R(r)Y_{l}^{m}(\phi, \theta)\) \(R(r)\) is the radial wavefunction and \(Y_{l}^{m}(\phi, \theta)\) is the angular wavefunction. Separating the angular variables in \(Y_{l}^{m}(\phi, \theta)\) gives \(Y_{l}^{m}(\phi, \theta)=\left[\dfrac{2l+1}{4\pi}\left(\dfrac{(l-\mid m \mid)!}{(l+\mid m \mid)!}\right)\right]^{\frac{1}{2}}P_l^{\mid m \mid}(cos(\theta))e^{im\phi}\) where \(P_l^{\mid m \mid}(cos(\theta))\) is a and is only in terms of the variable \(\theta\). The exponential function, which is only in terms of \(\phi\), determines the phase of the orbital. For the angular wavefunction, the square modulus gives the probability of finding the electron at a point in space on a ray described by \((\phi, \theta)\). The angular wavefunction describes the spherical harmonics of the electron's motion. Because orbitals are a cloud of the probability density of the electron, the square modulus of the angular wavefunction influences the direction and shape of the orbital. There are 3 quantum numbers defined by the Schrödinger wave equation. They are \(n\), \(l\), and \(m_{l}\). Each combination of these quantum numbers describe an orbital. Values for \(n\) come from from the radial wavefunction. \(n\) may be 1, 2, 3... Because they evolved from the separation of variables performed to solve the wavefunction, solutions to the angular wavefunction are quantized by the values for \(l\) and \(m_{l}\). Acceptable values for \(l\) are given by \(l=n-1\). The corresponding values for \(m_{l}\) are integers between \(-l\) and \(+l\). Orbitals descrbed by the same \(n\) and \(l\) values but different \(m_{l}\) values are degenerate, meaning that they are equal in energy but vary in their direction and, sometimes, shape. For \(p\) orbitals, \(l=1\), giving three \(m_{l}\) values and thus, 3 degenerate states. They are \(p_{x}\), \(p_{y}\)and \(p_{z}\). \(d\) orbitals have \(l=2\), giving 5 degenerate states. These are \(d_{xy}\), \(d_{xz}\), \(d_{yz}\), \(d_{z^2}\), \(d_{x^2-y^2}\). \(f\) orbitals have \(l=3\), giving a total of 7 degenerate states. 1. Which quantum numbers depend on the angular wavefunction? 2. Give the quantum numbers defined by the angular wavefunction for a \(d_{z^2}\) orbital. 3. Given that the spherical representation of the \(d_{x^2-y^2}\) orbital is \(r^2 sin^{2}(\theta) cos(2\phi)\), show that this matches the label for the orbital, \(x^2-y^2\). Hint: \(cos(2x)=cos^2(x)-sin^2(x)\) Solutions: 1. \(l\) and \(m_{l}\) 2. \(l=2\) and \(m_{l}=2,1,0,-1,-2\) 3. \(r^2 sin^{2}(\theta)cos(2\phi)=r^2 sin^{2}(\theta)(cos^2(\phi)-sin^2(\phi))\) \(r^2 sin^{2}(\theta)cos^2(\phi)-r^2 sin^{2}(\theta)sin^2(\phi)\) Since \(x=rsin(\theta)cos(\phi)\) and \(y=rsin(\theta)sin(\phi)\), \(r^2 sin^{2}(\theta) cos(2\phi)=x^2-y^2\) | 4,342 | 1,010 |
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Having now revised the basics of trends across and down the Periodic Table, we can use the concepts of Effective Nuclear Charge and Electronegativity to discuss the factors that contribute to the types of bonds formed between elements. Rules formulated by Kazimierz Fajans in 1923, can be used to predict whether a chemical bond is expected to be predominantly ionic or covalent, and depend on the relative charges and sizes of the cation and anion. . Although the bond in a compound like X Y may be considered to be 100% ionic, it will always have some degree of covalent character. When two oppositely charged ions (X and Y ) approach each other, the cation attracts electrons in the outermost shell of the anion but repels the positively charged nucleus. This results in a distortion, deformation or polarization of the anion. If the degree of polarization is quite small, an ionic bond is formed, while if the degree of polarization is large, a covalent bond results. The ability of a cation to distort an anion is known as its polarization power and the tendency of the anion to become polarized by the cation is known as its polarizability. The polarizing power and polarizability that enhances the formation of covalent bonds is favoured by the following factors: Reminder. Large cations are to be found on the bottom left of the periodic table and small anions on the top right. The greater the positive charge, the smaller the cation becomes and the is a measure of the charge to radius ratio. The cation charge increases (size decreases) and on the right, the anion size increases, both variations leading to an increase in the covalency. Thus covalency increases in the order: [Na Cl , NaCl] < [Mg 2(Cl) , MgCl ] < [Al 3(Cl) , AlCl ] and [Al 3(F) , AlF ] < [Al 3(Cl) , AlCl ] < [Al 3(Br) , AlBr ] Electronic configuration of the cation: for two cations of the same size and charge, the one with a pseudo noble-gas configuration (with 18 electrons in the outer-most shell) will be more polarizing than that with a noble gas configuration (with 8 electrons in the outermost shell). Thus zinc (II) chloride ( Zn(II) 1s 2s 2p 3s 3p 3d and Cl 1s 2s 2p 3s 3p ) is more covalent than magnesium chloride ( Mg(II) 1s 2s 2p ) despite the Zn ion (74 pm) and Mg ion (72 pm) having similar sizes and charges. From an MO perspective, the orbital overlap disperses the charge on each ion and so weakens the electrovalent forces throughout the solid, this can be used to explain the trend seen for the melting points of lithium halides. LiF = 870 °C, LiCl = 613 °C, LiBr = 547 °C, LiI = 446 °C It is found that the greater the possibility of polarization, the lower is the melting point and heat of sublimation and the greater is the solubility in non-polar solvents. The melting point of KCl is higher than that of AgCl though the crystal radii of Ag and K ions are almost the same. Solution: When the melting points of two compounds are compared, the one having the lower melting point is assumed to have the smaller degree of ionic character. In this case, both are chlorides, so the anion remains the same. The deciding factor must be the cation. (If the anions were different, then the answer could be affected by the variation of the anion.) Here the significant difference between the cations is in their electronic configurations. K = [Ar] and Ag =[Kr] 4d . This means a comparison needs to be made between a noble gas core and pseudo noble gas core, which as noted above holds that the pseudo noble gas would be the more polarizing. Based on Fajan's rules, it is expected that every ionic compound will have at least some amount of covalent character. The percentage of ionic character in a compound can be estimated from dipole moments. The bond dipole moment uses the idea of electric dipole moment to measure the polarity of a chemical bond within a molecule. It occurs whenever there is a separation of positive and negative charges. The bond dipole μ is given by: μ = δ d A bond dipole is modeled as +δ - δ- with a distance d between the partial charges. It is a vector, parallel to the bond axis and by convention points from minus to plus (note that many texts appear to ignore the convention and point from plus to minus). The SI unit for an electric dipole moment is the coulomb-meter, (C m). This is thought to produce values too large to be practical on the molecular scale so bond dipole moments are commonly measured in Debye, represented by the symbol, D. . This value arises from (1.602 x 10 * 1 x 10 ) / 3.336 x 10 where D = 3.336 x 10 C m (or 1 C m = 2.9979 x 10 D).
Typical dipole moments for simple diatomic molecules are in the range of 0 to 11 D (see Table below). The % ionic character = μ / μ (assuming 100% ionic bond) * 100 % From the Table below the observed dipole moment of KBr is given as 10.41 D, (3.473 x 10 coulomb metre), which being close to the upper level of 11 indicates that it is a highly polar molecule. The interatomic distance between K and Br is 282 pm. From this it is possible to calculate a theoretical dipole moment for the KBr molecule, assuming opposite charges of one fundamental unit located at each nucleus, and hence the percentage ionic character of KBr. Solution dipole moment μ = q * e * d coulomb metre
q = 1 for complete separation of unit charge
e = 1.602 x 10 C
d = 2.82 x 10 m for KBr (282 pm) Hence calculated μ = 1 * 1.602 x 10 * 2.82 x 10 = 4.518 x 10 Cm (13.54 D) The observed μ = 3.473 x 10 Cm (10.41 D) the % ionic character of KBr = 3.473 x 10 / 4.518 x 10 or 10.41 / 13.54 = 76.87% and the % covalent character is therefore about 23% (100 - 77). Given the observed dipole moment is 10.41 D (3.473 x 10 ) it is possible to estimate the charge distribution from the same equation by now solving for q:
Dipole moment μ = q * e * d Coulomb meters, but since q is no longer 1 we can substitute in values for μ and d to obtain an estimate for it.
q = μ /(e * d) = 3.473 x 10 / (1.602 x 10 * 2.82 x 10 ) thus q = 3.473 x 10 / (4.518 x 10 ) = 0.77 and the δ- and δ+ are -0.8 and +0.8 respectively. For HI, calculate the % of ionic character given a bond length = 161 pm and an observed dipole moment 0.44 D. Solution To calculate μ considering it as a 100% ionic bond μ = 1 * 1.602 x 10 * 1.61 x 10 / (3.336 x 10 ) = 7.73 D the % ionic character = 0.44/7.73 * 100 = 5.7% The calculated % ionic character is only 5.7% and the % covalent character is (100 - 5.7) = 94.3%. The ionic character arises from the polarizability and polarizing effects of H and I. Similarly, knowing the bond length and observed dipole moment of HCl, the % ionic character can be found to be 18%. Thus it can be seen that while HI is essentially covalent, HCl has significant ionic character. Note that by this simplistic definition, to achieve 100 % covalent character a compound must have an observed dipole moment of zero. Whilst not strictly true for heteronuclear molecules it does provide a simple qualitative method for predicting the bond character. It is possible to predict whether a given bond will be non-polar, polar covalent, or ionic based on the electronegativity difference, since the greater the difference, the more polar the bond. Linus Pauling proposed an empirical relationship which relates the percent ionic character in a bond to the electronegativity difference. percent ionic character= (1-e )* 100 This is shown as the curve in red below and is compared to the values for some diatomic molecules calculated from observed and calculated dipole moments. Intermolecular forces are the attractive forces between molecules without which all substances would be gases. The various types of these interactions span large differences in energy and for the halogens and interhalogens are generally quite small. The forces involved in these cases are called London dispersion forces (after Fritz Wolfgang London, 1900-1954). They are derived from momentary oscillations of electron charge in atoms and hence are present between all particles (atoms, ions and molecules). The ease with which the electron cloud of an atom can be distorted to become asymmetric is termed the molecule's polarizability. The greater the number of electrons an atom has, the farther the outer electrons will be from the nucleus, and the greater the chance for them to shift positions within the molecule. This means that larger nonpolar molecules tend to have stronger London dispersion forces. This is evident when considering the diatomic elements in Group 17, the Halogens. All of these homo-nuclear diatomic elements are nonpolar, covalently bonded molecules. Descending the group, fluorine and chlorine are gases, bromine is a liquid, and iodine is a solid. For nonpolar molecules, the farther you go down the group, the stronger the London dispersion forces. To picture how this occurs, compare the situation 1) where the electrons are evenly distributed and then consider 2) an instantaneous dipole that would arise from an uneven distribution of electrons on one side of the nucleus. When two molecules are close together, the instantaneous dipole of one molecule can induce a dipole in the second molecule. This results in synchronised motion of the electrons and an attraction between them. 3) when this effect is multiplied over numerous molecules the overall result is that the attraction keeps these molecules together, and for diiodine is sufficient to make this a solid. On average the electron cloud for molecules can be considered to be spherical in shape. When two non-polar molecules approach, attractions or repulsions between the electrons and nuclei can lead to distortions in their electron clouds (i.e. dipoles are induced). When more molecules interact these induced dipoles lead to intermolecular attraction. The changes seen in the variation of MP and BP for the dihalogens and binary interhalogens can be attributed to the increase in the London dispersion forces of attraction between the molecules. In general they increase with increasing atomic number. For the elements in the 2nd row, as the atomic number increases, the atomic radius of the elements decreases, the electronegativity increases, and the ionization energy increases.
The 2nd row has two metals (lithium and beryllium), making it the least metallic period and it has the most nonmetals, with four. The elements in the 2nd row often have the most extreme properties in their respective groups; for example, fluorine is the most reactive halogen, neon is the most inert noble gas, and lithium is the least reactive alkali metal.
These differences in properties with the subsequent rows are a result of: Apart from the 2nd row (ignoring H/He 1st row) the later rows all end with inert gases but these do not have completed quantum levels. The 2nd row elements in general can only use the 2s and 2p electrons for bonding restricting the total number of bonds to 4.
So N is not expected to have more than 4 bonds and 3 is common, while for P 5 and 6 bonded species are quite common.
Reactivity of metals and metalloids
For Lithium, compared to other alkali metals
Reaction with water:
Li reacts slowly with water at 25 °C
Na reacts violently and K in flames 2M(s) + 2H O(l) → 2M (aq) + 2OH + H (g) In general Li, Be, B, C, N, O, F are less reactive towards water than their heavier congeners.
Reaction with oxygen:
In conditions of excess oxygen, only Li forms a simple oxide, Li O. Other metals form peroxides and superoxides
Reaction with nitrogen:
Li reacts directly with N to form Li N
6Li(s) + N (g) → 2Li N(s).
No other alkali metal reacts with N
Solubility:
LiF, LiOH and Li CO are less soluble than the corresponding Na and K compounds
For Beryllium compared to the other alkaline earth metals:
With water:
All Group 2 metals except Be, react with water
M(s) + 2H O(l) → M (aq)+ 2OH (aq) + H (g)
With oxygen (air):
Be only reacts with air above 600 °C if it is finely powdered. The BeO that is formed is (other Group 2 oxides are basic).
Of the Group 2 elements only Be reacts with NaOH or KOH to liberate H and form [Be(OH) ] .
Li and Be are metals but are less conducting than the higher members of Group 1 and 2 elements due to their high IEs (electrons are close to nucleus).
Ionization of Boron to B requires a large input of energy and B adopts a covalent polymeric structure with semi-metallic properties.
The other elements of Group 14 become increasingly metallic as the group is descended due to the decrease in ionization energies.
Crystalline Boron is chemically inert - unaffected by boiling HCl and only slowly oxidized by hot concentrated HNO when finely powdered.
Covalent character
Li and Be are small and have strong polarizing abilities. Their compounds are more covalent than those of the heavier elements in their groups.
BeCl is covalent while MCl (M = Mg-Ba) are ionic. The conductivity of fused beryllium chloride is only 1/1000 that of sodium chloride under similar conditions.
Catenation
is the linkage of atoms of the same element into longer chains. Catenation occurs most readily in carbon, which forms covalent bonds with other carbon atoms to form longer chains and structures. This is the reason for the presence of the vast number of organic compounds in nature.
The ability of an element to catenate is primarily based on the bond energy of the element to itself, which decreases with more diffuse orbitals (those with higher azimuthal quantum number) overlapping to form the bond. Hence, carbon, with the least diffuse valence shell 2p orbital is capable of forming longer p-p sigma bonded chains of atoms than heavier elements which bond via higher valence shell orbitals.
Hetero-catenation is quite common in Inorganic Chemistry. Phosphates and silicates with P-O-P-O and Si-O-Si-O linkages are examples of this.
Multiple Bonds C, N and O are able to form multiple bonds (double and/or triple). In Group 14, C=C double bonds are stable (134 pm) but Si=Si double bonds (227 pm) are uncommon. The diagram below shows how multiple bonds are formed involving π overlap of 2p orbitals. By comparison the 3p orbitals of the corresponding third row elements Si, P, and S are more diffuse and the longer bond distances expected for these larger atoms would result in poor π overlap. Due to the high electron affinities and electronegativities of oxygen and fluorine, they tend to form strong ionic bonds with other elements. They even react with noble gases to form compounds such as XeO , XeO , XeF and XeF . In 1962 Neil Bartlett at the University of British Columbia reacted platinum hexafluoride and xenon, in an experiment that demonstrated the chemical reactivity of the noble gases. He discovered the mustard yellow compound, xenon hexafluoroplatinate, which is perhaps now best formulated as a mixture of species, [XeF ,PtF ] , [XeF ,Pt F ] , and [Xe F ] [PtF ] . A few hundred compounds of other noble gases have subsequently been discovered: in 1962 for radon, radon difluoride (RnF ), and in 1963 for krypton, krypton difluoride (KrF ). The first stable compound of argon was reported in 2000 when argon fluorohydride (HArF) was formed at a temperature of 40 K (-233.2 °C). Neutral compounds in which helium and neon are involved in chemical bonds have still not been formed. Noble gas compounds have already made an impact on our daily lives. XeF is a strong fluorinating agent and has been used to convert uracil to 5-fluorouracil, one of the first anti-tumor agents. | 15,629 | 1,013 |
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The Joule-Thomson effect is also known as the Joule-Kelvin effect. This effect is present in non ideal gasses, where a change in temperature occurs upon expansion. The Joule-Thomson coefficient is given by \[\mu_{\mathrm JT} = \left. \dfrac{\partial T}{\partial p} \right\vert_H\] where In terms of heat capacities one has \[\mu_{\mathrm JT} C_V = -\left. \dfrac{\partial E}{\partial V} \right \vert_T \] and \[\mu_{\mathrm JT} C_p = -\left. \dfrac{\partial H}{\partial p} \right \vert_T \] In terms of the second virial coefficient at zero pressure one has \[\mu_{\mathrm JT}\vert_{p=0} = ^0\!\!\phi = B_2(T) -T \dfrac{dB_2(T)}{dT}\] | 652 | 1,015 |
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is the study of the compounds that form between metals and , where a ligand is any molecule or ion that binds to the metal. A metal is the unit containing the metal bound to its ligands. For example, [PtCl (NH ) ] is the neutral metal complex where the Pt metal is bound to two Cl ligands and two NH ligands. If a complex is charged, it is called a (ex. [Pt(NH ) ] is a complex cation). A complex ion is stabilized by formation of a with ions of opposite charge (ex. [Pt(NH ) ]Cl ). It is convention to write the formula of a complex or complex ion inside of square brackets, while counterions are written outside of the brackets. In this convention, it is understood that ligands inside the brackets are bound directly to the metal ion, in the metal's (a.k.a inner coordination sphere). Ions written outside of the brackets are assumed to be in the , and they are not directly bound to the metal. A common metal complex is Ag(NH ) , formed when Ag ions are mixed with neutral ammonia molecules. \[Ag^+ + 2 NH_3 \rightarrow Ag(NH_3)_2^+\] A complex Ag(S O ) is formed between silver ions and negative thiosulfate ions: \[Ag^+ + 2 S_2O_3^{2-} \rightarrow Ag(S_2O_3)_2^{3-}\] The geometry and arrangement of ligands around the metal center affect the properties of a coordination compounds. Compounds with the same molecular formula can appear as with very different properties. Isomers are molecules that have identical chemical formulas, but have different arrangements of atoms in space. Isomers with different geometric arrangements of ligands are called . Isomers whose structures are mirror images of each other are called . The coordination chemistry was pioneered by Nobel Prize winner Alfred Werner (1866-1919). He received the Nobel Prize in 1913 for his coordination theory of transition metal-amine complexes. At the start of the 20th century, inorganic chemistry was not a prominant field until Werner studied the metal-amine complexes such as \([Co(NH_3)_6Cl_3]\). Werner recognized the existence of several forms of cobalt-ammonia chloride. These compounds have different color and other characteristics. The chemical formula has three chloride ions per mole, but the number of chloride ions that precipitate with Ag ions per formula is not always three. He thought only will form precipitate with silver ion. In the following table, the number below the is the number of ionized chloride ions per formula. To distinguish ionized chloride from the coordinated chloride, Werner formulated the and explained structure of the cobalt complexes. The structures of the complexes were proposed based on a coordination sphere of 6. The 6 ligands can be ammonia molecules or chloride ions. Two different structures were proposed for the last two compounds, the compound has two chloride ions on opposite vertices of an octahedral, whereas the the two chloride ions are adjacent to each other in the compound. The and compounds are known as geometric isomers. Other cobalt complexes studied by Werner are also interesting. It has been predicted that the complex Co(NH CH CH NH ) ClNH ] should exist in two forms, which are mirror images of each other. Werner isolated solids of the two forms, and structural studies confirmed his interpretations. The ligand NH CH CH NH is ethylenediamine (en) often represented by en. Sketch the structures of isomers Co(en) complex ion to show that they are mirror images of each other.
The images are shown on page 242 by Swaddle. If the triangular face of the end-amino group lie on the paper, you can draw lines to represent the en bidentate ligand. These lines will show that the two images are similar to the left-hand and right-hand screws. Structures of coordination compounds can be very complicated, and their names long because the ligands may already have long names. Knowing the rules of nomenclature not only enable you to understand what the complex is, but also let you give appropriate names to them. Often, several groups of the ligand are involved in a complex. The number of ligand molecules per complex is indicated by a Greek prefix: mono-, di- (or bis), tri-, tetra-, penta-, hexa, hepta-, octa-, nona-, (ennea-), deca- etc for 1, 2, 3, ... 10 etc. If the names of ligands already have one of these prefixes, the names are placed in parentheses. The prefices for the number of ligands become bis-, tris-, tetrakis, pentakis- etc. For neutral ligands, their names are not changed, except the following few: Normal names that will not change The last "e" in names of negative ions are changed to "o" in names of complexes. Sometimes "ide" is changed to "o". Note the following: The names of complexes start with the ligands, the anionic ones first, followed with neutral ligands and the metal. If the complex is negative, the name ends with "ate". At the very end are some Roman numerals representing the oxidation state of the metal. To give and remember all rules of nomenclature are hard to do. Pay attention to the names whenever you encounter any complexes is the way to learn. A bridging ligand is indicated by placing a - before its name. The - should be repeated for every bridging ligand. For example, (H3N)3Co(OH)3Co(NH3)3, Triamminecobalt(III)-m-trihydroxotriamminecobalt(III) Give the structural formula for chlorotriphenylphosphinepalladium(II)- m-dichlorochlorotriphenylphosphinepalladium(II). The structure is Name the complex: The name is Bis(ethylenediamine)cobalt(III)- - imido- -hydroxobis(ethylenediamine)cobalt(III) ion. When this compound dissolves in water, is the solution a conductor? What are the ions present in the solution of this compound? How many moles of chloride ions are present per mole of the compound? When potassiumtrichloroammineplatinate(II) dissolves in water, what ions are produced? What about chloropentaamminecobalt(III) chloride? Before modern structure determination methods were developed, the study of complexes were mostly done by chemical methods and deduction. The fact that CoCl2(NH3)4Cl has only two isomer suggests that | 6,096 | 1,016 |
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The Baeyer strain theory suggested that the larger cycloalkanes ring are difficult to synthesize because of angle strain associated with planar rings, as calculated in Table 12-3. We now know that, except for cyclopropane, none of the cycloalkanes have planar carbon rings and that the higher cycloalkanes have normal or nearly normal bond angles. The reason that the higher cycloalkanes are generally difficult to synthesize from open-chain compounds is not so much angle strain, as Baeyer hypothesized, but the low probability of having reactive groups on the two fairly remote ends of a long hydrocarbon chain come together to effect cyclization. Usually, coupling of reactive groups on the ends of molecules occurs in preference to cyclization, unless the reactions are carried out in very dilute solutions. This is called the technique for achieving ring formation when the ring-forming reaction has to compete with rapid inter-molecular reactions. With regard to conformations of the larger cycloalkanes, we first note that the chair form of cyclohexane is a “perfect” conformation for a cycloalkane. The \(\ce{C-C-C}\) bond angles are close to their normal values, all the adjacent hydrogens are staggered with respect to one another, and the 1,3-axial hydrogens are not close enough together to experience nonbonded repulsions. About the only qualification one could put on the ideality of the chair form is that the trans conformation of butane is somewhat more stable than the gauche conformation (Section 5-2), and that all of the \(\ce{C-C-C-C}\) segments of cyclohexane have the gauche arrangement. Arguing from this, J. Dale\(^6\) has suggested that large cycloalkane rings would tend to have trans \(\ce{C-C-C-C}\) segments to the degree possible and, indeed, cyclotetradecane seems to be most stable in a rectangular conformation with trans \(\ce{C-C-C-C}\) bond segments (Figure 12-16). This conformation has a number of possible substituent positions, but because only single isomers of monosubstituted cyclotetradecanes have been isolated, rapid equilibration of the various conformational isomers must occur. Other evidence indicates that the barrier to interconversion of these conformations is about \(7 \: \text{kcal mol}^{-1}\). With the cycloalkanes having 7 to 10 carbons, there are problems in trying to make either trans or gauche \(\ce{C-C-C-C}\) segments, because the sizes of these rings do not allow the proper bond angles or torsional angles, or else there are more or less serious nonbonded repulsions. Consequently each of these rings assumes a compromise conformation with some eclipsing, some nonbonded repulsions, and some angle distortions. Brief comments on some of these conformations follow. It will be useful to use molecular models to see the interactions involved. Possible conformations for cycloheptane include the “comfortable” appearing chair form, \(7\). However, this form has eclipsed hydrogens at \(\ce{C_4}\) and \(\ce{C_5}\) as well as nonbonded interactions between the axial-like hydrogens on \(\ce{C_3}\) and \(\ce{C_6}\). The best compromise conformation is achieved by a \(30^\text{o}\)-\(40^\text{o}\) rotation around the \(\ce{C_4-C_5}\) bond to relieve the eclipsing of the hydrogens. This spreads the interfering hydrogens at \(\ce{C_3}\) and \(\ce{C_6}\) and results in a somewhat less strained conformation called the . The twist chair, \(8\), is very flexible and probably only about \(3 \: \text{kcal mol}^{-1}\) of activation is required to interconvert the various possible monosubstituted cycloheptane conformations. There are several more or less reasonable looking cyclooctane conformations. After much research it now is clear that the favored conformation is the , \(9\), which is in equilibrium with a few tenths percent of the conformation, \(10\) : The activation energy for interconversion of these two forms is about \(10 \: \text{kcal mol}^{-1}\). The boat-chair conformation \(9\) is quite flexible and movement of its \(\ce{CH_2}\) groups between the various possible positions occurs with an activation energy of only about \(5 \: \text{kcal mol}^{-1}\). Several more or less reasonable conformations of cyclononane also can be developed, but the most favorable one is called the , which has three-fold symmetry (Figure 12-17). The activation energy for inversion of the ring is about \(6 \: \text{kcal mol}^{-1}\). The stable conformation of cyclodecane (Figure 12-18) is similar to that of cyclotetradecane (Figure 12-16). However, there are relatively short \(\ce{H} \cdot \cdot \cdot \cdot \ce{H}\) distances and the \(\ce{C-C-C}\) bond angles are somewhat distorted because of cross-ring hydrogen-hydrogen repulsions. The most stable position for a substituent on the cyclodecane ring is the one indicated in Figure 12-18. The least stable positions are those in which a substituent replaces any of the six hydrogens shown, because nonbonded interactions are particularly strong at these positions. The activation energy for interconversion of substituent positions is about \(6 \: \text{kcal mol}^{-1}\). Most stable conformation of cyclodecane; Dale and sawhorse representations. The shaded area in the sawhorse convention indicates substantial nonbonded \(\ce{H} \cdot \cdot \cdot \cdot \ce{H}\) interactions. \(^6\)Pronounced Dalluh. and (1977) | 5,367 | 1,017 |
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Colligative properties are important properties of as they describe how the properties of the will change as (or solutes) is (are) added. Before discussing these important properties, let us first review some definitions. Solutions can exist in solid (alloys of metals are an example of solid-phase solutions), liquid, or gaseous (aerosols are examples of gas-phase solutions) forms. For the most part, this discussion will focus on liquid-phase solutions. In general (and as will be discussed in Chapter 8 in more detail) a liquid will freeze when \[ \mu_{solid} \le \mu_{liquid} \nonumber \] As such, the freezing point of the solvent in a solution will be affected by anything that changes the chemical potential of the solvent. As it turns out, the chemical potential of the solvent is reduced by the presence of a solute. In a mixture, the chemical potential of component \(A\) can be calculated by \[\mu_A=\mu_A^o + RT \ln \chi_A \label{chemp} \] And because \(\chi_A\) is always less than (or equal to) 1, the chemical potential is always reduced by the addition of another component. The condition under which the solvent will freeze is \[ \mu_{A,solid} = \mu_{A,liquid} \nonumber \] where the chemical potential of the liquid is given by Equation \ref{chemp}, which rearrangement to \[\dfrac{ \mu_A -\mu_A^o}{RT} = \ln \chi_A \nonumber \] To evaluate the temperature dependence of the chemical potential, it is useful to consider the temperature derivative at constant pressure. \[ \left[ \dfrac{\partial}{\partial T} \left( \dfrac{\mu_A-\mu_A^o}{RT} \right) \right]_{p} = \left( \dfrac{\partial \ln \chi_A}{\partial T} \right)_p \nonumber \] \[- \dfrac{\mu_A - \mu_A^o}{RT^2} + \dfrac{1}{RT} \left[ \left( \dfrac{\partial \mu_A}{\partial T} \right)_p -\left( \dfrac{\partial \mu_A^o}{\partial T} \right)_p \right] =\left( \dfrac{\partial \ln \chi_A}{\partial T} \right)_p \label{bigeq} \] Recalling that \[ \mu = H = TS \nonumber \] and \[\left( \dfrac{\partial \mu}{\partial T} \right)_p =-S \nonumber \] Equation \ref{bigeq} becomes \[ - \dfrac{(H_A -TS_A - H_A^o + TS^o_A)}{RT^2} + \dfrac{1}{RT} \left[ -S_A + S_A^o\right] =\left( \dfrac{\partial \ln \chi_A}{\partial T} \right)_p \label{bigeq2} \] And noting that in the case of the solvent freezing, \(H_A^o\) is the enthalpy of the pure solvent in solid form, and \(H_A\) is the enthalpy of the solvent in the liquid solution. So \[ H_A^o - H_a = \Delta H_{fus} \nonumber \] Equation \ref{bigeq2} then becomes \[ \dfrac{\Delta H_{fus}}{RT^2} - \cancel{ \dfrac{-S_A + S_A^o}{RT}} + \cancel{\dfrac{-S_A + S_A^o}{RT}}=\left( \dfrac{\partial \ln \chi_A}{\partial T} \right)_p \nonumber \] or \[ \dfrac{\Delta H_{fus}}{RT^2} = \left( \dfrac{\partial \ln \chi_A}{\partial T} \right)_p \nonumber \] Separating the variables puts the equation into an integrable form. \[ \int_{T^o}^T \dfrac{\Delta H_{fus}}{RT^2} dT = \int d \ln \chi_A \label{int1} \] where \(T^{o}\) is the freezing point of the pure solvent and \(T\) is the temperature at which the solvent will begin to solidify in the solution. After integration of Equation \ref{int1}: \[ - \dfrac{\Delta H_{fus}}{R} \left( \dfrac{1}{T} - \dfrac{1}{T^{o}} \right) = \ln \chi_A \label{int3} \] This can be simplified further by noting that \[ \dfrac{1}{T} - \dfrac{1}{T^o} = \dfrac{T^o - T}{TT^o} = \dfrac{\Delta T}{TT^o} \nonumber \] where \(\Delta T\) is the difference between the freezing temperature of the pure solvent and that of the solvent in the solution. Also, for small deviations from the pure freezing point, \(TT^o\) can be replaced by the approximate value \((T^o)^2\). So the Equation \ref{int3} becomes \[ - \dfrac{\Delta H_{fus}}{R(T^o)^2} \Delta T = \ln \chi_A \label{int4} \] Further, for dilute solutions, for which \(\chi_A\) , the mole fraction of the solvent is very nearly 1, then \[ \ln \chi_A \approx -(1 -\chi_A) = -\chi_B \nonumber \] where \(\chi_B\) is the mole fraction of the solute. After a small bit of rearrangement, this results in an expression for freezing point depression of \[ \Delta T = \left( \dfrac{R(T^o)^2}{\Delta H_{fus}} \right) \chi_B \nonumber \] The first factor can be replaced by \(K_f\): \[\dfrac{R(T^o)^2}{\Delta H_{fus}} = K_f \nonumber \] which is the for the solvent. \(\Delta T\) gives the magnitude of the reduction of freezing point for the solution. Since \(\Delta H_{fus}\) and \(T^o\) are properties of the solvent, the freezing point depression property is independent of the solute and is a property based solely on the nature of the solvent. Further, since \(\chi_B\) was introduced as \((1 - \chi_A)\), it represents the sum of the mole fractions of all solutes present in the solution. It is important to keep in mind that for a real solution, freezing of the solvent changes the composition of the solution by decreasing the mole fraction of the solvent and increasing that of the solute. As such, the magnitude of \(\Delta T\) will change as the freezing process continually removes solvent from the liquid phase of the solution. The derivation of an expression describing boiling point elevation is similar to that for freezing point depression. In short, the introduction of a solute into a liquid solvent lowers the chemical potential of the solvent, cause it to favor the liquid phase over the vapor phase. As sch, the temperature must be increased to increase the chemical potential of the solvent in the liquid solution until it is equal to that of the vapor-phase solvent. The increase in the boiling point can be expressed as \[ \Delta T = K_b \chi_B \nonumber \] where \[\dfrac{R(T^o)^2}{\Delta H_{vap}} = K_b \nonumber \] is called the and, like the cryoscopic constant, is a property of the solvent that is independent of the solute or solutes. A very elegant derivation of the form of the models for freezing point depression and boiling point elevation has been shared by F. E. Schubert (Schubert, 1983). Cryoscopic and ebullioscopic constants are generally tabulated using molality as the unit of solute concentration rather than mole fraction. In this form, the equation for calculating the magnitude of the freezing point decrease or the boiling point increase is \[ \Delta T = K_f \,m \nonumber \] or \[ \Delta T = K_b \,m \nonumber \] where \(m\) is the concentration of the solute in moles per kg of solvent. Some values of \(K_f\) and \(K_b\) are shown in the table below. The boiling point of a solution of 3.00 g of an unknown compound in 25.0 g of CCl raises the boiling point to 81.5 °C. What is the molar mass of the compound? The approach here is to find the number of moles of solute in the solution. First, find the concentration of the solution: \[(85.5\, °C- 76.8\, °C) = \left( 5.02\, °C\,Kg/mol \right) m \nonumber \] \[ m= 0.936\, mol/kg \nonumber \] Using the number of kg of solvent, one finds the number for moles of solute: \[ \left( 0.936 \,mol/\cancel{kg} \right) (0.02\,\cancel{kg}) =0.0234 \, mol \nonumber \] The ratio of mass to moles yields the final answer: \[\dfrac{3.00 \,g}{0.0234} = 128 g/mol \nonumber \] For much the same reason as the lowering of freezing points and the elevation of boiling points for solvents into which a solute has been introduced, the vapor pressure of a volatile solvent will be decreased due to the introduction of a solute. The magnitude of this decrease can be quantified by examining the effect the solute has on the chemical potential of the solvent. In order to establish equilibrium between the solvent in the solution and the solvent in the vapor phase above the solution, the chemical potentials of the two phases must be equal. \[\mu_{vapor} = \mu_{solvent} \nonumber \] If the solute is not volatile, the vapor will be pure, so (assuming ideal behavior) \[\mu_{vap}^o + RT \ln \dfrac{p'}{p^o} = \mu_A^o + RT \ln \chi_A \label{eq3} \] Where \(p’\) is the vapor pressure of the solvent over the solution. Similarly, for the pure solvent in equilibrium with its vapor \[ \mu_A^o = \mu_{vap}^o + RT \ln \dfrac{p_A}{p^o} \label{eq4} \] where \(p^o\) is the standard pressure of 1 atm, and \(p_A\) is the vapor pressure of the pure solvent. Substituting Equation \ref{eq4} into Equation \ref{eq3} yields \[ \cancel{\mu_{vap}^o} + RT \ln \dfrac{p'}{p^o}= \left ( \cancel{\mu_{vap}^o} + RT \ln \dfrac{p_A}{p^o} \right) + RT \ln \chi_A \nonumber \] The terms for \(\mu_{vap}^o\) cancel, leaving \[ RT \ln \dfrac{p'}{p^o}= RT \ln \dfrac{p_A}{p^o} + RT \ln \chi_A \nonumber \] Subtracting \(RT \ln(P_A/P^o)\) from both side produces \[ RT \ln \dfrac{p'}{p^o} - RT \ln \dfrac{p_A}{p^o} = RT \ln \chi_A \nonumber \] which rearranges to \[RT \ln \dfrac{p'}{p_A} = RT \ln \chi_A \nonumber \] Dividing both sides by \(RT\) and then exponentiating yields \[ \dfrac{p'}{p_A} = \chi_A \nonumber \] or \[ p'=\chi_Ap_A \label{RL} \] This last result is Raoult’s Law. A more formal derivation would use the fugacities of the vapor phases, but would look essentially the same. Also, as in the case of freezing point depression and boiling point elevations, this derivation did not rely on the nature of the solute! However, unlike freezing point depression and boiling point elevation, this derivation did not rely on the solute being dilute, so the result should apply the entire range of concentrations of the solution. Consider a mixture of two volatile liquids A and B. The vapor pressure of pure A is 150 Torr at some temperature, and that of pure B is 300 Torr at the same temperature. What is the total vapor pressure above a mixture of these compounds with the mole fraction of B of 0.600. What is the mole fraction of B in the vapor that is in equilibrium with the liquid mixture? Using Raoult’s Law (Equation \ref{RL}) \[ p_A = (0.400)(150\, Toor) =60.0 \,Torr \nonumber \] \[ p_B = (0.600)(300\, Toor) =180.0 \,Torr \nonumber \] \[ p_{tot} = p_A + p_B = 240 \,Torr \nonumber \] To get the mole fractions in the gas phase, one can use Dalton’s Law of partial pressures. \[ \chi_A = \dfrac{ p_A}{p_{tot}} = \dfrac{60.0 \,Torr}{240\,Torr} = 0.250 \nonumber \] \[ \chi_B = \dfrac{ p_B}{p_{tot}} = \dfrac{180.0 \,Torr}{240\,Torr} = 0.750 \nonumber \] And, of course, it is also useful to note that the sum of the mole fractions is 1 (as it must be!) \[ \chi_A+\chi_B =1 \nonumber \] is a process by which solvent can pass through a semi-permeable membrane (a membrane through which solvent can pass, but not solute) from an area of low solute concentration to a region of high solute concentration. The is the pressure that when exerted on the region of high solute concentration will halt the process of osmosis. The nature of osmosis and the magnitude of the osmotic pressure can be understood by examining the chemical potential of a pure solvent and that of the solvent in a solution. The chemical potential of the solvent in the solution (before any extra pressure is applied) is given by \[ \mu_A = \mu_A^o + RT \ln \chi_A \nonumber \] And since x < 1, the chemical potential is of the solvent in a solution is always lower than that of the pure solvent. So, to prevent osmosis from occurring, something needs to be done to raise the chemical potential of the solvent in the solution. This can be accomplished by applying pressure to the solution. Specifically, the process of osmosis will stop when the chemical potential solvent in the solution is increased to the point of being equal to that of the pure solvent. The criterion, therefore, for osmosis to cease is \[ \mu_A^o(p) = \mu_A(\chi_b, +\pi) \nonumber \] To solve the problem to determine the magnitude of , the pressure dependence of the chemical potential is needed in addition to understanding the effect the solute has on lowering the chemical potential of the solvent in the solution. The magnitude, therefore, of the increase in chemical potential due to the application of excess pressure must be equal to the magnitude of the reduction of chemical potential by the reduced mole fraction of the solvent in the solution. We already know that the chemical potential of the solvent in the solution is reduced by an amount given by \[ \mu^o_A - \mu_A = RT \ln \chi_A \nonumber \] And the increase in chemical potential due to the application of excess pressure is given by \[ \mu(p+\pi) = \mu(p) + \int _{p}^{\pi} \left( \dfrac{\partial \mu}{\partial p} \right)_T dp \nonumber \] The integrals on the right can be evaluated by recognizing \[\left( \dfrac{\partial \mu}{\partial p} \right)_T = V \nonumber \] where \(V\) is the molar volume of the substance. Combining these expressions results in \[ -RT \ln \chi_A = \int_{p}^{p+\pi} V\,dp \nonumber \] If the molar volume of the solvent is independent of pressure (has a very small value of \(\kappa_T\) – which is the case for most liquids) the term on the right becomes. \[ \int_{p}^{\pi} V\,dP = \left. V p \right |_{p}^{p+\pi} = V\pi \nonumber \] Also, for values of \(\chi_A\) very close to 1 \[ \ln \chi_A \approx -(1- \chi_A) = - \chi_B \nonumber \] So, for dilute solutions \[ \chi_B RT = V\pi \nonumber \] Or after rearrangement \[ \pi \dfrac{\chi_B RT}{V} \nonumber \] again, where \(V\) is the molar volume of the solvent. And finally, since \(\chi_B/V\) is the concentration of the solute \(B\) for cases where \(n_B \ll n_A\). This allows one to write a simplified version of the expression which can be used in the case of very dilute solutions \[ \pi = [B]RT \nonumber \] When a pressure exceeding the osmotic pressure \(\pi\) is applied to the solution, the chemical potential of the solvent in the solution can be made to exceed that of the pure solvent on the other side of the membrane, causing reverse osmosis to occur. This is a very effective method, for example, for recovering pure water from a mixture such as a salt/water solution. Colligative properties are properties that depend on the of particles rather than their total mass. This implies that these properties can be used to measure molar mass. Colligative properties include: When we freeze a dilute solution the resulting frozen solvent is often quite a bit purer than the original solution. Let us consider this problem and make the following rather opposite assumptions about the solvent component: Under these two assumptions, we can consider the thermodynamic potentials of the solvent component (1) at the freezing point. They should be equal once equilibrium has been reached: \[μ_1^s = μ_1^{sln} \nonumber \] \[μ_1^s = μ_1^{liq*} + RT\ln\, a_1 \nonumber \] \[ \dfrac{μ_1^s - μ_1^{liq*} }{RT}=\ln \, a_1 \nonumber \] \[ \dfrac{-Δμ_1}{RT}=\ln \,a_1 \nonumber \] We can now apply by differentiation with respect to temperature: \[ \dfrac{\partial}{\partial\;T} \dfrac{Δμ_1}{T} = \dfrac{- Δ_{fus}H_{molar,1}}{T^2} \nonumber \] \[\dfrac{\partial [\ln\,a_1}{\partial T} = \dfrac{Δ_{fus}H_{molar,1}}{RT^2} \nonumber \] This means that we can actually use the quantity \(Δ_{fus}H_{molar,1}/RT^2\) to determine activities by integration, but usually Raoult Law is assumed valid: \[\ln a_1 \ln x_1 ≈ -x_2 \nonumber \] If we integrate \(Δ_{fus}H_{molar,1}/RT^2\) from the melting point of the pure solvent \(T_m^*\) to the actual melting point of the solution \(T_m\) we get: \[-x_2 = \dfrac{Δ_{fus}H_{molar,1}}{R} \left( \dfrac{1}{T_m^*}- \dfrac{1}{T_m} \right) = \dfrac{Δ_{fus}H_{molar,1}}{R} \left( \dfrac{T_m- T_m^*}{T_m^* T_m} \right) ≈ \dfrac{Δ_{fus}H_{molar,1}}{R} \left(\dfrac{ -ΔT}{T_m^{*2}} \right) \nonumber \] This is often rewritten in terms of molality as: \[ΔT = K_f.m \nonumber \] If K is known for the solvent we can add a number of grams of an unknown compound to the solvent measure the temperature depression, this tells us the molality. From molality and weight we can then calculate molar mass. Boiling point elevation is quite similar. | 15,773 | 1,018 |
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Mass spectrometry is an analytic method that employs ionization and mass analysis of compounds to determine the mass, formula and structure of the compound being analyzed. A mass analyzer is the component of the mass spectrometer that takes ionized masses and separates them based on charge to mass ratios and outputs them to the detector where they are detected and later converted to a digital output. There are six general types of mass analyzers that can be used for the separation of ions in a mass spectrometry. The DC bias will cause all the charged molecules to accelerate and move away from the center line, the rate being proportional to their charge to mass ratio. If their course goes off too far they will hit the metal rods or the sides of the container and be absorbed. So the DC bias acts like the magnetic field B of the mass spec and can be tuned to specific charge to mass ratios hitting the detector. The two sinusoidal electric fields at 90 orientation and 90 degrees degrees phase shift will cause an electric field which oscillates as a circle over time. So as the charged particles fly down toward the detector, they will be traveling in a spiral, the diameter of the spiral being determined by the charge to mass ratio of the molecule and the frequency and strength of the electric field. The combination of the DC bias and the circularly rotating electric field will be the charge particles will travel in a spiral which is curved. So by timing the peak of the curved spiral to coincide with the position of the detector at the end of the quadrupole, a great deal of selectivity to molecules charge to mass ratio can be obtained. TOF Analyzers separate ions by time without the use of an electric or magnetic field. In a crude sense, TOF is similar to chromatography, except there is no stationary/ mobile phase, instead the separation is based on the kinetic energy and velocity of the ions. Ions of the same charges have equal kinetic energies; kinetic energy of the ion in the flight tube is equal to the kinetic energy of the ion as it leaves the ion source: \[KE = \dfrac{mv^2}{2} = zV \label{1}\] The time of flight, or time it takes for the ion to travel the length of the flight tube is: \[T_f = \dfrac{L}{v} \label{2}\] Substituting Equation 1 for kinetic energy in Equation 2 for time of flight: \[T_f = L\sqrt{\dfrac{m}{z}}\sqrt{ \dfrac{1}{2 V}} \propto \sqrt{\dfrac{m}{z}} \label{3}\] During the analysis, \(L\), length of tube, the Voltage from the ion source \(V\) are held constant, which can be used to say that time of flight is directly proportional to the root of the mass to charge ratio. Unfortunately, at higher masses, resolution is difficult because flight time is longer. Also at high masses, not all of the ions of the same m/z values reach their ideal TOF velocities. To fix this problem, often a is added to the analyzer. The reflectron consists of a series of ring electrodes of very high voltage placed at the end of the flight tube. When an ion travels into the reflectron, it is reflected in the opposite direction due to the high voltage. The reflectron increases resolution by narrowing the broadband range of flight times for a single m/z value. Faster ions travel further into the reflectrons, and slower ions travel less into the reflector. This way both slow and fast ions, of the same m/z value, reach the detector at the same time rather then at different times, narrowing the bandwidth for the output signal. Similar to time of flight (TOF) analyzer mentioned earlier,in magnetic sector analyzers ions are accelerated through a flight tube, where the ions are separated by charge to mass ratios. The difference between magnetic sector and TOF is that a magnetic field is used to separate the ions. As moving charges enter a magnetic field, the charge is deflected to a circular motion of a unique radius in a direction perpendicular to the applied magnetic field. Ions in the magnetic field experience two equal forces; force due to the magnetic field and centripetal force. \[F_B= zvB =F_c= \dfrac{mv^2}{r} \label{4}\] The above equation can then be rearranged to give: \[v = \dfrac{Bzr}{m} \label{5}\] If this equation is substituted into the kinetic energy equation: \[KE= zV=\dfrac{mv^2}{2} \label{6}\] \[\dfrac{m}{z}=\dfrac{B^2r^2}{2V} \label{7}\] Basically the ions of a certain \(m/z\) value will have a unique path radius which can be determined if both magnetic field magnitude \(B\), and voltage difference \(V\) for region of acceleration are held constant. when similar ions pass through the magnetic field, they all will be deflected to the same degree and will all follow the same trajectory path. Those ions which are not selected by \(V\) and \(B\) values, will collide with either side of the flight tube wall or will not pass through the slit to the detector. Magnetic sector analyzers are used for mass focusing, they focus angular dispersions. Is similar to time of flight analyzer in that it separates the ions while in flight, but it separates using an electric field. Electrostatic sector analyzer consists of two curved plates of equal and opposite potential. As the ion travels through the electric field, it is deflected and the force on the ion due to the electric field is equal to the centripetal force on the ion. Here the ions of the same kinetic energy are focused, and ions of different kinetic energies are dispersed. \[KE = zV =\dfrac{mv^2}{2} \label{8}\] \[F_E= zE= F_c=\dfrac{mv^2}{R} \label{9}\] Electrostatic sector analyzers are energy focusers, where an ion beam is focused for energy. Electrostatic and magnetic sector analyzers when employed individually are single focusing instruments. However when both techniques are used together, it is called a double focusing instrument., because in this instrument both the energies and the angular dispersions are focused. This analyzer employs similar principles as the quadrupole analyzer mentioned above, it uses an electric field for the separation of the ions by mass to charge ratios. The analyzer is made with a ring electrode of a specific voltage and grounded end cap electrodes. The ions enter the area between the electrodes through one of the end caps. After entry, the electric field in the cavity due to the electrodes causes the ions of certain m/z values to orbit in the space. As the radio frequency voltage increases, heavier mass ion orbits become more stabilized and the light mass ions become less stabilized, causing them to collide with the wall, and eliminating the possibility of traveling to and being detected by the detector. The quadrupole ion trap usually runs a mass selective ejection, where selectively it ejects the trapped ions in order of in creasing mass by gradually increasing the applied radio frequency voltage. ICR is an ion trap that uses a magnetic field in order to trap ions into an orbit inside of it. In this analyzer there is no separation that occurs rather all the ions of a particular range are trapped inside, and an applied external electric field helps to generate a signal. As mentioned earlier, when a moving charge enters a magnetic field, it experiences a centripetal force making the ion orbit. Again the force on the ion due to the magnetic field is equal to the centripetal force on the ion. \[zvB=\dfrac{mv^2}{r} \label{10}\] Angular velocity of the ion perpendicular to the magnetic field can be substituted here \(w_c=v/r\) \[zB=mw_c \label{11}\] \[w_c=\dfrac{zB}{m} \label{12}\] Frequency of the orbit depends on the charge and mass of the ions, not the velocity. If the magnetic field is held constant, the charge to mass ratio of each ion can be determined by measuring the angular velocity \(\omega_c\). The relationship is that, at high \(w_c\), there is low m/z value, and at low \(\omega_c\), there is a high m/z value. Charges of opposite signs have the same angular velocity, the only difference is that they orbit in the opposite direction. To generate an electric signal from the trapped ions, a vary electric field is applied to the ion trap \[E=E_o \cos{(\omega_c t)} \label{13}\] When the \(\omega_c\) in the electric field matches the \(\omega_c\) of a certain ion, the ion absorbs energy making the velocity and orbiting radius of the ion increase. In this high energy orbit, as the ion oscillates between two plates, electrons accumulate at one of the plates over the other inducing an oscillating current, or current image. The current is directly proportional to the number of ions in the cell at a certain frequency. In a Fourier Transform ICR, all of the ions within the cell are excited simultaneously so that the current image is coupled with the image of all of the individual ion frequencies. is used to differential the summed signals to produce the desired results. | 8,831 | 1,019 |
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The first of the so-called "rare-earths" to be discovered, ytterbium takes its name from the Swedish village Ytterby (also the source of the names for terbium, erbium and yttrium). Discovery is credited to de Marignac in 1878. Initial identification was tediously made from the same mixture that most chemists of the time worked from: oxides of the lanthanides which gave rise to the term "rare-earth" due to its powdery consistency and often brownish color. But with the chemical separation techniques available at the time, it was very difficult to distinguish these similar elements. Even ytterbium itself turned out to hide another element. Lutetium was separated from it in 1907. Pure ytterbium is like most of the lanthanides: silvery and ductile, reacting slowly with air to form an oxide. Mostly obtained from monazite sand, ytterbium makes up about 0.03% of that mixture. | 900 | 1,020 |
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The structural theory of organic chemistry originated and developed from the concepts of valence and the tetrahedral carbon atom. It received powerful impetus from the electronic theory of bonding, as described in Chapter 6. We now express the structures of many organic compounds by simple bond diagrams which, when translated into three-dimensional models, are compatible with most observed molecular properties. Nonetheless, there are many situations for which ordinary structure theory is inadequate. An example is benzene ( ), which does not behave as would be expected if it were a cyclic polyene related to alkatrienes, such as \(\ce{CH_2=CH-CH=CH-CH=CH_2}\). There are many other substances that do not behave as predicted from the properties of simpler compounds. Some substances are more stable, some more reactive, some more acidic, some more basic, and so on, than we would have anticipated. In this chapter we shall look at the theories that explain some of these apparent anomalies. These theories will be based on quantum-mechanical arguments ( ). There are two popular approaches to the formulation of the structures and properties of organic compounds based on quantum mechanics - and methods. In the past, there has been great controversy as to which of these methods actually is more useful for qualitative purposes and, indeed, the adherents to one or the other could hardly even countenance suggestions of imperfections in their choice. Actually, neither is unequivocally better and one should know and use both - they are in fact more complementary than competitive. We have used the concepts of the resonance methods many times in previous chapters to explain the chemical behavior of compounds and to describe the structures of compounds that cannot be represented satisfactorily by a single valence-bond structure (e.g., benzene, ). We shall assume, therefore, that you are familiar with the qualitative ideas of resonance theory, and that you are aware that the so-called and are in fact synonymous. The further treatment given here emphasizes more directly the quantum-mechanical nature of valence-bond theory. The basis of molecular-orbital theory also is described and compared with valence-bond theory. First, however, we shall discuss general characteristics of simple covalent bonds that we would expect either theory to explain. and (1977) | 2,400 | 1,022 |
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In electronic spectroscopy, an atomic term symbol specifies a certain electronic state of an atom (usually a multi-electron one), by briefing the quantum numbers for the angular momenta of that atom. The form of an atomic term symbol implies coupling. Transitions between two different atomic states may be represented using their term symbols, to which certain rules apply. At the beginning, the spectroscopic notation for term symbols was derived from an obsolete system of categorizing spectral lines. In 1885, Johann Balmer, a Swiss mathematician, discovered the Balmer formula for a series of hydrogen emission lines. \[ \lambda=B \left(\dfrac{m^2}{m^2-4} \right)\] where Later it was extended by Johannes Rydberg and Walter Ritz. Yet this principle could hardly explain the discovery of fine structure, the splitting of spectral lines. In spectroscopy, spectral lines of alkali metals used to be divided into categories: sharp, principal, diffuse and fundamental, based on their fine structures. These categories, or "term series," then became associated with atomic energy levels along with the birth of the old quantum theory. The initials of those categories were employed to mark the atomic orbitals with respect to their azimuthal quantum numbers. The sequence of "s, p, d, f, g, h, i, k..." is known as the spectroscopic notation for atomic orbitals. By introducing spin as a nature of electrons, the fine structure of alkali spectra became further understood. The term "spin" was first used to describe the rotation of electrons. Later, although electrons have been proved unable to rotate, the word "spin" is reserved and used to describe the property of an electron that involves its intrinsic magnetism. LS coupling was first proposed by Henry Russell and Frederick Saunders in 1923. It perfectly explained the fine structures of hydrogen-like atomic spectra. The format of term symbols was developed in the Russell-Saunders coupling scheme. In the Russell-Saunders coupling scheme, term symbols are in the form of , where represents the total spin angular momentum, specifies the total orbital angular momentum, and refers to the total angular momentum. In a term symbol, is always an upper-case from the sequence "s, p, d, f, g, h, i, k...", wherein the first four letters stand for sharp, principal, diffuse and fundamental, and the rest follow in an alphabetical pattern. Note that the letter j is omitted. In today's physics, an electron in a spherically symmetric potential field can be described by four quantum numbers all together, which applies to hydrogen-like atoms only. Yet other atoms may undergo trivial approximations in order to fit in this description. Those quantum numbers each present a conserved property, such as the orbital angular momentum. They are sufficient to distinguish a particular electron within one atom. The term "angular momentum" describes the phenomenon that an electron distributes its position around the nucleus. Yet the underlying quantum mechanics is much more complicated than mere mechanical movement. Coupling of angular momenta was first introduced to explain the fine structures of atomic spectra. As for LS coupling, S, L, J and M are the four "good" quantum numbers to describe electronic states in lighter atoms. For heavier atoms, jj coupling is more applicable, where J, M , M and M are "good" quantum numbers. LS coupling, also known as Russell-Saunders coupling, assumes that the interaction between an electron's intrinsic angular momentum and its orbital angular momentum is small enough to be considered as an perturbation to the electronic Hamiltonian. Such interactions can be derived in a classical way. Let's suppose that the electron goes around the nucleus in a circular orbit, as in Bohr model. Set the electron's velocity to be . The electron experiences a magnetic field due to the relative movement of the nucleus \[ \mathbf{B}=\dfrac{1}{m_ec^2}(\mathbf{E} \times \mathbf{p})= \dfrac{Ze}{4 \pi \epsilon_0m_ec^2r^3} \mathbf{L} \label{2}\] while is the electric field at the electron due to the nucleus, the classical monumentum of the electron, and r the distance between the electron and the nucleus. The electron's spin brings a magnetic dipole moment \[ \mathbf \mu_s= \dfrac{-g_{se} \mathbf{s}}{2m_e}( \hat{ \mathbf{L}} \cdot \hat{ \mathbf{s}}) \label{3}\] where is the gyromagnetic ratio of an electron. Since the potential energy of the coulumbic attraction between the electron and nucleus is \[ V (r) = \dfrac{-Ze^2}{4πε_0r} \label{4}\] the interaction between and is \[ \hat{H}_{so}=\dfrac{g_s}{2m_e^2c^2} \dfrac{Ze^2}{4 \pi \epsilon_{_0}r^2}( \mathbf{L} \cdot \mathbf{s})=\dfrac{g_s}{2m_e^2c^2} \dfrac{ \partial{V}}{ \partial{r}}( \mathbf{L} \cdot \mathbf{s}) \label{5} \] After a correction due to centripetal acceleration , the interaction has a format of \[ \hat{H}_{so} = \dfrac{1}{2 \mu^2c^2} \dfrac{ \partial{V}}{ \partial{r}}( \mathbf{L} \cdot \mathbf{s}) \label{6} \] Thus the coupling energy is \[ \langle \psi_{nlm}| \hat{H}_{so}| \psi_{nlm} \rangle \label{7}\] In lighter atoms, the coupling energy is low enough be treated as a first-order perturbation to the total electronic Hamiltonian, hence LS coupling is applicable to them. For a single electron, the spin-orbit coupling angular momentum quantum number j has the following possible values = | - |, ..., + if the total angular momentum is defined as = + . The azimuthal counterpart of is , which can be a whole number in the range of [- ]. The first-order perturbation to the electronic energy can be deduced so \[ \dfrac{\hbar^2[j(j+1)-l(l+1)-s(s+1)]}{4\mu^2c^2}\int_0^\infty r^2dr \dfrac{1}{r} \dfrac{ \partial{V}}{\partial{r}}R_{nl}^2(r) \] Above is about the spin-orbit coupling of one electron. For many-electron atoms, the idea is similar. The coupling of angular momenta is \[ \mathbf{J}=\mathbf{L}+\mathbf{S} \label{9}\] thereby the total angular quantum number = | - |, ..., + where the total orbital quantum number \[ L=\sum\limits_i l_i \label{10}\] and the total spin quantum number \[ S=\sum \limits_{i}s_i \label{11}\] While is still the total angular momentum, and are the total orbital angular momentum and the total spin, respectively. The magnetic momentum due to is \[ \mathbf \mu_J=- \dfrac{g_Je \mathbf{J}}{2m_e} \label{12}\] wherein the Landé factor is \[ g_J=1+\dfrac{J(J+1)+S(S+1)-L(L+1)} {2J(J+1)} \label{13}\] supposing the gyromagnetic ratio of an electron is 2. For heavier atoms, the coupling between the total angular momenta of different electrons is more significant, causing the fine structures not to be "fine" any more. Therefore the coupling term can no more be considered as a perturbation to the electronic Hamiltonian, so that jj coupling is a better way to quantize the electron energy states and levels. For each electron, the quantum number = + . For the whole atom, the total angular momentum quantum number \[ J= \sum \limits_i j_i \label{14} \] Term symbols usually represent electronic states in the Russell-Saunders coupling scheme, where a typical atomic term symbol consists of the spin multiplicity, the symmetry label and the total angular momentum of the atom. They have the format of \[\large ^{2S+1}L_J\] such as , where = 1, L = 2, and = 2. Here is a commonly used method to determine term symbols for an electron configuration. It requires a table of possibilities of different "micro states," which happened to be called "Slater's table". Each row of the table represents a total magnetic quantum number, while each column does a total spin. Using this table we can pick out the possible electronic states easily since all terms are concentric rectangles on the table. The method of using a table to count possible "microstates" has been developed so long ago and honed by so many scientists and educators that it is hard to accredit a single person. Let's take the electronic configuration of d as an example. In the Slater's table, each cell contains the number of ways to assign the three electrons quantum numbers according to the M and M values. These assignments follow Pauli's exclusion law. The figure below shows an example to find out how many ways to assign quantum numbers to d electrons when M = 3 and M = -1/2. Now we start to subtract term symbols from this table. First there is a H state. And now it is subtracted from the table. And now is a F state. After being subtracted by F, the table becomes And now G. Now F. Here in the table are two D states. P. And the final deducted state is P. So in total the possible states for a d configuration are F, P, H, G, F, D, D and P. Taken into consideration, the possible states are: \[ ^4F_2\; ^4F_3 \; ^4F_4\; ^4P_0\; ^4P_1\; ^4P_2\; ^2H_{\frac{9}{2}} \; ^2H_{\frac{11}{2}} \; ^2G_{\frac{9}{2}} \; ^2G_{\frac{7}{2}} \; ^2F_{\frac{7}{2}}\; ^2F_{\frac{5}{2}} \; ^2D_{\frac{5}{2}} \; ^2D_{\frac{5}{2}} \; ^2D_{\frac{3}{2}} \; ^2D_{\frac{3}{2}}\; ^2P_{\frac{3}{2}} \; ^2P_{\frac{3}{2}} \] For lighter atoms before or among the first-row transition metals, this method works well. Another method is to use direct products in group theory to quickly work out possible term symbols for a certain electronic configuration. Basically, both electrons and holes are taken into consideration, which naturally results in the same term symbols for complementary configurations like p vs p . Electrons are categorized by spin, therefore divided into two categories, α and β, as are holes: α stands for +1, and βstands for -1, or vice versa. Term symbols of different possible configurations within one category are given. The term symbols for the total electronic configuration are derived from direct products of term symbols for different categories of electrons. For the p configuration, for example, the possible combinations of different categories are , , and . The first two combinations were assigned the partial term of . As and were given an symbol, the combination of them gives their direct product × = + [ ] + . The direct product for and is also × = + [ ] + . Considering the degeneracy, eventually the term symbols for p configuration are S, D and P. There is a specially modified version of this method for atoms with 2 unpaired electrons. The only step gives the direct product of the symmetries of the two orbitals. The degeneracy is still determined by . In general, states with a greater degeneracy have a lower energy. For one configuration, the level with the largest , which has the largest spin degeneracy, has the lowest energy. If two levels have the same value, then the one with the larger (and also the larger orbital degeneracy) have the lower energy. If the electrons in the subshell are fewer than half-filled, the ground state should have the smallest value of , otherwise the ground state has the greatest value of . Electrons of an atom may undergo certain transitions which may have strong or weak intensities. There are rules about which transitions should be strong and which should be weak. Usually an electronic transition is excited by heat or radiation. Electronic states can be interpreted by solutions of Schrödinger's equation. Those solutions have certain symmetries, which are a factor of whether transitions will be allowed or not. The transition may be triggered by an electric dipole momentum, a magnetic dipole momentum, and so on. These triggers are transition operators. The most common and usually most intense transitions occur in an electric dipolar field, so the selection rules are where a double dagger means not combinable. For jj coupling, only the third rule applies with an addition rule: Δ = 0; ±1. | 11,782 | 1,023 |
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Instructors can adopt existing LibreTexts texts or Remix them to quickly build course-specific resources to meet the needs of their students. Unlike traditional textbooks, LibreTexts’ web based origins allow powerful integration of advanced features and new access texts to improve postsecondary education at all levels of higher learning by developing an Open Access Resource environment. The project currently consists of 14 independently operating and interconnected libraries that are constantly being integrated. and are supported by the Department of Education Open Textbook Pilot 1525057, and 1413739. Unless otherwise noted, LibreTexts content is licensed by . Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not ). and are supported by the Department of Education Open Textbook Pilot Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. . . | 1,070 | 1,024 |
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Our next goal is to understand . Noncovalent interactions hold together the two strands DNA in the double helix, convert linear proteins to 3D structures that are necessary for enzyme activity, and are the basis for antibody-antigen association. More importantly, noncovalent interactions between water molecules are probably the feature of water that is most important for (the beginnings of life in the aqueous environment). Obviously, the topics of the next few sections, are of crucial importance to Biology. But in order to understand , we first need to develop a better understanding of the nature of bonds ranging from to . In other sections, chemical bonds are divided into two classes: covalent bonds, in which electrons are shared between atomic nuclei, and ionic bonds, in which electrons are transferred from one atom to the other. However, a sharp distinction between these two classes cannot be made. Unless both nuclei are the same (as in H ), an electron pair is never shared by both nuclei. There is thus some degree of electron transfer as well as electron sharing in most covalent bonds. On the other hand there is never a transfer of an electron from one nucleus to another in ionic compounds. The first nucleus always maintains some slight residual control over the transferred electron. are those in which electrons are shared equally between the two atoms involved. This can only happen for pairs of identical atoms. Iodine is a purple/black solid made up of \(\ce{I molecules, which should have a pure covalent bond by sharing 5p electrons. It's toxic and, in solution, used as a bacteriocide. The net charge on each atom is 0, meaning that the charge is the same as if it were an isolated \(\ce{I}\) atom. Another way of showing charge distribution is with an , where blue indicates positive charge and red indicates negative charge in regions of space around the two nuclei. Notice that the charge distribution is identical on the two iodines, although it is not uniformly distributed on either one (there's a positive region on the ends of the molecule, which probably results from electrons along the axis being drawn into the bond between the nuclei: It is curious that the iodine molecule, with no net charges on either atom, should attract other iodine molecules to make a solid. This exemplifies one kind of attraction important in biomolecules, the , and we'll discuss that a lot more later. It's not the most important kind of attraction, however. To see how stronger attractions between molecules arise, we need to see what happens when we change the \(\ce{I2}\) molecule slightly. Suppose we now change one \(\ce{I}\) atom to atoms in Group I: a \(\ce{Li}\) atom, \(\ce{Na}\) atom, and \(\ce{Cs}\) atom in succession. The products, lithium iodide (\(\ce{LiI}\)), sodium iodide (\(\ce{NaI}\)), and cesium iodide (\(\ce{CsI}\)) look like typical ionic compounds; they are all white crystalline solids. \(\ce{NaI}\) is used as a source of "iodine" (actually iodide) for "iodized salt", and looks just like \(\ce{NaCl}\). But the relatively low melting point of \(\ce{LiI}\) (459 C) is suggestive of covalent bonding. It is important to realize that all of these compounds exist as crystal lattices, not individual molecules, under ordinary conditions. The individual molecules that we're discussing are gas phase species, modeled in a vacuum. \(\ce{LiI}\) and its crystal lattice are shown here: The electrostatic potential surface confirms that there is sharing of electrons in \(\ce{LiF}\), because there is only a slight minimum in electron density between the atoms, and \(\ce{Li}\) has clearly distorted the spherical distribution of electrons on \(\ce{I}\), showing that electrons are shared. In a purely ionic compound, there would be virtually no electron density between the two electron clouds of the ions. We say that the small \(\ce{Li^{+}}\) ion distorts, or the large electron cloud of \(\ce{I^{-}}\). Large anions (negative ions) are easily polarized, while smaller ones, like F are much less polarizable because the electrons are held more tightly. We see that small cations (positive ions) like \(\ce{Li^{+}}\) are strong polarizers, while larger cations, like \(\ce{Na^{+}}\) or \(\ce{Cs^{+}}\) are less effective polarizers. Because \(\ce{Cs^{+}}\) is least effective in polarizing \(\ce{I^{-}}\), \(\ce{CsI}\) is the most ionic of the three. The electron cloud around the \(\ce{I^{-}}\) is almost spherical (undistorted), and there is a definite decrease in electron density in the region between \(\ce{Cs}\) and \(\ce{I}\). But there is still some sharing of electrons in \(\ce{CsI}\), because we do not see a region of zero electron density between two spherical ions. This is, in part, due to the fact that \(\ce{Cs}\) is large (near the bottom of Group I), so it is also slightly polarized by the iodine core. The extent of polarization in \(\ce{LiI}\) can be confirmed experimentally. An ion pair like \(\ce{LiI}\) has a negative end (\(\ce{I^{–}}\)) and a positive end (\(\ce{Li^{+}}\)). That is, it has two electrical “poles,” like the north and south magnetic poles of a magnet. The ion pair is therefore an electrical (literally “two poles“), and a quantity known as its dipole moment may be determined from experimental measurements. The \(μ\) is proportional to the size of the separated electrical charges \(Q\) and to the distance \(r\) between them: \[μ = Qr \label{1} \] In the \(\ce{LiI}\) ion pair the two nuclei are known to be separated by a distance of 239.2 pm. If the bond were completely ionic, there would be a net charge of –1.6021 × 10 C (the electronic charge) centered on the \(\ce{I}\) nucleus and a charge of +1.6021 × 10 C centered on the Li nucleus: The dipole moment would then be given (via Equation \ref{1}): \[\begin{align*} μ &= Qr \\[4pt] &= 1.6021 × 10^{–19} C × 239.2 × 10^{–12}\, m \\[4pt] &= 3.832 × 10^{–29}\, C\, m \end{align*} \nonumber \] The measured value of the dipole moment for the \(\ce{LiH}\) ion pair is 2.43× 10 C m, which is only about 64% of this value. This can only be because the negative charge is centered on the \(\ce{I}\) nucleus but shifted somewhat toward the \(\ce{Li^{+}}\) nucleus. This shift brings the opposite charges closer together, and the experimental dipole moment is smaller than would be expected. As the bond becomes less polarized, there is less electron sharing and the bond becomes more ionic. In the case of \(\ce{CsI}\), the charge is 0.822 e, so the dipole moment is 82% of the theoretical value for a totally ionic species. The bond distance is 270.0 pm, so the dipole moment is \[\begin{align*} μ &= Qr \\[4pt] &= 0.822\, e × 1.6021 × 10^{–19} C/e × 270.0 × 10^{–12} m \\[4pt] &= 3.56 × 10^{–29}\, C\, m \end{align*} \nonumber \] The polarization of the bond in \(\ce{LiI}\) gives it very different properties than the nonpolar \(\ce{I2}\). It's interesting that the blood/brain barrier allows nonpolar molecules, like \(\ce{O2}\) to pass freely, while more polar molecules may be prohibited. Ionic species, like the \(\ce{Li^{+}}\) and \(\ce{I^{–}}\) that result from dissolving \(\ce{LiI}\) in water, require special carrier-mediated transport mechanism which moderates the ion levels in the brain, even when plasma levels fluctuate significantly. | 7,330 | 1,025 |
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One of the most important properties of metallic elements is their ability to act as Lewis acids that form complexes with a variety of Lewis bases. A metal complex consists of a central metal atom or ion that is bonded to one or more (from the Latin ligare, meaning “to bind”), which are ions or molecules that contain one or more pairs of electrons that can be shared with the metal. Metal complexes can be neutral, such as Co(NH ) Cl ; positively charged, such as [Nd(H O) ] ; or negatively charged, such as [UF ] . Electrically charged metal complexes are sometimes called . A contains one or more metal complexes. Coordination compounds are important for at least three reasons. First, most of the elements in the periodic table are metals, and almost all metals form complexes, so metal complexes are a feature of the chemistry of more than half the elements. Second, many industrial catalysts are metal complexes, and such catalysts are steadily becoming more important as a way to control reactivity. For example, a mixture of a titanium complex and an organometallic compound of aluminum is the catalyst used to produce most of the polyethylene and polypropylene “plastic” items we use every day. Finally, transition-metal complexes are essential in biochemistry. Examples include hemoglobin, an iron complex that transports oxygen in our blood; cytochromes, iron complexes that transfer electrons in our cells; and complexes of Fe, Zn, Cu, and Mo that are crucial components of certain enzymes, the catalysts for all biological reactions. Coordination compounds have been known and used since antiquity; probably the oldest is the deep blue pigment called Prussian blue: \(\ce{KFe2(CN)6}\). The chemical nature of these substances, however, was unclear for a number of reasons. For example, many compounds called “double salts” were known, such as \(\ce{AlF3·3KF}\), \(\ce{Fe(CN)2·4KCN}\), and \(\ce{ZnCl2·2CsCl}\), which were combinations of simple salts in fixed and apparently arbitrary ratios. Why should \(\ce{AlF3·3KF}\) exist but not \(\ce{AlF3·4KF}\) or \(\ce{AlF3·2KF}\)? And why should a 3:1 KF:AlF3 mixture have different chemical and physical properties than either of its components? Similarly, adducts of metal salts with neutral molecules such as ammonia were also known—for example, \(\ce{CoCl3·6NH3}\), which was first prepared sometime before 1798. Like the double salts, the compositions of these adducts exhibited fixed and apparently arbitrary ratios of the components. For example, \(\ce{CoCl3·6NH3}\), \(\ce{CoCl3·5NH3}\), \(\ce{CoCl3·4NH3}\), and \(\ce{CoCl3·3NH3}\) were all known and had very different properties, but despite all attempts, chemists could not prepare \(\ce{CoCl3·2NH3}\) or \(\ce{CoCl3·NH3}\). Although the chemical composition of such compounds was readily established by existing analytical methods, their chemical nature was puzzling and highly controversial. The major problem was that what we now call valence (i.e., the oxidation state) and coordination number were thought to be identical. As a result, highly implausible (to modern eyes at least) structures were proposed for such compounds, including the “Chattanooga choo-choo” model for CoCl ·4NH shown here. The modern theory of coordination chemistry is based largely on the work of Alfred Werner (1866–1919; Nobel Prize in Chemistry in 1913). In a series of careful experiments carried out in the late 1880s and early 1890s, he examined the properties of several series of metal halide complexes with ammonia. For example, five different “adducts” of ammonia with PtCl were known at the time: PtCl ·nNH (n = 2–6). Some of Werner’s original data on these compounds are shown in Table \(\Page {1}\). The electrical conductivity of aqueous solutions of these compounds was roughly proportional to the number of ions formed per mole, while the number of chloride ions that could be precipitated as AgCl after adding Ag (aq) was a measure of the number of “free” chloride ions present. For example, Werner’s data on PtCl ·6NH in Table \(\Page {1}\) showed that all the chloride ions were present as free chloride. In contrast, PtCl ·2NH was a neutral molecule that contained no free chloride ions. Werner, the son of a factory worker, was born in Alsace. He developed an interest in chemistry at an early age, and he did his first independent research experiments at age 18. While doing his military service in southern Germany, he attended a series of chemistry lectures, and he subsequently received his PhD at the University of Zurich in Switzerland, where he was appointed professor of chemistry at age 29. He won the Nobel Prize in Chemistry in 1913 for his work on coordination compounds, which he performed as a graduate student and first presented at age 26. Apparently, Werner was so obsessed with solving the riddle of the structure of coordination compounds that his brain continued to work on the problem even while he was asleep. In 1891, when he was only 25, he woke up in the middle of the night and, in only a few hours, had laid the foundation for modern coordination chemistry. These data led Werner to postulate that metal ions have two different kinds of valence: (1) a primary valence ( ) that corresponds to the positive charge on the metal ion and (2) a secondary valence ( ) that is the total number of ligand-metal bonds bound to the metal ion. If \(\ce{Pt}\) had a primary valence of 4 and a secondary valence of 6, Werner could explain the properties of the \(\ce{PtCl4·NH3}\) adducts by the following reactions, where the metal complex is enclosed in square brackets: \[\begin{align*} \mathrm{[Pt(NH_3)_6]Cl_4} &\rightarrow \mathrm{[Pt(NH_3)_6]^{4+}(aq)+4Cl^-(aq)} \\[4pt]
\mathrm{[Pt(NH_3)_5Cl]Cl_3} &\rightarrow \mathrm{[Pt(NH_3)_5Cl]^{3+}(aq) +3Cl^-(aq)}\\[4pt]
\mathrm{[Pt(NH_3)_4Cl_2]Cl_2} &\rightarrow \mathrm{[Pt(NH_3)_4Cl_2]^{2+}(aq) +2Cl^-(aq)}\\[4pt]
\mathrm{[Pt(NH_3)_3Cl_3]Cl} &\rightarrow \mathrm{[Pt(NH_3)_3Cl_3]^+(aq) + Cl^-(aq)}\\[4pt]
\mathrm{[Pt(NH_3)_2Cl_4]} &\rightarrow \mathrm{[Pt(NH_3)_2Cl_4]^0(aq)} \end{align*} \label{23.9}\] Further work showed that the two missing members of the series—[Pt(NH )Cl ] and [PtCl ] —could be prepared as their mono- and dipotassium salts, respectively. Similar studies established coordination numbers of 6 for Co and Cr and 4 for Pt and Pd . Werner’s studies on the analogous Co complexes also allowed him to propose a structural model for metal complexes with a coordination number of 6. Thus he found that [Co(NH ) ]Cl (yellow) and [Co(NH ) Cl]Cl (purple) were 1:3 and 1:2 electrolytes. Unexpectedly, however, two different [Co(NH ) Cl ]Cl compounds were known: one was red, and the other was green (Figure \(\Page {1a}\)). Because both compounds had the same chemical composition and the same number of groups of the same kind attached to the same metal, there had to be something different about the arrangement of the ligands around the metal ion. Werner’s key insight was that the six ligands in [Co(NH ) Cl ]Cl had to be arranged at the vertices of an octahedron because that was the only structure consistent with the existence of two, and only two, arrangements of ligands (Figure \(\Page {1b}\). His conclusion was corroborated by the existence of only two different forms of the next compound in the series: Co(NH ) Cl . In Werner’s time, many complexes of the general formula MA B were known, but no more than two different compounds with the same composition had been prepared for any metal. To confirm Werner’s reasoning, calculate the maximum number of different structures that are possible for six-coordinate MA B complexes with each of the three most symmetrical possible structures: a hexagon, a trigonal prism, and an octahedron. What does the fact that no more than two forms of any MA B complex were known tell you about the three-dimensional structures of these complexes? : three possible structures and the number of different forms known for MA B complexes : number of different arrangements of ligands for MA B complex for each structure Sketch each structure, place a B ligand at one vertex, and see how many different positions are available for the second B ligand. The three regular six-coordinate structures are shown here, with each coordination position numbered so that we can keep track of the different arrangements of ligands. For each structure, all vertices are equivalent. We begin with a symmetrical MA complex and simply replace two of the A ligands in each structure to give an MA B complex: For the hexagon, we place the first B ligand at position 1. There are now three possible places for the second B ligand: at position 2 (or 6), position 3 (or 5), or position 4. These are the only possible arrangements. The (1, 2) and (1, 6) arrangements are chemically identical because the two B ligands are adjacent to each other. The (1, 3) and (1, 5) arrangements are also identical because in both cases the two B ligands are separated by an A ligand. Turning to the trigonal prism, we place the first B ligand at position 1. Again, there are three possible choices for the second B ligand: at position 2 or 3 on the same triangular face, position 4 (on the other triangular face but adjacent to 1), or position 5 or 6 (on the other triangular face but not adjacent to 1). The (1, 2) and (1, 3) arrangements are chemically identical, as are the (1, 5) and (1, 6) arrangements. In the octahedron, however, if we place the first B ligand at position 1, then we have only two choices for the second B ligand: at position 2 (or 3 or 4 or 5) or position 6. In the latter, the two B ligands are at opposite vertices of the octahedron, with the metal lying directly between them. Although there are four possible arrangements for the former, they are chemically identical because in all cases the two B ligands are adjacent to each other. The number of possible MA B arrangements for the three geometries is thus: hexagon, 3; trigonal prism, 3; and octahedron, 2. The fact that only two different forms were known for all MA B complexes that had been prepared suggested that the correct structure was the octahedron but did not prove it. For some reason one of the three arrangements possible for the other two structures could have been less stable or harder to prepare and had simply not yet been synthesized. When combined with analogous results for other types of complexes (e.g., MA B ), however, the data were best explained by an octahedral structure for six-coordinate metal complexes. Determine the maximum number of structures that are possible for a four-coordinate MA B complex with either a square planar or a tetrahedral symmetrical structure. square planar, 2; tetrahedral, 1 The coordination numbers of metal ions in metal complexes can range from 2 to at least 9. In general, the differences in energy between different arrangements of ligands are greatest for complexes with low coordination numbers and decrease as the coordination number increases. Usually only one or two structures are possible for complexes with low coordination numbers, whereas several different energetically equivalent structures are possible for complexes with high coordination numbers (n > 6). The following presents the most commonly encountered structures for coordination numbers 2–9. Many of these structures should be familiar to you from our discussion of the valence-shell electron-pair repulsion ( ) model because they correspond to the lowest-energy arrangements of n electron pairs around a central atom. Compounds with low coordination numbers exhibit the greatest differences in energy between different arrangements of ligands. Although it is rare for most metals, this coordination number is surprisingly common for d metal ions, especially Cu , Ag , Au , and Hg . An example is the [Au(CN) ] ion, which is used to extract gold from its ores. As expected based on VSEPR considerations, these complexes have the linear L–M–L structure shown here. Although it is also rare, this coordination number is encountered with d metal ions such as Cu and Hg . Among the few known examples is the HgI ion. Three-coordinate complexes almost always have the trigonal planar structure expected from the VSEPR model. Two common structures are observed for four-coordinate metal complexes: tetrahedral and square planar. The tetrahedral structure is observed for all four-coordinate complexes of nontransition metals, such as [BeF ] , and d ions, such as [ZnCl ] . It is also found for four-coordinate complexes of the first-row transition metals, especially those with halide ligands (e.g., [FeCl ] and [FeCl ] ). In contrast, square planar structures are routinely observed for four-coordinate complexes of second- and third-row transition metals with d electron configurations, such as Rh and Pd , and they are also encountered in some complexes of Ni and Cu . This coordination number is less common than 4 and 6, but it is still found frequently in two different structures: trigonal bipyramidal and square pyramidal. Because the energies of these structures are usually rather similar for most ligands, many five-coordinate complexes have distorted structures that lie somewhere between the two extremes. This coordination number is by far the most common. The six ligands are almost always at the vertices of an octahedron or a distorted octahedron. The only other six-coordinate structure is the trigonal prism, which is very uncommon in simple metal complexes. This relatively uncommon coordination number is generally encountered for only large metals (such as the second- and third-row transition metals, lanthanides, and actinides). At least three different structures are known, two of which are derived from an octahedron or a trigonal prism by adding a ligand to one face of the polyhedron to give a “capped” octahedron or trigonal prism. By far the most common, however, is the pentagonal bipyramid. This coordination number is relatively common for larger metal ions. The simplest structure is the cube, which is rare because it does not minimize interligand repulsive interactions. Common structures are the square antiprism and the dodecahedron, both of which can be generated from the cube. This coordination number is found in larger metal ions, and the most common structure is the tricapped trigonal prism, as in [Nd(H O) ] . Transition metals form metal complexes, polyatomic species in which a metal ion is bound to one or more ligands, which are groups bound to a metal ion. Complex ions are electrically charged metal complexes, and a coordination compound contains one or more metal complexes. Metal complexes with low coordination numbers generally have only one or two possible structures, whereas those with coordination numbers greater than six can have several different structures. Coordination numbers of two and three are common for d metal ions. Tetrahedral and square planar complexes have a coordination number of four; trigonal bipyramidal and square pyramidal complexes have a coordination number of five; and octahedral complexes have a coordination number of six. At least three structures are known for a coordination number of seven, which is generally found for only large metal ions. Coordination numbers of eight and nine are also found for larger metal ions. | 15,392 | 1,026 |
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(or just ) states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes. This law is a manifestation that enthalpy is a . Hess's Law is named after Russian Chemist and Doctor Germain Hess. Hess helped formulate the early principles of thermochemistry. His most famous paper, which was published in 1840, included his law on thermochemistry. Hess's law is due to enthalpy being a , which allows us to calculate the overall change in enthalpy by simply summing up the changes for each step of the way, until product is formed. All steps have to proceed at the same temperature and the equations for the individual steps must balance out. The principle underlying Hess's law does not just apply to Enthalpy and can be used to calculate other state functions like changes in and The heat of any reaction \(\Delta{H^°_f}\) for a specific reaction is equal to the sum of the heats of reaction for any set of reactions which in sum are equivalent to the overall reaction: (Although we have not considered the restriction, applicability of this law requires that all reactions considered proceed under similar conditions: we will consider all reactions to occur at constant pressure.) Hydrogen gas, which is of potential interest nationally as a clean fuel, can be generated by the reaction of carbon (coal) and water: \[C_{(s)} + 2 H_2O_{(g)} \rightarrow CO_{2\, (g)} + 2 H_{2\, (g)} \tag{2}\] Calorimetry reveals that this reaction requires the input of 90.1 kJ of heat for every mole of \(C_{(s)}\) consumed. By convention, when heat is absorbed during a reaction, we consider the quantity of heat to be a positive number: in chemical terms, \(q > 0\) for an endothermic reaction. When heat is evolved, the reaction is exothermic and \(q < 0\) by convention. It is interesting to ask where this input energy goes when the reaction occurs. One way to answer this question is to consider the fact that the reaction converts one fuel, \(C_{(s)}\), into another, \(H_{2(g)}\). To compare the energy available in each fuel, we can measure the heat evolved in the combustion of each fuel with one mole of oxygen gas. We observe that \[C_{(s)}+O_{2(g)} \rightarrow CO_{2(g)} \tag{3}\] produces \(393.5\, kJ\) for one mole of carbon burned; hence \(q=-393.5\, kJ\). The reaction \[2 H_{2(g)} + O_{2(g)} \rightarrow 2 H_2O_{(g)} \tag{4}\] produces 483.6 kJ for two moles of hydrogen gas burned, so q=-483.6 kJ. It is evident that more energy is available from combustion of the hydrogen fuel than from combustion of the carbon fuel, so it is not surprising that conversion of the carbon fuel to hydrogen fuel requires the input of energy. Of considerable importance is the observation that the heat input in equation [2], 90.1 kJ, is exactly equal to the difference between the heat evolved, -393.5 kJ, in the combustion of carbon and the heat evolved, -483.6 kJ, in the combustion of hydrogen. This is not a coincidence: if we take the combustion of carbon and add to it the reverse of the combustion of hydrogen, we get \[C_{(s)}+O_{2(g)} \rightarrow CO_{2(g)}\] \[2 H_2O_{(g)} \rightarrow 2 H_{2(g)} + O_{2(g)}\] \[C_{(s)} + O_{2(g)} + 2 H_2O_{(g)} \rightarrow CO_{2(g)} + 2 H_{2(g)} + O_{2(g)} \tag{5}\] Canceling the \(O_{2(g)}\) from both sides, since it is net neither a reactant nor product, equation [5] is equivalent to equation [2]. Thus, taking the combustion of carbon and "subtracting" the combustion of hydrogen (or more accurately, adding the reverse of the combustion of hydrogen) yields equation [2]. And, the heat of the combustion of carbon minus the heat of the combustion of hydrogen equals the heat of equation [2]. By studying many chemical reactions in this way, we discover that this result, known as Hess's Law, is general. A pictorial view of Hess's Law as applied to the heat of equation [2] is illustrative. In figure 1, the reactants C(s) + 2 H O(g) are placed together in a box, representing the state of the materials involved in the reaction prior to the reaction. The products CO (g) + 2 H (g) are placed together in a second box representing the state of the materials involved after the reaction. The reaction arrow connecting these boxes is labeled with the heat of this reaction. Now we take these same materials and place them in a third box containing C(s), O (g), and 2 H (g). This box is connected to the reactant and product boxes with reaction arrows, labeled by the heats of reaction in equation [3] and equation [4]. This picture of Hess's Law reveals that the heat of reaction along the "path" directly connecting the reactant state to the product state is exactly equal to the total heat of reaction along the alternative "path" connecting reactants to products via the intermediate state containing \(C_{(s)}\), \(O_{2(g)}\), and 2 \(H_{2(g)}\). A consequence of our observation of Hess's Law is therefore that the net heat evolved or absorbed during a reaction is independent of the path connecting the reactant to product (this statement is again subject to our restriction that all reactions in the alternative path must occur under constant pressure conditions). A slightly different view of figure 1 results from beginning at the reactant box and following a complete circuit through the other boxes leading back to the reactant box, summing the net heats of reaction as we go. We discover that the net heat transferred (again provided that all reactions occur under constant pressure) is exactly zero. This is a statement of the conservation of energy: the energy in the reactant state does not depend upon the processes which produced that state. Therefore, we cannot extract any energy from the reactants by a process which simply recreates the reactants. Were this not the case, we could endlessly produce unlimited quantities of energy by following the circuitous path which continually reproduces the initial reactants. By this reasoning, we can define an energy function whose value for the reactants is independent of how the reactant state was prepared. Likewise, the value of this energy function in the product state is independent of how the products are prepared. We choose this function, H, so that the change in the function, ΔH = H - H , is equal to the heat of reaction q under constant pressure conditions. H, which we call the enthalpy, is a state function, since its value depends only on the state of the materials under consideration, that is, the temperature, pressure and composition of these materials. The concept of a state function is somewhat analogous to the idea of elevation. Consider the difference in elevation between the first floor and the third floor of a building. This difference is independent of the path we choose to get from the first floor to the third floor. We can simply climb up two flights of stairs, or we can climb one flight of stairs, walk the length of the building, then walk a second flight of stairs. Or we can ride the elevator. We could even walk outside and have a crane lift us to the roof of the building, from which we climb down to the third floor. Each path produces exactly the same elevation gain, even though the distance traveled is significantly different from one path to the next. This is simply because the elevation is a "state function". Our elevation, standing on the third floor, is independent of how we got to the third floor, and the same is true of the first floor. Since the elevation thus a state function, the elevation gain is independent of the path. Now, the existence of an energy state function H is of considerable importance in calculating heats of reaction. Consider the prototypical reaction in subfigure 2.1, with reactants R being converted to products P. We wish to calculate the heat absorbed or released in this reaction, which is ΔH. Since H is a state function, we can follow any path from R to P and calculate ΔH along that path. In subfigure 2.2, we consider one such possible path, consisting of two reactions passing through an intermediate state containing all the atoms involved in the reaction, each in elemental form. This is a useful intermediate state since it can be used for any possible chemical reaction. For example, in figure 1, the atoms involved in the reaction are C, H, and O, each of which are represented in the intermediate state in elemental form. We can see in subfigure 2.2 that the ΔH for the overall reaction is now the difference between the ΔH in the formation of the products P from the elements and the ΔH in the formation of the reactants R from the elements. The ΔH values for formation of each material from the elements are thus of general utility in calculating ΔH for any reaction of interest. We therefore define the standard formation reaction for reactant R, as elements in standard state R and the heat involved in this reaction is the standard enthalpy of formation, designated by ΔH °. The subscript f, standing for "formation," indicates that the ΔH is for the reaction creating the material from the elements in standard state. The superscript ° indicates that the reactions occur under constant standard pressure conditions of 1 atm. From subfigure 2.2, we see that the heat of any reaction can be calculated from \[\Delta{H^°_f} = \Delta{H^°_{f,products}} -\Delta{H^°_{f,reactants}} \tag{6}\] Extensive tables of ΔH° values ( ) have been compiled that allows us to calculate with complete confidence the heat of reaction for any reaction of interest, even including hypothetical reactions which may be difficult to perform or impossibly slow to react. The enthalpy of a reaction does not depend on the elementary steps, but on the final state of the products and initial state of the reactants. Enthalpy is an extensive property and hence changes when the size of the sample changes. This means that the enthalpy of the reaction scales proportionally to the moles used in the reaction. For instance, in the following reaction, one can see that doubling the molar amounts simply doubles the enthalpy of the reaction. H (g) + 1/2O (g) → H O (g) ΔH° = -572 kJ 2H (g) + O (g) → 2H O (g) ΔH° = -1144kJ The sign of the reaction enthalpy changes when a process is reversed. H (g) + 1/2O (g) → H O (g) ΔH° = -572 kJ When switched: H O (g) → H (g) + 1/2O (g) ΔH° = +572 kJ Since enthalpy is a state function, it is path independent. Therefore, it does not matter what reactions one uses to obtain the final reaction. | 10,512 | 1,028 |
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The TLC pictured in this section shows elution of a TLC plate containing several samples of red food dye at different aqueous dilutions (0 = undiluted, 3 = 1 drop dye + 3 drops water, etc.). Draw a pencil line on a TLC plate \(\sim 1 \: \text{cm}\) from the bottom with a ruler, and mark the lanes. Don't put lanes too close to the edge or to each other. Spot a dilute sample on the pencil line of the correct lane, making very small spots ( in diameter). Rinse the spotter with a solvent (e.g. acetone) if going to use it for another sample. Place the sample in the TLC chamber with forceps, cap it, and leave it alone. Remove the plate when the solvent line is \(\sim 0.5 \: \text{cm}\) from the top. Immediately mark the solvent line with a pencil. Visualize if necessary. The components of a sample can appear as long streaks or "blobby" spots on a TLC plate if the samples are run at too high a concentration. For example, Figure 2.27b shows an eluted TLC plate containing five red food dye samples of different concentrations (lane 0 contains the dye in the concentration found at the grocery store; lane 3 contains 1 drop of dye diluted with 3 drops of water, etc.). After elution, the red and pink components from the undiluted dye (lane 0) streaked severely, as the TLC plate was "overloaded". When this happens proper equilibration between stationary and mobile phases does not occur. With further dilution (lane 12), the streaking disappeared and the spot shapes sharpened. If streaking is seen on a TLC plate, the sample should be diluted and run again. Figure 2.27c also demonstrates how dilution can improve the shape of a spot after elution. In this TLC, alkene and alkyne samples were spotted at somewhat high concentrations, while an improved dilution was used in Figure 2.27d. Note how the \(R_f\) appears to change in the two TLC plates. The more accurate \(R_f\) is of the diluted sample. Running TLC on concentrated samples gives inaccurate \(R_f\) values and may hide multiple spots. At times the solvent front may run unevenly on a TLC plate. This may occur if the plate was placed in the chamber at a slight tilt, if eluent splashed onto the plate during placement in the chamber, or if the chamber was jostled during elution. In cases where the front is dramatically different from one position to the other, the front should be measured for each of the plate (if calculating an \(R_f\)) instead of only once. Although capillary spotters for TLC can be purchased, some chemists prefer to create their own by stretching Pasteur pipettes (Figure 2.28). To make a spotter, hold a pipette by the edges while wearing thick gloves and position the middle of the pipette into the flame of a large Meker burner (Figure 2.28a). Warm the pipette until the glass becomes quite pliable (borosilicate glass softens at \(820^\text{o} \text{C}\),\(^5\) so this will take some time). Only rotate, but do not stretch the pipette at all while the pipette is in the flame. Then remove the pipette from the flame and immediately and rapidly pull the pipette to an arm's length. The thin sections can be broken into 6-12 inch segments and used for TLC. Some chemists scan, photograph, or electronically record their developed TLC plates, but it is much more common to copy a likeness of a TLC plate by hand into a laboratory notebook. It is important to copy a TLC plate "to scale", meaning the dimensions should be the same in the notebook as they are in actuality. To accomplish this, the TLC plate can be placed atop a notebook page and a rendering created beside it (Figure 2.29c). The back of the TLC plate should previously be wiped clean or else staining reagents may degrade the paper. It is important to copy the TLC plate as accurately as possible, drawing the spots exactly as they appear, even if they are streaky or blobby. All spots seen in a lane should be recorded, even if they are faint. Good record keeping means to record all observations, even if the importance is unknown; a faint, unexpected spot may become relevant at a later time. Several other notations should be made along with the sketch of the TLC plate (Figure 2.29d). The solvent system and identity of what was spotted in each lane must be recorded. For each spot, an \(R_f\) should be calculated (some chemists like to write the measurements on the TLC plate; see the plate in Figure 2.29d) along with notation of UV activity and stain color. If a spot changes appearance over time, as the orange spot in Figure 2.29b faded to light green over time (Figure 2.29d), the initial appearance should be recorded. \(^3\)The graphite in pencil will not travel with the eluent, but pen ink would. \(^4\)J. Sherman and B. Fried, , . \(^5\)G. L. Weissler, , 2\(^\text{nd}\) edition, , p. 315. | 4,790 | 1,029 |
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Credit for the synthesis of element 99 was given to the Berkeley team of Ghiorso et. al. in 1952 and the radioactive metal was named for Albert Einstein. The photo above shows about 0.3 micrograms of an unidentified Es compound. Evidence for the feasibility of the synthesis came from the debris of the first hydrogen bomb explosion in which U-238 was apparently transformed into U-253 which becomes Es-253 by the loss of seven beta particles. The laboratory synthesis of Es-253 begins with Pu-239 and goes through a five step process. The longest-lived isotope is Es-254 with a half-life of 276 days. | 621 | 1,030 |
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Cytochrome c occupies a prominent place in the mitochondrial electron-transport chain. Its water solubility, low molecular weight (12.4 kDa), stability, and ease of purification have allowed many experiments, which, when taken together, present a detailed picture of the structure and biological function of this electron carrier. X-ray structures of oxidized and reduced tuna cytochrome c are very similar; most of the differences are confined to changes in the orientations of the side chains of some surface-exposed amino acids and sub-Ångström adjustments of some groups in the protein interior. Upon reduction, the heme active site becomes slightly more ordered (Figure 6.33). Two-dimensional NMR studies confirm this interpretation of the x-ray data, and further establish that the crystal and solution structures of cytochrome c differ in only minor respects. Cytochrome c exhibits several pH-dependent conformational states. In particular, an alkaline transition with a pK ~ 9.1 has been observed for ferricytochrome c. This transition is believed to be associated with the dissociation of Met-80; the reduction potential decreases dramatically, and the 695-nm absorption band, associated with a sulfur → iron charge-transfer transition, disappears. The H NMR resonance due to ( H C-) Met-80 in deuterium-enriched ferricytochrome c disappears from its hyperfine-shifted upfield position without line broadening, and reappears coincident with the ( H C-)Met-65 resonance. In contrast, ferrocytochrome c maintains an ordered structure over the pH range 4 to 11. The heme iron in ferricytochrome c remains low-spin throughout this transition, and a new strong-field ligand must therefore replace Met-80. It has been suggested that an e-amino nitrogen of a nearby Lys provides the new donor atom, but this has not been confirmed. However, it is clear that reduction of ferricytochrome c at alkaline pH values below 11 causes a drastic conformational change at the heme site. The unknown sixth ligand must be displaced by Met-80 in order for the reduced protein to assume a structure similar to the one at neutral pH. This structural change is accompanied by a decrease in the rate of reduction of ferricytochrome c by hydrated electrons, as expected. How does the protein control the reduction potential of the iron center in cytochrome c? Factors that appear to play a role include the nature of the axial ligands, the stability and solvent accessibility of the heme crevice, and the hydrophobicities of the amino acids that line the heme crevice. These issues have been addressed theoretically and experimentally using cytochrome c variants engineered by protein semisynthesis or site-directed mutagenesis. Results for horse heart cytochrome c are set out in Table 6.6. Point mutations at either of positions 78 or 83 do not significantly alter E°'; however, the double mutant (Thr-78 → Asn-78; Tyr-83 → Phe-83) exhibits a substantially lower redox potential. Evidently, the results of such changes are not necessarily additive; great care must be taken in drawing conclusions about structure-function relations in engineered proteins. Finally, the ~310 mV difference between the values for the heme octapeptide and the native protein (the axial ligands are the same in both) provides a dramatic illustration of protein environmental effects on the redox potential: shielding the heme from the solvent is expected to stabilize Fe and therefore result in an increase in E°'. During the last fifteen years, much has been learned about the interaction of cytochrome c with its redox partners. Cytochrome c is a highly basic protein (pI = 10.05); lysine residues constitute most of the cationic amino acids. Despite the indication from the x-ray structures that only ~1 percent of the heme surface is solvent-exposed, the asymmetric distribution of surface charges, particularly a highly conserved ring of Lys residues surrounding the exposed edge of the heme crevice, led to the suggestion that electron-transfer reactions of cytochrome c (and other Class I cytochromes as well) occur via the exposed heme edge. Chemical modification of the surface Lys residues of cytochrome c has afforded opportunities to alter the properties of the surface \(\varepsilon\)-amino groups suspected to be involved in precursor complex formation. Margoliash and coworkers used a 4-carboxy-2,6-dinitrophenol (CDNP) modification of the Lys residues to map out the cytochrome c interaction domains with various transition-metal redox reagents and proteins. These experiments have shown that cytochrome c interacts with inorganic redox partners near the exposed heme edge. Numerous studies of cytochrome c with physiological reaction partners are in accord with electrostatic interactions featured in the model cytochrome c/cytochrome b complex discussed earlier. Similar types of interactions have been proposed for cytochrome c/flavodoxin and cytochrome c/cytochrome c peroxidase complexes. (Recent x-ray crystal structure work has shed new light on this problem.) Theoretical work additionally suggests that electrostatic forces exert torques on diffusing protein reactants that "steer" the proteins into a favorable docking geometry. However, the domains on cytochrome c for interaction with physiological redox partners are not identical, as Figure 6.34 illustrates. Reactions between cytochrome c and its physiological redox partners at low ionic strength generally are very fast, ~10 M s , even though the thermodynamic driving force may be as low as 20 mV, as it is for the reduction of cytochrome a in cytochrome c oxidase. Such rates are probably at the diffusioncontrolled limit for such protein-protein reactions. A more detailed understanding of these reactions will require studies that focus on the dynamical (rather than static) features of complexes of cytochrome c with other proteins. For example, there is evidence that a cytochrome c conformational change in the vicinity of the heme edge accompanies the formation of the complex with cytochrome c oxidase. Studies of the influence of geometry changes on activation energies are of particular importance in elucidating the mechanisms of protein-protein reactions. Intramolecular electron transfer in cytochrome c has been investigated by attaching photoactive Ru complexes to the protein surface. Ru(bpy) (CO ) (bpy = 2,2'-bipyridine) has been shown to react with surface His residues to yield, after addition of excess imidazole (im), Ru(bpy) (im)(His) . The protein-bound Ru complexes are luminescent, but the excited states (*Ru ) are rather short lived (\(\tau \leq\) 100 ns). When direct electron transfer from *Ru to the heme cannot compete with excited-state decay, electron-transfer quenchers (e.g., Ru(NH ) ) are added to the solution to intercept a small fraction (1-10%) of the excited molecules, yielding (with oxidative quenchers) Ru . If, before laser excitation of the Ru site, the heme is reduced, then the Fe to Ru reaction (k ) can be monitored by transient absorption spectroscopy. The k values for five different modified cytochromes have been reported: (Ru(His- 33), 2.6(3) x 10 ; Ru(His-39), 3.2(4) x 10 ; Ru(His-62), 1.0(2) x 10 ; Ru(His- 72), 9.0(3) x 10 ; and Ru(His-79), > 10 s ). According to Equation (6.27), rates become activationless when the reaction driving force (- \(\Delta\)G°) equals the reorganization energy (A), The driving force (0.74 eV) is approximately equal to the reorganization energy (0.8 eV) estimated for the Ru(bpy) (im)(His)-cyt c reactions. The activationless (maximum) rates (k ) are limited by H , where H is the electronic matrix element that couples the reactants and products at the transition state. Values of k and H for the Fe to Ru reactions are given in Table 6.7. 14.0 (11C) (1H) 13.9 (11C) (1H) 17.6 (7C) (1S) 20.6 (16C) (2H) Calculations that explicitly include the structure of the intervening medium have been particularly helpful in developing an understanding of distant electronic couplings. As discussed in Section IV.A, the couplings in proteins can be interpreted in terms of pathways comprised of covalent, H-bond, and through-space contacts. An algorithm has been developed that searches a protein structure for the best pathways coupling two redox sites (the pathways between the histidines (33, 39, 62, 72, 79) and the heme are shown in Figure 6.35). A given coupling pathway consisting of covalent bonds, H-bonds, and through-space jumps can be described in terms of an equivalent covalent pathway with an effective number of covalent bonds (n ). Multiplying the effective number of bonds by 1.4 Å/bond gives a-tunneling lengths (\(\sigma\)1) for the five pathways (Table 6.7) that correlate well with the maximum rates (one-bond limit set at 3 x 10 s ; slope of 0.71 Å ) (Figure 6.35). The 0.71 Å decay accords closely with related distance dependences for covalently coupled donor-acceptor molecules. Photosynthetic bacteria produce only one type of reaction center, unlike green plants (which produce two different kinds linked together in series), and are therefore the organisms of choice in photosynthetic electon-transfer research. As indicated in Section I.B, the original reaction center structure (Figure 6.15) lacked a quinone (Q ). Subsequent structures for reaction centers from other photosynthetic bacteria contain this quinone (Figure 6.36 See color plate section, page C-13.). The reaction center contains ten cofactors and three protein subunits. (Note that the structure contains a cytochrome subunit as well.) The cofactors are arrayed so that they nearly span the 40-Å-thick membrane (Figure 6.37 See color plate section, page C-13.). The iron atom is indicated by the red dot near the cytoplasmic side of the membrane (bottom). In spite of the near two-fold axis of symmetry, electron transfer proceeds along a pathway that is determined by the A branch. In particular, BChl and BPhe do not appear to play an important role in the electron transfers. It was demonstrated long ago that (BChl) is the primary electron donor and that ubiquinone (or metaquinone) is the ultimate electron acceptor. Transient flash photolysis experiments indicate that several electron-transfer steps occur in order to translocate the charge across the membrane (Figure 6.38). Curiously, the high-spin ferrous iron appears to play no functional role in the Q to Q electron transfer. In addition, the part played by BChl is not understood—it may act to promote reduction of BPhe via a superexchange mechanism. Cytochromes supply the reducing equivalents to reduce the special pair (BChl) . Estimated rate constants for the various electron-transfer steps, together with approximate reduction potentials, are displayed in Figure 6.39. For each step, the forward rate is orders of magnitude faster than the reverse reaction. The rapid rates suggest that attempts to obtain x-ray structures of intermediates (especially the early ones!) will not be successful. However, molecular dynamics methods are being explored in computer simulations of the structures of various intermediates. Within a few years we may begin to understand why the initial steps are so fast. | 11,277 | 1,034 |
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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 decomposition reaction of \(\ce{N_2O_4}\) to \(\ce{NO_2}\). Both the forward and reverse reactions for this system consist of a single elementary reaction, so the reaction rates are as follows: \[\text{forward rate} = k_f[\ce{N2O4}] \label{Eq1}\] and \[\text{reverse rate} = k_r[\ce{NO2}]^2 \label{Eq2}\] At equilibrium, the forward rate equals the reverse rate (definition of equilibrium): \[ k_f[\ce{N2O4}] = k_r[\ce{NO2}]^2 \label{Eq3}\] so \[\dfrac{k_f}{k_r}=\dfrac{[\ce{NO2}]^2}{[\ce{N2O4}]} \label{Eq4}\] The ratio of the rate constants gives us a new constant, the (\(K\)), which is defined as follows: \[K=\dfrac{k_f}{k_r} \label{Eq5}\] Hence there is a fundamental relationship between chemical kinetics and chemical equilibrium: under a given set of conditions, the composition of the equilibrium mixture is determined by the magnitudes of the rate constants for the forward and the reverse reactions. The equilibrium constant is equal to the rate constant for the forward reaction divided by the rate constant for the reverse reaction. Table \(\Page {1}\) lists the initial and equilibrium concentrations from five different experiments using the reaction system described by Equation \(\ref{Eq1}\). At equilibrium the magnitude of the quantity \([\ce{NO2}]^2/[\ce{N2O4}]\) is essentially the same for all five experiments. In fact, no matter what the initial concentrations of \(\ce{NO2}\) and \(\ce{N2O4}\) are, at equilibrium the quantity \([\ce{NO2}]^2/[\ce{N2O4}]\) will always be \(6.53 \pm 0.03 \times 10^{-3}\) at 25°C, which corresponds to the ratio of the rate constants for the forward and reverse reactions (Equation \ref{Eq5}). 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 any reversible reaction of the general form \[aA+bB \rightleftharpoons cC+dD \label{Eq6}\] where \(A\) and \(B\) are reactants, \(C\) and \(D\) are products, and \(a\), \(b\), \(c\), and \(d\) 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 and can be stated as follows: \[K=\dfrac{[C]^c[D]^d}{[A]^a[B]^b} \label{Eq7}\] where \(K\) is the equilibrium constant for the reaction. Equation \(\ref{Eq6}\) is called the equilibrium equation, and the right side of Equation \(\ref{Eq7}\) is called the equilibrium constant expression. The relationship shown in Equation \(\ref{Eq7}\) is true for any 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 \(K\) shown in Table \(\Page {2}\), 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 K greater than \(10^3\) 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 \(\ce{H_2}\) and \(\ce{Cl_2}\) to produce \(\ce{HCl}\), which has an equilibrium constant of \(1.6 \times 10^{33}\) at 300 K. Because \(\ce{H_2}\) is a good reductant and \(\ce{Cl_2}\) is a good oxidant, the reaction proceeds essentially to completion. In contrast, values of \(K\) less than \(10^{-3}\) 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 Table \(\Page {2}\) that equilibrium constants have no units, even though Equation \(\ref{Eq7}\) suggests that the units of concentration might not always cancel because the exponents may vary. As shown in Equation \(\ref{Eq8}\), the units of concentration cancel, which makes \(K\) unitless as well: \[ \dfrac{[A]_{measured}}{[A]_{standard\; state}}=\dfrac{\cancel{M}}{\cancel{M}} = \dfrac{\cancel{\frac{mol}{L}}}{\cancel{\frac{mol}{L}}} \label{Eq8}\] Many reactions have equilibrium constants between 1000 and 0.001 (\(10^3 \ge K \ge 10^{−3}\)), 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 \(\ce{HD}\): \[\ce{H2(g) + D2(g) \rightleftharpoons 2HD(g)} \label{Eq9}\] The equilibrium constant expression for this reaction is \[K= \dfrac{[HD]^2}{[H_2,D_2]}\] with \(K\) varying between 1.9 and 4 over a wide temperature range (100–1000 K). Thus an equilibrium mixture of \(\ce{H_2}\), \(\ce{D_2}\), and \(\ce{HD}\) contains significant concentrations of both product and reactants. Figure \(\Page {1}\) summarizes the relationship between the magnitude of \(K\) and the relative concentrations of reactants and products at equilibrium for a general reaction, written as reactants \rightleftharpoons products. Because there is a direct relationship between the kinetics of a reaction and the equilibrium concentrations of products and reactants (Equations \(\ref{Eq8}\) and \(\ref{Eq7}\)), when \(k_f \gg k_r\), \(K\) is a number, and the concentration of products at equilibrium predominate. This corresponds to an essentially irreversible reaction. Conversely, when \(k_f \ll k_r\), \(K\) is a very number, and the reaction produces almost no products as written. Systems for which \(k_f ≈ k_r\) have significant concentrations of both reactants and products at equilibrium. A large value of the equilibrium constant \(K\) 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 Equation \(\ref{Eq7}\). 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. : The only product is ammonia, which has a coefficient of 2. For the reactants, \(N_2\) has a coefficient of 1 and \(\ce{H2}\) has a coefficient of 3. The equilibrium constant expression is as follows: \[\dfrac{[NH_3]^2}{[N_2,H_2]^3}\nonumber\] The only product is carbon dioxide, which has a coefficient of 1. The reactants are \(CO\), with a coefficient of 1, and \(\ce{O_2}\), with a coefficient of \(\ce{1/2}\). Thus the equilibrium constant expression is as follows: \[\dfrac{[CO_2]}{[CO,O_2]^{1/2}}\nonumber\] This reaction is the reverse of the reaction in part b, with all coefficients multiplied by 2 to remove the fractional coefficient for \(\ce{O_2}\). The equilibrium constant expression is therefore the inverse of the expression in part b, with all exponents multiplied by 2: \[\dfrac{[CO]^2[O_2]}{[CO_2]^2}\nonumber\] Write the equilibrium constant expression for each reaction. \[K=\dfrac{[N_2,O_2]^{1/2}}{[N_2O]}\nonumber\] \[K=\dfrac{[CO_2]^{16}[H_2O]^{18}}{[C_8H_{18}]^2[O_2]^{25}}\nonumber\] \[K=\dfrac{[HI]^2}{[H_2,I_2]}\nonumber\] 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 \(K\) : 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. : Hydrogen and nitrogen react to form ammonia according to the following balanced chemical equation: \[3H_{2(g)}+N_{2(g)} \rightleftharpoons 2NH_{3(g)}\nonumber\] Values of the equilibrium constant at various temperatures were reported as 327°C, where \(K\) is smallest 25°C 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 Equation \(\ref{Eq6}\) in reverse, we obtain the following: \[cC+dD \rightleftharpoons aA+bB \label{Eq10}\] The corresponding equilibrium constant \(K′\) is as follows: \[K'=\dfrac{[A]^a[B]^b}{[C]^c[D]^d} \label{Eq11}\] This expression is the inverse of the expression for the original equilibrium constant, so \(K′ = 1/K\). 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 \(N_2O_4 \rightleftharpoons 2NO_2\) is as follows: \[K=\dfrac{[NO_2]^2}{[N_2O_4]} \label{Eq12}\] but for the opposite reaction, \(2 NO_2 \rightleftharpoons N_2O_4\), the equilibrium constant K′ is given by the inverse expression: \[K'=\dfrac{[N_2O_4]}{[NO_2]^2} \label{Eq13}\] Consider another example, the formation of water: \(2H_{2(g)}+O_{2(g)} \rightleftharpoons 2H_2O_{(g)}\). Because \(H_2\) is a good reductant and \(O_2\) is a good oxidant, this reaction has a very large equilibrium constant (\(K = 2.4 \times 10^{47}\) at 500 K). Consequently, the equilibrium constant for the reverse reaction, the decomposition of water to form \(O_2\) and \(H_2\), is very small: \(K′ = 1/K = 1/(2.4 \times 10^{47}) = 4.2 \times 10^{−48}\). 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 \(H_2\) and \(O_2\). 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 K″ is as follows: \[ K′′=\dfrac{[N_2O_4]^{1/2}}{[NO_2]} \label{Eq14}\] The values for K′ (Equation \(\ref{Eq13}\)) and K″ are related as follows: \[ K′′=(K')^{1/2}=\sqrt{K'} \label{Eq15}\] In general, if all the coefficients in a balanced chemical equation were subsequently multiplied by \(n\), then the new equilibrium constant is the original equilibrium constant raised to the \(n^{th}\) power. At 745 K, K is 0.118 for the following reaction: \[\ce{N2(g) + 3H2(g) \rightleftharpoons 2NH3(g)}\nonumber\] What is the equilibrium constant for each related reaction at 745 K? : balanced equilibrium equation, \(K\) at a given temperature, and equations of related reactions : values of \(K\) for related reactions : Write the equilibrium constant expression for the given reaction and for each related reaction. From these expressions, calculate \(K\) for each reaction. : The equilibrium constant expression for the given reaction of \(\ce{N2(g)}\) with \(\ce{H2(g)}\) to produce \(\ce{NH3(g)}\) at 745 K is as follows: \[K=\dfrac{[NH_3]^2}{[N_2,H_2]^3}=0.118\nonumber\] This reaction is the reverse of the one given, so its equilibrium constant expression is as follows: \[K'=\dfrac{1}{K}=\dfrac{[N_2,H_2]^3}{[NH_3]^2}=\dfrac{1}{0.118}=8.47\nonumber\] In this reaction, the stoichiometric coefficients of the given reaction are divided by 2, so the equilibrium constant is calculated as follows: \[K′′=\dfrac{[NH_3]}{[N_2]^{1/2}[H_2]^{3/2}}=K^{1/2}=\sqrt{K}=\sqrt{0.118} = 0.344\nonumber\] At 527°C, the equilibrium constant for the reaction \[\ce{2SO2(g) + O2(g) \rightleftharpoons 2SO3(g) }\nonumber\] is \(7.9 \times 10^4\). Calculate the equilibrium constant for the following reaction at the same temperature: \(3.6 \times 10^{−3}\) 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 \[\ce{aA + bB <=> cC + dD} \nonumber\] 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): \[K_p=\dfrac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b} \label{Eq16}\] Thus \(K_p\) for the decomposition of \(N_2O_4\) (Equation \(\ref{Eq1}\)) is as follows: \[K_p=\dfrac{(P_{NO_2})^2}{P_{N_2O_4}} \label{Eq17}\] 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 (\(T\)): \[K_p = K(RT)^{Δn} \label{Eq18}\] 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 \(\ref{Eq18}\), \(K_p = K\) 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, \[K_p = K(RT)^1 = KRT\] 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: \[N_{2(g)}+3H_{2(g)} \rightleftharpoons 2NH_{3(g)}\nonumber\] 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 \(\ref{Eq18}\) 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 \(\ref{Eq15}\), we have the following: \[K_p=K(RT)^{−2}=\dfrac{K}{(RT)^2}=\dfrac{0.118}{\{ [0.08206(L \cdot atm)/(mol \cdot K),745\; K]\}^2}=3.16 \times 10^{−5}\nonumber\] Because \(K_p\) is a unitless quantity, the answer is \( K_p = 3.16 \times 10^{−5}\). Calculate \(K_p\) for the reaction \[2SO_{2(g)}+O_{2(g)} \rightleftharpoons 2SO_{3(g)}\nonumber\] at 527°C, if \(K = 7.9 \times 10^4\) at this temperature. \(K_p = 1.2 \times 10^3\) When the products and reactants of a reaction at equilibrium in a single phase (e.g., liquid, gas or solids of different lattices), the system is a . In such situations, the concentrations of the reactants and products can vary over a wide range. In contrast, a system where the reactants and products are in two or more phase is called a (e.g, the reaction of a gas with a solid or liquid or two different solid lattices in co-existing). 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 activity of 1. (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: \[\ce{CO2(g) + C(s) <=> 2CO(g)} \label{Eq14.4.1}\] 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: \[K=\dfrac{[CO]^2}{[CO_2,C]} \label{Eq14.4.2}\] Because graphite is a solid, however, its molar concentration, determined from its density and molar mass, is essentially constant and has the following value: \[ [C] =\dfrac{2.26 \cancel{g}/{\cancel{cm^3}}}{12.01\; \cancel{g}/mol} \times 1000 \; \cancel{cm^3}/L = 188 \; mol/L = 188\;M \label{Eq14.4.3}\] We can rearrange Equation \(\ref{Eq14.4.3}\) so that the constant terms are on one side: \[ K[C]=K(188)=\dfrac{[CO]^2}{[CO_2]} \label{Eq14.4.4}\] Incorporating the constant value of \([C]\) into the equilibrium equation for the reaction in Equation \(\ref{Eq14.4.4}\), \[K'=\dfrac{[CO]^2}{[CO_2]} \label{Eq14.4.5}\] The equilibrium constant for this reaction can also be written in terms of the partial pressures of the gases: \[K_p=\dfrac{(P_{CO})^2}{P_{CO_2}} \label{Eq14.4.6}\] 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 \(\ref{Eq14.4.1}\) 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 \(\Page {2}\), 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 \[K=\dfrac{1}{[Cl_2]}\nonumber\] and \[K_p=\dfrac{1}{P_{Cl_2}}\nonumber\] 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: \[K=\dfrac{[H_2O]^4}{[H_2]^4}\nonumber\] and \[K_p=\dfrac{(P_{H_2O})^4}{(P_{H_2})^4}\nonumber\] Write the expressions for \(K\) and \(K_p\) for the following reactions. \(K = [CO_2]\) and \(K_p = P_{CO_2}\) \(K=\dfrac{[CO_2]^6[H_2O]^6}{[O_2]^6}\) and \(K_p=\dfrac{(P_{CO_2})^6(P_{H_2O})^6}{(P_{O_2})^6}\) 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. The ratio of the rate constants for the forward and reverse reactions at equilibrium is the equilibrium constant (K), 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 \(K\). For a system at equilibrium, the law of mass action relates \(K\) 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 equilibrium equation. The ratio is called the equilibrium constant expression. When a reaction is written in the reverse direction, \(K\) 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 (\(K_p\)) 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 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. | 22,881 | 1,035 |
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are those in which electrons are shared equally between the two atoms involved, as we saw earlier, where the iodine molecule was given as an example: It was also shown that replacing an I atom with a group I metal decreased the covalent nature of the bond, while increasing its percent ionic character. The molecule is changed from a poisonous and bactericidal substance to salt-like white crystalline solids which may be more or less toxic depending on the particular metal chosen. IWC replica Réplica de reloj But replacing one I atom in the purple solid I with another nonmetal also makes a significant difference. Replacing one of the iodine atoms with a hydrogen atom to make HI (hydrogen iodide) changes the chemistry significantly. HI is a colorless gas, and reacts with NaOH to give sodium iodide (used in iodized salt). Aqueous solution of HI are called hydroiodic acid, because HI dissolves extensively and readily in water to make acidic solutions by increasing the hydrogen ion (H ) concentration, while I is barely soluble in water. The polarity of the bond clearly has biological significance. The Jmol model and electrostatic potential surfaces differ from those of I in several ways. Charge is no longer equally distributed between the atoms; the I atom has an excess of about 0.05 electrons, on the average, over the number of electrons in the neutral atom, so it has a charge of -0.05e. Even with electron shielding, the highly positive iodine nucleus pulls electrons toward itself more than the single proton of the hydrogen nucleus attracts electrons. The H atom has lost 0.05 of an electron, so it has an electrostatic charge of +0.05 e. The molecule has two electrical "poles", and is called a . The bond, in which electrons are not equally shared, is called a . Electrostatic surface map for HI In the H—I bond the I has an excess of 0.05 electrons and hence has a negative charge. This situation is often indicated as follows: or H I The Greek letter δ (delta) is used here to indicate that electron transfer is not complete and that some sharing takes place. If the transfer had been complete, δ would have been 1.0. Because the Li—H bond is only negative at the one end and positive at the other, we often say that the bond is or , rather than 100 percent ionic. Elements in the upper right of the periodic table, which are small because the large nucleus contracts the valence shell, form much more polar bonds with H. For example, HF has a δ value of 0.43, compared with δ = 0.05 for HI. Lets see how to calculate the δ value for HI. The data for this example could be obtained from the Jmol model above, but there are a number of different methods for calculating charges, and the one used here may not be appropriate for this calculation. Nonetheless, the bond distance in HI can be measured by right clicking one the Jmol model, choosing "measurements", then clicking on the atoms in sequence. We'll use empirically (from experiment) measured values here: The dipole moment of the HI molecule is found to be 1.34 × 10 C m, while the H―I distance is 165.0 pm. Find the partial charge on the H and F atoms. Rearranging Eq. (1) from Polarizability, we have Thus the apparent charge on each end of the molecule is given by Since the charge on a single electron is 1.6021 × 10 C, we have So δ = 0.051. It is worth noting in the above example that the dipole moment measures the electrical imbalance of the and not just that of the H―I bonding pair. In the HI molecule there are four valence electron pairs, with the three lone pairs on the right of the I atom also contributing to the overall negative charge of I. | 3,674 | 1,036 |
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