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Learning Objectives
• To correlate the arrangement of atoms in the periodic table results in blocks corresponding to filling of the ns, np, nd, and nf orbitals
As you have learned, the electron configurations of the elements explain the otherwise peculiar shape of the periodic table. Although the table was originally organized on the basis of physical and chemical similarities between the elements within groups, these similarities are ultimately attributable to orbital energy levels and the Pauli principle, which cause the individual subshells to be filled in a particular order. As a result, the periodic table can be divided into “blocks” corresponding to the type of subshell that is being filled, as illustrated in Figure \(1\). For example, the two columns on the left, known as the s block, consist of elements in which the ns orbitals are being filled. The six columns on the right, elements in which the np orbitals are being filled, constitute the p block. In between are the 10 columns of the d block, elements in which the (n − 1)d orbitals are filled. At the bottom lie the 14 columns of the f block, elements in which the (n − 2)f orbitals are filled. Because two electrons can be accommodated per orbital, the number of columns in each block is the same as the maximum electron capacity of the subshell: 2 for ns, 6 for np, 10 for (n − 1)d, and 14 for (n − 2)f. Within each column, each element has the same valence electron configuration—for example, ns1 (group 1) or ns2np1 (group 13). As you will see, this is reflected in important similarities in the chemical reactivity and the bonding for the elements in each column.
Because each orbital can have a maximum of 2 electrons, there are 2 columns in the s block, 6 columns in the p block, 10 columns in the d block, and 14 columns in the f block.
Hydrogen and helium are placed somewhat arbitrarily. Although hydrogen is not an alkali metal, its 1s1 electron configuration suggests a similarity to lithium ([He]2s1) and the other elements in the first column. Although helium, with a filled ns subshell, should be similar chemically to other elements with an ns2 electron configuration, the closed principal shell dominates its chemistry, justifying its placement above neon on the right.
Example \(1\)
Use the periodic table to predict the valence electron configuration of all the elements of group 2 (beryllium, magnesium, calcium, strontium, barium, and radium).
Given: series of elements
Asked for: valence electron configurations
Strategy:
1. Identify the block in the periodic table to which the group 2 elements belong. Locate the nearest noble gas preceding each element and identify the principal quantum number of the valence shell of each element.
2. Write the valence electron configuration of each element by first indicating the filled inner shells using the symbol for the nearest preceding noble gas and then listing the principal quantum number of its valence shell, its valence orbitals, and the number of valence electrons in each orbital as superscripts.
Solution:
A The group 2 elements are in the s block of the periodic table, and as group 2 elements, they all have two valence electrons. Beginning with beryllium, we see that its nearest preceding noble gas is helium and that the principal quantum number of its valence shell is n = 2.
B Thus beryllium has an [He]s2 electron configuration. The next element down, magnesium, is expected to have exactly the same arrangement of electrons in the n = 3 principal shell: [Ne]s2. By extrapolation, we expect all the group 2 elements to have an ns2 electron configuration.
Exercise \(1\)
Use the periodic table to predict the characteristic valence electron configuration of the halogens in group 17.
Answer
All have an ns2np5 electron configuration, one electron short of a noble gas electron configuration. (Note that the heavier halogens also have filled (n − 1)d10 subshells, as well as an (n − 2)f14 subshell for Rn; these do not, however, affect their chemistry in any significant way.
Summary
The arrangement of atoms in the periodic table results in blocks corresponding to filling of the ns, np, nd, and nf orbitals to produce the distinctive chemical properties of the elements in the s block, p block, d block, and f block, respectively. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/06%3A_Electronic_Structure_of_Atoms/6.09%3A_Electron_Configurations_and_the_Periodic_Table.txt |
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
6.1: The Wave Nature of Light
Conceptual Problems
1. What are the characteristics of a wave? What is the relationship between electromagnetic radiation and wave energy?
2. At constant wavelength, what effect does increasing the frequency of a wave have on its speed? its amplitude?
3. List the following forms of electromagnetic radiation in order of increasing wavelength: x-rays, radio waves, infrared waves, microwaves, ultraviolet waves, visible waves, and gamma rays. List them in order of increasing frequency. Which has the highest energy?
4. A large industry is centered on developing skin-care products, such as suntan lotions and cosmetics, that cannot be penetrated by ultraviolet radiation. How does the wavelength of visible light compare with the wavelength of ultraviolet light? How does the energy of visible light compare with the energy of ultraviolet light? Why is this industry focused on blocking ultraviolet light rather than visible light?
Numerical Problems
1. The human eye is sensitive to what fraction of the electromagnetic spectrum, assuming a typical spectral range of 104 to 1020 Hz? If we came from the planet Krypton and had x-ray vision (i.e., if our eyes were sensitive to x-rays in addition to visible light), how would this fraction be changed?
2. What is the frequency in megahertz corresponding to each wavelength?
1. 755 m
2. 6.73 nm
3. 1.77 × 103 km
4. 9.88 Å
5. 3.7 × 10−10 m
3. What is the frequency in megahertz corresponding to each wavelength?
1. 5.8 × 10−7 m
2. 2.3 Å
3. 8.6 × 107 m
4. 6.2 mm
5. 3.7 nm
4. Line spectra are also observed for molecular species. Given the following characteristic wavelengths for each species, identify the spectral region (ultraviolet, visible, etc.) in which the following line spectra will occur. Given 1.00 mol of each compound and the wavelength of absorbed or emitted light, how much energy does this correspond to?
1. NH3, 1.0 × 10−2 m
2. CH3CH2OH, 9.0 μm
3. Mo atom, 7.1 Å
5. What is the speed of a wave in meters per second that has a wavelength of 1250 m and a frequency of 2.36 × 105 s−1?
6. A wave travels at 3.70 m/s with a frequency of 4.599 × 107 Hz and an amplitude of 1.0 m. What is its wavelength in nanometers?
7. An AM radio station broadcasts with a wavelength of 248.0 m. What is the broadcast frequency of the station in kilohertz? An AM station has a broadcast range of 92.6 MHz. What is the corresponding wavelength range in meters for this reception?
8. An FM radio station broadcasts with a wavelength of 3.21 m. What is the broadcast frequency of the station in megahertz? An FM radio typically has a broadcast range of 82–112 MHz. What is the corresponding wavelength range in meters for this reception?
9. A microwave oven operates at a frequency of approximately 2450 MHz. What is the corresponding wavelength? Water, with its polar molecules, absorbs electromagnetic radiation primarily in the infrared portion of the spectrum. Given this fact, why are microwave ovens used for cooking food?
6.2: Quantized Energy and Photons
Conceptual Problems
1. Describe the relationship between the energy of a photon and its frequency.
2. How was the ultraviolet catastrophe explained?
3. If electromagnetic radiation with a continuous range of frequencies above the threshold frequency of a metal is allowed to strike a metal surface, is the kinetic energy of the ejected electrons continuous or quantized? Explain your answer.
4. The vibrational energy of a plucked guitar string is said to be quantized. What do we mean by this? Are the sounds emitted from the 88 keys on a piano also quantized?
5. Which of the following exhibit quantized behavior: a human voice, the speed of a car, a harp, the colors of light, automobile tire sizes, waves from a speedboat?
Conceptual Answers
1. The energy of a photon is directly proportional to the frequency of the electromagnetic radiation.
2.
3.
4.
5. Quantized: harp, tire size, speedboat waves; continuous: human voice, colors of light, car speed.
Numerical Problems
1. What is the energy of a photon of light with each wavelength? To which region of the electromagnetic spectrum does each wavelength belong?
1. 4.33 × 105 m
2. 0.065 nm
3. 786 pm
2. How much energy is contained in each of the following? To which region of the electromagnetic spectrum does each wavelength belong?
1. 250 photons with a wavelength of 3.0 m
2. 4.2 × 106 photons with a wavelength of 92 μm
3. 1.78 × 1022 photons with a wavelength of 2.1 Å
3. A 6.023 x 1023 photons are found to have an energy of 225 kJ. What is the wavelength of the radiation?
4. Use the data in Table 2.1.1 to calculate how much more energetic a single gamma-ray photon is than a radio-wave photon. How many photons from a radio source operating at a frequency of 8 × 105 Hz would be required to provide the same amount of energy as a single gamma-ray photon with a frequency of 3 × 1019 Hz?
5. Use the data in Table 2.1.1 to calculate how much more energetic a single x-ray photon is than a photon of ultraviolet light.
6. A radio station has a transmitter that broadcasts at a frequency of 100.7 MHz with a power output of 50 kW. Given that 1 W = 1 J/s, how many photons are emitted by the transmitter each second?
Numerical Answers
1.
1. 4.59 × 10−31 J/photon, radio
2. 3.1 × 10−15 J/photon, gamma ray
3. 2.53 × 10−16 J/photon, gamma ray
2.
3. 532 nm
6.3: Line Spectra and the Bohr Model
Conceptual Problems
1. Is the spectrum of the light emitted by isolated atoms of an element discrete or continuous? How do these spectra differ from those obtained by heating a bulk sample of a solid element? Explain your answers.
2. Explain why each element has a characteristic emission and absorption spectra. If spectral emissions had been found to be continuous rather than discrete, what would have been the implications for Bohr’s model of the atom?
3. Explain the differences between a ground state and an excited state. Describe what happens in the spectrum of a species when an electron moves from a ground state to an excited state. What happens in the spectrum when the electron falls from an excited state to a ground state?
4. What phenomenon causes a neon sign to have a characteristic color? If the emission spectrum of an element is constant, why do some neon signs have more than one color?
5. How is light from a laser different from the light emitted by a light source such as a light bulb? Describe how a laser produces light.
Numerical Problems
1. Using a Bohr model and the transition from n = 2 to n = 3 in an atom with a single electron, describe the mathematical relationship between an emission spectrum and an absorption spectrum. What is the energy of this transition? What does the sign of the energy value represent in this case? What range of light is associated with this transition?
2. If a hydrogen atom is excited from an n = 1 state to an n = 3 state, how much energy does this correspond to? Is this an absorption or an emission? What is the wavelength of the photon involved in this process? To what region of the electromagnetic spectrum does this correspond?
3. The hydrogen atom emits a photon with a 486 nm wavelength, corresponding to an electron decaying from the n = 4 level to which level? What is the color of the emission?
4. An electron in a hydrogen atom can decay from the n = 3 level to n = 2 level. What is the color of the emitted light? What is the energy of this transition?
5. Calculate the wavelength and energy of the photon that gives rise to the third line in order of increasing energy in the Lyman series in the emission spectrum of hydrogen. In what region of the spectrum does this wavelength occur? Describe qualitatively what the absorption spectrum looks like.
6. The wavelength of one of the lines in the Lyman series of hydrogen is 121 nm. In what region of the spectrum does this occur? To which electronic transition does this correspond?
7. The emission spectrum of helium is shown. Estimate what change in energy (ΔE) gives rise to each line?
8. Removing an electron from solid potassium requires 222 kJ/mol. Would you expect to observe a photoelectric effect for potassium using a photon of blue light (λ = 485 nm)? What is the longest wavelength of energy capable of ejecting an electron from potassium? What is the corresponding color of light of this wavelength?
9. The binding energy of an electron is the energy needed to remove an electron from its lowest energy state. According to Bohr’s postulates, calculate the binding energy of an electron in a hydrogen atom. There are 6.02 x 1023 atoms in 1g of hydrogen atoms What wavelength in nanometers is required to remove such an electron from one hydrogen atom?
10. As a radio astronomer, you have observed spectral lines for hydrogen corresponding to a state with n = 320, and you would like to produce these lines in the laboratory. Is this feasible? Why or why not?
Numerical Answers
1. 656 nm; red light
2.
3. n = 2, blue-green light
4.
5. 97.2 nm, 2.04 × 10−18 J/photon, ultraviolet light, absorption spectrum is a single dark line at a wavelength of 97.2 nm
6.
7. Violet: 390 nm, 307 kJ/mol photons; Blue-purple: 440 nm, 272 kJ/mol photons; Blue-green: 500 nm, 239 kJ/mol photons; Orange: 580 nm, 206 kJ/mol photons; Red: 650 nm, 184 kJ/mol photons
8.
9. 1313 kJ/mol, λ ≤ 91.1 nm
6.4: The Wave Behavior of Matter
Conceptual Problems
1. Explain what is meant by each term and illustrate with a sketch:
1. standing wave
2. fundamental
3. overtone
4. node
2. How does Einstein’s theory of relativity illustrate the wave–particle duality of light? What properties of light can be explained by a wave model? What properties can be explained by a particle model?
3. In the modern theory of the electronic structure of the atom, which of de Broglie’s ideas have been retained? Which proved to be incorrect?
4. According to Bohr, what is the relationship between an atomic orbit and the energy of an electron in that orbit? Is Bohr’s model of the atom consistent with Heisenberg’s uncertainty principle? Explain your answer.
5. The development of ideas frequently builds on the work of predecessors. Complete the following chart by filling in the names of those responsible for each theory shown.
1.
2.
3.
4.
Numerical Problems
1. How much heat is generated by shining a carbon dioxide laser with a wavelength of 1.065 μm on a 68.95 kg sample of water if 1.000 mol of photons is absorbed and converted to heat? Is this enough heat to raise the temperature of the water by 4°C?
2. Show the mathematical relationship between energy and mass and between wavelength and mass. What is the effect of doubling the
1. mass of an object on its energy?
2. mass of an object on its wavelength?
3. frequency on its mass?
3. What is the de Broglie wavelength of a 39 g bullet traveling at 1020 m/s ± 10 m/s? What is the minimum uncertainty in the bullet’s position?
4. What is the de Broglie wavelength of a 6800 tn aircraft carrier traveling at 18 ± 0.1 knots (1 knot = 1.15 mi/h)? What is the minimum uncertainty in its position?
5. Calculate the mass of a particle if it is traveling at 2.2 × 106 m/s and has a frequency of 6.67 × 107 Hz. If the uncertainty in the velocity is known to be 0.1%, what is the minimum uncertainty in the position of the particle?
6. Determine the wavelength of a 2800 lb automobile traveling at 80 mi/h ± 3%. How does this compare with the diameter of the nucleus of an atom? You are standing 3 in. from the edge of the highway. What is the minimum uncertainty in the position of the automobile in inches?
Numerical Answers
1. E = 112.3 kJ, ΔT = 0.3893°C, over ten times more light is needed for a 4.0°C increase in temperature
2.
3. 1.7 × 10−35 m, uncertainty in position is ≥ 1.4 × 10−34 m
4.
5. 9.1 × 10−39 kg, uncertainty in position ≥ 2.6 m
6.5: Quantum Mechanics and Atomic Orbitals
Conceptual Problems
1. Why does an electron in an orbital with n = 1 in a hydrogen atom have lower energy than a free electron (n = ∞)?
2. What four variables are required to fully describe the position of any object in space? In quantum mechanics, one of these variables is not explicitly considered. Which one and why?
3. Chemists generally refer to the square of the wave function rather than to the wave function itself. Why?
4. Orbital energies of species with only one electron are defined by only one quantum number. Which one? In such a species, is the energy of an orbital with n = 2 greater than, less than, or equal to the energy of an orbital with n = 4? Justify your answer.
5. In each pair of subshells for a hydrogen atom, which has the higher energy? Give the principal and the azimuthal quantum number for each pair.
1. 1s, 2p
2. 2p, 2s
3. 2s, 3s
4. 3d, 4s
6. What is the relationship between the energy of an orbital and its average radius? If an electron made a transition from an orbital with an average radius of 846.4 pm to an orbital with an average radius of 476.1 pm, would an emission spectrum or an absorption spectrum be produced? Why?
7. In making a transition from an orbital with a principal quantum number of 4 to an orbital with a principal quantum number of 7, does the electron of a hydrogen atom emit or absorb a photon of energy? What would be the energy of the photon? To what region of the electromagnetic spectrum does this energy correspond?
8. What quantum number defines each of the following?
1. the overall shape of an orbital
2. the orientation of an electron with respect to a magnetic field
3. the orientation of an orbital in space
4. the average energy and distance of an electron from the nucleus
9. In an attempt to explain the properties of the elements, Niels Bohr initially proposed electronic structures for several elements with orbits holding a certain number of electrons, some of which are in the following table:
Element Number of Electrons Electrons in orbits with n =
4 3 2 1
H 1 1
He 2 2
Ne 10 8 2
Ar 18 8 8 2
Li 3 1 2
Na 11 1 8 2
K 19 1 8 8 2
Be 4 2 2
1. Draw the electron configuration of each atom based only on the information given in the table. What are the differences between Bohr’s initially proposed structures and those accepted today?
2. Using Bohr’s model, what are the implications for the reactivity of each element?
3. Give the actual electron configuration of each element in the table.
Numerical Problems
1. How many subshells are possible for n = 3? What are they?
2. How many subshells are possible for n = 5? What are they?
3. What value of l corresponds to a d subshell? How many orbitals are in this subshell?
4. What value of l corresponds to an f subshell? How many orbitals are in this subshell?
5. State the number of orbitals and electrons that can occupy each subshell.
1. 2s
2. 3p
3. 4d
4. 6f
6. State the number of orbitals and electrons that can occupy each subshell.
1. 1s
2. 4p
3. 5d
4. 4f
7. How many orbitals and subshells are found within the principal shell n = 6? How do these orbital energies compare with those for n = 4?
8. How many nodes would you expect a 4p orbital to have? A 5s orbital?
9. A p orbital is found to have one node in addition to the nodal plane that bisects the lobes. What would you predict to be the value of n? If an s orbital has two nodes, what is the value of n?
Numerical Answers
1. Three subshells, with l = 0 (s), l = 1 (p), and l = 2 (d).
2.
3. A d subshell has l = 2 and contains 5 orbitals.
4.
5.
1. 2 electrons; 1 orbital
2. 6 electrons; 3 orbitals
3. 10 electrons; 5 orbitals
4. 14 electrons; 7 orbitals
6.
7. A principal shell with n = 6 contains six subshells, with l = 0, 1, 2, 3, 4, and 5, respectively. These subshells contain 1, 3, 5, 7, 9, and 11 orbitals, respectively, for a total of 36 orbitals. The energies of the orbitals with n = 6 are higher than those of the corresponding orbitals with the same value of l for n = 4.
6.7: Many-Electron Atoms
Conceptual Problems
1. A set of four quantum numbers specifies each wave function. What information is given by each quantum number? What does the specified wave function describe?
2. List two pieces of evidence to support the statement that electrons have a spin.
3. The periodic table is divided into blocks. Identify each block and explain the principle behind the divisions. Which quantum number distinguishes the horizontal rows?
4. Identify the element with each ground state electron configuration.
1. [He]2s22p3
2. [Ar]4s23d1
3. [Kr]5s24d105p3
4. [Xe]6s24f 6
5. Identify the element with each ground state electron configuration.
1. [He]2s22p1
2. [Ar]4s23d8
3. [Kr]5s24d105p4
4. [Xe]6s2
6. Propose an explanation as to why the noble gases are inert.
Numerical Problems
1. How many magnetic quantum numbers are possible for a 4p subshell? A 3d subshell? How many orbitals are in these subshells?
2. How many magnetic quantum numbers are possible for a 6s subshell? A 4f subshell? How many orbitals does each subshell contain?
3. If l = 2 and ml = 2, give all the allowed combinations of the four quantum numbers (n, l, ml, ms) for electrons in the corresponding 3d subshell.
4. Give all the allowed combinations of the four quantum numbers (n, l, ml, ms) for electrons in a 4d subshell. How many electrons can the 4d orbital accommodate? How would this differ from a situation in which there were only three quantum numbers (n, l, m)?
5. Given the following sets of quantum numbers (n, l, ml, ms), identify each principal shell and subshell.
1. 1, 0, 0, ½
2. 2, 1, 0, ½
3. 3, 2, 0, ½
4. 4, 3, 3, ½
6. Is each set of quantum numbers allowed? Explain your answers.
1. n = 2; l = 1; ml = 2; ms = +½
2. n = 3, l = 0; ml = −1; ms = −½
3. n = 2; l = 2; ml = 1; ms = +½
4. n = 3; l = 2; ml = 2; ms = +½
7. List the set of quantum numbers for each electron in the valence shell of each element.
1. beryllium
2. xenon
3. lithium
4. fluorine
8. List the set of quantum numbers for each electron in the valence shell of each element.
1. carbon
2. magnesium
3. bromine
4. sulfur
9. Sketch the shape of the periodic table if there were three possible values of ms for each electron (+½, −½, and 0); assume that the Pauli principle is still valid.
10. Predict the shape of the periodic table if eight electrons could occupy the p subshell.
11. If the electron could only have spin +½, what would the periodic table look like?
12. If three electrons could occupy each s orbital, what would be the electron configuration of each species?
1. sodium
2. titanium
3. fluorine
4. calcium
13. If Hund’s rule were not followed and maximum pairing occurred, how many unpaired electrons would each species have? How do these numbers compare with the number found using Hund’s rule?
1. phosphorus
2. iodine
3. manganese
14. Write the electron configuration for each element in the ground state.
1. aluminum
2. calcium
3. sulfur
4. tin
5. nickel
6. tungsten
7. neodymium
8. americium
15. Write the electron configuration for each element in the ground state.
1. boron
2. rubidium
3. bromine
4. germanium
5. vanadium
6. palladium
7. bismuth
8. europium
16. Give the complete electron configuration for each element.
1. magnesium
2. potassium
3. titanium
4. selenium
5. iodine
6. uranium
7. germanium
17. Give the complete electron configuration for each element.
1. tin
2. copper
3. fluorine
4. hydrogen
5. thorium
6. yttrium
7. bismuth
18. Write the valence electron configuration for each element:
1. samarium
2. praseodymium
3. boron
4. cobalt
19. Using the Pauli exclusion principle and Hund’s rule, draw valence orbital diagrams for each element.
1. barium
2. neodymium
3. iodine
20. Using the Pauli exclusion principle and Hund’s rule, draw valence orbital diagrams for each element.
1. chlorine
2. silicon
3. scandium
21. How many unpaired electrons does each species contain?
1. lead
2. cesium
3. copper
4. silicon
5. selenium
22. How many unpaired electrons does each species contain?
1. helium
2. oxygen
3. bismuth
4. silver
5. boron
23. For each element, give the complete electron configuration, draw the valence electron configuration, and give the number of unpaired electrons present.
1. lithium
2. magnesium
3. silicon
4. cesium
5. lead
24. Use an orbital diagram to illustrate the aufbau principle, the Pauli exclusion principle, and Hund’s rule for each element.
1. carbon
2. sulfur
Numerical Answers
1. For a 4p subshell, n = 4 and l = 1. The allowed values of the magnetic quantum number, ml, are therefore +1, 0, −1, corresponding to three 4p orbitals. For a 3d subshell, n = 3 and l = 2. The allowed values of the magnetic quantum number, ml, are therefore +2, +1, 0, −1, −2, corresponding to five 3d orbitals. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/06%3A_Electronic_Structure_of_Atoms/6.E%3A_Electronic_Structure_of_Atoms_%28Exercises%29.txt |
Understanding the electronic structure of atoms requires an understanding of the properties of waves and electromagnetic radiation. A basic knowledge of the electronic structure of atoms requires an understanding of the properties of waves and electromagnetic radiation. A wave is a periodic oscillation by which energy is transmitted through space. All waves are periodic, repeating regularly in both space and time. Waves are characterized by several interrelated properties.
• Electronic structure: arrangement of electrons in atoms
• Electromagnetic radiation: aka radiant energy; form of energy that has wave characteristics and carries energy through space. All types of electromagnetic radiation move through a vacuum at a speed of 3.00 X 108 m/s (speed of light).
• Wavelength: the distance between identical points on successive waves
• Frequency: the number of complete wavelengths that pass a given point in 1s
$\nu λ = c \nonumber$
where $\nu$ = frequency, $λ$= wavelength, and $c$ = speed of light
Wavelength is expressed in units of length.
• Electromagnetic spectrum: various types of electromagnetic radiations arranged in order of increasing wavelength.
Frequency is expressed in Hertz (Hz), also denoted by s-1 or /s
Quantum Effects and Photons
• Quantum: smallest quantity of energy that can be emitted or absorbed as electromagnetic radiation
$E = hv \nonumber$
where $E$ = energy, $h$ = Planck’s constant, $\nu$ = frequency
Planck’s constant = 6.63 X 10-34 J•s
According to Planck’s theory, energy is always emitted or absorbed in whole-number multiples of hv, for example, hv, 2hv, 3hv, and so forth. We say that the allowed energies are quantized (that is, their values are restricted to certain quantities).
6.2: Quantized Energy and Photons
Blackbody radiation is the radiation emitted by hot objects and could not be explained with classical physics. Max Planck postulated that energy was quantized and may be emitted or absorbed only in integral multiples of a small unit of energy, known as a quantum. The energy of a quantum is proportional to the frequency of the radiation; the proportionality constant h is a fundamental constant (Planck’s constant). Albert Einstein used the quantization of energy to explain the photoelectric effect
• The photoelectric effect: when photons of sufficiently high energy strike a metal surface, electrons are emitted from the metal. The emitted electrons are drawn toward the other electrode, which is a positive terminal. As a result, current flows in a circuit.
• Photon: smallest increment (a quantum) of radiant energy; a photon of light with frequency v has an energy equal to hv.
• When a photon strikes the metal, its energy is transferred to an electron in the metal. A certain amount of energy is required for the electron to overcome the attractive forces that hold it within the metal. If the photons have less energy that this energy threshold, the electrons cannot escape from the metal surface. If a photon has sufficient energy, an electron is emitted. If the photon has more energy than necessary, the excess appears as kinetic energy of the emitted electron.
6.3: Line Spectra and the Bohr Model
There is an intimate connection between the atomic structure of an atom and its spectral characteristics. Most light is polychromatic and contains light of many wavelengths. Light that has only a single wavelength is monochromatic and is produced by devices called lasers, which use transitions between two atomic energy levels to produce light in a very narrow range of wavelengths. Atoms can also absorb light of certain energies, resulting in a transition from the ground state or a lower-energy e
Line Spectra
Radiation composed of a single wavelength is said to be monochromatic.
• Spectrum: distribution among various wavelengths of the radiant energy emitted or absorbed by an object
• Continuous spectrum: rainbow of colors, containing light of all wavelengths. Not all radiation sources produce a continuous spectrum
• Line spectrum: spectrum containing radiation of only specific wavelengths
Bohr’s Model
Electrons in a permitted orbit have a specific energy and are said to be in an "allowed" energy state. An electron in an allowed energy state will not radiate energy and therefore will not spiral into the nucleus.
$E_n = -R_H \dfrac{1}{n^2} \nonumber$
• RH = Rydberg constant: 2.18 X 10-18 J
• n = principal quantum number, corresponds to the different allowed orbits for the electron
All energies given by this equation will be negative. The lower (more negative) the energy is, the more stable the atom is. The lowest energy state is that for which n=1.
• Ground state: lowest energy state of an atom, n=1
• Excited state: when the electron is in higher energy orbit (less negative), n=2 or higher
If n becomes infinitely large (∞), the electron is completely separated from the nucleus:
$E_∞ = (-2.18 \times 10^{-18} J) \left(\dfrac{1}{∞^2}\right) = 0 \nonumber$
Thus, the state in which the electron is removed from the nucleus is the reference, or zero-energy, state of the hydrogen atom. It is important to remember that this zero-energy state is higher in energy than the states with negative energies
Electrons can change from one energy state to another by absorbing or emitting radiant energy. Radiant energy must be absorbed for an electron to move to a higher energy state, but is emitted when the electron moves to a lower energy state. .
$\Delta E = E_f – E_i \nonumber$
• If nf > ni , then ∆E is positive, radiant energy is absorbed
• If nf < ni, then ∆E is negative, radiant energy is emitted
6.4: The Wave Behavior of Matter
An electron possesses both particle and wave properties. Louis de Broglie showed that the wavelength of a particle is equal to Planck’s constant divided by the mass times the velocity of the particle. The electron in Bohr’s circular orbits could thus be described as a standing wave, one that does not move through space. Werner Heisenberg’s uncertainty principle states that it is impossible to precisely describe both the location and the speed of particles that exhibit wavelike behavior.
• Momentum: the product of the mass, m, and the velocity, v, of a particle
• Matter waves: term used to describe the wave characteristics of a particle
$λ = h / mv \nonumber$
λ = wavelength, h = Planck’s constant, m = mass, v = velocity
The Uncertainty Principle
• Uncertainty principle: theory first put forth by Heisenberg, states that is it impossible to determine both the exact momentum of the electron and its exact location.
6.5: Quantum Mechanics and Atomic Orbitals
There is a relationship between the motions of electrons in atoms and molecules and their energies that is described by quantum mechanics. Because of wave–particle duality, scientists must deal with the probability of an electron being at a particular point in space. To do so required the development of quantum mechanics, which uses wavefunctions to describe the mathematical relationship between the motion of electrons in atoms and molecules and their energies.
• Wave functions: represented by ψ, square of wave function, ψ2, provides information about an electron’s location when it is in an allowed energy state.
• Probability density: represented by ψ2, value that represents the probability that an electron will be found at a given point in space
• Electron density: the probability of finding and electron at any particular point in an atom. Equals ψ2.
Orbitals and Quantum Numbers
-Orbital: allowed energy state of an electron in the quantum-mechanical model of the atom; also used to describe the spatial distribution of an electron. Defined by the value of 3 quantum numbers; n, l, and ml.
value of 3 quantum numbers; n, l, and ml.
Value of l 0 1 2 3
Letter used s p d f
1. The principal quantum number, n, can have integral values of 1, 2, 3 and so forth. As n increases, the orbital becomes larger; the electron has a higher energy and is farther away from the nucleus.
2. The second quantum number, l, can have integral values from 0 to n – 1 for each value of n. This quantum number defines the shape of the orbital. Generally designated by the letters s, p, d, and f. These correspond to values ranging from 0 to 3.
3. The magnetic quantum number, ml, can have integral values between l and –l, including zero. This quantum number describes the orientation of the orbital in space.
Electron shell: collection of orbitals with the same value of n
Subshell: one or more orbitals with the same set of n and l values
1. Each shell is divided into the number of subshells equal to the principal quantum number, n, for that shell. The first shell consists of only the 1s subshell; the second shell consists of two subshells, 2s and 2p; the third of three subshell, 3s, 3p and 3d, and so forth.
2. Each subshell is divided into orbitals. Each s subshell consists of one orbital; each p subshell of three orbitals, each d subshell of five, and each f subshell of seven orbitals.
6.6: 3D Representation of Orbitals
Orbitals with l = 0 are s orbitals and are spherically symmetrical, with the greatest probability of finding the electron occurring at the nucleus. Orbitals with values of n > 1 and l = 0 contain one or more nodes. Orbitals with l = 1 are p orbitals and contain a nodal plane that includes the nucleus, giving rise to a dumbbell shape. Orbitals with l = 2 are d orbitals and have more complex shapes with at least two nodal surfaces. l = 3 orbitals are f orbitals, which are still more complex.
The s Orbitals: 1s orbital: most stable, spherically symmetric, figure indicates that the probability decreases as we move away from the nucleus. ALL s ORBITALS ARE SPHERICALLY SYMMETRIC.
• Nodal surfaces (nodes): intermediate regions where ψ2 goes to zero. The number of nodes increases with increasing value for the principal quantum number, n.
The p Orbitals: Electron density is concentrated on two sides of the nucleus, separated by a node at the nucleus. The orbitals of a given subshell have the same size and shape but differ from each other in orientation. The axis along which the orbital is oriented is not related to ml.
6.7: Many-Electron Atoms
In addition to the three quantum numbers (n, l, ml) dictated by quantum mechanics, a fourth quantum number is required to explain certain properties of atoms. This is the electron spin quantum number (ms), which can have values of +½ or −½ for any electron, corresponding to the two possible orientations of an electron in a magnetic field. This is important for chemistry because the Pauli exclusion principle implies that no orbital can contain more than two electrons (with opposite spin).
Although the shapes of the orbitals for many-electron atoms are the same as those for hydrogen, the presence of more than one electron greatly changes the energies of the orbitals. In hydrogen, the energy of an orbital depends only on its principal quantum number, however, in many-electron atoms, electron-electron repulsions cause different subshells to be at different energies
Effective Nuclear Charge
• Effective nuclear charge: net positive charge attracting electrons
$Z_{eff} = Z – S \nonumber$
• Zeff = effective nuclear charge
• Z = number of protons in the nucleus
• S = average number of electrons between the nucleus and electron in question
• Screening effect: effect of inner electrons in decreasing the nuclear charge experienced by outer electrons
Energies of Orbitals
The extent to which an electron will be screened by the other electrons depends on its electron distribution as we move outward from the nucleus.
• In a many-electron atom, for a given value of n, Zeff decreases with increasing value of l.
• In a many-electron atom, for a given value of n, the energy of an orbital increases with increasing value of l.
Degenerate: orbitals that have the same energy
Electron Spin and the Pauli Exclusion Principle
• Electron spin: property of the electron that makes it behave as though it were a tiny magnet. The electron behaves as if it were spinning on its axis; electron spin is quantized.
• Electron spin quantum number: denoted as ms. It can only have two possible values, +½ and –½, which we can interpret as indicating the two opposite directions in which the electron can spin.
• Pauli exclusion principle: states that no two electrons in an atom can have the same set of four quantum numbers n, l, ml, ms. This means that if we wish to put two electrons in an orbital and satisfy Pauli’s exclusion principle, our only choice is to assign different ms values to the electrons. Because there are only two values, we can conclude that an electron can hold a maximum of two electrons and they must have opposite spins.
6.8: Electron Configurations
Based on the Pauli principle and a knowledge of orbital energies obtained using hydrogen-like orbitals, it is possible to construct the periodic table by filling up the available orbitals beginning with the lowest-energy orbitals (the aufbau principle), which gives rise to a particular arrangement of electrons for each element (its electron configuration). Hund’s rule says that the lowest-energy arrangement of electrons is the one that places them in degenerate orbitals with parallel spins.
• Electron configuration: the way in which the electrons are distributed among the various orbitals. The most stable, or ground, electron configuration of an atom is that in which the electrons are in the lowest possible energy level
• Orbital diagram: representation of electron configuration in which each orbital is represented by a box and each electron by a half-arrow. A half-arrow pointing upward represents an electron with positive spin; one pointing downward represents an electron with a negative spin.
Writing Electron Configurations
• Hund’s rule: rule stating that electrons occupy degenerate orbitals in such a way as to maximize the number of electrons with the same spin. In other words, each orbital has one electron placed in it before paring of electron in orbitals occurs. Note that this rule applies to orbitals that are degenerate, which means that they have the same energy.
• Valence electrons: electrons in the outer shells
• Core electrons: electrons in the inner shells
• Transition elements: aka Transition metals; elements of the d orbitals
• Lanthanide elements: aka Rare-earth elements; 14 elements of the 4f orbitals, # 58-71
• Actinide elements: 14 elements of 5f orbitals, # 90-103. Most are not found in nature.
6.9: Electron Configurations and the Periodic Table
The arrangement of atoms in the periodic table results in blocks corresponding to filling of the ns, np, nd, and nf orbitals to produce the distinctive chemical properties of the elements in the s block, p block, d block, and f block, respectively.
• Main group elements: aka Representatives; s and p block elements
• F-block metals: 28 elements located below the table, f block elements | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/06%3A_Electronic_Structure_of_Atoms/6.S%3A_Electronic_Structure_of_Atoms_%28Summary%29.txt |
Last chapter, we presented the contemporary quantum mechanical model of the atom. In using this model to describe the electronic structures of the elements in order of increasing atomic number, we saw that periodic similarities in electron configuration correlate with periodic similarities in properties, which is the basis for the structure of the periodic table. For example, the noble gases have what is often called filled or closed-shell valence electron configurations. These closed shells are actually filled s and p subshells with a total of eight electrons, which are called octets; helium is an exception, with a closed 1s shell that has only two electrons. Because of their filled valence shells, the noble gases are generally unreactive. In contrast, the alkali metals have a single valence electron outside a closed shell and readily lose this electron to elements that require electrons to achieve an octet, such as the halogens. Thus because of their periodic similarities in electron configuration, atoms in the same column of the periodic table tend to form compounds with the same oxidation states and stoichiometries. Last chapter ended with the observation that, because all the elements in a column have the same valence electron configuration, the periodic table can be used to find the electron configuration of most of the elements at a glance.
In this chapter, we explore the relationship between the electron configurations of the elements, as reflected in their arrangement in the periodic table, and their physical and chemical properties. In particular, we focus on the similarities between elements in the same column and on the trends in properties that are observed across horizontal rows or down vertical columns. By the end of this chapter, your understanding of these trends and relationships will provide you with clues as to why argon is used in incandescent light bulbs, why coal and wood burst into flames when they come in contact with pure F2, why aluminum was discovered so late despite being the third most abundant element in Earth’s crust, and why lithium is commonly used in batteries. We begin by expanding on the brief discussion of the history of the periodic table and describing how it was created many years before electrons had even been discovered, much less discussed in terms of shells, subshells, orbitals, and electron spin.
• 7.1: Development of the Periodic Table
The periodic table arranges the elements according to their electron configurations, such that elements in the same column have the same valence electron configurations. Periodic variations in size and chemical properties are important factors in dictating the types of chemical reactions the elements undergo and the kinds of chemical compounds they form. The modern periodic table was based on empirical correlations of properties such as atomic mass.
• 7.2: Shielding and Effective Nuclear Charge
The calculation of orbital energies in atoms or ions with more than one electron (multielectron atoms or ions) is complicated by repulsive interactions between the electrons. The concept of electron shielding, in which intervening electrons act to reduce the positive nuclear charge experienced by an electron, allows the use of hydrogen-like orbitals and an effective nuclear charge to describe electron distributions in more complex atoms or ions.
• 7.3: Sizes of Atoms and Ions
Ionic radii share the same vertical trend as atomic radii, but the horizontal trends differ due to differences in ionic charges. A variety of methods have been established to measure the size of a single atom or ion. The covalent atomic radius (rcov) is half the internuclear distance in a molecule with two identical atoms bonded to each other, whereas the metallic atomic radius (rmet) is defined as half the distance between the nuclei of two adjacent atoms in a metallic element.
• 7.4: Ionization Energy
Generally, the first ionization energy and electronegativity values increase diagonally from the lower left of the periodic table to the upper right, and electron affinities become more negative across a row. The energy required to remove successive electrons from an atom increases steadily, with a substantial increase occurring with the removal of an electron from a filled inner shell.
• 7.5: Electron Affinities
The electron affinity (EA) of an element is the energy change that occurs when an electron is added to a gaseous atom to give an anion. In general, elements with the most negative electron affinities (the highest affinity for an added electron) are those with the smallest size and highest ionization energies and are located in the upper right corner of the periodic table.
• 7.6: Metals, Nonmetals, and Metalloids
The elements can be classified as metals, nonmetals, or metalloids.
• 7.7: Group Trends for the Active Metals
The elements within the same group of the periodic table tend to exhibit similar physical and chemical properties. Four major factors affect reactivity of metals: nuclear charge, atomic radius, shielding effect and sublevel arrangement (of electrons).
• 7.8: Group Trends for Selected Nonmetals
Non-metallic character is the ability to be reduced (be an oxidizing agent), form acidic hydroxides, form covalent compounds with non-metals. These characteristics increase with a larger nuclear charge and smaller radius, with no increase in shielding. The most active non-metal would be the one farthest up and to the right -- not including the noble gases (non-reactive.)
• 7.E: Periodic Properties of the Elements (Exercises)
Problems and selection solutions to the chapter.
• 7.S: Periodic Properties of the Elements (Summary)
Summary for "Chapter 7: Periodic Properties of the Elements" textmap for "Chemistry: The central Science".
07: Periodic Properties of the Elements
Learning Objectives
• To become familiar with the history of the periodic table.
The modern periodic table has evolved through a long history of attempts by chemists to arrange the elements according to their properties as an aid in predicting chemical behavior. One of the first to suggest such an arrangement was the German chemist Johannes Dobereiner (1780–1849), who noticed that many of the known elements could be grouped in triads (a set of three elements that have similar properties)—for example, chlorine, bromine, and iodine; or copper, silver, and gold. Dobereiner proposed that all elements could be grouped in such triads, but subsequent attempts to expand his concept were unsuccessful. We now know that portions of the periodic table—the d block in particular—contain triads of elements with substantial similarities. The middle three members of most of the other columns, such as sulfur, selenium, and tellurium in group 16 or aluminum, gallium, and indium in Group 13, also have remarkably similar chemistry.
By the mid-19th century, the atomic masses of many of the elements had been determined. The English chemist John Newlands (1838–1898), hypothesizing that the chemistry of the elements might be related to their masses, arranged the known elements in order of increasing atomic mass and discovered that every seventh element had similar properties (Figure $1$). Newlands therefore suggested that the elements could be classified into octaves. He described octaves as a group of seven elements which correspond to the horizontal rows in the main groups of today's periodic table. There were seven elements because the noble gases were not known at the time. Unfortunately, Newlands’s “law of octaves” did not seem to work for elements heavier than calcium, and his idea was publicly ridiculed. At one scientific meeting, Newlands was asked why he didn’t arrange the elements in alphabetical order instead of by atomic mass, since that would make just as much sense! Actually, Newlands was on the right track—with only a few exceptions, atomic mass does increase with atomic number, and similar properties occur every time a set of ns2np6 subshells is filled. Despite the fact that Newlands’s table had no logical place for the d-block elements, he was honored for his idea by the Royal Society of London in 1887.
John Newlands (1838–1898)
John Alexander Reina Newlands was an English chemist who worked on the development of the periodic table. He noticed that elemental properties repeated every seventh (or multiple of seven) element, as musical notes repeat every eighth note.
The periodic table achieved its modern form through the work of the German chemist Julius Lothar Meyer (1830–1895) and the Russian chemist Dimitri Mendeleev (1834–1907), both of whom focused on the relationships between atomic mass and various physical and chemical properties. In 1869, they independently proposed essentially identical arrangements of the elements. Meyer aligned the elements in his table according to periodic variations in simple atomic properties, such as “atomic volume” (Figure $2$), which he obtained by dividing the atomic mass (molar mass) in grams per mole by the density ($\rho$) of the element in grams per cubic centimeter. This property is equivalent to what is today defined as molar volume, the molar mass of an element divided by its density, (measured in cubic centimeters per mole):
$\dfrac{molar\; mass\left ( \cancel{g}/mol \right )}{density\left ( \cancel{g}/cm^{3} \right )}=molar\; volume\left ( cm^{3}/mol \right ) \label{7.1.1}$
As shown in Figure $2$, the alkali metals have the highest molar volumes of the solid elements. In Meyer’s plot of atomic volume versus atomic mass, the nonmetals occur on the rising portion of the graph, and metals occur at the peaks, in the valleys, and on the downslopes.
Dimitri Mendeleev (1834–1907)
When his family’s glass factory was destroyed by fire, Mendeleev moved to St. Petersburg, Russia, to study science. He became ill and was not expected to recover, but he finished his PhD with the help of his professors and fellow students.
In addition to the periodic table, another of Mendeleev’s contributions to science was an outstanding textbook, The Principles of Chemistry, which was used for many years. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/07%3A_Periodic_Properties_of_the_Elements/7.01%3A_Development_of_the_Periodic_Table.txt |
Learning Objectives
• To understand the basics of electron shielding and penetration
For an atom or an ion with only a single electron, we can calculate the potential energy by considering only the electrostatic attraction between the positively charged nucleus and the negatively charged electron. When more than one electron is present, however, the total energy of the atom or the ion depends not only on attractive electron-nucleus interactions but also on repulsive electron-electron interactions. When there are two electrons, the repulsive interactions depend on the positions of both electrons at a given instant, but because we cannot specify the exact positions of the electrons, it is impossible to exactly calculate the repulsive interactions. Consequently, we must use approximate methods to deal with the effect of electron-electron repulsions on orbital energies. These effects are the underlying basis for the periodic trends in elemental properties that we will explore in this chapter.
Electron Shielding and Effective Nuclear Charge
If an electron is far from the nucleus (i.e., if the distance $r$ between the nucleus and the electron is large), then at any given moment, many of the other electrons will be between that electron and the nucleus (Figure $1$). Hence the electrons will cancel a portion of the positive charge of the nucleus and thereby decrease the attractive interaction between it and the electron farther away. As a result, the electron farther away experiences an effective nuclear charge ($Z_{eff}$) that is less than the actual nuclear charge $Z$. This effect is called electron shielding.
As the distance between an electron and the nucleus approaches infinity, $Z_{eff}$ approaches a value of 1 because all the other ($Z − 1$) electrons in the neutral atom are, on the average, between it and the nucleus. If, on the other hand, an electron is very close to the nucleus, then at any given moment most of the other electrons are farther from the nucleus and do not shield the nuclear charge. At $r ≈ 0$, the positive charge experienced by an electron is approximately the full nuclear charge, or $Z_{eff} ≈ Z$. At intermediate values of $r$, the effective nuclear charge is somewhere between 1 and $Z$:
$1 ≤ Z_{eff} ≤ Z. \nonumber$
Notice that $Z_{eff} = Z$ only for hydrogen (Figure $2$).
Definition: Shielding
Shielding refers to the core electrons repelling the outer electrons, which lowers the effective charge of the nucleus on the outer electrons. Hence, the nucleus has "less grip" on the outer electrons insofar as it is shielded from them.
$Z_{eff}$ can be calculated by subtracting the magnitude of shielding from the total nuclear charge and the effective nuclear charge of an atom is given by the equation:
$Z_{eff}=Z-S \label{4}$
where $Z$ is the atomic number (number of protons in nucleus) and $S$ is the shielding constant and is approximated by number of electrons between the nucleus and the electron in question (the number of nonvalence electrons). The value of $Z_{eff}$ will provide information on how much of a charge an electron actually experiences.
We can see from Equation \ref{4} that the effective nuclear charge of an atom increases as the number of protons in an atom increases (Figure $2$). As we will discuss later on in the chapter, this phenomenon can explain the decrease in atomic radii we see as we go across the periodic table as electrons are held closer to the nucleus due to increase in number of protons and increase in effective nuclear charge.
The shielding constant can be estimated by totaling the screening by all nonvalence electrons ($n$) except the one in question.
$S = \sum_{i}^{n-1} S_i \label{2.6.0}$
where $S_i$ is the shielding of the ith electron.
Electrons that are shielded from the full charge of the nucleus experience an effective nuclear charge ($Z_{eff}$) of the nucleus, which is some degree less than the full nuclear charge an electron would feel in a hydrogen atom or hydrogenlike ion.
From Equations \ref{4} and \ref{2.6.0}, $Z_{eff}$ for a specific electron can be estimated is the shielding constants for that electron of all other electrons in species is known. A simple approximation is that all other non-valence electrons shield equally and fully:
$S_i=1 \label{simple}$
This crude approximation is demonstrated in Example $1$.
Example $1$: Fluorine, Neon, and Sodium
What is the effective attraction $Z_{eff}$ experienced by the valence electrons in the three isoelectronic species: the fluorine anion, the neutral neon atom, and sodium cation?
Solution
Each species has 10 electrons, and the number of nonvalence electrons is 2 (10 total electrons - 8 valence), but the effective nuclear charge varies because each has a different atomic number $A$. This is an application of Equations \ref{4} and \ref{2.6.0}. We use the simple assumption that all electrons shield equally and fully the valence electrons (Equation \ref{simple}).
The charge $Z$ of the nucleus of a fluorine atom is 9, but the valence electrons are screened appreciably by the core electrons (four electrons from the 1s and 2s orbitals) and partially by the 7 electrons in the 2p orbitals.
• $Z_\mathrm{eff}(\mathrm{F}^-) = 9 - 2 = 7+$
• $Z_\mathrm{eff}(\mathrm{Ne}) = 10 - 2 = 8+$
• $Z_\mathrm{eff}(\mathrm{Na}^+) = 11 - 2 = 9+$
So the sodium cation has the greatest effective nuclear charge. This also suggests that $\mathrm{Na}^+$ has the smallest radius of these species and that is correct.
Exercise $1$: Magnesium Species
What is the effective attraction $Z_{eff}$ experienced by the valence electrons in the magnesium anion, the neutral magnesium atom, and magnesium cation? Use the simple approximation for shielding constants. Compare your result for the magnesium atom to the more accurate value in Figure $2$ and proposed an origin for the difference.
Answer
• $Z_\mathrm{eff}(\ce{Mg}^{-}) = 12 - 10 = 2+$
• $Z_\mathrm{eff}(\ce{Mg}) = 12 - 10 = 2+$
• $Z_\mathrm{eff}(\ce{Mg}^{+}) = 12 - 10 = 2+$
Remember that the simple approximations in Equations \ref{2.6.0} and \ref{simple} suggest that valence electrons do not shield other valence electrons. Therefore, each of these species has the same number of non-valence electrons and Equation \ref{4} suggests the effective charge on each valence electron is identical for each of the three species.
This is not correct and a more complex model is needed to predict the experimental observed $Z_{eff}$ value. The ability of valence electrons to shield other valence electrons or in partial amounts (e.g., $S_i \neq 1$) is in violation of Equations \ref{2.6.0} and \ref{simple}. That fact that these approximations are poor is suggested by the experimental $Z_{eff}$ value shown in Figure $2$ for $\ce{Mg}$ of 3.2+. This is appreciably larger than the +2 estimated above, which means these simple approximations overestimate the total shielding constant $S$. A more sophisticated model is needed.
Electron Penetration
The approximation in Equation \ref{simple} is a good first order description of electron shielding, but the actual $Z_{eff}$ experienced by an electron in a given orbital depends not only on the spatial distribution of the electron in that orbital but also on the distribution of all the other electrons present. This leads to large differences in $Z_{eff}$ for different elements, as shown in Figure $2$ for the elements of the first three rows of the periodic table.
Penetration describes the proximity to which an electron can approach to the nucleus. In a multi-electron system, electron penetration is defined by an electron's relative electron density (probability density) near the nucleus of an atom (Figure $3$). Electrons in different orbitals have different electron densities around the nucleus. In other words, penetration depends on the shell ($n$) and subshell ($l$).
For example, a 1s electron (Figure $3$; purple curve) has greater electron density near the nucleus than a 2p electron (Figure $3$; red curve) and has a greater penetration. This related to the shielding constants since the 1s electrons are closer to the nucleus than a 2p electron, hence the 1s screens a 2p electron almost perfectly ($S=1$. However, the 2s electron has a lower shielding constant ($S<1$ because it can penetrate close to the nucleus in the small area of electron density within the first spherical node (Figure $3$; green curve). In this way the 2s electron can "avoid" some of the shielding effect of the inner 1s electron.
For the same shell value ($n$) the penetrating power of an electron follows this trend in subshells (Figure $3$):
$s > p > d \approx f. \label{better1}$
for different values of shell (n) and subshell (l), penetrating power of an electron follows this trend:
$\ce{1s > 2s > 2p > 3s > 3p > 4s > 3d > 4p > 5s > 4d > 5p > 6s > 4f ...} \label{better2}$
Definition: Penetration
Penetration describes the proximity of electrons in an orbital to the nucleus. Electrons that have greater penetration can get closer to the nucleus and effectively block out the charge from electrons that have less proximity.
Table $1$: Effective Nuclear Charges for Selected Atoms
Atom Sublevel Z Zeff
H 1s 1 1
He 1s 2 1.69
Li 1s, 2s 3 2.69, 1.28
Be 1s, 2s 4 3.68, 1.91
B 1s, 2s, 2p 5 4.68, 2.58, 2.42
F 1s, 2s, 2p 9 8.65, 5.13, 5.10
Na 1s, 2s, 2p, 3s 11 10.63, 6.57, 6.80, 2.51
Data from E. Clementi and D. L. Raimondi; The Journal of Chemical Physics 38, 2686 (1963).
Because of the effects of shielding and the different radial distributions of orbitals with the same value of n but different values of l, the different subshells are not degenerate in a multielectron atom. For a given value of n, the ns orbital is always lower in energy than the np orbitals, which are lower in energy than the nd orbitals, and so forth. As a result, some subshells with higher principal quantum numbers are actually lower in energy than subshells with a lower value of n; for example, the 4s orbital is lower in energy than the 3d orbitals for most atoms.
A Better Estimation of Shielding: Slater Rules
The concepts of electron shielding, orbital penetration and effective nuclear charge were introduced above, but we did so in a qualitative manner (e.g., Equations \ref{better1} and \ref{better2}). A more accurate model for estimating electron shielding and corresponding effective nuclear charge experienced is Slater's Rules. However, the application of these rules is outside the scope of this text.
Zeff and Electron Shielding: Zeff and Electron Shielding(opens in new window) [youtu.be]
Summary
The calculation of orbital energies in atoms or ions with more than one electron (multielectron atoms or ions) is complicated by repulsive interactions between the electrons. The concept of electron shielding, in which intervening electrons act to reduce the positive nuclear charge experienced by an electron, allows the use of hydrogen-like orbitals and an effective nuclear charge ($Z_{eff}$) to describe electron distributions in more complex atoms or ions. The degree to which orbitals with different values of l and the same value of n overlap or penetrate filled inner shells results in slightly different energies for different subshells in the same principal shell in most atoms. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/07%3A_Periodic_Properties_of_the_Elements/7.02%3A_Shielding_and_Effective_Nuclear_Charge.txt |
Learning Objectives
• To understand periodic trends in atomic radii.
• To predict relative ionic sizes within an isoelectronic series.
Although some people fall into the trap of visualizing atoms and ions as small, hard spheres similar to miniature table-tennis balls or marbles, the quantum mechanical model tells us that their shapes and boundaries are much less definite than those images suggest. As a result, atoms and ions cannot be said to have exact sizes; however, some atoms are larger or smaller than others, and this influences their chemistry. In this section, we discuss how atomic and ion “sizes” are defined and obtained.
Atomic Radii
Recall that the probability of finding an electron in the various available orbitals falls off slowly as the distance from the nucleus increases. This point is illustrated in Figure \(1\) which shows a plot of total electron density for all occupied orbitals for three noble gases as a function of their distance from the nucleus. Electron density diminishes gradually with increasing distance, which makes it impossible to draw a sharp line marking the boundary of an atom.
Figure \(1\) also shows that there are distinct peaks in the total electron density at particular distances and that these peaks occur at different distances from the nucleus for each element. Each peak in a given plot corresponds to the electron density in a given principal shell. Because helium has only one filled shell (n = 1), it shows only a single peak. In contrast, neon, with filled n = 1 and 2 principal shells, has two peaks. Argon, with filled n = 1, 2, and 3 principal shells, has three peaks. The peak for the filled n = 1 shell occurs at successively shorter distances for neon (Z = 10) and argon (Z = 18) because, with a greater number of protons, their nuclei are more positively charged than that of helium. Because the 1s2 shell is closest to the nucleus, its electrons are very poorly shielded by electrons in filled shells with larger values of n. Consequently, the two electrons in the n = 1 shell experience nearly the full nuclear charge, resulting in a strong electrostatic interaction between the electrons and the nucleus. The energy of the n = 1 shell also decreases tremendously (the filled 1s orbital becomes more stable) as the nuclear charge increases. For similar reasons, the filled n = 2 shell in argon is located closer to the nucleus and has a lower energy than the n = 2 shell in neon.
Figure \(1\) illustrates the difficulty of measuring the dimensions of an individual atom. Because distances between the nuclei in pairs of covalently bonded atoms can be measured quite precisely, however, chemists use these distances as a basis for describing the approximate sizes of atoms. For example, the internuclear distance in the diatomic Cl2 molecule is known to be 198 pm. We assign half of this distance to each chlorine atom, giving chlorine a covalent atomic radius (\(r_{cov}\)), which is half the distance between the nuclei of two like atoms joined by a covalent bond in the same molecule, of 99 pm or 0.99 Å (Figure \(\PageIndex{2a}\)). Atomic radii are often measured in angstroms (Å), a non-SI unit: 1 Å = 1 × 10−10 m = 100 pm.
In a similar approach, we can use the lengths of carbon–carbon single bonds in organic compounds, which are remarkably uniform at 154 pm, to assign a value of 77 pm as the covalent atomic radius for carbon. If these values do indeed reflect the actual sizes of the atoms, then we should be able to predict the lengths of covalent bonds formed between different elements by adding them. For example, we would predict a carbon–chlorine distance of 77 pm + 99 pm = 176 pm for a C–Cl bond, which is very close to the average value observed in many organochlorine compounds. A similar approach for measuring the size of ions is discussed later in this section.
Covalent atomic radii can be determined for most of the nonmetals, but how do chemists obtain atomic radii for elements that do not form covalent bonds? For these elements, a variety of other methods have been developed. With a metal, for example, the metallic atomic radius (\(r_{met}\)) is defined as half the distance between the nuclei of two adjacent metal atoms in the solid (Figure \(\PageIndex{2b}\)). For elements such as the noble gases, most of which form no stable compounds, we can use what is called the van der Waals atomic radius (\(r_{vdW}\)), which is half the internuclear distance between two nonbonded atoms in the solid (Figure \(\PageIndex{2c}\)). This is somewhat difficult for helium which does not form a solid at any temperature. An atom such as chlorine has both a covalent radius (the distance between the two atoms in a \(\ce{Cl2}\) molecule) and a van der Waals radius (the distance between two Cl atoms in different molecules in, for example, \(\ce{Cl2(s)}\) at low temperatures). These radii are generally not the same (Figure \(\PageIndex{2d}\)).
Periodic Trends in Atomic Radii
Because it is impossible to measure the sizes of both metallic and nonmetallic elements using any one method, chemists have developed a self-consistent way of calculating atomic radii using the quantum mechanical functions. Although the radii values obtained by such calculations are not identical to any of the experimentally measured sets of values, they do provide a way to compare the intrinsic sizes of all the elements and clearly show that atomic size varies in a periodic fashion (Figure \(3\)).
In the periodic table, atomic radii decrease from left to right across a row and increase from top to bottom down a column. Because of these two trends, the largest atoms are found in the lower left corner of the periodic table, and the smallest are found in the upper right corner (Figure \(4\)).
Trends in atomic size result from differences in the effective nuclear charges (\(Z_{eff}\)) experienced by electrons in the outermost orbitals of the elements. For all elements except H, the effective nuclear charge is always less than the actual nuclear charge because of shielding effects. The greater the effective nuclear charge, the more strongly the outermost electrons are attracted to the nucleus and the smaller the atomic radius.
Atomic radii decrease from left to right across a row and increase from top to bottom down a column.
The atoms in the second row of the periodic table (Li through Ne) illustrate the effect of electron shielding. All have a filled 1s2 inner shell, but as we go from left to right across the row, the nuclear charge increases from +3 to +10. Although electrons are being added to the 2s and 2p orbitals, electrons in the same principal shell are not very effective at shielding one another from the nuclear charge. Thus the single 2s electron in lithium experiences an effective nuclear charge of approximately +1 because the electrons in the filled 1s2 shell effectively neutralize two of the three positive charges in the nucleus. (More detailed calculations give a value of Zeff = +1.26 for Li.) In contrast, the two 2s electrons in beryllium do not shield each other very well, although the filled 1s2 shell effectively neutralizes two of the four positive charges in the nucleus. This means that the effective nuclear charge experienced by the 2s electrons in beryllium is between +1 and +2 (the calculated value is +1.66). Consequently, beryllium is significantly smaller than lithium. Similarly, as we proceed across the row, the increasing nuclear charge is not effectively neutralized by the electrons being added to the 2s and 2p orbitals. The result is a steady increase in the effective nuclear charge and a steady decrease in atomic size (Figure \(5\)).
The increase in atomic size going down a column is also due to electron shielding, but the situation is more complex because the principal quantum number n is not constant. As we saw in Chapter 2, the size of the orbitals increases as n increases, provided the nuclear charge remains the same. In group 1, for example, the size of the atoms increases substantially going down the column. It may at first seem reasonable to attribute this effect to the successive addition of electrons to ns orbitals with increasing values of n. However, it is important to remember that the radius of an orbital depends dramatically on the nuclear charge. As we go down the column of the group 1 elements, the principal quantum number n increases from 2 to 6, but the nuclear charge increases from +3 to +55!
As a consequence the radii of the lower electron orbitals in Cesium are much smaller than those in lithium and the electrons in those orbitals experience a much larger force of attraction to the nucleus. That force depends on the effective nuclear charge experienced by the the inner electrons. If the outermost electrons in cesium experienced the full nuclear charge of +55, a cesium atom would be very small indeed. In fact, the effective nuclear charge felt by the outermost electrons in cesium is much less than expected (6 rather than 55). This means that cesium, with a 6s1 valence electron configuration, is much larger than lithium, with a 2s1 valence electron configuration. The effective nuclear charge changes relatively little for electrons in the outermost, or valence shell, from lithium to cesium because electrons in filled inner shells are highly effective at shielding electrons in outer shells from the nuclear charge. Even though cesium has a nuclear charge of +55, it has 54 electrons in its filled 1s22s22p63s23p64s23d104p65s24d105p6 shells, abbreviated as [Xe]5s24d105p6, which effectively neutralize most of the 55 positive charges in the nucleus. The same dynamic is responsible for the steady increase in size observed as we go down the other columns of the periodic table. Irregularities can usually be explained by variations in effective nuclear charge.
Not all Electrons shield equally
Electrons in the same principal shell are not very effective at shielding one another from the nuclear charge, whereas electrons in filled inner shells are highly effective at shielding electrons in outer shells from the nuclear charge.
Example \(1\)
On the basis of their positions in the periodic table, arrange these elements in order of increasing atomic radius: aluminum, carbon, and silicon.
Given: three elements
Asked for: arrange in order of increasing atomic radius
Strategy:
1. Identify the location of the elements in the periodic table. Determine the relative sizes of elements located in the same column from their principal quantum number n. Then determine the order of elements in the same row from their effective nuclear charges. If the elements are not in the same column or row, use pairwise comparisons.
2. List the elements in order of increasing atomic radius.
Solution:
A These elements are not all in the same column or row, so we must use pairwise comparisons. Carbon and silicon are both in group 14 with carbon lying above, so carbon is smaller than silicon (C < Si). Aluminum and silicon are both in the third row with aluminum lying to the left, so silicon is smaller than aluminum (Si < Al) because its effective nuclear charge is greater.
B Combining the two inequalities gives the overall order: C < Si < Al.
Exercise \(1\)
On the basis of their positions in the periodic table, arrange these elements in order of increasing size: oxygen, phosphorus, potassium, and sulfur.
Answer
O < S < P < K
Atomic Radius: Atomic Radius, YouTube(opens in new window) [youtu.be]
Ionic Radii and Isoelectronic Series
An ion is formed when either one or more electrons are removed from a neutral atom to form a positive ion (cation) or when additional electrons attach themselves to neutral atoms to form a negative one (anion). The designations cation or anion come from the early experiments with electricity which found that positively charged particles were attracted to the negative pole of a battery, the cathode, while negatively charged ones were attracted to the positive pole, the anode.
Ionic compounds consist of regular repeating arrays of alternating positively charged cations and negatively charges anions. Although it is not possible to measure an ionic radius directly for the same reason it is not possible to directly measure an atom’s radius, it is possible to measure the distance between the nuclei of a cation and an adjacent anion in an ionic compound to determine the ionic radius (the radius of a cation or anion) of one or both. As illustrated in Figure \(6\), the internuclear distance corresponds to the sum of the radii of the cation and anion. A variety of methods have been developed to divide the experimentally measured distance proportionally between the smaller cation and larger anion. These methods produce sets of ionic radii that are internally consistent from one ionic compound to another, although each method gives slightly different values. For example, the radius of the Na+ ion is essentially the same in NaCl and Na2S, as long as the same method is used to measure it. Thus despite minor differences due to methodology, certain trends can be observed.
A comparison of ionic radii with atomic radii (Figure \(7\)) shows that a cation, having lost an electron, is always smaller than its parent neutral atom, and an anion, having gained an electron, is always larger than the parent neutral atom. When one or more electrons is removed from a neutral atom, two things happen: (1) repulsions between electrons in the same principal shell decrease because fewer electrons are present, and (2) the effective nuclear charge felt by the remaining electrons increases because there are fewer electrons to shield one another from the nucleus. Consequently, the size of the region of space occupied by electrons decreases and the ion shrinks (compare Li at 167 pm with Li+ at 76 pm). If different numbers of electrons can be removed to produce ions with different charges, the ion with the greatest positive charge is the smallest (compare Fe2+ at 78 pm with Fe3+ at 64.5 pm). Conversely, adding one or more electrons to a neutral atom causes electron–electron repulsions to increase and the effective nuclear charge to decrease, so the size of the probability region increases and the ion expands (compare F at 42 pm with F at 133 pm).
Cations are always smaller than the neutral atom and anions are always larger.
Because most elements form either a cation or an anion but not both, there are few opportunities to compare the sizes of a cation and an anion derived from the same neutral atom. A few compounds of sodium, however, contain the Na ion, allowing comparison of its size with that of the far more familiar Na+ ion, which is found in many compounds. The radius of sodium in each of its three known oxidation states is given in Table \(1\). All three species have a nuclear charge of +11, but they contain 10 (Na+), 11 (Na0), and 12 (Na) electrons. The Na+ ion is significantly smaller than the neutral Na atom because the 3s1 electron has been removed to give a closed shell with n = 2. The Na ion is larger than the parent Na atom because the additional electron produces a 3s2 valence electron configuration, while the nuclear charge remains the same.
Table \(1\): Experimentally Measured Values for the Radius of Sodium in Its Three Known Oxidation States
Na+ Na0 Na
Electron Configuration 1s22s22p6 1s22s22p63s1 1s22s22p63s2
Radius (pm) 102 154* 202
*The metallic radius measured for Na(s). †Source: M. J. Wagner and J. L. Dye, “Alkalides, Electrides, and Expanded Metals,” Annual Review of Materials Science 23 (1993): 225–253.
Ionic radii follow the same vertical trend as atomic radii; that is, for ions with the same charge, the ionic radius increases going down a column. The reason is the same as for atomic radii: shielding by filled inner shells produces little change in the effective nuclear charge felt by the outermost electrons. Again, principal shells with larger values of n lie at successively greater distances from the nucleus.
Because elements in different columns tend to form ions with different charges, it is not possible to compare ions of the same charge across a row of the periodic table. Instead, elements that are next to each other tend to form ions with the same number of electrons but with different overall charges because of their different atomic numbers. Such a set of species is known as an isoelectronic series. For example, the isoelectronic series of species with the neon closed-shell configuration (1s22s22p6) is shown in Table \(3\).
The sizes of the ions in this series decrease smoothly from N3− to Al3+. All six of the ions contain 10 electrons in the 1s, 2s, and 2p orbitals, but the nuclear charge varies from +7 (N) to +13 (Al). As the positive charge of the nucleus increases while the number of electrons remains the same, there is a greater electrostatic attraction between the electrons and the nucleus, which causes a decrease in radius. Consequently, the ion with the greatest nuclear charge (Al3+) is the smallest, and the ion with the smallest nuclear charge (N3−) is the largest. The neon atom in this isoelectronic series is not listed in Table \(3\), because neon forms no covalent or ionic compounds and hence its radius is difficult to measure.
Ion Radius (pm) Atomic Number
Table \(3\): Radius of Ions with the Neon Closed-Shell Electron Configuration. Source: R. D. Shannon, “Revised effective ionic radii and systematic studies of interatomic distances in halides and chalcogenides,” Acta Crystallographica 32, no. 5 (1976): 751–767.
N3− 146 7
O2− 140 8
F 133 9
Na+ 98 11
Mg2+ 79 12
Al3+ 57 13
Example \(2\)
Based on their positions in the periodic table, arrange these ions in order of increasing radius: Cl, K+, S2−, and Se2.
Given: four ions
Asked for: order by increasing radius
Strategy:
1. Determine which ions form an isoelectronic series. Of those ions, predict their relative sizes based on their nuclear charges. For ions that do not form an isoelectronic series, locate their positions in the periodic table.
2. Determine the relative sizes of the ions based on their principal quantum numbers n and their locations within a row.
Solution:
A We see that S and Cl are at the right of the third row, while K and Se are at the far left and right ends of the fourth row, respectively. K+, Cl, and S2− form an isoelectronic series with the [Ar] closed-shell electron configuration; that is, all three ions contain 18 electrons but have different nuclear charges. Because K+ has the greatest nuclear charge (Z = 19), its radius is smallest, and S2− with Z = 16 has the largest radius. Because selenium is directly below sulfur, we expect the Se2 ion to be even larger than S2−.
B The order must therefore be K+ < Cl < S2− < Se2.
Exercise \(2\)
Based on their positions in the periodic table, arrange these ions in order of increasing size: Br, Ca2+, Rb+, and Sr2+.
Answer
Ca2+ < Sr2+ < Rb+ < Br
Summary
Ionic radii share the same vertical trend as atomic radii, but the horizontal trends differ due to differences in ionic charges. A variety of methods have been established to measure the size of a single atom or ion. The covalent atomic radius (rcov) is half the internuclear distance in a molecule with two identical atoms bonded to each other, whereas the metallic atomic radius (rmet) is defined as half the distance between the nuclei of two adjacent atoms in a metallic element. The van der Waals radius (rvdW) of an element is half the internuclear distance between two nonbonded atoms in a solid. Atomic radii decrease from left to right across a row because of the increase in effective nuclear charge due to poor electron screening by other electrons in the same principal shell. Moreover, atomic radii increase from top to bottom down a column because the effective nuclear charge remains relatively constant as the principal quantum number increases. The ionic radii of cations and anions are always smaller or larger, respectively, than the parent atom due to changes in electron–electron repulsions, and the trends in ionic radius parallel those in atomic size. A comparison of the dimensions of atoms or ions that have the same number of electrons but different nuclear charges, called an isoelectronic series, shows a clear correlation between increasing nuclear charge and decreasing size. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/07%3A_Periodic_Properties_of_the_Elements/7.03%3A_Sizes_of_Atoms_and_Ions.txt |
Learning Objectives
• To correlate ionization energies with the chemistry of the elements
We have seen that when elements react, they often gain or lose enough electrons to achieve the valence electron configuration of the nearest noble gas. Why is this so? In this section, we develop a more quantitative approach to predicting such reactions by examining periodic trends in the energy changes that accompany ion formation.
Ionization Energies
Because atoms do not spontaneously lose electrons, energy is required to remove an electron from an atom to form a cation. Chemists define the ionization energy ($I$) of an element as the amount of energy needed to remove an electron from the gaseous atom $E$ in its ground state. $I$ is therefore the energy required for the reaction
$E_{(g)} \rightarrow E^+_{(g)} +e^- \;\;\ \text{energy required=I } \label{7.4.1}$
Because an input of energy is required, the ionization energy is always positive ($I > 0$) for the reaction as written in Equation $1$. Larger values of I mean that the electron is more tightly bound to the atom and harder to remove. Typical units for ionization energies are kilojoules/mole (kJ/mol) or electron volts (eV):
$1\; eV/atom = 96.49\; kJ/mol \nonumber$
If an atom possesses more than one electron, the amount of energy needed to remove successive electrons increases steadily. We can define a first ionization energy ($I_1$), a second ionization energy ($I_2$), and in general an nth ionization energy ($I_n$) according to the following reactions:
$\ce{E(g) \rightarrow E^+(g) +e^-} \;\;\ I_1=\text{1st ionization energy} \label{7.4.2}$
$\ce{E^{+}(g) \rightarrow E^{2+}(g) +e^-} \;\;\ I_2=\text{2nd ionization energy} \label{7.4.3}$
$\ce{E^{2+}(g) \rightarrow E^{3+}(g) +e^-} \;\;\ I_3=\text{3rd ionization energy} \label{7.4.4}$
Values for the ionization energies of $Li$ and $Be$ listed in Table $1$ show that successive ionization energies for an element increase as they go; that is, it takes more energy to remove the second electron from an atom than the first, and so forth. There are two reasons for this trend. First, the second electron is being removed from a positively charged species rather than a neutral one, so in accordance with Coulomb’s law, more energy is required. Second, removing the first electron reduces the repulsive forces among the remaining electrons, so the attraction of the remaining electrons to the nucleus is stronger.
Successive ionization energies for an element increase.
Table $1$: Ionization Energies (in kJ/mol) for Removing Successive Electrons from Li and Be. Source: Data from CRC Handbook of Chemistry and Physics (2004).
Reaction Electronic Transition $I$ Reaction Electronic Transition $I$
$\ce{Li (g)\rightarrow Li^+ (g) + e^-}$ $1s^22s^1 \rightarrow 1s^2$ I1 = 520.2 $\ce{Be (g) \rightarrow Be^+(g) + e^-}$ $1s^22s^2 \rightarrow 1s^22s^1$ I1 = 899.5
$\ce{Li^+(g) \rightarrow Li^{2+}(g) +e^-}$ $1s^2 \rightarrow 1s^1$ I2 = 7298.2 $\ce{Be^+(g) \rightarrow Be^{2+}(g) + e^-}$ $1s^22s^1 \rightarrow 1s^2$ I2 = 1757.1
$\ce{Li^{2+} (g) \rightarrow Li^{3+}(g) + e^-}$ $1s^1 \rightarrow 1s^0$ I3 = 11,815.0 $\ce{Be^{2+}(g) \rightarrow Be^{3+}(g) + e^-}$ $1s^2 \rightarrow 1s^1$ I3 = 14,848.8
$\ce{Be^{3+}(g) \rightarrow Be^{4+}(g) + e^-}$ $1s^1 \rightarrow 1s^0$ I4 = 21,006.6
The increase in successive ionization energies, however, is not linear, but increases drastically when removing electrons in lower $n$ orbitals closer to the nucleus. The most important consequence of the values listed in Table $1$ is that the chemistry of $\ce{Li}$ is dominated by the $\ce{Li^+}$ ion, while the chemistry of $\ce{Be}$ is dominated by the +2 oxidation state. The energy required to remove the second electron from $\ce{Li}$:
$\ce{Li^+(g) \rightarrow Li^{2+}(g) + e^-} \label{7.4.5}$
is more than 10 times greater than the energy needed to remove the first electron. Similarly, the energy required to remove the third electron from $\ce{Be}$:
$\ce{Be^{2+}(g) \rightarrow Be^{3+}(g) + e^-} \label{7.4.6}$
is about 15 times greater than the energy needed to remove the first electron and around 8 times greater than the energy required to remove the second electron. Both $\ce{Li^+}$ and $\ce{Be^{2+}}$ have 1s2 closed-shell configurations, and much more energy is required to remove an electron from the 1s2 core than from the 2s valence orbital of the same element. The chemical consequences are enormous: lithium (and all the alkali metals) forms compounds with the 1+ ion but not the 2+ or 3+ ions. Similarly, beryllium (and all the alkaline earth metals) forms compounds with the 2+ ion but not the 3+ or 4+ ions. The energy required to remove electrons from a filled core is prohibitively large and simply cannot be achieved in normal chemical reactions.
The energy required to remove electrons from a filled core is prohibitively large under normal reaction conditions.
Ionization Energy: Ionization Energy, YouTube(opens in new window) [youtu.be] (opens in new window)
Ionization Energies of s- and p-Block Elements
Ionization energies of the elements in the third row of the periodic table exhibit the same pattern as those of $Li$ and $Be$ (Table $2$): successive ionization energies increase steadily as electrons are removed from the valence orbitals (3s or 3p, in this case), followed by an especially large increase in ionization energy when electrons are removed from filled core levels as indicated by the bold diagonal line in Table $2$. Thus in the third row of the periodic table, the largest increase in ionization energy corresponds to removing the fourth electron from $Al$, the fifth electron from Si, and so forth—that is, removing an electron from an ion that has the valence electron configuration of the preceding noble gas. This pattern explains why the chemistry of the elements normally involves only valence electrons. Too much energy is required to either remove or share the inner electrons.
Table $2$: Successive Ionization Energies (in kJ/mol) for the Elements in the Third Row of the Periodic Table.Source: Data from CRC Handbook of Chemistry and Physics (2004).
Element $I_1$ $I_2$ $I_3$ $I_4$ $I_5$ $I_6$ $I_7$
*Inner-shell electron
Na 495.8 4562.4*
Mg 737.7 1450.7 7732.7
Al 577.4.4 1816.7 2744.8 11,577.4.4
Si 786.5 1577.1 3231.6 4355.5 16,090.6
P 1011.8 1907.4.4 2914.1 4963.6 6274.0 21,267.4.3
S 999.6 2251.8 3357 4556.2 7004.3 8495.8 27,107.4.3
Cl 1251.2 2297.7 3822 5158.6 6540 9362 11,018.2
Ar 1520.6 2665.9 3931 5771 7238 8781.0 11,995.3
Example $1$: Highest Fourth Ionization Energy
From their locations in the periodic table, predict which of these elements has the highest fourth ionization energy: B, C, or N.
Given: three elements
Asked for: element with highest fourth ionization energy
Strategy:
1. List the electron configuration of each element.
2. Determine whether electrons are being removed from a filled or partially filled valence shell. Predict which element has the highest fourth ionization energy, recognizing that the highest energy corresponds to the removal of electrons from a filled electron core.
Solution:
A These elements all lie in the second row of the periodic table and have the following electron configurations:
• B: [He]2s22p1
• C: [He]2s22p2
• N: [He]2s22p3
B The fourth ionization energy of an element ($I_4$) is defined as the energy required to remove the fourth electron:
$E^{3+}_{(g)} \rightarrow E^{4+}_{(g)} + e^- \nonumber$
Because carbon and nitrogen have four and five valence electrons, respectively, their fourth ionization energies correspond to removing an electron from a partially filled valence shell. The fourth ionization energy for boron, however, corresponds to removing an electron from the filled 1s2 subshell. This should require much more energy. The actual values are as follows: B, 25,026 kJ/mol; C, 6223 kJ/mol; and N, 7475 kJ/mol.
Exercise $1$: Lowest Second Ionization Energy
From their locations in the periodic table, predict which of these elements has the lowest second ionization energy: Sr, Rb, or Ar.
Answer
$\ce{Sr}$
The first column of data in Table $2$ shows that first ionization energies tend to increase across the third row of the periodic table. This is because the valence electrons do not screen each other very well, allowing the effective nuclear charge to increase steadily across the row. The valence electrons are therefore attracted more strongly to the nucleus, so atomic sizes decrease and ionization energies increase. These effects represent two sides of the same coin: stronger electrostatic interactions between the electrons and the nucleus further increase the energy required to remove the electrons.
However, the first ionization energy decreases at Al ([Ne]3s23p1) and at S ([Ne]3s23p4). The electron configurations of these "exceptions" provide the answer why. The electrons in aluminum’s filled 3s2 subshell are better at screening the 3p1 electron than they are at screening each other from the nuclear charge, so the s electrons penetrate closer to the nucleus than the p electron does and the p electron is more easily removed. The decrease at S occurs because the two electrons in the same p orbital repel each other. This makes the S atom slightly less stable than would otherwise be expected, as is true of all the group 16 elements.
The first ionization energies of the elements in the first six rows of the periodic table are plotted in Figure $1$ and are presented numerically and graphically in Figure $2$. These figures illustrate three important trends:
1. The changes seen in the second (Li to Ne), fourth (K to Kr), fifth (Rb to Xe), and sixth (Cs to Rn) rows of the s and p blocks follow a pattern similar to the pattern described for the third row of the periodic table. The transition metals are included in the fourth, fifth, and sixth rows, however, and the lanthanides are included in the sixth row. The first ionization energies of the transition metals are somewhat similar to one another, as are those of the lanthanides. Ionization energies increase from left to right across each row, with discrepancies occurring at ns2np1 (group 13), ns2np4 (group 16), and ns2(n − 1)d10 (group 12).
2. First ionization energies generally decrease down a column. Although the principal quantum number n increases down a column, filled inner shells are effective at screening the valence electrons, so there is a relatively small increase in the effective nuclear charge. Consequently, the atoms become larger as they acquire electrons. Valence electrons that are farther from the nucleus are less tightly bound, making them easier to remove, which causes ionization energies to decrease. A larger radius typically corresponds to a lower ionization energy.
3. Because of the first two trends, the elements that form positive ions most easily (have the lowest ionization energies) lie in the lower left corner of the periodic table, whereas those that are hardest to ionize lie in the upper right corner of the periodic table. Consequently, ionization energies generally increase diagonally from lower left (Cs) to upper right (He).
Generally, $I_1$ increases diagonally from the lower left of the periodic table to the upper right.
Gallium (Ga), which is the first element following the first row of transition metals, has the following electron configuration: [Ar]4s23d104p1. Its first ionization energy is significantly lower than that of the immediately preceding element, zinc, because the filled 3d10 subshell of gallium lies inside the 4p subshell, shielding the single 4p electron from the nucleus. Experiments have revealed something of even greater interest: the second and third electrons that are removed when gallium is ionized come from the 4s2 orbital, not the 3d10 subshell. The chemistry of gallium is dominated by the resulting Ga3+ ion, with its [Ar]3d10 electron configuration. This and similar electron configurations are particularly stable and are often encountered in the heavier p-block elements. They are sometimes referred to as pseudo noble gas configurations. In fact, for elements that exhibit these configurations, no chemical compounds are known in which electrons are removed from the (n − 1)d10 filled subshell.
Ionization Energies of Transition Metals & Lanthanides
As we noted, the first ionization energies of the transition metals and the lanthanides change very little across each row. Differences in their second and third ionization energies are also rather small, in sharp contrast to the pattern seen with the s- and p-block elements. The reason for these similarities is that the transition metals and the lanthanides form cations by losing the ns electrons before the (n − 1)d or (n − 2)f electrons, respectively. This means that transition metal cations have (n − 1)dn valence electron configurations, and lanthanide cations have (n − 2)fn valence electron configurations. Because the (n − 1)d and (n − 2)f shells are closer to the nucleus than the ns shell, the (n − 1)d and (n − 2)f electrons screen the ns electrons quite effectively, reducing the effective nuclear charge felt by the ns electrons. As Z increases, the increasing positive charge is largely canceled by the electrons added to the (n − 1)d or (n − 2)f orbitals.
That the ns electrons are removed before the (n − 1)d or (n − 2)f electrons may surprise you because the orbitals were filled in the reverse order. In fact, the ns, the (n − 1)d, and the (n − 2)f orbitals are so close to one another in energy, and interpenetrate one another so extensively, that very small changes in the effective nuclear charge can change the order of their energy levels. As the d orbitals are filled, the effective nuclear charge causes the 3d orbitals to be slightly lower in energy than the 4s orbitals. The [Ar]3d2 electron configuration of Ti2+ tells us that the 4s electrons of titanium are lost before the 3d electrons; this is confirmed by experiment. A similar pattern is seen with the lanthanides, producing cations with an (n − 2)fn valence electron configuration.
Because their first, second, and third ionization energies change so little across a row, these elements have important horizontal similarities in chemical properties in addition to the expected vertical similarities. For example, all the first-row transition metals except scandium form stable compounds as M2+ ions, whereas the lanthanides primarily form compounds in which they exist as M3+ ions.
Example $2$: Lowest First Ionization Energy
Use their locations in the periodic table to predict which element has the lowest first ionization energy: Ca, K, Mg, Na, Rb, or Sr.
Given: six elements
Asked for: element with lowest first ionization energy
Strategy:
Locate the elements in the periodic table. Based on trends in ionization energies across a row and down a column, identify the element with the lowest first ionization energy.
Solution:
These six elements form a rectangle in the two far-left columns of the periodic table. Because we know that ionization energies increase from left to right in a row and from bottom to top of a column, we can predict that the element at the bottom left of the rectangle will have the lowest first ionization energy: Rb.
Exercise $2$: Highest First Ionization Energy
Use their locations in the periodic table to predict which element has the highest first ionization energy: As, Bi, Ge, Pb, Sb, or Sn.
Answer
$\ce{As}$
Summary
The tendency of an element to lose electrons is one of the most important factors in determining the kind of compounds it forms. Periodic behavior is most evident for ionization energy (I), the energy required to remove an electron from a gaseous atom. The energy required to remove successive electrons from an atom increases steadily, with a substantial increase occurring with the removal of an electron from a filled inner shell. Consequently, only valence electrons can be removed in chemical reactions, leaving the filled inner shell intact. Ionization energies explain the common oxidation states observed for the elements. Ionization energies increase diagonally from the lower left of the periodic table to the upper right. Minor deviations from this trend can be explained in terms of particularly stable electronic configurations, called pseudo noble gas configurations, in either the parent atom or the resulting ion. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/07%3A_Periodic_Properties_of_the_Elements/7.04%3A_Ionization_Energy.txt |
Learning Objectives
• To master the concept of electron affinity as a measure of the energy required to add an electron to an atom or ion.
• To recognize the inverse relationship of ionization energies and electron affinities
The electron affinity ($EA$) of an element $E$ is defined as the energy change that occurs when an electron is added to a gaseous atom or ion:
$E_{(g)}+e^- \rightarrow E^-_{(g)} \;\;\; \text{energy change=}EA \label{7.5.1}$
Unlike ionization energies, which are always positive for a neutral atom because energy is required to remove an electron, electron affinities can be negative (energy is released when an electron is added), positive (energy must be added to the system to produce an anion), or zero (the process is energetically neutral). This sign convention is consistent with a negative value corresponded to the energy change for an exothermic process, which is one in which heat is released (Figure $1$).
The chlorine atom has the most negative electron affinity of any element, which means that more energy is released when an electron is added to a gaseous chlorine atom than to an atom of any other element:
$\ce{ Cl(g) + e^- \rightarrow Cl^- (g)} \;\;\; EA=-346\; kJ/mol \label{7.5.2}$
In contrast, beryllium does not form a stable anion, so its effective electron affinity is
$\ce{ Be(g) + e^- \rightarrow Be^- (g)} \;\;\; EA \ge 0 \label{7.5.3}$
Nitrogen is unique in that it has an electron affinity of approximately zero. Adding an electron neither releases nor requires a significant amount of energy:
$\ce{ N(g) + e^- \rightarrow N^- (g)} \;\;\; EA \approx 0 \label{7.5.4}$
Generally, electron affinities become more negative across a row of the periodic table.
In general, electron affinities of the main-group elements become less negative as we proceed down a column. This is because as n increases, the extra electrons enter orbitals that are increasingly far from the nucleus.
Atoms with the largest radii, which have the lowest ionization energies (affinity for their own valence electrons), also have the lowest affinity for an added electron. There are, however, two major exceptions to this trend:
1. The electron affinities of elements B through F in the second row of the periodic table are less negative than those of the elements immediately below them in the third row. Apparently, the increased electron–electron repulsions experienced by electrons confined to the relatively small 2p orbitals overcome the increased electron–nucleus attraction at short nuclear distances. Fluorine, therefore, has a lower affinity for an added electron than does chlorine. Consequently, the elements of the third row (n = 3) have the most negative electron affinities. Farther down a column, the attraction for an added electron decreases because the electron is entering an orbital more distant from the nucleus. Electron–electron repulsions also decrease because the valence electrons occupy a greater volume of space. These effects tend to cancel one another, so the changes in electron affinity within a family are much smaller than the changes in ionization energy.
2. The electron affinities of the alkaline earth metals become more negative from Be to Ba. The energy separation between the filled ns2 and the empty np subshells decreases with increasing n, so that formation of an anion from the heavier elements becomes energetically more favorable.
The equations for second and higher electron affinities are analogous to those for second and higher ionization energies:
$E_{(g)} + e^- \rightarrow E^-_{(g)} \;\;\; \text{energy change=}EA_1 \label{7.5.5}$
$E^-_{(g)} + e^- \rightarrow E^{2-}_{(g)} \;\;\; \text{energy change=}EA_2 \label{7.5.6}$
As we have seen, the first electron affinity can be greater than or equal to zero or negative, depending on the electron configuration of the atom. In contrast, the second electron affinity is always positive because the increased electron–electron repulsions in a dianion are far greater than the attraction of the nucleus for the extra electrons. For example, the first electron affinity of oxygen is −141 kJ/mol, but the second electron affinity is +744 kJ/mol:
$O_{(g)} + e^- \rightarrow O^-_{(g)} \;\;\; EA_1=-141 \;kJ/mol \label{7.5.7}$
$O^-_{(g)} + e^- \rightarrow O^{2-}_{(g)} \;\;\; EA_2=+744 \;kJ/mol \label{7.5.8}$
Thus the formation of a gaseous oxide ($O^{2−}$) ion is energetically quite unfavorable (estimated by adding both steps):
$O_{(g)} + 2e^- \rightarrow O^{2-}_{(g)} \;\;\; EA=+603 \;kJ/mol \label{7.5.9}$
Similarly, the formation of all common dianions (such as $S^{2−}$) or trianions (such as $P^{3−}$) is energetically unfavorable in the gas phase.
While first electron affinities can be negative, positive, or zero, second electron affinities are always positive.
Electron Affinity: Electron Affinity, YouTube(opens in new window) [youtu.be] (opens in new window)
If energy is required to form both positively charged cations and monatomic polyanions, why do ionic compounds such as $MgO$, $Na_2S$, and $Na_3P$ form at all? The key factor in the formation of stable ionic compounds is the favorable electrostatic interactions between the cations and the anions in the crystalline salt.
Example $1$: Contrasting Electron Affinities of Sb, Se, and Te
Based on their positions in the periodic table, which of Sb, Se, or Te would you predict to have the most negative electron affinity?
Given: three elements
Asked for: element with most negative electron affinity
Strategy:
1. Locate the elements in the periodic table. Use the trends in electron affinities going down a column for elements in the same group. Similarly, use the trends in electron affinities from left to right for elements in the same row.
2. Place the elements in order, listing the element with the most negative electron affinity first.
Solution:
A We know that electron affinities become less negative going down a column (except for the anomalously low electron affinities of the elements of the second row), so we can predict that the electron affinity of Se is more negative than that of Te. We also know that electron affinities become more negative from left to right across a row, and that the group 15 elements tend to have values that are less negative than expected. Because Sb is located to the left of Te and belongs to group 15, we predict that the electron affinity of Te is more negative than that of Sb. The overall order is Se < Te < Sb, so Se has the most negative electron affinity among the three elements.
Exercise $1$: Contrasting Electron Affinities of Rb, Sr, and Xe
Based on their positions in the periodic table, which of Rb, Sr, or Xe would you predict to most likely form a gaseous anion?
Answer
Rb
Summary
The electron affinity (EA) of an element is the energy change that occurs when an electron is added to a gaseous atom to give an anion. In general, elements with the most negative electron affinities (the highest affinity for an added electron) are those with the smallest size and highest ionization energies and are located in the upper right corner of the periodic table. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/07%3A_Periodic_Properties_of_the_Elements/7.05%3A_Electron_Affinities.txt |
Learning Objectives
• To understand the basic properties separating Metals from Nonmetals and Metalloids
An element is the simplest form of matter that cannot be split into simpler substances or built from simpler substances by any ordinary chemical or physical method. There are 118 elements known to us, out of which 92 are naturally occurring, while the rest have been prepared artificially. Elements are further classified into metals, non-metals, and metalloids based on their properties, which are correlated with their placement in the periodic table.
Metallic Elements Nonmetallic elements
Table $1$: Characteristic properties of metallic and non-metallic elements:
Distinguishing luster (shine) Non-lustrous, various colors
Malleable and ductile (flexible) as solids Brittle, hard or soft
Conduct heat and electricity Poor conductors
Metallic oxides are basic, ionic Nonmetallic oxides are acidic, covalent
Form cations in aqueous solution Form anions, oxyanions in aqueous solution
Metals
With the exception of hydrogen, all elements that form positive ions by losing electrons during chemical reactions are called metals. Thus metals are electropositive elements with relatively low ionization energies. They are characterized by bright luster, hardness, ability to resonate sound and are excellent conductors of heat and electricity. Metals are solids under normal conditions except for Mercury.
Physical Properties of Metals
Metals are lustrous, malleable, ductile, good conductors of heat and electricity. Other properties include:
• State: Metals are solids at room temperature with the exception of mercury, which is liquid at room temperature (Gallium is liquid on hot days).
• Luster: Metals have the quality of reflecting light from their surface and can be polished e.g., gold, silver and copper.
• Malleability: Metals have the ability to withstand hammering and can be made into thin sheets known as foils. For example, a sugar cube sized chunk of gold can be pounded into a thin sheet that will cover a football field.
• Ductility: Metals can be drawn into wires. For example, 100 g of silver can be drawn into a thin wire about 200 meters long.
• Hardness: All metals are hard except sodium and potassium, which are soft and can be cut with a knife.
• Valency: Metals typically have 1 to 3 electrons in the outermost shell of their atoms.
• Conduction: Metals are good conductors because they have free electrons. Silver and copper are the two best conductors of heat and electricity. Lead is the poorest conductor of heat. Bismuth, mercury and iron are also poor conductors
• Density: Metals have high density and are very heavy. Iridium and osmium have the highest densities whereas lithium has the lowest density.
• Melting and Boiling Points: Metals have high melting and boiling points. Tungsten has the highest melting and boiling points whereas mercury has the lowest. Sodium and potassium also have low melting points.
Chemical Properties of Metals
Metals are electropositive elements that generally form basic or amphoteric oxides with oxygen. Other chemical properties include:
• Electropositive Character: Metals tend to have low ionization energies, and typically lose electrons (i.e. are oxidized) when they undergo chemical reactions They normally do not accept electrons. For example:
• Alkali metals are always 1+ (lose the electron in s subshell)
• Alkaline earth metals are always 2+ (lose both electrons in s subshell)
• Transition metal ions do not follow an obvious pattern, 2+ is common (lose both electrons in s subshell), and 1+ and 3+ are also observed
$\ce{Na^0 \rightarrow Na^+ + e^{-}} \label{1.1}$
$\ce{Mg^0 \rightarrow Mg^{2+} + 2e^{-}} \label{1.2}$
$\ce{Al^0 \rightarrow Al^{3+} + 3e^{-}} \label{1.3}$
Compounds of metals with non-metals tend to be ionic in nature. Most metal oxides are basic oxides and dissolve in water to form metal hydroxides:
$\ce{Na2O(s) + H2O(l) \rightarrow 2NaOH(aq)}\label{1.4}$
$\ce{CaO(s) + H2O(l) \rightarrow Ca(OH)2(aq)} \label{1.5}$
Metal oxides exhibit their basic chemical nature by reacting with acids to form metal salts and water:
$\ce{MgO(s) + HCl(aq) \rightarrow MgCl2(aq) + H2O(l)} \label{1.6}$
$\ce{NiO(s) + H2SO4(aq) \rightarrow NiSO4(aq) + H2O(l)} \label{1.7}$
Example $1$
What is the chemical formula for aluminum oxide?
Solution
Al has a 3+ charge, the oxide ion is $O^{2-}$, thus $Al_2O_3$.
Example $2$
Would you expect it to be solid, liquid or gas at room temperature?
Solutions
Oxides of metals are characteristically solid at room temperature
Example $3$
Write the balanced chemical equation for the reaction of aluminum oxide with nitric acid:
Metal oxide + acid -> salt + water
$\ce{Al2O3(s) + 6HNO3(aq) \rightarrow 2Al(NO3)3(aq) + 3H2O(l)} \nonumber$
Nonmetals
Elements that tend to gain electrons to form anions during chemical reactions are called non-metals. These are electronegative elements with high ionization energies. They are non-lustrous, brittle and poor conductors of heat and electricity (except graphite). Non-metals can be gases, liquids or solids.
Physical Properties of Nonmetals
• Physical State: Most of the non-metals exist in two of the three states of matter at room temperature: gases (oxygen) and solids (carbon). Only bromine exists as a liquid at room temperature.
• Non-Malleable and Ductile: Non-metals are very brittle, and cannot be rolled into wires or pounded into sheets.
• Conduction: They are poor conductors of heat and electricity.
• Luster: These have no metallic luster and do not reflect light.
• Melting and Boiling Points: The melting points of non-metals are generally lower than metals, but are highly variable.
• Seven non-metals exist under standard conditions as diatomic molecules: $\ce{H2(g)}$, $\ce{N2(g)}$, $\ce{O2(g)}$, $\ce{F2(g)}$, $\ce{Cl2(g)}$, $\ce{Br2(l)}$, $\ce{I2(s)}$.
Chemical Properties of Nonmetals
Non-metals have a tendency to gain or share electrons with other atoms. They are electronegative in character. Nonmetals, when reacting with metals, tend to gain electrons (typically attaining noble gas electron configuration) and become anions:
$\ce{3Br2(l) + 2Al(s) \rightarrow 2AlBr3(s)} \nonumber$
Compounds composed entirely of nonmetals are covalent substances. They generally form acidic or neutral oxides with oxygen that that dissolve in water to form acids:
$\ce{CO2(g) + H2O(l)} \rightarrow \underset{\text{carbonic acid}}{\ce{H2CO3(aq)}} \nonumber$
As you may know, carbonated water is slightly acidic (carbonic acid).
Nonmetal oxides can combine with bases to form salts.
$\ce{CO2(g) + 2NaOH(aq) \rightarrow Na2CO3(aq) + H2O(l)} \nonumber$
Metalloids
Metalloids have properties intermediate between the metals and nonmetals. Metalloids are useful in the semiconductor industry. Metalloids are all solid at room temperature. They can form alloys with other metals. Some metalloids, such as silicon and germanium, can act as electrical conductors under the right conditions, thus they are called semiconductors. Silicon for example appears lustrous, but is not malleable nor ductile (it is brittle - a characteristic of some nonmetals). It is a much poorer conductor of heat and electricity than the metals. The physical properties of metalloids tend to be metallic, but their chemical properties tend to be non-metallic. The oxidation number of an element in this group can range from +5 to -2, depending on the group in which it is located.
Table $2$: Elements categorized into metals, non-metals and metalloids.
Metals Non-metals Metalloids
Gold Oxygen Silicon
Silver Carbon Boron
Copper Hydrogen Arsenic
Iron Nitrogen Antimony
Mercury Sulfur Germanium
Zinc Phosphorus
Trends in Metallic and Nonmetallic Character
Metallic character is strongest for the elements in the leftmost part of the periodic table, and tends to decrease as we move to the right in any period (nonmetallic character increases with increasing electronegativity and ionization energy values). Within any group of elements (columns), the metallic character increases from top to bottom (the electronegativity and ionization energy values generally decrease as we move down a group). This general trend is not necessarily observed with the transition metals. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/07%3A_Periodic_Properties_of_the_Elements/7.06%3A_Metals_Nonmetals_and_Metalloids.txt |
The elements within the same group of the periodic table tend to exhibit similar physical and chemical properties. Four major factors affect reactivity of metals: nuclear charge, atomic radius, shielding effect and sublevel arrangement (of electrons). Metal reactivity relates to ability to lose electrons (oxidize), form basic hydroxides, form ionic compounds with non-metals. In general, the bigger the atom, the greater the ability to lose electrons. The greater the shielding, the greater the ability to lose electrons. Therefore, metallic character increases going down the table, and decreases going across -- so the most active metal is towards the left and down.
Group 1: The Alkali Metals
The word "alkali" is derived from an Arabic word meaning "ashes". Many sodium and potassium compounds were isolated from wood ashes ($\ce{Na2CO3}$ and $\ce{K2CO3}$ are still occasionally referred to as "soda ash" and "potash"). In the alkali group, as we go down the group we have elements Lithium (Li), Sodium (Na), Potassium (K), Rubidium (Rb), Cesium (Cs) and Francium (Fr). Several physical properties of these elements are compared in Table $1$. These elements have all only one electron in their outermost shells. All the elements show metallic properties and have valence +1, hence they give up electron easily.
Table $1$: General Properties of Group I Metals
Element Electronic Configuration Melting Point (°C) Density (g/cm3) Atomic Radius Ionization Energy (kJ/mol)
Lithium $[He]2s^1$ 181 0.53 1.52 520
Sodium $[Ne]3s^1$ 98 0.97 1.86 496
Potassium $[Ar]4s^1$ 63 0.86 2.27 419
Rubidium $[Kr]5s^1$ 39 1.53 2.47 403
Cesium $[Xe]6s^1$ 28 1.88 2.65 376
As we move down the group (from Li to Fr), the following trends are observed (Table $1$):
• All have a single electron in an 's' valence orbital
• The melting point decreases
• The density increases
• The atomic radius increases
• The ionization energy decreases (first ionization energy)
The alkali metals have the lowest $I_1$ values of the elements
This represents the relative ease with which the lone electron in the outer 's' orbital can be removed.
The alkali metals are very reactive, readily losing 1 electron to form an ion with a 1+ charge:
$M \rightarrow M^+ + e- \nonumber$
Due to this reactivity, the alkali metals are found in nature only as compounds. The alkali metals combine directly with most nonmetals:
• React with hydrogen to form solid hydrides
$2M_{(s)} + H_{2(g)} \rightarrow 2MH(s) \nonumber$
(Note: hydrogen is present in the metal hydride as the hydride H- ion)
• React with sulfur to form solid sulfides
$2M_{(s)} + S_{(s)} \rightarrow M_2S_{(s)} \nonumber$
React with chlorine to form solid chlorides
$2M_{(s)} + Cl_{2(g)} \rightarrow 2MCl_{(s)} \nonumber$
Alkali metals react with water to produce hydrogen gas and alkali metal hydroxides; this is a very exothermic reaction (Figure $1$).
$2M_{(s)} + 2H_2O_{(l)} \rightarrow 2MOH_{(aq)} + H_{2(g)} \nonumber$
The reaction between alkali metals and oxygen is more complex:
• A common reaction is to form metal oxides which contain the O2- ion
$4Li_{(s)} + O_{2 (g)} \rightarrow \underbrace{2Li_2O_{(s)}}_{\text{lithium oxide}} \nonumber$
Other alkali metals can form metal peroxides (contains O22- ion)
$2Na(s) + O_{2 (g)} \rightarrow \underbrace{Na_2O_{2(s)}}_{\text{sodium peroxide}} \nonumber$
K, Rb and Cs can also form superoxides (O2- ion)
$K(s) + O_{2 (g)} \rightarrow \underbrace{KO_{2(s)}}_{\text{potassium superoxide}} \nonumber$
Colors via Absorption
The color of a chemical is produced when a valence electron in an atom is excited from one energy level to another by visible radiation. In this case, the particular frequency of light that excites the electron is absorbed. Thus, the remaining light that you see is white light devoid of one or more wavelengths (thus appearing colored). Alkali metals, having lost their outermost electrons, have no electrons that can be excited by visible radiation. Alkali metal salts and their aqueous solution are colorless unless they contain a colored anion.
Colors via Emission
When alkali metals are placed in a flame the ions are reduced (gain an electron) in the lower part of the flame. The electron is excited (jumps to a higher orbital) by the high temperature of the flame. When the excited electron falls back down to a lower orbital a photon is released. The transition of the valence electron of sodium from the 3p down to the 3s subshell results in release of a photon with a wavelength of 589 nm (yellow)
Flame colors:
• Lithium: crimson red
• Sodium: yellow
• Potassium: lilac
Group 2: The Alkaline Earth Metals
Compared with the alkali metals, the alkaline earth metals are typically harder, more dense, melt at a higher temperature. The first ionization energies ($I_1$) of the alkaline earth metals are not as low as the alkali metals. The alkaline earth metals are therefore less reactive than the alkali metals (Be and Mg are the least reactive of the alkaline earth metals). Several physical properties of these elements are compared in Table $2$.
Table $2$: General Properties of Group 2 Metals
Element Electronic Configuration Melting Point (°C) Density (g/cm3) Atomic Radius Ionization Energy (kJ/mol)
Beryllium $[He]2s^2$ 1278 1.85 1.52 899
Magnesium $[Ne]3s^2$ 649 1.74 1.60 738
Calcium $[Ar]4s^2$ 839 1.54 1.97 590
Strontium $[Kr]5s^2$ 769 2.54 2.15 549
Barium $[Xe]6s^2$ 725 3.51 2.17 503
Calcium, and elements below it, react readily with water at room temperature:
$Ca_{(s)} + 2H_2O_{(l)} \rightarrow Ca(OH)_{2(aq)} + H_{2(g)} \nonumber$
The tendency of the alkaline earths to lose their two valence electrons is demonstrated in the reactivity of Mg towards chlorine gas and oxygen:
$Mg_{(s)} + Cl_{2(g)} \rightarrow MgCl_{2(s)} \nonumber$
$2Mg_{(s)} + O_{2(g)} \rightarrow 2MgO_{(s)} \nonumber$
The 2+ ions of the alkaline earth metals have a noble gas like electron configuration and are thus form colorless or white compounds (unless the anion is itself colored). Flame colors:
• Calcium: brick red
• Strontium: crimson red
• Barium: green | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/07%3A_Periodic_Properties_of_the_Elements/7.07%3A_Group_Trends_for_the_Active_Metals.txt |
Learning Objectives
• To gain a descriptive understanding of the chemical properties of Hydrogen, the group 16, 17 and 18 elements.
Non-metallic character is the ability to be reduced (be an oxidizing agent), form acidic hydroxides, form covalent compounds with non-metals. These characteristics increase with a larger nuclear charge and smaller radius, with no increase in shielding. The most active non-metal would be the one farthest up and to the right -- not including the noble gases (non-reactive.)
Hydrogen
Hydrogen has a 1s1 electron configuration and is placed above the alkali metal group. Hydrogen is a non-metal, which occurs as a gas (H2) under normal conditions.
• Its ionization energy is considerably higher (due to lack of shielding, and thus higher $Z_{eff}$) than the rest of the Group 1 metals and is more like the nonmetals
• Hydrogen generally reacts with other nonmetals to form molecular compounds (typically highly exothermic)
• Hydrogen reacts with active metals to form metal hydrides which contain the H- hydride ion:
$2Na_{(s)} + H_{2(g)} \rightarrow 2NaH_{(s)} \label{7.8.1}$
• Hydrogen can also lose an electron to yield the aqueous $H^+_{(aq)}$ hydronium ion.
Group 16: The Oxygen Family
As we proceed down group 16 the elements become more metallic in nature:
• Oxygen is a gas, the rest are solids
• Oxygen, sulfur and selenium are nonmetals
• Tellurium is a metalloid with some metal properties
• Polonium is a metal
Oxygen can be found in two molecular forms, O2 and O3 (ozone). These two forms of oxygen are called allotropes (different forms of the same element in the same state)
$3O_{2(g)} \rightarrow 2O_{3(g)}\;\;\; \Delta H = 284.6\; kJ / mol \label{7.8.2}$
the reaction is endothermic, thus ozone is less stable that O2
Oxygen has a great tendency to attract electrons from other elements (i.e. to "oxidize" them)
• Oxygen in combination with metals is almost always present as the O2- ion (which has noble gas electronic configuration and is particularly stable)
• Two other oxygen anions are observed: peroxide (O22-) and superoxide (O2-)
Sulfur
Sulfur also exists in several allotropic forms, the most common stable allotrope is the yellow solid S8 (an 8 member ring of sulfur atoms). Like oxygen, sulfur has a tendency to gain electrons from other elements, and to form sulfides (which contain the S2- ion). This is particular true for the active metals:
$16Na_{(s)} + S_{8(s)} \rightarrow 8Na_2S_{(s)}\label{7.8.3}$
Note: most sulfur in nature is present as a metal-sulfur compound. Sulfur chemistry is more complex than that of oxygen.
Group 17: The Halogens
"Halogen" is derived from the Greek meaning "salt formers"
• Astatine is radioactive and rare, and some of its properties are unknown
• All the halogens are nonmetals
• Each element consists of diatomic molecules under standard conditions
Colors of diatomic halogens: (not flame colors)
• Fluorine: pale yellow
• Chlorine: yellow green
• Bromine: reddish brown
• Iodine: violet vapor
The halogens have some of the most negative electron affinities (i.e. large exothermic process in gaining an electron from another element)
$X_2 + 2e^- \rightarrow 2X^-\label{7.8.4}$
• Fluorine and chlorine are the most reactive halogens (highest electron affinities). Fluorine will remove electrons from almost any substance (including several of the noble gases from Group 18).
Note
The chemistry of the halogens is dominated by their tendency to gain electrons from other elements (forming a halide ion)
In 1992, 22.3 billion pounds of chlorine was produced. Both chlorine and sodium can be produced by electrolysis of molten sodium chloride (table salt). The electricity is used to strip electrons from chloride ions and transfer them to sodium ions to produce chlorine gas and solid sodium metal
Chlorine reacts slowly with water to produce hydrochloric acid and hypochlorous acid:
$Cl_{2(g)} + H_2O_{(l)} \rightarrow HCl_{(aq)} + HOCl_{(aq)}\label{7.8}.5$
Hypochlorous acid is a disinfectant, thus chlorine is a useful addition to swimming pool water
The halogens react with most metals to form ionic halides:
$Cl_{2(g)} + 2Na_{(s)} \rightarrow 2NaCl_{(s)}\label{7.8.6}$
Group 18: The Noble Gases
• Nonmetals
• Gases at room temperature
• monoatomic
• completely filled 's' and 'p' subshells
• large first ionization energy, but this decreases somewhat as we move down the group
Rn is highly radioactive and some of its properties are unknown
They are exceptionally unreactive. It was reasoned that if any of these were reactive, they would most likely be Rn, Xe or Kr where the first ionization energies were lower.
Note
In order to react, they would have to be combined with an element which had a high tendency to remove electrons from other atoms. Such as fluorine.
Compounds of noble gases to date:
$XeF_2$ $XeF_4$ $XeF_6$
only one compound with Kr has been made
$KrF_2$
No compounds observed with He, Ne, or Ar; they are truly inert gases. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/07%3A_Periodic_Properties_of_the_Elements/7.08%3A_Group_Trends_for_Selected_Nonmetals.txt |
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
7.1: Development of the Periodic Table
Conceptual Problems
1. Johannes Dobereiner is credited with developing the concept of chemical triads. Which of the group 15 elements would you expect to compose a triad? Would you expect B, Al, and Ga to act as a triad? Justify your answers.
2. Despite the fact that Dobereiner, Newlands, Meyer, and Mendeleev all contributed to the development of the modern periodic table, Mendeleev is credited with its origin. Why was Mendeleev’s periodic table accepted so rapidly?
3. How did Moseley’s contribution to the development of the periodic table explain the location of the noble gases?
4. The eka- naming scheme devised by Mendeleev was used to describe undiscovered elements.
1. Use this naming method to predict the atomic number of eka-mercury, eka-astatine, eka-thallium, and eka-hafnium.
2. Using the eka-prefix, identify the elements with these atomic numbers: 79, 40, 51, 117, and 121.
Numerical Problem
1. Based on the data given, complete the table.
Species Molar Mass (g/mol) Density (g/cm3) Molar Volume (cm3/mol)
A 40.078 25.85
B 39.09 0.856
C 32.065 16.35
D 1.823 16.98
E 26.98 9.992
F 22.98 0.968
Plot molar volume versus molar mass for these substances. According to Meyer, which would be considered metals and which would be considered nonmetals?
Numerical Answer
1. Species Molar Mass (g/mol) Density (g/cm3) Molar Volume (cm3/mol)
A 40.078 1.550 25.85
B 39.09 0.856 45.67
C 32.065 1.961 16.35
D 30.95 1.823 16.98
E 26.98 2.700 9.992
F 22.98 0.968 23.7
2. Meyer found that the alkali metals had the highest molar volumes, and that molar volumes decreased steadily with increasing atomic mass, then leveled off, and finally rose again. The elements located on the rising portion of a plot of molar volume versus molar mass were typically nonmetals. If we look at the plot of the data in the table, we can immediately identify those elements with the largest molar volumes (A, B, F) as metals located on the left side of the periodic table. The element with the smallest molar volume (E) is aluminum. The plot shows that the subsequent elements (C, D) have molar volumes that are larger than that of E, but smaller than those of A and B. Thus, C and D are most likely to be nonmetals (which is the case: C = sulfur, D = phosphorus).
7.2: Effective Nuclear Charge
Conceptual Problems
1. What happens to the energy of a given orbital as the nuclear charge Z of a species increases? In a multielectron atom and for a given nuclear charge, the Zeff experienced by an electron depends on its value of l. Why?
2. The electron density of a particular atom is divided into two general regions. Name these two regions and describe what each represents.
3. As the principal quantum number increases, the energy difference between successive energy levels decreases. Why? What would happen to the electron configurations of the transition metals if this decrease did not occur?
4. Describe the relationship between electron shielding and Zeff on the outermost electrons of an atom. Predict how chemical reactivity is affected by a decreased effective nuclear charge.
5. If a given atom or ion has a single electron in each of the following subshells, which electron is easier to remove?
• 2s, 3s
• 3p, 4d
• 2p, 1s
• 3d, 4s
7.3: Sizes of Atoms and Ions
Conceptual Problems
1. The electrons of the 1s shell have a stronger electrostatic attraction to the nucleus than electrons in the 2s shell. Give two reasons for this.
2. Predict whether Na or Cl has the more stable 1s2 shell and explain your rationale.
3. Arrange K, F, Ba, Pb, B, and I in order of decreasing atomic radius.
4. Arrange Ag, Pt, Mg, C, Cu, and Si in order of increasing atomic radius.
5. Using the periodic table, arrange Li, Ga, Ba, Cl, and Ni in order of increasing atomic radius.
6. Element M is a metal that forms compounds of the type MX2, MX3, and MX4, where X is a halogen. What is the expected trend in the ionic radius of M in these compounds? Arrange these compounds in order of decreasing ionic radius of M.
7. The atomic radii of Na and Cl are 190 and 79 pm, respectively, but the distance between sodium and chlorine in NaCl is 282 pm. Explain this discrepancy.
8. Are shielding effects on the atomic radius more pronounced across a row or down a group? Why?
9. What two factors influence the size of an ion relative to the size of its parent atom? Would you expect the ionic radius of S2− to be the same in both MgS and Na2S? Why or why not?
10. Arrange Br, Al3+, Sr2+, F, O2−, and I in order of increasing ionic radius.
11. Arrange P3−, N3−, Cl, In3+, and S2− in order of decreasing ionic radius.
12. How is an isoelectronic series different from a series of ions with the same charge? Do the cations in magnesium, strontium, and potassium sulfate form an isoelectronic series? Why or why not?
13. What isoelectronic series arises from fluorine, nitrogen, magnesium, and carbon? Arrange the ions in this series by
1. increasing nuclear charge.
2. increasing size.
14. What would be the charge and electron configuration of an ion formed from calcium that is isoelectronic with
1. a chloride ion?
2. Ar+?
Conceptual Answers
1. The 1s shell is closer to the nucleus and therefore experiences a greater electrostatic attraction. In addition, the electrons in the 2s subshell are shielded by the filled 1s2 shell, which further decreases the electrostatic attraction to the nucleus.
2.
3. Ba > K > Pb > I > B > F
4.
5.
6.
7. The sum of the calculated atomic radii of sodium and chlorine atoms is 253 pm. The sodium cation is significantly smaller than a neutral sodium atom (102 versus 154 pm), due to the loss of the single electron in the 3s orbital. Conversely, the chloride ion is much larger than a neutral chlorine atom (181 versus 99 pm), because the added electron results in greatly increased electron–electron repulsions within the filled n = 3 principal shell. Thus, transferring an electron from sodium to chlorine decreases the radius of sodium by about 50%, but causes the radius of chlorine to almost double. The net effect is that the distance between a sodium ion and a chloride ion in NaCl is greater than the sum of the atomic radii of the neutral atoms.
Numerical Problems
1. Plot the ionic charge versus ionic radius using the following data for Mo: Mo3+, 69 pm; Mo4+, 65 pm; and Mo5+, 61 pm. Then use this plot to predict the ionic radius of Mo6+. Is the observed trend consistent with the general trends discussed in the chapter? Why or why not?
2. Internuclear distances for selected ionic compounds are given in the following table.
1. If the ionic radius of Li+ is 76 pm, what is the ionic radius of each of the anions?
LiF LiCl LiBr LiI
Distance (pm) 209 257 272 296
• What is the ionic radius of Na+?
NaF NaCl NaBr NaI
Distance (pm) 235 282 298 322
• Arrange the gaseous species Mg2+, P3−, Br, S2−, F, and N3− in order of increasing radius and justify your decisions.
7.4: Ionization Energy
Conceptual Problems
1. Identify each statement as either true or false and explain your reasoning.
1. Ionization energies increase with atomic radius.
2. Ionization energies decrease down a group.
3. Ionization energies increase with an increase in the magnitude of the electron affinity.
4. Ionization energies decrease diagonally across the periodic table from He to Cs.
5. Ionization energies depend on electron configuration.
6. Ionization energies decrease across a row.
2. Based on electronic configurations, explain why the first ionization energies of the group 16 elements are lower than those of the group 15 elements, which is contrary to the general trend.
3. The first through third ionization energies do not vary greatly across the lanthanides. Why? How does the effective nuclear charge experienced by the ns electron change when going from left to right (with increasing atomic number) in this series?
4. Most of the first row transition metals can form at least two stable cations, for example iron(II) and iron(III). In contrast, scandium and zinc each form only a single cation, the Sc3+ and Zn2+ ions, respectively. Use the electron configuration of these elements to provide an explanation.
5. Of the elements Nd, Al, and Ar, which will readily form(s) +3 ions? Why?
6. Orbital energies can reverse when an element is ionized. Of the ions B3+, Ga3+, Pr3+, Cr3+, and As3+, in which would you expect this reversal to occur? Explain your reasoning.
7. The periodic trends in electron affinities are not as regular as periodic trends in ionization energies, even though the processes are essentially the converse of one another. Why are there so many more exceptions to the trends in electron affinities compared to ionization energies?
8. Elements lying on a lower right to upper left diagonal line cannot be arranged in order of increasing electronegativity according to where they occur in the periodic table. Why?
9. Why do ionic compounds form, if energy is required to form gaseous cations?
10. Why is Pauling’s definition of electronegativity considered to be somewhat limited?
11. Based on their positions in the periodic table, arrange Sb, O, P, Mo, K, and H in order of increasing electronegativity.
12. Based on their positions in the periodic table, arrange V, F, B, In, Na, and S in order of decreasing electronegativity.
Conceptual Answers
5. Both Al and Nd will form a cation with a +3 charge. Aluminum is in Group 13, and loss of all three valence electrons will produce the Al3+ ion with a noble gas configuration. Neodymium is a lanthanide, and all of the lanthanides tend to form +3 ions because the ionization potentials do not vary greatly across the row, and a +3 charge can be achieved with many oxidants.
11. K < Mo ≈ Sb < P ≈ H < O
Numerical Problems
1. The following table gives values of the first and third ionization energies for selected elements:
Number of Electrons Element I1 (E → E+ + e, kJ/mol) Element I3 (E2+ → E3+ + e, kJ/mol)
11 Na 495.9 Al 2744.8
12 Mg 737.8 Si 3231.6
13 Al 577.6 P 2914.1
14 Si 786.6 S 3357
15 P 1011.9 Cl 3822
16 S 999.6 Ar 3931
17 Cl 1251.2 K 4419.6
18 Ar 1520.6 Ca 4912.4
Plot the ionization energies versus the number of electrons. Explain why the slopes of the I1 and I3 plots are different, even though the species in each row of the table have the same electron configurations.
2. Would you expect the third ionization energy of iron, corresponding to the removal of an electron from a gaseous Fe2+ ion, to be larger or smaller than the fourth ionization energy, corresponding to the removal of an electron from a gaseous Fe3+ ion? Why? How would these ionization energies compare to the first ionization energy of Ca?
3. Which would you expect to have the highest first ionization energy: Mg, Al, or Si? Which would you expect to have the highest third ionization energy. Why?
4. Use the values of the first ionization energies given in Figure 7.11 to construct plots of first ionization energy versus atomic number for (a) boron through oxygen in the second period; and (b) oxygen through tellurium in group 16. Which plot shows more variation? Explain the reason for the variation in first ionization energies for this group of elements.
5. Arrange Ga, In, and Zn in order of increasing first ionization energies. Would the order be the same for second and third ionization energies? Explain your reasoning.
6. Arrange each set of elements in order of increasing magnitude of electron affinity.
1. Pb, Bi, and Te
2. Na, K, and Rb
3. P, C, and Ge
7. Arrange each set of elements in order of decreasing magnitude of electron affinity.
1. As, Bi, and N
2. O, F, and Ar
3. Cs, Ba, and Rb
8. Of the species F, O, Al3+, and Li+, which has the highest electron affinity? Explain your reasoning.
9. Of the species O, N2−, Hg2+, and H+, which has the highest electron affinity? Which has the lowest electron affinity? Justify your answers.
10. The Mulliken electronegativity of element A is 542 kJ/mol. If the electron affinity of A is −72 kJ/mol, what is the first ionization energy of element A? Use the data in the following table as a guideline to decide if A is a metal, a nonmetal, or a semimetal. If 1 g of A contains 4.85 × 1021 molecules, what is the identity of element A?
Na Al Si S Cl
EA (kJ/mol) −59.6 −41.8 −134.1 −200.4 −348.6
I (kJ/mol) 495.8 577.5 786.5 999.6 1251.2
11. Based on their valence electron configurations, classify the following elements as either electrical insulators, electrical conductors, or substances with intermediate conductivity: S, Ba, Fe, Al, Te, Be, O, C, P, Sc, W, Na, B, and Rb.
12. Using the data in Problem 10, what conclusions can you draw with regard to the relationship between electronegativity and electrical properties? Estimate the approximate electronegativity of a pure element that is very dense, lustrous, and malleable.
13. Of the elements Al, Mg, O2, Ti, I2, and H2, which, if any, would you expect to be a good reductant? Explain your reasoning.
14. Of the elements Zn, B, Li, Se, Co, and Br2, which if any, would you expect to be a good oxidant? Explain your reasoning.
15. Determine whether each species is a good oxidant, a good reductant, or neither.
1. Ba
2. Mo
3. Al
4. Ni
5. O2
6. Xe
16. Determine whether each species is a good oxidant, a good reductant, or neither.
1. Ir
2. Cs
3. Be
4. B
5. N
6. Po
7. Ne
17. Of the species I2, O, Zn, Sn2+, and K+, choose which you would expect to be a good oxidant. Then justify your answer.
18. Based on the valence electron configuration of the noble gases, would you expect them to have positive or negative electron affinities? What does this imply about their most likely oxidation states? their reactivity?
Numerical Answers
1. The general features of both plots are roughly the same, with a small peak at 12 electrons and an essentially level region from 15–16 electrons. The slope of the I3 plot is about twice as large as the slope of the I1 plot, however, because the I3 values correspond to removing an electron from an ion with a +2 charge rather than a neutral atom. The greater charge increases the effect of the steady rise in effective nuclear charge across the row.
2.
3. Electron configurations: Mg, 1s22s22p63s2; Al, 1s22s22p63s23p1; Si, 1s22s22p63s23p2; First ionization energies increase across the row due to a steady increase in effective nuclear charge; thus, Si has the highest first ionization energy. The third ionization energy corresponds to removal of a 3s electron for Al and Si, but for Mg it involves removing a 2p electron from a filled inner shell; consequently, the third ionization energy of Mg is the highest.
4.
5.
6.
1. Bi > As > N
2. F > O >> Ar
3. Rb > Cs > Ba
7.
8. Hg2+ > H+ > O > N2−; Hg2+ has the highest positive charge plus a relatively low energy vacant set of orbitals (the 6p subshell) to accommodate an added electron, giving it the greatest electron affinity; N2− has a greater negative charge than O, so electron–electron repulsions will cause its electron affinity to be even lower (more negative) than that of O.
9.
10. insulators: S, O, C (diamond), P; conductors: Ba, Fe, Al, C (graphite), Be, Sc, W, Na, Rb; Te and B are semimetals and semiconductors.
11.
12. Mg, Al, Ti, and H2
13.
1. reductant
2. neither
3. reductant
4. reductant
5. oxidant
6. neither
14.
15. I2 is the best oxidant, with a moderately strong tendency to accept an electron to form the I ion, with a closed shell electron configuration. O would probably also be an oxidant, with a tendency to add an electron to form salts containing the oxide ion, O2−. Zn and Sn2+ are all reductants, while K+ has no tendency to act as an oxidant or a reductant.
7.5: Electron Affinities
see above question to tease out | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/07%3A_Periodic_Properties_of_the_Elements/7.E%3A_Periodic_Properties_of_the_Elements_%28Exercises%29.txt |
7.1: Development of the Periodic Table
• 1869: Dmitri Mendeleev (Russia) and Lothar Meyer (Germany) publish nearly identical schemes for classifying the elements
• Both had arranged the elements in order of increasing atomic weights
• Mendeleev predicted the existence and properties of Germanium (Ge) and Gallium (Ga). When they were discovered, their properties were found to closely match those predicted by Mendeleev
• 1913: Henry Moseley discovers concept of atomic numbers. Observed that frequencies of Xrays were different for each element. Was able to arrange these frequencies in order by assigning each element a unique whole number, which he called the atomic number
7.2: Effective Nuclear Charge
• Orbitals with the same value of n are referred to as a shell
• When looking at a radial electrondensity graph, the maxima (peaks) designate the areas with the higher probabilities of finding electrons
• The 1s shell in Argon is closer to the nucleus than that of Helium. This is because both 1s shells are the only ones not shielded from the nucleus by other electrons. However, since Argon has a stronger nuclear charge, it can attract its electrons more easily
• radial electron density – probability of finding the electron at a particular distance from the nucleus
• as nuclear charge increases, 1s shells pulled closer to nucleus
7.3: Sizes of Atoms and Ions
Atomic radius: radius of an atom of a given element
1. Within each group, the atomic radius tends to increase going from top to bottom
2. Within each period, the atomic radius tends to decrease moving left to right
Two factors determine the size of the outermost orbital
1. Its principal quantum number (as it increases, the size of the orbital increases)
2. The effective nuclear charge (reduces the size of the orbital by pulling electrons closer)
Atomic Sizes
• atomic radius – estimation of radius of atoms
• estimate by assuming that atoms are spheres that touch each other
• atomic radius increases down a group, decrease down row
• atomic radius affected by principal quantum number and effective nuclear charge
• increase principal quantum number, increases size of orbital
• increase effective nuclear charge, reduces size or orbital
7.4: Ionization Energy
Ionization energy: energy required to remove an electron from a gaseous atom when the atom is in its ground state.
• first ionization energy – energy needed to remove the first electron
• second ionization energy – energy needed to remove second electron
• the greater ionization energy, harder it is to remove electrons
• The first ionization energy, I1, is the energy needed to remove the first electron. The second ionization energy, I2, removes the second electron, and so on. The HIGHER the ionization energy, the MORE DIFFICULT it is to remove an electron. Every element exhibits a large increase in ionization energy when electrons are removed from its noblegas core. The inner electrons are too tightly bound to the nucleus to be lost from the atom or even shared with another atom
$I_1< I_2 < I_3 \nonumber$
• Positive nuclear charge remains same, number electrons decreases -> effective nuclear charge increases
• Greater effective nuclear charge, greater energy required to remove electron
• Sharp increase in ionization energy when inner-shell electron removed
• Only outer most electrons involved in sharing and transfer of electrons in bonding and reactions
Periodic Trends in Ionization Energies
1. Within each period, I1 generally increases with increasing atomic number. The alkali metals show the lowest ionization energy in each row, and the noble gases the highest
2. Within each group, ionization energy generally decreases with increasing atomic number.
• Energy needed to remove electrons from outer shell depends on nuclear charge and average distance between of the electron from the nucleus
• As we move across a period, there is both an increase in effective nuclear charge and a decrease in atomic radius, causing the ionization energy to increase
• As we move down a column, the atomic radius increases, while the effective charge remains essentially constant. Thus, the attraction between the nucleus and the electron decreases in this direction, causing the ionization energy to decrease.
7.5: Electron Affinities
• Electron affinity: energy change that occurs when an electron is added to a gaseous atom or ion
• For most neutral atoms and for all positively charged ions, energy is released when an electron is added
• The greater the attraction between the species and the added electron, the more exothermic the process
• On the other hand, adding an electron to anions and to some atoms makes them unstable.
• The general trend for electron affinity is to become increasingly negative (stronger binding of electrons) as we move across each period toward the halogens. The halogens have the most negative electron affinities since they are one electron short of noblegas configuration
• Electron affinities do not change greatly as we go down a group
7.6: Metals, Nonmetals, and Metalloids
Metallic Elements and Nonmetallic Elements
Metallic elements Nonmetallic elements
Distinguishing luster Nonlustrous; various colors
Malleable and ductile as solids Solids are usually brittle; may be hard or soft
Good thermal and electrical conductivity Poor conductors of heat and electricity
Most metallic oxides are basic, ionic solids Most nonmetallic oxides are molecular, acidic compounds
Exist in aqueous solution mainly as cations Exist in aqueous solution mainly as anions or oxyanions
Metals
• See properties above. All are solids at room temperature except mercury, which is a liquid.
• Metals tend to have low ionization energies and are consequently oxidized (lose electrons) when they undergo chemical reaction.
• Compounds of metals with nonmetals tend to be ionic substances
• Most metals are basic oxides (oxide that either reacts with water to form a base or reacts with an acid to form a salt and water)
Nonmetals
• Generally have lower melting points than those of metals.
• Seven nonmetals exist as diatomic molecules under ordinary conditions (H2, N2, O2, F2, Cl2, Br2, I2). Bromine is liquid; Iodine is a volatile solid, the rest are gases. The remaining nonmetals are solids.
• Nonmetals, in reacting with metals, tend to gain electrons and become anions
• Compound composed entirely of nonmetals are molecular substances
• Most nonmetals are acidic oxides (oxide that forms acid when added to water; soluble nonmetal oxides are acidic oxides)
Metalloids
• Have properties intermediate between those of metals and nonmetals. May have some metallic properties but lack others.
• Several metalloids are electrical semiconductors
Trends in Metallic and Nonmetallic Character
Metallic character: the extent to which an element exhibits the physical and chemical properties characteristic of metals, for example, luster, malleability, ductility, and good thermal and electrical conductivity.
1. Metallic character is strongest for the elements in the leftmost part of the periodic table and tends to decrease as we move to the right in any period
2. Within any group of representative elements, the metallic character increases progressively from top to bottom
7.7: Group Trends for the Active Metals
Group 1: The Alkali Metals
• Metallic properties; silvery, metallic luster, high thermal and electrical conductivity
• Low densities and melting points; increasing atomic radius and decreasing ionization energy as we move down the family
• Lowest I1 values of the elements, which reflects the ease with which their outer electrons can be removed. This makes the alkali metals very reactive, readily losing one electron.
• In the hydrides of the alkali metals (LiH, NaH, and so forth), hydrogen is present as H, called the hydride ion.
• The alkali metals react vigorously with water, producing hydrogen gas and solutions of alkali metal hydroxides (very exothermic).
Group 2: The Alkaline Earth Metals
• Solids, typical metallic properties, denser and harder than alkali metals (also melt at higher temperatures)
• Less reactive than alkali metals (have slightly higher ionization energy)
• Ease with which the elements lose their electrons increases as we move down the family
7.8: Group Trends for Selected Nonmetals
Hydrogen
• Hydrogen is a nonmetal, which occurs as a diatomic gas under normal conditions
• Its ionization energy is considerably higher (due to lack of shielding, and thus higher Zeff) thatn the metals and is more like the nonmetals
• Hydrogen generally reacts with other nonmetals to form molecular compounds (typically highly exothermic)
• Hydrogen reacts with active metals to form metal hydrides that contain the H hydride ion
• Hydrogen can also lose an electron to yield the aqueous H+ ion
Group 16: The Oxygen Family
• As we proceed down group 6A, the increase in metallic character is clearly evident
• Allotropes: different forms of the same element in the same state (ie: O2 and O3)
• Oxygen has a great tendency to attract electrons from other elements ("oxidize" them)
• Oxygen in combination with metals is almost always present as the oxide, O2 ion, which has noblegas configuration and is particularly stable
• Sulfur is the second most important element in the 6A group. It also exists in several allotropic, the most common and stable being S8.
• Sulfur also has a great tendency to gain electrons from other elements to form sulfides
• Because Sulfur is below Oxygen, its tendency to form sulfide anions is not is not as great as that of Oxygen to form oxide ions
Group 17: The Halogens
• All halogens are typical nonmetals. Their melting and boiling points increase with increasing atomic numbers
• Fluorine and Chlorine are gases at room temperature, Bromine is a liquid, and Iodine is a solid
• The halogens have among the most negative electron affinities; they have a great tendency to gain electrons.
• Fluorine and Chlorine are the most reactive. Fluorine removes electrons from almost any substance, even water, and usually very exothermically.
• Halogens react with most metals to form ionic halides and with Hydrogen to form gaseous hydrogen halides.
Group 18: The Noble Gases
• All noble gases are monoatomic (single atom)
• All have large first ionization energies, with a decrease as we go down the group
• Until the early 1960’s, they were called inert gases because they were thought to be incapable of forming chemical compounds.
• Only noblegas compounds known today are: XeF2, XeF4, XeF6, and KrF2 | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/07%3A_Periodic_Properties_of_the_Elements/7.S%3A_Periodic_Properties_of_the_Elements_%28Summary%29.txt |
We described the relationship between the chemical properties and reactivity of an element and its position in the periodic table. In this chapter and the following chapter, we describe the interactions that hold atoms together in chemical substances, and we examine the factors that determine how the atoms of a substance are arranged in space. Our goal is to understand how the properties of the component atoms in a chemical compound determine the structure and reactivity of the compound. The properties described previously were properties of isolated atoms, yet most of the substances in our world consist of atoms held together in molecules, ionic compounds, or metallic solids. The properties of these substances depend on not only the characteristics of the component atoms but also how those atoms are bonded to one another.
What you learn in this chapter about chemical bonding and molecular structure will help you understand how different substances with the same atoms can have vastly different physical and chemical properties. For example, oxygen gas (O2) is essential for life, yet ozone (O3) is toxic to cells, although as you learned previously, ozone in the upper atmosphere shields us from harmful ultraviolet light. Moreover, you saw that diamond is a hard, transparent solid that is a gemstone; graphite is a soft, black solid that is a lubricant; and fullerenes are molecular species with carbon cage structures—yet all of these are composed of carbon. As you learn about bonding, you will also discover why, although carbon and silicon both have ns2np2 valence electron configurations and form dioxides, CO2 is normally a gas that condenses into the volatile molecular solid known as dry ice, whereas SiO2 is a nonvolatile solid with a network structure that can take several forms, including beach sand and quartz crystals.
• 8.1: Chemical Bonds, Lewis Symbols, and the Octet Rule
Lewis dot symbols can be used to predict the number of bonds formed by most elements in their compounds. Lewis electron dot symbols, which consist of the chemical symbol for an element surrounded by dots that represent its valence electrons, grouped into pairs often placed above, below, and to the left and right of the symbol. The structures reflect the fact that the elements in period 2 and beyond tend to gain, lose, or share electrons to reach a total of 8 valence electrons in their compounds.
• 8.2: Ionic Bonding
The amount of energy needed to separate a gaseous ion pair is its bond energy. The formation of ionic compounds are usually extremely exothermic. The strength of the electrostatic attraction between ions with opposite charges is directly proportional to the magnitude of the charges on the ions and inversely proportional to the internuclear distance.
• 8.3: Covalent Bonding
The strength of a covalent bond depends on the overlap between the valence orbitals of the bonded atoms. Bond order is the number of electron pairs that hold two atoms together. Single bonds have a bond order of one, and multiple bonds with bond orders of two (a double bond) and three (a triple bond) are quite common. In closely related compounds with bonds between the same kinds of atoms, the bond with the highest bond order is both the shortest and the strongest.
• 8.4: Bond Polarity and Electronegativity
Bond polarity and ionic character increase with an increasing difference in electronegativity. The electronegativity (χ) of an element is the relative ability of an atom to attract electrons to itself in a chemical compound and increases diagonally from the lower left of the periodic table to the upper right. The Pauling electronegativity scale is based on measurements of the strengths of covalent bonds between different atoms, whereas the Mulliken electronegativity of an element is the average
• 8.5: Drawing Lewis Structures
Lewis dot symbols provide a simple rationalization of why elements form compounds with the observed stoichiometries. A plot of the overall energy of a covalent bond as a function of internuclear distance is identical to a plot of an ionic pair because both result from attractive and repulsive forces between charged entities. Lewis structures are an attempt to rationalize why certain stoichiometries are commonly observed for the elements of particular families.
• 8.6: Resonance Structures
Some molecules have two or more chemically equivalent Lewis electron structures, called resonance structures. Resonance is a mental exercise and method within the Valence Bond Theory of bonding that describes the delocalization of electrons within molecules. These structures are written with a double-headed arrow between them, indicating that none of the Lewis structures accurately describes the bonding but that the actual structure is an average of the individual resonance structures.
• 8.7: Exceptions to the Octet Rule
Following the Octet Rule for Lewis Dot Structures leads to the most accurate depictions of stable molecular and atomic structures and because of this we always want to use the octet rule when drawing Lewis Dot Structures. There are three exceptions: (1) When there are an odd number of valence electrons, (2) When there are too few valence electrons, and (3) when there are too many valence electrons
• 8.8: Strength of Covalent Bonds
Bond order is the number of electron pairs that hold two atoms together. Single bonds have a bond order of one, and multiple bonds with bond orders of two (a double bond) and three (a triple bond) are quite common. The bond with the highest bond order is both the shortest and the strongest. In bonds with the same bond order between different atoms, trends are observed that, with few exceptions, result in the strongest single bonds being formed between the smallest atoms.
• 8.E: Basic Concepts of Chemical Bonding (Exercises)
Problems and select solutions to the chapter.
• 8.S: Basic Concepts of Chemical Bonding (Summary)
A summary of the key concepts in this chapter of the Textmap created for "Chemistry: The Central Science" by Brown et al.
Thumbnail: Non-polar covalent bonds in methane (\(\ce{CH4}\)). The Lewis structure shows electrons shared between C and H atoms. (CC BY-sa 2.5; DynaBlast via Wikipedia).
08: Basic Concepts of Chemical Bonding
Learning Objectives
• To use Lewis electron dot symbols to predict the number of bonds an element will form.
Why are some substances chemically bonded molecules and others are an association of ions? The answer to this question depends upon the electronic structures of the atoms and nature of the chemical forces within the compounds. Although there are no sharply defined boundaries, chemical bonds are typically classified into three main types: ionic bonds, covalent bonds, and metallic bonds. In this chapter, each type of bond wil be discussed and the general properties found in typical substances in which the bond type occurs
1. Ionic bonds results from electrostatic forces that exist between ions of opposite charge. These bonds typically involves a metal with a nonmetal
2. Covalent bonds result from the sharing of electrons between two atoms. The bonds typically involves one nonmetallic element with another
3. Metallic bonds These bonds are found in solid metals (copper, iron, aluminum) with each metal bonded to several neighboring groups and bonding electrons free to move throughout the 3-dimensional structure.
Each bond classification is discussed in detail in subsequent sections of the chapter. Let's look at the preferred arrangements of electrons in atoms when they form chemical compounds.
Lewis Symbols
At the beginning of the 20th century, the American chemist G. N. Lewis (1875–1946) devised a system of symbols—now called Lewis electron dot symbols (often shortened to Lewis dot symbols) that can be used for predicting the number of bonds formed by most elements in their compounds. Each Lewis dot symbol consists of the chemical symbol for an element surrounded by dots that represent its valence electrons.
Lewis Dot symbols:
• convenient representation of valence electrons
• allows you to keep track of valence electrons during bond formation
• consists of the chemical symbol for the element plus a dot for each valence electron
To write an element’s Lewis dot symbol, we place dots representing its valence electrons, one at a time, around the element’s chemical symbol. Up to four dots are placed above, below, to the left, and to the right of the symbol (in any order, as long as elements with four or fewer valence electrons have no more than one dot in each position). The next dots, for elements with more than four valence electrons, are again distributed one at a time, each paired with one of the first four. For example, the electron configuration for atomic sulfur is [Ne]3s23p4, thus there are six valence electrons. Its Lewis symbol would therefore be:
Fluorine, for example, with the electron configuration [He]2s22p5, has seven valence electrons, so its Lewis dot symbol is constructed as follows:
Lewis used the unpaired dots to predict the number of bonds that an element will form in a compound. Consider the symbol for nitrogen in Figure 8.1.2. The Lewis dot symbol explains why nitrogen, with three unpaired valence electrons, tends to form compounds in which it shares the unpaired electrons to form three bonds. Boron, which also has three unpaired valence electrons in its Lewis dot symbol, also tends to form compounds with three bonds, whereas carbon, with four unpaired valence electrons in its Lewis dot symbol, tends to share all of its unpaired valence electrons by forming compounds in which it has four bonds.
The Octet Rule
In 1904, Richard Abegg formulated what is now known as Abegg's rule, which states that the difference between the maximum positive and negative valences of an element is frequently eight. This rule was used later in 1916 when Gilbert N. Lewis formulated the "octet rule" in his cubical atom theory. The octet rule refers to the tendency of atoms to prefer to have eight electrons in the valence shell. When atoms have fewer than eight electrons, they tend to react and form more stable compounds. Atoms will react to get in the most stable state possible. A complete octet is very stable because all orbitals will be full. Atoms with greater stability have less energy, so a reaction that increases the stability of the atoms will release energy in the form of heat or light ;reactions that decrease stability must absorb energy, getting colder.
When discussing the octet rule, we do not consider d or f electrons. Only the s and p electrons are involved in the octet rule, making it a useful rule for the main group elements (elements not in the transition metal or inner-transition metal blocks); an octet in these atoms corresponds to an electron configurations ending with s2p6.
Definition: Octet Rule
A stable arrangement is attended when the atom is surrounded by eight electrons. This octet can be made up by own electrons and some electrons which are shared. Thus, an atom continues to form bonds until an octet of electrons is made. This is known as octet rule by Lewis.
1. Normally two electrons pairs up and forms a bond, e.g., \(\ce{H_2}\)
2. For most atoms there will be a maximum of eight electrons in the valence shell (octet structure), e.g., \(\ce{CH_4}\)
The other tendency of atoms is to maintain a neutral charge. Only the noble gases (the elements on the right-most column of the periodic table) have zero charge with filled valence octets. All of the other elements have a charge when they have eight electrons all to themselves. The result of these two guiding principles is the explanation for much of the reactivity and bonding that is observed within atoms: atoms seek to share electrons in a way that minimizes charge while fulfilling an octet in the valence shell.
The noble gases rarely form compounds. They have the most stable configuration (full octet, no charge), so they have no reason to react and change their configuration. All other elements attempt to gain, lose, or share electrons to achieve a noble gas configuration.
Example \(1\): Salt
The formula for table salt is NaCl. It is the result of Na+ ions and Cl- ions bonding together. If sodium metal and chlorine gas mix under the right conditions, they will form salt. The sodium loses an electron, and the chlorine gains that electron. In the process, a great amount of light and heat is released. The resulting salt is mostly unreactive — it is stable. It will not undergo any explosive reactions, unlike the sodium and chlorine that it is made of. Why?
Solution
Referring to the octet rule, atoms attempt to get a noble gas electron configuration, which is eight valence electrons. Sodium has one valence electron, so giving it up would result in the same electron configuration as neon. Chlorine has seven valence electrons, so if it takes one it will have eight (an octet). Chlorine has the electron configuration of argon when it gains an electron.
The octet rule could have been satisfied if chlorine gave up all seven of its valence electrons and sodium took them. In that case, both would have the electron configurations of noble gasses, with a full valence shell. However, their charges would be much higher. It would be Na7- and Cl7+, which is much less stable than Na+ and Cl-. Atoms are more stable when they have no charge, or a small charge.
Lewis dot symbols can also be used to represent the ions in ionic compounds. The reaction of cesium with fluorine, for example, to produce the ionic compound CsF can be written as follows:
No dots are shown on Cs+ in the product because cesium has lost its single valence electron to fluorine. The transfer of this electron produces the Cs+ ion, which has the valence electron configuration of Xe, and the F ion, which has a total of eight valence electrons (an octet) and the Ne electron configuration. This description is consistent with the statement that among the main group elements, ions in simple binary ionic compounds generally have the electron configurations of the nearest noble gas. The charge of each ion is written in the product, and the anion and its electrons are enclosed in brackets. This notation emphasizes that the ions are associated electrostatically; no electrons are shared between the two elements.
Atoms often gain, lose, or share electrons to achieve the same number of electrons as the noble gas closest to them in the periodic table.
As you might expect for such a qualitative approach to bonding, there are exceptions to the octet rule, which we describe elsewhere. These include molecules in which one or more atoms contain fewer or more than eight electrons.
Summary
Lewis dot symbols can be used to predict the number of bonds formed by most elements in their compounds. One convenient way to predict the number and basic arrangement of bonds in compounds is by using Lewis electron dot symbols, which consist of the chemical symbol for an element surrounded by dots that represent its valence electrons, grouped into pairs often placed above, below, and to the left and right of the symbol. The structures reflect the fact that the elements in period 2 and beyond tend to gain, lose, or share electrons to reach a total of eight valence electrons in their compounds, the so-called octet rule. Hydrogen, with only two valence electrons, does not obey the octet rule.
Contributors and Attributions
• Wikipedia
• National Programme on Technology Enhanced Learning (India) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/08%3A_Basic_Concepts_of_Chemical_Bonding/8.01%3A_Chemical_Bonds_Lewis_Symbols_and_the_Octet_Rule.txt |
Learning Objectives
• To describe the characteristics of ionic bonding.
• To quantitatively describe the energetic factors involved in the formation of an ionic bond.
Ions are atoms or molecules which are electrically charged. Cations are positively charged and anions carry a negative charge. Ions form when atoms gain or lose electrons. Since electrons are negatively charged, an atom that loses one or more electrons will become positively charged; an atom that gains one or more electrons becomes negatively charged. Ionic bonding is the attraction between positively- and negatively-charged ions. These oppositely charged ions attract each other to form ionic networks (or lattices). Electrostatics explains why this happens: opposite charges attract and like charges repel. When many ions attract each other, they form large, ordered, crystal lattices in which each ion is surrounded by ions of the opposite charge. Generally, when metals react with non-metals, electrons are transferred from the metals to the non-metals. The metals form positively-charged ions and the non-metals form negatively-charged ions.
Generating Ionic Bonds
Ionic bonds form when metals and non-metals chemically react. By definition, a metal is relatively stable if it loses electrons to form a complete valence shell and becomes positively charged. Likewise, a non-metal becomes stable by gaining electrons to complete its valence shell and become negatively charged. When metals and non-metals react, the metals lose electrons by transferring them to the non-metals, which gain them. Consequently, ions are formed, which instantly attract each other—ionic bonding.
In the overall ionic compound, positive and negative charges must be balanced, because electrons cannot be created or destroyed, only transferred. Thus, the total number of electrons lost by the cationic species must equal the total number of electrons gained by the anionic species.
Example $1$: Sodium Chloride
For example, in the reaction of Na (sodium) and Cl (chlorine), each Cl atom takes one electron from a Na atom. Therefore each Na becomes a Na+ cation and each Cl atom becomes a Cl- anion. Due to their opposite charges, they attract each other to form an ionic lattice. The formula (ratio of positive to negative ions) in the lattice is $\ce{NaCl}$.
$\ce{2Na (s) + Cl 2(g) \rightarrow 2NaCl (s)} \nonumber$
These ions are arranged in solid NaCl in a regular three-dimensional arrangement (or lattice):
NaCl lattice. (left) 3-D structure and (right) simple 2D slice through lattes. Images used with permission from Wikipedia and Mike Blaber.
The chlorine has a high affinity for electrons, and the sodium has a low ionization energy. Thus the chlorine gains an electron from the sodium atom. This can be represented using ewis dot symbols (here we will consider one chlorine atom, rather than Cl2):
The arrow indicates the transfer of the electron from sodium to chlorine to form the Na+ metal ion and the Cl- chloride ion. Each ion now has an octet of electrons in its valence shell:
• Na+: 2s22p6
• Cl-: 3s23p6
Energetics of Ionic Bond Formation
Ionic bonds are formed when positively and negatively charged ions are held together by electrostatic forces. Consider a single pair of ions, one cation and one anion. How strong will the force of their attraction be? According to Coulomb's Law, the energy of the electrostatic attraction ($E$) between two charged particles is proportional to the magnitude of the charges and inversely proportional to the internuclear distance between the particles ($r$):
$E \propto \dfrac{Q_{1}Q_{2}}{r} \label{Eq1a}$
$E = k\dfrac{Q_{1}Q_{2}}{r} \label{Eq1b}$
where each ion’s charge is represented by the symbol Q. The proportionality constant k is equal to 2.31 × 10−28 J·m. This value of k includes the charge of a single electron (1.6022 × 10−19 C) for each ion. The equation can also be written using the charge of each ion, expressed in coulombs (C), incorporated in the constant. In this case, the proportionality constant, k, equals 8.999 × 109 J·m/C2. In the example given, Q1 = +1(1.6022 × 10−19 C) and Q2 = −1(1.6022 × 10−19 C). If Q1 and Q2 have opposite signs (as in NaCl, for example, where Q1 is +1 for Na+ and Q2 is −1 for Cl), then E is negative, which means that energy is released when oppositely charged ions are brought together from an infinite distance to form an isolated ion pair.
Energy is always released when a bond is formed and correspondingly, it always requires energy to break a bond.
As shown by the green curve in the lower half of Figure $1$, the maximum energy would be released when the ions are infinitely close to each other, at r = 0. Because ions occupy space and have a structure with the positive nucleus being surrounded by electrons, however, they cannot be infinitely close together. At very short distances, repulsive electron–electron interactions between electrons on adjacent ions become stronger than the attractive interactions between ions with opposite charges, as shown by the red curve in the upper half of Figure $1$. The total energy of the system is a balance between the attractive and repulsive interactions. The purple curve in Figure $1$ shows that the total energy of the system reaches a minimum at r0, the point where the electrostatic repulsions and attractions are exactly balanced. This distance is the same as the experimentally measured bond distance.
Consider the energy released when a gaseous $Na^+$ ion and a gaseous $Cl^-$ ion are brought together from r = ∞ to r = r0. Given that the observed gas-phase internuclear distance is 236 pm, the energy change associated with the formation of an ion pair from an $Na^+_{(g)}$ ion and a $Cl^-_{(g)}$ ion is as follows:
\begin{align*} E &= k\dfrac{Q_{1}Q_{2}}{r_{0}} \[4pt] &= (2.31 \times {10^{ - 28}}\rm{J}\cdot \cancel{m} ) \left( \dfrac{( + 1)( - 1)}{236\; \cancel{pm} \times 10^{ - 12} \cancel{m/pm}} \right) \[4pt] &= - 9.79 \times 10^{ - 19}\; J/ion\; pair \label{Eq2} \end{align*}
The negative value indicates that energy is released. Our convention is that if a chemical process provides energy to the outside world, the energy change is negative. If it requires energy, the energy change is positive. To calculate the energy change in the formation of a mole of NaCl pairs, we need to multiply the energy per ion pair by Avogadro’s number:
$E=\left ( -9.79 \times 10^{ - 19}\; J/ \cancel{ion pair} \right )\left ( 6.022 \times 10^{ 23}\; \cancel{ion\; pair}/mol\right )=-589\; kJ/mol \label{Eq3}$
This is the energy released when 1 mol of gaseous ion pairs is formed, not when 1 mol of positive and negative ions condenses to form a crystalline lattice. Because of long-range interactions in the lattice structure, this energy does not correspond directly to the lattice energy of the crystalline solid. However, the large negative value indicates that bringing positive and negative ions together is energetically very favorable, whether an ion pair or a crystalline lattice is formed.
We summarize the important points about ionic bonding:
• At r0, the ions are more stable (have a lower potential energy) than they are at an infinite internuclear distance. When oppositely charged ions are brought together from r = ∞ to r = r0, the energy of the system is lowered (energy is released).
• Because of the low potential energy at r0, energy must be added to the system to separate the ions. The amount of energy needed is the bond energy.
• The energy of the system reaches a minimum at a particular internuclear distance (the bond distance).
Example $2$: LiF
Calculate the amount of energy released when 1 mol of gaseous Li+F ion pairs is formed from the separated ions. The observed internuclear distance in the gas phase is 156 pm.
Given: cation and anion, amount, and internuclear distance
Asked for: energy released from formation of gaseous ion pairs
Strategy:
Substitute the appropriate values into Equation $\ref{Eq1b}$ to obtain the energy released in the formation of a single ion pair and then multiply this value by Avogadro’s number to obtain the energy released per mole.
Solution:
Inserting the values for Li+F into Equation $\ref{Eq1b}$ (where Q1 = +1, Q2 = −1, and r = 156 pm), we find that the energy associated with the formation of a single pair of Li+F ions is
\begin{align*} E &=k \dfrac{Q_1Q_2}{r_0} \[4pt] &=\left(2.31 \times 10^{−28} J⋅\cancel{m} \right) \left(\dfrac{\text{(+1)(−1)}}{156\; pm \times 10^{−12} \cancel{m/pm}} \right)\[4pt] &=−1.48 \times 10^{−18} \end{align*} \nonumber
Then the energy released per mole of Li+F ion pairs is
\begin{align*} E&= \left(−1.48 \times 10^{−18} J/ \cancel{\text{ion pair}}\right) \left(6.022 \times 10^{23} \cancel{\text{ion pair}}/mol\right)\[4pt] &−891 \;kJ/mol \end{align*} \nonumber
Because Li+ and F are smaller than Na+ and Cl (see Section 7.3), the internuclear distance in LiF is shorter than in NaCl. Consequently, in accordance with Equation $\ref{Eq1b}$, much more energy is released when 1 mol of gaseous Li+F ion pairs is formed (−891 kJ/mol) than when 1 mol of gaseous Na+Cl ion pairs is formed (−589 kJ/mol).
Exercise $2$: Magnesium oxide
Calculate the amount of energy released when 1 mol of gaseous $\ce{MgO}$ ion pairs is formed from the separated ions. The internuclear distance in the gas phase is 175 pm.
Answer
−3180 kJ/mol = −3.18 × 103 kJ/mol
Electron Configuration of Ions
How does the energy released in lattice formation compare to the energy required to strip away a second electron from the Na+ ion? Since the Na+ ion has a noble gas electron configuration, stripping away the next electron from this stable arrangement would require more energy than what is released during lattice formation (Sodium I2 = 4,560 kJ/mol). Thus, sodium is present in ionic compounds as Na+ and not Na2+. Likewise, adding an electron to fill a valence shell (and achieve noble gas electron configuration) is exothermic or only slightly endothermic. To add an additional electron into a new subshell requires tremendous energy - more than the lattice energy. Thus, we find Cl- in ionic compounds, but not Cl2-.
Table $1$: Lattice energies range from around 700 kJ/mol to 4000 kJ/mol:
Compound Lattice Energy (kJ/mol)
LiF 1024
LiI 744
NaF 911
NaCl 788
NaI 693
KF 815
KBr 682
KI 641
MgF2 2910
SrCl2 2130
MgO 3938
This amount of energy can compensate for values as large as I3 for valence electrons (i.e. can strip away up to 3 valence electrons). Because most transition metals would require the removal of more than 3 electrons to attain a noble gas core, they are not found in ionic compounds with a noble gas core. A transition metal always loses electrons first from the higher 's' subshell, before losing from the underlying 'd' subshell. (The remaining electrons in the unfilled d subshell are the reason for the bright colors observed in many transition metal compounds!) For example, iron ions will not form a noble gas core:
• Fe: [Ar]4s23d6
• Fe2+: [Ar] 3d6
• Fe3+: [Ar] 3d5
Some metal ions can form a pseudo noble gas core (and be colorless), for example:
• Ag: [Kr]5s14d10 Ag+ [Kr]4d10 Compound: AgCl
• Cd: [Kr]5s24d10 Cd2+ [Kr]4d10 Compound: CdS
The valence electrons do not adhere to the "octet rule" in this case (a limitation of the usefulness of this rule). Note: The silver and cadmium atoms lost the 5s electrons in achieving the ionic state.
When a positive ion is formed from an atom, electrons are always lost first from the subshell with the largest principle quantum number
Polyatomic Ions
Not all ionic compounds are formed from only two elements. Many polyatomic ions exist, in which two or more atoms are bound together by covalent bonds. They form a stable grouping which carries a charge (positive or negative). The group of atoms as a whole acts as a charged species in forming an ionic compound with an oppositely charged ion. Polyatomic ions may be either positive or negative, for example:
• NH4+ (ammonium) = cation
• SO42- (sulfate) = anion
The principles of ionic bonding with polyatomic ions are the same as those with monatomic ions. Oppositely charged ions come together to form a crystalline lattice, releasing a lattice energy. Based on the shapes and charges of the polyatomic ions, these compounds may form crystalline lattices with interesting and complex structures.
Summary
The amount of energy needed to separate a gaseous ion pair is its bond energy. The formation of ionic compounds are usually extremely exothermic. The strength of the electrostatic attraction between ions with opposite charges is directly proportional to the magnitude of the charges on the ions and inversely proportional to the internuclear distance. The total energy of the system is a balance between the repulsive interactions between electrons on adjacent ions and the attractive interactions between ions with opposite charges. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/08%3A_Basic_Concepts_of_Chemical_Bonding/8.02%3A_Ionic_Bonding.txt |
Learning Objectives
• To understand the relationship between bond order, bond length, and bond energy.
In proposing his theory that octets can be completed by two atoms sharing electron pairs, Lewis provided scientists with the first description of covalent bonding. In this section, we expand on this and describe some of the properties of covalent bonds. The general properties of Ionic substances are:
• usually brittle
• high melting point
• organized into an ordered lattice of atoms, which can be cleaved along a smooth line
However, the vast majority of chemical substances are not ionic in nature. G.N. Lewis reasoned that an atom might attain a noble gas electron configuration by sharing electrons.
A chemical bond formed by sharing a pair of electrons is called a covalent bond
Lewis Structures
Lewis structures (also known as Lewis dot diagrams, electron dot diagrams, Lewis dot formulas, Lewis dot structures, and electron dot structures) are diagrams that show the bonding between atoms of a molecule and the lone pairs of electrons that may exist in the molecule. Lewis structures show each atom and its position in the structure of the molecule using its chemical symbol. Lines are drawn between atoms that are bonded to one another (pairs of dots can be used instead of lines). Excess electrons that form lone pairs are represented as pairs of dots, and are placed next to the atoms.
The diatomic hydrogen molecule (H2) is the simplest model of a covalent bond, and is represented in Lewis structures as:
The shared pair of electrons provides each hydrogen atom with two electrons in its valence shell (the 1s) orbital. In a sense, each hydrogen atoms has the electron configuration of the noble gas helium When two chlorine atoms covalently bond to form $Cl_2$, the following sharing of electrons occurs:
Each chlorine atom shared the bonding pair of electrons and achieves the electron configuration of the noble gas argon. In Lewis structures the bonding pair of electrons is usually displayed as a line, and the unshared electrons as dots:
The shared electrons are not located in a fixed position between the nuclei. In the case of the $H_2$ compound, the electron density is concentrated between the two nuclei:
The two atoms are bound into the $H_2$ molecule mainly due to the attraction of the positively charged nuclei for the negatively charged electron cloud located between them. Examples of hydride compounds of the above elements (covalent bonds with hydrogen:
Multiple bonds
The sharing of a pair of electrons represents a single covalent bond, usually just referred to as a single bond. However, in many molecules atoms attain complete octets by sharing more than one pair of electrons between them:
• Two electron pairs shared a double bond
• Three electron pairs shared a triple bond
Because each nitrogen contains 5 valence electrons, they need to share 3 pairs to each achieve a valence octet. N2 is fairly inert, due to the strong triple bond between the two nitrogen atoms and the N - N bond distance in N2 is 1.10 Å (fairly short). From a study of various Nitrogen containing compounds bond distance as a function of bond type can be summarized as follows:
• $\ce{N-N}$: 1.47Å
• $\ce{N=N}$: 1.24Å
• $\ce{N:=N}$:1.10Å
For the nonmetals (and the 's' block metals) the number of valence electrons is equal to the group number:
nonmetals (and the 's' block metals) the number of valence electrons is equal to the group number
Element Group Valence electrons Bonds needed to form valence octet
aF 17 7 1
O 16 6 2
N 15 5 3
C 14 4 4
Thus, the Lewis bonds successfully describe the covalent interactions between various nonmetal elements. When we draw Lewis structures, we place one, two, or three pairs of electrons between adjacent atoms. In the Lewis bonding model, the number of electron pairs that hold two atoms together is called the bond order. For a single bond, such as the C–C bond in H3C–CH3, the bond order is one. For a double bond (such as H2C=CH2), the bond order is two. For a triple bond, such as HC≡CH, the bond order is three.
When analogous bonds in similar compounds are compared, bond length decreases as bond order increases. The bond length data in Table $1$, for example, show that the C–C distance in H3C–CH3 (153.5 pm) is longer than the distance in H2C=CH2 (133.9 pm), which in turn is longer than that in HC≡CH (120.3 pm). Additionally, as noted in Section 8.5, molecules or ions whose bonding must be described using resonance structures usually have bond distances that are intermediate between those of single and double bonds, as we demonstrated with the C–C distances in benzene. The relationship between bond length and bond order is not linear, however. A double bond is not half as long as a single bond, and the length of a C=C bond is not the average of the lengths of C≡C and C–C bonds. Nevertheless, as bond orders increase, bond lengths generally decrease.
Table $1$: Bond Lengths and Bond Dissociation Energies for Bonds with Different Bond Orders in Selected Gas-Phase Molecules at 298 K
Compound Bond Order Bond Length (pm) Bond Dissociation Energy (kJ/mol) Compound Bond Order Bond Length (pm) Bond Dissociation Energy (kJ/mol)
H3C–CH3 1 153.5 376 H3C–NH2 1 147.1 331
H2C=CH2 2 133.9 728 H2C=NH 2 127.3 644
HC≡CH 3 120.3 965 HC≡N 3 115.3 937
H2N–NH2 1 144.9 275.3 H3C–OH 1 142.5 377
HN=NH 2 125.2 456 H2C=O 2 120.8 732
N≡N 3 109.8 945.3 O=C=O 2 116.0 799
HO–OH 1 147.5 213 C≡O 3 112.8 1076.5
O=O 2 120.7 498.4
Sources: Data from CRC Handbook of Chemistry and Physics (2004); Lange’s Handbook of Chemistry (2005); CCCBDB [cccbdb.nist.gov].
As a general rule, the distance between bonded atoms decreases as the number of shared electron pairs increases
The Relationship between Bond Order & Bond Energy
As shown in Table $1$, triple bonds between like atoms are shorter than double bonds, and because more energy is required to completely break all three bonds than to completely break two, a triple bond is also stronger than a double bond. Similarly, double bonds between like atoms are stronger and shorter than single bonds. Bonds of the same order between different atoms show a wide range of bond energies, however.
Table $1$: Average Bond Energies (kJ/mol) for Commonly Encountered Bonds at 273 K
Single Bonds Multiple Bonds
H–H 432 C–C 346 N–N ≈167 O–O ≈142 F–F 155 C=C 602
H–C 411 C–Si 318 N–O 201 O–F 190 F–Cl 249 C≡C 835
H–Si 318 C–N 305 N–F 283 O–Cl 218 F–Br 249 C=N 615
H–N 386 C–O 358 N–Cl 313 O–Br 201 F–I 278 C≡N 887
H–P ≈322 C–S 272 N–Br 243 O–I 201 Cl–Cl 240 C=O 749
H–O 459 C–F 485 P–P 201 S–S 226 Cl–Br 216 C≡O 1072
H–S 363 C–Cl 327 S–F 284 Cl–I 208 N=N 418
H–F 565 C–Br 285 S–Cl 255 Br–Br 190 N≡N 942
H–Cl 428 C–I 213 S–Br 218 Br–I 175 N=O 607
H–Br 362 Si–Si 222 I–I 149 O=O 494
H–I 295 Si–O 452 S=O 532
Source: Data from J. E. Huheey, E. A. Keiter, and R. L. Keiter, Inorganic Chemistry, 4th ed. (1993).
Table $2$ lists the average values for some commonly encountered bonds. Although the values shown vary widely, we can observe four trends:
1. Bonds between hydrogen and atoms in the same column of the periodic table decrease in strength as we go down the column. Thus an H–F bond is stronger than an H–I bond, H–C is stronger than H–Si, H–N is stronger than H–P, H–O is stronger than H–S, and so forth. The reason for this is that the region of space in which electrons are shared between two atoms becomes proportionally smaller as one of the atoms becomes larger (part (a) in Figure 8.3.1).
2. Bonds between like atoms usually become weaker as we go down a column (important exceptions are noted later). For example, the C–C single bond is stronger than the Si–Si single bond, which is stronger than the Ge–Ge bond, and so forth. As two bonded atoms become larger, the region between them occupied by bonding electrons becomes proportionally smaller, as illustrated in part (b) in Figure 8.3.1. Noteworthy exceptions are single bonds between the period 2 atoms of groups 15, 16, and 17 (i.e., N, O, F), which are unusually weak compared with single bonds between their larger congeners. It is likely that the N–N, O–O, and F–F single bonds are weaker than might be expected due to strong repulsive interactions between lone pairs of electrons on adjacent atoms. The trend in bond energies for the halogens is therefore$\ce{Cl\bond{-}Cl > Br\bond{-}Br > F\bond{-}F > I–I} \nonumber$ Similar effects are also seen for the O–O versus S–S and for N–N versus P–P single bonds.
Bonds between hydrogen and atoms in a given column in the periodic table are weaker down the column; bonds between like atoms usually become weaker down a column.
1. Because elements in periods 3 and 4 rarely form multiple bonds with themselves, their multiple bond energies are not accurately known. Nonetheless, they are presumed to be significantly weaker than multiple bonds between lighter atoms of the same families. Compounds containing an Si=Si double bond, for example, have only recently been prepared, whereas compounds containing C=C double bonds are one of the best-studied and most important classes of organic compounds.
2. Multiple bonds between carbon, oxygen, or nitrogen and a period 3 element such as phosphorus or sulfur tend to be unusually strong. In fact, multiple bonds of this type dominate the chemistry of the period 3 elements of groups 15 and 16. Multiple bonds to phosphorus or sulfur occur as a result of d-orbital interactions, e..g, for the SO42 ion. In contrast, silicon in group 14 has little tendency to form discrete silicon–oxygen double bonds. Consequently, SiO2 has a three-dimensional network structure in which each silicon atom forms four Si–O single bonds, which makes the physical and chemical properties of SiO2 very different from those of CO2.
Bond strengths increase as bond order increases, while bond distances decrease.
The Relationship between Molecular Structure and Bond Energy
Bond energy is defined as the energy required to break a particular bond in a molecule in the gas phase. Its value depends on not only the identity of the bonded atoms but also their environment. Thus the bond energy of a C–H single bond is not the same in all organic compounds. For example, the energy required to break a C–H bond in methane varies by as much as 25% depending on how many other bonds in the molecule have already been broken (Table $2$); that is, the C–H bond energy depends on its molecular environment. Except for diatomic molecules, the bond energies listed in Table $2$ are average values for all bonds of a given type in a range of molecules. Even so, they are not likely to differ from the actual value of a given bond by more than about 10%.
Table $3$: Energies for the Dissociation of Successive C–H Bonds in Methane
Reaction D (kJ/mol)
CH4(g) → CH3(g) + H(g) 439
CH3(g) → CH2(g) + H(g) 462
CH2(g) → CH(g) + H(g) 424
CH(g) → C(g) + H(g) 338
Source: Data from CRC Handbook of Chemistry and Physics (2004).
We can estimate the enthalpy change for a chemical reaction by adding together the average energies of the bonds broken in the reactants and the average energies of the bonds formed in the products and then calculating the difference between the two. If the bonds formed in the products are stronger than those broken in the reactants, then energy will be released in the reaction (ΔHrxn < 0):
$ΔH_{rxn} \approx \sum{\text{(bond energies of bonds broken)}}−\sum{\text{(bond energies of bonds formed)}} \label{8.3.1}$
The ≈ sign is used because we are adding together average bond energies; hence this approach does not give exact values for ΔHrxn.
Let’s consider the reaction of 1 mol of n-heptane (C7H16) with oxygen gas to give carbon dioxide and water. This is one reaction that occurs during the combustion of gasoline:
$CH_3(CH_2)_5CH_{3(l)} + 11 O_{2(g)} \rightarrow 7 CO_{2(g)} + 8 H_2O_{(g)} \label{8.3.2}$
In this reaction, 6 C–C bonds, 16 C–H bonds, and 11 O=O bonds are broken per mole of n-heptane, while 14 C=O bonds (two for each CO2) and 16 O–H bonds (two for each H2O) are formed. The energy changes can be tabulated as follows:
Bonds Broken (kJ/mol) and Bonds Formed (kJ/mol)
Bonds Broken (kJ/mol) Bonds Formed (kJ/mol)
6 C–C 346 × 6 = 2076 14 C=O 799 × 14 = 11,186
16 C–H 411 × 16 = 6576 16 O–H 459 × 16 = 7344
11 O=O 494 × 11 = 5434 Total = 18,530
Total = 14,086
The bonds in the products are stronger than the bonds in the reactants by about 4444 kJ/mol. This means that $ΔH_{rxn}$ is approximately −4444 kJ/mol, and the reaction is highly exothermic (which is not too surprising for a combustion reaction).
If we compare this approximation with the value obtained from measured $ΔH_f^o$ values ($ΔH_{rxn} = −481\;7 kJ/mol$), we find a discrepancy of only about 8%, less than the 10% typically encountered. Chemists find this method useful for calculating approximate enthalpies of reaction for molecules whose actual $ΔH^ο_f$ values are unknown. These approximations can be important for predicting whether a reaction is exothermic or endothermic—and to what degree.
Example $1$
The compound RDX (Research Development Explosive) is a more powerful explosive than dynamite and is used by the military. When detonated, it produces gaseous products and heat according to the following reaction. Use the approximate bond energies in Table 8.3.2 to estimate the $ΔH_{rxn}$ per mole of RDX.
Given: chemical reaction, structure of reactant, and Table 8.3.2.
Asked for: $ΔH_{rxn}$ per mole
Strategy:
1. List the types of bonds broken in RDX, along with the bond energy required to break each type. Multiply the number of each type by the energy required to break one bond of that type and then add together the energies. Repeat this procedure for the bonds formed in the reaction.
2. Use Equation 8.3.1 to calculate the amount of energy consumed or released in the reaction (ΔHrxn).
Solution:
We must add together the energies of the bonds in the reactants and compare that quantity with the sum of the energies of the bonds in the products. A nitro group (–NO2) can be viewed as having one N–O single bond and one N=O double bond, as follows:
In fact, however, both N–O distances are usually the same because of the presence of two equivalent resonance structures.
A We can organize our data by constructing a table:
Bonds Broken (kJ/mol)
Bonds Broken (kJ/mol) Bonds Broken (kJ/mol)
6 C–H 411 × 6 = 2466 6 C=O 799 × 6 = 4794
3 N–N 167 × 3 = 501 6 O–H 459 × 6 = 2754
3 N–O 201 × 3 = 603 Total = 10,374
3 N=O 607 × 3 = 1821
1.5 O=O 494 × 1.5 = 741
Total = 7962
B From Equation 8.3.1, we have
$ΔH_{rxn} \approx \sum{\text{(bond energies of bonds broken)}}−\sum{\text{(bond energies of bonds formed)}} \nonumber$
$= 7962 \; kJ/mol − 10,374 \; kJ/mol \nonumber$
\[=−2412 \;kJ/mol]
Thus this reaction is also highly exothermic.
Exercise $1$
The molecule HCFC-142b, a hydrochlorofluorocarbon used in place of chlorofluorocarbons (CFCs) such as the Freons, can be prepared by adding HCl to 1,1-difluoroethylene:
Use tabulated bond energies to calculate $ΔH_{rxn}$.
Answer
−54 kJ/mol
Summary
The strength of a covalent bond depends on the overlap between the valence orbitals of the bonded atoms. Bond order is the number of electron pairs that hold two atoms together. Single bonds have a bond order of one, and multiple bonds with bond orders of two (a double bond) and three (a triple bond) are quite common. In closely related compounds with bonds between the same kinds of atoms, the bond with the highest bond order is both the shortest and the strongest. In bonds with the same bond order between different atoms, trends are observed that, with few exceptions, result in the strongest single bonds being formed between the smallest atoms. Tabulated values of average bond energies can be used to calculate the enthalpy change of many chemical reactions. If the bonds in the products are stronger than those in the reactants, the reaction is exothermic and vice versa.
• Wikipedia | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/08%3A_Basic_Concepts_of_Chemical_Bonding/8.03%3A_Covalent_Bonding.txt |
Learning Objectives
• To define electronegativity and bond polarity
• To calculate the percent ionic character of a covalent polar bond
The electron pairs shared between two atoms are not necessarily shared equally. For example, while the bonding electron pair is shared equally in the covalent bond in $Cl_2$, in $NaCl$ the 3s electron is stripped from the Na atom and is incorporated into the electronic structure of the Cl atom - and the compound is most accurately described as consisting of individual $Na^+$ and $Cl^-$ ions (ionic bonding). For most covalent substances, their bond character falls between these two extremes. As demonstrated below, the bond polarity is a useful concept for describing the sharing of electrons between atoms within a covalent bond:
• A nonpolar covalent bond is one in which the electrons are shared equally between two atoms.
• A polar covalent bond is one in which one atom has a greater attraction for the electrons than the other atom. If this relative attraction is great enough, then the bond is an ionic bond.
Electronegativity
The elements with the highest ionization energies are generally those with the most negative electron affinities, which are located toward the upper right corner of the periodic table. Conversely, the elements with the lowest ionization energies are generally those with the least negative electron affinities and are located in the lower left corner of the periodic table.
Because the tendency of an element to gain or lose electrons is so important in determining its chemistry, various methods have been developed to quantitatively describe this tendency. The most important method uses a measurement called electronegativity (represented by the Greek letter chi, χ, pronounced “ky” as in “sky”), defined as the relative ability of an atom to attract electrons to itself in a chemical compound. Elements with high electronegativities tend to acquire electrons in chemical reactions and are found in the upper right corner of the periodic table. Elements with low electronegativities tend to lose electrons in chemical reactions and are found in the lower left corner of the periodic table.
Unlike ionization energy or electron affinity, the electronegativity of an atom is not a simple, fixed property that can be directly measured in a single experiment. In fact, an atom’s electronegativity should depend to some extent on its chemical environment because the properties of an atom are influenced by its neighbors in a chemical compound. Nevertheless, when different methods for measuring the electronegativity of an atom are compared, they all tend to assign similar relative values to a given element. For example, all scales predict that fluorine has the highest electronegativity and cesium the lowest of the stable elements, which suggests that all the methods are measuring the same fundamental property.
Electronegativity is defined as the ability of an atom in a particular molecule to attract electrons to itself. The greater the value, the greater the attractiveness for electrons.
Electronegativity is a function of:
1. the atom's ionization energy (how strongly the atom holds on to its own electrons) and
2. the atom's electron affinity (how strongly the atom attracts other electrons).
Both of these are properties of the isolated atom. An element will be highly electronegative if it has a large (negative) electron affinity and a high ionization energy (always endothermic, or positive for neutral atoms). Thus, it will attract electrons from other atoms and resist having its own electrons attracted away.
The Pauling Electronegativity Scale
The original electronegativity scale, developed in the 1930s by Linus Pauling (1901– 1994) was based on measurements of the strengths of covalent bonds between different elements. Pauling arbitrarily set the electronegativity of fluorine at 4.0 (although today it has been refined to 3.98), thereby creating a scale in which all elements have values between 0 and 4.0.
Periodic variations in Pauling’s electronegativity values are illustrated in Figures $1$ and $2$. If we ignore the inert gases and elements for which no stable isotopes are known, we see that fluorine ($\chi = 3.98$) is the most electronegative element and cesium is the least electronegative nonradioactive element ($\chi = 0.79$). Because electronegativities generally increase diagonally from the lower left to the upper right of the periodic table, elements lying on diagonal lines running from upper left to lower right tend to have comparable values (e.g., O and Cl and N, S, and Br).
Figure $2$: Pauling Electronegativity Values of the s-, p-, d-, and f-Block Elements. Values for most of the actinides are approximate. Elements for which no data are available are shown in gray. Source: Data from L. Pauling, The Nature of the Chemical Bond, 3rd ed. (1960).
Linus Pauling (1901-1994)
When he was nine, Pauling’s father died, and his mother tried to convince him to quit school to support the family. He did not quit school, but was later denied a high school degree, and had to work several jobs to put himself through college. Pauling would go on to become one of the most influential chemists of the century if not all time. He won two Nobel Prizes, one for chemistry in 1954 and one for peace in 1962.
Pauling’s method is limited by the fact that many elements do not form stable covalent compounds with other elements; hence their electronegativities cannot be measured by his method. Other definitions have since been developed that address this problem, e.g., the Mulliken, Allred-Rochow, and Allen electronegativity scales. The Mulliken electronegativity of an element is the average of its first ionization energy and the absolute value of its electron affinity, showing the relationship between electronegativity and these other periodic properties.
Electronegativity Differences between Metals and Nonmetals
An element’s electronegativity provides us with a single value that we can use to characterize the chemistry of an element. Elements with a high electronegativity (χ ≥ 2.2 in Figure $2$) have very negative affinities and large ionization potentials, so they are generally nonmetals and electrical insulators that tend to gain electrons in chemical reactions (i.e., they are oxidants). In contrast, elements with a low electronegativity ($\chi \le 1.8$) have electron affinities that have either positive or small negative values and small ionization potentials, so they are generally metals and good electrical conductors that tend to lose their valence electrons in chemical reactions (i.e., they are reductants). In between the metals and nonmetals, along the heavy diagonal line running from B to At is a group of elements with intermediate electronegativities (χ ~ 2.0). These are the metalloids (or semimetals), elements that have some of the chemical properties of both nonmetals and metals. The distinction between metals and nonmetals is one of the most fundamental we can make in categorizing the elements and predicting their chemical behavior. Figure $3$ shows the strong correlation between electronegativity values, metallic versus nonmetallic character, and location in the periodic table.
Electronegativity values increase from lower left to upper right in the periodic table.
The rules for assigning oxidation states(opens in new window) are based on the relative electronegativities of the elements; the more electronegative element in a binary compound is assigned a negative oxidation state. As we shall see, electronegativity values are also used to predict bond energies, bond polarities, and the kinds of reactions that compounds undergo.
Example $1$: Increasing Electronegativity
On the basis of their positions in the periodic table, arrange Cl, Se, Si, and Sr in order of increasing electronegativity and classify each as a metal, a nonmetal, or a metalloid.
Given: four elements
Asked for: order by increasing electronegativity and classification
Strategy:
1. Locate the elements in the periodic table. From their diagonal positions from lower left to upper right, predict their relative electronegativities.
2. Arrange the elements in order of increasing electronegativity.
3. Classify each element as a metal, a nonmetal, or a metalloid according to its location about the diagonal belt of metalloids running from B to At.
Solution:
A Electronegativity increases from lower left to upper right in the periodic table (Figure 8.4.2). Because Sr lies far to the left of the other elements given, we can predict that it will have the lowest electronegativity. Because Cl lies above and to the right of Se, we can predict that χCl > χSe. Because Si is located farther from the upper right corner than Se or Cl, its electronegativity should be lower than those of Se and Cl but greater than that of Sr. B The overall order is therefore χSr < χSi < χSe < χCl.
C To classify the elements, we note that Sr lies well to the left of the diagonal belt of metalloids running from B to At; while Se and Cl lie to the right and Si lies in the middle. We can predict that Sr is a metal, Si is a metalloid, and Se and Cl are nonmetals.
Exercise $1$
On the basis of their positions in the periodic table, arrange Ge, N, O, Rb, and Zr in order of increasing electronegativity and classify each as a metal, a nonmetal, or a metalloid.
Answer
Rb < Zr < Ge < N < O; metals (Rb, Zr); metalloid (Ge); nonmetal (N, O)
Percent Ionic Character of a Covalent polar bond
The two idealized extremes of chemical bonding: (1) ionic bonding—in which one or more electrons are transferred completely from one atom to another, and the resulting ions are held together by purely electrostatic forces—and (2) covalent bonding, in which electrons are shared equally between two atoms. Most compounds, however, have polar covalent bonds, which means that electrons are shared unequally between the bonded atoms. Figure $4$ compares the electron distribution in a polar covalent bond with those in an ideally covalent and an ideally ionic bond. Recall that a lowercase Greek delta ($\delta$) is used to indicate that a bonded atom possesses a partial positive charge, indicated by $\delta^+$, or a partial negative charge, indicated by $\delta^-$, and a bond between two atoms that possess partial charges is a polar bond.
Bond Polarity
The polarity of a bond—the extent to which it is polar—is determined largely by the relative electronegativities of the bonded atoms. Electronegativity (χ) was defined as the ability of an atom in a molecule or an ion to attract electrons to itself. Thus there is a direct correlation between electronegativity and bond polarity. A bond is nonpolar if the bonded atoms have equal electronegativities. If the electronegativities of the bonded atoms are not equal, however, the bond is polarized toward the more electronegative atom. A bond in which the electronegativity of B (χB) is greater than the electronegativity of A (χA), for example, is indicated with the partial negative charge on the more electronegative atom:
$\begin{matrix} _{less\; electronegative}& & _{more\; electronegative}\ A\; \; &-& B\; \; \; \; \ ^{\delta ^{+}} & & ^{\delta ^{-}} \end{matrix} \label{8.4.1}$
One way of estimating the ionic character of a bond—that is, the magnitude of the charge separation in a polar covalent bond—is to calculate the difference in electronegativity between the two atoms: Δχ = χB − χA.
To predict the polarity of the bonds in Cl2, HCl, and NaCl, for example, we look at the electronegativities of the relevant atoms: χCl = 3.16, χH = 2.20, and χNa = 0.93. Cl2 must be nonpolar because the electronegativity difference (Δχ) is zero; hence the two chlorine atoms share the bonding electrons equally. In NaCl, Δχ is 2.23. This high value is typical of an ionic compound (Δχ ≥ ≈1.5) and means that the valence electron of sodium has been completely transferred to chlorine to form Na+ and Cl ions. In HCl, however, Δχ is only 0.96. The bonding electrons are more strongly attracted to the more electronegative chlorine atom, and so the charge distribution is
$\begin{matrix} _{\delta ^{+}}& & _{\delta ^{-}}\ H\; \; &-& Cl \end{matrix} \nonumber$
Remember that electronegativities are difficult to measure precisely and different definitions produce slightly different numbers. In practice, the polarity of a bond is usually estimated rather than calculated.
Bond polarity and ionic character increase with an increasing difference in electronegativity.
As with bond energies, the electronegativity of an atom depends to some extent on its chemical environment. It is therefore unlikely that the reported electronegativities of a chlorine atom in NaCl, Cl2, ClF5, and HClO4 would be exactly the same.
Dipole Moments
The asymmetrical charge distribution in a polar substance such as HCl produces a dipole moment where $Qr$ in meters (m). is abbreviated by the Greek letter mu (µ). The dipole moment is defined as the product of the partial charge Q on the bonded atoms and the distance r between the partial charges:
$\mu=Qr \label{8.4.2}$
where Q is measured in coulombs (C) and r in meters. The unit for dipole moments is the debye (D):
$1\; D = 3.3356\times 10^{-30}\; C\cdot ·m \label{8.4.3}$
When a molecule with a dipole moment is placed in an electric field, it tends to orient itself with the electric field because of its asymmetrical charge distribution (Figure $4$).
We can measure the partial charges on the atoms in a molecule such as HCl using Equation $\ref{8.4.2}$. If the bonding in HCl were purely ionic, an electron would be transferred from H to Cl, so there would be a full +1 charge on the H atom and a full −1 charge on the Cl atom. The dipole moment of HCl is 1.109 D, as determined by measuring the extent of its alignment in an electric field, and the reported gas-phase H–Cl distance is 127.5 pm. Hence the charge on each atom is
$Q=\dfrac{\mu }{r} =1.109\;\cancel{D}\left ( \dfrac{3.3356\times 10^{-30}\; C\cdot \cancel{m}}{1\; \cancel{D}} \right )\left ( \dfrac{1}{127.8\; \cancel{pm}} \right )\left ( \dfrac{1\; \cancel{pm}}{10^{-12\;} \cancel{m}} \right )=2.901\times 10^{-20}\;C \label{8.4.4}$
By dividing this calculated value by the charge on a single electron (1.6022 × 10−19 C), we find that the electron distribution in HCl is asymmetric and that effectively it appears that there is a net negative charge on the Cl of about −0.18, effectively corresponding to about 0.18 e. This certainly does not mean that there is a fraction of an electron on the Cl atom, but that the distribution of electron probability favors the Cl atom side of the molecule by about this amount.
$\dfrac{2.901\times 10^{-20}\; \cancel{C}}{1.6022\times 10^{-19}\; \cancel{C}}=0.1811\;e^{-} \label{8.4.5}$
To form a neutral compound, the charge on the H atom must be equal but opposite. Thus the measured dipole moment of HCl indicates that the H–Cl bond has approximately 18% ionic character (0.1811 × 100), or 82% covalent character. Instead of writing HCl as
$\begin{matrix} _{\delta ^{+}}& & _{\delta ^{-}}\ H\; \; &-& Cl \end{matrix} \nonumber$
we can therefore indicate the charge separation quantitatively as
$\begin{matrix} _{0.18\delta ^{+}}& & _{0.18\delta ^{-}}\ H\; \; &-& Cl \end{matrix} \nonumber$
Our calculated results are in agreement with the electronegativity difference between hydrogen and chlorine χH = 2.20; χCl = 3.16, χCl − χH = 0.96), a value well within the range for polar covalent bonds. We indicate the dipole moment by writing an arrow above the molecule. Mathematically, dipole moments are vectors, and they possess both a magnitude and a direction. The dipole moment of a molecule is the vector sum of the dipoles of the individual bonds. In HCl, for example, the dipole moment is indicated as follows:
The arrow shows the direction of electron flow by pointing toward the more electronegative atom.
The charge on the atoms of many substances in the gas phase can be calculated using measured dipole moments and bond distances. Figure $6$ shows a plot of the percent ionic character versus the difference in electronegativity of the bonded atoms for several substances. According to the graph, the bonding in species such as NaCl(g) and CsF(g) is substantially less than 100% ionic in character. As the gas condenses into a solid, however, dipole–dipole interactions between polarized species increase the charge separations. In the crystal, therefore, an electron is transferred from the metal to the nonmetal, and these substances behave like classic ionic compounds. The data in Figure $6$ show that diatomic species with an electronegativity difference of less than 1.5 are less than 50% ionic in character, which is consistent with our earlier description of these species as containing polar covalent bonds. The use of dipole moments to determine the ionic character of a polar bond is illustrated in Example $2$.
Example $2$
In the gas phase, NaCl has a dipole moment of 9.001 D and an Na–Cl distance of 236.1 pm. Calculate the percent ionic character in NaCl.
Given: chemical species, dipole moment, and internuclear distance
Asked for: percent ionic character
Strategy:
A Compute the charge on each atom using the information given and Equation $\ref{8.4.2}$.
B Find the percent ionic character from the ratio of the actual charge to the charge of a single electron.
Solution:
A The charge on each atom is given by
$Q=\dfrac{\mu }{r} =9.001\;\cancel{D}\left ( \dfrac{3.3356\times 10^{-30}\; C\cdot \cancel{m}}{1\; \cancel{D}} \right )\left ( \dfrac{1}{236.1\; \cancel{pm}} \right )\left ( \dfrac{1\; \cancel{pm}}{10^{-12\;} \cancel{m}} \right )=1.272\times 10^{-19}\;C \nonumber$
Thus NaCl behaves as if it had charges of 1.272 × 10−19 C on each atom separated by 236.1 pm.
B The percent ionic character is given by the ratio of the actual charge to the charge of a single electron (the charge expected for the complete transfer of one electron):
$\% \; ionic\; character=\left ( \dfrac{1.272\times 10^{-19}\; \cancel{C}}{1.6022\times 10^{-19}\; \cancel{C}} \right )\left ( 100 \right )=79.39\%\simeq 79\% \nonumber$
Exercise $2$
In the gas phase, silver chloride (AgCl) has a dipole moment of 6.08 D and an Ag–Cl distance of 228.1 pm. What is the percent ionic character in silver chloride?
Answer
55.5%
Summary
Bond polarity and ionic character increase with an increasing difference in electronegativity. The electronegativity (χ) of an element is the relative ability of an atom to attract electrons to itself in a chemical compound and increases diagonally from the lower left of the periodic table to the upper right. The Pauling electronegativity scale is based on measurements of the strengths of covalent bonds between different atoms, whereas the Mulliken electronegativity of an element is the average of its first ionization energy and the absolute value of its electron affinity. Elements with a high electronegativity are generally nonmetals and electrical insulators and tend to behave as oxidants in chemical reactions. Conversely, elements with a low electronegativity are generally metals and good electrical conductors and tend to behave as reductants in chemical reactions.
Compounds with polar covalent bonds have electrons that are shared unequally between the bonded atoms. The polarity of such a bond is determined largely by the relative electronegativites of the bonded atoms. The asymmetrical charge distribution in a polar substance produces a dipole moment, which is the product of the partial charges on the bonded atoms and the distance between them. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/08%3A_Basic_Concepts_of_Chemical_Bonding/8.04%3A_Bond_Polarity_and_Electronegativity.txt |
Learning Objectives
• To use Lewis dot symbols to explain the stoichiometry of a compound
We begin our discussion of the relationship between structure and bonding in covalent compounds by describing the interaction between two identical neutral atoms—for example, the H2 molecule, which contains a purely covalent bond. Each hydrogen atom in H2 contains one electron and one proton, with the electron attracted to the proton by electrostatic forces. As the two hydrogen atoms are brought together, additional interactions must be considered (Figure $1$):
• The electrons in the two atoms repel each other because they have the same charge (
• The electrons in the two atoms repel each other because they have the same charge (E > 0).
• Similarly, the protons in adjacent atoms repel each other (E > 0).
• The electron in one atom is attracted to the oppositely charged proton in the other atom and vice versa (E < 0). Recall that it is impossible to specify precisely the position of the electron in either hydrogen atom. Hence the quantum mechanical probability distributions must be used.
A plot of the potential energy of the system as a function of the internuclear distance (Figure $2$) shows that the system becomes more stable (the energy of the system decreases) as two hydrogen atoms move toward each other from r = ∞, until the energy reaches a minimum at r = r0 (the observed internuclear distance in H2 is 74 pm). Thus at intermediate distances, proton–electron attractive interactions dominate, but as the distance becomes very short, electron–electron and proton–proton repulsive interactions cause the energy of the system to increase rapidly. Notice the similarity between Figures $1$ and $2$, which described a system containing two oppositely charged ions. The shapes of the energy versus distance curves in the two figures are similar because they both result from attractive and repulsive forces between charged entities.
At long distances, both attractive and repulsive interactions are small. As the distance between the atoms decreases, the attractive electron–proton interactions dominate, and the energy of the system decreases. At the observed bond distance, the repulsive electron–electron and proton–proton interactions just balance the attractive interactions, preventing a further decrease in the internuclear distance. At very short internuclear distances, the repulsive interactions dominate, making the system less stable than the isolated atoms.
Using Lewis Dot Symbols to Describe Covalent Bonding
The valence electron configurations of the constituent atoms of a covalent compound are important factors in determining its structure, stoichiometry, and properties. For example, chlorine, with seven valence electrons, is one electron short of an octet. If two chlorine atoms share their unpaired electrons by making a covalent bond and forming Cl2, they can each complete their valence shell:
Each chlorine atom now has an octet. The electron pair being shared by the atoms is called a bonding pair; the other three pairs of electrons on each chlorine atom are called lone pairs. Lone pairs are not involved in covalent bonding. If both electrons in a covalent bond come from the same atom, the bond is called a coordinate covalent bond. Examples of this type of bonding are presented in Section 8.6 when we discuss atoms with less than an octet of electrons.
We can illustrate the formation of a water molecule from two hydrogen atoms and an oxygen atom using Lewis dot symbols:
The structure on the right is the Lewis electron structure, or Lewis structure, for H2O. With two bonding pairs and two lone pairs, the oxygen atom has now completed its octet. Moreover, by sharing a bonding pair with oxygen, each hydrogen atom now has a full valence shell of two electrons. Chemists usually indicate a bonding pair by a single line, as shown here for our two examples:
The following procedure can be used to construct Lewis electron structures for more complex molecules and ions:
1. Arrange the atoms to show specific connections. When there is a central atom, it is usually the least electronegative element in the compound. Chemists usually list this central atom first in the chemical formula (as in CCl4 and CO32, which both have C as the central atom), which is another clue to the compound’s structure. Hydrogen and the halogens are almost always connected to only one other atom, so they are usually terminal rather than central.
2. Determine the total number of valence electrons in the molecule or ion. Add together the valence electrons from each atom. (Recall that the number of valence electrons is indicated by the position of the element in the periodic table.) If the species is a polyatomic ion, remember to add or subtract the number of electrons necessary to give the total charge on the ion. For CO32, for example, we add two electrons to the total because of the −2 charge.
3. Place a bonding pair of electrons between each pair of adjacent atoms to give a single bond. In H2O, for example, there is a bonding pair of electrons between oxygen and each hydrogen.
4. Beginning with the terminal atoms, add enough electrons to each atom to give each atom an octet (two for hydrogen). These electrons will usually be lone pairs.
5. If any electrons are left over, place them on the central atom. We will explain later that some atoms are able to accommodate more than eight electrons.
6. If the central atom has fewer electrons than an octet, use lone pairs from terminal atoms to form multiple (double or triple) bonds to the central atom to achieve an octet. This will not change the number of electrons on the terminal atoms.
Now let’s apply this procedure to some particular compounds, beginning with one we have already discussed.
The central atom is usually the least electronegative element in the molecule or ion; hydrogen and the halogens are usually terminal.
The $H_2O$ Molecule
1. Because H atoms are almost always terminal, the arrangement within the molecule must be HOH.
2. Each H atom (group 1) has 1 valence electron, and the O atom (group 16) has 6 valence electrons, for a total of 8 valence electrons.
3. Placing one bonding pair of electrons between the O atom and each H atom gives H:O:H, with 4 electrons left over.
4. Each H atom has a full valence shell of 2 electrons.
5. Adding the remaining 4 electrons to the oxygen (as two lone pairs) gives the following structure:
This is the Lewis structure we drew earlier. Because it gives oxygen an octet and each hydrogen two electrons, we do not need to use step 6.
The $OCl^−$ Ion
1. With only two atoms in the molecule, there is no central atom.
2. Oxygen (group 16) has 6 valence electrons, and chlorine (group 17) has 7 valence electrons; we must add one more for the negative charge on the ion, giving a total of 14 valence electrons.
3. Placing a bonding pair of electrons between O and Cl gives O:Cl, with 12 electrons left over.
4. If we place six electrons (as three lone pairs) on each atom, we obtain the following structure:
Each atom now has an octet of electrons, so steps 5 and 6 are not needed. The Lewis electron structure is drawn within brackets as is customary for an ion, with the overall charge indicated outside the brackets, and the bonding pair of electrons is indicated by a solid line. OCl is the hypochlorite ion, the active ingredient in chlorine laundry bleach and swimming pool disinfectant.
The $CH_2O$ Molecule
1. Because carbon is less electronegative than oxygen and hydrogen is normally terminal, C must be the central atom. One possible arrangement is as follows:
2. Each hydrogen atom (group 1) has one valence electron, carbon (group 14) has 4 valence electrons, and oxygen (group 16) has 6 valence electrons, for a total of [(2)(1) + 4 + 6] = 12 valence electrons.
3. Placing a bonding pair of electrons between each pair of bonded atoms gives the following:
Six electrons are used, and 6 are left over.
4. Adding all 6 remaining electrons to oxygen (as three lone pairs) gives the following:
Although oxygen now has an octet and each hydrogen has 2 electrons, carbon has only 6 electrons.
5. There are no electrons left to place on the central atom.
6. To give carbon an octet of electrons, we use one of the lone pairs of electrons on oxygen to form a carbon–oxygen double bond:
Both the oxygen and the carbon now have an octet of electrons, so this is an acceptable Lewis electron structure. The O has two bonding pairs and two lone pairs, and C has four bonding pairs. This is the structure of formaldehyde, which is used in embalming fluid.
An alternative structure can be drawn with one H bonded to O. Formal charges, discussed later in this section, suggest that such a structure is less stable than that shown previously.
Example $1$
Write the Lewis electron structure for each species.
1. NCl3
2. S22
3. NOCl
Given: chemical species
Asked for: Lewis electron structures
Strategy:
Use the six-step procedure to write the Lewis electron structure for each species.
Solution:
1. Nitrogen is less electronegative than chlorine, and halogen atoms are usually terminal, so nitrogen is the central atom. The nitrogen atom (group 15) has 5 valence electrons and each chlorine atom (group 17) has 7 valence electrons, for a total of 26 valence electrons. Using 2 electrons for each N–Cl bond and adding three lone pairs to each Cl account for (3 × 2) + (3 × 2 × 3) = 24 electrons. Rule 5 leads us to place the remaining 2 electrons on the central N:
Nitrogen trichloride is an unstable oily liquid once used to bleach flour; this use is now prohibited in the United States.
2. In a diatomic molecule or ion, we do not need to worry about a central atom. Each sulfur atom (group 16) contains 6 valence electrons, and we need to add 2 electrons for the −2 charge, giving a total of 14 valence electrons. Using 2 electrons for the S–S bond, we arrange the remaining 12 electrons as three lone pairs on each sulfur, giving each S atom an octet of electrons:
3. Because nitrogen is less electronegative than oxygen or chlorine, it is the central atom. The N atom (group 15) has 5 valence electrons, the O atom (group 16) has 6 valence electrons, and the Cl atom (group 17) has 7 valence electrons, giving a total of 18 valence electrons. Placing one bonding pair of electrons between each pair of bonded atoms uses 4 electrons and gives the following:
Adding three lone pairs each to oxygen and to chlorine uses 12 more electrons, leaving 2 electrons to place as a lone pair on nitrogen:
Because this Lewis structure has only 6 electrons around the central nitrogen, a lone pair of electrons on a terminal atom must be used to form a bonding pair. We could use a lone pair on either O or Cl. Because we have seen many structures in which O forms a double bond but none with a double bond to Cl, it is reasonable to select a lone pair from O to give the following:
All atoms now have octet configurations. This is the Lewis electron structure of nitrosyl chloride, a highly corrosive, reddish-orange gas.
Exercise $1$
Write Lewis electron structures for CO2 and SCl2, a vile-smelling, unstable red liquid that is used in the manufacture of rubber.
Using Lewis Electron Structures to Explain Stoichiometry
Lewis dot symbols provide a simple rationalization of why elements form compounds with the observed stoichiometries. In the Lewis model, the number of bonds formed by an element in a neutral compound is the same as the number of unpaired electrons it must share with other atoms to complete its octet of electrons. For the elements of Group 17 (the halogens), this number is one; for the elements of Group 16 (the chalcogens), it is two; for Group 15 elements, three; and for Group 14 elements four. These requirements are illustrated by the following Lewis structures for the hydrides of the lightest members of each group:
Elements may form multiple bonds to complete an octet. In ethylene, for example, each carbon contributes two electrons to the double bond, giving each carbon an octet (two electrons/bond × four bonds = eight electrons). Neutral structures with fewer or more bonds exist, but they are unusual and violate the octet rule.
Allotropes of an element can have very different physical and chemical properties because of different three-dimensional arrangements of the atoms; the number of bonds formed by the component atoms, however, is always the same. As noted at the beginning of the chapter, diamond is a hard, transparent solid; graphite is a soft, black solid; and the fullerenes have open cage structures. Despite these differences, the carbon atoms in all three allotropes form four bonds, in accordance with the octet rule.
Lewis structures explain why the elements of groups 14–17 form neutral compounds with four, three, two, and one bonded atom(s), respectively.
Elemental phosphorus also exists in three forms: white phosphorus, a toxic, waxy substance that initially glows and then spontaneously ignites on contact with air; red phosphorus, an amorphous substance that is used commercially in safety matches, fireworks, and smoke bombs; and black phosphorus, an unreactive crystalline solid with a texture similar to graphite (Figure $3$). Nonetheless, the phosphorus atoms in all three forms obey the octet rule and form three bonds per phosphorus atom.
Formal Charges
It is sometimes possible to write more than one Lewis structure for a substance that does not violate the octet rule, as we saw for CH2O, but not every Lewis structure may be equally reasonable. In these situations, we can choose the most stable Lewis structure by considering the formal charge on the atoms, which is the difference between the number of valence electrons in the free atom and the number assigned to it in the Lewis electron structure. The formal charge is a way of computing the charge distribution within a Lewis structure; the sum of the formal charges on the atoms within a molecule or an ion must equal the overall charge on the molecule or ion. A formal charge does not represent a true charge on an atom in a covalent bond but is simply used to predict the most likely structure when a compound has more than one valid Lewis structure.
To calculate formal charges, we assign electrons in the molecule to individual atoms according to these rules:
• Nonbonding electrons are assigned to the atom on which they are located.
• Bonding electrons are divided equally between the bonded atoms.
For each atom, we then compute a formal charge:
$\begin{matrix} formal\; charge= & valence\; e^{-}- & \left ( non-bonding\; e^{-}+\frac{bonding\;e^{-}}{2} \right )\ & ^{\left ( free\; atom \right )} & ^{\left ( atom\; in\; Lewis\; structure \right )} \end{matrix} \label{8.5.1}$ (atom in Lewis structure)
To illustrate this method, let’s calculate the formal charge on the atoms in ammonia (NH3) whose Lewis electron structure is as follows:
A neutral nitrogen atom has five valence electrons (it is in group 15). From its Lewis electron structure, the nitrogen atom in ammonia has one lone pair and shares three bonding pairs with hydrogen atoms, so nitrogen itself is assigned a total of five electrons [2 nonbonding e + (6 bonding e/2)]. Substituting into Equation $\ref{8.5.2}$, we obtain
$formal\; charge\left ( N \right )=5\; valence\; e^{-}-\left ( 2\; non-bonding\; e^{-} +\dfrac{6\; bonding\; e^{-}}{2} \right )=0 \label{8.5.2}$
A neutral hydrogen atom has one valence electron. Each hydrogen atom in the molecule shares one pair of bonding electrons and is therefore assigned one electron [0 nonbonding e + (2 bonding e/2)]. Using Equation $\ref{8.5.2}$ to calculate the formal charge on hydrogen, we obtain
$formal\; charge\left ( H \right )=1\; valence\; e^{-}-\left ( 0\; non-bonding\; e^{-} +\dfrac{2\; bonding\; e^{-}}{2} \right )=0 \label{8.5.3}$
The hydrogen atoms in ammonia have the same number of electrons as neutral hydrogen atoms, and so their formal charge is also zero. Adding together the formal charges should give us the overall charge on the molecule or ion. In this example, the nitrogen and each hydrogen has a formal charge of zero. When summed the overall charge is zero, which is consistent with the overall charge on the NH3 molecule.
An atom, molecule, or ion has a formal charge of zero if it has the number of bonds that is typical for that species.
Typically, the structure with the most charges on the atoms closest to zero is the more stable Lewis structure. In cases where there are positive or negative formal charges on various atoms, stable structures generally have negative formal charges on the more electronegative atoms and positive formal charges on the less electronegative atoms. The next example further demonstrates how to calculate formal charges.
Example $2$: The Ammonium Ion
Calculate the formal charges on each atom in the NH4+ ion.
Given: chemical species
Asked for: formal charges
Strategy:
Identify the number of valence electrons in each atom in the NH4+ ion. Use the Lewis electron structure of NH4+ to identify the number of bonding and nonbonding electrons associated with each atom and then use Equation $\ref{8.5.2}$ to calculate the formal charge on each atom.
Solution:
The Lewis electron structure for the NH4+ ion is as follows:
The nitrogen atom shares four bonding pairs of electrons, and a neutral nitrogen atom has five valence electrons. Using Equation $\ref{8.5.1}$, the formal charge on the nitrogen atom is therefore
$formal\; charge\left ( N \right )=5-\left ( 0+\dfrac{8}{2} \right )=0 \nonumber$
Each hydrogen atom in has one bonding pair. The formal charge on each hydrogen atom is therefore
$formal\; charge\left ( H \right )=1-\left ( 0+\dfrac{2}{2} \right )=0 \nonumber$
The formal charges on the atoms in the NH4+ ion are thus
Adding together the formal charges on the atoms should give us the total charge on the molecule or ion. In this case, the sum of the formal charges is 0 + 1 + 0 + 0 + 0 = +1.
Exercise $2$
Write the formal charges on all atoms in BH4.
Answer
If an atom in a molecule or ion has the number of bonds that is typical for that atom (e.g., four bonds for carbon), its formal charge is zero. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/08%3A_Basic_Concepts_of_Chemical_Bonding/8.05%3A_Drawing_Lewis_Structures.txt |
Learning Objectives
• To understand the concept of resonance.
Resonance structures are a set of two or more Lewis Structures that collectively describe the electronic bonding of a single polyatomic species including fractional bonds and fractional charges. Resonance structures are capable of describing delocalized electrons that cannot be expressed by a single Lewis formula with an integral number of covalent bonds.
Sometimes one Lewis Structure is not Enough
Sometimes, even when formal charges are considered, the bonding in some molecules or ions cannot be described by a single Lewis structure. Resonance is a way of describing delocalized electrons within certain molecules or polyatomic ions where the bonding cannot be expressed by a single Lewis formula. A molecule or ion with such delocalized electrons is represented by several contributing structures (also called resonance structures or canonical forms). Such is the case for ozone (\(\ce{O3}\)), an allotrope of oxygen with a V-shaped structure and an O–O–O angle of 117.5°.
Ozone (\(O_3\))
1. We know that ozone has a V-shaped structure, so one O atom is central:
2. Each O atom has 6 valence electrons, for a total of 18 valence electrons.
3. Assigning one bonding pair of electrons to each oxygen–oxygen bond gives
with 14 electrons left over.
4. If we place three lone pairs of electrons on each terminal oxygen, we obtain
and have 2 electrons left over.
5. At this point, both terminal oxygen atoms have octets of electrons. We therefore place the last 2 electrons on the central atom:
6. The central oxygen has only 6 electrons. We must convert one lone pair on a terminal oxygen atom to a bonding pair of electrons—but which one? Depending on which one we choose, we obtain either
Which is correct? In fact, neither is correct. Both predict one O–O single bond and one O=O double bond. As you will learn, if the bonds were of different types (one single and one double, for example), they would have different lengths. It turns out, however, that both O–O bond distances are identical, 127.2 pm, which is shorter than a typical O–O single bond (148 pm) and longer than the O=O double bond in O2 (120.7 pm).
Equivalent Lewis dot structures, such as those of ozone, are called resonance structures. The position of the atoms is the same in the various resonance structures of a compound, but the position of the electrons is different. Double-headed arrows link the different resonance structures of a compound:
The double-headed arrow indicates that the actual electronic structure is an average of those shown, not that the molecule oscillates between the two structures.
When it is possible to write more than one equivalent resonance structure for a molecule or ion, the actual structure is the average of the resonance structures.
The Carbonate (\(CO_3^{2−} \)) Ion
Like ozone, the electronic structure of the carbonate ion cannot be described by a single Lewis electron structure. Unlike O3, though, the actual structure of CO32 is an average of three resonance structures.
1. Because carbon is the least electronegative element, we place it in the central position:
2. Carbon has 4 valence electrons, each oxygen has 6 valence electrons, and there are 2 more for the −2 charge. This gives 4 + (3 × 6) + 2 = 24 valence electrons.
3. Six electrons are used to form three bonding pairs between the oxygen atoms and the carbon:
4. We divide the remaining 18 electrons equally among the three oxygen atoms by placing three lone pairs on each and indicating the −2 charge:
5. No electrons are left for the central atom.
6. At this point, the carbon atom has only 6 valence electrons, so we must take one lone pair from an oxygen and use it to form a carbon–oxygen double bond. In this case, however, there are three possible choices:
As with ozone, none of these structures describes the bonding exactly. Each predicts one carbon–oxygen double bond and two carbon–oxygen single bonds, but experimentally all C–O bond lengths are identical. We can write resonance structures (in this case, three of them) for the carbonate ion:
The actual structure is an average of these three resonance structures.
The Nitrate (\(NO_3^-\)) ion
1. Count up the valence electrons: (1*5) + (3*6) + 1(ion) = 24 electrons
2. Draw the bond connectivities:
3. Add octet electrons to the atoms bonded to the center atom:
4. Place any leftover electrons (24-24 = 0) on the center atom:
5. Does the central atom have an octet?
• NO, it has 6 electrons
• Add a multiple bond (first try a double bond) to see if the central atom can achieve an octet:
6. Does the central atom have an octet?
• YES
• Are there possible resonance structures? YES
Note: We would expect that the bond lengths in the \(\ce{NO_3^{-}}\) ion to be somewhat shorter than a single bond.
Example \(1\): Benzene
Benzene is a common organic solvent that was previously used in gasoline; it is no longer used for this purpose, however, because it is now known to be a carcinogen. The benzene molecule (\(\ce{C6H6}\)) consists of a regular hexagon of carbon atoms, each of which is also bonded to a hydrogen atom. Use resonance structures to describe the bonding in benzene.
Given: molecular formula and molecular geometry
Asked for: resonance structures
Strategy:
1. Draw a structure for benzene illustrating the bonded atoms. Then calculate the number of valence electrons used in this drawing.
2. Subtract this number from the total number of valence electrons in benzene and then locate the remaining electrons such that each atom in the structure reaches an octet.
3. Draw the resonance structures for benzene.
Solution:
A Each hydrogen atom contributes 1 valence electron, and each carbon atom contributes 4 valence electrons, for a total of (6 × 1) + (6 × 4) = 30 valence electrons. If we place a single bonding electron pair between each pair of carbon atoms and between each carbon and a hydrogen atom, we obtain the following:
Each carbon atom in this structure has only 6 electrons and has a formal charge of +1, but we have used only 24 of the 30 valence electrons.
B If the 6 remaining electrons are uniformly distributed pairwise on alternate carbon atoms, we obtain the following:
Three carbon atoms now have an octet configuration and a formal charge of −1, while three carbon atoms have only 6 electrons and a formal charge of +1. We can convert each lone pair to a bonding electron pair, which gives each atom an octet of electrons and a formal charge of 0, by making three C=C double bonds.
C There are, however, two ways to do this:
Each structure has alternating double and single bonds, but experimentation shows that each carbon–carbon bond in benzene is identical, with bond lengths (139.9 pm) intermediate between those typically found for a C–C single bond (154 pm) and a C=C double bond (134 pm). We can describe the bonding in benzene using the two resonance structures, but the actual electronic structure is an average of the two. The existence of multiple resonance structures for aromatic hydrocarbons like benzene is often indicated by drawing either a circle or dashed lines inside the hexagon:
Exercise \(1\): Nitrite Ion
The sodium salt of nitrite is used to relieve muscle spasms. Draw two resonance structures for the nitrite ion (NO2).
Answer
Resonance structures are particularly common in oxoanions of the p-block elements, such as sulfate and phosphate, and in aromatic hydrocarbons, such as benzene and naphthalene.
Warning
If several reasonable resonance forms for a molecule exists, the "actual electronic structure" of the molecule will probably be intermediate between all the forms that you can draw. The classic example is benzene in Example \(1\). One would expect the double bonds to be shorter than the single bonds, but if one overlays the two structures, you see that one structure has a single bond where the other structure has a double bond. The best measurements that we can make of benzene do not show two bond lengths - instead, they show that the bond length is intermediate between the two resonance structures.
Resonance structures is a mechanism that allows us to use all of the possible resonance structures to try to predict what the actual form of the molecule would be. Single bonds, double bonds, triple bonds, +1 charges, -1 charges, these are our limitations in explaining the structures, and the true forms can be in between - a carbon-carbon bond could be mostly single bond with a little bit of double bond character and a partial negative charge, for example.
Summary
Some molecules have two or more chemically equivalent Lewis electron structures, called resonance structures. Resonance is a mental exercise and method within the Valence Bond Theory of bonding that describes the delocalization of electrons within molecules. These structures are written with a double-headed arrow between them, indicating that none of the Lewis structures accurately describes the bonding but that the actual structure is an average of the individual resonance structures. Resonance structures are used when one Lewis structure for a single molecule cannot fully describe the bonding that takes place between neighboring atoms relative to the empirical data for the actual bond lengths between those atoms. The net sum of valid resonance structures is defined as a resonance hybrid, which represents the overall delocalization of electrons within the molecule. A molecule that has several resonance structures is more stable than one with fewer. Some resonance structures are more favorable than others. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/08%3A_Basic_Concepts_of_Chemical_Bonding/8.06%3A_Resonance_Structures.txt |
Learning Objectives
• To assign a Lewis dot symbol to elements not having an octet of electrons in their compounds.
Three cases can be constructed that do not follow the octet rule, and as such, they are known as the exceptions to the octet rule. Following the Octet Rule for Lewis Dot Structures leads to the most accurate depictions of stable molecular and atomic structures and because of this we always want to use the octet rule when drawing Lewis Dot Structures. However, it is hard to imagine that one rule could be followed by all molecules. There is always an exception, and in this case, three exceptions:
1. When there are an odd number of valence electrons
2. When there are too few valence electrons
3. When there are too many valence electrons
Exception 1: Species with Odd Numbers of Electrons
The first exception to the Octet Rule is when there are an odd number of valence electrons. An example of this would be Nitrogen (II) Oxide also called nitric oxide ($\ce{NO}$. Nitrogen has 5 valence electrons while Oxygen has 6. The total would be 11 valence electrons to be used. The Octet Rule for this molecule is fulfilled in the above example, however that is with 10 valence electrons. The last one does not know where to go. The lone electron is called an unpaired electron. But where should the unpaired electron go? The unpaired electron is usually placed in the Lewis Dot Structure so that each element in the structure will have the lowest formal charge possible. The formal charge is the perceived charge on an individual atom in a molecule when atoms do not contribute equal numbers of electrons to the bonds they participate in.
No formal charge at all is the most ideal situation. An example of a stable molecule with an odd number of valence electrons would be nitric oxide. nitric oxide has 11 valence electrons. If you need more information about formal charges, see Lewis Structures. If we were to imagine nitric oxide had ten valence electrons we would come up with the Lewis Structure (Figure $1$):
Let's look at the formal charges of Figure $2$ based on this Lewis structure. Nitrogen normally has five valence electrons. In Figure $1$, it has two lone pair electrons and it participates in two bonds (a double bond) with oxygen. This results in nitrogen having a formal charge of +1. Oxygen normally has six valence electrons. In Figure $1$, oxygen has four lone pair electrons and it participates in two bonds with nitrogen. Oxygen therefore has a formal charge of 0. The overall molecule here has a formal charge of +1 (+1 for nitrogen, 0 for oxygen. +1 + 0 = +1). However, if we add the eleventh electron to nitrogen (because we want the molecule to have the lowest total formal charge), it will bring both the nitrogen and the molecule's overall charges to zero, the most ideal formal charge situation. That is exactly what is done to get the correct Lewis structure for nitric oxide (Figure $2$):
Free Radicals
There are actually very few stable molecules with odd numbers of electrons that exist, since that unpaired electron is willing to react with other unpaired electrons. Most odd electron species are highly reactive, which we call Free Radicals. Because of their instability, free radicals bond to atoms in which they can take an electron from in order to become stable, making them very chemically reactive. Radicals are found as both reactants and products, but generally react to form more stable molecules as soon as they can. In order to emphasize the existence of the unpaired electron, radicals are denoted with a dot in front of their chemical symbol as with $\cdot OH$, the hydroxyl radical. An example of a radical you may by familiar with already is the gaseous chlorine atom, denoted $\cdot Cl$. Interestingly, an odd Number of Valence Electrons will result in the molecule being paramagnetic.
Exception 2: Incomplete Octets
The second exception to the Octet Rule is when there are too few valence electrons that results in an incomplete Octet. There are even more occasions where the octet rule does not give the most correct depiction of a molecule or ion. This is also the case with incomplete octets. Species with incomplete octets are pretty rare and generally are only found in some beryllium, aluminum, and boron compounds including the boron hydrides. Let's take a look at one such hydride, BH3 (Borane).
If one were to make a Lewis structure for BH3 following the basic strategies for drawing Lewis structures, one would probably come up with this structure (Figure $2$):
The problem with this structure is that boron has an incomplete octet; it only has six electrons around it. Hydrogen atoms can naturally only have only 2 electrons in their outermost shell (their version of an octet), and as such there are no spare electrons to form a double bond with boron. One might surmise that the failure of this structure to form complete octets must mean that this bond should be ionic instead of covalent. However, boron has an electronegativity that is very similar to hydrogen, meaning there is likely very little ionic character in the hydrogen to boron bonds, and as such this Lewis structure, though it does not fulfill the octet rule, is likely the best structure possible for depicting BH3 with Lewis theory. One of the things that may account for BH3's incomplete octet is that it is commonly a transitory species, formed temporarily in reactions that involve multiple steps.
Let's take a look at another incomplete octet situation dealing with boron, BF3 (Boron trifluorine). Like with BH3, the initial drawing of a Lewis structure of BF3 will form a structure where boron has only six electrons around it (Figure $4$).
If you look Figure $4$, you can see that the fluorine atoms possess extra lone pairs that they can use to make additional bonds with boron, and you might think that all you have to do is make one lone pair into a bond and the structure will be correct. If we add one double bond between boron and one of the fluorines we get the following Lewis Structure (Figure $5$):
Each fluorine has eight electrons, and the boron atom has eight as well! Each atom has a perfect octet, right? Not so fast. We must examine the formal charges of this structure. The fluorine that shares a double bond with boron has six electrons around it (four from its two lone pairs of electrons and one each from its two bonds with boron). This is one less electron than the number of valence electrons it would have naturally (Group Seven elements have seven valence electrons), so it has a formal charge of +1. The two flourines that share single bonds with boron have seven electrons around them (six from their three lone pairs and one from their single bonds with boron). This is the same amount as the number of valence electrons they would have on their own, so they both have a formal charge of zero. Finally, boron has four electrons around it (one from each of its four bonds shared with fluorine). This is one more electron than the number of valence electrons that boron would have on its own, and as such boron has a formal charge of -1.
This structure is supported by the fact that the experimentally determined bond length of the boron to fluorine bonds in BF3 is less than what would be typical for a single bond (see Bond Order and Lengths). However, this structure contradicts one of the major rules of formal charges: Negative formal charges are supposed to be found on the more electronegative atom(s) in a bond, but in the structure depicted in Figure $5$, a positive formal charge is found on fluorine, which not only is the most electronegative element in the structure, but the most electronegative element in the entire periodic table ($\chi=4.0$). Boron on the other hand, with the much lower electronegativity of 2.0, has the negative formal charge in this structure. This formal charge-electronegativity disagreement makes this double-bonded structure impossible.
However the large electronegativity difference here, as opposed to in BH3, signifies significant polar bonds between boron and fluorine, which means there is a high ionic character to this molecule. This suggests the possibility of a semi-ionic structure such as seen in Figure $6$:
None of these three structures is the "correct" structure in this instance. The most "correct" structure is most likely a resonance of all three structures: the one with the incomplete octet (Figure $4$), the one with the double bond (Figure $5$), and the one with the ionic bond (Figure $6$). The most contributing structure is probably the incomplete octet structure (due to Figure $5$ being basically impossible and Figure $6$ not matching up with the behavior and properties of BF3). As you can see even when other possibilities exist, incomplete octets may best portray a molecular structure.
As a side note, it is important to note that BF3 frequently bonds with a F- ion in order to form BF4- rather than staying as BF3. This structure completes boron's octet and it is more common in nature. This exemplifies the fact that incomplete octets are rare, and other configurations are typically more favorable, including bonding with additional ions as in the case of BF3 .
Example $1$: $NF_3$
Draw the Lewis structure for boron trifluoride (BF3).
Solution
1. Add electrons (3*7) + 3 = 24
2. Draw connectivities:
3. Add octets to outer atoms:
4. Add extra electrons (24-24=0) to central atom:
5. Does central electron have octet?
• NO. It has 6 electrons
• Add a multiple bond (double bond) to see if central atom can achieve an octet:
6. The central Boron now has an octet (there would be three resonance Lewis structures)
However...
• In this structure with a double bond the fluorine atom is sharing extra electrons with the boron.
• The fluorine would have a '+' partial charge, and the boron a '-' partial charge, this is inconsistent with the electronegativities of fluorine and boron.
• Thus, the structure of BF3, with single bonds, and 6 valence electrons around the central boron is the most likely structure
BF3 reacts strongly with compounds which have an unshared pair of electrons which can be used to form a bond with the boron:
Exception 3: Expanded Valence Shells
More common than incomplete octets are expanded octets where the central atom in a Lewis structure has more than eight electrons in its valence shell. In expanded octets, the central atom can have ten electrons, or even twelve. Molecules with expanded octets involve highly electronegative terminal atoms, and a nonmetal central atom found in the third period or below, which those terminal atoms bond to. For example, $PCl_5$ is a legitimate compound (whereas $NCl_5$) is not:
Expanded valence shells are observed only for elements in period 3 (i.e. n=3) and beyond
The 'octet' rule is based upon available ns and np orbitals for valence electrons (2 electrons in the s orbitals, and 6 in the p orbitals). Beginning with the n=3 principle quantum number, the d orbitals become available (l=2). The orbital diagram for the valence shell of phosphorous is:
Hence, the third period elements occasionally exceed the octet rule by using their empty d orbitals to accommodate additional electrons. Size is also an important consideration:
• The larger the central atom, the larger the number of electrons which can surround it
• Expanded valence shells occur most often when the central atom is bonded to small electronegative atoms, such as F, Cl and O.
There is currently much scientific exploration and inquiry into the reason why expanded valence shells are found. The top area of interest is figuring out where the extra pair(s) of electrons are found. Many chemists think that there is not a very large energy difference between the 3p and 3d orbitals, and as such it is plausible for extra electrons to easily fill the 3d orbital when an expanded octet is more favorable than having a complete octet. This matter is still under hot debate, however and there is even debate as to what makes an expanded octet more favorable than a configuration that follows the octet rule.
One of the situations where expanded octet structures are treated as more favorable than Lewis structures that follow the octet rule is when the formal charges in the expanded octet structure are smaller than in a structure that adheres to the octet rule, or when there are less formal charges in the expanded octet than in the structure a structure that adheres to the octet rule.
Example $2$: The $SO_4^{-2}$ ion
Such is the case for the sulfate ion, SO4-2. A strict adherence to the octet rule forms the following Lewis structure:
If we look at the formal charges on this molecule, we can see that all of the oxygen atoms have seven electrons around them (six from the three lone pairs and one from the bond with sulfur). This is one more electron than the number of valence electrons then they would have normally, and as such each of the oxygens in this structure has a formal charge of -1. Sulfur has four electrons around it in this structure (one from each of its four bonds) which is two electrons more than the number of valence electrons it would have normally, and as such it carries a formal charge of +2.
If instead we made a structure for the sulfate ion with an expanded octet, it would look like this:
Looking at the formal charges for this structure, the sulfur ion has six electrons around it (one from each of its bonds). This is the same amount as the number of valence electrons it would have naturally. This leaves sulfur with a formal charge of zero. The two oxygens that have double bonds to sulfur have six electrons each around them (four from the two lone pairs and one each from the two bonds with sulfur). This is the same amount of electrons as the number of valence electrons that oxygen atoms have on their own, and as such both of these oxygen atoms have a formal charge of zero. The two oxygens with the single bonds to sulfur have seven electrons around them in this structure (six from the three lone pairs and one from the bond to sulfur). That is one electron more than the number of valence electrons that oxygen would have on its own, and as such those two oxygens carry a formal charge of -1. Remember that with formal charges, the goal is to keep the formal charges (or the difference between the formal charges of each atom) as small as possible. The number of and values of the formal charges on this structure (-1 and 0 (difference of 1) in Figure $12$, as opposed to +2 and -1 (difference of 3) in Figure $12$) is significantly lower than on the structure that follows the octet rule, and as such an expanded octet is plausible, and even preferred to a normal octet, in this case.
Example $3$: The $ICl_4^-$ Ion
Draw the Lewis structure for $ICl_4^-$ ion.
Solution
1. Count up the valence electrons: 7+(4*7)+1 = 36 electrons
2. Draw the connectivities:
3. Add octet of electrons to outer atoms:
4. Add extra electrons (36-32=4) to central atom:
5. The ICl4- ion thus has 12 valence electrons around the central Iodine (in the 5d orbitals)
Expanded Lewis structures are also plausible depictions of molecules when experimentally determined bond lengths suggest partial double bond characters even when single bonds would already fully fill the octet of the central atom. Despite the cases for expanded octets, as mentioned for incomplete octets, it is important to keep in mind that, in general, the octet rule applies.
Summary
Following the Octet Rule for Lewis Dot Structures leads to the most accurate depictions of stable molecular and atomic structures and because of this we always want to use the octet rule when drawing Lewis Dot Structures. There are three exceptions: (1) When there are an odd number of valence electrons, (2) When there are too few valence electrons, and (3) when there are too many valence electrons | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/08%3A_Basic_Concepts_of_Chemical_Bonding/8.07%3A_Exceptions_to_the_Octet_Rule.txt |
Learning Objectives
• The define Bond-dissociation energy (bond energy)
• To correlate bond strength with bond length
• To define and used average bond energies
In proposing his theory that octets can be completed by two atoms sharing electron pairs, Lewis provided scientists with the first description of covalent bonding. In this section, we expand on this and describe some of the properties of covalent bonds. The stability of a molecule is a function of the strength of the covalent bonds holding the atoms together.
The Relationship between Bond Order and Bond Energy
Triple bonds between like atoms are shorter than double bonds, and because more energy is required to completely break all three bonds than to completely break two, a triple bond is also stronger than a double bond. Similarly, double bonds between like atoms are stronger and shorter than single bonds. Bonds of the same order between different atoms show a wide range of bond energies, however. Table $1$ lists the average values for some commonly encountered bonds. Although the values shown vary widely, we can observe four trends:
Table $1$: Average Bond Energies (kJ/mol) for Commonly Encountered Bonds at 273 K
Single Bonds Multiple Bonds
H–H 432 C–C 346 N–N ≈167 O–O ≈142 F–F 155 C=C 602
H–C 411 C–Si 318 N–O 201 O–F 190 F–Cl 249 C≡C 835
H–Si 318 C–N 305 N–F 283 O–Cl 218 F–Br 249 C=N 615
H–N 386 C–O 358 N–Cl 313 O–Br 201 F–I 278 C≡N 887
H–P ≈322 C–S 272 N–Br 243 O–I 201 Cl–Cl 240 C=O 749
H–O 459 C–F 485 P–P 201 S–S 226 Cl–Br 216 C≡O 1072
H–S 363 C–Cl 327 S–F 284 Cl–I 208 N=N 418
H–F 565 C–Br 285 S–Cl 255 Br–Br 190 N≡N 942
H–Cl 428 C–I 213 S–Br 218 Br–I 175 N=O 607
H–Br 362 Si–Si 222 I–I 149 O=O 494
H–I 295 Si–O 452 S=O 532
Source: Data from J. E. Huheey, E. A. Keiter, and R. L. Keiter, Inorganic Chemistry, 4th ed. (1993).
1. Bonds between hydrogen and atoms in the same column of the periodic table decrease in strength as we go down the column. Thus an H–F bond is stronger than an H–I bond, H–C is stronger than H–Si, H–N is stronger than H–P, H–O is stronger than H–S, and so forth. The reason for this is that the region of space in which electrons are shared between two atoms becomes proportionally smaller as one of the atoms becomes larger (part (a) in Figure 8.11).
2. Bonds between like atoms usually become weaker as we go down a column (important exceptions are noted later). For example, the C–C single bond is stronger than the Si–Si single bond, which is stronger than the Ge–Ge bond, and so forth. As two bonded atoms become larger, the region between them occupied by bonding electrons becomes proportionally smaller, as illustrated in part (b) in Figure 8.11. Noteworthy exceptions are single bonds between the period 2 atoms of groups 15, 16, and 17 (i.e., N, O, F), which are unusually weak compared with single bonds between their larger congeners. It is likely that the N–N, O–O, and F–F single bonds are weaker than might be expected due to strong repulsive interactions between lone pairs of electrons on adjacent atoms. The trend in bond energies for the halogens is therefore $\ce{Cl–Cl > Br–Br > F–F > I–I} \nonumber$ Similar effects are also seen for the O–O versus S–S and for N–N versus P–P single bonds.
Bonds between hydrogen and atoms in a given column in the periodic table are weaker down the column; bonds between like atoms usually become weaker down a column.
1. Because elements in periods 3 and 4 rarely form multiple bonds with themselves, their multiple bond energies are not accurately known. Nonetheless, they are presumed to be significantly weaker than multiple bonds between lighter atoms of the same families. Compounds containing an Si=Si double bond, for example, have only recently been prepared, whereas compounds containing C=C double bonds are one of the best-studied and most important classes of organic compounds.
1. Multiple bonds between carbon, oxygen, or nitrogen and a period 3 element such as phosphorus or sulfur tend to be unusually strong. In fact, multiple bonds of this type dominate the chemistry of the period 3 elements of groups 15 and 16. Multiple bonds to phosphorus or sulfur occur as a result of d-orbital interactions, as we discussed for the SO42 ion in Section 8.6. In contrast, silicon in group 14 has little tendency to form discrete silicon–oxygen double bonds. Consequently, SiO2 has a three-dimensional network structure in which each silicon atom forms four Si–O single bonds, which makes the physical and chemical properties of SiO2 very different from those of CO2.
Bond strengths increase as bond order increases, while bond distances decrease.
The Relationship between Molecular Structure and Bond Energy
Bond energy is defined as the energy required to break a particular bond in a molecule in the gas phase. Its value depends on not only the identity of the bonded atoms but also their environment. Thus the bond energy of a C–H single bond is not the same in all organic compounds. For example, the energy required to break a C–H bond in methane varies by as much as 25% depending on how many other bonds in the molecule have already been broken (Table $2$); that is, the C–H bond energy depends on its molecular environment. Except for diatomic molecules, the bond energies listed in Table $1$ are average values for all bonds of a given type in a range of molecules. Even so, they are not likely to differ from the actual value of a given bond by more than about 10%.
Table $2$: Energies for the Dissociation of Successive C–H Bonds in Methane. Source: Data from CRC Handbook of Chemistry and Physics (2004).
Reaction D (kJ/mol)
CH4(g) → CH3(g) + H(g) 439
CH3(g) → CH2(g) + H(g) 462
CH2(g) → CH(g) + H(g) 424
CH(g) → C(g) + H(g) 338
We can estimate the enthalpy change for a chemical reaction by adding together the average energies of the bonds broken in the reactants and the average energies of the bonds formed in the products and then calculating the difference between the two. If the bonds formed in the products are stronger than those broken in the reactants, then energy will be released in the reaction ($ΔH_{rxn} < 0$):
$ΔH_{rxn} \approx \sum{\text{(bond energies of bonds broken)}}−\sum{\text{(bond energies of bonds formed)}} \label{$1$}$
The ≈ sign is used because we are adding together average bond energies; hence this approach does not give exact values for ΔHrxn.
Let’s consider the reaction of 1 mol of n-heptane (C7H16) with oxygen gas to give carbon dioxide and water. This is one reaction that occurs during the combustion of gasoline:
$\ce{CH3(CH2)5CH3(l) + 11 O2(g) \rightarrow 7 CO2(g) + 8 H2O(g)} \label{$2$}$
In this reaction, 6 C–C bonds, 16 C–H bonds, and 11 O=O bonds are broken per mole of n-heptane, while 14 C=O bonds (two for each CO2) and 16 O–H bonds (two for each H2O) are formed. The energy changes can be tabulated as follows:
Binds Broken (kJ/mol) and Bonds Formed (kJ/mol)
Bonds Broken (kJ/mol) Bonds Formed (kJ/mol)
6 C–C 346 × 6 = 2076 14 C=O 799 × 14 = 11,186
16 C–H 411 × 16 = 6576 16 O–H 459 × 16 = 7344
11 O=O 494 × 11 = 5434 Total = 18,530
Total = 14,086
The bonds in the products are stronger than the bonds in the reactants by about 4444 kJ/mol. This means that $ΔH_{rxn}$ is approximately −4444 kJ/mol, and the reaction is highly exothermic (which is not too surprising for a combustion reaction).
If we compare this approximation with the value obtained from measured $ΔH_f^o$ values ($ΔH_{rxn} = −481\;7 kJ/mol$), we find a discrepancy of only about 8%, less than the 10% typically encountered. Chemists find this method useful for calculating approximate enthalpies of reaction for molecules whose actual $ΔH^ο_f$ values are unknown. These approximations can be important for predicting whether a reaction is exothermic or endothermic—and to what degree.
Example $1$: Explosives
The compound RDX (Research Development Explosive) is a more powerful explosive than dynamite and is used by the military. When detonated, it produces gaseous products and heat according to the following reaction. Use the approximate bond energies in Table $1$ to estimate the $ΔH_{rxn}$ per mole of RDX.
Given: chemical reaction, structure of reactant, and Table $1$.
Asked for: $ΔH_{rxn}$ per mole
Strategy:
1. List the types of bonds broken in RDX, along with the bond energy required to break each type. Multiply the number of each type by the energy required to break one bond of that type and then add together the energies. Repeat this procedure for the bonds formed in the reaction.
2. Use Equation $1$ to calculate the amount of energy consumed or released in the reaction (ΔHrxn).
Solution:
We must add together the energies of the bonds in the reactants and compare that quantity with the sum of the energies of the bonds in the products. A nitro group (–NO2) can be viewed as having one N–O single bond and one N=O double bond, as follows:
In fact, however, both N–O distances are usually the same because of the presence of two equivalent resonance structures.
A We can organize our data by constructing a table:
Bonds Broken (kJ/mol)
Bonds Broken (kJ/mol) Bonds Broken (kJ/mol)
6 C–H 411 × 6 = 2466 6 C=O 799 × 6 = 4794
3 N–N 167 × 3 = 501 6 O–H 459 × 6 = 2754
3 N–O 201 × 3 = 603 Total = 10,374
3 N=O 607 × 3 = 1821
1.5 O=O 494 × 1.5 = 741
Total = 7962
B From Equation $1$, we have
\begin{align*} ΔH_{rxn} &\approx \sum{\text{(bond energies of bonds broken)}}−\sum{\text{(bond energies of bonds formed)}} \[4pt] &= 7962 \; kJ/mol − 10,374 \; kJ/mol \[4pt] &=−2412 \;kJ/mol \end{align*} \nonumber
Thus this reaction is also highly exothermic
Exercise $1$: Freon
The molecule HCFC-142b is a hydrochlorofluorocarbon that is used in place of chlorofluorocarbons (CFCs) such as the Freons and can be prepared by adding HCl to 1,1-difluoroethylene:
HCL to produce CH3Cf2Cl and HCFC142b." data-cke-saved-src="/@api/deki/files/129585/imageedit_31_2777381278.png" src="/@api/deki/files/129585/imageedit_31_2777381278.png" data-quail-id="371">
Use tabulated bond energies to calculate $ΔH_{rxn}$.
Answer
−54 kJ/mol
Bond Dissociation Energy
Bond Dissociation Energy (also referred to as Bond energy) is the enthalpy change ($\Delta H$, heat input) required to break a bond (in 1 mole of a gaseous substance)
What about when we have a compound which is not a diatomic molecule? Consider the dissociation of methane:
There are four equivalent C-H bonds, thus we can that the dissociation energy for a single C-H bond would be:
\begin{align*} D(C-H) &= (1660/4)\, kJ/mol \[4pt] &= 415 \,kJ/mol \end{align*} \nonumber
The bond energy for a given bond is influenced by the rest of the molecule. However, this is a relatively small effect (suggesting that bonding electrons are localized between the bonding atoms). Thus, the bond energy for most bonds varies little from the average bonding energy for that type of bond
Bond energy is always a positive value - it takes energy to break a covalent bond (conversely energy is released during bond formation)
Bond (kJ/mol)
Table $4$: Average bond energies:
C-F 485
C-Cl 328
C-Br 276
C-I 240
C-C 348
C-N 293
C-O 358
C-F 485
C-C 348
C=C 614
C=C 839
The more stable a molecule (i.e. the stronger the bonds) the less likely the molecule is to undergo a chemical reaction.
Bond Energies and the Enthalpy of Reactions
If we know which bonds are broken and which bonds are made during a chemical reaction, we can estimate the enthalpy change of the reaction ($\Delta H_{rxn}$) even if we do not know the enthalpies of formation (($\Delta H_{f}^o$)for the reactants and products:
$\Delta H = \sum \text{bond energies of broken bonds} - \sum \text{bond energies of formed bonds} \label{8.8.3}$
Example $2$: Chlorination of Methane
What is the enthalpy of reaction between 1 mol of chlorine and 1 mol methane?
Solution
We use Equation \ref{8.8.3}, which requires tabulating bonds broken and formed.
• Bonds broken: 1 mol of Cl-Cl bonds, 1 mol of C-H bonds
• Bonds formed: 1 mol of H-Cl bonds, 1 mol of C-Cl bonds
\begin{align*} \Delta H &= [D(Cl-Cl) + D(C-H)] - [D(H-Cl)+D(C-Cl)] \[4pt] &= [242 kJ + 413 kJ] - [431 kJ + 328 kJ] \[4pt] &= -104 \,kJ \end{align*} \nonumber
Thus, the reaction is exothermic (because the bonds in the products are stronger than the bonds in the reactants)
Example $3$: Combustion of Ethane
What is the enthalpy of reaction for the combustion of 1 mol of ethane?
Solution
We use Equation \ref{8.8.3}, which requires tabulating bonds broken and formed.
• bonds broken: 6 moles C-H bonds, 1 mol C-C bonds, 7/2 moles of O=O bonds
• bonds formed: 4 moles C=O bonds, 6 moles O-H bonds
\begin{align*} \Delta {H} &= [(6 \times 413) + (348) + (\frac{7}{2} \times 495)] - [(4 \times 799) + (6 \times 463)] \[4pt] &= 4558 - 5974 \[4pt] &= -1416\; kJ \end{align*} \nonumber
Therefor the reaction is exothermic.
Table $5$: Bond strength and bond length
Bond Bond Energy (kJ/mol) Bond Length (Å)
C-C 348 1.54
C=C 614 1.34
C=C 839 1.
As the number of bonds between two atoms increases, the bond grows shorter and stronger
Summary
Bond order is the number of electron pairs that hold two atoms together. Single bonds have a bond order of one, and multiple bonds with bond orders of two (a double bond) and three (a triple bond) are quite common. In closely related compounds with bonds between the same kinds of atoms, the bond with the highest bond order is both the shortest and the strongest. In bonds with the same bond order between different atoms, trends are observed that, with few exceptions, result in the strongest single bonds being formed between the smallest atoms. Tabulated values of average bond energies can be used to calculate the enthalpy change of many chemical reactions. If the bonds in the products are stronger than those in the reactants, the reaction is exothermic and vice versa.
The breakage and formation of bonds is similar to a relationship: you can either get married or divorced and it is more favorable to be married.
• Energy is always released to make bonds, which is why the enthalpy change for breaking bonds is always positive.
• Energy is always required to break bonds. Atoms are much happier when they are "married" and release energy because it is easier and more stable to be in a relationship (e.g., to generate octet electronic configurations). The enthalpy change is always negative because the system is releasing energy when forming bond. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/08%3A_Basic_Concepts_of_Chemical_Bonding/8.08%3A_Strength_of_Covalent_Bonds.txt |
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
8.1: CHEMICAL BONDS, LEWIS SYMBOLS AND THE OCTET RULE
Conceptual Problems
1. The Lewis electron system is a simplified approach for understanding bonding in covalent and ionic compounds. Why do chemists still find it useful?
2. Is a Lewis dot symbol an exact representation of the valence electrons in an atom or ion? Explain your answer.
3. How can the Lewis electron dot system help to predict the stoichiometry of a compound and its chemical and physical properties?
4. How is a Lewis dot symbol consistent with the quantum mechanical model of the atom? How is it different?
Conceptual Answer
3. Lewis dot symbols allow us to predict the number of bonds atoms will form, and therefore the stoichiometry of a compound. The Lewis structure of a compound also indicates the presence or absence of lone pairs of electrons, which provides information on the compound’s chemical reactivity and physical properties.
8.2: IONIC BONDING
Conceptual Problems
1. Describe the differences in behavior between NaOH and CH3OH in aqueous solution. Which solution would be a better conductor of electricity? Explain your reasoning.
2. What is the relationship between the strength of the electrostatic attraction between oppositely charged ions and the distance between the ions? How does the strength of the electrostatic interactions change as the size of the ions increases?
3. Which will result in the release of more energy: the interaction of a gaseous sodium ion with a gaseous oxide ion or the interaction of a gaseous sodium ion with a gaseous bromide ion? Why?
4. Which will result in the release of more energy: the interaction of a gaseous chloride ion with a gaseous sodium ion or a gaseous potassium ion? Explain your answer.
5. What are the predominant interactions when oppositely charged ions are
1. far apart?
2. at internuclear distances close to r0?
3. very close together (at a distance that is less than the sum of the ionic radii)?
6. Several factors contribute to the stability of ionic compounds. Describe one type of interaction that destabilizes ionic compounds. Describe the interactions that stabilize ionic compounds.
7. What is the relationship between the electrostatic attractive energy between charged particles and the distance between the particles?
Conceptual Answer
1.
2.
3. The interaction of a sodium ion and an oxide ion. The electrostatic attraction energy between ions of opposite charge is directly proportional to the charge on each ion (Q1 and Q2 in Equation 9.1). Thus, more energy is released as the charge on the ions increases (assuming the internuclear distance does not increase substantially). A sodium ion has a +1 charge; an oxide ion, a −2 charge; and a bromide ion, a −1 charge. For the interaction of a sodium ion with an oxide ion, Q1 = +1 and Q2 = −2, whereas for the interaction of a sodium ion with a bromide ion, Q1 = +1 and Q2 = −1. The larger value of Q1 × Q2 for the sodium ion–oxide ion interaction means it will release more energy.
Numerical Problems
1. How does the energy of the electrostatic interaction between ions with charges +1 and −1 compare to the interaction between ions with charges +3 and −1 if the distance between the ions is the same in both cases? How does this compare with the magnitude of the interaction between ions with +3 and −3 charges?
2. How many grams of gaseous MgCl2 are needed to give the same electrostatic attractive energy as 0.5 mol of gaseous LiCl? The ionic radii are Li+ = 76 pm, Mg+2 = 72 pm, and Cl = 181 pm.
3. Sketch a diagram showing the relationship between potential energy and internuclear distance (from r = ∞ to r = 0) for the interaction of a bromide ion and a potassium ion to form gaseous KBr. Explain why the energy of the system increases as the distance between the ions decreases from r = r0 to r = 0.
4. Calculate the magnitude of the electrostatic attractive energy (E, in kilojoules) for 85.0 g of gaseous SrS ion pairs. The observed internuclear distance in the gas phase is 244.05 pm.
5. What is the electrostatic attractive energy (E, in kilojoules) for 130 g of gaseous HgI2? The internuclear distance is 255.3 pm.
Numerical Answers
1. According to Equation 9.1, in the first case Q1Q2 = (+1)(−1) = −1; in the second case, Q1Q2 = (+3)(−1) = −3. Thus, E will be three times larger for the +3/−1 ions. For +3/−3 ions, Q1Q2 = (+3)(−3) = −9, so E will be nine times larger than for the +1/−1 ions.
2.
3.
At r < r0, the energy of the system increases due to electron–electron repulsions between the overlapping electron distributions on adjacent ions. At very short internuclear distances, electrostatic repulsions between adjacent nuclei also become important.
8.3: COVALENT BONDING
Conceptual Problems
1. Which would you expect to be stronger—an S–S bond or an Se–Se bond? Why?
2. Which element—nitrogen, phosphorus, or arsenic—will form the strongest multiple bond with oxygen? Why?
3. Why do multiple bonds between oxygen and period 3 elements tend to be unusually strong?
4. What can bond energies tell you about reactivity?
5. Bond energies are typically reported as average values for a range of bonds in a molecule rather than as specific values for a single bond? Why?
6. If the bonds in the products are weaker than those in the reactants, is a reaction exothermic or endothermic? Explain your answer.
7. A student presumed that because heat was required to initiate a particular reaction, the reaction product would be stable. Instead, the product exploded. What information might have allowed the student to predict this outcome?
Numerical Problems
1. What is the bond order about the central atom(s) of hydrazine (N2H4), nitrogen, and diimide (N2H2)? Draw Lewis electron structures for each compound and then arrange these compounds in order of increasing N–N bond distance. Which of these compounds would you expect to have the largest N–N bond energy? Explain your answer.
2. What is the carbon–carbon bond order in ethylene (C2H4), BrH2CCH2Br, and FCCH? Arrange the compounds in order of increasing C–C bond distance. Which would you expect to have the largest C–C bond energy? Why?
3. From each pair of elements, select the one with the greater bond strength? Explain your choice in each case.
1. P–P, Sb–Sb
2. Cl–Cl, I–I
3. O–O, Se–Se
4. S–S, Cl–Cl
5. Al–Cl, B–Cl
4. From each pair of elements, select the one with the greater bond strength? Explain your choice in each case.
1. Te–Te, S–S
2. C–H, Ge–H
3. Si–Si, P–P
4. Cl–Cl, F–F
5. Ga–H, Al–H
5. Approximately how much energy per mole is required to completely dissociate acetone [(CH3)2CO] and urea [(NH2)2CO] into their constituent atoms?
6. Approximately how much energy per mole is required to completely dissociate ethanol, formaldehyde, and hydrazine into their constituent atoms?
7. Is the reaction of diimine (N2H2) with oxygen to produce nitrogen and water exothermic or endothermic? Quantify your answer.
Numerical Answer
1. N2H4, bond order 1; N2H2, bond order 2; N2, bond order 3; N–N bond distance: N2 < N2H2 < N2H4; Largest bond energy: N2; Highest bond order correlates with strongest and shortest bond.
8.4: BOND POLARITY AND ELECRONEGATIVITY
Conceptual Problems
1. Why do ionic compounds such as KI exhibit substantially less than 100% ionic character in the gas phase?
2. Of the compounds LiI and LiF, which would you expect to behave more like a classical ionic compound? Which would have the greater dipole moment in the gas phase? Explain your answers.
Numerical Problems
1. Predict whether each compound is purely covalent, purely ionic, or polar covalent.
1. RbCl
2. S8
3. TiCl2
4. SbCl3
5. LiI
6. Br2
2. Based on relative electronegativities, classify the bonding in each compound as ionic, covalent, or polar covalent. Indicate the direction of the bond dipole for each polar covalent bond.
1. NO
2. HF
3. MgO
4. AlCl3
5. SiO2
6. the C=O bond in acetone
7. O3
3. Based on relative electronegativities, classify the bonding in each compound as ionic, covalent, or polar covalent. Indicate the direction of the bond dipole for each polar covalent bond.
1. NaBr
2. OF2
3. BCl3
4. the S–S bond in CH3CH2SSCH2CH3
5. the C–Cl bond in CH2Cl2
6. the O–H bond in CH3OH
7. PtCl42
4. Classify each species as having 0%–40% ionic character, 40%–60% ionic character, or 60%–100% ionic character based on the type of bonding you would expect. Justify your reasoning.
1. CaO
2. S8
3. AlBr3
4. ICl
5. Na2S
6. SiO2
7. LiBr
5. If the bond distance in HCl (dipole moment = 1.109 D) were double the actual value of 127.46 pm, what would be the effect on the charge localized on each atom? What would be the percent negative charge on Cl? At the actual bond distance, how would doubling the charge on each atom affect the dipole moment? Would this represent more ionic or covalent character?
6. Calculate the percent ionic character of HF (dipole moment = 1.826 D) if the H–F bond distance is 92 pm.
7. Calculate the percent ionic character of CO (dipole moment = 0.110 D) if the C–O distance is 113 pm.
8. Calculate the percent ionic character of PbS and PbO in the gas phase, given the following information: for PbS, r = 228.69 pm and µ = 3.59 D; for PbO, r = 192.18 pm and µ = 4.64 D. Would you classify these compounds as having covalent or polar covalent bonds in the solid state?
8.5: DRAWING LEWIS STRUCTURES
Conceptual Problems
1. Compare and contrast covalent and ionic compounds with regard to
1. volatility.
2. melting point.
3. electrical conductivity.
4. physical appearance.
2. What are the similarities between plots of the overall energy versus internuclear distance for an ionic compound and a covalent compound? Why are the plots so similar?
3. Which atom do you expect to be the central atom in each of the following species?
1. SO42
2. NH4+
3. BCl3
4. SO2Cl2
4. Which atom is the central atom in each of the following species?
1. PCl3
2. CHCl3
3. SO2
4. IF3
5. What is the relationship between the number of bonds typically formed by the period 2 elements in groups 14, 15, and 16 and their Lewis electron structures?
6. Although formal charges do not represent actual charges on atoms in molecules or ions, they are still useful. Why?
Numerical Problems
1. Give the electron configuration and the Lewis dot symbol for the following. How many more electrons can each atom accommodate?
1. Se
2. Kr
3. Li
4. Sr
5. H
2. Give the electron configuration and the Lewis dot symbol for the following. How many more electrons can each atom accommodate?
1. Na
2. Br
3. Ne
4. C
5. Ga
3. Based on Lewis dot symbols, predict the preferred oxidation state of Be, F, B, and Cs.
4. Based on Lewis dot symbols, predict the preferred oxidation state of Br, Rb, O, Si, and Sr.
5. Based on Lewis dot symbols, predict how many bonds gallium, silicon, and selenium will form in their neutral compounds.
6. Determine the total number of valence electrons in the following.
1. Cr
2. Cu+
3. NO+
4. XeF2
5. Br2
6. CH2Cl2
7. NO3
8. H3O+
7. Determine the total number of valence electrons in the following.
1. Ag
2. Pt2+
3. H2S
4. OH
5. I2
6. CH4
7. SO42
8. NH4+.
8. Draw Lewis electron structures for the following.
1. F2
2. SO2
3. AlCl4
4. SO32
5. BrCl
6. XeF4
7. NO+
8. PCl3
9. Draw Lewis electron structures for the following.
1. Br2
2. CH3Br
3. SO42
4. O2
5. S22
6. BF3
10. Draw Lewis electron structures for CO2, NO2, SO2, and NO2+. From your diagram, predict which pair(s) of compounds have similar electronic structures.
11. Write Lewis dot symbols for each pair of elements. For a reaction between each pair of elements, predict which element is the oxidant, which element is the reductant, and the final stoichiometry of the compound formed.
1. K, S
2. Sr, Br
3. Al, O
4. Mg, Cl
12. Write Lewis dot symbols for each pair of elements. For a reaction between each pair of elements, predict which element is the oxidant, which element is the reductant, and the final stoichiometry of the compound formed.
1. Li, F
2. Cs, Br
3. Ca, Cl
4. B, F
13. Use Lewis dot symbols to predict whether ICl and NO4 are chemically reasonable formulas.
14. Draw a plausible Lewis electron structure for a compound with the molecular formula Cl3PO.
15. Draw a plausible Lewis electron structure for a compound with the molecular formula CH4O.
16. While reviewing her notes, a student noticed that she had drawn the following structure in her notebook for acetic acid:
Why is this structure not feasible? Draw an acceptable Lewis structure for acetic acid. Show the formal charges of all nonhydrogen atoms in both the correct and incorrect structures.
17. A student proposed the following Lewis structure shown for acetaldehyde.
Why is this structure not feasible? Draw an acceptable Lewis structure for acetaldehyde. Show the formal charges of all nonhydrogen atoms in both the correct and incorrect structures.
18. Draw the most likely structure for HCN based on formal charges, showing the formal charge on each atom in your structure. Does this compound have any plausible resonance structures? If so, draw one.
19. Draw the most plausible Lewis structure for NO3. Does this ion have any other resonance structures? Draw at least one other Lewis structure for the nitrate ion that is not plausible based on formal charges.
20. At least two Lewis structures can be drawn for BCl3. Using arguments based on formal charges, explain why the most likely structure is the one with three B–Cl single bonds.
21. Using arguments based on formal charges, explain why the most feasible Lewis structure for SO42 has two sulfur–oxygen double bonds.
22. At least two distinct Lewis structures can be drawn for N3. Use arguments based on formal charges to explain why the most likely structure contains a nitrogen–nitrogen double bond.
23. Is H–O–N=O a reasonable structure for the compound HNO2? Justify your answer using Lewis electron dot structures.
24. Is H–O=C–H a reasonable structure for a compound with the formula CH2O? Use Lewis electron dot structures to justify your answer.
25. Explain why the following Lewis structure for SO32 is or is not reasonable.
Numerical Answers
1.
1. [Ar]4s23d104p4
Selenium can accommodate two more electrons, giving the Se2 ion.
2. [Ar]4s23d104p6
Krypton has a closed shell electron configuration, so it cannot accommodate any additional electrons.
3. 1s22s1
Lithium can accommodate one additional electron in its 2s orbital, giving the Li ion.
4. [Kr]5s2
Strontium has a filled 5s subshell, and additional electrons would have to be placed in an orbital with a higher energy. Thus strontium has no tendency to accept an additional electron.
5. 1s1
Hydrogen can accommodate one additional electron in its 1s orbital, giving the H ion.
2.
3. Be2+, F, B3+, Cs+
4.
5.
6.
7.
1. 11
2. 8
3. 8
4. 8
5. 14
6. 8
7. 32
8. 8
8.
9.
10.
11.
1. K is the reductant; S is the oxidant. The final stoichiometry is K2S.
2. Sr is the reductant; Br is the oxidant. The final stoichiometry is SrBr2.
3. Al is the reductant; O is the oxidant. The final stoichiometry is Al2O3.
4. Mg is the reductant; Cl is the oxidant. The final stoichiometry is MgCl2.
12.
13.
14.
15. The only structure that gives both oxygen and carbon an octet of electrons is the following:
16.
17. The student’s proposed structure has two flaws: the hydrogen atom with the double bond has four valence electrons (H can only accommodate two electrons), and the carbon bound to oxygen only has six valence electrons (it should have an octet). An acceptable Lewis structure is
The formal charges on the correct and incorrect structures are as follows:
18.
19. The most plausible Lewis structure for NO3 is:
There are three equivalent resonance structures for nitrate (only one is shown), in which nitrogen is doubly bonded to one of the three oxygens. In each resonance structure, the formal charge of N is +1; for each singly bonded O, it is −1; and for the doubly bonded oxygen, it is 0.
The following is an example of a Lewis structure that is not plausible:
This structure nitrogen has six bonds (nitrogen can form only four bonds) and a formal charge of –1.
20.
21. With four S–O single bonds, each oxygen in SO42 has a formal charge of −1, and the central sulfur has a formal charge of +2. With two S=O double bonds, only two oxygens have a formal charge of –1, and sulfur has a formal charge of zero. Lewis structures that minimize formal charges tend to be lowest in energy, making the Lewis structure with two S=O double bonds the most probable.
22.
23. Yes. This is a reasonable Lewis structure, because the formal charge on all atoms is zero, and each atom (except H) has an octet of electrons.
8.6: RESONANCE STRUCTURES
Conceptual Problems
1. Why are resonance structures important?
2. In what types of compounds are resonance structures particularly common?
3. True or False, The picture below is a resonance structure?
Numerical Problems
1. Draw all the resonance structures for each ion.
1. HSO4
2. HSO3
1. Draw the Lewis Dot Structure for SO42- and all possible resonance structures. Which of the following resonance structure is not favored among the Lewis Structures? Explain why. Assign Formal Charges.
2. Draw the Lewis Dot Structure for CH3COO- and all possible resonance structures. Assign Formal Charges. Choose the most favorable Lewis Structure.
3. Draw the Lewis Dot Structure for HPO32- and all possible resonance structures. Assign Formal Charges.
4. Draw the Lewis Dot Structure for CHO21- and all possible resonance structures. Assign Formal Charges.
5. Draw the Resonance Hybrid Structure for PO43-.
6. Draw the Resonance Hybrid Structure for NO3-.
Numerical Answers
1. False, because the electrons were not moved around, only the atoms (this violates the Resonance Structure Rules).
3. Below are the all Lewis dot structure with formal charges (in red) for Sulfate (SO42-). There isn't a most favorable resonance of the Sulfate ion because they are all identical in charge and there is no change in Electronegativity between the Oxygen atoms.
4. Below is the resonance for CH3COO-, formal charges are displayed in red. The Lewis Structure with the most formal charges is not desirable, because we want the Lewis Structure with the least formal charge.
5. The resonance for HPO32-, and the formal charges (in red).
6. The resonance for CHO21-, and the formal charges (in red).
7. The resonance hybrid for PO43-, hybrid bonds are in red.
8. The resonance hybrid for NO3-, hybrid bonds are in red.
8.7: EXCEPTIONS TO THE OCTET RULE
Conceptual Problems
1. What regions of the periodic table contain elements that frequently form molecules with an odd number of electrons? Explain your answer.
2. How can atoms expand their valence shell? What is the relationship between an expanded valence shell and the stability of an ion or a molecule?
3. What elements are known to form compounds with less than an octet of electrons? Why do electron-deficient compounds form?
4. List three elements that form compounds that do not obey the octet rule. Describe the factors that are responsible for the stability of these compounds.
Numerical Problems
1. What is the major weakness of the Lewis system in predicting the electron structures of PCl6 and other species containing atoms from period 3 and beyond?
2. The compound aluminum trichloride consists of Al2Cl6 molecules with the following structure (lone pairs of electrons removed for clarity):
Does this structure satisfy the octet rule? What is the formal charge on each atom? Given the chemical similarity between aluminum and boron, what is a plausible explanation for the fact that aluminum trichloride forms a dimeric structure rather than the monomeric trigonal planar structure of BCl3?
3. Draw Lewis electron structures for ClO4, IF5, SeCl4, and SbF5.
4. Draw Lewis electron structures for ICl3, Cl3PO, Cl2SO, and AsF6.
5. Draw plausible Lewis structures for the phosphate ion, including resonance structures. What is the formal charge on each atom in your structures?
6. Draw an acceptable Lewis structure for PCl5, a compound used in manufacturing a form of cellulose. What is the formal charge of the central atom? What is the oxidation number of the central atom?
7. Using Lewis structures, draw all of the resonance structures for the BrO3 ion.
8. Draw an acceptable Lewis structure for xenon trioxide (XeO3), including all resonance structures.
Numerical Answers
1.
2.
3. ClO4 (one of four equivalent resonance structures)
4.
5.
The formal charge on phosphorus is 0, while three oxygen atoms have a formal charge of −1 and one has a formal charge of zero.
6.
Practice Problems
1. Draw the Lewis structure for the molecule I3-.
2. Draw the molecule ClF3.
3. The central atom for an expanded octet must have an atomic number larger than what?
4. Draw the Lewis structure for the molecule NO2.
5. Which Lewis structure is more likely?
or
Practice Answers
1.
2.
3. 10 (Sodium and higher)
4.
5. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/08%3A_Basic_Concepts_of_Chemical_Bonding/8.E%3A_Basic_Concepts_of_Chemical_Bonding_%28Exercises%29.txt |
8.1: CHEMICALBONDS, LEWISSYMBOLSANDTHEOCTETRULE
Lewis dot symbols can be used to predict the number of bonds formed by most elements in their compounds. Lewis electron dot symbols, which consist of the chemical symbol for an element surrounded by dots that represent its valence electrons, grouped into pairs often placed above, below, and to the left and right of the symbol. The structures reflect the fact that the elements in period 2 and beyond tend to gain, lose, or share electrons to reach a total of 8 valence electrons in their compounds.
• Ionic bond: bond formed on the basis of electrostatic forces that exist between oppositely charged ions. The ions are formed from atoms by transfer of one or more electrons
• Covalent bond: bond formed between two or more atoms by a sharing of electrons
• Metallic bond: bonding in which the bonding electrons are relatively free to move throughout the 3D structure
• Electron dot symbols: aka Lewis symbols; simple and convenient way of showing the valence electrons of atoms and keeping tack of them in the course of bond formation
• The number of valence electrons of any representative element is the same as the column number of the element in the periodic table
• Octet rule: atoms tend to lose or gain electrons until they are surrounded by 8 valence electrons
8.2: IONICBONDING
The amount of energy needed to separate a gaseous ion pair is its bond energy. Forming ionic compounds is usually extremely exothermic. The strength of the electrostatic attraction between ions with opposite charges is directly proportional to the magnitude of the charges on the ions and inversely proportional to the internuclear distance.
Energies of Ionic Bond Formation
The formation of ionic compounds is very exothermic
Removing an electron from an atom, such as Na, is endothermic because energy needs to be used to overcome the attractive forces within the atom. Adding an electron is the opposite process and releases lots of energy
The principal reason that ionic compounds are stable is the attraction between ions of unlike charge. This attraction draws the ions together, releasing energy and causing the ions to form a solid array (lattice)
Lattice energy: energy required to separate completely a mole of a solid ionic compounds into its gaseous ions
Large values of lattice energy mean that the ions are strongly attracted to one another
Energy released by the attraction between the ions of unlike charges more than makes up for the endothermic nature of ionization energies, making the formation of ionic compounds an exothermic process
$E = k \dfrac{Q_1Q_2}{ d} \nonumber$
• E = potential energy of two interacting charged particles
• Q1 and Q2 = charges on the particles
• D= distance between the particles
• K = constant; 8.99 X 109 Jm/C2
For a given arrangement of ions, the lattice energy increases as the charges of ions increase and as their radii decrease. The magnitude of lattice energies depends primarily on the ionic charges because ionic radii do not vary over a wide range.
Electron Configurations of Ions
Many ions tend to have noble gas electron configurations. This is why Na can have a +1 charge, but not a +2 one. Once an ion has reached noble gas configuration, it wants to stay there.
• Na: 1s2 2s2 2p6 3s1 = [Ne] 3s1
• Na+: 1s2 2s2 2p6 = [Ne]
• Na2+: 1s2 2s2 2p5
Similarly, addition of electrons to nonmetals is either exothermic or slightly endothermic as long as electrons are being added to the valence shell. Further addition of electrons requires tremendous amounts of energy; more than is available form the lattice energy
• Cl: 1s2 2s2 2p6 3s2 3p5 = [Ne] 3s2 3p5
• Cl: 1s2 2s2 2p6 3s2 3p6 = [Ar]
• Cl2: 1s2 2s2 2p6 3s2 3p6 4s1 = [Ar]
The lattice energies of ionic compounds are generally large enough to compensate for the loss of up to only 3 electrons from atoms. Thus we find cations only having charges of +1, +2, or +3.
Because most transition metals have more than 3 electrons beyond a noble gas core, attainment of a noble gas configuration for these ions is usually impossible.
When a positive ion is formed from an atom, electrons are always lost first from the subshell with the largest value of n. Thus, a transition metal always loses the outer s electrons before it loses electrons from the underlying d subshell.
8.3 Sizes of Ions
Sizes of ions are important in determining both the way in which the ions pack in a solid and the lattice energy of the solid. It is also a major factor governing the properties of ions in solution
The size of an atom depends on its nuclear charge, the number of electrons it possesses, and the orbitals in which the outer-shell electrons reside
Positive ions are formed by removing 1 or more electrons from the outermost region of the atom. Thus, the formation of a cation not only vacates the most spatially extended orbitals, it also decreases the total electron-electron repulsions. Hence, cations are smaller than the original atoms from which they came.
The opposite happens when speaking of negative ions. An added electron increases electron-electron repulsions and causes the electrons to spread out more in space.
For ions of the same charge, size increases as we go down a group
8.3: COVALENT BONDING
The strength of a covalent bond depends on the overlap between the valence orbitals of the bonded atoms. Bond order is the number of electron pairs that hold two atoms together. Single bonds have a bond order of one, and multiple bonds with bond orders of two (a double bond) and three (a triple bond) are quite common. In closely related compounds with bonds between the same kinds of atoms, the bond with the highest bond order is both the shortest and the strongest.
Ionic substances are usually brittle with high melting points. They are usually crystalline, meaning that they have flat surfaces that make characteristic angles with one another.
• Covalent bond: chemical bond formed by sharing a pair of electrons
• Lewis structure: structure that represents bonding using dots for unpaired electrons and lines for bonds
For nonmetals, the number of valence electrons is the same as the group number
Knowing this, we can predict that an element in Group 7A would need one covalent bond in order to get an octet, an element in Group 6A would need two, and so on.
Multiple Bonds
• Single bond: sharing of one pair of electrons, one covalent bond
• Double bond: two shared electrons
Distance between bonded atoms decreases as the number of shared electron pairs increases
8.4: BONDPOLARITY AND ELECRONEGATIVITY
Bond polarity and ionic character increase with an increasing difference in electronegativity. The electronegativity (χ) of an element is the relative ability of an atom to attract electrons to itself in a chemical compound and increases diagonally from the lower left of the periodic table to the upper right. The Pauling electronegativity scale is based on measurements of the strengths of covalent bonds between different atoms, whereas the Mulliken electronegativity of an element is the average
• Bond polarity: measure of how equally the electrons are shared between the atoms in a chemical bond
• Nonpolar bond: one in which the electrons are shared equally between the two atoms
• Polar covalent bond: one of the atoms exerts a greater attraction for the electron than the other
Electronegativity
Used to estimate whether a bond will be nonpolar, polar covalent, or ionic
Electronegativity: ability of an atom in a molecule to attract electrons to itself
An atom with a very negative electron affinity and high ionization energy will both attract electrons from other atoms and resist having its electrons attracted away; it will be highly electronegative
Highest electronegativity = 4.0 (Fluorine), lowest = 0.7 (Cesium)
Electronegativity increases form left to right, and usually decreases with increasing atomic number in any one group
Electronegativity and Bond Polarity
Differences in electronegativities:
Nonpolar = 0 – 0.4
Polar covalent = 0.4 – 1.6
Ionic = > 1.6 (> 50% = ionic)
δ+ and δ : "delta sign"; represent partial positive and negative charge. The atom with the δ is the more electronegative one
8.5: DRAWING LEWIS STRUCTURES
Lewis dot symbols provide a simple rationalization of why elements form compounds with the observed stoichiometries. A plot of the overall energy of a covalent bond as a function of internuclear distance is identical to a plot of an ionic pair because both result from attractive and repulsive forces between charged entities. Lewis structures are an attempt to rationalize why certain stoichiometries are commonly observed for the elements of particular families.
1. Sum the valence electrons from all atoms. For an anion, add an electron to the total negative charge. For a cation, subtract an electron.
2. Write the symbols for the atoms to show which atoms are attached to which, and connect them with a single bond.
3. Complete the octets of the atoms bonded to the central atom.
4. Place any leftover electrons on the central atom, even if doing so results in more than an octet
5. If there are not enough electrons to give the central atom an octet, try multiple bonds
8.6: RESONANCE STRUCTURES
Some molecules have two or more chemically equivalent Lewis electron structures, called resonance structures. Resonance is a mental exercise and method within the Valence Bond Theory of bonding that describes the delocalization of electrons within molecules. These structures are written with a double-headed arrow between them, indicating that none of the Lewis structures accurately describes the bonding but that the actual structure is an average of the individual resonance structures.
Resonance structures (resonance forms) are individual Lewis structures in cases where two or more Lewis structures are equally good descriptions of a single molecule. If a molecule (or ion) has two or more resonance structures, the molecule is a blend of these structures. The molecule does not oscillate rapidly between two or more different forms.
8.7: EXCEPTIONSTO THE OCTET RULE
Following the Octet Rule for Lewis Dot Structures leads to the most accurate depictions of stable molecular and atomic structures and because of this we always want to use the octet rule when drawing Lewis Dot Structures. There are three exceptions: (1) When there are an odd number of valence electrons, (2) When there are too few valence electrons, and (3) when there are too many valence electrons
1. Molecules with an odd number of electrons
2. Molecules in which an atom has less than an octet
3. Molecules in which an atom has more than an octet
Odd Number of Electrons
In a few molecules, such as ClO2, NO, and NO2, the number of electrons is odd. In NO for example, there are 5+6 = 11 valence electron. Hence, complete pairing of these electrons is impossible and an octet around each atom cannot be achieved.
Less Than an Octet
Second type of exception occurs when there are fewer than eight electrons around an atom in a molecule or ion. Relatively rare situation; most often encountered in compounds of Boron and Beryllium. For example, let’s consider Boron Trifluoride, BF3
There are 6 electrons around the Boron atom. We can form a double bond between Boron and any of the 3 Fluorine atoms (3 possible resonance structures)
However, by doing so, we forced a Fluorine atom to share additional electrons with Boron. This would make the F atom to have a +1 charge, and the Boron atom to have a –1 charge, which is extremely unfavorable.
We then conclude that the structures containing the double bonds are less important than the one illustrated on the right. Since in this case Boron has only 6 valence electrons, it will react violently with molecules that have an unshared pair of electrons.
More Than an Octet
The octet rule works as well as it does because the representative elements usually employ only an ns and three np valence shell orbitals in bonding, and these hold eight electrons.
Because elements of the second period have only 2s and 2p orbitals, they cannot have more than an octet of electrons in their valence shells. However, from the third period on, the elements have unfilled nd orbitals that can be used in bonding.
Size also plays an important role in determining whether an atom can accommodate more than eight electrons. The larger the central atom, the larger the number of atoms that can surround it. The size of the surrounding atoms is also important. Expanded valence shells occur most often when the central atom is bonded to the smallest and most electronegative atoms.
8.8: STRENGTHSOF COVALENT BONDS
Bond order is the number of electron pairs that hold two atoms together. Single bonds have a bond order of one, and multiple bonds with bond orders of two (a double bond) and three (a triple bond) are quite common. The bond with the highest bond order is both the shortest and the strongest. In bonds with the same bond order between different atoms, trends are observed that, with few exceptions, result in the strongest single bonds being formed between the smallest atoms.
Bonddissociation energy: aka bond energy; enthalpy change, ΔH, required to break a particular bond in a mole of gaseous substance.
For polyatomic molecules, we must often utilize average bond energies.
Bond energy is always positive, the greater the bond energy, the stronger the bond
A molecule with strong bonds generally has less tendency to undergo chemical change than does one with weak bonds
Bond Energies and the Enthalpy of Reactions
ΔH = Σ (bond energies of bonds broken) – Σ (bond energies of bonds formed)
If ΔH > 0, the reaction is endothermic
If ΔH < 0, the reaction is exothermic
Cl – Cl (g) + H – CH3 (g) → H – Cl (g) + Cl – CH3 (g)
Bonds broken: 1 mol Cl – Cl, 1 mol C – H
Bonds made: 1 mol H – Cl, 1 mol C – Cl
$∆H = [D (Cl – Cl) + D(C – H)] [D (H – Cl) + D (Cl – Cl)] \nonumber$
= (242 kJ + 413 kJ) – (431kJ + 328 kJ)
= 104 kJ
Bond Strength and Bond Length
As the number of bonds between a given element increase, the bond energy increases and the bond length decreases. Hence, the atoms are held more tightly and closely together. In general, as the number of bonds between two atoms increases, the bond grows shorter and stronger.
8.10 Oxidation Numbers
Oxidation Numbers: aka Oxidation states; a positive or negative whole number assigned to an element in a molecule or ion on the basis of a set of normal rules; to some degree it reflects the positive or negative character of an atom
Oxidation numbers do NOT correspond to real charges on the atoms, EXCEPT in the special case of simple ionic substances
1. The oxidation form of an element in its elemental form is zero.
2. The oxidation number of a monoatomic ion is the same as its charge. For example, the oxidation number of sodium in Na+ is +1, and that of sulfur in S2 is –2
3. In binary compounds (those with two different elements), the element with greater electronegativity is assigned a negative oxidation number equal to its charge in simple ionic compounds the element. For example, consider the oxidation state of Cl in PCl3. Cl is more electronegative than P. In its simple ionic compounds, Cl appears as the ion Cl. Thus, in PCl3, Cl is assigned an oxidation number of –1.
4. The sum of the oxidation numbers equals zero for an electrically neutral compound and equals the overall charge of the ionic species. For example, PCl3 is a neutral molecule. Thus, the sum of oxidation number of the P and Cl atoms must equal zero Because the oxidation number of each Cl in this compound is –1 (rule 3), the oxidation number of P must be +3.
Group 1A elements are +1, Group 2A elements are +2, and Aluminum is +3.
The most electronegative element, F, is always found in the –1 oxidation state. Oxygen is usually in the –2 state, however, it can be –1 in peroxides.
Hydrogen has an oxidation number of +1 when it is bonded to a more electronegative element (most nonmetals), and of –1 when bonded to less electronegative elements (most metals)
Oxidation Numbers and Nomenclature
Name of the less electronegative element is given first, followed by the name of the more electronegative element modified to have an –ide ending
Compounds of metals in higher oxidation states tend to be molecular rather than ionic | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/08%3A_Basic_Concepts_of_Chemical_Bonding/8.S%3A_Basic_Concepts_of_Chemical_Bonding_%28Summary%29.txt |
• 9.1: Molecular Shapes
The Lewis electron-pair approach described previously can be used to predict the number and types of bonds between the atoms in a substance, and it indicates which atoms have lone pairs of electrons. This approach gives no information about the actual arrangement of atoms in space, however.
• 9.2: The VSEPR Model
The VSEPR model can predict the structure of nearly any molecule or polyatomic ion in which the central atom is a nonmetal, as well as the structures of many molecules and polyatomic ions with a central metal atom. The VSEPR model is not a theory; it does not attempt to explain observations. Instead, it is a counting procedure that accurately predicts the three-dimensional structures of a large number of compounds, which cannot be predicted using the Lewis electron-pair approach.
• 9.3: Molecular Shape and Molecular Polarity
Compounds with polar covalent bonds have electrons that are shared unequally between the bonded atoms. The polarity of such a bond is determined largely by the relative electronegativites of the bonded atoms. The asymmetrical charge distribution in a polar substance produces a dipole moment, which is the product of the partial charges on the bonded atoms and the distance between them.
• 9.4: Covalent Bonding and Orbital Overlap
A more sophisticated treatment of bonding is needed for systems such as these. In this section, we present a quantum mechanical description of bonding, in which bonding electrons are viewed as being localized between the nuclei of the bonded atoms. The overlap of bonding orbitals is substantially increased through a process called hybridization, which results in the formation of stronger bonds.
• 9.5: Hybrid Orbitals
The localized valence bonding theory uses a process called hybridization, in which atomic orbitals that are similar in energy but not equivalent are combined mathematically to produce sets of equivalent orbitals that are properly oriented to form bonds. These new combinations are called hybrid atomic orbitals because they are produced by combining (hybridizing) two or more atomic orbitals from the same atom.
• 9.6: Multiple Bonds
To describe the bonding in more complex molecules with multiple bonds, we can use an approach that uses hybrid atomic orbitals to describe the $\sigma$ bonding and molecular orbitals to describe the $\pi$ bonding. In this approach, unhybridized np orbitals on atoms bonded to one another are allowed to interact to produce bonding, antibonding, or nonbonding combinations.
• 9.7: Molecular Orbitals
A molecular orbital is an allowed spatial distribution of electrons in a molecule that is associated with a particular orbital energy. Unlike an atomic orbital, which is centered on a single atom, a molecular orbital extends over all the atoms in a molecule or ion. Hence the molecular orbital theory of bonding is a delocalized approach. Molecular orbitals are constructed using linear combinations of atomic orbitals (LCAOs), which describe overlapping atomic orbitals.
• 9.8: Second-Row Diatomic Molecules
Molecular orbital energy-level diagrams for diatomic molecules can be created if the electron configuration of the parent atoms is known, following a few simple rules. Most important, the number of molecular orbitals in a molecule is the same as the number of atomic orbitals that interact. The difference between bonding and antibonding molecular orbital combinations is proportional to the overlap of the parent orbitals and decreases as the energy difference between the atomic orbitals increases.
• 9.E: Exercises
Problems and select solutions to the Chapter.
• 9.S: Molecular Geometry and Bonding Theories (Summary)
Thumbnail: Covalent bonding in ethane (CC BY-SA-NC 3.0; Anonymous)
09: Molecular Geometry and Bonding Theories
The Lewis electron-pair approach described previously can be used to predict the number and types of bonds between the atoms in a substance, and it indicates which atoms have lone pairs of electrons. This approach gives no information about the actual arrangement of atoms in space, however.
Molecular Geometry
The specific three dimensional arrangement of atoms in molecules is referred to as molecular geometry. We also define molecular geometry as the positions of the atomic nuclei in a molecule. There are various instrumental techniques such as X-Ray crystallography and other experimental techniques which can be used to tell us where the atoms are located in a molecule. Using advanced techniques, very complicated structures for proteins, enzymes, DNA, and RNA have been determined. Molecular geometry is associated with the chemistry of vision, smell, taste, drug reactions, and enzyme controlled reactions to name a few.
Example \(1\): Carbon Tetrachloride
The Lewis structure of carbon tetrachloride provides information about connectivities, provides information about valence orbitals, and provides information about bond character.
However, the Lewis structure provides no information about the shape of the molecule, which is defined by the bond angles and the bond lengths. For carbon tetrachloride, each C-Cl bond length is 1.78Å and each Cl-C-Cl bond angle is 109.5°. Hence, carbon tetrachloride is tetrahedral in structure:
Molecular Geometries of \(AB_n\) molecules
Molecular geometry is associated with the specific orientation of bonding atoms. A careful analysis of electron distributions in orbitals will usually result in correct molecular geometry determinations. In addition, the simple writing of Lewis diagrams can also provide important clues for the determination of molecular geometry. Molecular shapes, or geometries, are critical to molecular recognition and function. Table \(1\) shows some examples of geometries where a central atom \(A\) is bonded to two or more \(X\) atoms. As indicated in several of the geometries below, non-bonding electrons \(E\) can strongly influence the molecular geometry of the molecule; this is discussed in more details in VSEPR Model" data-cke-saved-href="/Bookshelves/General_Chemistry/Map:_Chemistry_-_The_Central_Science_(Brown_et_al.)/09:_Molecular_Geometry_and_Bonding_Theories/9.02:_The_VSEPR_Model" href="/Bookshelves/General_Chemistry/Map:_Chemistry_-_The_Central_Science_(Brown_et_al.)/09:_Molecular_Geometry_and_Bonding_Theories/9.02:_The_VSEPR_Model" data-quail-id="26">Section 9.2.
Table \(1\):
6 5 4 3 2
AX6
octahedral
AX5
trigonal bipyramidal
AX4
tetrahedral
AX3
trigonal planar
AX2
linear
1 lone pair of electrons
AX5E
square pyramidal
AX4E
distorted tetrahedron
AX3E
pyramidal
AX2E
nonlinear
AXE
linear
2 lone pairs of electrons
AX4E2
square planar
AX3E2
T-shaped
AX2E2
bent
AXE2
linear
These structures can generally be predicted, when A is a nonmetal, using the "valence-shell electron-pair repulsion model (VSEPR) discussed in the next section.
Contributors and Attributions
• Robyn Rindge (Class of '98) who now works for PDI Dreamworks (look for his name in the credits of Shrek2.). Robyn drew these rotating molecules using Infini-D (MetaCreations).
• Paul Groves, chemistry teacher at South Pasadena High School and Chemmy Bear(opens in new window) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/09%3A_Molecular_Geometry_and_Bonding_Theories/9.01%3A_Molecular_Shapes.txt |
Learning Objectives
• To use the VSEPR model to predict molecular geometries.
• To predict whether a molecule has a dipole moment.
The Lewis electron-pair approach can be used to predict the number and types of bonds between the atoms in a substance, and it indicates which atoms have lone pairs of electrons. This approach gives no information about the actual arrangement of atoms in space, however. We continue our discussion of structure and bonding by introducing the valence-shell electron-pair repulsion (VSEPR) model (pronounced “vesper”), which can be used to predict the shapes of many molecules and polyatomic ions. Keep in mind, however, that the VSEPR model, like any model, is a limited representation of reality; the model provides no information about bond lengths or the presence of multiple bonds.
The VSEPR Model
The VSEPR model can predict the structure of nearly any molecule or polyatomic ion in which the central atom is a nonmetal, as well as the structures of many molecules and polyatomic ions with a central metal atom. The premise of the VSEPR theory is that electron pairs located in bonds and lone pairs repel each other and will therefore adopt the geometry that places electron pairs as far apart from each other as possible. This theory is very simplistic and does not account for the subtleties of orbital interactions that influence molecular shapes; however, the simple VSEPR counting procedure accurately predicts the three-dimensional structures of a large number of compounds, which cannot be predicted using the Lewis electron-pair approach.
We can use the VSEPR model to predict the geometry of most polyatomic molecules and ions by focusing only on the number of electron pairs around the central atom, ignoring all other valence electrons present. According to this model, valence electrons in the Lewis structure form groups, which may consist of a single bond, a double bond, a triple bond, a lone pair of electrons, or even a single unpaired electron, which in the VSEPR model is counted as a lone pair. Because electrons repel each other electrostatically, the most stable arrangement of electron groups (i.e., the one with the lowest energy) is the one that minimizes repulsions. Groups are positioned around the central atom in a way that produces the molecular structure with the lowest energy, as illustrated in Figures \(1\) and \(2\).
In the VSEPR model, the molecule or polyatomic ion is given an AXmEn designation, where A is the central atom, X is a bonded atom, E is a nonbonding valence electron group (usually a lone pair of electrons), and m and n are integers. Each group around the central atom is designated as a bonding pair (BP) or lone (nonbonding) pair (LP). From the BP and LP interactions we can predict both the relative positions of the atoms and the angles between the bonds, called the bond angles. Using this information, we can describe the molecular geometry, the arrangement of the bonded atoms in a molecule or polyatomic ion.
VESPR Produce to predict Molecular geometry
This VESPR procedure is summarized as follows:
1. Draw the Lewis electron structure of the molecule or polyatomic ion.
2. Determine the electron group arrangement around the central atom that minimizes repulsions.
3. Assign an AXmEn designation; then identify the LP–LP, LP–BP, or BP–BP interactions and predict deviations from ideal bond angles.
4. Describe the molecular geometry.
We will illustrate the use of this procedure with several examples, beginning with atoms with two electron groups. In our discussion we will refer to Figure \(2\) and Figure \(3\), which summarize the common molecular geometries and idealized bond angles of molecules and ions with two to six electron groups.
Two Electron Groups
Our first example is a molecule with two bonded atoms and no lone pairs of electrons, \(BeH_2\).
AX2 Molecules: BeH2
1. The central atom, beryllium, contributes two valence electrons, and each hydrogen atom contributes one. The Lewis electron structure is
3. Both groups around the central atom are bonding pairs (BP). Thus BeH2 is designated as AX2.
4. From Figure \(3\) we see that with two bonding pairs, the molecular geometry that minimizes repulsions in BeH2 is linear.
AX2 Molecules: CO2
1. The central atom, carbon, contributes four valence electrons, and each oxygen atom contributes six. The Lewis electron structure is
2. The carbon atom forms two double bonds. Each double bond is a group, so there are two electron groups around the central atom. Like BeH2, the arrangement that minimizes repulsions places the groups 180° apart.
3. Once again, both groups around the central atom are bonding pairs (BP), so CO2 is designated as AX2.
4. VSEPR only recognizes groups around the central atom. Thus the lone pairs on the oxygen atoms do not influence the molecular geometry. With two bonding pairs on the central atom and no lone pairs, the molecular geometry of CO2 is linear (Figure \(3\)). The structure of \(\ce{CO2}\) is shown in Figure \(1\).
Three Electron Groups
AX3 Molecules: BCl3
1. The central atom, boron, contributes three valence electrons, and each chlorine atom contributes seven valence electrons. The Lewis electron structure is
3. All electron groups are bonding pairs (BP), so the structure is designated as AX3.
4. From Figure \(3\) we see that with three bonding pairs around the central atom, the molecular geometry of BCl3 is trigonal planar, as shown in Figure \(2\).
AX3 Molecules: CO32−
1. The central atom, carbon, has four valence electrons, and each oxygen atom has six valence electrons. As you learned previously, the Lewis electron structure of one of three resonance forms is represented as
3. All electron groups are bonding pairs (BP). With three bonding groups around the central atom, the structure is designated as AX3.
4. We see from Figure \(3\) that the molecular geometry of CO32 is trigonal planar with bond angles of 120°.
In our next example we encounter the effects of lone pairs and multiple bonds on molecular geometry for the first time.
AX2E Molecules: SO2
1. The central atom, sulfur, has 6 valence electrons, as does each oxygen atom. With 18 valence electrons, the Lewis electron structure is shown below.
3. There are two bonding pairs and one lone pair, so the structure is designated as AX2E. This designation has a total of three electron pairs, two X and one E. Because a lone pair is not shared by two nuclei, it occupies more space near the central atom than a bonding pair (Figure \(4\)). Thus bonding pairs and lone pairs repel each other electrostatically in the order BP–BP < LP–BP < LP–LP. In SO2, we have one BP–BP interaction and two LP–BP interactions.
4. The molecular geometry is described only by the positions of the nuclei, not by the positions of the lone pairs. Thus with two nuclei and one lone pair the shape is bent, or V shaped, which can be viewed as a trigonal planar arrangement with a missing vertex (Figures \(2\) and \(3\)). The O-S-O bond angle is expected to be less than 120° because of the extra space taken up by the lone pair.
As with SO2, this composite model of electron distribution and negative electrostatic potential in ammonia shows that a lone pair of electrons occupies a larger region of space around the nitrogen atom than does a bonding pair of electrons that is shared with a hydrogen atom.
Like lone pairs of electrons, multiple bonds occupy more space around the central atom than a single bond, which can cause other bond angles to be somewhat smaller than expected. This is because a multiple bond has a higher electron density than a single bond, so its electrons occupy more space than those of a single bond. For example, in a molecule such as CH2O (AX3), whose structure is shown below, the double bond repels the single bonds more strongly than the single bonds repel each other. This causes a deviation from ideal geometry (an H–C–H bond angle of 116.5° rather than 120°).
Four Electron Groups
One of the limitations of Lewis structures is that they depict molecules and ions in only two dimensions. With four electron groups, we must learn to show molecules and ions in three dimensions.
AX4 Molecules: CH4
1. The central atom, carbon, contributes four valence electrons, and each hydrogen atom has one valence electron, so the full Lewis electron structure is
2. There are four electron groups around the central atom. As shown in Figure \(2\), repulsions are minimized by placing the groups in the corners of a tetrahedron with bond angles of 109.5°.
3. All electron groups are bonding pairs, so the structure is designated as AX4.
4. With four bonding pairs, the molecular geometry of methane is tetrahedral (Figure \(3\)).
AX3E Molecules: NH3
1. In ammonia, the central atom, nitrogen, has five valence electrons and each hydrogen donates one valence electron, producing the Lewis electron structure
2. There are four electron groups around nitrogen, three bonding pairs and one lone pair. Repulsions are minimized by directing each hydrogen atom and the lone pair to the corners of a tetrahedron.
3. With three bonding pairs and one lone pair, the structure is designated as AX3E. This designation has a total of four electron pairs, three X and one E. We expect the LP–BP interactions to cause the bonding pair angles to deviate significantly from the angles of a perfect tetrahedron.
4. There are three nuclei and one lone pair, so the molecular geometry is trigonal pyramidal. In essence, this is a tetrahedron with a vertex missing (Figure \(3\)). However, the H–N–H bond angles are less than the ideal angle of 109.5° because of LP–BP repulsions (Figure \(3\) and Figure \(4\)).
AX2E2 Molecules: H2O
1. Oxygen has six valence electrons and each hydrogen has one valence electron, producing the Lewis electron structure
3. With two bonding pairs and two lone pairs, the structure is designated as AX2E2 with a total of four electron pairs. Due to LP–LP, LP–BP, and BP–BP interactions, we expect a significant deviation from idealized tetrahedral angles.
4. With two hydrogen atoms and two lone pairs of electrons, the structure has significant lone pair interactions. There are two nuclei about the central atom, so the molecular shape is bent, or V shaped, with an H–O–H angle that is even less than the H–N–H angles in NH3, as we would expect because of the presence of two lone pairs of electrons on the central atom rather than one. This molecular shape is essentially a tetrahedron with two missing vertices.
Five Electron Groups
In previous examples it did not matter where we placed the electron groups because all positions were equivalent. In some cases, however, the positions are not equivalent. We encounter this situation for the first time with five electron groups.
AX5 Molecules: PCl5
1. Phosphorus has five valence electrons and each chlorine has seven valence electrons, so the Lewis electron structure of PCl5 is
3. All electron groups are bonding pairs, so the structure is designated as AX5. There are no lone pair interactions.
4. The molecular geometry of PCl5 is trigonal bipyramidal, as shown in Figure \(3\). The molecule has three atoms in a plane in equatorial positions and two atoms above and below the plane in axial positions. The three equatorial positions are separated by 120° from one another, and the two axial positions are at 90° to the equatorial plane. The axial and equatorial positions are not chemically equivalent, as we will see in our next example.
AX4E Molecules: SF4
1. The sulfur atom has six valence electrons and each fluorine has seven valence electrons, so the Lewis electron structure is
With an expanded valence, this species is an exception to the octet rule.
2. There are five groups around sulfur, four bonding pairs and one lone pair. With five electron groups, the lowest energy arrangement is a trigonal bipyramid, as shown in Figure \(2\).
3. We designate SF4 as AX4E; it has a total of five electron pairs. However, because the axial and equatorial positions are not chemically equivalent, where do we place the lone pair? If we place the lone pair in the axial position, we have three LP–BP repulsions at 90°. If we place it in the equatorial position, we have two 90° LP–BP repulsions at 90°. With fewer 90° LP–BP repulsions, we can predict that the structure with the lone pair of electrons in the equatorial position is more stable than the one with the lone pair in the axial position. We also expect a deviation from ideal geometry because a lone pair of electrons occupies more space than a bonding pair.
At 90°, the two electron pairs share a relatively large region of space, which leads to strong repulsive electron–electron interactions.
4. With four nuclei and one lone pair of electrons, the molecular structure is based on a trigonal bipyramid with a missing equatorial vertex; it is described as a seesaw. The Faxial–S–Faxial angle is 173° rather than 180° because of the lone pair of electrons in the equatorial plane.
AX3E2 Molecules: BrF3
1. The bromine atom has seven valence electrons, and each fluorine has seven valence electrons, so the Lewis electron structure is
Once again, we have a compound that is an exception to the octet rule.
2. There are five groups around the central atom, three bonding pairs and two lone pairs. We again direct the groups toward the vertices of a trigonal bipyramid.
3. With three bonding pairs and two lone pairs, the structural designation is AX3E2 with a total of five electron pairs. Because the axial and equatorial positions are not equivalent, we must decide how to arrange the groups to minimize repulsions. If we place both lone pairs in the axial positions, we have six LP–BP repulsions at 90°. If both are in the equatorial positions, we have four LP–BP repulsions at 90°. If one lone pair is axial and the other equatorial, we have one LP–LP repulsion at 90° and three LP–BP repulsions at 90°:
Structure (c) can be eliminated because it has a LP–LP interaction at 90°. Structure (b), with fewer LP–BP repulsions at 90° than (a), is lower in energy. However, we predict a deviation in bond angles because of the presence of the two lone pairs of electrons.
4. The three nuclei in BrF3 determine its molecular structure, which is described as T shaped. This is essentially a trigonal bipyramid that is missing two equatorial vertices. The Faxial–Br–Faxial angle is 172°, less than 180° because of LP–BP repulsions (Figure \(2\).1).
Because lone pairs occupy more space around the central atom than bonding pairs, electrostatic repulsions are more important for lone pairs than for bonding pairs.
AX2E3 Molecules: I3−
1. Each iodine atom contributes seven electrons and the negative charge one, so the Lewis electron structure is
2. There are five electron groups about the central atom in I3, two bonding pairs and three lone pairs. To minimize repulsions, the groups are directed to the corners of a trigonal bipyramid.
3. With two bonding pairs and three lone pairs, I3 has a total of five electron pairs and is designated as AX2E3. We must now decide how to arrange the lone pairs of electrons in a trigonal bipyramid in a way that minimizes repulsions. Placing them in the axial positions eliminates 90° LP–LP repulsions and minimizes the number of 90° LP–BP repulsions.
The three lone pairs of electrons have equivalent interactions with the three iodine atoms, so we do not expect any deviations in bonding angles.
4. With three nuclei and three lone pairs of electrons, the molecular geometry of I3 is linear. This can be described as a trigonal bipyramid with three equatorial vertices missing. The ion has an I–I–I angle of 180°, as expected.
Six Electron Groups
Six electron groups form an octahedron, a polyhedron made of identical equilateral triangles and six identical vertices (Figure \(2\).)
AX6 Molecules: SF6
1. The central atom, sulfur, contributes six valence electrons, and each fluorine atom has seven valence electrons, so the Lewis electron structure is
With an expanded valence, this species is an exception to the octet rule.
2. There are six electron groups around the central atom, each a bonding pair. We see from Figure \(2\) that the geometry that minimizes repulsions is octahedral.
3. With only bonding pairs, SF6 is designated as AX6. All positions are chemically equivalent, so all electronic interactions are equivalent.
4. There are six nuclei, so the molecular geometry of SF6 is octahedral.
AX5E Molecules: BrF5
1. The central atom, bromine, has seven valence electrons, as does each fluorine, so the Lewis electron structure is
With its expanded valence, this species is an exception to the octet rule.
2. There are six electron groups around the Br, five bonding pairs and one lone pair. Placing five F atoms around Br while minimizing BP–BP and LP–BP repulsions gives the following structure:
3. With five bonding pairs and one lone pair, BrF5 is designated as AX5E; it has a total of six electron pairs. The BrF5 structure has four fluorine atoms in a plane in an equatorial position and one fluorine atom and the lone pair of electrons in the axial positions. We expect all Faxial–Br–Fequatorial angles to be less than 90° because of the lone pair of electrons, which occupies more space than the bonding electron pairs.
4. With five nuclei surrounding the central atom, the molecular structure is based on an octahedron with a vertex missing. This molecular structure is square pyramidal. The Faxial–B–Fequatorial angles are 85.1°, less than 90° because of LP–BP repulsions.
AX4E2 Molecules: ICl4−
1. The central atom, iodine, contributes seven electrons. Each chlorine contributes seven, and there is a single negative charge. The Lewis electron structure is
2. There are six electron groups around the central atom, four bonding pairs and two lone pairs. The structure that minimizes LP–LP, LP–BP, and BP–BP repulsions is
3. ICl4 is designated as AX4E2 and has a total of six electron pairs. Although there are lone pairs of electrons, with four bonding electron pairs in the equatorial plane and the lone pairs of electrons in the axial positions, all LP–BP repulsions are the same. Therefore, we do not expect any deviation in the Cl–I–Cl bond angles.
4. With five nuclei, the ICl4− ion forms a molecular structure that is square planar, an octahedron with two opposite vertices missing.
The relationship between the number of electron groups around a central atom, the number of lone pairs of electrons, and the molecular geometry is summarized in Figure \(6\).
Figure \(6\): Overview of Molecular Geometries
Example \(1\)
Using the VSEPR model, predict the molecular geometry of each molecule or ion.
1. PF5 (phosphorus pentafluoride, a catalyst used in certain organic reactions)
2. H3O+ (hydronium ion)
Given: two chemical species
Asked for: molecular geometry
Strategy:
1. Draw the Lewis electron structure of the molecule or polyatomic ion.
2. Determine the electron group arrangement around the central atom that minimizes repulsions.
3. Assign an AXmEn designation; then identify the LP–LP, LP–BP, or BP–BP interactions and predict deviations in bond angles.
4. Describe the molecular geometry.
Solution:
1. A The central atom, P, has five valence electrons and each fluorine has seven valence electrons, so the Lewis structure of PF5 is
C All electron groups are bonding pairs, so PF5 is designated as AX5. Notice that this gives a total of five electron pairs. With no lone pair repulsions, we do not expect any bond angles to deviate from the ideal.
D The PF5 molecule has five nuclei and no lone pairs of electrons, so its molecular geometry is trigonal bipyramidal.
2. A The central atom, O, has six valence electrons, and each H atom contributes one valence electron. Subtracting one electron for the positive charge gives a total of eight valence electrons, so the Lewis electron structure is
B There are four electron groups around oxygen, three bonding pairs and one lone pair. Like NH3, repulsions are minimized by directing each hydrogen atom and the lone pair to the corners of a tetrahedron.
C With three bonding pairs and one lone pair, the structure is designated as AX3E and has a total of four electron pairs (three X and one E). We expect the LP–BP interactions to cause the bonding pair angles to deviate significantly from the angles of a perfect tetrahedron.
D There are three nuclei and one lone pair, so the molecular geometry is trigonal pyramidal, in essence a tetrahedron missing a vertex. However, the H–O–H bond angles are less than the ideal angle of 109.5° because of LP–BP repulsions:
Exercise \(1\)
Using the VSEPR model, predict the molecular geometry of each molecule or ion.
1. XeO3
2. PF6
3. NO2+
Answer a
trigonal pyramidal
Answer b
octahedral
Answer c
linear
Example \(2\)
Predict the molecular geometry of each molecule.
1. XeF2
2. SnCl2
Given: two chemical compounds
Asked for: molecular geometry
Strategy:
Use the strategy given in Example\(1\).
Solution:
1. A Xenon contributes eight electrons and each fluorine seven valence electrons, so the Lewis electron structure is
B There are five electron groups around the central atom, two bonding pairs and three lone pairs. Repulsions are minimized by placing the groups in the corners of a trigonal bipyramid.
C From B, XeF2 is designated as AX2E3 and has a total of five electron pairs (two X and three E). With three lone pairs about the central atom, we can arrange the two F atoms in three possible ways: both F atoms can be axial, one can be axial and one equatorial, or both can be equatorial:
The structure with the lowest energy is the one that minimizes LP–LP repulsions. Both (b) and (c) have two 90° LP–LP interactions, whereas structure (a) has none. Thus both F atoms are in the axial positions, like the two iodine atoms around the central iodine in I3. All LP–BP interactions are equivalent, so we do not expect a deviation from an ideal 180° in the F–Xe–F bond angle.
D With two nuclei about the central atom, the molecular geometry of XeF2 is linear. It is a trigonal bipyramid with three missing equatorial vertices.
2. A The tin atom donates 4 valence electrons and each chlorine atom donates 7 valence electrons. With 18 valence electrons, the Lewis electron structure is
B There are three electron groups around the central atom, two bonding groups and one lone pair of electrons. To minimize repulsions the three groups are initially placed at 120° angles from each other.
C From B we designate SnCl2 as AX2E. It has a total of three electron pairs, two X and one E. Because the lone pair of electrons occupies more space than the bonding pairs, we expect a decrease in the Cl–Sn–Cl bond angle due to increased LP–BP repulsions.
D With two nuclei around the central atom and one lone pair of electrons, the molecular geometry of SnCl2 is bent, like SO2, but with a Cl–Sn–Cl bond angle of 95°. The molecular geometry can be described as a trigonal planar arrangement with one vertex missing.
Exercise \(2\)
Predict the molecular geometry of each molecule.
1. SO3
2. XeF4
Answer a
trigonal planar
Answer b
square planar
Molecules with No Single Central Atom
The VSEPR model can be used to predict the structure of somewhat more complex molecules with no single central atom by treating them as linked AXmEn fragments. We will demonstrate with methyl isocyanate (CH3–N=C=O), a volatile and highly toxic molecule that is used to produce the pesticide Sevin. In 1984, large quantities of Sevin were accidentally released in Bhopal, India, when water leaked into storage tanks. The resulting highly exothermic reaction caused a rapid increase in pressure that ruptured the tanks, releasing large amounts of methyl isocyanate that killed approximately 3800 people and wholly or partially disabled about 50,000 others. In addition, there was significant damage to livestock and crops.
We can treat methyl isocyanate as linked AXmEn fragments beginning with the carbon atom at the left, which is connected to three H atoms and one N atom by single bonds. The four bonds around carbon mean that it must be surrounded by four bonding electron pairs in a configuration similar to AX4. We can therefore predict the CH3–N portion of the molecule to be roughly tetrahedral, similar to methane:
The nitrogen atom is connected to one carbon by a single bond and to the other carbon by a double bond, producing a total of three bonds, C–N=C. For nitrogen to have an octet of electrons, it must also have a lone pair:
Because multiple bonds are not shown in the VSEPR model, the nitrogen is effectively surrounded by three electron pairs. Thus according to the VSEPR model, the C–N=C fragment should be bent with an angle less than 120°.
The carbon in the –N=C=O fragment is doubly bonded to both nitrogen and oxygen, which in the VSEPR model gives carbon a total of two electron pairs. The N=C=O angle should therefore be 180°, or linear. The three fragments combine to give the following structure:
Certain patterns are seen in the structures of moderately complex molecules. For example, carbon atoms with four bonds (such as the carbon on the left in methyl isocyanate) are generally tetrahedral. Similarly, the carbon atom on the right has two double bonds that are similar to those in CO2, so its geometry, like that of CO2, is linear. Recognizing similarities to simpler molecules will help you predict the molecular geometries of more complex molecules.
Example \(3\)
Use the VSEPR model to predict the molecular geometry of propyne (H3C–C≡CH), a gas with some anesthetic properties.
Given: chemical compound
Asked for: molecular geometry
Strategy:
Count the number of electron groups around each carbon, recognizing that in the VSEPR model, a multiple bond counts as a single group. Use Figure \(3\) to determine the molecular geometry around each carbon atom and then deduce the structure of the molecule as a whole.
Solution:
Because the carbon atom on the left is bonded to four other atoms, we know that it is approximately tetrahedral. The next two carbon atoms share a triple bond, and each has an additional single bond. Because a multiple bond is counted as a single bond in the VSEPR model, each carbon atom behaves as if it had two electron groups. This means that both of these carbons are linear, with C–C≡C and C≡C–H angles of 180°.
Exercise \(3\)
Predict the geometry of allene (H2C=C=CH2), a compound with narcotic properties that is used to make more complex organic molecules.
Answer
The terminal carbon atoms are trigonal planar, the central carbon is linear, and the C–C–C angle is 180°.
Molecular Dipole Moments
You previously learned how to calculate the dipole moments of simple diatomic molecules. In more complex molecules with polar covalent bonds, the three-dimensional geometry and the compound’s symmetry determine whether there is a net dipole moment. Mathematically, dipole moments are vectors; they possess both a magnitude and a direction. The dipole moment of a molecule is therefore the vector sum of the dipole moments of the individual bonds in the molecule. If the individual bond dipole moments cancel one another, there is no net dipole moment. Such is the case for CO2, a linear molecule (Figure \(\PageIndex{8a}\)). Each C–O bond in CO2 is polar, yet experiments show that the CO2 molecule has no dipole moment. Because the two C–O bond dipoles in CO2 are equal in magnitude and oriented at 180° to each other, they cancel. As a result, the CO2 molecule has no net dipole moment even though it has a substantial separation of charge. In contrast, the H2O molecule is not linear (Figure \(\PageIndex{8b}\)); it is bent in three-dimensional space, so the dipole moments do not cancel each other. Thus a molecule such as H2O has a net dipole moment. We expect the concentration of negative charge to be on the oxygen, the more electronegative atom, and positive charge on the two hydrogens. This charge polarization allows H2O to hydrogen-bond to other polarized or charged species, including other water molecules.
Other examples of molecules with polar bonds are shown in Figure \(9\). In molecular geometries that are highly symmetrical (most notably tetrahedral and square planar, trigonal bipyramidal, and octahedral), individual bond dipole moments completely cancel, and there is no net dipole moment. Although a molecule like CHCl3 is best described as tetrahedral, the atoms bonded to carbon are not identical. Consequently, the bond dipole moments cannot cancel one another, and the molecule has a dipole moment. Due to the arrangement of the bonds in molecules that have V-shaped, trigonal pyramidal, seesaw, T-shaped, and square pyramidal geometries, the bond dipole moments cannot cancel one another. Consequently, molecules with these geometries always have a nonzero dipole moment. Molecules with asymmetrical charge distributions have a net dipole moment.
Example \(4\)
Which molecule(s) has a net dipole moment?
1. \(\ce{H2S}\)
2. \(\ce{NHF2}\)
3. \(\ce{BF3}\)
Given: three chemical compounds
Asked for: net dipole moment
Strategy:
For each three-dimensional molecular geometry, predict whether the bond dipoles cancel. If they do not, then the molecule has a net dipole moment.
Solution:
1. The total number of electrons around the central atom, S, is eight, which gives four electron pairs. Two of these electron pairs are bonding pairs and two are lone pairs, so the molecular geometry of \(\ce{H2S}\) is bent (Figure \(6\)). The bond dipoles cannot cancel one another, so the molecule has a net dipole moment.
2. Difluoroamine has a trigonal pyramidal molecular geometry. Because there is one hydrogen and two fluorines, and because of the lone pair of electrons on nitrogen, the molecule is not symmetrical, and the bond dipoles of NHF2 cannot cancel one another. This means that NHF2 has a net dipole moment. We expect polarization from the two fluorine atoms, the most electronegative atoms in the periodic table, to have a greater affect on the net dipole moment than polarization from the lone pair of electrons on nitrogen.
3. The molecular geometry of BF3 is trigonal planar. Because all the B–F bonds are equal and the molecule is highly symmetrical, the dipoles cancel one another in three-dimensional space. Thus BF3 has a net dipole moment of zero:
Exercise \(4\)
Which molecule(s) has a net dipole moment?
• \(\ce{CH3Cl}\)
• \(\ce{SO3}\)
• \(\ce{XeO3}\)
Answer
\(\ce{CH3Cl}\) and \(\ce{XeO3}\)
Summary
Lewis electron structures give no information about molecular geometry, the arrangement of bonded atoms in a molecule or polyatomic ion, which is crucial to understanding the chemistry of a molecule. The valence-shell electron-pair repulsion (VSEPR) model allows us to predict which of the possible structures is actually observed in most cases. It is based on the assumption that pairs of electrons occupy space, and the lowest-energy structure is the one that minimizes electron pair–electron pair repulsions. In the VSEPR model, the molecule or polyatomic ion is given an AXmEn designation, where A is the central atom, X is a bonded atom, E is a nonbonding valence electron group (usually a lone pair of electrons), and m and n are integers. Each group around the central atom is designated as a bonding pair (BP) or lone (nonbonding) pair (LP). From the BP and LP interactions we can predict both the relative positions of the atoms and the angles between the bonds, called the bond angles. From this we can describe the molecular geometry. The VSEPR model can be used to predict the shapes of many molecules and polyatomic ions, but it gives no information about bond lengths and the presence of multiple bonds. A combination of VSEPR and a bonding model, such as Lewis electron structures, is necessary to understand the presence of multiple bonds.
Molecules with polar covalent bonds can have a dipole moment, an asymmetrical distribution of charge that results in a tendency for molecules to align themselves in an applied electric field. Any diatomic molecule with a polar covalent bond has a dipole moment, but in polyatomic molecules, the presence or absence of a net dipole moment depends on the structure. For some highly symmetrical structures, the individual bond dipole moments cancel one another, giving a dipole moment of zero. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/09%3A_Molecular_Geometry_and_Bonding_Theories/9.02%3A_The_VSEPR_Model.txt |
Learning Objectives
• To calculate the percent ionic character of a covalent polar bond
Previously, we described the two idealized extremes of chemical bonding:
• ionic bonding—in which one or more electrons are transferred completely from one atom to another, and the resulting ions are held together by purely electrostatic forces—and
• covalent bonding, in which electrons are shared equally between two atoms.
Most compounds, however, have polar covalent bonds, which means that electrons are shared unequally between the bonded atoms. Figure $1$ compares the electron distribution in a polar covalent bond with those in an ideally covalent and an ideally ionic bond. Recall that a lowercase Greek delta ($\delta$) is used to indicate that a bonded atom possesses a partial positive charge, indicated by $\delta^+$, or a partial negative charge, indicated by $\delta^-$, and a bond between two atoms that possess partial charges is a polar bond.
Bond Polarity
The polarity of a bond—the extent to which it is polar—is determined largely by the relative electronegativities of the bonded atoms. Electronegativity ($\chi$) was defined as the ability of an atom in a molecule or an ion to attract electrons to itself. Thus there is a direct correlation between electronegativity and bond polarity. A bond is nonpolar if the bonded atoms have equal electronegativities. If the electronegativities of the bonded atoms are not equal, however, the bond is polarized toward the more electronegative atom. A bond in which the electronegativity of B ($\chi_B$) is greater than the electronegativity of A ($\chi_A$), for example, is indicated with the partial negative charge on the more electronegative atom:
$\begin{matrix} _{less\; electronegative}& & _{more\; electronegative}\ A\; &-& B\; \; \ ^{\delta ^{+}} & & ^{\delta ^{-}} \end{matrix} \label{9.3.1}$
One way of estimating the ionic character of a bond—that is, the magnitude of the charge separation in a polar covalent bond—is to calculate the difference in electronegativity between the two atoms:
$Δ\chi = \chi_B − \chi_A. \nonumber$
To predict the polarity of the bonds in Cl2, HCl, and NaCl, for example, we look at the electronegativities of the relevant atoms (Table A2): $\chi_{Cl} = 3.16$, $\chi_H = 2.20$, and $\chi_{Na} = 0.93$. $\ce{Cl2}$ must be nonpolar because the electronegativity difference ($Δ\chi$) is zero; hence the two chlorine atoms share the bonding electrons equally. In NaCl, $Δ\chi$ is 2.23. This high value is typical of an ionic compound ($Δ\chi ≥ ≈1.5$) and means that the valence electron of sodium has been completely transferred to chlorine to form Na+ and Cl ions. In HCl, however, $Δ\chi$ is only 0.96. The bonding electrons are more strongly attracted to the more electronegative chlorine atom, and so the charge distribution is
$\begin{matrix} _{\delta ^{+}}& & _{\delta ^{-}}\ H\; \; &-& Cl \end{matrix} \nonumber$
Remember that electronegativities are difficult to measure precisely and different definitions produce slightly different numbers. In practice, the polarity of a bond is usually estimated rather than calculated.
Bond polarity and ionic character increase with an increasing difference in electronegativity.
As with bond energies, the electronegativity of an atom depends to some extent on its chemical environment. It is therefore unlikely that the reported electronegativities of a chlorine atom in $\ce{NaCl}$, $\ce{Cl2}$, $\ce{ClF5}$, and $\ce{HClO4}$ would be exactly the same.
Dipole Moments
The asymmetrical charge distribution in a polar substance such as $\ce{HCl}$ produces a dipole moment where $Qr$ in meters ($m$). is abbreviated by the Greek letter mu ($\mu$). The dipole moment is defined as the product of the partial charge $Q$ on the bonded atoms and the distance $r$ between the partial charges:
$\mu=Qr \label{9.3.2}$
where $Q$ is measured in coulombs ($C$) and $r$ in meters. The unit for dipole moments is the debye (D):
$1\; D = 3.3356\times 10^{-30}\; C\cdot m \label{9.7.3}$
When a molecule with a dipole moment is placed in an electric field, it tends to orient itself with the electric field because of its asymmetrical charge distribution (Figure $2$).
We can measure the partial charges on the atoms in a molecule such as $\ce{HCl}$ using Equation \ref{9.3.2}. If the bonding in $\ce{HCl}$ were purely ionic, an electron would be transferred from H to Cl, so there would be a full +1 charge on the H atom and a full −1 charge on the Cl atom. The dipole moment of $\ce{HCl}$ is 1.109 D, as determined by measuring the extent of its alignment in an electric field, and the reported gas-phase H–Cl distance is 127.5 pm. Hence the charge on each atom is
\begin{align*} Q &=\dfrac{\mu }{r} \[4pt] &=1.109\;\cancel{D}\left ( \dfrac{3.3356\times 10^{-30}\; C\cdot \cancel{m}}{1\; \cancel{D}} \right )\left ( \dfrac{1}{127.8\; \cancel{pm}} \right )\left ( \dfrac{1\; \cancel{pm}}{10^{-12\;} \cancel{m}} \right ) \[4pt] &=2.901\times 10^{-20}\;C \label{9.7.4} \end{align*}
By dividing this calculated value by the charge on a single electron (1.6022 × 10−19 C), we find that the electron distribution in$\ce{HCl}$ is asymmetric and that effectively it appears that there is a net negative charge on the $\ce{Cl}$ of about −0.18, effectively corresponding to about 0.18 e. This certainly does not mean that there is a fraction of an electron on the $\ce{Cl}$ atom, but that the distribution of electron probability favors the Cl atom side of the molecule by about this amount.
$\dfrac{2.901\times 10^{-20}\; \cancel{C}}{1.6022\times 10^{-19}\; \cancel{C}}=0.1811\;e^{-} \label{9.7.5}$
To form a neutral compound, the charge on the $\ce{H}$ atom must be equal but opposite. Thus the measured dipole moment of $\ce{HCl}$ indicates that the H–Cl bond has approximately 18% ionic character (0.1811 × 100), or 82% covalent character. Instead of writing$\ce{HCl}$ as
$\begin{matrix} _{\delta ^{+}}& & _{\delta ^{-}}\ H\; \; &-& Cl \end{matrix}$
we can therefore indicate the charge separation quantitatively as
$\begin{matrix} _{0.18\delta ^{+}}& & _{0.18\delta ^{-}}\ H\; \; &-& Cl \end{matrix}$
Our calculated results are in agreement with the electronegativity difference between hydrogen and chlorine ($\chi_H = 2.20$ and $\chi_{Cl} = 3.16$) so
$\chi_{Cl} − \chi_H = 0.96 \nonumber$
This is a value well within the range for polar covalent bonds. We indicate the dipole moment by writing an arrow above the molecule.Mathematically, dipole moments are vectors, and they possess both a magnitude and a direction. The dipole moment of a molecule is the vector sum of the dipoles of the individual bonds. In HCl, for example, the dipole moment is indicated as follows:
The arrow shows the direction of electron flow by pointing toward the more electronegative atom.
A warning about Dipole Moment arrows
As the figure above shows, we represent dipole moments by an arrow with a length proportional to $μ$ and pointing from the positive charge to the negative charge. However, the opposite convention is still widely used especially among physicists.
The charge on the atoms of many substances in the gas phase can be calculated using measured dipole moments and bond distances. Figure $2$ shows a plot of the percent ionic character versus the difference in electronegativity of the bonded atoms for several substances. According to the graph, the bonding in species such as $NaCl_{(g)}$ and $CsF_{(g)}$ is substantially less than 100% ionic in character. As the gas condenses into a solid, however, dipole–dipole interactions between polarized species increase the charge separations. In the crystal, therefore, an electron is transferred from the metal to the nonmetal, and these substances behave like classic ionic compounds. The data in Figure $2$ show that diatomic species with an electronegativity difference of less than 1.5 are less than 50% ionic in character, which is consistent with our earlier description of these species as containing polar covalent bonds. The use of dipole moments to determine the ionic character of a polar bond is illustrated in Example $1$.
Example $1$: percent ionic character
In the gas phase, NaCl has a dipole moment of 9.001 D and an Na–Cl distance of 236.1 pm. Calculate the percent ionic character in NaCl.
Given: chemical species, dipole moment, and internuclear distance
Asked for: percent ionic character
Strategy:
1. Compute the charge on each atom using the information given and Equation $\ref{9.3.2}$
2. Find the percent ionic character from the ratio of the actual charge to the charge of a single electron.
Solution:
A The charge on each atom is given by
\begin{align*} Q &=\dfrac{\mu }{r} \[4pt] &=9.001\;\cancel{D}\left ( \dfrac{3.3356\times 10^{-30}\; C\cdot \cancel{m}}{1\; \cancel{D}} \right )\left ( \dfrac{1}{236.1\; \cancel{pm}} \right )\left ( \dfrac{1\; \cancel{pm}}{10^{-12\;} \cancel{m}} \right ) \[4pt] &=1.272\times 10^{-19}\;C \end{align*} \nonumber
Thus NaCl behaves as if it had charges of 1.272 × 10−19 C on each atom separated by 236.1 pm.
B The percent ionic character is given by the ratio of the actual charge to the charge of a single electron (the charge expected for the complete transfer of one electron):
\begin{align*} \text{ ionic character} &=\left ( \dfrac{1.272\times 10^{-19}\; \cancel{C}}{1.6022\times 10^{-19}\; \cancel{C}} \right )\left ( 100 \right ) \[4pt] &=79.39\%\simeq 79\% \end{align*} \nonumber
Exercise $1$
In the gas phase, silver chloride ($\ce{AgCl}$) has a dipole moment of 6.08 D and an Ag–Cl distance of 228.1 pm. What is the percent ionic character in silver chloride?
Answer
55.5%
Summary
Bond polarity and ionic character increase with an increasing difference in electronegativity.
$\mu = Qr \nonumber$
Compounds with polar covalent bonds have electrons that are shared unequally between the bonded atoms. The polarity of such a bond is determined largely by the relative electronegativites of the bonded atoms. The asymmetrical charge distribution in a polar substance produces a dipole moment, which is the product of the partial charges on the bonded atoms and the distance between them. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/09%3A_Molecular_Geometry_and_Bonding_Theories/9.03%3A_Molecular_Shape_and_Molecular_Polarity.txt |
Learning Objectives
• To describe the bonding in simple compounds using valence bond theory.
Although the VSEPR model is a simple and useful method for qualitatively predicting the structures of a wide range of compounds, it is not infallible. It predicts, for example, that H2S and PH3 should have structures similar to those of \(\ce{H2O}\) and \(\ce{NH3}\), respectively. In fact, structural studies have shown that the H–S–H and H–P–H angles are more than 12° smaller than the corresponding bond angles in \(\ce{H2O}\) and \(\ce{NH3}\). More disturbing, the VSEPR model predicts that the simple group 2 halides (MX2), which have four valence electrons, should all have linear X–M–X geometries. Instead, many of these species, including \(\ce{SrF2}\) and \(\ce{BaF2}\), are significantly bent. A more sophisticated treatment of bonding is needed for systems such as these. In this section, we present a quantum mechanical description of bonding, in which bonding electrons are viewed as being localized between the nuclei of the bonded atoms. The overlap of bonding orbitals is substantially increased through a process called hybridization, which results in the formation of stronger bonds.
Introduction
As we have talked about using Lewis structures to depict the bonding in covalent compounds, we have been very vague in our language about the actual nature of the chemical bonds themselves. We know that a covalent bond involves the ‘sharing’ of a pair of electrons between two atoms - but how does this happen, and how does it lead to the formation of a bond holding the two atoms together?
The valence bond theory is introduced to describe bonding in covalent molecules. In this model, bonds are considered to form from the overlapping of two atomic orbitals on different atoms, each orbital containing a single electron. In looking at simple inorganic molecules such as H2 or HF, our present understanding of s and p atomic orbitals will suffice. To explain the bonding in organic molecules, however, we will need to introduce the concept of hybrid orbitals.
Example: The H2 molecule
The simplest case to consider is the hydrogen molecule, H2. When we say that the two electrons from each of the hydrogen atoms are shared to form a covalent bond between the two atoms, what we mean in valence bond theory terms is that the two spherical 1s orbitals overlap, allowing the two electrons to form a pair within the two overlapping orbitals. In simple terms, we can say that both electrons now spend more time between the two nuclei and thus hold the atoms together. As we will see, the situation is not quite so simple as that, because the electron pair must still obey quantum mechanics - that is, the two electrons must now occupy a shared orbital space. This will be the essential principle of valence bond theory.
How far apart are the two nuclei? That is a very important issue to consider. If they are too far apart, their respective 1s orbitals cannot overlap, and thus no covalent bond can form - they are still just two separate hydrogen atoms. As they move closer and closer together, orbital overlap begins to occur, and a bond begins to form. This lowers the potential energy of the system, as new, attractive positive-negative electrostatic interactions become possible between the nucleus of one atom and the electron of the second. However, something else is happening at the same time: as the atoms get closer, the repulsive positive-positive interaction between the two nuclei also begins to increase.
At first this repulsion is more than offset by the attraction between nuclei and electrons, but at a certain point, as the nuclei get even closer, the repulsive forces begin to overcome the attractive forces, and the potential energy of the system rises quickly. When the two nuclei are ‘too close’, we have a very unstable, high-energy situation. There is a defined optimal distance between the nuclei in which the potential energy is at a minimum, meaning that the combined attractive and repulsive forces add up to the greatest overall attractive force. This optimal internuclear distance is the bond length. For the H2 molecule, this distance is 74 x 10-12 meters, or 0.74 Å (Å means angstrom, or 10-10 meters). Likewise, the difference in potential energy between the lowest state (at the optimal internuclear distance) and the state where the two atoms are completely separated is called the bond energy. For the hydrogen molecule, this energy is equal to about 104 kcal/mol.
Every covalent bond in a given molecule has a characteristic length and strength. In general, carbon-carbon single bonds are about 1.5 Å long (Å means angstrom, or 10-10 meters) while carbon-carbon double bonds are about 1.3 Å, carbon-oxygen double bonds are about 1.2 Å, and carbon-hydrogen bonds are in the range of 1.0 – 1.1 Å. Most covalent bonds in organic molecules range in strength from just under 100 kcal/mole (for a carbon-hydrogen bond in ethane, for example) up to nearly 200 kcal/mole. You can refer to tables in reference books such as the CRC Handbook of Chemistry and Physics for extensive lists of bond lengths, strengths, and many other data for specific organic compounds.
Balls and Springs
Although we tend to talk about "bond length" as a specific distance, it is not accurate to picture covalent bonds as rigid sticks of unchanging length - rather, it is better to picture them as springs which have a defined length when relaxed, but which can be compressed, extended, and bent. This ‘springy’ picture of covalent bonds will become very important, when we study the analytical technique known as infrared (IR) spectroscopy.
Valence Bond Theory: A Localized Bonding Approach
You learned that as two hydrogen atoms approach each other from an infinite distance, the energy of the system reaches a minimum. This region of minimum energy in the energy diagram corresponds to the formation of a covalent bond between the two atoms at an H–H distance of 74 pm (Figure \(1\)). According to quantum mechanics, bonds form between atoms because their atomic orbitals overlap, with each region of overlap accommodating a maximum of two electrons with opposite spin, in accordance with the Pauli principle. In this case, a bond forms between the two hydrogen atoms when the singly occupied 1s atomic orbital of one hydrogen atom overlaps with the singly occupied 1s atomic orbital of a second hydrogen atom. Electron density between the nuclei is increased because of this orbital overlap and results in a localized electron-pair bond (Figure \(1\)).
Although Lewis and VSEPR structures also contain localized electron-pair bonds, neither description uses an atomic orbital approach to predict the stability of the bond. Doing so forms the basis for a description of chemical bonding known as valence bond theory, which is built on two assumptions:
1. The strength of a covalent bond is proportional to the amount of overlap between atomic orbitals; that is, the greater the overlap, the more stable the bond.
2. An atom can use different combinations of atomic orbitals to maximize the overlap of orbitals used by bonded atoms.
Figure \(2\) shows an electron-pair bond formed by the overlap of two ns atomic orbitals, two np atomic orbitals, and an ns and an np orbital where n = 2. Notice that bonding overlap occurs when the interacting atomic orbitals have the correct orientation (are "pointing at" each other) and are in phase (represented by colors in Figure \(2\) ).
Maximum overlap occurs between orbitals with the same spatial orientation and similar energies.
Let’s examine the bonds in BeH2, for example. According to the VSEPR model, BeH2 is a linear compound with four valence electrons and two Be–H bonds. Its bonding can also be described using an atomic orbital approach. Beryllium has a 1s22s2 electron configuration, and each H atom has a 1s1 electron configuration. Because the Be atom has a filled 2s subshell, however, it has no singly occupied orbitals available to overlap with the singly occupied 1s orbitals on the H atoms. If a singly occupied 1s orbital on hydrogen were to overlap with a filled 2s orbital on beryllium, the resulting bonding orbital would contain three electrons, but the maximum allowed by quantum mechanics is two. How then is beryllium able to bond to two hydrogen atoms? One way would be to add enough energy to excite one of its 2s electrons into an empty 2p orbital and reverse its spin, in a process called promotion:
In this excited state, the Be atom would have two singly occupied atomic orbitals (the 2s and one of the 2p orbitals), each of which could overlap with a singly occupied 1s orbital of an H atom to form an electron-pair bond. Although this would produce \(\ce{BeH2}\), the two Be–H bonds would not be equivalent: the 1s orbital of one hydrogen atom would overlap with a Be 2s orbital, and the 1s orbital of the other hydrogen atom would overlap with an orbital of a different energy, a Be 2p orbital. Experimental evidence indicates, however, that the two Be–H bonds have identical energies. To resolve this discrepancy and explain how molecules such as \(\ce{BeH2}\) form, scientists developed the concept of hybridization. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/09%3A_Molecular_Geometry_and_Bonding_Theories/9.04%3A_Covalent_Bonding_and_Orbital_Overlap.txt |
The localized valence bond theory uses a process called hybridization, in which atomic orbitals that are similar in energy but not equivalent are combined mathematically to produce sets of equivalent orbitals that are properly oriented to form bonds. These new combinations are called hybrid atomic orbitals because they are produced by combining (hybridizing) two or more atomic orbitals from the same atom.
Hybridization of s and p Orbitals
In BeH2, we can generate two equivalent orbitals by combining the 2s orbital of beryllium and any one of the three degenerate 2p orbitals. By taking the sum and the difference of Be 2s and 2pz atomic orbitals, for example, we produce two new orbitals with major and minor lobes oriented along the z-axes, as shown in Figure $1$.
Because the difference A − B can also be written as A + (−B), in Figure $2$ and subsequent figures we have reversed the phase(s) of the orbital being subtracted, which is the same as multiplying it by −1 and adding. This gives us Equation \ref{9.5.1b}, where the value $\frac{1}{\sqrt{2}}$ is needed mathematically to indicate that the 2s and 2p orbitals contribute equally to each hybrid orbital.
$sp = \dfrac{1}{\sqrt{2}} (2s + 2p_z) \label{9.5.1a}$
and
$sp = \dfrac{1}{\sqrt{2}} (2s - 2p_z) \label{9.5.1b}$
The nucleus resides just inside the minor lobe of each orbital. In this case, the new orbitals are called sp hybrids because they are formed from one s and one p orbital. The two new orbitals are equivalent in energy, and their energy is between the energy values associated with pure s and p orbitals, as illustrated in this diagram:
Because both promotion and hybridization require an input of energy, the formation of a set of singly occupied hybrid atomic orbitals is energetically uphill. The overall process of forming a compound with hybrid orbitals will be energetically favorable only if the amount of energy released by the formation of covalent bonds is greater than the amount of energy used to form the hybrid orbitals (Figure $4$). As we will see, some compounds are highly unstable or do not exist because the amount of energy required to form hybrid orbitals is greater than the amount of energy that would be released by the formation of additional bonds.
The concept of hybridization also explains why boron, with a 2s22p1 valence electron configuration, forms three bonds with fluorine to produce BF3, as predicted by the Lewis and VSEPR approaches. With only a single unpaired electron in its ground state, boron should form only a single covalent bond. By the promotion of one of its 2s electrons to an unoccupied 2p orbital, however, followed by the hybridization of the three singly occupied orbitals (the 2s and two 2p orbitals), boron acquires a set of three equivalent hybrid orbitals with one electron each, as shown here:
Looking at the 2s22p2 valence electron configuration of carbon, we might expect carbon to use its two unpaired 2p electrons to form compounds with only two covalent bonds. We know, however, that carbon typically forms compounds with four covalent bonds. We can explain this apparent discrepancy by the hybridization of the 2s orbital and the three 2p orbitals on carbon to give a set of four degenerate sp3 (“s-p-three” or “s-p-cubed”) hybrid orbitals, each with a single electron:
In addition to explaining why some elements form more bonds than would be expected based on their valence electron configurations, and why the bonds formed are equal in energy, valence bond theory explains why these compounds are so stable: the amount of energy released increases with the number of bonds formed. In the case of carbon, for example, much more energy is released in the formation of four bonds than two, so compounds of carbon with four bonds tend to be more stable than those with only two. Carbon does form compounds with only two covalent bonds (such as CH2 or CF2), but these species are highly reactive, unstable intermediates that only form in certain chemical reactions.
Valence bond theory explains the number of bonds formed in a compound and the relative bond strengths.
The bonding in molecules such as NH3 or H2O, which have lone pairs on the central atom, can also be described in terms of hybrid atomic orbitals. In NH3, for example, N, with a 2s22p3 valence electron configuration, can hybridize its 2s and 2p orbitals to produce four sp3 hybrid orbitals. Placing five valence electrons in the four hybrid orbitals, we obtain three that are singly occupied and one with a pair of electrons:
The three singly occupied sp3 lobes can form bonds with three H atoms, while the fourth orbital accommodates the lone pair of electrons. Similarly, H2O has an sp3 hybridized oxygen atom that uses two singly occupied sp3 lobes to bond to two H atoms, and two to accommodate the two lone pairs predicted by the VSEPR model. Such descriptions explain the approximately tetrahedral distribution of electron pairs on the central atom in NH3 and H2O. Unfortunately, however, recent experimental evidence indicates that in NH3 and H2O, the hybridized orbitals are not entirely equivalent in energy, making this bonding model an active area of research.
Example $1$
Use the VSEPR model to predict the number of electron pairs and molecular geometry in each compound and then describe the hybridization and bonding of all atoms except hydrogen.
1. H2S
2. CHCl3
Given: two chemical compounds
Asked for: number of electron pairs and molecular geometry, hybridization, and bonding
Strategy:
1. Using the VSEPR approach to determine the number of electron pairs and the molecular geometry of the molecule.
2. From the valence electron configuration of the central atom, predict the number and type of hybrid orbitals that can be produced. Fill these hybrid orbitals with the total number of valence electrons around the central atom and describe the hybridization.
Solution:
1. A H2S has four electron pairs around the sulfur atom with two bonded atoms, so the VSEPR model predicts a molecular geometry that is bent, or V shaped. B Sulfur has a 3s23p4 valence electron configuration with six electrons, but by hybridizing its 3s and 3p orbitals, it can produce four sp3 hybrids. If the six valence electrons are placed in these orbitals, two have electron pairs and two are singly occupied. The two sp3 hybrid orbitals that are singly occupied are used to form S–H bonds, whereas the other two have lone pairs of electrons. Together, the four sp3 hybrid orbitals produce an approximately tetrahedral arrangement of electron pairs, which agrees with the molecular geometry predicted by the VSEPR model.
2. A The CHCl3 molecule has four valence electrons around the central atom. In the VSEPR model, the carbon atom has four electron pairs, and the molecular geometry is tetrahedral. B Carbon has a 2s22p2 valence electron configuration. By hybridizing its 2s and 2p orbitals, it can form four sp3 hybridized orbitals that are equal in energy. Eight electrons around the central atom (four from C, one from H, and one from each of the three Cl atoms) fill three sp3 hybrid orbitals to form C–Cl bonds, and one forms a C–H bond. Similarly, the Cl atoms, with seven electrons each in their 3s and 3p valence subshells, can be viewed as sp3 hybridized. Each Cl atom uses a singly occupied sp3 hybrid orbital to form a C–Cl bond and three hybrid orbitals to accommodate lone pairs.
Exercise $1$
Use the VSEPR model to predict the number of electron pairs and molecular geometry in each compound and then describe the hybridization and bonding of all atoms except hydrogen.
1. the BF4 ion
2. hydrazine (H2N–NH2)
Answer a
B is sp3 hybridized; F is also sp3 hybridized so it can accommodate one B–F bond and three lone pairs. The molecular geometry is tetrahedral.
Answer b
Each N atom is sp3 hybridized and uses one sp3 hybrid orbital to form the N–N bond, two to form N–H bonds, and one to accommodate a lone pair. The molecular geometry about each N is trigonal pyramidal.
The number of hybrid orbitals used by the central atom is the same as the number of electron pairs around the central atom.
Hybridization Using d Orbitals
Hybridization is not restricted to the ns and np atomic orbitals. The bonding in compounds with central atoms in the period 3 and below can also be described using hybrid atomic orbitals. In these cases, the central atom can use its valence (n − 1)d orbitals as well as its ns and np orbitals to form hybrid atomic orbitals, which allows it to accommodate five or more bonded atoms (as in PF5 and SF6). Using the ns orbital, all three np orbitals, and one (n − 1)d orbital gives a set of five sp3d hybrid orbitals that point toward the vertices of a trigonal bipyramid (part (a) in Figure $7$). In this case, the five hybrid orbitals are not all equivalent: three form a triangular array oriented at 120° angles, and the other two are oriented at 90° to the first three and at 180° to each other.
Similarly, the combination of the ns orbital, all three np orbitals, and two nd orbitals gives a set of six equivalent sp3d2 hybrid orbitals oriented toward the vertices of an octahedron (part (b) in Figure 9.5.6). In the VSEPR model, PF5 and SF6 are predicted to be trigonal bipyramidal and octahedral, respectively, which agrees with a valence bond description in which sp3d or sp3d2 hybrid orbitals are used for bonding.
Example $2$
What is the hybridization of the central atom in each species? Describe the bonding in each species.
1. XeF4
2. SO42
3. SF4
Given: three chemical species
Asked for: hybridization of the central atom
Strategy:
1. Determine the geometry of the molecule using the strategy in Example $1$. From the valence electron configuration of the central atom and the number of electron pairs, determine the hybridization.
2. Place the total number of electrons around the central atom in the hybrid orbitals and describe the bonding.
Solution:
1. A Using the VSEPR model, we find that Xe in XeF4 forms four bonds and has two lone pairs, so its structure is square planar and it has six electron pairs. The six electron pairs form an octahedral arrangement, so the Xe must be sp3d2 hybridized. B With 12 electrons around Xe, four of the six sp3d2 hybrid orbitals form Xe–F bonds, and two are occupied by lone pairs of electrons.
2. A The S in the SO42 ion has four electron pairs and has four bonded atoms, so the structure is tetrahedral. The sulfur must be sp3 hybridized to generate four S–O bonds. B Filling the sp3 hybrid orbitals with eight electrons from four bonds produces four filled sp3 hybrid orbitals.
3. A The S atom in SF4 contains five electron pairs and four bonded atoms. The molecule has a seesaw structure with one lone pair:
To accommodate five electron pairs, the sulfur atom must be sp3d hybridized. B Filling these orbitals with 10 electrons gives four sp3d hybrid orbitals forming S–F bonds and one with a lone pair of electrons.
Exercise $2$
What is the hybridization of the central atom in each species? Describe the bonding.
1. PCl4+
2. BrF3
3. SiF62
Answer a
sp3 with four P–Cl bonds
Answer a
sp3d with three Br–F bonds and two lone pairs
Answer a
sp3d2 with six Si–F bonds
Hybridization using d orbitals allows chemists to explain the structures and properties of many molecules and ions. Like most such models, however, it is not universally accepted. Nonetheless, it does explain a fundamental difference between the chemistry of the elements in the period 2 (C, N, and O) and those in period 3 and below (such as Si, P, and S).
Period 2 elements do not form compounds in which the central atom is covalently bonded to five or more atoms, although such compounds are common for the heavier elements. Thus whereas carbon and silicon both form tetrafluorides (CF4 and SiF4), only SiF4 reacts with F to give a stable hexafluoro dianion, SiF62. Because there are no 2d atomic orbitals, the formation of octahedral CF62 would require hybrid orbitals created from 2s, 2p, and 3d atomic orbitals. The 3d orbitals of carbon are so high in energy that the amount of energy needed to form a set of sp3d2 hybrid orbitals cannot be equaled by the energy released in the formation of two additional C–F bonds. These additional bonds are expected to be weak because the carbon atom (and other atoms in period 2) is so small that it cannot accommodate five or six F atoms at normal C–F bond lengths due to repulsions between electrons on adjacent fluorine atoms. Perhaps not surprisingly, then, species such as CF62 have never been prepared.
Example $3$: $\ce{OF4}$
What is the hybridization of the oxygen atom in OF4? Is OF4 likely to exist?
Given: chemical compound
Asked for: hybridization and stability
Strategy:
1. Predict the geometry of OF4 using the VSEPR model.
2. From the number of electron pairs around O in OF4, predict the hybridization of O. Compare the number of hybrid orbitals with the number of electron pairs to decide whether the molecule is likely to exist.
Solution:
A The VSEPR model predicts that OF4 will have five electron pairs, resulting in a trigonal bipyramidal geometry with four bonding pairs and one lone pair. B To accommodate five electron pairs, the O atom would have to be sp3d hybridized. The only d orbital available for forming a set of sp3d hybrid orbitals is a 3d orbital, which is much higher in energy than the 2s and 2p valence orbitals of oxygen. As a result, the OF4 molecule is unlikely to exist. In fact, it has not been detected.
Exercise $3$
What is the hybridization of the boron atom in $BF_6^{3−}$? Is this ion likely to exist?
Answer a
sp3d2 hybridization; no
Summary
Hybridization increases the overlap of bonding orbitals and explains the molecular geometries of many species whose geometry cannot be explained using a VSEPR approach. The localized bonding model (called valence bond theory) assumes that covalent bonds are formed when atomic orbitals overlap and that the strength of a covalent bond is proportional to the amount of overlap. It also assumes that atoms use combinations of atomic orbitals (hybrids) to maximize the overlap with adjacent atoms. The formation of hybrid atomic orbitals can be viewed as occurring via promotion of an electron from a filled ns2 subshell to an empty np or (n − 1)d valence orbital, followed by hybridization, the combination of the orbitals to give a new set of (usually) equivalent orbitals that are oriented properly to form bonds. The combination of an ns and an np orbital gives rise to two equivalent sp hybrids oriented at 180°, whereas the combination of an ns and two or three np orbitals produces three equivalent sp2 hybrids or four equivalent sp3 hybrids, respectively. The bonding in molecules with more than an octet of electrons around a central atom can be explained by invoking the participation of one or two (n − 1)d orbitals to give sets of five sp3d or six sp3d2 hybrid orbitals, capable of forming five or six bonds, respectively. The spatial orientation of the hybrid atomic orbitals is consistent with the geometries predicted using the VSEPR model. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/09%3A_Molecular_Geometry_and_Bonding_Theories/9.05%3A_Hybrid_Orbitals.txt |
Learning Objectives
• To explain resonance structures using molecular orbitals.
So far in our valence bond orbital descriptions we have not dealt with polyatomic systems with multiple bonds. To do so, we can use an approach in which we describe $\sigma$ bonding using localized electron-pair bonds formed by hybrid atomic orbitals, and $\pi$ bonding using molecular orbitals formed by unhybridized np atomic orbitals.
Multiple Bonding
We begin our discussion by considering the bonding in ethylene (C2H4). Experimentally, we know that the H–C–H and H–C–C angles in ethylene are approximately 120°. This angle suggests that the carbon atoms are sp2 hybridized, which means that a singly occupied sp2 orbital on one carbon overlaps with a singly occupied s orbital on each H and a singly occupied sp2 lobe on the other C. The sp2 hybridization can be represented as follows:
When considering multiple bonds, we can introduce the more sophisticated molecular orbital theory to better understand how orbital overlap creates bonding orbitals. Recall that atomic orbitals represent electron wavefunctions; this implies that "orbital overlap" actually involves the combination of those wavefunctions, which can occur via both constructive and destructive interference. Thus, when the two singly occupied 2pz orbitals in ethylene overlap, they actually create both a $\pi$ bonding orbital (constructive combination) and a $\pi$* antibonding orbital (destructive combination), which produces the energy-level diagram shown in Figure $2$. With the formation of a $\pi$ bonding orbital, electron density increases in the plane between the carbon nuclei. Electrons occupying this orbital lower the potential energy of the combination and tend to hold the two nuclei together (i.e., they form a bond.) The $\pi$* orbital lies outside the internuclear region and has a nodal plane perpendicular to the internuclear axis; electrons in this orbital would tend to push the nuclei apart, so it is called an antibonding orbital. Because each 2pz orbital has a single electron, there are only two electrons, enough to fill only the bonding ($\pi$) level, leaving the $\pi$* orbital empty. Consequently, the C–C bond in ethylene consists of a $\sigma$ bond and a $\pi$ bond, which together give a C=C double bond. Our model is supported by the facts that the measured carbon–carbon bond is shorter than that in ethane (133.9 pm versus 153.5 pm) and the bond is stronger (728 kJ/mol versus 376 kJ/mol in ethane). The two CH2 fragments are coplanar, which maximizes the overlap of the two singly occupied 2pz orbitals.
Triple bonds, as in acetylene (C2H2), can also be explained using a combination of hybrid atomic orbitals and molecular orbitals. The four atoms of acetylene are collinear, which suggests that each carbon is sp hybridized. If one sp lobe on each carbon atom is used to form a C–C $\sigma$ bond and one is used to form the C–H $\sigma$ bond, then each carbon will still have two unhybridized 2p orbitals (a 2px,y pair), each with one electron (part (a) in Figure $3$).
In complex molecules, hybrid orbitals and valence bond theory can be used to describe $\sigma$ bonding, and unhybridized $\pi$ orbitals and molecular orbital theory can be used to describe $\pi$ bonding.
Example $1$
Describe the bonding in HCN using a combination of hybrid atomic orbitals and molecular orbitals. The HCN molecule is linear.
Given: chemical compound and molecular geometry
Asked for: bonding description using hybrid atomic orbitals and molecular orbitals
Strategy:
1. From the geometry given, predict the hybridization in HCN. Use the hybrid orbitals to form the $\sigma$-bonded framework of the molecule and determine the number of valence electrons that are used for $\sigma$ bonding.
2. Determine the number of remaining valence electrons. Use any remaining unhybridized p orbitals to form $\pi$ and $\pi$* orbitals.
3. Fill the orbitals with the remaining electrons in order of increasing energy. Describe the bonding in HCN.
Solution:
A Because HCN is a linear molecule, it is likely that the bonding can be described in terms of sp hybridization at carbon. Because the nitrogen atom can also be described as sp hybridized, we can use one sp hybrid on each atom to form a C–N $\sigma$ bond. This leaves one sp hybrid on each atom to either bond to hydrogen (C) or hold a lone pair of electrons (N). Of 10 valence electrons (5 from N, 4 from C, and 1 from H), 4 are used for $\sigma$ bonding:
B We are now left with 2 electrons on N (5 valence electrons minus 1 bonding electron minus 2 electrons in the lone pair) and 2 electrons on C (4 valence electrons minus 2 bonding electrons). We have two unhybridized 2p atomic orbitals left on carbon and two on nitrogen, each occupied by a single electron. These four 2p atomic orbitals can be combined to give four molecular orbitals: two $\pi$ (bonding) orbitals and two $\pi$* (antibonding) orbitals. C With 4 electrons available, only the $\pi$ orbitals are filled. The overall result is a triple bond (1 $\sigma$ and 2 $\pi$) between C and N.
Exercise $1$
Describe the bonding in formaldehyde (H2C=O), a trigonal planar molecule, using a combination of hybrid atomic orbitals and molecular orbitals.
Answer
• $\sigma$-bonding framework: Carbon and oxygen are sp2 hybridized. Two sp2 hybrid orbitals on oxygen have lone pairs, two sp2 hybrid orbitals on carbon form C–H bonds, and one sp2 hybrid orbital on C and O forms a C–O $\sigma$ bond.
• $\pi$ bonding: Unhybridized, singly occupied 2p atomic orbitals on carbon and oxygen interact to form $\pi$ (bonding) and $\pi$* (antibonding) molecular orbitals. With two electrons, only the $\pi$ (bonding) orbital is occupied.
Molecular Orbitals and Resonance Structures
Resonance structures can be used to describe the bonding in molecules such as ozone (O3) and the nitrite ion (NO2). Ozone can be represented by either of these Lewis electron structures:
Although the VSEPR model correctly predicts that both species are bent, it gives no information about their bond orders.
Experimental evidence indicates that ozone has a bond angle of 117.5°. Because this angle is close to 120°, it is likely that the central oxygen atom in ozone is trigonal planar and sp2 hybridized. If we assume that the terminal oxygen atoms are also sp2 hybridized, then we obtain the $\sigma$-bonded framework shown in Figure $4$. Two of the three sp2 lobes on the central O are used to form O–O sigma bonds, and the third has a lone pair of electrons. Each terminal oxygen atom has two lone pairs of electrons that are also in sp2 lobes. In addition, each oxygen atom has one unhybridized 2p orbital perpendicular to the molecular plane. The $\sigma$ bonds and lone pairs account for a total of 14 electrons (five lone pairs and two $\sigma$ bonds, each containing 2 electrons). Each oxygen atom in ozone has 6 valence electrons, so O3 has a total of 18 valence electrons. Subtracting 14 electrons from the total gives us 4 electrons that must occupy the three unhybridized 2p orbitals.
With a molecular orbital approach to describe the $\pi$ bonding, three 2p atomic orbitals give us three molecular orbitals, as shown in Figure $5$. One of the molecular orbitals is a $\pi$ bonding molecular orbital, which is shown as a banana-shaped region of electron density above and below the molecular plane. This region has no nodes perpendicular to the O3 plane. The molecular orbital with the highest energy has two nodes that bisect the O–O $\sigma$ bonds; it is a $\pi$* antibonding orbital. The third molecular orbital contains a single node that is perpendicular to the O3 plane and passes through the central O atom; because the orbital nodes do not directly touch, this is a nonbonding molecular orbital. Because electrons in nonbonding orbitals are neither bonding nor antibonding, they are ignored in calculating bond orders.
We can now place the remaining four electrons in the three energy levels shown in Figure $5$, thereby filling the $\pi$ bonding and the nonbonding levels. The result is a single $\pi$ bond holding three oxygen atoms together, or $½ \pi$ bond per O–O. We therefore predict the overall O–O bond order to be $½ \pi$ bond plus 1 $\sigma$ bond), just as predicted using resonance structures. The molecular orbital description, however, makes it more clear that resonance really means that electrons are delocalized over all three atoms at once. The molecular orbital approach also shows that the $\pi$ nonbonding orbital is localized on the terminal O atoms, which suggests that they are more electron rich than the central O atom (corresponding to the "extra" lone pair seen on one of the terminal O atoms in the Lewis resonance structures). The reactivity of ozone is consistent with the predicted charge localization.
Resonance structures are a crude way of describing molecular orbitals that extend over more than two atoms.
Example $2$
Describe the bonding in the nitrite ion in terms of a combination of hybrid atomic orbitals and molecular orbitals. Lewis dot structures and the VSEPR model predict that the NO2 ion is bent.
Given: chemical species and molecular geometry
Asked for: bonding description using hybrid atomic orbitals and molecular orbitals
Strategy:
1. Calculate the number of valence electrons in NO2. From the structure, predict the type of atomic orbital hybridization in the ion.
2. Predict the number and type of molecular orbitals that form during bonding. Use valence electrons to fill these orbitals and then calculate the number of electrons that remain.
3. If there are unhybridized orbitals, place the remaining electrons in these orbitals in order of increasing energy. Calculate the bond order and describe the bonding.
Solution:
A The lone pair of electrons on nitrogen and a bent structure suggest that the bonding in NO2 is similar to the bonding in ozone. This conclusion is supported by the fact that nitrite also contains 18 valence electrons (5 from N and 6 from each O, plus 1 for the −1 charge). The bent structure implies that the nitrogen is sp2 hybridized.
B If we assume that the oxygen atoms are sp2 hybridized as well, then we can use two sp2 hybrid orbitals on each oxygen and one sp2 hybrid orbital on nitrogen to accommodate the five lone pairs of electrons. Two sp2 hybrid orbitals on nitrogen form $\sigma$ bonds with the remaining sp2 hybrid orbital on each oxygen. The $\sigma$ bonds and lone pairs account for 14 electrons. We are left with three unhybridized 2p orbitals, one on each atom, perpendicular to the plane of the molecule, and 4 electrons. Just as with ozone, these three 2p orbitals interact to form bonding, nonbonding, and antibonding $\pi$ molecular orbitals. The bonding molecular orbital is spread over the nitrogen and both oxygen atoms.
C Placing 4 electrons in the energy-level diagram fills both the bonding and nonbonding molecular orbitals and gives a $\pi$ bond order of 1/2 per N–O bond. The overall N–O bond order is $1\;\frac{1}{2}$, consistent with a resonance structure.
Exercise$2$
Describe the bonding in the formate ion (HCO2), in terms of a combination of hybrid atomic orbitals and molecular orbitals.
Answer
Like nitrite, formate is a planar polyatomic ion with 18 valence electrons. The $\sigma$ bonding framework can be described in terms of sp2 hybridized carbon and oxygen, which account for 14 electrons. The three unhybridized 2p orbitals (on C and both O atoms) form three $\pi$ molecular orbitals, and the remaining 4 electrons occupy both the bonding and nonbonding $\pi$ molecular orbitals. The overall C–O bond order is therefore $frac{3}{2}$
The Chemistry of Vision
Hydrocarbons in which two or more carbon–carbon double bonds are directly linked by carbon–carbon single bonds (called conjugated structures) are generally more stable than expected because of resonance. Because the double bonds are close enough to interact electronically with one another, the $\pi$ electrons are shared over all the carbon atoms, as illustrated for 1,3-butadiene in Figure $6$. As the number of interacting atomic orbitals increases, the number of molecular orbitals increases, the energy spacing between molecular orbitals decreases, and the systems become more stable (Figure $7$). Thus as a chain of alternating double and single bonds becomes longer, the energy required to excite an electron from the highest-energy occupied (bonding) orbital to the lowest-energy unoccupied (antibonding) orbital decreases. If the chain is long enough, the amount of energy required to excite an electron corresponds to the energy of visible light. For example, vitamin A is yellow because its chain of five alternating double bonds is able to absorb violet light. Many of the colors we associate with dyes result from this same phenomenon; most dyes are organic compounds with alternating double bonds.
As the number of atomic orbitals increases, the difference in energy between the resulting molecular orbital energy levels decreases, which allows light of lower energy to be absorbed. As a result, organic compounds with long chains of carbon atoms and alternating single and double bonds tend to become more deeply colored as the number of double bonds increases.
As the number of interacting atomic orbitals increases, the energy separation between the resulting molecular orbitals steadily decreases.
A derivative of vitamin A called retinal is used by the human eye to detect light and has a structure with alternating C=C double bonds. When visible light strikes retinal, the energy separation between the molecular orbitals is sufficiently close that the energy absorbed corresponds to the energy required to change one double bond in the molecule from cis, where like groups are on the same side of the double bond, to trans, where they are on opposite sides, initiating a process that causes a signal to be sent to the brain. If this mechanism is defective, we lose our vision in dim light. Once again, a molecular orbital approach to bonding explains a process that cannot be explained using any of the other approaches we have described.
Summary
Polyatomic systems with multiple bonds can be described using hybrid atomic orbitals for $\sigma$ bonding and molecular orbitals to describe $\pi$ bonding. To describe the bonding in more complex molecules with multiple bonds, we can use an approach that uses hybrid atomic orbitals to describe the $\sigma$ bonding and molecular orbitals to describe the $\pi$ bonding. In this approach, unhybridized np orbitals on atoms bonded to one another are allowed to interact to produce bonding, antibonding, or nonbonding combinations. For $\pi$ bonds between two atoms (as in ethylene or acetylene), the resulting molecular orbitals are virtually identical to the $\pi$ molecular orbitals in diatomic molecules such as O2 and N2. Applying the same approach to $\pi$ bonding between three or four atoms requires combining three or four unhybridized np orbitals on adjacent atoms to generate $\pi$ bonding, antibonding, and nonbonding molecular orbitals extending over all of the atoms. Filling the resulting energy-level diagram with the appropriate number of electrons explains the bonding in molecules or ions that previously required the use of resonance structures in the Lewis electron-pair approach. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/09%3A_Molecular_Geometry_and_Bonding_Theories/9.06%3A_Multiple_Bonds.txt |
Learning Objectives
• To use molecular orbital theory to predict bond order
None of the approaches we have described so far can adequately explain why some compounds are colored and others are not, why some substances with unpaired electrons are stable, and why others are effective semiconductors. These approaches also cannot describe the nature of resonance. Such limitations led to the development of a new approach to bonding in which electrons are not viewed as being localized between the nuclei of bonded atoms but are instead delocalized throughout the entire molecule. Just as with the valence bond theory, the approach we are about to discuss is based on a quantum mechanical model.
Previously, we described the electrons in isolated atoms as having certain spatial distributions, called orbitals, each with a particular orbital energy. Just as the positions and energies of electrons in atoms can be described in terms of atomic orbitals (AOs), the positions and energies of electrons in molecules can be described in terms of molecular orbitals (MOs) A particular spatial distribution of electrons in a molecule that is associated with a particular orbital energy.—a spatial distribution of electrons in a molecule that is associated with a particular orbital energy. As the name suggests, molecular orbitals are not localized on a single atom but extend over the entire molecule. Consequently, the molecular orbital approach, called molecular orbital theory is a delocalized approach to bonding.
Molecular Orbital Theory: A Delocalized Bonding Approach
Although the molecular orbital theory is computationally demanding, the principles on which it is based are similar to those we used to determine electron configurations for atoms. The key difference is that in molecular orbitals, the electrons are allowed to interact with more than one atomic nucleus at a time. Just as with atomic orbitals, we create an energy-level diagram by listing the molecular orbitals in order of increasing energy. We then fill the orbitals with the required number of valence electrons according to the Pauli principle. This means that each molecular orbital can accommodate a maximum of two electrons with opposite spins.
Molecular Orbitals Involving Only ns Atomic Orbitals
We begin our discussion of molecular orbitals with the simplest molecule, H2, formed from two isolated hydrogen atoms, each with a 1s1 electron configuration. As we explained in Chapter 9, electrons can behave like waves. In the molecular orbital approach, the overlapping atomic orbitals are described by mathematical equations called wave functions. The 1s atomic orbitals on the two hydrogen atoms interact to form two new molecular orbitals, one produced by taking the sum of the two H 1s wave functions, and the other produced by taking their difference:
$\begin{matrix} MO(1)= & AO(atom\; A) & +& AO(atomB) \ MO(1)= & AO(atom\; A) & -&AO(atomB) \end{matrix} \label{9.7.1}$
The molecular orbitals created from Equation $\ref{9.7.1}$ are called linear combinations of atomic orbitals (LCAOs) Molecular orbitals created from the sum and the difference of two wave functions (atomic orbitals). A molecule must have as many molecular orbitals as there are atomic orbitals.
A molecule must have as many molecular orbitals as there are atomic orbitals.
Adding two atomic orbitals corresponds to constructive interference between two waves, thus reinforcing their intensity; the internuclear electron probability density is increased. The molecular orbital corresponding to the sum of the two H 1s orbitals is called a σ1s combination (pronounced “sigma one ess”) (part (a) and part (b) in Figure $1$). In a sigma (σ) orbital, (i.e., a bonding molecular orbital in which the electron density along the internuclear axis and between the nuclei has cylindrical symmetry), the electron density along the internuclear axis and between the nuclei has cylindrical symmetry; that is, all cross-sections perpendicular to the internuclear axis are circles. The subscript 1s denotes the atomic orbitals from which the molecular orbital was derived: The ≈ sign is used rather than an = sign because we are ignoring certain constants that are not important to our argument.
$\sigma _{1s} \approx 1s\left ( A \right ) + 1s\left ( B \right ) \label{9.7.2}$
Conversely, subtracting one atomic orbital from another corresponds to destructive interference between two waves, which reduces their intensity and causes a decrease in the internuclear electron probability density (part (c) and part (d) in Figure $1$). The resulting pattern contains a node where the electron density is zero. The molecular orbital corresponding to the difference is called $\sigma _{1s}^{*}$ (“sigma one ess star”). In a sigma star (σ*) orbital An antibonding molecular orbital in which there is a region of zero electron probability (a nodal plane) perpendicular to the internuclear axis., there is a region of zero electron probability, a nodal plane, perpendicular to the internuclear axis:
$\sigma _{1s}^{\star } \approx 1s\left ( A \right ) - 1s\left ( B \right ) \label{9.7.3}$
The electron density in the σ1s molecular orbital is greatest between the two positively charged nuclei, and the resulting electron–nucleus electrostatic attractions reduce repulsions between the nuclei. Thus the σ1s orbital represents a bonding molecular orbital. A molecular orbital that forms when atomic orbitals or orbital lobes with the same sign interact to give increased electron probability between the nuclei due to constructive reinforcement of the wave functions. In contrast, electrons in the $\sigma _{1s}^{\star }$ orbital are generally found in the space outside the internuclear region. Because this allows the positively charged nuclei to repel one another, the $\sigma _{1s}^{\star }$ orbital is an antibonding molecular orbital (a molecular orbital that forms when atomic orbitals or orbital lobes of opposite sign interact to give decreased electron probability between the nuclei due to destructive reinforcement of the wave functions).
Antibonding orbitals contain a node perpendicular to the internuclear axis; bonding orbitals do not.
Energy-Level Diagrams
Because electrons in the σ1s orbital interact simultaneously with both nuclei, they have a lower energy than electrons that interact with only one nucleus. This means that the σ1s molecular orbital has a lower energy than either of the hydrogen 1s atomic orbitals. Conversely, electrons in the $\sigma _{1s}^{\star }$ orbital interact with only one hydrogen nucleus at a time. In addition, they are farther away from the nucleus than they were in the parent hydrogen 1s atomic orbitals. Consequently, the $\sigma _{1s}^{\star }$ molecular orbital has a higher energy than either of the hydrogen 1s atomic orbitals. The σ1s (bonding) molecular orbital is stabilized relative to the 1s atomic orbitals, and the $\sigma _{1s}^{\star }$ (antibonding) molecular orbital is destabilized. The relative energy levels of these orbitals are shown in the energy-level diagram (a schematic drawing that compares the energies of the molecular orbitals (bonding, antibonding, and nonbonding) with the energies of the parent atomic orbitals) in Figure $2$
A bonding molecular orbital is always lower in energy (more stable) than the component atomic orbitals, whereas an antibonding molecular orbital is always higher in energy (less stable).
To describe the bonding in a homonuclear diatomic molecule (a molecule that consists of two atoms of the same element) such as H2, we use molecular orbitals; that is, for a molecule in which two identical atoms interact, we insert the total number of valence electrons into the energy-level diagram (Figure $2$). We fill the orbitals according to the Pauli principle and Hund’s rule: each orbital can accommodate a maximum of two electrons with opposite spins, and the orbitals are filled in order of increasing energy. Because each H atom contributes one valence electron, the resulting two electrons are exactly enough to fill the σ1s bonding molecular orbital. The two electrons enter an orbital whose energy is lower than that of the parent atomic orbitals, so the H2 molecule is more stable than the two isolated hydrogen atoms. Thus molecular orbital theory correctly predicts that H2 is a stable molecule. Because bonds form when electrons are concentrated in the space between nuclei, this approach is also consistent with our earlier discussion of electron-pair bonds.
Bond Order in Molecular Orbital Theory
In the Lewis electron structures, the number of electron pairs holding two atoms together was called the bond order. In the molecular orbital approach, bond order One-half the net number of bonding electrons in a molecule. is defined as one-half the net number of bonding electrons:
$bond\; order=\dfrac{number\; of \; bonding\; electrons-number\; of \; antibonding\; electrons}{2} \label{9.7.4}$
To calculate the bond order of H2, we see from Figure $2$ that the σ1s (bonding) molecular orbital contains two electrons, while the $\sigma _{1s}^{\star }$ (antibonding) molecular orbital is empty. The bond order of H2 is therefore
$\dfrac{2-0}{2}=1 \label{9.7.5}$
This result corresponds to the single covalent bond predicted by Lewis dot symbols. Thus molecular orbital theory and the Lewis electron-pair approach agree that a single bond containing two electrons has a bond order of 1. Double and triple bonds contain four or six electrons, respectively, and correspond to bond orders of 2 and 3. We can use energy-level diagrams such as the one in Figure $2$ to describe the bonding in other pairs of atoms and ions where n = 1, such as the H2+ ion, the He2+ ion, and the He2 molecule. Again, we fill the lowest-energy molecular orbitals first while being sure not to violate the Pauli principle or Hund’s rule.
Figure $\PageIndex{3a}$ shows the energy-level diagram for the H2+ ion, which contains two protons and only one electron. The single electron occupies the σ1s bonding molecular orbital, giving a (σ1s)1 electron configuration. The number of electrons in an orbital is indicated by a superscript. In this case, the bond order is
$\dfrac{1-0}{2}=1/2 \nonumber$
Because the bond order is greater than zero, the H2+ ion should be more stable than an isolated H atom and a proton. We can therefore use a molecular orbital energy-level diagram and the calculated bond order to predict the relative stability of species such as H2+. With a bond order of only 1/2 the bond in H2+ should be weaker than in the H2 molecule, and the H–H bond should be longer. As shown in Table $1$, these predictions agree with the experimental data.
Figure $\PageIndex{3b}$ is the molecular orbital energy-level diagram for He2+. This ion has a total of three valence electrons. Because the first two electrons completely fill the σ1s molecular orbital, the Pauli principle states that the third electron must be in the $\sigma _{1s}^{\star}$ antibonding orbital, giving a $\left (\sigma _{1s} \right )^{2}\left (\sigma _{1s}^{\star } \right )^{1}$ electron configuration. This electron configuration gives a bond order of
$\dfrac{2-1}{2}=1/2 \nonumber$
As with H2+, the He2+ ion should be stable, but the He–He bond should be weaker and longer than in H2. In fact, the He2+ ion can be prepared, and its properties are consistent with our predictions (Table $1$).
Table $1$: Molecular Orbital Electron Configurations, Bond Orders, Bond Lengths, and Bond Energies for some Simple Homonuclear Diatomic Molecules and Ions
Molecule or Ion Electron Configuration Bond Order Bond Length (pm) Bond Energy (kJ/mol)
H2+ 1s)1 1/2 106 269
H2 1s)2 1 74 436
He2+ $\left (\sigma _{1s} \right )^{2}\left (\sigma _{1s}^{\star } \right )^{1}$ 1/2 108 251
He2 $\left (\sigma _{1s} \right )^{2}\left (\sigma _{1s}^{\star } \right )^{2}$ 0 not observed not observed
Finally, we examine the He2 molecule, formed from two He atoms with 1s2 electron configurations. Figure $\PageIndex{3c}$ is the molecular orbital energy-level diagram for He2. With a total of four valence electrons, both the σ1s bonding and $\sigma _{1s}^{\star }$ antibonding orbitals must contain two electrons. This gives a $\left (\sigma _{1s} \right )^{2}\left (\sigma _{1s}^{\star } \right )^{2}$ electron configuration, with a predicted bond order of (2 − 2) ÷ 2 = 0, which indicates that the He2 molecule has no net bond and is not a stable species. Experiments show that the He2 molecule is actually less stable than two isolated He atoms due to unfavorable electron–electron and nucleus–nucleus interactions.
In molecular orbital theory, electrons in antibonding orbitals effectively cancel the stabilization resulting from electrons in bonding orbitals. Consequently, any system that has equal numbers of bonding and antibonding electrons will have a bond order of 0, and it is predicted to be unstable and therefore not to exist in nature. In contrast to Lewis electron structures and the valence bond approach, molecular orbital theory is able to accommodate systems with an odd number of electrons, such as the H2+ ion.
In contrast to Lewis electron structures and the valence bond approach, molecular orbital theory can accommodate systems with an odd number of electrons.
Example $1$
Use a molecular orbital energy-level diagram, such as those in Figure $2$, to predict the bond order in the He22+ ion. Is this a stable species?
Given: chemical species
Asked for: molecular orbital energy-level diagram, bond order, and stability
Strategy:
1. Combine the two He valence atomic orbitals to produce bonding and antibonding molecular orbital
2. s. Draw the molecular orbital energy-level diagram for the system.
3. Determine the total number of valence electrons in the He22+ ion. Fill the molecular orbitals in the energy-level diagram beginning with the orbital with the lowest energy. Be sure to obey the Pauli principle and Hund’s rule while doing so.
4. Calculate the bond order and predict whether the species is stable.
Solution:
A Two He 1s atomic orbitals combine to give two molecular orbitals: a σ1s bonding orbital at lower energy than the atomic orbitals and a $\sigma _{1s}^{\star }$ antibonding orbital at higher energy. The bonding in any diatomic molecule with two He atoms can be described using the following molecular orbital diagram:
B The He22+ ion has only two valence electrons (two from each He atom minus two for the +2 charge). We can also view He22+ as being formed from two He+ ions, each of which has a single valence electron in the 1s atomic orbital. We can now fill the molecular orbital diagram:
The two electrons occupy the lowest-energy molecular orbital, which is the bonding (σ1s) orbital, giving a (σ1s)2 electron configuration. To avoid violating the Pauli principle, the electron spins must be paired. C So the bond order is
$\dfrac{2-0}{2} =1 \nonumber$
He22+ is therefore predicted to contain a single He–He bond. Thus it should be a stable species.
Exercise $1$
Use a molecular orbital energy-level diagram to predict the valence-electron configuration and bond order of the H22 ion. Is this a stable species?
Answer
H22 has a valence electron configuration of $\left (\sigma _{1s} \right )^{2}\left (\sigma _{1s}^{\star } \right )^{2}$ with a bond order of 0. It is therefore predicted to be unstable.
So far, our discussion of molecular orbitals has been confined to the interaction of valence orbitals, which tend to lie farthest from the nucleus. When two atoms are close enough for their valence orbitals to overlap significantly, the filled inner electron shells are largely unperturbed; hence they do not need to be considered in a molecular orbital scheme. Also, when the inner orbitals are completely filled, they contain exactly enough electrons to completely fill both the bonding and antibonding molecular orbitals that arise from their interaction. Thus the interaction of filled shells always gives a bond order of 0, so filled shells are not a factor when predicting the stability of a species. This means that we can focus our attention on the molecular orbitals derived from valence atomic orbitals.
A molecular orbital diagram that can be applied to any homonuclear diatomic molecule with two identical alkali metal atoms (Li2 and Cs2, for example) is shown in part (a) in Figure $4$, where M represents the metal atom. Only two energy levels are important for describing the valence electron molecular orbitals of these species: a σns bonding molecular orbital and a σ*ns antibonding molecular orbital. Because each alkali metal (M) has an ns1 valence electron configuration, the M2 molecule has two valence electrons that fill the σns bonding orbital. As a result, a bond order of 1 is predicted for all homonuclear diatomic species formed from the alkali metals (Li2, Na2, K2, Rb2, and Cs2). The general features of these M2 diagrams are identical to the diagram for the H2 molecule in Figure $4$. Experimentally, all are found to be stable in the gas phase, and some are even stable in solution.
Similarly, the molecular orbital diagrams for homonuclear diatomic compounds of the alkaline earth metals (such as Be2), in which each metal atom has an ns2 valence electron configuration, resemble the diagram for the He2 molecule in part (c) in Figure $2$. As shown in part (b) in Figure $4$, this is indeed the case. All the homonuclear alkaline earth diatomic molecules have four valence electrons, which fill both the σns bonding orbital and the σns* antibonding orbital and give a bond order of 0. Thus Be2, Mg2, Ca2, Sr2, and Ba2 are all expected to be unstable, in agreement with experimental data.In the solid state, however, all the alkali metals and the alkaline earth metals exist as extended lattices held together by metallic bonding. At low temperatures, $Be_2$ is stable.
Example $2$
Use a qualitative molecular orbital energy-level diagram to predict the valence electron configuration, bond order, and likely existence of the Na2 ion.
Given: chemical species
Asked for: molecular orbital energy-level diagram, valence electron configuration, bond order, and stability
Strategy:
1. Combine the two sodium valence atomic orbitals to produce bonding and antibonding molecular orbitals. Draw the molecular orbital energy-level diagram for this system.
2. Determine the total number of valence electrons in the Na2 ion. Fill the molecular orbitals in the energy-level diagram beginning with the orbital with the lowest energy. Be sure to obey the Pauli principle and Hund’s rule while doing so.
3. Calculate the bond order and predict whether the species is stable.
Solution:
A Because sodium has a [Ne]3s1 electron configuration, the molecular orbital energy-level diagram is qualitatively identical to the diagram for the interaction of two 1s atomic orbitals. B The Na2 ion has a total of three valence electrons (one from each Na atom and one for the negative charge), resulting in a filled σ3s molecular orbital, a half-filled σ3s* and a $\left ( \sigma _{3s} \right )^{2}\left ( \sigma _{3s}^{\star } \right )^{1}$ electron configuration.
C The bond order is (2-1)÷2=1/2 With a fractional bond order, we predict that the Na2 ion exists but is highly reactive.
Exercise $2$
Use a qualitative molecular orbital energy-level diagram to predict the valence electron configuration, bond order, and likely existence of the Ca2+ ion.
Answer
Ca2+ has a $\left ( \sigma _{4s} \right )^{2}\left ( \sigma _{4s}^{\star } \right )^{1}$ electron configurations and a bond order of 1/2 and should exist.
Molecular Orbitals Formed from ns and np Atomic Orbitals
Atomic orbitals other than ns orbitals can also interact to form molecular orbitals. Because individual p, d, and f orbitals are not spherically symmetrical, however, we need to define a coordinate system so we know which lobes are interacting in three-dimensional space. Recall that for each np subshell, for example, there are npx, npy, and npz orbitals. All have the same energy and are therefore degenerate, but they have different spatial orientations.
$\sigma _{np_{z}}=np_{z}\left ( A \right )-np_{z}\left ( B \right ) \label{9.7.6}$
Just as with ns orbitals, we can form molecular orbitals from np orbitals by taking their mathematical sum and difference. When two positive lobes with the appropriate spatial orientation overlap, as illustrated for two npz atomic orbitals in part (a) in Figure $5$, it is the mathematical difference of their wave functions that results in constructive interference, which in turn increases the electron probability density between the two atoms. The difference therefore corresponds to a molecular orbital called a $\sigma _{np_{z}}$ bonding molecular orbital because, just as with the σ orbitals discussed previously, it is symmetrical about the internuclear axis (in this case, the z-axis):
$\sigma _{np_{z}}=np_{z}\left ( A \right )-np_{z}\left ( B \right ) \label{9.7.7A}$
The other possible combination of the two npz orbitals is the mathematical sum:
$\sigma _{np_{z}}=np_{z}\left ( A \right )+np_{z}\left ( B \right ) \label{9.7.7}$
In this combination, shown in part (b) in Figure $5$, the positive lobe of one npz atomic orbital overlaps the negative lobe of the other, leading to destructive interference of the two waves and creating a node between the two atoms. Hence this is an antibonding molecular orbital. Because it, too, is symmetrical about the internuclear axis, this molecular orbital is called a $\sigma _{np_{z}}=np_{z}\left ( A \right )-np_{z}\left ( B \right )$ antibonding molecular orbital. Whenever orbitals combine, the bonding combination is always lower in energy (more stable) than the atomic orbitals from which it was derived, and the antibonding combination is higher in energy (less stable).
Overlap of atomic orbital lobes with the same sign produces a bonding molecular orbital, regardless of whether it corresponds to the sum or the difference of the atomic orbitals.
The remaining p orbitals on each of the two atoms, npx and npy, do not point directly toward each other. Instead, they are perpendicular to the internuclear axis. If we arbitrarily label the axes as shown in Figure $6$, we see that we have two pairs of np orbitals: the two npx orbitals lying in the plane of the page, and two npy orbitals perpendicular to the plane. Although these two pairs are equivalent in energy, the npx orbital on one atom can interact with only the npx orbital on the other, and the npy orbital on one atom can interact with only the npy on the other. These interactions are side-to-side rather than the head-to-head interactions characteristic of σ orbitals. Each pair of overlapping atomic orbitals again forms two molecular orbitals: one corresponds to the arithmetic sum of the two atomic orbitals and one to the difference. The sum of these side-to-side interactions increases the electron probability in the region above and below a line connecting the nuclei, so it is a bonding molecular orbital that is called a pi (π) orbital (a bonding molecular orbital formed from the side-to-side interactions of two or more parallel np atomic orbitals). The difference results in the overlap of orbital lobes with opposite signs, which produces a nodal plane perpendicular to the internuclear axis; hence it is an antibonding molecular orbital, called a pi star (π*) orbital An antibonding molecular orbital formed from the difference of the side-to-side interactions of two or more parallel np atomic orbitals, creating a nodal plane perpendicular to the internuclear axis..
$\pi _{np_{x}}=np_{x}\left ( A \right )+np_{x}\left ( B \right ) \label{9.7.8}$
$\pi ^{\star }_{np_{x}}=np_{x}\left ( A \right )-np_{x}\left ( B \right ) \label{9.7.9}$
The two npy orbitals can also combine using side-to-side interactions to produce a bonding $\pi _{np_{y}}$ molecular orbital and an antibonding $\pi _{np_{y}}^{\star }$ molecular orbital. Because the npx and npy atomic orbitals interact in the same way (side-to-side) and have the same energy, the $\pi _{np_{x}}$ and $\pi _{np_{y}}$molecular orbitals are a degenerate pair, as are the $\pi _{np_{x}}^{\star }$ and $\pi _{np_{y}}^{\star }$ molecular orbitals.
Figure $7$ is an energy-level diagram that can be applied to two identical interacting atoms that have three np atomic orbitals each. There are six degenerate p atomic orbitals (three from each atom) that combine to form six molecular orbitals, three bonding and three antibonding. The bonding molecular orbitals are lower in energy than the atomic orbitals because of the increased stability associated with the formation of a bond. Conversely, the antibonding molecular orbitals are higher in energy, as shown. The energy difference between the σ and σ* molecular orbitals is significantly greater than the difference between the two π and π* sets. The reason for this is that the atomic orbital overlap and thus the strength of the interaction are greater for a σ bond than a π bond, which means that the σ molecular orbital is more stable (lower in energy) than the π molecular orbitals.
Although many combinations of atomic orbitals form molecular orbitals, we will discuss only one other interaction: an ns atomic orbital on one atom with an npz atomic orbital on another. As shown in Figure $8$, the sum of the two atomic wave functions (ns + npz) produces a σ bonding molecular orbital. Their difference (nsnpz) produces a σ* antibonding molecular orbital, which has a nodal plane of zero probability density perpendicular to the internuclear axis.
Summary
Molecular orbital theory, a delocalized approach to bonding, can often explain a compound’s color, why a compound with unpaired electrons is stable, semiconductor behavior, and resonance, none of which can be explained using a localized approach. A molecular orbital (MO) is an allowed spatial distribution of electrons in a molecule that is associated with a particular orbital energy. Unlike an atomic orbital (AO), which is centered on a single atom, a molecular orbital extends over all the atoms in a molecule or ion. Hence the molecular orbital theory of bonding is a delocalized approach. Molecular orbitals are constructed using linear combinations of atomic orbitals (LCAOs), which are usually the mathematical sums and differences of wave functions that describe overlapping atomic orbitals. Atomic orbitals interact to form three types of molecular orbitals.
A completely bonding molecular orbital contains no nodes (regions of zero electron probability) perpendicular to the internuclear axis, whereas a completely antibonding molecular orbital contains at least one node perpendicular to the internuclear axis. A sigma (σ) orbital (bonding) or a sigma star (σ*) orbital (antibonding) is symmetrical about the internuclear axis. Hence all cross-sections perpendicular to that axis are circular. Both a pi (π) orbital (bonding) and a pi star (π*) orbital (antibonding) possess a nodal plane that contains the nuclei, with electron density localized on both sides of the plane.
The energies of the molecular orbitals versus those of the parent atomic orbitals can be shown schematically in an energy-level diagram. The electron configuration of a molecule is shown by placing the correct number of electrons in the appropriate energy-level diagram, starting with the lowest-energy orbital and obeying the Pauli principle; that is, placing only two electrons with opposite spin in each orbital. From the completed energy-level diagram, we can calculate the bond order, defined as one-half the net number of bonding electrons. In bond orders, electrons in antibonding molecular orbitals cancel electrons in bonding molecular orbitals, while electrons in nonbonding orbitals have no effect and are not counted. Bond orders of 1, 2, and 3 correspond to single, double, and triple bonds, respectively. Molecules with predicted bond orders of 0 are generally less stable than the isolated atoms and do not normally exist.
Contributors and Attributions
1. Orbitals or orbital lobes with the same sign interact to give increased electron probability along the plane of the internuclear axis because of constructive reinforcement of the wave functions. Consequently, electrons in such molecular orbitals help to hold the positively charged nuclei together. Such orbitals are bonding molecular orbitals, and they are always lower in energy than the parent atomic orbitals.
2. Orbitals or orbital lobes with opposite signs interact to give decreased electron probability density between the nuclei because of destructive interference of the wave functions. Consequently, electrons in such molecular orbitals are primarily located outside the internuclear region, leading to increased repulsions between the positively charged nuclei. These orbitals are called antibonding molecular orbitals, and they are always higher in energy than the parent atomic orbitals.
3. Some atomic orbitals interact only very weakly, and the resulting molecular orbitals give essentially no change in the electron probability density between the nuclei. Hence electrons in such orbitals have no effect on the bonding in a molecule or ion. These orbitals are nonbonding molecular orbitals, and they have approximately the same energy as the parent atomic orbitals. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/09%3A_Molecular_Geometry_and_Bonding_Theories/9.07%3A_Molecular_Orbitals.txt |
Learning Objectives
• To apply Molecular Orbital Theory to the diatomic homonuclear molecule from the elements in the second period.
If we combine the splitting schemes for the 2s and 2p orbitals, we can predict bond order in all of the diatomic molecules and ions composed of elements in the first complete row of the periodic table. Remember that only the valence orbitals of the atoms need be considered; as we saw in the cases of lithium hydride and dilithium, the inner orbitals remain tightly bound and retain their localized atomic character. We now describe examples of systems involving period 2 homonuclear diatomic molecules, such as N2, O2, and F2.
four key points to remember when drawing molecular orbital diagrams:
1. The number of molecular orbitals produced is the same as the number of atomic orbitals used to create them (the "law of conservation of orbitals").
2. As the overlap between two atomic orbitals increases, the difference in energy between the resulting bonding and antibonding molecular orbitals increases.
3. When two atomic orbitals combine to form a pair of molecular orbitals, the bonding molecular orbital is stabilized about as much as the antibonding molecular orbital is destabilized.
4. The interaction between atomic orbitals is greatest when they have the same energy.
We illustrate how to use these points by constructing a molecular orbital energy-level diagram for F2. We use the diagram in part (a) in Figure $1$; the n = 1 orbitals (σ1s and σ1s*) are located well below those of the n = 2 level and are not shown. As illustrated in the diagram, the σ2s and σ2s* molecular orbitals are much lower in energy than the molecular orbitals derived from the 2p atomic orbitals because of the large difference in energy between the 2s and 2p atomic orbitals of fluorine. The lowest-energy molecular orbital derived from the three 2p orbitals on each F is $\sigma _{2p_{z}}$ and the next most stable are the two degenerate orbitals, $\pi _{2p_{x}}$ and $\pi _{2p_{y}}$. For each bonding orbital in the diagram, there is an antibonding orbital, and the antibonding orbital is destabilized by about as much as the corresponding bonding orbital is stabilized. As a result, the $\sigma ^{\star }_{2p_{z}}$ orbital is higher in energy than either of the degenerate $\pi _{2p_{x}}^{\star }$ and $\pi _{2p_{y}}^{\star }$ orbitals. We can now fill the orbitals, beginning with the one that is lowest in energy.
Each fluorine has 7 valence electrons, so there are a total of 14 valence electrons in the F2 molecule. Starting at the lowest energy level, the electrons are placed in the orbitals according to the Pauli principle and Hund’s rule. Two electrons each fill the σ2s and σ2s* orbitals, 2 fill the $\sigma _{2p_{z}}$ orbital, 4 fill the two degenerate π orbitals, and 4 fill the two degenerate π* orbitals, for a total of 14 electrons. To determine what type of bonding the molecular orbital approach predicts F2 to have, we must calculate the bond order. According to our diagram, there are 8 bonding electrons and 6 antibonding electrons, giving a bond order of (8 − 6) ÷ 2 = 1. Thus F2 is predicted to have a stable F–F single bond, in agreement with experimental data.
We now turn to a molecular orbital description of the bonding in O2. It so happens that the molecular orbital description of this molecule provided an explanation for a long-standing puzzle that could not be explained using other bonding models. To obtain the molecular orbital energy-level diagram for O2, we need to place 12 valence electrons (6 from each O atom) in the energy-level diagram shown in part (b) in Figure $1$. We again fill the orbitals according to Hund’s rule and the Pauli principle, beginning with the orbital that is lowest in energy. Two electrons each are needed to fill the σ2s and σ2s* orbitals, 2 more to fill the $\sigma _{2p_{z}}$ orbital, and 4 to fill the degenerate $\pi _{2p_{x}}^{\star }$ and $\pi _{2p_{y}}^{\star }$ orbitals. According to Hund’s rule, the last 2 electrons must be placed in separate π* orbitals with their spins parallel, giving two unpaired electrons. This leads to a predicted bond order of (8 − 4) ÷ 2 = 2, which corresponds to a double bond, in agreement with experimental data (Table 4.5): the O–O bond length is 120.7 pm, and the bond energy is 498.4 kJ/mol at 298 K.
None of the other bonding models can predict the presence of two unpaired electrons in O2. Chemists had long wondered why, unlike most other substances, liquid O2 is attracted into a magnetic field. As shown in Figure $2$, it actually remains suspended between the poles of a magnet until the liquid boils away. The only way to explain this behavior was for O2 to have unpaired electrons, making it paramagnetic, exactly as predicted by molecular orbital theory. This result was one of the earliest triumphs of molecular orbital theory over the other bonding approaches we have discussed.
The magnetic properties of O2 are not just a laboratory curiosity; they are absolutely crucial to the existence of life. Because Earth’s atmosphere contains 20% oxygen, all organic compounds, including those that compose our body tissues, should react rapidly with air to form H2O, CO2, and N2 in an exothermic reaction. Fortunately for us, however, this reaction is very, very slow. The reason for the unexpected stability of organic compounds in an oxygen atmosphere is that virtually all organic compounds, as well as H2O, CO2, and N2, have only paired electrons, whereas oxygen has two unpaired electrons. Thus the reaction of O2 with organic compounds to give H2O, CO2, and N2 would require that at least one of the electrons on O2 change its spin during the reaction. This would require a large input of energy, an obstacle that chemists call a spin barrier. Consequently, reactions of this type are usually exceedingly slow. If they were not so slow, all organic substances, including this book and you, would disappear in a puff of smoke!
For period 2 diatomic molecules to the left of N2 in the periodic table, a slightly different molecular orbital energy-level diagram is needed because the $\sigma _{2p_{z}}$ molecular orbital is slightly higher in energy than the degenerate $\pi ^{\star }_{np_{x}}$ and $\pi ^{\star }_{np_{y}}$ orbitals. The difference in energy between the 2s and 2p atomic orbitals increases from Li2 to F2 due to increasing nuclear charge and poor screening of the 2s electrons by electrons in the 2p subshell. The bonding interaction between the 2s orbital on one atom and the 2pz orbital on the other is most important when the two orbitals have similar energies. This interaction decreases the energy of the σ2s orbital and increases the energy of the $\sigma _{2p_{z}}$ orbital. Thus for Li2, Be2, B2, C2, and N2, the $\sigma _{2p_{z}}$ orbital is higher in energy than the $\sigma _{3p_{z}}$ orbitals, as shown in Figure $3$ Experimentally, it is found that the energy gap between the ns and np atomic orbitals increases as the nuclear charge increases (Figure $3$ ). Thus for example, the $\sigma _{2p_{z}}$ molecular orbital is at a lower energy than the $\pi _{2p_{x,y}}$ pair.
Completing the diagram for N2 in the same manner as demonstrated previously, we find that the 10 valence electrons result in 8 bonding electrons and 2 antibonding electrons, for a predicted bond order of 3, a triple bond. Experimental data show that the N–N bond is significantly shorter than the F–F bond (109.8 pm in N2 versus 141.2 pm in F2), and the bond energy is much greater for N2 than for F2 (945.3 kJ/mol versus 158.8 kJ/mol, respectively). Thus the N2 bond is much shorter and stronger than the F2 bond, consistent with what we would expect when comparing a triple bond with a single bond.
Example $3$: Diatomic Sulfur
Use a qualitative molecular orbital energy-level diagram to predict the electron configuration, the bond order, and the number of unpaired electrons in S2, a bright blue gas at high temperatures.
Given: chemical species
Asked for: molecular orbital energy-level diagram, bond order, and number of unpaired electrons
Strategy:
1. Write the valence electron configuration of sulfur and determine the type of molecular orbitals formed in S2. Predict the relative energies of the molecular orbitals based on how close in energy the valence atomic orbitals are to one another.
2. Draw the molecular orbital energy-level diagram for this system and determine the total number of valence electrons in S2.
3. Fill the molecular orbitals in order of increasing energy, being sure to obey the Pauli principle and Hund’s rule.
4. Calculate the bond order and describe the bonding.
Solution:
A Sulfur has a [Ne]3s23p4 valence electron configuration. To create a molecular orbital energy-level diagram similar to those in Figure $1$ and Figure $3$, we need to know how close in energy the 3s and 3p atomic orbitals are because their energy separation will determine whether the $\pi _{3p_{x,y}}$ or the $\sigma _{3p_{z}}$> molecular orbital is higher in energy. Because the nsnp energy gap increases as the nuclear charge increases (Figure $3$), the $\sigma _{3p_{z}}$ molecular orbital will be lower in energy than the $\pi _{3p_{x,y}}$ pair.
B The molecular orbital energy-level diagram is as follows:
Each sulfur atom contributes 6 valence electrons, for a total of 12 valence electrons.
C Ten valence electrons are used to fill the orbitals through $\pi _{3p_{x}}$ and $\pi _{3p_{y}}$, leaving 2 electrons to occupy the degenerate $\pi ^{\star }_{3p_{x}}$ and $\pi ^{\star }_{3p_{y}}$ pair. From Hund’s rule, the remaining 2 electrons must occupy these orbitals separately with their spins aligned. With the numbers of electrons written as superscripts, the electron configuration of S2 is $\left ( \sigma _{3s} \right )^{2}\left ( \sigma ^{\star }_{3s} \right )^{2}\left ( \sigma _{3p_{z}} \right )^{2}\left ( \pi _{3p_{x,y}} \right )^{4}\left ( \pi _{3p ^{\star }_{x,y}} \right )^{2}$ with 2 unpaired electrons. The bond order is (8 − 4) ÷ 2 = 2, so we predict an S=S double bond.
Exercise $3$: The Peroxide Ion
Use a qualitative molecular orbital energy-level diagram to predict the electron configuration, the bond order, and the number of unpaired electrons in the peroxide ion (O22).
Answer
$\left ( \sigma _{2s} \right )^{2}\left ( \sigma ^{\star }_{2s} \right )^{2}\left ( \sigma _{2p_{z}} \right )^{2}\left ( \pi _{2p_{x,y}} \right )^{4}\left ( \pi _{2p ^{\star }_{x,y}} \right )^{4}$ bond order of 1; no unpaired electrons
Molecular Orbitals for Heteronuclear Diatomic Molecules
Diatomic molecules with two different atoms are called heteronuclear diatomic molecules. When two nonidentical atoms interact to form a chemical bond, the interacting atomic orbitals do not have the same energy. If, for example, element B is more electronegative than element A (χB > χA), the net result is a “skewed” molecular orbital energy-level diagram, such as the one shown for a hypothetical A–B molecule in Figure $4$. The atomic orbitals of element B are uniformly lower in energy than the corresponding atomic orbitals of element A because of the enhanced stability of the electrons in element B. The molecular orbitals are no longer symmetrical, and the energies of the bonding molecular orbitals are more similar to those of the atomic orbitals of B. Hence the electron density of bonding electrons is likely to be closer to the more electronegative atom. In this way, molecular orbital theory can describe a polar covalent bond.
A molecular orbital energy-level diagram is always skewed toward the more electronegative atom.
An Odd Number of Valence Electrons: NO
Nitric oxide (NO) is an example of a heteronuclear diatomic molecule. The reaction of O2 with N2 at high temperatures in internal combustion engines forms nitric oxide, which undergoes a complex reaction with O2 to produce NO2, which in turn is responsible for the brown color we associate with air pollution. Recently, however, nitric oxide has also been recognized to be a vital biological messenger involved in regulating blood pressure and long-term memory in mammals.
Because NO has an odd number of valence electrons (5 from nitrogen and 6 from oxygen, for a total of 11), its bonding and properties cannot be successfully explained by either the Lewis electron-pair approach or valence bond theory. The molecular orbital energy-level diagram for NO (Figure $13$) shows that the general pattern is similar to that for the O2 molecule (Figure $11$). Because 10 electrons are sufficient to fill all the bonding molecular orbitals derived from 2p atomic orbitals, the 11th electron must occupy one of the degenerate π* orbitals. The predicted bond order for NO is therefore (8-3) ÷ 2 = 2 1/2 . Experimental data, showing an N–O bond length of 115 pm and N–O bond energy of 631 kJ/mol, are consistent with this description. These values lie between those of the N2 and O2 molecules, which have triple and double bonds, respectively. As we stated earlier, molecular orbital theory can therefore explain the bonding in molecules with an odd number of electrons, such as NO, whereas Lewis electron structures cannot.
Note that electronic structure studies show the ground state configuration of $\ce{NO}$ to be $\left ( \sigma _{2s} \right)^{2}\left ( \sigma ^{\star }_{2s} \right)^{2}\left ( \pi _{2p_{x,y}} \right)^{4} \left ( \sigma _{2p_{z}} \right)^{2} \left ( \pi _{2p ^{\star }_{x,y}} \right)^{1}$ in order of increasing energy. Hence, the $\pi _{2p_{x,y}}$ orbitals are lower in energy than the $\sigma _{2p_{z}}$ orbital. This is because the $\ce{NO}$ molecule is near the transition of flipping energies levels observed in homonuclear diatomics where the sigma bond drops below the pi bond (Figure $11$).
Molecular orbital theory can also tell us something about the chemistry of $NO$. As indicated in the energy-level diagram in Figure $13$, NO has a single electron in a relatively high-energy molecular orbital. We might therefore expect it to have similar reactivity as alkali metals such as Li and Na with their single valence electrons. In fact, $NO$ is easily oxidized to the $NO^+$ cation, which is isoelectronic with $N_2$ and has a bond order of 3, corresponding to an N≡O triple bond.
Nonbonding Molecular Orbitals
Molecular orbital theory is also able to explain the presence of lone pairs of electrons. Consider, for example, the HCl molecule, whose Lewis electron structure has three lone pairs of electrons on the chlorine atom. Using the molecular orbital approach to describe the bonding in HCl, we can see from Figure $6$ that the 1s orbital of atomic hydrogen is closest in energy to the 3p orbitals of chlorine. Consequently, the filled Cl 3s atomic orbital is not involved in bonding to any appreciable extent, and the only important interactions are those between the H 1s and Cl 3p orbitals. Of the three p orbitals, only one, designated as 3pz, can interact with the H 1s orbital. The 3px and 3py atomic orbitals have no net overlap with the 1s orbital on hydrogen, so they are not involved in bonding. Because the energies of the Cl 3s, 3px, and 3py orbitals do not change when HCl forms, they are called nonbonding molecular orbitals. A nonbonding molecular orbital occupied by a pair of electrons is the molecular orbital equivalent of a lone pair of electrons. By definition, electrons in nonbonding orbitals have no effect on bond order, so they are not counted in the calculation of bond order. Thus the predicted bond order of HCl is (2 − 0) ÷ 2 = 1. Because the σ bonding molecular orbital is closer in energy to the Cl 3pz than to the H 1s atomic orbital, the electrons in the σ orbital are concentrated closer to the chlorine atom than to hydrogen. A molecular orbital approach to bonding can therefore be used to describe the polarization of the H–Cl bond to give $H^{\delta +} -- Cl^{\delta -}$.
Electrons in nonbonding molecular orbitals have no effect on bond order.
Example $4$: The Cyanide Ion
Use a “skewed” molecular orbital energy-level diagram like the one in Figure $4$ to describe the bonding in the cyanide ion (CN). What is the bond order?
Given: chemical species
Asked for: “skewed” molecular orbital energy-level diagram, bonding description, and bond order
Strategy:
1. Calculate the total number of valence electrons in CN. Then place these electrons in a molecular orbital energy-level diagram like Figure $4$ in order of increasing energy. Be sure to obey the Pauli principle and Hund’s rule while doing so.
2. Calculate the bond order and describe the bonding in CN.
Solution:
A The CN ion has a total of 10 valence electrons: 4 from C, 5 from N, and 1 for the −1 charge. Placing these electrons in an energy-level diagram like Figure $4$ fills the five lowest-energy orbitals, as shown here:
Because $\chi_N > \chi_C$, the atomic orbitals of N (on the right) are lower in energy than those of C.
B The resulting valence electron configuration gives a predicted bond order of (8 − 2) ÷ 2 = 3, indicating that the CN ion has a triple bond, analogous to that in N2.
Exercise $4$: The Hypochlorite Ion
Use a qualitative molecular orbital energy-level diagram to describe the bonding in the hypochlorite ion (OCl). What is the bond order?
Answer
All molecular orbitals except the highest-energy σ* are filled, giving a bond order of 1.
Although the molecular orbital approach reveals a great deal about the bonding in a given molecule, the procedure quickly becomes computationally intensive for molecules of even moderate complexity. Furthermore, because the computed molecular orbitals extend over the entire molecule, they are often difficult to represent in a way that is easy to visualize. Therefore we do not use a pure molecular orbital approach to describe the bonding in molecules or ions with more than two atoms. Instead, we use a valence bond approach and a molecular orbital approach to explain, among other things, the concept of resonance, which cannot adequately be explained using other methods.
Summary
Molecular orbital energy-level diagrams for diatomic molecules can be created if the electron configuration of the parent atoms is known, following a few simple rules. Most important, the number of molecular orbitals in a molecule is the same as the number of atomic orbitals that interact. The difference between bonding and antibonding molecular orbital combinations is proportional to the overlap of the parent orbitals and decreases as the energy difference between the parent atomic orbitals increases. With such an approach, the electronic structures of virtually all commonly encountered homonuclear diatomic molecules, molecules with two identical atoms, can be understood. The molecular orbital approach correctly predicts that the O2 molecule has two unpaired electrons and hence is attracted into a magnetic field. In contrast, most substances have only paired electrons. A similar procedure can be applied to molecules with two dissimilar atoms, called heteronuclear diatomic molecules, using a molecular orbital energy-level diagram that is skewed or tilted toward the more electronegative element. Molecular orbital theory is able to describe the bonding in a molecule with an odd number of electrons such as NO and even to predict something about its chemistry. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/09%3A_Molecular_Geometry_and_Bonding_Theories/9.08%3A_Second-Row_Diatomic_Molecules.txt |
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
9.2: THE VSEPR MODEL
Conceptual Problems
1. What is the main difference between the VSEPR model and Lewis electron structures?
2. What are the differences between molecular geometry and Lewis electron structures? Can two molecules with the same Lewis electron structures have different molecular geometries? Can two molecules with the same molecular geometry have different Lewis electron structures? In each case, support your answer with an example.
3. How does the VSEPR model deal with the presence of multiple bonds?
4. Three molecules have the following generic formulas: AX2, AX2E, and AX2E2. Predict the molecular geometry of each, and arrange them in order of increasing X–A–X angle.
5. Which has the smaller angles around the central atom—H2S or SiH4? Why? Do the Lewis electron structures of these molecules predict which has the smaller angle?
6. Discuss in your own words why lone pairs of electrons occupy more space than bonding pairs. How does the presence of lone pairs affect molecular geometry?
• When using VSEPR to predict molecular geometry, the importance of repulsions between electron pairs decreases in the following order: LP–LP, LP–BP, BP–BP. Explain this order. Draw structures of real molecules that separately show each of these interactions.
• How do multiple bonds affect molecular geometry? Does a multiple bond take up more or less space around an atom than a single bond? a lone pair?
• Straight-chain alkanes do not have linear structures but are “kinked.” Using n-hexane as an example, explain why this is so. Compare the geometry of 1-hexene to that of n-hexane.
• How is molecular geometry related to the presence or absence of a molecular dipole moment?
• How are molecular geometry and dipole moments related to physical properties such as melting point and boiling point?
• What two features of a molecule’s structure and bonding are required for a molecule to be considered polar? Is COF2 likely to have a significant dipole moment? Explain your answer.
• When a chemist says that a molecule is polar, what does this mean? What are the general physical properties of polar molecules?
• Use the VSPER model and your knowledge of bonding and dipole moments to predict which molecules will be liquids or solids at room temperature and which will be gases. Explain your rationale for each choice. Justify your answers.
1. CH3Cl
2. PCl3
3. CO
4. SF6
5. IF5
6. CH3OCH3
7. CCl3H
8. H3COH
• The idealized molecular geometry of BrF5 is square pyramidal, with one lone pair. What effect does the lone pair have on the actual molecular geometry of BrF5? If LP–BP repulsions were weaker than BP–BP repulsions, what would be the effect on the molecular geometry of BrF5?
• Which has the smallest bond angle around the central atom—H2S, H2Se, or H2Te? the largest? Justify your answers.
• Which of these molecular geometries always results in a molecule with a net dipole moment: linear, bent, trigonal planar, tetrahedral, seesaw, trigonal pyramidal, square pyramidal, and octahedral? For the geometries that do not always produce a net dipole moment, what factor(s) will result in a net dipole moment?
Conceptual Answers
1.
2.
3. To a first approximation, the VSEPR model assumes that multiple bonds and single bonds have the same effect on electron pair geometry and molecular geometry; in other words, VSEPR treats multiple bonds like single bonds. Only when considering fine points of molecular structure does VSEPR recognize that multiple bonds occupy more space around the central atom than single bonds.
4.
5.
6.
7.
8.
9.
10.
11. Physical properties like boiling point and melting point depend upon the existence and magnitude of the dipole moment of a molecule. In general, molecules that have substantial dipole moments are likely to exhibit greater intermolecular interactions, resulting in higher melting points and boiling points.
12.
13. The term “polar” is generally used to mean that a molecule has an asymmetrical structure and contains polar bonds. The resulting dipole moment causes the substance to have a higher boiling or melting point than a nonpolar substance.
Numerical Problems
1. Give the number of electron groups around the central atom and the molecular geometry for each molecule. Classify the electron groups in each species as bonding pairs or lone pairs.
1. BF3
2. PCl3
3. XeF2
4. AlCl4
5. CH2Cl2
2. Give the number of electron groups around the central atom and the molecular geometry for each species. Classify the electron groups in each species as bonding pairs or lone pairs.
1. ICl3
2. CCl3+
3. H2Te
4. XeF4
5. NH4+
3. Give the number of electron groups around the central atom and the molecular geometry for each molecule. For structures that are not linear, draw three-dimensional representations, clearly showing the positions of the lone pairs of electrons.
1. HCl
2. NF3
3. ICl2+
4. N3
5. H3O+
4. Give the number of electron groups around the central atom and the molecular geometry for each molecule. For structures that are not linear, draw three-dimensional representations, clearly showing the positions of the lone pairs of electrons.
1. SO3
2. NH2
3. NO3
4. I3
5. OF2
5. What is the molecular geometry of ClF3? Draw a three-dimensional representation of its structure and explain the effect of any lone pairs on the idealized geometry.
6. Predict the molecular geometry of each of the following.
1. ICl3
2. AsF5
3. NO2
4. TeCl4
7. Predict whether each molecule has a net dipole moment. Justify your answers and indicate the direction of any bond dipoles.
1. NO
2. HF
3. PCl3
4. CO2
5. SO2
6. SF4
8. Predict whether each molecule has a net dipole moment. Justify your answers and indicate the direction of any bond dipoles.
1. OF2
2. BCl3
3. CH2Cl2
4. TeF4
5. CH3OH
6. XeO4
9. Of the molecules Cl2C=Cl2, IF3, and SF6, which has a net dipole moment? Explain your reasoning.
10. Of the molecules SO3, XeF4, and H2C=Cl2, which has a net dipole moment? Explain your reasoning.
Numerical Answers
1. trigonal planar (all electron groups are bonding pairs)
2. tetrahedral (one lone pair on P)
3. trigonal bipyramidal (three lone pairs on Xe)
4. tetrahedral (all electron groups on Al are bonding pairs)
5. tetrahedral (all electron groups on C are bonding pairs)
1.
2.
1. four electron groups, linear molecular geometry
2. four electron groups, pyramidal molecular geometry
3. four electron groups, bent molecular geometry
4. two electron groups, linear molecular geometry
5. four electron groups, pyramidal molecular geometry
3.
4. The idealized geometry is T shaped, but the two lone pairs of electrons on Cl will distort the structure, making the F–Cl–F angle less than 180°.
5.
6.
7.
8. Cl2C=CCl2: Although the C–Cl bonds are rather polar, the individual bond dipoles cancel one another in this symmetrical structure, and Cl2C=CCl2 does not have a net dipole moment.
IF3: In this structure, the individual I–F bond dipoles cannot cancel one another, giving IF3 a net dipole moment.
SF6: The S–F bonds are quite polar, but the individual bond dipoles cancel one another in an octahedral structure. Thus, SF6 has no net dipole moment.
9.3: MOLECULAR SHAPE AND MOLECULAR POLARITY
Conceptual Problems
1. Why do ionic compounds such as KI exhibit substantially less than 100% ionic character in the gas phase?
2. Of the compounds LiI and LiF, which would you expect to behave more like a classical ionic compound? Which would have the greater dipole moment in the gas phase? Explain your answers.
Numerical Problems
1. Predict whether each compound is purely covalent, purely ionic, or polar covalent.
1. RbCl
2. S8
3. TiCl2
4. SbCl3
5. LiI
6. Br2
2. Based on relative electronegativities, classify the bonding in each compound as ionic, covalent, or polar covalent. Indicate the direction of the bond dipole for each polar covalent bond.
1. NO
2. HF
3. MgO
4. AlCl3
5. SiO2
6. the C=O bond in acetone
7. O3
3. Based on relative electronegativities, classify the bonding in each compound as ionic, covalent, or polar covalent. Indicate the direction of the bond dipole for each polar covalent bond.
1. NaBr
2. OF2
3. BCl3
4. the S–S bond in CH3CH2SSCH2CH3
5. the C–Cl bond in CH2Cl2
6. the O–H bond in CH3OH
7. PtCl42−
4. Classify each species as having 0%–40% ionic character, 40%–60% ionic character, or 60%–100% ionic character based on the type of bonding you would expect. Justify your reasoning.
1. CaO
2. S8
3. AlBr3
4. ICl
5. Na2S
6. SiO2
7. LiBr
5. If the bond distance in HCl (dipole moment = 1.109 D) were double the actual value of 127.46 pm, what would be the effect on the charge localized on each atom? What would be the percent negative charge on Cl? At the actual bond distance, how would doubling the charge on each atom affect the dipole moment? Would this represent more ionic or covalent character?
6. Calculate the percent ionic character of HF (dipole moment = 1.826 D) if the H–F bond distance is 92 pm.
7. Calculate the percent ionic character of CO (dipole moment = 0.110 D) if the C–O distance is 113 pm.
8. Calculate the percent ionic character of PbS and PbO in the gas phase, given the following information: for PbS, r = 228.69 pm and µ = 3.59 D; for PbO, r = 192.18 pm and µ = 4.64 D. Would you classify these compounds as having covalent or polar covalent bonds in the solid state?
9.5: HYBRID ORBITALS
Conceptual Problems
1. Arrange sp, sp3, and sp2 in order of increasing strength of the bond formed to a hydrogen atom. Explain your reasoning.
2. What atomic orbitals are combined to form sp3, sp, sp3d2, and sp3d? What is the maximum number of electron-pair bonds that can be formed using each set of hybrid orbitals?
3. Why is it incorrect to say that an atom with sp2 hybridization will form only three bonds? The carbon atom in the carbonate anion is sp2 hybridized. How many bonds to carbon are present in the carbonate ion? Which orbitals on carbon are used to form each bond?
4. If hybridization did not occur, how many bonds would N, O, C, and B form in a neutral molecule, and what would be the approximate molecular geometry?
5. How are hybridization and molecular geometry related? Which has a stronger correlation—molecular geometry and hybridization or Lewis structures and hybridization?
6. In the valence bond approach to bonding in BeF2, which step(s) require(s) an energy input, and which release(s) energy?
7. The energies of hybrid orbitals are intermediate between the energies of the atomic orbitals from which they are formed. Why?
8. How are lone pairs on the central atom treated using hybrid orbitals?
9. Because nitrogen bonds to only three hydrogen atoms in ammonia, why doesn’t the nitrogen atom use sp2 hybrid orbitals instead of sp3 hybrids?
10. Using arguments based on orbital hybridization, explain why the CCl62− ion does not exist.
11. Species such as NF52− and OF42− are unknown. If 3d atomic orbitals were much lower energy, low enough to be involved in hybrid orbital formation, what effect would this have on the stability of such species? Why? What molecular geometry, electron-pair geometry, and hybridization would be expected for each molecule?
Numerical Problems
1. Draw an energy-level diagram showing promotion and hybridization to describe the bonding in CH3. How does your diagram compare with that for methane? What is the molecular geometry?
2. Draw an energy-level diagram showing promotion and hybridization to describe the bonding in CH3+. How does your diagram compare with that for methane? What is the molecular geometry?
3. Draw the molecular structure, including any lone pairs on the central atom, state the hybridization of the central atom, and determine the molecular geometry for each molecule.
1. BBr3
2. PCl3
3. NO3
4. Draw the molecular structure, including any lone pairs on the central atom, state the hybridization of the central atom, and determine the molecular geometry for each species.
1. AsBr3
2. CF3+
3. H2O
5. What is the hybridization of the central atom in each of the following?
1. CF4
2. CCl22−
3. IO3
4. SiH4
6. What is the hybridization of the central atom in each of the following?
1. CCl3+
2. CBr2O
3. CO32−
4. IBr2
7. What is the hybridization of the central atom in PF6? Is this ion likely to exist? Why or why not? What would be the shape of the molecule?
8. What is the hybridization of the central atom in SF5? Is this ion likely to exist? Why or why not? What would be the shape of the molecule?
Numerical Answers
1.
The promotion and hybridization process is exactly the same as shown for CH4 in the chapter. The only difference is that the C atom uses the four singly occupied sp3 hybrid orbitals to form electron-pair bonds with only three H atoms, and an electron is added to the fourth hybrid orbital to give a charge of 1–. The electron-pair geometry is tetrahedral, but the molecular geometry is pyramidal, as in NH3.
2.
3.
1.
sp2, trigonal planar
2.
sp3, pyramidal
3.
sp2, trigonal planar
4.
5. The central atoms in CF4, CCl22–, IO3, and SiH4 are all sp3 hybridized.
6.
7. The phosphorus atom in the PF6 ion is sp3d2 hybridized, and the ion is octahedral. The PF6 ion is isoelectronic with SF6 and has essentially the same structure. It should therefore be a stable species.
9.6: MULTIPLE BONDS
Conceptual Problems
1. What information is obtained by using the molecular orbital approach to bonding in O3 that is not obtained using the VSEPR model? Can this information be obtained using a Lewis electron-pair approach?
2. How is resonance explained using the molecular orbital approach?
3. Indicate what information can be obtained by each method:
Lewis Electron Structures VSEPR Model Valence Bond Theory Molecular Orbital Theory
Geometry
Resonance
Orbital Hybridization
Reactivity
Expanded Valences
Bond Order
Numerical Problems
1. Using both a hybrid atomic orbital and molecular orbital approaches, describe the bonding in \(BCl_3\) and \(CS_3^{2−}\).
2. Use both a hybrid atomic orbital and molecular orbital approaches to describe the bonding in \(CO_2\) and \(N_3^−\). | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/09%3A_Molecular_Geometry_and_Bonding_Theories/9.E%3A_Exercises.txt |
9.1: Molecular Shapes
The Lewis electron-pair approach described previously can be used to predict the number and types of bonds between the atoms in a substance, and it indicates which atoms have lone pairs of electrons. This approach gives no information about the actual arrangement of atoms in space, however.
• Bond angles: angles made by the lines joining the nuclei of the atoms in a molecule
• Bond angles determine overall shape of molecule
VSEPR Model" data-cke-saved-href="/Bookshelves/General_Chemistry/Map:_Chemistry_-_The_Central_Science_(Brown_et_al.)/09:_Molecular_Geometry_and_Bonding_Theories/9.02:_The_VSEPR_Model" href="/Bookshelves/General_Chemistry/Map:_Chemistry_-_The_Central_Science_(Brown_et_al.)/09:_Molecular_Geometry_and_Bonding_Theories/9.02:_The_VSEPR_Model" data-quail-id="12">9.2: The VSEPR Model
The VSEPR model can predict the structure of nearly any molecule or polyatomic ion in which the central atom is a nonmetal, as well as the structures of many molecules and polyatomic ions with a central metal atom. The VSEPR model is not a theory; it does not attempt to explain observations. Instead, it is a counting procedure that accurately predicts the three-dimensional structures of a large number of compounds, which cannot be predicted using the Lewis electron-pair approach.
The Valence-Shell Electron Pair-Repulsion Model
• Valence-shell electron-pair repulsion (VSEPR) model: model that accounts for the geometric arrangements of shared and unshared electron pairs around a central atom in terms of the repulsions between electron pairs
• Electrons repel one another; hence, the best arrangement of a given number of electron pairs is the one that minimizes the repulsions among them.
Predicting Molecular Geometries
• Two types of valence-shell electron pairs; bonding pairs and nonbonding pairs.
• Electron-pair geometry: 3-D arrangement of electron pairs around an atom according to the VSEPR model
• Molecular geometry: the arrangement in space of the atoms of a molecule
When describing the shapes of molecules, we always give the molecular geometry rather than the electron-pair geometry. Molecular geometries only include BONDING pairs. For example, NH3 is a tetrahedral according to electron-pair geometry because it has 3 bonds and an unshared pair of electron. However, according to molecular geometry it is a trigonal pyramidal because it has only 3 bonds.
How to predict molecular geometries using VSEPR model:
1. Sketch the Lewis dot structure of the molecule or ion
2. Count the total number of electron pairs around the central atom and arrange them in the way that minimizes electron-pair repulsions
3. Describe the molecular geometry in terms of the angular arrangement of the bonding pairs
Four or Fewer Valence-Shell Electron Pairs
A double or triple bond has essentially the same effect on bond angles as a single bond and is therefore counted as one bonding pair (that is, one electron region) when predicting geometry
The Effect of Nonbonding Electrons and Multiple Bonds on Bond Angles
Bond angles decrease as the number of nonbonding electron pairs increases. Bonding pairs are attracted by the two nuclei. By contrast, nonbonding electrons move under the attractive influence of only one nucleus and thus spread out more in space.
• Nonbonding electron pairs exert greater repulsive forces on adjacent electron pairs and thus tend to compress the angles between the bonding pairs
• Electrons in multiple bonds, like nonbonding pairs, exert a greater repulsive force on adjacent electron pairs than do single bonds.
Geometries of Molecules with Expanded Valence Shells
• When the central atom of a molecule or ion is from the third period of the periodic table and beyond (has d subshell) that atom may have more than four electron pairs around it
• The most stable electron-pair geometry for five electron pairs is the Trigonal bipyramid.
• Trigonal bipyramid geometries have two distinct types of electron pairs. They are axial pairs and equatorial pairs. In axial position, an electron pair is situated 90° from the three equatorial pairs. In an equatorial position, an electron pairs is situated 120° from the other two equatorial pairs and 90° from the two axial pairs.
• Because nonbonding pairs exert larger repulsions than bonding pairs, they always occupy the equatorial positions
• The most stable electron-pair geometry for six electron pairs is the octahedron.
• In an octahedron, the angles are either 90° or 180°, and all six positions are equivalent
Molecules with No Single Central Atom
Consider acetic acid molecule
• Notice that the leftmost C has four bonded pairs around it. Hence, it must be a tetrahedral.
• The central C has 3 bonding pairs around it (counting the double bond as a single pair), hence, it is trigonal planar
• O has two bonding pairs, and two nonbonding ones. Hence, it is bent.
9.3: Molecular Shape and Molecular Polarity
Compounds with polar covalent bonds have electrons that are shared unequally between the bonded atoms. The polarity of such a bond is determined largely by the relative electronegativites of the bonded atoms. The asymmetrical charge distribution in a polar substance produces a dipole moment, which is the product of the partial charges on the bonded atoms and the distance between them.
Polar: in a molecule, if its centers of negative and positive charge don’t coincide, aka dipole
• If there are no charges on the opposite ends of a molecule, or if the charges have the same sign, the molecule does not have a dipole moment and is therefore nonpolar. Any diatomic molecule with a polar bond is a polar molecule
• Polar molecules align themselves with respect to one another. The negative end of a polar molecule is attracted to a positive ion, the positive end is attracted to a negative ion
• Dipole moment: measures the degree of polarity of a given molecule. The dipole moment, $\mu$, is the product of the charge at either end of the dipole, $Q$, times the distance, $r$, between the charges.
$\mu =Qr \nonumber$
Therefore, the dipole moment increases as the quantity of charge that is separated increases and as the distance between the positive and negative centers increases. Dipole moments are generally reported in debyes (D), a unit equaling 3.33 X 10-30 C-m.
The Polarity of Polyatomic Molecules
• The polarity of a molecule containing more than two atoms depends on both the polarities of the bonds and the geometry of the molecule
• For each polar bond in a molecule, we can consider the bond dipole, that is, the dipole moment due only to the two atoms bonded together. We must then ask what the overall dipole moment results from adding up the individual bond dipoles. Let’s use CO2 (linear) as an example
• In CO2, each C – O bond is polar, and because they are identical, the bond dipoles are equal in magnitude.
• The fact that both C – O bonds are polar doesn’t necessarily mean the whole CO2 molecule is
• Bond dipoles and dipole moments are vector quantities; they have both a magnitude and direction. When calculating the overall dipole moment, both the magnitude and direction of the bond dipoles must be considered in this sum of vectors
• The two bond dipoles in CO2, although equal in magnitude, are opposite in direction, which makes them cancel each other out. Hence, CO2 has an overall dipole moment of zero. Even though the oxygen atoms in CO2 carry a partial negative charge, the geometry of the molecule dictates the overall dipole moment.
• Now, let’s consider H2O. Both of its bonds are identical once again, but the geometry is bent instead of linear. Hence, the bonds do not cancel each other out, and H2O has a dipole moment
• Molecules in which the central is symmetrically surrounded by identical atoms (BF3, CCl4) are nonpolar.
• For ABn molecules in which ALL of the B atoms are the same, certain symmetrical geometries – linear (AB2), trigonal planar (AB3), tetrahedral and square planar (AB4), trigonal bipyramidal (AB5), and octahedral (AB6) – must lead to nonpolar molecules, regardless of how polar the individual bonds are.
9.4: Covalent Bonding and Orbital Overlap
A more sophisticated treatment of bonding is needed for systems such as these. In this section, we present a quantum mechanical description of bonding, in which bonding electrons are viewed as being localized between the nuclei of the bonded atoms. The overlap of bonding orbitals is substantially increased through a process called hybridization, which results in the formation of stronger bonds.
Valence bond theory: model of chemical bonding in which an electron-pair bond is formed between two atoms by the overlap of orbitals on the two atoms
• Covalent bonding occurs when atoms share electrons. This increases the electron density between the two nuclei, and is visualized as occurring when a valence atomic orbital of one atom merges with that of another atom
• As the distance between the atoms decreases, their overlap increases. Because of the resultant increase in electron density between the nuclei, the potential energy of the system decreases, increasing the strength of the bond.
• The observed bond length is the distance at which the attractive forces between unlike charges (electrons and nuclei) are balanced by the repulsive forces between like charges (electron-electron and nucleus-nucleus)
9.5: Hybrid Orbitals
The localized valence bonding theory uses a process called hybridization, in which atomic orbitals that are similar in energy but not equivalent are combined mathematically to produce sets of equivalent orbitals that are properly oriented to form bonds. These new combinations are called hybrid atomic orbitals because they are produced by combining (hybridizing) two or more atomic orbitals from the same atom.
sp Hybrid Orbitals
• Use the salt BeF2 as an example. Its geometry is linear, however it is now hard to determine which orbitals on the Be atom overlap with those of the F atoms. The electron configuration of Be is 1s2 2s2. Hence, Be in its ground state has no unpaired electrons and is unable to form bonds
• Be’s only option is to "promote" one of its two electrons in the 2s shell to the 2p shell. It will then have a configuration of 1s2 2s1 2p1. Since the 2p orbital is at a higher energy level than the 2s orbital, this promotion requires energy
• The Be atom now has two unpaired electrons (2s and 2p) and would be able to form two bonds, which will not be similar. The two new orbitals are identical in shape (two lobes; one much larger than the other) but are oriented in different directions. Two hybrid orbitals have been created
• Hybrid orbitals: orbitals that result from the mixing of different kinds of atomic orbitals on the same atom. For example and sp3 hybrid results from the mixing, or hybridizing, of one s orbital and three p orbitals
• Hybridization: mixing of different types of atomic orbitals to produce a set of equivalent hybrid orbitals
• In this case, we have hybridized one s and one p orbital, so we call each hybrid an sp hybrid orbital. For the Be atom of BeF2, the orbital diagram is written as follows: 1s2 2sp2
• Hybrid orbitals have one large lobe and can therefore be directed at surrounding atoms more effectively, forming stronger bonds. The energy released by such bonds enables the promotion of a 2s electron to the 2p level.
sp2 and sp3 Hybrid Orbitals
• Whenever we mix a certain number of atomic orbitals, we get the same number of hybrid orbitals. Each of these hybrid orbitals is equivalent to the others but points in a different direction.
• Consider BF3. B has an electron configuration of 2s2 2p1. It promotes one of its 2s electron to the 2p orbital to have a configuration of 2s1 2p2. Then, it hybridizes into 3 sp2 hybrid orbitals
The same process can apply to the creation of sp3 hybrid orbitals
Hybridization Involving d Orbitals
Same process as sp2 orbitals. For example, mixing one s orbital, three p orbitals, and two d orbitals gives six sp3d2 hybrid orbitals, which are directed at the vertices of an octahedron
Section Summary
The molecular geometry needs to be known first in order to use the concept of hybridization, then we can use the following rules:
1. Draw the Lewis structure for the molecule or ion
2. Determine the electron-pair geometry using the VSEPR model
3. Specify the hybrid orbitals needed to accommodate the electron pairs based on their geometrical arrangement
9.6: Multiple Bonds
To describe the bonding in more complex molecules with multiple bonds, we can use an approach that uses hybrid atomic orbitals to describe the σσ bonding and molecular orbitals to describe the $π$ bonding. In this approach, unhybridized np orbitals on atoms bonded to one another are allowed to interact to produce bonding, antibonding, or nonbonding combinations.
In each covalent bond we have considered so far, the electron density is concentrated symmetrically about the line that connects the two nuclei (internuclear axis). This axis passes through the middle of the overlap region
• Sigma (σ) bond: covalent bond in which electron density is concentrated along the internuclear axis
• Pi (π) bond: covalent bond in which electron density is concentrated above and below the internuclear axis. Pi bonds are generally weaker than sigma bonds (less overlap)
Usually, single bonds are $\sigma$ bonds. A double bond has one $\sigma$ bond and one $\pi$ bond. A triple bond has one $\sigma$ bond and two $\pi$ bonds.
Consider C2H4 as an example. After hybridization, C has 3 sp2 hybrid orbitals and one electron in the remaining unhybridized 2p orbital. The unhybridized 2p electron is directly perpendicular to the plane that contains the three sp2 hybrid orbitals
• The C – H $\sigma$ bond is formed by the overlap of a sp2 hybrid orbital with the 1s orbital of hydrogen. Hence, the bond uses two electrons
• The C – C $\sigma$ bond is formed by the overlap of two sp2 hybrid orbitals, one on each carbon atom, and requires two more electrons. Overall, C2H4 has 12 valence electrons, 10 of which are used in the formation of the bonds
• The remaining two valence electrons reside in the unhybridized 2p orbitals, one electron on each of the atoms
• The 2p orbitals can overlap with one another in a sideways fashion. This will lead to the electron density being concentrated above and below the C – C bond axis, hence being a $\pi$ bond
Because $\pi$ bonds require that portions of a molecule be planar, they can introduce rigidity into molecules (strongly affects the properties of substances)
Although $\pi$ bonds can be formed from d orbitals, we will only consider those formed by p orbitals. They can only form if unhybridized p orbitals are present on the bonded atoms. Only atoms having sp or sp2 hybridization can be involved in $\pi$ bonding. Double and triple bonds are more common in molecules with small atoms
Delocalized Bonding
• σ and $\pi$ electrons are localized when they are associated totally with two atoms forming the bond.
• In some molecules, particularly those with more than one resonance form, we cannot accurately describe the bonding as localized. In order to deal with this, we "smear out" the bonds (delocalize them) so that it fits both resonance structures as best as possible
Delocalized electrons: electrons that are spread over a number of atoms in a molecule rather than localized between a pair of atoms
Section Summary
1. Every pair of bonded atoms shares one or more pairs of electrons. In every bond, at least one pair of electrons is localized in the space between the atoms, in a $\sigma$ bond. The appropriate set of hybrid orbitals used to form the $\sigma$ bonds between an atom and its neighbors is determined by the observed geometry of the molecule
2. The electrons in $\sigma$ bonds are localized in the region between two bonded atoms and do not make a significant contributions to the bonding between any other two atoms
3. When atoms share more than one pair of electrons, the additional pairs are in $\pi$ bonds. The centers of chare density in a $\pi$ bond lie above an below the bond axis.
4. Molecules with two or more resonance forms can have $\pi$ bonds that extend over more than two bonded atoms. Electrons in $\pi$ bonds that extend over more than two bonded atoms are said to be delocalized
9.7: Molecular Orbitals
A molecular orbital is an allowed spatial distribution of electrons in a molecule that is associated with a particular orbital energy. Unlike an atomic orbital, which is centered on a single atom, a molecular orbital extends over all the atoms in a molecule or ion. Hence the molecular orbital theory of bonding is a delocalized approach. Molecular orbitals are constructed using linear combinations of atomic orbitals (LCAOs), which describe overlapping atomic orbitals.
• Molecular orbital theory: theory that predicts that electrons in molecules exist in allowed energy states.
• Molecular orbitals: allowed state for an electron in a molecule. Molecular orbitals are very similar to atomic orbitals. Both hold a maximum of two electrons (with opposite spins), have definite energies, and their electron-density distribution can be visualized by using contour representations. However, molecular orbitals are associated with the entire molecule
The Hydrogen Molecule
Consider H2 as an example. Whenever two atomic orbitals overlap, two molecular orbitals form. Thus, the overlap of the 1s orbitals of two hydrogen atoms to form H2 produces two molecular orbitals.
Bonding molecular orbital: molecular orbital in which the electron density is concentrated in the internuclear region. The energy of a bonding molecular orbital is lower than the energy of the separate atomic orbitals from which it forms.
The bonding molecular orbital results from summing the two atomic orbitals so that the atomic orbital wave functions enhance each other in the bond region. Because an electron in this molecular orbital is strongly attracted to both nuclei, the electron is more stable (lower energy) than it is in the 1s orbital of hydrogen. Because it concentrates electron density between both nuclei, the bonding molecular orbital holds the atoms together in a covalent bond.
Antibonding molecular orbital: molecular orbital in which electron density is concentrated outside the region between the two nuclei of bonded atoms. Such orbitals, designated as σ* or π*, are less stable (of higher energy) than bonding molecular orbitals
• An electron in the antibonding molecular orbital is repelled from the bonding region and is therefore less stable (higher energy) than it is in the 1s orbital of the hydrogen atom
• Sigma (σ) molecular orbital: molecular orbital that centers the electron density about an imaginary line passing through two nuclei
• Energy level diagram: aka Molecular orbital diagram; diagram that shows the energies of molecular orbitals relative to the atomic orbitals from which they are derived
• Electrons occupying a bonding molecular orbital are called bonding electrons
Bond Order
• Bond order: related to the stability of a covalent bond.
$\text{Bond order} = \dfrac{1}{2} (\text{number of bonding electrons – number of antibonding electrons})\nonumber$
A bond order of 1 represents a single bond, a bond order of 2 represents a double bond, and a bond order of 3 represents a triple bond. A bond order of 0 means that no bonds exist
9.8: Second-Row Diatomic Molecules
Molecular orbital energy-level diagrams for diatomic molecules can be created if the electron configuration of the parent atoms is known, following a few simple rules. Most important, the number of molecular orbitals in a molecule is the same as the number of atomic orbitals that interact. The difference between bonding and antibonding molecular orbital combinations is proportional to the overlap of the parent orbitals and decreases as the energy difference between the atomic orbitals increases.
Material here applies to homonuclear diatomic molecules (those composed of identical atoms)
1. The number of atomic orbitals formed equals the number of atomic orbitals combined
2. Atomic orbitals combine most effectively with other atomic orbitals of similar energy
3. The effectiveness with which two atomic orbitals combine is proportional to their overlap with one another; that is, as the overlap increases, the bonding orbital is lowered in energy, and the antibonding orbital is raised in energy
4. Each molecular orbital can accommodate at most two electrons, with their spins paired (Pauli exclusion principle)
5. When molecular orbitals have the same energy, one electron enters each orbital (with parallel spins) before spin pairing occurs (Hund’s rule)
Molecular Orbitals for Li2 and Be2
• Lithium has 1s and 2s orbitals, but both are extremely different. According to the Lewis structure of Li2, there is a single Li – Li bond.
• Because the 1s and 2s orbitals are so different in energy, it is likely that the 1s orbitals interact with each other. (See rule 2)
• This would create a much larger overlap between the 2s orbitals (see figure)
• As a result, the energy seperation between the σ2s and σ*2s is greater than that for the 1s-based orbitals.
• Each Li atom has 3 electrons; so six electrons must be placed in the molecular orbitals of Li2
• The electrons occupy the σ1s , σ*1s , and σ2s orbitals (not the σ*2s orbital). There are 4 bonding orbitals and 2 nonbonding ones, so the bond order equals ½ (4 – 2) = 1. The molecule has a single bond, in accord with its Lewis structure.
Filled atomic subshells usually do not contribute significantly to bonding in molecule formation
In Be2, each Be has 4 electrons, for a total of eight. When we place them in molecular orbitals, we fill the σ1s , σ*1s , σ2s , and σ*2s orbitals. There is an equal number of bonding and antibonding electrons, so the bond order is 0. Thus, Be2 doesn’t exist.
Molecular Orbitals from 2p Atomic Orbitals
• z axis is internuclear axis (in this example). 2pz orbitals face each other in a "head-to-head" fashion along the z-axis. They combine like the s orbitals, with one combination concentrating electron density between both nuclei, and the other excluding the electron density from the bonding region. In each of these combinations, the electron density lies along the internuclear axis. Hence, they are σ molecular orbitals: σ2p and σ*2p
• The other 2p orbitals overlap in a sideways fashion and thus concentrate electron density on opposite sides of the internuclear axis. Molecular orbitals of this type are called pi molecular orbitals . We get two π2p molecular orbitals by combining 2px and 2py molecular orbitals. The resulting π2p molecular orbitals are degenerate (same energy). Likewise, two degenerate π*2p antibonding molecular orbitals are formed.
• The 2pz orbitals on two atoms point directly at one another, hence, having a greater overlap than the 2px and 2py orbitals. From rule 3, we can expect the σ2p molecular orbital to be lower in energy (more stable) tha n the π2p molecular orbitals.
Electron Configurations for B2 Through F2
• The elements Boron through Fluorine all have 2s and 2p valence electrons
• In the energy diagram, the σ2p molecular orbital is at higher energy than the π2p molecular orbitals, contrary to what we would expect from rule 3.
• This is because of interaction between the σ2s and σ2p molecular orbitals. It results in the σ2p molecular orbital being pushed upward in energy to the point where it is above the π2p orbi tals, and the σ2s molecular orbital is pushed down in energy. (The interaction between σ 2p and σ 2s molecular orbitals decreases as we move from left to right in the second period. As a result, the σ 2p is higher in energy than the π 2p orbitals for B2, C2, and N2. For O2, F2, and Ne2 , the σ 2p orbital is lower in energy than the π 2p orbitals)
• Boron has 3 valence electrons. Thus, for B2, we must place 6 electrons in molecular orbitals. Four of these occupy the σ2s and σ*2s molecular orbitals, leading to no net bonding. The last two electrons are put in the π2p bonding molecular orbitals; one electron is put in each π2p molecular orbital with the same spin (rule 5). Therefore, B2 has a bond order of 1.
Electron Configurations and Molecular Properties
• Molecules with one or more unpaired electrons are attracted into a magnetic field. The more unpaired electrons, the stronger the force of the attraction
• Paramagnetism: property that a substance possesses if it contains one or more unpaired electrons. A paramagnetic substance is drawn into a magnetic field
• Diamagnetism: type of magnetism that causes a substance with no unpaired electrons to be weakly repelled from a magnetic field.
• Diamagnetism is a much weaker effect than paramagnetism
• As bond orders increase, bond distances decrease and bond-dissociation energies increase.
• Molecules with the same bond order do not have the same bond distances and bond-dissociation energies. There are other factors involved, such as nuclear charges and the extend of orbital overlap.
9 | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/09%3A_Molecular_Geometry_and_Bonding_Theories/9.S%3A_Molecular_Geometry_and_Bonding_Theories_%28Summary%29.txt |
Previously, we focused on the microscopic properties of matter—the properties of individual atoms, ions, and molecules—and how the electronic structures of atoms and ions determine the stoichiometry and three-dimensional geometry of the compounds they form. We will now focus on macroscopic properties—the behavior of aggregates with large numbers of atoms, ions, or molecules. An understanding of macroscopic properties is central to an understanding of chemistry. Why, for example, are many substances gases under normal pressures and temperatures (1.0 atm, 25°C), whereas others are liquids or solids? We will examine each form of matter—gases, liquids, and solids—as well as the nature of the forces, such as hydrogen bonding and electrostatic interactions, that hold molecular and ionic compounds together in these three states.
In this chapter, we explore the relationships among pressure, temperature, volume, and the amount of gases. You will learn how to use these relationships to describe the physical behavior of a sample of both a pure gaseous substance and mixtures of gases. By the end of this chapter, your understanding of the gas laws and the model used to explain the behavior of gases will allow you to explain how straws and hot-air balloons work, why hand pumps cannot be used in wells beyond a certain depth, why helium-filled balloons deflate so rapidly, and how a gas can be liquefied for use in preserving biological tissue.
• 10.1: Characteristics of Gases
Bulk matter can exist in three states: gas, liquid, and solid. Gases have the lowest density of the three, are highly compressible, and fill their containers completely. Elements that exist as gases at room temperature and pressure are clustered on the right side of the periodic table; they occur as either monatomic gases (the noble gases) or diatomic molecules (some halogens, N₂, O₂).
• 10.2: Pressure
Pressure is defined as the force exerted per unit area; it can be measured using a barometer or manometer. Four quantities must be known for a complete physical description of a sample of a gas: temperature, volume, amount, and pressure. Pressure is force per unit area of surface; the SI unit for pressure is the pascal (Pa), defined as 1 newton per square meter (N/m²). The pressure exerted by an object is proportional to the force it exerts and inversely proportional to the area.
• 10.3: The Gas Laws
The volume of a gas is inversely proportional to its pressure and directly proportional to its temperature and the amount of gas. Boyle showed that the volume of a sample of a gas is inversely proportional to pressure (Boyle’s law), Charles and Gay-Lussac demonstrated that the volume of a gas is directly proportional to its temperature at constant pressure (Charles’s law), and Avogadro showed that the volume of a gas is directly proportional to the number of moles of gas (Avogadro’s law).
• 10.4: The Ideal Gas Equation
The empirical relationships among the volume, the temperature, the pressure, and the amount of a gas can be combined into the ideal gas law, PV = nRT. The proportionality constant, R, is called the gas constant. The ideal gas law describes the behavior of an ideal gas, a hypothetical substance whose behavior can be explained quantitatively by the ideal gas law and the kinetic molecular theory of gases. Standard temperature and pressure (STP) is 0°C and 1 atm.
• 10.5: Further Applications of the Ideal-Gas Equations
The relationship between the amounts of products and reactants in a chemical reaction can be expressed in units of moles or masses of pure substances, of volumes of solutions, or of volumes of gaseous substances. The ideal gas law can be used to calculate the volume of gaseous products or reactants as needed. In the laboratory, gases produced in a reaction are often collected by the displacement of water from filled vessels; the amount of gas can be calculated from the volume of water displaced.
• 10.6: Gas Mixtures and Partial Pressures
The pressure exerted by each gas in a gas mixture is independent of the pressure exerted by all other gases present. Consequently, the total pressure exerted by a mixture of gases is the sum of the partial pressures of the components (Dalton’s law of partial pressures). The amount of gas in a mixture may be described by its partial pressure or its mole fraction. In a mixture, the partial pressure of each gas is the product of the total pressure and the mole fraction.
• 10.7: Kinetic-Molecular Theory
The behavior of ideal gases is explained by the kinetic molecular theory of gases. Molecular motion, which leads to collisions between molecules and the container walls, explains pressure, and the large intermolecular distances in gases explain their high compressibility. Although all gases have the same average kinetic energy at a given temperature, they do not all possess the same root mean square speed. The actual values of speed and kinetic energy are not the same for all gas particles.
• 10.8: Molecular Effusion and Diffusion
Diffusion is the gradual mixing of gases to form a sample of uniform composition even in the absence of mechanical agitation. In contrast, effusion is the escape of a gas from a container through a tiny opening into an evacuated space. The rate of effusion of a gas is inversely proportional to the square root of its molar mass (Graham’s law), a relationship that closely approximates the rate of diffusion. As a result, light gases tend to diffuse and effuse much more rapidly than heavier gases.
• 10.9: Real Gases - Deviations from Ideal Behavior
No real gas exhibits ideal gas behavior, although many real gases approximate it over a range of conditions. Gases most closely approximate ideal gas behavior at high temperatures and low pressures. Deviations from ideal gas law behavior can be described by the van der Waals equation, which includes empirical constants to correct for the actual volume of the gaseous molecules and quantify the reduction in pressure due to intermolecular attractive forces.
• 10.E: Exercises
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
• 10.S: Gases (Summary)
Thumbnail: Motion of gas molecules. The randomized thermal vibrations of fundamental particles such as atoms and molecules—gives a substance its “kinetic temperature.” Here, the size of helium atoms relative to their spacing is shown to scale under 1950 atmospheres of pressure. (CC BY-SA 3.0; Greg L).
10: Gases
Learning Objectives
• To describe the characteristics of a gas.
The three common phases (or states) of matter are gases, liquids, and solids. Gases have the lowest density of the three, are highly compressible, and completely fill any container in which they are placed. Gases behave this way because their intermolecular forces are relatively weak, so their molecules are constantly moving independently of the other molecules present. Solids, in contrast, are relatively dense, rigid, and incompressible because their intermolecular forces are so strong that the molecules are essentially locked in place. Liquids are relatively dense and incompressible, like solids, but they flow readily to adapt to the shape of their containers, like gases. We can therefore conclude that the sum of the intermolecular forces in liquids are between those of gases and solids. Figure \(1\) compares the three states of matter and illustrates the differences at the molecular level.
The state of a given substance depends strongly on conditions. For example, H2O is commonly found in all three states: solid ice, liquid water, and water vapor (its gaseous form). Under most conditions, we encounter water as the liquid that is essential for life; we drink it, cook with it, and bathe in it. When the temperature is cold enough to transform the liquid to ice, we can ski or skate on it, pack it into a snowball or snow cone, and even build dwellings with it. Water vapor (the term vapor refers to the gaseous form of a substance that is a liquid or a solid under normal conditions so nitrogen (N2) and oxygen (O2) are referred to as gases, but gaseous water in the atmosphere is called water vapor) is a component of the air we breathe, and it is produced whenever we heat water for cooking food or making coffee or tea. Water vapor at temperatures greater than 100°C is called steam. Steam is used to drive large machinery, including turbines that generate electricity. The properties of the three states of water are summarized in Table 10.1.
Table \(1\): Properties of Water at 1.0 atm
Temperature State Density (g/cm3)
≤0°C solid (ice) 0.9167 (at 0.0°C)
0°C–100°C liquid (water) 0.9997 (at 4.0°C)
≥100°C vapor (steam) 0.005476 (at 127°C)
The geometric structure and the physical and chemical properties of atoms, ions, and molecules usually do not depend on their physical state; the individual water molecules in ice, liquid water, and steam, for example, are all identical. In contrast, the macroscopic properties of a substance depend strongly on its physical state, which is determined by intermolecular forces and conditions such as temperature and pressure.
Figure \(2\) shows the locations in the periodic table of those elements that are commonly found in the gaseous, liquid, and solid states. Except for hydrogen, the elements that occur naturally as gases are on the right side of the periodic table. Of these, all the noble gases (group 18) are monatomic gases, whereas the other gaseous elements are diatomic molecules (H2, N2, O2, F2, and Cl2). Oxygen can also form a second allotrope, the highly reactive triatomic molecule ozone (O3), which is also a gas. In contrast, bromine (as Br2) and mercury (Hg) are liquids under normal conditions (25°C and 1.0 atm, commonly referred to as “room temperature and pressure”). Gallium (Ga), which melts at only 29.76°C, can be converted to a liquid simply by holding a container of it in your hand or keeping it in a non-air-conditioned room on a hot summer day. The rest of the elements are all solids under normal conditions.
All of the gaseous elements (other than the monatomic noble gases) are molecules. Within the same group (1, 15, 16 and 17), the lightest elements are gases. All gaseous substances are characterized by weak interactions between the constituent molecules or atoms.
Summary
Bulk matter can exist in three states: gas, liquid, and solid. Gases have the lowest density of the three, are highly compressible, and fill their containers completely. Elements that exist as gases at room temperature and pressure are clustered on the right side of the periodic table; they occur as either monatomic gases (the noble gases) or diatomic molecules (some halogens, N2, O2). | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/10%3A_Gases/10.01%3A_Characteristics_of_Gases.txt |
Learning Objectives
• to describe and measure the pressure of a gas.
At the macroscopic level, a complete physical description of a sample of a gas requires four quantities:
• temperature (expressed in kelvins),
• volume (expressed in liters),
• amount (expressed in moles), and
• pressure (in atmospheres).
As we demonstrated below, these variables are not independent (i.e., they cannot be arbitrarily be varied). If we know the values of any three of these quantities, we can calculate the fourth and thereby obtain a full physical description of the gas. Temperature, volume, and amount have been discussed in previous chapters. We now discuss pressure and its units of measurement.
Units of Pressure
Any object, whether it is your computer, a person, or a sample of gas, exerts a force on any surface with which it comes in contact. The air in a balloon, for example, exerts a force against the interior surface of the balloon, and a liquid injected into a mold exerts a force against the interior surface of the mold, just as a chair exerts a force against the floor because of its mass and the effects of gravity. If the air in a balloon is heated, the increased kinetic energy of the gas eventually causes the balloon to burst because of the increased pressure ($P$) of the gas, the force ($F$) per unit area ($A$) of surface:
$P=\dfrac{\rm Force}{\rm Area}=\dfrac{F}{A}\label{10.2.1}$
Pressure is dependent on both the force exerted and the size of the area to which the force is applied. We know from Equation $\ref{10.2.1}$ that applying the same force to a smaller area produces a higher pressure. When we use a hose to wash a car, for example, we can increase the pressure of the water by reducing the size of the opening of the hose with a thumb.
The units of pressure are derived from the units used to measure force and area. The SI unit for pressure, derived from the SI units for force (newtons) and area (square meters), is the newton per square meter ($N/m^2$), which is called the Pascal (Pa), after the French mathematician Blaise Pascal (1623–1662):
$\rm 1 \;Pa=1\;N/m^2 \label{10.2.2}$
Example $1$
Assuming a paperback book has a mass of 2.00 kg, a length of 27.0 cm, a width of 21.0 cm, and a thickness of 4.5 cm, what pressure does it exert on a surface if it is
1. lying flat?
2. standing on edge in a bookcase?
Given: mass and dimensions of object
Asked for: pressure
Strategy:
1. Calculate the force exerted by the book and then compute the area that is in contact with a surface.
2. Substitute these two values into Equation $\ref{10.2.1}$ to find the pressure exerted on the surface in each orientation.
Solution:
The force exerted by the book does not depend on its orientation. Recall that the force exerted by an object is F = ma, where m is its mass and a is its acceleration. In Earth’s gravitational field, the acceleration is due to gravity (9.8067 m/s2 at Earth’s surface). In SI units, the force exerted by the book is therefore
$F = ma = 2.00 \;\rm kg\times 9.8067 \dfrac{\rm m}{\rm s^2} = 19.6 \dfrac{\rm kg·m}{\rm s^2} = 19.6\;\rm N \nonumber$
A We calculated the force as 19.6 N. When the book is lying flat, the area is
$A=\rm0.270 \;m\times0.210 \;m= 0.0567 \;m^2. \nonumber$
B The pressure exerted by the text lying flat is thus
$P=\dfrac{F}{A}=\dfrac{19.6\;\rm N}{0.0567\;\rm m^2}=3.46\times10^2 \rm Pa \nonumber$
A If the book is standing on its end, the force remains the same, but the area decreases:
$\rm A=\rm21.0 \;cm\times4.5 \;cm = 0.210 \;m\times0.045 \;m = 9.5 \times 10^{−3} \;\rm m^2 \nonumber$
B The pressure exerted by the text lying flat is thus
$P=\dfrac{19.6\;\rm N}{9.5\times10^{-3}\;\rm m^2}=2.06\times10^3 \;\rm Pa \nonumber$
Exercise $1$
What pressure does a 60.0 kg student exert on the floor
1. when standing flat-footed in the laboratory in a pair of tennis shoes (the surface area of the soles is approximately 180 cm2)?
2. as she steps heel-first onto a dance floor wearing high-heeled shoes (the area of the heel = 1.0 cm2)?
Answer a
3.27 × 104 Pa
Answer b
5.9 × 106 Pa
Barometric Pressure
Just as we exert pressure on a surface because of gravity, so does our atmosphere. We live at the bottom of an ocean of gases that becomes progressively less dense with increasing altitude. Approximately 99% of the mass of the atmosphere lies within 30 km of Earth’s surface (Figure $1$). Every point on Earth’s surface experiences a net pressure called barometric pressure. The pressure exerted by the atmosphere is considerable: a 1 m2 column, measured from sea level to the top of the atmosphere, has a mass of about 10,000 kg, which gives a pressure of about 101 kPa:
Barometric pressure can be measured using a barometer, a device invented in 1643 by one of Galileo’s students, Evangelista Torricelli (1608–1647). A barometer may be constructed from a long glass tube that is closed at one end. It is filled with mercury and placed upside down in a dish of mercury without allowing any air to enter the tube. Some of the mercury will run out of the tube, but a relatively tall column remains inside (Figure $2$). Why doesn’t all the mercury run out? Gravity is certainly exerting a downward force on the mercury in the tube, but it is opposed by the pressure of the atmosphere pushing down on the surface of the mercury in the dish, which has the net effect of pushing the mercury up into the tube. Because there is no air above the mercury inside the tube in a properly filled barometer (it contains a vacuum), there is no pressure pushing down on the column. Thus the mercury runs out of the tube until the pressure exerted by the mercury column itself exactly balances the pressure of the atmosphere. The pressure exerted by the mercury column can be expressed as:
\begin{align} P&=\dfrac{F}{A} \[4pt] &= \dfrac{mg}{A} \[4pt] &= \dfrac{\rho V\cdot g}{A} \[4pt] &= \dfrac{ \rho \cdot Ah\cdot g}{A} \[4pt] &= \rho gh \end{align} \nonumber
with
• $g$ is the gravitational acceleration,
• $m$ is the mass,
• $\rho$ is the density,
• $V$ is the volume,
• $A$ is the bottom area, and
• $h$ is height of the mercury column.
Under normal weather conditions at sea level, the two forces are balanced when the top of the mercury column is approximately 760 mm above the level of the mercury in the dish, as shown in Figure $2$. This value varies with meteorological conditions and altitude. In Denver, Colorado, for example, at an elevation of about 1 mile, or 1609 m (5280 ft), the height of the mercury column is 630 mm rather than 760 mm.
Mercury barometers have been used to measure barometric pressure for so long that they have their own unit for pressure: the millimeter of mercury (mmHg), often called the torr, after Torricelli. Standard barometric pressure is the barometric pressure required to support a column of mercury exactly 760 mm tall; this pressure is also referred to as 1 atmosphere (atm). These units are also related to the pascal:
\begin{align} \rm 1\; atm &= 760 \; mmHg \[4pt] &= 760 \; torr \[4pt] &= 1.01325 \times 10^5 \; Pa \[4pt] &= 101.325 \; kPa\label{10.2.3} \end{align}
Thus a pressure of 1 atm equals 760 mmHg exactly.
We are so accustomed to living under this pressure that we never notice it. Instead, what we notice are changes in the pressure, such as when our ears pop in fast elevators in skyscrapers or in airplanes during rapid changes in altitude. We make use of barometric pressure in many ways. We can use a drinking straw because sucking on it removes air and thereby reduces the pressure inside the straw. The barometric pressure pushing down on the liquid in the glass then forces the liquid up the straw.
Example $2$: Barometric Pressure
One of the authors visited Rocky Mountain National Park several years ago. After departing from an airport at sea level in the eastern United States, he arrived in Denver (altitude 5280 ft), rented a car, and drove to the top of the highway outside Estes Park (elevation 14,000 ft). He noticed that even slight exertion was very difficult at this altitude, where the barometric pressure is only 454 mmHg. Convert this pressure to
1. atmospheres (atm).
2. bar.
Given: pressure in millimeters of mercury
Asked for: pressure in atmospheres and bar
Strategy:
Use the conversion factors in Equation $\ref{10.2.3}$ to convert from millimeters of mercury to atmospheres and kilopascals.
Solution:
From Equation $\ref{10.2.3}$, we have 1 atm = 760 mmHg = 101.325 kPa. The pressure at 14,000 ft in atm is thus
\begin{align} P &=\rm 454 \;mmHg\times\dfrac{1\;atm}{760\;mmHg} \[4pt] &= 0.597\;atm \nonumber \end{align} \nonumber
The pressure in bar is given by
\begin{align} P&=\rm 0.597\;atm\times\dfrac{1.01325\;bar}{1\;atm}\[4pt] &= 0.605\;bar \nonumber \end{align} \nonumber
Exercise $2$: Barometric Pressure
Mt. Everest, at 29,028 ft above sea level, is the world’s tallest mountain. The normal barometric pressure at this altitude is about 0.308 atm. Convert this pressure to
1. millimeters of mercury.
2. bar.
Answer a
234 mmHg;
Answer b
0.312 bar
Manometers
Barometers measure barometric pressure, but manometers measure the pressures of samples of gases contained in an apparatus. The key feature of a manometer is a U-shaped tube containing mercury (or occasionally another nonvolatile liquid). A closed-end manometer is shown schematically in part (a) in Figure $3$. When the bulb contains no gas (i.e., when its interior is a near vacuum), the heights of the two columns of mercury are the same because the space above the mercury on the left is a near vacuum (it contains only traces of mercury vapor). If a gas is released into the bulb on the right, it will exert a pressure on the mercury in the right column, and the two columns of mercury will no longer be the same height. The difference between the heights of the two columns is equal to the pressure of the gas.
If the tube is open to the atmosphere instead of closed, as in the open-end manometer shown in part (b) in Figure $3$, then the two columns of mercury have the same height only if the gas in the bulb has a pressure equal to the barometric pressure. If the gas in the bulb has a higher pressure, the mercury in the open tube will be forced up by the gas pushing down on the mercury in the other arm of the U-shaped tube. The pressure of the gas in the bulb is therefore the sum of the barometric pressure (measured with a barometer) and the difference in the heights of the two columns. If the gas in the bulb has a pressure less than that of the atmosphere, then the height of the mercury will be greater in the arm attached to the bulb. In this case, the pressure of the gas in the bulb is the barometric pressure minus the difference in the heights of the two columns.
Example $3$
Suppose you want to construct a closed-end manometer to measure gas pressures in the range 0.000–0.200 atm. Because of the toxicity of mercury, you decide to use water rather than mercury. How tall a column of water do you need? (The density of water is 1.00 g/cm3; the density of mercury is 13.53 g/cm3.)
Given: pressure range and densities of water and mercury
Asked for: column height
Strategy:
1. Calculate the height of a column of mercury corresponding to 0.200 atm in millimeters of mercury. This is the height needed for a mercury-filled column.
2. From the given densities, use a proportion to compute the height needed for a water-filled column.
Solution:
A In millimeters of mercury, a gas pressure of 0.200 atm is
$P=\rm 0.200\;atm\times\dfrac{760\;mmHg}{1\;atm}=152\;mmHg \nonumber$
Using a mercury manometer, you would need a mercury column at least 152 mm high.
B Because water is less dense than mercury, you need a taller column of water to achieve the same pressure as a given column of mercury. The height needed for a water-filled column corresponding to a pressure of 0.200 atm is proportional to the ratio of the density of mercury to the density of water
$P=d_{\rm wat}gh_{\rm wat}=d_{\rm Hg}gh_{\rm Hg} \nonumber$
$h_{\rm wat}=h_{\rm Hg}\times\dfrac{d_{\rm Hg}}{g_{\rm wat}}=\rm152\;mm\times\dfrac{13.53\;g/cm^3}{1.00\;g/cm^3}=2070\;mm \nonumber$
The answer makes sense: it takes a taller column of a less dense liquid to achieve the same pressure.
Exercise $3$
Suppose you want to design a barometer to measure barometric pressure in an environment that is always hotter than 30°C. To avoid using mercury, you decide to use gallium, which melts at 29.76°C; the density of liquid gallium at 25°C is 6.114 g/cm3. How tall a column of gallium do you need if P = 1.00 atm?
Answer
1.68 m
The answer to Example $3$ also tells us the maximum depth of a farmer’s well if a simple suction pump will be used to get the water out. The 1.00 atm corresponds to a column height of
\begin{align} h_{\rm wat} &=h_{\rm Hg}\times\dfrac{d_{\rm Hg}}{g_{\rm wat}} \nonumber \[4pt] &=\rm760\;mm\times\dfrac{13.53\;g/cm^3}{1.00\;g/cm^3} \nonumber \[4pt] &= 1.03\times10^4\;mm \nonumber \[4pt] &= 10.3\;m \nonumber \end{align} \nonumber
A suction pump is just a more sophisticated version of a straw: it creates a vacuum above a liquid and relies on barometric pressure to force the liquid up a tube. If 1 atm pressure corresponds to a 10.3 m (33.8 ft) column of water, then it is physically impossible for barometric pressure to raise the water in a well higher than this. Until electric pumps were invented to push water mechanically from greater depths, this factor greatly limited where people could live because obtaining water from wells deeper than about 33 ft was difficult.
Summary
Pressure is defined as the force exerted per unit area; it can be measured using a barometer or manometer. Four quantities must be known for a complete physical description of a sample of a gas: temperature, volume, amount, and pressure. Pressure is force per unit area of surface; the SI unit for pressure is the pascal (Pa), defined as 1 newton per square meter (N/m2). The pressure exerted by an object is proportional to the force it exerts and inversely proportional to the area on which the force is exerted. The pressure exerted by Earth’s atmosphere, called barometric pressure, is about 101 kPa or 14.7 lb/in.2 at sea level. barometric pressure can be measured with a barometer, a closed, inverted tube filled with mercury. The height of the mercury column is proportional to barometric pressure, which is often reported in units of millimeters of mercury (mmHg), also called torr. Standard barometric pressure, the pressure required to support a column of mercury 760 mm tall, is yet another unit of pressure: 1 atmosphere (atm). A manometer is an apparatus used to measure the pressure of a sample of a gas. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/10%3A_Gases/10.02%3A_Pressure.txt |
Learning Objectives
• To understand the relationships among pressure, temperature, volume, and the amount of a gas.
Early scientists explored the relationships among the pressure of a gas (P) and its temperature (T), volume (V), and amount (n) by holding two of the four variables constant (amount and temperature, for example), varying a third (such as pressure), and measuring the effect of the change on the fourth (in this case, volume). The history of their discoveries provides several excellent examples of the scientific method.
The Relationship between Pressure and Volume: Boyle's Law
As the pressure on a gas increases, the volume of the gas decreases because the gas particles are forced closer together. Conversely, as the pressure on a gas decreases, the gas volume increases because the gas particles can now move farther apart. Weather balloons get larger as they rise through the atmosphere to regions of lower pressure because the volume of the gas has increased; that is, the atmospheric gas exerts less pressure on the surface of the balloon, so the interior gas expands until the internal and external pressures are equal.
The Irish chemist Robert Boyle (1627–1691) carried out some of the earliest experiments that determined the quantitative relationship between the pressure and the volume of a gas. Boyle used a J-shaped tube partially filled with mercury, as shown in Figure $1$. In these experiments, a small amount of a gas or air is trapped above the mercury column, and its volume is measured at atmospheric pressure and constant temperature. More mercury is then poured into the open arm to increase the pressure on the gas sample. The pressure on the gas is atmospheric pressure plus the difference in the heights of the mercury columns, and the resulting volume is measured. This process is repeated until either there is no more room in the open arm or the volume of the gas is too small to be measured accurately. Data such as those from one of Boyle’s own experiments may be plotted in several ways (Figure $2$). A simple plot of $V$ versus $P$ gives a curve called a hyperbola and reveals an inverse relationship between pressure and volume: as the pressure is doubled, the volume decreases by a factor of two. This relationship between the two quantities is described as follows:
$PV = \rm constant \label{10.3.1}$
Dividing both sides by $P$ gives an equation illustrating the inverse relationship between $P$ and $V$:
$V=\dfrac{\rm const.}{P} = {\rm const.}\left(\dfrac{1}{P}\right) \label{10.3.2}$
or
$V \propto \dfrac{1}{P} \label{10.3.3}$
where the ∝ symbol is read “is proportional to.” A plot of V versus 1/P is thus a straight line whose slope is equal to the constant in Equations $\ref{10.3.1}$ and $\ref{10.3.3}$. Dividing both sides of Equation $\ref{10.3.1}$ by V instead of P gives a similar relationship between P and 1/V. The numerical value of the constant depends on the amount of gas used in the experiment and on the temperature at which the experiments are carried out. This relationship between pressure and volume is known as Boyle’s law, after its discoverer, and can be stated as follows: At constant temperature, the volume of a fixed amount of a gas is inversely proportional to its pressure. This law in practice is shown in Figure $2$.
At constant temperature, the volume of a fixed amount of a gas is inversely proportional to its pressure
The Relationship between Temperature and Volume: Charles's Law
Hot air rises, which is why hot-air balloons ascend through the atmosphere and why warm air collects near the ceiling and cooler air collects at ground level. Because of this behavior, heating registers are placed on or near the floor, and vents for air-conditioning are placed on or near the ceiling. The fundamental reason for this behavior is that gases expand when they are heated. Because the same amount of substance now occupies a greater volume, hot air is less dense than cold air. The substance with the lower density—in this case hot air—rises through the substance with the higher density, the cooler air.
The first experiments to quantify the relationship between the temperature and the volume of a gas were carried out in 1783 by an avid balloonist, the French chemist Jacques Alexandre César Charles (1746–1823). Charles’s initial experiments showed that a plot of the volume of a given sample of gas versus temperature (in degrees Celsius) at constant pressure is a straight line. Similar but more precise studies were carried out by another balloon enthusiast, the Frenchman Joseph-Louis Gay-Lussac (1778–1850), who showed that a plot of V versus T was a straight line that could be extrapolated to a point at zero volume, a theoretical condition now known to correspond to −273.15°C (Figure $3$).A sample of gas cannot really have a volume of zero because any sample of matter must have some volume. Furthermore, at 1 atm pressure all gases liquefy at temperatures well above −273.15°C. Note from part (a) in Figure $3$ that the slope of the plot of V versus T varies for the same gas at different pressures but that the intercept remains constant at −273.15°C. Similarly, as shown in part (b) in Figure $3$, plots of V versus T for different amounts of varied gases are straight lines with different slopes but the same intercept on the T axis.
The significance of the invariant T intercept in plots of V versus T was recognized in 1848 by the British physicist William Thomson (1824–1907), later named Lord Kelvin. He postulated that −273.15°C was the lowest possible temperature that could theoretically be achieved, for which he coined the term absolute zero (0 K).
We can state Charles’s and Gay-Lussac’s findings in simple terms: At constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature (in kelvins). This relationship, illustrated in part (b) in Figure $3$ is often referred to as Charles’s law and is stated mathematically as
$V ={\rm const.}\; T \label{10.3.4}$
or
$V \propto T \label{10.3.5}$
with temperature expressed in kelvins, not in degrees Celsius. Charles’s law is valid for virtually all gases at temperatures well above their boiling points.
The Relationship between Amount and Volume: Avogadro's Law
We can demonstrate the relationship between the volume and the amount of a gas by filling a balloon; as we add more gas, the balloon gets larger. The specific quantitative relationship was discovered by the Italian chemist Amedeo Avogadro, who recognized the importance of Gay-Lussac’s work on combining volumes of gases. In 1811, Avogadro postulated that, at the same temperature and pressure, equal volumes of gases contain the same number of gaseous particles (Figure $4$). This is the historic “Avogadro’s hypothesis.”
A logical corollary to Avogadro's hypothesis (sometimes called Avogadro’s law) describes the relationship between the volume and the amount of a gas: At constant temperature and pressure, the volume of a sample of gas is directly proportional to the number of moles of gas in the sample. Stated mathematically,
$V ={\rm const.} \; (n) \label{10.3.6}$
or
$V \propto.n \text{@ constant T and P} \label{10.3.7}$
This relationship is valid for most gases at relatively low pressures, but deviations from strict linearity are observed at elevated pressures.
For a sample of gas,
• V increases as P decreases (and vice versa)
• V increases as T increases (and vice versa)
• V increases as n increases (and vice versa)
The relationships among the volume of a gas and its pressure, temperature, and amount are summarized in Figure $5$. Volume increases with increasing temperature or amount, but decreases with increasing pressure.
Summary
The volume of a gas is inversely proportional to its pressure and directly proportional to its temperature and the amount of gas. Boyle showed that the volume of a sample of a gas is inversely proportional to its pressure (Boyle’s law), Charles and Gay-Lussac demonstrated that the volume of a gas is directly proportional to its temperature (in kelvins) at constant pressure (Charles’s law), and Avogadro postulated that the volume of a gas is directly proportional to the number of moles of gas present (Avogadro’s law). Plots of the volume of gases versus temperature extrapolate to zero volume at −273.15°C, which is absolute zero (0 K), the lowest temperature possible. Charles’s law implies that the volume of a gas is directly proportional to its absolute temperature. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/10%3A_Gases/10.03%3A_The_Gas_Laws.txt |
Learning Objectives
• Derive the ideal gas law from the constituent gas laws
• To use the ideal gas law to describe the behavior of a gas.
In this module, the relationship between pressure, temperature, volume, and amount of a gas are described and how these relationships can be combined to give a general expression that describes the behavior of a gas.
Deriving the Ideal Gas Law
Any set of relationships between a single quantity (such as $V$) and several other variables ($P$, $T$, and $n$) can be combined into a single expression that describes all the relationships simultaneously. The three individual expressions were derived previously:
• Boyle’s law
$V \propto \dfrac{1}{P} \;\; \text{@ constant n and T} \nonumber$
• Charles’s law
$V \propto T \;\; \text{@ constant n and P} \nonumber$
• Avogadro’s law
$V \propto n \;\; \text{@ constant T and P} \nonumber$
Combining these three expressions gives
$V \propto \dfrac{nT}{P} \label{10.4.1}$
which shows that the volume of a gas is proportional to the number of moles and the temperature and inversely proportional to the pressure. This expression can also be written as
$V= {\rm Cons.} \left( \dfrac{nT}{P} \right) \label{10.4.2}$
By convention, the proportionality constant in Equation $\ref{10.4.1}$ is called the gas constant, which is represented by the letter $R$. Inserting R into Equation $\ref{10.4.2}$ gives
$V = \dfrac{RnT}{P} = \dfrac{nRT}{P} \label{10.4.3}$
Clearing the fractions by multiplying both sides of Equation $\ref{10.4.4}$ by $P$ gives
$PV = nRT \label{10.4.4}$
This equation is known as the ideal gas law.
An ideal gas is defined as a hypothetical gaseous substance whose behavior is independent of attractive and repulsive forces and can be completely described by the ideal gas law. In reality, there is no such thing as an ideal gas, but an ideal gas is a useful conceptual model that allows us to understand how gases respond to changing conditions. As we shall see, under many conditions, most real gases exhibit behavior that closely approximates that of an ideal gas. The ideal gas law can therefore be used to predict the behavior of real gases under most conditions. The ideal gas law does not work well at very low temperatures or very high pressures, where deviations from ideal behavior are most commonly observed.
Significant deviations from ideal gas behavior commonly occur at low temperatures and very high pressures.
Before we can use the ideal gas law, however, we need to know the value of the gas constant R. Its form depends on the units used for the other quantities in the expression. If V is expressed in liters (L), P in atmospheres (atm), T in kelvins (K), and n in moles (mol), then
$R = 0.08206 \dfrac{\rm L\cdot atm}{\rm K\cdot mol} \label{10.4.5}$
Because the product PV has the units of energy, R can also have units of J/(K•mol):
$R = 8.3145 \dfrac{\rm J}{\rm K\cdot mol}\label{10.4.6}$
Standard Conditions of Temperature and Pressure
Scientists have chosen a particular set of conditions to use as a reference: 0°C (273.15 K) and $\rm1\; bar = 100 \;kPa = 10^5\;Pa$ pressure, referred to as standard temperature and pressure (STP).
$\text{STP:} \hspace{2cm} T=273.15\;{\rm K}\text{ and }P=\rm 1\;bar=10^5\;Pa \nonumber$
Please note that STP was defined differently in the past. The old definition was based on a standard pressure of 1 atm.
We can calculate the volume of 1.000 mol of an ideal gas under standard conditions using the variant of the ideal gas law given in Equation $\ref{10.4.4}$:
$V=\dfrac{nRT}{P}\label{10.4.7}$
Thus the volume of 1 mol of an ideal gas is 22.71 L at STP and 22.41 L at 0°C and 1 atm, approximately equivalent to the volume of three basketballs. The molar volumes of several real gases at 0°C and 1 atm are given in Table 10.3, which shows that the deviations from ideal gas behavior are quite small. Thus the ideal gas law does a good job of approximating the behavior of real gases at 0°C and 1 atm. The relationships described in Section 10.3 as Boyle’s, Charles’s, and Avogadro’s laws are simply special cases of the ideal gas law in which two of the four parameters (P, V, T, and n) are held fixed.
Table $1$: Molar Volumes of Selected Gases at 0°C and 1 atm
Gas Molar Volume (L)
He 22.434
Ar 22.397
H2 22.433
N2 22.402
O2 22.397
CO2 22.260
NH3 22.079
Applying the Ideal Gas Law
The ideal gas law allows us to calculate the value of the fourth variable for a gaseous sample if we know the values of any three of the four variables (P, V, T, and n). It also allows us to predict the final state of a sample of a gas (i.e., its final temperature, pressure, volume, and amount) following any changes in conditions if the parameters (P, V, T, and n) are specified for an initial state. Some applications are illustrated in the following examples. The approach used throughout is always to start with the same equation—the ideal gas law—and then determine which quantities are given and which need to be calculated. Let’s begin with simple cases in which we are given three of the four parameters needed for a complete physical description of a gaseous sample.
Example $1$
The balloon that Charles used for his initial flight in 1783 was destroyed, but we can estimate that its volume was 31,150 L (1100 ft3), given the dimensions recorded at the time. If the temperature at ground level was 86°F (30°C) and the atmospheric pressure was 745 mmHg, how many moles of hydrogen gas were needed to fill the balloon?
Given: volume, temperature, and pressure
Asked for: amount of gas
Strategy:
1. Solve the ideal gas law for the unknown quantity, in this case n.
2. Make sure that all quantities are given in units that are compatible with the units of the gas constant. If necessary, convert them to the appropriate units, insert them into the equation you have derived, and then calculate the number of moles of hydrogen gas needed.
Solution:
A We are given values for P, T, and V and asked to calculate n. If we solve the ideal gas law (Equation $\ref{10.4.4}$) for $n$, we obtain
$\rm745\;mmHg\times\dfrac{1\;atm}{760\;mmHg}=0.980\;atm \nonumber$
B P and T are given in units that are not compatible with the units of the gas constant [R = 0.08206 (L•atm)/(K•mol)]. We must therefore convert the temperature to kelvins and the pressure to atmospheres:
$T=273+30=303{\rm K}\nonumber$
Substituting these values into the expression we derived for n, we obtain
\begin{align*} n &=\dfrac{PV}{RT} \[4pt] &=\rm\dfrac{0.980\;atm\times31150\;L}{0.08206\dfrac{atm\cdot L}{\rm mol\cdot K}\times 303\;K} \[4pt] &=1.23\times10^3\;mol \end{align*} \nonumber
Exercise $1$
Suppose that an “empty” aerosol spray-paint can has a volume of 0.406 L and contains 0.025 mol of a propellant gas such as CO2. What is the pressure of the gas at 25°C?
Answer
1.5 atm
In Example $1$, we were given three of the four parameters needed to describe a gas under a particular set of conditions, and we were asked to calculate the fourth. We can also use the ideal gas law to calculate the effect of changes in any of the specified conditions on any of the other parameters, as shown in Example $5$.
General Gas Equation
When a gas is described under two different conditions, the ideal gas equation must be applied twice - to an initial condition and a final condition. This is:
$\begin{array}{cc}\text{Initial condition }(i) & \text{Final condition} (f) \ P_iV_i=n_iRT_i & P_fV_f=n_fRT_f\end{array} \nonumber$
Both equations can be rearranged to give:
$R=\dfrac{P_iV_i}{n_iT_i} \hspace{1cm} R=\dfrac{P_fV_f}{n_fT_f} \nonumber$
The two equations are equal to each other since each is equal to the same constant $R$. Therefore, we have:
$\dfrac{P_iV_i}{n_iT_i}=\dfrac{P_fV_f}{n_fT_f}\label{10.4.8}$
The equation is called the general gas equation. The equation is particularly useful when one or two of the gas properties are held constant between the two conditions. In such cases, the equation can be simplified by eliminating these constant gas properties.
Example $2$
Suppose that Charles had changed his plans and carried out his initial flight not in August but on a cold day in January, when the temperature at ground level was −10°C (14°F). How large a balloon would he have needed to contain the same amount of hydrogen gas at the same pressure as in Example $1$?
Given: temperature, pressure, amount, and volume in August; temperature in January
Asked for: volume in January
Strategy:
1. Use the results from Example $1$ for August as the initial conditions and then calculate the change in volume due to the change in temperature from 30°C to −10°C. Begin by constructing a table showing the initial and final conditions.
2. Simplify the general gas equation by eliminating the quantities that are held constant between the initial and final conditions, in this case $P$ and $n$.
3. Solve for the unknown parameter.
Solution:
A To see exactly which parameters have changed and which are constant, prepare a table of the initial and final conditions:
Solution to Example 10.4.2
Initial (August) Final (January)
$T_i=30\, °C = 303\, K$ $T_f=−10\,°C = 263\, K$
$P_i= 0.980 \, atm$ $P_f= 0.980\, atm$
$n_i=1.23 × 10^3\, mol$ $n_f= 1.23 × 10^3\, mol$
$V_i=31150\, L$ $V_f=?$
B Both $n$ and $P$ are the same in both cases ($n_i=n_f,P_i=P_f$). Therefore, Equation \ref{10.4.8} can be simplified to:
$\dfrac{V_i}{T_i}=\dfrac{V_f}{T_f} \nonumber$
This is the relationship first noted by Charles.
C Solving the equation for $V_f$, we get:
\begin{align*} V_f &=V_i\times\dfrac{T_f}{T_i} \[4pt] &=\rm31150\;L\times\dfrac{263\;K}{303\;K} \[4pt] &=2.70\times10^4\;L \end{align*} \nonumber
It is important to check your answer to be sure that it makes sense, just in case you have accidentally inverted a quantity or multiplied rather than divided. In this case, the temperature of the gas decreases. Because we know that gas volume decreases with decreasing temperature, the final volume must be less than the initial volume, so the answer makes sense. We could have calculated the new volume by plugging all the given numbers into the ideal gas law, but it is generally much easier and faster to focus on only the quantities that change.
Exercise $2$
At a laboratory party, a helium-filled balloon with a volume of 2.00 L at 22°C is dropped into a large container of liquid nitrogen (T = −196°C). What is the final volume of the gas in the balloon?
Answer
0.52 L
Example $1$ illustrates the relationship originally observed by Charles. We could work through similar examples illustrating the inverse relationship between pressure and volume noted by Boyle (PV = constant) and the relationship between volume and amount observed by Avogadro (V/n = constant). We will not do so, however, because it is more important to note that the historically important gas laws are only special cases of the ideal gas law in which two quantities are varied while the other two remain fixed. The method used in Example $1$ can be applied in any such case, as we demonstrate in Example $2$ (which also shows why heating a closed container of a gas, such as a butane lighter cartridge or an aerosol can, may cause an explosion).
Example $3$
Aerosol cans are prominently labeled with a warning such as “Do not incinerate this container when empty.” Assume that you did not notice this warning and tossed the “empty” aerosol can in Exercise $1$ (0.025 mol in 0.406 L, initially at 25°C and 1.5 atm internal pressure) into a fire at 750°C. What would be the pressure inside the can (if it did not explode)?
Given: initial volume, amount, temperature, and pressure; final temperature
Asked for: final pressure
Strategy:
Follow the strategy outlined in Example $2$.
Solution:
Prepare a table to determine which parameters change and which are held constant:
Solution to Example 10.4.3
Initial Final
$V_i=0.406\;\rm L$ $V_f=0.406\;\rm L$
$n_i=0.025\;\rm mol$ $n_f=0.025\;\rm mol$
$T_i=\rm25\;^\circ C=298\;K$ $T_i=\rm750\;^\circ C=1023\;K$
$P_i=1.5\;\rm atm$ $P_f=?$
Both $V$ and $n$ are the same in both cases ($V_i=V_f,n_i=n_f$). Therefore, Equation can be simplified to:
$\dfrac{P_i}{T_i} = \dfrac{P_f}{T_f} \nonumber$
By solving the equation for $P_f$, we get:
\begin{align*} P_f &=P_i\times\dfrac{T_f}{T_i} \[4pt] &=\rm1.5\;atm\times\dfrac{1023\;K}{298\;K} \[4pt] &=5.1\;atm \end{align*} \nonumber
This pressure is more than enough to rupture a thin sheet metal container and cause an explosion!
Exercise $3$
Suppose that a fire extinguisher, filled with CO2 to a pressure of 20.0 atm at 21°C at the factory, is accidentally left in the sun in a closed automobile in Tucson, Arizona, in July. The interior temperature of the car rises to 160°F (71.1°C). What is the internal pressure in the fire extinguisher?
Answer
23.4 atm
In Examples $1$ and $2$, two of the four parameters (P, V, T, and n) were fixed while one was allowed to vary, and we were interested in the effect on the value of the fourth. In fact, we often encounter cases where two of the variables P, V, and T are allowed to vary for a given sample of gas (hence n is constant), and we are interested in the change in the value of the third under the new conditions.
Example $4$
We saw in Example $1$ that Charles used a balloon with a volume of 31,150 L for his initial ascent and that the balloon contained 1.23 × 103 mol of H2 gas initially at 30°C and 745 mmHg. Suppose that Gay-Lussac had also used this balloon for his record-breaking ascent to 23,000 ft and that the pressure and temperature at that altitude were 312 mmHg and −30°C, respectively. To what volume would the balloon have had to expand to hold the same amount of hydrogen gas at the higher altitude?
Given: initial pressure, temperature, amount, and volume; final pressure and temperature
Asked for: final volume
Strategy:
Follow the strategy outlined in Example $3$.
Solution:
Begin by setting up a table of the two sets of conditions:
Solution to Example 10.4.4
Initial Final
$P_i=745\;\rm mmHg=0.980\;atm$ $P_f=312\;\rm mmHg=0.411\;atm$
$T_i=\rm30\;^\circ C=303\;K$ $T_f=\rm750-30\;^\circ C=243\;K$
$n_i=\rm1.2\times10^3\;mol$ $n_i=\rm1.2\times10^3\;mol$
$V_i=\rm31150\;L$ $V_f=?$
By eliminating the constant property ($n$) of the gas, Equation $\ref{10.4.8}$ is simplified to:
$\dfrac{P_iV_i}{T_i}=\dfrac{P_fV_f}{T_f} \nonumber$
By solving the equation for $V_f$, we get:
\begin{align*} V_f &=V_i\times\dfrac{P_i}{P_f}\dfrac{T_f}{T_i} \[4pt] &=\rm3.115\times10^4\;L\times\dfrac{0.980\;atm}{0.411\;atm}\dfrac{243\;K}{303\;K} \[4pt] &=5.96\times10^4\;L \end{align*} \nonumber
Does this answer make sense? Two opposing factors are at work in this problem: decreasing the pressure tends to increase the volume of the gas, while decreasing the temperature tends to decrease the volume of the gas. Which do we expect to predominate? The pressure drops by more than a factor of two, while the absolute temperature drops by only about 20%. Because the volume of a gas sample is directly proportional to both T and 1/P, the variable that changes the most will have the greatest effect on V. In this case, the effect of decreasing pressure predominates, and we expect the volume of the gas to increase, as we found in our calculation.
We could also have solved this problem by solving the ideal gas law for V and then substituting the relevant parameters for an altitude of 23,000 ft:
Except for a difference caused by rounding to the last significant figure, this is the same result we obtained previously. There is often more than one “right” way to solve chemical problems.
Exercise $4$
A steel cylinder of compressed argon with a volume of 0.400 L was filled to a pressure of 145 atm at 10°C. At 1.00 atm pressure and 25°C, how many 15.0 mL incandescent light bulbs could be filled from this cylinder? (Hint: find the number of moles of argon in each container.)
Answer
4.07 × 103
Using the Ideal Gas Law to Calculate Gas Densities and Molar Masses
The ideal gas law can also be used to calculate molar masses of gases from experimentally measured gas densities. To see how this is possible, we first rearrange the ideal gas law to obtain
$\dfrac{n}{V}=\dfrac{P}{RT}\label{10.4.9}$
The left side has the units of moles per unit volume (mol/L). The number of moles of a substance equals its mass ($m$, in grams) divided by its molar mass ($M$, in grams per mole):
$n=\dfrac{m}{M}\label{10.4.10}$
Substituting this expression for $n$ into Equation $\ref{10.4.9}$ gives
$\dfrac{m}{MV}=\dfrac{P}{RT}\label{10.4.11}$
Because $m/V$ is the density $d$ of a substance, we can replace $m/V$ by $d$ and rearrange to give
$\rho=\dfrac{m}{V}=\dfrac{MP}{RT}\label{10.4.12}$
The distance between particles in gases is large compared to the size of the particles, so their densities are much lower than the densities of liquids and solids. Consequently, gas density is usually measured in grams per liter (g/L) rather than grams per milliliter (g/mL).
Example $5$
Calculate the density of butane at 25°C and a pressure of 750 mmHg.
Given: compound, temperature, and pressure
Asked for: density
Strategy:
1. Calculate the molar mass of butane and convert all quantities to appropriate units for the value of the gas constant.
2. Substitute these values into Equation $\ref{10.4.12}$ to obtain the density.
Solution:
A The molar mass of butane (C4H10) is
$M=(4)(12.011) + (10)(1.0079) = 58.123 \rm g/mol \nonumber$
Using 0.08206 (L•atm)/(K•mol) for R means that we need to convert the temperature from degrees Celsius to kelvins (T = 25 + 273 = 298 K) and the pressure from millimeters of mercury to atmospheres:
$P=\rm750\;mmHg\times\dfrac{1\;atm}{760\;mmHg}=0.987\;atm \nonumber$
B Substituting these values into Equation $\ref{10.4.12}$ gives
$\rho=\rm\dfrac{58.123\;g/mol\times0.987\;atm}{0.08206\dfrac{L\cdot atm}{K\cdot mol}\times298\;K}=2.35\;g/L \nonumber$
Exercise $5$: Density of Radon
Radon (Rn) is a radioactive gas formed by the decay of naturally occurring uranium in rocks such as granite. It tends to collect in the basements of houses and poses a significant health risk if present in indoor air. Many states now require that houses be tested for radon before they are sold. Calculate the density of radon at 1.00 atm pressure and 20°C and compare it with the density of nitrogen gas, which constitutes 80% of the atmosphere, under the same conditions to see why radon is found in basements rather than in attics.
Answer
radon, 9.23 g/L; N2, 1.17 g/L
A common use of Equation $\ref{10.4.12}$ is to determine the molar mass of an unknown gas by measuring its density at a known temperature and pressure. This method is particularly useful in identifying a gas that has been produced in a reaction, and it is not difficult to carry out. A flask or glass bulb of known volume is carefully dried, evacuated, sealed, and weighed empty. It is then filled with a sample of a gas at a known temperature and pressure and reweighed. The difference in mass between the two readings is the mass of the gas. The volume of the flask is usually determined by weighing the flask when empty and when filled with a liquid of known density such as water. The use of density measurements to calculate molar masses is illustrated in Example $6$.
Example $6$
The reaction of a copper penny with nitric acid results in the formation of a red-brown gaseous compound containing nitrogen and oxygen. A sample of the gas at a pressure of 727 mmHg and a temperature of 18°C weighs 0.289 g in a flask with a volume of 157.0 mL. Calculate the molar mass of the gas and suggest a reasonable chemical formula for the compound.
Given: pressure, temperature, mass, and volume
Asked for: molar mass and chemical formula
Strategy:
1. Solve Equation $\ref{10.4.12}$ for the molar mass of the gas and then calculate the density of the gas from the information given.
2. Convert all known quantities to the appropriate units for the gas constant being used. Substitute the known values into your equation and solve for the molar mass.
3. Propose a reasonable empirical formula using the atomic masses of nitrogen and oxygen and the calculated molar mass of the gas.
Solution:
A Solving Equation $\ref{10.4.12}$ for the molar mass gives
$M=\dfrac{mRT}{PV}=\dfrac{dRT}{P} \nonumber$
Density is the mass of the gas divided by its volume:
$\rho=\dfrac{m}{V}=\dfrac{0.289\rm g}{0.157\rm L}=1.84 \rm g/L\nonumber$
B We must convert the other quantities to the appropriate units before inserting them into the equation:
$T=18+273=291 K\nonumber$
$P=727 \, mmHg \times \dfrac{1\rm atm} {760\rm mmHg} =0.957\rm atm \nonumber$
The molar mass of the unknown gas is thus
$M=\rm\dfrac{1.84\;g/L\times0.08206\dfrac{L\cdot atm}{K\cdot mol}\times291\;K}{0.957\;atm}=45.9 g/mol\nonumber$
C The atomic masses of N and O are approximately 14 and 16, respectively, so we can construct a list showing the masses of possible combinations:
$M({\rm NO})=14 + 16=30 \rm\; g/mol\nonumber$
$M({\rm N_2O})=(2)(14)+16=44 \rm\;g/mol\nonumber$
$M({\rm NO_2})=14+(2)(16)=46 \rm\;g/mol\nonumber$
The most likely choice is NO2 which is in agreement with the data. The red-brown color of smog also results from the presence of NO2 gas.
Exercise $6$
You are in charge of interpreting the data from an unmanned space probe that has just landed on Venus and sent back a report on its atmosphere. The data are as follows: pressure, 90 atm; temperature, 557°C; density, 58 g/L. The major constituent of the atmosphere (>95%) is carbon. Calculate the molar mass of the major gas present and identify it.
Answer
44 g/mol; $CO_2$
Summary
The ideal gas law is derived from empirical relationships among the pressure, the volume, the temperature, and the number of moles of a gas; it can be used to calculate any of the four properties if the other three are known.
Ideal gas equation: $PV = nRT$,
where $R = 0.08206 \dfrac{\rm L\cdot atm}{\rm K\cdot mol}=8.3145 \dfrac{\rm J}{\rm K\cdot mol}$
General gas equation: $\dfrac{P_iV_i}{n_iT_i}=\dfrac{P_fV_f}{n_fT_f}$
Density of a gas: $\rho=\dfrac{MP}{RT}$
The empirical relationships among the volume, the temperature, the pressure, and the amount of a gas can be combined into the ideal gas law, PV = nRT. The proportionality constant, R, is called the gas constant and has the value 0.08206 (L•atm)/(K•mol), 8.3145 J/(K•mol), or 1.9872 cal/(K•mol), depending on the units used. The ideal gas law describes the behavior of an ideal gas, a hypothetical substance whose behavior can be explained quantitatively by the ideal gas law and the kinetic molecular theory of gases. Standard temperature and pressure (STP) is 0°C and 1 atm. The volume of 1 mol of an ideal gas at STP is 22.41 L, the standard molar volume. All of the empirical gas relationships are special cases of the ideal gas law in which two of the four parameters are held constant. The ideal gas law allows us to calculate the value of the fourth quantity (P, V, T, or n) needed to describe a gaseous sample when the others are known and also predict the value of these quantities following a change in conditions if the original conditions (values of P, V, T, and n) are known. The ideal gas law can also be used to calculate the density of a gas if its molar mass is known or, conversely, the molar mass of an unknown gas sample if its density is measured. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/10%3A_Gases/10.04%3A_The_Ideal_Gas_Equation.txt |
Learning Objectives
• To relate the amount of gas consumed or released in a chemical reaction to the stoichiometry of the reaction.
• To understand how the ideal gas equation and the stoichiometry of a reaction can be used to calculate the volume of gas produced or consumed in a reaction.
With the ideal gas law, we can use the relationship between the amounts of gases (in moles) and their volumes (in liters) to calculate the stoichiometry of reactions involving gases, if the pressure and temperature are known. This is important for several reasons. Many reactions that are carried out in the laboratory involve the formation or reaction of a gas, so chemists must be able to quantitatively treat gaseous products and reactants as readily as they quantitatively treat solids or solutions. Furthermore, many, if not most, industrially important reactions are carried out in the gas phase for practical reasons. Gases mix readily, are easily heated or cooled, and can be transferred from one place to another in a manufacturing facility via simple pumps and plumbing.
Gas Densities and Molar Mass
The ideal-gas equation can be manipulated to solve a variety of different types of problems. For example, the density, $\rho$, of a gas, depends on the number of gas molecules in a constant volume. To determine this value, we rearrange the ideal gas equation to
$\dfrac{n}{V}=\dfrac{P}{RT}\label{10.5.1}$
Density of a gas is generally expressed in g/L (mass over volume). Multiplication of the left and right sides of Equation \ref{10.5.1} by the molar mass in g/mol ($M$) of the gas gives
$\rho= \dfrac{g}{L}=\dfrac{PM}{RT} \label{10.5.2}$
This allows us to determine the density of a gas when we know the molar mass, or vice versa.
The density of a gas INCREASES with increasing pressure and DECREASES with increasing temperature
Example $1$
What is the density of nitrogen gas ($\ce{N_2}$) at 248.0 Torr and 18º C?
Step 1: Write down your given information
• P = 248.0 Torr
• V = ?
• n = ?
• R = 0.0820574 L•atm•mol-1 K-1
• T = 18º C
Step 2: Convert as necessary.
$(248 \; \rm{Torr}) \times \dfrac{1 \; \rm{atm}}{760 \; \rm{Torr}} = 0.3263 \; \rm{atm} \nonumber$
$18\,^oC + 273 = 291 K\nonumber$
Step 3: This one is tricky. We need to manipulate the Ideal Gas Equation to incorporate density into the equation.
Write down all known equations:
$PV = nRT \nonumber$
$\rho=\dfrac{m}{V} \nonumber$
where $\rho$ is density, $m$ is mass, and $V$ is volume.
$m=M \times n \nonumber$
where $M$ is molar mass and $n$ is the number of moles.
Now take the definition of density (Equation \ref{10.5.1})
$\rho=\dfrac{m}{V} \nonumber$
Keeping in mind $m=M \times n$...replace $(M \times n)$ for $mass$ within the density formula.
\begin{align*} \rho &=\dfrac{M \times n}{V} \[4pt] \dfrac{\rho}{M} &= \dfrac{n}{V} \end{align*} \nonumber
Now manipulate the Ideal Gas Equation
\begin{align*} PV &= nRT \[4pt] \dfrac{n}{V} &= \dfrac{P}{RT} \end{align*} \nonumber
$(n/V)$ is in both equations.
\begin{align*} \dfrac{n}{V} &= \dfrac{\rho}{M} \[4pt] &= \dfrac{P}{RT} \end{align*} \nonumber
Now combine them please.
$\dfrac{\rho}{M} = \dfrac{P}{RT}\nonumber$
Isolate density.
$\rho = \dfrac{PM}{RT} \nonumber$
Step 4: Now plug in the information you have.
\begin{align*} \rho &= \dfrac{PM}{RT} \[4pt] &= \dfrac{(0.3263\; \rm{atm})(2*14.01 \; \rm{g/mol})}{(0.08206\, L\, atm/K mol)(291 \; \rm{K})} \[4pt] &= 0.3828 \; g/L \end{align*} \nonumber
An example of varying density for a useful purpose is the hot air balloon, which consists of a bag (called the envelope) that is capable of containing heated air. As the air in the envelope is heated, it becomes less dense than the surrounding cooler air (Equation $\ref{10.5.2}$), which is has enough lifting power (due to buoyancy) to cause the balloon to float and rise into the air. Constant heating of the air is required to keep the balloon aloft. As the air in the balloon cools, it contracts, allowing outside cool air to enter, and the density increases. When this is carefully controlled by the pilot, the balloon can land as gently as it rose.
Determining Gas Volumes in Chemical Reactions
The ideal gas law can be used to calculate volume of gases consumed or produced. The ideal-gas equation frequently is used to interconvert between volumes and molar amounts in chemical equations.
Example $\PageIndex{2A}$
What volume of carbon dioxide gas is produced at STP by the decomposition of 0.150 g $\ce{CaCO_3}$ via the equation:
$\ce{CaCO3(s) \rightarrow CaO(s) + CO2(g)} \nonumber$
Solution
Begin by converting the mass of calcium carbonate to moles.
$\dfrac{0.150\;g}{100.1\;g/mol} = 0.00150\; mol \nonumber$
The stoichiometry of the reaction dictates that the number of moles $\ce{CaCO_3}$ decomposed equals the number of moles $\ce{CO2}$ produced. Use the ideal-gas equation to convert moles of $\ce{CO2}$ to a volume.
\begin{align*} V &= \dfrac{nRT}{PR} \[4pt] &= \dfrac{(0.00150\;mol)\left( 0.08206\; \frac{L \cdot atm}{mol \cdot K} \right) ( 273.15\;K)}{1\;atm} \[4pt] &= 0.0336\;L \; or \; 33.6\;mL \end{align*} \nonumber
Example $\PageIndex{2B}$
A 3.00 L container is filled with $\ce{Ne(g)}$ at 770 mmHg at 27oC. A $0.633\;\rm{g}$ sample of $\ce{CO2}$ vapor is then added.
• What is the partial pressure of $\ce{CO2}$ and $\ce{Ne}$ in atm?
• What is the total pressure in the container in atm?
Solution
Step 1: Write down all given information, and convert as necessary.
Before:
• $P = 770\,mmHg \rigtharrow 1.01 \,atm \nonumber \] • \(V = 3.00\,L$
• $n_{\ce{Ne}} = ?$
• $T = 27^o C \rightarrow 300\; K$
Other Unknowns: $n_{\ce{CO2}}$= ?
$n_{CO_2} = 0.633\; \rm{g} \;CO_2 \times \dfrac{1 \; \rm{mol}}{44\; \rm{g}} = 0.0144\; \rm{mol} \; CO_2 \nonumber$
Step 2: After writing down all your given information, find the unknown moles of $\ce{Ne}$.
\begin{align*} n_{Ne} &= \dfrac{PV}{RT} \[4pt] &= \dfrac{(1.01\; \rm{atm})(3.00\; \rm{L})}{(0.08206\;atm\;L/mol\;K)(300\; \rm{K})} \[4pt] &= 0.123 \; \rm{mol} \end{align*} \nonumber
Because the pressure of the container before the $\ce{CO2}$ was added contained only $\ce{Ne}$, that is your partial pressure of $Ne$. After converting it to atm, you have already answered part of the question!
$P_{Ne} = 1.01\; \rm{atm} \nonumber$
Step 3: Now that have pressure for $\ce{Ne}$, you must find the partial pressure for $CO_2$. Use the ideal gas equation.
$\dfrac{P_{Ne}\cancel{V}}{n_{Ne}\cancel{RT}} = \dfrac{P_{CO_2}\cancel{V}}{n_{CO_2}\cancel{RT}} \nonumber$
but because both gases share the same Volume ($V$) and Temperature ($T$) and since the Gas Constant ($R$) is constants, all three terms cancel.
\begin{align*} \dfrac{P}{n_{Ne}} &= \dfrac{P}{n_{CO_2}} \[4pt] \dfrac{1.01 \; \rm{atm}}{0.123\; \rm{mol} \;Ne} &= \dfrac{P_{CO_2}}{0.0144\; \rm{mol} \;CO_2} \[4pt] P_{CO_2} &= 0.118 \; \rm{atm} \end{align*} \nonumber
This is the partial pressure $\ce{CO_2}$.
Step 4: Now find total pressure.
\begin{align*} P_{total} &= P_{Ne} + P_{CO_2} \[4pt] &= 1.01 \; \rm{atm} + 0.118\; \rm{atm} \[4pt] &= 1.128\; \rm{atm} \[4pt] &\approx 1.13\; \rm{atm} \; \text{(with appropriate significant figures)} \end{align*} \nonumber
Example $\PageIndex{2C}$: Sulfuric Acid
Sulfuric acid, the industrial chemical produced in greatest quantity (almost 45 million tons per year in the United States alone), is prepared by the combustion of sulfur in air to give $\ce{SO2}$, followed by the reaction of $\ce{SO2}$ with $\ce{O2}$ in the presence of a catalyst to give $\ce{SO3}$, which reacts with water to give $\ce{H2SO4}$. The overall chemical equation is as follows:
$\ce {2S(s) + 3O2(g) + 2H2O(l) \rightarrow 2H2SO4(aq)} \nonumber$
What volume of O2 (in liters) at 22°C and 745 mmHg pressure is required to produce 1.00 ton (907.18 kg) of H2SO4?
Given: reaction, temperature, pressure, and mass of one product
Asked for: volume of gaseous reactant
Strategy:
A Calculate the number of moles of H2SO4 in 1.00 ton. From the stoichiometric coefficients in the balanced chemical equation, calculate the number of moles of $\ce{O2}$ required.
B Use the ideal gas law to determine the volume of $\ce{O2}$ required under the given conditions. Be sure that all quantities are expressed in the appropriate units.
Solution:
mass of $\ce{H2SO4}$ → moles $\ce{H2SO4}$ → moles $\ce{O2}$ → liters $\ce{O2}$
A We begin by calculating the number of moles of H2SO4 in 1.00 ton:
$\rm\dfrac{907.18\times10^3\;g\;H_2SO_4}{(2\times1.008+32.06+4\times16.00)\;g/mol}=9250\;mol\;H_2SO_4 \nonumber$
We next calculate the number of moles of $\ce{O2}$ required:
$\rm9250\;mol\;H_2SO_4\times\dfrac{3mol\; O_2}{2mol\;H_2SO_4}=1.389\times10^4\;mol\;O_2 \nonumber$
B After converting all quantities to the appropriate units, we can use the ideal gas law to calculate the volume of O2:
\begin{align*} V&=\dfrac{nRT}{P} \[4pt] &=\rm\dfrac{1.389\times10^4\;mol\times0.08206\dfrac{L\cdot atm}{mol\cdot K}\times(273+22)\;K}{745\;mmHg\times\dfrac{1\;atm}{760\;mmHg}} \[4pt] &=3.43\times10^5\;L \end{align*} \nonumber
The answer means that more than 300,000 L of oxygen gas are needed to produce 1 ton of sulfuric acid. These numbers may give you some appreciation for the magnitude of the engineering and plumbing problems faced in industrial chemistry.
Exercise $2$
Charles used a balloon containing approximately 31,150 L of $\ce{H2}$ for his initial flight in 1783. The hydrogen gas was produced by the reaction of metallic iron with dilute hydrochloric acid according to the following balanced chemical equation:
$\ce{ Fe(s) + 2 HCl(aq) \rightarrow H2(g) + FeCl2(aq)} \nonumber$
How much iron (in kilograms) was needed to produce this volume of $\ce{H2}$ if the temperature were 30°C and the atmospheric pressure was 745 mmHg?
Answer
68.6 kg of Fe (approximately 150 lb)
Example $3$: Emergency Air bags
Sodium azide ($\ce{NaN_3}$) decomposes to form sodium metal and nitrogen gas according to the following balanced chemical equation:
$\ce{ 2NaN3 \rightarrow 2Na(s) + 3N2(g)} \nonumber$
This reaction is used to inflate the air bags that cushion passengers during automobile collisions. The reaction is initiated in air bags by an electrical impulse and results in the rapid evolution of gas. If the $\ce{N_2}$ gas that results from the decomposition of a 5.00 g sample of $\ce{NaN_3}$ could be collected by displacing water from an inverted flask, what volume of gas would be produced at 21°C and 762 mmHg?
Given: reaction, mass of compound, temperature, and pressure
Asked for: volume of nitrogen gas produced
Strategy:
A Calculate the number of moles of $\ce{N_2}$ gas produced. From the data in Table S3, determine the partial pressure of $\ce{N_2}$ gas in the flask.
B Use the ideal gas law to find the volume of $\ce{N_2}$ gas produced.
Solution:
A Because we know the mass of the reactant and the stoichiometry of the reaction, our first step is to calculate the number of moles of $\ce{N_2}$ gas produced:
$\rm\dfrac{5.00\;g\;NaN_3}{(22.99+3\times14.01)\;g/mol}\times\dfrac{3mol\;N_2}{2mol\;NaN_3}=0.115\;mol\; N_2 \nonumber$
The pressure given (762 mmHg) is the total pressure in the flask, which is the sum of the pressures due to the N2 gas and the water vapor present. Table S3 tells us that the vapor pressure of water is 18.65 mmHg at 21°C (294 K), so the partial pressure of the $\ce{N_2}$ gas in the flask is only
\begin{align*} \rm(762 − 18.65)\;mmHg \times\dfrac{1\;atm}{760\;mmHg} &= 743.4\; \cancel{mmHg} \times\dfrac{1\;atm}{760\;\cancel{mmHg}} \[4pt] &= 0.978\; atm. \end{align*} \nonumber
B Solving the ideal gas law for V and substituting the other quantities (in the appropriate units), we get
$V=\dfrac{nRT}{P}=\rm\dfrac{0.115\;mol\times0.08206\dfrac{atm\cdot L}{mol\cdot K}\times294\;K}{0.978\;atm}=2.84\;L \nonumber$
Exercise$3$
A 1.00 g sample of zinc metal is added to a solution of dilute hydrochloric acid. It dissolves to produce $\ce{H2}$ gas according to the equation
$\ce{ Zn(s) + 2 HCl(aq) → H2(g) + ZnCl2(aq)}. \nonumber$
The resulting H2 gas is collected in a water-filled bottle at 30°C and an atmospheric pressure of 760 mmHg. What volume does it occupy?
Answer
0.397 L
Summary
The relationship between the amounts of products and reactants in a chemical reaction can be expressed in units of moles or masses of pure substances, of volumes of solutions, or of volumes of gaseous substances. The ideal gas law can be used to calculate the volume of gaseous products or reactants as needed. In the laboratory, gases produced in a reaction are often collected by the displacement of water from filled vessels; the amount of gas can then be calculated from the volume of water displaced and the atmospheric pressure. A gas collected in such a way is not pure, however, but contains a significant amount of water vapor. The measured pressure must therefore be corrected for the vapor pressure of water, which depends strongly on the temperature. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/10%3A_Gases/10.05%3A_Further_Applications_of_the_Ideal-Gas_Equations.txt |
Learning Objectives
• To determine the contribution of each component gas to the total pressure of a mixture of gases
In our use of the ideal gas law thus far, we have focused entirely on the properties of pure gases with only a single chemical species. But what happens when two or more gases are mixed? In this section, we describe how to determine the contribution of each gas present to the total pressure of the mixture.
Partial Pressures
The ideal gas law assumes that all gases behave identically and that their behavior is independent of attractive and repulsive forces. If volume and temperature are held constant, the ideal gas equation can be rearranged to show that the pressure of a sample of gas is directly proportional to the number of moles of gas present:
$P=n \left(\dfrac{RT}{V}\right) = n \times \rm const. \label{10.6.1}$
Nothing in the equation depends on the nature of the gas—only the amount.
With this assumption, let’s suppose we have a mixture of two ideal gases that are present in equal amounts. What is the total pressure of the mixture? Because the pressure depends on only the total number of particles of gas present, the total pressure of the mixture will simply be twice the pressure of either component. More generally, the total pressure exerted by a mixture of gases at a given temperature and volume is the sum of the pressures exerted by each gas alone. Furthermore, if we know the volume, the temperature, and the number of moles of each gas in a mixture, then we can calculate the pressure exerted by each gas individually, which is its partial pressure, the pressure the gas would exert if it were the only one present (at the same temperature and volume).
To summarize, the total pressure exerted by a mixture of gases is the sum of the partial pressures of component gases. This law was first discovered by John Dalton, the father of the atomic theory of matter. It is now known as Dalton’s law of partial pressures. We can write it mathematically as
\begin{align} P_{tot} &= P_1+P_2+P_3+P_4 \ldots \[4pt] &= \sum_{i=1}^n{P_i} \label{10.6.2} \end{align}
where $P_{tot}$ is the total pressure and the other terms are the partial pressures of the individual gases (up to $n$ component gases).
For a mixture of two ideal gases, $A$ and $B$, we can write an expression for the total pressure:
\begin{align} P_{tot} &=P_A+P_B \[4pt] &=n_A\left(\dfrac{RT}{V}\right) + n_B\left(\dfrac{RT}{V}\right) \[4pt] &=(n_A+n_B)\left(\dfrac{RT}{V}\right) \label{10.6.3} \end{align}
More generally, for a mixture of $n$ component gases, the total pressure is given by
\begin{align} P_{tot} &=(P_1+P_2+P_3+ \; \cdots +P_n)\left(\dfrac{RT}{V}\right)\label{10.6.2a} \[4pt] &=\sum_{i=1}^n{P_i}\left(\dfrac{RT}{V}\right)\label{10.6.2b} \end{align}
Equation $\ref{10.6.2b}$ restates Equation $\ref{10.6.3}$ in a more general form and makes it explicitly clear that, at constant temperature and volume, the pressure exerted by a gas depends on only the total number of moles of gas present, whether the gas is a single chemical species or a mixture of dozens or even hundreds of gaseous species. For Equation $\ref{10.6.2b}$ to be valid, the identity of the particles present cannot have an effect. Thus an ideal gas must be one whose properties are not affected by either the size of the particles or their intermolecular interactions because both will vary from one gas to another. The calculation of total and partial pressures for mixtures of gases is illustrated in Example $1$.
Example $1$: The Bends
Deep-sea divers must use special gas mixtures in their tanks, rather than compressed air, to avoid serious problems, most notably a condition called “the bends.” At depths of about 350 ft, divers are subject to a pressure of approximately 10 atm. A typical gas cylinder used for such depths contains 51.2 g of $O_2$ and 326.4 g of He and has a volume of 10.0 L. What is the partial pressure of each gas at 20.00°C, and what is the total pressure in the cylinder at this temperature?
Given: masses of components, total volume, and temperature
Asked for: partial pressures and total pressure
Strategy:
1. Calculate the number of moles of $\ce{He}$ and $\ce{O_2}$ present.
2. Use the ideal gas law to calculate the partial pressure of each gas. Then add together the partial pressures to obtain the total pressure of the gaseous mixture.
Solution:
A The number of moles of $\ce{He}$ is
$n_{\rm He}=\rm\dfrac{326.4\;g}{4.003\;g/mol}=81.54\;mol \nonumber$
The number of moles of $\ce{O_2}$ is
$n_{\rm O_2}=\rm \dfrac{51.2\;g}{32.00\;g/mol}=1.60\;mol \nonumber$
B We can now use the ideal gas law to calculate the partial pressure of each:
$P_{\rm He}=\dfrac{n_{\rm He}RT}{V}=\rm\dfrac{81.54\;mol\times0.08206\;\dfrac{atm\cdot L}{mol\cdot K}\times293.15\;K}{10.0\;L}=196.2\;atm \nonumber$
$P_{\rm O_2}=\dfrac{n_{\rm O_2} RT}{V}=\rm\dfrac{1.60\;mol\times0.08206\;\dfrac{atm\cdot L}{mol\cdot K}\times293.15\;K}{10.0\;L}=3.85\;atm \nonumber$
The total pressure is the sum of the two partial pressures:
$P_{\rm tot}=P_{\rm He}+P_{\rm O_2}=\rm(196.2+3.85)\;atm=200.1\;atm \nonumber$
Exercise $1$
A cylinder of compressed natural gas has a volume of 20.0 L and contains 1813 g of methane and 336 g of ethane. Calculate the partial pressure of each gas at 22.0°C and the total pressure in the cylinder.
Answer
$P_{CH_4}=137 \; atm$; $P_{C_2H_6}=13.4\; atm$; $P_{tot}=151\; atm$
Mole Fractions of Gas Mixtures
The composition of a gas mixture can be described by the mole fractions of the gases present. The mole fraction ($\chi$) of any component of a mixture is the ratio of the number of moles of that component to the total number of moles of all the species present in the mixture ($n_{tot}$):
$\chi_A=\dfrac{\text{moles of A}}{\text{total moles}}= \dfrac{n_A}{n_{tot}} =\dfrac{n_A}{n_A+n_B+\cdots}\label{10.6.5}$
The mole fraction is a dimensionless quantity between 0 and 1. If $\chi_A = 1.0$, then the sample is pure $A$, not a mixture. If $\chi_A = 0$, then no $A$ is present in the mixture. The sum of the mole fractions of all the components present must equal 1.
To see how mole fractions can help us understand the properties of gas mixtures, let’s evaluate the ratio of the pressure of a gas $A$ to the total pressure of a gas mixture that contains $A$. We can use the ideal gas law to describe the pressures of both gas $A$ and the mixture: $P_A = n_ART/V$ and $P_{tot} = n_tRT/V$. The ratio of the two is thus
$\dfrac{P_A}{P_{tot}}=\dfrac{n_ART/V}{n_{tot}RT/V} = \dfrac{n_A}{n_{tot}}=\chi_A \label{10.6.6}$
Rearranging this equation gives
$P_A = \chi_AP_{tot} \label{10.6.7}$
That is, the partial pressure of any gas in a mixture is the total pressure multiplied by the mole fraction of that gas. This conclusion is a direct result of the ideal gas law, which assumes that all gas particles behave ideally. Consequently, the pressure of a gas in a mixture depends on only the percentage of particles in the mixture that are of that type, not their specific physical or chemical properties. By volume, Earth’s atmosphere is about 78% $N_2$, 21% $O_2$, and 0.9% $Ar$, with trace amounts of gases such as $CO_2$, $H_2O$, and others. This means that 78% of the particles present in the atmosphere are $N_2$; hence the mole fraction of $N_2$ is 78%/100% = 0.78. Similarly, the mole fractions of $O_2$ and $Ar$ are 0.21 and 0.009, respectively. Using Equation \ref{10.6.7}, we therefore know that the partial pressure of N2 is 0.78 atm (assuming an atmospheric pressure of exactly 760 mmHg) and, similarly, the partial pressures of $O_2$ and $Ar$ are 0.21 and 0.009 atm, respectively.
Example $2$: Exhaling Composition
We have just calculated the partial pressures of the major gases in the air we inhale. Experiments that measure the composition of the air we exhale yield different results, however. The following table gives the measured pressures of the major gases in both inhaled and exhaled air. Calculate the mole fractions of the gases in exhaled air.
the mole fractions of the gases
Inhaled Air / mmHg Exhaled Air / mmHg
$P_{\rm N_2}$ 597 568
$P_{\rm O_2}$ 158 116
$P_{\rm H_2O}$ 0.3 28
$P_{\rm CO_2}$ 5 48
$P_{\rm Ar}$ 8 8
$P_{tot}$ 767 767
Given: pressures of gases in inhaled and exhaled air
Asked for: mole fractions of gases in exhaled air
Strategy:
Calculate the mole fraction of each gas using Equation $\ref{10.6.7}$.
Solution:
The mole fraction of any gas $A$ is given by
$\chi_A=\dfrac{P_A}{P_{tot}} \nonumber$
where $P_A$ is the partial pressure of $A$ and $P_{tot}$ is the total pressure. For example, the mole fraction of $\ce{CO_2}$ is given as:
$\chi_{\rm CO_2}=\rm\dfrac{48\;mmHg}{767\;mmHg}=0.063 \nonumber$
The following table gives the values of $\chi_A$ for the gases in the exhaled air.
Solutions to Example 10.6.2
Gas Mole Fraction
${\rm N_2}$ 0.741
${\rm O_2}$ 0.151
${\rm H_2O}$ 0.037
${\rm CO_2}$ 0.063
${\rm Ar}$ 0.010
Exercise $2$
Venus is an inhospitable place, with a surface temperature of 560°C and a surface pressure of 90 atm. The atmosphere consists of about 96% CO2 and 3% N2, with trace amounts of other gases, including water, sulfur dioxide, and sulfuric acid. Calculate the partial pressures of CO2 and N2.
Answer
$P_{\rm CO_2}=\rm86\; atm \nonumber$
$P_{\rm N_2}=\rm2.7\;atm \nonumber$
Summary
The partial pressure of each gas in a mixture is proportional to its mole fraction. The pressure exerted by each gas in a gas mixture (its partial pressure) is independent of the pressure exerted by all other gases present. Consequently, the total pressure exerted by a mixture of gases is the sum of the partial pressures of the components (Dalton’s law of partial pressures). The amount of gas present in a mixture may be described by its partial pressure or its mole fraction. The mole fraction of any component of a mixture is the ratio of the number of moles of that substance to the total number of moles of all substances present. In a mixture of gases, the partial pressure of each gas is the product of the total pressure and the mole fraction of that gas. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/10%3A_Gases/10.06%3A_Gas_Mixtures_and_Partial_Pressures.txt |
Learning Objectives
• To understand the significance of the kinetic molecular theory of gases.
The laws that describe the behavior of gases were well established long before anyone had developed a coherent model of the properties of gases. In this section, we introduce a theory that describes why gases behave the way they do. The theory we introduce can also be used to derive laws such as the ideal gas law from fundamental principles and the properties of individual particles.
A Molecular Description
The kinetic molecular theory of gases explains the laws that describe the behavior of gases. Developed during the mid-19th century by several physicists, including the Austrian Ludwig Boltzmann (1844–1906), the German Rudolf Clausius (1822–1888), and the Englishman James Clerk Maxwell (1831–1879), who is also known for his contributions to electricity and magnetism, this theory is based on the properties of individual particles as defined for an ideal gas and the fundamental concepts of physics. Thus the kinetic molecular theory of gases provides a molecular explanation for observations that led to the development of the ideal gas law. The kinetic molecular theory of gases is based on the following five postulates:
five postulates of Kinetic Molecular Theory
1. A gas is composed of a large number of particles called molecules (whether monatomic or polyatomic) that are in constant random motion.
2. Because the distance between gas molecules is much greater than the size of the molecules, the volume of the molecules is negligible.
3. Intermolecular interactions, whether repulsive or attractive, are so weak that they are also negligible.
4. Gas molecules collide with one another and with the walls of the container, but these collisions are perfectly elastic; that is, they do not change the average kinetic energy of the molecules.
5. The average kinetic energy of the molecules of any gas depends on only the temperature, and at a given temperature, all gaseous molecules have exactly the same average kinetic energy.
Although the molecules of real gases have nonzero volumes and exert both attractive and repulsive forces on one another, for the moment we will focus on how the kinetic molecular theory of gases relates to the properties of gases we have been discussing. In the following sections, we explain how this theory must be modified to account for the behavior of real gases.
Postulates 1 and 4 state that gas molecules are in constant motion and collide frequently with the walls of their containers. The collision of molecules with their container walls results in a momentum transfer (impulse) from molecules to the walls (Figure $2$).
The momentum transfer to the wall perpendicular to $x$ axis as a molecule with an initial velocity $u_x$ in $x$ direction hits is expressed as:
$\Delta p_x=2mu_x \label{10.7.1}$
The collision frequency, a number of collisions of the molecules to the wall per unit area and per second, increases with the molecular speed and the number of molecules per unit volume.
$f\propto (u_x) \times \Big(\dfrac{N}{V}\Big) \label{10.7.2}$
The pressure the gas exerts on the wall is expressed as the product of impulse and the collision frequency.
$P\propto (2mu_x)\times(u_x)\times\Big(\dfrac{N}{V}\Big)\propto \Big(\dfrac{N}{V}\Big)mu_x^2 \label{10.7.3}$
At any instant, however, the molecules in a gas sample are traveling at different speed. Therefore, we must replace $u_x^2$ in the expression above with the average value of $u_x^2$, which is denoted by $\overline{u_x^2}$. The overbar designates the average value over all molecules.
The exact expression for pressure is given as :
$P=\dfrac{N}{V}m\overline{u_x^2} \label{10.7.4}$
Finally, we must consider that there is nothing special about $x$ direction. We should expect that
$\overline{u_x^2}= \overline{u_y^2}=\overline{u_z^2}=\dfrac{1}{3}\overline{u^2}. \nonumber$
Here the quantity $\overline{u^2}$ is called the mean-square speed defined as the average value of square-speed ($u^2$) over all molecules. Since
$u^2=u_x^2+u_y^2+u_z^2 \nonumber$
for each molecule, then
$\overline{u^2}=\overline{u_x^2}+\overline{u_y^2}+\overline{u_z^2}. \nonumber$
By substituting $\dfrac{1}{3}\overline{u^2}$ for $\overline{u_x^2}$ in the expression above, we can get the final expression for the pressure:
$P=\dfrac{1}{3}\dfrac{N}{V}m\overline{u^2} \label{10.7.5}$
Because volumes and intermolecular interactions are negligible, postulates 2 and 3 state that all gaseous particles behave identically, regardless of the chemical nature of their component molecules. This is the essence of the ideal gas law, which treats all gases as collections of particles that are identical in all respects except mass. Postulate 2 also explains why it is relatively easy to compress a gas; you simply decrease the distance between the gas molecules.
Postulate 5 provides a molecular explanation for the temperature of a gas. Postulate 5 refers to the average translational kinetic energy of the molecules of a gas $(\overline{e_K})$, which can be represented as and states that at a given Kelvin temperature $(T)$, all gases have the same value of
$\overline{e_K}=\dfrac{1}{2}m\overline{u^2}=\dfrac{3}{2}\dfrac{R}{N_A}T \label{10.7.6}$
where $N_A$ is the Avogadro's constant. The total translational kinetic energy of 1 mole of molecules can be obtained by multiplying the equation by $N_A$:
$N_A\overline{e_K}=\dfrac{1}{2}M\overline{u^2}=\dfrac{3}{2}RT \label{10.7.7}$
where $M$ is the molar mass of the gas molecules and is related to the molecular mass by $M=N_Am$. By rearranging the equation, we can get the relationship between the root-mean square speed ($u_{\rm rms}$) and the temperature. The rms speed ($u_{\rm rms}$) is the square root of the sum of the squared speeds divided by the number of particles:
$u_{\rm rms}=\sqrt{\overline{u^2}}=\sqrt{\dfrac{u_1^2+u_2^2+\cdots u_N^2}{N}} \label{10.7.8}$
where $N$ is the number of particles and $u_i$ is the speed of particle $i$.
The relationship between $u_{\rm rms}$ and the temperature is given by:
$u_{\rm rms}=\sqrt{\dfrac{3RT}{M}} \label{10.7.9}$
In Equation $\ref{10.7.9}$, $u_{\rm rms}$ has units of meters per second; consequently, the units of molar mass $M$ are kilograms per mole, temperature $T$ is expressed in kelvins, and the ideal gas constant $R$ has the value 8.3145 J/(K•mol). Equation $\ref{10.7.9}$ shows that $u_{\rm rms}$ of a gas is proportional to the square root of its Kelvin temperature and inversely proportional to the square root of its molar mass. The root mean-square speed of a gas increase with increasing temperature. At a given temperature, heavier gas molecules have slower speeds than do lighter ones.
The rms speed and the average speed do not differ greatly (typically by less than 10%). The distinction is important, however, because the rms speed is the speed of a gas particle that has average kinetic energy. Particles of different gases at the same temperature have the same average kinetic energy, not the same average speed. In contrast, the most probable speed (vp) is the speed at which the greatest number of particles is moving. If the average kinetic energy of the particles of a gas increases linearly with increasing temperature, then Equation $\ref{10.7.8}$ tells us that the rms speed must also increase with temperature because the mass of the particles is constant. At higher temperatures, therefore, the molecules of a gas move more rapidly than at lower temperatures, and vp increases.
At a given temperature, all gaseous particles have the same average kinetic energy but not the same average speed.
Example $1$
The speeds of eight particles were found to be 1.0, 4.0, 4.0, 6.0, 6.0, 6.0, 8.0, and 10.0 m/s. Calculate their average speed ($v_{\rm av}$) root mean square speed ($v_{\rm rms}$), and most probable speed ($v_{\rm m}$).
Given: particle speeds
Asked for: average speed ($v_{\rm av}$), root mean square speed ($v_{\rm rms}$), and most probable speed ($v_{\rm m}$)
Strategy:
Use Equation $\ref{10.7.6}$ to calculate the average speed and Equation $\ref{10.7.8}$ to calculate the rms speed. Find the most probable speed by determining the speed at which the greatest number of particles is moving.
Solution:
The average speed is the sum of the speeds divided by the number of particles:
$v_{\rm av}=\rm\dfrac{(1.0+4.0+4.0+6.0+6.0+6.0+8.0+10.0)\;m/s}{8}=5.6\;m/s \nonumber$
The rms speed is the square root of the sum of the squared speeds divided by the number of particles:
$v_{\rm rms}=\rm\sqrt{\dfrac{(1.0^2+4.0^2+4.0^2+6.0^2+6.0^2+6.0^2+8.0^2+10.0^2)\;m^2/s^2}{8}}=6.2\;m/s \nonumber$
The most probable speed is the speed at which the greatest number of particles is moving. Of the eight particles, three have speeds of 6.0 m/s, two have speeds of 4.0 m/s, and the other three particles have different speeds. Hence $v_{\rm m}=6.0$ m/s. The $v_{\rm rms}$ of the particles, which is related to the average kinetic energy, is greater than their average speed.
Boltzmann Distributions
At any given time, what fraction of the molecules in a particular sample has a given speed? Some of the molecules will be moving more slowly than average, and some will be moving faster than average, but how many in each situation? Answers to questions such as these can have a substantial effect on the amount of product formed during a chemical reaction. This problem was solved mathematically by Maxwell in 1866; he used statistical analysis to obtain an equation that describes the distribution of molecular speeds at a given temperature. Typical curves showing the distributions of speeds of molecules at several temperatures are displayed in Figure $3$. Increasing the temperature has two effects. First, the peak of the curve moves to the right because the most probable speed increases. Second, the curve becomes broader because of the increased spread of the speeds. Thus increased temperature increases the value of the most probable speed but decreases the relative number of molecules that have that speed. Although the mathematics behind curves such as those in Figure $3$ were first worked out by Maxwell, the curves are almost universally referred to as Boltzmann distributions, after one of the other major figures responsible for the kinetic molecular theory of gases.
The Relationships among Pressure, Volume, and Temperature
We now describe how the kinetic molecular theory of gases explains some of the important relationships we have discussed previously.
• Pressure versus Volume: At constant temperature, the kinetic energy of the molecules of a gas and hence the rms speed remain unchanged. If a given gas sample is allowed to occupy a larger volume, then the speed of the molecules does not change, but the density of the gas (number of particles per unit volume) decreases, and the average distance between the molecules increases. Hence the molecules must, on average, travel farther between collisions. They therefore collide with one another and with the walls of their containers less often, leading to a decrease in pressure. Conversely, increasing the pressure forces the molecules closer together and increases the density, until the collective impact of the collisions of the molecules with the container walls just balances the applied pressure.
• Volume versus Temperature: Raising the temperature of a gas increases the average kinetic energy and therefore the rms speed (and the average speed) of the gas molecules. Hence as the temperature increases, the molecules collide with the walls of their containers more frequently and with greater force. This increases the pressure, unless the volume increases to reduce the pressure, as we have just seen. Thus an increase in temperature must be offset by an increase in volume for the net impact (pressure) of the gas molecules on the container walls to remain unchanged.
• Pressure of Gas Mixtures: Postulate 3 of the kinetic molecular theory of gases states that gas molecules exert no attractive or repulsive forces on one another. If the gaseous molecules do not interact, then the presence of one gas in a gas mixture will have no effect on the pressure exerted by another, and Dalton’s law of partial pressures holds.
Example $2$
The temperature of a 4.75 L container of N2 gas is increased from 0°C to 117°C. What is the qualitative effect of this change on the
1. average kinetic energy of the N2 molecules?
2. rms speed of the N2 molecules?
3. average speed of the N2 molecules?
4. impact of each N2 molecule on the wall of the container during a collision with the wall?
5. total number of collisions per second of N2 molecules with the walls of the entire container?
6. number of collisions per second of N2 molecules with each square centimeter of the container wall?
7. pressure of the N2 gas?
Given: temperatures and volume
Asked for: effect of increase in temperature
Strategy:
Use the relationships among pressure, volume, and temperature to predict the qualitative effect of an increase in the temperature of the gas.
Solution:
1. Increasing the temperature increases the average kinetic energy of the N2 molecules.
2. An increase in average kinetic energy can be due only to an increase in the rms speed of the gas particles.
3. If the rms speed of the N2 molecules increases, the average speed also increases.
4. If, on average, the particles are moving faster, then they strike the container walls with more energy.
5. Because the particles are moving faster, they collide with the walls of the container more often per unit time.
6. The number of collisions per second of N2 molecules with each square centimeter of container wall increases because the total number of collisions has increased, but the volume occupied by the gas and hence the total area of the walls are unchanged.
7. The pressure exerted by the N2 gas increases when the temperature is increased at constant volume, as predicted by the ideal gas law.
Exercise $2$
A sample of helium gas is confined in a cylinder with a gas-tight sliding piston. The initial volume is 1.34 L, and the temperature is 22°C. The piston is moved to allow the gas to expand to 2.12 L at constant temperature. What is the qualitative effect of this change on the
1. average kinetic energy of the He atoms?
2. rms speed of the He atoms?
3. average speed of the He atoms?
4. impact of each He atom on the wall of the container during a collision with the wall?
5. total number of collisions per second of He atoms with the walls of the entire container?
6. number of collisions per second of He atoms with each square centimeter of the container wall?
7. pressure of the He gas?
Answer a
no change
Answer b
no change
Answer c
no change
Answer d
no change
Answer e
decreases
Answer f
decreases
Answer g
decreases
Summary
• The kinetic molecular theory of gases provides a molecular explanation for the observations that led to the development of the ideal gas law.
• Average kinetic energy:$\overline{e_K}=\dfrac{1}{2}m{u_{\rm rms}}^2=\dfrac{3}{2}\dfrac{R}{N_A}T n\nonumber$
• Root mean square speed: $u_{\rm rms}=\sqrt{\dfrac{u_1^2+u_2^2+\cdots u_N^2}{N}} \nonumber$
• Kinetic molecular theory of gases: $u_{\rm rms}=\sqrt{\dfrac{3RT}{M}} \nonumber$
The behavior of ideal gases is explained by the kinetic molecular theory of gases. Molecular motion, which leads to collisions between molecules and the container walls, explains pressure, and the large intermolecular distances in gases explain their high compressibility. Although all gases have the same average kinetic energy at a given temperature, they do not all possess the same root mean square (rms) speed (vrms). The actual values of speed and kinetic energy are not the same for all particles of a gas but are given by a Boltzmann distribution, in which some molecules have higher or lower speeds (and kinetic energies) than average. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/10%3A_Gases/10.07%3A_Kinetic-Molecular_Theory.txt |
Learning Objectives
• To understand the significance of the kinetic molecular theory of gases
We now describe how the kinetic molecular theory of gases explains some of the important relationships we have discussed previously.
Diffusion and Effusion
As you have learned, the molecules of a gas are not stationary but in constant and random motion. If someone opens a bottle of perfume in the next room, for example, you are likely to be aware of it soon. Your sense of smell relies on molecules of the aromatic substance coming into contact with specialized olfactory cells in your nasal passages, which contain specific receptors (protein molecules) that recognize the substance. How do the molecules responsible for the aroma get from the perfume bottle to your nose? You might think that they are blown by drafts, but, in fact, molecules can move from one place to another even in a draft-free environment.
Diffusion is the gradual mixing of gases due to the motion of their component particles even in the absence of mechanical agitation such as stirring. The result is a gas mixture with uniform composition. Diffusion is also a property of the particles in liquids and liquid solutions and, to a lesser extent, of solids and solid solutions. The related process, effusion, is the escape of gaseous molecules through a small (usually microscopic) hole, such as a hole in a balloon, into an evacuated space.
The phenomenon of effusion had been known for thousands of years, but it was not until the early 19th century that quantitative experiments related the rate of effusion to molecular properties. The rate of effusion of a gaseous substance is inversely proportional to the square root of its molar mass. This relationship is referred to as Graham’s law, after the Scottish chemist Thomas Graham (1805–1869). The ratio of the effusion rates of two gases is the square root of the inverse ratio of their molar masses:
$\dfrac{\text{rate of effusion A}}{\text{rate of effusion B}}=\sqrt{\dfrac{M_B}{M_A}} \label{10.8.1}$
Helium (M = 4.00 g/mol) effuses much more rapidly than ethylene oxide (M = 44.0 g/mol). Because helium is less dense than air, helium-filled balloons “float” at the end of a tethering string. Unfortunately, rubber balloons filled with helium soon lose their buoyancy along with much of their volume. In contrast, rubber balloons filled with air tend to retain their shape and volume for a much longer time. Because helium has a molar mass of 4.00 g/mol, whereas air has an average molar mass of about 29 g/mol, pure helium effuses through the microscopic pores in the rubber balloon $\sqrt{\dfrac{29}{4.00}}=2.7$ times faster than air. For this reason, high-quality helium-filled balloons are usually made of Mylar, a dense, strong, opaque material with a high molecular mass that forms films that have many fewer pores than rubber. Hence, mylar balloons can retain their helium for days.
At a given temperature, heavier molecules move more slowly than lighter molecules.
Example $1$
During World War II, scientists working on the first atomic bomb were faced with the challenge of finding a way to obtain large amounts of $\ce{^{235}U}$. Naturally occurring uranium is only 0.720% $\ce{^{235}U}$, whereas most of the rest (99.275%) is $\ce{^{238}U}$, which is not fissionable (i.e., it will not break apart to release nuclear energy) and also actually poisons the fission process. Because both isotopes of uranium have the same reactivity, they cannot be separated chemically. Instead, a process of gaseous effusion was developed using the volatile compound $UF_6$ (boiling point = 56°C).
1. Calculate the ratio of the rates of effusion of 235UF6 and 238UF6 for a single step in which UF6 is allowed to pass through a porous barrier. (The atomic mass of 235U is 235.04, and the atomic mass of 238U is 238.05.)
2. If n identical successive separation steps are used, the overall separation is given by the separation in a single step (in this case, the ratio of effusion rates) raised to the nth power. How many effusion steps are needed to obtain 99.0% pure 235UF6?
Given: isotopic content of naturally occurring uranium and atomic masses of 235U and 238U
Asked for: ratio of rates of effusion and number of effusion steps needed to obtain 99.0% pure 235UF6
Strategy:
1. Calculate the molar masses of 235UF6 and 238UF6, and then use Graham’s law to determine the ratio of the effusion rates. Use this value to determine the isotopic content of 235UF6 after a single effusion step.
2. Divide the final purity by the initial purity to obtain a value for the number of separation steps needed to achieve the desired purity. Use a logarithmic expression to compute the number of separation steps required.
Solution:
A The first step is to calculate the molar mass of UF6 containing 235U and 238U. Luckily for the success of the separation method, fluorine consists of a single isotope of atomic mass 18.998. The molar mass of 235UF6 is
234.04 + (6)(18.998) = 349.03 g/mol
The molar mass of 238UF6 is
238.05 + (6)(18.998) = 352.04 g/mol
The difference is only 3.01 g/mol (less than 1%). The ratio of the effusion rates can be calculated from Graham’s law using Equation $\ref{10.8.1}$: $\rm\dfrac{\text{rate }^{235}UF_6}{\text{rate }^{238}UF_6}=\sqrt{\dfrac{352.04\;g/mol}{349.03\;g/mol}}=1.0043 \nonumber$
B To obtain 99.0% pure 235UF6 requires many steps. We can set up an equation that relates the initial and final purity to the number of times the separation process is repeated:
final purity = (initial purity)(separation)n
In this case, 0.990 = (0.00720)(1.0043)n, which can be rearranged to give
$1.0043^n=\dfrac{0.990}{0.00720}=137.50 \nonumber$
Taking the logarithm of both sides gives
\begin{align} n\ln(1.0043)&=\ln(137.50) \[4pt] n &=\dfrac{\ln(137.50)}{\ln(1.0043)} \[4pt]&=1148 \end{align} \nonumber
Thus at least a thousand effusion steps are necessary to obtain highly enriched 235U. Below is a small part of a system that is used to prepare enriched uranium on a large scale.
Exercise $1$
Helium consists of two isotopes: 3He (natural abundance = 0.000134%) and 4He (natural abundance = 99.999866%). Their atomic masses are 3.01603 and 4.00260, respectively. Helium-3 has unique physical properties and is used in the study of ultralow temperatures. It is separated from the more abundant 4He by a process of gaseous effusion.
1. Calculate the ratio of the effusion rates of 3He and 4He and thus the enrichment possible in a single effusion step.
2. How many effusion steps are necessary to yield 99.0% pure 3He?
Answer a
ratio of effusion rates = 1.15200; one step gives 0.000154% 3He
Answer b
96 steps
Rates of Diffusion or Effusion
Graham’s law is an empirical relationship that states that the ratio of the rates of diffusion or effusion of two gases is the square root of the inverse ratio of their molar masses. The relationship is based on the postulate that all gases at the same temperature have the same average kinetic energy. We can write the expression for the average kinetic energy of two gases with different molar masses:
$KE=\dfrac{1}{2}\dfrac{M_{\rm A}}{N_A}v_{\rm rms,A}^2=\dfrac{1}{2}\dfrac{M_{\rm B}}{N_A}v_{\rm rms,B}^2\label{10.8.2}$
Multiplying both sides by 2 and rearranging give
$\dfrac{v_{\rm rms, B}^2}{v_{\rm rms,A}^2}=\dfrac{M_{\rm A}}{M_{\rm B}}\label{10.8.3}$
Taking the square root of both sides gives
$\dfrac{v_{\rm rms, B}}{v_{\rm rms,A}}=\sqrt{\dfrac{M_{\rm A}}{M_{\rm B}}}\label{10.8.4}$
Thus the rate at which a molecule, or a mole of molecules, diffuses or effuses is directly related to the speed at which it moves. Equation $\ref{10.8.4}$ shows that Graham’s law is a direct consequence of the fact that gaseous molecules at the same temperature have the same average kinetic energy.
Typically, gaseous molecules have a speed of hundreds of meters per second (hundreds of miles per hour). The effect of molar mass on these speeds is dramatic, as illustrated in Figure $3$ for some common gases. Because all gases have the same average kinetic energy, according to the Boltzmann distribution, molecules with lower masses, such as hydrogen and helium, have a wider distribution of speeds.
The lightest gases have a wider distribution of speeds and the highest average speeds.
Molecules with lower masses have a wider distribution of speeds and a higher average speed.
Gas molecules do not diffuse nearly as rapidly as their very high speeds might suggest. If molecules actually moved through a room at hundreds of miles per hour, we would detect odors faster than we hear sound. Instead, it can take several minutes for us to detect an aroma because molecules are traveling in a medium with other gas molecules. Because gas molecules collide as often as 1010 times per second, changing direction and speed with each collision, they do not diffuse across a room in a straight line, as illustrated schematically in Figure $4$.
The average distance traveled by a molecule between collisions is the mean free path. The denser the gas, the shorter the mean free path; conversely, as density decreases, the mean free path becomes longer because collisions occur less frequently. At 1 atm pressure and 25°C, for example, an oxygen or nitrogen molecule in the atmosphere travels only about 6.0 × 10−8 m (60 nm) between collisions. In the upper atmosphere at about 100 km altitude, where gas density is much lower, the mean free path is about 10 cm; in space between galaxies, it can be as long as 1 × 1010 m (about 6 million miles).
The denser the gas, the shorter the mean free path.
Example $2$
Calculate the rms speed of a sample -butene (C4H8) at 20°C.
Given: compound and temperature
Asked for: rms speed
Strategy:
Calculate the molar mass of cis-2-butene. Be certain that all quantities are expressed in the appropriate units and then use Equation 10.8.5 to calculate the rms speed of the gas.
Solution:
To use Equation 10.8.4, we need to calculate the molar mass of cis-2-butene and make sure that each quantity is expressed in the appropriate units. Butene is C4H8, so its molar mass is 56.11 g/mol. Thus
\begin{align} u_{\rm rms} &= \sqrt{\dfrac{3RT}{M}} \[4pt] &=\rm\sqrt{\dfrac{3\times8.3145\;\dfrac{J}{K\cdot mol}\times(20+273)\;K}{56.11\times10^{-3}\;kg}}\[4pt] &=361\;m/s \end{align} \nonumber
or approximately 810 mi/h.
Exercise $1$
Calculate the rms speed of a sample of radon gas at 23°C.
Answer
$1.82 \times 10^2\; m/s$ (about 410 mi/h)
The kinetic molecular theory of gases demonstrates how a successful theory can explain previously observed empirical relationships (laws) in an intuitively satisfying way. Unfortunately, the actual gases that we encounter are not “ideal,” although their behavior usually approximates that of an ideal gas.
Summary
• The kinetic molecular theory of gases provides a molecular explanation for the observations that led to the development of the ideal gas law.
• Average kinetic energy:$\overline{e_K}=\dfrac{1}{2}m{u_{\rm rms}}^2=\dfrac{3}{2}\dfrac{R}{N_A}T, \nonumber$
• Root mean square speed: $u_{\rm rms}=\sqrt{\dfrac{u_1^2+u_2^2+\cdots u_N^2}{N}}, \nonumber$
• Kinetic molecular theory of gases: $u_{\rm rms}=\sqrt{\dfrac{3RT}{M}}. \nonumber$
• Graham’s law for effusion: $\dfrac{v_{\rm rms, B}}{v_{\rm rms,A}}=\sqrt{\dfrac{M_{\rm A}}{M_{\rm B}}} \nonumber$
Diffusion is the gradual mixing of gases to form a sample of uniform composition even in the absence of mechanical agitation. In contrast, effusion is the escape of a gas from a container through a tiny opening into an evacuated space. The rate of effusion of a gas is inversely proportional to the square root of its molar mass (Graham’s law), a relationship that closely approximates the rate of diffusion. As a result, light gases tend to diffuse and effuse much more rapidly than heavier gases. The mean free path of a molecule is the average distance it travels between collisions. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/10%3A_Gases/10.08%3A_Molecular_Effusion_and_Diffusion.txt |
Learning Objectives
• To recognize the differences between the behavior of an ideal gas and a real gas
• To understand how molecular volumes and intermolecular attractions cause the properties of real gases to deviate from those predicted by the ideal gas law.
The postulates of the kinetic molecular theory of gases ignore both the volume occupied by the molecules of a gas and all interactions between molecules, whether attractive or repulsive. In reality, however, all gases have nonzero molecular volumes. Furthermore, the molecules of real gases interact with one another in ways that depend on the structure of the molecules and therefore differ for each gaseous substance. In this section, we consider the properties of real gases and how and why they differ from the predictions of the ideal gas law. We also examine liquefaction, a key property of real gases that is not predicted by the kinetic molecular theory of gases.
Pressure, Volume, and Temperature Relationships in Real Gases
For an ideal gas, a plot of $PV/nRT$ versus $P$ gives a horizontal line with an intercept of 1 on the $PV/nRT$ axis. Real gases, however, show significant deviations from the behavior expected for an ideal gas, particularly at high pressures (Figure $\PageIndex{1a}$). Only at relatively low pressures (less than 1 atm) do real gases approximate ideal gas behavior (Figure $\PageIndex{1b}$).
Real gases also approach ideal gas behavior more closely at higher temperatures, as shown in Figure $2$ for $N_2$. Why do real gases behave so differently from ideal gases at high pressures and low temperatures? Under these conditions, the two basic assumptions behind the ideal gas law—namely, that gas molecules have negligible volume and that intermolecular interactions are negligible—are no longer valid.
Because the molecules of an ideal gas are assumed to have zero volume, the volume available to them for motion is always the same as the volume of the container. In contrast, the molecules of a real gas have small but measurable volumes. At low pressures, the gaseous molecules are relatively far apart, but as the pressure of the gas increases, the intermolecular distances become smaller and smaller (Figure $3$). As a result, the volume occupied by the molecules becomes significant compared with the volume of the container. Consequently, the total volume occupied by the gas is greater than the volume predicted by the ideal gas law. Thus at very high pressures, the experimentally measured value of PV/nRT is greater than the value predicted by the ideal gas law.
Moreover, all molecules are attracted to one another by a combination of forces. These forces become particularly important for gases at low temperatures and high pressures, where intermolecular distances are shorter. Attractions between molecules reduce the number of collisions with the container wall, an effect that becomes more pronounced as the number of attractive interactions increases. Because the average distance between molecules decreases, the pressure exerted by the gas on the container wall decreases, and the observed pressure is less than expected (Figure $4$). Thus as shown in Figure $2$, at low temperatures, the ratio of $PV/nRT$ is lower than predicted for an ideal gas, an effect that becomes particularly evident for complex gases and for simple gases at low temperatures. At very high pressures, the effect of nonzero molecular volume predominates. The competition between these effects is responsible for the minimum observed in the $PV/nRT$ versus $P$ plot for many gases.
Nonzero molecular volume makes the actual volume greater than predicted at high pressures; intermolecular attractions make the pressure less than predicted.
At high temperatures, the molecules have sufficient kinetic energy to overcome intermolecular attractive forces, and the effects of nonzero molecular volume predominate. Conversely, as the temperature is lowered, the kinetic energy of the gas molecules decreases. Eventually, a point is reached where the molecules can no longer overcome the intermolecular attractive forces, and the gas liquefies (condenses to a liquid).
The van der Waals Equation
The Dutch physicist Johannes van der Waals (1837–1923; Nobel Prize in Physics, 1910) modified the ideal gas law to describe the behavior of real gases by explicitly including the effects of molecular size and intermolecular forces. In his description of gas behavior, the so-called van der Waals equation,
$\underbrace{ \left(P + \dfrac{an^2}{V^2}\right)}_{\text{Pressure Term}} \overbrace{(V − nb)}^{\text{Pressure Term}} =nRT \label{10.9.1}$
a and b are empirical constants that are different for each gas. The values of $a$ and $b$ are listed in Table $1$ for several common gases.
Table $1$:: van der Waals Constants for Some Common Gases (see Table A8 for more complete list)
Gas a ((L2·atm)/mol2) b (L/mol)
He 0.03410 0.0238
Ne 0.205 0.0167
Ar 1.337 0.032
H2 0.2420 0.0265
N2 1.352 0.0387
O2 1.364 0.0319
Cl2 6.260 0.0542
NH3 4.170 0.0371
CH4 2.273 0.0430
CO2 3.610 0.0429
The pressure term in Equation $\ref{10.9.1}$ corrects for intermolecular attractive forces that tend to reduce the pressure from that predicted by the ideal gas law. Here, $n^2/V^2$ represents the concentration of the gas ($n/V$) squared because it takes two particles to engage in the pairwise intermolecular interactions of the type shown in Figure $4$. The volume term corrects for the volume occupied by the gaseous molecules.
The correction for volume is negative, but the correction for pressure is positive to reflect the effect of each factor on V and P, respectively. Because nonzero molecular volumes produce a measured volume that is larger than that predicted by the ideal gas law, we must subtract the molecular volumes to obtain the actual volume available. Conversely, attractive intermolecular forces produce a pressure that is less than that expected based on the ideal gas law, so the $an^2/V^2$ term must be added to the measured pressure to correct for these effects.
Example $1$
You are in charge of the manufacture of cylinders of compressed gas at a small company. Your company president would like to offer a 4.00 L cylinder containing 500 g of chlorine in the new catalog. The cylinders you have on hand have a rupture pressure of 40 atm. Use both the ideal gas law and the van der Waals equation to calculate the pressure in a cylinder at 25°C. Is this cylinder likely to be safe against sudden rupture (which would be disastrous and certainly result in lawsuits because chlorine gas is highly toxic)?
Given: volume of cylinder, mass of compound, pressure, and temperature
Asked for: safety
Strategy:
A Use the molar mass of chlorine to calculate the amount of chlorine in the cylinder. Then calculate the pressure of the gas using the ideal gas law.
B Obtain a and b values for Cl2 from Table $1$. Use the van der Waals equation ($\ref{10.9.1}$) to solve for the pressure of the gas. Based on the value obtained, predict whether the cylinder is likely to be safe against sudden rupture.
Solution:
A We begin by calculating the amount of chlorine in the cylinder using the molar mass of chlorine (70.906 g/mol):
\begin{align} n &=\dfrac{m}{M} \[4pt] &= \rm\dfrac{500\;g}{70.906\;g/mol} \[4pt] &=7.052\;mol\nonumber \end{align} \nonumber
Using the ideal gas law and the temperature in kelvin (298 K), we calculate the pressure:
\begin{align} P &=\dfrac{nRT}{V} \[4pt] &=\rm\dfrac{7.052\;mol\times 0.08206\dfrac{L\cdot atm}{mol\cdot K}\times298\;K}{4.00\;L} \[4pt] &= 43.1\;atm \end{align} \nonumber
If chlorine behaves like an ideal gas, you have a real problem!
B Now let’s use the van der Waals equation with the a and b values for Cl2 from Table $1$. Solving for $P$ gives
\begin{align}P&=\dfrac{nRT}{V-nb}-\dfrac{an^2}{V^2}\&=\rm\dfrac{7.052\;mol\times0.08206\dfrac{L\cdot atm}{mol\cdot K}\times298\;K}{4.00\;L-7.052\;mol\times0.0542\dfrac{L}{mol}}-\dfrac{6.260\dfrac{L^2atm}{mol^2}\times(7.052\;mol)^2}{(4.00\;L)^2}\&=\rm28.2\;atm\end{align} \nonumber
This pressure is well within the safety limits of the cylinder. The ideal gas law predicts a pressure 15 atm higher than that of the van der Waals equation.
Exercise $1$
A 10.0 L cylinder contains 500 g of methane. Calculate its pressure to two significant figures at 27°C using the
1. ideal gas law.
2. van der Waals equation.
Answer a
77 atm
Answer b
67 atm
Liquefaction of Gases
Liquefaction of gases is the condensation of gases into a liquid form, which is neither anticipated nor explained by the kinetic molecular theory of gases. Both the theory and the ideal gas law predict that gases compressed to very high pressures and cooled to very low temperatures should still behave like gases, albeit cold, dense ones. As gases are compressed and cooled, however, they invariably condense to form liquids, although very low temperatures are needed to liquefy light elements such as helium (for He, 4.2 K at 1 atm pressure).
Liquefaction can be viewed as an extreme deviation from ideal gas behavior. It occurs when the molecules of a gas are cooled to the point where they no longer possess sufficient kinetic energy to overcome intermolecular attractive forces. The precise combination of temperature and pressure needed to liquefy a gas depends strongly on its molar mass and structure, with heavier and more complex molecules usually liquefying at higher temperatures. In general, substances with large van der Waals $a$ coefficients are relatively easy to liquefy because large a coefficients indicate relatively strong intermolecular attractive interactions. Conversely, small molecules with only light elements have small a coefficients, indicating weak intermolecular interactions, and they are relatively difficult to liquefy. Gas liquefaction is used on a massive scale to separate O2, N2, Ar, Ne, Kr, and Xe. After a sample of air is liquefied, the mixture is warmed, and the gases are separated according to their boiling points.
A large value of a in the van der Waals equation indicates the presence of relatively strong intermolecular attractive interactions.
The ultracold liquids formed from the liquefaction of gases are called cryogenic liquids, from the Greek kryo, meaning “cold,” and genes, meaning “producing.” They have applications as refrigerants in both industry and biology. For example, under carefully controlled conditions, the very cold temperatures afforded by liquefied gases such as nitrogen (boiling point = 77 K at 1 atm) can preserve biological materials, such as semen for the artificial insemination of cows and other farm animals. These liquids can also be used in a specialized type of surgery called cryosurgery, which selectively destroys tissues with a minimal loss of blood by the use of extreme cold.
Moreover, the liquefaction of gases is tremendously important in the storage and shipment of fossil fuels (Figure $5$). Liquefied natural gas (LNG) and liquefied petroleum gas (LPG) are liquefied forms of hydrocarbons produced from natural gas or petroleum reserves. LNG consists mostly of methane, with small amounts of heavier hydrocarbons; it is prepared by cooling natural gas to below about −162°C. It can be stored in double-walled, vacuum-insulated containers at or slightly above atmospheric pressure. Because LNG occupies only about 1/600 the volume of natural gas, it is easier and more economical to transport. LPG is typically a mixture of propane, propene, butane, and butenes and is primarily used as a fuel for home heating. It is also used as a feedstock for chemical plants and as an inexpensive and relatively nonpolluting fuel for some automobiles.
Summary
No real gas exhibits ideal gas behavior, although many real gases approximate it over a range of conditions. Deviations from ideal gas behavior can be seen in plots of PV/nRT versus P at a given temperature; for an ideal gas, PV/nRT versus P = 1 under all conditions. At high pressures, most real gases exhibit larger PV/nRT values than predicted by the ideal gas law, whereas at low pressures, most real gases exhibit PV/nRT values close to those predicted by the ideal gas law. Gases most closely approximate ideal gas behavior at high temperatures and low pressures. Deviations from ideal gas law behavior can be described by the van der Waals equation, which includes empirical constants to correct for the actual volume of the gaseous molecules and quantify the reduction in pressure due to intermolecular attractive forces. If the temperature of a gas is decreased sufficiently, liquefaction occurs, in which the gas condenses into a liquid form. Liquefied gases have many commercial applications, including the transport of large amounts of gases in small volumes and the uses of ultracold cryogenic liquids. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/10%3A_Gases/10.09%3A_Real_Gases_-_Deviations_from_Ideal_Behavior.txt |
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These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
10.1: Characteristics of Gases
Q10.1.1
Explain the differences between the microscopic and the macroscopic properties of matter. Is the boiling point of a compound a microscopic or macroscopic property? molecular mass? Why?
Q10.1.2
How do the microscopic properties of matter influence the macroscopic properties? Can you relate molecular mass to boiling point? Why or why not?
Q10.1.3
For a substance that has gas, liquid, and solid phases, arrange these phases in order of increasing density.
1. strength of intermolecular interactions.
2. compressibility.
3. molecular motion.
4. order in the arrangement of the molecules or atoms.
Q10.1.4
Which elements of the periodic table exist as gases at room temperature and pressure? Of these, which are diatomic molecules and which are monatomic? Which elements are liquids at room temperature and pressure? Which portion of the periodic table contains elements whose binary hydrides are most likely gases at room temperature?
Q10.2.1
What four quantities must be known to completely describe a sample of a gas? What units are commonly used for each quantity?
Q10.2.2
If the applied force is constant, how does the pressure exerted by an object change as the area on which the force is exerted decreases? In the real world, how does this relationship apply to the ease of driving a small nail versus a large nail?
Q10.2.3
As the force on a fixed area increases, does the pressure increase or decrease? With this in mind, would you expect a heavy person to need smaller or larger snowshoes than a lighter person? Explain.
Q10.2.4
What do we mean by barometric pressure? Is the barometric pressure at the summit of Mt. Rainier greater than or less than the pressure in Miami, Florida? Why?
Q10.2.5
Which has the highest barometric pressure—a cave in the Himalayas, a mine in South Africa, or a beach house in Florida? Which has the lowest?
Q10.2.6
Mars has an average barometric pressure of 0.007 atm. Would it be easier or harder to drink liquid from a straw on Mars than on Earth? Explain your answer.
Q10.2.7
Is the pressure exerted by a 1.0 kg mass on a 2.0 m2 area greater than or less than the pressure exerted by a 1.0 kg mass on a 1.0 m2 area? What is the difference, if any, between the pressure of the atmosphere exerted on a 1.0 m2 piston and a 2.0 m2 piston?
Q10.2.8
If you used water in a barometer instead of mercury, what would be the major difference in the instrument?
Q10.2.9
Calculate the pressure in pascals and in atmospheres exerted by a carton of milk that weighs 1.5 kg and has a base of 7.0 cm × 7.0 cm. If the carton were lying on its side (height = 25 cm), would it exert more or less pressure? Explain your reasoning.
Q10.2.10
If barometric pressure at sea level is 1.0 × 105 Pa, what is the mass of air in kilograms above a 1.0 cm2 area of your skin as you lie on the beach? If barometric pressure is 8.2 × 104 Pa on a mountaintop, what is the mass of air in kilograms above a 4.0 cm2 patch of skin?
Q10.2.11
Complete the following table:
atm kPa mmHg torr
1.40
723
43.2
Q10.2.12
The SI unit of pressure is the pascal, which is equal to 1 N/m2. Show how the product of the mass of an object and the acceleration due to gravity result in a force that, when exerted on a given area, leads to a pressure in the correct SI units. What mass in kilograms applied to a 1.0 cm2 area is required to produce a pressure of
1. 1.0 atm?
2. 1.0 torr?
3. 1 mmHg?
4. 1 kPa?
Q10.2.13
If you constructed a manometer to measure gas pressures over the range 0.60–1.40 atm using the liquids given in the following table, how tall a column would you need for each liquid? The density of mercury is 13.5 g/cm3. Based on your results, explain why mercury is still used in barometers, despite its toxicity.
Liquid Density (20°C) Column Height (m)
isopropanol 0.785
coconut oil 0.924
glycerine 1.259
10.3: The Gas Laws
Q10.3.1
Sketch a graph of the volume of a gas versus the pressure on the gas. What would the graph of V versus P look like if volume was directly proportional to pressure?
Q10.3.2
What properties of a gas are described by Boyle’s law, Charles’s law, and Avogadro’s law? In each law, what quantities are held constant? Why does the constant in Boyle’s law depend on the amount of gas used and the temperature at which the experiments are carried out?
Q10.3.3
Use Charles’s law to explain why cooler air sinks.
Q10.3.4
Use Boyle’s law to explain why it is dangerous to heat even a small quantity of water in a sealed container.
Q10.3.5
A 1.00 mol sample of gas at 25°C and 1.0 atm has an initial volume of 22.4 L. Calculate the results of each change, assuming all the other conditions remain constant.
1. The pressure is changed to 85.7 mmHg. How many milliliters does the gas occupy?
2. The volume is reduced to 275 mL. What is the pressure in millimeters of mercury?
3. The pressure is increased to 25.3 atm. What is the temperature in degrees Celsius?
4. The sample is heated to 30°C. What is the volume in liters?
5. The sample is compressed to 1255 mL, and the pressure is increased to 2555 torr. What is the temperature of the gas in kelvins?
S10.3.5
1. 1.99 × 105 mL
2. 6.19 × 104 mmHg
3. 7270°C
4. 22.8 L
5. 51.4 K
Q10.3.6
A 1.00 mol sample of gas is at 300 K and 4.11 atm. What is the volume of the gas under these conditions? The sample is compressed to 6.0 atm at constant temperature, giving a volume of 3.99 L. Is this result consistent with Boyle’s law?
10.4: The Ideal Gas Equation
Q10.4.1
For an ideal gas, is volume directly proportional or inversely proportional to temperature? What is the volume of an ideal gas at absolute zero?
Q10.4.2
What is meant by STP? If a gas is at STP, what further information is required to completely describe the state of the gas?
Q10.4.3
For a given amount of a gas, the volume, temperature, and pressure under any one set of conditions are related to the volume, the temperature, and the pressure under any other set of conditions by the equation
\[ \dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2} \]
Q10.4.4
Derive this equation from the ideal gas law. At constant temperature, this equation reduces to one of the gas laws discussed previosuly; which one? At constant pressure, this equation reduces to one of the laws discussed in Section 10.3"; which one?
Q10.4.5
Predict the effect of each change on one variable if the other variables are held constant.
1. If the number of moles of gas increases, what is the effect on the temperature of the gas?
2. If the temperature of a gas decreases, what is the effect on the pressure of the gas?
3. If the volume of a gas increases, what is the effect on the temperature of the gas?
4. If the pressure of a gas increases, what is the effect on the number of moles of the gas?
Q10.4.6
What would the ideal gas law be if the following were true?
1. volume were proportional to pressure
2. temperature were proportional to amount
3. pressure were inversely proportional to temperature
4. volume were inversely proportional to temperature
5. both pressure and volume were inversely proportional to temperature
Given the following initial and final values, what additional information is needed to solve the problem using the ideal gas law?
Given Solve for
V1, T1, T2, n1 n 2
P1, P2, T2, n2 n 1
T1, T2 V 2
P1, n1 P 2
Q10.4.7
Given the following information and using the ideal gas law, what equation would you use to solve the problem?
Given Solve for
P1, P2, T1 T 2
V1, n1, n2 V 2
T1, T2, V1, V2, n2 n 1
Q10.4.8
Using the ideal gas law as a starting point, derive the relationship between the density of a gas and its molar mass. Which would you expect to be denser—nitrogen or oxygen? Why does radon gas accumulate in basements and mine shafts?
Q10.4.9
Use the ideal gas law to derive an equation that relates the remaining variables for a sample of an ideal gas if the following are held constant.
1. amount and volume
2. pressure and amount
3. temperature and volume
4. temperature and amount
5. pressure and temperature
S10.4.9
1. P/T = constant
2. V/T = constant (Charles’ law)
3. P/n = constant
4. PV = constant (Boyle’s law)
5. V/n = constant (Avogadro’s law)
Q10.4.10
Tennis balls that are made for Denver, Colorado, feel soft and do not bounce well at lower altitudes. Use the ideal gas law to explain this observation. Will a tennis ball designed to be used at sea level be harder or softer and bounce better or worse at higher altitudes?
Q10.4.11
Calculate the number of moles in each sample at STP.
1. 1580 mL of NO2
2. 847 cm3 of HCl
3. 4.792 L of H2
4. a 15.0 cm × 6.7 cm × 7.5 cm container of ethane
S10.4.11
1. 7.05 × 10−2 mol
2. 3.78 × 10−2 mol
3. 0.2138 mol
4. 3.4 × 10−2 mol
Q10.4.12
Calculate the number of moles in each sample at STP.
1. 2200 cm3 of CO2
2. 1200 cm3 of N2
3. 3800 mL of SO2
4. 13.75 L of NH3
Q10.4.13
Calculate the mass of each sample at STP.
1. 36 mL of HI
2. 550 L of H2S
3. 1380 cm3 of CH4
1. 0.21 g HI;
2. 840 g H2S;
3. 0.988 g CH4
Q10.4.14
Calculate the mass of each sample at STP.
1. 3.2 L of N2O
2. 65 cm3 of Cl2
3. 3600 mL of HBr
Q10.4.15
Calculate the volume in liters of each sample at STP.
1. 1.68 g of Kr
2. 2.97 kg of propane (C3H8)
3. 0.643 mg of (CH3)2O
S10.4.15
1. 0.449 L Kr
2. 1510 L C3H8
3. 3.13 × 10−4 L (CH3)2O
Q10.4.16
Calculate the volume in liters of each sample at STP.
1. 3.2 g of Xe
2. 465 mg of CS2
3. 5.34 kg of acetylene (C2H2)
Q10.4.17
Calculate the volume of each gas at STP.
1. 1.7 L at 28°C and 96.4 kPa
2. 38.0 mL at 17°C and 103.4 torr
3. 650 mL at −15°C and 723 mmHg
Q10.4.18
Calculate the volume of each gas at STP.
1. 2.30 L at 23°C and 740 mmHg
2. 320 mL at 13°C and 97.2 kPa
3. 100.5 mL at 35°C and 1.4 atm
1. 1.5 L
2. 4.87 mL
3. 650 mL
Q10.4.19
A 8.60 L tank of nitrogen gas at a pressure of 455 mmHg is connected to an empty tank with a volume of 5.35 L. What is the final pressure in the system after the valve connecting the two tanks is opened? Assume that the temperature is constant.
281 mmHg
Q10.4.20
At constant temperature, what pressure in atmospheres is needed to compress 14.2 L of gas initially at 25.2 atm to a volume of 12.4 L? What pressure is needed to compress 27.8 L of gas to 20.6 L under similar conditions?
Q10.4.21
One method for preparing hydrogen gas is to pass HCl gas over hot aluminum; the other product of the reaction is AlCl3. If you wanted to use this reaction to fill a balloon with a volume of 28,500 L at sea level and a temperature of 78°F, what mass of aluminum would you need? What volume of HCl at STP would you need?
S10.4.21
20.9 kg Al, 5.20 × 104 L HCl
Q10.4.21
An 3.50 g sample of acetylene is burned in excess oxygen according to the following reaction:
\[\ce{2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(l)}\]
At STP, what volume of CO2(g) is produced?
Q10.4.22
Calculate the density of ethylene (C2H4) under each set of conditions.
1. 7.8 g at 0.89 atm and 26°C
2. 6.3 mol at 102.6 kPa and 38°C
3. 9.8 g at 3.1 atm and −45°C
1. 1.0 g/L
2. 1.1 g/L
3. 4.6 g/L
Q10.4.23
Determine the density of O2 under each set of conditions.
1. 42 g at 1.1 atm and 25°C
2. 0.87 mol at 820 mmHg and 45°C
3. 16.7 g at 2.4 atm and 67°C
Q10.4.24
At 140°C, the pressure of a diatomic gas in a 3.0 L flask is 635 kPa. The mass of the gas is 88.7 g. What is the most likely identity of the gas?
Q10.4.25
What volume must a balloon have to hold 6.20 kg of H2 for an ascent from sea level to an elevation of 20,320 ft, where the temperature is −37°C and the pressure is 369 mmHg?
Q10.4.26
What must be the volume of a balloon that can hold 313.0 g of helium gas and ascend from sea level to an elevation of 1.5 km, where the temperature is 10.0°C and the pressure is 635.4 mmHg?
2174 L
Q10.4.27
A typical automobile tire is inflated to a pressure of 28.0 lb/in.2 Assume that the tire is inflated when the air temperature is 20°C; the car is then driven at high speeds, which increases the temperature of the tire to 43°C. What is the pressure in the tire? If the volume of the tire had increased by 8% at the higher temperature, what would the pressure be?
Q10.4.28
The average respiratory rate for adult humans is 20 breaths per minute. If each breath has a volume of 310 mL of air at 20°C and 0.997 atm, how many moles of air does a person inhale each day? If the density of air is 1.19 kg/m3, what is the average molecular mass of air?
Q10.4.29
Kerosene has a self-ignition temperature of 255°C. It is a common accelerant used by arsonists, but its presence is easily detected in fire debris by a variety of methods. If a 1.0 L glass bottle containing a mixture of air and kerosene vapor at an initial pressure of 1 atm and an initial temperature of 23°C is pressurized, at what pressure would the kerosene vapor ignite?
10.5: Further Applications of the Ideal-Gas Equations
Q10.5.1
Why are so many industrially important reactions carried out in the gas phase?
Q10.5.2
The volume of gas produced during a chemical reaction can be measured by collecting the gas in an inverted container filled with water. The gas forces water out of the container, and the volume of liquid displaced is a measure of the volume of gas. What additional information must be considered to determine the number of moles of gas produced? The volume of some gases cannot be measured using this method. What property of a gas precludes the use of this method?
Q10.5.3
Equal masses of two solid compounds (A and B) are placed in separate sealed flasks filled with air at 1 atm and heated to 50°C for 10 hours. After cooling to room temperature, the pressure in the flask containing A was 1.5 atm. In contrast, the pressure in the flask containing B was 0.87 atm. Suggest an explanation for these observations. Would the masses of samples A and B still be equal after the experiment? Why or why not?
Q10.5.4
Balance each chemical equation and then determine the volume of the indicated reactant at STP that is required for complete reaction. Assuming complete reaction, what is the volume of the products?
1. SO2(g) + O2(g) → SO3(g) given 2.4 mol of O2
2. H2(g) + Cl2(g) → HCl(g) given 0.78 g of H2
3. C2H6(g) + O2(g) → CO2(g) + H2O(g) given 1.91 mol of O2
Q10.5.5
During the smelting of iron, carbon reacts with oxygen to produce carbon monoxide, which then reacts with iron(III) oxide to produce iron metal and carbon dioxide. If 1.82 L of CO2 at STP is produced,
1. what mass of CO is consumed?
2. what volume of CO at STP is consumed?
3. how much O2 (in liters) at STP is used?
4. what mass of carbon is consumed?
5. how much iron metal (in grams) is produced?
Q10.5.6
Complete decomposition of a sample of potassium chlorate produced 1.34 g of potassium chloride and oxygen gas.
1. What is the mass of KClO3 in the original sample?
2. What mass of oxygen is produced?
3. What is the volume of oxygen produced at STP?
S10.5.6
1. 2.20 g KClO3
2. 0.863 g O2
3. 604 mL O2
Q10.5.7
The combustion of a 100.0 mg sample of an herbicide in excess oxygen produced 83.16 mL of CO2 and 72.9 mL of H2O vapor at STP. A separate analysis showed that the sample contained 16.44 mg of chlorine. If the sample is known to contain only C, H, Cl, and N, determine the percent composition and the empirical formula of the herbicide.
Q10.5.8
The combustion of a 300.0 mg sample of an antidepressant in excess oxygen produced 326 mL of CO2 and 164 mL of H2O vapor at STP. A separate analysis showed that the sample contained 23.28% oxygen. If the sample is known to contain only C, H, O, and N, determine the percent composition and the empirical formula of the antidepressant.
S10.5.8
Percent composition: 58.3% C, 4.93% H, 23.28% O, and 13.5% N; empirical formula: C10H10O3N2
10.6: Gas Mixtures and Partial Pressures
Q10.6.1
Dalton’s law of partial pressures makes one key assumption about the nature of the intermolecular interactions in a mixture of gases. What is it?
Q10.6.2
What is the relationship between the partial pressure of a gas and its mole fraction in a mixture?
Q10.6.3
What is the partial pressure of each gas if the following amounts of substances are placed in a 25.0 L container at 25°C? What is the total pressure of each mixture?
1. 1.570 mol of CH4 and 0.870 mol of CO2
2. 2.63 g of CO and 1.24 g of NO2
3. 1.78 kg of CH3Cl and 0.92 kg of SO2
Q10.6.4
What is the partial pressure of each gas in the following 3.0 L mixtures at 37°C as well as the total pressure?
1. 0.128 mol of SO2 and 0.098 mol of methane (CH4)
2. 3.40 g of acetylene (C2H2) and 1.54 g of He
3. 0.267 g of NO, 4.3 g of Ar, and 0.872 g of SO2
Q10.6.5
In a mixture of helium, oxygen, and methane in a 2.00 L container, the partial pressures of He and O2 are 13.6 kPa and 29.2 kPa, respectively, and the total pressure inside the container is 95.4 kPa. What is the partial pressure of methane? If the methane is ignited to initiate its combustion with oxygen and the system is then cooled to the original temperature of 30°C, what is the final pressure inside the container (in kilopascals)?
S10.6.5
52.6 kPa, 66.2 kPa
Q10.6.6
A 2.00 L flask originally contains 1.00 g of ethane (C2H6) and 32.0 g of oxygen at 21°C. During ignition, the ethane reacts completely with oxygen to produce CO2 and water vapor, and the temperature of the flask increases to 200°C. Determine the total pressure and the partial pressure of each gas before and after the reaction.
Q10.6.7
If a 20.0 L cylinder at 19°C is charged with 5.0 g each of sulfur dioxide and oxygen, what is the partial pressure of each gas? The sulfur dioxide is ignited in the oxygen to produce sulfur trioxide gas, and the mixture is allowed to cool to 19°C at constant pressure. What is the final volume of the cylinder? What is the partial pressure of each gas in the piston?
Q10.6.8
The highest point on the continent of Europe is Mt. Elbrus in Russia, with an elevation of 18,476 ft. The highest point on the continent of South America is Mt. Aconcagua in Argentina, with an elevation of 22,841 ft.
1. The following table shows the variation of atmospheric pressure with elevation. Use the data in the table to construct a plot of pressure versus elevation.
Elevation (km) Pressure in Summer (mmHg) Pressure in Winter (mmHg)
0.0 760.0 760.0
1.0 674.8 670.6
1.5 635.4 629.6
2.0 598.0 590.8
3.0 528.9 519.7
5.0 410.6 398.7
7.0 314.9 301.6
9.0 237.8 224.1
2. Use your graph to estimate the pressures in millimeters of mercury during the summer and the winter at the top of both mountains in both atmospheres and kilopascals.
3. Given that air is 20.95% O2 by volume, what is the partial pressure of oxygen in atmospheres during the summer at each location?
10.8: Molecular Effusion and Diffusion
Q10.8.1
Which of the following processes represents effusion, and which represents diffusion?
1. helium escaping from a hole in a balloon
2. vapor escaping from the surface of a liquid
3. gas escaping through a membrane
Q10.8.2
Which postulate of the kinetic molecular theory of gases most readily explains the observation that a helium-filled balloon is round?
Q10.8.3
Why is it relatively easy to compress a gas? How does the compressibility of a gas compare with that of a liquid? A solid? Why? Which of the postulates of the kinetic molecular theory of gases most readily explains these observations?
Q10.8.4
What happens to the average kinetic energy of a gas if the rms speed of its particles increases by a factor of 2? How is the rms speed different from the average speed?
Q10.8.5
Which gas—radon or helium—has a higher average kinetic energy at 100°C? Which has a higher average speed? Why? Which postulate of the kinetic molecular theory of gases most readily supports your answer?
Q10.8.6
What is the relationship between the average speed of a gas particle and the temperature of the gas? What happens to the distribution of molecular speeds if the temperature of a gas is increased? Decreased?
Q10.8.7
Qualitatively explain the relationship between the number of collisions of gas particles with the walls of a container and the pressure of a gas. How does increasing the temperature affect the number of collisions?
Q10.8.8
What happens to the average kinetic energy of a gas at constant temperature if the
1. volume of the gas is increased?
2. pressure of the gas is increased?
Q10.8.9
What happens to the density of a gas at constant temperature if the
1. volume of the gas is increased?
2. pressure of the gas is increased?
Q10.8.10
Use the kinetic molecular theory of gases to describe how a decrease in volume produces an increase in pressure at constant temperature. Similarly, explain how a decrease in temperature leads to a decrease in volume at constant pressure.
Q10.8.11
Graham’s law is valid only if the two gases are at the same temperature. Why?
Q10.8.12
If we lived in a helium atmosphere rather than in air, would we detect odors more or less rapidly than we do now? Explain your reasoning. Would we detect odors more or less rapidly at sea level or at high altitude? Why?
Q10.8.13
At a given temperature, what is the ratio of the rms speed of the atoms of Ar gas to the rms speed of molecules of H2 gas?
S10.8.13
At any temperature, the rms speed of hydrogen is 4.45 times that of argon.
Q10.8.14
At a given temperature, what is the ratio of the rms speed of molecules of CO gas to the rms speed of molecules of H2S gas?
Q10.8.15
What is the ratio of the rms speeds of argon and oxygen at any temperature? Which diffuses more rapidly?
Q10.8.16
What is the ratio of the rms speeds of Kr and NO at any temperature? Which diffuses more rapidly?
Q10.8.17
Deuterium (D) and tritium (T) are heavy isotopes of hydrogen. Tritium has an atomic mass of 3.016 amu and has a natural abundance of 0.000138%. The effusion of hydrogen gas (containing a mixture of H2, HD, and HT molecules) through a porous membrane can be used to obtain samples of hydrogen that are enriched in tritium. How many membrane passes are necessary to give a sample of hydrogen gas in which 1% of the hydrogen molecules are HT?
Q10.8.18
Samples of HBr gas and NH3 gas are placed at opposite ends of a 1 m tube. If the two gases are allowed to diffuse through the tube toward one another, at what distance from each end of the tube will the gases meet and form solid NH4Br?
10.9: Real Gases: Deviations from Ideal Behavior
Q10.9.1
What factors cause deviations from ideal gas behavior? Use a sketch to explain your answer based on interactions at the molecular level.
Q10.9.2
Explain the effect of nonzero atomic volume on the ideal gas law at high pressure. Draw a typical graph of volume versus 1/P for an ideal gas and a real gas.
Q10.9.3
For an ideal gas, the product of pressure and volume should be constant, regardless of the pressure. Experimental data for methane, however, show that the value of PV decreases significantly over the pressure range 0 to 120 atm at 0°C. The decrease in PV over the same pressure range is much smaller at 100°C. Explain why PV decreases with increasing temperature. Why is the decrease less significant at higher temperatures.
Q10.9.4
What is the effect of intermolecular forces on the liquefaction of a gas? At constant pressure and volume, does it become easier or harder to liquefy a gas as its temperature increases? Explain your reasoning. What is the effect of increasing the pressure on the liquefaction temperature?
Q10.9.5
Describe qualitatively what a and b, the two empirical constants in the van der Waals equation, represent.
Q10.9.6
In the van der Waals equation, why is the term that corrects for volume negative and the term that corrects for pressure positive? Why is n/V squared?
Q10.9.7
Liquefaction of a gas depends strongly on two factors. What are they? As temperature is decreased, which gas will liquefy first—ammonia, methane, or carbon monoxide? Why?
Q10.9.8
What is a cryogenic liquid? Describe three uses of cryogenic liquids.
Q10.9.9
Air consists primarily of O2, N2, Ar, Ne, Kr, and Xe. Use the concepts discussed in this chapter to propose two methods by which air can be separated into its components. Which component of air will be isolated first?
Q10.9.10
How can gas liquefaction facilitate the storage and transport of fossil fuels? What are the potential drawbacks to these methods?
Q10.9.11
The van der Waals constants for xenon are a = 4.19 (L2·atm)/mol2 and b = 0.0510 L/mol. If a 0.250 mol sample of xenon in a container with a volume of 3.65 L is cooled to −90°C, what is the pressure of the sample assuming ideal gas behavior? What would be the actual pressure under these conditions?
Q10.9.12
The van der Waals constants for water vapor are a = 5.46 (L2·atm)/mol2 and b = 0.0305 L/mol. If a 20.0 g sample of water in a container with a volume of 5.0 L is heated to 120°C, what is the pressure of the sample assuming ideal gas behavior? What would be the actual pressure under these conditions? | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/10%3A_Gases/10.E%3A_Exercises.txt |
10.1: Characteristics of Gases
Bulk matter can exist in three states: gas, liquid, and solid. Gases have the lowest density of the three, are highly compressible, and fill their containers completely. Elements that exist as gases at room temperature and pressure are clustered on the right side of the periodic table; they occur as either monatomic gases (the noble gases) or diatomic molecules (some halogens, N₂, O₂).
• Gases expand spontaneously to fill containers in which they are held, equaling their volume. Consequently, they are highly compressible.
• Gases form homogeneous mixtures with each other regardless of the identities or relative proportions of the component gases
• The characteristic properties of gases arise because the individual molecules are relatively far apart, hence, acting largely as though they were alone.
10.2: Pressure
Pressure is defined as the force exerted per unit area; it can be measured using a barometer or manometer. Four quantities must be known for a complete physical description of a sample of a gas: temperature, volume, amount, and pressure. Pressure is force per unit area of surface; the SI unit for pressure is the pascal (Pa), defined as 1 newton per square meter (N/m²). The pressure exerted by an object is proportional to the force it exerts and inversely proportional to the area.
• Pressure, P, is the Force, F, that acts on a given Area, A:
$P = F / A \nonumber$
Atmospheric Pressure and the Barometer
• The force, F, exerted by any object is the product of its mass, m, times its acceleration, a: F = ma
• SI unit for force is kg-m/s2 and is called the Newton (N)
• SI unit of pressure is N/m2, called a Pascal
• Standard atmospheric pressure: defined as 760 torr (760 mm Hg), or, in SI units, 101.325 kPa
• Atmosphere: unit of pressure equal to 760 torr; 1 atm = 101.325 kPa
10.3: The Gas Laws
The volume of a gas is inversely proportional to its pressure and directly proportional to its temperature and the amount of gas. Boyle showed that the volume of a sample of a gas is inversely proportional to pressure (Boyle’s law), Charles and Gay-Lussac demonstrated that the volume of a gas is directly proportional to its temperature at constant pressure (Charles’s law), and Avogadro showed that the volume of a gas is directly proportional to the number of moles of gas (Avogadro’s law).
The Pressure-Volume Relationship: Boyle’s Law
• Boyle’s law states that the volume of a fixed quantity of gas maintained at constant temperature is inversely proportional to the pressure. When two measurements are inversely proportional, one gets smaller as the other one gets larger.
$PV = constant \nonumber$
where $P$ = pressure, $V$ = volume
The Temperature-Volume Relationship: Charles’s Law
• Charles’s law: states that the volume of a fixed amount of gas maintained at constant pressure is directly proportional to its absolute temperature. Thus, as the pressure gets higher, so does the temperature.
$\dfrac{V}{T} = constant \nonumber$
where $V$ = volume, $T$ = Temperature
The Quantity-Volume Relationship: Avogadro’s Law
• Law of combining volumes: at a given pressure and temperature, the volumes of gases that react with one another are in the ratios of small whole numbers. (ie: 2H2 + O2 = 2H2O)
• Avogadro’s hypothesis: equal volumes of gases at the same temperature and pressure contain equal number of molecules
• Avogadro’s law: The volume of a gas maintained at constant temperature and pressure is directly proportional to the number of moles of the gas
$V = constant \times n \nonumber$
where V = volume, n = number of moles
10.4: The Ideal Gas Equation
The empirical relationships among the volume, the temperature, the pressure, and the amount of a gas can be combined into the ideal gas law, PV = nRT. The proportionality constant, R, is called the gas constant. The ideal gas law describes the behavior of an ideal gas, a hypothetical substance whose behavior can be explained quantitatively by the ideal gas law and the kinetic molecular theory of gases. Standard temperature and pressure (STP) is 0°C and 1 atm.
Ideal-gas equation
$PV = nRT \nonumber$
where P = pressure, V = volume, n = number of moles, R = gas constant, T = Temperature (always expressed on absolute-temperature scale, usually Kelvin)
The Gas constant ($R$) is the constant of proportionality in the ideal-gas equation. Some values of R are given below
Some values of R
Units Numerical value
L-atm/mol-K 0.08206
Cal/mol-K 1.987
J/mol-K 8.314
M3-Pa/mol-K 8.314
L-torr/mol-K 62.36
Standard temperature and pressure (STP): 0°C and 1 atm. 1 mol of gas at STP has a volume of 22.41 L (molar volume)
10.5: Further Applications of the Ideal-Gas Equations
The relationship between the amounts of products and reactants in a chemical reaction can be expressed in units of moles or masses of pure substances, of volumes of solutions, or of volumes of gaseous substances. The ideal gas law can be used to calculate the volume of gaseous products or reactants as needed. In the laboratory, gases produced in a reaction are often collected by the displacement of water from filled vessels; the amount of gas can be calculated from the volume of water displaced.
Density of a gas (\*\rho\) = density, M = molar mass):
$\rho = \dfrac{PM}{ RT} \nonumber$
Molar mass of a gas:
$M = \dfrac{ \rho RT }{ P} \nonumber$
10.6: Gas Mixtures and Partial Pressures
The pressure exerted by each gas in a gas mixture is independent of the pressure exerted by all other gases present. Consequently, the total pressure exerted by a mixture of gases is the sum of the partial pressures of the components (Dalton’s law of partial pressures). The amount of gas in a mixture may be described by its partial pressure or its mole fraction. In a mixture, the partial pressure of each gas is the product of the total pressure and the mole fraction.
• Partial pressure: the pressure exerted by a particular gas in a mixture
• Dalton’s law of partial pressures: law stating that the total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone
The total pressure at constant temperature and volume is determined by the total number of moles of gas present, whether that total represents just one substance or a mixture
Partial Pressures and Mole Fractions
• Mole fraction: the ratio of the number of one component of a mixture to the total moles of all components; abbreviated $\chi$, with a subscript to identify the components.
The partial pressure of a gas in a mixture is its mole fraction times the total pressure
10.7: Kinetic-Molecular Theory
The behavior of ideal gases is explained by the kinetic molecular theory of gases. Molecular motion, which leads to collisions between molecules and the container walls, explains pressure, and the large intermolecular distances in gases explain their high compressibility. Although all gases have the same average kinetic energy at a given temperature, they do not all possess the same root mean square speed. The actual values of speed and kinetic energy are not the same for all gas particles.
Kinetic-molecular theory: set of assumptions about the nature of gases. These assumptions, when translated into mathematical form, yield the ideal-gas equation
1. Gases consist of large numbers of molecules that are in continuous, random motion
2. The volume of all the molecules of the gas is negligible compared to the total volume in which the gas is contained
3. Attractive and repulsive forces between gas molecules are negligible
4. Energy can be transferred between molecules during collisions, but the average kinetic energy of the molecules does not change with time, as long as the temperature of the gas remains constant
5. The average kinetic energy of the molecules is proportional to absolute temperature. At any given temperature, the molecules of all gases have the same average kinetic energy
The pressure of a gas is caused by collisions of molecules with the walls of the container. The magnitude of the pressure is determined by how often and how "hard" the molecules strike the walls.
If two different gases are at the same temperature, they have the same average kinetic energy. If the temperature of a gas is doubled, its kinetic energy also doubles. Hence, molecular motion increases with increasing temperature.
• Root-mean-square (rms) speed: the square root of the squared speeds of the gas molecules in a gas sample. This quantity is the speed of a molecule possessing average kinetic energy.
The rms speed is important because the average kinetic energy of the gas molecules, $ε$, is related directly to $u^2$:
$ε = \dfrac{1}{2}mu^2 \nonumber$
where $m$ is the mass of the molecule
Because mass doesn’t change with temperature, the rms speed (and also the average speed) of molecules must increase as temperature increases
Applications to the Gas Laws
1. Effect of a volume increase at constant temperature: If the volume is increased, the molecules must move a longer distance between collisions. Consequently, there are fewer collisions per unit time with the container walls, and pressure decreases.
2. Effect of a temperature increase at constant volume: An increases in temperature means an increase in the average kinetic energy of the molecules. If there is no change in volume, there will be more collisions with the walls per unit time. Furthermore, the molecules strike harder, hence explaining how the observed pressure increases.
10.8: Molecular Effusion and Diffusion
Diffusion is the gradual mixing of gases to form a sample of uniform composition even in the absence of mechanical agitation. In contrast, effusion is the escape of a gas from a container through a tiny opening into an evacuated space. The rate of effusion of a gas is inversely proportional to the square root of its molar mass (Graham’s law), a relationship that closely approximates the rate of diffusion. As a result, light gases tend to diffuse and effuse much more rapidly than heavier gases.
A gas composed of light gas particles will have the same average kinetic energy as one composed of much heavier particles, provided that the two gases are at the same temperature. The mass, $m$, of the particles in the lighter gas is smaller that that in the heavier gas. Consequently, the particles of the lighter gas must have a higher rms speed, $u$, than the heavier one:
$u =\sqrt{\dfrac{3RT}{ M}} \nonumber$
Since M is in the denominator, the less massive the gas molecules, the higher the rms speed
• Effusion: the escape of a gas through an orifice or hole. The rate of effusion depends on the molecular mass of the gas.
• Diffusion: the spreading of one substance through a space or through another substance
Graham’s Law of Effusion
• Graham’s law: law stating that the rate of effusion of a gas is inversely proportional to the square root of its molecular weight
$\dfrac{r_1}{ r_2} = \sqrt{\dfrac{ M_2}{ M_1}} \nonumber$
where $r$ is the rate of effusion
• The rate of effusion is also directly proportional to the rms speed of the molecules. This is because the only way for the molecule to escape is to "collide" with the opening. Hence, the faster the molecules are moving, the greater the likelihood that a molecule will hit the opening and effuse.
Diffusion and Mean Free Path
• Diffusion, like effusion, is faster for light molecules than for heavy ones. The diffusion of gases is much slower than molecular speeds because of molecular collisions. Because of these collisions, the direction of motion of a gas molecule is constantly changing, making this a slow process.
• Mean free path: average distance traveled by a molecule between collisions. The higher the density of a gas, the smaller the mean free path
10.9: Real Gases - Deviations from Ideal Behavior
No real gas exhibits ideal gas behavior, although many real gases approximate it over a range of conditions. Gases most closely approximate ideal gas behavior at high temperatures and low pressures. Deviations from ideal gas law behavior can be described by the van der Waals equation, which includes empirical constants to correct for the actual volume of the gaseous molecules and quantify the reduction in pressure due to intermolecular attractive forces.
The ideal gas equation may be rearranged as follows to understand deviations from ideal-gas behavior:
$\dfrac{PV}{RT} = n \nonumber$
• For a mole of ideal gas (n = 1), the quantity PV / RT = 1 at all pressures. However, real gases do not behave in such a way. At high pressures, the deviation is very high, however it is less with lower pressures. In general, the deviations from idea behavior increase as temperature decreases, becoming significant near the temperature at which the gas is converted into a liquid.
• Basic assumptions in the kinetic molecular theory suggest that molecules of an ideal gas occupy no space and have no attraction for each other. Real molecules, however, do have finite volumes, and they do attract one another.
• Also, if the volume of the container in which the gas is contained is large, the molecules have plenty of free space, and do not take much of the container’s volume itself. However, as pressure increases, the gas molecules occupy a much larger fraction of the container’s volume.
• In addition, the attractive forces between molecules come into play at short distances, as when molecules are crowded together at high pressures. Because of these attractive forces, the impact of a given molecule with the wall of the container is lessened.
• Temperature determines how effective attractive forces between gas molecules are. As the gas is cooled, the average kinetic energy decreases, while intermolecular attractions remain constant.
The Van der Walls Equation
According to the ideal gas equation:
$\underbrace{P = \dfrac{nRT}{ V}}_{\text{ideal gas}} \nonumber$
According to Van der Waals:
$P = \dfrac{nRT}{ V – nb} – \dfrac{n^2a}{V^2} \nonumber$
Correction for volume of molecules – Correction for molecular attractions
• The Van der Waals constant b is a measure of the actual volume occupied by a mole of gas molecules; b has units of L/mol.
• The Van der Walls constant a has units of L2-atm/mol2. The magnitude of a reflects how strongly the gas molecules attract each other
Van der Waals equation
$\left[ P + \left(\dfrac{n^2a}{ V^2}\right) \right] (V – nb) = nRT \nonumber$
The Van der Waals constants a and b are different for each gas. The values of these constants generally increase with an increase in mass of the molecule and with an increase in the complexity of its structures. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/10%3A_Gases/10.S%3A_Gases_%28Summary%29.txt |
The physical properties of a substance depends upon its physical state. Water vapor, liquid water and ice all have the same chemical properties, but their physical properties are considerably different. In general Covalent bonds determine: molecular shape, bond energies, chemical properties, while intermolecular forces (non-covalent bonds) influence the physical properties of liquids and solids. The kinetic molecular theory of gases described in Chapter 10 gives a reasonably accurate description of the behavior of gases. A similar model can be applied to liquids, but it must take into account the nonzero volumes of particles and the presence of strong intermolecular attractive forces.
• 11.1: A Molecular Comparison of Gases, Liquids, and Solids
The state of a substance depends on the balance between the kinetic energy of the individual particles (molecules or atoms) and the intermolecular forces. The kinetic energy keeps the molecules apart and moving around, and is a function of the temperature of the substance and the intermolecular forces try to draw the particles together.
• 11.2: Intermolecular Forces
Molecules in liquids are held to other molecules by intermolecular interactions, which are weaker than the intramolecular interactions that hold molecules and polyatomic ions together. The three major types of intermolecular interactions are dipole–dipole interactions, London dispersion forces (these two are often referred to collectively as van der Waals forces), and hydrogen bonds.
• 11.3: Some Properties of Liquids
Surface tension, capillary action, and viscosity are unique properties of liquids that depend on the nature of intermolecular interactions. Surface tension is the energy required to increase the surface area of a liquid. Surfactants are molecules that reduce the surface tension of polar liquids like water. Capillary action is the phenomenon in which liquids rise up into a narrow tube called a capillary. The viscosity of a liquid is its resistance to flow.
• 11.4: Phase Changes
Fusion, vaporization, and sublimation are endothermic processes, whereas freezing, condensation, and deposition are exothermic processes. Changes of state are examples of phase changes, or phase transitions. All phase changes are accompanied by changes in the energy of a system. Changes from a more-ordered state to a less-ordered state (such as a liquid to a gas) are endothermic. Changes from a less-ordered state to a more-ordered state (such as a liquid to a solid) are always exothermic.
• 11.5: Vapor Pressure
Because the molecules of a liquid are in constant motion and possess a wide range of kinetic energies, at any moment some fraction of them has enough energy to escape from the surface of the liquid to enter the gas or vapor phase. This process, called vaporization or evaporation, generates a vapor pressure above the liquid. Molecules in the gas phase can collide with the liquid surface and reenter the liquid via condensation. Eventually, a steady state or dynamic equilibrium is reached.
• 11.6: Phase Diagrams
The states of matter exhibited by a substance under different temperatures and pressures can be summarized graphically in a phase diagram, which is a plot of pressure versus temperature. Phase diagrams contain discrete regions corresponding to the solid, liquid, and gas phases. The solid and liquid regions are separated by the melting curve of the substance, and the liquid and gas regions are separated by its vapor pressure curve, which ends at the critical point.
• 11.7: Structure of Solids
A crystalline solid can be represented by its unit cell, which is the smallest identical unit that when stacked together produces the characteristic three-dimensional structure. Solids are characterized by an extended three-dimensional arrangement of atoms, ions, or molecules in which the components are generally locked into their positions. The components can be arranged in a regular repeating three-dimensional array. The smallest repeating unit of a crystal lattice is the unit cell.
• 11.8: Bonding in Solids
The major types of solids are ionic, molecular, covalent, and metallic. Ionic solids consist of positively and negatively charged ions held together by electrostatic forces; the strength of the bonding is reflected in the lattice energy. Ionic solids tend to have high melting points and are rather hard. Molecular solids are held together by relatively weak forces, such as dipole–dipole interactions, hydrogen bonds, and London dispersion forces. Metallic solids have unusual properties.
• 11.E: Liquids and Intermolecular Forces (Exercises)
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al.
• 11.S: Liquids and Intermolecular Forces (Summary)
This is the summary Module for the chapter "Liquids and Intermolecular Forces" in the Brown et al. General Chemistry Textmap.
Thumbnail: A water drop. (CC BY 2.0; José Manuel Suárez).
11: Liquids and Intermolecular Forces
Learning Objectives
• To be familiar with the kinetic molecular description of liquids.
The physical properties of a substance depends upon its physical state. Water vapor, liquid water and ice all have the same chemical properties, but their physical properties are considerably different. In general covalent bonds determine: molecular shape, bond energies, chemical properties, while intermolecular forces (non-covalent bonds) influence the physical properties of liquids and solids. The kinetic molecular theory of gases gives a reasonably accurate description of the behavior of gases. A similar model can be applied to liquids, but it must take into account the nonzero volumes of particles and the presence of strong intermolecular attractive forces.
The state of a substance depends on the balance between the kinetic energy of the individual particles (molecules or atoms) and the intermolecular forces. The kinetic energy keeps the molecules apart and moving around, and is a function of the temperature of the substance. The intermolecular forces are attractive forces that try to draw the particles together (Figure \(2\)). A discussed previously, gasses are very sensitive to temperatures and pressure. However, these also affect liquids and solids too. Heating and cooling can change the kinetic energy of the particles in a substance, and so, we can change the physical state of a substance by heating or cooling it. Increasing the pressure on a substance forces the molecules closer together, which increases the strength of intermolecular forces.
Below is an overview of the general properties of the three different phases of matter.
Properties of Gases
• A collection of widely separated molecules
• The kinetic energy of the molecules is greater than any attractive forces between the molecules
• The lack of any significant attractive force between molecules allows a gas to expand to fill its container
• If attractive forces become large enough, then the gases exhibit non-ideal behavior
Properties of Liquids
• The intermolecular attractive forces are strong enough to hold molecules close together
• Liquids are more dense and less compressible than gasses
• Liquids have a definite volume, independent of the size and shape of their container
• The attractive forces are not strong enough, however, to keep neighboring molecules in a fixed position and molecules are free to move past or slide over one another
Thus, liquids can be poured and assume the shape of their containers.
Properties of Solids
• The intermolecular forces between neighboring molecules are strong enough to keep them locked in position
• Solids (like liquids) are not very compressible due to the lack of space between molecules
• If the molecules in a solid adopt a highly ordered packing arrangement, the structures are said to be crystalline
Due to the strong intermolecular forces between neighboring molecules, solids are rigid.
• Cooling a gas may change the state to a liquid
• Cooling a liquid may change the state to a solid
• Increasing the pressure on a gas may change the state to a liquid
• Increasing the pressure on a liquid may change the state to a solid
Physical Properties of Liquids
In a gas, the distance between molecules, whether monatomic or polyatomic, is very large compared with the size of the molecules; thus gases have a low density and are highly compressible. In contrast, the molecules in liquids are very close together, with essentially no empty space between them. As in gases, however, the molecules in liquids are in constant motion, and their kinetic energy (and hence their speed) depends on their temperature. We begin our discussion by examining some of the characteristic properties of liquids to see how each is consistent with a modified kinetic molecular description.
The properties of liquids can be explained using a modified version of the kinetic molecular theory of gases described previously. This model explains the higher density, greater order, and lower compressibility of liquids versus gases; the thermal expansion of liquids; why they diffuse; and why they adopt the shape (but not the volume) of their containers. A kinetic molecular description of liquids must take into account both the nonzero volumes of particles and the presence of strong intermolecular attractive forces. Solids and liquids have particles that are fairly close to one another, and are thus called "condensed phases" to distinguish them from gases
• Density: The molecules of a liquid are packed relatively close together. Consequently, liquids are much denser than gases. The density of a liquid is typically about the same as the density of the solid state of the substance. Densities of liquids are therefore more commonly measured in units of grams per cubic centimeter (g/cm3) or grams per milliliter (g/mL) than in grams per liter (g/L), the unit commonly used for gases.
• Molecular Order: Liquids exhibit short-range order because strong intermolecular attractive forces cause the molecules to pack together rather tightly. Because of their higher kinetic energy compared to the molecules in a solid, however, the molecules in a liquid move rapidly with respect to one another. Thus unlike the ions in the ionic solids, the molecules in liquids are not arranged in a repeating three-dimensional array. Unlike the molecules in gases, however, the arrangement of the molecules in a liquid is not completely random.
• Compressibility: Liquids have so little empty space between their component molecules that they cannot be readily compressed. Compression would force the atoms on adjacent molecules to occupy the same region of space.
• Thermal Expansion: The intermolecular forces in liquids are strong enough to keep them from expanding significantly when heated (typically only a few percent over a 100°C temperature range). Thus the volumes of liquids are somewhat fixed. Notice from Table S1 (with a shorten version in Table \(1\)) that the density of water, for example, changes by only about 3% over a 90-degree temperature range.
Table \(1\): The Density of Water at Various Temperatures
T (°C) Density (g/cm3)
0 0.99984
30 0.99565
60 0.98320
90 0.96535
• Diffusion: Molecules in liquids diffuse because they are in constant motion. A molecule in a liquid cannot move far before colliding with another molecule, however, so the mean free path in liquids is very short, and the rate of diffusion is much slower than in gases.
• Fluidity: Liquids can flow, adjusting to the shape of their containers, because their molecules are free to move. This freedom of motion and their close spacing allow the molecules in a liquid to move rapidly into the openings left by other molecules, in turn generating more openings, and so forth (Figure \(3\)). | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/11%3A_Liquids_and_Intermolecular_Forces/11.01%3A_A_Molecular_Comparison_of_Gases_Liquids_and_Solids.txt |
Learning Objectives
• To describe the intermolecular forces in liquids.
The properties of liquids are intermediate between those of gases and solids, but are more similar to solids. In contrast to intramolecular forces, such as the covalent bonds that hold atoms together in molecules and polyatomic ions, intermolecular forces hold molecules together in a liquid or solid. Intermolecular forces are generally much weaker than covalent bonds. For example, it requires 927 kJ to overcome the intramolecular forces and break both O–H bonds in 1 mol of water, but it takes only about 41 kJ to overcome the intermolecular attractions and convert 1 mol of liquid water to water vapor at 100°C. (Despite this seemingly low value, the intermolecular forces in liquid water are among the strongest such forces known!) Given the large difference in the strengths of intra- and intermolecular forces, changes between the solid, liquid, and gaseous states almost invariably occur for molecular substances without breaking covalent bonds.
The properties of liquids are intermediate between those of gases and solids, but are more similar to solids.
Intermolecular forces determine bulk properties, such as the melting points of solids and the boiling points of liquids. Liquids boil when the molecules have enough thermal energy to overcome the intermolecular attractive forces that hold them together, thereby forming bubbles of vapor within the liquid. Similarly, solids melt when the molecules acquire enough thermal energy to overcome the intermolecular forces that lock them into place in the solid.
Intermolecular forces are electrostatic in nature; that is, they arise from the interaction between positively and negatively charged species. Like covalent and ionic bonds, intermolecular interactions are the sum of both attractive and repulsive components. Because electrostatic interactions fall off rapidly with increasing distance between molecules, intermolecular interactions are most important for solids and liquids, where the molecules are close together. These interactions become important for gases only at very high pressures, where they are responsible for the observed deviations from the ideal gas law at high pressures.
In this section, we explicitly consider three kinds of intermolecular interactions. There are two additional types of electrostatic interaction that you are already familiar with: the ion–ion interactions that are responsible for ionic bonding, and the ion–dipole interactions that occur when ionic substances dissolve in a polar substance such as water. The first two are often described collectively as van der Waals forces.
Dipole–Dipole Interactions
Polar covalent bonds behave as if the bonded atoms have localized fractional charges that are equal but opposite (i.e., the two bonded atoms generate a dipole). If the structure of a molecule is such that the individual bond dipoles do not cancel one another, then the molecule has a net dipole moment. Molecules with net dipole moments tend to align themselves so that the positive end of one dipole is near the negative end of another and vice versa, as shown in Figure \(\PageIndex{1a}\).
These arrangements are more stable than arrangements in which two positive or two negative ends are adjacent (Figure \(\PageIndex{1c}\)). Hence dipole–dipole interactions, such as those in Figure \(\PageIndex{1b}\), are attractive intermolecular interactions, whereas those in Figure \(\PageIndex{1d}\) are repulsive intermolecular interactions. Because molecules in a liquid move freely and continuously, molecules always experience both attractive and repulsive dipole–dipole interactions simultaneously, as shown in Figure \(2\). On average, however, the attractive interactions dominate.
Because each end of a dipole possesses only a fraction of the charge of an electron, dipole–dipole interactions are substantially weaker than the interactions between two ions, each of which has a charge of at least ±1, or between a dipole and an ion, in which one of the species has at least a full positive or negative charge. In addition, the attractive interaction between dipoles falls off much more rapidly with increasing distance than do the ion–ion interactions. Recall that the attractive energy between two ions is proportional to 1/r, where r is the distance between the ions. Doubling the distance (r → 2r) decreases the attractive energy by one-half. In contrast, the energy of the interaction of two dipoles is proportional to 1/r3, so doubling the distance between the dipoles decreases the strength of the interaction by 23, or 8-fold. Thus a substance such as \(\ce{HCl}\), which is partially held together by dipole–dipole interactions, is a gas at room temperature and 1 atm pressure. Conversely, \(\ce{NaCl}\), which is held together by interionic interactions, is a high-melting-point solid. Within a series of compounds of similar molar mass, the strength of the intermolecular interactions increases as the dipole moment of the molecules increases, as shown in Table \(1\).
Table \(1\): Relationships Between the Dipole Moment and the Boiling Point for Organic Compounds of Similar Molar Mass
Compound Molar Mass (g/mol) Dipole Moment (D) Boiling Point (K)
C3H6 (cyclopropane) 42 0 240
CH3OCH3 (dimethyl ether) 46 1.30 248
CH3CN (acetonitrile) 41 3.9 355
The attractive energy between two ions is proportional to 1/r, whereas the attractive energy between two dipoles is proportional to 1/r6.
Video Discussing Dipole Intermolecular Forces. Source: Dipole Intermolecular Force, YouTube(opens in new window) [youtu.be]
Example \(1\)
Arrange ethyl methyl ether (CH3OCH2CH3), 2-methylpropane [isobutane, (CH3)2CHCH3], and acetone (CH3COCH3) in order of increasing boiling points. Their structures are as follows:
Given: compounds.
Asked for: order of increasing boiling points.
Strategy:
Compare the molar masses and the polarities of the compounds. Compounds with higher molar masses and that are polar will have the highest boiling points.
Solution:
The three compounds have essentially the same molar mass (58–60 g/mol), so we must look at differences in polarity to predict the strength of the intermolecular dipole–dipole interactions and thus the boiling points of the compounds.
The first compound, 2-methylpropane, contains only C–H bonds, which are not very polar because C and H have similar electronegativities. It should therefore have a very small (but nonzero) dipole moment and a very low boiling point.
Ethyl methyl ether has a structure similar to H2O; it contains two polar C–O single bonds oriented at about a 109° angle to each other, in addition to relatively nonpolar C–H bonds. As a result, the C–O bond dipoles partially reinforce one another and generate a significant dipole moment that should give a moderately high boiling point.
Acetone contains a polar C=O double bond oriented at about 120° to two methyl groups with nonpolar C–H bonds. The C–O bond dipole therefore corresponds to the molecular dipole, which should result in both a rather large dipole moment and a high boiling point.
Thus we predict the following order of boiling points:
2-methylpropane < ethyl methyl ether < acetone
This result is in good agreement with the actual data: 2-methylpropane, boiling point = −11.7°C, and the dipole moment (μ) = 0.13 D; methyl ethyl ether, boiling point = 7.4°C and μ = 1.17 D; acetone, boiling point = 56.1°C and μ = 2.88 D.
Exercise \(1\)
Arrange carbon tetrafluoride (CF4), ethyl methyl sulfide (CH3SC2H5), dimethyl sulfoxide [(CH3)2S=O], and 2-methylbutane [isopentane, (CH3)2CHCH2CH3] in order of decreasing boiling points.
Answer
dimethyl sulfoxide (boiling point = 189.9°C) > ethyl methyl sulfide (boiling point = 67°C) > 2-methylbutane (boiling point = 27.8°C) > carbon tetrafluoride (boiling point = −128°C)
London Dispersion Forces
Thus far, we have considered only interactions between polar molecules. Other factors must be considered to explain why many nonpolar molecules, such as bromine, benzene, and hexane, are liquids at room temperature; why others, such as iodine and naphthalene, are solids. Even the noble gases can be liquefied or solidified at low temperatures, high pressures, or both (Table \(2\)).
What kind of attractive forces can exist between nonpolar molecules or atoms? This question was answered by Fritz London (1900–1954), a German physicist who later worked in the United States. In 1930, London proposed that temporary fluctuations in the electron distributions within atoms and nonpolar molecules could result in the formation of short-lived instantaneous dipole moments, which produce attractive forces called London dispersion forces between otherwise nonpolar substances.
Table \(2\): Normal Melting and Boiling Points of Some Elements and Nonpolar Compounds
Substance Molar Mass (g/mol) Melting Point (°C) Boiling Point (°C)
Ar 40 −189.4 −185.9
Xe 131 −111.8 −108.1
N2 28 −210 −195.8
O2 32 −218.8 −183.0
F2 38 −219.7 −188.1
I2 254 113.7 184.4
CH4 16 −182.5 −161.5
Consider a pair of adjacent He atoms, for example. On average, the two electrons in each He atom are uniformly distributed around the nucleus. Because the electrons are in constant motion, however, their distribution in one atom is likely to be asymmetrical at any given instant, resulting in an instantaneous dipole moment. As shown in part (a) in Figure \(3\), the instantaneous dipole moment on one atom can interact with the electrons in an adjacent atom, pulling them toward the positive end of the instantaneous dipole or repelling them from the negative end. The net effect is that the first atom causes the temporary formation of a dipole, called an induced dipole, in the second. Interactions between these temporary dipoles cause atoms to be attracted to one another. These attractive interactions are weak and fall off rapidly with increasing distance. London was able to show with quantum mechanics that the attractive energy between molecules due to temporary dipole–induced dipole interactions falls off as 1/r6. Doubling the distance therefore decreases the attractive energy by 26, or 64-fold.
Instantaneous dipole–induced dipole interactions between nonpolar molecules can produce intermolecular attractions just as they produce interatomic attractions in monatomic substances like Xe. This effect, illustrated for two H2 molecules in part (b) in Figure \(3\), tends to become more pronounced as atomic and molecular masses increase (Table \(2\)). For example, Xe boils at −108.1°C, whereas He boils at −269°C. The reason for this trend is that the strength of London dispersion forces is related to the ease with which the electron distribution in a given atom can be perturbed. In small atoms such as He, the two 1s electrons are held close to the nucleus in a very small volume, and electron–electron repulsions are strong enough to prevent significant asymmetry in their distribution. In larger atoms such as Xe, however, the outer electrons are much less strongly attracted to the nucleus because of filled intervening shells. As a result, it is relatively easy to temporarily deform the electron distribution to generate an instantaneous or induced dipole. The ease of deformation of the electron distribution in an atom or molecule is called its polarizability. Because the electron distribution is more easily perturbed in large, heavy species than in small, light species, we say that heavier substances tend to be much more polarizable than lighter ones.
For similar substances, London dispersion forces get stronger with increasing molecular size.
The polarizability of a substance also determines how it interacts with ions and species that possess permanent dipoles. Thus, London dispersion forces are responsible for the general trend toward higher boiling points with increased molecular mass and greater surface area in a homologous series of compounds, such as the alkanes (part (a) in Figure \(4\)). The strengths of London dispersion forces also depend significantly on molecular shape because shape determines how much of one molecule can interact with its neighboring molecules at any given time. For example, part (b) in Figure \(4\) shows 2,2-dimethylpropane (neopentane) and n-pentane, both of which have the empirical formula C5H12. Neopentane is almost spherical, with a small surface area for intermolecular interactions, whereas n-pentane has an extended conformation that enables it to come into close contact with other n-pentane molecules. As a result, the boiling point of neopentane (9.5°C) is more than 25°C lower than the boiling point of n-pentane (36.1°C).
All molecules, whether polar or nonpolar, are attracted to one another by London dispersion forces in addition to any other attractive forces that may be present. In general, however, dipole–dipole interactions in small polar molecules are significantly stronger than London dispersion forces, so the former predominate.
Video Discussing London/Dispersion Intermolecular Forces. Source: Dispersion Intermolecular Force, YouTube(opens in new window) [youtu.be]
Example \(2\)
Arrange n-butane, propane, 2-methylpropane [isobutene, (CH3)2CHCH3], and n-pentane in order of increasing boiling points.
Given: compounds
Asked for: order of increasing boiling points
Strategy:
Determine the intermolecular forces in the compounds, and then arrange the compounds according to the strength of those forces. The substance with the weakest forces will have the lowest boiling point.
Solution:
The four compounds are alkanes and nonpolar, so London dispersion forces are the only important intermolecular forces. These forces are generally stronger with increasing molecular mass, so propane should have the lowest boiling point and n-pentane should have the highest, with the two butane isomers falling in between. Of the two butane isomers, 2-methylpropane is more compact, and n-butane has the more extended shape. Consequently, we expect intermolecular interactions for n-butane to be stronger due to its larger surface area, resulting in a higher boiling point. The overall order is thus as follows, with actual boiling points in parentheses: propane (−42.1°C) < 2-methylpropane (−11.7°C) < n-butane (−0.5°C) < n-pentane (36.1°C).
Exercise \(2\)
Arrange GeH4, SiCl4, SiH4, CH4, and GeCl4 in order of decreasing boiling points.
Answer
GeCl4 (87°C) > SiCl4 (57.6°C) > GeH4 (−88.5°C) > SiH4 (−111.8°C) > CH4 (−161°C)
Hydrogen Bonds
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 London dispersion forces dominate, as illustrated for the covalent hydrides of elements of groups 14–17 in Figure \(5\). 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 H2Te and H2Se 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 hydrogen bonds, as shown for ice in Figure \(6\). A hydrogen bond is usually indicated by a dotted line between the hydrogen atom attached to O, N, or F (the hydrogen bond donor) and the atom that has the lone pair of electrons (the hydrogen bond acceptor). 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 not 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, cage like 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 both a hydrogen bond donor and 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.
Video Discussing Hydrogen Bonding Intermolecular Forces. Source: Hydrogen Bonding Intermolecular Force, YouTube(opens in new window) [youtu.be]
Example \(3\)
Considering CH3OH, C2H6, Xe, and (CH3)3N, which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures.
Given: compounds
Asked for: formation of hydrogen bonds and structure
Strategy:
1. Identify the compounds with a hydrogen atom attached to O, N, or F. These are likely to be able to act as hydrogen bond donors.
2. Of the compounds that can act as hydrogen bond donors, identify those that also contain lone pairs of electrons, which allow them to be hydrogen bond acceptors. If a substance is both a hydrogen donor and a hydrogen bond acceptor, draw a structure showing the hydrogen bonding.
Solution:
A. Of the species listed, xenon (Xe), ethane (C2H6), and trimethylamine [(CH3)3N] do not contain a hydrogen atom attached to O, N, or F; hence they cannot act as hydrogen bond donors.
B. The one compound that can act as a hydrogen bond donor, methanol (CH3OH), 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:
Exercise \(3\)
Considering CH3CO2H, (CH3)3N, NH3, and CH3F, which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures.
Answer
CH3CO2H and 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 HF can form only two hydrogen bonds at a time as can, on average, pure liquid NH3. 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 four hydrogen bonds at a time.
Example \(4\): Buckyballs
Arrange C60 (buckminsterfullerene, which has a cage structure), NaCl, He, Ar, and N2O in order of increasing boiling points.
Given: compounds.
Asked for: order of increasing boiling points.
Strategy:
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.
Solution
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 N2O have very similar molar masses (40 and 44 g/mol, respectively), but N2O is polar while Ar is not. Consequently, N2O should have a higher boiling point. A C60 molecule is nonpolar, but its molar mass is 720 g/mol, much greater than that of Ar or N2O. Because the boiling points of nonpolar substances increase rapidly with molecular mass, C60 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) < N2O (−88.5°C) < C60 (>280°C) < NaCl (1465°C).
Exercise \(4\)
Arrange 2,4-dimethylheptane, Ne, CS2, Cl2, and KBr in order of decreasing boiling points.
Answer
KBr (1435°C) > 2,4-dimethylheptane (132.9°C) > CS2 (46.6°C) > Cl2 (−34.6°C) > Ne (−246°C)
Example \(5\)
Identify the most significant intermolecular force in each substance.
1. C3H8
2. CH3OH
3. H2S
Solution
1. Although C–H bonds are polar, they are only minimally polar. The most significant intermolecular force for this substance would be dispersion forces.
2. This molecule has an H atom bonded to an O atom, so it will experience hydrogen bonding.
3. Although this molecule does not experience hydrogen bonding, the Lewis electron dot diagram and VSEPR indicate that it is bent, so it has a permanent dipole. The most significant force in this substance is dipole-dipole interaction.
Exercise \(6\)
Identify the most significant intermolecular force in each substance.
1. HF
2. HCl
Answer a
hydrogen bonding
Answer b
dipole-dipole interactions
Summary
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 van der Waals forces), and hydrogen bonds. Dipole–dipole interactions 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/r3, where r is the distance between dipoles. London dispersion forces are due to the formation of instantaneous dipole moments 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 induced dipole in adjacent molecules; their energy falls off as 1/r6. Larger atoms tend to be more polarizable than smaller ones, because their outer electrons are less tightly bound and are therefore more easily perturbed. Hydrogen bonds 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 hydrogen bond donor) can interact strongly with a lone pair of electrons of a partially negatively charged O, N, or F atom on adjacent molecules (the hydrogen bond acceptor). Because of strong O⋅⋅⋅H hydrogen bonding between water molecules, water has an unusually high boiling point, and ice has an open, cage like structure that is less dense than liquid water. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/11%3A_Liquids_and_Intermolecular_Forces/11.02%3A_Intermolecular_Forces.txt |
Learning Objectives
• To describe the unique properties of liquids.
Although you have been introduced to some of the interactions that hold molecules together in a liquid, we have not yet discussed the consequences of those interactions for the bulk properties of liquids. We now turn our attention to three unique properties of liquids that intimately depend on the nature of intermolecular interactions:
• surface tension,
• capillary action, and
• viscosity.
Surface Tension
If liquids tend to adopt the shapes of their containers, then why do small amounts of water on a freshly waxed car form raised droplets instead of a thin, continuous film? The answer lies in a property called surface tension, which depends on intermolecular forces. Surface tension is the energy required to increase the surface area of a liquid by a unit amount and varies greatly from liquid to liquid based on the nature of the intermolecular forces, e.g., water with hydrogen bonds has a surface tension of 7.29 x 10-2 J/m2 (at 20°C), while mercury with metallic bonds has as surface tension that is 15 times higher: 4.86 x 10-1 J/m2 (at 20°C).
Figure \(1\) presents a microscopic view of a liquid droplet. A typical molecule in the interior of the droplet is surrounded by other molecules that exert attractive forces from all directions. Consequently, there is no net force on the molecule that would cause it to move in a particular direction. In contrast, a molecule on the surface experiences a net attraction toward the drop because there are no molecules on the outside to balance the forces exerted by adjacent molecules in the interior. Because a sphere has the smallest possible surface area for a given volume, intermolecular attractive interactions between water molecules cause the droplet to adopt a spherical shape. This maximizes the number of attractive interactions and minimizes the number of water molecules at the surface. Hence raindrops are almost spherical, and drops of water on a waxed (nonpolar) surface, which does not interact strongly with water, form round beads. A dirty car is covered with a mixture of substances, some of which are polar. Attractive interactions between the polar substances and water cause the water to spread out into a thin film instead of forming beads.
The same phenomenon holds molecules together at the surface of a bulk sample of water, almost as if they formed a skin. When filling a glass with water, the glass can be overfilled so that the level of the liquid actually extends above the rim. Similarly, a sewing needle or a paper clip can be placed on the surface of a glass of water where it “floats,” even though steel is much denser than water. Many insects take advantage of this property to walk on the surface of puddles or ponds without sinking. This is even observable in the zero gravity conditions of space as shown in Figure \(2\) (and more so in the video link) where water wrung from a wet towel continues to float along the towel's surface!
Such phenomena are manifestations of surface tension, which is defined as the energy required to increase the surface area of a liquid by a specific amount. Surface tension is therefore measured as energy per unit area, such as joules per square meter (J/m2) or dyne per centimeter (dyn/cm), where 1 dyn = 1 × 10−5 N. The values of the surface tension of some representative liquids are listed in Table \(1\). Note the correlation between the surface tension of a liquid and the strength of the intermolecular forces: the stronger the intermolecular forces, the higher the surface tension. For example, water, with its strong intermolecular hydrogen bonding, has one of the highest surface tension values of any liquid, whereas low-boiling-point organic molecules, which have relatively weak intermolecular forces, have much lower surface tensions. Mercury is an apparent anomaly, but its very high surface tension is due to the presence of strong metallic bonding.
Table \(1\): Surface Tension, Viscosity, Vapor Pressure (at 25°C Unless Otherwise Indicated), and Normal Boiling Points of Common Liquids
Substance Surface Tension (× 10−3 J/m2) Viscosity (mPa•s) Vapor Pressure (mmHg) Normal Boiling Point (°C)
Organic Compounds
diethyl ether 17 0.22 531 34.6
n-hexane 18 0.30 149 68.7
acetone 23 0.31 227 56.5
ethanol 22 1.07 59 78.3
ethylene glycol 48 16.1 ~0.08 198.9
Liquid Elements
bromine 41 0.94 218 58.8
mercury 486 1.53 0.0020 357
Water
0°C 75.6 1.79 4.6
20°C 72.8 1.00 17.5
60°C 66.2 0.47 149
100°C 58.9 0.28 760
Adding soaps and detergents that disrupt the intermolecular attractions between adjacent water molecules can reduce the surface tension of water. Because they affect the surface properties of a liquid, soaps and detergents are called surface-active agents, or surfactants. In the 1960s, US Navy researchers developed a method of fighting fires aboard aircraft carriers using “foams,” which are aqueous solutions of fluorinated surfactants. The surfactants reduce the surface tension of water below that of fuel, so the fluorinated solution is able to spread across the burning surface and extinguish the fire. Such foams are now used universally to fight large-scale fires of organic liquids.
Capillary Action
Intermolecular forces also cause a phenomenon called capillary action, which is the tendency of a polar liquid to rise against gravity into a small-diameter tube (a capillary), as shown in Figure \(3\). When a glass capillary is is placed in liquid water, water rises up into the capillary. The height to which the water rises depends on the diameter of the tube and the temperature of the water but not on the angle at which the tube enters the water. The smaller the diameter, the higher the liquid rises.
• Cohesive forces bind molecules of the same type together
• Adhesive forces bind a substance to a surface
Capillary action is the net result of two opposing sets of forces: cohesive forces, which are the intermolecular forces that hold a liquid together, and adhesive forces, which are the attractive forces between a liquid and the substance that composes the capillary. Water has both strong adhesion to glass, which contains polar SiOH groups, and strong intermolecular cohesion. When a glass capillary is put into water, the surface tension due to cohesive forces constricts the surface area of water within the tube, while adhesion between the water and the glass creates an upward force that maximizes the amount of glass surface in contact with the water. If the adhesive forces are stronger than the cohesive forces, as is the case for water, then the liquid in the capillary rises to the level where the downward force of gravity exactly balances this upward force. If, however, the cohesive forces are stronger than the adhesive forces, as is the case for mercury and glass, the liquid pulls itself down into the capillary below the surface of the bulk liquid to minimize contact with the glass (Figure \(4\)). The upper surface of a liquid in a tube is called the meniscus, and the shape of the meniscus depends on the relative strengths of the cohesive and adhesive forces. In liquids such as water, the meniscus is concave; in liquids such as mercury, however, which have very strong cohesive forces and weak adhesion to glass, the meniscus is convex (Figure \(4\)).
Polar substances are drawn up a glass capillary and generally have a concave meniscus.
Fluids and nutrients are transported up the stems of plants or the trunks of trees by capillary action. Plants contain tiny rigid tubes composed of cellulose, to which water has strong adhesion. Because of the strong adhesive forces, nutrients can be transported from the roots to the tops of trees that are more than 50 m tall. Cotton towels are also made of cellulose; they absorb water because the tiny tubes act like capillaries and “wick” the water away from your skin. The moisture is absorbed by the entire fabric, not just the layer in contact with your body.
Viscosity
Viscosity (η) is the resistance of a liquid to flow. Some liquids, such as gasoline, ethanol, and water, flow very readily and hence have a low viscosity. Others, such as motor oil, molasses, and maple syrup, flow very slowly and have a high viscosity. The two most common methods for evaluating the viscosity of a liquid are (1) to measure the time it takes for a quantity of liquid to flow through a narrow vertical tube and (2) to measure the time it takes steel balls to fall through a given volume of the liquid. The higher the viscosity, the slower the liquid flows through the tube and the steel balls fall. Viscosity is expressed in units of the poise (mPa•s); the higher the number, the higher the viscosity. The viscosities of some representative liquids are listed in Table 11.3.1 and show a correlation between viscosity and intermolecular forces. Because a liquid can flow only if the molecules can move past one another with minimal resistance, strong intermolecular attractive forces make it more difficult for molecules to move with respect to one another. The addition of a second hydroxyl group to ethanol, for example, which produces ethylene glycol (HOCH2CH2OH), increases the viscosity 15-fold. This effect is due to the increased number of hydrogen bonds that can form between hydroxyl groups in adjacent molecules, resulting in dramatically stronger intermolecular attractive forces.
There is also a correlation between viscosity and molecular shape. Liquids consisting of long, flexible molecules tend to have higher viscosities than those composed of more spherical or shorter-chain molecules. The longer the molecules, the easier it is for them to become “tangled” with one another, making it more difficult for them to move past one another. London dispersion forces also increase with chain length. Due to a combination of these two effects, long-chain hydrocarbons (such as motor oils) are highly viscous.
Viscosity increases as intermolecular interactions or molecular size increases.
Video Discussing Surface Tension and Viscosity. Video Link: Surface Tension, Viscosity, & Melting Point, YouTube(opens in new window) [youtu.be]
Application: Motor Oils
Motor oils and other lubricants demonstrate the practical importance of controlling viscosity. The oil in an automobile engine must effectively lubricate under a wide range of conditions, from subzero starting temperatures to the 200°C that oil can reach in an engine in the heat of the Mojave Desert in August. Viscosity decreases rapidly with increasing temperatures because the kinetic energy of the molecules increases, and higher kinetic energy enables the molecules to overcome the attractive forces that prevent the liquid from flowing. As a result, an oil that is thin enough to be a good lubricant in a cold engine will become too “thin” (have too low a viscosity) to be effective at high temperatures.
The viscosity of motor oils is described by an SAE (Society of Automotive Engineers) rating ranging from SAE 5 to SAE 50 for engine oils: the lower the number, the lower the viscosity (Figure \(5\)). So-called single-grade oils can cause major problems. If they are viscous enough to work at high operating temperatures (SAE 50, for example), then at low temperatures, they can be so viscous that a car is difficult to start or an engine is not properly lubricated. Consequently, most modern oils are multigrade, with designations such as SAE 20W/50 (a grade used in high-performance sports cars), in which case the oil has the viscosity of an SAE 20 oil at subzero temperatures (hence the W for winter) and the viscosity of an SAE 50 oil at high temperatures. These properties are achieved by a careful blend of additives that modulate the intermolecular interactions in the oil, thereby controlling the temperature dependence of the viscosity. Many of the commercially available oil additives “for improved engine performance” are highly viscous materials that increase the viscosity and effective SAE rating of the oil, but overusing these additives can cause the same problems experienced with highly viscous single-grade oils.
Example \(1\)
Based on the nature and strength of the intermolecular cohesive forces and the probable nature of the liquid–glass adhesive forces, predict what will happen when a glass capillary is put into a beaker of SAE 20 motor oil. Will the oil be pulled up into the tube by capillary action or pushed down below the surface of the liquid in the beaker? What will be the shape of the meniscus (convex or concave)? (Hint: the surface of glass is lined with Si–OH groups.)
Given: substance and composition of the glass surface
Asked for: behavior of oil and the shape of meniscus
Strategy:
1. Identify the cohesive forces in the motor oil.
2. Determine whether the forces interact with the surface of glass. From the strength of this interaction, predict the behavior of the oil and the shape of the meniscus.
Solution
A Motor oil is a nonpolar liquid consisting largely of hydrocarbon chains. The cohesive forces responsible for its high boiling point are almost solely London dispersion forces between the hydrocarbon chains.
B Such a liquid cannot form strong interactions with the polar Si–OH groups of glass, so the surface of the oil inside the capillary will be lower than the level of the liquid in the beaker. The oil will have a convex meniscus similar to that of mercury.
Exercise \(1\)
Predict what will happen when a glass capillary is put into a beaker of ethylene glycol. Will the ethylene glycol be pulled up into the tube by capillary action or pushed down below the surface of the liquid in the beaker? What will be the shape of the meniscus (convex or concave)?
Answer
Capillary action will pull the ethylene glycol up into the capillary. The meniscus will be concave.
Summary
Surface tension, capillary action, and viscosity are unique properties of liquids that depend on the nature of intermolecular interactions. Surface tension is the energy required to increase the surface area of a liquid by a given amount. The stronger the intermolecular interactions, the greater the surface tension. Surfactants are molecules, such as soaps and detergents, that reduce the surface tension of polar liquids like water. Capillary action is the phenomenon in which liquids rise up into a narrow tube called a capillary. It results when cohesive forces, the intermolecular forces in the liquid, are weaker than adhesive forces, the attraction between a liquid and the surface of the capillary. The shape of the meniscus, the upper surface of a liquid in a tube, also reflects the balance between adhesive and cohesive forces. The viscosity of a liquid is its resistance to flow. Liquids that have strong intermolecular forces tend to have high viscosities. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/11%3A_Liquids_and_Intermolecular_Forces/11.03%3A_Some_Properties_of_Liquids.txt |
Learning Objectives
• To calculate the energy changes that accompany phase changes.
We take advantage of changes between the gas, liquid, and solid states to cool a drink with ice cubes (solid to liquid), cool our bodies by perspiration (liquid to gas), and cool food inside a refrigerator (gas to liquid and vice versa). We use dry ice, which is solid CO2, as a refrigerant (solid to gas), and we make artificial snow for skiing and snowboarding by transforming a liquid to a solid. In this section, we examine what happens when any of the three forms of matter is converted to either of the other two. These changes of state are often called phase changes. The six most common phase changes are shown in Figure $1$.
Energy Changes That Accompany Phase Changes
Phase changes are always accompanied by a change in the energy of a system. For example, converting a liquid, in which the molecules are close together, to a gas, in which the molecules are, on average, far apart, requires an input of energy (heat) to give the molecules enough kinetic energy to allow them to overcome the intermolecular attractive forces. The stronger the attractive forces, the more energy is needed to overcome them. Solids, which are highly ordered, have the strongest intermolecular interactions, whereas gases, which are very disordered, have the weakest. Thus any transition from a more ordered to a less ordered state (solid to liquid, liquid to gas, or solid to gas) requires an input of energy; it is endothermic. Conversely, any transition from a less ordered to a more ordered state (liquid to solid, gas to liquid, or gas to solid) releases energy; it is exothermic. The energy change associated with each common phase change is shown in Figure $1$.
ΔH is positive for any transition from a more ordered to a less ordered state and negative for a transition from a less ordered to a more ordered state.
Previously, we defined the enthalpy changes associated with various chemical and physical processes. The melting points and molar enthalpies of fusion ($ΔH_{fus}$), the energy required to convert from a solid to a liquid, a process known as fusion (or melting), as well as the normal boiling points and enthalpies of vaporization ($ΔH_{vap}$) of selected compounds are listed in Table $1$.
Table $1$: Melting and Boiling Points and Enthalpies of Fusion and Vaporization for Selected Substances. Values given under 1 atm. of external pressure.
Substance Melting Point (°C) ΔHfus (kJ/mol) Boiling Point (°C) ΔHvap (kJ/mol)
N2 −210.0 0.71 −195.8 5.6
HCl −114.2 2.00 −85.1 16.2
Br2 −7.2 10.6 58.8 30.0
CCl4 −22.6 2.56 76.8 29.8
CH3CH2OH (ethanol) −114.1 4.93 78.3 38.6
CH3(CH2)4CH3 (n-hexane) −95.4 13.1 68.7 28.9
H2O 0 6.01 100 40.7
Na 97.8 2.6 883 97.4
NaF 996 33.4 1704 176.1
The substances with the highest melting points usually have the highest enthalpies of fusion; they tend to be ionic compounds that are held together by very strong electrostatic interactions. Substances with high boiling points are those with strong intermolecular interactions that must be overcome to convert a liquid to a gas, resulting in high enthalpies of vaporization. The enthalpy of vaporization of a given substance is much greater than its enthalpy of fusion because it takes more energy to completely separate molecules (conversion from a liquid to a gas) than to enable them only to move past one another freely (conversion from a solid to a liquid).
Less energy is needed to allow molecules to move past each other than to separate them totally.
The direct conversion of a solid to a gas, without an intervening liquid phase, is called sublimation. The amount of energy required to sublime 1 mol of a pure solid is the enthalpy of sublimation (ΔHsub). Common substances that sublime at standard temperature and pressure (STP; 0°C, 1 atm) include CO2 (dry ice); iodine (Figure $2$); naphthalene, a substance used to protect woolen clothing against moths; and 1,4-dichlorobenzene. As shown in Figure $1$, the enthalpy of sublimation of a substance is the sum of its enthalpies of fusion and vaporization provided all values are at the same T; this is an application of Hess’s law.
$ΔH_{sub} =ΔH_{fus} +ΔH_{vap} \label{Eq1}$
Fusion, vaporization, and sublimation are endothermic processes; they occur only with the absorption of heat. Anyone who has ever stepped out of a swimming pool on a cool, breezy day has felt the heat loss that accompanies the evaporation of water from the skin. Our bodies use this same phenomenon to maintain a constant temperature: we perspire continuously, even when at rest, losing about 600 mL of water daily by evaporation from the skin. We also lose about 400 mL of water as water vapor in the air we exhale, which also contributes to cooling. Refrigerators and air-conditioners operate on a similar principle: heat is absorbed from the object or area to be cooled and used to vaporize a low-boiling-point liquid, such as ammonia or the chlorofluorocarbons (CFCs) and the hydrofluorocarbons (HCFCs). The vapor is then transported to a different location and compressed, thus releasing and dissipating the heat. Likewise, ice cubes efficiently cool a drink not because of their low temperature but because heat is required to convert ice at 0°C to liquid water at 0°C.
Temperature Curves
The processes on the right side of Figure $1$—freezing, condensation, and deposition, which are the reverse of fusion, sublimation, and vaporization—are exothermic. Thus heat pumps that use refrigerants are essentially air-conditioners running in reverse. Heat from the environment is used to vaporize the refrigerant, which is then condensed to a liquid in coils within a house to provide heat. The energy changes that occur during phase changes can be quantified by using a heating or cooling curve.
Heating Curves
Figure $3$ shows a heating curve, a plot of temperature versus heating time, for a 75 g sample of water. The sample is initially ice at 1 atm and −23°C; as heat is added, the temperature of the ice increases linearly with time. The slope of the line depends on both the mass of the ice and the specific heat (Cs) of ice, which is the number of joules required to raise the temperature of 1 g of ice by 1°C. As the temperature of the ice increases, the water molecules in the ice crystal absorb more and more energy and vibrate more vigorously. At the melting point, they have enough kinetic energy to overcome attractive forces and move with respect to one another. As more heat is added, the temperature of the system does not increase further but remains constant at 0°C until all the ice has melted. Once all the ice has been converted to liquid water, the temperature of the water again begins to increase. Now, however, the temperature increases more slowly than before because the specific heat capacity of water is greater than that of ice. When the temperature of the water reaches 100°C, the water begins to boil. Here, too, the temperature remains constant at 100°C until all the water has been converted to steam. At this point, the temperature again begins to rise, but at a faster rate than seen in the other phases because the heat capacity of steam is less than that of ice or water.
Thus the temperature of a system does not change during a phase change. In this example, as long as even a tiny amount of ice is present, the temperature of the system remains at 0°C during the melting process, and as long as even a small amount of liquid water is present, the temperature of the system remains at 100°C during the boiling process. The rate at which heat is added does not affect the temperature of the ice/water or water/steam mixture because the added heat is being used exclusively to overcome the attractive forces that hold the more condensed phase together. Many cooks think that food will cook faster if the heat is turned up higher so that the water boils more rapidly. Instead, the pot of water will boil to dryness sooner, but the temperature of the water does not depend on how vigorously it boils.
The temperature of a sample does not change during a phase change.
If heat is added at a constant rate, as in Figure $3$, then the length of the horizontal lines, which represents the time during which the temperature does not change, is directly proportional to the magnitude of the enthalpies associated with the phase changes. In Figure $3$, the horizontal line at 100°C is much longer than the line at 0°C because the enthalpy of vaporization of water is several times greater than the enthalpy of fusion.
A superheated liquid is a sample of a liquid at the temperature and pressure at which it should be a gas. Superheated liquids are not stable; the liquid will eventually boil, sometimes violently. The phenomenon of superheating causes “bumping” when a liquid is heated in the laboratory. When a test tube containing water is heated over a Bunsen burner, for example, one portion of the liquid can easily become too hot. When the superheated liquid converts to a gas, it can push or “bump” the rest of the liquid out of the test tube. Placing a stirring rod or a small piece of ceramic (a “boiling chip”) in the test tube allows bubbles of vapor to form on the surface of the object so the liquid boils instead of becoming superheated. Superheating is the reason a liquid heated in a smooth cup in a microwave oven may not boil until the cup is moved, when the motion of the cup allows bubbles to form.
Cooling Curves
The cooling curve, a plot of temperature versus cooling time, in Figure $4$ plots temperature versus time as a 75 g sample of steam, initially at 1 atm and 200°C, is cooled. Although we might expect the cooling curve to be the mirror image of the heating curve in Figure $3$, the cooling curve is not an identical mirror image. As heat is removed from the steam, the temperature falls until it reaches 100°C. At this temperature, the steam begins to condense to liquid water. No further temperature change occurs until all the steam is converted to the liquid; then the temperature again decreases as the water is cooled. We might expect to reach another plateau at 0°C, where the water is converted to ice; in reality, however, this does not always occur. Instead, the temperature often drops below the freezing point for some time, as shown by the little dip in the cooling curve below 0°C. This region corresponds to an unstable form of the liquid, a supercooled liquid. If the liquid is allowed to stand, if cooling is continued, or if a small crystal of the solid phase is added (a seed crystal), the supercooled liquid will convert to a solid, sometimes quite suddenly. As the water freezes, the temperature increases slightly due to the heat evolved during the freezing process and then holds constant at the melting point as the rest of the water freezes. Subsequently, the temperature of the ice decreases again as more heat is removed from the system.
Supercooling effects have a huge impact on Earth’s climate. For example, supercooling of water droplets in clouds can prevent the clouds from releasing precipitation over regions that are persistently arid as a result. Clouds consist of tiny droplets of water, which in principle should be dense enough to fall as rain. In fact, however, the droplets must aggregate to reach a certain size before they can fall to the ground. Usually a small particle (a nucleus) is required for the droplets to aggregate; the nucleus can be a dust particle, an ice crystal, or a particle of silver iodide dispersed in a cloud during seeding (a method of inducing rain). Unfortunately, the small droplets of water generally remain as a supercooled liquid down to about −10°C, rather than freezing into ice crystals that are more suitable nuclei for raindrop formation. One approach to producing rainfall from an existing cloud is to cool the water droplets so that they crystallize to provide nuclei around which raindrops can grow. This is best done by dispersing small granules of solid CO2 (dry ice) into the cloud from an airplane. Solid CO2 sublimes directly to the gas at pressures of 1 atm or lower, and the enthalpy of sublimation is substantial (25.3 kJ/mol). As the CO2 sublimes, it absorbs heat from the cloud, often with the desired results.
A Video Discussing the Thermodynamics of Phase Changes. Video Source: The Thermodynamics of Phase Changes, YouTube(opens in new window) [youtu.be]
Example $1$: Cooling Tea
If a 50.0 g ice cube at 0.0°C is added to 500 mL of tea at 20.0°C, what is the temperature of the tea when the ice cube has just melted? Assume that no heat is transferred to or from the surroundings. The density of water (and iced tea) is 1.00 g/mL over the range 0°C–20°C, the specific heats of liquid water and ice are 4.184 J/(g•°C) and 2.062 J/(g•°C), respectively, and the enthalpy of fusion of ice is 6.01 kJ/mol.
Given: mass, volume, initial temperature, density, specific heats, and $ΔH_{fus}$
Asked for: final temperature
Strategy
Substitute the given values into the general equation relating heat gained (by the ice) to heat lost (by the tea) to obtain the final temperature of the mixture.
Solution
When two substances or objects at different temperatures are brought into contact, heat will flow from the warmer one to the cooler. The amount of heat that flows is given by
$q=mC_sΔT \nonumber$
where $q$ is heat, $m$ is mass, $C_s$ is the specific heat, and $ΔT$ is the temperature change. Eventually, the temperatures of the two substances will become equal at a value somewhere between their initial temperatures. Calculating the temperature of iced tea after adding an ice cube is slightly more complicated. The general equation relating heat gained and heat lost is still valid, but in this case we also have to take into account the amount of heat required to melt the ice cube from ice at 0.0°C to liquid water at 0.0°C.
The amount of heat gained by the ice cube as it melts is determined by its enthalpy of fusion in kJ/mol:
$q=nΔH_{fus} \nonumber$
For our 50.0 g ice cube:
\begin{align*} q_{ice} &= 50.0 g⋅\dfrac{1\: mol}{18.02\:g}⋅6.01\: kJ/mol \[4pt] &= 16.7\, kJ \end{align*} \nonumber
Thus, when the ice cube has just melted, it has absorbed 16.7 kJ of heat from the tea. We can then substitute this value into the first equation to determine the change in temperature of the tea:
$q_{tea} = - 16,700 J = 500 mL⋅\dfrac{1.00\: g}{1\: mL}⋅4.184 J/(g•°C) ΔT \nonumber$
$ΔT = - 7.98 °C = T_f - T_i \nonumber$
$T_f = 12.02 °C \nonumber$
This would be the temperature of the tea when the ice cube has just finished melting; however, this leaves the melted ice still at 0.0°C. We might more practically want to know what the final temperature of the mixture of tea will be once the melted ice has come to thermal equilibrium with the tea. To determine this, we can add one more step to the calculation by plugging in to the general equation relating heat gained and heat lost again:
\begin{align*} q_{ice} &= - q_{tea} \[4pt] q_{ice} &= m_{ice}C_sΔT = 50.0g⋅4.184 J/(g•°C)⋅(T_f - 0.0°C) \[4pt] &= 209.2 J/°C⋅T_f \end{align*} \nonumber
$q_{tea} = m_{tea}C_sΔT = 500g⋅4.184 J/(g•°C)⋅(T_f - 12.02°C) = 2092 J/°C⋅T_f - 25,150 J \nonumber$
$209.2 J/°C⋅T_f = - 2092 J/°C⋅T_f + 25,150 J \nonumber$
$2301.2 J/°C⋅T_f = 25,150 J \nonumber$
$T_f = 10.9 °C \nonumber$
The final temperature is in between the initial temperatures of the tea (12.02 °C) and the melted ice (0.0 °C), so this answer makes sense. In this example, the tea loses much more heat in melting the ice than in mixing with the cold water, showing the importance of accounting for the heat of phase changes!
Exercise $1$: Death by Freezing
Suppose you are overtaken by a blizzard while ski touring and you take refuge in a tent. You are thirsty, but you forgot to bring liquid water. You have a choice of eating a few handfuls of snow (say 400 g) at −5.0°C immediately to quench your thirst or setting up your propane stove, melting the snow, and heating the water to body temperature before drinking it. You recall that the survival guide you leafed through at the hotel said something about not eating snow, but you cannot remember why—after all, it’s just frozen water. To understand the guide’s recommendation, calculate the amount of heat that your body will have to supply to bring 400 g of snow at −5.0°C to your body’s internal temperature of 37°C. Use the data in Example $1$
Answer
200 kJ (4.1 kJ to bring the ice from −5.0°C to 0.0°C, 133.6 kJ to melt the ice at 0.0°C, and 61.9 kJ to bring the water from 0.0°C to 37°C), which is energy that would not have been expended had you first melted the snow.
Summary
Fusion, vaporization, and sublimation are endothermic processes, whereas freezing, condensation, and deposition are exothermic processes. Changes of state are examples of phase changes, or phase transitions. All phase changes are accompanied by changes in the energy of a system. Changes from a more-ordered state to a less-ordered state (such as a liquid to a gas) are endothermic. Changes from a less-ordered state to a more-ordered state (such as a liquid to a solid) are always exothermic. The conversion of a solid to a liquid is called fusion (or melting). The energy required to melt 1 mol of a substance is its enthalpy of fusion (ΔHfus). The energy change required to vaporize 1 mol of a substance is the enthalpy of vaporization (ΔHvap). The direct conversion of a solid to a gas is sublimation. The amount of energy needed to sublime 1 mol of a substance is its enthalpy of sublimation (ΔHsub) and is the sum of the enthalpies of fusion and vaporization. Plots of the temperature of a substance versus heat added or versus heating time at a constant rate of heating are called heating curves. Heating curves relate temperature changes to phase transitions. A superheated liquid, a liquid at a temperature and pressure at which it should be a gas, is not stable. A cooling curve is not exactly the reverse of the heating curve because many liquids do not freeze at the expected temperature. Instead, they form a supercooled liquid, a metastable liquid phase that exists below the normal melting point. Supercooled liquids usually crystallize on standing, or adding a seed crystal of the same or another substance can induce crystallization. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/11%3A_Liquids_and_Intermolecular_Forces/11.04%3A_Phase_Changes.txt |
Learning Objectives
• To know how and why the vapor pressure of a liquid varies with temperature.
• To understand that the equilibrium vapor pressure of a liquid depends on the temperature and the intermolecular forces present.
• To understand that the relationship between pressure, enthalpy of vaporization, and temperature is given by the Clausius-Clapeyron equation.
Nearly all of us have heated a pan of water with the lid in place and shortly thereafter heard the sounds of the lid rattling and hot water spilling onto the stovetop. When a liquid is heated, its molecules obtain sufficient kinetic energy to overcome the forces holding them in the liquid and they escape into the gaseous phase. By doing so, they generate a population of molecules in the vapor phase above the liquid that produces a pressure—the vapor pressure of the liquid. In the situation we described, enough pressure was generated to move the lid, which allowed the vapor to escape. If the vapor is contained in a sealed vessel, however, such as an unvented flask, and the vapor pressure becomes too high, the flask will explode (as many students have unfortunately discovered). In this section, we describe vapor pressure in more detail and explain how to quantitatively determine the vapor pressure of a liquid.
Evaporation and Condensation
Because the molecules of a liquid are in constant motion, we can plot the fraction of molecules with a given kinetic energy (KE) against their kinetic energy to obtain the kinetic energy distribution of the molecules in the liquid (Figure $1$), just as we did for a gas. As for gases, increasing the temperature increases both the average kinetic energy of the particles in a liquid and the range of kinetic energy of the individual molecules. If we assume that a minimum amount of energy ($E_0$) is needed to overcome the intermolecular attractive forces that hold a liquid together, then some fraction of molecules in the liquid always has a kinetic energy greater than $E_0$. The fraction of molecules with a kinetic energy greater than this minimum value increases with increasing temperature. Any molecule with a kinetic energy greater than $E_0$ has enough energy to overcome the forces holding it in the liquid and escape into the vapor phase. Before it can do so, however, a molecule must also be at the surface of the liquid, where it is physically possible for it to leave the liquid surface; that is, only molecules at the surface can undergo evaporation (or vaporization), where molecules gain sufficient energy to enter a gaseous state above a liquid’s surface, thereby creating a vapor pressure.
To understand the causes of vapor pressure, consider the apparatus shown in Figure $2$. When a liquid is introduced into an evacuated chamber (part (a) in Figure $2$), the initial pressure above the liquid is approximately zero because there are as yet no molecules in the vapor phase. Some molecules at the surface, however, will have sufficient kinetic energy to escape from the liquid and form a vapor, thus increasing the pressure inside the container. As long as the temperature of the liquid is held constant, the fraction of molecules with $KE > E_0$ will not change, and the rate at which molecules escape from the liquid into the vapor phase will depend only on the surface area of the liquid phase.
As soon as some vapor has formed, a fraction of the molecules in the vapor phase will collide with the surface of the liquid and reenter the liquid phase in a process known as condensation (part (b) in Figure $2$). As the number of molecules in the vapor phase increases, the number of collisions between vapor-phase molecules and the surface will also increase. Eventually, a steady state will be reached in which exactly as many molecules per unit time leave the surface of the liquid (vaporize) as collide with it (condense). At this point, the pressure over the liquid stops increasing and remains constant at a particular value that is characteristic of the liquid at a given temperature. The rates of evaporation and condensation over time for a system such as this are shown graphically in Figure $3$.
Equilibrium Vapor Pressure
Two opposing processes (such as evaporation and condensation) that occur at the same rate and thus produce no net change in a system, constitute a dynamic equilibrium. In the case of a liquid enclosed in a chamber, the molecules continuously evaporate and condense, but the amounts of liquid and vapor do not change with time. The pressure exerted by a vapor in dynamic equilibrium with a liquid is the equilibrium vapor pressure of the liquid.
If a liquid is in an open container, however, most of the molecules that escape into the vapor phase will not collide with the surface of the liquid and return to the liquid phase. Instead, they will diffuse through the gas phase away from the container, and an equilibrium will never be established. Under these conditions, the liquid will continue to evaporate until it has “disappeared.” The speed with which this occurs depends on the vapor pressure of the liquid and the temperature. Volatile liquids have relatively high vapor pressures and tend to evaporate readily; nonvolatile liquids have low vapor pressures and evaporate more slowly. Although the dividing line between volatile and nonvolatile liquids is not clear-cut, as a general guideline, we can say that substances with vapor pressures greater than that of water (Figure $4$) are relatively volatile, whereas those with vapor pressures less than that of water are relatively nonvolatile. Thus diethyl ether (ethyl ether), acetone, and gasoline are volatile, but mercury, ethylene glycol, and motor oil are nonvolatile.
The equilibrium vapor pressure of a substance at a particular temperature is a characteristic of the material, like its molecular mass, melting point, and boiling point. It does not depend on the amount of liquid as long as at least a tiny amount of liquid is present in equilibrium with the vapor. The equilibrium vapor pressure does, however, depend very strongly on the temperature and the intermolecular forces present, as shown for several substances in Figure $4$. Molecules that can hydrogen bond, such as ethylene glycol, have a much lower equilibrium vapor pressure than those that cannot, such as octane. The nonlinear increase in vapor pressure with increasing temperature is much steeper than the increase in pressure expected for an ideal gas over the corresponding temperature range. The temperature dependence is so strong because the vapor pressure depends on the fraction of molecules that have a kinetic energy greater than that needed to escape from the liquid, and this fraction increases exponentially with temperature. As a result, sealed containers of volatile liquids are potential bombs if subjected to large increases in temperature. The gas tanks on automobiles are vented, for example, so that a car won’t explode when parked in the sun. Similarly, the small cans (1–5 gallons) used to transport gasoline are required by law to have a pop-off pressure release.
Volatile substances have low boiling points and relatively weak intermolecular interactions; nonvolatile substances have high boiling points and relatively strong intermolecular interactions.
A Video Discussing Vapor Pressure and Boiling Points. Video Source: Vapor Pressure & Boiling Point(opens in new window) [youtu.be]
The exponential rise in vapor pressure with increasing temperature in Figure $4$ allows us to use natural logarithms to express the nonlinear relationship as a linear one.
$\boxed{\ln P =\dfrac{-\Delta H_{vap}}{R}\left ( \dfrac{1}{T} \right) + C} \label{Eq1}$
where
• $\ln P$ is the natural logarithm of the vapor pressure,
• $ΔH_{vap}$ is the enthalpy of vaporization,
• $R$ is the universal gas constant [8.314 J/(mol•K)],
• $T$ is the temperature in kelvins, and
• $C$ is the y-intercept, which is a constant for any given line.
Plotting $\ln P$ versus the inverse of the absolute temperature ($1/T$) is a straight line with a slope of −ΔHvap/R. Equation $\ref{Eq1}$, called the Clausius–Clapeyron Equation, can be used to calculate the $ΔH_{vap}$ of a liquid from its measured vapor pressure at two or more temperatures. The simplest way to determine $ΔH_{vap}$ is to measure the vapor pressure of a liquid at two temperatures and insert the values of $P$ and $T$ for these points into Equation $\ref{Eq2}$, which is derived from the Clausius–Clapeyron equation:
$\ln\left ( \dfrac{P_{1}}{P_{2}} \right)=\dfrac{-\Delta H_{vap}}{R}\left ( \dfrac{1}{T_{1}}-\dfrac{1}{T_{2}} \right) \label{Eq2}$
Conversely, if we know ΔHvap and the vapor pressure $P_1$ at any temperature $T_1$, we can use Equation $\ref{Eq2}$ to calculate the vapor pressure $P_2$ at any other temperature $T_2$, as shown in Example $1$.
A Video Discussing the Clausius-Clapeyron Equation. Video Link: The Clausius-Clapeyron Equation(opens in new window) [youtu.be]
Example $1$: Vapor Pressure of Mercury
The experimentally measured vapor pressures of liquid Hg at four temperatures are listed in the following table:
experimentally measured vapor pressures of liquid Hg at four temperatures
T (°C) 80.0 100 120 140
P (torr) 0.0888 0.2729 0.7457 1.845
From these data, calculate the enthalpy of vaporization (ΔHvap) of mercury and predict the vapor pressure of the liquid at 160°C. (Safety note: mercury is highly toxic; when it is spilled, its vapor pressure generates hazardous levels of mercury vapor.)
Given: vapor pressures at four temperatures
Asked for: ΔHvap of mercury and vapor pressure at 160°C
Strategy:
1. Use Equation $\ref{Eq2}$ to obtain ΔHvap directly from two pairs of values in the table, making sure to convert all values to the appropriate units.
2. Substitute the calculated value of ΔHvap into Equation $\ref{Eq2}$ to obtain the unknown pressure (P2).
Solution:
A The table gives the measured vapor pressures of liquid Hg for four temperatures. Although one way to proceed would be to plot the data using Equation $\ref{Eq1}$ and find the value of ΔHvap from the slope of the line, an alternative approach is to use Equation $\ref{Eq2}$ to obtain ΔHvap directly from two pairs of values listed in the table, assuming no errors in our measurement. We therefore select two sets of values from the table and convert the temperatures from degrees Celsius to kelvin because the equation requires absolute temperatures. Substituting the values measured at 80.0°C (T1) and 120.0°C (T2) into Equation $\ref{Eq2}$ gives
\begin{align*} \ln \left ( \dfrac{0.7457 \; \cancel{Torr}}{0.0888 \; \cancel{Torr}} \right) &=\dfrac{-\Delta H_{vap}}{8.314 \; J/mol\cdot K}\left ( \dfrac{1}{\left ( 120+273 \right)K}-\dfrac{1}{\left ( 80.0+273 \right)K} \right) \[4pt] \ln\left ( 8.398 \right) &=\dfrac{-\Delta H_{vap}}{8.314 \; J/mol\cdot \cancel{K}}\left ( -2.88\times 10^{-4} \; \cancel{K^{-1}} \right) \[4pt] 2.13 &=-\Delta H_{vap} \left ( -3.46 \times 10^{-4} \right) J^{-1}\cdot mol \[4pt] \Delta H_{vap} &=61,400 \; J/mol = 61.4 \; kJ/mol \end{align*} \nonumber
B We can now use this value of ΔHvap to calculate the vapor pressure of the liquid (P2) at 160.0°C (T2):
$\ln\left ( \dfrac{P_{2} }{0.0888 \; torr} \right)=\dfrac{-61,400 \; \cancel{J/mol}}{8.314 \; \cancel{J/mol} \; K^{-1}}\left ( \dfrac{1}{\left ( 160+273 \right)K}-\dfrac{1}{\left ( 80.0+273 \right) K} \right) \nonumber$
Using the relationship $e^{\ln x} = x$, we have
\begin{align*} \ln \left ( \dfrac{P_{2} }{0.0888 \; Torr} \right) &=3.86 \[4pt] \dfrac{P_{2} }{0.0888 \; Torr} &=e^{3.86} = 47.5 \[4pt] P_{2} &= 4.21 Torr \end{align*} \nonumber
At 160°C, liquid Hg has a vapor pressure of 4.21 torr, substantially greater than the pressure at 80.0°C, as we would expect.
Exercise $1$: Vapor Pressure of Nickel
The vapor pressure of liquid nickel at 1606°C is 0.100 torr, whereas at 1805°C, its vapor pressure is 1.000 torr. At what temperature does the liquid have a vapor pressure of 2.500 torr?
Answer
1896°C
Boiling Points
As the temperature of a liquid increases, the vapor pressure of the liquid increases until it equals the external pressure, or the atmospheric pressure in the case of an open container. Bubbles of vapor begin to form throughout the liquid, and the liquid begins to boil. The temperature at which a liquid boils at exactly 1 atm pressure is the normal boiling point of the liquid. For water, the normal boiling point is exactly 100°C. The normal boiling points of the other liquids in Figure $4$ are represented by the points at which the vapor pressure curves cross the line corresponding to a pressure of 1 atm. Although we usually cite the normal boiling point of a liquid, the actual boiling point depends on the pressure. At a pressure greater than 1 atm, water boils at a temperature greater than 100°C because the increased pressure forces vapor molecules above the surface to condense. Hence the molecules must have greater kinetic energy to escape from the surface. Conversely, at pressures less than 1 atm, water boils below 100°C.
Table $1$: The Boiling Points of Water at Various Locations on Earth
Place Altitude above Sea Level (ft) Atmospheric Pressure (mmHg) Boiling Point of Water (°C)
Mt. Everest, Nepal/Tibet 29,028 240 70
Bogota, Colombia 11,490 495 88
Denver, Colorado 5280 633 95
Washington, DC 25 759 100
Dead Sea, Israel/Jordan −1312 799 101.4
Typical variations in atmospheric pressure at sea level are relatively small, causing only minor changes in the boiling point of water. For example, the highest recorded atmospheric pressure at sea level is 813 mmHg, recorded during a Siberian winter; the lowest sea-level pressure ever measured was 658 mmHg in a Pacific typhoon. At these pressures, the boiling point of water changes minimally, to 102°C and 96°C, respectively. At high altitudes, on the other hand, the dependence of the boiling point of water on pressure becomes significant. Table $1$ lists the boiling points of water at several locations with different altitudes. At an elevation of only 5000 ft, for example, the boiling point of water is already lower than the lowest ever recorded at sea level. The lower boiling point of water has major consequences for cooking everything from soft-boiled eggs (a “three-minute egg” may well take four or more minutes in the Rockies and even longer in the Himalayas) to cakes (cake mixes are often sold with separate high-altitude instructions). Conversely, pressure cookers, which have a seal that allows the pressure inside them to exceed 1 atm, are used to cook food more rapidly by raising the boiling point of water and thus the temperature at which the food is being cooked.
As pressure increases, the boiling point of a liquid increases and vice versa.
Example $2$: Boiling Mercury
Use Figure $4$ to estimate the following.
1. the boiling point of water in a pressure cooker operating at 1000 mmHg
2. the pressure required for mercury to boil at 250°C
Given: Data in Figure $4$, pressure, and boiling point
Asked for: corresponding boiling point and pressure
Strategy:
1. To estimate the boiling point of water at 1000 mmHg, refer to Figure $4$ and find the point where the vapor pressure curve of water intersects the line corresponding to a pressure of 1000 mmHg.
2. To estimate the pressure required for mercury to boil at 250°C, find the point where the vapor pressure curve of mercury intersects the line corresponding to a temperature of 250°C.
Solution:
1. A The vapor pressure curve of water intersects the P = 1000 mmHg line at about 110°C; this is therefore the boiling point of water at 1000 mmHg.
2. B The vertical line corresponding to 250°C intersects the vapor pressure curve of mercury at P ≈ 75 mmHg. Hence this is the pressure required for mercury to boil at 250°C.
Exercise $2$: Boiling Ethlyene Glycol
Ethylene glycol is an organic compound primarily used as a raw material in the manufacture of polyester fibers and fabric industry, and polyethylene terephthalate resins (PET) used in bottling. Use the data in Figure $4$ to estimate the following.
1. the normal boiling point of ethylene glycol
2. the pressure required for diethyl ether to boil at 20°C.
Answer a
200°C
Answer b
450 mmHg
Summary
Because the molecules of a liquid are in constant motion and possess a wide range of kinetic energies, at any moment some fraction of them has enough energy to escape from the surface of the liquid to enter the gas or vapor phase. This process, called vaporization or evaporation, generates a vapor pressure above the liquid. Molecules in the gas phase can collide with the liquid surface and reenter the liquid via condensation. Eventually, a steady state is reached in which the number of molecules evaporating and condensing per unit time is the same, and the system is in a state of dynamic equilibrium. Under these conditions, a liquid exhibits a characteristic equilibrium vapor pressure that depends only on the temperature. We can express the nonlinear relationship between vapor pressure and temperature as a linear relationship using the Clausius–Clapeyron equation. This equation can be used to calculate the enthalpy of vaporization of a liquid from its measured vapor pressure at two or more temperatures. Volatile liquids are liquids with high vapor pressures, which tend to evaporate readily from an open container; nonvolatile liquids have low vapor pressures. When the vapor pressure equals the external pressure, bubbles of vapor form within the liquid, and it boils. The temperature at which a substance boils at a pressure of 1 atm is its normal boiling point. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/11%3A_Liquids_and_Intermolecular_Forces/11.05%3A_Vapor_Pressure.txt |
Learning Objectives
• To understand the basics of a one-component phase diagram as a function of temperature and pressure in a closed system.
• To be able to identify the triple point, the critical point, and four regions: solid, liquid, gas, and a supercritical fluid.
The state exhibited by a given sample of matter depends on the identity, temperature, and pressure of the sample. A phase diagram is a graphic summary of the physical state of a substance as a function of temperature and pressure in a closed system.
Introduction
A typical phase diagram consists of discrete regions that represent the different phases exhibited by a substance (Figure $1$). Each region corresponds to the range of combinations of temperature and pressure over which that phase is stable. The combination of high pressure and low temperature (upper left of Figure $1$) corresponds to the solid phase, whereas the gas phase is favored at high temperature and low pressure (lower right). The combination of high temperature and high pressure (upper right) corresponds to a supercritical fluid.
The solid phase is favored at low temperature and high pressure; the gas phase is favored at high temperature and low pressure.
The lines in a phase diagram correspond to the combinations of temperature and pressure at which two phases can coexist in equilibrium. In Figure $1$, the line that connects points A and D separates the solid and liquid phases and shows how the melting point of a solid varies with pressure. The solid and liquid phases are in equilibrium all along this line; crossing the line horizontally corresponds to melting or freezing. The line that connects points A and B is the vapor pressure curve of the liquid, which we discussed in Section 11.5. It ends at the critical point, beyond which the substance exists as a supercritical fluid. The line that connects points A and C is the vapor pressure curve of the solid phase. Along this line, the solid is in equilibrium with the vapor phase through sublimation and deposition. Finally, point A, where the solid/liquid, liquid/gas, and solid/gas lines intersect, is the triple point; it is the only combination of temperature and pressure at which all three phases (solid, liquid, and gas) are in equilibrium and can therefore exist simultaneously. Because no more than three phases can ever coexist, a phase diagram can never have more than three lines intersecting at a single point.
Remember that a phase diagram, such as the one in Figure $1$, is for a single pure substance in a closed system, not for a liquid in an open beaker in contact with air at 1 atm pressure. In practice, however, the conclusions reached about the behavior of a substance in a closed system can usually be extrapolated to an open system without a great deal of error.
The Phase Diagram of Water
Figure $2$ shows the phase diagram of water and illustrates that the triple point of water occurs at 0.01°C and 0.00604 atm (4.59 mmHg). Far more reproducible than the melting point of ice, which depends on the amount of dissolved air and the atmospheric pressure, the triple point (273.16 K) is used to define the absolute (Kelvin) temperature scale. The triple point also represents the lowest pressure at which a liquid phase can exist in equilibrium with the solid or vapor. At pressures less than 0.00604 atm, therefore, ice does not melt to a liquid as the temperature increases; the solid sublimes directly to water vapor. Sublimation of water at low temperature and pressure can be used to “freeze-dry” foods and beverages. The food or beverage is first cooled to subzero temperatures and placed in a container in which the pressure is maintained below 0.00604 atm. Then, as the temperature is increased, the water sublimes, leaving the dehydrated food (such as that used by backpackers or astronauts) or the powdered beverage (as with freeze-dried coffee).
The phase diagram for water illustrated in Figure $\PageIndex{2b}$ shows the boundary between ice and water on an expanded scale. The melting curve of ice slopes up and slightly to the left rather than up and to the right as in Figure $1$; that is, the melting point of ice decreases with increasing pressure; at 100 MPa (987 atm), ice melts at −9°C. Water behaves this way because it is one of the few known substances for which the crystalline solid is less dense than the liquid (others include antimony and bismuth). Increasing the pressure of ice that is in equilibrium with water at 0°C and 1 atm tends to push some of the molecules closer together, thus decreasing the volume of the sample. The decrease in volume (and corresponding increase in density) is smaller for a solid or a liquid than for a gas, but it is sufficient to melt some of the ice.
In Figure $\PageIndex{2b}$ point A is located at P = 1 atm and T = −1.0°C, within the solid (ice) region of the phase diagram. As the pressure increases to 150 atm while the temperature remains the same, the line from point A crosses the ice/water boundary to point B, which lies in the liquid water region. Consequently, applying a pressure of 150 atm will melt ice at −1.0°C. We have already indicated that the pressure dependence of the melting point of water is of vital importance. If the solid/liquid boundary in the phase diagram of water were to slant up and to the right rather than to the left, ice would be denser than water, ice cubes would sink, water pipes would not burst when they freeze, and antifreeze would be unnecessary in automobile engines.
Ice Skating: An Incorrect Hypothesis of Phase Transitions
Until recently, many textbooks described ice skating as being possible because the pressure generated by the skater’s blade is high enough to melt the ice under the blade, thereby creating a lubricating layer of liquid water that enables the blade to slide across the ice. Although this explanation is intuitively satisfying, it is incorrect, as we can show by a simple calculation.
Recall that pressure (P) is the force (F) applied per unit area (A):
$P=\dfrac{F}{A} \nonumber$
To calculate the pressure an ice skater exerts on the ice, we need to calculate only the force exerted and the area of the skate blade. If we assume a 75.0 kg (165 lb) skater, then the force exerted by the skater on the ice due to gravity is
$F = mg \nonumber$
where m is the mass and g is the acceleration due to Earth’s gravity (9.81 m/s2). Thus the force is
$F = (75.0\; kg)(9.81\; m/s^2) = 736\; (kg•m)/s^2 = 736 N \nonumber$
If we assume that the skate blades are 2.0 mm wide and 25 cm long, then the area of the bottom of each blade is
$A = (2.0 \times 10^{−3}\; m)(25 \times 10^{−2}\; m) = 5.0 \times 10^{−4} m^2 \nonumber$
If the skater is gliding on one foot, the pressure exerted on the ice is
$P= \dfrac{736\;N}{5.0 \times 10^{-4} \; m^2} = 1.5 \times 10^6 \; N/m^2 = 1.5 \times 10^6\; Pa =15 \; atm \nonumber$
The pressure is much lower than the pressure needed to decrease the melting point of ice by even 1°C, and experience indicates that it is possible to skate even when the temperature is well below freezing. Thus pressure-induced melting of the ice cannot explain the low friction that enables skaters (and hockey pucks) to glide. Recent research indicates that the surface of ice, where the ordered array of water molecules meets the air, consists of one or more layers of almost liquid water. These layers, together with melting induced by friction as a skater pushes forward, appear to account for both the ease with which a skater glides and the fact that skating becomes more difficult below about −7°C, when the number of lubricating surface water layers decreases.
Example $1$: Water
Referring to the phase diagram of water in Figure $2$:
1. predict the physical form of a sample of water at 400°C and 150 atm.
2. describe the changes that occur as the sample in part (a) is slowly allowed to cool to −50°C at a constant pressure of 150 atm.
Given: phase diagram, temperature, and pressure
Asked for: physical form and physical changes
Strategy:
1. Identify the region of the phase diagram corresponding to the initial conditions and identify the phase that exists in this region.
2. Draw a line corresponding to the given pressure. Move along that line in the appropriate direction (in this case cooling) and describe the phase changes.
Solution:
1. A Locate the starting point on the phase diagram in part (a) in Figure $2$. The initial conditions correspond to point A, which lies in the region of the phase diagram representing water vapor. Thus water at T = 400°C and P = 150 atm is a gas.
2. B Cooling the sample at constant pressure corresponds to moving left along the horizontal line in part (a) in Figure $2$. At about 340°C (point B), we cross the vapor pressure curve, at which point water vapor will begin to condense and the sample will consist of a mixture of vapor and liquid. When all of the vapor has condensed, the temperature drops further, and we enter the region corresponding to liquid water (indicated by point C). Further cooling brings us to the melting curve, the line that separates the liquid and solid phases at a little below 0°C (point D), at which point the sample will consist of a mixture of liquid and solid water (ice). When all of the water has frozen, cooling the sample to −50°C takes us along the horizontal line to point E, which lies within the region corresponding to solid water. At P = 150 atm and T = −50°C, therefore, the sample is solid ice.
Exercise $2$
Referring to the phase diagram of water in Figure $2$, predict the physical form of a sample of water at −0.0050°C as the pressure is gradually increased from 1.0 mmHg to 218 atm.
Answer
The sample is initially a gas, condenses to a solid as the pressure increases, and then melts when the pressure is increased further to give a liquid.
The Phase Diagram of Carbon Dioxide
In contrast to the phase diagram of water, the phase diagram of CO2 (Figure $3$) has a more typical melting curve, sloping up and to the right. The triple point is −56.6°C and 5.11 atm, which means that liquid CO2 cannot exist at pressures lower than 5.11 atm. At 1 atm, therefore, solid CO2 sublimes directly to the vapor while maintaining a temperature of −78.5°C, the normal sublimation temperature. Solid CO2 is generally known as dry ice because it is a cold solid with no liquid phase observed when it is warmed.
Also notice the critical point at 30.98°C and 72.79 atm. Supercritical carbon dioxide is emerging as a natural refrigerant, making it a low carbon (and thus a more environmentally friendly) solution for domestic heat pumps.
The Critical Point
As the phase diagrams above demonstrate, a combination of high pressure and low temperature allows gases to be liquefied. As we increase the temperature of a gas, liquefaction becomes more and more difficult because higher and higher pressures are required to overcome the increased kinetic energy of the molecules. In fact, for every substance, there is some temperature above which the gas can no longer be liquefied, regardless of pressure. This temperature is the critical temperature (Tc), the highest temperature at which a substance can exist as a liquid. Above the critical temperature, the molecules have too much kinetic energy for the intermolecular attractive forces to hold them together in a separate liquid phase. Instead, the substance forms a single phase that completely occupies the volume of the container. Substances with strong intermolecular forces tend to form a liquid phase over a very large temperature range and therefore have high critical temperatures. Conversely, substances with weak intermolecular interactions have relatively low critical temperatures. Each substance also has a critical pressure (Pc), the minimum pressure needed to liquefy it at the critical temperature. The combination of critical temperature and critical pressure is called the critical point. The critical temperatures and pressures of several common substances are listed in Figure $1$.
Figure $1$: Critical Temperatures and Pressures of Some Simple Substances
Substance Tc (°C) Pc (atm)
NH3 132.4 113.5
CO2 31.0 73.8
CH3CH2OH (ethanol) 240.9 61.4
He −267.96 2.27
Hg 1477 1587
CH4 −82.6 46.0
N2 −146.9 33.9
H2O 374.0 217.7
High-boiling-point, nonvolatile liquids have high critical temperatures and vice versa.
Supercritical Fluids
A Video Discussing Phase Diagrams. Video Source: Phase Diagrams(opens in new window) [youtu.be] | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/11%3A_Liquids_and_Intermolecular_Forces/11.06%3A_Phase_Diagrams.txt |
Learning Objectives
• To know the characteristic properties of crystalline and amorphous solids.
• To recognize the unit cell of a crystalline solid.
• To calculate the density of a solid given its unit cell.
Crystalline solids have regular ordered arrays of components held together by uniform intermolecular forces, whereas the components of amorphous solids are not arranged in regular arrays. With few exceptions, the particles that compose a solid material, whether ionic, molecular, covalent, or metallic, are held in place by strong attractive forces between them. When we discuss solids, therefore, we consider the positions of the atoms, molecules, or ions, which are essentially fixed in space, rather than their motions (which are more important in liquids and gases). The constituents of a solid can be arranged in two general ways: they can form a regular repeating three-dimensional structure called a crystal lattice, thus producing a crystalline solid, or they can aggregate with no particular order, in which case they form an amorphous solid (from the Greek ámorphos, meaning “shapeless”).
Crystalline solids, or crystals, have distinctive internal structures that in turn lead to distinctive flat surfaces, or faces. The faces intersect at angles that are characteristic of the substance. When exposed to x-rays, each structure also produces a distinctive pattern that can be used to identify the material. The characteristic angles do not depend on the size of the crystal; they reflect the regular repeating arrangement of the component atoms, molecules, or ions in space. When an ionic crystal is cleaved (Figure $2$, for example, repulsive interactions cause it to break along fixed planes to produce new faces that intersect at the same angles as those in the original crystal. In a covalent solid such as a cut diamond, the angles at which the faces meet are also not arbitrary but are determined by the arrangement of the carbon atoms in the crystal.
Crystals tend to have relatively sharp, well-defined melting points because all the component atoms, molecules, or ions are the same distance from the same number and type of neighbors; that is, the regularity of the crystalline lattice creates local environments that are the same. Thus the intermolecular forces holding the solid together are uniform, and the same amount of thermal energy is needed to break every interaction simultaneously.
Amorphous solids have two characteristic properties. When cleaved or broken, they produce fragments with irregular, often curved surfaces; and they have poorly defined patterns when exposed to x-rays because their components are not arranged in a regular array. An amorphous, translucent solid is called a glass. Almost any substance can solidify in amorphous form if the liquid phase is cooled rapidly enough. Some solids, however, are intrinsically amorphous, because either their components cannot fit together well enough to form a stable crystalline lattice or they contain impurities that disrupt the lattice. For example, although the chemical composition and the basic structural units of a quartz crystal and quartz glass are the same—both are SiO2 and both consist of linked SiO4 tetrahedra—the arrangements of the atoms in space are not. Crystalline quartz contains a highly ordered arrangement of silicon and oxygen atoms, but in quartz glass the atoms are arranged almost randomly. When molten SiO2 is cooled rapidly (4 K/min), it forms quartz glass, whereas the large, perfect quartz crystals sold in mineral shops have had cooling times of thousands of years. In contrast, aluminum crystallizes much more rapidly. Amorphous aluminum forms only when the liquid is cooled at the extraordinary rate of 4 × 1013 K/s, which prevents the atoms from arranging themselves into a regular array.
In an amorphous solid, the local environment, including both the distances to neighboring units and the numbers of neighbors, varies throughout the material. Different amounts of thermal energy are needed to overcome these different interactions. Consequently, amorphous solids tend to soften slowly over a wide temperature range rather than having a well-defined melting point like a crystalline solid. If an amorphous solid is maintained at a temperature just below its melting point for long periods of time, the component molecules, atoms, or ions can gradually rearrange into a more highly ordered crystalline form.
Crystals have sharp, well-defined melting points; amorphous solids do not.
Crystals
Because a crystalline solid consists of repeating patterns of its components in three dimensions (a crystal lattice), we can represent the entire crystal by drawing the structure of the smallest identical units that, when stacked together, form the crystal. This basic repeating unit is called a unit cell. For example, the unit cell of a sheet of identical postage stamps is a single stamp, and the unit cell of a stack of bricks is a single brick. In this section, we describe the arrangements of atoms in various unit cells.
Unit cells are easiest to visualize in two dimensions. In many cases, more than one unit cell can be used to represent a given structure, as shown for the Escher drawing in the chapter opener and for a two-dimensional crystal lattice in Figure $4$. Usually the smallest unit cell that completely describes the order is chosen. The only requirement for a valid unit cell is that repeating it in space must produce the regular lattice. Thus the unit cell in Figure $\PageIndex{4d}$ is not a valid choice because repeating it in space does not produce the desired lattice (there are triangular holes). The concept of unit cells is extended to a three-dimensional lattice in the schematic drawing in Figure $5$.
The Unit Cell
There are seven fundamentally different kinds of unit cells, which differ in the relative lengths of the edges and the angles between them (Figure $6$). Each unit cell has six sides, and each side is a parallelogram. We focus primarily on the cubic unit cells, in which all sides have the same length and all angles are 90°, but the concepts that we introduce also apply to substances whose unit cells are not cubic.
If the cubic unit cell consists of eight component atoms, molecules, or ions located at the corners of the cube, then it is called simple cubic (Figure $\PageIndex{7a}$). If the unit cell also contains an identical component in the center of the cube, then it is body-centered cubic (bcc) ($\PageIndex{7b}$). If there are components in the center of each face in addition to those at the corners of the cube, then the unit cell is face-centered cubic (fcc) (Figure $\PageIndex{7c}$).
As indicated in Figure $7$, a solid consists of a large number of unit cells arrayed in three dimensions. Any intensive property of the bulk material, such as its density, must therefore also be related to its unit cell. Because density is the mass of substance per unit volume, we can calculate the density of the bulk material from the density of a single unit cell. To do this, we need to know the size of the unit cell (to obtain its volume), the molar mass of its components, and the number of components per unit cell. When we count atoms or ions in a unit cell, however, those lying on a face, an edge, or a corner contribute to more than one unit cell, as shown in Figure $7$. For example, an atom that lies on a face of a unit cell is shared by two adjacent unit cells and is therefore counted as ${1\over 2}$ atom per unit cell. Similarly, an atom that lies on the edge of a unit cell is shared by four adjacent unit cells, so it contributes ${1 \over 4}$ atom to each. An atom at a corner of a unit cell is shared by all eight adjacent unit cells and therefore contributes ${1 \over 8}$ atom to each. The statement that atoms lying on an edge or a corner of a unit cell count as ${1 \over 4}$ or ${1 \over 8}$ atom per unit cell, respectively, is true for all unit cells except the hexagonal one, in which three unit cells share each vertical edge and six share each corner (Figure $7$:), leading to values of ${1 \over 3}$ and ${1 \over 6}$ atom per unit cell, respectively, for atoms in these positions. In contrast, atoms that lie entirely within a unit cell, such as the atom in the center of a body-centered cubic unit cell, belong to only that one unit cell.
For all unit cells except hexagonal, atoms on the faces contribute ${1\over 2}$ atom to each unit cell, atoms on the edges contribute ${1 \over 4}$ atom to each unit cell, and atoms on the corners contribute ${1 \over 8}$ atom to each unit cell.
Example $1$: The Unit Cell for Gold
Metallic gold has a face-centered cubic unit cell ($\PageIndex{7c}$). How many Au atoms are in each unit cell?
Given: unit cell
Asked for: number of atoms per unit cell
Strategy
Using Figure $7$, identify the positions of the Au atoms in a face-centered cubic unit cell and then determine how much each Au atom contributes to the unit cell. Add the contributions of all the Au atoms to obtain the total number of Au atoms in a unit cell.
Solution
As shown in Figure $7$, a face-centered cubic unit cell has eight atoms at the corners of the cube and six atoms on the faces. Because atoms on a face are shared by two unit cells, each counts as ${1 \over 2}$ atom per unit cell, giving $6\times{1 \over 2}=3$ Au atoms per unit cell. Atoms on a corner are shared by eight unit cells and hence contribute only ${1\over 8}$ atom per unit cell, giving $8\times{1\over 8}=1$ Au atom per unit cell. The total number of Au atoms in each unit cell is thus $3 + 1 = 4$.
Exercise $1$: Unit Cell for Iron
Metallic iron has a body-centered cubic unit cell (Figure $\PageIndex{7b}$). How many Fe atoms are in each unit cell?
Answer
two
Now that we know how to count atoms in unit cells, we can use unit cells to calculate the densities of simple compounds. Note, however, that we are assuming a solid consists of a perfect regular array of unit cells, whereas real substances contain impurities and defects that affect many of their bulk properties, including density. Consequently, the results of our calculations will be close but not necessarily identical to the experimentally obtained values.
Example $2$: Density of Iron
Calculate the density of metallic iron, which has a body-centered cubic unit cell (Figure $\PageIndex{7b}$) with an edge length of 286.6 pm.
Given: unit cell and edge length
Asked for: density
Strategy:
1. Determine the number of iron atoms per unit cell.
2. Calculate the mass of iron atoms in the unit cell from the molar mass and Avogadro’s number. Then divide the mass by the volume of the cell.
Solution:
A We know from Example $1$ that each unit cell of metallic iron contains two Fe atoms.
B The molar mass of iron is 55.85 g/mol. Because density is mass per unit volume, we need to calculate the mass of the iron atoms in the unit cell from the molar mass and Avogadro’s number and then divide the mass by the volume of the cell (making sure to use suitable units to get density in g/cm3):
$\textit{mass of Fe} =\left ( 2 \; \cancel{atoms} \; Fe \right )\left ( \dfrac{ 1 \; \cancel{mol}}{6.022\times 10^{23} \; \cancel{atoms}} \right )\left ( \dfrac{55.85 \; g}{\cancel{mol}} \right ) =1.855\times 10^{-22} \; g \nonumber$
$volume=\left [ \left ( 286.6 \; pm \right )\left ( \dfrac{10^{-12 }\; \cancel{m}}{\cancel{pm}} \right )\left ( \dfrac{10^{2} \; cm}{\cancel{m}} \right ) \right ] =2.345\times 10^{-23} \; cm^{3} \nonumber$
$density = \dfrac{1.855\times 10^{-22} \; g}{2.345\times 10^{-23} \; cm^{3}} = 7.880 g/cm^{3} \nonumber$
This result compares well with the tabulated experimental value of 7.874 g/cm3.
Exercise $2$: Density of Gold
Calculate the density of gold, which has a face-centered cubic unit cell (Figure $\PageIndex{7c}$) with an edge length of 407.8 pm.
Answer
19.29 g/cm3
Packing of Spheres
Our discussion of the three-dimensional structures of solids has considered only substances in which all the components are identical. As we shall see, such substances can be viewed as consisting of identical spheres packed together in space; the way the components are packed together produces the different unit cells. Most of the substances with structures of this type are metals.
Simple Cubic Structure
The arrangement of the atoms in a solid that has a simple cubic unit cell was shown in Figure $\PageIndex{5a}$. Each atom in the lattice has only six nearest neighbors in an octahedral arrangement. Consequently, the simple cubic lattice is an inefficient way to pack atoms together in space: only 52% of the total space is filled by the atoms. The only element that crystallizes in a simple cubic unit cell is polonium. Simple cubic unit cells are, however, common among binary ionic compounds, where each cation is surrounded by six anions and vice versa (Figure $8$).
Body-Centered Cubic Structure
The body-centered cubic unit cell is a more efficient way to pack spheres together and is much more common among pure elements. Each atom has eight nearest neighbors in the unit cell, and 68% of the volume is occupied by the atoms. As shown in Figure $8$, the body-centered cubic structure consists of a single layer of spheres in contact with each other and aligned so that their centers are at the corners of a square; a second layer of spheres occupies the square-shaped “holes” above the spheres in the first layer. The third layer of spheres occupies the square holes formed by the second layer, so that each lies directly above a sphere in the first layer, and so forth. All the alkali metals, barium, radium, and several of the transition metals have body-centered cubic structures.
Hexagonal Close-Packed and Cubic Close-Packed Structures
The most efficient way to pack spheres is the close-packed arrangement, which has two variants. A single layer of close-packed spheres is shown in Figure $\PageIndex{6a}$. Each sphere is surrounded by six others in the same plane to produce a hexagonal arrangement. Above any set of seven spheres are six depressions arranged in a hexagon. In principle, all six sites are the same, and any one of them could be occupied by an atom in the next layer. Actually, however, these six sites can be divided into two sets, labeled B and C in Figure $\PageIndex{9a}$. Sites B and C differ because as soon as we place a sphere at a B position, we can no longer place a sphere in any of the three C positions adjacent to A and vice versa.
If we place the second layer of spheres at the B positions in Figure $\PageIndex{9a}$, we obtain the two-layered structure shown in Figure $\PageIndex{9b}$. There are now two alternatives for placing the first atom of the third layer: we can place it directly over one of the atoms in the first layer (an A position) or at one of the C positions, corresponding to the positions that we did not use for the atoms in the first or second layers (Figure $\PageIndex{9c}$). If we choose the first arrangement and repeat the pattern in succeeding layers, the positions of the atoms alternate from layer to layer in the pattern ABABAB…, resulting in a hexagonal close-packed (hcp) structure (Figure $\PageIndex{9a}$). If we choose the second arrangement and repeat the pattern indefinitely, the positions of the atoms alternate as ABCABC…, giving a cubic close-packed (ccp) structure (Figure $\PageIndex{9b}$). Because the ccp structure contains hexagonally packed layers, it does not look particularly cubic. As shown in Figure $\PageIndex{9b}$, however, simply rotating the structure reveals its cubic nature, which is identical to a fcc structure. The hcp and ccp structures differ only in the way their layers are stacked. Both structures have an overall packing efficiency of 74%, and in both each atom has 12 nearest neighbors (6 in the same plane plus 3 in each of the planes immediately above and below).
Table $1$ compares the packing efficiency and the number of nearest neighbors for the different cubic and close-packed structures; the number of nearest neighbors is called the coordination number. Most metals have hcp, ccp, or bcc structures, although several metals exhibit both hcp and ccp structures, depending on temperature and pressure.
Table $1$: Properties of the Common Structures of Metals
Structure Percentage of Space Occupied by Atoms Coordination Number
simple cubic 52 6
body-centered cubic 68 8
hexagonal close packed 74 12
cubic close packed (identical to face-centered cubic) 74 12
Summary
A crystalline solid can be represented by its unit cell, which is the smallest identical unit that when stacked together produces the characteristic three-dimensional structure. Solids are characterized by an extended three-dimensional arrangement of atoms, ions, or molecules in which the components are generally locked into their positions. The components can be arranged in a regular repeating three-dimensional array (a crystal lattice), which results in a crystalline solid, or more or less randomly to produce an amorphous solid. Crystalline solids have well-defined edges and faces, diffract x-rays, and tend to have sharp melting points. In contrast, amorphous solids have irregular or curved surfaces, do not give well-resolved x-ray diffraction patterns, and melt over a wide range of temperatures.
The smallest repeating unit of a crystal lattice is the unit cell. The simple cubic unit cell contains only eight atoms, molecules, or ions at the corners of a cube. A body-centered cubic (bcc) unit cell contains one additional component in the center of the cube. A face-centered cubic (fcc) unit cell contains a component in the center of each face in addition to those at the corners of the cube. Simple cubic and bcc arrangements fill only 52% and 68% of the available space with atoms, respectively. The hexagonal close-packed (hcp) structure has an ABABAB… repeating arrangement, and the cubic close-packed (ccp) structure has an ABCABC… repeating pattern; the latter is identical to an fcc lattice. The hcp and ccp arrangements fill 74% of the available space and have a coordination number of 12 for each atom in the lattice, the number of nearest neighbors. The simple cubic and bcc lattices have coordination numbers of 6 and 8, respectively. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/11%3A_Liquids_and_Intermolecular_Forces/11.07%3A_Structure_of_Solids.txt |
Learning Objectives
• To understand the correlation between bonding and the properties of solids.
• To classify solids as ionic, molecular, covalent (network), or metallic, where the general order of increasing strength of interactions.
Crystalline solids fall into one of four categories. All four categories involve packing discrete molecules or atoms into a lattice or repeating array, though network solids are a special case. The categories are distinguished by the nature of the interactions holding the discrete molecules or atoms together. Based on the nature of the forces that hold the component atoms, molecules, or ions together, solids may be formally classified as ionic, molecular, covalent (network), or metallic. The variation in the relative strengths of these four types of interactions correlates nicely with their wide variation in properties.
Table $1$: Solids may be formally classified as ionic, molecular, covalent (network), or metallic
Type of Solid Interaction Properties Examples
Ionic Ionic High Melting Point, Brittle, Hard NaCl, MgO
Molecular Hydrogen Bonding,
Dipole-Dipole,
London Dispersion
Low Melting Point, Nonconducting H2, CO2
Metallic Metallic Bonding Variable Hardness and Melting Point (depending upon strength of metallic bonding), Conducting Fe, Mg
Network Covalent Bonding High Melting Point, Hard, Nonconducting C (diamond),
SiO2 (quartz)
In ionic and molecular solids, there are no chemical bonds between the molecules, atoms, or ions. The solid consists of discrete chemical species held together by intermolecular forces that are electrostatic or Coulombic in nature. This behavior is most obvious for an ionic solid such as $NaCl$, where the positively charged Na+ ions are attracted to the negatively charged $Cl^-$ ions. Even in the absence of ions, however, electrostatic forces are operational. For polar molecules such as $CH_2Cl_2$, the positively charged region of one molecular is attracted to the negatively charged region of another molecule (dipole-dipole interactions). For a nonpolar molecule such as $CO_2$, which has no permanent dipole moment, the random motion of electrons gives rise to temporary polarity (a temporary dipole moment). Electrostatic attractions between two temporarily polarized molecules are called London Dispersion Forces.
Hydrogen bonding is a term describing an attractive interaction between a hydrogen atom from a molecule or a molecular fragment X–H in which X is more electronegative than H, and an atom or a group of atoms in the same or a different molecule, in which there is evidence of bond formation. (See the IUPAC Provisional Recommendation on the definition of a hydrogen bond.) Dots are employed to indicate the presence of a hydrogen bond: X–H•••Y. The attractive interaction in a hydrogen bond typically has a strong electrostatic contribution, but dispersion forces and weak covalent bonding are also present.
In metallic solids and network solids, however, chemical bonds hold the individual chemical subunits together. The crystal is essential a single, macroscopic molecule with continuous chemical bonding throughout the entire structure. In metallic solids, the valence electrons are no longer exclusively associated with a single atom. Instead these electrons exist in molecular orbitals that are delocalized over many atoms, producing an electronic band structure. The metallic crystal essentially consists of a set of metal cations in a sea of electrons. This type of chemical bonding is called metallic bonding.
Ionic Solids
You learned previously that an ionic solid consists of positively and negatively charged ions held together by electrostatic forces. The strength of the attractive forces depends on the charge and size of the ions that compose the lattice and determines many of the physical properties of the crystal.
The lattice energy (i.e., the energy required to separate 1 mol of a crystalline ionic solid into its component ions in the gas phase) is directly proportional to the product of the ionic charges and inversely proportional to the sum of the radii of the ions. For example, NaF and CaO both crystallize in the face-centered cubic (fcc) sodium chloride structure, and the sizes of their component ions are about the same: Na+ (102 pm) versus Ca2+ (100 pm), and F (133 pm) versus O2− (140 pm). Because of the higher charge on the ions in CaO, however, the lattice energy of CaO is almost four times greater than that of NaF (3401 kJ/mol versus 923 kJ/mol). The forces that hold Ca and O together in CaO are much stronger than those that hold Na and F together in NaF, so the heat of fusion of CaO is almost twice that of NaF (59 kJ/mol versus 33.4 kJ/mol), and the melting point of CaO is 2927°C versus 996°C for NaF. In both cases, however, the values are large; that is, simple ionic compounds have high melting points and are relatively hard (and brittle) solids.
Molecular Solids
Molecular solids consist of atoms or molecules held to each other by dipole–dipole interactions, London dispersion forces, or hydrogen bonds, or any combination of these. The arrangement of the molecules in solid benzene is as follows:
Because the intermolecular interactions in a molecular solid are relatively weak compared with ionic and covalent bonds, molecular solids tend to be soft, low melting, and easily vaporized ($ΔH_{fus}$ and $ΔH_{vap}$ are low). For similar substances, the strength of the London dispersion forces increases smoothly with increasing molecular mass. For example, the melting points of benzene (C6H6), naphthalene (C10H8), and anthracene (C14H10), with one, two, and three fused aromatic rings, are 5.5°C, 80.2°C, and 215°C, respectively. The enthalpies of fusion also increase smoothly within the series: benzene (9.95 kJ/mol) < naphthalene (19.1 kJ/mol) < anthracene (28.8 kJ/mol). If the molecules have shapes that cannot pack together efficiently in the crystal, however, then the melting points and the enthalpies of fusion tend to be unexpectedly low because the molecules are unable to arrange themselves to optimize intermolecular interactions. Thus toluene (C6H5CH3) and m-xylene [m-C6H4(CH3)2] have melting points of −95°C and −48°C, respectively, which are significantly lower than the melting point of the lighter but more symmetrical analog, benzene.
Self-healing rubber is an example of a molecular solid with the potential for significant commercial applications. The material can stretch, but when snapped into pieces it can bond back together again through reestablishment of its hydrogen-bonding network without showing any sign of weakness. Among other applications, it is being studied for its use in adhesives and bicycle tires that will self-heal.
Covalent Network Solids
Covalent solids are formed by networks or chains of atoms or molecules held together by covalent bonds. A perfect single crystal of a covalent solid is therefore a single giant molecule. For example, the structure of diamond, shown in part (a) in Figure $1$, consists of sp3 hybridized carbon atoms, each bonded to four other carbon atoms in a tetrahedral array to create a giant network. The carbon atoms form six-membered rings.
The unit cell of diamond can be described as an fcc array of carbon atoms with four additional carbon atoms inserted into four of the tetrahedral holes. It thus has the zinc blende structure described in Section 12.3, except that in zinc blende the atoms that compose the fcc array are sulfur and the atoms in the tetrahedral holes are zinc. Elemental silicon has the same structure, as does silicon carbide (SiC), which has alternating C and Si atoms. The structure of crystalline quartz (SiO2), shown in Section 12.1, can be viewed as being derived from the structure of silicon by inserting an oxygen atom between each pair of silicon atoms.
All compounds with the diamond and related structures are hard, high-melting-point solids that are not easily deformed. Instead, they tend to shatter when subjected to large stresses, and they usually do not conduct electricity very well. In fact, diamond (melting point = 3500°C at 63.5 atm) is one of the hardest substances known, and silicon carbide (melting point = 2986°C) is used commercially as an abrasive in sandpaper and grinding wheels. It is difficult to deform or melt these and related compounds because strong covalent (C–C or Si–Si) or polar covalent (Si–C or Si–O) bonds must be broken, which requires a large input of energy.
Other covalent solids have very different structures. For example, graphite, the other common allotrope of carbon, has the structure shown in part (b) in Figure $1$. It contains planar networks of six-membered rings of sp2 hybridized carbon atoms in which each carbon is bonded to three others. This leaves a single electron in an unhybridized 2pz orbital that can be used to form C=C double bonds, resulting in a ring with alternating double and single bonds. Because of its resonance structures, the bonding in graphite is best viewed as consisting of a network of C–C single bonds with one-third of a π bond holding the carbons together, similar to the bonding in benzene.
To completely describe the bonding in graphite, we need a molecular orbital approach similar to the one used for benzene in Chapter 9. In fact, the C–C distance in graphite (141.5 pm) is slightly longer than the distance in benzene (139.5 pm), consistent with a net carbon–carbon bond order of 1.33. In graphite, the two-dimensional planes of carbon atoms are stacked to form a three-dimensional solid; only London dispersion forces hold the layers together. As a result, graphite exhibits properties typical of both covalent and molecular solids. Due to strong covalent bonding within the layers, graphite has a very high melting point, as expected for a covalent solid (it actually sublimes at about 3915°C). It is also very soft; the layers can easily slide past one another because of the weak interlayer interactions. Consequently, graphite is used as a lubricant and as the “lead” in pencils; the friction between graphite and a piece of paper is sufficient to leave a thin layer of carbon on the paper. Graphite is unusual among covalent solids in that its electrical conductivity is very high parallel to the planes of carbon atoms because of delocalized C–C π bonding. Finally, graphite is black because it contains an immense number of alternating double bonds, which results in a very small energy difference between the individual molecular orbitals. Thus light of virtually all wavelengths is absorbed. Diamond, on the other hand, is colorless when pure because it has no delocalized electrons. Table $2$ compares the strengths of the intermolecular and intramolecular interactions for three covalent solids, showing the comparative weakness of the interlayer interactions.
Table $2$: A Comparison of Intermolecular (ΔHsub) and Intramolecular Interactions
Substance ΔHsub (kJ/mol) Average Bond Energy (kJ/mol)
phosphorus (s) 58.98 201
sulfur (s) 64.22 226
iodine (s) 62.42 149
Carbon: An example of an Covalent Network Solid
In network solids, conventional chemical bonds hold the chemical subunits together. The bonding between chemical subunits, however, is identical to that within the subunits, resulting in a continuous network of chemical bonds. One common examples of network solids are diamond (a form of pure carbon) Carbon exists as a pure element at room temperature in three different forms: graphite (the most stable form), diamond, and fullerene.
Diamonds
The structure of diamond is shown at the right in a "ball-and-stick" format. The balls represent the carbon atoms and the sticks represent a covalent bond. Be aware that in the "ball-and-stick" representation the size of the balls do not accurately represent the size of carbon atoms. In addition, a single stick is drawn to represent a covalent bond irrespective of whether the bond is a single, double, or triple bond or requires resonance structures to represent. In the diamond structure, all bonds are single covalent bonds ($\sigma$ bonds). The "space-filling" format is an alternate representation that displays atoms as spheres with a radius equal to the van der Waals radius, thus providing a better sense of the size of the atoms.
Notice that diamond is a network solid. The entire solid is an "endless" repetition of carbon atoms bonded to each other by covalent bonds. (In the display at the right, the structure is truncated to fit in the display area.)
Questions to consider
• What is the bonding geometry around each carbon?
• What is the hybridization of carbon in diamond?
• The diamond structure consists of a repeating series of rings. How many carbon atoms are in a ring?
• Diamond are renowned for its hardness. Explain why this property is expected on the basis of the structure of diamond.
Graphite
The most stable form of carbon is graphite. Graphite consists of sheets of carbon atoms covalently bonded together. These sheets are then stacked to form graphite. Figure $3$ shows a ball-and-stick representation of graphite with sheets that extended "indefinitely" in the xy plane, but the structure has been truncated for display purposed. Graphite may also be regarded as a network solid, even though there is no bonding in the z direction. Each layer, however, is an "endless" bonded network of carbon atoms.
Questions to consider
• What is the bonding geometry around each carbon?
• What is the hybridization of carbon in graphite?
• The a layer of the graphite structure consists of a repeating series of rings. How many carbon atoms are in a ring?
• What force holds the carbon sheets together in graphite?
• Graphite is very slippery and is often used in lubricants. Explain why this property is expected on the basis of the structure of graphite.
• The slipperiness of graphite is enhanced by the introduction of impurities. Where would such impurities be located and why would they make graphite a better lubricant?
Fullerenes
Until the mid 1980's, pure carbon was thought to exist in two forms: graphite and diamond. The discovery of C60 molecules in interstellar dust in 1985 added a third form to this list. The existence of C60, which resembles a soccer ball, had been hypothesized by theoreticians for many years. In the late 1980's synthetic methods were developed for the synthesis of C60, and the ready availability of this form of carbon led to extensive research into its properties.
The C60 molecule (Figure $4$; left), is called buckminsterfullerene, though the shorter name fullerene is often used. The name is a tribute to the American architect R. Buckminster Fuller, who is famous for designing and constructing geodesic domes which bear a close similarity to the structure of C60. As is evident from the display, C60 is a sphere composed of six-member and five-member carbon rings. These balls are sometimes fondly referred to as "Bucky balls". It should be noted that fullerenes are an entire class of pure carbon compounds rather than a single compound. A distorted sphere containing more than 60 carbon atoms have also been found, and it is also possible to create long tubes (Figure $4$; right). All of these substances are pure carbon.
Questions to Consider
• What is the bonding geometry around each carbon? (Note that this geometry is distorted in $C_{60}$.)
• What is the hybridization of carbon in fullerene?
• A single crystal of C60 falls into which class of crystalline solids?
• It has been hypothesized that C60 would make a good lubricant. Why might C60 make a good lubricant?
Metallic Solids
Metallic solids such as crystals of copper, aluminum, and iron are formed by metal atoms Figure $5$. The structure of metallic crystals is often described as a uniform distribution of atomic nuclei within a “sea” of delocalized electrons. The atoms within such a metallic solid are held together by a unique force known as metallic bonding that gives rise to many useful and varied bulk properties. All exhibit high thermal and electrical conductivity, metallic luster, and malleability. Many are very hard and quite strong. Because of their malleability (the ability to deform under pressure or hammering), they do not shatter and, therefore, make useful construction materials.
Metals are characterized by their ability to reflect light, called luster, their high electrical and thermal conductivity, their high heat capacity, and their malleability and ductility. Every lattice point in a pure metallic element is occupied by an atom of the same metal. The packing efficiency in metallic crystals tends to be high, so the resulting metallic solids are dense, with each atom having as many as 12 nearest neighbors.
Bonding in metallic solids is quite different from the bonding in the other kinds of solids we have discussed. Because all the atoms are the same, there can be no ionic bonding, yet metals always contain too few electrons or valence orbitals to form covalent bonds with each of their neighbors. Instead, the valence electrons are delocalized throughout the crystal, providing a strong cohesive force that holds the metal atoms together.
Valence electrons in a metallic solid are delocalized, providing a strong cohesive force that holds the atoms together.
The strength of metallic bonds varies dramatically. For example, cesium melts at 28.4°C, and mercury is a liquid at room temperature, whereas tungsten melts at 3680°C. Metallic bonds tend to be weakest for elements that have nearly empty (as in Cs) or nearly full (Hg) valence subshells, and strongest for elements with approximately half-filled valence shells (as in W). As a result, the melting points of the metals increase to a maximum around group 6 and then decrease again from left to right across the d block. Other properties related to the strength of metallic bonds, such as enthalpies of fusion, boiling points, and hardness, have similar periodic trends.
A somewhat oversimplified way to describe the bonding in a metallic crystal is to depict the crystal as consisting of positively charged nuclei in an electron sea (Figure $6$). In this model, the valence electrons are not tightly bound to any one atom but are distributed uniformly throughout the structure. Very little energy is needed to remove electrons from a solid metal because they are not bound to a single nucleus. When an electrical potential is applied, the electrons can migrate through the solid toward the positive electrode, thus producing high electrical conductivity. The ease with which metals can be deformed under pressure is attributed to the ability of the metal ions to change positions within the electron sea without breaking any specific bonds. The transfer of energy through the solid by successive collisions between the metal ions also explains the high thermal conductivity of metals. This model does not, however, explain many of the other properties of metals, such as their metallic luster and the observed trends in bond strength as reflected in melting points or enthalpies of fusion. Some general properties of the four major classes of solids are summarized in Table $2$.
Table $2$: Properties of the Major Classes of Solids
Ionic Solids Molecular Solids Covalent Solids Metallic Solids
*Many exceptions exist. For example, graphite has a relatively high electrical conductivity within the carbon planes, and diamond has the highest thermal conductivity of any known substance.
poor conductors of heat and electricity poor conductors of heat and electricity poor conductors of heat and electricity* good conductors of heat and electricity
relatively high melting point low melting point high melting point melting points depend strongly on electron configuration
hard but brittle; shatter under stress soft very hard and brittle easily deformed under stress; ductile and malleable
relatively dense low density low density usually high density
dull surface dull surface dull surface lustrous
The general order of increasing strength of interactions in a solid is:
molecular solids < ionic solids ≈ metallic solids < covalent solids
Example $1$
Classify Ge, RbI, C6(CH3)6, and Zn as ionic, molecular, covalent, or metallic solids and arrange them in order of increasing melting points.
Given: compounds
Asked for: classification and order of melting points
Strategy:
1. Locate the component element(s) in the periodic table. Based on their positions, predict whether each solid is ionic, molecular, covalent, or metallic.
2. Arrange the solids in order of increasing melting points based on your classification, beginning with molecular solids.
Solution:
A Germanium lies in the p block just under Si, along the diagonal line of semimetallic elements, which suggests that elemental Ge is likely to have the same structure as Si (the diamond structure). Thus Ge is probably a covalent solid. RbI contains a metal from group 1 and a nonmetal from group 17, so it is an ionic solid containing Rb+ and I ions. The compound C6(CH3)6 is a hydrocarbon (hexamethylbenzene), which consists of isolated molecules that stack to form a molecular solid with no covalent bonds between them. Zn is a d-block element, so it is a metallic solid.
B Arranging these substances in order of increasing melting points is straightforward, with one exception. We expect C6(CH3)6 to have the lowest melting point and Ge to have the highest melting point, with RbI somewhere in between. The melting points of metals, however, are difficult to predict based on the models presented thus far. Because Zn has a filled valence shell, it should not have a particularly high melting point, so a reasonable guess is C6(CH3)6 < Zn ~ RbI < Ge. The actual melting points are C6(CH3)6, 166°C; Zn, 419°C; RbI, 642°C; and Ge, 938°C. This agrees with our prediction.
Exercise $1$
Classify C60, BaBr2, GaAs, and AgZn as ionic, covalent, molecular, or metallic solids and then arrange them in order of increasing melting points.
Answer
C60 (molecular) < AgZn (metallic) ~ BaBr2 (ionic) < GaAs (covalent). The actual melting points are C60, about 300°C; AgZn, about 700°C; BaBr2, 856°C; and GaAs, 1238°C.
Summary
The major types of solids are ionic, molecular, covalent, and metallic. Ionic solids consist of positively and negatively charged ions held together by electrostatic forces; the strength of the bonding is reflected in the lattice energy. Ionic solids tend to have high melting points and are rather hard. Molecular solids are held together by relatively weak forces, such as dipole–dipole interactions, hydrogen bonds, and London dispersion forces. As a result, they tend to be rather soft and have low melting points, which depend on their molecular structure. Covalent solids consist of two- or three-dimensional networks of atoms held together by covalent bonds; they tend to be very hard and have high melting points. Metallic solids have unusual properties: in addition to having high thermal and electrical conductivity and being malleable and ductile, they exhibit luster, a shiny surface that reflects light. An alloy is a mixture of metals that has bulk metallic properties different from those of its constituent elements. Alloys can be formed by substituting one metal atom for another of similar size in the lattice (substitutional alloys), by inserting smaller atoms into holes in the metal lattice (interstitial alloys), or by a combination of both. Although the elemental composition of most alloys can vary over wide ranges, certain metals combine in only fixed proportions to form intermetallic compounds with unique properties. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/11%3A_Liquids_and_Intermolecular_Forces/11.08%3A_Bonding_in_Solids.txt |
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
11.1: A Molecular Comparison of Gases, Liquids, and Solids
Q11.1.1
A liquid, unlike a gas, is virtually incompressible. Explain what this means using macroscopic and microscopic descriptions. What general physical properties do liquids share with solids? What properties do liquids share with gases?
Q11.1.2
Using a kinetic molecular approach, discuss the differences and similarities between liquids and gases with regard to
1. thermal expansion.
2. fluidity.
3. diffusion.
Q11.1.3
How must the ideal gas law be altered to apply the kinetic molecular theory of gases to liquids? Explain.
Q11.1.4
Why are the root mean square speeds of molecules in liquids less than the root mean square speeds of molecules in gases?
11.2: Intermolecular Forces
Q11.2.1
What is the main difference between intramolecular interactions and intermolecular interactions? Which is typically stronger? How are changes of state affected by these different kinds of interactions?
Q11.2.2
Describe the three major kinds of intermolecular interactions discussed in this chapter and their major features. The hydrogen bond is actually an example of one of the other two types of interaction. Identify the kind of interaction that includes hydrogen bonds and explain why hydrogen bonds fall into this category.
Q11.2.3
Which are stronger—dipole–dipole interactions or London dispersion forces? Which are likely to be more important in a molecule with heavy atoms? Explain your answers.
Q11.2.4
Explain why hydrogen bonds are unusually strong compared to other dipole–dipole interactions. How does the strength of hydrogen bonds compare with the strength of covalent bonds?
Q11.2.5
Liquid water is essential for life as we know it, but based on its molecular mass, water should be a gas under standard conditions. Why is water a liquid rather than a gas under standard conditions?
Q11.2.6
Describe the effect of polarity, molecular mass, and hydrogen bonding on the melting point and boiling point of a substance.
Q11.2.7
Why are intermolecular interactions more important for liquids and solids than for gases? Under what conditions must these interactions be considered for gases?
Q11.2.8
Using acetic acid as an example, illustrate both attractive and repulsive intermolecular interactions. How does the boiling point of a substance depend on the magnitude of the repulsive intermolecular interactions?
Q11.2.9
In group 17, elemental fluorine and chlorine are gases, whereas bromine is a liquid and iodine is a solid. Why?
Q11.2.10
The boiling points of the anhydrous hydrogen halides are as follows: HF, 19°C; HCl, −85°C; HBr, −67°C; and HI, −34°C. Explain any trends in the data, as well as any deviations from that trend.
Q11.2.11
Identify the most important intermolecular interaction in each of the following.
1. SO2
2. HF
3. CO2
4. CCl4
5. CH2Cl2
Q11.2.12
Identify the most important intermolecular interaction in each of the following.
1. LiF
2. I2
3. ICl
4. NH3
5. NH2Cl
Q11.2.13
Would you expect London dispersion forces to be more important for Xe or Ne? Why? (The atomic radius of Ne is 38 pm, whereas that of Xe is 108 pm.)
Q11.2.14
Arrange Kr, Cl2, H2, N2, Ne, and O2 in order of increasing polarizability. Explain your reasoning.
Q11.2.15
Both water and methanol have anomalously high boiling points due to hydrogen bonding, but the boiling point of water is greater than that of methanol despite its lower molecular mass. Why? Draw the structures of these two compounds, including any lone pairs, and indicate potential hydrogen bonds.
Q11.2.16
The structures of ethanol, ethylene glycol, and glycerin are as follows:
Arrange these compounds in order of increasing boiling point. Explain your rationale.
Q11.2.17
Do you expect the boiling point of H2S to be higher or lower than that of H2O? Justify your answer.
Q11.2.18
Ammonia (NH3), methylamine (CH3NH2), and ethylamine (CH3CH2NH2) are gases at room temperature, while propylamine (CH3CH2CH2NH2) is a liquid at room temperature. Explain these observations.
Q11.2.19
Why is it not advisable to freeze a sealed glass bottle that is completely filled with water? Use both macroscopic and microscopic models to explain your answer. Is a similar consideration required for a bottle containing pure ethanol? Why or why not?
Q11.2.20
Which compound in the following pairs will have the higher boiling point? Explain your reasoning.
• NH3 or PH3
• ethylene glycol (HOCH2CH2OH) or ethanol
• 2,2-dimethylpropanol [CH3C(CH3)2CH2OH] or n-butanol (CH3CH2CH2CH2OH)
Q11.2.21
Some recipes call for vigorous boiling, while others call for gentle simmering. What is the difference in the temperature of the cooking liquid between boiling and simmering? What is the difference in energy input?
Q11.2.22
Use the melting of a metal such as lead to explain the process of melting in terms of what is happening at the molecular level. As a piece of lead melts, the temperature of the metal remains constant, even though energy is being added continuously. Why?
Q11.2.23
How does the O–H distance in a hydrogen bond in liquid water compare with the O–H distance in the covalent O–H bond in the H2O molecule? What effect does this have on the structure and density of ice?
Q11.2.24
Explain why the hydrogen bonds in liquid HF are stronger than the corresponding intermolecular H⋅⋅⋅I interactions in liquid HI.
1. In which substance are the individual hydrogen bonds stronger: HF or H2O? Explain your reasoning.
2. For which substance will hydrogen bonding have the greater effect on the boiling point: HF or H2O? Explain your reasoning.
Answers
1. Water is a liquid under standard conditions because of its unique ability to form four strong hydrogen bonds per molecule.
2. As the atomic mass of the halogens increases, so does the number of electrons and the average distance of those electrons from the nucleus. Larger atoms with more electrons are more easily polarized than smaller atoms, and the increase in polarizability with atomic number increases the strength of London dispersion forces. These intermolecular interactions are strong enough to favor the condensed states for bromine and iodine under normal conditions of temperature and pressure.
1. The V-shaped SO2 molecule has a large dipole moment due to the polar S=O bonds, so dipole–dipole interactions will be most important.
2. The H–F bond is highly polar, and the fluorine atom has three lone pairs of electrons to act as hydrogen bond acceptors; hydrogen bonding will be most important.
3. Although the C=O bonds are polar, this linear molecule has no net dipole moment; hence, London dispersion forces are most important.
4. This is a symmetrical molecule that has no net dipole moment, and the Cl atoms are relatively polarizable; thus, London dispersion forces will dominate.
5. This molecule has a small dipole moment, as well as polarizable Cl atoms. In such a case, dipole–dipole interactions and London dispersion forces are often comparable in magnitude.
3. Water has two polar O–H bonds with H atoms that can act as hydrogen bond donors, plus two lone pairs of electrons that can act as hydrogen bond acceptors, giving a net of four hydrogen bonds per H2O molecule. Although methanol also has two lone pairs of electrons on oxygen that can act as hydrogen bond acceptors, it only has one O–H bond with an H atom that can act as a hydrogen bond donor. Consequently, methanol can only form two hydrogen bonds per molecule on average, versus four for water. Hydrogen bonding therefore has a much greater effect on the boiling point of water.
4. Vigorous boiling causes more water molecule to escape into the vapor phase, but does not affect the temperature of the liquid. Vigorous boiling requires a higher energy input than does gentle simmering.
11.3: Some Properties of Liquids
Conceptual Problems
1. Why is a water droplet round?
2. How is the environment of molecules on the surface of a liquid droplet different from that of molecules in the interior of the droplet? How is this difference related to the concept of surface tension?
3. Explain the role of intermolecular and intramolecular forces in surface tension.
4. A mosquito is able to walk across water without sinking, but if a few drops of detergent are added to the water, the insect will sink. Why?
5. Explain how soaps or surfactants decrease the surface tension of a liquid. How does the meniscus of an aqueous solution in a capillary change if a surfactant is added? Illustrate your answer with a diagram.
6. Of CH2Cl2, hexane, and ethanol, which has the lowest viscosity? Which has the highest surface tension? Explain your reasoning in each case.
7. At 25°C, cyclohexanol has a surface tension of 32.92 mN/m2, whereas the surface tension of cyclohexanone, which is very similar chemically, is only 25.45 mN/m2. Why is the surface tension of cyclohexanone so much less than that of cyclohexanol?
8. What is the relationship between
1. surface tension and temperature?
2. viscosity and temperature?
Explain your answers in terms of a microscopic picture.
9. What two opposing forces are responsible for capillary action? How do these forces determine the shape of the meniscus?
10. Which of the following liquids will have a concave meniscus in a glass capillary? Explain your reasoning.
1. pentane
2. diethylene glycol (HOCH2CH2OCH2CH2OH)
3. carbon tetrachloride
11. How does viscosity depend on molecular shape? What molecular features make liquids highly viscous?
Conceptual Answers
1. Adding a soap or a surfactant to water disrupts the attractive intermolecular interactions between water molecules, thereby decreasing the surface tension. Because water is a polar molecule, one would expect that a soap or a surfactant would also disrupt the attractive interactions responsible for adhesion of water to the surface of a glass capillary. As shown in the sketch, this would decrease the height of the water column inside the capillary, as well as making the meniscus less concave.
2. As the structures indicate, cyclohexanol is a polar substance that can engage in hydrogen bonding, much like methanol or ethanol; consequently, it is expected to have a higher surface tension due to stronger intermolecular interactions.
3. Cohesive forces are the intermolecular forces that hold the molecules of the liquid together, while adhesive forces are the attractive forces between the molecules of the liquid and the walls of the capillary. If the adhesive forces are stronger than the cohesive forces, the liquid is pulled up into the capillary and the meniscus is concave. Conversely, if the cohesive forces are stronger than the adhesive forces, the level of the liquid inside the capillary will be lower than the level outside the capillary, and the meniscus will be convex.
4. Viscous substances often consist of molecules that are much longer than they are wide and whose structures are often rather flexible. As a result, the molecules tend to become tangled with one another (much like overcooked spaghetti), which decreases the rate at which they can move through the liquid.
Numerical Problems
1. The viscosities of five liquids at 25°C are given in the following table. Explain the observed trends in viscosity.
Compound Molecular Formula Viscosity (mPa•s)
benzene C6H6 0.604
aniline C6H5NH2 3.847
1,2-dichloroethane C2H4Cl2 0.779
heptane C7H16 0.357
1-heptanol C7H15OH 5.810
• The following table gives values for the viscosity, boiling point, and surface tension of four substances. Examine these data carefully to see whether the data for each compound are internally consistent and point out any obvious errors or inconsistencies. Explain your reasoning.
Compound Viscosity (mPa•s at 20°C) Boiling Point (°C) Surface Tension (dyn/cm at 25°C)
A 0.41 61 27.16
B 0.55 65 22.55
C 0.92 105 36.76
D 0.59 110 28.53
• Surface tension data (in dyn/cm) for propanoic acid (C3H6O2), and 2-propanol (C3H8O), as a function of temperature, are given in the following table. Plot the data for each compound and explain the differences between the two graphs. Based on these data, which molecule is more polar?
Compound 25°C 50°C 75°C
propanoic acid 26.20 23.72 21.23
2-propanol 20.93 18.96 16.98
Numerical Answer
3. The plots of surface tension versus temperature for propionic acid and isopropanol have essentially the same slope, but at all temperatures the surface tension of propionic acid is about 30% greater than for isopropanol. Because surface tension is a measure of the cohesive forces in a liquid, these data suggest that the cohesive forces for propionic acid are significantly greater than for isopropanol. Both substances consist of polar molecules with similar molecular masses, and the most important intermolecular interactions are likely to be dipole–dipole interactions. Consequently, these data suggest that propionic acid is more polar than isopropanol.
11.4: Phase Changes
Conceptual Problems
1. In extremely cold climates, snow can disappear with no evidence of its melting. How can this happen? What change(s) in state are taking place? Would you expect this phenomenon to be more common at high or low altitudes? Explain your answer.
2. Why do car manufacturers recommend that an automobile should not be left standing in subzero temperatures if its radiator contains only water? Car manufacturers also warn car owners that they should check the fluid level in a radiator only when the engine is cool. What is the basis for this warning? What is likely to happen if it is ignored?
3. Use Hess’s law and a thermochemical cycle to show that, for any solid, the enthalpy of sublimation is equal to the sum of the enthalpy of fusion of the solid and the enthalpy of vaporization of the resulting liquid.
4. Three distinct processes occur when an ice cube at −10°C is used to cool a glass of water at 20°C. What are they? Which causes the greatest temperature change in the water?
5. When frost forms on a piece of glass, crystals of ice are deposited from water vapor in the air. How is this process related to sublimation? Describe the energy changes that take place as the water vapor is converted to frost.
6. What phase changes are involved in each process? Which processes are exothermic, and which are endothermic?
1. ice melting
2. distillation
3. condensation forming on a window
4. the use of dry ice to create a cloud for a theatrical production
7. What phase changes are involved in each process? Which processes are exothermic, and which are endothermic?
1. evaporation of methanol
2. crystallization
3. liquefaction of natural gas
4. the use of naphthalene crystals to repel moths
8. Why do substances with high enthalpies of fusion tend to have high melting points?
9. Why is the enthalpy of vaporization of a compound invariably much larger than its enthalpy of fusion?
10. What is the opposite of fusion, sublimation, and condensation? Describe the phase change in each pair of opposing processes and state whether each phase change is exothermic or endothermic.
11. Draw a typical heating curve (temperature versus amount of heat added at a constant rate) for conversion of a solid to a liquid and then to a gas. What causes some regions of the plot to have a positive slope? What is happening in the regions of the plot where the curve is horizontal, meaning that the temperature does not change even though heat is being added?
12. If you know the mass of a sample of a substance, how could you use a heating curve to calculate the specific heat of the substance, as well as the change in enthalpy associated with a phase change?
13. Draw the heating curve for a liquid that has been superheated. How does this differ from a normal heating curve for a liquid? Draw the cooling curve for a liquid that has been supercooled. How does this differ from a normal cooling curve for a liquid?
Conceptual Answers
1. When snow disappears without melting, it must be subliming directly from the solid state to the vapor state. The rate at which this will occur depends solely on the partial pressure of water, not on the total pressure due to other gases. Consequently, altitude (and changes in atmospheric pressure) will not affect the rate of sublimation directly.
3 The general equations and enthalpy changes for the changes of state involved in converting a solid to a gas are:
$\text{solid} \rightarrow \text{liquid}: \Delta H_{fus}$
$\text{liquid} \rightarrow \text{gas}: \Delta H_{vap}$
$\text{solid} \rightarrow \text{gas}: \Delta{H_{sub}}= \Delta{H_{fus}} + \Delta{H_{vap}}$
The relationship between these enthalpy changes is shown schematically in the thermochemical cycle below:
1. The formation of frost on a surface is an example of deposition, which is the reverse of sublimation. The change in enthalpy for deposition is equal in magnitude, but opposite in sign, to ΔHsub, which is a positive number: ΔHsub = ΔHfus + ΔHvap.
1. liquid + heat → vapor: endothermic
2. liquid → solid + heat: exothermic
3. gas → liquid + heat: exothermic
4. solid + heat → vapor: endothermic
2. The enthalpy of vaporization is larger than the enthalpy of fusion because vaporization requires the addition of enough energy to disrupt all intermolecular interactions and create a gas in which the molecules move essentially independently. In contrast, fusion requires much less energy, because the intermolecular interactions in a liquid and a solid are similar in magnitude in all condensed phases. Fusion requires only enough energy to overcome the intermolecular interactions that lock molecules in place in a lattice, thereby allowing them to move more freely.
3. The portions of the curve with a positive slope correspond to heating a single phase, while the horizontal portions of the curve correspond to phase changes. During a phase change, the temperature of the system does not change, because the added heat is melting the solid at its melting point or evaporating the liquid at its boiling point.
4. A superheated liquid exists temporarily as liquid with a temperature above the normal boiling point of the liquid. When a supercooled liquid boils, the temperature drops as the liquid is converted to vapor.
Conversely, a supercooled liquid exists temporarily as a liquid with a temperature lower than the normal melting point of the solid. As shown below, when a supercooled liquid crystallizes, the temperature increases as the liquid is converted to a solid.
Numerical Problems
1. The density of oxygen at 1 atm and various temperatures is given in the following table. Plot the data and use your graph to predict the normal boiling point of oxygen.
T (K) 60 70 80 90 100 120 140
d (mol/L) 40.1 38.6 37.2 35.6 0.123 0.102 0.087
2. The density of propane at 1 atm and various temperatures is given in the following table. Plot the data and use your graph to predict the normal boiling point of propane.
T (K) 100 125 150 175 200 225 250 275
d (mol/L) 16.3 15.7 15.0 14.4 13.8 13.2 0.049 0.044
• Draw the cooling curve for a sample of the vapor of a compound that has a melting point of 34°C and a boiling point of 77°C as it is cooled from 100°C to 0°C.
• Propionic acid has a melting point of −20.8°C and a boiling point of 141°C. Draw a heating curve showing the temperature versus time as heat is added at a constant rate to show the behavior of a sample of propionic acid as it is heated from −50°C to its boiling point. What happens above 141°C?
• A 0.542 g sample of I2 requires 96.1 J of energy to be converted to vapor. What is the enthalpy of sublimation of I2?
• A 2.0 L sample of gas at 210°C and 0.762 atm condenses to give 1.20 mL of liquid, and 476 J of heat is released during the process. What is the enthalpy of vaporization of the compound?
• One fuel used for jet engines and rockets is aluminum borohydride [Al(BH4)3], a liquid that readily reacts with water to produce hydrogen. The liquid has a boiling point of 44.5°C. How much energy is needed to vaporize 1.0 kg of aluminum borohydride at 20°C, given a ΔHvap of 30 kJ/mol and a molar heat capacity (Cp) of 194.6 J/(mol•K)?
• How much energy is released when freezing 100.0 g of dimethyl disulfide (C2H6S2) initially at 20°C? Use the following information: melting point = −84.7°C, ΔHfus = 9.19 kJ/mol, Cp = 118.1 J/(mol•K).
The following four problems use the following information (the subscript p indicates measurements taken at constant pressure): ΔHfus(H2O) = 6.01 kJ/mol, ΔHvap(H2O) = 40.66 kJ/mol, Cp(s)(crystalline H2O) = 38.02 J/(mol•K), Cp(l)(liquid H2O) = 75.35 J/(mol•K), and Cp(g)(H2O gas) = 33.60 J/(mol•K).
1. How much heat is released in the conversion of 1.00 L of steam at 21.9 atm and 200°C to ice at −6.0°C and 1 atm?
2. How much heat must be applied to convert a 1.00 g piece of ice at −10°C to steam at 120°C?
3. How many grams of boiling water must be added to a glass with 25.0 g of ice at −3°C to obtain a liquid with a temperature of 45°C?
4. How many grams of ice at −5.0°C must be added to 150.0 g of water at 22°C to give a final temperature of 15°C?
Numerical Answers
1. The transition from a liquid to a gaseous phase is accompanied by a drastic decrease in density. According to the data in the table and the plot, the boiling point of liquid oxygen is between 90 and 100 K (actually 90.2 K).
2. 45.0 kJ/mol
3. 488 kJ
1. 32.6 kJ
2. 57 g
11.5: Vapor Pressure
Conceptual Problems
1. What is the relationship between the boiling point, vapor pressure, and temperature of a substance and atmospheric pressure?
2. What is the difference between a volatile liquid and a nonvolatile liquid? Suppose that two liquid substances have the same molecular mass, but one is volatile and the other is nonvolatile. What differences in the molecular structures of the two substances could account for the differences in volatility?
3. An “old wives’ tale” states that applying ethanol to the wrists of a child with a very high fever will help to reduce the fever because blood vessels in the wrists are close to the skin. Is there a scientific basis for this recommendation? Would water be as effective as ethanol?
4. Why is the air over a strip of grass significantly cooler than the air over a sandy beach only a few feet away?
5. If gasoline is allowed to sit in an open container, it often feels much colder than the surrounding air. Explain this observation. Describe the flow of heat into or out of the system, as well as any transfer of mass that occurs. Would the temperature of a sealed can of gasoline be higher, lower, or the same as that of the open can? Explain your answer.
6. What is the relationship between the vapor pressure of a liquid and
1. its temperature?
2. the surface area of the liquid?
3. the pressure of other gases on the liquid?
4. its viscosity?
7. At 25°C, benzene has a vapor pressure of 12.5 kPa, whereas the vapor pressure of acetic acid is 2.1 kPa. Which is more volatile? Based on the intermolecular interactions in the two liquids, explain why acetic acid has the lower vapor pressure.
Numerical Problems
1. Acetylene (C2H2), which is used for industrial welding, is transported in pressurized cylinders. Its vapor pressure at various temperatures is given in the following table. Plot the data and use your graph to estimate the vapor pressure of acetylene at 293 K. Then use your graph to determine the value of ΔHvap for acetylene. How much energy is required to vaporize 2.00 g of acetylene at 250 K?
T (K) 145 155 175 200 225 250 300
P (mmHg) 1.3 7.8 32.2 190 579 1370 5093
2. The following table gives the vapor pressure of water at various temperatures. Plot the data and use your graph to estimate the vapor pressure of water at 25°C and at 75°C. What is the vapor pressure of water at 110°C? Use these data to determine the value of ΔHvap for water.
T (°C) 0 10 30 50 60 80 100
P (mmHg) 4.6 9.2 31.8 92.6 150 355 760
3. The ΔHvap of carbon tetrachloride is 29.8 kJ/mol, and its normal boiling point is 76.8°C. What is its boiling point at 0.100 atm?
4. The normal boiling point of sodium is 883°C. If ΔHvap is 97.4 kJ/mol, what is the vapor pressure (in millimeters of mercury) of liquid sodium at 300°C?
5. An unknown liquid has a vapor pressure of 0.860 atm at 63.7°C and a vapor pressure of 0.330 atm at 35.1°C. Use the data in Table 11.6 in Section 11.5 to identify the liquid.
6. An unknown liquid has a boiling point of 75.8°C at 0.910 atm and a boiling point of 57.2°C at 0.430 atm. Use the data in Table 11.6 in Section 11.5 to identify the liquid.
7. If the vapor pressure of a liquid is 0.850 atm at 20°C and 0.897 atm at 25°C, what is the normal boiling point of the liquid?
8. If the vapor pressure of a liquid is 0.799 atm at 99.0°C and 0.842 atm at 111°C, what is the normal boiling point of the liquid?
9. The vapor pressure of liquid SO2 is 33.4 torr at −63.4°C and 100.0 torr at −47.7 K.
1. What is the ΔHvap of SO2?
2. What is its vapor pressure at −24.5 K?
3. At what temperature is the vapor pressure equal to 220 torr?
10. The vapor pressure of CO2 at various temperatures is given in the following table:
T (°C) −120 −110 −100 −90
P (torr) 9.81 34.63 104.81 279.5
1. What is ΔHvap over this temperature range?
2. What is the vapor pressure of CO2 at −70°C?
3. At what temperature does CO2 have a vapor pressure of 310 torr?
Numerical Answers
1. vapor pressure at 273 K is 3050 mmHg; ΔHvap = 18.7 kJ/mol, 1.44 kJ
2. 12.5°C
3. ΔHvap = 28.9 kJ/mol, n-hexane
4. ΔHvap = 7.81 kJ/mol, 36°C
11.6: Phase Diagrams
Conceptual Problems
1. A phase diagram is a graphic representation of the stable phase of a substance at any combination of temperature and pressure. What do the lines separating different regions in a phase diagram indicate? What information does the slope of a line in a phase diagram convey about the physical properties of the phases it separates? Can a phase diagram have more than one point where three lines intersect?
2. If the slope of the line corresponding to the solid/liquid boundary in the phase diagram of water were positive rather than negative, what would be the effect on aquatic life during periods of subzero temperatures? Explain your answer.
Conceptual Answer
1. The lines in a phase diagram represent boundaries between different phases; at any combination of temperature and pressure that lies on a line, two phases are in equilibrium. It is physically impossible for more than three phases to coexist at any combination of temperature and pressure, but in principle there can be more than one triple point in a phase diagram. The slope of the line separating two phases depends upon their relative densities. For example, if the solid–liquid line slopes up and to the right, the liquid is less dense than the solid, while if it slopes up and to the left, the liquid is denser than the solid.
Numerical Problems
1. Naphthalene (C10H8) is the key ingredient in mothballs. It has normal melting and boiling points of 81°C and 218°C, respectively. The triple point of naphthalene is 80°C at 1000 Pa. Use these data to construct a phase diagram for naphthalene and label all the regions of your diagram.
2. Argon is an inert gas used in welding. It has normal boiling and freezing points of 87.3 K and 83.8 K, respectively. The triple point of argon is 83.8 K at 0.68 atm. Use these data to construct a phase diagram for argon and label all the regions of your diagram.
11.7: Structure of Solids
Conceptual Problems
1. Compare the solid and liquid states in terms of
a. rigidity of structure.
b. long-range order.
c. short-range order.
2. How do amorphous solids differ from crystalline solids in each characteristic? Which of the two types of solid is most similar to a liquid?
a. rigidity of structure
b. long-range order
c. short-range order
3. Why is the arrangement of the constituent atoms or molecules more important in determining the properties of a solid than a liquid or a gas?
4. Why are the structures of solids usually described in terms of the positions of the constituent atoms rather than their motion?
5. What physical characteristics distinguish a crystalline solid from an amorphous solid? Describe at least two ways to determine experimentally whether a material is crystalline or amorphous.
6. Explain why each characteristic would or would not favor the formation of an amorphous solid.
a. slow cooling of pure molten material
b. impurities in the liquid from which the solid is formed
c. weak intermolecular attractive forces
7. A student obtained a solid product in a laboratory synthesis. To verify the identity of the solid, she measured its melting point and found that the material melted over a 12°C range. After it had cooled, she measured the melting point of the same sample again and found that this time the solid had a sharp melting point at the temperature that is characteristic of the desired product. Why were the two melting points different? What was responsible for the change in the melting point?
Conceptual Answers II
3. The arrangement of the atoms or molecules is more important in determining the properties of a solid because of the greater persistent long-range order of solids. Gases and liquids cannot readily be described by the spatial arrangement of their components because rapid molecular motion and rearrangement defines many of the properties of liquids and gases.
7. The initial solid contained the desired compound in an amorphous state, as indicated by the wide temperature range over which melting occurred. Slow cooling of the liquid caused it to crystallize, as evidenced by the sharp second melting point observed at the expected temperature.
Conceptual Problems II
1. Why is it valid to represent the structure of a crystalline solid by the structure of its unit cell? What are the most important constraints in selecting a unit cell?
2. All unit cell structures have six sides. Can crystals of a solid have more than six sides? Explain your answer.
3. Explain how the intensive properties of a material are reflected in the unit cell. Are all the properties of a bulk material the same as those of its unit cell? Explain your answer.
4. The experimentally measured density of a bulk material is slightly higher than expected based on the structure of the pure material. Propose two explanations for this observation.
5. The experimentally determined density of a material is lower than expected based on the arrangement of the atoms in the unit cell, the formula mass, and the size of the atoms. What conclusion(s) can you draw about the material?
6. Only one element (polonium) crystallizes with a simple cubic unit cell. Why is polonium the only example of an element with this structure?
7. What is meant by the term coordination number in the structure of a solid? How does the coordination number depend on the structure of the metal?
8. Arrange the three types of cubic unit cells in order of increasing packing efficiency. What is the difference in packing efficiency between the hcp structure and the ccp structure?
9. The structures of many metals depend on pressure and temperature. Which structure—bcc or hcp—would be more likely in a given metal at very high pressures? Explain your reasoning.
10. A metal has two crystalline phases. The transition temperature, the temperature at which one phase is converted to the other, is 95°C at 1 atm and 135°C at 1000 atm. Sketch a phase diagram for this substance. The metal is known to have either a ccp structure or a simple cubic structure. Label the regions in your diagram appropriately and justify your selection for the structure of each phase.
Numerical Problems II
1. Metallic rhodium has an fcc unit cell. How many atoms of rhodium does each unit cell contain?
2. Chromium has a structure with two atoms per unit cell. Is the structure of this metal simple cubic, bcc, fcc, or hcp?
3. The density of nickel is 8.908 g/cm3. If the metallic radius of nickel is 125 pm, what is the structure of metallic nickel?
4. The density of tungsten is 19.3 g/cm3. If the metallic radius of tungsten is 139 pm, what is the structure of metallic tungsten?
5. An element has a density of 10.25 g/cm3 and a metallic radius of 136.3 pm. The metal crystallizes in a bcc lattice. Identify the element.
6. A 21.64 g sample of a nonreactive metal is placed in a flask containing 12.00 mL of water; the final volume is 13.81 mL. If the length of the edge of the unit cell is 387 pm and the metallic radius is 137 pm, determine the packing arrangement and identify the element.
7. A sample of an alkali metal that has a bcc unit cell is found to have a mass of 1.000 g and a volume of 1.0298 cm3. When the metal reacts with excess water, the reaction produces 539.29 mL of hydrogen gas at 0.980 atm and 23°C. Identify the metal, determine the unit cell dimensions, and give the approximate size of the atom in picometers.
8. A sample of an alkaline earth metal that has a bcc unit cell is found to have a mass 5.000 g and a volume of 1.392 cm3. Complete reaction with chlorine gas requires 848.3 mL of chlorine gas at 1.050 atm and 25°C. Identify the metal, determine the unit cell dimensions, and give the approximate size of the atom in picometers.
9. Lithium crystallizes in a bcc structure with an edge length of 3.509 Å. Calculate its density. What is the approximate metallic radius of lithium in picometers?
10. Vanadium is used in the manufacture of rust-resistant vanadium steel. It forms bcc crystals with a density of 6.11 g/cm3 at 18.7°C. What is the length of the edge of the unit cell? What is the approximate metallic radius of the vanadium in picometers?
11. A simple cubic cell contains one metal atom with a metallic radius of 100 pm.
1. Determine the volume of the atom(s) contained in one unit cell [the volume of a sphere = (${4 \over 3}$)πr3].
2. What is the length of one edge of the unit cell? (Hint: there is no empty space between atoms.)
3. Calculate the volume of the unit cell.
4. Determine the packing efficiency for this structure.
5. Use the steps in Problem 11 to calculate the packing efficiency for a bcc unit cell with a metallic radius of 1.00 Å.
Numerical Answers
1. four
3. fcc
5. molybdenum
7. sodium, unit cell edge = 428 pm, r = 185 pm
9. d = 0.5335 g/cm3, r =151.9 pm
11.8: Bonding in Solids
Conceptual Problems
1. Four vials labeled A–D contain sucrose, zinc, quartz, and sodium chloride, although not necessarily in that order. The following table summarizes the results of the series of analyses you have performed on the contents:
A B C D
Melting Point high high high low
Thermal Conductivity poor poor good poor
Electrical Conductivity in Solid State moderate poor high poor
Electrical Conductivity in Liquid State high poor high poor
Hardness hard hard soft soft
Luster none none high none
Match each vial with its contents.
2. Do ionic solids generally have higher or lower melting points than molecular solids? Why? Do ionic solids generally have higher or lower melting points than covalent solids? Explain your reasoning.
3. The strength of London dispersion forces in molecular solids tends to increase with molecular mass, causing a smooth increase in melting points. Some molecular solids, however, have significantly lower melting points than predicted by their molecular masses. Why?
4. Suppose you want to synthesize a solid that is both heat resistant and a good electrical conductor. What specific types of bonding and molecular interactions would you want in your starting materials?
5. Explain the differences between an interstitial alloy and a substitutional alloy. Given an alloy in which the identity of one metallic element is known, how could you determine whether it is a substitutional alloy or an interstitial alloy?
6. How are intermetallic compounds different from interstitial alloys or substitutional alloys?
Conceptual Answers
1.
a. NaCl, ionic solid
b. quartz, covalent solid
c. zinc, metal
d. sucrose, molecular solid
5. In a substitutional alloy, the impurity atoms are similar in size and chemical properties to the atoms of the host lattice; consequently, they simply replace some of the metal atoms in the normal lattice and do not greatly perturb the structure and physical properties. In an interstitial alloy, the impurity atoms are generally much smaller, have very different chemical properties, and occupy holes between the larger metal atoms. Because interstitial impurities form covalent bonds to the metal atoms in the host lattice, they tend to have a large effect on the mechanical properties of the metal, making it harder, less ductile, and more brittle. Comparing the mechanical properties of an alloy with those of the parent metal could be used to decide whether the alloy were a substitutional or interstitial alloy.
Numerical Problems
1. Will the melting point of lanthanum(III) oxide be higher or lower than that of ferrous bromide? The relevant ionic radii are as follows: La3+, 104 pm; O2−, 132 pm; Fe2+, 83 pm; and Br, 196 pm. Explain your reasoning.
2. Draw a graph showing the relationship between the electrical conductivity of metallic silver and temperature.
3. Which has the higher melting point? Explain your reasoning in each case.
a. Os or Hf
b. SnO2 or ZrO2
c. Al2O3 or SiO2
4. Draw a graph showing the relationship between the electrical conductivity of a typical semiconductor and temperature.
Numerical Answer
3.
a. Osmium has a higher melting point, due to more valence electrons for metallic bonding.
b. Zirconium oxide has a higher melting point, because it has greater ionic character.
c. Aluminum oxide has a higher melting point, again because it has greater ionic character. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/11%3A_Liquids_and_Intermolecular_Forces/11.E%3A_Liquids_and_Intermolecular_Forces_%28Exercises%29.txt |
• intermolecular forces – forces that exist between molecules
11.1: A Molecular Comparison of Gases, Liquids, and Solids
The state of a substance depends on the balance between the kinetic energy of the individual particles (molecules or atoms) and the intermolecular forces. The kinetic energy keeps the molecules apart and moving around, and is a function of the temperature of the substance and the intermolecular forces try to draw the particles together.
• gases
• average kinetic energy of the molecules is larger than average energy of attractions between molecules
• lack of strong attractive forces allows gases to expand
• liquids
• denser than gases
• have a definite volume
• attractive forces not strong enough to keep molecules from moving allowing liquids to hold shape of container
• solids
• intermolecular forces hold molecules together and keep them from moving
• not very compressible
• crystalline – solids with highly ordered structures
Gas, Liquid and Solid
Gas assumes both the volume and shape of container is compressible diffusion within a gas occurs rapidly flows readily
Liquid Assumes the shape of the portion of the container it occupies Does not expand to fill container Is virtually incompressible Diffusion within a liquid occurs slowly Flows readily
Solid Retains its own shape and volume Is virtually incompressible Diffusion within a solid occurs extremely slowly Does not flow
• state of substance depends on balance between the kinetic energies of the particles and interparticle energies of attraction
• kinetic energies depends on temperature and tend to keep particles apart and moving
• interparticle attractions draw particles together
• condensed phases – liquids and solids because particles are close together compared to gases
• increase temperature forces molecules to be closer together ® increase in strength of intermolecular forces
11.2: Intermolecular Forces
Molecules in liquids are held to other molecules by intermolecular interactions, which are weaker than the intramolecular interactions that hold molecules and polyatomic ions together. The three major types of intermolecular interactions are dipole–dipole interactions, London dispersion forces (these two are often referred to collectively as van der Waals forces), and hydrogen bonds.
• intermolecular forces weaker than ionic or covalent bonds
• many properties of liquids reflect strengths of intermolecular forces
• three types of intermolecular forces: dipole-dipole forces, London dispersion forces, and hydrogen-bonding forces
• also called van der Waals forces
• less than 15% as strong as covalent or ionic bonds
• electrostatic in nature, involves attractions between positive and negative species
11.2.1 Ion-Dipole Forces
• Ion-Dipole Force – exists between an ion and partial charge at one end of a polar molecule
• Polar molecules are dipoles
• magnitude of attraction increases as either the charge of ion or magnitude of dipole moment increases
11.2.2 Dipole-Dipole Forces
• dipole-dipole force – exists between neutral polar molecules
• effective only when polar molecules are very close together
• weaker than ion-dipole forces
• for molecules of approximately equal mass and size, the strengths of intermolecular attractions increase with increasing polarity
• increase dipole moment ® increase boiling point
11.2.3 London Dispersion Forces
• interparticle forces that exist between nonpolar atoms or molecules
• motion of electrons can create an instantaneous dipole moment
• molecules have to be very close together
• polarizability – ease in which the charge distribution in a molecule can be distorted
• greater polarizability ® more easily electron cloud can be distorted to give momentary dipole
• larger molecules have greater polarizability
• London dispersion forces increase with increasing molecular size
• Dispersion forces increase in strength with increasing molecular weight
• Molecular shape affects intermolecular attractions
• greater surface contact ® greater boiling point and London dispersion forces
• dispersion forces operate between all molecules
• comparing relative strengths of intermolecular attractions:
• 1) comparable molecular weights and shapes = equal dispersion forces
• differences in magnitudes of attractive forces due to differences in strengths of dipole-dipole attractions
• most polar molecule has strongest attractions
• 2) differing molecular weights = dispersion forces tend to be the decisive ones
• differences in magnitudes of attractive forces associated with differences in molecular weights
• most massive molecular has strongest attractions
11.2.4 Hydrogen Bonding
• hydrogen bonding – special type of intermolecular attraction that exists between the hydrogen atom in a polar bond and an unshared electron pair on a nearby electronegative ion or atom
• hydrogen bond with F, N, and O is polar
• density of ice is lower than that of liquid water
• when water freezes the molecules assume the ordered open arrangement ® makes ice less dense than water
• a given mass of ice has a greater volume than the same mass of water
• structure of ice allows the maximum number of hydrogen bonding interactions to exist
11.2.5 Comparing Intermolecular Forces
• dispersion forces found in all substances
• strengths of forces increase with increases molecular weight and also depend on shape
• dipole-dipole forces add to effect of dispersion forces and found in polar molecules
• hydrogen bonds tend to be strongest intermolecular force
11.3: Some Properties of Liquids
Surface tension, capillary action, and viscosity are unique properties of liquids that depend on the nature of intermolecular interactions. Surface tension is the energy required to increase the surface area of a liquid. Surfactants are molecules that reduce the surface tension of polar liquids like water. Capillary action is the phenomenon in which liquids rise up into a narrow tube called a capillary. The viscosity of a liquid is its resistance to flow.
• two properties of liquids: viscosity and surface tension
11.3.1 Viscosity
• viscosity – resistance of a liquid to flow
• the greater the viscosity the more slowly the liquid flows
• measured by timing how long it takes a certain amount of liquid to flow through a thin tube under gravitational forces
• can also be measured by how long it takes steel spheres to fall through the liquid
• viscosity related to ease with which individual molecules of liquid can move with respect to one another
• depends on attractive forces between molecules, and whether structural features exist to cause molecules to be entangled
• viscosity decreases with increasing temperature
11.3.2 Surface Tension
• surface tension – energy required to increase the surface area of a liquid by a unit amount
• surface tension of water at 20° C is 7.29 x 10-2 J/m2
• 7.29 x 10-2 J/m2 must be supplied to increase surface area of a given amount of water by 1 m2
• cohesive forces – intermolecular forces that bind similar molecules
• adhesive forces – intermolecular forces that bind a substance to a surface
• capillary action – rise of liquids up very narrow tubes
11.4: Phase Changes
Fusion, vaporization, and sublimation are endothermic processes, whereas freezing, condensation, and deposition are exothermic processes. Changes of state are examples of phase changes, or phase transitions. All phase changes are accompanied by changes in the energy of a system. Changes from a more-ordered state to a less-ordered state (such as a liquid to a gas) are endothermic. Changes from a less-ordered state to a more-ordered state (such as a liquid to a solid) are always exothermic.
11.4.1 Energy Changes Accompanying Phase Changes
• phase changes require energy
• phase changes to less ordered state requires energy
• melting process of solid called fusion
• heat of fusion – enthalpy change of melting a solid
• D Hfus water = 6.01 kJ/mol
• heat of vaporization – heat needed for vaporization of liquid
• D Hvap water = 40.67 kJ/mol
• melting, vaporization, and sublimation are endothermic
• freezing, condensation, and deposition are exothermic
11.4.2 Heating Curves
• heating curve – graph of temperature of system versus the amount of heat added
• used to calculate enthalpy changes
• supercooled water – when water if cooled to a temperature below 0° C
11.4.3 Critical Temperature and Pressure
• critical temperature – highest temperature at which a substance can exist as a liquid
• critical pressure – pressure required to bring about liquefaction at critical temperature
• the greater the intermolecular attractive forces, the more readily gases liquefy ® higher critical temperature
• cannot liquefy a gas by applying pressure if gas is above critical temperature
11.5: Vapor Pressure
Because the molecules of a liquid are in constant motion and possess a wide range of kinetic energies, at any moment some fraction of them has enough energy to escape from the surface of the liquid to enter the gas or vapor phase. This process, called vaporization or evaporation, generates a vapor pressure above the liquid. Molecules in the gas phase can collide with the liquid surface and reenter the liquid via condensation. Eventually, a steady state or dynamic equilibrium is reached.
vapor pressure – measures tendency of a liquid to evaporate
11.5.1 Explaining Vapor Pressure on the Molecular Level
• dynamic equilibrium – condition when two opposing processes are occurring simultaneously at equal rates
• vapor pressure of a liquid is the pressure exerted by its vapor when the liquid and vapor states are in dynamic equilibrium
11.5.2 Volatility, Vapor Pressure, and Temperature
• volatile – liquids that evaporate readily
• vapor pressure increases with increasing temperature
11.5.3 Vapor Pressure and Boiling Point
• liquids boil when its vapor pressure equals the external pressure acting on the surface of the liquid
• temperature of boiling increase with increasing external pressure
• normal boiling point – boiling point of a liquid at 1 atm
• higher pressures cause water to boil at higher temperatures
11.6: Phase Diagrams
The states of matter exhibited by a substance under different temperatures and pressures can be summarized graphically in a phase diagram, which is a plot of pressure versus temperature. Phase diagrams contain discrete regions corresponding to the solid, liquid, and gas phases. The solid and liquid regions are separated by the melting curve of the substance, and the liquid and gas regions are separated by its vapor pressure curve, which ends at the critical point.
• phase diagrams – graphical way to summarize conditions under which equilibria exist between the different states of matter
• three important curves:
• 1) vapor pressure curve of liquid
• shows equilibrium of liquid and gas phases
• normal boiling point = point on curve where pressure at 1 atm
• 2) variation in vapor pressure of solid at it sublimes at different temperatures
• 3) change in melting point of solid with increasing pressure
• slopes right as pressure increases
• higher temperatures needed to melt solids at higher pressures
• melting point of solid identical to freezing point
• differ only in temperature direction from which phase change is approached
• melting point at 1 atm is the normal melting point
• triple point – point at which all three phases are at equilibrium
• gas phase stable at low pressures and high temperatures
• solid phase stable at low temperatures and high pressures
• liquid phase – stable between gas and solids
11.6.1 the Phase diagrams of H2O and CO2
• melting point of CO2 increases with increasing pressure
• melting point of H2O decreases with increasing pressure
• triple point of H2O (0.0098° C and 4.58 torr) at lower pressure than CO2 (-56.4° C and 5.11 atm)
• solid CO2 does not melt but sublimes
• CO2 does not have a normal melting point but a normal sublimation point
• CO2 absorbs energy at ordinary temperatures
11.7: Structure of Solids
A crystalline solid can be represented by its unit cell, which is the smallest identical unit that when stacked together produces the characteristic three-dimensional structure. Solids are characterized by an extended three-dimensional arrangement of atoms, ions, or molecules in which the components are generally locked into their positions. The components can be arranged in a regular repeating three-dimensional array. The smallest repeating unit of a crystal lattice is the unit cell.
• crystalline solid – solid whose atoms, ion, or molecules are ordered in well-defined arrangements
• flat surfaces or faces that make definite angles
• regular shapes
• amorphous solid – solid whose particles have no orderly structure
• lack well-defined faces and shapes
• mixtures of molecules that do not stack together well
• large, complicated molecules
• intermolecular forces vary in strength
• does not melt at a specific temperature but soften over a temperature range
11.7.1 Unit Cell
• unit cell – repeating unit of a solid
• crystal lattice – three-dimensional array of points, each representing an identical environment within the crystal
• three types of cubic unit cell: primitive cubic, body-centered cubic, and face-centered cubic
• primitive cubic – lattice points at corners only
• body-centered cubic – lattice points at corners and center
• face-centered cubic – lattice points at center of each face and at each corner
11.7.2 The Crystal structure of Sodium Chloride
• total cation-to-anion ratio of a unit cell must be the same as that for entire crystal
11.7.3 Close Packing of Spheres
• structures of crystalline solids are those that bring particles in closest contact to maximize the attractive forces
• most particles that make up solids are spherical
• two forms of close packing: cubic close packing and hexagonal close packing
• hexagonal close packing – spheres of the third layer that are placed in line with those of the first layer
• coordination number – number of particles immediately surrounding a particle in the crystal structure
• both forms of close packing have coordination number of 12
11.8: Bonding in Solids
The major types of solids are ionic, molecular, covalent, and metallic. Ionic solids consist of positively and negatively charged ions held together by electrostatic forces; the strength of the bonding is reflected in the lattice energy. Ionic solids tend to have high melting points and are rather hard. Molecular solids are held together by relatively weak forces, such as dipole–dipole interactions, hydrogen bonds, and London dispersion forces. Metallic solids have unusual properties.
Major Types of Solid
Type of Solid Forms of Unit Particles Forces Between Particles Properties Examples
Molecular Atoms of molecules London dispersion, dipole-dipole forces, hydrogen bonds Fairly soft, low to moderately high melting point, poor thermal and electrical conduction Argon, methane, sucrose, dry ice
Covalent-network Atoms connected in a network of covalent bonds Covalent bonds Very hard, very high melting point, often poor thermal and electrical conduction Diamond, quartz
Ionic Positive and negative ions Electrostatic attractions Hard and brittle, high melting point, poor thermal and electrical conduction Typical salts
Metallic atoms Metallic bonds Soft to very hard, low to very high melting point, excellent thermal and electrical conduction, malleable and ductile All metallic elements
11.8.1 Molecular Solids
• molecular solids – atoms or molecules held together by intermolecular forces
• soft, low melting points
• gases or liquids at room temperature from molecular solids at low temperature
• properties depends on strengths of forces and ability of molecules to pack efficiently in three dimensions
• intermolecular forces that depend on close contact are not as effective
11.8.2 Covalent-Network Solids
• covalent-network solids – atoms held together in large networks or chains by covalent bonds
• hard, high melting points
11.8.3 Ionic Solids
• ionic solids – ions held together by ionic bonds
• strength depends on charges of ions
• structure of ionic solids depends on charges and relative sizes of ions
11.8.4 Metallic Solids
• metallic solids – metal atoms
• usually have hexagonal close-packed, cubic close-packed, or body-centered-cubic structures
• each atom has 8 or 12 adjacent atoms
• bonding due to valence electrons that are delocalized throughout entire solid
• strength of bonding increases as number of electrons available for bonding increases
• mobility of electrons make metallic solids good conductors of heat and electricity | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/11%3A_Liquids_and_Intermolecular_Forces/11.S%3A_Liquids_and_Intermolecular_Forces_%28Summary%29.txt |
Thumbnail: The alpha form of solid polonium. (Public Domain; Cadmium).
12: Solids and Modern Materials
It can often be quite straightforward to tell materials classes apart by look or feel. Metals are usually more reflective or 'metallic' looking, ceramics are commonly matte and polymers may be shiny or matte, but are typically less dense than either metals or ceramics. Composites may be harder to immediately identify, but the surface may appear non-uniform and/or sectioning the sample may reveal fibres or particles.
It is useful to note that taking a cross-section can often be helpful in identifying materials used in components as the internal material and/or microstructure may differ from that at the edge.
Some materials may not be quite so easily identified simply from appearance and texture, but considering the factors mentioned can help to narrow down the possibilities. The table below gives a rough guideline to the kinds of properties you would expect from each class, and a few examples:
Properties expected from each class
Class Common Properties Examples
Metal Hard, ductile and conduct heat and electricity Copper (wires), stainless steel (cutlery)
Polymer Widely variable, often soft and flexible Polystyrene (cups), polycarbonate (CDs), polyethylene (plastic bags)
Ceramic Hard, brittle, resistant to corrosion, electrically non-conductive Concrete (buildings), PZT (piezoelectric used in lighters and ultrasonic transducers), porcelain (vases, teacups)
The tree-diagram below shows an overview of a variety of materials that might be encountered:
Coatings
Many components may have some kind of coating; a covering of another material designed to improve the surface qualities of the item. The improvement could be for many reasons including: corrosion resistance, appearance, adhesion, wear resistance and scratch resistance.
Different kinds of coating will have different processing methods. It is often possible to deduce the method from the composition (of the coating and the bulk of the component) and the shape of the component.
Modern Materials
• To become familiar with the properties of some contemporary materials.
In addition to polymers, other materials, such as ceramics, high-strength alloys, and composites, play a major role in almost every aspect of our lives. Until relatively recently, steel was used for any application that required an especially strong and durable material, such as bridges, automobiles, airplanes, golf clubs, and tennis rackets. In the last 15 to 20 years, however, graphite or boron fiber golf clubs and tennis rackets have made wood and steel obsolete for these items. Likewise, a modern jet engine now is largely composed of Ti and Ni by weight rather than steel (Table 12.9.1). The percentage of iron in wings and fuselages is similarly low, which indicates the extent to which other materials have supplanted steel. The Chevrolet Corvette introduced in 1953 was considered unusual because its body was constructed of fiberglass, a composite material, rather than steel; by 1992, Jaguar fabricated an all-aluminum limited-edition vehicle. In fact, the current models of many automobiles have engines that are made mostly of aluminum rather than steel. In this section, we describe some of the chemistry behind three classes of contemporary materials: ceramics, superalloys, and composites.
Table 12.9.1 The Approximate Elemental Composition of a Modern Jet Engine
Element Percentage by Mass
titanium 38
nickel 37
chromium 12
cobalt 6
aluminum 3
niobium 1
tantalum 0.025
Ceramics
A ceramic is any nonmetallic, inorganic solid that is strong enough for use in structural applications. Traditional ceramics, which are based on metal silicates or aluminosilicates, are the materials used to make pottery, china, bricks, and concrete. Modern ceramics contain a much wider range of components and can be classified as either ceramic oxides, which are based on metal oxides such as alumina (Al2O3), zirconia (ZrO2), and beryllia (BeO), or nonoxide ceramics, which are based on metal carbides such as silicon carbide (carborundum, SiC) and tungsten carbide (WC), or nitrides like silicon nitride (Si3N4) and boron nitride (BN).
All modern ceramics are hard, lightweight, and stable at very high temperatures. Unfortunately, however, they are also rather brittle, tending to crack or break under stresses that would cause metals to bend or dent. Thus a major challenge for materials scientists is to take advantage of the desirable properties of ceramics, such as their thermal and oxidative stability, chemical inertness, and toughness, while finding ways to decrease their brittleness to use them in new applications. Few metals can be used in jet engines, for example, because most lose mechanical strength and react with oxygen at the very high operating temperatures inside the engines (approximately 2000°C). In contrast, ceramic oxides such as Al2O3 cannot react with oxygen regardless of the temperature because aluminum is already in its highest possible oxidation state (Al3+). Even nonoxide ceramics such as silicon and boron nitrides and silicon carbide are essentially unreactive in air up to about 1500°C. Producing a high-strength ceramic for service use involves a process called sintering, which fuses the grains into a dense and strong material (Figure 12.9.2).
Figure 12.9.2: Sintering These photos show the effects of sintering magnesium oxide grains: (a) the microstructure before sintering; (b) the microstructure of the ceramic after sintering for two hours at 1250°C; and (c) the microstructure after sintering for two hours at 1450°C. During the sintering process, the grains fuse, forming a dense and strong material.
Ceramics are hard, lightweight, and able to withstand high temperatures, but they are also brittle.
One of the most widely used raw materials for making ceramics is clay. Clay minerals consist of hydrated alumina (Al2O3) and silica (SiO2) that have a broad range of impurities, including barium, calcium, sodium, potassium, and iron. Although the structures of clay minerals are complicated, they all contain layers of metal atoms linked by oxygen atoms. Water molecules fit between the layers to form a thin film of water. When hydrated, clays can be easily molded, but during high-temperature heat treatment, called firing, a dense and strong ceramic is produced.
Because ceramics are so hard, they are easily contaminated by the material used to grind them. In fact, the ceramic often grinds the metal surface of the mill almost as fast as the mill grinds the ceramic! The sol-gel process was developed to address this problem. In this process, a water-soluble precursor species, usually a metal or semimetal alkoxide [M(OR)n] undergoes a hydrolysis reaction to form a cloudy aqueous dispersion called a sol. The sol contains particles of the metal or semimetal hydroxide [M(OH)n], which are typically 1–100 nm in diameter. As the reaction proceeds, molecules of water are eliminated from between the M(OH)n units in a condensation reaction, and the particles fuse together, producing oxide bridges, M–O–M. Eventually, the particles become linked in a three-dimensional network that causes the solution to form a gel, similar to a gelatin dessert. Heating the gel to 200°C–500°C causes more water to be eliminated, thus forming small particles of metal oxide that can be amazingly uniform in size. This chemistry starts with highly pure SiCl4 and proceeds via the following reactions starting with the alkoxide formation
$SiCl_{4}\left ( s \right )+4CH_{3}CH_{2}OH\left ( l \right )+4NH_{3}\left ( g \right ){\rightarrow}SiO\left ( OCH_{2}CH_{3} \right )_{4}\left ( s \right )+4NH_{4}Sl\left ( s \right ) \label{12.8.1}$
and then the hydrolysis of the alkoxide
$SiO\left ( OCH_{2}CH_{3} \right )_{4}\left ( s \right )+4H_{2}O\left ( l \right ) {\rightarrow} \left ( HO \right )_{3}Si-OH\left ( s \right )+ 4CH_{3}CH_{2}OH\left ( aq \right )$
ending with the condensation
$\left ( HO_{3} \right )Si-OH\left ( s \right )+nHO-Si\left ( OH \right )_{3}\left ( s \right )\rightarrow \left ( HO_{3} \right )Si\left ( -O-Si\left ( OH \right )_{3} \right )_{n}\left ( s \right )+nH_{2}O\left ( l \right )$
Nature uses the same process to create opal gemstones.
Superalloys
Superalloys are high-strength alloys, often with a complex composition, that are used in systems requiring mechanical strength, high surface stability (minimal flaking or pitting), and resistance to high temperatures. The aerospace industry, for example, requires materials that have high strength-to-weight ratios to improve the fuel efficiency of advanced propulsion systems, and these systems must operate safely at temperatures greater than 1000°C.
Superalloys are used in systems requiring mechanical strength, minimal flaking or pitting, and high-temperature resistance.
Although most superalloys are based on nickel, cobalt, or iron, other metals are used as well. Pure nickel or cobalt is relatively easily oxidized, but adding small amounts of other metals (Al, Co, Cr, Mo, Nb, Ti, and W) results in an alloy that has superior properties. Consequently, most of the internal parts of modern gas turbine jet engines are now made of superalloys based on either nickel (used in blades and disks) or cobalt (used in vanes, combustion chamber liners, and afterburners). The cobalt-based superalloys are not as strong as the nickel-based ones, but they have excellent corrosion resistance at high temperatures.
Other alloys, such as aluminum–lithium and alloys based on titanium, also have applications in the aerospace industry. Because aluminum–lithium alloys are lighter, stiffer, and more resistant to fatigue at high temperatures than aluminum itself, they are used in engine parts and in the metal “skins” that cover wings and bodies. Titanium’s high strength, corrosion resistance, and lightweight properties are equally desirable for applications where minimizing weight is important (as in airplanes). Unfortunately, however, metallic titanium reacts rapidly with air at high temperatures to form TiN and TiO2. The welding of titanium or any similar processes must therefore be carried out in an argon or inert gas atmosphere, which adds significantly to the cost. Initially, titanium and its alloys were primarily used in military applications, but more recently, they have been used as components of the airframes of commercial planes, in ship structures, and in biological implants.
Composite Materials
Composite materials have at least two distinct components: the matrix (which constitutes the bulk of the material) and fibers or granules that are embedded within the matrix and limit the growth of cracks by pinning defects in the bulk material (Figure 12.9.3). The resulting material is stronger, tougher, stiffer, and more resistant to corrosion than either component alone. Composites are thus the nanometer-scale equivalent of reinforced concrete, in which steel rods greatly increase the mechanical strength of the cement matrix, and are extensively used in the aircraft industry, among others. For example, the Boeing 777 is 9% composites by weight, whereas the newly developed Boeing 787 is 50% composites by weight. Not only does the use of composite materials reduce the weight of the aircraft, and therefore its fuel consumption, but it also allows new design concepts because composites can be molded. Moreover, by using composites in the Boeing 787 multiple functions can be integrated into a single system, such as acoustic damping, thermal regulation, and the electrical system.
Three distinct types of composite material are generally recognized, distinguished by the nature of the matrix. These are polymer-matrix composites, metal-matrix composites, and ceramic-matrix composites.
Figure 12.9.3: Some Possible Arrangements of Fibers in Fiber-Reinforced Composite Materials The arrangements shown range from discontinuous and randomly oriented to continuous and aligned. The fibers limit the growth of cracks by pinning defects within the matrix.
Composites are stronger, tougher, stiffer, and more resistant to corrosion than their components alone.
Fiberglass is a polymer-matrix composite that consists of glass fibers embedded in a polymer, forming tapes that are then arranged in layers impregnated with epoxy. The result is a strong, stiff, lightweight material that is resistant to chemical degradation. It is not strong enough, however, to resist cracking or puncturing on impact. Stronger, stiffer polymer-matrix composites contain fibers of carbon (graphite), boron, or polyamides such as Kevlar. High-tech tennis rackets and golf clubs as well as the skins of modern military aircraft such as the “stealth” F-117A fighters and B-2 bombers are made from both carbon fiber–epoxy and boron fiber–epoxy composites. Compared with metals, these materials are 25%–50% lighter and thus reduce operating costs. Similarly, the space shuttle payload bay doors and panels are made of a carbon fiber–epoxy composite. The structure of the Boeing 787 has been described as essentially one giant macromolecule, where everything is fastened through cross-linked chemical bonds reinforced with carbon fiber.
Metal-matrix composites consist of metals or metal alloys reinforced with fibers. They offer significant advantages for high-temperature applications but pose major manufacturing challenges. For example, obtaining a uniform distribution and alignment of the reinforcing fibers can be difficult, and because organic polymers cannot survive the high temperatures of molten metals, only fibers composed of boron, carbon, or ceramic (such as silicon carbide) can be used. Aluminum alloys reinforced with boron fibers are used in the aerospace industry, where their strength and lightweight properties make up for their relatively high cost. The skins of hypersonic aircraft and structural units in the space shuttle are made of metal-matrix composites.
Ceramic-matrix composites contain ceramic fibers in a ceramic matrix material. A typical example is alumina reinforced with silicon carbide fibers. Combining the two very high-melting-point materials results in a composite that has excellent thermal stability, great strength, and corrosion resistance, while the SiC fibers reduce brittleness and cracking. Consequently, these materials are used in very high-temperature applications, such as the leading edge of wings of hypersonic airplanes and jet engine parts. They are also used in the protective ceramic tiles on the space shuttle, which contain short fibers of pure SiO2 mixed with fibers of an aluminum–boron–silicate ceramic. These tiles are excellent thermal insulators and extremely light (their density is only about 0.2 g/cm3). Although their surface reaches a temperature of about 1250°C during reentry into Earth’s atmosphere, the temperature of the underlying aluminum alloy skin stays below 200°C.
Example $1$:
An engineer is tasked with designing a jet ski hull. What material is most suited to this application? Why?
Given: design objective
Asked for: most suitable material
Strategy:
Determine under what conditions the design will be used. Then decide what type of material is most appropriate.
Solution:
A jet ski hull must be lightweight to maximize speed and fuel efficiency. Because of its use in a marine environment, it must also be resistant to impact and corrosion. A ceramic material provides rigidity but is brittle and therefore tends to break or crack under stress, such as when it impacts waves at high speeds. Superalloys provide strength and stability, but a superalloy is probably too heavy for this application. Depending on the selection of metals, it might not be resistant to corrosion in a marine environment either. Composite materials, however, provide strength, stiffness, and corrosion resistance; they are also lightweight materials. This is not a high-temperature application, so we do not need a metal-matrix composite or a ceramic-matrix composite. The best choice of material is a polymer-matrix composite with Kevlar fibers to increase the strength of the composite on impact.
Exercise $1$
In designing a new generation of space shuttle, National Aeronautics and Space Administration (NASA) engineers are considering thermal-protection devices to protect the skin of the craft. Among the materials being considered are titanium- or nickel-based alloys and silicon-carbide ceramic reinforced with carbon fibers. Why are these materials suitable for this application?
Answer: Ti- or Ni-based alloys have a high strength-to-weight ratio, resist corrosion, and are safe at high temperatures. Reinforced ceramic is lightweight; has high thermal and oxidative stability; and is chemically inert, tough, and impact resistant.
Summary
Ceramics are nonmetallic, inorganic solids that are typically strong; they have high melting points but are brittle. The two major classes of modern ceramics are ceramic oxides and nonoxide ceramics, which are composed of nonmetal carbides or nitrides. The production of ceramics generally involves pressing a powder of the material into the desired shape and sintering at a temperature just below its melting point. The necessary fine powders of ceramic oxides with uniformly sized particles can be produced by the sol-gel process. Superalloys are new metal phases based on cobalt, nickel, or iron that exhibit unusually high temperature stability and resistance to oxidation. Composite materials consist of at least two phases: a matrix that constitutes the bulk of the material and fibers or granules that act as a reinforcement. Polymer-matrix composites have reinforcing fibers embedded in a polymer matrix. Metal-matrix composites have a metal matrix and fibers of boron, graphite, or ceramic. Ceramic-matrix composites use reinforcing fibers, usually also ceramic, to make the matrix phase less brittle.
Key Takeaway
• Materials that have contemporary applications include ceramics, high-strength alloys, and composites, whose properties can be modified as needed.
Conceptual Problems
1. Can a compound based on titanium oxide qualify as a ceramic material? Explain your answer.
2. What features make ceramic materials attractive for use under extreme conditions? What are some potential drawbacks of ceramics?
3. How do composite materials differ from the other classes of materials discussed in this chapter? What advantages do composites have versus other materials?
4. How does the matrix control the properties of a composite material? What is the role of an additive in determining the properties of a composite material?
• Anonymous | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/12%3A_Solids_and_Modern_Materials/12.01%3A_Classes_of_Materials.txt |
The most popular building material in the world today, concrete was first developed and exploited by the ancient Romans, who used it to create monumental public spaces such as aqueducts, churches, baths, and the Colosseum. Although concrete is a durable material with many building applications, the design of Roman concrete structures reinforced Roman ideas about social status and imperial power. This chapter explores the rich history of concrete and its legacy in the modern world, touching upon the role of concrete in ancient Rome, today’s technical advances in concrete construction, and concrete’s environmental drawbacks. The chapter also examines how concrete construction is shaped by societal ideals today, just as it was by societal ideas in ancient Rome.
Pourable, moldable, durable, waterproof, and relatively easy and inexpensive to manufacture, concrete is the world’s most popular building material. We live, work, and play on and in buildings and roads constructed from it. Architects exploit its properties to create artistic tours de force as well as utilitarian monuments
Origin of Concrete
As far back as the sixth millennium (6000–5000 BCE), the ancient Mesopotamians knew that heating calcium carbonate, a substance occurring naturally in limestone rocks, creates a new substance, known today as quicklime, in a process described chemically as CACO₃ + heat(1000° C) =CO₂ + CaO. This chemical reaction releases carbon dioxide into the atmosphere (more on this later). The resulting material, when mixed with water, bonds to other surfaces. The early residents of Çatalhöyük, an ancient city in modern-day Turkey used this substance to coat their walls, providing a surface for painted decoration.
Modern Concrete
Despite more highly mechanized manufacturing techniques, the process of creating concrete still relies on the basic chemical reaction exploited by the Romans, the reaction that releases carbon dioxide, a major pollutant, into the atmosphere. The world’s yearly production of concrete has been rising steadily and today is over four billion tons. Concrete production is responsible for nearly eight percent of the anthropogenic (human-made) greenhouse gases released into the air. By some estimates, the external climate and health damages caused by concrete production amount to approximately 74 percent of the value of the industry itself, putting the external costs of concrete higher than natural gas and oil and only slightly less than coal. As even more concrete is produced, the amount of carbon dioxide will also rise unless we develop smarter, greener methods of production. Concrete manufacturers recognize this problem, but any solution will have to be cost effective for worldwide adoption to take place.
The first step in making concrete is to produce cement as discussed in the video below.
An additional cost of manufacturing concrete is the need for sand as a component of the finished product. Today, sand is becoming an increasingly rare and sought-after commodity. As the Romans knew, only sand worn by water (river or ocean sand), not sand that has been exposed to the elements (desert sand), is suitable. While there are problems are connected with new construction, older concrete structures pose other issues. Most modern concrete is reinforced by metal bars that lead to eventual cracking as the metal expands and contracts. Can we recycle ruined concrete buildings? How can we stabilize and repair buildings? Engineers are working to develop new technologies that can sense imminent structural issues before a bridge or building collapses. To prevent damage in new construction, engineers developed Smartcrete, a form of concrete that can repair itself. New methods of concrete construction such as Ductal, which requires no metal, and the use of cloth as a framing material are also potential answers to this problem.
Although some striking modern architectural monuments have been built from concrete—the Guggenheim Museum in New York, for example—many consider modern concrete stark and ugly because of the many utilitarian buildings constructed from it. Because, unlike the Romans, we can make concrete with a smooth surface, it is not necessary for us to cover it with other materials.
New concrete technologies continue to emerge. Among the possibilities is concrete laid by robots. Acknowledging the close link between buildings and social structure allows us to wonder if there might be hidden costs to this technology. What types of workforce changes would occur if machines took over this aspect of building construction? Would using robots free humans to do other things or would it merely eliminate a large category of jobs?
Future of Concrete
Building construction remains the primary use of concrete today. Might there not be other uses for such a versatile material beyond architecture? Architects, engineers, and others are beginning to address this question. For example, kitchen designers are using concrete for countertops, taking advantage of its durability, cost effectiveness, modern appearance, and ability to resist water. People are even considering how ancient Roman concrete might be used to address sea level rise associated with global warming.
To encourage thinking about a common material in a different light, the American Society for Civil Engineering sponsors an annual concrete canoe contest. This challenge forces students to broaden their ideas about possible applications for this common material. Engineering students from across the US attempt to build and race a concrete canoe. They are judged not only on the results of the race, but also on their design concept. Concrete is certainly not the first material that comes to mind when thinking about canoes, although it is waterproof. But a concrete canoe suggests that if we think beyond the limits imposed on the use of concrete by our societal worldview and historical traditions, we may be able to find newer and more effective ways to use this versatile material.
Contributed By
Excerpted from a longer piece by Mary Ann Eaverly, University of Florida | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/12%3A_Solids_and_Modern_Materials/12.02%3A_Materials_for_Structure/12.2.01%3A_Concrete.txt |
Learning Objectives
• Outline the general approach for the metallurgy of iron into steel
The early application of iron to the manufacture of tools and weapons was possible because of the wide distribution of iron ores and the ease with which iron compounds in the ores could be reduced by carbon. For a long time, charcoal was the form of carbon used in the reduction process. The production and use of iron became much more widespread about 1620, when coke was introduced as the reducing agent. Coke is a form of carbon formed by heating coal in the absence of air to remove impurities.
The overall reaction for the production of iron in a blast furnace is as follows:
$\mathrm{Fe_2O_3(s) +3C(s)\xrightarrow{\Delta}2Fe(l) +3CO(g)} \label{23.2.3}$
The actual reductant is CO, which reduces Fe2O3 to give Fe(l) and CO2(g) (Equation $\ref{23.2.3}$); the CO2 is then reduced back to CO by reaction with excess carbon. As the ore, lime, and coke drop into the furnace (Figure $1$), any silicate minerals in the ore react with the lime to produce a low-melting mixture of calcium silicates called slag, which floats on top of the molten iron. Molten iron is then allowed to run out the bottom of the furnace, leaving the slag behind. Originally, the iron was collected in pools called pigs, which is the origin of the name pig iron.
The first step in the metallurgy of iron is usually roasting the ore (heating the ore in air) to remove water, decomposing carbonates into oxides, and converting sulfides into oxides. The oxides are then reduced in a blast furnace that is 80–100 feet high and about 25 feet in diameter (Figure $2$) in which the roasted ore, coke, and limestone (impure CaCO3) are introduced continuously into the top. Molten iron and slag are withdrawn at the bottom. The entire stock in a furnace may weigh several hundred tons.
Near the bottom of a furnace are nozzles through which preheated air is blown into the furnace. As soon as the air enters, the coke in the region of the nozzles is oxidized to carbon dioxide with the liberation of a great deal of heat. The hot carbon dioxide passes upward through the overlying layer of white-hot coke, where it is reduced to carbon monoxide:
$\ce{CO2}(g)+\ce{C}(s)⟶\ce{2CO}(g) \nonumber$
The carbon monoxide serves as the reducing agent in the upper regions of the furnace. The individual reactions are indicated in Figure $2$. The iron oxides are reduced in the upper region of the furnace. In the middle region, limestone (calcium carbonate) decomposes, and the resulting calcium oxide combines with silica and silicates in the ore to form slag. The slag is mostly calcium silicate and contains most of the commercially unimportant components of the ore:
$\ce{CaO}(s)+\ce{SiO2}(s)⟶\ce{CaSiO3}(l) \nonumber$
Just below the middle of the furnace, the temperature is high enough to melt both the iron and the slag. They collect in layers at the bottom of the furnace; the less dense slag floats on the iron and protects it from oxidation. Several times a day, the slag and molten iron are withdrawn from the furnace. The iron is transferred to casting machines or to a steelmaking plant (Figure $3$).
Steel
Much of the iron produced is refined and converted into steel. Steel is made from iron by removing impurities and adding substances such as manganese, chromium, nickel, tungsten, molybdenum, and vanadium to produce alloys with properties that make the material suitable for specific uses. Most steels also contain small but definite percentages of carbon (0.04%–2.5%). However, a large part of the carbon contained in iron must be removed in the manufacture of steel; otherwise, the excess carbon would make the iron brittle. However, there is not just one substance called steel - they are a family of alloys of iron with carbon or various metals.
Impurities in the iron from the Blast Furnace include carbon, sulfur, phosphorus and silicon, which have to be removed.
• Removal of sulfur: Sulfur has to be removed first in a separate process. Magnesium powder is blown through the molten iron and the sulfur reacts with it to form magnesium sulfide. This forms a slag on top of the iron and can be removed. $Mg + S \rightarrow MgS \label{127}$
• Removal of carbon: The still impure molten iron is mixed with scrap iron (from recycling) and oxygen is blown on to the mixture. The oxygen reacts with the remaining impurities to form various oxides. The carbon forms carbon monoxide. Since this is a gas it removes itself from the iron! This carbon monoxide can be cleaned and used as a fuel gas.
• Removal of other elements: Elements like phosphorus and silicon react with the oxygen to form acidic oxides. These are removed using quicklime (calcium oxide) which is added to the furnace during the oxygen blow. They react to form compounds such as calcium silicate or calcium phosphate which form a slag on top of the iron.
Table $1$: Special Steels
iron mixed with special properties uses include
stainless steel chromium and nickel resists corrosion cutlery, cooking utensils, kitchen sinks, industrial equipment for food and drink processing
titanium steel titanium withstands high temperatures gas turbines, spacecraft
manganese steel manganese very hard rock-breaking machinery, some railway track (e.g. points), military helmets
Cast iron has already been mentioned above. This section deals with the types of iron and steel which are produced as a result of the steel-making process.
• Wrought iron: If all the carbon is removed from the iron to give high purity iron, it is known as wrought iron. Wrought iron is quite soft and easily worked and has little structural strength. It was once used to make decorative gates and railings, but these days mild steel is normally used instead.
• Mild steel: Mild steel is iron containing up to about 0.25% of carbon. The presence of the carbon makes the steel stronger and harder than pure iron. The higher the percentage of carbon, the harder the steel becomes. Mild steel is used for lots of things - nails, wire, car bodies, ship building, girders and bridges amongst others.
• High carbon steel: High carbon steel contains up to about 1.5% of carbon. The presence of the extra carbon makes it very hard, but it also makes it more brittle. High carbon steel is used for cutting tools and masonry nails (nails designed to be driven into concrete blocks or brickwork without bending). High carbon steel tends to fracture rather than bend if mistreated.
• Special steels: These are iron alloyed with other metals (Table $1$).
Video: You can watch an animation of steelmaking that walks you through the process (Steelmaking animation, YouTube [www.youtube.com]).
12.2.03: Aluminum
Aluminum is easily oxidized, and so its ore, Al2O3, is difficult to reduce. In fact water is reduced rather than Al3+(aq), and so electrolysis must be carried out in a molten salt. Even this is difficult because the melting point of Al2O3 is above 2000°C—a temperature which is very difficult to maintain.
The first successful method for reducing Al2O3 is the one still used today. It was developed in the United States in 1886 by Charles Hall (1863 to 1914), who was then 23 years old and fresh out of Oberlin College. Hall realized that if Al2O3 were dissolved in another molten salt, the melting point of the mixture would be lower than for either pure substance. The substance Hall used was cryolite, Na3AlF6, in which the Al2O3 can be dissolved at just over 1000°C.
The electrolytic cell used for the Hall process. (Figure $1$) consists of a steel box lined with graphite. This contains the molten Na3AlF6 and Al2O3 and also serves as the cathode. The anode is a large cylinder of carbon. Passage of electrical current maintains the high temperature of the cell and causes the following half-equations to occur:
$\text{Al}^{3+} + \text{3}e^{-} \rightarrow \text{Al}(l) \label{1}$
$\text{2O}^{2-} + \text{C}(s) \rightarrow \text{CO}_2(g) + \text{4}e^{-}\label{2}$
Since the carbon anode is consumed by the oxidation half-equation, it must be replaced periodically. The video below shows the process
Aluminum production requires vast quantities of electrical energy, both to maintain the high temperature and to cause half-equations $\ref{1}$ and $\ref{2}$ to occur. Currently about 5 percent of the total electrical energy produced in the United States goes into the Hall process. Much of this energy comes from combustion of fossil fuels and hence consumes a valuable, nonrenewable resource. Since Al is protected from oxidation back to Al2O3 by a surface coating of oxide, it is a prime candidate for recycling, as well as for applications such as house siding, where it is expected to remain for a long time. Throwing away aluminum beverage cans, on the other hand, is a tremendous waste of energy.
Several other easily oxidized metals are currently produced by electrolysis, but not in such large quantities as Al. Mg is obtained by electrolyzing molten MgCl2 which is derived from seawater, and Na and Ca are produced together from a molten mixture of NaCl and CaCl2.
Contributions and Attributions
This page is licensed under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Ed Vitz, John W. Moore, Justin Shorb, Xavier Prat-Resina, Tim Wendorff, & Adam Hahn. A detailed versioning history of the edits to source content is available upon request. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/12%3A_Solids_and_Modern_Materials/12.02%3A_Materials_for_Structure/12.2.02%3A_Steel.txt |
Learning Objectives
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially important that you know the precise meanings of all the bold terms in the context of this topic.
• Aside from their high molar masses, how do synthetic polymers differ from ordinary molecular solids?
• Polymers can be classified according to their chemical composition, their physical properties, and their general application. For each of these three categories, name two examples that that might be considered when adapting a polymer to a particular end-use.
• and a thermoset, and comment on the molecular basis for their different properties, including crystallinity.
• Describe the two general methods of polymer synthesis.
• Name two kinds each of commercially-important synthetic thermoplastics and thermosets, and specify some of their principal uses.
• Name two kinds of commercially-important natural polymers.
• Describe some of the concerns and sources of small-molecule release from polymers.
• What are some of the problems connected with recycling or re-use of polymeric materials?
Plastics and natural materials such as rubber or cellulose are composed of very large molecules called polymers. Polymers are constructed from relatively small molecular fragments known as monomers that are joined together. Wool, cotton, silk, wood and leather are examples of natural polymers that have been known and used since ancient times. This group includes biopolymers such as proteins and carbohydrates that are constituents of all living organisms.
Synthetic polymers, which includes the large group known as plastics, came into prominence in the early twentieth century. Chemists' ability to engineer them to yield a desired set of properties (strength, stiffness, density, heat resistance, electrical conductivity) has greatly expanded the many roles they play in the modern industrial economy. This Module deals mostly with synthetic polymers, but will include a synopsis of some of the more important natural polymers. It will close with a summary of some of the very significant environmental problems created by the wide use of plastics.
Polymers and "pure substances"
Let's begin by looking at an artificial polymer that is known to everyone in the form of flexible, transparent plastic bags: polyethylene. It is also the simplest polymer, consisting of random-length (but generally very long) chains made up of two-carbon units.
You will notice some "fuzziness" in the way that the polyethylene structures are represented above. The squiggly lines at the ends of the long structure indicate that the same pattern extends indefinitely. The more compact notation on the right shows the minimal repeating unit enclosed in brackets overprinted with a dash; this means the same thing and is the preferred way of depicting polymer structures.
In most areas of chemistry, a "pure substance" has a definite structure, molar mass, and properties. It turns out, however, that few polymeric substances are uniform in this way. This is especially the case with synthetic polymers, whose molecular weights cover a range of values, as may the sequence, orientation, and connectivity of the individual monomers. So most synthetic polymers are really mixtures rather than pure substances in the ordinary chemical sense of the term. Their molecular weights are typically distributed over a wide range.
Figure \(1\): Polymers
Don't be misled by chemical formulas that depict polymers such as polyethylene as reasonably straight chains of substituted carbon atoms. Free rotation around C—C bonds allows long polymer molecules to curl up and and tangle very much like spaghetti (Figure \(2\)). Thus polymers generally form amorphous solids. There are, however, ways in which certain polymers can be partially oriented.
Classification of polymers
Polymers can be classified in ways that reflect their chemical makeup, or perhaps more importantly, their properties and applications. Many of these factors are strongly interdependent, and most are discussed in much more detail in subsequent sections of this page.
Chemistry
• Nature of the monomeric units
• Average chain length and molecular weight
• Homopolymers (one kind of monomeric unit) or copolymers;
• Chain topology: how the monomeric units are connected
• Presence or absence of cross-branching
• Method of polymerization
Properties
• Density
• Thermal properties — can they soften or melt when heated?
• Degree of crystallinity
• Physical properties such as hardness, strength, machineability.
• Solubility, permeability to gases
Applications
• molded and formed objects ("plastics")
• sheets and films
• elastomers (i.e., elastic polymers such as rubber)
• adhesives
• coatings, paints, inks
• fibers and yarns
Physical properties of polymers
The physical properties of a polymer such as its strength and flexibility depend on:
• chain length - in general, the longer the chains the stronger the polymer;
• side groups - polar side groups (including those that lead to hydrogen bonding) give stronger attraction between polymer chains, making the polymer stronger;
• branching - straight, unbranched chains can pack together more closely than highly branched chains, giving polymers that have higher density, are more crystalline and therefore stronger;
• cross-linking - if polymer chains are linked together extensively by covalent bonds, the polymer is harder and more difficult to melt.
Amorphous and crystalline polymers
The spaghetti-like entanglements of polymer molecules tend to produce amorphous solids, but it often happens that some parts can become sufficiently aligned to produce a region exhibiting crystal-like order, so it is not uncommon for some polymeric solids to consist of a random mixture of amorphous and crystalline regions. As might be expected, shorter and less-branched polymer chains can more easily organize themselves into ordered layers than have can long chains. Hydrogen-bonding between adjacent chains also helps, and is very important in fiber-forming polymers both synthetic (Nylon 6.6) and natural (cotton cellulose).
Pure crystalline solids have definite melting points, but polymers, if they melt at all, exhibit a more complex behavior. At low temperatures, the tangled polymer chains tend to behave as rigid glasses. For example, the natural polymer that we call rubber becomes hard and brittle when cooled to liquid nitrogen temperature. Many synthetic polymers remain in this state to well above room temperature.
The melting of a crystalline compound corresponds to a sudden loss of long-range order; this is the fundamental reason that such solids exhibit definite melting points, and it is why there is no intermediate form between the liquid and the solid states. In amorphous solids there is no long-range order, so there is no melting point in the usual sense. Such solids simply become less and less viscous as the temperature is raised.
In some polymers (known as thermoplastics) there is a fairly definite softening point that is observed when the thermal kinetic energy becomes high enough to allow internal rotation to occur within the bonds and to allow the individual molecules to slide independently of their neighbors, thus rendering them more flexible and deformable. This defines the glass transition temperature tg .
Depending on the degree of crystallinity, there will be a higher temperature, the melting point tm , at which the crystalline regions come apart and the material becomes a viscous liquid. Such liquids can easily be injected into molds to manufacture objects of various shapes, or extruded into sheets or fibers. Other polymers (generally those that are highly cross-linked) do not melt at all; these are known as thermosets. If they are to be made into molded objects, the polymerization reaction must take place within the molds — a far more complicated process. About 20% of the commercially-produced polymers are thermosets; the remainder are thermoplastics.
2 Thermoplastic polymer structures
Homopolymers and heteropolymers
Copolymerization is an invaluable tool for "tuning" polymers so that they have the right combination of properties for an application. For example, homopolymeric polystyrene is a rigid and very brittle transparent thermoplastic with a glass transition temperature of 97°C. Copolymerizing it with acrylonitrile yields an alternating "SAN" copolymer in which tg is raised to 107°, making it useable for transparent drink containers.
A polymer that is composed of identical monomeric units (as is polyethylene) is called a homopolymer. Heteropolymers are built up from more than one type of monomer. Artificial heteropolymers are more commonly known as copolymers.
Chain topology
Polymers may also be classified as straight-chained or branched, leading to forms such as these:
The monomers can be joined end-to-end, and they can also be cross-linked to provide a harder material:
If the cross-links are fairly long and flexible, adjacent chains can move with respect to each other, producing an elastic polymer or elastomer
Chain configuration and tacticity
In a linear polymer such as polyethylene, rotations around carbon-carbon single bonds can allow the chains to bend or curl up in various ways, resulting in the spaghetti-like mixture of these different conformations we alluded to above. But if one of the hydrogen atoms is replaced by some other entity such as a methyl group, the relative orientations of the individual monomer units that make up a linear section of any carbon chain becomes an important characteristic of the polymer.
Cis-trans isomerism occurs because rotation around carbon-carbon double bonds is not possible — unlike the case for single bonds. Any pair of unlike substituents attached to the two carbons is permanently locked into being on the same side (cis) or opposite sides (trans) of the double bond.
If the carbon chain contains double bonds, then cis-trans isomerism becomes possible, giving rise to two different possible configurations (known as diastereomers) at each unit of the chain. This seemingly small variable can profoundly affect the nature of the polymer. For example, the latex in natural rubber is made mostly of cis-polyisoprene, whereas the trans isomer (known as gutta percha latex) has very different (and generally inferior) properties.
Chirality
The tetrahedral nature of carbon bonding has an important consequence that is not revealed by simple two-dimensional structural formulas: atoms attached to the carbon can be on one side or on the other, and these will not be geometrically equivalent if all four of the groups attached to a single carbon atom are different. Such carbons (and the groups attached to them) are said to be chiral, and can exist in two different three-dimensional forms known as enantiomers.
For an individual carbon atom in a polymer chain, two of its attached groups will ordinarily be the chain segments on either side of the carbon. If the two remaining groups are different (say one hydrogen and the other methyl), then the above conditions are satisfied and this part of the chain can give rise to two enantiomeric forms.
A chain that can be represented as (in which the orange and green circles represent different groups) will have multiple chiral centers, giving rise to a huge number of possible enantiomers. In practice, it is usually sufficient to classify chiral polymers into the following three classes of stereoregularity, usually referred to as tacticity.
The tacticity of a polymer chain can have a major influence on its properties. Atactic polymers, for example, being more disordered, cannot crystallize.
One of the major breakthroughs in polymer chemistry occurred in the early 1950s when the German chemist Karl Ziegler discovered a group of catalysts that could efficiently polymerize ethylene. At about the same time, Giulio Natta (Italian) made the first isotactic (and crystalline) polyethylene. The Ziegler-Natta catalysts revolutionized polymer chemistry by making it possible to control the stereoregularity of these giant molecules. The two shared the 1963 Nobel Prize in Chemistry
3 How polymers are made
Polymers are made by joining small molecules into large ones. But most of these monomeric molecules are perfectly stable as they are, so chemists have devised two general methods to make them react with each other, building up the backbone chain as the reaction proceeds.
Condensation-elimination polymerization
This method (also known as step-growth) requires that the monomers possess two or more kinds of functional groups that are able to react with each other in such a way that parts of these groups combine to form a small molecule (often H2O) which is eliminated from the two pieces. The now-empty bonding positions on the two monomers can then join together .
This occurs, for example, in the synthesis of the Nylon family of polymers in which the eliminated H2O molecule comes from the hydroxyl group of the acid and one of the amino hydrogens:
Note that the monomeric units that make up the polymer are not identical with the starting components.
Addition polymerization
Addition or chain-growth polymerization involves the rearrangement of bonds within the monomer in such a way that the monomers link up directly with each other:
In order to make this happen, a chemically active molecule (called an initiator) is needed to start what is known as a chain reaction. The manufacture of polyethylene is a very common example of such a process. It employs a free-radical initiator that donates its unpaired electron to the monomer, making the latter highly reactive and able to form a bond with another monomer at this site.
In theory, only a single chain-initiation process needs to take place, and the chain-propagation step then repeats itself indefinitely, but in practice multiple initiation steps are required, and eventually two radicals react (chain termination) to bring the polymerization to a halt.
As with all polymerizations, chains having a range of molecular weights are produced, and this range can be altered by controlling the pressure and temperature of the process.
4 Gallery of common synthetic polymers
Thermoplastics
Note: the left panels below show the polymer name and synonyms, structural formula, glass transition temperature, melting point/decomposition temperature, and (where applicable) the resin identification symbol used to facilitate recycling.
Polycarbonate (Lexan®)
Tg = 145°C, Tm = 225°C.
This polymer was discovered independently in Germany and the U.S. in 1953. Lexan is exceptionally hard and strong; we see it most commonly in the form of compact disks. It was once widely used in water bottles, but concerns about leaching of unreacted monomer (bisphenol-A) has largely suppressed this market.
Polyethylene terephthalate (PET, Mylar)
Tg = 76°C, Tm = 250°C.
Thin and very strong films of this material are made by drawing out the molten polymer in both directions, thus orienting the molecules into a highly crystalline state that becomes "locked-in" on cooling. Its many applications include food packaging (in foil-laminated drink containers and microwaveable frozen-food containers), overhead-projector film, weather balloons, and as aluminum-coated reflective material in spacecraft and other applications.
Nylon (a polyamide)
Tg = 50°C, Tm = 255°C.
Nylon has a fascinating history, both scientific and cultural. It was invented by DuPont chemist Wallace Carothers (1896-1937). The common form Nylon 6.6 has six carbon atoms in both parts of its chain; there are several other kinds. Notice that the two copolymer sub-units are held together by peptide bonds, the same kinds that join amino acids into proteins.
Nylon 6.6 has good abrasion resistance and is self-lubricating, which makes it a good engineering material. It is also widely used as a fiber in carpeting, clothing, and tire cord.
For an interesting account of the development of Nylon, see Enough for One Liftetime: Wallace Carothers, Inventor of Nylonby Ann Gaines (1971)
Polyacrylonitrile (Orlon, Acrilan, "acrylic" fiber)
Tg = 85°C, Tm = 318°C.
Used in the form of fibers in rugs, blankets, and clothing, especially cashmere-like sweaters. The fabric is very soft, but tends to "pill" — i.e., produce fuzz-like blobs. Owing to its low glass transition temperature, it requires careful treatment in cleaning and ironing.
Polyethylene
Tg = –78°C, Tm = 100°C.
LDPE
HDPE
Control of polymerization by means of catalysts and additives has led to a large variety of materials based on polyethylene that exhibit differences in densities, degrees of chain branching and crystallinity, and cross-linking. Some major types are low-density (LDPE), linear low density (LLDPE), high-density (HDPE).
LDPE was the first commercial form (1933) and is used mostly for ordinary "plastic bags", but also for food containers and in six-pack soda can rings. Its low density is due to long-chain branching that inhibits close packing. LLDPE has less branching; its greater toughness allows its use in those annoyingly-thin plastic bags often found in food markets.
A "very low density" form (VLDPE) with extensive short-chain branching is now used for plastic stretch wrap (replacing the original component of Saran Wrap) and in flexible tubing.
HDPE has mostly straight chains and is therefore stronger. It is widely used in milk jugs and similar containers, garbage containers, and as an "engineering plastic" for machine parts.
Polymethylmethacrylate (Plexiglass, Lucite, Perspex)
Tg = 114°C, Tm = 130-140°C.
This clear, colorless polymer is widely used in place of glass, where its greater impact resistance, lighter weight, and machineability are advantages. It is normally copolymerized with other substances to improve its properties. Aircraft windows, plastic signs, and lighting panels are very common applications. Its compatibility with human tissues has led to various medical applications, such as replacement lenses for cataract patients.
Polypropylene
Tg = –10°C, Tm = 173°C.
PP
Polypropylene is used alone or as a copolymer, usually with with ethylene. These polymers have an exceptionally wide range of uses — rope, binder covers, plastic bottles, staple yarns, non-woven fabrics, electric kettles. When uncolored, it is translucent but not transparent. Its resistance to fatigue makes it useful for food containers and their lids, and flip-top lids on bottled products such as ketchup.
polystyrene
Tg = 95°C, Tm = 240°C.
PS
Polystyrene is transparent but rather brittle, and yellows under uv light.
Widely used for inexpensive packaging materials and "take-out trays", foam "packaging peanuts", CD cases, foam-walled drink cups, and other thin-walled and moldable parts.
polyvinyl acetate
Tg = 30°C
PVA is too soft and low-melting to be used by itself; it is commonly employed as a water-based emulsion in paints, wood glue and other adhesives.
polyvinyl chloride ("vinyl", "PVC")
Tg = 85°C, Tm = 240°C.
PVC
This is one of the world's most widely used polymers. By itself it is quite rigid and used in construction materials such as pipes, house siding, flooring. Addition of plasticizers make it soft and flexible for use in upholstery, electrical insulation, shower curtains and waterproof fabrics. There is some effort being made to phase out this polymer owing to environmental concerns (see below).
Synthetic rubbers
Neoprene (polychloroprene)
Tg = –70°C
Polybutadiene Tg < –90°C
Neoprene, invented in 1930, was the first mass-produced synthetic rubber. It is used for such things as roofing membranes and wet suits.
Polybutadiene substitutes a hydrogen for the chlorine; it is the major component (usually admixed with other rubbers) of tires. Synthetic rubbers played a crucial role in World War II
SBS (styrene-butadiene-styrene) rubber is a block copolymer whose special durability makes it valued for tire treads.
Polytetrafluroethylene (Teflon, PTFE)
Decomposes above 350°C.
This highly-crystalline fluorocarbon is exceptionally inert to chemicals and solvents. Water and oils do not wet it, which accounts for its use in cooking ware and other anti-stick applications, including personal care products.
These properties — non-adhesion to other materials, non-wetability, and very low coefficient of friction ("slipperyness") — have their origin in the highly electronegative nature of fluorine whose atoms partly shield the carbon chain. Fluorine's outer electrons are so strongly attracted to its nucleus that they are less available to participate in London (dispersion force) interactions.
Polyaramid (Kevlar)
Sublimation temperature 450°C.
Kevlar is known for its ability to be spun into fibers that have five times the tensile strength of steel. It was first used in the 1970s to replace steel tire cords. Bullet-proof vests are one of it more colorful uses, but other applications include boat hulls, drum heads, sports equipment, and as a replacement for asbestos in brake pads. It is often combined with carbon or glass fibers in composite materials.
The high tensile strength is due in part to the extensive hydrogen bonding between adjacent chains.
Kevlar also has the distinction of having been invented by a woman chemist, Stephanie Kwolek.
Thermosets
The thermoplastic materials described above are chains based on relatively simple monomeric units having varying degrees of polymerization, branching, bending, cross-linking and crystallinity, but with each molecular chain being a discrete unit. In thermosets, the concept of an individual molecular unit is largely lost; the material becomes more like a gigantic extended molecule of its own — hence the lack of anything like a glass transition temperature or a melting point.
These properties have their origins in the nature of the monomers used to produce them. The most important feature is the presence of multiple reactive sites that are able to form what amount to cross-links at every center. The phenolic resins, typified by the reaction of phenol with formaldehyde, illustrate the multiplicity of linkages that can be built.
Phenolic resins
These are made by condensing one or more types of phenols (hydroxy-substituted benzene rings) with formaldehyde, as illustrated above. This was the first commercialized synthetic molding plastic. It was developed in 1907-1909 by the Belgian chemist Leo Baekeland, hence the common name bakelite. The brown material (usually bulked up with wood powder) was valued for its electrical insulating properties (light fixtures, outlets and other wiring devices) as well as for consumer items prior to the mid-century. Since that time, more recently developed polymers have largely displaced these uses. Phenolics are still extensively used as adhesives in plywood manufacture, and for making paints and varnishes.
Urea resins
Condensation of formaldehyde with urea yields lighter-colored and less expensive materials than phenolics. The major use if urea-formaldehyde resins is in bonding wood particles into particle board. Other uses are as baked-on enamel coatings for kitchen appliances and to coat cotton and rayon fibers to impart wrinkle- water-, and stain-resistance to the finished fabrics.
Melamine resins
Melamine, with even more amino (–NH2) groups than urea, reacts with formaldehyde to form colorless solids that are harder then urea resins. The are most widely encountered in dinner-ware (plastic plates, cups and serving bowls) and in plastic laminates such as Formica.
Alkyd-polyester resins
An ester is the product of the reaction of an organic acid with an alcohol, so polyesters result when multifunctional acids such as phthalic acid react with polyhydric alcohols such as glycerol. The term alkyd derives from the two words alcohol and acid.
Alkyd resins were first made by Berzelius in 1847, and they were first commercialized as Glyptal (glycerine + phthalic acid) varnishes for the paint industry in 1902.
The later development of other polyesters greatly expanded their uses into a wide variety of fibers and molded products, ranging from clothing fabrics and pillow fillings to glass-reinforced plastics.
Epoxy resins
This large and industrially-important group of resins typically starts by condensing bisphenol-A with epichlorohydrin in the presence of a catalyst. (The -epiprefix refers to the epoxide group in which an oxygen atom that bridges two carbons.) These resins are usually combined with others to produce the desired properties. Epoxies are especially valued as glues and adhesives, as their setting does not depend on evaporation and the setting time can be varied over a wide range. In the two-part resins commonly sold for home use, the unpolymerized mixture and the hardener catalyst are packaged separately for mixing just prior to use. In some formulations the polymerization is initiated by heat ("heat curing"). Epoxy dental fillings are cured by irradiation with uv light.
Polyurethanes
Organic isocyanates R–NCO react with multifunctional alcohols to form polymeric carbamates, commonly referred to as polyurethanes. Their major use is in plastic foams for thermal insulation and upholstery, but a very large number of other applications, including paints and varnishes and plastic wheels used in fork-lift trucks, shopping carts and skateboards.
Silicones
Polysiloxanes (–Si–O–Si-) are the most important of the small class inorganic polymers. The commercial silicone polymers usually contained attached organic side groups that aid to cross-linking. Silicones can be made in a wide variety of forms; those having lower molecular weights are liquids, while the more highly polymerized materials are rubbery solids. These polymers have a similarly wide variety of applications: lubricants, caulking materials and sealants, medical implants, non-stick cookware coatings, hair-conditioners and other personal-care products.
Natural Polymers
Polymers derived from plants have been essential components of human existence for thousands of years. In this survey we will look at only those that have major industrial uses, so we will not be discussing the very important biopolymers proteins and nucleic acids.
Polysaccharides
Polysaccharides are polymers of sugars; they play essential roles in energy storage, signaling, and as structural components in all living organisms. The only ones we will be concerned with here are those composed of glucose, the most important of the six-carbon hexoses. Glucose serves as the primary fuel of most organisms.
Glucose, however, is highly soluble and cannot be easily stored, so organisms make polymeric forms of glucose to set aside as reserve storage, from which glucose molecules can be withdrawn as needed.
Glycogen
In humans and higher animals, the reserve storage polymer is glycogen. It consists of roughly 60,000 glucose units in a highly branched configuration. Glycogen is made mostly in the liver under the influence of the hormone insulin which triggers a process in which digested glucose is polymerized and stored mostly in that organ. A few hours after a meal, the glucose content of the blood begins to fall, and glycogen begins to be broken down in order to maintain the body's required glucose level.
Starch
In plants, these glucose-polymer reserves are known as starch. Starch granules are stored in seeds or tubers to provide glucose for the energy needs of newly-germinated plants, and in the twigs of deciduous plants to tide them over during the winter when photosynthesis (the process in which glucose is synthesized from CO2 and H2O) does not take place. The starches in food grains such as rice and wheat, and in tubers such as potatoes, are a major nutritional source for humans.
Plant starches are mixtures of two principal forms, amylose and amylopectin. Amylose is a largely-unbranched polymer of 500 to 20,000 glucose molecules that curls up into a helical form that is stabilized by internal hydrogen bonding. Amylopectin is a much larger polymer having up to two million glucose residues arranged into branches of 20 to 30 units.
Cellulose and its derivatives
Cellulose is the most abundant organic compound on the earth. Extensive hydrogen bonding between the chains causes native cellulose to be about 70% crystalline. It also raises the melting point (>280°C) to above its combustion temperature. The structures of starch and cellulose appear to be very similar; in the latter, every other glucose molecule is "upside-down". But the consequences of this are far-reaching; starch can dissolve in water and can be digested by higher animals including humans, whereas cellulose is insoluble and undigestible. Cellulose serves as the principal structural component of green plants and (along with lignin) in wood.
Cotton is one of the purest forms of cellulose and has been cultivated since ancient times. Its ability to absorb water (which increases its strength) makes cotton fabrics especially useful for clothing in very warm climates.
Cotton also serves (along with treated wood pulp) as the source the industrial production of cellulose-derived materials which were the first "plastic" materials of commercial importance.
• Nitrocellulose was developed in the latter part of the 19th Century. It is prepared by treating cotton with nitric acid, which reacts with the hydroxyl groups in the cellulose chain. It was first used to make molded objects the first material used for a photographic film base by Eastman Kodak. Its extreme flammability posed considerable danger in movie theaters, and its spontaneous slow decomposition over time had seriously degraded many early films before they were transferred to more stable media. Nitrocellulose was also used as an explosive and propellant, for which applications it is known as guncotton.
• Cellulose acetate was developed in the early 1900s and became the first artificial fiber that was woven into fabrics that became prized for their lustrous appearance and wearing comfort. Kodak developed it as a "safety film" base in the 1930's to replace nitrocellulose, but it did not come into wide use for this purpose until 1948. A few years later, is became the base material for magnetic recording tape.
• Viscose is the general term for "regenerated" forms of cellulose made from solutions of the polymer in certain strong solvents. When extruded into a thin film it becomes cellophane which has been used as a food wrapping since 1912 and is the base for transparent adhesive tapes such as Scotch Tape. Viscose solutions extruded through a spinneret produce fibers known as rayon. Rayon (right) was the first "artificial silk" and has been used for tire cord, apparel, and carpets. It was popular for womens' stockings before Nylon became available for this purpose.
Rubber
A variety of plants produce a sap consisting of a colloidal dispersion of cis-polyisoprene. This milky fluid is especially abundant in the rubber tree (Hevea), from which it drips when the bark is wounded. After collection, the latex is coagulated to obtain the solid rubber. Natural rubber is thermoplastic, with a glass transition temperature of –70°C.
cis-polyisoprene
Raw natural rubber tends to be sticky when warm and brittle when cold, so it was little more than a novelty material when first introduced to Europe around 1770. It did not become generally useful until the mid-nineteenth century when Charles Goodyear found that heating it with sulfur — a process he called vulcanization — could greatly improve its properties.
Why does a rubber band heat up when it is stretched, and why does it spontaneously snap back? It all has to do with entropy.
Vulcanization creates disulfide cross-links that prevent the polyisoprene chains from sliding over each other. The degree of cross-linking can be controlled to produce a rubber having the desired elasticity and hardness. More recently, other kinds of chemical treatment (such as epoxidation) have been developed to produce rubbers for special purposes.
Better things for better living... through chemistry" is a famous commercial slogan that captured the attitude of the public around 1940 when synthetic polymers were beginning to make a major impact in people's lives. What was not realized at the time, however, were some of the problems these materials would create as their uses multiplied and the world became more wary of "chemicals". (DuPont dropped the "through chemistry" part in 1982.)
Small-molecule release
Many kinds of polymers contain small molecules — either unreacted monomers, or substances specifically added (plasticizers, uv absorbers, flame retardants, etc.) to modify their properties. Many of these smaller molecules are able to diffuse through the material and be released into any liquid or air in contact with the plastic — and eventually into the aquatic environment. Those that are used for building materials (in mobile homes, for example) can build up in closed environments and contribute to indoor air pollution.
Residual monomer
Formation of long polymer chains is a complicated and somewhat random process that is never perfectly stoichiometric. It is therefore not uncommon for some unreacted monomer to remain in the finished product. Some of these monomers, such as formaldehyde, styrene (from polystyrene, including polystyrene foam food take-out containers), vinyl chloride, and bisphenol-A (from polycarbonates) are known carcinogens. Although there is little evidence that the small quantities that diffuse into the air or leach out into fluids pose a quantifiable health risk, people are understandably reluctant to tolerate these exposures, and public policy is gradually beginning to regulate them.
Perfluorooctanoic acid (PFOA), the monomer from which Teflon is made, has been the subject of a 2004 lawsuit against a DuPont factory that contaminated groundwater. Small amounts of PFOA have been detected in gaseous emissions from hot fluorocarbon products.
Plasticizers
These substances are compounded into certain types of plastics to render them more flexible by lowering the glass transition temperature. They accomplish this by taking up space between the polymer chains and acting as lubricants to enable the chains to more readily slip over each other. Many (but not all) are small enough to be diffusible and a potential source of health problems.
Polyvinyl chloride polymers are one of the most widely-plasticized types, and the odors often associated with flexible vinyl materials such as garden hoses, waterbeds, cheap shower curtains, raincoats and upholstery are testament to their ability to migrate into the environment.
The well-known "new car smell" is largely due to plasticizer release from upholstery and internal trim.
There is now an active movement to develop non-diffusible and "green" plasticizers that do not present these dangers.
Endocrine disrupters
Plastics-related compounds are not the only kind of endocrine disrupters found in the environment. Others include pesticide and fungicide residues, and industrial chemicals such as polychlorinated biphenols (PCBs).
To complicate matters even further, many of these small molecules have been found to be physiologically active owing to their ability to mimic the action of hormones or other signaling molecules, probably by fitting into and binding with the specialized receptor sites present in many tissues. The evidence that many of these chemicals are able to act in this way at the cellular level is fairly clear, but there is still some dispute whether many of these pose actual health risks to adult humans at the relatively low concentrations in which they commonly occur in the environment.
There is, however, some concern about the effects of these substances on non-adults and especially on fetuses, given that endocrines are intimately connected with sexual differentiation and neurological development which continues up through the late teens.
Decomposition products
Most commonly-used polymers are not readily biodegradable, particularly under the anaerobic conditions of most landfills. And what decomposition does occur will combine with rainwater to form leachates that can contaminate nearby streams and groundwater supplies. Partial photodecomposition, initiated by exposure to sunlight, is a more likely long-term fate for exposed plastics, resulting in tiny broken-up fragments. Many of these materials are less dense than seawater, and once they enter the oceans through coastal sewage outfalls or from marine vessel wastes, they tend to remain there indefinitely.
Open burning of polymeric materials containing chlorine (polyvinyl chloride, for example) is known to release compounds such as dioxins that persist in the environment. Incineration under the right conditions can effectively eliminate this hazard.
Disposed products containing fluorocarbons (Teflon-coated ware, some personal-care, waterproofing and anti-stick materials) break down into perfluorooctane sulfonate which has been shown to damage aquatic animals.
Hazards to animals
There are two general types of hazards that polymers can introduce into the aquatic environment. One of these relates to the release of small molecules that act as hormone disrupters as described above. It is well established that small aquatic animals such as fish are being seriously affected by such substances in many rivers and estuarine systems, but details of the sources and identities of these molecules have not been identified. One confounding factor is the release of sewage water containing human birth-control drugs (which have a feminizing effect on sexual development) into many waterways.
The other hazard relates to pieces of plastic waste that aquatic animals mistake for food or become entangled in.
This plastic bag (probably mistaken for a jellyfish, the sea turtle's only food) cannot be regurgitated and leads to intestinal blockage and a slow death.
Remains of an albatross that mistook bits of plastic junk for food
These dangers occur throughout the ocean, but are greatly accentuated in regions known as gyres. These are regions of the ocean in which a combination of ocean currents drives permanent vortices that tend to collect and concentrate floating materials. The most notorious of these are the Great Pacific Gyres that have accumulated astounding quantities of plastic waste.
Recycling
The huge quantity (one estimate is 108 metric tons per year) of plastic materials produced for consumer and industrial use has created a gigantic problem of what to do with plastic waste which is difficult to incinerate safely and which, being largely non-biodegradable, threatens to overwhelm the capacity of landfills. An additional consideration is that de novo production most of the major polymers consumes non-renewable hydrocarbon resources.
Plastic water bottles (left) present a special recycling problem because of their widespread use in away-from-home locations.
Plastics recycling has become a major industry, greatly aided by enlightened trash management policies in the major developed nations. However, it is plagued with some special problems of its own:
• Recycling is only profitable when there is a market for the regenerated material. Such markets vary with the economic cycle (they practically disappeared during the recession that commenced in 2008.)
• The energy-related costs of collecting and transporting plastic waste, and especially of processing it for re-use, are frequently the deciding factor in assessing the practicability of recycling.
• Collection of plastic wastes from diverse sources and locations and their transport to processing centers consumes energy and presents numerous operational problems.
• Most recycling processes are optimized for particular classes of polymers. The diversity of plastic types necessitates their separation into different waste streams — usually requiring manual (i.e., low-cost) labor. This in turn encourages shipment of these wastes to low-wage countries, thus reducing the availability of recycled materials in the countries in which the plastics originated.
Some of the major recycling processes include
• Thermal decomposition processes that can accommodate mixed kinds of plastics and render them into fuel oil, but the large inputs of energy they require have been a problem.
• A very small number of condensation polymers can be depolymerized so that the monomers can be recovered and re-used.
• Thermopolymers can be melted and pelletized, but those of widely differing types must be treated separately to avoid incompatability problems.
• Thermosets are usually shredded and used as filler material in recycled thermopolymers.
In order to facilitate efficient recycling, a set of seven resin idenfication codes has been established (the seventh, not shown below, is "other").
These codes are stamped on the bottoms of many containers of widely-distributed products. Not all categories are accepted by all local recycling authorities, so residents need to be informed about which kinds should be placed in recycling containers and which should be combined with ordinary trash.
Tire recycling
The large number of rubber tires that are disposed of, together with the increasing reluctance of landfills to accept them, has stimulated considerable innovation in the re-use of this material, especially in the construction industry. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/12%3A_Solids_and_Modern_Materials/12.02%3A_Materials_for_Structure/12.2.04%3A_Polymers.txt |
• 12.3.1: Anti-Cancer Drugs I
• 12.3.2: Anti-Cancer Drugs II
• 12.3.3: Antidepressants
Antidepressant drugs are used to restore mentally depressed patients to an improved mental status. Depression results from a deficiency of norepinephrine at receptors in the brain. Mechanisms that increase their effective concentration at the receptor sites should alleviate depression.
• 12.3.4: Barbiturates and Benzodiazepines
Barbiturates are central nervous system depressants and are similar, in many ways, to the depressant effects of alcohol.
• 12.3.5: Drugs Acting Upon the Central Nervous System
Chemical influences are capable of producing a myriad of effects on the activity and function of the central nervous system. Since our knowledge of different regions of brain function and the neurotransmitters in the brain is limited, the explanations for the mechanisms of drug action may be vague. The known neurotransmitters are: acetylcholine which is involved with memory and learning; norepinephrine which is involved with mania-depression and emotions.
• 12.3.6: Hallucinogenic Drugs
Hallucinogenic agents, also called psychomimetic agents, are capable of producing hallucinations, sensory illusions and bizarre thoughts. The primary effect of these compounds is to consistently alter thought and sensory perceptions.
• 12.3.7: Local Anesthetics
Local anesthetics are agents that reversibly block the generation and conduction of nerve impulses along a nerve fiber. They depress impulses from sensory nerves of the skin, surfaces of mucosa, and muscles to the central nervous system. These agents are widely used in surgery, dentistry, and ophthalmology to block transmission of impulses in peripheral nerve endings.
• 12.3.8: Misc Antibiotics
Antibiotics are specific chemical substances derived from or produced by living organisms that are capable of inhibiting the life processes of other organisms.
• 12.3.9: Narcotic Analgesic Drugs
Narcotic agents are potent analgesics which are effective for the relief of severe pain. Analgesics are selective central nervous system depressants used to relieve pain. The term analgesic means "without pain". Even in therapeutic doses, narcotic analgesics can cause respiratory depression, nausea, and drowsiness.
• 12.3.10: Penicillin
The penicillins were the first antibiotics discovered as natural products from the mold Penicillium.
• 12.3.11: Sulfa Drugs
Sulfonamides are synthetic antimicrobial agents with a wide spectrum encompassing most gram-positive and many gram-negative organisms. These drugs were the first efficient treatment to be employed systematically for the prevention and cure of bacterial infections.
12.03: Materials for Medicine
The available anticancer drugs have distinct mechanisms of action which may vary in their effects on different types of normal and cancer cells. A single "cure" for cancer has proved elusive since there is not a single type of cancer but as many as 100 different types of cancer. In addition, there are very few demonstrable biochemical differences between cancerous cells and normal cells. For this reason the effectiveness of many anticancer drugs is limited by their toxicity to normal rapidly growing cells in the intestinal and bone marrow areas. A final problem is that cancerous cells which are initially suppressed by a specific drug may develop a resistance to that drug. For this reason cancer chemotherapy may consist of using several drugs in combination for varying lengths of time.
Introduction
Chemotherapy drugs, are sometimes feared because of a patient's concern about toxic effects. Their role is to slow and hopefully halt the growth and spread of a cancer. There are three goals associated with the use of the most commonly-used anticancer agents.
1. Damage the DNA of the affected cancer cells.
2. Inhibit the synthesis of new DNA strands to stop the cell from replicating, because the replication of the cell is what allows the tumor to grow.
3. Stop mitosis or the actual splitting of the original cell into two new cells. Stopping mitosis stops cell division (replication) of the cancer and may ultimately halt the progression of the cancer.
Unfortunately, the majority of drugs currently on the market are not specific, which leads to the many common side effects associated with cancer chemotherapy. Because the common approach of all chemotherapy is to decrease the growth rate (cell division) of the cancer cells, the side effects are seen in bodily systems that naturally have a rapid turnover of cells iincluding skin, hair, gastrointestinal, and bone marrow. These healthy, normal cells, also end up damaged by the chemotherapy program.
Categories of Chemotherapy Drugs
In general, chemotherapy agents can be divided into three main categories based on their mechanism of action.
• Stop the synthesis of pre DNA molecule building blocks
These agents work in a number of different ways. DNA building blocks are folic acid, heterocyclic bases, and nucleotides, which are made naturally within cells. All of these agents work to block some step in the formation of nucleotides or deoxyribonucleotides (necessary for making DNA). When these steps are blocked, the nucleotides, which arethe building blocks of DNA and RNA, can not be synthesized. Thus the cells can not replicate because they can nnot make DNA without the nucleotides. Examples of drugs in this class include 1) methotrexate (Abitrexate®),2) fluorouracil (Adrucil®), 3) hydroxyurea (Hydrea®), and 4) mercaptopurine (Purinethol®).
• Directly damage the DNA in the nucleus of the cell
These agents chemically damage DNA and RNA. They disrupt replication of the DNA and either totally halt replication or cause the manufacture of nonsense DNA or RNA (i.e. the new DNA or RNA does not code for anything useful). Examples of drugs in this class include cisplatin (Platinol®) and 7) antibiotics - daunorubicin (Cerubidine®), doxorubicin (Adriamycin®), and etoposide (VePesid®).
• Effect the synthesis or breakdown of the mitotic spindles
Mitotic spindles serve as molecular railroads with "North and South Poles" in the cell when a cell starts to divide itself into two new cells. These spindles are very important because they help to split the newly copied DNA such that a copy goes to each of the two new cells during cell division. These drugs disrupt the formation of these spindles and therefore interrupt cell division. Examples of drugs in this class of 8) miotic disrupters include: Vinblastine (Velban®), Vincristine (Oncovin®) and Pacitaxel (Taxol®).
Enzyme L-asparaginase
In the 1950's a biochemical difference in metabolism related to the amino acid asparagine was found. Normal cells apparently can synthesize asparagine while leukemia cells cannot. If leukemia cells are deprived of asparagine, they will eventually die. In an almost unrecognized and parallel discovery, it was found that blood serum from guinea pigs and other South American rodents had antileukemia properties. The enzyme L-asparaginase was eventually identified as the anticancer agent. L-asparaginase was isolated and tested successfully on human leukemias. Eventually the enzyme asparaginase was also found and isolated from the bacteria, E. coli.
If the enzyme L-asparaginase is given to humans, various types of leukemias can be controlled. Tumor cells, more specifically lymphatic tumor cells, require huge amounts of asparagines to keep up with their rapid, malignant growth. This means they use both asparagine from the diet as well as what they can make themselves (which is limited) to satisfy their large asparagines demand. L-asparaginase is an enzyme that destroys asparagine external to the cell. Normal cells are able to make all the asparagine they need internally whereas tumor cells become depleted rapidly and die.The enzyme converts asparagine in the blood into aspartic acid by a deamination reaction. The leukemia cells are thus deprived of their supply of asparagine and will die.
Methotrexate
Methotrexate inhibits folic acid reductase which is responsible for the conversion of folic acid to tetrahydrofolic acid. At two stages in the biosynthesis of purines (adenine and guanine) and at one stage in the synthesis of pyrimidines (thymine, cytosine, and uracil), one-carbon transfer reactions occur which require specific coenzymes synthesized in the cell from tetrahydrofolic acid.
Tetrahydrofolic acid itself is synthesized in the cell from folic acid with the help of an enzyme, folic acid reductase. Methotrexate looks a lot like folic acid to the enzyme, so it binds to it thinking that it is folic acid. In fact, methotrexate looks so good to the enzyme that it binds to it quite strongly and inhibits the enzyme. Thus, DNA synthesis cannot proceed because the coenzymes needed for one-carbon transfer reactions are not produced from tetrahydrofolic acid because there is no tetrahydrofolic acid. Again, without DNA, no cell division.
5-Fluorouracil
5-Fluorouracil (5-FU; Adrucil®, Fluorouracil, Efudex®, Fluoroplex®) is an effective pyrimidine antimetabolite. Fluorouracil is synthesized into the nucleotide, 5-fluoro-2-deoxyuridine. This product acts as an antimetabolite by inhibiting the synthesis of 2-deoxythymidine because the carbon - fluorine bond is extremely stable and prevents the addition of a methyl group in the 5-position. The failure to synthesize the thymidine nucleotide results in little or no production of DNA. Two other similar drugs include: gemcitabine (Gemzar®) and arabinosylcytosine (araC). They all work through similar mechanisms.
Contributors
Charles Ophardt (Professor Emeritus, Elmhurst College); Virtual Chembook | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/12%3A_Solids_and_Modern_Materials/12.03%3A_Materials_for_Medicine/12.3.01%3A_Anti-Cancer_Drugs_I.txt |
Hydroxyurea
Hydroxyurea blocks an enzyme which converts the cytosine nucleotide into the deoxy derivative. In addition, DNA synthesis is further inhibited because hydroxyurea blocks the incorporation of the thymidine nucleotide into the DNA strand.
Mercaptopurine
Mercaptopurine, a chemical analog of the purine adenine, inhibits the biosynthesis of adenine nucleotides by acting as an antimetabolite. In the body, 6-MP is converted to the corresponding ribonucleotide. 6-MP ribonucleotide is a potent inhibitor of the conversion of a compound called inosinic acid to adenine Without adenine, DNA cannot be synthesized. 6-MP also works by being incorporated into nucleic acids as thioguanosine, rendering the resulting nucleic acids (DNA, RNA) unable to direct proper protein synthesis.
Thioguanine
Thioguanine is an antimetabolite in the synthesis of guanine nucleotides.
Alkylating Agents
Alkylating agents involve reactions with guanine in DNA. These drugs add methyl or other alkyl groups onto molecules where they do not belong. This in turn inhibits their correct utilization by base pairing and causes a miscoding of DNA.
1. In the first mechanism an alkylating agent attaches alkyl groups to DNA bases. This alteration results in the DNA being fragmented by repair enzymes in their attempts to replace the alkylated bases.
2. A second mechanism by which alkylating agents cause DNA damage is the formation of cross-bridges, bonds between atoms in the DNA. In this process, two bases are linked together by an alkylating agent that has two DNA binding sites. Cross-linking prevents DNA from being separated for synthesis or transcription.
3. The third mechanism of action of alkylating agents causes the mispairing of the nucleotides leading to mutations.
There are six groups of alkylating agents: nitrogen mustards; ethylenimes; alkylsulfonates; triazenes; piperazines; and nitrosureas. Cyclosporamide is a classical example of the role of the host metabolism in the activation of an alkylating agent and is one or the most widely used agents of this class. It was hoped that the cancer cells might posses enzymes capable of accomplishing the cleavage, thus resulting in the selective production of an activated nitrogen mustard in the malignant cells. Compare the top and bottom structures in the graphic on the left.
Antibiotics
A number of antibiotics such as anthracyclines, dactinomycin, bleomycin, adriamycin, mithramycin, bind to DNA and inactivate it. Thus the synthesis of RNA is prevented. General properties of these drugs include: interaction with DNA in a variety of different ways including intercalation (squeezing between the base pairs), DNA strand breakage and inhibition with the enzyme topoisomerase II. Most of these compounds have been isolated from natural sources and antibiotics. However, they lack the specificity of the antimicrobial antibiotics and thus produce significant toxicity.
The anthracyclines are among the most important antitumor drugs available. Doxorubicin is widely used for the treatment of several solid tumors while daunorubicin and idarubicin are used exclusively for the treatment of leukemia. These agents have a number of important effects including: intercalating (squeezing between the base pairs) with DNA affecting many functions of the DNA including DNA and RNA synthesis. Breakage of the DNA strand can also occur by inhibition of the enzyme topoisomerase II.
Dactinomycin (Actinomycin D)
At low concentrations dactinomycin inhibits DNA directed RNA synthesis and at higher concentrations DNA synthesis is also inhibited. All types of RNA are affected, but ribosomal RNA is more sensitive. Dactinomycin binds to double stranded DNA , permitting RNA chain initiation but blocking chain elongation. Binding to the DNA depends on the presence of guanine.
Mitotic Disrupters
Plant alkaloids like vincristine prevent cell division, or mitosis. There are several phases of mitosis, one of which is the metaphase. During metaphase, the cell pulls duplicated DNA chromosomes to either side of the parent cell in structures called "spindles". These spindles ensure that each new cell gets a full set of DNA. Spindles are microtubular fibers formed with the help of the protein "tubulin". Vincristine binds to tubulin, thus preventing the formation of spindles and cell division.
Taxol
Paclitaxel (taxol) was first isolated from the from the bark of the Pacific Yew (Taxus brevifolia). Docetaxel is a more potent analog that is produced semisynthetically. In contrast to other microtubule antagonists, taxol disrupts the equilibrium between free tubulin and mircrotubules by shifting it in the direction of assembly, rather than disassembly. As a result, taxol treatment causes both the stabilization of microtubules and the formation of abnormal bundles of microtubules. The net effect is still the disruption of mitosis.
Mechanism of Intercalating Agents
Intercalating agents wedge between bases along the DNA. The intercalated drug molecules affect the structure of the DNA, preventing polymerase and other DNA binding proteins from functioning properly. The result is prevention of DNA synthesis, inhibition of transcription and induction of mutations. Examples include: Carboplatin and Cisplatin.
These related drugs covalently bind to DNA with preferential binding to the N-7 position of guanine and adenine. They are able to bind to two different sites on DNA producing cross-links, either intrastrand (within the same DNA molecule which results in inhibition of DNA synthesis and transcription.
Contributors
Charles Ophardt (Professor Emeritus, Elmhurst College); Virtual Chembook | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/12%3A_Solids_and_Modern_Materials/12.03%3A_Materials_for_Medicine/12.3.02%3A_Anti-Cancer_Drugs_II.txt |
Antidepressant drugs are used to restore mentally depressed patients to an improved mental status. Depression results from a deficiency of norepinephrine at receptors in the brain. Mechanisms that increase their effective concentration at the receptor sites should alleviate depression. Antidepressant drugs act by one or more of the following stimulation type mechanisms:
1. Increase release of norepinephrine: Amphetamines and electroconvulsive therapy act by this mechanism. Amphetamines mimic norepinephrine.
2. Prevent inactivation of norepinephrine:Monoamine oxidase (MAO) inhibitors are thought to act as antidepressant agents in part by preventing the breakdown and inactivation of norepinephrine.
3. Prevent the re uptake of norepinephrine:The action of norepinephrine at the receptor site is terminated by the re uptake of norepinephrine by the neuron from which it was originally released.
Tricyclic Antidepressants
The tricyclic antidepressants are the most effective drugs presently available for the treatment of depression. These act by increasing the release of norepinephrine. Amphetamine and cocaine can also act in this manner. Imipramine, amitriptylin, and other closely related drugs are among the drugs currently most widely used for the treatment of major depression.
• imipramine (Tofranil)
• desipramine (Norpramin)
The activity of the tricyclic drugs depends on the central ring of seven or eight atoms which confers an angled or twisted conformation. The side chain must have at least 2 carbons although 3 appear to be better. The amine group may be either tertiary or secondary. All tricyclic antidepressants block the re-uptake of norepinephrine at nerve terminals. However, the potency and selectivity for the inhibition of the uptake of norepinephrine, serotonin, and dopamine vary greatly among the agents. The tertiary amine tricyclics seem to inhibit the serotonin uptake pump, whereas the secondary amine ones seem better in switching off the NE pump. For instance, imipramine is a potent and selective blocker of serotonin transport, while desipramine inhibits the uptake of norepinephrine.
Serotonin
Serotonin (5-hydroxytryptamine or 5-HT) is a monoamine neurotransmitter found in cardiovascular tissue, in endothelial cells, in blood cells, and in the central nervous system. The role of serotonin in neurological function is diverse, and there is little doubt that serotonin is an important CNS neurotransmitter. Although some of the serotonin is metabolized by monoamine oxidase, most of the serotonin released into the post-synaptic space is removed by the neuron through a re-uptake mechanism inhibited by the tricyclic antidepressants and the newer, more selective antidepressant re uptake inhibitors such as fluoxetine and sertraline.
Selective Serotonin Reuptake Inhibitors
In recent years, selective serotonin reuptake inhibitors have been introduced for the treatment of depression. Prozac is the most famous drug in this class. Clomiprimine, fluoxetine (Prozac), sertraline and paroxetine selectively block the re uptake of serotonin, thereby increasing the levels of serotonin in the central nervous system. Note the similarities and differences between the tricyclic antidepressants and the selective serotonin re uptake inhibitors. Clomipramine has been useful in the treatment of obsessive-compulsive disorders.
Monoamine Oxidase Inhibitors
Monoamine oxidase (MAO) causes the oxidative deamination of norephinephrine, serotonin, and other amines. This oxidation is the method of reducing the concentration of the neurotransmitter after it has sent the signal at the receptor site. A drug which inhibits this enzyme has the effect of increasing the concentration of the norepinephrine which in turn causes a stimulation effect. Most MAO inhibitors are hydrazine derivatives. Hydrazine is highly reactive and may form a strong covalent bond with MAO with consequent inhibition for up to 5 days.
These drugs are less effective and produce more side effects than the tricyclic antidepressants. For example, they lower blood pressure and were at one time used to treat hypertension. Their use in psychiatry has also become very limited as the tricyclic antidepressants have come to dominate the treatment of depression and allied conditions. Thus, MAOIs are used most often when tricyclic antidepressants give unsatisfactory results.
Phenelzine is the hydrazine analog of phenylethylamine, a substrate of MAO. This and several other MAOIs, such as isocarboxazide, are structurally related to amphetamine and were synthesized in an attempt to enhance central stimulant properties.
• phenelzine (Nardil)
• isocarboxazid (Marplan)
Contributors
Charles Ophardt (Professor Emeritus, Elmhurst College); Virtual Chembook | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/12%3A_Solids_and_Modern_Materials/12.03%3A_Materials_for_Medicine/12.3.03%3A_Antidepressants.txt |
Hypnotic and sedative drugs are non-selective, general depressants of the central nervous system. If the dose is relatively low, a sedative action results in a reduction in restlessness and emotional tension. A larger dose of the same drug produces a hypnotic sleep inducing effect. As the dosage is increased further, the result is anesthesia or death if the dosage is sufficiently high.
Introduction
The barbiturates once enjoyed a long period of extensive use as sedative-hypnotic drugs; however, except for a few specialized uses, they have been largely replaced by the much safer benzodiazepines. Barbiturates are central nervous system depressants and are similar, in many ways, to the depressant effects of alcohol. To date, there are about 2,500 derivatives of barbituric acid of which only 15 are used medically. The first barbiturate was synthesized from barbituric acid in 1864. The original use of barbiturates was to replace drugs such as opiates, bromides, and alcohol to induce sleep.
The hyponotic and sedative effects produced by barbiturates are usually ascribed to their interference of nerve transmission to the cortex. Various theories for the action of barbiturates include: changes in ion movements across the cell membrane; interactions with cholinergic and non cholinergic receptor sites; impairment of biochemical reactions which provide energy; and depression of selected areas of the brain. The structures of the barbiturates can be related to the duration of effective action. Although over 2000 derivatives of barbituric acid have been synthesized only about a dozen are currently used. All of the barbiturates are related to the structure of barbituric acid shown below.
The duration of effect depends mainly on the alkyl groups attached to carbon # 5 which confer lipid solubility to the drug. The duration of effective action decreases as the total number of carbons at C # 5 increases. To be more specific, a long effect is achieved by a short chain and/or phenyl group. A short duration effect occurs when there are the most carbons and branches in the alkyl chains
Benzodiazepines
The term benzodiazepine refers to the portion of the structure composed of a benzene ring (A) fused to a seven-membered diazepine ring (B). However, since all of the important benzodiazepines contain a aryl substituent ring C) and a 1, 4-diazepine ring, the term has come to mean the aryl-1,4-benzodiazepines. There are several useful benzodiazepines available: chlordiazepoxide (Librium) and diazapam (Valium).
The actions of benzodiazepines are a result of increased activation of receptors by gamma-aminobutyric acid (GABA). Benzodiazepine receptors are located on the alpha subunit of the GABA receptor located almost exclusively on postsynaptic nerve endings in the CNS (especially cerebral cortex). Benzodiazepines enhance the GABA transmitter in the opening of postsynaptic chloride channels which leads to hyperpolarization of cell membranes. That is, they "bend" the receptor slightly so that GABA molecules attach to and activate their receptors more effectively and more often.
Contributors
Charles Ophardt (Professor Emeritus, Elmhurst College); Virtual Chembook
12.3.05: Drugs Acting Upon the Central Nervous System
The central nervous system directs the functions of all tissues of the body. The peripheral nervous system receives thousands of sensory inputs and transmits them to the brain via the spinal cord. The brain processes this incoming information and discards 99% as unimportant. After sensory information has been evaluated, selected areas of the central nervous system initiate nerve impulses to organs or tissue to make an appropriate response.
Chemical influences are capable of producing a myriad of effects on the activity and function of the central nervous system. Since our knowledge of different regions of brain function and the neurotransmitters in the brain is limited, the explanations for the mechanisms of drug action may be vague. The known neurotransmitters are: acetylcholine which is involved with memory and learning; norepinephrine which is involved with mania-depression and emotions; and serotonin which is involved with biological rhythms, sleep, emotion, and pain.
Central Nervous System Stimulants
Stimulants are drugs that exert their action through excitation of the central nervous system. Psychic stimulants include caffeine, cocaine, and various amphetamines. These drugs are used to enhance mental alertness and reduce drowsiness and fatigue. However, increasing the dosage of caffeine above 200 mg (about 2 cups of coffee) does not increase mental performance but may increase nervousness, irritability, tremors, and headache. Heavy coffee drinkers become psychically dependent upon caffeine. If caffeine is withheld, a person may experience mild withdrawal symptoms characterized by irritability, nervousness, and headache.
Caffeine and the chemically related xanthines, theophylline and theobromine, decrease in the order given in their stimulatory action. They may be included in some over-the-counter drugs. The action of caffeine is to block adenosine receptors as an antagonist. As caffeine has a similar structure to the adenosine group. This means that caffeine will fit adenosine receptors as well as adenosine itself. It inhibits the release of neurotransmitters from presynaptic sites but works in concert with norepinephrine or angiotensin to augment their actions. Antagonism of adenosine receptors by caffeine would appear to promote neurotransmitter release, thus explaining the stimulatory effects of caffeine.
Amphetamines
The stimulation caused by amphetamines is caused by excessive release of norepinephrine from storage sites in the peripheral nervous system. It is not known whether the same action occurs in the central nervous system. Two other theories for their action are that they are degraded slower than norepinephrine or that they could act on serotonin receptor sites. Therapeutic doses of amphetamine elevate mood, reduce feelings of fatigue and hunger, facilitate powers of concentration, and increase the desire and capacity to carry out work. They induce exhilarating feelings of power, strength, energy, self-assertion, focus and enhanced motivation. The need to sleep or eat is diminished.
Levoamphetamine (Benzedrine), dextroamphetamine (Dexedrine), and methamphetamine (Methedrine) are collectively referred to as amphetamines. Benzedrine is a mixture of both the dextro and levoamphetamine isomers. The dextro isomer is several times more potent than the levo isomer.
The misuse and abuse of amphetamines is a significant problem which may include the house wife taking diet pills, athletes desiring an improved performance, the truck driver driving non-stop coast to-coast, or a student cramming all night for an exam. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/12%3A_Solids_and_Modern_Materials/12.03%3A_Materials_for_Medicine/12.3.04%3A_Barbiturates_and_Benzodiazepines.txt |
Hallucinogenic agents, also called psychomimetic agents, are capable of producing hallucinations, sensory illusions and bizarre thoughts. The primary effect of these compounds is to consistently alter thought and sensory perceptions. Some of these drugs are used in medicine to produce model psychoses as aids in psychotherapy. Another purpose is to investigate the relationship of mind, brain, and biochemistry with the purpose of elucidating mental diseases such as schizophrenia.
Introduction
A large body of evidence links the action of hallucinogenic agents to effects at serotonin receptor sites in the central nervous system. Whether the receptor site is stimulated or blocked is not exactly known. The serotonin receptor site may consist of three polar or ionic areas to complement the structure of serotonin as shown in the graphic on the left.
Mescalin and Psilocybin
The drugs shown in the graphic can be isolated from natural sources: lysergic acid amide from morning glory seeds, psilocybin from the "magic mushroom", Psilocybe mexicana. The hallucinogenic molecules fit into the same receptors as the neuro-transmitter, and over-stimulate them, leading to false signals being created.
Mescaline is isolated from a peyote cactus. The natives of Central America first made use of these drugs in religious ceremonies, believing the vivid, colorful hallucinations had religious significance. The Aztecs even had professional mystics and prophets who achieved their inspiration by eating the mescaline-containing peyote cactus (Lophophora williamsii). Indeed, the cactus was so important to the Aztecs that they named it teo-nancacyl, or "God's Flesh". This plant was said to have been distributed to the guests at the coronation of Montezuma to make the ceremony seem even more spectacular.
Lysergic acid diethylamide (LSD)
LSD is one of the most powerful hallucinogenic drugs known. LSD stimulates centers of the sympathetic nervous system in the midbrain, which leads to pupillary dilation, increase in body temperature, and rise in the blood-sugar level. LSD also has a serotonin-blocking effect. The hallucinogenic effects of lysergic acid diethylamide (LSD) are also the result of the complex interactions of the drug with both the serotoninergic and dopaminergic systems.
During the first hour after ingestion, the user may experience visual changes with extreme changes in mood. The user may also suffer impaired depth and time perception, with distorted perception of the size and shape of objects, movements, color, sound, touch and the user's own body image.
Serotonin
Serotonin (5-hydroxytryptamine or 5-HT) is a monoamine neurotransmitter found in cardiovascular tissue, in endothelial cells, in blood cells, and in the central nervous system. The role of serotonin in neurological function is diverse, and there is little doubt that serotonin is an important CNS neurotransmitter. Although some of the serotonin is metabolized by monoamine oxidase, most of the serotonin released into the post-synaptic space is removed by the neuron through a re uptake mechanism inhibited by the tricyclic antidepressants and the newer, more selective antidepressant re uptake inhibitors such as fluoxetine and sertraline.
• en.Wikipedia.org/wiki/Lysergi...d_diethylamide
• en.Wikipedia.org/wiki/Serotonin
12.3.07: Local Anesthetics
Unlike other drugs which act in the region of the synapse, local anesthetics are agents that reversibly block the generation and conduction of nerve impulses along a nerve fiber. They depress impulses from sensory nerves of the skin, surfaces of mucosa, and muscles to the central nervous system. These agents are widely used in surgery, dentistry, and ophthalmology to block transmission of impulses in peripheral nerve endings.
Introduction
Most local anesthetics can be represented by the following general formula. In both the official chemical name and the proprietary name, a local anesthetic drug can be recognized by the "-caine" ending. The ester linkage can also be an amide linkage. The most recent research indicates that the local anesthetic binds to a phospholipid in the nerve membrane and inhibits the ability of the phospholipid to bind Ca+2 ions.
Practically all of the free-base forms of the drugs are liquids. For this reason most of these drugs are used as salts (chloride, sulfate, etc.) which are water soluble, odorless, and crystalline solids. As esters these drugs are easily hydrolyzed with consequent loss of activity. The amide form of the drug is more stable and resistant to hydrolysis.
Benzocaine and Lidocaine
Two local anesthtics are shown below. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/12%3A_Solids_and_Modern_Materials/12.03%3A_Materials_for_Medicine/12.3.06%3A_Hallucinogenic_Drugs.txt |
Antibiotics are specific chemical substances derived from or produced by living organisms that are capable of inhibiting the life processes of other organisms. The first antibiotics were isolated from microorganisms but some are now obtained from higher plants and animals. Over 3,000 antibiotics have been identified but only a few dozen are used in medicine. Antibiotics are the most widely prescribed class of drugs comprising 12% of the prescriptions in the United States.
Macrolides
Macrolides are products of actinomycetes (soil bacteria) or semi-synthetic derivatives of them. Erythromycin is an orally effective antibiotic discovered in 1952 in the metabolic products of a strain of Streptocyces erythreus, originally obtained from a soil sample.
Erythromycin and other macrolide antibiotics inhibit protein synthesis by binding to the 23S rRNA molecule (in the 50S subunit) of the bacterial ribosome blocking the exit of the growing peptide chain. of sensitive microorganisms. (Humans do not have 50 S ribosomal subunits, but have ribosomes composed of 40 S and 60 S subunits). Certain resistant microorganisms with mutational changes in components of this subunit of the ribosome fail to bind the drug. The association between erythromycin and the ribosome is reversible and takes place only when the 50 S subunit is free from tRNA molecules bearing nascent peptide chains. Gram-positive bacteria accumulate about 100 times more erythromycin than do gram-negative microorganisms. The non ionized from of the drug is considerably more permeable to cells, and this probably explains the increased antimicrobial activity that is observed in alkaline pH.
Tetracyclines
Tetracyclines have the broadest spectrum of antimicrobial activity. These may include: Aureomycin, Terramycin, and Panmycin. Four fused 6-membered rings, as shown in the figure below, form the basic structure from which the various tetracyclines are made. The various derivatives are different at one or more of four sites on the rigid, planar ring structure. The classical tetracyclines were derived from Streptomyces spp., but the newer derivatives are semisynthetic as is generally true for newer members of other drug groups.
Tetracyclines inhibit bacterial protein synthesis by blocking the attachment of the transfer RNA-amino acid to the ribosome. More precisely they are inhibitors of the codon-anticodon interaction. Tetracyclines can also inhibit protein synthesis in the host, but are less likely to reach the concentration required because eukaryotic cells do not have a tetracycline uptake mechanism.
Streptomycin
Streptomycin is effective against gram-negative bacteria, although it is also used in the treatment of tuberculosis. Streptomycin binds to the 30S ribosome and changes its shape so that it and inhibits protein synthesis by causing a misreading of messenger RNA information.
Chloramphenicol
Chloromycetin is also a broad spectrum antibiotic that possesses activity similar to the tetracylines. At present, it is the only antibiotic prepared synthetically. It is reserved for treatment of serious infections because it is potentially highly toxic to bone marrow cells. It inhibits protein synthesis by attaching to the ribosome and interferes with the formation of peptide bonds between amino acids. It behaves as an antimetabolite for the essential amino acid phenylalanine at ribosomal binding sites.
Table 1: Some clinically important antibiotics
Antibiotic Producer organism Activity Site or mode of action
Penicillin Penicillium chrysogenum Gram-positive bacteria Wall synthesis
Cephalosporin Cephalosporium acremonium Broad spectrum Wall synthesis
Griseofulvin Penicillium griseofulvum Dermatophytic fungi Microtubules
Bacitracin Bacillus subtilis Gram-positive bacteria Wall synthesis
Polymyxin B Bacillus polymyxa Gram-negative bacteria Cell membrane
Amphotericin B Streptomyces nodosus Fungi Cell membrane
Erythromycin Streptomyces erythreus Gram-positive bacteria Protein synthesis
Neomycin Streptomyces fradiae Broad spectrum Protein synthesis
Streptomycin Streptomyces griseus Gram-negative bacteria Protein synthesis
Tetracycline Streptomyces rimosus Broad spectrum Protein synthesis
Vancomycin Streptomyces orientalis Gram-positive bacteria Protein synthesis
Gentamicin Micromonospora purpurea Broad spectrum Protein synthesis
Rifamycin Streptomyces mediterranei Tuberculosis Protein synthesis
Contributors
Charles Ophardt (Professor Emeritus, Elmhurst College); Virtual Chembook | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/12%3A_Solids_and_Modern_Materials/12.03%3A_Materials_for_Medicine/12.3.08%3A_Misc_Antibiotics.txt |
Narcotic agents are potent analgesics which are effective for the relief of severe pain. Analgesics are selective central nervous system depressants used to relieve pain. The term analgesic means "without pain". Even in therapeutic doses, narcotic analgesics can cause respiratory depression, nausea, and drowsiness. Long term administration produces tolerance, psychic, and physical dependence called addiction.
Introduction
Narcotic agents may be classified into four categories:
1. Morphine and codeine - natural alkaloids of opium.
2. Synthetic derivatives of morphine such as heroin.
3. Synthetic agents which resemble the morphine structure.
4. Narcotic antagonists which are used as antidotes for overdoses of narcotic analgesics.
The main pharmacological action of analgesics is on the cerebrum and medulla of the central nervous system. Another effect is on the smooth muscle and glandular secretions of the respiratory and gastro-intestinal tract. The precise mechanism of action is unknown although the narcotics appear to interact with specific receptor sites to interfere with pain impulses.
Receptor Site
A schematic for an analgesic receptor site may look as shown in the graphic below with morphine. Three areas are needed: a flat areas to accommodate a flat nonpolar aromatic ring, a cavity to accept another series of rings perpendicular, and an anionic site for polar interaction of the amine group.
Enkephalins
Recently investigators have discovered two compounds in the brain called enkephalins which resemble morphine in structure. Each one is a peptide composed of 5 amino acids and differ only in the last amino acid. The peptide sequences are: tyr-gly-gly-phe-leu and tyr-gly-gly-phe-met. Molecular models show that the structures of the enkephalins has some similarities with morphine. The main feature in common appears to be the aromatic ring with the -OH group attached (tyr). Methadone and other similar analgesics have 2 aromatic rings which would be similar to the enkephalins (tyr and phe).
Analgesics may relieve pain by preventing the release of acetylcholine. Enkephalin molecules are released from a nerve cell and bind to analgesic receptor sites on the nerve cell sending the impulse. The binding of enkephalin or morphine-like drugs changes the shape of the nerve sending the impulse in such a fashion as to prevent the cell from releasing acetylcholine. As a result, the pain impulse cannot be transmitted and the brain does not preceive pain.
Morphine and Codeine
Morphine exerts a narcotic action manifested by analgesia, drowsiness, changes in mood, and mental clouding. The major medical action of morphine sought in the CNS is analgesia. Opiates suppress the "cough center" which is also located in the brain stem, the medulla. Such an action is thought to underlie the use of opiate narcotics as cough suppressants. Codeine appears to be particularly effective in this action and is widely used for this purpose.
Narcotic analgesics cause an addictive physical dependence. If the drug is discontinued, withdrawal symptoms are experienced. Although the reasons for addiction and withdrawal symptoms are not completely known, recent experiments have provided some information. A nucleotide known as cyclicadenosine monophosphate (cAMP) is synthesized with the aid of the enzyme adenylate cyclase. Enkephalin and morphine-like drugs inhibit this enzyme and thus decrease the amount of cAMP in the cells. In order to compensate for the decreased cAMP, the cells synthesize more enzyme in an attempt to produce more cAMP. Since more enzyme has been produced, more morphine is required as an inhibitor to keep the cAMP at a low level. This cycle repeats itself causing an increase in the tolerance level and increasing the amounts of morphine required. If morphine is suddenly withheld, withdrawal symptoms are probably caused by a high concentration of cAMP since the synthesizing enzyme, adenylate cyclase, is no longer being inhibited.
Morphine and codeine are contained in opium from the poppy (Papaver Somniterum) plant found in Turkey, Mexico, Southeast Asia, China, and India. This plant is 3-4 feet tall with 5-8 egg shaped capsules on top. Ten days after the poppy blooms in June, incisions are made in the capsules permitting a milky fluid to ooze out. The following day the gummy mass (now brown) is carefully scraped off and pressed into cakes of raw opium to dry.
Opium contains over 20 compounds but only morphine (10%) and codeine (0.5%) are of any importance. Morphine is extracted from the opium and isolated in a relatively pure form. Since codeine is in such low concentration, it is synthesized from morphine by an ether-type methylation of an alcohol group. Codeine has only a fraction of the potency compared to morphine. It is used with aspirin and as a cough suppressant.
Heroin
Heroin is synthesized from morphine by a relatively simple esterification reaction of two alcohol (phenol) groups with acetic anhydride (equivalent to acetic acid). Heroin is much more potent than morphine but without the respiratory depression effect. A possible reason may be that heroin passes the blood-brain barrier much more rapidly than morphine. Once in the brain, the heroin is hydrolyzed to morphine which is responsible for its activity.
Synthetic narcotic analgesics include meperidine and methadone. Meperidine is the most common subsitute for morphine. It exerts several pharmacological effects: analgesic, local anesthetic, and mild antihistamine. This multiple activity may be explained by its structural resemblance to morphine, atropine, and histamine.
Methadone is more active and more toxic than morphine. It can be used for the relief of may types of pain. In addition it is used as a narcotic substitute in addiction treatment because it prevents morphine abstinence syndrome. Methadone was synthesized by German chemists during Wold War II when the United States and our allies cut off their opium supply. And it is difficult to fight a war without analgesics so the Germans went to work and synthesized a number of medications in use today, including demerol and darvon which is structurally simular to methadone. And before we go further lets clear up another myth. Methadone, or dolophine was not named after Adolf Hitler. The "dol" in dolophine comes from the latin root "dolor." The female name Dolores is derived from it and the term dol is used in pain research to measure pain e.g., one dol is 1 unit of pain.
Even methadone, which looks strikingly different from other opioid agonists, has steric forces which produce a configuration that closely resembles that of other opiates. See the graphic on the left and the top graphic on this page. In other words, steric forces bend the molecule of methadone into the correct configuration to fit into the opiate receptor. When you take methadone it first must be metabolized in the liver to a product that your body can use. Excess methadone is also stored in the liver and blood stream and this is how methadone works its 'time release trick' and last for 24 hours or more. Once in the blood stream metabolized methadone is slowly passed to the brain when it is needed to fill opiate receptors. Methadone is the only effective treatment for heroin addiction. It works to smooth the ups and down of heroin craving and allows the person to function normally.
Narcotic Antagonists
Narcotic Antagonists prevent or abolish excessive respiratory depression caused by the administration of morphine or related compounds. They act by competing for the same analgesic receptor sites. They are structurally related to morphine with the exception of the group attached to nitrogen.
Nalorphine precipitates withdrawal symptoms and produces behavioral disturbances in addition to the antogonism action. Naloxane is a pure antagonist with no morphine like effects. It blocks the euphoric effect of heroin when given before heroin. Naltrexone became clinically available in 1985 as a new narcotic antagonist. Its actions resemble those of naloxone, but naltrexone is well is well absorbed orally and is long acting, necessitating only a dose of 50 to 100 mg. Therefore, it is useful in narcotic treatment programs where it is desired to maintain an individual on chronic therapy with a narcotic antagonist. In individuals taking naltrexone, subsequent injection of an opiate will produce little or no effect. Naltrexone appears to be particularly effective for the treatment of narcotic dependence in addicts who have more to gain by being drug-free rather than drug dependant
Contributors and Attributions
• Charles Ophardt, Professor Emeritus, Elmhurst College; Virtual Chembook
• Poppies image from: leda.lycaeum.org | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/12%3A_Solids_and_Modern_Materials/12.03%3A_Materials_for_Medicine/12.3.09%3A_Narcotic_Analgesic_Drugs.txt |
The penicillins were the first antibiotics discovered as natural products from the mold Penicillium.
Introduction
In 1928, Sir Alexander Fleming, professor of bacteriology at St. Mary's Hospital in London, was culturing Staphylococcus aureus. He noticed zones of inhibition where mold spores were growing. He named the mold Penicillium rubrum. It was determined that a secretion of the mold was effective against Gram-positive bacteria.
Figure 1: Beta Lactam Structure
Penicillins as well as cephalosporins are called beta-lactam antibiotics and are characterized by three fundamental structural requirements: the fused beta-lactam structure (shown in the blue and red rings, a free carboxyl acid group (shown in red bottom right), and one or more substituted amino acid side chains (shown in black). The lactam structure can also be viewed as the covalent bonding of pieces of two amino acids - cysteine (blue) and valine (red).
Penicillin-G where R = an ethyl pheny group, is the most potent of all penicillin derivatives. It has several shortcomings and is effective only against gram-positive bacteria. It may be broken down in the stomach by gastric acids and is poorly and irregularly absorbed into the blood stream. In addition many disease producing staphylococci are able to produce an enzyme capable of inactivating penicillin-G. Various semisynthetic derivatives have been produced which overcome these shortcomings.
Powerful electron-attracting groups attached to the amino acid side chain such as in phenethicillin prevent acid attack. A bulky group attached to the amino acid side chain provides steric hindrance which interferes with the enzyme attachment which would deactivate the pencillins i.e. methicillin. Refer to Table 2 for the structures. Finally if the polar character is increased as in ampicillin or carbenicillin, there is a greater activity against Gram-negative bacteria.
Penicillin Mode of Action
All penicillin derivatives produce their bacteriocidal effects by inhibition of bacterial cell wall synthesis. Specifically, the cross linking of peptides on the mucosaccharide chains is prevented. If cell walls are improperly made cell walls allow water to flow into the cell causing it to burst. Resemblances between a segment of penicillin structure and the backbone of a peptide chain have been used to explain the mechanism of action of beta-lactam antibiotics. The structures of a beta-lactam antibiotic and a peptide are shown on the left for comparison. Follow the trace of the red oxygens and blue nitrogen atoms.
Gram-positive bacteria possess a thick cell wall composed of a cellulose-like structural sugar polymer covalently bound to short peptide units in layers.The polysaccharide portion of the peptidoglycan structure is made of repeating units of N-acetylglucosamine linked b-1,4 to N-acetylmuramic acid (NAG-NAM). The peptide varies, but begins with L-Ala and ends with D-Ala. In the middle is a dibasic amino acid, diaminopimelate (DAP). DAP (orange) provides a linkage to the D-Ala (gray) residue on an adjacent peptide.
The bacterial cell wall synthesis is completed when a cross link between two peptide chains attached to polysaccharide backbones is formed. The cross linking is catalyzed by the enzyme transpeptidase. First the terminal alanine from each peptide is hydrolyzed and secondly one alanine is joined to lysine through an amide bond.
Peptidoglycan image courtesy of the University of Texas-Houston Medical School
Penicillin binds at the active site of the transpeptidase enzyme that cross-links the peptidoglycan strands. It does this by mimicking the D-alanyl-D-alanine residues that would normally bind to this site. Penicillin irreversibly inhibits the enzyme transpeptidase by reacting with a serine residue in the transpeptidase. This reaction is irreversible and so the growth of the bacterial cell wall is inhibited. Since mammal cells do not have the same type of cell walls, penicillin specifically inhibits only bacterial cell wall synthesis.
Bacterial Resistance
As early as the 1940s, bacteria began to combat the effectiveness of penicillin. Penicillinases (or beta-lactamases) are enzymes produced by structurally susceptable bacteria which renders penicillin useless by hydrolysing the peptide bond in the beta-lactam ring of the nucleus. Penicillinase is a response of bacterial adaptation to its adverse environment, namely the presence of a substance which inhibits its growth. Many other antibiotics are also rendered ineffective because of this same type of resistance.
Severe Allergic Shock
It is estimated that between 300-500 people die each year from penicillin-induced anaphylaxis, a severe allergic shock reaction to penicillin. In afflicted individuals, the beta-lactam ring binds to serum proteins, initiating an IgE-mediated inflammatory response. Penicillin and ala-ala peptide - Chime in new window
Cephalosporins
Cephalosporins are the second major group of beta-lactam antibiotics. They differ from penicillins by having the beta-lactam ring as a 6 member ring. The other difference, which is more significant from a medicinal chemistry stand point, is the existence of a functional group (R) at position 3 of the fused ring system. This now allows for molecular variations to effect changes in properties by diversifying the groups at position 3.
The first member of the newer series of beta-lactams was isolated in 1956 from extracts of Cephalosporium acremonium, a sewer fungus. Like penicillin, cephalosporins are valuable because of their low toxicity and their broad spectrum of action against various diseases. In this way, cephalosporin is very similar to penicillin. Cephalosporins are one of the most widely used antibiotics, and economically speaking, has about 29% of the antibiotic market. The cephalosporins are possibly the single most important group of antibiotics today and are equal in importance to penicillin.
The structure and mode of action of the cephalosporins are similar to that of penicillin. They affect bacterial growth by inhibiting cell wall synthesis, in Gram-positive and negative bacteria. Some brand names include: cefachlor, cefadroxil, cefoxitin, ceftriaxone. Cephalexin
12.3.11: Sulfa Drugs
Sulfonamides are synthetic antimicrobial agents with a wide spectrum encompassing most gram-positive and many gram-negative organisms. These drugs were the first efficient treatment to be employed systematically for the prevention and cure of bacterial infections.
Contributors and Attributions
Charles Ophardt (Professor Emeritus, Elmhurst College); Virtual Chembook | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/12%3A_Solids_and_Modern_Materials/12.03%3A_Materials_for_Medicine/12.3.10%3A_Penicillin.txt |
• 12.4.1: Bonding in Electronic Materials
Bonding in metals and semiconductors can be described using band theory, in which a set of molecular orbitals is generated that extends throughout the solid. The primary learning objective of this Module is to describe the electrical properties of solid using band theory.
• 12.4.2: Silicon Devices
The synthesis and purification of bulk polycrystalline semiconductor material represents the first step towards the commercial fabrication of an electronic device. This polycrystalline material is then used as the raw material for the formation of single crystal material that is processed to semiconductor wafers.
12.04: Materials for Electronics
Bonding in metals and semiconductors can be described using band theory, in which a set of molecular orbitals is generated that extends throughout the solid. The primary learning objective of this Module is to describe the electrical properties of solid using band theory.
Band Theory
To explain the observed properties of metals, a more sophisticated approach is needed than the electron-sea model commonly described. The molecular orbital theory used to explain the delocalized π bonding in polyatomic ions and molecules such as NO2, ozone, and 1,3-butadiene can be adapted to accommodate the much higher number of atomic orbitals that interact with one another simultaneously in metals.
In a 1 mol sample of a metal, there can be more than 1024 orbital interactions to consider. In our molecular orbital description of metals, however, we begin by considering a simple one-dimensional example: a linear arrangement of n metal atoms, each containing a single electron in an s orbital. We use this example to describe an approach to metallic bonding called band theory, which assumes that the valence orbitals of the atoms in a solid interact, generating a set of molecular orbitals that extend throughout the solid.
One-Dimensional Systems
If the distance between the metal atoms is short enough for the orbitals to interact, they produce bonding, antibonding, and nonbonding molecular orbitals. The left portion of Figure \(1\) shows the pattern of molecular orbitals that results from the interaction of ns orbitals as n increases from 2 to 5.
As we saw previously, the lowest-energy orbital is the completely bonding molecular orbital, whereas the highest-energy orbital is the completely antibonding molecular orbital. Molecular orbitals of intermediate energy have fewer nodes than the totally antibonding molecular orbital. The energy separation between adjacent orbitals decreases as the number of interacting orbitals increases. For n = 30, there are still discrete, well-resolved energy levels, but as n increases from 30 to a number close to Avogadro’s number, the spacing between adjacent energy levels becomes almost infinitely small. The result is essentially a continuum of energy levels, as shown on the right in Figure \(1\), each of which corresponds to a particular molecular orbital extending throughout the linear array of metal atoms. The levels that are lowest in energy correspond to mostly bonding combinations of atomic orbitals, those highest in energy correspond to mostly antibonding combinations, and those in the middle correspond to essentially nonbonding combinations.
The continuous set of allowed energy levels shown on the right in Figure \(1\) is called an energy band. The difference in energy between the highest and lowest energy levels is the bandwidth and is proportional to the strength of the interaction between orbitals on adjacent atoms: the stronger the interaction, the larger the bandwidth. Because the band contains as many energy levels as molecular orbitals, and the number of molecular orbitals is the same as the number of interacting atomic orbitals, the band in Figure \(1\) contains n energy levels corresponding to the combining of s orbitals from n metal atoms. Each of the original s orbitals could contain a maximum of two electrons, so the band can accommodate a total of 2n electrons. Recall, however, that each of the metal atoms we started with contained only a single electron in each s orbital, so there are only n electrons to place in the band. Just as with atomic orbitals or molecular orbitals, the electrons occupy the lowest energy levels available. Consequently, only the lower half of the band is filled. This corresponds to filling all of the bonding molecular orbitals in the linear array of metal atoms and results in the strongest possible bonding.
Multidimensional Systems
The previous example was a one-dimensional array of atoms that had only s orbitals. To extrapolate to two- or three-dimensional systems and atoms with electrons in p and d orbitals is straightforward in principle, even though in practice the mathematics becomes more complex, and the resulting molecular orbitals are more difficult to visualize. The resulting energy-level diagrams are essentially the same as the diagram of the one-dimensional example in Figure \(1\), with the following exception: they contain as many bands as there are different types of interacting orbitals. Because different atomic orbitals interact differently, each band will have a different bandwidth and will be centered at a different energy, corresponding to the energy of the parent atomic orbital of an isolated atom.
Band Gap
Because the 1s, 2s, and 2p orbitals of a period 3 atom are filled core levels, they do not interact strongly with the corresponding orbitals on adjacent atoms. Hence they form rather narrow bands that are well separated in energy (Figure \(2\)). These bands are completely filled (both the bonding and antibonding levels are completely populated), so they do not make a net contribution to bonding in the solid. The energy difference between the highest level of one band and the lowest level of the next is the band gap. It represents a set of forbidden energies that do not correspond to any allowed combinations of atomic orbitals.
Because they extend farther from the nucleus, the valence orbitals of adjacent atoms (3s and 3p in Figure \(2\)) interact much more strongly with one another than do the filled core levels; as a result, the valence bands have a larger bandwidth. In fact, the bands derived from the 3s and 3p atomic orbitals are wider than the energy gap between them, so the result is overlapping bands. These have molecular orbitals derived from two or more valence orbitals with similar energies. As the valence band is filled with one, two, or three electrons per atom for Na, Mg, and Al, respectively, the combined band that arises from the overlap of the 3s and 3p bands is also filling up; it has a total capacity of eight electrons per atom (two electrons for each 3s orbital and six electrons for each set of 3p orbitals). With Na, therefore, which has one valence electron, the combined valence band is one-eighth filled; with Mg (two valence electrons), it is one-fourth filled; and with Al, it is three-eighths filled, as indicated in Figure \(2\). The partially filled valence band is absolutely crucial for explaining metallic behavior because it guarantees that there are unoccupied energy levels at an infinitesimally small energy above the highest occupied level.
Band theory can explain virtually all the properties of metals. Metals conduct electricity, for example, because only a very small amount of energy is required to excite an electron from a filled level to an empty one, where it is free to migrate rapidly throughout the crystal in response to an applied electric field. Similarly, metals have high heat capacities (as you no doubt remember from the last time a doctor or a nurse placed a stethoscope on your skin) because the electrons in the valence band can absorb thermal energy by being excited to the low-lying empty energy levels. Finally, metals are lustrous because light of various wavelengths can be absorbed, causing the valence electrons to be excited into any of the empty energy levels above the highest occupied level. When the electrons decay back to low-lying empty levels, they emit light of different wavelengths. Because electrons can be excited from many different filled levels in a metallic solid and can then decay back to any of many empty levels, light of varying wavelengths is absorbed and reemitted, which results in the characteristic shiny appearance that we associate with metals.
Requirements for Metallic Behavior
For a solid to exhibit metallic behavior,
• it must have a set of delocalized orbitals forming a band of allowed energy levels, and
• the resulting band must be partially filled (10%–90%) with electrons.
Without a set of delocalized orbitals, there is no pathway by which electrons can move through the solid.
Band theory explains the correlation between the valence electron configuration of a metal and the strength of metallic bonding. The valence electrons of transition metals occupy either their valence ns, (n − 1)d, and np orbitals (with a total capacity of 18 electrons per metal atom) or their ns and (n − 1)d orbitals (a total capacity of 12 electrons per metal atom). These atomic orbitals are close enough in energy that the derived bands overlap, so the valence electrons are not confined to a specific orbital. Metals with 6 to 9 valence electrons (which correspond to groups 6–9) are those most likely to fill the valence bands approximately halfway. Those electrons therefore occupy the highest possible number of bonding levels, while the number of antibonding levels occupied is minimal. Not coincidentally, the elements of these groups exhibit physical properties consistent with the presence of the strongest metallic bonding, such as very high melting points.
Insulators
In contrast to metals, electrical insulators are materials that conduct electricity poorly because their valence bands are full. The energy gap between the highest filled levels and the lowest empty levels is so large that the empty levels are inaccessible: thermal energy cannot excite an electron from a filled level to an empty one. The valence-band structure of diamond, for example, is shown in Figure \(\PageIndex{3a}\). Because diamond has only 4 bonded neighbors rather than the 6 to 12 typical of metals, the carbon 2s and 2p orbitals combine to form two bands in the solid, with the one at lower energy representing bonding molecular orbitals and the one at higher energy representing antibonding molecular orbitals. Each band can accommodate four electrons per atom, so only the lower band is occupied. Because the energy gap between the filled band and the empty band is very large (530 kJ/mol), at normal temperatures thermal energy cannot excite electrons from the filled level into the empty band. Thus there is no pathway by which electrons can move through the solid, so diamond has one of the lowest electrical conductivities known.
Semiconductors
What if the difference in energy between the highest occupied level and the lowest empty level is intermediate between those of electrical conductors and insulators? This is the case for silicon and germanium, which have the same structure as diamond. Because Si–Si and Ge–Ge bonds are substantially weaker than C–C bonds, the energy gap between the filled and empty bands becomes much smaller as we go down group 14 (part (b) and part (c) of Figure \(2\)). Consequently, thermal energy is able to excite a small number of electrons from the filled valence band of Si and Ge into the empty band above it, which is called the conduction band.
Exciting electrons from the filled valence band to the empty conduction band causes an increase in electrical conductivity for two reasons:
1. The electrons in the previously vacant conduction band are free to migrate through the crystal in response to an applied electric field.
2. Excitation of an electron from the valence band produces a “hole” in the valence band that is equivalent to a positive charge. The hole in the valence band can migrate through the crystal in the direction opposite that of the electron in the conduction band by means of a “bucket brigade” mechanism in which an adjacent electron fills the hole, thus generating a hole where the second electron had been, and so forth.
Consequently, Si is a much better electrical conductor than diamond, and Ge is even better, although both are still much poorer conductors than a typical metal (Figure \(4\)).
Substances such as Si and Ge that have conductivities between those of metals and insulators are called semiconductors. Many binary compounds of the main group elements exhibit semiconducting behavior similar to that of Si and Ge. For example, gallium arsenide (GaAs) is isoelectronic with Ge and has the same crystalline structure, with alternating Ga and As atoms; not surprisingly, it is also a semiconductor. The electronic structure of semiconductors is compared with the structures of metals and insulators in Figure \(5\).
Temperature and Conductivity
Because thermal energy can excite electrons across the band gap in a semiconductor, increasing the temperature increases the number of electrons that have sufficient kinetic energy to be promoted into the conduction band. The electrical conductivity of a semiconductor therefore increases rapidly with increasing temperature, in contrast to the behavior of a purely metallic crystal. In a metal, as an electron travels through the crystal in response to an applied electrical potential, it cannot travel very far before it encounters and collides with a metal nucleus. The more often such encounters occur, the slower the net motion of the electron through the crystal, and the lower the conductivity. As the temperature of the solid increases, the metal atoms in the lattice acquire more and more kinetic energy. Because their positions are fixed in the lattice, however, the increased kinetic energy increases only the extent to which they vibrate about their fixed positions. At higher temperatures, therefore, the metal nuclei collide with the mobile electrons more frequently and with greater energy, thus decreasing the conductivity. This effect is, however, substantially smaller than the increase in conductivity with temperature exhibited by semiconductors. For example, the conductivity of a tungsten wire decreases by a factor of only about two over the temperature range 750–1500 K, whereas the conductivity of silicon increases approximately 100-fold over the same temperature range. These trends are illustrated in Figure \(6\).
n- and p-Type Semiconductors
Doping is a process used to tune the electrical properties of commercial semiconductors by deliberately introducing small amounts of impurities. If an impurity contains more valence electrons than the atoms of the host lattice (e.g., when small amounts of a group 15 atom are introduced into a crystal of a group 14 element), then the doped solid has more electrons available to conduct current than the pure host has. As shown in Figure \(\PageIndex{7a}\), adding an impurity such as phosphorus to a silicon crystal creates occasional electron-rich sites in the lattice. The electronic energy of these sites lies between those of the filled valence band and the empty conduction band but closer to the conduction band. Because the atoms that were introduced are surrounded by host atoms, and the electrons associated with the impurity are close in energy to the conduction band, those extra electrons are relatively easily excited into the empty conduction band of the host. Such a substance is called an n-type semiconductor, with the n indicating that the added charge carriers are negative (they are electrons).
If the impurity atoms contain fewer valence electrons than the atoms of the host (e.g., when small amounts of a group 13 atom are introduced into a crystal of a group 14 element), then the doped solid has fewer electrons than the pure host. Perhaps unexpectedly, this also results in increased conductivity because the impurity atoms generate holes in the valence band. As shown in Figure \(\PageIndex{7b}\), adding an impurity such as gallium to a silicon crystal creates isolated electron-deficient sites in the host lattice. The electronic energy of these empty sites also lies between those of the filled valence band and the empty conduction band of the host but much closer to the filled valence band. It is therefore relatively easy to excite electrons from the valence band of the host to the isolated impurity atoms, thus forming holes in the valence band. This kind of substance is called a p-type semiconductor, with the p standing for positive charge carrier (i.e., a hole). Holes in what was a filled band are just as effective as electrons in an empty band at conducting electricity.
The electrical conductivity of a semiconductor is roughly proportional to the number of charge carriers, so doping is a precise way to adjust the conductivity of a semiconductor over a wide range. The entire semiconductor industry is built on methods for preparing samples of Si, Ge, or GaAs doped with precise amounts of desired impurities and assembling silicon chips and other complex devices with junctions between n- and p-type semiconductors in varying numbers and arrangements.
Because silicon does not stand up well to temperatures above approximately 100°C, scientists have been interested in developing semiconductors made from diamonds, a more thermally stable material. A new method has been developed based on vapor deposition, in which a gaseous mixture is heated to a high temperature to produce carbon that then condenses on a diamond kernel. This is the same method now used to create cultured diamonds, which are indistinguishable from natural diamonds. The diamonds are heated to more than 2000°C under high pressure to harden them even further. Doping the diamonds with boron has produced p-type semiconductors, whereas doping them with boron and deuterium achieves n-type behavior. Because of their thermal stability, diamond semiconductors have potential uses as microprocessors in high-voltage applications.
Example \(1\)
A crystalline solid has the following band structure, with the purple areas representing regions occupied by electrons. The lower band is completely occupied by electrons, and the upper level is about one-third filled with electrons.
1. Predict the electrical properties of this solid.
2. What would happen to the electrical properties if all of the electrons were removed from the upper band? Would you use a chemical oxidant or reductant to effect this change?
3. What would happen to the electrical properties if enough electrons were added to completely fill the upper band? Would you use a chemical oxidant or reductant to effect this change?
Given: band structure
Asked for: variations in electrical properties with conditions
Strategy:
1. Based on the occupancy of the lower and upper bands, predict whether the substance will be an electrical conductor. Then predict how its conductivity will change with temperature.
2. After all the electrons are removed from the upper band, predict how the band gap would affect the electrical properties of the material. Determine whether you would use a chemical oxidant or reductant to remove electrons from the upper band.
3. Predict the effect of a filled upper band on the electrical properties of the solid. Then decide whether you would use an oxidant or a reductant to fill the upper band.
Solution:
1. The material has a partially filled band, which is critical for metallic behavior. The solid will therefore behave like a metal, with high electrical conductivity that decreases slightly with increasing temperature.
2. Removing all of the electrons from the partially filled upper band would create a solid with a filled lower band and an empty upper band, separated by an energy gap. If the band gap is large, the material will be an electrical insulator. If the gap is relatively small, the substance will be a semiconductor whose electrical conductivity increases rapidly with increasing temperature. Removing the electrons would require an oxidant because oxidants accept electrons.
3. Adding enough electrons to completely fill the upper band would produce an electrical insulator. Without another empty band relatively close in energy above the filled band, semiconductor behavior would be impossible. Adding electrons to the solid would require a reductant because reductants are electron donors.
Exercise \(1\)
A substance has the following band structure, in which the lower band is half-filled with electrons (purple area) and the upper band is empty.
1. Predict the electrical properties of the solid.
2. What would happen to the electrical properties if all of the electrons were removed from the lower band? Would you use a chemical oxidant or reductant to effect this change?
3. What would happen to the electrical properties if enough electrons were added to completely fill the lower band? Would you use a chemical oxidant or reductant to effect this change?
Answer:
1. The solid has a partially filled band, so it has the electrical properties of a conductor.
2. Removing all of the electrons from the lower band would produce an electrical insulator with two empty bands. An oxidant is required.
3. Adding enough electrons to completely fill the lower level would result in an electrical insulator if the energy gap between the upper and lower bands is relatively large, or a semiconductor if the band gap is relatively small. A reductant is required.
• Metallic behavior requires a set of delocalized orbitals and a band of allowed energy levels that is partially occupied.
• The electrical conductivity of a semiconductor increases with increasing temperature, whereas the electrical conductivity of a metal decreases with increasing temperature.
• n-Type semiconductors are negative charge carriers; the impurity has more valence electrons than the host. p-Type semiconductors are positive charge carriers; the impurity has fewer valence electrons than the host.
Summary
Band theory assumes that the valence orbitals of the atoms in a solid interact to generate a set of molecular orbitals that extend throughout the solid; the continuous set of allowed energy levels is an energy band. The difference in energy between the highest and lowest allowed levels within a given band is the bandwidth, and the difference in energy between the highest level of one band and the lowest level of the band above it is the band gap. If the width of adjacent bands is larger than the energy gap between them, overlapping bands result, in which molecular orbitals derived from two or more kinds of valence orbitals have similar energies. Metallic properties depend on a partially occupied band corresponding to a set of molecular orbitals that extend throughout the solid to form a band of energy levels. If a solid has a filled valence band with a relatively low-lying empty band above it (a conduction band), then electrons can be excited by thermal energy from the filled band into the vacant band where they can then migrate through the crystal, resulting in electrical conductivity. Electrical insulators are poor conductors because their valence bands are full. Semiconductors have electrical conductivities intermediate between those of insulators and metals. The electrical conductivity of semiconductors increases rapidly with increasing temperature, whereas the electrical conductivity of metals decreases slowly with increasing temperature. The properties of semiconductors can be modified by doping, or introducing impurities. Adding an element with more valence electrons than the atoms of the host populates the conduction band, resulting in an n-type semiconductor with increased electrical conductivity. Adding an element with fewer valence electrons than the atoms of the host generates holes in the valence band, resulting in a p-type semiconductor that also exhibits increased electrical conductivity. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/12%3A_Solids_and_Modern_Materials/12.04%3A_Materials_for_Electronics/12.4.01%3A_Bonding_in_Electronic_Materials.txt |
Introduction
The synthesis and purification of bulk polycrystalline semiconductor material represents the first step towards the commercial fabrication of an electronic device. This polycrystalline material is then used as the raw material for the formation of single crystal material that is processed to semiconductor wafers. The strong influence on the electric characteristics of a semiconductors exhibited by small amounts of some impurities requires that the bulk raw material be of very high purity (> 99.9999%). Although some level of purification is possible during the crystallization process it is important to use as high a purity starting material as possible.
Following oxygen (46%), silicon (L. silicis flint) is the most abundant element in the earth's crust (28%). However, silicon does not occur in its elemental form, but as its oxide (SiO2) or as silicates. Sand, quartz, amethyst, agate, flint, and opal are some of the forms in which the oxide appears. Granite, hornblende, asbestos, feldspar, clay and mica, etc. are a few of the numerous silicate minerals. With such boundless supplies of the raw material, the costs associated with the production of bulk silicon is not one of abstraction and conversion of the oxide(s), but of purification of the crude elemental silicon. While 98% elemental silicon, known as metallurgical-grade silicon (MGS), is readily produced on a large scale, the requirements of extreme purity for electronic device fabrication require additional purification steps in order to produce electronic-grade silicon (EGS). Electronic-grade silicon is also known as semiconductor-grade silicon (SGS). In order for the purity levels to be acceptable for subsequent crystal growth and device fabrication, EGS must have carbon and oxygen impurity levels less than a few parts per million (ppm), and metal impurities at the parts per billion (ppb) range or lower. Table $1$ and Table $2$ give typical impurity concentrations in MGS and EGS, respectively. Besides the purity, the production cost and the specifications must meet the industry desires.
Table $1$: Typical impurity concentrations found in metallurgical-grade silicon (MGS).
Element Concentration (ppm) Element Concentration (ppm)
aluminum 1000-4350 manganese 50-120
boron 40-60 molybdenum < 20
calcium 245-500 nickel 10-105
chromium 50-200 phosphorus 20-50
copper 15-45 titanium 140-300
iron 1550-6500 vanadium 50-250
magnesium 10-50 zirconium 20
Table $2$: Typical impurity concentrations found in electronic-grade silicon (EGS).
Element Concentration (ppb) Element Concentration (ppb)
arsenic < 0.001 gold < 0.00001
antimony < 0.001 iron 0.1-1.0
boron ≤ 0.1 nickel 0.1-0.5
carbon 100-1000 oxygen 100-400
chromium < 0.01 phosphorus ≤ 0.3
cobalt 0.001 silver 0.001
copper 0.1 zinc < 0.1
Metallurgical-grade silicon (MGS)
The typical source material for commercial production of elemental silicon is quartzite gravel; a relatively pure form of sand (SiO2). The first step in the synthesis of silicon is the melting and reduction of the silica in a submerged-electrode arc furnace. An example of which is shown schematically in Figure $1$, along with the appropriate chemical reactions. A mixture of quartzite gravel and carbon are heated to high temperatures (ca. 1800 °C) in the furnace. The carbon bed consists of a mixture of coal, coke, and wood chips. The latter providing the necessary porosity such that the gases created during the reaction (SiO and CO) are able to flow through the bed.
The overall reduction reaction of SiO2 is expressed in (7.10.1), however, the reaction sequence is more complex than this overall reaction implies, and involves the formation of SiC and SiO intermediates. The initial reaction between molten SiO2 and C, (7.10.2), takes place in the arc between adjacent electrodes, where the local temperature can exceed 2000 °C. The SiO and CO thus generated flow to cooler zones in the furnace where SiC is formed, (7.10.3), or higher in the bed where they reform SiO2 and C, (7.10.2). The SiC reacts with molten SiO2, (7.10.4), producing the desired silicon along with SiO and CO. The molten silicon formed is drawn-off from the furnace and solidified.
$\text{SiO}_2\text{(liquid) + 2 C(solid)} \rightarrow \text{Si(liquid) + 2 CO(gas)} \nonumber$
$\text{SiO}_2\text{ + 2 C} \xrightleftharpoons[\text{<1600 °C}]{\text{>1700 °C}} \text{SiO + CO} \nonumber$
$\text{SiO + 2C} \rightarrow \text{SiC + CO (1600 - 1700 °C)} \nonumber$
$\text{SiC _ SiO}_2 \rightarrow \text{Si + SiO + CO} \nonumber$
The as-produced MGS is approximately 98-99% pure, with the major impurities being aluminum and iron (Table $1$), however, obtaining low levels of boron impurities is of particular importance, because it is difficult to remove and serves as a dopant for silicon. The drawbacks of the above process are that it is energy and raw material intensive. It is estimated that the production of one metric ton (1,000 kg) of MGS requires 2500 - 2700 kg quartzite, 600 kg charcoal, 600 - 700 kg coal or coke, 300 - 500 kg wood chips, and 500,000 kWh of electric power. Currently, approximately 500,000 metric tons of MGS are produced per year, worldwide. Most of the production (ca. 70%) is used for metallurgical applications (e.g., aluminum-silicon alloys are commonly used for automotive engine blocks) from whence its name is derived. Applications in a variety of chemical products such as silicone resins account for about 30%, and only 1% or less of the total production of MGS is used in the manufacturing of high-purity EGS for the electronics industry. The current worldwide consumption of EGS is approximately 5 x 106 kg per year.
Electronic-grade silicon (EGS)
Electronic-grade silicon (EGS) is a polycrystalline material of exceptionally high purity and is the raw material for the growth of single-crystal silicon. EGS is one of the purest materials commonly available, see Table $2$. The formation of EGS from MGS is accomplished through chemical purification processes. The basic concept of which involves the conversion of MGS to a volatile silicon compound, which is purified by distillation, and subsequently decomposed to re-form elemental silicon of higher purity (i.e., EGS). Irrespective of the purification route employed, the first step is physical pulverization of MGS followed by its conversion to the volatile silicon compounds.
A number of compounds, such as monosilane (SiH4), dichlorosilane (SiH2Cl2), trichlorosilane (SiHCl3), and silicon tetrachloride (SiCl4), have been considered as chemical intermediates. Among these, SiHCl3 has been used predominantly as the intermediate compound for subsequent EGS formation, although SiH4 is used to a lesser extent. Silicon tetrachloride and its lower chlorinated derivatives are used for the chemical vapor deposition (CVD) growth of Si and SiO2. The boiling points of silane and its chlorinated products (Table $3$) are such that they are conveniently separated from each other by fractional distillation.
Table $3$: Boiling points of silane and chlorosilanes at 760 mmHg (1 atmosphere).
Compound Boiling point (°C)
SiH4 -112.3
SiH3Cl -30.4
SiH2Cl2 8.3
SiHCl3 31.5
SiCl4 57.6
The reasons for the predominant use of SiHCl3 in the synthesis of EGS are as follows:
1. SiHCl3 can be easily formed by the reaction of anhydrous hydrogen chloride with MGS at reasonably low temperatures (200 - 400 °C);
2. it is liquid at room temperature so that purification can be accomplished using standard distillation techniques;
3. it is easily handled and if dry can be stored in carbon steel tanks;
4. its liquid is easily vaporized and, when mixed with hydrogen it can be transported in steel lines without corrosion;
5. it can be reduced at atmospheric pressure in the presence of hydrogen;
6. its deposition can take place on heated silicon, thus eliminating contact with any foreign surfaces that may contaminate the resulting silicon; and
7. it reacts at lower temperatures (1000 - 1200 °C) and at faster rates than does SiCl4.
Chlorosilane (Seimens) process
Trichlorosilane is synthesized by heating powdered MGS with anhydrous hydrogen chloride (HCl) at around 300 °C in a fluidized-bed reactor, (7.10.5).
$\text{Si(solid_ + 3 HCl(gas)} \xrightleftharpoons[\text{>900 °C}]{\text{ca. 300 °C}} \text{SiHCl}_3\text{(vapor) + H}_2\text{(gas)} \nonumber$
Since the reaction is actually an equilibrium and the formation of SiHCl3 highly exothermic, efficient removal of generated heat is essential to assure a maximum yield of SiHCl3. While the stoichiometric reaction is that shown in (7.10.5), a mixture of chlorinated silanes is actually prepared which must be separated by fractional distillation, along with the chlorides of any impurities. In particular iron, aluminum, and boron are removed as FeCl3 (b.p. = 316 °C), AlCl3 (m.p. = 190 °C subl.), and BCl3 (b.p. = 12.65 °C), respectively. Fractional distillation of SiHCl3 from these impurity halides result in greatly increased purity with a concentration of electrically active impurities of less than 1 ppb.
EGS is prepared from purified SiHCl3 in a chemical vapor deposition (CVD) process similar to the epitaxial growth of Si. The high-purity SiHCl3 is vaporized, diluted with high-purity hydrogen, and introduced into the Seimens deposition reactor, shown schematically in Figure $2$. Within the reactor, thin silicon rods called slim rods (ca. 4 mm diameter) are supported by graphite electrodes. Resistance heating of the slim rods causes the decomposition of the SiHCl3 to yield silicon, as described by the reverse reaction shown in (7.10.5).
The shift in the equilibrium from forming SiHCl3 from Si at low temperature, to forming Si from SiHCl3 at high temperature is as a consequence of the temperature dependence, (7.10.6), of the equilibrium constant, (7.10.7) where ρ = partial pressure, for (7.10.5). Since the formation of SiHCl3 is exothermic, i.e., ΔH < 0, an increase in the temperature causes the partial pressure of SiHCl3 to decrease. Thus, the Siemens process is typically run at ca. 1100 °C, while the reverse fluidized bed process is carried out at 300 °C.
$\text{lnK}_{\text{p}} \text{ = } \dfrac{ \text{-}\Delta\text{H}}{\text{RT}} \nonumber$
$\text{K}_{\text{p}} \text{ = } \dfrac{^{\rho}\text{SiHCl}_3 \text{ } ^{\rho}\text{H}_2}{^{\rho}\text{HCl}} \nonumber$
The slim rods act as a nucleation point for the deposition of silicon, and the resulting polycrystalline rod consists of columnar grains of silicon (polysilicon) grown perpendicular to the rod axis. Growth occurs at less than 1 mm per hour, and after deposition for 200 to 300 hours high-purity (EGS) polysilicon rods of 150 - 200 mm in diameter are produced. For subsequent float-zone refining the polysilicon EGS rods are cut into long cylindrical rods. Alternatively, the as-formed polysilicon rods are broken into chunks for single crystal growth processes, for example Czochralski melt growth.
In addition to the formation of silicon, the HCl coproduct reacts with the SiHCl3 reactant to form silicon tetrachloride (SiCl4) and hydrogen as major byproducts of the process, (7.10.8). This reaction represents a major disadvantage with the Seimens process: poor efficiency of silicon and chlorine consumption. Typically, only 30% of the silicon introduced into CVD reactor is converted into high-purity polysilicon.
$\text{HCl + SiHCl}_3 \rightarrow \text{SiCl}_4\text{ + H}_2 \nonumber$
In order to improve efficiency the HCl, SiCl4, H2, and unreacted SiHCl3 are separated and recovered for recycling. Figure $3$ illustrates the entire chlorosilane process starting with MGS and including the recycling of the reaction byproducts to achieve high overall process efficiency. As a consequence, the production cost of high-purity EGS depends on the commercial usefulness of the byproduct, SiCl4. Additional disadvantages of the Seimens process are derived from its relatively small batch size, slow growth rate, and high power consumption. These issues have lead to the investigation of alternative cost efficient routes to EGS.
Silane process
An alternative process for the production of EGS that has begun to receive commercial attention is the pyrolysis of silane (SiH4). The advantages of producing EGS from SiH4 instead of SiHCl3 are potentially lower costs associated with lower reaction temperatures, and less harmful byproducts. Silane decomposes < 900 °C to give silicon and hydrogen, (7.10.9).
$\text{SiH}_4\text{(vapor)} \rightarrow \text{Si(solid) + 2 H}_2\text{(gas)} \nonumber$
Silane may be prepared by a number of routes, each having advantages with respect to purity and production cost. The simplest process involves the direct reaction of MGS powders with magnesium at 500 °C in a hydrogen atmosphere, to form magnesium silicide (Mg2Si). The magnesium silicide is then reacted with ammonium chloride in liquid ammonia below 0 °C, (7.10.10).
$\text{Mg}_2\text{Si + 4 NH}_4\text{Cl} \rightarrow \text{SiH}_4\text{ + 2 MgCl}_2\text{ + 5 NH}_3 \nonumber$
This process is ideally suited to the removal of boron impurities (a p-type dopant in Si), because the diborane (B2H6) produced during the reaction forms the Lewis acid-base complex, H3B(NH3), whose volatility is sufficiently lower than SiH4, allowing for the purification of the latter. It is possible to prepare EGS with a boron content of ≤ 20 ppt using SiH4 synthesized in this manner. However, phosphorus (another dopant) in the form of PH3 may be present as a contaminant requiring subsequent purification of the SiH4.
Alternative routes to SiH4 involve the chemical reduction of SiCl4 by either lithium hydride, (7.10.11), lithium aluminum hydride, (7.10.12), or via hydrogenation in the presence of elemental silicon, (7.10.13) - (7.10.16). The hydride reduction reactions may be carried-out on relatively large scales (ca. 50 kg), but only batch processes. In contrast, Union Carbide has adapted the hydrogenation to a continuous process, involving disproportionation reactions of chlorosilanes, (7.10.14) - (7.10.16), and the fractional distillation of silane, Table $3$.
$\text{SiCl}_4\text{ + LiH} \rightarrow \text{SiH}_4\text{ + 4 LiCl} \nonumber$
$\text{SiCl}_4\text{ + LiAlH}_4 \rightarrow \text {SiH}_4 \text{ + LiCl + AlCl}_4 \nonumber$
$\text{SiCl}_4\text{ + 2 H}_2\text{ + Si(98%)} \rightarrow \text{4 SiHCl}_3 \nonumber$
$\text{2 SiHCl}_3 \rightarrow \text{SiH}_2\text{Cl}_2\text{ + 2 SiHCl}_3 \nonumber$
$\text{3 SiH}_2\text{Cl}_2 \rightarrow \text{SiH}_2\text{Cl + 2 SiHCl}_3 \nonumber$
$\text{2 SiH}_3\text{Cl} \rightarrow \text{SiH}_4\text{ + SiH}_2\text{Cl}_2 \nonumber$
Pyrolysis of silane on resistively heated polysilicon filaments at 700 - 800 °C yields polycrystalline EGS. As noted above, the EGS formed has remarkably low boron impurities compared with material prepared from trichlorosilane. Moreover, the resulting EGS is less contaminated with transition metals from the reactor container because SiH4 decomposition does not cause as much of a corrosion problem as is observed with halide precursor compounds.
Granular polysilicon deposition
Both the chlorosilane (Seimens) and silane processes result in the formation of rods of EGS. However, there has been increased interest in the formation of granular polycrystalline EGS. This process was developed in 1980’s, and relies on the decomposition of SiH4 in a fluidized-bed deposition reactor to produce free-flowing granular polysilicon.
Tiny silicon particles are fluidized in a SiH4/H2 flow, and act as seed crystal onto which polysilicon deposits to form free-flowing spherical particles. The size distribution of the particles thus formed is over the range from 0.1 to 1.5 mm in diameter with an average particle size of 0.7 mm. The fluidized-bed seed particles are originally made by grinding EGS in a ball (or hammer) mill and leaching the product with acid, hydrogen peroxide, and water. This process is time-consuming and costly, and tended to introduce undesirable impurities from the metal grinders. In a new method, large EGS particles are fired at each other by a high-speed stream of inert gas and the collision breaks them down into particles of suitable size for a fluidized bed. This process has the main advantage that it introduces no foreign materials and requires no leaching or other post purification.
The fluidized-bed reactors are much more efficient than traditional rod reactors as a consequence of the greater surface area available during CVD growth of silicon. It has been suggested that fluidized-bed reactors require 1/5 to 1/10 the energy, and half the capital cost of the traditional process. The quality of fluidized-bed polysilicon has proven to be equivalent to polysilicon produced by the conventional methods. Moreover, granular EGS in a free-flowing form, and with high bulk density, enables crystal growers to obtain the high, reproducible production yields out of each crystal growth run. For example, in the Czochralski crystal growth process, crucibles can be quickly and easily filled to uniform loading with granular EGS, which typically exceed those of randomly stacked polysilicon chunks produced by the Siemens silane process.
Zone refining
The technique of zone refining is used to purify solid materials and is commonly employed in metallurgical refining. In the case of silicon may be used to obtain the desired ultimate purity of EGS, which has already been purified by chemical processes. Zone refining was invented by Pfann, and makes use of the fact that the equilibrium solubility of any impurity (e.g., Al) is different in the solid and liquid phases of a material (e.g., Si). For the dilute solutions, as is observed in EGS silicon, an equilibrium segregation coefficient (k0) is defined by k0 = Cs/Cl, where Cs and Cl are the equilibrium concentrations of the impurity in the solid and liquid near the interface, respectively.
If k0 is less than 1 then the impurities are left in the melt as the molten zone is moved along the material. In a practical sense a molten zone is established in a solid rod. The zone is then moved along the rod from left to right. If k0 < 1 then the frozen part left on the trailing edge of the moving molten zone will be purer than the material that melts in on the right-side leading edge of the moving molten zone. Consequently the solid to the left of the molten zone is purer than the solid on the right. At the completion of the first pass the impurities become concentrated to the right of the solid sample. Repetition of the process allows for purification to exceptionally high levels. Table $4$ lists the equilibrium segregation coefficients for common impurity and dopant elements in silicon; it should be noted that they are all less than 1.
Table $4$: Segregation coefficients for common impurity and dopant elements in silicon.
Element k0 Element k0
aluminum 0.002 iron 8 x 10-6
boron 0.8 oxygen 0.25
carbon 0.07 phosphorus 0.35
copper 4 x 10-6 antimony 0.023
Bibliography
• K. G. Baraclough, K. G., in The Chemistry of the Semiconductor Industry, Eds. S. J. Moss and A. Ledwith, Blackie and Sons, Glasgow, Scotland (1987).
• L. D. Crossman and J. A. Baker, Semiconductor Silicon 1977, Electrochem. Soc., Princeton, New Jersey (1977).
• W. C. O’Mara, Ed. Handbook of Semiconductor Silicon Technology, Noyes Pub., New Jersey (1990).
• W. G. Pfann, Zone Melting, John Wiley & Sons, New York, (1966).
• F. Shimura, Semiconductor Silicon Crystal Technology, Academic Press (1989). | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/12%3A_Solids_and_Modern_Materials/12.04%3A_Materials_for_Electronics/12.4.02%3A_Silicon_Devices.txt |
can be widely defined as an . An can be considered to have a random arrangement of , such as observed in a gas, but more realistically can considered to only lack long-range order such as those found in crystalline solids. is clear (optically transparent) silica which is composed largely of silicon dioxide (SiO). The definition of does not restrict either the composition or the optical properties of the material, implying a wide variety of different materials that are considered . In fact, theoretically, any crystalline solid that can be brought to a liquid state, can be forced into an state through rapid solidification via extraordinary cooling rates. This is easily observable by recognizing that quartz, a very common crystalline solid, has the same composition as silica (SiO) but was cooled slowly enough to form long-range order. Non-silica glasses, in particular metallic glasses, can obtain unique electric, optical, or thermal properties from their crystalline counterparts through glassification. Non-metallic glasses can obtain similarly unique properties by adjusting elemental compositions and introducing dopants. has been desirable for its optical properties since silica is transparent in the visible spectrum. Although there are crystalline materials that are similarly transparent (quartz), they have several properties which make them undesirable as optical media in many cases; though each grain may be transparent, grain boundaries reflect and/or scatter light in poly crystalline materials; unless cut along specific planes, the faces of crystals are forced to conform to a rigid geometric order which may also scatter light. Silica is not restricted by these constraints and has properties making it even more desirable both in terms of clarity and malleability; being unrestricted by a defined internal structure, the surface of is molecularly smooth, bound only by , even along curved faces. This is particularly important since most optical instruments (microscopes, telescopes, and eyeglasses) require smooth curved surfaces. isn’t the only the fits the profile for optical functions. Acrylic , poly(methyl methacrylate), is a polymer, fitting the description of by being an . Acrylic has a very similar refractive index to silica (~1.5) and is physically lighter, softer, and more shatter-resistant than . Polycarbonates is another class of optically transparent polymers. It has even more desirable physical properties than acrylic in terms of strength and impact resistance. , in particular the index of refraction, can be modified by doping the . Doping the with low density materials such as boron can lower the index of refraction. Similarly, doping the material with higher density dopants, such as oxides of lead, titanium, barium or zirconium, can drastically increase the index of refraction. Glasses made or doped with germanium or phosphates are vitally important in the field of. filters are often made of chalcogenide , composed of two non-oxygen group 16 elements (sulfur, selenium, tellurium) and one group 14 element (silicon, tin, lead) or group 15 element (phosphorus, arsenic, antimony, bismuth) in its most simple form. Often the size of the allow for leniency to create more complex amorphous chalcogenide glasses involving several different elements in compositions that only roughly match the two to one ratio of silica. These materials transmit only in the IR or near IR range, appearing black or faintly blue, while still having similar malleability of silica . Example IR filter chalcogenide glasses include AsSe, GeAsSe, and GeAsSeTe. is a well-known insulator, having a resistivity on the order of 10 ohm m. is especially desirable in the field of for its insulating properties; in device fabrication, is deposited between metals or as very thin insulators. Doped , such as phosphosilicate (phosphorus doped) or borophosphosilicate (boron and phosphorus doped) are often used instead of pure silica for their lower melting temperatures and increased planarization (forming smooth flat surfaces). The addition of fluorine, a highly electronegative element, into the can lower the and thus the dielectric constant of the , making it more desirable for integrated circuits. The addition of large , such as lead (II) oxide, can reduce the mobility of other significantly increasing the resistivity of the material. It’s possible to give silica a level of electrical conductivity by dissolving small alkali metal into the , which have high mobility’s at increasing temperatures. is commonly considered to be very susceptible to thermal shock and breaks or cracks easily when suddenly changing temperatures. This is true for the cheapest and most common form of which is soda-lime-silica , which contains roughly 30% sodium oxide (NaO), lime (CaO), magnesia (MgO) and alumina (AlO). Soda-lime has a of thermal expansion of 93.5E-7 cm/cm.°C, which describes the relative increase in size per degree Celsius change in temperature. Pure or nearly pure (~96%) silica has a small of thermal expansion of 7.5E-7 cm/cm.°C due to the homogeneity of the solid. Silica also has a significantly higher melting temperature, combined with the purity requirements, this makes it more expensive to produce. Borosilicate (~13% BO) is a very common form of used in cookware for its thermal shock resistance, having a of 35E-7 cm/cm.°C. Unlike pure silica , borosilicate is cheaper to produce, having a lower melting temperature and having less stringent purity requirements. Corning has developed glasses with ultra-low thermal expansion coefficients with values on the order of 10 cm/cm.°C. does play a huge role in thermal insulation in the form of fiber and wool. wool involves the production of very thin strands on soda-lime to form a low density packing material. wool can achieve higher specific heats than either or water on the order of 7 J/g.K. requires massive cooling rates, on the order of 10 K/s. These cooling rates can be achieved by a variety of methods including:
12.06: Materials for Nanotechnology
Inorganic nanomaterials, (e.g. quantum dots, nanowires and nanorods) because of their interesting optical and electrical properties, could be used in optoelectronics. Furthermore, the optical and electronic properties of nanomaterials which depend on their size and shape can be tuned via synthetic techniques. There are the possibilities to use those materials in organic material based optoelectronic devices such as Organic solar cells, OLEDs etc. The operating principles of such devices are governed by photoinduced processes like electron transfer and energy transfer. The performance of the devices depends on the efficiency of the photoinduced process responsible for their functioning. Therefore, better understanding of those photoinduced processes in organic/inorganic nanomaterial composite systems is necessary in order to use them in organic optoelectronic devices.
Nanoparticles or nanocrystals made of metals, semiconductors, or oxides are of particular interest for their mechanical, electrical, magnetic, optical, chemical and other properties. Nanoparticles have been used as quantum dots and as chemical catalysts such as nanomaterial-based catalysts. Recently, a range of nanoparticles are extensively investigated for biomedical applications including tissue engineering, drug delivery, biosensor.
Nanoparticles are of great scientific interest as they are effectively a bridge between bulk materials and atomic or molecular structures. A bulk material should have constant physical properties regardless of its size, but at the nano-scale this is often not the case. Size-dependent properties are observed such as quantum confinement in semiconductor particles, surface plasmon resonance in some metal particles and superparamagnetism in magnetic materials.
Nanoparticles exhibit a number of special properties relative to bulk material. For example, the bending of bulk copper (wire, ribbon, etc.) occurs with movement of copper atoms/clusters at about the 50 nm scale. Copper nanoparticles smaller than 50 nm are considered super hard materials that do not exhibit the same malleability and ductility as bulk copper. The change in properties is not always desirable. Ferroelectric materials smaller than 10 nm can switch their magnetisation direction using room temperature thermal energy, thus making them useless for memory storage. Suspensions of nanoparticles are possible because the interaction of the particle surface with the solvent is strong enough to overcome differences in density, which usually result in a material either sinking or floating in a liquid. Nanoparticles often have unexpected visual properties because they are small enough to confine their electrons and produce quantum effects. For example gold nanoparticles appear deep red to black in solution.
The often very high surface area to volume ratio of nanoparticles provides a tremendous driving force for diffusion, especially at elevated temperatures. Sintering is possible at lower temperatures and over shorter durations than for larger particles. This theoretically does not affect the density of the final product, though flow difficulties and the tendency of nanoparticles to agglomerate do complicate matters. The surface effects of nanoparticles also reduces the incipient melting temperature.
• Wikipedia | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/12%3A_Solids_and_Modern_Materials/12.05%3A_Materials_for_Optics.txt |
In all solutions, whether gaseous, liquid, or solid, the substance present in the greatest amount is the solvent, and the substance or substances present in lesser amounts are the solute(s). The solute does not have to be in the same physical state as the solvent, but the physical state of the solvent usually determines the state of the solution. As long as the solute and solvent combine to give a homogeneous solution, the solute is said to be soluble in the solvent.
• 13.1: The Solution Process
Solutions are homogeneous mixtures of two or more substances whose components are uniformly distributed on a microscopic scale. The component present in the greatest amount is the solvent, and the components present in lesser amounts are the solute(s). The formation of a solution from a solute and a solvent is a physical process, not a chemical one. Substances that are miscible, such as gases, form a single phase in all proportions when mixed. Substances that form separate phases are immiscible.
• 13.2: Saturated Solutions and Solubility
The solubility of a substance is the maximum amount of a solute that can dissolve in a given quantity of solvent; it depends on the chemical nature of both the solute and the solvent and on the temperature and pressure. When a solution contains the maximum amount of solute that can dissolve under a given set of conditions, it is a saturated solution. Otherwise, it is unsaturated. Supersaturated solutions, which contain more dissolved solute than allowed under particular conditions, are unstable.
• 13.3: Factors Affecting Solubility
The solubility of most substances depends strongly on the temperature and, in the case of gases, on the pressure. The solubility of most solid or liquid solutes increases with increasing temperature. The components of a mixture can often be separated using fractional crystallization, which separates compounds according to their solubilities. The solubility of a gas decreases with increasing temperature. Henry’s law describes the relationship between the pressure and the solubility of a gas.
• 13.4: Ways of Expressing Concentration
Different units are used to express the concentrations of a solution depending on the application. The concentration of a solution is the quantity of solute in a given quantity of solution. It can be expressed in several ways.
• 13.5: Colligative Properties
Colligative properties of a solution depend on only the total number of dissolved particles in solution, not on their chemical identity. Colligative properties include vapor pressure, boiling point, freezing point, and osmotic pressure. The addition of a nonvolatile solute (one without a measurable vapor pressure) decreases the vapor pressure of the solvent. The vapor pressure of the solution is proportional to the mole fraction of solvent in the solution, a relationship known as Raoult’s law.
• 13.6: Colloids
A colloid can be classified as a sol, a dispersion of solid particles in a liquid or solid; a gel, a semisolid sol in which all of the liquid phase has been absorbed by the solid particles; an aerosol, a dispersion of solid or liquid particles in a gas; or an emulsion, a dispersion of one liquid phase in another. A colloid can be distinguished from a true solution by its ability to scatter a beam of light, known as the Tyndall effect.
• 13.E: Properties of Solutions (Exercises)
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al.
• 13.S: Properties of Solutions (Summary)
A summary of the key concepts in this chapter of the Textmap created for "Chemistry: The Central Science" by Brown et al.
Thumbnail: Nile red solution. Image use with permission (CC BY-SA 3.0; Armin Kübelbeck).
13: Properties of Solutions
Learning Objectives
• To understand how enthalpy and entropy changes affect solution formation.
• To use the magnitude of the changes in both enthalpy and entropy to predict whether a given solute–solvent combination will spontaneously form a solution.
In all solutions, whether gaseous, liquid, or solid, the substance present in the greatest amount is the solvent, and the substance or substances present in lesser amounts are the solute(s). The solute does not have to be in the same physical state as the solvent, but the physical state of the solvent usually determines the state of the solution. As long as the solute and solvent combine to give a homogeneous solution, the solute is said to be soluble in the solvent. Table $1$ lists some common examples of gaseous, liquid, and solid solutions and identifies the physical states of the solute and solvent in each.
Table $1$: Types of Solutions
Solution Solute Solvent Examples
gas gas gas air, natural gas
liquid gas liquid seltzer water ($CO_2$ gas in water)
liquid liquid liquid alcoholic beverage (ethanol in water), gasoline
liquid solid liquid tea, salt water
solid gas solid $H_2$ in Pd (used for $H_2$ storage)
solid solid liquid mercury in silver or gold (amalgam often used in dentistry)
solid solid solid alloys and other "solid solutions"
Forming a Solution
The formation of a solution from a solute and a solvent is a physical process, not a chemical one. That is, both solute and solvent can be recovered in chemically unchanged forms using appropriate separation methods. For example, solid zinc nitrate dissolves in water to form an aqueous solution of zinc nitrate:
$\ce{Zn(NO3)2(s) + H2O(l) \rightarrow Zn^{2+}(aq) + 2NO^{-}3(aq)} \label{13.1.1}$
Because $Zn(NO_3)_2$ can be recovered easily by evaporating the water, this is a physical process. In contrast, metallic zinc appears to dissolve in aqueous hydrochloric acid. In fact, the two substances undergo a chemical reaction to form an aqueous solution of zinc chloride with evolution of hydrogen gas:
$\ce{ Zn(s) + 2H^{+}(aq) + 2Cl^{-}(aq) \rightarrow Zn^{2+}(aq) + 2Cl^{-}(aq) + H2(g)} \label{13.1.2}$
When the solution evaporates, we do not recover metallic zinc, so we cannot say that metallic zinc is soluble in aqueous hydrochloric acid because it is chemically transformed when it dissolves. The dissolution of a solute in a solvent to form a solution does not involve a chemical transformation (that it is a physical change).
Dissolution of a solute in a solvent to form a solution does not involve a chemical transformation.
Substances that form a single homogeneous phase in all proportions are said to be completely miscible in one another. Ethanol and water are miscible, just as mixtures of gases are miscible. If two substances are essentially insoluble in each other, such as oil and water, they are immiscible. Examples of gaseous solutions that we have already discussed include Earth’s atmosphere.
The Role of Enthalpy in Solution Formation
Energy is required to overcome the intermolecular interactions in a solute, which can be supplied only by the new interactions that occur in the solution, when each solute particle is surrounded by particles of the solvent in a process called solvation (or hydration when the solvent is water). Thus all of the solute–solute interactions and many of the solvent–solvent interactions must be disrupted for a solution to form. In this section, we describe the role of enthalpy in this process.
Because enthalpy is a state function, we can use a thermochemical cycle to analyze the energetics of solution formation. The process occurs in three discrete steps, indicated by $ΔH_1$, $ΔH_2$, and $ΔH_3$ in Figure $2$. The overall enthalpy change in the formation of the solution ($\Delta H_{soln}$) is the sum of the enthalpy changes in the three steps:
$\Delta H_{soln} = \Delta H_1 + \Delta H_2 + \Delta H_3 \label{13.1.3}$
When a solvent is added to a solution, steps 1 and 2 are both endothermic because energy is required to overcome the intermolecular interactions in the solvent ($\Delta H_1$) and the solute ($\Delta H_2$). Because $ΔH$ is positive for both steps 1 and 2, the solute–solvent interactions ($\Delta H_3$) must be stronger than the solute–solute and solvent–solvent interactions they replace in order for the dissolution process to be exothermic ($\Delta H_{soln} < 0$). When the solute is an ionic solid, $ΔH_2$ corresponds to the lattice energy that must be overcome to form a solution. The higher the charge of the ions in an ionic solid, the higher the lattice energy. Consequently, solids that have very high lattice energies, such as $MgO$ (−3791 kJ/mol), are generally insoluble in all solvents.
A positive value for $ΔH_{soln}$ does not mean that a solution will not form. Whether a given process, including formation of a solution, occurs spontaneously depends on whether the total energy of the system is lowered as a result. Enthalpy is only one of the contributing factors. A high $ΔH_{soln}$ is usually an indication that the substance is not very soluble. Instant cold packs used to treat athletic injuries, for example, take advantage of the large positive $ΔH_{soln}$ of ammonium nitrate during dissolution (+25.7 kJ/mol), which produces temperatures less than 0°C (Figure $3$).
Entropy and Solution Formation
The enthalpy change that accompanies a process is important because processes that release substantial amounts of energy tend to occur spontaneously. A second property of any system, its entropy, is also important in helping us determine whether a given process occurs spontaneously. We will discuss entropy in more detail elsewhere, but for now we can state that entropy ($S$) is a thermodynamic property of all substances that is proportional to their degree of disorder. A perfect crystal at 0 K, whose atoms are regularly arranged in a perfect lattice and are motionless, has an entropy of zero. In contrast, gases have large positive entropies because their molecules are highly disordered and in constant motion at high speeds.
The formation of a solution disperses molecules, atoms, or ions of one kind throughout a second substance, which generally increases the disorder and results in an increase in the entropy of the system. Thus entropic factors almost always favor formation of a solution. In contrast, a change in enthalpy may or may not favor solution formation. The London dispersion forces that hold cyclohexane and n-hexane together in pure liquids, for example, are similar in nature and strength. Consequently, $ΔH_{soln}$ should be approximately zero, as is observed experimentally. Mixing equal amounts of the two liquids, however, produces a solution in which the n-hexane and cyclohexane molecules are uniformly distributed over approximately twice the initial volume. In this case, the driving force for solution formation is not a negative $ΔH_{soln}$ but rather the increase in entropy due to the increased disorder in the mixture. All spontaneous processes with $ΔH \ge 0$ are characterized by an increase in entropy. In other cases, such as mixing oil with water, salt with gasoline, or sugar with hexane, the enthalpy of solution is large and positive, and the increase in entropy resulting from solution formation is not enough to overcome it. Thus in these cases a solution does not form.
All spontaneous processes with ΔH ≥ 0 are characterized by an increase in entropy.
Table $2$ summarizes how enthalpic factors affect solution formation for four general cases. The column on the far right uses the relative magnitudes of the enthalpic contributions to predict whether a solution will form from each of the four. Keep in mind that in each case entropy favors solution formation. In two of the cases the enthalpy of solution is expected to be relatively small and can be either positive or negative. Thus the entropic contribution dominates, and we expect a solution to form readily. In the other two cases the enthalpy of solution is expected to be large and positive. The entropic contribution, though favorable, is usually too small to overcome the unfavorable enthalpy term. Hence we expect that a solution will not form readily.
Table $2$: Relative Changes in Enthalpies for Different Solute–Solvent Combinations*
$ΔH_1$ (separation of solvent molecules) $ΔH_2$ (separation of solute particles) $ΔH_3$ (solute–solvent interactions) $ΔH_{soln}$ ($ΔH_1$ + $ΔH_2$ +$ΔH_3$) Result of Mixing Solute and Solvent†
large; positive large; positive large; negative small; positive or negative solution will usually form
small; positive large; positive small; negative large; positive solution will not form
large; positive small; positive small; negative large; positive solution will not form
small; positive small; positive small; negative small; positive or negative solution will usually form
*$ΔH_1$, $ΔH_2$, and $ΔH_3$ refer to the processes indicated in the thermochemical cycle shown in Figure $2$.
In all four cases, entropy increases.
In contrast to liquid solutions, the intermolecular interactions in gases are weak (they are considered to be nonexistent in ideal gases). Hence mixing gases is usually a thermally neutral process ($ΔH_{soln} \approx 0$), and the entropic factor due to the increase in disorder is dominant (Figure $4$). Consequently, all gases dissolve readily in one another in all proportions to form solutions.
Example $1$
Considering $\ce{LiCl}$, benzoic acid ($\ce{C6H5CO2H}$), and naphthalene, which will be most soluble and which will be least soluble in water?
Given: three compounds
Asked for: relative solubilities in water
Strategy: Assess the relative magnitude of the enthalpy change for each step in the process shown in Figure $2$. Then use Table $2$ to predict the solubility of each compound in water and arrange them in order of decreasing solubility.
Solution:
The first substance, $\ce{LiCl}$, is an ionic compound, so a great deal of energy is required to separate its anions and cations and overcome the lattice energy (ΔH2 is far greater than zero in Equation $\ref{13.1.1}$). Because water is a polar substance, the interactions between both Li+ and Cl ions and water should be favorable and strong. Thus we expect $ΔH_3$ to be far less than zero, making LiCl soluble in water. In contrast, naphthalene is a nonpolar compound, with only London dispersion forces holding the molecules together in the solid state. We therefore expect $ΔH_2$ to be small and positive. We also expect the interaction between polar water molecules and nonpolar naphthalene molecules to be weak $ΔH_3 \approx 0$. Hence we do not expect naphthalene to be very soluble in water, if at all. Benzoic acid has a polar carboxylic acid group and a nonpolar aromatic ring. We therefore expect that the energy required to separate solute molecules (ΔH2) will be greater than for naphthalene and less than for LiCl. The strength of the interaction of benzoic acid with water should also be intermediate between those of LiCl and naphthalene. Hence benzoic acid is expected to be more soluble in water than naphthalene but less soluble than $\ce{LiCl}$. We thus predict $\ce{LiCl}$ to be the most soluble in water and naphthalene to be the least soluble.
Exercise $1$
Considering ammonium chloride, cyclohexane, and ethylene glycol ($HOCH_2CH_2OH$), which will be most soluble and which will be least soluble in benzene?
Answer
The most soluble is cyclohexane; the least soluble is ammonium chloride.
Summary
Solutions are homogeneous mixtures of two or more substances whose components are uniformly distributed on a microscopic scale. The component present in the greatest amount is the solvent, and the components present in lesser amounts are the solute(s). The formation of a solution from a solute and a solvent is a physical process, not a chemical one. Substances that are miscible, such as gases, form a single phase in all proportions when mixed. Substances that form separate phases are immiscible. Solvation is the process in which solute particles are surrounded by solvent molecules. When the solvent is water, the process is called hydration. The overall enthalpy change that accompanies the formation of a solution, $ΔH_{soln}$, is the sum of the enthalpy change for breaking the intermolecular interactions in both the solvent and the solute and the enthalpy change for the formation of new solute–solvent interactions. Exothermic ($ΔH_{soln} < 0$) processes favor solution formation. In addition, the change in entropy, the degree of disorder of the system, must be considered when predicting whether a solution will form. An increase in entropy (a decrease in order) favors dissolution. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/13%3A_Properties_of_Solutions/13.01%3A_The_Solution_Process.txt |
Learning Objectives
• To understand the relationship between solubility and molecular structure.
• To demonstrate how the strength of intramolecular bonding determines the solubility of a solute in a given solvent.
When a solute dissolves, its individual atoms, molecules, or ions interact with the solvent, become solvated, and are able to diffuse independently throughout the solution (Figure $\PageIndex{1a}$). This is not, however, a unidirectional process. If the molecule or ion happens to collide with the surface of a particle of the undissolved solute, it may adhere to the particle in a process called crystallization. Dissolution and crystallization continue as long as excess solid is present, resulting in a dynamic equilibrium analogous to the equilibrium that maintains the vapor pressure of a liquid. We can represent these opposing processes as follows:
$solute + solvent \underset{crystallization}{\stackrel{dissolution}{\longrightleftharpoons}} solution \nonumber$
Although the terms precipitation and crystallization are both used to describe the separation of solid solute from a solution, crystallization refers to the formation of a solid with a well-defined crystalline structure, whereas precipitation refers to the formation of any solid phase, often one with very small particles.
Factors Affecting Solubility
The maximum amount of a solute that can dissolve in a solvent at a specified temperature and pressure is its solubility. Solubility is often expressed as the mass of solute per volume (g/L) or mass of solute per mass of solvent (g/g), or as the moles of solute per volume (mol/L). Even for very soluble substances, however, there is usually a limit to how much solute can dissolve in a given quantity of solvent. In general, the solubility of a substance depends on not only the energetic factors we have discussed but also the temperature and, for gases, the pressure. At 20°C, for example, 177 g of NaI, 91.2 g of NaBr, 35.9 g of NaCl, and only 4.1 g of NaF dissolve in 100 g of water. At 70°C, however, the solubilities increase to 295 g of NaI, 119 g of NaBr, 37.5 g of NaCl, and 4.8 g of NaF. As you learned in Chapter 12, the lattice energies of the sodium halides increase from NaI to NaF. The fact that the solubilities decrease as the lattice energy increases suggests that the $ΔH_2$ term in Figure 13.1 dominates for this series of compounds.
A solution with the maximum possible amount of solute is saturated. If a solution contains less than the maximum amount of solute, it is unsaturated. When a solution is saturated and excess solute is present, the rate of dissolution is exactly equal to the rate of crystallization (Figure $\PageIndex{1b}$). Using the value just stated, a saturated aqueous solution of NaCl, for example, contains 35.9 g of NaCl per 100 mL of water at 20°C. We can prepare a homogeneous saturated solution by adding excess solute (in this case, greater than 35.9 g of NaCl) to the solvent (water), stirring until the maximum possible amount of solute has dissolved, and then removing undissolved solute by filtration.
The solubility of most solids increases with increasing temperature.
Because the solubility of most solids increases with increasing temperature, a saturated solution that was prepared at a higher temperature usually contains more dissolved solute than it would contain at a lower temperature. When the solution is cooled, it can therefore become supersaturated (Figure $\PageIndex{1c}$). Like a supercooled or superheated liquid, a supersaturated solution is unstable. Consequently, adding a small particle of the solute, a seed crystal, will usually cause the excess solute to rapidly precipitate or crystallize, sometimes with spectacular results. The rate of crystallization in Equation $\ref{13.2.1}$ is greater than the rate of dissolution, so crystals or a precipitate form (Figure $\PageIndex{1d}$). In contrast, adding a seed crystal to a saturated solution reestablishes the dynamic equilibrium, and the net quantity of dissolved solute no longer changes.
Video $1$: hot ice (sodium acetate) beautiful science experiment. watered-down sodium acetate trihydrate. Needle crystal is truly wonderful structures
Because crystallization is the reverse of dissolution, a substance that requires an input of heat to form a solution ($ΔH_{soln} > 0$) releases that heat when it crystallizes from solution ($ΔH_{crys} < 0$). The amount of heat released is proportional to the amount of solute that exceeds its solubility. Two substances that have a positive enthalpy of solution are sodium thiosulfate ($Na_2S_2O_3$) and sodium acetate ($CH_3CO_2Na$), both of which are used in commercial hot packs, small bags of supersaturated solutions used to warm hands (see Figure 13.1.3).
Interactions in Liquid Solutions
The interactions that determine the solubility of a substance in a liquid depend largely on the chemical nature of the solute (such as whether it is ionic or molecular) rather than on its physical state (solid, liquid, or gas). We will first describe the general case of forming a solution of a molecular species in a liquid solvent and then describe the formation of a solution of an ionic compound.
Solutions of Molecular Substances in Liquids
The London dispersion forces, dipole–dipole interactions, and hydrogen bonds that hold molecules to other molecules are generally weak. Even so, energy is required to disrupt these interactions. As described in Section 13.1, unless some of that energy is recovered in the formation of new, favorable solute–solvent interactions, the increase in entropy on solution formation is not enough for a solution to form.
For solutions of gases in liquids, we can safely ignore the energy required to separate the solute molecules ($ΔH_2 = 0$) because the molecules are already separated. Thus we need to consider only the energy required to separate the solvent molecules ($ΔH_1$) and the energy released by new solute–solvent interactions ($ΔH_3$). Nonpolar gases such as $N_2$, $O_2$, and $Ar$ have no dipole moment and cannot engage in dipole–dipole interactions or hydrogen bonding. Consequently, the only way they can interact with a solvent is by means of London dispersion forces, which may be weaker than the solvent–solvent interactions in a polar solvent. It is not surprising, then, that nonpolar gases are most soluble in nonpolar solvents. In this case, $ΔH_1$ and $ΔH_3$ are both small and of similar magnitude. In contrast, for a solution of a nonpolar gas in a polar solvent, $ΔH_1$ is far greater than $ΔH_3$. As a result, nonpolar gases are less soluble in polar solvents than in nonpolar solvents. For example, the concentration of $N_2$ in a saturated solution of $N_2$ in water, a polar solvent, is only $7.07 \times 10^{-4}\; M$ compared with $4.5 \times 10^{-3}\; M$ for a saturated solution of $N_2$ in benzene, a nonpolar solvent.
The solubilities of nonpolar gases in water generally increase as the molecular mass of the gas increases, as shown in Table $1$. This is precisely the trend expected: as the gas molecules become larger, the strength of the solvent–solute interactions due to London dispersion forces increases, approaching the strength of the solvent–solvent interactions.
Table $1$: Solubilities of Selected Gases in Water at 20°C and 1 atm Pressure
Gas Solubility (M) × 10−4
He 3.90
Ne 4.65
Ar 15.2
Kr 27.9
Xe 50.2
H2 8.06
N2 7.07
CO 10.6
O2 13.9
N2O 281
CH4 15.5
Virtually all common organic liquids, whether polar or not, are miscible. The strengths of the intermolecular attractions are comparable; thus the enthalpy of solution is expected to be small ($ΔH_{soln} \approx 0$), and the increase in entropy drives the formation of a solution. If the predominant intermolecular interactions in two liquids are very different from one another, however, they may be immiscible. For example, organic liquids such as benzene, hexane, $CCl_4$, and $CS_2$ (S=C=S) are nonpolar and have no ability to act as hydrogen bond donors or acceptors with hydrogen-bonding solvents such as $H_2O$, $HF$, and $NH_3$; hence they are immiscible in these solvents. When shaken with water, they form separate phases or layers separated by an interface (Figure $2$), the region between the two layers.
Just because two liquids are immiscible, however, does not mean that they are completely insoluble in each other. For example, 188 mg of benzene dissolves in 100 mL of water at 23.5°C. Adding more benzene results in the separation of an upper layer consisting of benzene with a small amount of dissolved water (the solubility of water in benzene is only 178 mg/100 mL of benzene). The solubilities of simple alcohols in water are given in Table $2$.
Table $2$: Solubilities of Straight-Chain Organic Alcohols in Water at 20°C
Alcohol Solubility (mol/100 g of $H_2O$)
methanol completely miscible
ethanol completely miscible
n-propanol completely miscible
n-butanol 0.11
n-pentanol 0.030
n-hexanol 0.0058
n-heptanol 0.0008
Only the three lightest alcohols (methanol, ethanol, and n-propanol) are completely miscible with water. As the molecular mass of the alcohol increases, so does the proportion of hydrocarbon in the molecule. Correspondingly, the importance of hydrogen bonding and dipole–dipole interactions in the pure alcohol decreases, while the importance of London dispersion forces increases, which leads to progressively fewer favorable electrostatic interactions with water. Organic liquids such as acetone, ethanol, and tetrahydrofuran are sufficiently polar to be completely miscible with water yet sufficiently nonpolar to be completely miscible with all organic solvents.
The same principles govern the solubilities of molecular solids in liquids. For example, elemental sulfur is a solid consisting of cyclic $S_8$ molecules that have no dipole moment. Because the $S_8$ rings in solid sulfur are held to other rings by London dispersion forces, elemental sulfur is insoluble in water. It is, however, soluble in nonpolar solvents that have comparable London dispersion forces, such as $CS_2$ (23 g/100 mL). In contrast, glucose contains five –OH groups that can form hydrogen bonds. Consequently, glucose is very soluble in water (91 g/120 mL of water) but essentially insoluble in nonpolar solvents such as benzene. The structure of one isomer of glucose is shown here.
Low-molecular-mass hydrocarbons with highly electronegative and polarizable halogen atoms, such as chloroform ($CHCl_3$) and methylene chloride ($CH_2Cl_2$), have both significant dipole moments and relatively strong London dispersion forces. These hydrocarbons are therefore powerful solvents for a wide range of polar and nonpolar compounds. Naphthalene, which is nonpolar, and phenol ($C_6H_5OH$), which is polar, are very soluble in chloroform. In contrast, the solubility of ionic compounds is largely determined not by the polarity of the solvent but rather by its dielectric constant, a measure of its ability to separate ions in solution, as you will soon see.
Example $1$
Identify the most important solute–solvent interactions in each solution.
1. iodine in benzene solvent
2. aniline ($\ce{C_6H_5NH_2}$) in dichloromethane ($\ce{CH_2Cl_2}$) solvent
1. iodine in water solvent
Given: components of solutions
Asked for: predominant solute–solvent interactions
Strategy:
Identify all possible intermolecular interactions for both the solute and the solvent: London dispersion forces, dipole–dipole interactions, or hydrogen bonding. Determine which is likely to be the most important factor in solution formation.
Solution
1. Benzene and $\ce{I2}$ are both nonpolar molecules. The only possible attractive forces are London dispersion forces.
2. Aniline is a polar molecule with a dipole moment of 1.6 D and has an $\ce{–NH_2}$ group that can act as a hydrogen bond donor. Dichloromethane is also polar with a 1.5 D dipole moment, but it has no obvious hydrogen bond acceptor. Therefore, the most important interactions between aniline and $CH_2Cl_2$ are likely to be dipole-dipole interactions.
3. Water is a highly polar molecule that engages in extensive hydrogen bonding, whereas $I_2$ is a nonpolar molecule that cannot act as a hydrogen bond donor or acceptor. The slight solubility of $\ce{I_2}$ in water ($1.3 \times 10^{-3}\; mol/L$ at 25°C) is due to London dispersion forces.
Exercise $1$
Identify the most important interactions in each solution:
1. ethylene glycol ($HOCH_2CH_2OH$) in acetone
2. acetonitrile ($\ce{CH_3C≡N}$) in acetone
3. n-hexane in benzene
Answer a
hydrogen bonding
Answer b
London interactions
Answer c
London dispersion forces
Hydrophilic and Hydrophobic Solutes
A solute can be classified as hydrophilic (literally, “water loving”), meaning that it has an electrostatic attraction to water, or hydrophobic (“water fearing”), meaning that it repels water. A hydrophilic substance is polar and often contains O–H or N–H groups that can form hydrogen bonds to water. For example, glucose with its five O–H groups is hydrophilic. In contrast, a hydrophobic substance may be polar but usually contains C–H bonds that do not interact favorably with water, as is the case with naphthalene and n-octane. Hydrophilic substances tend to be very soluble in water and other strongly polar solvents, whereas hydrophobic substances are essentially insoluble in water and soluble in nonpolar solvents such as benzene and cyclohexane.
The difference between hydrophilic and hydrophobic substances has substantial consequences in biological systems. For example, vitamins can be classified as either fat soluble or water soluble. Fat-soluble vitamins, such as vitamin A, are mostly nonpolar, hydrophobic molecules. As a result, they tend to be absorbed into fatty tissues and stored there. In contrast, water-soluble vitamins, such as vitamin C, are polar, hydrophilic molecules that circulate in the blood and intracellular fluids, which are primarily aqueous. Water-soluble vitamins are therefore excreted much more rapidly from the body and must be replenished in our daily diet. A comparison of the chemical structures of vitamin A and vitamin C quickly reveals why one is hydrophobic and the other hydrophilic.
Because water-soluble vitamins are rapidly excreted, the risk of consuming them in excess is relatively small. Eating a dozen oranges a day is likely to make you tired of oranges long before you suffer any ill effects due to their high vitamin C content. In contrast, fat-soluble vitamins constitute a significant health hazard when consumed in large amounts. For example, the livers of polar bears and other large animals that live in cold climates contain large amounts of vitamin A, which have occasionally proven fatal to humans who have eaten them.
Example $2$
The following substances are essential components of the human diet:
Using what you know of hydrophilic and hydrophobic solutes, classify each as water soluble or fat soluble and predict which are likely to be required in the diet on a daily basis.
1. arginine
2. pantothenic acid
3. oleic acid
Given: chemical structures
Asked for: classification as water soluble or fat soluble; dietary requirement
Strategy:
Based on the structure of each compound, decide whether it is hydrophilic or hydrophobic. If it is hydrophilic, it is likely to be required on a daily basis.
Solution:
1. Arginine is a highly polar molecule with two positively charged groups and one negatively charged group, all of which can form hydrogen bonds with water. As a result, it is hydrophilic and required in our daily diet.
2. Although pantothenic acid contains a hydrophobic hydrocarbon portion, it also contains several polar functional groups ($\ce{–OH}$ and $\ce{–CO_2H}$) that should interact strongly with water. It is therefore likely to be water soluble and required in the diet. (In fact, pantothenic acid is almost always a component of multiple-vitamin tablets.)
3. Oleic acid is a hydrophobic molecule with a single polar group at one end. It should be fat soluble and not required daily.
Exercise $2$
These compounds are consumed by humans: caffeine, acetaminophen, and vitamin D. Identify each as primarily hydrophilic (water soluble) or hydrophobic (fat soluble), and predict whether each is likely to be excreted from the body rapidly or slowly.
Answer
Caffeine and acetaminophen are water soluble and rapidly excreted, whereas vitamin D is fat soluble and slowly excreted
Solid Solutions
Solutions are not limited to gases and liquids; solid solutions also exist. For example, amalgams, which are usually solids, are solutions of metals in liquid mercury. Because most metals are soluble in mercury, amalgams are used in gold mining, dentistry, and many other applications. A major difficulty when mining gold is separating very small particles of pure gold from tons of crushed rock. One way to accomplish this is to agitate a suspension of the crushed rock with liquid mercury, which dissolves the gold (as well as any metallic silver that might be present). The very dense liquid gold–mercury amalgam is then isolated and the mercury distilled away.
An alloy is a solid or liquid solution that consists of one or more elements in a metallic matrix. A solid alloy has a single homogeneous phase in which the crystal structure of the solvent remains unchanged by the presence of the solute. Thus the microstructure of the alloy is uniform throughout the sample. Examples are substitutional and interstitial alloys such as brass or solder. Liquid alloys include sodium/potassium and gold/mercury. In contrast, a partial alloy solution has two or more phases that can be homogeneous in the distribution of the components, but the microstructures of the two phases are not the same. As a liquid solution of lead and tin is cooled, for example, different crystalline phases form at different cooling temperatures. Alloys usually have properties that differ from those of the component elements.
Network solids such as diamond, graphite, and $\ce{SiO_2}$ are insoluble in all solvents with which they do not react chemically. The covalent bonds that hold the network or lattice together are simply too strong to be broken under normal conditions. They are certainly much stronger than any conceivable combination of intermolecular interactions that might occur in solution. Most metals are insoluble in virtually all solvents for the same reason: the delocalized metallic bonding is much stronger than any favorable metal atom–solvent interactions. Many metals react with solutions such as aqueous acids or bases to produce a solution. However, as we saw in Section 13.1, in these instances the metal undergoes a chemical transformation that cannot be reversed by simply removing the solvent.
Solids with very strong intermolecular bonding tend to be insoluble.
Solubilities of Ionic Substances in Liquids
Previously, you were introduced to guidelines for predicting the solubility of ionic compounds in water. Ionic substances are generally most soluble in polar solvents; the higher the lattice energy, the more polar the solvent must be to overcome the lattice energy and dissolve the substance. Because of its high polarity, water is the most common solvent for ionic compounds. Many ionic compounds are soluble in other polar solvents, however, such as liquid ammonia, liquid hydrogen fluoride, and methanol. Because all these solvents consist of molecules that have relatively large dipole moments, they can interact favorably with the dissolved ions.
The ion–dipole interactions between $\ce{Li^{+}}$ ions and acetone molecules in a solution of LiCl in acetone are shown in Figure $3$. The energetically favorable $\ce{Li^{+}}$–acetone interactions make $ΔH_3$ sufficiently negative to overcome the positive $ΔH_1$ and $ΔH_2$. Because the dipole moment of acetone (2.88 D), and thus its polarity, is actually larger than that of water (1.85 D), one might even expect that LiCl would be more soluble in acetone than in water. In fact, the opposite is true: 83 g of LiCl dissolve in 100 mL of water at 20°C, but only about 4.1 g of $\ce{LiCl}$ dissolve in 100 mL of acetone. This apparent contradiction arises from the fact that the dipole moment is a property of a single molecule in the gas phase. A more useful measure of the ability of a solvent to dissolve ionic compounds is its dielectric constant (ε), which is the ability of a bulk substance to decrease the electrostatic forces between two charged particles. By definition, the dielectric constant of a vacuum is 1. In essence, a solvent with a high dielectric constant causes the charged particles to behave as if they have been moved farther apart. At 25°C, the dielectric constant of water is 80.1, one of the highest known, and that of acetone is only 21.0. Hence water is better able to decrease the electrostatic attraction between $\ce{Li^{+}}$ and $\ce{Cl^{-}}$ ions, so $\ce{LiCl}$ is more soluble in water than in acetone. This behavior is in contrast to that of molecular substances, for which polarity is the dominant factor governing solubility.
A solvent’s dielectric constant is the most useful measure of its ability to dissolve ionic compounds. A solvent’s polarity is the dominant factor in dissolving molecular substances.
It is also possible to dissolve ionic compounds in organic solvents using crown ethers, cyclic compounds with the general formula $(OCH_2CH_2)_n$. Crown ethers are named using both the total number of atoms in the ring and the number of oxygen atoms. Thus 18-crown-6 is an 18-membered ring with six oxygen atoms (Figure $\PageIndex{1a}$). The cavity in the center of the crown ether molecule is lined with oxygen atoms and is large enough to be occupied by a cation, such as $K^+$. The cation is stabilized by interacting with lone pairs of electrons on the surrounding oxygen atoms. Thus crown ethers solvate cations inside a hydrophilic cavity, whereas the outer shell, consisting of C–H bonds, is hydrophobic. Crown ethers are useful for dissolving ionic substances such as $KMnO_4$ in organic solvents such as isopropanol $[(CH_3)_2CHOH]$ (Figure $5$). The availability of crown ethers with cavities of different sizes allows specific cations to be solvated with a high degree of selectivity.
Cryptands (from the Greek kryptós, meaning “hidden”) are compounds that can completely surround a cation with lone pairs of electrons on oxygen and nitrogen atoms (Figure $\PageIndex{4b}$). The number in the name of the cryptand is the number of oxygen atoms in each strand of the molecule. Like crown ethers, cryptands can be used to prepare solutions of ionic compounds in solvents that are otherwise too nonpolar to dissolve them.
Summary
The solubility of a substance is the maximum amount of a solute that can dissolve in a given quantity of solvent; it depends on the chemical nature of both the solute and the solvent and on the temperature and pressure. When a solution contains the maximum amount of solute that can dissolve under a given set of conditions, it is a saturated solution. Otherwise, it is unsaturated. Supersaturated solutions, which contain more dissolved solute than allowed under particular conditions, are not stable; the addition of a seed crystal, a small particle of solute, will usually cause the excess solute to crystallize. A system in which crystallization and dissolution occur at the same rate is in dynamic equilibrium. The solubility of a substance in a liquid is determined by intermolecular interactions, which also determine whether two liquids are miscible. Solutes can be classified as hydrophilic (water loving) or hydrophobic (water fearing). Vitamins with hydrophilic structures are water soluble, whereas those with hydrophobic structures are fat soluble. Many metals dissolve in liquid mercury to form amalgams. Covalent network solids and most metals are insoluble in nearly all solvents. The solubility of ionic compounds is largely determined by the dielectric constant (ε) of the solvent, a measure of its ability to decrease the electrostatic forces between charged particles. Solutions of many ionic compounds in organic solvents can be dissolved using crown ethers, cyclic polyethers large enough to accommodate a metal ion in the center, or cryptands, compounds that completely surround a cation. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/13%3A_Properties_of_Solutions/13.02%3A_Saturated_Solutions_and_Solubility.txt |
Learning Objectives
• To understand the relationship among temperature, pressure, and solubility.
• The understand that the solubility of a solid may increase or decrease with increasing temperature,
• To understand that the solubility of a gas decreases with an increase in temperature and a decrease in pressure.
Experimentally it is found that the solubility of most compounds depends strongly on temperature and, if a gas, on pressure as well. As we shall see, the ability to manipulate the solubility by changing the temperature and pressure has several important consequences.
Effect of Temperature on the Solubility of Solids
Figure $1$ shows plots of the solubilities of several organic and inorganic compounds in water as a function of temperature. Although the solubility of a solid generally increases with increasing temperature, there is no simple relationship between the structure of a substance and the temperature dependence of its solubility. Many compounds (such as glucose and $\ce{CH_3CO_2Na}$) exhibit a dramatic increase in solubility with increasing temperature. Others (such as $\ce{NaCl}$ and $\ce{K_2SO_4}$) exhibit little variation, and still others (such as $\ce{Li_2SO_4}$) become less soluble with increasing temperature.
Notice in particular the curves for $\ce{NH4NO3}$ and $\ce{CaCl2}$. The dissolution of ammonium nitrate in water is endothermic ($ΔH_{soln} = +25.7\; kJ/mol$), whereas the dissolution of calcium chloride is exothermic ($ΔH_{soln} = −68.2 \;kJ/mol$), yet Figure $1$ shows that the solubility of both compounds increases sharply with increasing temperature. In fact, the magnitudes of the changes in both enthalpy and entropy for dissolution are temperature dependent. Because the solubility of a compound is ultimately determined by relatively small differences between large numbers, there is generally no good way to predict how the solubility will vary with temperature.
The variation of solubility with temperature has been measured for a wide range of compounds, and the results are published in many standard reference books. Chemists are often able to use this information to separate the components of a mixture by fractional crystallization, the separation of compounds on the basis of their solubilities in a given solvent. For example, if we have a mixture of 150 g of sodium acetate ($\ce{CH_3CO_2Na}$) and 50 g of $\ce{KBr}$, we can separate the two compounds by dissolving the mixture in 100 g of water at 80°C and then cooling the solution slowly to 0°C. According to the temperature curves in Figure $1$, both compounds dissolve in water at 80°C, and all 50 g of $\ce{KBr}$ remains in solution at 0°C. Only about 36 g of $\ce{CH3CO2Na}$ are soluble in 100 g of water at 0°C, however, so approximately 114 g (150 g − 36 g) of $\ce{CH_3CO_2Na}$ crystallizes out on cooling. The crystals can then be separated by filtration. Thus fractional crystallization allows us to recover about 75% of the original $\ce{CH_3CO_2Na}$ in essentially pure form in only one step.
Fractional crystallization is a common technique for purifying compounds as diverse as those shown in Figure $1$ and from antibiotics to enzymes. For the technique to work properly, the compound of interest must be more soluble at high temperature than at low temperature, so that lowering the temperature causes it to crystallize out of solution. In addition, the impurities must be more soluble than the compound of interest (as was $\ce{KBr}$ in this example) and preferably present in relatively small amounts.
Effect of Temperature on the Solubility of Gases
The solubility of gases in liquids decreases with increasing temperature, as shown in Figure $2$. Attractive intermolecular interactions in the gas phase are essentially zero for most substances. When a gas dissolves, it does so because its molecules interact with solvent molecules. Because heat is released when these new attractive interactions form, dissolving most gases in liquids is an exothermic process ($ΔH_{soln} < 0$). Conversely, adding heat to the solution provides thermal energy that overcomes the attractive forces between the gas and the solvent molecules, thereby decreasing the solubility of the gas. The phenomenon is similar to that involved in the increase in the vapor pressure of a pure liquid with increasing temperature. In the case of vapor pressure, however, it is attractive forces between solvent molecules that are being overcome by the added thermal energy when the temperature is increased.
The decrease in the solubilities of gases at higher temperatures has both practical and environmental implications. Anyone who routinely boils water in a teapot or electric kettle knows that a white or gray deposit builds up on the inside and must eventually be removed. The same phenomenon occurs on a much larger scale in the giant boilers used to supply hot water or steam for industrial applications, where it is called “boiler scale,” a deposit that can seriously decrease the capacity of hot water pipes (Figure $3$). The problem is not a uniquely modern one: aqueducts that were built by the Romans 2000 years ago to carry cold water from alpine regions to warmer, drier regions in southern France were clogged by similar deposits. The chemistry behind the formation of these deposits is moderately complex and will be described elsewhere, but the driving force is the loss of dissolved $\ce{CO2}$ from solution. Hard water contains dissolved $\ce{Ca^{2+}}$ and $\ce{HCO3^{-}}$ (bicarbonate) ions. Calcium bicarbonate ($\ce{Ca(HCO3)2}$ is rather soluble in water, but calcium carbonate ($\ce{CaCO3}$) is quite insoluble. A solution of bicarbonate ions can react to form carbon dioxide, carbonate ion, and water:
$\ce{2HCO3^{-}(aq) -> CO3^{2-}(aq) + H2O(l) + CO2(aq)} \label{13.9}$
Heating the solution decreases the solubility of $\ce{CO2}$, which escapes into the gas phase above the solution. In the presence of calcium ions, the carbonate ions precipitate as insoluble calcium carbonate, the major component of boiler scale.
Thermal Pollution
In thermal pollution, lake or river water that is used to cool an industrial reactor or a power plant is returned to the environment at a higher temperature than normal. Because of the reduced solubility of $\ce{O2}$ at higher temperatures (Figure $2$), the warmer water contains less dissolved oxygen than the water did when it entered the plant. Fish and other aquatic organisms that need dissolved oxygen to live can literally suffocate if the oxygen concentration of their habitat is too low. Because the warm, oxygen-depleted water is less dense, it tends to float on top of the cooler, denser, more oxygen-rich water in the lake or river, forming a barrier that prevents atmospheric oxygen from dissolving. Eventually even deep lakes can be suffocated if the problem is not corrected. Additionally, most fish and other nonmammalian aquatic organisms are cold-blooded, which means that their body temperature is the same as the temperature of their environment. Temperatures substantially greater than the normal range can lead to severe stress or even death. Cooling systems for power plants and other facilities must be designed to minimize any adverse effects on the temperatures of surrounding bodies of water.
A similar effect is seen in the rising temperatures of bodies of water such as the k0oi89Chesapeake Bay, the largest estuary in North America, where \lobal warming has been implicated as the cause. For each 1.5°C that the bay’s water warms, the capacity of water to dissolve oxygen decreases by about 1.1%. Many marine species that are at the southern limit of their distributions have shifted their populations farther north. In 2005, the eelgrass, which forms an important nursery habitat for fish and shellfish, disappeared from much of the bay following record high water temperatures. Presumably, decreased oxygen levels decreased populations of clams and other filter feeders, which then decreased light transmission to allow the eelsgrass to grow. The complex relationships in ecosystems such as the Chesapeake Bay are especially sensitive to temperature fluctuations that cause a deterioration of habitat quality.
Effect of Pressure on the Solubility of Gases: Henry’s Law
External pressure has very little effect on the solubility of liquids and solids. In contrast, the solubility of gases increases as the partial pressure of the gas above a solution increases. This point is illustrated in Figure $4$, which shows the effect of increased pressure on the dynamic equilibrium that is established between the dissolved gas molecules in solution and the molecules in the gas phase above the solution. Because the concentration of molecules in the gas phase increases with increasing pressure, the concentration of dissolved gas molecules in the solution at equilibrium is also higher at higher pressures.
The relationship between pressure and the solubility of a gas is described quantitatively by Henry’s law, which is named for its discoverer, the English physician and chemist, William Henry (1775–1836):
$C = kP \label{13.3.1}$
where
• $C$ is the concentration of dissolved gas at equilibrium,
• $P$ is the partial pressure of the gas, and
• $k$ is the Henry’s law constant, which must be determined experimentally for each combination of gas, solvent, and temperature.
Although the gas concentration may be expressed in any convenient units, we will use molarity exclusively. The units of the Henry’s law constant are therefore mol/(L·atm) = M/atm. Values of the Henry’s law constants for solutions of several gases in water at 20°C are listed in Table $1$.
As the data in Table $1$ demonstrate, the concentration of a dissolved gas in water at a given pressure depends strongly on its physical properties. For a series of related substances, London dispersion forces increase as molecular mass increases. Thus among the Group 18 elements, the Henry’s law constants increase smoothly from $\ce{He}$ to $\ce{Ne}$ to $\ce{Ar}$.
Table $1$: Henry’s Law Constants for Selected Gases in Water at 20°C
Gas Henry’s Law Constant [mol/(L·atm)] × 10−4
$\ce{He}$ 3.9
$\ce{Ne}$ 4.7
$\ce{Ar}$ 15
$\ce{H_2}$ 8.1
$\ce{N_2}$ 7.1
$\ce{O_2}$ 14
$\ce{CO_2}$ 392
Oxygen is Especially Soluble
Nitrogen and oxygen are the two most prominent gases in the Earth’s atmosphere and they share many similar physical properties. However, as Table $1$ shows, $\ce{O2}$ is twice as soluble in water as $\ce{N2}$. Many factors contribute to solubility including the nature of the intermolecular forces at play. For a details discussion, see "The O2/N2 Ratio Gas Solubility Mystery" by Rubin Battino and Paul G. Seybold (J. Chem. Eng. Data 2011, 56, 5036–5044),
Gases that react chemically with water, such as $\ce{HCl}$ and the other hydrogen halides, $\ce{H2S}$, and $\ce{NH3}$, do not obey Henry’s law; all of these gases are much more soluble than predicted by Henry’s law. For example, $\ce{HCl}$ reacts with water to give $\ce{H^{+}(aq)}$ and $\ce{Cl^{-}(aq)}$, not dissolved $\ce{HCl}$ molecules, and its dissociation into ions results in a much higher solubility than expected for a neutral molecule.
Gases that react with water do not obey Henry’s law.
Henry’s law has important applications. For example, bubbles of $\ce{CO2}$ form as soon as a carbonated beverage is opened because the drink was bottled under $\ce{CO2}$ at a pressure greater than 1 atm. When the bottle is opened, the pressure of $\ce{CO2}$ above the solution drops rapidly, and some of the dissolved gas escapes from the solution as bubbles. Henry’s law also explains why scuba divers have to be careful to ascend to the surface slowly after a dive if they are breathing compressed air. At the higher pressures under water, more N2 from the air dissolves in the diver’s internal fluids. If the diver ascends too quickly, the rapid pressure change causes small bubbles of $\ce{N2}$ to form throughout the body, a condition known as “the bends.” These bubbles can block the flow of blood through the small blood vessels, causing great pain and even proving fatal in some cases.
Due to the low Henry’s law constant for $\ce{O2}$ in water, the levels of dissolved oxygen in water are too low to support the energy needs of multicellular organisms, including humans. To increase the $\ce{O2}$ concentration in internal fluids, organisms synthesize highly soluble carrier molecules that bind $\ce{O2}$ reversibly. For example, human red blood cells contain a protein called hemoglobin that specifically binds $\ce{O2}$ and facilitates its transport from the lungs to the tissues, where it is used to oxidize food molecules to provide energy. The concentration of hemoglobin in normal blood is about 2.2 mM, and each hemoglobin molecule can bind four $\ce{O2}$ molecules. Although the concentration of dissolved $\ce{O2}$ in blood serum at 37°C (normal body temperature) is only 0.010 mM, the total dissolved $\ce{O2}$ concentration is 8.8 mM, almost a thousand times greater than would be possible without hemoglobin. Synthetic oxygen carriers based on fluorinated alkanes have been developed for use as an emergency replacement for whole blood. Unlike donated blood, these “blood substitutes” do not require refrigeration and have a long shelf life. Their very high Henry’s law constants for $\ce{O2}$ result in dissolved oxygen concentrations comparable to those in normal blood.
A Video Discussing Henry's Law. Video Link: Henry's Law (The Solubility of Gases in Solvents), YouTube(opens in new window) [youtu.be]
Example $1$
The Henry’s law constant for $\ce{O2}$ in water at 25°C is $1.27 \times 10^{-3} M/atm$, and the mole fraction of $\ce{O2}$ in the atmosphere is 0.21. Calculate the solubility of $\ce{O2}$ in water at 25°C at an atmospheric pressure of 1.00 atm.
Given: Henry’s law constant, mole fraction of $\ce{O2}$, and pressure
Asked for: solubility
Strategy:
1. Use Dalton’s law of partial pressures to calculate the partial pressure of oxygen. (For more information about Dalton’s law of partial pressures)
2. Use Henry’s law to calculate the solubility, expressed as the concentration of dissolved gas.
Solution:
A According to Dalton’s law, the partial pressure of $\ce{O2}$ is proportional to the mole fraction of $\ce{O2}$:
\begin{align*} P_A &= \chi_A P_t \[4pt] &= (0.21)(1.00\; atm) \[4pt] &= 0.21\; atm \end{align*} \nonumber
B From Henry’s law, the concentration of dissolved oxygen under these conditions is
\begin{align*} [\ce{CO2}] &= k P_{\ce{O2}} \[4pt] &=(1.27 \times 10^{-3}\; M/\cancel{atm}) (0.21\; \cancel{atm}) \[4pt] &=2.7 \times 10^{-4}\; M \end{align*} \nonumber
Exercise $1$
To understand why soft drinks “fizz” and then go “flat” after being opened, calculate the concentration of dissolved $\ce{CO2}$ in a soft drink:
1. bottled under a pressure of 5.0 atm of $\ce{CO2}$
2. in equilibrium with the normal partial pressure of $\ce{CO_2}$ in the atmosphere (approximately $3 \times 10^{-4} \;atm$). The Henry’s law constant for $\ce{CO2}$ in water at 25°C is $3.4 \times 10^{-2}\; M/atm$.
Answer a
$0.17 M$
Answer b
$1 \times 10^{-5} M$
Summary
The solubility of most substances depends strongly on the temperature and, in the case of gases, on the pressure. The solubility of most solid or liquid solutes increases with increasing temperature. The components of a mixture can often be separated using fractional crystallization, which separates compounds according to their solubilities. The solubility of a gas decreases with increasing temperature. Henry’s law describes the relationship between the pressure and the solubility of a gas. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/13%3A_Properties_of_Solutions/13.03%3A_Factors_Affecting_Solubility.txt |
Learning Objectives
• To describe the concentration of a solution in the way that is most appropriate for a particular problem or application.
• To be familiar with the different units used to express the concentrations of a solution.
There are several different ways to quantitatively describe the concentration of a solution. For example, molarity is a useful way to describe solution concentrations for reactions that are carried out in solution. Mole fractions are used not only to describe gas concentrations but also to determine the vapor pressures of mixtures of similar liquids. Example $1$ reviews the methods for calculating the molarity and mole fraction of a solution when the masses of its components are known.
Example $1$: Molarity and Mole Fraction
Commercial vinegar is essentially a solution of acetic acid in water. A bottle of vinegar has 3.78 g of acetic acid per 100.0 g of solution. Assume that the density of the solution is 1.00 g/mL.
1. What is its molarity?
2. What is its mole fraction?
Given: mass of substance and mass and density of solution
Asked for: molarity and mole fraction
Strategy:
1. Calculate the number of moles of acetic acid in the sample. Then calculate the number of liters of solution from its mass and density. Use these results to determine the molarity of the solution.
2. Determine the mass of the water in the sample and calculate the number of moles of water. Then determine the mole fraction of acetic acid by dividing the number of moles of acetic acid by the total number of moles of substances in the sample.
Solution:
A: The molarity is the number of moles of acetic acid per liter of solution. We can calculate the number of moles of acetic acid as its mass divided by its molar mass.
\begin{align*} \text{moles } \ce{CH_3CO_2H} &=\dfrac{3.78\; \cancel{\ce{g}}\; \ce{CH_3CO_2H}}{60.05\; \cancel{\ce{g}}/\ce{mol}} \[4pt] &=0.0629 \; \ce{mol} \end{align*} \nonumber
The volume of the solution equals its mass divided by its density.
\begin{align*} \text{volume} &=\dfrac{\text{mass}}{\text{density}} \[4pt] &=\dfrac{100.0\; \cancel{\ce{g}}\; \text{solution}}{1.00\; \cancel{\ce{g}}/\ce{mL}}=100\; mL\nonumber \end{align*} \nonumber
Then calculate the molarity directly.
\begin{align*} \text{molarity of } \ce{CH_3CO_2H} &=\dfrac{\text{moles } \ce{CH3CO2H} }{\text{liter solution}} \[4pt] &=\dfrac{0.0629\; mol\; \ce{CH_3CO_2H}}{(100\; \cancel{\ce{mL}})(1\; L/1000\; \cancel{\ce{mL}})}=0.629\; M \; \ce{CH_3CO_2H} \end{align*} \nonumber
This result makes intuitive sense. If 100.0 g of aqueous solution (equal to 100 mL) contains 3.78 g of acetic acid, then 1 L of solution will contain 37.8 g of acetic acid, which is a little more than $\ce{ 1/2}$ mole. Keep in mind, though, that the mass and volume of a solution are related by its density; concentrated aqueous solutions often have densities greater than 1.00 g/mL.
B: To calculate the mole fraction of acetic acid in the solution, we need to know the number of moles of both acetic acid and water. The number of moles of acetic acid is 0.0629 mol, as calculated in part (a). We know that 100.0 g of vinegar contains 3.78 g of acetic acid; hence the solution also contains (100.0 g − 3.78 g) = 96.2 g of water. We have
$moles\; \ce{H_2O}=\dfrac{96.2\; \cancel{\ce{g}}\; \ce{H_2O}}{18.02\; \cancel{\ce{g}}/mol}=5.34\; mol\; \ce{H_2O}\nonumber$
The mole fraction $\chi$ of acetic acid is the ratio of the number of moles of acetic acid to the total number of moles of substances present:
\begin{align*} \chi_{\ce{CH3CO2H}} &=\dfrac{moles\; \ce{CH_3CO_2H}}{moles \; \ce{CH_3CO_2H} + moles\; \ce{H_2O}} \[4pt] &=\dfrac{0.0629\; mol}{0.0629 \;mol + 5.34\; mol} \[4pt] &=0.0116=1.16 \times 10^{−2} \end{align*} \nonumber
This answer makes sense, too. There are approximately 100 times as many moles of water as moles of acetic acid, so the ratio should be approximately 0.01.
Exercise $1$: Molarity and Mole Fraction
A solution of $\ce{HCl}$ gas dissolved in water (sold commercially as “muriatic acid,” a solution used to clean masonry surfaces) has 20.22 g of $\ce{HCl}$ per 100.0 g of solution, and its density is 1.10 g/mL.
1. What is its molarity?
2. What is its mole fraction?
Answer a
6.10 M HCl
Answer b
$\chi_{HCl} = 0.111$
The concentration of a solution can also be described by its molality (m), the number of moles of solute per kilogram of solvent:
$\text{molality (m)} =\dfrac{\text{moles solute}}{\text{kilogram solvent}} \label{Eq1}$
Molality, therefore, has the same numerator as molarity (the number of moles of solute) but a different denominator (kilogram of solvent rather than liter of solution). For dilute aqueous solutions, the molality and molarity are nearly the same because dilute solutions are mostly solvent. Thus because the density of water under standard conditions is very close to 1.0 g/mL, the volume of 1.0 kg of $H_2O$ under these conditions is very close to 1.0 L, and a 0.50 M solution of $KBr$ in water, for example, has approximately the same concentration as a 0.50 m solution.
Another common way of describing concentration is as the ratio of the mass of the solute to the total mass of the solution. The result can be expressed as mass percentage, parts per million (ppm), or parts per billion (ppb):
\begin{align} \text{mass percentage}&=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 100 \label{Eq2} \[4pt] \text{parts per million (ppm)} &=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 10^{6} \label{Eq3} \[4pt] \text{parts per billion (ppb)}&=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 10^{9} \label{Eq4} \end{align}
In the health sciences, the concentration of a solution is often expressed as parts per thousand (ppt), indicated as a proportion. For example, adrenalin, the hormone produced in high-stress situations, is available in a 1:1000 solution, or one gram of adrenalin per 1000 g of solution.
The labels on bottles of commercial reagents often describe the contents in terms of mass percentage. Sulfuric acid, for example, is sold as a 95% aqueous solution, or 95 g of $\ce{H_2SO_4}$ per 100 g of solution. Parts per million and parts per billion are used to describe concentrations of highly dilute solutions. These measurements correspond to milligrams and micrograms of solute per kilogram of solution, respectively. For dilute aqueous solutions, this is equal to milligrams and micrograms of solute per liter of solution (assuming a density of 1.0 g/mL).
Example $2$: Molarity and Mass
Several years ago, millions of bottles of mineral water were contaminated with benzene at ppm levels. This incident received a great deal of attention because the lethal concentration of benzene in rats is 3.8 ppm. A 250 mL sample of mineral water has 12.7 ppm of benzene. Because the contaminated mineral water is a very dilute aqueous solution, we can assume that its density is approximately 1.00 g/mL.
1. What is the molarity of the solution?
2. What is the mass of benzene in the sample?
Given: volume of sample, solute concentration, and density of solution
Asked for: molarity of solute and mass of solute in 250 mL
Strategy:
1. Use the concentration of the solute in parts per million to calculate the molarity.
2. Use the concentration of the solute in parts per million to calculate the mass of the solute in the specified volume of solution.
Solution:
a. A To calculate the molarity of benzene, we need to determine the number of moles of benzene in 1 L of solution. We know that the solution contains 12.7 ppm of benzene. Because 12.7 ppm is equivalent to 12.7 mg/1000 g of solution and the density of the solution is 1.00 g/mL, the solution contains 12.7 mg of benzene per liter (1000 mL). The molarity is therefore
\begin{align*} \text{molarity}&=\dfrac{\text{moles}}{\text{liter solution}} \[4pt] &=\dfrac{(12.7\; \cancel{mg}) \left(\frac{1\; \cancel{g}}{1000\; \cancel{mg}}\right)\left(\frac{1\; mol}{78.114\; \cancel{g}}\right)}{1.00\; L} \[4pt] &=1.63 \times 10^{-4} M\end{align*} \nonumber
b. B We are given that there are 12.7 mg of benzene per 1000 g of solution, which is equal to 12.7 mg/L of solution. Hence the mass of benzene in 250 mL (250 g) of solution is
\begin{align*} \text{mass of benzene} &=\dfrac{(12.7\; mg\; \text{benzene})(250\; \cancel{mL})}{1000\; \cancel{mL}} \[4pt] &=3.18\; mg \[4pt] &=3.18 \times 10^{-3}\; g\; \text{benzene} \end{align*} \nonumber
Exercise $2$: Molarity of Lead Solution
The maximum allowable concentration of lead in drinking water is 9.0 ppb.
1. What is the molarity of $\ce{Pb^{2+}}$ in a 9.0 ppb aqueous solution?
2. Use your calculated concentration to determine how many grams of $\ce{Pb^{2+}}$ are in an 8 oz glass of water.
Answer a
4.3 × 10−8 M
Answer b
2 × 10−6 g
How do chemists decide which units of concentration to use for a particular application? Although molarity is commonly used to express concentrations for reactions in solution or for titrations, it does have one drawback—molarity is the number of moles of solute divided by the volume of the solution, and the volume of a solution depends on its density, which is a function of temperature. Because volumetric glassware is calibrated at a particular temperature, typically 20°C, the molarity may differ from the original value by several percent if a solution is prepared or used at a significantly different temperature, such as 40°C or 0°C. For many applications this may not be a problem, but for precise work these errors can become important. In contrast, mole fraction, molality, and mass percentage depend on only the masses of the solute and solvent, which are independent of temperature.
Mole fraction is not very useful for experiments that involve quantitative reactions, but it is convenient for calculating the partial pressure of gases in mixtures, as discussed previously. Mole fractions are also useful for calculating the vapor pressures of certain types of solutions. Molality is particularly useful for determining how properties such as the freezing or boiling point of a solution vary with solute concentration. Because mass percentage and parts per million or billion are simply different ways of expressing the ratio of the mass of a solute to the mass of the solution, they enable us to express the concentration of a substance even when the molecular mass of the substance is unknown. Units of ppb or ppm are also used to express very low concentrations, such as those of residual impurities in foods or of pollutants in environmental studies.
Table $1$ summarizes the different units of concentration and typical applications for each. When the molar mass of the solute and the density of the solution are known, it becomes relatively easy with practice to convert among the units of concentration we have discussed, as illustrated in Example $3$.
Table $1$: Different Units for Expressing the Concentrations of Solutions*
Unit Definition Application
*The molarity of a solution is temperature dependent, but the other units shown in this table are independent of temperature.
molarity (M) moles of solute/liter of solution (mol/L) Used for quantitative reactions in solution and titrations; mass and molecular mass of solute and volume of solution are known.
mole fraction ($\chi$) moles of solute/total moles present (mol/mol) Used for partial pressures of gases and vapor pressures of some solutions; mass and molecular mass of each component are known.
molality (m) moles of solute/kg of solvent (mol/kg) Used in determining how colligative properties vary with solute concentration; masses and molecular mass of solute are known.
mass percentage (%) [mass of solute (g)/mass of solution (g)] × 100 Useful when masses are known but molecular masses are unknown.
parts per thousand (ppt) [mass of solute/mass of solution] × 103 (g solute/kg solution) Used in the health sciences, ratio solutions are typically expressed as a proportion, such as 1:1000.
parts per million (ppm) [mass of solute/mass of solution] × 106 (mg solute/kg solution) Used for trace quantities; masses are known but molecular masses may be unknown.
parts per billion (ppb) [mass of solute/mass of solution] × 109 (µg solute/kg solution) Used for trace quantities; masses are known but molecular masses may be unknown.
Example $3$: Vodka
Vodka is essentially a solution of ethanol in water. Typical vodka is sold as “80 proof,” which means that it contains 40.0% ethanol by volume. The density of pure ethanol is 0.789 g/mL at 20°C. If we assume that the volume of the solution is the sum of the volumes of the components (which is not strictly correct), calculate the following for the ethanol in 80-proof vodka.
1. the mass percentage
2. the mole fraction
3. the molarity
4. the molality
Given: volume percent and density
Asked for: mass percentage, mole fraction, molarity, and molality
Strategy:
1. Use the density of the solute to calculate the mass of the solute in 100.0 mL of solution. Calculate the mass of water in 100.0 mL of solution.
2. Determine the mass percentage of solute by dividing the mass of ethanol by the mass of the solution and multiplying by 100.
3. Convert grams of solute and solvent to moles of solute and solvent. Calculate the mole fraction of solute by dividing the moles of solute by the total number of moles of substances present in solution.
4. Calculate the molarity of the solution: moles of solute per liter of solution. Determine the molality of the solution by dividing the number of moles of solute by the kilograms of solvent.
Solution:
The key to this problem is to use the density of pure ethanol to determine the mass of ethanol ($CH_3CH_2OH$), abbreviated as EtOH, in a given volume of solution. We can then calculate the number of moles of ethanol and the concentration of ethanol in any of the required units. A Because we are given a percentage by volume, we assume that we have 100.0 mL of solution. The volume of ethanol will thus be 40.0% of 100.0 mL, or 40.0 mL of ethanol, and the volume of water will be 60.0% of 100.0 mL, or 60.0 mL of water. The mass of ethanol is obtained from its density:
$mass\; of\; EtOH=(40.0\; \cancel{mL})\left(\dfrac{0.789\; g}{\cancel{mL}}\right)=31.6\; g\; EtOH\nonumber$
If we assume the density of water is 1.00 g/mL, the mass of water is 60.0 g. We now have all the information we need to calculate the concentration of ethanol in the solution.
B The mass percentage of ethanol is the ratio of the mass of ethanol to the total mass of the solution, expressed as a percentage:
\begin{align*} \%EtOH &=\left(\dfrac{mass\; of\; EtOH}{mass\; of\; solution}\right)(100) \[4pt] &=\left(\dfrac{31.6\; \cancel{g}\; EtOH}{31.6\; \cancel{g} \;EtOH +60.0\; \cancel{g} \; H_2O} \right)(100) \[4pt]&= 34.5\%\end{align*} \nonumber
C The mole fraction of ethanol is the ratio of the number of moles of ethanol to the total number of moles of substances in the solution. Because 40.0 mL of ethanol has a mass of 31.6 g, we can use the molar mass of ethanol (46.07 g/mol) to determine the number of moles of ethanol in 40.0 mL:
\begin{align*} moles\; \ce{CH_3CH_2OH}&=(31.6\; \cancel{g\; \ce{CH_3CH_2OH}}) \left(\dfrac{1\; mol}{46.07\; \cancel{g\; \ce{CH_3CH_2OH}}}\right) \[4pt] &=0.686 \;mol\; \ce{CH_3CH_2OH} \end{align*} \nonumber
Similarly, the number of moles of water is
$moles \;\ce{H_2O}=(60.0\; \cancel{g \; \ce{H_2O}}) \left(\dfrac{1 \;mol\; \ce{H_2O}}{18.02\; \cancel{g\; \ce{H_2O}}}\right)=3.33\; mol\; \ce{H_2O}\nonumber$
The mole fraction of ethanol is thus
$\chi_{\ce{CH_3CH_2OH}}=\dfrac{0.686\; \cancel{mol}}{0.686\; \cancel{mol} + 3.33\;\cancel{ mol}}=0.171\nonumber$
D The molarity of the solution is the number of moles of ethanol per liter of solution. We already know the number of moles of ethanol per 100.0 mL of solution, so the molarity is
$M_{\ce{CH_3CH_2OH}}=\left(\dfrac{0.686\; mol}{100\; \cancel{mL}}\right)\left(\dfrac{1000\; \cancel{mL}}{L}\right)=6.86 \;M\nonumber$
The molality of the solution is the number of moles of ethanol per kilogram of solvent. Because we know the number of moles of ethanol in 60.0 g of water, the calculation is again straightforward:
$m_{\ce{CH_3CH_2OH}}=\left(\dfrac{0.686\; mol\; EtOH}{60.0\; \cancel{g}\; H_2O } \right) \left(\dfrac{1000\; \cancel{g}}{kg}\right)=\dfrac{11.4\; mol\; EtOH}{kg\; H_2O}=11.4 \;m\nonumber$
Exercise $3$: Toluene/Benzene Solution
A solution is prepared by mixing 100.0 mL of toluene with 300.0 mL of benzene. The densities of toluene and benzene are 0.867 g/mL and 0.874 g/mL, respectively. Assume that the volume of the solution is the sum of the volumes of the components. Calculate the following for toluene.
1. mass percentage
2. mole fraction
3. molarity
4. molality
Answer a
mass percentage toluene = 24.8%
Answer b
$\chi_{toluene} = 0.219$
Answer c
2.35 M toluene
Answer d
3.59 m toluene
A Video Discussing Different Measures of Concentration. Video Link: Measures of Concentration, YouTube (opens in new window) [youtu.be]
A Video Discussing how to Convert Measures of Concentration. Video Link: Converting Units of Concentration, YouTube(opens in new window) [youtu.be]
Summary
Different units are used to express the concentrations of a solution depending on the application. The concentration of a solution is the quantity of solute in a given quantity of solution. It can be expressed in several ways: molarity (moles of solute per liter of solution); mole fraction, the ratio of the number of moles of solute to the total number of moles of substances present; mass percentage, the ratio of the mass of the solute to the mass of the solution times 100; parts per thousand (ppt), grams of solute per kilogram of solution; parts per million (ppm), milligrams of solute per kilogram of solution; parts per billion (ppb), micrograms of solute per kilogram of solution; and molality (m), the number of moles of solute per kilogram of solvent. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/13%3A_Properties_of_Solutions/13.04%3A_Ways_of_Expressing_Concentration.txt |
Learning Objectives
• To describe the relationship between solute concentration and the physical properties of a solution.
• To understand that the total number of nonvolatile solute particles determines the decrease in vapor pressure, increase in boiling point, and decrease in freezing point of a solution versus the pure solvent.
Many of the physical properties of solutions differ significantly from those of the pure substances discussed in earlier chapters, and these differences have important consequences. For example, the limited temperature range of liquid water (0°C–100°C) severely limits its use. Aqueous solutions have both a lower freezing point and a higher boiling point than pure water. Probably one of the most familiar applications of this phenomenon is the addition of ethylene glycol (“antifreeze”) to the water in an automobile radiator. This solute lowers the freezing point of the water, preventing the engine from cracking in very cold weather from the expansion of pure water on freezing. Antifreeze also enables the cooling system to operate at temperatures greater than 100°C without generating enough pressure to explode.
Changes in the freezing point and boiling point of a solution depend primarily on the number of solute particles present rather than the kind of particles. Such properties of solutions are called colligative properties (from the Latin colligatus, meaning “bound together” as in a quantity). As we will see, the vapor pressure and osmotic pressure of solutions are also colligative properties.
Counting concentrations
When we determine the number of particles in a solution, it is important to remember that not all solutions with the same molarity contain the same concentration of solute particles. Consider, for example, 0.01 M aqueous solutions of sucrose, $\ce{NaCl}$, and $\ce{CaCl_2}$. Because sucrose dissolves to give a solution of neutral molecules, the concentration of solute particles in a 0.01 M sucrose solution is 0.01 M. In contrast, both $\ce{NaCl}$ and $\ce{CaCl_2}$ are ionic compounds that dissociate in water to yield solvated ions. As a result, a 0.01 M aqueous solution of $\ce{NaCl}$ contains 0.01 M $\ce{Na^{+}}$ ions and 0.01 M $\ce{Cl^{−}}$ ions, for a total particle concentration of 0.02 M. Similarly, the $\ce{CaCl_2}$ solution contains 0.01 M $\ce{Ca^{2+}}$ ions and 0.02 M $\ce{Cl^{−}}$ ions, for a total particle concentration of 0.03 M. These values are correct for dilute solutions, where the dissociation of the compounds to form separately solvated ions is complete.
At higher concentrations (typically >1 M), especially with salts of small, highly charged ions (such as $\ce{Mg^{2+}}$ or $\ce{Al^{3+}}$), or in solutions with less polar solvents, dissociation to give separate ions is often incomplete. The sum of the concentrations of the dissolved solute particles dictates the physical properties of a solution. In the following discussion, we must therefore keep the chemical nature of the solute firmly in mind.
Vapor Pressure of Solutions and Raoult’s Law
Adding a nonvolatile solute, one whose vapor pressure is too low to measure readily, to a volatile solvent decreases the vapor pressure of the solvent. We can understand this phenomenon qualitatively by examining Figure $1$, which is a schematic diagram of the surface of a solution of glucose in water. In an aqueous solution of glucose, a portion of the surface area is occupied by nonvolatile glucose molecules rather than by volatile water molecules. As a result, fewer water molecules can enter the vapor phase per unit time, even though the surface water molecules have the same kinetic energy distribution as they would in pure water. At the same time, the rate at which water molecules in the vapor phase collide with the surface and reenter the solution is unaffected. The net effect is to shift the dynamic equilibrium between water in the vapor and the liquid phases, decreasing the vapor pressure of the solution compared with the vapor pressure of the pure solvent.
Figure $2$ shows two beakers, one containing pure water and one containing an aqueous glucose solution, in a sealed chamber. We can view the system as having two competing equilibria: water vapor will condense in both beakers at the same rate, but water molecules will evaporate more slowly from the glucose solution because fewer water molecules are at the surface. Eventually all of the water will evaporate from the beaker containing the liquid with the higher vapor pressure (pure water) and condense in the beaker containing the liquid with the lower vapor pressure (the glucose solution). If the system consisted of only a beaker of water inside a sealed container, equilibrium between the liquid and vapor would be achieved rather rapidly, and the amount of liquid water in the beaker would remain constant.
If the particles of a solute are essentially the same size as those of the solvent and both solute and solvent have roughly equal probabilities of being at the surface of the solution, then the effect of a solute on the vapor pressure of the solvent is proportional to the number of sites occupied by solute particles at the surface of the solution. Doubling the concentration of a given solute causes twice as many surface sites to be occupied by solute molecules, resulting in twice the decrease in vapor pressure. The relationship between solution composition and vapor pressure is therefore
$P_A=\chi_AP^0_A \label{13.5.1}$
where $P_A$ is the vapor pressure of component A of the solution (in this case the solvent), $\chi_A$ is the mole fraction of $A$ in solution, and $P^0_A$ is the vapor pressure of pure $A$. Equation $\ref{13.5.1}$ is known as Raoult’s law, after the French chemist who developed it. If the solution contains only a single nonvolatile solute (B), then $\chi_A + \chi_B = 1$, and we can substitute $\chi_A = 1 − \chi_B$ to obtain
\begin{align} P_A &=(1−\chi_B)P^0_A \[4pt] &=P^0_A−\chi_BP^0_A \label{13.5.2} \end{align}
Rearranging and defining $ΔP_A=P^0_A−P_A$, we obtain a relationship between the decrease in vapor pressure and the mole fraction of nonvolatile solute:
\begin{align} P^0_A−P_A &=ΔP_A \[4pt] &=\chi_BP^0_A \label{13.5.3} \end{align}
We can solve vapor pressure problems in either of two ways: by using Equation $\ref{13.5.1}$ to calculate the actual vapor pressure above a solution of a nonvolatile solute, or by using Equation $\ref{13.5.3}$ to calculate the decrease in vapor pressure caused by a specified amount of a nonvolatile solute.
Example $1$: Anti-Freeze
Ethylene glycol ($\ce{HOCH_2CH_2OH}$), the major ingredient in commercial automotive antifreeze, increases the boiling point of radiator fluid by lowering its vapor pressure. At 100°C, the vapor pressure of pure water is 760 mmHg. Calculate the vapor pressure of an aqueous solution containing 30.2% ethylene glycol by mass, a concentration commonly used in climates that do not get extremely cold in winter.
Given: identity of solute, percentage by mass, and vapor pressure of pure solvent
Asked for: vapor pressure of solution
Strategy:
1. Calculate the number of moles of ethylene glycol in an arbitrary quantity of water, and then calculate the mole fraction of water.
2. Use Raoult’s law to calculate the vapor pressure of the solution.
Solution:
A A 30.2% solution of ethylene glycol contains 302 g of ethylene glycol per kilogram of solution; the remainder (698 g) is water. To use Raoult’s law to calculate the vapor pressure of the solution, we must know the mole fraction of water. Thus we must first calculate the number of moles of both ethylene glycol (EG) and water present:
$moles \;EG=(302 \;\cancel{g}) \left( \dfrac{1\; mol}{62.07\; \cancel{g}} \right)=4.87\; mol\; EG \nonumber$
$moles \; H_2O=(698 \;\cancel{g}) \left( \dfrac{1\; mol}{18.02\; \cancel{g}} \right)=38.7\; mol\; H_2O \nonumber$
The mole fraction of water is thus
$\chi_{H_2O}=\dfrac{38.7\; \cancel{mol} \; H_2O}{38.7\; \cancel{mol}\; H_2O +4.87 \cancel{mol}\; EG} =0.888 \nonumber$
B From Raoult’s law (Equation $\ref{13.5.1}$), the vapor pressure of the solution is
\begin{align*} P_{H_2O} &=(\chi_{H2_O})(P^0_{H_2O}) \[4pt] &=(0.888)(760\; mmHg) =675 \;mmHg \end{align*} \nonumber
Alternatively, we could solve this problem by calculating the mole fraction of ethylene glycol and then using Equation $\ref{13.5.3}$ to calculate the resulting decrease in vapor pressure:
$\chi_{EG}=\dfrac{4.87\; mol\; EG}{4.87\; mol\; EG+38.7\; mol\; H_2O}=0.112 \nonumber$
$ΔP_{H2_O}=(\chi_{EG})(P^0_{H_2O})=(0.112)(760\; mmHg)=85.1\; mmHg \nonumber$
$P_{H_2O}=P^0_{H_2O}−ΔP_{H_2O}=760\; mmHg−85.1\; mmHg=675\; mmHg \nonumber$
The same result is obtained using either method.
Exercise $1$
Seawater is an approximately 3.0% aqueous solution of $\ce{NaCl}$ by mass with about 0.5% of other salts by mass. Calculate the decrease in the vapor pressure of water at 25°C caused by this concentration of $\ce{NaCl}$, remembering that 1 mol of $\ce{NaCl}$ produces 2 mol of solute particles. The vapor pressure of pure water at 25°C is 23.8 mmHg.
Answer
0.45 mmHg. This may seem like a small amount, but it constitutes about a 2% decrease in the vapor pressure of water and accounts in part for the higher humidity in the north-central United States near the Great Lakes, which are freshwater lakes. The decrease therefore has important implications for climate modeling.
Even when a solute is volatile, meaning that it has a measurable vapor pressure, we can still use Raoult’s law. In this case, we calculate the vapor pressure of each component separately. The total vapor pressure of the solution ($P_{tot}$) is the sum of the vapor pressures of the components:
$P_{tot}=P_A+P_B=\chi_AP^0_A+\chi_BP^0_B \label{13.5.4}$
Because $\chi_B = 1 − \chi_A$ for a two-component system,
$P_{tot}=\chi_AP^0_A+(1−\chi_A)P^0_B \label{13.5.5}$
Thus we need to specify the mole fraction of only one of the components in a two-component system. Consider, for example, the vapor pressure of solutions of benzene and toluene of various compositions. At 20°C, the vapor pressures of pure benzene and toluene are 74.7 and 22.3 mmHg, respectively. The vapor pressure of benzene in a benzene–toluene solution is
$P_{C_6H_6}=\chi_{C_6H_6}P^0_{C_6H_6} \label{13.5.6}$
and the vapor pressure of toluene in the solution is
$P{C_6H_5CH_3}=\chi_{C_6H_5CH_3}P^0_{C_6H_5CH3} \label{13.5.7}$
Equations $\ref{13.5.6}$ and $\ref{13.5.7}$ are both in the form of the equation for a straight line: $y = mx + b$, where $b = 0$. Plots of the vapor pressures of both components versus the mole fractions are therefore straight lines that pass through the origin, as shown in Figure $3$. Furthermore, a plot of the total vapor pressure of the solution versus the mole fraction is a straight line that represents the sum of the vapor pressures of the pure components. Thus the vapor pressure of the solution is always greater than the vapor pressure of either component.
A solution of two volatile components that behaves like the solution in Figure $3$, which is defined as a solution that obeys Raoult’s law. Like an ideal gas, an ideal solution is a hypothetical system whose properties can be described in terms of a simple model. Mixtures of benzene and toluene approximate an ideal solution because the intermolecular forces in the two pure liquids are almost identical in both kind and magnitude. Consequently, the change in enthalpy on solution formation is essentially zero ($ΔH_{soln} ≈ 0$), which is one of the defining characteristics of an ideal solution.
Ideal solutions and ideal gases are both simple models that ignore intermolecular interactions.
Most real solutions, however, do not obey Raoult’s law precisely, just as most real gases do not obey the ideal gas law exactly. Real solutions generally deviate from Raoult’s law because the intermolecular interactions between the two components A and B differ. We can distinguish between two general kinds of behavior, depending on whether the intermolecular interactions between molecules A and B are stronger or weaker than the A–A and B–B interactions in the pure components. If the A–B interactions are stronger than the A–A and B–B interactions, each component of the solution exhibits a lower vapor pressure than expected for an ideal solution, as does the solution as a whole. The favorable A–B interactions effectively stabilize the solution compared with the vapor. This kind of behavior is called a negative deviation from Raoult’s law. Systems stabilized by hydrogen bonding between two molecules, such as acetone and ethanol, exhibit negative deviations from Raoult’s law. Conversely, if the A–B interactions are weaker than the A–A and B–B interactions yet the entropy increase is enough to allow the solution to form, both A and B have an increased tendency to escape from the solution into the vapor phase. The result is a higher vapor pressure than expected for an ideal solution, producing a positive deviation from Raoult’s law. In a solution of $\ce{CCl_4}$ and methanol, for example, the nonpolar $\ce{CCl_4}$ molecules interrupt the extensive hydrogen bonding network in methanol, and the lighter methanol molecules have weaker London dispersion forces than the heavier $\ce{CCl_4}$ molecules. Consequently, solutions of $\ce{CCl_4}$ and methanol exhibit positive deviations from Raoult’s law.
Example $2$
For each system, compare the intermolecular interactions in the pure liquids and in the solution to decide whether the vapor pressure will be greater than that predicted by Raoult’s law (positive deviation), approximately equal to that predicted by Raoult’s law (an ideal solution), or less than the pressure predicted by Raoult’s law (negative deviation).
1. cyclohexane and ethanol
2. methanol and acetone
3. n-hexane and isooctane (2,2,4-trimethylpentane)
Given: identity of pure liquids
Asked for: predicted deviation from Raoult’s law (Equation \ref{13.5.1})
Strategy:
Identify whether each liquid is polar or nonpolar, and then predict the type of intermolecular interactions that occur in solution.
Solution:
1. Liquid ethanol contains an extensive hydrogen bonding network, and cyclohexane is nonpolar. Because the cyclohexane molecules cannot interact favorably with the polar ethanol molecules, they will disrupt the hydrogen bonding. Hence the A–B interactions will be weaker than the A–A and B–B interactions, leading to a higher vapor pressure than predicted by Raoult’s law (a positive deviation).
2. Methanol contains an extensive hydrogen bonding network, but in this case the polar acetone molecules create A–B interactions that are stronger than the A–A or B–B interactions, leading to a negative enthalpy of solution and a lower vapor pressure than predicted by Raoult’s law (a negative deviation).
3. Hexane and isooctane are both nonpolar molecules (isooctane actually has a very small dipole moment, but it is so small that it can be ignored). Hence the predominant intermolecular forces in both liquids are London dispersion forces. We expect the A–B interactions to be comparable in strength to the A–A and B–B interactions, leading to a vapor pressure in good agreement with that predicted by Raoult’s law (an ideal solution).
Exercise $2$
For each system, compare the intermolecular interactions in the pure liquids with those in the solution to decide whether the vapor pressure will be greater than that predicted by Raoult’s law (positive deviation), approximately equal to that predicted by Raoult’s law (an ideal solution), or less than the pressure predicted by Raoult’s law (negative deviation):
1. benzene and n-hexane
2. ethylene glycol and $\ce{CCl_4}$
3. acetic acid and n-propanol
Answer a
approximately equal
Answer b
positive deviation (vapor pressure greater than predicted)
Answer c
negative deviation (vapor pressure less than predicted)
A Video Discussing Roult's Law. Video Link: Introduction to the Vapor Pressure of a Solution (Raoult's Law), YouTube(opens in new window) [youtu.be]
A Video Discussing How to find the Vapor Pressure of a Solution. Video Link: Finding the Vapor Pressure of a Solution (Nonionic-Nonvolatile Solute), YouTube(opens in new window) [youtu.be]
Boiling Point Elevation
Recall that the normal boiling point of a substance is the temperature at which the vapor pressure equals 1 atm. If a nonvolatile solute lowers the vapor pressure of a solvent, it must also affect the boiling point. Because the vapor pressure of the solution at a given temperature is less than the vapor pressure of the pure solvent, achieving a vapor pressure of 1 atm for the solution requires a higher temperature than the normal boiling point of the solvent. Thus the boiling point of a solution is always greater than that of the pure solvent. We can see why this must be true by comparing the phase diagram for an aqueous solution with the phase diagram for pure water (Figure $4$). The vapor pressure of the solution is less than that of pure water at all temperatures. Consequently, the liquid–vapor curve for the solution crosses the horizontal line corresponding to P = 1 atm at a higher temperature than does the curve for pure water.
Figure $4$: Phase Diagrams of Pure Water and an Aqueous Solution of a Nonvolatile Solute. The vaporization curve for the solution lies below the curve for pure water at all temperatures, which results in an increase in the boiling point and a decrease in the freezing point of the solution.
The boiling point of a solution with a nonvolatile solute is always greater than the boiling point of the pure solvent.
The magnitude of the increase in the boiling point is related to the magnitude of the decrease in the vapor pressure. As we have just discussed, the decrease in the vapor pressure is proportional to the concentration of the solute in the solution. Hence the magnitude of the increase in the boiling point must also be proportional to the concentration of the solute (Figure $5$).
We can define the boiling point elevation ($ΔT_b$) as the difference between the boiling points of the solution and the pure solvent:
$ΔT_b=T_b−T^0_b \label{13.5.8}$
where $T_b$ is the boiling point of the solution and $T^0_b$ is the boiling point of the pure solvent. We can express the relationship between $ΔT_b$ and concentration as follows
$ΔT_b = mK_b \label{13.5.9}$
where m is the concentration of the solute expressed in molality, and $K_b$ is the molal boiling point elevation constant of the solvent, which has units of °C/m. Table $1$ lists characteristic $K_b$ values for several commonly used solvents. For relatively dilute solutions, the magnitude of both properties is proportional to the solute concentration.
Table $1$: Boiling Point Elevation Constants (Kb) and Freezing Point Depression Constants (Kf) for Some Solvents
Solvent Boiling Point (°C) Kb (°C/m) Freezing Point (°C) Kf (°C/m)
acetic acid 117.90 3.22 16.64 3.63
benzene 80.09 2.64 5.49 5.07
d-(+)-camphor 207.4 4.91 178.8 37.8
carbon disulfide 46.2 2.42 −112.1 3.74
carbon tetrachloride 76.8 5.26 −22.62 31.4
chloroform 61.17 3.80 −63.41 4.60
nitrobenzene 210.8 5.24 5.70 6.87
water 100.00 0.51 0.00 1.86
The concentration of the solute is typically expressed as molality rather than mole fraction or molarity for two reasons. First, because the density of a solution changes with temperature, the value of molarity also varies with temperature. If the boiling point depends on the solute concentration, then by definition the system is not maintained at a constant temperature. Second, molality and mole fraction are proportional for relatively dilute solutions, but molality has a larger numerical value (a mole fraction can be only between zero and one). Using molality allows us to eliminate nonsignificant zeros.
According to Table $1$, the molal boiling point elevation constant for water is 0.51°C/m. Thus a 1.00 m aqueous solution of a nonvolatile molecular solute such as glucose or sucrose will have an increase in boiling point of 0.51°C, to give a boiling point of 100.51°C at 1.00 atm. The increase in the boiling point of a 1.00 m aqueous $\ce{NaCl}$ solution will be approximately twice as large as that of the glucose or sucrose solution because 1 mol of $\ce{NaCl}$ produces 2 mol of dissolved ions. Hence a 1.00 m $\ce{NaCl}$ solution will have a boiling point of about 101.02°C.
Example $3$
In Example $1$, we calculated that the vapor pressure of a 30.2% aqueous solution of ethylene glycol at 100°C is 85.1 mmHg less than the vapor pressure of pure water. We stated (without offering proof) that this should result in a higher boiling point for the solution compared with pure water. Now that we have seen why this assertion is correct, calculate the boiling point of the aqueous ethylene glycol solution.
Given: composition of solution
Asked for: boiling point
Strategy:
Calculate the molality of ethylene glycol in the 30.2% solution. Then use Equation $\ref{13.5.9}$ to calculate the increase in boiling point.
Solution:
From Example $1$, we know that a 30.2% solution of ethylene glycol in water contains 302 g of ethylene glycol (4.87 mol) per 698 g of water. The molality of the solution is thus
$\text{molality of ethylene glycol}= \left(\dfrac{4.87 \;mol}{698 \; \cancel{g} \;H_2O} \right) \left(\dfrac{1000\; \cancel{g}}{1 \;kg} \right)=6.98\, m \nonumber$
From Equation $\ref{13.5.9}$, the increase in boiling point is therefore
$ΔT_b=m K_b=(6.98 \cancel{m})(0.51°C/\cancel{m})=3.6°C \nonumber$
The boiling point of the solution is thus predicted to be 104°C. With a solute concentration of almost 7 m, however, the assumption of a dilute solution used to obtain Equation $\ref{13.5.9}$ may not be valid.
Exercise $3$
Assume that a tablespoon (5.00 g) of $\ce{NaCl}$ is added to 2.00 L of water at 20.0°C, which is then brought to a boil to cook spaghetti. At what temperature will the water boil?
Answer
100.04°C, or 100°C to three significant figures. (Recall that 1 mol of $\ce{NaCl}$ produces 2 mol of dissolved particles. The small increase in temperature means that adding salt to the water used to cook pasta has essentially no effect on the cooking time.)
A Video Discussing Boiling Point Elevation and Freezing Point Depression. Video Link: Boiling Point Elevation and Freezing Point Depression, YouTube(opens in new window) [youtu.be] (opens in new window)
Freezing Point Depression
The phase diagram in Figure $4$ shows that dissolving a nonvolatile solute in water not only raises the boiling point of the water but also lowers its freezing point. The solid–liquid curve for the solution crosses the line corresponding to $P = 1\,atm$ at a lower temperature than the curve for pure water.
We can understand this result by imagining that we have a sample of water at the normal freezing point temperature, where there is a dynamic equilibrium between solid and liquid. Water molecules are continuously colliding with the ice surface and entering the solid phase at the same rate that water molecules are leaving the surface of the ice and entering the liquid phase. If we dissolve a nonvolatile solute such as glucose in the liquid, the dissolved glucose molecules will reduce the number of collisions per unit time between water molecules and the ice surface because some of the molecules colliding with the ice will be glucose. Glucose, though, has a very different structure than water, and it cannot fit into the ice lattice. Consequently, the presence of glucose molecules in the solution can only decrease the rate at which water molecules in the liquid collide with the ice surface and solidify. Meanwhile, the rate at which the water molecules leave the surface of the ice and enter the liquid phase is unchanged. The net effect is to cause the ice to melt. The only way to reestablish a dynamic equilibrium between solid and liquid water is to lower the temperature of the system, which decreases the rate at which water molecules leave the surface of the ice crystals until it equals the rate at which water molecules in the solution collide with the ice.
By analogy to our treatment of boiling point elevation,the freezing point depression ($ΔT_f$) is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution:
$ΔT_f=T^0_f−T_f \label{13.5.10}$
where $T^0_f$ is the freezing point of the pure solvent and $T_f$ is the freezing point of the solution.
The order of the terms is reversed compared with Equation $\ref{13.5.8}$ to express the freezing point depression as a positive number. The relationship between $ΔT_f$ and the solute concentration is given by an equation analogous to Equation $\ref{13.5.9}$:
$ΔT_f = mK_f \label{13.5.11}$
where $m$ is the molality of the solution and $K_f$ is the molal freezing point depression constant for the solvent (in units of °C/m).
Like $K_b$, each solvent has a characteristic value of $K_f$ (see Table $1$). Freezing point depression depends on the total number of dissolved nonvolatile solute particles, just as with boiling point elevation. Thus an aqueous $\ce{NaCl}$ solution has twice as large a freezing point depression as a glucose solution of the same molality.
People who live in cold climates use freezing point depression to their advantage in many ways. For example, salt is used to melt ice and snow on roads and sidewalks, ethylene glycol is added to engine coolant water to prevent an automobile engine from being destroyed, and methanol is added to windshield washer fluid to prevent the fluid from freezing.
The decrease in vapor pressure, increase in boiling point, and decrease in freezing point of a solution versus a pure liquid all depend on the total number of dissolved nonvolatile solute particles.
Example $4$: Salting the Roads
In colder regions of the United States, $\ce{NaCl}$ or $\ce{CaCl_2}$ is often sprinkled on icy roads in winter to melt the ice and make driving safer. Use the data in the Figure below to estimate the concentrations of two saturated solutions at 0°C, one of $\ce{NaCl}$ and one of $\ce{CaCl_2}$, and calculate the freezing points of both solutions to see which salt is likely to be more effective at melting ice.
Given: solubilities of two compounds
Asked for: concentrations and freezing points
Strategy:
1. Estimate the solubility of each salt in 100 g of water from the figure. Determine the number of moles of each in 100 g and calculate the molalities.
2. Determine the concentrations of the dissolved salts in the solutions. Substitute these values into Equation $\ref{13.5.11}$ to calculate the freezing point depressions of the solutions.
Solution:
A From Figure above, we can estimate the solubilities of $\ce{NaCl}$ and $\ce{CaCl_2}$ to be about 36 g and 60 g, respectively, per 100 g of water at 0°C. The corresponding concentrations in molality are
$m_{NaCl}=\left(\dfrac{36 \; \cancel{g \;NaCl}}{100 \;\cancel{g} \;H_2O}\right)\left(\dfrac{1\; mol\; NaCl}{58.44\; \cancel{ g\; NaCl}}\right)\left(\dfrac{1000\; \cancel{g}}{1\; kg}\right)=6.2\; m \nonumber$
$m_{CaCl_2}=\left(\dfrac{60\; \cancel{g\; CaCl_2}}{100\;\cancel{g}\; H_2O}\right)\left(\dfrac{1\; mol\; CaCl_2}{110.98\; \cancel{g\; CaCl_2}}\right)\left(\dfrac{1000 \;\cancel{g}}{1 kg}\right)=5.4\; m \nonumber$
The lower formula mass of $\ce{NaCl}$ more than compensates for its lower solubility, resulting in a saturated solution that has a slightly higher concentration than $CaCl_2$.
B Because these salts are ionic compounds that dissociate in water to yield two and three ions per formula unit of $\ce{NaCl}$ and $CaCl_2$, respectively, the actual concentrations of the dissolved species in the two saturated solutions are 2 × 6.2 m = 12 m for $\ce{NaCl}$ and 3 × 5.4 m = 16 m for $CaCl_2$. The resulting freezing point depressions can be calculated using Equation $\ref{13.5.11}$:
$\ce{NaCl}: ΔT_f=mK_f=(12\; \cancel{m})(1.86°C/\cancel{m})=22°C \nonumber$
$\ce{CaCl2}: ΔT_f=mK_f=(16\;\cancel{m})(1.86°C/\cancel{m})=30°C \nonumber$
Because the freezing point of pure water is 0°C, the actual freezing points of the solutions are −22°C and −30°C, respectively. Note that $CaCl_2$ is substantially more effective at lowering the freezing point of water because its solutions contain three ions per formula unit. In fact, $CaCl_2$ is the salt usually sold for home use, and it is also often used on highways.
Because the solubilities of both salts decrease with decreasing temperature, the freezing point can be depressed by only a certain amount, regardless of how much salt is spread on an icy road. If the temperature is significantly below the minimum temperature at which one of these salts will cause ice to melt (say −35°C), there is no point in using salt until it gets warmer
Exercise $4$
Calculate the freezing point of the 30.2% solution of ethylene glycol in water whose vapor pressure and boiling point we calculated in Examples $5$ and $5$.
Answer
−13.0°C
Example $5$
Arrange these aqueous solutions in order of decreasing freezing points: 0.1 m $KCl$, 0.1 m glucose, 0.1 m SrCl2, 0.1 m ethylene glycol, 0.1 m benzoic acid, and 0.1 m HCl.
Given: molalities of six solutions
Asked for: relative freezing points
Strategy:
1. Identify each solute as a strong, weak, or nonelectrolyte, and use this information to determine the number of solute particles produced.
2. Multiply this number by the concentration of the solution to obtain the effective concentration of solute particles. The solution with the highest effective concentration of solute particles has the largest freezing point depression.
Solution:
A Because the molal concentrations of all six solutions are the same, we must focus on which of the substances are strong electrolytes, which are weak electrolytes, and which are nonelectrolytes to determine the actual numbers of particles in solution. $\ce{KCl}$, $\ce{SrCl_2}$, and $\ce{HCl}$ are strong electrolytes, producing two, three, and two ions per formula unit, respectively. Benzoic acid is a weak electrolyte (approximately one particle per molecule), and glucose and ethylene glycol are both nonelectrolytes (one particle per molecule).
B The molalities of the solutions in terms of the total particles of solute are: $\ce{KCl}$ and $\ce{HCl}$, 0.2 m; $SrCl_2$, 0.3 m; glucose and ethylene glycol, 0.1 m; and benzoic acid, 0.1–0.2 m. Because the magnitude of the decrease in freezing point is proportional to the concentration of dissolved particles, the order of freezing points of the solutions is: glucose and ethylene glycol (highest freezing point, smallest freezing point depression) > benzoic acid > $\ce{HCl}$ = $\ce{KCl}$ > $\ce{SrCl_2}$.
Exercise $5$
Arrange these aqueous solutions in order of increasing freezing points: 0.2 m $\ce{NaCl}$, 0.3 m acetic acid, 0.1 m $\ce{CaCl_2}$, and 0.2 m sucrose.
Answer
0.2 m $\ce{NaCl}$ (lowest freezing point) < 0.3 m acetic acid ≈ 0.1 m $\ce{CaCl_2}$ < 0.2 m sucrose (highest freezing point)
Colligative properties can also be used to determine the molar mass of an unknown compound. One method that can be carried out in the laboratory with minimal equipment is to measure the freezing point of a solution with a known mass of solute. This method is accurate for dilute solutions (≤1% by mass) because changes in the freezing point are usually large enough to measure accurately and precisely. By comparing $K_b$ and $K_f$ values in Table $1$, we see that changes in the boiling point are smaller than changes in the freezing point for a given solvent. Boiling point elevations are thus more difficult to measure precisely. For this reason, freezing point depression is more commonly used to determine molar mass than is boiling point elevation. Because of its very large value of $K_f$ (37.8°C/m), d-(+)-camphor (Table $1$) is often used to determine the molar mass of organic compounds by this method.
Example $6$: Sulfur
A 7.08 g sample of elemental sulfur is dissolved in 75.0 g of $CS_2$ to create a solution whose freezing point is −113.5°C. Use these data to calculate the molar mass of elemental sulfur and thus the formula of the dissolved $\ce{S_n}$ molecules (i.e., what is the value of $n$?).
Given: masses of solute and solvent and freezing point
Asked for: molar mass and number of $\ce{S}$ atoms per molecule
Strategy:
1. Use Equation $\ref{13.5.10}$, the measured freezing point of the solution, and the freezing point of $CS_2$ from Table $1$ to calculate the freezing point depression. Then use Equation $\ref{13.5.11}$ and the value of $K_f$ from Table $1$ to calculate the molality of the solution.
2. From the calculated molality, determine the number of moles of solute present.
3. Use the mass and number of moles of the solute to calculate the molar mass of sulfur in solution. Divide the result by the molar mass of atomic sulfur to obtain $n$, the number of sulfur atoms per mole of dissolved sulfur.
Solution:
A The first step is to calculate the freezing point depression using Equation $\ref{13.5.10}$:
$ΔT_f=T^0_f−T_f=−112.1°C−(−113.5°C)=1.4°C \nonumber$
Then Equation $\ref{13.5.11}$ gives
$m=\dfrac{ΔT_f}{K_f}=\dfrac{1.4° \cancel{C}}{3.74° \cancel{C}/m}=0.37\;m \nonumber$
B The total number of moles of solute present in the solution is
$\text{moles solute}=\left(\dfrac{0.37 mol}{\cancel{kg}}\right) (75.0\; g) \left(\dfrac{1 kg}{1000\; g}\right)=0.028 \;mol \nonumber$
C We now know that 0.708 g of elemental sulfur corresponds to 0.028 mol of solute. The molar mass of dissolved sulfur is thus
$\text{molar mass}=\dfrac{7.08\; g}{0.028\; mol}=260\; g/mol \nonumber$
The molar mass of atomic sulfur is 32 g/mol, so there must be 260/32 = 8.1 sulfur atoms per mole, corresponding to a formula of $\ce{S_8}$.
Exercise $6$
One of the byproducts formed during the synthesis of $C_{60}$ is a deep red solid containing only carbon. A solution of 205 mg of this compound in 10.0 g of $CCl_4$ has a freezing point of −23.38°C. What are the molar mass and most probable formula of the substance?
Answer
847 g/mol; $\ce{C_{70}}$
A Video Discussing how to find the Molecular Weight of an Unknown using Colligative Properties. Video Link: Finding the Molecular Weight of an Unknown using Colligative Properties, YouTube(opens in new window) [youtu.be]
Osmotic Pressure
Osmotic pressure is a colligative property of solutions that is observed using a semipermeable membrane, a barrier with pores small enough to allow solvent molecules to pass through but not solute molecules or ions. The net flow of solvent through a semipermeable membrane is called osmosis (from the Greek osmós, meaning “push”). The direction of net solvent flow is always from the side with the lower concentration of solute to the side with the higher concentration.
Osmosis can be demonstrated using a U-tube like the one shown in Figure $6$, which contains pure water in the left arm and a dilute aqueous solution of glucose in the right arm. A net flow of water through the membrane occurs until the levels in the arms eventually stop changing, which indicates that equilibrium has been reached. The osmotic pressure ($\Pi$) of the glucose solution is the difference in the pressure between the two sides, in this case the heights of the two columns. Although the semipermeable membrane allows water molecules to flow through in either direction, the rate of flow is not the same in both directions because the concentration of water is not the same in the two arms. The net flow of water through the membrane can be prevented by applying a pressure to the right arm that is equal to the osmotic pressure of the glucose solution.
Just as with any other colligative property, the osmotic pressure of a solution depends on the concentration of dissolved solute particles. Osmotic pressure obeys a law that resembles the ideal gas equation:
$\Pi=\dfrac{nRT}{V}=MRT \label{13.5.12}$
where $M$ is the number of moles of solute per unit volume of solution (i.e., the molarity of the solution), $R$ is the ideal gas constant, and $T$ is the absolute temperature.
As shown in Example $7$, osmotic pressures tend to be quite high, even for rather dilute solutions.
Example $7$
When placed in a concentrated salt solution, certain yeasts are able to produce high internal concentrations of glycerol to counteract the osmotic pressure of the surrounding medium. Suppose that the yeast cells are placed in an aqueous solution containing 4.0% $\ce{NaCl}$ by mass; the solution density is 1.02 g/mL at 25°C.
1. Calculate the osmotic pressure of a 4.0% aqueous $\ce{NaCl}$ solution at 25°C.
2. If the normal osmotic pressure inside a yeast cell is 7.3 atm, corresponding to a total concentration of dissolved particles of 0.30 M, what concentration of glycerol must the cells synthesize to exactly balance the external osmotic pressure at 25°C?
Given: concentration, density, and temperature of $\ce{NaCl}$ solution; internal osmotic pressure of cell
Asked for: osmotic pressure of $\ce{NaCl}$ solution and concentration of glycerol needed
Strategy:
1. Calculate the molarity of the $\ce{NaCl}$ solution using the formula mass of the solute and the density of the solution. Then calculate the total concentration of dissolved particles.
2. Use Equation \ref{13.5.12} to calculate the osmotic pressure of the solution.
3. Subtract the normal osmotic pressure of the cells from the osmotic pressure of the salt solution to obtain the additional pressure needed to balance the two. Use Equation \ref{13.5.12} to calculate the molarity of glycerol needed to create this osmotic pressure.
Solution:
A The solution contains 4.0 g of $\ce{NaCl}$ per 100 g of solution. Using the formula mass of $\ce{NaCl}$ (58.44 g/mol) and the density of the solution (1.02 g/mL), we can calculate the molarity:
\begin{align*} M_{NaCl} &=\dfrac{moles\; NaCl}{\text{liter solution}} \[4pt] &=\left(\dfrac{4.0 \; \cancel{g} \;NaCl}{58.44\; \cancel{g}/mol\; NaCl}\right)\left(\dfrac{1}{100\; \cancel{g \;solution}}\right)\left(\dfrac{1.02\; \cancel{g\; solution}}{1.00\; \cancel{mL}\; solution}\right)\left(\dfrac{1000\; \cancel{mL}}{1\; L}\right) \[4pt] &= 0.70\; M\; \ce{NaCl} \end{align*} \nonumber
Because 1 mol of $\ce{NaCl}$ produces 2 mol of particles in solution, the total concentration of dissolved particles in the solution is (2)(0.70 M) = 1.4 M.
B Now we can use Equation \ref{13.5.12} to calculate the osmotic pressure of the solution:
\begin{align*} \Pi &=MRT \[4pt] &=(1.4 \;mol/L)\left[ 0.0821\; (L⋅atm)/(K⋅mol) \right ] (298\; K)\[4pt] &=34 \;atm\end{align*} \nonumber
C If the yeast cells are to exactly balance the external osmotic pressure, they must produce enough glycerol to give an additional internal pressure of (34 atm − 7.3 atm) = 27 atm. Glycerol is a nonelectrolyte, so we can solve Equation \ref{13.5.12} for the molarity corresponding to this osmotic pressure:
\begin{align*} M&=\dfrac{\Pi}{RT}\[4pt] &=\dfrac{27\; \cancel{atm}}{[0.0821(L⋅\cancel{atm})/(\cancel{K}⋅mol)] (298 \;\cancel{K})}\[4pt] &=1.1 \;M \;\text{glycerol} \end{align*} \nonumber
In solving this problem, we could also have recognized that the only way the osmotic pressures can be the same inside the cells and in the solution is if the concentrations of dissolved particles are the same. We are given that the normal concentration of dissolved particles in the cells is 0.3 M, and we have calculated that the $\ce{NaCl}$ solution is effectively 1.4 M in dissolved particles. The yeast cells must therefore synthesize enough glycerol to increase the internal concentration of dissolved particles from 0.3 M to 1.4 M—that is, an additional 1.1 M concentration of glycerol.
Exercise $7$
Assume that the fluids inside a sausage are approximately 0.80 M in dissolved particles due to the salt and sodium nitrite used to prepare them. Calculate the osmotic pressure inside the sausage at 100°C to learn why experienced cooks pierce the semipermeable skin of sausages before boiling them.
Answer
24 atm
Because of the large magnitude of osmotic pressures, osmosis is extraordinarily important in biochemistry, biology, and medicine. Virtually every barrier that separates an organism or cell from its environment acts like a semipermeable membrane, permitting the flow of water but not solutes. The same is true of the compartments inside an organism or cell. Some specialized barriers, such as those in your kidneys, are slightly more permeable and use a related process called dialysis, which permits both water and small molecules to pass through but not large molecules such as proteins.
The same principle has long been used to preserve fruits and their essential vitamins over the long winter. High concentrations of sugar are used in jams and jellies not for sweetness alone but because they greatly increase the osmotic pressure. Thus any bacteria not killed in the cooking process are dehydrated, which keeps them from multiplying in an otherwise rich medium for bacterial growth. A similar process using salt prevents bacteria from growing in ham, bacon, salt pork, salt cod, and other preserved meats. The effect of osmotic pressure is dramatically illustrated in Figure $7$, which shows what happens when red blood cells are placed in a solution whose osmotic pressure is much lower or much higher than the internal pressure of the cells.
In addition to capillary action, trees use osmotic pressure to transport water and other nutrients from the roots to the upper branches. Evaporation of water from the leaves results in a local increase in the salt concentration, which generates an osmotic pressure that pulls water up the trunk of the tree to the leaves.
Finally, a process called reverse osmosis can be used to produce pure water from seawater. As shown schematically in Figure $8$, applying high pressure to seawater forces water molecules to flow through a semipermeable membrane that separates pure water from the solution, leaving the dissolved salt behind. Large-scale desalinization plants that can produce hundreds of thousands of gallons of freshwater per day are common in the desert lands of the Middle East, where they supply a large proportion of the freshwater needed by the population. Similar facilities are now being used to supply freshwater in southern California. Small, hand-operated reverse osmosis units can produce approximately 5 L of freshwater per hour, enough to keep 25 people alive, and are now standard equipment on US Navy lifeboats.
A Video Discussing Osmotic Pressure. Video Link: Osmotic Pressure, YouTube(opens in new window) [youtu.be] (opens in new window)
Colligative Properties of Electrolyte Solutions
Thus far we have assumed that we could simply multiply the molar concentration of a solute by the number of ions per formula unit to obtain the actual concentration of dissolved particles in an electrolyte solution. We have used this simple model to predict such properties as freezing points, melting points, vapor pressure, and osmotic pressure. If this model were perfectly correct, we would expect the freezing point depression of a 0.10 m solution of sodium chloride, with 2 mol of ions per mole of $\ce{NaCl}$ in solution, to be exactly twice that of a 0.10 m solution of glucose, with only 1 mol of molecules per mole of glucose in solution. In reality, this is not always the case. Instead, the observed change in freezing points for 0.10 m aqueous solutions of $\ce{NaCl}$ and $\ce{KCl}$ are significantly less than expected (−0.348°C and −0.344°C, respectively, rather than −0.372°C), which suggests that fewer particles than we expected are present in solution.
The relationship between the actual number of moles of solute added to form a solution and the apparent number as determined by colligative properties is called the van’t Hoff factor ($i$) and is defined as follows:
$i=\dfrac{\text{apparent number of particles in solution}}{\text{ number of moles of solute dissolved}} \label{13.5.13}$
Named for Jacobus Hendricus van’t Hoff (1852–1911), a Dutch chemistry professor at the University of Amsterdam who won the first Nobel Prize in Chemistry (1901) for his work on thermodynamics and solutions.
As the solute concentration increases, the van’t Hoff factor decreases.
The van’t Hoff factor is therefore a measure of a deviation from ideal behavior. The lower the van’t Hoff factor, the greater the deviation. As the data in Table $2$ show, the van’t Hoff factors for ionic compounds are somewhat lower than expected; that is, their solutions apparently contain fewer particles than predicted by the number of ions per formula unit. As the concentration of the solute increases, the van’t Hoff factor decreases because ionic compounds generally do not totally dissociate in aqueous solution.
Table $2$: van’t Hoff Factors for 0.0500 M Aqueous Solutions of Selected Compounds at 25°C
Compound i (measured) i (ideal)
glucose 1.0 1.0
sucrose 1.0 1.0
$NaCl$ 1.9 2.0
$HCl$ 1.9 2.0
$MgCl_2$ 2.7 3.0
$FeCl_3$ 3.4 4.0
$Ca(NO_3)_2$ 2.5 3.0
$AlCl_3$ 3.2 4.0
$MgSO_4$ 1.4 2.0
Instead, some of the ions exist as ion pairs, a cation and an anion that for a brief time are associated with each other without an intervening shell of water molecules (Figure $9$). Each of these temporary units behaves like a single dissolved particle until it dissociates. Highly charged ions such as $Mg^{2+}$, $Al^{3+}$, $SO_4^{2−}$, and $PO_4^{3−}$ have a greater tendency to form ion pairs because of their strong electrostatic interactions. The actual number of solvated ions present in a solution can be determined by measuring a colligative property at several solute concentrations.
Example $8$: Iron Chloride in Water
A 0.0500 M aqueous solution of $FeCl_3$ has an osmotic pressure of 4.15 atm at 25°C. Calculate the van’t Hoff factor $i$ for the solution.
Given: solute concentration, osmotic pressure, and temperature
Asked for: van’t Hoff factor
Strategy:
1. Use Equation $\ref{13.5.12}$ to calculate the expected osmotic pressure of the solution based on the effective concentration of dissolved particles in the solvent.
2. Calculate the ratio of the observed osmotic pressure to the expected value. Multiply this number by the number of ions of solute per formula unit, and then use Equation $\ref{13.5.13}$ to calculate the van’t Hoff factor.
Solution:
A If $\ce{FeCl_3}$ dissociated completely in aqueous solution, it would produce four ions per formula unit [Fe3+(aq) plus 3Cl−(aq)] for an effective concentration of dissolved particles of 4 × 0.0500 M = 0.200 M. The osmotic pressure would be
\begin{align*} \Pi &=MRT \[4pt] &=(0.200 \;mol/L) \left[0.0821\;(L⋅atm)/(K⋅mol) \right] (298\; K)=4.89\; atm \end{align*} \nonumber
B The observed osmotic pressure is only 4.15 atm, presumably due to ion pair formation. The ratio of the observed osmotic pressure to the calculated value is 4.15 atm/4.89 atm = 0.849, which indicates that the solution contains (0.849)(4) = 3.40 particles per mole of $\ce{FeCl_3}$ dissolved. Alternatively, we can calculate the observed particle concentration from the osmotic pressure of 4.15 atm:
$4.15\; atm=M \left( \dfrac{0.0821 \;(L⋅atm)}{(K⋅mol)}\right) (298 \;K) \nonumber$
or after rearranging
$M = 0.170 mol \nonumber$
The ratio of this value to the expected value of 0.200 M is 0.170 M/0.200 M = 0.850, which again gives us (0.850)(4) = 3.40 particles per mole of $\ce{FeCl_3}$ dissolved. From Equation $\ref{13.5.13}$, the van’t Hoff factor for the solution is
$i=\dfrac{\text{3.40 particles observed}}{\text{1 formula unit}\; \ce{FeCl_3}}=3.40 \nonumber$
Exercise $8$: Magnesium Chloride in Water
Calculate the van’t Hoff factor for a 0.050 m aqueous solution of $MgCl_2$ that has a measured freezing point of −0.25°C.
Answer
2.7 (versus an ideal value of 3).
A Video Discussing the Colligative Properties in Solutions. Video Link: Colligative Properties in Solutions, YouTube(opens in new window) [youtu.be]
Summary
The colligative properties of a solution depend on only the total number of dissolved particles in solution, not on their chemical identity. Colligative properties include vapor pressure, boiling point, freezing point, and osmotic pressure. The addition of a nonvolatile solute (one without a measurable vapor pressure) decreases the vapor pressure of the solvent. The vapor pressure of the solution is proportional to the mole fraction of solvent in the solution, a relationship known as Raoult’s law. Solutions that obey Raoult’s law are called ideal solutions. Most real solutions exhibit positive or negative deviations from Raoult’s law. The boiling point elevation ($ΔT_b$) and freezing point depression ($ΔT_f$) of a solution are defined as the differences between the boiling and freezing points, respectively, of the solution and the pure solvent. Both are proportional to the molality of the solute. When a solution and a pure solvent are separated by a semipermeable membrane, a barrier that allows solvent molecules but not solute molecules to pass through, the flow of solvent in opposing directions is unequal and produces an osmotic pressure, which is the difference in pressure between the two sides of the membrane. Osmosis is the net flow of solvent through such a membrane due to different solute concentrations. Dialysis uses a semipermeable membrane with pores that allow only small solute molecules and solvent molecules to pass through. In more concentrated solutions, or in solutions of salts with highly charged ions, the cations and anions can associate to form ion pairs, which decreases their effect on the colligative properties of the solution. The extent of ion pair formation is given by the van’t Hoff factor (i), the ratio of the apparent number of particles in solution to the number predicted by the stoichiometry of the salt.
• Henry’s law: $C = kP \nonumber$
• Raoult’s law: $P_A=\chi_AP^0_A \nonumber$
• vapor pressure lowering: $P^0_A−P_A=ΔP_A=\chi_BP^0_A \nonumber$
• vapor pressure of a system containing two volatile components: $P_{tot}=\chi_AP^0_A+(1−\chi_A)P^0_B \nonumber$
• boiling point elevation: $ΔT_b = mK_b \nonumber$
• freezing point depression: $ΔT_f = mK_f \nonumber$
• osmotic pressure: $\Pi=nRTV=MRT \nonumber$
• van ’t Hoff factor: $i=\dfrac{\text{apparent number of particles in solution}}{\text{ number of moles of solute dissolved}} \nonumber$ | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/13%3A_Properties_of_Solutions/13.05%3A_Colligative_Properties.txt |
Learning Objectives
• To distinguish between true solutions and solutions with aggregate particles.
Suspensions and colloids are two common types of mixtures whose properties are in many ways intermediate between those of true solutions and heterogeneous mixtures. A suspension is a heterogeneous mixture of particles with diameters of about 1 µm (1000 nm) that are distributed throughout a second phase. Common suspensions include paint, blood, and hot chocolate, which are solid particles in a liquid, and aerosol sprays, which are liquid particles in a gas. If the suspension is allowed to stand, the two phases will separate, which is why paints must be thoroughly stirred or shaken before use. A colloid is also a heterogeneous mixture, but the particles of a colloid are typically smaller than those of a suspension, generally in the range of 2 to about 500 nm in diameter. Colloids include fog and clouds (liquid particles in a gas), milk (solid particles in a liquid), and butter (solid particles in a solid). Other colloids are used industrially as catalysts. Unlike in a suspension, the particles in a colloid do not separate into two phases on standing. The only combination of substances that cannot produce a suspension or a colloid is a mixture of two gases because their particles are so small that they always form true solutions. The properties of suspensions, colloids, and solutions are summarized in Table \(1\).
Table \(1\): Properties of Liquid Solutions, Colloids, and Suspensions
Type of Mixture Approximate Size of Particles (nm) Characteristic Properties Examples
solution < 2 not filterable; does not separate on standing; does not scatter visible light air, white wine, gasoline, salt water
colloid 2–500 scatters visible light; translucent or opaque; not filterable; does not separate on standing smoke, fog, ink, milk, butter, cheese
suspension 500–1000 cloudy or opaque; filterable; separates on standing muddy water, hot cocoa, blood, paint
Colloids and Suspensions
Colloids were first characterized in about 1860 by Thomas Graham, who also gave us Graham’s law of diffusion and effusion. Although some substances, such as starch, gelatin, and glue, appear to dissolve in water to produce solutions, Graham found that they diffuse very slowly or not at all compared with solutions of substances such as salt and sugar. Graham coined the word colloid (from the Greek kólla, meaning “glue”) to describe these substances, as well as the words sol and gel to describe certain types of colloids in which all of the solvent has been absorbed by the solid particles, thus preventing the mixture from flowing readily, as we see in Jell-O. Two other important types of colloids are aerosols, which are dispersions of solid or liquid particles in a gas, and emulsions, which are dispersions of one liquid in another liquid with which it is immiscible.
Colloids share many properties with solutions. For example, the particles in both are invisible without a powerful microscope, do not settle on standing, and pass through most filters. However, the particles in a colloid scatter a beam of visible light, a phenomenon known as the Tyndall effect,The effect is named after its discoverer, John Tyndall, an English physicist (1820–1893). whereas the particles of a solution do not. The Tyndall effect is responsible for the way the beams from automobile headlights are clearly visible from the side on a foggy night but cannot be seen from the side on a clear night. It is also responsible for the colored rays of light seen in many sunsets, where the sun’s light is scattered by water droplets and dust particles high in the atmosphere. An example of the Tyndall effect is shown in Figure \(1\).
Although colloids and suspensions can have particles similar in size, the two differ in stability: the particles of a colloid remain dispersed indefinitely unless the temperature or chemical composition of the dispersing medium is changed. The chemical explanation for the stability of colloids depends on whether the colloidal particles are hydrophilic or hydrophobic.
Most proteins, including those responsible for the properties of gelatin and glue, are hydrophilic because their exterior surface is largely covered with polar or charged groups. Starch, a long-branched polymer of glucose molecules, is also hydrophilic. A hydrophilic colloid particle interacts strongly with water, resulting in a shell of tightly bound water molecules that prevents the particles from aggregating when they collide. Heating such a colloid can cause aggregation because the particles collide with greater energy and disrupt the protective shell of solvent. Moreover, heat causes protein structures to unfold, exposing previously buried hydrophobic groups that can now interact with other hydrophobic groups and cause the particles to aggregate and precipitate from solution. When an egg is boiled, for example, the egg white, which is primarily a colloidal suspension of a protein called albumin, unfolds and exposes its hydrophobic groups, which aggregate and cause the albumin to precipitate as a white solid.
In some cases, a stable colloid can be transformed to an aggregated suspension by a minor chemical modification. Consider, for example, the behavior of hemoglobin, a major component of red blood cells. Hemoglobin molecules normally form a colloidal suspension inside red blood cells, which typically have a “donut” shape and are easily deformed, allowing them to squeeze through the capillaries to deliver oxygen to tissues. In a common inherited disease called sickle-cell anemia, one of the amino acids in hemoglobin that has a hydrophilic carboxylic acid side chain (glutamate) is replaced by another amino acid that has a hydrophobic side chain (valine). Under some conditions, the abnormal hemoglobin molecules can aggregate to form long, rigid fibers that cause the red blood cells to deform, adopting a characteristic sickle shape that prevents them from passing through the capillaries (Figure \(2\)). The reduction in blood flow results in severe cramps, swollen joints, and liver damage. Until recently, many patients with sickle-cell anemia died before the age of 30 from infection, blood clots, or heart or kidney failure, although individuals with the sickle-cell genetic trait are more resistant to malaria than are those with “normal” hemoglobin.
Aggregation and precipitation can also result when the outer, charged layer of a particle is neutralized by ions with the opposite charge. In inland waterways, clay particles, which have a charged surface, form a colloidal suspension. High salt concentrations in seawater neutralize the charge on the particles, causing them to precipitate and form land at the mouths of large rivers, as seen in the satellite view in Figure \(3\). Charge neutralization is also an important strategy for precipitating solid particles from gaseous colloids such as smoke, and it is widely used to reduce particulate emissions from power plants that burn fossil fuels.
Emulsions
Emulsions are colloids formed by the dispersion of a hydrophobic liquid in water, thereby bringing two mutually insoluble liquids, such as oil and water, in close contact. Various agents have been developed to stabilize emulsions, the most successful being molecules that combine a relatively long hydrophobic “tail” with a hydrophilic “head”:
Examples of such emulsifying agents include soaps, which are salts of long-chain carboxylic acids, such as sodium stearate \(\ce{[CH_3(CH_2)_{16}CO_2−Na^{+}]}\), and detergents, such as sodium dodecyl sulfate \(\ce{[CH_3(CH_2)_{11}OSO_3−Na^{+}]}\), whose structures are as follows:
When you wash your laundry, the hydrophobic tails of soaps and detergents interact with hydrophobic particles of dirt or grease through dispersion forces, dissolving in the interior of the hydrophobic particle. The hydrophilic group is then exposed at the surface of the particle, which enables it to interact with water through ion–dipole forces and hydrogen bonding. This causes the particles of dirt or grease to disperse in the wash water and allows them to be removed by rinsing. Similar agents are used in the food industry to stabilize emulsions such as mayonnaise.
A related mechanism allows us to absorb and digest the fats in buttered popcorn and French fries. To solubilize the fats so that they can be absorbed, the gall bladder secretes a fluid called bile into the small intestine. Bile contains a variety of bile salts, detergent-like molecules that emulsify the fats.
Micelles
Detergents and soaps are surprisingly soluble in water in spite of their hydrophobic tails. The reason for their solubility is that they do not, in fact, form simple solutions. Instead, above a certain concentration they spontaneously form micelles, which are spherical or cylindrical aggregates that minimize contact between the hydrophobic tails and water. In a micelle, only the hydrophilic heads are in direct contact with water, and the hydrophobic tails are in the interior of the aggregate (Figure \(\PageIndex{4a}\)).
A large class of biological molecules called phospholipids consists of detergent-like molecules with a hydrophilic head and two hydrophobic tails, as can be seen in the molecule of phosphatidylcholine. The additional tail results in a cylindrical shape that prevents phospholipids from forming a spherical micelle. Consequently, phospholipids form bilayers, extended sheets consisting of a double layer of molecules. As shown in Figure \(\PageIndex{4b}\), the hydrophobic tails are in the center of the bilayer, where they are not in contact with water, and the hydrophilic heads are on the two surfaces, in contact with the surrounding aqueous solution.
A cell membrane is essentially a mixture of phospholipids that form a phospholipid bilayer. One definition of a cell is a collection of molecules surrounded by a phospholipid bilayer that is capable of reproducing itself. The simplest cells are bacteria, which consist of only a single compartment surrounded by a single membrane. Animal and plant cells are much more complex, however, and contain many different kinds of compartments, each surrounded by a membrane and able to carry out specialized tasks.
Summary
A suspension is a heterogeneous mixture of particles of one substance distributed throughout a second phase; the dispersed particles separate from the dispersing phase on standing. In contrast, the particles in a colloid are smaller and do not separate on standing. A colloid can be classified as a sol, a dispersion of solid particles in a liquid or solid; a gel, a semisolid sol in which all of the liquid phase has been absorbed by the solid particles; an aerosol, a dispersion of solid or liquid particles in a gas; or an emulsion, a dispersion of one liquid phase in another. A colloid can be distinguished from a true solution by its ability to scatter a beam of light, known as the Tyndall effect. Hydrophilic colloids contain an outer shell of groups that interact favorably with water, whereas hydrophobic colloids have an outer surface with little affinity for water. Emulsions are prepared by dispersing a hydrophobic liquid in water. In the absence of a dispersed hydrophobic liquid phase, solutions of detergents in water form organized spherical aggregates called micelles. Phospholipids are a class of detergent-like molecules that have two hydrophobic tails attached to a hydrophilic head. A bilayer is a two-dimensional sheet consisting of a double layer of phospholipid molecules arranged tail to tail with a hydrophobic interior and a hydrophilic exterior. Cells are collections of molecules that are surrounded by a phospholipid bilayer called a cell membrane and are able to reproduce themselves. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/13%3A_Properties_of_Solutions/13.06%3A_Colloids.txt |
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
13.1: The Solution Process
Conceptual Problems
1. Classify each of the following as a heterogeneous mixture or homogeneous mixture. Explain your rationale in each case.
1. aqueous ammonia
2. liquid decongestant
3. vinegar
4. seawater
5. gasoline
6. fog
2. Solutions and heterogeneous mixtures are at the extreme ends of the solubility scale. Name one type of mixture that is intermediate on this scale. How are the properties of the mixture you have chosen different from those of a solution or a heterogeneous mixture?
1. a naphthalene mothball dissolving in benzene
2. a sample of a common drain cleaner that has a mixture of $NaOH$ crystals and $Al$ chunks dissolving in water to give $H_2$ gas and an aqueous solution of $Na^+$, $OH^−$, and $Al^{3+}$ ions
3. an iron ship anchor slowly dissolving in seawater
4. sodium metal dissolving in liquid ammonia
5. a sugar cube dissolving in a cup of hot tea
6. $SO_3$ gas dissolving in water to produce sulfuric acid
7. calcium oxide dissolving in water to produce a basic solution
8. metallic gold dissolving in a small quantity of liquid mercury
1. Classify each process as simple dissolution or a chemical reaction.
2. Classify each process as simple dissolution or a chemical reaction.
3. You notice that a gas is evolved as you are dissolving a solid in a liquid. Will you be able to recover your original solid by evaporation? Why or why not?
4. Why is heat evolved when sodium hydroxide pellets are dissolved in water? Does this process correspond to simple dissolution or a chemical reaction? Justify your answer.
5. Which process(es) is the simple formation of a solution, and which process(es) involves a chemical reaction?
1. mixing an aqueous solution of NaOH with an aqueous solution of HCl
2. bubbling HCl gas through water
3. adding iodine crystals to CCl4
4. adding sodium metal to ethanol to produce sodium ethoxide (C2H5ONa+) and hydrogen gas
1. Using thermochemical arguments, explain why some substances that do not form a solution at room temperature will form a solution when heated. Explain why a solution can form even when $ΔH_{soln}$ is positive.
2. If you wanted to formulate a new compound that could be used in an instant cold pack, would you select a compound with a positive or negative value of $ΔH_{soln}$ in water? Justify your answer.
3. Why is entropy the dominant factor in the formation of solutions of two or more gases? Is it possible for two gases to be immiscible? Why or why not?
Conceptual Answers
1. Homogeneous mixtures: aqueous ammonia, liquid decongestant, vinegar, and gasoline. Heterogeneous mixtures: seawater and fog.
2. All are chemical reactions except dissolving iodine crystals in $CCl_4$.
13.2: Saturated Solutions and Solubility
Conceptual Problems
1. If a compound is only slightly soluble in a particular solvent, what are the relative strengths of the solvent–solvent and solute–solute interactions versus the solute–solvent interactions?
2. Predict whether each of the following sets of conditions favors formation of a solution:
Intermolecular Attractive Forces (Solute) Intermolecular Attractive Forces (Solvent) $ΔH_{soln}$
London dispersion hydrogen bonding slightly positive
dipole–dipole hydrogen bonding very negative
ionic dipole–dipole slightly positive
ionic London dispersion positive
1. Arrange the following liquids in order of increasing solubility in water: t-butanol [(CH3)3COH], benzene, ammonia, and heptane. Justify your answer.
2. Which compound in each pair will be more soluble in water? Explain your reasoning in each case.
1. toluene ($C_7H_8$) or ethyl ether ($C_2H_5OC_2H_5$)
2. chloroform ($CHCl_3$) or acetone ($CH_3COCH_3$)
3. carbon tetrachloride ($CCl_4$) or tetrahydrofuran ($C_4H_8O$)
4. $CaCl_2$ or $CH_2Cl_2$
3. Which compound in each pair will be more soluble in benzene? Explain your reasoning in each case.
1. cyclohexane or methanol
2. $I_2$ or $MgCl_2$
3. methylene chloride ($CH_2Cl_2$) or acetic acid
4. Two water-insoluble compounds—n-decylamine $[CH_3(CH_2)_9NH_2]$ and n-decane—can be separated by the following procedure: The compounds are dissolved in a solvent such as toluene that is immiscible with water. When adding an aqueous $HCl$ solution to the mixture and stirring vigorously, the $HCl$ reacts with one of the compounds to produce a salt. When the stirring is stopped and the mixture is allowed to stand, two layers are formed. At this point, each layer contains only one of the two original compounds. After the layers are separated, adding aqueous $NaOH$ to the aqueous layer liberates one of the original compounds, which can then be removed by stirring with a second portion of toluene to extract it from the water.
1. Identify the compound that is present in each layer following the addition of HCl. Explain your reasoning.
2. How can the original compounds be recovered from the toluene solution?
5. Bromine and iodine are both soluble in $CCl_4$, but bromine is much more soluble. Why?
6. A solution is made by mixing 50.0 mL of liquid A with 75.0 mL of liquid B. Which is the solute, and which is the solvent? Is it valid to assume that the volume of the resulting solution will be 125 mL? Explain your answer.
7. The compounds NaI, NaBr, and NaCl are far more soluble in water than NaF, a substance that is used to fluoridate drinking water. In fact, at 25°C the solubility of NaI is 184 g/100 mL of water, versus only 4.2 g/100 mL of water for NaF. Why is sodium iodide so much more soluble in water? Do you expect KCl to be more soluble or less soluble in water than NaCl?
8. When water is mixed with a solvent with which it is immiscible, the two liquids usually form two separate layers. If the density of the nonaqueous solvent is 1.75 g/mL at room temperature, sketch the appearance of the heterogeneous mixture in a beaker and label which layer is which. If you were not sure of the density and the identity of the other liquid, how might you be able to identify which is the aqueous layer?
9. When two liquids are immiscible, the addition of a third liquid can occasionally be used to induce the formation of a homogeneous solution containing all three.
1. Ethylene glycol ($HOCH_2CH_2OH$) and hexane are immiscible, but adding acetone $[(CH_3)_2CO]$ produces a homogeneous solution. Why does adding a third solvent produce a homogeneous solution?
2. Methanol and n-hexane are immiscible. Which of the following solvents would you add to create a homogeneous solution—water, n-butanol, or cyclohexane? Justify your choice.
10. Some proponents of vitamin therapy for combating illness encourage the consumption of large amounts of fat-soluble vitamins. Why can this be dangerous? Would it be as dangerous to consume large amounts of water-soluble vitamins? Why or why not?
11. Why are most metals insoluble in virtually all solvents?
12. Because sodium reacts violently with water, it is difficult to weigh out small quantities of sodium metal for a reaction due to its rapid reaction with small amounts of moisture in the air. Would a Na/Hg amalgam be as sensitive to moisture as metallic sodium? Why or why not? A Na/K alloy is a liquid at room temperature. Will it be more or less sensitive to moisture than solid Na or K?
13. Dental amalgams often contain high concentrations of Hg, which is highly toxic. Why isn’t dental amalgam toxic?
14. Arrange 2,2,3-trimethylpentane, 1-propanol, toluene ($C_7H_8$), and dimethyl sulfoxide [$\ce{(CH_3)_2S=O}$] in order of increasing dipole moment. Explain your reasoning.
15. Arrange acetone, chloroform, cyclohexane, and 2-butanol in order of increasing dielectric constant. Explain your reasoning.
16. Dissolving a white crystalline compound in ethanol gave a blue solution. Evaporating the ethanol from the solution gave a bluish-crystalline product, which slowly transformed into the original white solid on standing in the air for several days. Explain what happened. How does the mass of the initial bluish solid compare with the mass of the white solid finally recovered?
17. You have been asked to develop a new drug that could be used to bind Fe3+ ions in patients who suffer from iron toxicity, allowing the bound iron to be excreted in the urine. Would you consider a crown ether or a cryptand to be a reasonable candidate for such a drug? Explain your answer.
18. Describe two different situations in which fractional crystallization will not work as a separation technique when attempting to isolate a single compound from a mixture.
19. You have been given a mixture of two compounds—A and B—and have been told to isolate pure A. You know that pure A has a lower solubility than pure B and that the solubilities of both A and B increase with temperature. Outline a procedure to isolate pure A. If B had the lower solubility, could you use the same procedure to isolate pure A? Why or why not?
Conceptual Answers
7. London dispersion forces increase with increasing atomic mass. Iodine is a solid while bromine is a liquid due to the greater intermolecular interactions between the heavier iodine atoms. Iodine is less soluble than bromine in virtually all solvents because it requires more energy to separate $I_2$ molecules than $Br_2$ molecules.
11.
1. A third solvent with intermediate polarity and/or dielectric constant can effectively dissolve both of the immiscible solvents, creating a single liquid phase.
2. n-butanol—it is intermediate in polarity between methanol and n-hexane, while water is more polar than either and cyclohexane is comparable to n-hexane.
15. In dental amalgam, the mercury atoms are locked in a solid phase that does not undergo corrosion under physiological conditions; hence, the mercury atoms cannot readily diffuse to the surface where they could vaporize or undergo chemical reaction.
21. Dissolve the mixture of A and B in a solvent in which they are both soluble when hot and relatively insoluble when cold, filter off any undissolved B, and cool slowly. Pure A should crystallize, while B stays in solution. If B were less soluble, it would be impossible to obtain pure A by this method in a single step, because some of the less soluble compound (B) will always be present in the solid that crystallizes from solution.
13.3: Factors Affecting Solubility
Conceptual Problems
1. Use the kinetic molecular theory of gases discussed in Chapter 10 to explain why the solubility of virtually all gases in liquids decreases with increasing temperature.
2. An industrial plant uses water from a nearby stream to cool its reactor and returns the water to the stream after use. Over a period of time, dead fish start to appear downstream from the plant, but there is no evidence for any leaks of potentially toxic chemicals into the stream. What other factor might be causing the fish to die?
3. One manufacturer’s instructions for setting up an aquarium specify that if boiled water is used, the water must be cooled to room temperature and allowed to stand overnight before fish are added. Why is it necessary for the water to stand?
4. Using a carbonated beverage as an example, discuss the effect of temperature on the “fizz.” How does the “foaminess” of a carbonated beverage differ between Los Angeles, California, and Denver, Colorado?
5. A common laboratory technique for degassing a solvent is to place it in a flask that is sealed to the atmosphere and then evacuate the flask to remove any gases above the liquid. Why is this procedure effective? Why does the temperature of the solvent usually decrease substantially during this process?
Conceptual Answers
1. When water is boiled, all of the dissolved oxygen and nitrogen are removed. When the water is cooled to room temperature, it initially contains very little dissolved oxygen. Allowing the water to stand overnight allows oxygen in the air to dissolve, so that the fish will not suffocate.
1. Evacuating the flask to remove gases decreases the partial pressure of oxygen above the solution. According to Henry’s law, the solubility of any gas decreases as its partial pressure above the solution decreases. Consequently, dissolved oxygen escapes from solution into the gas phase, where it is removed by the vacuum pump. Filling the flask with nitrogen gas and repeating this process several times effectively removes almost all of the dissolved oxygen. The temperature of the solvent decreases because some solvent evaporates as well during this process. The heat that is required to evaporate some of the liquid is initially removed from the rest of the solvent, decreasing its temperature.
Numerical Problems
1. The solubility of $CO_2$ in water at 0°C and 1 atm is 0.335 g/100 g of $H_2O$. At 20°C and 1 atm, the solubility of $CO_2$ in water is 0.169 g/100 g of $H_2O$.
1. What volume of $CO_2$ would be released by warming 750 g of water saturated with $CO_2$ from 0°C to 20°C?
2. What is the value of the Henry’s law constant for $CO_2$ under each set of conditions?
1. The solubility of $O_2$ in 100 g of $H_2O$ at varying temperatures and a pressure of 1 atm is given in the following table:
Solubility (g) Temperature (°C)
0.0069 0
0.0054 10
0.0043 20
1. What is the value of the Henry’s law constant at each temperature?
2. Does Henry’s law constant increase or decrease with increasing temperature?
3. At what partial pressure of $O_2$ would the concentration of $O_2$ in water at 0°C be the same as the concentration in water at 20°C at a partial pressure of 1 atm?
4. Assuming that air is 20% $O_2$ by volume, at what atmospheric pressure would the $O_2$ concentration be the same at 20°C as it is at atmospheric pressure and 0°C?
Numerical Answer
1. 0.678 L $CO_2$
2. $k_{0°C} = 7.61 \times 10^{-2}\; M/atm$, $k_{20°C} = 3.84 \times 10^{-2}\; M/atm$
13.4: Ways of Expressing Concentration
Conceptual Problems
1. Does the molality have the same numerical value as the molarity for a highly concentrated aqueous solution of fructose ($C_6H_{12}O_6$) (approximately 3.2 M)? Why or why not?
2. Explain why the molality and molarity of an aqueous solution are not always numerically identical. Will the difference between the two be greater for a dilute or a concentrated solution? Explain your answer.
3. Under what conditions are molality and molarity likely to be equal? Is the difference between the two greater when water is the solvent or when the solvent is not water? Why?
4. What is the key difference between using mole fraction or molality versus molarity to describe the concentration of a solution? Which unit(s) of concentration is most appropriate for experiments that must be carried out at several different temperatures?
5. An experiment that relies on very strict control of the reaction stoichiometry calls for adding 50.0 mL of a 0.95 M solution of A to 225 mL of a 1.01 M solution of B, followed by heating for 1 h at 60°C. To save time, a student decided to heat solution B to 60°C before measuring out 225 mL of solution B, transferring it to the flask containing solution A, and proceeding normally. This change in procedure caused the yield of product to be less than usual. How could such an apparently minor change in procedure have resulted in a decrease in the yield?
Numerical Problems
1. Complete the following table for aqueous solutions of the compounds given.
Compound Molarity (M) Solution Density (g/mL) Mole Fraction (X)
$H_2SO_4$ 18.0 1.84
$CH_3COOH$ 1.00 $7.21 \times 10^{−3}$
$KOH$ 3.60 1.16
1. Complete the following table for each compound given.
Compound Mass (g) Volume of Solution (mL) Molarity (M)
$Na_2SO_4$ 7.80 225
$KNO_3$ 125 1.27
$NaO_2CCH_3$ 18.64 0.95
1. How would you prepare 100.0 mL of an aqueous solution with 0.40 M KI? a solution with 0.65 M $NaCN$?
2. Calculate the molality of a solution with 775 mg of $NaCl$ in 500.0 g of water. Do you expect the molarity to be the same as the molality? Why or why not?
3. What is the molarity of each solution?
1. 12.8 g of glucose ($C_6H_{12}O_{6}$) in water, total volume 150.0 mL
2. 9.2 g of $Na_3PO_4$ in water, total volume 200.0 mL
3. 843 mg of I2 in $EtOH$, total volume 150.0 mL
1. A medication used to treat abnormal heart rhythms is labeled “Procainamide 0.5 g/250 cc.” Express this concentration in parts per thousand.
2. Meperidine is a medication used for pain relief. A bottle of meperidine is labeled as 50 mg/mL. Express this concentration in parts per thousand.
3. An aqueous solution that is 4.61% NaOH by mass has a density of 1.06 g/mL. Calculate the molarity of the solution, the mole fraction of $NaOH$, and the molality of the solution.
4. A solution of concentrated phosphoric acid contains 85.0% $H_3PO_4$ by mass and has a density of 1.684 g/mL. Calculate the following.
1. the molarity of the solution
2. the mole fraction of $H_3PO_4$
3. the molality of the solution
1. A solution of commercial concentrated nitric acid is 16 M $HNO3$ and has a density of 1.42 g/mL. What is the percentage of $HNO3$ in the solution by mass? What is the molality?
2. A commercial aqueous ammonia solution contains 28.0% $NH3$ by mass and has a density of 0.899 g/mL. Calculate the following.
1. the molarity
2. the mole fraction
3. Concentrated, or glacial, acetic acid is pure acetic acid and has a density of 1.053 g/mL. It is widely used in organic syntheses, in the manufacture of rayon and plastics, as a preservative in foods, and occasionally to treat warts. What volume of glacial acetic acid is required to prepare 5.00 L of a 1.75 M solution of acetic acid in ethanol?
4. Solutions of sodium carbonate decahydrate, also known as washing soda, are used as skin cleansers. The solubility of this compound in cold water is 21.52 g/100 mL. If a saturated solution has a density of 1.20 g/mL, what is its molarity? What is the mole fraction of sodium carbonate decahydrate in this solution?
5. Hydrogen peroxide ($H_2O_2$) is usually sold over the counter as an aqueous solution that is 3% by mass. Assuming a solution density of 1.01 g/mL, what is the molarity of hydrogen peroxide? What is the molar concentration of a solution that is 30% hydrogen peroxide by mass (density = 1.112 g/mL)? How would you prepare 100.0 mL of a 3% solution from the 30% solution?
6. Determine the concentration of a solution with 825 mg of $Na_2HPO_4$ dissolved in 450.0 mL of $H_2O$ at 20°C in molarity, molality, mole fraction, and parts per million. Assume that the density of the solution is the same as that of water. Which unit of concentration is most convenient for calculating vapor pressure changes? Why?
7. How many moles of $Cl^−$ are there in 25.0 mL of a 0.15 M $CaCl_2$ solution?
8. How many moles of Na+ are there in 25.0 g of a $1.33 \times 10^{-3}\; m \;Na_2HPO_4$ solution? What is the sodium concentration of this solution in ppb?
9. How many grams of copper are there in 30.0 mL of a 0.100 M $CuSO_4$ solution?
10. How many grams of nitrate ion are there in 75.0 g of a $1.75 \times 10^{−4} m$ $Pb(NO_3)_2$ solution? What is the nitrate concentration of the solution in ppb?
11. How many milliliters of a 0.750 M solution of $K_2CrO_4$ are required to deliver 250 mg of chromate ion?
12. How many milliliters of a $1.95 \times 10^{−6} M$ solution of $Ag_3PO_4$ are required to deliver 14.0 mg of $Ag^+$?
13. Iron reacts with bromine according to the following equation:$2Fe_{(s)} + 3Br_{2(aq)} \rightarrow 2FeBr_{3(aq)}$
14. How many milliliters of a $5.0 \times 10^{−2} M$ solution of bromine in water are required to react completely with 750.0 mg of iron metal?
15. Aluminum reacts with $HCl$ according to the following equation: $2Al_{(s)} + 6HCl_{(aq)} \rightarrow 2AlCl_{3(aq)} + 3H_{2(g)}$
16. If 25.0 mL of a solution of $HCl$ in water is required to react completely with 1.05 g of aluminum metal, what is the molarity of the HCl solution?
17. The precipitation of silver chloride is a diagnostic test for the presence of chloride ion. If 25.0 mL of 0.175 M $AgNO_3$ are required to completely precipitate the chloride ions from 10.0 mL of an $NaCl$ solution, what was the original concentration of $NaCl$?
18. Barium sulfate is virtually insoluble. If a 10.0 mL solution of 0.333 M $Ba(NO_3)_2$ is stirred with 40.0 mL of a 0.100 M $Na_2SO_4$, how many grams of barium sulfate will precipitate? Which reactant is present in excess? What is its final concentration?
13.5: Colligative Properties
Conceptual Problems
1. Why does the vapor pressure of a solvent decrease when adding a nonvolatile solute?
2. Does seawater boil at the same temperature as distilled water? If not, which has the higher boiling point? Explain your answer.
3. Which will be more soluble in benzene—$O_2$ or $HCl$? Will $H_2S$ or $HCl$ be more soluble in water? Explain your reasoning in each case.
4. Will the vapor pressure of a solution of hexane and heptane have an ideal vapor pressure curve (i.e., obey Raoult’s law)? Explain your answer. What properties of two liquids determine whether a solution of the two will exhibit an ideal behavior?
5. Predict whether the following mixtures will exhibit negative, zero, or positive deviations from Raoult’s law. Explain your reasoning in each case.
• carbon tetrachloride and heptane
• methanol and tetrahydrofuran (C4H8O)
• acetone [(CH3)2C=CO] and dichloromethane
• hexane and methanol
6. Why are deviations from the ideal behavior predicted by Raoult’s law more common for solutions of liquids than are deviations from the ideal behavior predicted by the ideal gas law for solutions of gases?
7. Boiling point elevation is proportional to the molal concentration of the solute. Is it also proportional to the molar concentration of the solution? Why or why not?
8. Many packaged foods in sealed bags are cooked by placing the bag in boiling water. How could you reduce the time required to cook the contents of the bag using this cooking method?
9. If the costs per kilogram of ethylene glycol and of ethanol were the same, which would be the more cost-effective antifreeze?
10. Many people get thirsty after eating foods such as ice cream or potato chips that have a high sugar or salt content, respectively. Suggest an explanation for this phenomenon.
11. When two aqueous solutions with identical concentrations are separated by a semipermeable membrane, no net movement of water occurs. What happens when a solute that cannot penetrate the membrane is added to one of the solutions? Why?
12. A solution injected into blood vessels must have an electrolyte concentration nearly identical to that found in blood plasma. Why? What would happen if red blood cells were placed in distilled water? What would happen to red blood cells if they were placed in a solution that had twice the electrolyte concentration of blood plasma?
13. If you were stranded on a desert island, why would drinking seawater lead to an increased rate of dehydration, eventually causing you to die of thirst?
14. What is the relationship between the van’t Hoff factor for a compound and its lattice energy?
Numerical Problems
1. Hemoglobin is the protein that is responsible for the red color of blood and for transporting oxygen from the lungs to the tissues. A solution with 11.2 mg of hemoglobin per mL has an osmotic pressure of 2.9 mmHg at 5°C. What is the molecular mass of hemoglobin?
2. To determine the molar mass of the antifreeze protein from the Arctic right-eye flounder, the osmotic pressure of a solution containing 13.2 mg of protein per mL was measured and found to be 21.2 mmHg at 10°C. What is the molar mass of the protein?
3. What is the osmotic pressure at 21.0°C of 13.5 mL of a solution with 1.77 g of sucrose (\)C_{12}H_{22}O_{11})\)? A solution of $NaNO_3$ is generated by dissolving 1.25 g of NaNO3 in enough water to give a final volume of 25.0 mL. What is the osmotic pressure of this sample at 25.0°C?
4. Which would have the lower vapor pressure—an aqueous solution that is 0.12 M in glucose or one that is 0.12 M in $CaCl_2$? Why?
5. What is the total particle concentration expected for each aqueous solution? Which would produce the highest osmotic pressure?
1. 0.35 M KBr
2. 0.11 M MgSO4
3. 0.26 M $MgCl_2$
4. 0.24 M glucose (C6H12O6)
1. The boiling point of an aqueous solution of sodium chloride is 100.37°C. What is the molality of the solution? How many grams of $NaCl$ are present in 125 g of the solution?
2. Calculate the boiling point of a solution of sugar prepared by dissolving 8.4 g of glucose ($C_6H_{12}O_6$) in 250 g of water.
3. At 37°C, the vapor pressure of 300.0 g of water was reduced from 0.062 atm to 0.058 atm by the addition of NaBr. How many grams of NaBr were added?
4. How many grams of KCl must be added to reduce the vapor pressure of 500.0 g of $H_2O$ from 17.5 mmHg to 16.0 mmHg at 20.0°C?
5. How much $NaCl$ would you have to add to 2.0 L of water at a mountain lodge at an elevation of 7350 ft, where the pressure is 0.78 atm and the boiling point of water is 94°C, to get the water to boil at the same temperature as in New Orleans, Louisiana, where the pressure is 1.00 atm?
6. You have three solutions with the following compositions: 12.5 g of KCl in 250 mL of water, 12.5 g of glucose in 400 mL of water, and 12.5 g of $MgCl_2$ in 350 mL of water. Which will have the highest boiling point?
7. Assuming the price per kilogram is the same, which is a better salt to use for deicing wintry roads—$NaCl$ or $MgCl_2$? Why? Would magnesium chloride be an effective deicer at a temperature of −8°C?
8. How many grams of $KNO_3$ must be added to water to produce the same boiling point elevation as a solution of 2.03 g of $MgCl_2$ in a total volume of 120.0 mL of solution, assuming complete dissociation? If the van’t Hoff factor for $MgCl_2$ at this concentration is 2.73, how much $KNO_3$ would be needed?
9. Calculate the quantity of each compound that would need to be added to lower the freezing point of 500.0 mL of water by 1.0°C: KBr, ethylene glycol, $MgBr_2$, ethanol. Assume that the density of water is 1.00 g/cm3.
10. The melting point depression of biphenyl (melting point = 69.0°C) can be used to determine the molecular mass of organic compounds. A mixture of 100.0 g of biphenyl and 2.67 g of naphthalene ($C_{10}H_8$) has a melting point of 68.50°C. If a mixture of 1.00 g of an unknown compound with 100.0 g of biphenyl has a melting point of 68.86°C, what is the molar mass of the unknown compound?
11. Four solutions of urea in water were prepared, with concentrations of 0.32 m, 0.55 m, 1.52 m, and 3.16 m. The freezing points of these solutions were found to be −0.595°C, −1.02°C, −2.72°C, and −5.71°C, respectively. Graphically determine the freezing point depression constant for water. A fifth solution made by dissolving 6.22 g of urea in 250.0 g of water has a freezing point of −0.75°C. Use these data to determine the molar mass of urea.
12. The term osmolarity has been used to describe the total solute concentration of a solution (generally water), where 1 osmole (Osm) is equal to 1 mol of an ideal, nonionizing molecule.
1. What is the osmolarity of a 1.5 M solution of glucose? a 1.5 M solution of $NaCl$? a 1.5 M solution of $CaCl_2$?
2. What is the relationship between osmolarity and the concentration of water?
3. What would be the direction of flow of water through a semipermeable membrane separating a 0.1 M solution of $NaCl$ and a 0.1 M solution of $CaCl_2$?
13. At 40°C, the vapor pressures of pure $CCl_4$ and cyclohexane are 0.2807 atm and 0.2429 atm, respectively. Assuming ideal behavior, what is the vapor pressure of a solution with a $CCl_4$ mole fraction of 0.475? What is the mole fraction of cyclohexane in the vapor phase? The boiling points of $CCl_4$ and cyclohexane are 76.8°C and 80.7°C, respectively.
14. A benzene/toluene solution with a mole fraction of benzene of 0.6589 boils at 88°C at 1 atm. The vapor pressures of pure benzene and toluene at this temperature are 1.259 atm and 0.4993 atm, respectively. What is the composition of the vapor above the boiling solution at this temperature?
15. Plot the vapor pressure of the solution versus composition for the system $CCl_4:CH_3CN$ at 45°C, given the following experimental data:
$X_{CCl_4}(liquid)$ 0.035 0.375 0.605 0.961
$X_{CCl_4}(vapor)$ 0.180 0.543 0.594 0.800
Total P (atm) 0.326 0.480 0.488 0.414
• Source: Adapted from L. Brown and W. Foch, Australian Journal of Chemistry, 9 (1956): 180.
• Does your diagram show behavior characteristic of an ideal solution? Explain your answer.
Numerical Answers
1. $6.7 \times 10^4\; amu$
3. 9.24 atm
5. The $CaCl_2$ solution will have a lower vapor pressure, because it contains three times as many particles as the glucose solution.
7. 0.36 m $NaCl$, 2.6 g $NaCl$
9. 60 g NaBr
11. 700 g $NaCl$
13. $MgCl_2$ produces three particles in solution versus two for $NaCl$, so the same molal concentration of $MgCl_2$ will produce a 50% greater freezing point depression than for $NaCl$. Nonetheless, the molar mass of $MgCl_2$ is 95.3 g/mol versus 48.45 g/mol for $NaCl$. Consequently, a solution containing 1 g $NaCl$ per 1000 g $H_2O$ will produce a freezing point depression of 0.064°C versus 0.059°C for a solution containing 1 g $MgCl_2$ per 1000 g $H_2O$. Thus, given equal cost per gram, $NaCl$ is more effective. Yes, $MgCl_2$ would be effective at −8°C; a 1.43 m solution (136 g per 1000 g H2O) would be required.
16.
kf = 1.81(°C•kg)/mol, molecular mass of urea = 60.0 g/mol
13.6: Colloids
Conceptual Problems
1. How does a colloid differ from a suspension? Which has a greater effect on solvent properties, such as vapor pressure?
2. Is homogenized milk a colloid or a suspension? Is human plasma a colloid or a suspension? Justify your answers.
3. How would you separate the components of an emulsion of fat dispersed in an aqueous solution of sodium chloride? | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/13%3A_Properties_of_Solutions/13.E%3A_Properties_of_Solutions_%28Exercises%29.txt |
These is a summary of key concepts of the chapter in the Textmap created for "Chemistry: The Central Science" by Brown et al.
13.1: The Solution Process
• interaction between solute and solvent molecules
• hydration – solvation when solvent is water
13.1.1 Energy Changes and Solution Formation
• overall enthalpy change in formation of a solution
• $\Delta H_{\mathit{soln}}=\Delta H_1+\Delta H_2+\Delta H_3$
• $\Delta H_1=$ separation of solute molecules
• $\Delta H_1=$ separation of solvent molecules
• $\Delta H_3=$ formation of solute-solvent interactions
• separation of solute particles is endothermic
• separation of solvent is endothermic
• third is exothermic
• formation of solution can be either exothermic or endothermic
• exothermic processes are spontaneous
• solution will not form if enthalpy is too endothermic
• H3 has to be comparable to H1 + H2
• Ionic substances cannot dissolve in nonpolar liquids
• Polar liquids do not form solutions with nonpolar liquids
13.1.2 Solution Formation, Spontaneity, and Disorder
• two nonpolar substances dissolve in one another
• attractive forces = London dispersion forces
• two factors in processes that are spontaneous: energy and disorder
• processes in which the energy content of the system decreases tend to occur spontaneously
• exothermic
• processes in which the disorder of the system increases tend to occur spontaneously
• solutions will form unless solute-solute or solvent-solvent interactions too strong relative to solute-solvent interactions
13.1.3 Solution Formation and Chemical Reactions
• distinguish between physical process of solution formation from chemical process that leads to a solution
13.2: Saturated Solutions and Solubility
• crystallization – reverse process of solution
• dynamic equilibrium – when equilibrium exists between process of solution and crystallization
• solute said to be saturated
• solubility – amount of solute needed to saturate a solution
• unsaturated – when there isn’t enough solute to saturate a solution
• supersaturated – when there is more solute than needed to saturate a solution
• for most salts crystallization of excess solute is exothermic
13.3: Factors Affecting Solubility
Solute-Solvent Interactions
• solubility increases with increasing molar mass
• London dispersion forces increase with increasing size and mass of gas molecules
• Miscible – pairs of liquids that mix in all proportions
• Immiscible – opposite of miscible
• Hydrogen-bonding interactions between solute and solvent leads to high solubility
• Substances with similar intermolecular attractive forces tend to be soluble in one another
• "like dissolves like"
Pressure Effects
• solubility of a gas in any solvent increases as pressure of gas over solvent increases
• relationship between pressure and solubility: Henry’s Law:
• $C_g = kP_g$
• Cg solubility of gas in solution phase (usually expressed as molarity), Pg partial pressure of gas over solution, k is proportionality constant (Henry’s Law constant)
• Henry’s law constant different fore each solute-solvent pair, and temperature
Temperature Effects
• solubility of most solid solutes in water increases as temperature of solution increase
• solubility of gases in water decreases with increasing temperature
• decreases solubility of O2 in water as temperature increases in one the effects of thermal pollution
13.4: Ways of Expressing Concentration
• dilute and concentrated used to describe solution qualitatively
• mass percentage of component in solution:
$\displaystyle\textit{mass% of component} = \frac{\textit{mass of component in soln}}{\textit{total mass of soln}}\times 100 \nonumber$
• very dilute solutions expressed in parts per million (ppm)
$\displaystyle\textit{ppm of component} = \frac{\textit{mass of component in soln}}{\textit{total mass of soln}}\times 10^6 \nonumber$
• 1ppm = 1g solute for each (106) grams of solution or 1mg solute per kg solution
• 1ppm = 1mg solute/L solution
• 1 ppb = 1g of solute/109 grams of solution or 1 m g solute/ L of solution
13.4.1 Mole Fraction, Molarity, and Molality
$\displaystyle\textit{mole fraction of component} = \frac{\textit{moles of component}}{\textit{total moles of all components}} \nonumber$
• sum of mole fractions of all components of solution must equal one
• $\displaystyle\textit{molarity} = \frac{\textit{moles solute}}{\textit{liters soln}}$
• $\displaystyle\textit{molality} = \frac{\textit{moles solute}}{\textit{kilograms of solvent}}$
• molality goes not vary with temperature
• molarity changes with temperature because of expansion and contraction of solution
13.5: Colligative Properties
Colligative properties are physical properties that depend on quantity
Lowering the Vapor Pressure
• vapor pressure over pure solvent higher than that over solution
• vapor pressure needed to obtain equilibrium of pure solvent higher than that of solution
Raoult’s Law
• Raoult’s law: $P_A = X_AP_{A}^°$
• PA = vapor pressure of solution, XA = mole fraction of solvent, PA° = vapor pressure of the pure solvent
• Ideal solution – solution that obeys Raoult’s law
• Solute concentration is low, solute and solvent have similar molecular sizes and similar types of intermolecular attractions
Boiling-Point Elevation
• normal boiling point of pure liquid is the temperature at which pressure is 1 atm
• addition of a nonvolatile solute lowers vapor pressure of solution
• $\Delta T_b=K_b m$
• Kb = molal boiling-point-elevation constant
• Depends only on solvent
• boiling point elevation proportional to number of solute particles present in given quantity of solution
Freezing-Point Depression
• freezing point of solution is temperature at which the first crystals of pure solvent form in equilibrium
• freezing point of solution lower than pure liquid
• freezing point directly proportional to the molality of the solute:
• $\Delta T_f=K_f m$
• Kf = molal freezing-point-depression constant
Osmosis
• semipermeable – membranes that allow passage of some molecules and not others
• osmosis – the net movement of solvent molecules from the less concentrated solution to the more concentrated solution
• net movement of solvent always toward the solution with the higher solute concentration
• osmotic pressure – pressure needed to prevent osmosis, p
• $\pi = \left(\frac{n}{V}\right)RT=MRT$
• M = molarity of solution
• if solutions identical osmosis will not occur and said to be isotonic
• if one solution lower osmotic pressure = hypotonic, the solution that has higher osmotic pressure = hypertonic
• crenation = when cells shrivel up from the loss of water
• hemolysis = when cells rupture due to to much water
Determination of Molar Mass
• colligative properties can be used to find molar mass
13.6: Colloids
Colloidal dispersions (colloids) are intermediate types of dispersions or suspensions
• intermediate solutions between solutions and heterogeneous mixtures
• colloids can be gases, liquids, or solids
• colloid particles have size between 10 - 2000Å
• tyndall effect – scattering of light by colloids
Hydrophilic and Hydrophobic Colloids
• hydrophilic – colloids in which the dispersion medium is water
• hydrophobic – colloids not dispersed in water
• hydrophobic colloids have to be stabilized before being put in water
• natural lack of affinity for water causes separation
• can be stabilized by the adsorption of ions on the surface
• adsorped ions interact with water
• can also be stabilized by presence of other hydrophilic groups on surface
Removal of Colloidal Particles
• coagulation – enlarging colloidal particles
• heating or adding an electrolyte to mixture
• heating increases number of collisions and particles stick together increasing their size
• electrolytes causes neutralization of the surface charges of the particles which remove the electrostatic repulsion
• dialysis – use of semipermeable membranes to filter out colloidal particles | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/13%3A_Properties_of_Solutions/13.S%3A_Properties_of_Solutions_%28Summary%29.txt |
This chapter will present a quantitative description of when the chemical composition of a system is not constant with time. Chemical kinetics is the study of reaction rates, the changes in the concentrations of reactants and products with time. With a discussion of chemical kinetics, the reaction rates or the changes in the concentrations of reactants and products with time are studied. The techniques you are about to learn will enable you to describe the speed of many such changes and predict how the composition of each system will change in response to changing conditions. As you learn about the factors that affect reaction rates, the methods chemists use for reporting and calculating those rates, and the clues that reaction rates provide about events at the molecular level.
• 14.1: Factors that Affect Reaction Rates
There are many factors that influence the reaction rates of chemical reactions include the concentration of reactants, temperature, the physical state of reactants and their dispersion, the solvent, and the presence of a catalyst.
• 14.2: Reaction Rates
In this Module, the quantitative determination of a reaction rate is demonstrated. Reaction rates can be determined over particular time intervals or at a given point in time. A rate law describes the relationship between reactant rates and reactant concentrations. Reaction rates are reported as either the average rate over a period of time or as the instantaneous rate at a single time. Reaction rates can be determined over particular time intervals or at a given point in time.
• 14.3: Concentration and Rates (Differential Rate Laws)
The rate law for a reaction is a mathematical relationship between the reaction rate and the concentrations of species in solution. Rate laws can be expressed either as a differential rate law, describing the change in reactant or product concentrations as a function of time, or as an integrated rate law, describing the actual concentrations of reactants or products as a function of time. The rate constant and reaction order are extracted directly from the rate law.
• 14.4: The Change of Concentration with Time (Integrated Rate Laws)
The reaction rate of a zeroth-order reaction is independent of the concentration of the reactants. The reaction rate of a first-order reaction is directly proportional to the concentration of one reactant. The reaction rate of a simple second-order reaction is proportional to the square of the concentration of one reactant. Knowing the rate law of a reaction gives clues to the reaction mechanism.
• 14.5: Temperature and Rate
A minimum energy (activation energy,Ea) is required for a collision between molecules to result in a chemical reaction. Plots of potential energy for a system versus the reaction coordinate show an energy barrier that must be overcome for the reaction to occur. The arrangement of atoms at the highest point of this barrier is the activated complex, or transition state, of the reaction. At a given temperature, the higher the Ea, the slower the reaction.
• 14.6: Reaction Mechanisms
A balanced chemical reaction does not necessarily reveal either the individual elementary reactions by which a reaction occurs or its rate law. A reaction mechanism is the microscopic path by which reactants are transformed into products. Each step is an elementary reaction. Species that are formed in one step and consumed in another are intermediates. Each elementary reaction can be described in terms of its molecularity. The slowest step in a reaction mechanism is the rate-determining step.
• 14.7: Catalysis
Catalysts participate in a chemical reaction and increase its rate. They do not appear in the reaction’s net equation and are not consumed during the reaction. Catalysts allow a reaction to proceed via a pathway that has a lower activation energy than the uncatalyzed reaction. In heterogeneous catalysis, catalysts provide a surface to which reactants bind in a process of adsorption. In homogeneous catalysis, catalysts are in the same phase as the reactants. Enzymes are biological catalysts.
• 14.E: Exercises
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. Complementary General Chemistry question banks can also be found for other Textmaps.
• 14.S: Chemical Kinetics (Summary)
A summary of the key concepts in this chapter of the Textmap created for "Chemistry: The Central Science" by Brown et al.
14: Chemical Kinetics
Learning Objectives
• To gain a general overview of some of the important factors that manipulate chemical reaction rates.
There are many factors that influence the reaction rates of chemical reactions include the concentration of reactants, temperature, the physical state of reactants and their dispersion, the solvent, and the presence of a catalyst.
Although a balanced chemical equation for a reaction describes the quantitative relationships between the amounts of reactants present and the amounts of products that can be formed, it gives us no information about whether or how fast a given reaction will occur. This information is obtained by studying the chemical kinetics of a reaction, which depend on various factors: reactant concentrations, temperature, physical states and surface areas of reactants, and solvent and catalyst properties if either are present. By studying the kinetics of a reaction, chemists gain insights into how to control reaction conditions to achieve a desired outcome.
Concentration Effects
Two substances cannot possibly react with each other unless their constituent particles (molecules, atoms, or ions) come into contact. If there is no contact, the reaction rate will be zero. Conversely, the more reactant particles that collide per unit time, the more often a reaction between them can occur. Consequently, the reaction rate usually increases as the concentration of the reactants increases.
Temperature Effects
Increasing the temperature of a system increases the average kinetic energy of its constituent particles. As the average kinetic energy increases, the particles move faster and collide more frequently per unit time and possess greater energy when they collide. Both of these factors increase the reaction rate. Hence the reaction rate of virtually all reactions increases with increasing temperature. Conversely, the reaction rate of virtually all reactions decreases with decreasing temperature. For example, refrigeration retards the rate of growth of bacteria in foods by decreasing the reaction rates of biochemical reactions that enable bacteria to reproduce.
In systems where more than one reaction is possible, the same reactants can produce different products under different reaction conditions. For example, in the presence of dilute sulfuric acid and at temperatures around 100°C, ethanol is converted to diethyl ether:
$\mathrm{2CH_3CH_2OH}\xrightarrow{\mathrm{H_2SO_4}}\mathrm{CH_3CH_2OCH_2CH_3}+\mathrm{H_2O} \label{14.1.1}$
At 180°C, however, a completely different reaction occurs, which produces ethylene as the major product:
$\mathrm{CH_3CH_2OH}\xrightarrow{\mathrm{H_2SO_4}}\mathrm{C_2H_4}+\mathrm{H_2O} \label{14.1.2}$
Phase and Surface Area Effects
When two reactants are in the same fluid phase, their particles collide more frequently than when one or both reactants are solids (or when they are in different fluids that do not mix). If the reactants are uniformly dispersed in a single homogeneous solution, then the number of collisions per unit time depends on concentration and temperature, as we have just seen. If the reaction is heterogeneous, however, the reactants are in two different phases, and collisions between the reactants can occur only at interfaces between phases. The number of collisions between reactants per unit time is substantially reduced relative to the homogeneous case, and, hence, so is the reaction rate. The reaction rate of a heterogeneous reaction depends on the surface area of the more condensed phase.
Automobile engines use surface area effects to increase reaction rates. Gasoline is injected into each cylinder, where it combusts on ignition by a spark from the spark plug. The gasoline is injected in the form of microscopic droplets because in that form it has a much larger surface area and can burn much more rapidly than if it were fed into the cylinder as a stream. Similarly, a pile of finely divided flour burns slowly (or not at all), but spraying finely divided flour into a flame produces a vigorous reaction.
Solvent Effects
The nature of the solvent can also affect the reaction rates of solute particles. For example, a sodium acetate solution reacts with methyl iodide in an exchange reaction to give methyl acetate and sodium iodide.
$CH_3CO_2Na_{(soln)} + CH_3I_{(l)} \rightarrow CH_3CO_2CH_{3\; (soln)} + NaI_{(soln)} \label{14.1.3}$
This reaction occurs 10 million times more rapidly in the organic solvent dimethylformamide [DMF; (CH3)2NCHO] than it does in methanol (CH3OH). Although both are organic solvents with similar dielectric constants (36.7 for DMF versus 32.6 for methanol), methanol is able to hydrogen bond with acetate ions, whereas DMF cannot. Hydrogen bonding reduces the reactivity of the oxygen atoms in the acetate ion.
Solvent viscosity is also important in determining reaction rates. In highly viscous solvents, dissolved particles diffuse much more slowly than in less viscous solvents and can collide less frequently per unit time. Thus the reaction rates of most reactions decrease rapidly with increasing solvent viscosity.
Catalyst Effects
A catalyst is a substance that participates in a chemical reaction and increases the reaction rate without undergoing a net chemical change itself. Consider, for example, the decomposition of hydrogen peroxide in the presence and absence of different catalysts. Because most catalysts are highly selective, they often determine the product of a reaction by accelerating only one of several possible reactions that could occur.
Most of the bulk chemicals produced in industry are formed with catalyzed reactions. Recent estimates indicate that about 30% of the gross national product of the United States and other industrialized nations relies either directly or indirectly on the use of catalysts. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/14%3A_Chemical_Kinetics/14.01%3A_Factors_that_Affect_Reaction_Rates.txt |
Learning Objectives
• To determine the reaction rate of a reaction.
Reaction rates are usually expressed as the concentration of reactant consumed or the concentration of product formed per unit time. The units are thus moles per liter per unit time, written as M/s, M/min, or M/h. To measure reaction rates, chemists initiate the reaction, measure the concentration of the reactant or product at different times as the reaction progresses, perhaps plot the concentration as a function of time on a graph, and then calculate the change in the concentration per unit time.
The progress of a simple reaction (A → B) is shown in Figure $1$; the beakers are snapshots of the composition of the solution at 10 s intervals. The number of molecules of reactant (A) and product (B) are plotted as a function of time in the graph. Each point in the graph corresponds to one beaker in Figure $1$. The reaction rate is the change in the concentration of either the reactant or the product over a period of time. The concentration of A decreases with time, while the concentration of B increases with time.
$\textrm{rate}=\dfrac{\Delta [\textrm B]}{\Delta t}=-\dfrac{\Delta [\textrm A]}{\Delta t} \label{Eq1}$
Square brackets indicate molar concentrations, and the capital Greek delta (Δ) means “change in.” Because chemists follow the convention of expressing all reaction rates as positive numbers, however, a negative sign is inserted in front of Δ[A]/Δt to convert that expression to a positive number. The reaction rate calculated for the reaction A → B using Equation $\ref{Eq1}$ is different for each interval (this is not true for every reaction, as shown below). A greater change occurs in [A] and [B] during the first 10 s interval, for example, than during the last, meaning that the reaction rate is greatest at first.
Reaction rates generally decrease with time as reactant concentrations decrease.
A Video Discussing Average Reaction Rates. Video Link: Introduction to Chemical Reaction Kinetics(opens in new window) [youtu.be] (opens in new window)
Determining the Reaction Rate of Hydrolysis of Aspirin
We can use Equation $\ref{Eq1}$ to determine the reaction rate of hydrolysis of aspirin, probably the most commonly used drug in the world (more than 25,000,000 kg are produced annually worldwide). Aspirin (acetylsalicylic acid) reacts with water (such as water in body fluids) to give salicylic acid and acetic acid, as shown in Figure $2$.
Because salicylic acid is the actual substance that relieves pain and reduces fever and inflammation, a great deal of research has focused on understanding this reaction and the factors that affect its rate. Data for the hydrolysis of a sample of aspirin are in Table $1$ and are shown in the graph in Figure $3$.
Table $1$: Data for Aspirin Hydrolysis in Aqueous Solution at pH 7.0 and 37°C*
Time (h) [Aspirin] (M) [Salicylic Acid] (M)
*The reaction at pH 7.0 is very slow. It is much faster under acidic conditions, such as those found in the stomach.
0 5.55 × 10−3 0
2.0 5.51 × 10−3 0.040 × 10−3
5.0 5.45 × 10−3 0.10 × 10−3
10 5.35 × 10−3 0.20 × 10−3
20 5.15 × 10−3 0.40 × 10−3
30 4.96 × 10−3 0.59 × 10−3
40 4.78 × 10−3 0.77 × 10−3
50 4.61 × 10−3 0.94 × 10−3
100 3.83 × 10−3 1.72 × 10−3
200 2.64 × 10−3 2.91 × 10−3
300 1.82 × 10−3 3.73 × 10−3
The data in Table $1$ were obtained by removing samples of the reaction mixture at the indicated times and analyzing them for the concentrations of the reactant (aspirin) and one of the products (salicylic acid).
The average reaction rate for a given time interval can be calculated from the concentrations of either the reactant or one of the products at the beginning of the interval (time = t0) and at the end of the interval (t1). Using salicylic acid, the reaction rate for the interval between t = 0 h and t = 2.0 h (recall that change is always calculated as final minus initial) is calculated as follows:
\begin{align*}\textrm{rate}_{(t=0-2.0\textrm{ h})}&=\frac{[\textrm{salicyclic acid}]_2-[\textrm{salicyclic acid}]_0}{\textrm{2.0 h}-\textrm{0 h}} \&=\frac{0.040\times10^{-3}\textrm{ M}-0\textrm{ M}}{\textrm{2.0 h}}=2.0\times10^{-5}\textrm{ M/h} \end{align*} \nonumber
The reaction rate can also be calculated from the concentrations of aspirin at the beginning and the end of the same interval, remembering to insert a negative sign, because its concentration decreases:
\begin{align*}\textrm{rate}_{(t=0-2.0\textrm{ h})}&=-\dfrac{[\textrm{aspirin}]_2-[\textrm{aspirin}]_0}{\mathrm{2.0\,h-0\,h}} \&=-\dfrac{(5.51\times10^{-3}\textrm{ M})-(5.55\times10^{-3}\textrm{ M})}{\textrm{2.0 h}} \&=2\times10^{-5}\textrm{ M/h}\end{align*} \nonumber
If the reaction rate is calculated during the last interval given in Table $1$(the interval between 200 h and 300 h after the start of the reaction), the reaction rate is significantly slower than it was during the first interval (t = 0–2.0 h):
\begin{align*}\textrm{rate}_{(t=200-300\textrm{h})}&=\dfrac{[\textrm{salicyclic acid}]_{300}-[\textrm{salicyclic acid}]_{200}}{\mathrm{300\,h-200\,h}} \&=-\dfrac{(3.73\times10^{-3}\textrm{ M})-(2.91\times10^{-3}\textrm{ M})}{\textrm{100 h}} \&=8.2\times10^{-6}\textrm{ M/h}\end{align*} \nonumber
Calculating the Reaction Rate of Fermentation of Sucrose
In the preceding example, the stoichiometric coefficients in the balanced chemical equation are the same for all reactants and products; that is, the reactants and products all have the coefficient 1. Consider a reaction in which the coefficients are not all the same, the fermentation of sucrose to ethanol and carbon dioxide:
$\underset{\textrm{sucrose}}{\mathrm{C_{12}H_{22}O_{11}(aq)}}+\mathrm{H_2O(l)}\rightarrow\mathrm{4C_2H_5OH(aq)}+4\mathrm{CO_2(g)} \label{Eq2}$
The coefficients indicate that the reaction produces four molecules of ethanol and four molecules of carbon dioxide for every one molecule of sucrose consumed. As before, the reaction rate can be found from the change in the concentration of any reactant or product. In this particular case, however, a chemist would probably use the concentration of either sucrose or ethanol because gases are usually measured as volumes and, as explained in Chapter 10, the volume of CO2 gas formed depends on the total volume of the solution being studied and the solubility of the gas in the solution, not just the concentration of sucrose. The coefficients in the balanced chemical equation tell us that the reaction rate at which ethanol is formed is always four times faster than the reaction rate at which sucrose is consumed:
$\dfrac{\Delta[\mathrm{C_2H_5OH}]}{\Delta t}=-\dfrac{4\Delta[\textrm{sucrose}]}{\Delta t} \label{Eq3}$
The concentration of the reactant—in this case sucrose—decreases with time, so the value of Δ[sucrose] is negative. Consequently, a minus sign is inserted in front of Δ[sucrose] in Equation $\ref{Eq3}$ so the rate of change of the sucrose concentration is expressed as a positive value. Conversely, the ethanol concentration increases with time, so its rate of change is automatically expressed as a positive value.
Often the reaction rate is expressed in terms of the reactant or product with the smallest coefficient in the balanced chemical equation. The smallest coefficient in the sucrose fermentation reaction (Equation $\ref{Eq2}$) corresponds to sucrose, so the reaction rate is generally defined as follows:
$\textrm{rate}=-\dfrac{\Delta[\textrm{sucrose}]}{\Delta t}=\dfrac{1}{4}\left (\dfrac{\Delta[\mathrm{C_2H_5OH}]}{\Delta t} \right ) \label{Eq4}$
Example $1$: Decomposition Reaction I
Consider the thermal decomposition of gaseous N2O5 to NO2 and O2 via the following equation:
$\mathrm{2N_2O_5(g)}\xrightarrow{\,\Delta\,}\mathrm{4NO_2(g)}+\mathrm{O_2(g)} \nonumber$
Write expressions for the reaction rate in terms of the rates of change in the concentrations of the reactant and each product with time.
Given: balanced chemical equation
Asked for: reaction rate expressions
Strategy:
1. Choose the species in the equation that has the smallest coefficient. Then write an expression for the rate of change of that species with time.
2. For the remaining species in the equation, use molar ratios to obtain equivalent expressions for the reaction rate.
Solution
A Because O2 has the smallest coefficient in the balanced chemical equation for the reaction, define the reaction rate as the rate of change in the concentration of O2 and write that expression.
B The balanced chemical equation shows that 2 mol of N2O5 must decompose for each 1 mol of O2 produced and that 4 mol of NO2 are produced for every 1 mol of O2 produced. The molar ratios of O2 to N2O5 and to NO2 are thus 1:2 and 1:4, respectively. This means that the rate of change of [N2O5] and [NO2] must be divided by its stoichiometric coefficient to obtain equivalent expressions for the reaction rate. For example, because NO2 is produced at four times the rate of O2, the rate of production of NO2 is divided by 4. The reaction rate expressions are as follows:
$\textrm{rate}=\dfrac{\Delta[\mathrm O_2]}{\Delta t}=\dfrac{\Delta[\mathrm{NO_2}]}{4\Delta t}=-\dfrac{\Delta[\mathrm{N_2O_5}]}{2\Delta t}$
Exercise $1$: Contact Process I
The contact process is used in the manufacture of sulfuric acid. A key step in this process is the reaction of $SO_2$ with $O_2$ to produce $SO_3$.
$2SO_{2(g)} + O_{2(g)} \rightarrow 2SO_{3(g)} \nonumber$
Write expressions for the reaction rate in terms of the rate of change of the concentration of each species.
Answer
$\textrm{rate}=-\dfrac{\Delta[\mathrm O_2]}{\Delta t}=-\dfrac{\Delta[\mathrm{SO_2}]}{2\Delta t}=\dfrac{\Delta[\mathrm{SO_3}]}{2\Delta t}$
Instantaneous Rates of Reaction
The instantaneous rate of a reaction is the reaction rate at any given point in time. As the period of time used to calculate an average rate of a reaction becomes shorter and shorter, the average rate approaches the instantaneous rate. Comparing this to calculus, the instantaneous rate of a reaction at a given time corresponds to the slope of a line tangent to the concentration-versus-time curve at that point—that is, the derivative of concentration with respect to time.
The distinction between the instantaneous and average rates of a reaction is similar to the distinction between the actual speed of a car at any given time on a trip and the average speed of the car for the entire trip. Although the car may travel for an extended period at 65 mph on an interstate highway during a long trip, there may be times when it travels only 25 mph in construction zones or 0 mph if you stop for meals or gas. The average speed on the trip may be only 50 mph, whereas the instantaneous speed on the interstate at a given moment may be 65 mph. Whether the car can be stopped in time to avoid an accident depends on its instantaneous speed, not its average speed. There are important differences between the speed of a car during a trip and the speed of a chemical reaction, however. The speed of a car may vary unpredictably over the length of a trip, and the initial part of a trip is often one of the slowest. In a chemical reaction, the initial interval typically has the fastest rate (though this is not always the case), and the reaction rate generally changes smoothly over time.
Chemical kinetics generally focuses on one particular instantaneous rate, which is the initial reaction rate, t = 0. Initial rates are determined by measuring the reaction rate at various times and then extrapolating a plot of rate versus time to t = 0.
Example $2$: Decomposition Reaction II
Using the reaction shown in Example $1$, calculate the reaction rate from the following data taken at 56°C:
$2N_2O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)} \nonumber$
calculate the reaction rate from the following data taken at 56°C:
Time (s) [N2O5] (M) [NO2] (M) [O2] (M)
240 0.0388 0.0314 0.00792
600 0.0197 0.0699 0.0175
Given: balanced chemical equation and concentrations at specific times
Asked for: reaction rate
Strategy:
1. Using the equations in Example $1$, subtract the initial concentration of a species from its final concentration and substitute that value into the equation for that species.
2. Substitute the value for the time interval into the equation. Make sure your units are consistent.
Solution
A Calculate the reaction rate in the interval between t1 = 240 s and t2 = 600 s. From Example $1$, the reaction rate can be evaluated using any of three expressions:
$\textrm{rate}=\dfrac{\Delta[\mathrm O_2]}{\Delta t}=\dfrac{\Delta[\mathrm{NO_2}]}{4\Delta t}=-\dfrac{\Delta[\mathrm{N_2O_5}]}{2\Delta t} \nonumber$
Subtracting the initial concentration from the final concentration of N2O5 and inserting the corresponding time interval into the rate expression for N2O5,
$\textrm{rate}=-\dfrac{\Delta[\mathrm{N_2O_5}]}{2\Delta t}=-\dfrac{[\mathrm{N_2O_5}]_{600}-[\mathrm{N_2O_5}]_{240}}{2(600\textrm{ s}-240\textrm{ s})} \nonumber$
B Substituting actual values into the expression,
$\textrm{rate}=-\dfrac{\mathrm{\mathrm{0.0197\;M-0.0388\;M}}}{2(360\textrm{ s})}=2.65\times10^{-5} \textrm{ M/s}$
Similarly, NO2 can be used to calculate the reaction rate:
$\textrm{rate}=\dfrac{\Delta[\mathrm{NO_2}]}{4\Delta t}=\dfrac{[\mathrm{NO_2}]_{600}-[\mathrm{NO_2}]_{240}}{4(\mathrm{600\;s-240\;s})}=\dfrac{\mathrm{0.0699\;M-0.0314\;M}}{4(\mathrm{360\;s})}=\mathrm{2.67\times10^{-5}\;M/s} \nonumber$
Allowing for experimental error, this is the same rate obtained using the data for N2O5. The data for O2 can also be used:
$\textrm{rate}=\dfrac{\Delta[\mathrm{O_2}]}{\Delta t}=\dfrac{[\mathrm{O_2}]_{600}-[\mathrm{O_2}]_{240}}{\mathrm{600\;s-240\;s}}=\dfrac{\mathrm{0.0175\;M-0.00792\;M}}{\mathrm{360\;s}}=\mathrm{2.66\times10^{-5}\;M/s} \nonumber$
Again, this is the same value obtained from the N2O5 and NO2 data. Thus, the reaction rate does not depend on which reactant or product is used to measure it.
Exercise $2$: Contact Process II
Using the data in the following table, calculate the reaction rate of $SO_2(g)$ with $O_2(g)$ to give $SO_3(g)$.
$2SO_{2(g)} + O_{2(g)} \rightarrow 2SO_{3(g)} \nonumber$
calculate the reaction rate of $SO_2(g)$ with $O_2(g)$ to give $SO_3(g)$.
Time (s) [SO2] (M) [O2] (M) [SO3] (M)
300 0.0270 0.0500 0.0072
720 0.0194 0.0462 0.0148
Answer:
9.0 × 10−6 M/s
Summary
In this Module, the quantitative determination of a reaction rate is demonstrated. Reaction rates can be determined over particular time intervals or at a given point in time. A rate law describes the relationship between reactant rates and reactant concentrations. Reaction rates are reported as either the average rate over a period of time or as the instantaneous rate at a single time. Reaction rates can be determined over particular time intervals or at a given point in time.
• General definition of rate for A → B: $\textrm{rate}=\frac{\Delta [\textrm B]}{\Delta t}=-\frac{\Delta [\textrm A]}{\Delta t} \nonumber$ | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/14%3A_Chemical_Kinetics/14.02%3A_Reaction_Rates.txt |
Learning Objectives
• To understand the meaning of the rate law.
The factors that affect the reaction rate of a chemical reaction, which may determine whether a desired product is formed. In this section, we will show you how to quantitatively determine the reaction rate.
Rate Laws
Typically, reaction rates decrease with time because reactant concentrations decrease as reactants are converted to products. Reaction rates generally increase when reactant concentrations are increased. This section examines mathematical expressions called rate laws, which describe the relationships between reactant rates and reactant concentrations. Rate laws are mathematical descriptions of experimentally verifiable data.
Rate laws may be written from either of two different but related perspectives. A differential rate law expresses the reaction rate in terms of changes in the concentration of one or more reactants (Δ[R]) over a specific time interval (Δt). In contrast, an integrated rate law describes the reaction rate in terms of the initial concentration ([R]0) and the measured concentration of one or more reactants ([R]) after a given amount of time (t); integrated rate laws are discussed in more detail later. The integrated rate law is derived by using calculus to integrate the differential rate law. Whether using a differential rate law or integrated rate law, always make sure that the rate law gives the proper units for the reaction rate, usually moles per liter per second (M/s).
Reaction Orders
For a reaction with the general equation:
$aA + bB \rightarrow cC + dD \label{14.3.1}$
the experimentally determined rate law usually has the following form:
$\text{rate} = k[A]^m[B]^n \label{14.3.2}$
The proportionality constant (k) is called the rate constant, and its value is characteristic of the reaction and the reaction conditions. A given reaction has a particular rate constant value under a given set of conditions, such as temperature, pressure, and solvent; varying the temperature or the solvent usually changes the value of the rate constant. The numerical value of k, however, does not change as the reaction progresses under a given set of conditions.
Under a given set of conditions, the value of the rate constant does not change as the reaction progresses.
The reaction rate thus depends on the rate constant for the given set of reaction conditions and the concentration of A and B raised to the powers m and n, respectively. The values of m and n are derived from experimental measurements of the changes in reactant concentrations over time and indicate the reaction order, the degree to which the reaction rate depends on the concentration of each reactant; m and n need not be integers. For example, Equation $\ref{14.3.2}$ tells us that Equation $\ref{14.3.1}$ is mth order in reactant A and nth order in reactant B. It is important to remember that n and m are not related to the stoichiometric coefficients a and b in the balanced chemical equation and must be determined experimentally. The overall reaction order is the sum of all the exponents in the rate law: m + n.
The orders of the reactions (e.g. n and m) are not related to the stoichiometric coefficients in the balanced chemical (e.g., a and b).
To illustrate how chemists interpret a differential rate law, consider the experimentally derived rate law for the hydrolysis of t-butyl bromide in 70% aqueous acetone.
This reaction produces t-butanol according to the following equation:
$(CH_3)_3CBr_{(soln)} + H_2O_{(soln)} \rightarrow (CH_3)_3COH_{(soln)} + HBr_{(soln)} \label{14.3.3}$
Combining the rate expression in Equation $\ref{14.3.2}$ with the definition of average reaction rate
$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t} \nonumber$
gives a general expression for the differential rate law:
$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm A]^m[\textrm B]^n \label{14.3.4}$
Inserting the identities of the reactants into Equation $\ref{14.3.4}$ gives the following expression for the differential rate law for the reaction:
$\textrm{rate}=-\dfrac{\Delta[\mathrm{(CH_3)_3CBr}]}{\Delta t}=k[\mathrm{(CH_3)_3CBr}]^m[\mathrm{H_2O}]^n \label{14.3.5}$
Experiments to determine the rate law for the hydrolysis of t-butyl bromide show that the reaction rate is directly proportional to the concentration of (CH3)3CBr but is independent of the concentration of water. Therefore, m and n in Equation $\ref{14.3.4}$ are 1 and 0, respectively, and,
$\text{rate} = k[(CH_3)_3CBr]^1[H_2O]^0 = k[(CH_3)_3CBr] \label{14.3.6}$
Because the exponent for the reactant is 1, the reaction is first order in (CH3)3CBr. It is zeroth order in water because the exponent for [H2O] is 0. (Recall that anything raised to the zeroth power equals 1.) Thus, the overall reaction order is 1 + 0 = 1. The reaction orders state in practical terms that doubling the concentration of (CH3)3CBr doubles the reaction rate of the hydrolysis reaction, halving the concentration of (CH3)3CBr halves the reaction rate, and so on. Conversely, increasing or decreasing the concentration of water has no effect on the reaction rate. (Again, when working with rate laws, there is no simple correlation between the stoichiometry of the reaction and the rate law. The values of k, m, and n in the rate law must be determined experimentally.) Experimental data show that k has the value 5.15 × 10−4 s−1 at 25°C. The rate constant has units of reciprocal seconds (s−1) because the reaction rate is defined in units of concentration per unit time (M/s). The units of a rate constant depend on the rate law for a particular reaction.
Under conditions identical to those for the t-butyl bromide reaction, the experimentally derived differential rate law for the hydrolysis of methyl bromide (CH3Br) is as follows:
$\textrm{rate}=-\dfrac{\Delta[\mathrm{CH_3Br}]}{\Delta t}=k'[\mathrm{CH_3Br}] \label{14.3.7}$
This reaction also has an overall reaction order of 1, but the rate constant in Equation $\ref{14.3.7}$ is approximately 106 times smaller than that for t-butyl bromide. Thus, methyl bromide hydrolyzes about 1 million times more slowly than t-butyl bromide, and this information tells chemists how the reactions differ on a molecular level.
Frequently, changes in reaction conditions also produce changes in a rate law. In fact, chemists often alter reaction conditions to study the mechanics of a reaction. For example, when t-butyl bromide is hydrolyzed in an aqueous acetone solution containing OH ions rather than in aqueous acetone alone, the differential rate law for the hydrolysis reaction does not change. In contrast, for methyl bromide, the differential rate law becomes
$\text{rate} =k″[CH_3Br][OH^−] \nonumber$
with an overall reaction order of 2. Although the two reactions proceed similarly in neutral solution, they proceed very differently in the presence of a base, providing clues as to how the reactions differ on a molecular level.
Example $1$: Writing Rate Laws from Reaction Orders
An experiment shows that the reaction of nitrogen dioxide with carbon monoxide:
$\ce{NO2}(g)+\ce{CO}(g)⟶\ce{NO}(g)+\ce{CO2}(g) \nonumber$
is second order in NO2 and zero order in CO at 100 °C. What is the rate law for the reaction?
Solution
The reaction will have the form:
$\ce{rate}=k[\ce{NO2}]^m[\ce{CO}]^n \nonumber$
The reaction is second order in NO2; thus m = 2. The reaction is zero order in CO; thus n = 0. The rate law is:
$\ce{rate}=k[\ce{NO2}]^2[\ce{CO}]^0=k[\ce{NO2}]^2 \nonumber$
Remember that a number raised to the zero power is equal to 1, thus [CO]0 = 1, which is why we can simply drop the concentration of CO from the rate equation: the rate of reaction is solely dependent on the concentration of NO2. When we consider rate mechanisms later in this chapter, we will explain how a reactant’s concentration can have no effect on a reaction despite being involved in the reaction.
Exercise $\PageIndex{1A}$
The rate law for the reaction:
$\ce{H2}(g)+\ce{2NO}(g)⟶\ce{N2O}(g)+\ce{H2O}(g) \nonumber$
has been experimentally determined to be $rate = k[NO]^2[H_2]$. What are the orders with respect to each reactant, and what is the overall order of the reaction?
Answer
• order in NO = 2
• order in H2 = 1
• overall order = 3
Exercise $\PageIndex{1B}$
In a transesterification reaction, a triglyceride reacts with an alcohol to form an ester and glycerol. Many students learn about the reaction between methanol (CH3OH) and ethyl acetate (CH3CH2OCOCH3) as a sample reaction before studying the chemical reactions that produce biodiesel:
$\ce{CH3OH + CH3CH2OCOCH3 ⟶ CH3OCOCH3 + CH3CH2OH} \nonumber$
The rate law for the reaction between methanol and ethyl acetate is, under certain conditions, experimentally determined to be:
$\ce{rate}=k[\ce{CH3OH}] \nonumber$
What is the order of reaction with respect to methanol and ethyl acetate, and what is the overall order of reaction?
Answer
• order in CH3OH = 1
• order in CH3CH2OCOCH3 = 0
• overall order = 1
Example $2$: Differential Rate Laws
Below are three reactions and their experimentally determined differential rate laws. For each reaction, give the units of the rate constant, give the reaction order with respect to each reactant, give the overall reaction order, and predict what happens to the reaction rate when the concentration of the first species in each chemical equation is doubled.
1. $\mathrm{2HI(g)}\xrightarrow{\textrm{Pt}}\mathrm{H_2(g)}+\mathrm{I_2(g)} \ \textrm{rate}=-\frac{1}{2}\left (\frac{\Delta[\mathrm{HI}]}{\Delta t} \right )=k[\textrm{HI}]^2 \nonumber$
2. $\mathrm{2N_2O(g)}\xrightarrow{\Delta}\mathrm{2N_2(g)}+\mathrm{O_2(g)} \ \textrm{rate}=-\frac{1}{2}\left (\frac{\Delta[\mathrm{N_2O}]}{\Delta t} \right )=k \nonumber$
3. $\mathrm{cyclopropane(g)}\rightarrow\mathrm{propane(g)} \ \textrm{rate}=-\frac{\Delta[\mathrm{cyclopropane}]}{\Delta t}=k[\mathrm{cyclopropane}] \nonumber$
Given: balanced chemical equations and differential rate laws
Asked for: units of rate constant, reaction orders, and effect of doubling reactant concentration
Strategy:
1. Express the reaction rate as moles per liter per second [mol/(L·s), or M/s]. Then determine the units of each chemical species in the rate law. Divide the units for the reaction rate by the units for all species in the rate law to obtain the units for the rate constant.
2. Identify the exponent of each species in the rate law to determine the reaction order with respect to that species. Add all exponents to obtain the overall reaction order.
3. Use the mathematical relationships as expressed in the rate law to determine the effect of doubling the concentration of a single species on the reaction rate.
Solution
1. A [HI]2 will give units of (moles per liter)2. For the reaction rate to have units of moles per liter per second, the rate constant must have reciprocal units [1/(M·s)]:
$k\textrm M^2=\dfrac{\textrm M}{\textrm s}k=\dfrac{\textrm{M/s}}{\textrm M^2}=\dfrac{1}{\mathrm{M\cdot s}}=\mathrm{M^{-1}\cdot s^{-1}} \nonumber$
B The exponent in the rate law is 2, so the reaction is second order in HI. Because HI is the only reactant and the only species that appears in the rate law, the reaction is also second order overall.
C If the concentration of HI is doubled, the reaction rate will increase from k[HI]02 to k(2[HI])02 = 4k[HI]02. The reaction rate will therefore quadruple.
1. A Because no concentration term appears in the rate law, the rate constant must have M/s units for the reaction rate to have M/s units.
B The rate law tells us that the reaction rate is constant and independent of the N2O concentration. That is, the reaction is zeroth order in N2O and zeroth order overall.
C Because the reaction rate is independent of the N2O concentration, doubling the concentration will have no effect on the reaction rate.
1. A The rate law contains only one concentration term raised to the first power. Hence the rate constant must have units of reciprocal seconds (s−1) to have units of moles per liter per second for the reaction rate: M·s−1 = M/s.
B The only concentration in the rate law is that of cyclopropane, and its exponent is 1. This means that the reaction is first order in cyclopropane. Cyclopropane is the only species that appears in the rate law, so the reaction is also first order overall.
C Doubling the initial cyclopropane concentration will increase the reaction rate from k[cyclopropane]0 to 2k[cyclopropane]0. This doubles the reaction rate.
Exercise $2$
Given the following two reactions and their experimentally determined differential rate laws: determine the units of the rate constant if time is in seconds, determine the reaction order with respect to each reactant, give the overall reaction order, and predict what will happen to the reaction rate when the concentration of the first species in each equation is doubled.
1. \begin{align}\textrm{CH}_3\textrm N\textrm{=NCH}_3\textrm{(g)}\rightarrow\mathrm{C_2H_6(g)}+\mathrm{N_2(g)}\hspace{5mm}&\textrm{rate}=-\frac{\Delta[\textrm{CH}_3\textrm N\textrm{=NCH}_3]}{\Delta t} \&=k[\textrm{CH}_3\textrm N\textrm{=NCH}_3]\end{align}
2. \begin{align}\mathrm{2NO_2(g)}+\mathrm{F_2(g)}\rightarrow\mathrm{2NO_2F(g)}\hspace{5mm}&\textrm{rate}=-\frac{\Delta[\mathrm{F_2}]}{\Delta t}=-\frac{1}{2}\left ( \frac{\Delta[\mathrm{NO_2}]}{\Delta t} \right ) \&=k[\mathrm{NO_2}][\mathrm{F_2}]\end{align}
Answer a
s−1; first order in CH3N=NCH3; first order overall; doubling [CH3N=NCH3] will double the reaction rate.
Answer b
M−1·s−1; first order in NO2, first order in F2; second order overall; doubling [NO2] will double the reaction rate.
Determining the Rate Law of a Reaction
The number of fundamentally different mechanisms (sets of steps in a reaction) is actually rather small compared to the large number of chemical reactions that can occur. Thus understanding reaction mechanisms can simplify what might seem to be a confusing variety of chemical reactions. The first step in discovering the reaction mechanism is to determine the reaction’s rate law. This can be done by designing experiments that measure the concentration(s) of one or more reactants or products as a function of time. For the reaction $A + B \rightarrow products$, for example, we need to determine k and the exponents m and n in the following equation:
$\text{rate} = k[A]^m[B]^n \label{14.4.11}$
To do this, we might keep the initial concentration of B constant while varying the initial concentration of A and calculating the initial reaction rate. This information would permit us to deduce the reaction order with respect to A. Similarly, we could determine the reaction order with respect to B by studying the initial reaction rate when the initial concentration of A is kept constant while the initial concentration of B is varied. In earlier examples, we determined the reaction order with respect to a given reactant by comparing the different rates obtained when only the concentration of the reactant in question was changed. An alternative way of determining reaction orders is to set up a proportion using the rate laws for two different experiments. Rate data for a hypothetical reaction of the type $A + B \rightarrow products$ are given in Table $1$.
Table $1$: Rate Data for a Hypothetical Reaction of the Form $A + B \rightarrow products$
Experiment [A] (M) [B] (M) Initial Rate (M/min)
1 0.50 0.50 8.5 × 10−3
2 0.75 0.50 19 × 10−3
3 1.00 0.50 34 × 10−3
4 0.50 0.75 8.5 × 10−3
5 0.50 1.00 8.5 × 10−3
The general rate law for the reaction is given in Equation $\ref{14.4.11}$. We can obtain m or n directly by using a proportion of the rate laws for two experiments in which the concentration of one reactant is the same, such as Experiments 1 and 3 in Table $3$.
$\dfrac{\mathrm{rate_1}}{\mathrm{rate_3}}=\dfrac{k[\textrm A_1]^m[\textrm B_1]^n}{k[\textrm A_3]^m[\textrm B_3]^n} \nonumber$
Inserting the appropriate values from Table $3$,
$\dfrac{8.5\times10^{-3}\textrm{ M/min}}{34\times10^{-3}\textrm{ M/min}}=\dfrac{k[\textrm{0.50 M}]^m[\textrm{0.50 M}]^n}{k[\textrm{1.00 M}]^m[\textrm{0.50 M}]^n} \nonumber$
Because 1.00 to any power is 1, [1.00 M]m = 1.00 M. We can cancel like terms to give 0.25 = [0.50]m, which can also be written as 1/4 = [1/2]m. Thus we can conclude that m = 2 and that the reaction is second order in A. By selecting two experiments in which the concentration of B is the same, we were able to solve for m.
Conversely, by selecting two experiments in which the concentration of A is the same (e.g., Experiments 5 and 1), we can solve for n.
$\dfrac{\mathrm{rate_1}}{\mathrm{rate_5}}=\dfrac{k[\mathrm{A_1}]^m[\mathrm{B_1}]^n}{k[\mathrm{A_5}]^m[\mathrm{B_5}]^n}$
Substituting the appropriate values from Table $3$,
$\dfrac{8.5\times10^{-3}\textrm{ M/min}}{8.5\times10^{-3}\textrm{ M/min}}=\dfrac{k[\textrm{0.50 M}]^m[\textrm{0.50 M}]^n}{k[\textrm{0.50 M}]^m[\textrm{1.00 M}]^n} \nonumber$
Canceling leaves 1.0 = [0.50]n, which gives $n = 0$; that is, the reaction is zeroth order in $B$. The experimentally determined rate law is therefore
rate = k[A]2[B]0 = k[A]2
We can now calculate the rate constant by inserting the data from any row of Table $3$ into the experimentally determined rate law and solving for $k$. Using Experiment 2, we obtain
19 × 10−3 M/min = k(0.75 M)2
3.4 × 10−2 M−1·min−1 = k
You should verify that using data from any other row of Table $1$ gives the same rate constant. This must be true as long as the experimental conditions, such as temperature and solvent, are the same.
Example $3$
Nitric oxide is produced in the body by several different enzymes and acts as a signal that controls blood pressure, long-term memory, and other critical functions. The major route for removing NO from biological fluids is via reaction with $O_2$ to give $NO_2$, which then reacts rapidly with water to give nitrous acid and nitric acid:
These reactions are important in maintaining steady levels of NO. The following table lists kinetics data for the reaction of NO with O2 at 25°C:
$2NO(g) + O_2(g) \rightarrow 2NO_2(g) \nonumber$
Determine the rate law for the reaction and calculate the rate constant.
rate law for the reaction and calculate the rate constant.
Experiment [NO]0 (M) [O2]0 (M) Initial Rate (M/s)
1 0.0235 0.0125 7.98 × 10−3
2 0.0235 0.0250 15.9 × 10−3
3 0.0470 0.0125 32.0 × 10−3
4 0.0470 0.0250 63.5 × 10−3
Given: balanced chemical equation, initial concentrations, and initial rates
Asked for: rate law and rate constant
Strategy:
1. Compare the changes in initial concentrations with the corresponding changes in rates of reaction to determine the reaction order for each species. Write the rate law for the reaction.
2. Using data from any experiment, substitute appropriate values into the rate law. Solve the rate equation for k.
Solution
A Comparing Experiments 1 and 2 shows that as [O2] is doubled at a constant value of [NO2], the reaction rate approximately doubles. Thus the reaction rate is proportional to [O2]1, so the reaction is first order in O2. Comparing Experiments 1 and 3 shows that the reaction rate essentially quadruples when [NO] is doubled and [O2] is held constant. That is, the reaction rate is proportional to [NO]2, which indicates that the reaction is second order in NO. Using these relationships, we can write the rate law for the reaction:
rate = k[NO]2[O2]
B The data in any row can be used to calculate the rate constant. Using Experiment 1, for example, gives
$k=\dfrac{\textrm{rate}}{[\mathrm{NO}]^2[\mathrm{O_2}]}=\dfrac{7.98\times10^{-3}\textrm{ M/s}}{(0.0235\textrm{ M})^2(0.0125\textrm{ M})}=1.16\times10^3\;\mathrm{ M^{-2}\cdot s^{-1}} \nonumber$
Alternatively, using Experiment 2 gives
$k=\dfrac{\textrm{rate}}{[\mathrm{NO}]^2[\mathrm{O_2}]}=\dfrac{15.9\times10^{-3}\textrm{ M/s}}{(0.0235\textrm{ M})^2(0.0250\textrm{ M})}=1.15\times10^3\;\mathrm{ M^{-2}\cdot s^{-1}} \nonumber$
The difference is minor and associated with significant digits and likely experimental error in making the table.
The overall reaction order $(m + n) = 3$, so this is a third-order reaction whose rate is determined by three reactants. The units of the rate constant become more complex as the overall reaction order increases.
Exercise $3$
The peroxydisulfate ion (S2O82) is a potent oxidizing agent that reacts rapidly with iodide ion in water:
$S_2O^{2−}_{8(aq)} + 3I^−_{(aq)} \rightarrow 2SO^{2−}_{4(aq)} + I^−_{3(aq)} \nonumber$
The following table lists kinetics data for this reaction at 25°C. Determine the rate law and calculate the rate constant.
kinetics data for this reaction at 25°C.
Experiment [S2O82]0 (M) [I]0 (M) Initial Rate (M/s)
1 0.27 0.38 2.05
2 0.40 0.38 3.06
3 0.40 0.22 1.76
Answer:
rate = k[S2O82][I]; k = 20 M−1·s−1
A Video Discussing Initial Rates and Rate Law Expressions. Video Link: Initial Rates and Rate Law Expressions(opens in new window) [youtu.be]
Summary
The rate law for a reaction is a mathematical relationship between the reaction rate and the concentrations of species in solution. Rate laws can be expressed either as a differential rate law, describing the change in reactant or product concentrations as a function of time, or as an integrated rate law, describing the actual concentrations of reactants or products as a function of time. The rate constant (k) of a rate law is a constant of proportionality between the reaction rate and the reactant concentration. The exponent to which a concentration is raised in a rate law indicates the reaction order, the degree to which the reaction rate depends on the concentration of a particular reactant. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/14%3A_Chemical_Kinetics/14.03%3A_Concentration_and_Rates_%28Differential_Rate_Laws%29.txt |
Learning Objectives
• To apply rate laws to zeroth, first and second order reactions.
Either the differential rate law or the integrated rate law can be used to determine the reaction order from experimental data. Often, the exponents in the rate law are the positive integers: 1 and 2 or even 0. Thus the reactions are zeroth, first, or second order in each reactant. The common patterns used to identify the reaction order are described in this section, where we focus on characteristic types of differential and integrated rate laws and how to determine the reaction order from experimental data. The learning objective of this Module is to know how to determine the reaction order from experimental data.
Zeroth-Order Reactions
A zeroth-order reaction is one whose rate is independent of concentration; its differential rate law is
$\text{rate} = k. \nonumber$
We refer to these reactions as zeroth order because we could also write their rate in a form such that the exponent of the reactant in the rate law is 0:
$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm{reactant}]^0=k(1)=k \label{14.4.1}$
Because rate is independent of reactant concentration, a graph of the concentration of any reactant as a function of time is a straight line with a slope of $−k$. The value of $k$ is negative because the concentration of the reactant decreases with time. Conversely, a graph of the concentration of any product as a function of time is a straight line with a slope of $k$, a positive value.
The integrated rate law for a zeroth-order reaction also produces a straight line and has the general form
$[A] = [A]_0 − kt \label{14.4.2}$
where $[A]_0$ is the initial concentration of reactant $A$. Equation \ref{14.4.2} has the form of the algebraic equation for a straight line,
$y = mx + b, \nonumber$
with $y = [A]$, $mx = −kt$, and $b = [A]_0$.)
Units
In a zeroth-order reaction, the rate constant must have the same units as the reaction rate, typically moles per liter per second.
Although it may seem counterintuitive for the reaction rate to be independent of the reactant concentration(s), such reactions are rather common. They occur most often when the reaction rate is determined by available surface area. An example is the decomposition of N2O on a platinum (Pt) surface to produce N2 and O2, which occurs at temperatures ranging from 200°C to 400°C:
$\mathrm{2N_2O(g)}\xrightarrow{\textrm{Pt}}\mathrm{2N_2(g)}+\mathrm{O_2(g)} \label{14.4.3}$
Without a platinum surface, the reaction requires temperatures greater than 700°C, but between 200°C and 400°C, the only factor that determines how rapidly N2O decomposes is the amount of Pt surface available (not the amount of Pt). As long as there is enough N2O to react with the entire Pt surface, doubling or quadrupling the N2O concentration will have no effect on the reaction rate. At very low concentrations of N2O, where there are not enough molecules present to occupy the entire available Pt surface, the reaction rate is dependent on the N2O concentration. The reaction rate is as follows:
$\textrm{rate}=-\dfrac{1}{2}\left (\dfrac{\Delta[\mathrm{N_2O}]}{\Delta t} \right )=\dfrac{1}{2}\left (\dfrac{\Delta[\mathrm{N_2}]}{\Delta t} \right )=\dfrac{\Delta[\mathrm{O_2}]}{\Delta t}=k[\mathrm{N_2O}]^0=k \label{14.4.4}$
Thus the rate at which N2O is consumed and the rates at which N2 and O2 are produced are independent of concentration. As shown in Figure $2$, the change in the concentrations of all species with time is linear. Most important, the exponent (0) corresponding to the N2O concentration in the experimentally derived rate law is not the same as the reactant’s stoichiometric coefficient in the balanced chemical equation (2). For this reaction, as for all others, the rate law must be determined experimentally.
A zeroth-order reaction that takes place in the human liver is the oxidation of ethanol (from alcoholic beverages) to acetaldehyde, catalyzed by the enzyme alcohol dehydrogenase. At high ethanol concentrations, this reaction is also a zeroth-order reaction. The overall reaction equation is
where \ce{NAD^{+}}\) (nicotinamide adenine dinucleotide) and $\ce{NADH}$ (reduced nicotinamide adenine dinucleotide) are the oxidized and reduced forms, respectively, of a species used by all organisms to transport electrons. When an alcoholic beverage is consumed, the ethanol is rapidly absorbed into the blood. Its concentration then decreases at a constant rate until it reaches zero (Figure $\PageIndex{3a}$). An average 70 kg person typically takes about 2.5 h to oxidize the 15 mL of ethanol contained in a single 12 oz can of beer, a 5 oz glass of wine, or a shot of distilled spirits (such as whiskey or brandy). The actual rate, however, varies a great deal from person to person, depending on body size and the amount of alcohol dehydrogenase in the liver. The reaction rate does not increase if a greater quantity of alcohol is consumed over the same period of time because the reaction rate is determined only by the amount of enzyme present in the liver. Contrary to popular belief, the caffeine in coffee is ineffective at catalyzing the oxidation of ethanol. When the ethanol has been completely oxidized and its concentration drops to essentially zero, the rate of oxidation also drops rapidly (part (b) in Figure $3$).
These examples illustrate two important points:
1. In a zeroth-order reaction, the reaction rate does not depend on the reactant concentration.
2. A linear change in concentration with time is a clear indication of a zeroth-order reaction.
First-Order Reactions
In a first-order reaction, the reaction rate is directly proportional to the concentration of one of the reactants. First-order reactions often have the general form A → products. The differential rate for a first-order reaction is as follows:
$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm A] \label{14.4.5}$
If the concentration of A is doubled, the reaction rate doubles; if the concentration of A is increased by a factor of 10, the reaction rate increases by a factor of 10, and so forth. Because the units of the reaction rate are always moles per liter per second, the units of a first-order rate constant are reciprocal seconds (s−1).
The integrated rate law for a first-order reaction can be written in two different ways: one using exponents and one using logarithms. The exponential form is as follows:
$[A] = [A]_0e^{−kt} \label{14.4.6}$
where $[A]_0$ is the initial concentration of reactant $A$ at $t = 0$; $k$ is the rate constant; and e is the base of the natural logarithms, which has the value 2.718 to three decimal places. Recall that an integrated rate law gives the relationship between reactant concentration and time. Equation $\ref{14.4.6}$ predicts that the concentration of A will decrease in a smooth exponential curve over time. By taking the natural logarithm of each side of Equation $\ref{14.4.6}$ and rearranging, we obtain an alternative logarithmic expression of the relationship between the concentration of $A$ and $t$:
$\ln[A] = \ln[A]_0 − kt \label{14.4.7}$
Because Equation $\ref{14.4.7}$ has the form of the algebraic equation for a straight line,
$y = mx + b, \nonumber$
with $y = \ln[A]$ and $b = \ln[A]_0$, a plot of $\ln[A]$ versus $t$ for a first-order reaction should give a straight line with a slope of $−k$ and an intercept of $\ln[A]_0$. Either the differential rate law (Equation $\ref{14.4.5}$) or the integrated rate law (Equation $\ref{14.4.7}$) can be used to determine whether a particular reaction is first order.
First-order reactions are very common. One reaction that exhibits apparent first-order kinetics is the hydrolysis of the anticancer drug cisplatin. Cisplatin, the first “inorganic” anticancer drug to be discovered, is unique in its ability to cause complete remission of the relatively rare, but deadly cancers of the reproductive organs in young adults. The structures of cisplatin and its hydrolysis product are as follows:
Both platinum compounds have four groups arranged in a square plane around a Pt(II) ion. The reaction shown in Figure $5$ is important because cisplatin, the form in which the drug is administered, is not the form in which the drug is active. Instead, at least one chloride ion must be replaced by water to produce a species that reacts with deoxyribonucleic acid (DNA) to prevent cell division and tumor growth. Consequently, the kinetics of the reaction in Figure $4$ have been studied extensively to find ways of maximizing the concentration of the active species.
If a plot of reactant concentration versus time is not linear but a plot of the natural logarithm of reactant concentration versus time is linear, then the reaction is first order.
The rate law and reaction order of the hydrolysis of cisplatin are determined from experimental data, such as those displayed in Table $1$. The table lists initial rate data for four experiments in which the reaction was run at pH 7.0 and 25°C but with different initial concentrations of cisplatin.
Table $1$: Rates of Hydrolysis of Cisplatin as a Function of Concentration at pH 7.0 and 25°C
Experiment [Cisplatin]0 (M) Initial Rate (M/min)
1 0.0060 9.0 × 10−6
2 0.012 1.8 × 10−5
3 0.024 3.6 × 10−5
4 0.030 4.5 × 10−5
Because the reaction rate increases with increasing cisplatin concentration, we know this cannot be a zeroth-order reaction. Comparing Experiments 1 and 2 in Table $1$ shows that the reaction rate doubles [(1.8 × 10−5 M/min) ÷ (9.0 × 10−6 M/min) = 2.0] when the concentration of cisplatin is doubled (from 0.0060 M to 0.012 M). Similarly, comparing Experiments 1 and 4 shows that the reaction rate increases by a factor of 5 [(4.5 × 10−5 M/min) ÷ (9.0 × 10−6 M/min) = 5.0] when the concentration of cisplatin is increased by a factor of 5 (from 0.0060 M to 0.030 M). Because the reaction rate is directly proportional to the concentration of the reactant, the exponent of the cisplatin concentration in the rate law must be 1, so the rate law is rate = k[cisplatin]1. Thus the reaction is first order. Knowing this, we can calculate the rate constant using the differential rate law for a first-order reaction and the data in any row of Table $1$. For example, substituting the values for Experiment 3 into Equation $\ref{14.4.5}$,
3.6 × 10−5 M/min = k(0.024 M)
1.5 × 10−3 min−1 = k
Knowing the rate constant for the hydrolysis of cisplatin and the rate constants for subsequent reactions that produce species that are highly toxic enables hospital pharmacists to provide patients with solutions that contain only the desired form of the drug.
Example $1$
At high temperatures, ethyl chloride produces HCl and ethylene by the following reaction:
$\ce{CH_3CH_2Cl(g) ->[\Delta] HCl(g) + C_2H_4(g)} \nonumber$
Using the rate data for the reaction at 650°C presented in the following table, calculate the reaction order with respect to the concentration of ethyl chloride and determine the rate constant for the reaction.
data for the reaction at 650°C
Experiment [CH3CH2Cl]0 (M) Initial Rate (M/s)
1 0.010 1.6 × 10−8
2 0.015 2.4 × 10−8
3 0.030 4.8 × 10−8
4 0.040 6.4 × 10−8
Given: balanced chemical equation, initial concentrations of reactant, and initial rates of reaction
Asked for: reaction order and rate constant
Strategy:
1. Compare the data from two experiments to determine the effect on the reaction rate of changing the concentration of a species.
2. Compare the observed effect with behaviors characteristic of zeroth- and first-order reactions to determine the reaction order. Write the rate law for the reaction.
C Use measured concentrations and rate data from any of the experiments to find the rate constant.
Solution
The reaction order with respect to ethyl chloride is determined by examining the effect of changes in the ethyl chloride concentration on the reaction rate.
A Comparing Experiments 2 and 3 shows that doubling the concentration doubles the reaction rate, so the reaction rate is proportional to [CH3CH2Cl]. Similarly, comparing Experiments 1 and 4 shows that quadrupling the concentration quadruples the reaction rate, again indicating that the reaction rate is directly proportional to [CH3CH2Cl].
B This behavior is characteristic of a first-order reaction, for which the rate law is rate = k[CH3CH2Cl].
C We can calculate the rate constant (k) using any row in the table. Selecting Experiment 1 gives the following:
1.60 × 10−8 M/s = k(0.010 M)
1.6 × 10−6 s−1 = k
Exercise $1$
Sulfuryl chloride (SO2Cl2) decomposes to SO2 and Cl2 by the following reaction:
$SO_2Cl_2(g) → SO_2(g) + Cl_2(g) \nonumber$
Data for the reaction at 320°C are listed in the following table. Calculate the reaction order with regard to sulfuryl chloride and determine the rate constant for the reaction.
Data for the reaction at 320°C
Experiment [SO2Cl2]0 (M) Initial Rate (M/s)
1 0.0050 1.10 × 10−7
2 0.0075 1.65 × 10−7
3 0.0100 2.20 × 10−7
4 0.0125 2.75 × 10−7
Answer
first order; k = 2.2 × 10−5 s−1
We can also use the integrated rate law to determine the reaction rate for the hydrolysis of cisplatin. To do this, we examine the change in the concentration of the reactant or the product as a function of time at a single initial cisplatin concentration. Figure $\PageIndex{6a}$ shows plots for a solution that originally contained 0.0100 M cisplatin and was maintained at pH 7 and 25°C.
The concentration of cisplatin decreases smoothly with time, and the concentration of chloride ion increases in a similar way. When we plot the natural logarithm of the concentration of cisplatin versus time, we obtain the plot shown in part (b) in Figure $6$. The straight line is consistent with the behavior of a system that obeys a first-order rate law. We can use any two points on the line to calculate the slope of the line, which gives us the rate constant for the reaction. Thus taking the points from part (a) in Figure $6$ for t = 100 min ([cisplatin] = 0.0086 M) and t = 1000 min ([cisplatin] = 0.0022 M),
\begin{align*}\textrm{slope}&=\dfrac{\ln [\textrm{cisplatin}]_{1000}-\ln [\textrm{cisplatin}]_{100}}{\mathrm{1000\;min-100\;min}} \[4pt] -k&=\dfrac{\ln 0.0022-\ln 0.0086}{\mathrm{1000\;min-100\;min}}=\dfrac{-6.12-(-4.76)}{\mathrm{900\;min}}=-1.51\times10^{-3}\;\mathrm{min^{-1}} \[4pt] k&=1.5\times10^{-3}\;\mathrm{min^{-1}}\end{align*} \nonumber
The slope is negative because we are calculating the rate of disappearance of cisplatin. Also, the rate constant has units of min−1 because the times plotted on the horizontal axes in parts (a) and (b) in Figure $6$ are in minutes rather than seconds.
The reaction order and the magnitude of the rate constant we obtain using the integrated rate law are exactly the same as those we calculated earlier using the differential rate law. This must be true if the experiments were carried out under the same conditions.
Video Example Using the First-Order Integrated Rate Law Equation:
Example $2$
If a sample of ethyl chloride with an initial concentration of 0.0200 M is heated at 650°C, what is the concentration of ethyl chloride after 10 h? How many hours at 650°C must elapse for the concentration to decrease to 0.0050 M (k = 1.6 × 10−6 s−1) ?
Given: initial concentration, rate constant, and time interval
Asked for: concentration at specified time and time required to obtain particular concentration
Strategy:
1. Substitute values for the initial concentration ([A]0) and the calculated rate constant for the reaction (k) into the integrated rate law for a first-order reaction. Calculate the concentration ([A]) at the given time t.
2. Given a concentration [A], solve the integrated rate law for time t.
Solution
The exponential form of the integrated rate law for a first-order reaction (Equation $\ref{14.4.6}$) is [A] = [A]0ekt.
A Having been given the initial concentration of ethyl chloride ([A]0) and having the rate constant of k = 1.6 × 10−6 s−1, we can use the rate law to calculate the concentration of the reactant at a given time t. Substituting the known values into the integrated rate law,
\begin{align*}[\mathrm{CH_3CH_2Cl}]_{\mathrm{10\;h}}&=[\mathrm{CH_3CH_2Cl}]_0e^{-kt} \[4pt] &=\textrm{0.0200 M}(e^{\large{-(1.6\times10^{-6}\textrm{ s}^{-1})[(10\textrm{ h})(60\textrm{ min/h})(60\textrm{ s/min})]}}) \[4pt] &=0.0189\textrm{ M} \nonumber\end{align*} \nonumber
We could also have used the logarithmic form of the integrated rate law (Equation $\ref{14.4.7}$):
\begin{align*}\ln[\mathrm{CH_3CH_2Cl}]_{\textrm{10 h}}&=\ln [\mathrm{CH_3CH_2Cl}]_0-kt \[4pt] &=\ln 0.0200-(1.6\times10^{-6}\textrm{ s}^{-1})[(\textrm{10 h})(\textrm{60 min/h})(\textrm{60 s/min})] \[4pt] &=-3.912-0.0576=-3.970 \nonumber \[4pt] [\mathrm{CH_3CH_2Cl}]_{\textrm{10 h}}&=e^{-3.970}\textrm{ M} \nonumber \[4pt] &=0.0189\textrm{ M} \nonumber\end{align*} \nonumber
B To calculate the amount of time required to reach a given concentration, we must solve the integrated rate law for $t$. Equation $\ref{14.4.7}$ gives the following:
\begin{align*}\ln[\mathrm{CH_3CH_2Cl}]_t &=\ln[\mathrm{CH_3CH_2Cl}]_0-kt \[4pt] kt &=\ln[\mathrm{CH_3CH_2Cl}]_0-\ln[\mathrm{CH_3CH_2Cl}]_t=\ln\dfrac{[\mathrm{CH_3CH_2Cl}]_0}{[\mathrm{CH_3CH_2Cl}]_t} \[4pt] t &=\dfrac{1}{k}\left (\ln\dfrac{[\mathrm{CH_3CH_2Cl}]_0}{[\mathrm{CH_3CH_2Cl}]_t} \right )=\dfrac{1}{1.6\times10^{-6}\textrm{ s}^{-1}}\left(\ln \dfrac{0.0200\textrm{ M}}{0.0050\textrm{ M}}\right) \[4pt] &=\dfrac{\ln 4.0}{1.6\times10^{-6}\textrm{ s}^{-1}}=8.7\times10^5\textrm{ s}=240\textrm{ h}=2.4\times10^2\textrm{ h} \nonumber \end{align*} \nonumber
Exercise $2$
In the exercise in Example $1$, you found that the decomposition of sulfuryl chloride ($\ce{SO2Cl2}$) is first order, and you calculated the rate constant at 320°C.
1. Use the form(s) of the integrated rate law to find the amount of $\ce{SO2Cl2}$ that remains after 20 h if a sample with an original concentration of 0.123 M is heated at 320°C.
2. How long would it take for 90% of the SO2Cl2 to decompose?
Answer a
0.0252 M
Answer b
29 h
Second-Order Reactions
The simplest kind of second-order reaction is one whose rate is proportional to the square of the concentration of one reactant. These generally have the form
$\ce{2A → products.}\nonumber$
A second kind of second-order reaction has a reaction rate that is proportional to the product of the concentrations of two reactants. Such reactions generally have the form A + B → products. An example of the former is a dimerization reaction, in which two smaller molecules, each called a monomer, combine to form a larger molecule (a dimer).
The differential rate law for the simplest second-order reaction in which 2A → products is as follows:
$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{2\Delta t}=k[\textrm A]^2 \label{14.4.8}$
Consequently, doubling the concentration of A quadruples the reaction rate. For the units of the reaction rate to be moles per liter per second (M/s), the units of a second-order rate constant must be the inverse (M−1·s−1). Because the units of molarity are expressed as mol/L, the unit of the rate constant can also be written as L(mol·s).
For the reaction 2A → products, the following integrated rate law describes the concentration of the reactant at a given time:
$\dfrac{1}{[\textrm A]}=\dfrac{1}{[\textrm A]_0}+kt \label{14.4.9}$
Because Equation $\ref{14.4.9}$ has the form of an algebraic equation for a straight line, y = mx + b, with y = 1/[A] and b = 1/[A]0, a plot of 1/[A] versus t for a simple second-order reaction is a straight line with a slope of k and an intercept of 1/[A]0.
Second-order reactions generally have the form 2A → products or A + B → products.
Video Discussing the Second-Order Integrated Rate Law Equation: Second-Order Integrated Rate Law Equation(opens in new window) [youtu.be]
Simple second-order reactions are common. In addition to dimerization reactions, two other examples are the decomposition of NO2 to NO and O2 and the decomposition of HI to I2 and H2. Most examples involve simple inorganic molecules, but there are organic examples as well. We can follow the progress of the reaction described in the following paragraph by monitoring the decrease in the intensity of the red color of the reaction mixture.
Many cyclic organic compounds that contain two carbon–carbon double bonds undergo a dimerization reaction to give complex structures. One example is as follows:
Figure $7$
For simplicity, we will refer to this reactant and product as “monomer” and “dimer,” respectively. The systematic name of the monomer is 2,5-dimethyl-3,4-diphenylcyclopentadienone. The systematic name of the dimer is the name of the monomer followed by “dimer.” Because the monomers are the same, the general equation for this reaction is 2A → product. This reaction represents an important class of organic reactions used in the pharmaceutical industry to prepare complex carbon skeletons for the synthesis of drugs. Like the first-order reactions studied previously, it can be analyzed using either the differential rate law (Equation $\ref{14.4.8}$) or the integrated rate law (Equation $\ref{14.4.9}$).
Table $2$: Rates of Reaction as a Function of Monomer Concentration for an Initial Monomer Concentration of 0.0054 M
Time (min) [Monomer] (M) Instantaneous Rate (M/min)
10 0.0044 8.0 × 10−5
26 0.0034 5.0 × 10−5
44 0.0027 3.1 × 10−5
70 0.0020 1.8 × 10−5
120 0.0014 8.0 × 10−6
To determine the differential rate law for the reaction, we need data on how the reaction rate varies as a function of monomer concentrations, which are provided in Table $2$. From the data, we see that the reaction rate is not independent of the monomer concentration, so this is not a zeroth-order reaction. We also see that the reaction rate is not proportional to the monomer concentration, so the reaction is not first order. Comparing the data in the second and fourth rows shows that the reaction rate decreases by a factor of 2.8 when the monomer concentration decreases by a factor of 1.7:
$\dfrac{5.0\times10^{-5}\textrm{ M/min}}{1.8\times10^{-5}\textrm{ M/min}}=2.8\hspace{5mm}\textrm{and}\hspace{5mm}\dfrac{3.4\times10^{-3}\textrm{ M}}{2.0\times10^{-3} \textrm{ M}}=1.7 \nonumber$
Because (1.7)2 = 2.9 ≈ 2.8, the reaction rate is approximately proportional to the square of the monomer concentration.
rate ∝ [monomer]2
This means that the reaction is second order in the monomer. Using Equation $\ref{14.4.8}$ and the data from any row in Table $2$, we can calculate the rate constant. Substituting values at time 10 min, for example, gives the following:
\begin{align}\textrm{rate}&=k[\textrm A]^2 \8.0\times10^{-5}\textrm{ M/min}&=k(4.4\times10^{-3}\textrm{ M})^2 \4.1 \textrm{ M}^{-1}\cdot \textrm{min}^{-1}&=k\end{align} \nonumber
We can also determine the reaction order using the integrated rate law. To do so, we use the decrease in the concentration of the monomer as a function of time for a single reaction, plotted in Figure $\PageIndex{8a}$. The measurements show that the concentration of the monomer (initially 5.4 × 10−3 M) decreases with increasing time. This graph also shows that the reaction rate decreases smoothly with increasing time. According to the integrated rate law for a second-order reaction, a plot of 1/[monomer] versus t should be a straight line, as shown in Figure $\PageIndex{8b}$. Any pair of points on the line can be used to calculate the slope, which is the second-order rate constant. In this example, k = 4.1 M−1·min−1, which is consistent with the result obtained using the differential rate equation. Although in this example the stoichiometric coefficient is the same as the reaction order, this is not always the case. The reaction order must always be determined experimentally.
For two or more reactions of the same order, the reaction with the largest rate constant is the fastest. Because the units of the rate constants for zeroth-, first-, and second-order reactions are different, however, we cannot compare the magnitudes of rate constants for reactions that have different orders.
Example $3$
At high temperatures, nitrogen dioxide decomposes to nitric oxide and oxygen.
$\mathrm{2NO_2(g)}\xrightarrow{\Delta}\mathrm{2NO(g)}+\mathrm{O_2(g)} \nonumber$
Experimental data for the reaction at 300°C and four initial concentrations of NO2 are listed in the following table:
Experimental data for the reaction at 300°C and four initial concentrations of NO2
Experiment [NO2]0 (M) Initial Rate (M/s)
1 0.015 1.22 × 10−4
2 0.010 5.40 × 10−5
3 0.0080 3.46 × 10−5
4 0.0050 1.35 × 10−5
Determine the reaction order and the rate constant.
Given: balanced chemical equation, initial concentrations, and initial rates
Asked for: reaction order and rate constant
Strategy:
1. From the experiments, compare the changes in the initial reaction rates with the corresponding changes in the initial concentrations. Determine whether the changes are characteristic of zeroth-, first-, or second-order reactions.
2. Determine the appropriate rate law. Using this rate law and data from any experiment, solve for the rate constant (k).
Solution
A We can determine the reaction order with respect to nitrogen dioxide by comparing the changes in NO2 concentrations with the corresponding reaction rates. Comparing Experiments 2 and 4, for example, shows that doubling the concentration quadruples the reaction rate [(5.40 × 10−5) ÷ (1.35 × 10−5) = 4.0], which means that the reaction rate is proportional to [NO2]2. Similarly, comparing Experiments 1 and 4 shows that tripling the concentration increases the reaction rate by a factor of 9, again indicating that the reaction rate is proportional to [NO2]2. This behavior is characteristic of a second-order reaction.
B We have rate = k[NO2]2. We can calculate the rate constant (k) using data from any experiment in the table. Selecting Experiment 2, for example, gives the following:
\begin{align*}\textrm{rate}&=k[\mathrm{NO_2}]^2 \5.40\times10^{-5}\textrm{ M/s}&=k(\mathrm{\mathrm{0.010\;M}})^2 \0.54\mathrm{\;M^{-1}\cdot s^{-1}}&=k\end{align*} \nonumber
Exercise $3$
When the highly reactive species HO2 forms in the atmosphere, one important reaction that then removes it from the atmosphere is as follows:
$2HO_{2(g)} \rightarrow H_2O_{2(g)} + O_{2(g)} \nonumber$
The kinetics of this reaction have been studied in the laboratory, and some initial rate data at 25°C are listed in the following table:
Some initial rate data at 25°C
Experiment [HO2]0 (M) Initial Rate (M/s)
1 1.1 × 10−8 1.7 × 10−7
2 2.5 × 10−8 8.8 × 10−7
3 3.4 × 10−8 1.6 × 10−6
4 5.0 × 10−8 3.5 × 10−6
Determine the reaction order and the rate constant.
Answer
second order in HO2; k = 1.4 × 109 M−1·s−1
If a plot of reactant concentration versus time is not linear, but a plot of 1/(reactant concentration) versus time is linear, then the reaction is second order.
Example $4$
If a flask that initially contains 0.056 M NO2 is heated at 300°C, what will be the concentration of NO2 after 1.0 h? How long will it take for the concentration of NO2 to decrease to 10% of the initial concentration? Use the integrated rate law for a second-order reaction (Equation \ref{14.4.9}) and the rate constant calculated above.
Given: balanced chemical equation, rate constant, time interval, and initial concentration
Asked for: final concentration and time required to reach specified concentration
Strategy:
1. Given k, t, and [A]0, use the integrated rate law for a second-order reaction to calculate [A].
2. Setting [A] equal to 1/10 of [A]0, use the same equation to solve for $t$.
Solution
A We know k and [NO2]0, and we are asked to determine [NO2] at t = 1 h (3600 s). Substituting the appropriate values into Equation \ref{14.4.9},
\begin{align*}\dfrac{1}{[\mathrm{NO_2}]_{3600}}&=\dfrac{1}{[\mathrm{NO_2}]_0}+kt \[4pt] &=\dfrac{1}{0.056\textrm{ M}}+[(0.54 \mathrm{\;M^{-1}\cdot s^{-1}})(3600\textrm{ s})] \[4pt] &=2.0\times10^3\textrm{ M}^{-1}\end{align*} \nonumber
Thus [NO2]3600 = 5.1 × 10−4 M.
B In this case, we know k and [NO2]0, and we are asked to calculate at what time [NO2] = 0.1[NO2]0 = 0.1(0.056 M) = 0.0056 M. To do this, we solve Equation $\ref{14.4.9}$ for t, using the concentrations given.
\begin{align*} t &=\dfrac{(1/[\mathrm{NO_2}])-(1/[\mathrm{NO_2}]_0)}{k} \[4pt] &=\dfrac{(1/0.0056 \textrm{ M})-(1/0.056\textrm{ M})}{0.54 \;\mathrm{M^{-1}\cdot s^{-1}}} \[4pt] &=3.0\times10^2\textrm{ s}=5.0\textrm{ min} \end{align*} \nonumber
NO2 decomposes very rapidly; under these conditions, the reaction is 90% complete in only 5.0 min.
Exercise $4$
In the previous exercise, you calculated the rate constant for the decomposition of HO2 as k = 1.4 × 109 M−1·s−1. This high rate constant means that HO2 decomposes rapidly under the reaction conditions given in the problem. In fact, the HO2 molecule is so reactive that it is virtually impossible to obtain in high concentrations. Given a 0.0010 M sample of HO2, calculate the concentration of HO2 that remains after 1.0 h at 25°C. How long will it take for 90% of the HO2 to decompose? Use the integrated rate law for a second-order reaction (Equation $\ref{14.4.9}$) and the rate constant calculated in the exercise in Example $3$.
Answer
2.0 × 10−13 M; 6.4 × 10−6 s
In addition to the simple second-order reaction and rate law we have just described, another very common second-order reaction has the general form $A + B \rightarrow products$, in which the reaction is first order in $A$ and first order in $B$. The differential rate law for this reaction is as follows:
$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=-\dfrac{\Delta[\textrm B]}{\Delta t}=k[\textrm A][\textrm B] \nonumber$
Because the reaction is first order both in A and in B, it has an overall reaction order of 2. (The integrated rate law for this reaction is rather complex, so we will not describe it.) We can recognize second-order reactions of this sort because the reaction rate is proportional to the concentrations of each reactant.
Summary
The reaction rate of a zeroth-order reaction is independent of the concentration of the reactants. The reaction rate of a first-order reaction is directly proportional to the concentration of one reactant. The reaction rate of a simple second-order reaction is proportional to the square of the concentration of one reactant. Knowing the rate law of a reaction gives clues to the reaction mechanism.
• zeroth-order reaction: $\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k \nonumber$ $[A] = [A]_0 − kt \nonumber$
• first-order reaction: $\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm A] \nonumber$ $[A] = [A]_0e^{−kt} \nonumber$ $\ln[A] = \ln[A]_0 − kt \nonumber$
• second-order reaction: $\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm A]^2 \nonumber$ $\dfrac{1}{[\textrm A]}=\dfrac{1}{[\textrm A]_0}+kt \nonumber$ | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/14%3A_Chemical_Kinetics/14.04%3A_The_Change_of_Concentration_with_Time_%28Integrated_Rate_Laws%29.txt |
Learning Objectives
• To understand why and how chemical reactions occur.
It is possible to use kinetics studies of a chemical system, such as the effect of changes in reactant concentrations, to deduce events that occur on a microscopic scale, such as collisions between individual particles. Such studies have led to the collision model of chemical kinetics, which is a useful tool for understanding the behavior of reacting chemical species. The collision model explains why chemical reactions often occur more rapidly at higher temperatures. For example, the reaction rates of many reactions that occur at room temperature approximately double with a temperature increase of only 10°C. In this section, we will use the collision model to analyze this relationship between temperature and reaction rates. Before delving into the relationship between temperature and reaction rate, we must discuss three microscopic factors that influence the observed macroscopic reaction rates.
Microscopic Factor 1: Collisional Frequency
Central to collision model is that a chemical reaction can occur only when the reactant molecules, atoms, or ions collide. Hence, the observed rate is influence by the frequency of collisions between the reactants. The collisional frequency is the average rate in which two reactants collide for a given system and is used to express the average number of collisions per unit of time in a defined system. While deriving the collisional frequency ($Z_{AB}$) between two species in a gas is straightforward, it is beyond the scope of this text and the equation for collisional frequency of $A$ and $B$ is the following:
$Z_{AB} = N_{A}N_{B}\left(r_{A} + r_{B}\right)^2\sqrt{ \dfrac{8\pi k_{B}T}{\mu_{AB}}} \label{freq}$
with
• $N_A$ and $N_B$ are the numbers of $A$ and $B$ molecules in the system, respectively
• $r_a$ and $r_b$ are the radii of molecule $A$ and $B$, respectively
• $k_B$ is the Boltzmann constant $k_B$ =1.380 x 10-23 Joules Kelvin
• $T$ is the temperature in Kelvin
• $\mu_{AB}$ is calculated via $\mu_{AB} = \frac{m_Am_B}{m_A + m_B}$
The specifics of Equation \ref{freq} are not important for this conversation, but it is important to identify that $Z_{AB}$ increases with increasing density (i.e., increasing $N_A$ and $N_B$), with increasing reactant size ($r_a$ and $r_b$), with increasing velocities (predicted via Kinetic Molecular Theory), and with increasing temperature (although weakly because of the square root function).
A Video Discussing Collision Theory of Kinetics: Collusion Theory of Kinetics (opens in new window) [youtu.be]
Microscopic Factor 2: Activation Energy
Previously, we discussed the kinetic molecular theory of gases, which showed that the average kinetic energy of the particles of a gas increases with increasing temperature. Because the speed of a particle is proportional to the square root of its kinetic energy, increasing the temperature will also increase the number of collisions between molecules per unit time. What the kinetic molecular theory of gases does not explain is why the reaction rate of most reactions approximately doubles with a 10°C temperature increase. This result is surprisingly large considering that a 10°C increase in the temperature of a gas from 300 K to 310 K increases the kinetic energy of the particles by only about 4%, leading to an increase in molecular speed of only about 2% and a correspondingly small increase in the number of bimolecular collisions per unit time.
The collision model of chemical kinetics explains this behavior by introducing the concept of activation energy ($E_a$). We will define this concept using the reaction of $\ce{NO}$ with ozone, which plays an important role in the depletion of ozone in the ozone layer:
$\ce{NO(g) + O_3(g) \rightarrow NO_2(g) + O_2(g)} \nonumber$
Increasing the temperature from 200 K to 350 K causes the rate constant for this particular reaction to increase by a factor of more than 10, whereas the increase in the frequency of bimolecular collisions over this temperature range is only 30%. Thus something other than an increase in the collision rate must be affecting the reaction rate.
Experimental rate law for this reaction is
$\text{rate} = k [\ce{NO}][\ce{O3}] \nonumber$
and is used to identify how the reaction rate (not the rate constant) vares with concentration. The rate constant, however, does vary with temperature. Figure $1$ shows a plot of the rate constant of the reaction of $\ce{NO}$ with $\ce{O3}$ at various temperatures. The relationship is not linear but instead resembles the relationships seen in graphs of vapor pressure versus temperature (e.g, the Clausius-Claperyon equation). In all three cases, the shape of the plots results from a distribution of kinetic energy over a population of particles (electrons in the case of conductivity; molecules in the case of vapor pressure; and molecules, atoms, or ions in the case of reaction rates). Only a fraction of the particles have sufficient energy to overcome an energy barrier.
In the case of vapor pressure, particles must overcome an energy barrier to escape from the liquid phase to the gas phase. This barrier corresponds to the energy of the intermolecular forces that hold the molecules together in the liquid. In conductivity, the barrier is the energy gap between the filled and empty bands. In chemical reactions, the energy barrier corresponds to the amount of energy the particles must have to react when they collide. This energy threshold, called the activation energy, was first postulated in 1888 by the Swedish chemist Svante Arrhenius (1859–1927; Nobel Prize in Chemistry 1903). It is the minimum amount of energy needed for a reaction to occur. Reacting molecules must have enough energy to overcome electrostatic repulsion, and a minimum amount of energy is required to break chemical bonds so that new ones may be formed. Molecules that collide with less than the threshold energy bounce off one another chemically unchanged, with only their direction of travel and their speed altered by the collision. Molecules that are able to overcome the energy barrier are able to react and form an arrangement of atoms called the activated complex or the transition state of the reaction. The activated complex is not a reaction intermediate; it does not last long enough to be detected readily.
Any phenomenon that depends on the distribution of thermal energy in a population of particles has a nonlinear temperature dependence.
We can graph the energy of a reaction by plotting the potential energy of the system as the reaction progresses. Figure $2$ shows a plot for the NO–O3 system, in which the vertical axis is potential energy and the horizontal axis is the reaction coordinate, which indicates the progress of the reaction with time. The activated complex is shown in brackets with an asterisk. The overall change in potential energy for the reaction ($ΔE$) is negative, which means that the reaction releases energy. (In this case, $ΔE$ is −200.8 kJ/mol.) To react, however, the molecules must overcome the energy barrier to reaction ($E_a$ is 9.6 kJ/mol). That is, 9.6 kJ/mol must be put into the system as the activation energy. Below this threshold, the particles do not have enough energy for the reaction to occur.
Figure $\PageIndex{3a}$ illustrates the general situation in which the products have a lower potential energy than the reactants. In contrast, Figure $\PageIndex{3b}$ illustrates the case in which the products have a higher potential energy than the reactants, so the overall reaction requires an input of energy; that is, it is energetically uphill, and $ΔE > 0$. Although the energy changes that result from a reaction can be positive, negative, or even zero, in most cases an energy barrier must be overcome before a reaction can occur. This means that the activation energy is almost always positive; there is a class of reactions called barrierless reactions, but those are discussed elsewhere.
For similar reactions under comparable conditions, the one with the smallest Ea will occur most rapidly.
Whereas $ΔE$ is related to the tendency of a reaction to occur spontaneously, $E_a$ gives us information about the reaction rate and how rapidly the reaction rate changes with temperature. For two similar reactions under comparable conditions, the reaction with the smallest $E_a$ will occur more rapidly.
Figure $4$ shows both the kinetic energy distributions and a potential energy diagram for a reaction. The shaded areas show that at the lower temperature (300 K), only a small fraction of molecules collide with kinetic energy greater than Ea; however, at the higher temperature (500 K) a much larger fraction of molecules collide with kinetic energy greater than Ea. Consequently, the reaction rate is much slower at the lower temperature because only a relatively few molecules collide with enough energy to overcome the potential energy barrier.
Video Discussing Transition State Theory: Transition State Theory(opens in new window) [youtu.be]
Microscopic Factor 3: Sterics
Even when the energy of collisions between two reactant species is greater than $E_a$, most collisions do not produce a reaction. The probability of a reaction occurring depends not only on the collision energy but also on the spatial orientation of the molecules when they collide. For $\ce{NO}$ and $\ce{O3}$ to produce $\ce{NO2}$ and $\ce{O2}$, a terminal oxygen atom of $\ce{O3}$ must collide with the nitrogen atom of $\ce{NO}$ at an angle that allows $\ce{O3}$ to transfer an oxygen atom to $\ce{NO}$ to produce $\ce{NO2}$ (Figure $4$). All other collisions produce no reaction. Because fewer than 1% of all possible orientations of $\ce{NO}$ and $\ce{O3}$ result in a reaction at kinetic energies greater than $E_a$, most collisions of $\ce{NO}$ and $\ce{O3}$ are unproductive. The fraction of orientations that result in a reaction is called the steric factor ($\rho$) and its value can range from $\rho=0$ (no orientations of molecules result in reaction) to $\rho=1$ (all orientations result in reaction).
Macroscopic Behavior: The Arrhenius Equation
The collision model explains why most collisions between molecules do not result in a chemical reaction. For example, nitrogen and oxygen molecules in a single liter of air at room temperature and 1 atm of pressure collide about 1030 times per second. If every collision produced two molecules of $\ce{NO}$, the atmosphere would have been converted to $\ce{NO}$ and then $\ce{NO2}$ a long time ago. Instead, in most collisions, the molecules simply bounce off one another without reacting, much as marbles bounce off each other when they collide.
For an $A + B$ elementary reaction, all three microscopic factors discussed above that affect the reaction rate can be summarized in a single relationship:
$\text{rate} = (\text{collision frequency}) \times (\text{steric factor}) \times (\text{fraction of collisions with } E > E_a ) \nonumber$
where
$\text{rate} = k[A][B] \label{14.5.2}$
Arrhenius used these relationships to arrive at an equation that relates the magnitude of the rate constant for a reaction to the temperature, the activation energy, and the constant, $A$, called the frequency factor:
$k=Ae^{-E_{\Large a}/RT} \label{14.5.3}$
The frequency factor is used to convert concentrations to collisions per second (scaled by the steric factor). Because the frequency of collisions depends on the temperature, $A$ is actually not constant (Equation \ref{freq}). Instead, $A$ increases slightly with temperature as the increased kinetic energy of molecules at higher temperatures causes them to move slightly faster and thus undergo more collisions per unit time.
Equation $\ref{14.5.3}$ is known as the Arrhenius equation and summarizes the collision model of chemical kinetics, where $T$ is the absolute temperature (in K) and R is the ideal gas constant [8.314 J/(K·mol)]. $E_a$ indicates the sensitivity of the reaction to changes in temperature. The reaction rate with a large $E_a$ increases rapidly with increasing temperature, whereas the reaction rate with a smaller $E_a$ increases much more slowly with increasing temperature.
If we know the reaction rate at various temperatures, we can use the Arrhenius equation to calculate the activation energy. Taking the natural logarithm of both sides of Equation $\ref{14.5.3}$,
\begin{align} \ln k &=\ln A+\left(-\dfrac{E_{\textrm a}}{RT}\right) \[4pt] &=\ln A+\left[\left(-\dfrac{E_{\textrm a}}{R}\right)\left(\dfrac{1}{T}\right)\right] \label{14.5.4} \end{align}
Equation $\ref{14.5.4}$ is the equation of a straight line,
$y = mx + b \nonumber$
where $y = \ln k$ and $x = 1/T$. This means that a plot of $\ln k$ versus $1/T$ is a straight line with a slope of $−E_a/R$ and an intercept of $\ln A$. In fact, we need to measure the reaction rate at only two temperatures to estimate $E_a$.
Knowing the $E_a$ at one temperature allows us to predict the reaction rate at other temperatures. This is important in cooking and food preservation, for example, as well as in controlling industrial reactions to prevent potential disasters. The procedure for determining $E_a$ from reaction rates measured at several temperatures is illustrated in Example $1$.
A Video Discussing The Arrhenius Equation: The Arrhenius Equation(opens in new window) [youtu.be]
Example $1$: Chirping Tree Crickets
Many people believe that the rate of a tree cricket’s chirping is related to temperature. To see whether this is true, biologists have carried out accurate measurements of the rate of tree cricket chirping ($f$) as a function of temperature ($T$). Use the data in the following table, along with the graph of ln[chirping rate] versus $1/T$ to calculate $E_a$ for the biochemical reaction that controls cricket chirping. Then predict the chirping rate on a very hot evening, when the temperature is 308 K (35°C, or 95°F).
Chirping Tree Crickets Frequency Table
Frequency (f; chirps/min) ln f T (K) 1/T (K)
200 5.30 299 3.34 × 10−3
179 5.19 298 3.36 × 10−3
158 5.06 296 3.38 × 10−3
141 4.95 294 3.40 × 10−3
126 4.84 293 3.41 × 10−3
112 4.72 292 3.42 × 10−3
100 4.61 290 3.45 × 10−3
89 4.49 289 3.46 × 10−3
79 4.37 287 3.48 × 10−3
Given: chirping rate at various temperatures
Asked for: activation energy and chirping rate at specified temperature
Strategy:
1. From the plot of $\ln f$ versus $1/T$, calculate the slope of the line (−Ea/R) and then solve for the activation energy.
2. Express Equation \ref{14.5.4} in terms of k1 and T1 and then in terms of k2 and T2.
3. Subtract the two equations; rearrange the result to describe k2/k1 in terms of T2 and T1.
4. Using measured data from the table, solve the equation to obtain the ratio k2/k1. Using the value listed in the table for k1, solve for k2.
Solution
A If cricket chirping is controlled by a reaction that obeys the Arrhenius equation, then a plot of $\ln f$ versus $1/T$ should give a straight line (Figure $6$).
Also, the slope of the plot of $\ln f$ versus $1/T$ should be equal to $−E_a/R$. We can use the two endpoints in Figure $6$ to estimate the slope:
\begin{align*}\textrm{slope}&=\dfrac{\Delta\ln f}{\Delta(1/T)} \[4pt] &=\dfrac{5.30-4.37}{3.34\times10^{-3}\textrm{ K}^{-1}-3.48\times10^{-3}\textrm{ K}^{-1}} \[4pt] &=\dfrac{0.93}{-0.14\times10^{-3}\textrm{ K}^{-1}} \[4pt] &=-6.6\times10^3\textrm{ K}\end{align*} \nonumber
A computer best-fit line through all the points has a slope of −6.67 × 103 K, so our estimate is very close. We now use it to solve for the activation energy:
\begin{align*} E_{\textrm a} &=-(\textrm{slope})(R) \[4pt] &=-(-6.6\times10^3\textrm{ K})\left(\dfrac{8.314 \textrm{ J}}{\mathrm{K\cdot mol}}\right)\left(\dfrac{\textrm{1 KJ}}{\textrm{1000 J}}\right) \[4pt] &=\dfrac{\textrm{55 kJ}}{\textrm{mol}} \end{align*} \nonumber
B If the activation energy of a reaction and the rate constant at one temperature are known, then we can calculate the reaction rate at any other temperature. We can use Equation \ref{14.5.4} to express the known rate constant ($k_1$) at the first temperature ($T_1$) as follows:
$\ln k_1=\ln A-\dfrac{E_{\textrm a}}{RT_1} \nonumber$
Similarly, we can express the unknown rate constant ($k_2$) at the second temperature ($T_2$) as follows:
$\ln k_2=\ln A-\dfrac{E_{\textrm a}}{RT_2} \nonumber$
C These two equations contain four known quantities (Ea, T1, T2, and k1) and two unknowns (A and k2). We can eliminate A by subtracting the first equation from the second:
\begin{align*} \ln k_2-\ln k_1 &=\left(\ln A-\dfrac{E_{\textrm a}}{RT_2}\right)-\left(\ln A-\dfrac{E_{\textrm a}}{RT_1}\right) \[4pt] &=-\dfrac{E_{\textrm a}}{RT_2}+\dfrac{E_{\textrm a}}{RT_1} \end{align*} \nonumber
Then
$\ln \dfrac{k_2}{k_1}=\dfrac{E_{\textrm a}}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right) \nonumber$
D To obtain the best prediction of chirping rate at 308 K (T2), we try to choose for T1 and k1 the measured rate constant and corresponding temperature in the data table that is closest to the best-fit line in the graph. Choosing data for T1 = 296 K, where f = 158, and using the $E_a$ calculated previously,
\begin{align*} \ln\dfrac{k_{T_2}}{k_{T_1}} &=\dfrac{E_{\textrm a}}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right) \[4pt] &=\dfrac{55\textrm{ kJ/mol}}{8.314\textrm{ J}/(\mathrm{K\cdot mol})}\left(\dfrac{1000\textrm{ J}}{\textrm{1 kJ}}\right)\left(\dfrac{1}{296 \textrm{ K}}-\dfrac{1}{\textrm{308 K}}\right) \[4pt] &=0.87 \end{align*} \nonumber
Thus k308/k296 = 2.4 and k308 = (2.4)(158) = 380, and the chirping rate on a night when the temperature is 308 K is predicted to be 380 chirps per minute.
Exercise $\PageIndex{1A}$
The equation for the decomposition of $\ce{NO2}$ to $\ce{NO}$ and $\ce{O2}$ is second order in $\ce{NO2}$:
$\ce{2NO2(g) → 2NO(g) + O2(g)} \nonumber$
Data for the reaction rate as a function of temperature are listed in the following table. Calculate $E_a$ for the reaction and the rate constant at 700 K.
Data for the reaction rate as a function of temperature
T (K) k (M−1·s−1)
592 522
603 755
627 1700
652 4020
656 5030
Answer
$E_a$ = 114 kJ/mol; k700= 18,600 M−1·s−1 = 1.86 × 104 M−1·s−1.
Exercise $\PageIndex{1B}$
What $E_a$ results in a doubling of the reaction rate with a 10°C increase in temperature from 20° to 30°C?
Answer
about 51 kJ/mol
A Video Discussing Graphing Using the Arrhenius Equation: Graphing Using the Arrhenius Equation (opens in new window) [youtu.be] (opens in new window)
Summary
For a chemical reaction to occur, an energy threshold must be overcome, and the reacting species must also have the correct spatial orientation. The Arrhenius equation is $k=Ae^{-E_{\Large a}/RT}$. A minimum energy (activation energy,v$E_a$) is required for a collision between molecules to result in a chemical reaction. Plots of potential energy for a system versus the reaction coordinate show an energy barrier that must be overcome for the reaction to occur. The arrangement of atoms at the highest point of this barrier is the activated complex, or transition state, of the reaction. At a given temperature, the higher the Ea, the slower the reaction. The fraction of orientations that result in a reaction is the steric factor. The frequency factor, steric factor, and activation energy are related to the rate constant in the Arrhenius equation: $k=Ae^{-E_{\Large a}/RT}$. A plot of the natural logarithm of k versus 1/T is a straight line with a slope of −Ea/R. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/14%3A_Chemical_Kinetics/14.05%3A_Temperature_and_Rate.txt |
Learning Objectives
• To determine the individual steps of a simple reaction.
One of the major reasons for studying chemical kinetics is to use measurements of the macroscopic properties of a system, such as the rate of change in the concentration of reactants or products with time, to discover the sequence of events that occur at the molecular level during a reaction. This molecular description is the mechanism of the reaction; it describes how individual atoms, ions, or molecules interact to form particular products. The stepwise changes are collectively called the reaction mechanism.
In an internal combustion engine, for example, isooctane reacts with oxygen to give carbon dioxide and water:
$\ce{2C8H18 (l) + 25O2(g) -> 16CO2 (g) + 18H2O(g)} \label{14.6.1}$
For this reaction to occur in a single step, 25 dioxygen molecules and 2 isooctane molecules would have to collide simultaneously and be converted to 34 molecules of product, which is very unlikely. It is more likely that a complex series of reactions takes place in a stepwise fashion. Each individual reaction, which is called an elementary reaction, involves one, two, or (rarely) three atoms, molecules, or ions. The overall sequence of elementary reactions is the mechanism of the reaction. The sum of the individual steps, or elementary reactions, in the mechanism must give the balanced chemical equation for the overall reaction.
The overall sequence of elementary reactions is the mechanism of the reaction.
Molecularity and the Rate-Determining Step
To demonstrate how the analysis of elementary reactions helps us determine the overall reaction mechanism, we will examine the much simpler reaction of carbon monoxide with nitrogen dioxide.
$\ce{NO2(g) + CO(g) -> NO(g) + CO2 (g)} \label{14.6.2}$
From the balanced chemical equation, one might expect the reaction to occur via a collision of one molecule of $\ce{NO2}$ with a molecule of $\ce{CO}$ that results in the transfer of an oxygen atom from nitrogen to carbon. The experimentally determined rate law for the reaction, however, is as follows:
$rate = k[\ce{NO2}]^2 \label{14.6.3}$
The fact that the reaction is second order in $[\ce{NO2}]$ and independent of $[\ce{CO}]$ tells us that it does not occur by the simple collision model outlined previously. If it did, its predicted rate law would be
$rate = k[\ce{NO2}][\ce{CO}]. \nonumber$
The following two-step mechanism is consistent with the rate law if step 1 is much slower than step 2:
two-step mechanism
$\textrm{step 1}$ $\mathrm{NO_2}+\mathrm{NO_2}\xrightarrow{\textrm{slow}}\mathrm{NO_3}+\textrm{NO}$ $\textrm{elementary reaction}$
$\textrm{step 2}$ $\underline{\mathrm{NO_3}+\mathrm{CO}\rightarrow\mathrm{NO_2}+\mathrm{CO_2}}$ $\textrm{elementary reaction}$
$\textrm{sum}$ $\mathrm{NO_2}+\mathrm{CO}\rightarrow\mathrm{NO}+\mathrm{CO_2}$ $\textrm{overall reaction}$
According to this mechanism, the overall reaction occurs in two steps, or elementary reactions. Summing steps 1 and 2 and canceling on both sides of the equation gives the overall balanced chemical equation for the reaction. The $\ce{NO3}$ molecule is an intermediate in the reaction, a species that does not appear in the balanced chemical equation for the overall reaction. It is formed as a product of the first step but is consumed in the second step.
The sum of the elementary reactions in a reaction mechanism must give the overall balanced chemical equation of the reaction.
Using Molecularity to Describe a Rate Law
The molecularity of an elementary reaction is the number of molecules that collide during that step in the mechanism. If there is only a single reactant molecule in an elementary reaction, that step is designated as unimolecular; if there are two reactant molecules, it is bimolecular; and if there are three reactant molecules (a relatively rare situation), it is termolecular. Elementary reactions that involve the simultaneous collision of more than three molecules are highly improbable and have never been observed experimentally. (To understand why, try to make three or more marbles or pool balls collide with one another simultaneously!)
Writing the rate law for an elementary reaction is straightforward because we know how many molecules must collide simultaneously for the elementary reaction to occur; hence the order of the elementary reaction is the same as its molecularity (Table $1$). In contrast, the rate law for the reaction cannot be determined from the balanced chemical equation for the overall reaction. The general rate law for a unimolecular elementary reaction (A → products) is
$rate = k[A]. \nonumber$
For bimolecular reactions, the reaction rate depends on the number of collisions per unit time, which is proportional to the product of the concentrations of the reactants, as shown in Figure $1$. For a bimolecular elementary reaction of the form A + B → products, the general rate law is
$rate = k[A][B]. \nonumber$
Table $1$: Common Types of Elementary Reactions and Their Rate Laws
Elementary Reaction Molecularity Rate Law Reaction Order
A → products unimolecular rate = k[A] first
2A → products bimolecular rate = k[A]2 second
A + B → products bimolecular rate = k[A][B] second
2A + B → products termolecular rate = k[A]2[B] third
A + B + C → products termolecular rate = k[A][B][C] third
For elementary reactions, the order of the elementary reaction is the same as its molecularity. In contrast, the rate law cannot be determined from the balanced chemical equation for the overall reaction (unless it is a single step mechanism and is therefore also an elementary step).
Identifying the Rate-Determining Step
Note the important difference between writing rate laws for elementary reactions and the balanced chemical equation of the overall reaction. Because the balanced chemical equation does not necessarily reveal the individual elementary reactions by which the reaction occurs, we cannot obtain the rate law for a reaction from the overall balanced chemical equation alone. In fact, it is the rate law for the slowest overall reaction, which is the same as the rate law for the slowest step in the reaction mechanism, the rate-determining step, that must give the experimentally determined rate law for the overall reaction.This statement is true if one step is substantially slower than all the others, typically by a factor of 10 or more. If two or more slow steps have comparable rates, the experimentally determined rate laws can become complex. Our discussion is limited to reactions in which one step can be identified as being substantially slower than any other. The reason for this is that any process that occurs through a sequence of steps can take place no faster than the slowest step in the sequence. In an automotive assembly line, for example, a component cannot be used faster than it is produced. Similarly, blood pressure is regulated by the flow of blood through the smallest passages, the capillaries. Because movement through capillaries constitutes the rate-determining step in blood flow, blood pressure can be regulated by medications that cause the capillaries to contract or dilate. A chemical reaction that occurs via a series of elementary reactions can take place no faster than the slowest step in the series of reactions.
Look at the rate laws for each elementary reaction in our example as well as for the overall reaction.
rate laws for each elementary reaction in our example as well as for the overall reaction.
$\textrm{step 1}$ $\mathrm{NO_2}+\mathrm{NO_2}\xrightarrow{\mathrm{k_1}}\mathrm{NO_3}+\textrm{NO}$ $\textrm{rate}=k_1[\mathrm{NO_2}]^2\textrm{ (predicted)}$
$\textrm{step 2}$ $\underline{\mathrm{NO_3}+\mathrm{CO}\xrightarrow{k_2}\mathrm{NO_2}+\mathrm{CO_2}}$ $\textrm{rate}=k_2[\mathrm{NO_3}][\mathrm{CO}]\textrm{ (predicted)}$
$\textrm{sum}$ $\mathrm{NO_2}+\mathrm{CO}\xrightarrow{k}\mathrm{NO}+\mathrm{CO_2}$ $\textrm{rate}=k[\mathrm{NO_2}]^2\textrm{ (observed)}$
The experimentally determined rate law for the reaction of $NO_2$ with $CO$ is the same as the predicted rate law for step 1. This tells us that the first elementary reaction is the rate-determining step, so $k$ for the overall reaction must equal $k_1$. That is, NO3 is formed slowly in step 1, but once it is formed, it reacts very rapidly with CO in step 2.
Sometimes chemists are able to propose two or more mechanisms that are consistent with the available data. If a proposed mechanism predicts the wrong experimental rate law, however, the mechanism must be incorrect.
Example $1$: A Reaction with an Intermediate
In an alternative mechanism for the reaction of $\ce{NO2}$ with $\ce{CO}$ with $\ce{N2O4}$ appearing as an intermediate.
alternative mechanism for the reaction of $\ce{NO2}$ with $\ce{CO}$ with $\ce{N2O4}$ appearing as an intermediate.
$\textrm{step 1}$ $\mathrm{NO_2}+\mathrm{NO_2}\xrightarrow{k_1}\mathrm{N_2O_4}$
$\textrm{step 2}$ $\underline{\mathrm{N_2O_4}+\mathrm{CO}\xrightarrow{k_2}\mathrm{NO}+\mathrm{NO_2}+\mathrm{CO_2}}$
$\textrm{sum}$ $\mathrm{NO_2}+\mathrm{CO}\rightarrow\mathrm{NO}+\mathrm{CO_2}$
Write the rate law for each elementary reaction. Is this mechanism consistent with the experimentally determined rate law (rate = k[NO2]2)?
Given: elementary reactions
Asked for: rate law for each elementary reaction and overall rate law
Strategy:
1. Determine the rate law for each elementary reaction in the reaction.
2. Determine which rate law corresponds to the experimentally determined rate law for the reaction. This rate law is the one for the rate-determining step.
Solution
A The rate law for step 1 is rate = k1[NO2]2; for step 2, it is rate = k2[N2O4][CO].
B If step 1 is slow (and therefore the rate-determining step), then the overall rate law for the reaction will be the same: rate = k1[NO2]2. This is the same as the experimentally determined rate law. Hence this mechanism, with N2O4 as an intermediate, and the one described previously, with NO3 as an intermediate, are kinetically indistinguishable. In this case, further experiments are needed to distinguish between them. For example, the researcher could try to detect the proposed intermediates, NO3 and N2O4, directly.
Exercise $1$
Iodine monochloride ($\ce{ICl}$) reacts with $\ce{H2}$ as follows:
$\ce{2ICl(l) + H_2(g) \rightarrow 2HCl(g) + I_2(s)} \nonumber$
The experimentally determined rate law is $rate = k[\ce{ICl}][\ce{H2}]$. Write a two-step mechanism for this reaction using only bimolecular elementary reactions and show that it is consistent with the experimental rate law. (Hint: $\ce{HI}$ is an intermediate.)
Answer
Solutions to Exercise 14.6.1
$\textrm{step 1}$ $\mathrm{ICl}+\mathrm{H_2}\xrightarrow{k_1}\mathrm{HCl}+\mathrm{HI}$ $\mathrm{rate}=k_1[\mathrm{ICl}][\mathrm{H_2}]\,(\textrm{slow})$
$\textrm{step 2}$ $\underline{\mathrm{HI}+\mathrm{ICl}\xrightarrow{k_2}\mathrm{HCl}+\mathrm{I_2}}$ $\mathrm{rate}=k_2[\mathrm{HI}][\mathrm{ICl}]\,(\textrm{fast})$
$\textrm{sum}$ $\mathrm{2ICl}+\mathrm{H_2}\rightarrow\mathrm{2HCl}+\mathrm{I_2}$
This mechanism is consistent with the experimental rate law if the first step is the rate-determining step.
Example $2$ : Nitrogen Oxide Reacting with Molecular Hydrogen
Assume the reaction between $\ce{NO}$ and $\ce{H_2}$ occurs via a three-step process:
the reaction between $\ce{NO}$ and $\ce{H_2}$ occurs via a three-step process
$\textrm{step 1}$ $\mathrm{NO}+\mathrm{NO}\xrightarrow{k_1}\mathrm{N_2O_2}$ $\textrm{(fast)}$
$\textrm{step 2}$ $\mathrm{N_2O_2}+\mathrm{H_2}\xrightarrow{k_2}\mathrm{N_2O}+\mathrm{H_2O}$ $\textrm{(slow)}$
$\textrm{step 3}$ $\mathrm{N_2O}+\mathrm{H_2}\xrightarrow{k_3}\mathrm{N_2}+\mathrm{H_2O}$ $\textrm{(fast)}$
Write the rate law for each elementary reaction, write the balanced chemical equation for the overall reaction, and identify the rate-determining step. Is the rate law for the rate-determining step consistent with the experimentally derived rate law for the overall reaction:
$\text{rate} = k[\ce{NO}]^2[\ce{H_2}]? \tag{observed}$
Answer
• Step 1: $rate = k_1[\ce{NO}]^2$
• Step 2: $rate = k_2[\ce{N_2O_2}][\ce{H_2}]$
• Step 3: $rate = k_3[\ce{N_2O}][\ce{H_2}]$
The overall reaction is then
$\ce{2NO(g) + 2H_2(g) -> N2(g) + 2H2O(g)} \nonumber$
• Rate Determining Step : #2
• Yes, because the rate of formation of $[\ce{N_2O_2}] = k_1[\ce{NO}]^2$. Substituting $k_1[\ce{NO}]^2$ for $[\ce{N_2O_2}]$ in the rate law for step 2 gives the experimentally derived rate law for the overall chemical reaction, where $k = k_1k_2$.
Reaction Mechanism (Slow step followed by fast step): Reaction Mechanism (Slow step Followed by Fast Step)(opens in new window) [youtu.be] (opens in new window)
Summary
A balanced chemical reaction does not necessarily reveal either the individual elementary reactions by which a reaction occurs or its rate law. A reaction mechanism is the microscopic path by which reactants are transformed into products. Each step is an elementary reaction. Species that are formed in one step and consumed in another are intermediates. Each elementary reaction can be described in terms of its molecularity, the number of molecules that collide in that step. The slowest step in a reaction mechanism is the rate-determining step. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/14%3A_Chemical_Kinetics/14.06%3A_Reaction_Mechanisms.txt |
Learning Objectives
• To understand how catalysts increase the reaction rate and the selectivity of chemical reactions.
Catalysts are substances that increase the reaction rate of a chemical reaction without being consumed in the process. A catalyst, therefore, does not appear in the overall stoichiometry of the reaction it catalyzes, but it must appear in at least one of the elementary reactions in the mechanism for the catalyzed reaction. The catalyzed pathway has a lower Ea, but the net change in energy that results from the reaction (the difference between the energy of the reactants and the energy of the products) is not affected by the presence of a catalyst (Figure $1$). Nevertheless, because of its lower Ea, the reaction rate of a catalyzed reaction is faster than the reaction rate of the uncatalyzed reaction at the same temperature. Because a catalyst decreases the height of the energy barrier, its presence increases the reaction rates of both the forward and the reverse reactions by the same amount. In this section, we will examine the three major classes of catalysts: heterogeneous catalysts, homogeneous catalysts, and enzymes.
A catalyst affects Ea, not ΔE.
Heterogeneous Catalysis
In heterogeneous catalysis, the catalyst is in a different phase from the reactants. At least one of the reactants interacts with the solid surface in a physical process called adsorption in such a way that a chemical bond in the reactant becomes weak and then breaks. Poisons are substances that bind irreversibly to catalysts, preventing reactants from adsorbing and thus reducing or destroying the catalyst’s efficiency.
An example of heterogeneous catalysis is the interaction of hydrogen gas with the surface of a metal, such as Ni, Pd, or Pt. As shown in part (a) in Figure $2$, the hydrogen–hydrogen bonds break and produce individual adsorbed hydrogen atoms on the surface of the metal. Because the adsorbed atoms can move around on the surface, two hydrogen atoms can collide and form a molecule of hydrogen gas that can then leave the surface in the reverse process, called desorption. Adsorbed H atoms on a metal surface are substantially more reactive than a hydrogen molecule. Because the relatively strong H–H bond (dissociation energy = 432 kJ/mol) has already been broken, the energy barrier for most reactions of H2 is substantially lower on the catalyst surface.
Figure $2$ shows a process called hydrogenation, in which hydrogen atoms are added to the double bond of an alkene, such as ethylene, to give a product that contains C–C single bonds, in this case ethane. Hydrogenation is used in the food industry to convert vegetable oils, which consist of long chains of alkenes, to more commercially valuable solid derivatives that contain alkyl chains. Hydrogenation of some of the double bonds in polyunsaturated vegetable oils, for example, produces margarine, a product with a melting point, texture, and other physical properties similar to those of butter.
Several important examples of industrial heterogeneous catalytic reactions are in Table $1$. Although the mechanisms of these reactions are considerably more complex than the simple hydrogenation reaction described here, they all involve adsorption of the reactants onto a solid catalytic surface, chemical reaction of the adsorbed species (sometimes via a number of intermediate species), and finally desorption of the products from the surface.
Table $1$: Some Commercially Important Reactions that Employ Heterogeneous Catalysts
Commercial Process Catalyst Initial Reaction Final Commercial Product
contact process V2O5 or Pt 2SO2 + O2 → 2SO3 H2SO4
Haber process Fe, K2O, Al2O3 N2 + 3H2 → 2NH3 NH3
Ostwald process Pt and Rh 4NH3 + 5O2 → 4NO + 6H2O HNO3
water–gas shift reaction Fe, Cr2O3, or Cu CO + H2O → CO2 + H2 H2 for NH3, CH3OH, and other fuels
steam reforming Ni CH4 + H2O → CO + 3H2 H2
methanol synthesis ZnO and Cr2O3 CO + 2H2 → CH3OH CH3OH
Sohio process bismuth phosphomolybdate $\mathrm{CH}_2\textrm{=CHCH}_3+\mathrm{NH_3}+\mathrm{\frac{3}{2}O_2}\rightarrow\mathrm{CH_2}\textrm{=CHCN}+\mathrm{3H_2O}$ $\underset{\textrm{acrylonitrile}}{\mathrm{CH_2}\textrm{=CHCN}}$
catalytic hydrogenation Ni, Pd, or Pt RCH=CHR′ + H2 → RCH2—CH2R′ partially hydrogenated oils for margarine, and so forth
Homogeneous Catalysis
In homogeneous catalysis, the catalyst is in the same phase as the reactant(s). The number of collisions between reactants and catalyst is at a maximum because the catalyst is uniformly dispersed throughout the reaction mixture. Many homogeneous catalysts in industry are transition metal compounds (Table $2$), but recovering these expensive catalysts from solution has been a major challenge. As an added barrier to their widespread commercial use, many homogeneous catalysts can be used only at relatively low temperatures, and even then they tend to decompose slowly in solution. Despite these problems, a number of commercially viable processes have been developed in recent years. High-density polyethylene and polypropylene are produced by homogeneous catalysis.
Table $2$: Some Commercially Important Reactions that Employ Homogeneous Catalysts
Commercial Process Catalyst Reactants Final Product
Union Carbide [Rh(CO)2I2] CO + CH3OH CH3CO2H
hydroperoxide process Mo(VI) complexes CH3CH=CH2 + R–O–O–H
hydroformylation Rh/PR3 complexes RCH=CH2 + CO + H2 RCH2CH2CHO
adiponitrile process Ni/PR3complexes 2HCN + CH2=CHCH=CH2 NCCH2CH2CH2CH2CN used to synthesize nylon
olefin polymerization (RC5H5)2ZrCl2 CH2=CH2 –(CH2CH2–)n: high-density polyethylene
Enzymes
Enzymes, catalysts that occur naturally in living organisms, are almost all protein molecules with typical molecular masses of 20,000–100,000 amu. Some are homogeneous catalysts that react in aqueous solution within a cellular compartment of an organism. Others are heterogeneous catalysts embedded within the membranes that separate cells and cellular compartments from their surroundings. The reactant in an enzyme-catalyzed reaction is called a substrate.
Because enzymes can increase reaction rates by enormous factors (up to 1017 times the uncatalyzed rate) and tend to be very specific, typically producing only a single product in quantitative yield, they are the focus of active research. At the same time, enzymes are usually expensive to obtain, they often cease functioning at temperatures greater than 37 °C, have limited stability in solution, and have such high specificity that they are confined to turning one particular set of reactants into one particular product. This means that separate processes using different enzymes must be developed for chemically similar reactions, which is time-consuming and expensive. Thus far, enzymes have found only limited industrial applications, although they are used as ingredients in laundry detergents, contact lens cleaners, and meat tenderizers. The enzymes in these applications tend to be proteases, which are able to cleave the amide bonds that hold amino acids together in proteins. Meat tenderizers, for example, contain a protease called papain, which is isolated from papaya juice. It cleaves some of the long, fibrous protein molecules that make inexpensive cuts of beef tough, producing a piece of meat that is more tender. Some insects, like the bombadier beetle, carry an enzyme capable of catalyzing the decomposition of hydrogen peroxide to water (Figure $3$).
Enzyme inhibitors cause a decrease in the reaction rate of an enzyme-catalyzed reaction by binding to a specific portion of an enzyme and thus slowing or preventing a reaction from occurring. Irreversible inhibitors are therefore the equivalent of poisons in heterogeneous catalysis. One of the oldest and most widely used commercial enzyme inhibitors is aspirin, which selectively inhibits one of the enzymes involved in the synthesis of molecules that trigger inflammation. The design and synthesis of related molecules that are more effective, more selective, and less toxic than aspirin are important objectives of biomedical research.
Summary
Catalysts participate in a chemical reaction and increase its rate. They do not appear in the reaction’s net equation and are not consumed during the reaction. Catalysts allow a reaction to proceed via a pathway that has a lower activation energy than the uncatalyzed reaction. In heterogeneous catalysis, catalysts provide a surface to which reactants bind in a process of adsorption. In homogeneous catalysis, catalysts are in the same phase as the reactants. Enzymes are biological catalysts that produce large increases in reaction rates and tend to be specific for certain reactants and products. The reactant in an enzyme-catalyzed reaction is called a substrate. Enzyme inhibitors cause a decrease in the reaction rate of an enzyme-catalyzed reaction. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/14%3A_Chemical_Kinetics/14.07%3A_Catalysis.txt |
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
14.1: Factors that Affect Reaction Rates
Q14.1.1
What information can you obtain by studying the chemical kinetics of a reaction? Does a balanced chemical equation provide the same information? Why or why not?
S14.1.1
Kinetics gives information on the reaction rate and reaction mechanism; the balanced chemical equation gives only the stoichiometry of the reaction.
Q14.1.2
If you were tasked with determining whether to proceed with a particular reaction in an industrial facility, why would studying the chemical kinetics of the reaction be important to you?
S14.1.2
Studying chemical kinetics determines whether to proceed with a reaction as it measures the rate of a reaction. Reactions conducted in an industrial facility mix compounds together, heating and stirring them for a while, before moving to the next process. When the compounds are then moved to the next phase of the process it is important to know how long to hold the reaction at one stage before continuing, to make sure a reaction has finished before starting the next one
Q14.1.3
What is the relationship between each of the following factors and the reaction rate: reactant concentration, temperature of the reaction, physical properties of the reactants, physical and chemical properties of the solvent, and the presence of a catalyst?
S14.1.3
Reactant concentration:
Reaction rate and the concentration of reactants have a direct relationship; as concentration goes up, so does reaction rate. This is because a higher concentration causes the reactants to have a higher probability to collide with a another reactant particle, possibly inducing the chemical reaction. This comes from the Collision theory.
Reaction temperature:
Reaction rate and the reaction temperature also have a direct relationship; as temperature goes up, so does reaction rate. A higher temperature makes the particles move at a faster speed so when two or more reactant particles collide, they collide with more energy and are more likely to reach the activation energy threshold, starting a chemical reaction.
Physical properties of the reactants:
If the reactants have the same physical properties, they are more likely to react, increasing the reaction rate, because the reactants mix and are more likely to collide another one of the reactants.
Physical and chemical properties of the solvent:
The properties of the solvent also affect the rates of a reaction. High viscosity means that particles more slowly than in low viscosity solvents, so reaction rates are slower in high viscosity solvents than in low viscosity solvents. This indicates an inverse relationship between viscosity and reaction rate.
Presence of a catalyst:
The chemical definition of a catalyst in a substance that increases the rate of a reaction without being used up in the reaction. Therefore, if a catalyst is present, reaction rate increases.
Q14.1.4
A slurry is a mixture of a finely divided solid with a liquid in which it is only sparingly soluble. As you prepare a reaction, you notice that one of your reactants forms a slurry with the solvent, rather than a solution. What effect will this have on the reaction rate? What steps can you take to try to solve the problem?
Q14.1.5
Why does the reaction rate of virtually all reactions increase with an increase in temperature? If you were to make a glass of sweetened iced tea the old-fashioned way, by adding sugar and ice cubes to a glass of hot tea, which would you add first?
S14.1.5
Increasing the temperature increases the average kinetic energy of molecules and ions, causing them to collide more frequently and with greater energy, which increases the reaction rate. First dissolve sugar in the hot tea, and then add the ice.
Q14.1.6
In a typical laboratory setting, a reaction is carried out in a ventilated hood with air circulation provided by outside air. A student noticed that a reaction that gave a high yield of a product in the winter gave a low yield of that same product in the summer, even though his technique did not change and the reagents and concentrations used were identical. What is a plausible explanation for the different yields?
Q14.1.7
A very active area of chemical research involves the development of solubilized catalysts that are not made inactive during the reaction process. Such catalysts are expected to increase reaction rates significantly relative to the same reaction run in the presence of a heterogeneous catalyst. What is the reason for anticipating that the relative rate will increase?
Q14.1.8
Water has a dielectric constant more than two times greater than that of methanol (80.1 for H2O and 33.0 for CH3OH). Which would be your solvent of choice for a substitution reaction between an ionic compound and a polar reagent, both of which are soluble in either methanol or water? Why?
14.2: Reaction Rates
Q14.2.1
Explain why the reaction rate is generally fastest at early time intervals. For the second-order A + B → C, what would the plot of the concentration of C versus time look like during the course of the reaction?
S14.2.1
Reactant concentrations are highest at the beginning of a reaction. The plot of [C] versus t is a curve with a slope that becomes steadily less positive.
Q14.2.2
Explain the differences between a differential rate law and an integrated rate law. What two components do they have in common? Which form is preferred for obtaining a reaction order and a rate constant? Why?
Q14.2.3
Diffusion-controlled reactions have rates that are determined only by the reaction rate at which two reactant molecules can diffuse together. These reactions are rapid, with second-order rate constants typically on the order of 1010 L/(mol·s). Are the reactions expected to be faster or slower in solvents that have a low viscosity? Why? Consider the reactions H3O+ + OH → 2H2O and H3O+ + N(CH3)3 → H2O + HN(CH3)3+ in aqueous solution. Which would have the higher rate constant? Why?
S14.2.3
Faster in a less viscous solvent because the rate of diffusion is higher; the H3O+/OH reaction is faster due to the decreased relative size of reactants and the higher electrostatic attraction between the reactants.
Q14.2.4
What information can be obtained from the reaction order? What correlation does the reaction order have with the stoichiometry of the overall equation?
Q14.2.5
During the hydrolysis reaction A + H2O → B + C, the concentration of A decreases much more rapidly in a polar solvent than in a nonpolar solvent. How would this effect be reflected in the overall reaction order?
Q14.2.6
The reaction rate of a particular reaction in which A and B react to make C is as follows:
$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=\dfrac{1}{2}\left ( \dfrac{\Delta[\textrm C]}{\Delta t} \right )$
Write a reaction equation that is consistent with this rate law. What is the rate expression with respect to time if 2A are converted to 3C?
Q14.2.7
While commuting to work, a person drove for 12 min at 35 mph, then stopped at an intersection for 2 min, continued the commute at 50 mph for 28 min, drove slowly through traffic at 38 mph for 18 min, and then spent 1 min pulling into a parking space at 3 mph. What was the average rate of the commute? What was the instantaneous rate at 13 min? at 28 min?
Q14.2.8
Why do most studies of chemical reactions use the initial rates of reaction to generate a rate law? How is this initial rate determined? Given the following data, what is the reaction order? Estimate.
Time (s) [A] (M)
120 0.158
240 0.089
360 0.062
Q14.2.9
Predict how the reaction rate will be affected by doubling the concentration of the first species in each equation.
1. C2H5I → C2H4 + HI: rate = k[C2H5I]
2. SO + O2 → SO2 + O: rate = k[SO][O2]
3. 2CH3 → C2H6: rate = k[CH3]2
4. ClOO → Cl + O2: rate = k
Q14.2.10
Cleavage of C2H6 to produce two CH3· radicals is a gas-phase reaction that occurs at 700°C. This reaction is first order, with k = 5.46 × 10−4 s−1. How long will it take for the reaction to go to 15% completion? to 50% completion?
298 s; 1270 s
Q14.2.11
Three chemical processes occur at an altitude of approximately 100 km in Earth’s atmosphere.
$\mathrm{N_2^+}+\mathrm{O_2}\xrightarrow{k_1}\mathrm{N_2}+\mathrm{O_2^+}$
$\mathrm{O_2^+}+\mathrm{O}\xrightarrow{k_2}\mathrm{O_2}+\mathrm{O^+}$
$(\mathrm{O^+}+\mathrm{N_2}\xrightarrow{k_3}\mathrm{NO^+}+\mathrm{N}$
Write a rate law for each elementary reaction. If the rate law for the overall reaction were found to be rate = k[N2+][O2], which one of the steps is rate limiting?
Q14.2.12
The oxidation of aqueous iodide by arsenic acid to give I3 and arsenous acid proceeds via the following reaction:
$\mathrm{H_3AsO_4(aq)}+\mathrm{3I^-(aq)}+\mathrm{2H^+(aq)}\overset{k_{\textrm f}}{\underset{k_{\textrm r}}{\rightleftharpoons}}\mathrm{H_3AsO_3(aq)}+\mathrm{I_3^-(aq)}+\mathrm{H_2O(l)}$
Write an expression for the initial rate of decrease of [I3], Δ[I3]/Δt. When the reaction rate of the forward reaction is equal to that of the reverse reaction: kf/kr = [H3AsO3][I3]/[H3AsO4][I]3[H+]2. Based on this information, what can you say about the nature of the rate-determining steps for the reverse and the forward reactions?
14.4: The Change of Concentration with Time
Q14.4.1
What are the characteristics of a zeroth-order reaction? Experimentally, how would you determine whether a reaction is zeroth order?
Q14.4.2
Predict whether the following reactions are zeroth order and explain your reasoning.
1. a substitution reaction of an alcohol with HCl to form an alkyl halide and water
2. catalytic hydrogenation of an alkene
3. hydrolysis of an alkyl halide to an alcohol
4. enzymatic conversion of nitrate to nitrite in a soil bacterium
Q14.4.3
In a first-order reaction, what is the advantage of using the integrated rate law expressed in natural logarithms over the rate law expressed in exponential form?
Q14.4.4
If the reaction rate is directly proportional to the concentration of a reactant, what does this tell you about (a) the reaction order with respect to the reactant and (b) the overall reaction order?
Q14.4.5
The reaction of NO with O2 is found to be second order with respect to NO and first order with respect to O2. What is the overall reaction order? What is the effect of doubling the concentration of each reagent on the reaction rate?
Q14.4.6
Iodide reduces Fe(III) according to the following reaction:
$2Fe^{3+}(soln) + 2I^−(soln) → 2Fe^{2+}(soln) + I_2(soln)$
Experimentally, it was found that doubling the concentration of Fe(III) doubled the reaction rate, and doubling the iodide concentration increased the reaction rate by a factor of 4. What is the reaction order with respect to each species? What is the overall rate law? What is the overall reaction order?
S14.4.6
First order in Fe3+; second order in I; third order overall; rate = k[Fe3+][I]2.
Q14.4.7
Benzoyl peroxide is a medication used to treat acne. Its rate of thermal decomposition at several concentrations was determined experimentally, and the data were tabulated as follows:
Experiment [Benzoyl Peroxide]0 (M) Initial Rate (M/s)
1 1.00 2.22 × 10−4
2 0.70 1.64 × 10−4
3 0.50 1.12 × 10−4
4 0.25 0.59 × 10−4
What is the reaction order with respect to benzoyl peroxide? What is the rate law for this reaction?
S14.4.7
The general rate law is: rate = k [Benzoyl Peroxide]m. In order to find the reaction order with respect to benzoyl peroxide, divide two rate laws and solve for m:
$\dfrac{rate_2}{rate_1}=\dfrac{k_2}{k_1} \left ( \dfrac{[Benzoyl Peroxide]_2}{[Benzoyl Peroxide]_1} \right )^m$
$\dfrac{1.64 × 10^-4 \dfrac{M}{s}}{2.22 × 10^-4 \dfrac{M}{s}}=( \dfrac{0.070 M}{1.00 M})^m$
$0.738=(0.7)^m$
$m=0.85$
rate law: rate = k [Benzoyl Peroxide]0.85
Q14.4.8
1-Bromopropane is a colorless liquid that reacts with S2O32− according to the following reaction:
$C_3H_7Br + S_2O_3^{2−} → C_3H_7S_2O_3^− + Br^−$
The reaction is first order in 1-bromopropane and first order in S2O32−, with a rate constant of 8.05 × 10−4 M−1·s−1. If you began a reaction with 40 mmol/100 mL of C3H7Br and an equivalent concentration of S2O32−, what would the initial reaction rate be? If you were to decrease the concentration of each reactant to 20 mmol/100 mL, what would the initial reaction rate be?
S14.4.8
1.29 × 10−4 M/s; 3.22 × 10−5 M/s
Q14.4.9
The experimental rate law for the reaction 3A + 2B → C + D was found to be Δ[C]/Δt = k[A]2[B] for an overall reaction that is third order. Because graphical analysis is difficult beyond second-order reactions, explain the procedure for determining the rate law experimentally.
14.5: Temperature and Rate
Q14.5.1
Although an increase in temperature results in an increase in kinetic energy, this increase in kinetic energy is not sufficient to explain the relationship between temperature and reaction rates. How does the activation energy relate to the chemical kinetics of a reaction? Why does an increase in temperature increase the reaction rate despite the fact that the average kinetic energy is still less than the activation energy?
S14.5.1
Activation energy is required for a collision between molecules to result in a chemical reaction, as well it is related to the rate due to the Arrhenius equation
Activation energy is the threshold of energy needed in order for a reaction to occur. Reactant particles must collide with enough energy to be able to break chemical bonds, that will then allow the creation of new bonds. If the particles do not have enough energy, when they collide they will simply bounce off one another.
k=Ae(-Ea/RT). The increase in temperature increases the rate of reaction despite the fact the average kinetic is less than the activation energy since it increase the rate of collision between molecules.
With an increase in temperature, there is a greater distribution of kinetic energy among reactant particles. This increase of temperature allows the rate in which particles collide with one another to increase. Although the average kinetic energy is still lower than the activation energy, the increase of collisions among particles increases the chance of particles that contain enough energy to overcome the energy barrier to collide. Thus, the reaction rate increases due to this increased rate of collisions.
Additionally, there is a higher amount of molecules that have sufficient kinetic energy to overcome the energy barrier, despite the average energy of all these molecules still being lower than the activation energy.
Q14.5.2
For any given reaction, what is the relationship between the activation energy and each of the following?
1. electrostatic repulsions
2. bond formation in the activated complex
3. the nature of the activated complex
S14.5.2
1.) Electrostatic repulsion: Electrostatic repulsion is the unfavorable interaction between two species of like charge. Activation energy is the minimum amount of energy needed for a reaction to occur. Reacting molecules must have enough energy to overcome electrostatic repulsion, and a minimum amount of energy is required to break chemical bonds so that new ones may be formed.
2.) Bond formation in the activated complex: An activated complex is an intermediate state that is formed during the conversion of reactants into products. Bond breaking can increase activation energy as breaking bonds requires energy.
3.) Nature of the activated complex: The activation energy of a chemical reaction is the difference between the energy of the activated complex and the energy of the reactants. If the structure has a high steric hindrance the activation energy will be higher.
Q14.5.3
If you are concerned with whether a reaction will occur rapidly, why would you be more interested in knowing the magnitude of the activation energy than the change in potential energy for the reaction?
Q14.5.4
The product C in the reaction A + B → C + D can be separated easily from the reaction mixture. You have been given pure A and pure B and are told to determine the activation energy for this reaction to determine whether the reaction is suitable for the industrial synthesis of C. How would you do this? Why do you need to know the magnitude of the activation energy to make a decision about feasibility?
Q14.5.5
Above Ea, molecules collide with enough energy to overcome the energy barrier for a reaction. Is it possible for a reaction to occur at a temperature less than that needed to reach Ea? Explain your answer.
Q14.5.6
What is the relationship between A, Ea, and T? How does an increase in A affect the reaction rate?
Q14.5.7
Of two highly exothermic reactions with different values of Ea, which would need to be monitored more carefully: the one with the smaller value or the one with the higher value? Why?
Q14.5.8
What happens to the approximate rate of a reaction when the temperature of the reaction is increased from 20°C to 30°C? What happens to the reaction rate when the temperature is raised to 70°C? For a given reaction at room temperature (20°C), what is the shape of a plot of reaction rate versus temperature as the temperature is increased to 70°C?
S14.5.8
The reaction rate will approximately double: 20°C to 30°C, the reaction rate increases by about 21 = 2; 20°C to 70°C, the reaction rate increases by about 25 = 32-fold. A plot of reaction rate versus temperature will give an exponential increase: rate ∝ 2ΔT/10.
Q14.5.9
Acetaldehyde, used in silvering mirrors and some perfumes, undergoes a second-order decomposition between 700 and 840 K. From the data in the following table, would you say that acetaldehyde follows the general rule that each 10 K increase in temperature doubles the reaction rate?
T (K) k (M−1·s−1)
720 0.024
740 0.051
760 0.105
800 0.519
Q14.5.10
Bromoethane reacts with hydroxide ion in water to produce ethanol. The activation energy for this reaction is 90 kJ/mol. If the reaction rate is 3.6 × 10−5 M/s at 25°C, what would the reaction rate be at the following temperatures?
1. 15°C
2. 30°C
3. 45°C
Q14.5.11
An enzyme-catalyzed reaction has an activation energy of 15 kcal/mol. How would the value of the rate constant differ between 20°C and 30°C? If the enzyme reduced the $E_a$ from 25 kcal/mol to 15 kcal/mol, by what factor has the enzyme increased the reaction rate at each temperature?
S14.5.11
1. 1.0 × 10−5 M/s
2. 6.6 × 10−5 M/s
3. 3.5 × 10−4 M/s
Q14.5.12
The data in the following table are the rate constants as a function of temperature for the dimerization of 1,3-butadiene. What is the activation energy for this reaction?
T (K) k (M−1·min−1)
529 1.4
560 3.7
600 25
645 82
100 kJ/mol
Q14.5.13
The reaction rate at 25°C is 1.0 × 10−4 M/s. Increasing the temperature to 75°C causes the reaction rate to increase to 7.0 × 10−2 M/s. Estimate $E_a$ for this process. If $E_a$ were 25 kJ/mol and the reaction rate at 25°C is 1.0 × 10−4 M/s, what would be the reaction rate at 75°C?
14.6: Reaction Mechanisms
Q14.6.1
How does the term molecularity relate to elementary reactions? How does it relate to the overall balanced chemical equation?
Q14.6.2
What is the relationship between the reaction order and the molecularity of a reaction? What is the relationship between the reaction order and the balanced chemical equation?
Q14.6.3
When you determine the rate law for a given reaction, why is it valid to assume that the concentration of an intermediate does not change with time during the course of the reaction?
Q14.6.3
If you know the rate law for an overall reaction, how would you determine which elementary reaction is rate determining? If an intermediate is contained in the rate-determining step, how can the experimentally determined rate law for the reaction be derived from this step?
Q14.6.4
Give the rate-determining step for each case.
1. Traffic is backed up on a highway because two lanes merge into one.
2. Gas flows from a pressurized cylinder fitted with a gas regulator and then is bubbled through a solution.
3. A document containing text and graphics is downloaded from the Internet.
Q14.6.5
Before being sent on an assignment, an aging James Bond was sent off to a health farm where part of the program’s focus was to purge his body of radicals. Why was this goal considered important to his health?
S14.6.5
Free radicals are uncharged molecules with an unpaired valence electron. The reason these are so dangerous is because they like to grab electrons from other atoms to fill their own outer shell. This allows them to impair protein function because free radicals readily oxidize proteins and cell membrane which could lead to a loss of function. It was important to purge James Bond of radicals because radicals set off chain reactions of continuously pulling electrons from molecules, which in turn, can damage cells in the body.
Q14.6.6
Cyclopropane, a mild anesthetic, rearranges to propylene via a collision that produces and destroys an energized species. The important steps in this rearrangement are as follows:
where M is any molecule, including cyclopropane. Only those cyclopropane molecules with sufficient energy (denoted with an asterisk) can rearrange to propylene. Which step determines the rate constant of the overall reaction?
S14.6.6
When we are trying to decide which step is a rate-determining step, you have to look at which step is the slowest of them all, since we have no knowledge of the time scale for either of these steps we have to look elsewhere to determine the rate-determining step.
In our specific problem, we do not know what exact molecules we have however we do see that the first step is an equilibrium since it has a double-headed arrow with two k values one that is the reciprocal of the other. Since it is an equilibrium we can reasonably assume that energized propane is rapidly created which would lead us to the conclusion that the second step is the slower of the two steps and is, therefore, the rate-determining step. So, the second step (with k2) determines the rate constant of the overall reaction.
Q14.6.7
Above approximately 500 K, the reaction between NO2 and CO to produce CO2 and NO follows the second-order rate law Δ[CO2]/Δt = k[NO2][CO]. At lower temperatures, however, the rate law is Δ[CO2]/Δt = k′[NO2]2, for which it is known that NO3 is an intermediate in the mechanism. Propose a complete low-temperature mechanism for the reaction based on this rate law. Which step is the slowest?
S14.6.7
Given that NO3 is an intermediate in the low temperature reaction mechanism, we automatically know two things: 1) NO3 won't show up in the final overall reaction, and thus, 2) NO3 will be in the products of the first reaction of the mechanism and in the reactants of the second reaction. We can also assume that since we're given an intermediate from the problem, this is the only intermediate (so we won't have to dream up any other compounds that might exist in the series of reactions.) These things being said, the lower temperature reaction mechanism will look like this:
$2NO_2(g) \rightarrow NO_3(g) + NO(g) \tag{1}$
$CO(g) + NO_3(g) \rightarrow CO_2(g) + NO_2(g) \tag{2}$
And the overall reaction will look like this (notice how NO3 is not present):
$CO(g) + NO_2(g) \rightarrow CO_2(g) + NO(g) \tag{overall reaction}$
Now that we have the reaction mechanism written out, we can go about determining which step is the slowest. It would be pretty tricky to do this if we weren't given any further information, however, we know two more things: 1) the high-temperature mechanism's rate = k[NO2][CO] meaning that it took place in one step (given that the overall reaction is also equal to this rate) and 2) the low-temperature mechanism's rate = k′[NO2]2. These two things being said, we've both confirmed that our proposed low-temperature reaction mechanism is in fact two steps, and that we have a means to find which step is slower.
By using the "guess-and-check" method we can label each step reaction one at a time as the "slow reaction" and see if the rate matches up with the rate given to us.
Let's first try the 2nd reaction. (see above)
By using rate laws we can determine that the rate of the reaction must be in terms of its reactants, which follows:
$\textrm{k}\textrm{[CO][NO}_3\textrm{]}$
... But wait! We can't have the overall reaction rate in terms of an intermediate.
By looking at the 1st reaction, we can determine that we can sub in "[NO2] /[NO]" for "[NO3]" since by writing the full reaction rate of the first step and solving for [NO3] this is equivalent. So, we now have the overall rate of the mechanism as the following given the 2nd reaction is the "slow reaction:"
$\textrm{k}\dfrac{\textrm{[CO][NO}_2\textrm{]}^2}{\textrm{[NO}\textrm{]}}$
Note that NO2 is raised to the second power to account for the stoichiometry of the balanced reaction.
So clearly the 2nd reaction isn't the slow reaction since the rate is not equivalent to what we were given!
Let's check the rate of the 1st reaction now...
$\textrm{k}\textrm{[NO}_2\textrm{]}^2$
What do you know... the rates are equal!
We have now confirmed that the 1st reaction is the slow reaction equation, since its rate is equivalent to the overall reaction rate.
Q14.6.8
Nitramide (O2NNH2) decomposes in aqueous solution to N2O and H2O. What is the experimental rate law (Δ[N2O]/Δt) for the decomposition of nitramide if the mechanism for the decomposition is as follows?
$\mathrm{O_2NNH_2}\overset{k_1}{\underset{k_{-1}}{\rightleftharpoons}}\mathrm{O_2NNH^-}+\mathrm{H^+}$ $(\textrm{fast})$
$\mathrm{O_2NNH^-}\xrightarrow{k_2}\mathrm{N_2O}+\mathrm{OH^-}$ $(\textrm{slow})$
$\mathrm{H^+}+\mathrm{OH^-}\xrightarrow{k_3}\mathrm{H_2O}$ $(\textrm{fast})$
Assume that the rates of the forward and reverse reactions in the first equation are equal.
S14.6.8
We know that the slowest step of the reaction is the rate determining step, since it usually has the highest activation energy requirement. As a result, the slowest step of the reaction is the experimental rate law we are looking for.
Note: since the slowest step is the rate determining step, that usually means there is some intermediate in between. Intermediates should NEVER be a part of the rate law mechanism.
$rate = rate_2 = {k_2}{[O_2NNH^{-}]}$
Since Nitramide is an intermediate, we must find some way to substitute it. To solve that problem, we look for where Nitramide is produced and consumed. We see that
$rate_1 = {k_1}{[O_2NNH_2]}$
$rate_{-1} = {k_{-1}}{[O_2NNH^-]}{[H^+]}$
Since these two rates produce and consume the same amount of $O_2NNH^{-}$ over the same time period, we can set them equal to each other and solve for the intermediate
$rate_1 = rate_{-1}$
${k_1}{[O_2NNH_2]} = {k_{-1}}{[O_2NNH^-]}{[H^+]}$
${[O_2NNH^-]} = \frac{k_1{[O_2NNH_2]}}{k_{-1}{[H^+]}}$
Substituting this equation back into our original equation gives us
$rate=rate_2=k_2\frac{k_1[O_2NNH_2]}{k_{-1}[H^+]}$
With all of the rate constants (k), we can clean up our equation a little bit by saying
$k = \frac{k_2{k_1}}{k_{-1}}$
Leaving us with
$rate = \frac{k[O_2NNH_2]}{[H^+]}$
A14.6.8
$\textrm{rate}=k_2\dfrac{k_1[\mathrm{O_2NNH_2}]}{k_{-1}[\mathrm{H^+}]}=k\dfrac{[\mathrm{O_2NNH_2}]}{[\mathrm{H^+}]}$
Q14.6.9
The following reactions are given:
$\mathrm{A+B}\overset{k_1}{\underset{k_{-1}}{\rightleftharpoons}}\mathrm{C+D}$
$\mathrm{D+E}\xrightarrow{k_2}\mathrm F$
What is the relationship between the relative magnitudes of $k_{−1}$ and $k_2$ if these reactions have the following rate law?
$\dfrac{Δ[F]}{Δt} = k\dfrac{[A][B][E]}{[C]}$
How does the magnitude of $k_1$ compare to that of $k_2$? Under what conditions would you expect the rate law to be
$\dfrac{Δ[F]}{Δt} =k′[A][B]?$
Assume that the rates of the forward and reverse reactions in the first equation are equal.
S14.6.9
First, because we have broken the equations down into elementary steps we can write the rate laws for each step.
Step1:
$A+B\xrightarrow[]{k_{1}} C+D$
$rate=k_{1}[A][B]$
Step 2:
$C+D \xrightarrow[]{k_{-1}} A+B$
$rate=k_{-1}[C][D]$
Step 3:
$D+E \xrightarrow[]{k_{2}} F$
$rate=k_{2}[D][E]$
If we add a these steps together we see that we get overall reaction
$A+B+E \rightarrow C+F$
we can see that [D] is an intermediate and $k_{1}=k_{-1}$.
Since we are not told which steps are fast or slow we need to use Steady State Approximation.
If the second step is the slower step (k-1>>k2) then our rate determining step would be
$rate=k_{2}[D][E]$
Since we can only write rate laws in terms of products and reactants we have to rewrite this so that we are not including an intermediate.
Assume: rate of [D] formation = rate of its disappearance
$k_{1}[A][B]=k_{-1}[C][D]+k_{2}[D][E]$
$k_{1}[A][B]=[D](k_{-1}[C]+k_{2}[E])$
Solving for [D] we find that
$[D]= \frac{k_{1}[A][B]}{(k_{-1}[C]+k_{2}[E])}$
now we can use this to substitute the intermediate [D] in the rate law to get an appropriate rate law.
$rate=\frac{k_{2} k_{1}[A][B][E]}{(k_{-1}[C]+k_{2}[E])}$
because we had already established k-1>>k2 we can assume that
$k_{-1}[C]+k_{2}[E]\approx k_{-1}[C]$
this would give us the observed rate law
$\frac{\Delta [F]}{\Delta t}=\frac{k_{2}k_{1}[A][B][E]}{k_{-1}[C]}$
to make this clearer we can set
$k=\frac{(k_{2})(k_{1})}{(k_{-1})}$
and we can then simplify it down to
$\dfrac{Δ[F]}{Δt} = k\dfrac{[A][B][E]}{[C]}$
we can see that all of these rate constants are related by this ratio k2k1/k-1. Since k2 is our rate determining step k-1>>k2 and since k1=k-1 then we can see that k1>>k2.
We would expect the rate law to be
$\dfrac{Δ[F]}{Δt} =k′[A][B]?$
if the rate determining step aka the slowest step is that corresponding to (step 1) since the rate law for this step is k1[A][B] and this is the exact the same as the rate law that we were given.
14.7: Catalysis
Q14.7.1
What effect does a catalyst have on the activation energy of a reaction? What effect does it have on the frequency factor (A)? What effect does it have on the change in potential energy for the reaction?
S14.7.1
A catalyst lowers the activation energy of a reaction. Some catalysts can also orient the reactants and thereby increase the frequency factor. Catalysts have no effect on the change in potential energy for a reaction.
Q14.7.2
How is it possible to affect the product distribution of a reaction by using a catalyst?
Q14.7.3
A heterogeneous catalyst works by interacting with a reactant in a process called adsorption. What occurs during this process? Explain how this can lower the activation energy.
S14.7.3
In adsorption, a reactant binds tightly to a surface. Because intermolecular interactions between the surface and the reactant weaken or break bonds in the reactant, its reactivity is increased, and the activation energy for a reaction is often decreased.
Q14.7.4
What effect does increasing the surface area of a heterogeneous catalyst have on a reaction? Does increasing the surface area affect the activation energy? Explain your answer.
Q14.7.5
Identify the differences between a heterogeneous catalyst and a homogeneous catalyst in terms of the following.
1. ease of recovery
2. collision frequency
3. temperature sensitivity
4. cost
S14.7.4
1. Heterogeneous catalysts are easier to recover.
2. Collision frequency is greater for homogeneous catalysts.
3. Homogeneous catalysts are often more sensitive to temperature.
4. Homogeneous catalysts are often more expensive.
Q14.7.6
An area of intensive chemical research involves the development of homogeneous catalysts, even though homogeneous catalysts generally have a number of operational difficulties. Propose one or two reasons why a homogenous catalyst may be preferred.
S14.7.6
Catalysts are compounds that, when added to chemical reactions, reduce the activation energy and increase the reaction rate. The amount of a catalyst does not change during a reaction, as it is not consumed as part of the reaction process. Catalysts lower the energy required to reach the transition state of the reaction, allowing more molecular interactions to achieve that state. However, catalysts do not affect the degree to which a reaction progresses. In other words, though catalysts affect reaction kinetics, the equilibrium state remains unaffected.
Catalysts can be classified into two types: homogenous and heterogeneous. Homogenous catalysts are those which exist in the same phase (gas or liquid) as the reactants, while heterogeneous catalysts are not in the same phase as the reactants. Typically, heterogeneous catalysis involves the use of solid catalysts placed in a liquid reaction mixture. For this question, we will be discussing homogenous catalysts.
Most of the times, homogeneous catalysis involves the introduction of an aqueous phase catalyst into an aqueous solution of reactants. One reason why homogeneous catalysts are preferred over heterogeneous catalysts because homogeneous catalysts mix well in chemical reactions in comparison to heterogeneous catalysts. However, homogeneous catalyst is often irrecoverable after the reaction has run to completion.
Q14.7.7
Consider the following reaction between cerium(IV) and thallium(I) ions:
$\ce{2Ce^{4+} + Tl^+ → 2Ce^{3+} + Tl^{3+}}$
This reaction is slow, but Mn2+ catalyzes it, as shown in the following mechanism:
$\ce{Ce^{4+} + Mn^{2+} → Ce^{3+} + Mn^{3+}}$
$\ce{Ce^{4+} + Mn^{3+} → Ce^{3+} + Mn^{4+}}$
$\ce{Mn^{4+} + Tl^{+ }→ Tl^{3+} + Mn^{2+}}$
In what way does $Mn^{2+}$ increase the reaction rate?
S14.7.7
The Mn2+ ion donates two electrons to Ce4+, one at a time, and then accepts two electrons from Tl+. Because Mn can exist in three oxidation states separated by one electron, it is able to couple one-electron and two-electron transfer reactions.
Q14.7.8
The text identifies several factors that limit the industrial applications of enzymes. Still, there is keen interest in understanding how enzymes work for designing catalysts for industrial applications. Why?
S14.7.8
Enzymes are expensive to make, denature and fail at certain temperatures, are not that stable in a solution, and are very specific to the reaction it was made for. However, scientists can use the observations from enzymes to create catalysts that are more effective in aiding the reaction and cost less to produce. Overall, catalysts still play a large part in lowering the activation energy for reactions. Creating new catalysts can help in the improvement of areas such as medical, ecological, and even commercial products.
Q14.7.9
Most enzymes have an optimal pH range; however, care must be taken when determining pH effects on enzyme activity. A decrease in activity could be due to the effects of changes in pH on groups at the catalytic center or to the effects on groups located elsewhere in the enzyme. Both examples are observed in chymotrypsin, a digestive enzyme that is a protease that hydrolyzes polypeptide chains. Explain how a change in pH could affect the catalytic activity due to (a) effects at the catalytic center and (b) effects elsewhere in the enzyme. (Hint: remember that enzymes are composed of functional amino acids.)
Q14.7.10
At some point during an enzymatic reaction, the concentration of the activated complex, called an enzyme–substrate complex ($ES$), and other intermediates involved in the reaction is nearly constant. When a single substrate is involved, the reaction can be represented by the following sequence of equations:
$\text {enzyme (E) + substrate (S)} \rightleftharpoons \text{enzyme-substrate complex (ES)} \rightleftharpoons \text{enzyme (E) + product (P)}$
This can also be shown as follows:
$E + S \underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}} ES \underset{k_{-2}}{\stackrel{k_{2}}{\rightleftharpoons}} E+P$
Using molar concentrations and rate constants, write an expression for the rate of disappearance of the enzyme–substrate complex. Typically, enzyme concentrations are small, and substrate concentrations are high. If you were determining the rate law by varying the substrate concentrations under these conditions, what would be your apparent reaction order?
S14.7.10
$\dfrac{Δ[ES]}{Δt}=−(k_2+k_{−1})[ES]+k_1[E][S]+k_{−2}[E][P] \approx 0$; zeroth order in substrate.
Q14.7.11
A particular reaction was found to proceed via the following mechanism:
• $A + B → C + D$ (slow)
• $2C → E$ (fast)
• $E + A → B + F$ (fast)
What is the overall reaction? Is this reaction catalytic, and if so, what species is the catalyst? Identify the intermediates
Q14.7.12
A particular reaction has two accessible pathways (A and B), each of which favors conversion of X to a different product (Y and Z, respectively). Under uncatalyzed conditions pathway A is favored, but in the presence of a catalyst pathway B is favored. Pathway B is reversible, whereas pathway A is not. Which product is favored in the presence of a catalyst? without a catalyst? Draw a diagram illustrating what is occurring with and without the catalyst.
S14.7.12
In both cases, the product of pathway A is favored. All of the Z produced in the catalyzed reversible pathway B will eventually be converted to X as X is converted irreversibly to Y by pathway A.
Q14.7.13
The kinetics of an enzyme-catalyzed reaction can be analyzed by plotting the reaction rate versus the substrate concentration. This type of analysis is referred to as a Michaelis–Menten treatment. At low substrate concentrations, the plot shows behavior characteristic of first-order kinetics, but at very high substrate concentrations, the behavior shows zeroth-order kinetics. Explain this phenomenon. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/14%3A_Chemical_Kinetics/14.E%3A_Exercises.txt |
These is a summary of key concepts of the chapter in the Textmap created for "Chemistry: The Central Science" by Brown et al.
14.1: Factors that Affect Reaction Rates
chemical kinetics – area of chemistry dealing with speeds/rates of reactions
• rates of reactions affected by four factors
1. concentrations of reactants
2. temperature at which reaction occurs
3. presence of a catalyst
4. surface area of solid or liquid reactants and/or catalysts
14.2: Reaction Rates
• reaction rate – speed of a chemical reaction
$\displaystyle \textit{average rate} = \frac{\textit{change #moles B}}{\textit{change in time}}= \frac{\Delta\textit{moles B}}{\Delta t}\textit{ if }A \to B \nonumber$
$\Delta\textit{moles B} = \textit{moles B at final time}- \textit{moles B at initial time} \nonumber$
$\displaystyle \textit{average rate} = -\frac{\Delta\textit{moles A}}{\Delta t}\textit{ if }A \to B \nonumber$
14.2.1 Rates in Terms of Concentrations
• rate calculated in units of M/s
• brackets around a substance indicate the concentration
• instantaneous rate – rate at a particular time
• instantaneous rate obtained from the straight line tangent that touches the curve at a specific point
• slopes give instantaneous rates
• instantaneous rate also referred to as the rate
14.2.2 Reaction Rates and Stoichiometry
• for the irreversible reaction $aA+bB\to cC+dD$
$\displaystyle\textit{rate} = -\frac{1}{a}\frac{\Delta [A]}{\Delta t} = -\frac{1}{b}\frac{\Delta [B]}{\Delta t} = \frac{1}{c}\frac{\Delta [C]}{\Delta t} = \frac{1}{d}\frac{\Delta [D]}{\Delta t} \nonumber$
14.3: Concentration and Rate
• equation used only if C and D only substances formed
• Rate = k[A][B]
• Rate law – expression that shows that rate depends on concentrations of reactants
• k = rate constant
14.3.1 Reaction Order
• Rate = k[reactant 1]m[reactant 2]n
• m, n are called reaction orders
• m+n, overall reaction order
• reaction orders do not have to correspond with coefficients in balanced equation
• values of reaction order determined experimentally
• reaction order can be fractional or negative
14.3.2 Units of Rates Constants
• units of rate constant depend on overall reaction order of rate law
• for reaction of second order overall
• units of rate = (units of rate constant)(units of concentration)2
• units of rate constant = M-1s-1
14.3.3 Using Initial Rates to Determine Rate Laws
• zero order – no change in rate when concentration changed
• first order – change in concentration gives proportional changes in rate
• second order – change in concentration changes rate by the square of the concentration change, such as 22 or 32, etc…
• rate constant does not depend on concentration
14.4: The Change of Concentration with Time
• rate laws can be converted into equations that give concentrations of reactants or products
14.4.1 First-Order Reactions
$\textit{rate} = -\frac{\Delta [A]}{\Delta t} = k[A] \nonumber$
and in integral form:
$\ln[A]_t - \ln[A]_0 =-kt \nonumber$
or
$\ln\frac{[A]_t}{[A]_0} = -kt \nonumber$
$\ln[A]_t = - kt + \ln[A]_0 \nonumber$
• corresponds to a straight line with $y = mx + b$
• equations used to determine:
1. concentration of reactant remaining at any time
2. time required for given fraction of sample to react
3. time required for reactant concentration to reach a certain level
14.3.2 Half-Life
• half-life of first order reaction
$\displaystyle t_{\frac{1}{2}} = -\frac{\ln\frac{1}{2}}{k} = \frac{0.693}{k} \nonumber$
• half-life – time required for concentration of reactant to drop to one-half of initial value
• $t_{1/2}$ of first order independent of initial concentrations
• half-life same at any given time of reaction
• in first order reaction – concentrations of reactant decreases by ½ in each series of regularly spaced time intervals
14.3.3 Second-Order Reactions
• rate depends on reactant concentration raised to second power or concentrations of two different reactants each raised to first power
$\text{Rate} = k[A]^2 \nonumber$
$\displaystyle\frac{1}{[A]_t} = kt + \frac{1}{[A]_0} \nonumber$
$\displaystyle\textit{half life} = t_{\frac{1}{2}} = \frac{1}{k[A]_0} \nonumber$
• half life dependent on initial concentration of reactant
14.5: Temperature and Rate
• rate constant must increase with increasing temperature, thus increasing the rate of reaction
14.5.1 The Collision Model
• collision model – molecules must collide to react
• greater frequency of collisions the greater the reaction rate
• for most reactions only a small fraction of collisions leads to a reaction
14.5.2 Activation Energy
• Svante August Arrhenius
• Molecules must have a minimum amount of energy to react
• Energy comes from kinetic energy of collisions
• Kinetic energy used to break bonds
• Activation energy, Ea – minimum energy required to initiate a chemical reaction
• Activated complex or transition state – atoms at the top of the energy barrier
• Rate depends on temperature and Ea
• Lower Ea means faster reaction
• Reactions occur when collisions between molecules occur with enough energy and proper orientation
14.5.3 The Arrhenius Equation
• reaction rate data:
• theArrhenius Equation:
$\displaystyle k = A e^{\frac{-E_a}{RT}} \nonumber$
• $k$ = rate constant, $E_a$ = activation energy, $R$ = gas constant (8.314 J/(mol K)), $T$ = absolute temperature, $A$ = frequency factor
• $A$ relates to frequency of collisions, favorable orientations
$\displaystyle \ln k = -\frac{E_a}{RT} + \ln A \nonumber$
• the $\ln k$ vs. $1/t$ graph (also known as an Arrhenius plot) has a slope $–E_a/R$ and the y-intercept $\ln A$
• for two temperatures:
$\displaystyle \ln \frac{k_1}{k_2} = \frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) \nonumber$
• used to calculate rate constant, $k_1$ and $T_1$
14.6: Reaction Mechanisms
• reaction mechanism – process by which a reaction occurs
14.6.1 Elementary Steps
• elementary steps – each step in a reaction
• molecularity – if only one molecule involved in step
• unimolecular – if only one molecule involved in step
• bimolecular – elementary step involving collision of two reactant molecules
• termolecular – elementary step involving simultaneous collision of three molecules
• elementary steps in multi-step mechanism must always add to give chemical equation of overall process
• intermediate – product formed in one step and consumed in a later step
14.6.2 Rate Laws of Elementary Steps
• if reaction is known to be an elementary step then the rate law is known
• rate of unimolecular step is first order (Rate = k[A])
• rate of bimolecular steps is second order (Rate = k[A][B])
• first order in [A] and [B]
• if double [A] than number of collisions of A and B will double
14.6.3 Rate Laws of Multi-step Mechanisms
• rate-determining step – slowest elementary step
• determines rate law of overall reaction
14.6.4 Mechanisms with an Initial Slow Step vs. Mechanisms with an Initial Fast Step
• intermediates are usually unstable, in low concentration, and difficult to isolate
• when a fast step precedes a slow one, solve for concentration of intermediate by assuming that equilibrium is established in fast step
14.7: Catalysis
• catalyst – substance that changes speed of chemical reaction without undergoing a permanent chemical change
14.7.1 Homogeneous Catalysis
• homogeneous catalyst – catalyst that is present in same phase as reacting molecule
• catalysts alter Ea or A
• generally catalysts lowers overall Ea for chemical reaction
• catalysts provides a different mechanism for reaction
14.7.2 Heterogeneous Catalysis
• exists in different phase from reactants
• initial step in heterogeneous catalyst is adsorption
• adsorption – binding of molecules to surface
• adsorption occurs because ions/atoms at surface of solid extremely reactive
14.7.3 Enzymes
• biological catalysts
• large protein molecules with molecular weights 10,000 – 1 million amu
• catalase – enzyme in blood and liver that decomposes hydrogen peroxide into water and oxygen
• substrates – substances that undergo reaction at the active site
• lock-and-key model – substrate molecules bind specifically to the active site
• enzyme-substrate complex – combination of enzyme and substrate
• binding between enzyme and substrate involves intermolecular forces (dipole-dipole, hydrogen bonding, and London dispersion forces)
• product from reaction leaves enzyme allowing for another substrate to enter enzyme
• enzyme inhibitors – molecules that bind strongly to enzymes
• turnover number – number of catalyzed reactions occurring at a particular active site
• large turnover numbers = low activation energies | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/14%3A_Chemical_Kinetics/14.S%3A_Chemical_Kinetics_%28Summary%29.txt |
We introduced the concept of equilibrium in Chapter 11, where you learned that a liquid and a vapor are in equilibrium when the number of molecules evaporating from the surface of the liquid per unit time is the same as the number of molecules condensing from the vapor phase. Vapor pressure is an example of a physical equilibrium because only the physical form of the substance changes. Similarly, in Chapter 13, we discussed saturated solutions, another example of a physical equilibrium, in which the rate of dissolution of a solute is the same as the rate at which it crystallizes from solution.
In this chapter, we describe the methods chemists use to quantitatively describe the composition of chemical systems at equilibrium, and we discuss how factors such as temperature and pressure influence the equilibrium composition. As you study these concepts, you will also learn how urban smog forms and how reaction conditions can be altered to produce H2 rather than the combustion products CO2 and H2O from the methane in natural gas. You will discover how to control the composition of the gases emitted in automobile exhaust and how synthetic polymers such as the polyacrylonitrile used in sweaters and carpets are produced on an industrial scale.
• 15.1: The Concept of 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.
• 15.2: The Equilibrium Constant
The law of mass action describes a system at equilibrium in terms of the concentrations of the products and the reactants. For a system involving one or more gases, either the molar concentrations of the gases or their partial pressures can be used. The equilibrium constant can be defined in terms of forward and reverse rate constants via the law of mass action.
• 15.3: Interpreting and Working with Equilibrium Constants
The magnitude of the equilibrium constant, K, indicates the extent to which a reaction will proceed: If K is a large number, it means that the equilibrium concentration of the products is large. In this case, the reaction as written will proceed to the right (resulting in an increase in the concentration of products) If K is a small number, it means that the equilibrium concentration of the reactants is large.
• 15.4: Heterogeneous Equilibria
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.
• 15.5: Calculating Equilibrium Constants
Various methods can be used to solve the two fundamental types of equilibrium problems: (1) those in which we calculate the concentrations of reactants and products at equilibrium and (2) those in which we use the equilibrium constant and the initial concentrations of reactants to determine the composition of the equilibrium mixture. When an equilibrium constant is calculated from equilibrium concentrations, concentrations or partial pressures are use into the equilibrium constant expression.
• 15.6: Applications of Equilibrium Constants
The reaction Quotient (\(Q\) or \(Q_p\)) has the same form as the equilibrium constant expression, but it is derived from concentrations obtained at any time. When a reaction system is at equilibrium, \(Q = K\). Graphs derived by plotting a few equilibrium concentrations for a system at a given temperature and pressure can be used to predict the direction in which a reaction will proceed. Points that do not lie on the line or curve are nonequilibrium states.
• 15.7: Le Chatelier's Principle
Systems at equilibrium can be disturbed by changes to temperature, concentration, and, in some cases, volume and pressure; volume and pressure changes will disturb equilibrium if the number of moles of gas is different on the reactant and product sides of the reaction. The system's response to these disturbances is described by Le Châtelier's principle: The system will respond in a way that counteracts the disturbance. Adding a catalyst affects the reaction rates but does not alter equilibrium.
• 15.E: Exercises
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al.
• 15.S: Chemical Equilibrium (Summary)
15: Chemical Equilibrium
Learning Objectives
• To understand what is meant by chemical equilibrium.
In the last chapter, we discussed the principles of chemical kinetics, which deal with the rate of change, or how quickly a given chemical reaction occurs. We now turn our attention to the extent 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 reversible 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 chemical equilibrium, 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 $1$). 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 $2$ 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 $\PageIndex{2a}$), in accordance with the stoichiometry of the reaction, we reach exactly the same equilibrium composition (Figure $\PageIndex{2b}$). Thus equilibrium can be approached from either direction in a chemical reaction.
Figure $3$ 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.
Example $1$
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?
Given: three reaction systems
Asked for: relative time to reach chemical equilibrium
Strategy:
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.
Solution:
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.
Exercise $1$
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?
Answer
system 2
A Video Introduction to Dynamic Equilibrium: Introduction to Dynamic Equilibrium(opens in new window) [youtu.be]
Summary
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. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/15%3A_Chemical_Equilibrium/15.01%3A_The_Concept_of_Equilibrium.txt |
Learning Objectives
• To know the relationship between the equilibrium constant and the rate constants for the forward and reverse reactions.
• To write an equilibrium constant expression for any reaction.
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 $\ce{NO2}$. Both the forward and reverse reactions for this system consist of a single elementary reaction, so the reaction rates are as follows:
$\text{forward rate} = k_f[\ce{N2O4}] \nonumber$
and
$\text{reverse rate} = k_r[\ce{NO2}]^2\nonumber$
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 equilibrium constant ($K$), 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.
A Video for Determining the Equilibrium Expression: Determining the Equilibrium Expression(opens in new window) [youtu.be]
Table $1$ lists the initial and equilibrium concentrations from five different experiments using the reaction system described by Equation $\ref{Eq3}$. 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. 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.
Table $1$: Initial and Equilibrium Concentrations for $\ce{NO2}:\ce{N2O4}$ Mixtures at 25°C
Initial Concentrations Concentrations at Equilibrium
Experiment [$\ce{N2O4}$] (M) [$\ce{NO2}$] (M) [$\ce{N2O4}$] (M) [$\ce{NO2}$] (M) $K = [\ce{NO2}]^2/[\ce{N2O4}]$
1 0.0500 0.0000 0.0417 0.0165 $6.54 \times 10^{-3}$
2 0.0000 0.1000 0.0417 0.0165 $6.54 \times 10^{-3}$
3 0.0750 0.0000 0.0647 0.0206 $6.56 \times 10^{-3}$
4 0.0000 0.0750 0.0304 0.0141 $6.54 \times 10^{-3}$
5 0.0250 0.0750 0.0532 0.0186 $6.50 \times 10^{-3}$
Developing an Equilibrium Constant Expression
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 law of mass action (or law of chemical equilibrium) 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 $2$, 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 $H_2$ and $Cl_2$ to produce $HCl$, which has an equilibrium constant of $1.6 \times 10^{33}$ at 300 K. Because $H_2$ is a good reductant and $Cl_2$ 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.
Table $2$: Equilibrium Constants for Selected Reactions*
Reaction Temperature (K) Equilibrium Constant (K)
*Equilibrium constants vary with temperature. The K values shown are for systems at the indicated temperatures.
$S_{(s)}+O_{2(g)} \rightleftharpoons SO_{2(g)}$ 300 $4.4 \times 10^{53}$
$2H_{2(g)}+O_{2(g)} \rightleftharpoons 2H2O_{(g)}$ 500 $2.4 \times 10^{47}$
$H_{2(g)}+Cl_{2(g)} \rightleftharpoons 2HCl_{(g)}$ 300 $1.6 \times 10^{33}$
$H_{2(g)}+Br_{2(g)} \rightleftharpoons 2HBr_{(g)}$ 300 $4.1 \times 10^{18}$
$2NO_{(g)}+O_{2(g)} \rightleftharpoons 2NO_{2(g)}$ 300 $4.2 \times 10^{13}$
$3H_{2(g)}+N_{2(g)} \rightleftharpoons 2NH_{3(g)}$ 300 $2.7 \times 10^{8}$
$H_{2(g)}+D_{2(g)} \rightleftharpoons 2HD_{(g)}$ 100 $1.92$
$H_{2(g)}+I_{2(g)} \rightleftharpoons 2HI_{(g)}$ 300 $2.9 \times 10^{-1}$
$I_{2(g)} \rightleftharpoons 2I_{(g)}$ 800 $4.6 \times 10^{-7}$
$Br_{2(g)} \rightleftharpoons 2Br_{(g)}$ 1000 $4.0 \times 10^{-7}$
$Cl_{2(g)} \rightleftharpoons 2Cl_{(g)}$ 1000 $1.8 \times 10^{-9}$
$F_{2(g)} \rightleftharpoons 2F_{(g)}$ 500 $7.4 \times 10^{-13}$
Effective vs. True Concentrations
You will also notice in Table $2$ 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. In fact, equilibrium constants are calculated using “effective concentrations,” or activities, of reactants and products, which are the ratios of the measured concentrations to a standard state of 1 M. As shown in Equation $\ref{Eq8}$, the units of concentration cancel, which makes $K$ unitless as well:
$\dfrac{[A]_{measured}}{[A]_{standard\; state}}=\dfrac{\cancel{M}}{\cancel{M}} = \dfrac{\cancel{\frac{mol}{L}}}{\cancel{\frac{mol}{L}}} \label{Eq8}$
Because equilibrium constants are calculated using “effective concentrations” relative to a standard state of 1 M, values of K are unitless.
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 $\ce{HD}$:
$\ce{H2(g) + D2(g) <=> 2HD(g)} \nonumber$
The equilibrium constant expression for this reaction is
$K= \dfrac{[HD]^2}{[H_2][D_2]} \nonumber$
with $K$ varying between 1.9 and 4 over a wide temperature range (100–1000 K). Thus an equilibrium mixture of $H_2$, $D_2$, and $HD$ contains significant concentrations of both product and reactants.
Figure $3$ 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 large 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 small 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.
Example $1$: Equilibrium Constant Expressions
Write the equilibrium constant expression for each reaction.
• $\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$
• $\ce{CO(g) + 1/2 O2(g) <=> CO2(g)}$
• $\ce{2CO2(g) <=> 2CO(g) + O2(g)}$
Given: balanced chemical equations
Asked for: equilibrium constant expressions
Strategy:
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.
Solution:
The only product is ammonia, which has a coefficient of 2. For the reactants, $\ce{N2}$ has a coefficient of 1 and $\ce{H2}$ has a coefficient of 3. The equilibrium constant expression is as follows:
$\dfrac{[\ce{NH3}]^2}{[\ce{N2}][\ce{H2}]^3} \nonumber$
The only product is carbon dioxide, which has a coefficient of 1. The reactants are $\ce{CO}$, with a coefficient of 1, and $\ce{O2}$, with a coefficient of $\frac{1}{2}$. Thus the equilibrium constant expression is as follows:
$\dfrac{[\ce{CO2}]}{[\ce{CO}][\ce{O2}]^{1/2}} \nonumber$
This reaction is the reverse of the reaction in part b, with all coefficients multiplied by 2 to remove the fractional coefficient for $\ce{O2}$. The equilibrium constant expression is therefore the inverse of the expression in part b, with all exponents multiplied by 2
$\dfrac{[\ce{CO}]^2[\ce{O2}]}{[\ce{CO2}]^2} \nonumber$
Exercise $1$
Write the equilibrium constant expression for each reaction.
1. $\ce{N2O(g) <=> N2(g) + 1/2O2(g)}$
2. $\ce{2C8H18(g) + 25O2(g) <=> 16CO2(g) + 18H2O(g)}$
3. $\ce{H2(g) + I2(g) <=> 2HI(g)}$
Answer a
$K=\dfrac{[N_2][O_2]^{1/2}}{[N_2O]}$
Answer b
$K=\dfrac{[CO_2]^{16}[H_2O]^{18}}{[C_8H_{18}]^2[O_2]^{25}}$
Answer c
$K=\dfrac{[HI]^2}{[H_2][I_2]}$
Example $2$
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.
1. $H_{2(g)}+I_{2(g)} \rightleftharpoons 2HI_{(g)}\;\;\; K_{(700K)}=54$
2. $2CO_{2(g)} \rightleftharpoons 2CO_{(g)}+O_{2(g)}\;\;\; K_{(1200K)}=3.1 \times 10^{-18}$
3. $PCl_{5(g)} \rightleftharpoons PCl_{3(g)}+Cl_{2(g)}\;\;\; K_{(613K)}=97$
4. $2O_{3(g)} \rightleftharpoons 3O_{2(g)} \;\;\; K_{(298 K)}=5.9 \times 10^{55}$
Given: systems and values of $K$
Asked for: composition of systems at equilibrium
Strategy:
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.
Solution:
1. Only system 4 has $K \gg 10^3$, so at equilibrium it will consist of essentially only products.
2. System 2 has $K \ll 10^{-3}$, so the reactants have little tendency to form products under the conditions specified; thus, at equilibrium the system will contain essentially only reactants.
3. Both systems 1 and 3 have equilibrium constants in the range $10^3 \ge K \ge 10^{-3}$, indicating that the equilibrium mixtures will contain appreciable amounts of both products and reactants.
Exercise $2$
Hydrogen and nitrogen react to form ammonia according to the following balanced chemical equation:
$\ce{3H2(g) + N2(g) <=> 2NH3(g)} \nonumber$
Values of the equilibrium constant at various temperatures were reported as
• $K_{25°C} = 3.3 \times 10^8$,
• $K_{177°C} = 2.6 \times 10^3$, and
• $K_{327°C} = 4.1$.
1. At which temperature would you expect to find the highest proportion of $H_2$ and $N_2$ in the equilibrium mixture?
2. Assuming that the reaction rates are fast enough so that equilibrium is reached quickly, at what temperature would you design a commercial reactor to operate to maximize the yield of ammonia?
Answer a
327°C, where $K$ is smallest
Answer b
25°C
Video which Discusses What Does K Tell us About a Reaction?: What Does K Tell us About a Reaction?(opens in new window) [youtu.be]
Variations in the Form of the Equilibrium Constant Expression
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 $\ce{N2O4 <=> 2NO2}$ is as follows:
$K=\dfrac{[\ce{NO2}]^2}{[\ce{N2O4}]} \label{Eq12}$
but for the opposite reaction, $2 NO_2 \rightleftharpoons N_2O_4$, the equilibrium constant K′ is given by the inverse expression:
$K'=\dfrac{[\ce{N2O4}]}{[\ce{NO2}]^2} \label{Eq13}$
Consider another example, the formation of water:
$\ce{2H2(g) + O2(g) <=> 2H2O(g)}. \nonumber$
Because $\ce{H2}$ is a good reductant and $\ce{O2}$ 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 $\ce{O2}$ and $\ce{H2}$, 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 $\ce{H2}$ and $\ce{O2}$.
The equilibrium constant for a reaction written in reverse is the inverse 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
$\ce{2NO2 <=> N2O4} \nonumber$
as
$\ce{NO2 <=> 1/2 N2O4} \nonumber$
with the equilibrium constant K″ is as follows:
$K′′=\dfrac{[\ce{N2O4}]^{1/2}}{[\ce{NO2}]} \label{Eq14}$
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.
A Video Discussing Relationships Involving Equilibrium Constants: Relationships Involving Equilibrium Constants(opens in new window) [youtu.be] (opens in new window)
Example $3$: The Haber Process
At 745 K, K is 0.118 for the following reaction:
$\ce{N2(g) + 3H2(g) <=> 2NH3(g)} \nonumber$
What is the equilibrium constant for each related reaction at 745 K?
1. $\ce{2NH3(g) <=> N2(g) + 3H2(g)}$
2. $\ce{1/2 N2(g) + 3/2 H2(g) <=> NH3(g)}$
Given: balanced equilibrium equation, $K$ at a given temperature, and equations of related reactions
Asked for: values of $K$ for related reactions
Strategy:
Write the equilibrium constant expression for the given reaction and for each related reaction. From these expressions, calculate $K$ for each reaction.
Solution:
The equilibrium constant expression for the given reaction of $N_{2(g)}$ with $H_{2(g)}$ to produce $NH_{3(g)}$ 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:
$K′′=\dfrac{[NH_3]}{[N_2]^{1/2}[H_2]^{3/2}}=K^{1/2}=\sqrt{K}=\sqrt{0.118} = 0.344 \nonumber$
Exercise
At 527°C, the equilibrium constant for the reaction
$\ce{2SO2(g) + O2(g) <=> 2SO3(g)} \nonumber$
is $7.9 \times 10^4$. Calculate the equilibrium constant for the following reaction at the same temperature:
$\ce{SO3(g) <=> SO2(g) + 1/2 O2(g)} \nonumber$
Answer
$3.6 \times 10^{-3}$
Equilibrium Constant Expressions for Systems that Contain Gases
For reactions that involve species in solution, the concentrations used in equilibrium calculations are usually expressed in moles/liter. For gases, however, the concentrations are usually expressed in terms of partial pressures rather than molarity, where the standard state is 1 atm of pressure. The symbol $K_p$ is used to denote equilibrium constants calculated from partial pressures. For the general reaction $aA+bB \rightleftharpoons cC+dD$, in which all the components are gases, the equilibrium constant expression can be written as the ratio of the partial pressures of the products and reactants (each raised to its coefficient in the chemical equation):
$K_p=\dfrac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b} \label{Eq16}$
Thus $K_p$ for the decomposition of $N_2O_4$ (Equation 15.1) 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 fugacity, just as activity is the effective concentration.
Because partial pressures are usually expressed in atmospheres or mmHg, the molar concentration of a gas and its partial pressure do not have the same numerical value. Consequently, the numerical values of $K$ and $K_p$ are usually different. They are, however, related by the ideal gas constant ($R$) and the absolute temperature ($T$):
$\color{red} K_p = K(RT)^{Δn} \label{Eq18}$
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,
$K_p = K(RT)^1 = KRT \nonumber$
Example $4$: The Haber Process (again)
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:
$\ce{N2(g) + 3H2(g) <=> 2NH3(g)} \nonumber$
What is $K_p$ for this reaction at the same temperature?
Given: equilibrium equation, equilibrium constant, and temperature
Asked for: $K_p$
Strategy:
Use the coefficients in the balanced chemical equation to calculate $Δn$. Then use Equation $\ref{Eq18}$ to calculate $K$ from $K_p$.
Solution:
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:
\begin{align*} K_p &=K(RT)^{-2} \[4pt] &=\dfrac{K}{(RT)^2} \[4pt] &=\dfrac{0.118}{\{ [0.08206(L \cdot atm)/(mol \cdot K)][745\; K]\}^2} \[4pt] &=3.16 \times 10^{-5} \end{align*} \nonumber
Because $K_p$ is a unitless quantity, the answer is $K_p = 3.16 \times 10^{-5}$.
Exercise $4$
Calculate $K_p$ for the reaction
$\ce{2SO2(g) + O2(g) \rightleftharpoons 2SO3(g)} \nonumber$
at 527°C, if $K = 7.9 \times 10^4$ at this temperature.
Answer
$K_p = 1.2 \times 10^3$
Video Discussing Converting Kc to Kp: Converting Kc to Kp(opens in new window) [youtu.be]
Equilibrium Constant Expressions for the Sums of Reactions
Chemists frequently need to know the equilibrium constant for a reaction that has not been previously studied. In such cases, the desired reaction can often be written as the sum of other reactions for which the equilibrium constants are known. The equilibrium constant for the unknown reaction can then be calculated from the tabulated values for the other reactions.
To illustrate this procedure, let’s consider the reaction of $\ce{N2}$ with $\ce{O2}$ to give $\ce{NO2}$. This reaction is an important source of the $\ce{NO2}$ that gives urban smog its typical brown color. The reaction normally occurs in two distinct steps. In the first reaction (step 1), $\ce{N2}$ reacts with $\ce{O2}$ at the high temperatures inside an internal combustion engine to give $\ce{NO}$. The released $\ce{NO}$ then reacts with additional $\ce{O2}$ to give $\ce{NO2}$ (step 2). The equilibrium constant for each reaction at 100°C is also given.
$\ce{N2(g) + O2(g) <=> 2NO(g)}\;\; K_1=2.0 \times 10^{-25} \label{step 1}$
$\ce{2NO(g) + O2(g) <=> 2NO2(g)}\;\;\;K_2=6.4 \times 10^9 \label{step 2}$
Summing reactions (step 1) and (step 2) gives the overall reaction of $N_2$ with $O_2$:
$\ce{N2(g) + 2O2(g) <=> 2NO2(g)} \;\;\;K_3=? \label{overall reaction 3}$
The equilibrium constant expressions for the reactions are as follows:
$K_1=\dfrac{[NO]^2}{[N_2][O_2]}\;\;\; K_2=\dfrac{[NO_2]^2}{[NO]^2[O_2]}\;\;\; K_3=\dfrac{[NO_2]^2}{[N_2][O_2]^2} \nonumber$
What is the relationship between $K_1$, $K_2$, and $K_3$, all at 100°C? The expression for $K_1$ has $[NO]^2$ in the numerator, the expression for $K_2$ has $[NO]^2$ in the denominator, and $[NO]^2$ does not appear in the expression for $K_3$. Multiplying $K_1$ by $K_2$ and canceling the $[NO]^2$ terms,
$K_1K_2=\dfrac{\cancel{[NO]^2}}{[N_2][O_2]} \times \dfrac{[NO_2]^2}{\cancel{[NO]^2}[O_2]}=\dfrac{[NO_2]^2}{[N_2][O_2]^2}=K_3 \nonumber$
Thus the product of the equilibrium constant expressions for $K_1$ and $K_2$ is the same as the equilibrium constant expression for $K_3$:
$K_3 = K_1K_2 = (2.0 \times 10^{-25})(6.4 \times 10^9) = 1.3 \times 10^{-15} \nonumber$
The equilibrium constant for a reaction that is the sum of two or more reactions is equal to the product of the equilibrium constants for the individual reactions. In contrast, recall that according to Hess’s Law, $ΔH$ for the sum of two or more reactions is the sum of the ΔH values for the individual reactions.
To determine $K$ for a reaction that is the sum of two or more reactions, add the reactions but multiply the equilibrium constants.
Example $6$
The following reactions occur at 1200°C:
1. $CO_{(g)}+3H_{2(g)} \rightleftharpoons CH_{4(g)}+H_2O_{(g)} \;\;\;K_1=9.17 \times 10^{-2}$
2. $CH_{4(g)}+2H_2S_{(g)} \rightleftharpoons CS_{2(g)}+4H_{2(g})\;\;\; K_2=3.3 \times 10^4$
Calculate the equilibrium constant for the following reaction at the same temperature.
1. $CO_{(g)}+2H_2S_{(g)} \rightleftharpoons CS_{2(g)}+H_2O_{(g)}+H_{2(g)}\;\;\; K_3=?$
Given: two balanced equilibrium equations, values of $K$, and an equilibrium equation for the overall reaction
Asked for: equilibrium constant for the overall reaction
Strategy:
Arrange the equations so that their sum produces the overall equation. If an equation had to be reversed, invert the value of $K$ for that equation. Calculate $K$ for the overall equation by multiplying the equilibrium constants for the individual equations.
Solution:
The key to solving this problem is to recognize that reaction 3 is the sum of reactions 1 and 2:
$CO_{(g)}+ \cancel{3H_{2(g)}} \rightleftharpoons \cancel{CH_{4(g)}} + H_2O_{(g)} \nonumber$
$\cancel{CH_{4(g)}} +2H_2S_{(g)} \rightleftharpoons CS_{2(g)} + \cancel{3H_{2(g)}} + H_{2(g)} \nonumber$
$CO_{(g)} + 3H_{2(g)} \rightleftharpoons CS_{2(g)}+H_2O_{(g)}+H_{2(g)} \nonumber$
The values for $K_1$ and $K_2$ are given, so it is straightforward to calculate $K_3$:
$K_3 = K_1K_2 = (9.17 \times 10^{-2})(3.3 \times 10^4) = 3.03 \times 10^3 \nonumber$
Exercise $6$
In the first of two steps in the industrial synthesis of sulfuric acid, elemental sulfur reacts with oxygen to produce sulfur dioxide. In the second step, sulfur dioxide reacts with additional oxygen to form sulfur trioxide. The reaction for each step is shown, as is the value of the corresponding equilibrium constant at 25°C. Calculate the equilibrium constant for the overall reaction at this same temperature.
1. $\frac{1}{8}S_{8(s)}+O_{2(g)} \rightleftharpoons SO_{2(g)}\;\;\; K_1=4.4 \times 10^{53}$
2. $SO_{2(g)}+\frac{1}{2}O_{2(g)} \rightleftharpoons SO_{3(g)}\;\;\; K_2=2.6 \times 10^{12}$
3. $\frac{1}{8}S_{8(s)}+\frac{3}{2}O_{2(g)} \rightleftharpoons SO_{3(g)}\;\;\; K_3=?$
Answer
$K_3 = 1.1 \times 10^{66}$
Summary
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 equilibrium system that contains products and reactants in a single phase is a homogeneous equilibrium; a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibrium. When a reaction can be expressed as the sum of two or more reactions, its equilibrium constant is equal to the product of the equilibrium constants for the individual reactions.
• The law of mass action describes a system at equilibrium in terms of the concentrations of the products and the reactants.
• For a system involving one or more gases, either the molar concentrations of the gases or their partial pressures can be used.
• Definition of equilibrium constant in terms of forward and reverse rate constants: $K=\dfrac{k_f}{k_r} \nonumber$
• Equilibrium constant expression (law of mass action): $K=\dfrac{[C]^c[D]^d}{[A]^a[B]^b} \nonumber$
• Equilibrium constant expression for reactions involving gases using partial pressures: $K_p=\dfrac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b} \nonumber$
• Relationship between $K_p$ and $K$: $K_p = K(RT)^{Δn} \nonumber$ | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/15%3A_Chemical_Equilibrium/15.02%3A_The_Equilibrium_Constant.txt |
Learning Objectives
• How to figure out in which direction a reaction will go (i.e. towards making product, or more reactant)
• Calculating equilibrium concentrations. This may involve knowing equilibrium values for some of the reactants and products and determining the concentration of an unknown.
• Alternatively, we may be provided with the starting concentrations of reactants and products and may be asked to find the equilibrium concentrations
Your ability to interpret the numerical value of a quantity in terms of what it means in a practical sense is an essential part of developing a working understanding of Chemistry. This is particularly the case for equilibrium constants, whose values span the entire range of the positive numbers. Although there is no explicit rule, for most practical purposes you can say that equilibrium constants within the range of roughly 0.01 to 100 indicate that a chemically significant amount of all components of the reaction system will be present in an equilibrium mixture and that the reaction will be incomplete or “reversible”.
As an equilibrium constant approaches the limits of zero or infinity, the reaction can be increasingly characterized as a one-way process; we say it is “complete” or “irreversible”. The latter term must of course not be taken literally; the Le Chatelier principle still applies (especially insofar as temperature is concerned), but addition or removal of reactants or products will have less effect.
Kinetically Hindered Reactions
Although it is by no means a general rule, it frequently happens that reactions having very large equilibrium constants are kinetically hindered, often to the extent that the reaction essentially does not take place.
The examples in the following table are intended to show that numbers (values of K), no matter how dull they may look, do have practical consequences!
Table $1$: Examples of Reversible Reactions
Reaction
K
remarks
$N_{2(g)} + O_{2(g)} \rightleftharpoons 2 NO_{(g)}$ $5 \times 10^{–31}$ at 25°C,
0.0013 at 2100°C
These two very different values of K illustrate very nicely why reducing combustion-chamber temperatures in automobile engines is environmentally friendly.
$3 H_{2(g)} + N_{2(g)} \rightleftharpoons 2 NH_{3(g)}$ $7 \times 10^5$ at 25°C,
56 at 1300°C
See the discussion of this reaction in the section on the Haber process.
$H_{2(g)} \rightleftharpoons 2 H_{(g)}$ $10^{–36}$ at 25°C,
$6 \times 10^{–5}$ at 5000°
Dissociation of any stable molecule into its atoms is endothermic. This means that all molecules will decompose at sufficiently high temperatures.
$H_2O_{(g)} \rightleftharpoons H_{2(g)} + ½ O_{2(g)}$ $8 \times 10^{–41}$ at 25°C You won’t find water a very good source of oxygen gas at ordinary temperatures!
$CH_3COOH_{(l)} \rightleftharpoons 2 H_2O_{(l)} + 2 C_{(s)}$
$K_c = 10^{13}$ at 25°C This tells us that acetic acid has a great tendency to decompose to carbon, but nobody has ever found graphite (or diamonds!) forming in a bottle of vinegar. A good example of a super kinetically-hindered reaction!
Do Equilibrium Constants have Units?
The equilibrium expression for the synthesis of ammonia
$\ce{ 3 H2(g) + N2(g) -> 2 NH3(g)} \label{15.4.1}$
can be expressed as
$K_p =\dfrac{P^2_{NH_3}}{P_{N_2}P^3_{H_2}} \label{15.4.2}$
or
$K_c = \dfrac{[NH_3]^2}{[N_2] [H_2]^3} \label{15.4.3}$
so $K_p$ for this process would appear to have units of atm–2, and $K_c$ would be expressed in mol–2 L2. And yet these quantities are often represented as being dimensionless. Which is correct? The answer is that both forms are acceptable. There are some situations (which you will encounter later) in which K’s must be considered dimensionless, but in simply quoting the value of an equilibrium constant it is permissible to include the units, and this may even be useful in order to remove any doubt about the units of the individual terms in equilibrium expressions containing both pressure and concentration terms. In carrying out your own calculations, however, there is rarely any real need to show the units.
Strictly speaking, equilibrium expressions do not have units because the concentration or pressure terms that go into them are really ratios having the forms (n mol L–1)/(1 mol L–1) or (n atm)/(1 atm) in which the unit quantity in the denominator refers to the standard state of the substance; thus the units always cancel out.
Strictly speaking, equilibrium expressions do not have units.
For substances that are liquids or solids, the standard state is just the concentration of the substance within the liquid or solid, so for something like $\ce{CaF(s)}$, the term going into the equilibrium expression is $[\ce{CaF2}]/[\ce{CaF2}]$ which cancels to unity; this is the reason we do not need to include terms for solid or liquid phases in equilibrium expressions. The subject of standard states would take us beyond where we need to be at this point in the course, so we will simply say that the concept is made necessary by the fact that energy, which ultimately governs chemical change, is always relative to some arbitrarily defined zero value which, for chemical substances, is the standard state.
Summary
The magnitude of the equilibrium constant, $K$, indicates the extent to which a reaction will proceed:
• If $K$ is a large number, it means that the equilibrium concentration of the products is large. In this case, the reaction as written will proceed to the right (resulting in an increase in the concentration of products)
• If $K$ is a small number, it means that the equilibrium concentration of the reactants is large. In this case, the reaction as written will proceed to the left (resulting in an increase in the concentration of reactants)
Knowing the value of the equilibrium constant, $K$, will allow us to determine: (1) he direction a reaction will proceed to achieve equilibrium and (2) the ratios of the concentrations of reactants and products when equilibrium is reached | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/15%3A_Chemical_Equilibrium/15.03%3A_Interpreting_and_Working_with_Equilibrium_Constants.txt |
Learning Objectives
• To understand how different phases affect equilibria.
When the products and reactants of an equilibrium reaction form a single phase, whether gas or liquid, the system is a homogeneous equilibrium. In such situations, the concentrations of the reactants and products can vary over a wide range. In contrast, a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibrium, such as the reaction of a gas with a solid or liquid.
As noted in the previous section, the equilibrium constant expression is actually a ratio of activities. To simplify the calculations in general chemistry courses, the activity of each substance in the reaction is often approximated using a ratio of the molarity of a substance compared to the standard state of that substance. For substances that are liquids or solids, the standard state is just the concentration of the substance within the liquid or solid. Because the molar concentrations of pure liquids and solids normally do not vary greatly with temperature, the ratio of the molarity to the standard state for substances that are liquids or solids always has a value of 1. For example, for a compound such as CaF2(s), the term going into the equilibrium expression is [CaF2]/[CaF2] which cancels to unity. Thus, when the activities of the solids and liquids (including solvents) are incorporated into the equilibrium expression, they do not change the value.
Consider the following reaction, which is used in the final firing of some types of pottery to produce brilliant metallic glazes:
$\ce{CO2(g) + C(s) \rightleftharpoons 2CO(g)} \label{Eq14.4.1}$
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:
$K=\dfrac{a_{\ce{CO}}^2}{a_{\ce{CO2}}a_{C}}=\dfrac{[\ce{CO}]^2}{[\ce{CO2}][1]}=\dfrac{[\ce{CO}]^2}{[\ce{CO_2}]}\label{Eq14.4.2}$
The equilibrium constant for this reaction can also be written in terms of the partial pressures of the gases:
$K_p=\dfrac{(P_{CO})^2}{P_{CO_2}} \label{Eq14.4.3}$
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 activities 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 $\ce{CO}$ and $\ce{CO_2}$, 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 $1$, it does not matter whether 1 g or 100 g of solid carbon is present; in either case, the composition of the gaseous components of the system will be the same at equilibrium.
Example $1$
Write each expression for $K$, incorporating all constants, and $K_p$ for the following equilibrium reactions.
1. $\ce{PCl3(l) + Cl2(g) <=> PCl5(s)}$
2. $\ce{Fe3O4(s) + 4H2(g) <=> 3Fe(s) + 4H2O(g)}$
Given: balanced equilibrium equations.
Asked for: expressions for $K$ and $K_p$.
Strategy:
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.
Solution
This reaction contains a pure solid ($PCl_5$) and a pure liquid ($PCl_3$). Their activities are equal to 1, so when incorporated into the equilibrium constant expression, they do not change the value. So
$K=\dfrac{1}{(1)[Cl_2]} \nonumber$
and
$K_p=\dfrac{1}{(1)P_{Cl_2}} \nonumber$
This reaction contains two pure solids ($Fe_3O_4$ and $Fe$), which are each assigned a value of 1 in the equilibrium constant expressions:
$K=\dfrac{(1)[H_2O]^4}{(1)[H_2]^4} \nonumber$
and
$K_p=\dfrac{(1)(P_{H_2O})^4}{(1)(P_{H_2})^4} \nonumber$
Exercise $1$
Write the expressions for $K$ and $K_p$ for the following reactions.
1. $\ce{CaCO3(s) <=> CaO(s) + CO2(g)}$
2. $\underset{glucose}{\ce{C6H12O6(s)}} + \ce{6O2(g) <=> 6CO2(g) + 6H2O(g)}$
Answer a
$K = [\ce{CO_2}]$ and $K_p = P_{\ce{CO_2}}$
Answer b
$K=\dfrac{[CO_2]^6[H_2O]^6}{[O_2]^6}$ and $K_p=\dfrac{(P_{CO_2})^6(P_{H_2O})^6}{(P_{O_2})^6}$
For reactions carried out in solution, the solvent is assumed to be pure, and therefore is assigned an activity equal to 1 in the equilibrium constant expression. The activities of the solutes are approximated by their molarities. The result is that the equilibrium constant expressions appear to only depend upon the concentrations of the solutes.
The activities of pure solids, pure liquids, and solvents are defined as having a value of '1'. Often, it is said that these activities are "left out" of equilibrium constant expressions. This is an unfortunate use of words. The activities are not "left out" of equilibrium constant expressions. Rather, because they have a value of '1', they do not change the value of the equilibrium constant when they are multiplied together with the other terms. The activities of the solutes are approximated by their molarities.
Summary
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.
Contributors and Attributions
• Anonymous
• Modified by Tom Neils (Grand Rapids Community College) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/15%3A_Chemical_Equilibrium/15.04%3A_Heterogeneous_Equilibria.txt |
Learning Objectives
• To solve quantitative problems involving chemical equilibriums.
There are two fundamental kinds of equilibrium problems:
1. those in which we are given the concentrations of the reactants and the products at equilibrium (or, more often, information that allows us to calculate these concentrations), and we are asked to calculate the equilibrium constant for the reaction; and
2. those in which we are given the equilibrium constant and the initial concentrations of reactants, and we are asked to calculate the concentration of one or more substances at equilibrium. In this section, we describe methods for solving both kinds of problems.
Calculating an Equilibrium Constant from Equilibrium Concentrations
We saw in the exercise in Example 6 in Section 15.2 that the equilibrium constant for the decomposition of $CaCO_{3(s)}$ to $CaO_{(s)}$ and $CO_{2(g)}$ is $K = [CO_2]$. At 800°C, the concentration of $CO_2$ in equilibrium with solid $CaCO_3$ and $CaO$ is $2.5 \times 10^{-3}\; M$. Thus K at 800°C is $2.5 \times 10^{-3}$. (Remember that equilibrium constants are unitless.)
A more complex example of this type of problem is the conversion of n-butane, an additive used to increase the volatility of gasoline, into isobutane (2-methylpropane).
This reaction can be written as follows:
$\ce{n-butane_{(g)} \rightleftharpoons isobutane_{(g)}} \label{Eq1}$
and the equilibrium constant $K = [\text{isobutane}]/[\text{n-butane}]$. At equilibrium, a mixture of n-butane and isobutane at room temperature was found to contain 0.041 M isobutane and 0.016 M n-butane. Substituting these concentrations into the equilibrium constant expression,
$K=\dfrac{[\textit{isobutane}]}{[\textit{n-butane}]}=0.041\; M = 2.6 \label{Eq2}$
Thus the equilibrium constant for the reaction as written is 2.6.
Example $1$
The reaction between gaseous sulfur dioxide and oxygen is a key step in the industrial synthesis of sulfuric acid:
$2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}\nonumber$
A mixture of $SO_2$ and $O_2$ was maintained at 800 K until the system reached equilibrium. The equilibrium mixture contained
• $5.0 \times 10^{-2}\; M\; SO_3$,
• $3.5 \times 10^{-3}\; M\; O_2$, and
• $3.0 \times 10^{-3}\; M\; SO_2$.
Calculate $K$ and $K_p$ at this temperature.
Given: balanced equilibrium equation and composition of equilibrium mixture
Asked for: equilibrium constant
Strategy
Write the equilibrium constant expression for the reaction. Then substitute the appropriate equilibrium concentrations into this equation to obtain $K$.
Solution
Substituting the appropriate equilibrium concentrations into the equilibrium constant expression,
$K=\dfrac{[SO_3]^2}{[SO_2]^2[O_2]}=\dfrac{(5.0 \times 10^{-2})^2}{(3.0 \times 10^{-3})^2(3.5 \times 10^{-3})}=7.9 \times 10^4 \nonumber$
To solve for $K_p$, we use the relationship derived previously
$K_p = K(RT)^{Δn} \nonumber$
where $Δn = 2 − 3 = −1$:
$K_p=K(RT)^{Δn}\nonumber$
$K_p=7.9 \times 10^4 [(0.08206\; L⋅atm/mol⋅K)(800 K)]^{−1}\nonumber$
$K_p=1.2 \times 10^3\nonumber$
Exercise $1$
Hydrogen gas and iodine react to form hydrogen iodide via the reaction
$2 NOCl_{(g)} \leftrightharpoons 2NO_{(g)}+Cl_{2(g)}\nonumber$
A mixture of $H_2$ and $I_2$ was maintained at 740 K until the system reached equilibrium. The equilibrium mixture contained
• $1.37\times 10^{−2}\; M\; HI$,
• $6.47 \times 10^{−3}\; M\; H_2$, and
• $5.94 \times 10^{-4}\; M\; I_2$.
Calculate $K$ and $K_p$ for this reaction.
Answer
$K = 48.8$ and $K_p = 48.8$
Chemists are not often given the concentrations of all the substances, and they are not likely to measure the equilibrium concentrations of all the relevant substances for a particular system. In such cases, we can obtain the equilibrium concentrations from the initial concentrations of the reactants and the balanced chemical equation for the reaction, as long as the equilibrium concentration of one of the substances is known. Example $2$ shows one way to do this.
Example $2$
A 1.00 mol sample of $NOCl$ was placed in a 2.00 L reactor and heated to 227°C until the system reached equilibrium. The contents of the reactor were then analyzed and found to contain 0.056 mol of $Cl_2$. Calculate $K$ at this temperature. The equation for the decomposition of $NOCl$ to $NO$ and $Cl_2$ is as follows:
$2 NOCl_{(g)} \rightleftharpoons 2NO_{(g)}+Cl_{2(g)}\nonumber$
Given: balanced equilibrium equation, amount of reactant, volume, and amount of one product at equilibrium
Asked for: $K$
Strategy:
1. Write the equilibrium constant expression for the reaction. Construct a table showing the initial concentrations, the changes in concentrations, and the final concentrations (as initial concentrations plus changes in concentrations).
2. Calculate all possible initial concentrations from the data given and insert them in the table.
3. Use the coefficients in the balanced chemical equation to obtain the changes in concentration of all other substances in the reaction. Insert those concentration changes in the table.
4. Obtain the final concentrations by summing the columns. Calculate the equilibrium constant for the reaction.
Solution
A The first step in any such problem is to balance the chemical equation for the reaction (if it is not already balanced) and use it to derive the equilibrium constant expression. In this case, the equation is already balanced, and the equilibrium constant expression is as follows:
$K=\dfrac{[NO]^2[Cl_2]}{[NOCl]^2}\nonumber$
To obtain the concentrations of $NOCl$, $NO$, and $Cl_2$ at equilibrium, we construct a table showing what is known and what needs to be calculated. We begin by writing the balanced chemical equation at the top of the table, followed by three lines corresponding to the initial concentrations, the changes in concentrations required to get from the initial to the final state, and the final concentrations.
$2 NOCl_{(g)} \rightleftharpoons 2NO_{(g)}+Cl_{2(g)}\nonumber$
ICE $[NOCl]$ $[NO]$ $[Cl_2]$
Initial
Change
Final
B Initially, the system contains 1.00 mol of $NOCl$ in a 2.00 L container. Thus $[NOCl]_i = 1.00\; mol/2.00\; L = 0.500\; M$. The initial concentrations of $NO$ and $Cl_2$ are $0\; M$ because initially no products are present. Moreover, we are told that at equilibrium the system contains 0.056 mol of $Cl_2$ in a 2.00 L container, so $[Cl_2]_f = 0.056 \;mol/2.00 \;L = 0.028\; M$. We insert these values into the following table:
$2 NOCl_{(g)} \rightleftharpoons 2NO_{(g)}+Cl_{2(g)}\nonumber$
ICE $[NOCl]$ $[NO]$ $[Cl_2]$
Initial 0.500 0 0
Change
Final 0.028
C We use the stoichiometric relationships given in the balanced chemical equation to find the change in the concentration of $Cl_2$, the substance for which initial and final concentrations are known:
$Δ[Cl_2] = 0.028 \;M_{(final)} − 0.00\; M_{(initial)}] = +0.028\; M\nonumber$
According to the coefficients in the balanced chemical equation, 2 mol of $NO$ are produced for every 1 mol of $Cl_2$, so the change in the $NO$ concentration is as follows:
$Δ[NO]=\left(\dfrac{0.028\; \cancel{mol \;Cl_2}}{ L}\right)\left(\dfrac{2\; mol\; NO}{1 \cancel{\;mol \;Cl_2}}\right)=0.056\; M\nonumber$
Similarly, 2 mol of $NOCl$ are consumed for every 1 mol of $Cl_2$ produced, so the change in the $NOCl$ concentration is as follows:
$Δ[NOCl]= \left(\dfrac{0.028\; \cancel{mol\; Cl_2}}{L}\right) \left(\dfrac{−2\; mol \;NOCl}{1\; \cancel{mol\; Cl_2}} \right) = -0.056 \;M\nonumber$
We insert these values into our table:
$2 NOCl_{(g)} \rightleftharpoons 2NO_{(g)}+Cl_{2(g)}\nonumber$
ICE $[NOCl]$ $[NO]$ $[Cl_2]$
Initial 0.500 0 0
Change −0.056 +0.056 +0.028
Final 0.028
D We sum the numbers in the $[NOCl]$ and $[NO]$ columns to obtain the final concentrations of $NO$ and $NOCl$:
$[NO]_f = 0.000\; M + 0.056 \;M = 0.056\; M\nonumber$
$[NOCl]_f = 0.500\; M + (−0.056\; M) = 0.444 M\nonumber$
We can now complete the table:
$2 NOCl_{(g)} \rightleftharpoons 2NO_{(g)}+Cl_{2(g)}\nonumber$
ICE $[NOCl] \([NO]$ $[Cl_2]$
initial 0.500 0 0
change −0.056 +0.056 +0.028
final 0.444 0.056 0.028
We can now calculate the equilibrium constant for the reaction:
$K=\dfrac{[NO]^2[Cl_2]}{[NOCl]^2}=\dfrac{(0.056)^2(0.028)}{(0.444)^2}=4.5 \times 10^{−4}\nonumber$
Exercise $2$
The German chemist Fritz Haber (1868–1934; Nobel Prize in Chemistry 1918) was able to synthesize ammonia ($NH_3$) by reacting $0.1248\; M \;H_2$ and $0.0416\; M \;N_2$ at about 500°C. At equilibrium, the mixture contained 0.00272 M $NH_3$. What is $K$ for the reaction
$N_2+3H_2 \rightleftharpoons 2NH_3\nonumber$
at this temperature? What is $K_p$?
Answer
$K = 0.105$ and $K_p = 2.61 \times 10^{-5}$
A Video Disucssing Using ICE Tables to find Kc: Using ICE Tables to find Kc(opens in new window) [youtu.be]
Calculating Equilibrium Concentrations from the Equilibrium Constant
To describe how to calculate equilibrium concentrations from an equilibrium constant, we first consider a system that contains only a single product and a single reactant, the conversion of n-butane to isobutane (Equation $\ref{Eq1}$), for which K = 2.6 at 25°C. If we begin with a 1.00 M sample of n-butane, we can determine the concentration of n-butane and isobutane at equilibrium by constructing a table showing what is known and what needs to be calculated, just as we did in Example $2$.
$\text{n-butane}_{(g)} \rightleftharpoons \text{isobutane}_{(g)} \nonumber$
ICE $[\text{n-butane}_{(g)} ]$ $[\text{isobutane}_{(g)}]$
Initial
Change
Final
The initial concentrations of the reactant and product are both known: [n-butane]i = 1.00 M and [isobutane]i = 0 M. We need to calculate the equilibrium concentrations of both n-butane and isobutane. Because it is generally difficult to calculate final concentrations directly, we focus on the change in the concentrations of the substances between the initial and the final (equilibrium) conditions. If, for example, we define the change in the concentration of isobutane (Δ[isobutane]) as $+x$, then the change in the concentration of n-butane is Δ[n-butane] = $−x$. This is because the balanced chemical equation for the reaction tells us that 1 mol of n-butane is consumed for every 1 mol of isobutane produced. We can then express the final concentrations in terms of the initial concentrations and the changes they have undergone.
$\text{n-butane}_{(g)} \rightleftharpoons \text{isobutane}_{(g)} \nonumber$
ICE $[\text{n-butane}_{(g)} ]$ $[\text{isobutane}_{(g)}]$
Initial 1.00 0
Change $−x$ $+x$
Final $(1.00 − x)$ $(0 + x) = x$
Substituting the expressions for the final concentrations of n-butane and isobutane from the table into the equilibrium equation,
$K=\dfrac{[\text{isobutane}]}{[\text{n-butane}]}=\dfrac{x}{1.00−x}=2.6 \nonumber$
Rearranging and solving for $x$,
$x=2.6(1.00−x)= 2.6−2.6x \nonumber$
$x+2.6x= 2.6 \nonumber$
$x=0.72 \nonumber$
We obtain the final concentrations by substituting this $x$ value into the expressions for the final concentrations of n-butane and isobutane listed in the table:
$[\text{n-butane}]_f = (1.00 − x) M = (1.00 − 0.72) M = 0.28\; M \nonumber$
$[\text{isobutane}]_f = (0.00 + x) M = (0.00 + 0.72) M = 0.72\; M \nonumber$
We can check the results by substituting them back into the equilibrium constant expression to see whether they give the same $K$ that we used in the calculation:
$K=\dfrac{[\text{isobutane}]}{[\text{n-butane}]}=\left(\dfrac{0.72\; \cancel{M}}{0.28\;\cancel{M}}\right) =2.6 \nonumber$
This is the same $K$ we were given, so we can be confident of our results.
Example $3$ illustrates a common type of equilibrium problem that you are likely to encounter.
Example $3$: The water–gas shift reaction
The water–gas shift reaction is important in several chemical processes, such as the production of H2 for fuel cells. This reaction can be written as follows:
$H_{2(g)}+CO_{2(g)} \rightleftharpoons H_2O_{(g)}+CO_{(g)}\nonumber$
$K = 0.106$ at 700 K. If a mixture of gases that initially contains 0.0150 M $H_2$ and 0.0150 M $CO_2$ is allowed to equilibrate at 700 K, what are the final concentrations of all substances present?
Given: balanced equilibrium equation, $K$, and initial concentrations
Asked for: final concentrations
Strategy:
1. Construct a table showing what is known and what needs to be calculated. Define $x$ as the change in the concentration of one substance. Then use the reaction stoichiometry to express the changes in the concentrations of the other substances in terms of $x$. From the values in the table, calculate the final concentrations.
2. Write the equilibrium equation for the reaction. Substitute appropriate values from the ICE table to obtain $x$.
3. Calculate the final concentrations of all species present. Check your answers by substituting these values into the equilibrium constant expression to obtain $K$.
Solution
A The initial concentrations of the reactants are $[H_2]_i = [CO_2]_i = 0.0150\; M$. Just as before, we will focus on the change in the concentrations of the various substances between the initial and final states. If we define the change in the concentration of $H_2O$ as $x$, then $Δ[H_2O] = +x$. We can use the stoichiometry of the reaction to express the changes in the concentrations of the other substances in terms of $x$. For example, 1 mol of $CO$ is produced for every 1 mol of $H_2O$, so the change in the $CO$ concentration can be expressed as $Δ[CO] = +x$. Similarly, for every 1 mol of $H_2O$ produced, 1 mol each of $H_2$ and $CO_2$ are consumed, so the change in the concentration of the reactants is $Δ[H_2] = Δ[CO_2] = −x$. We enter the values in the following table and calculate the final concentrations.
$H_{2(g)}+CO_{2(g)} \rightleftharpoons H_2O_{(g)}+CO_{(g)}\nonumber$
ICE $[H_2]$ $[CO_2]$ $[H_2O]$ $[CO]$
Initial 0.0150 0.0150 0 0
Change $−x$ $−x$ $+x$ $+x$
Final $(0.0150 − x)$ $(0.0150 − x)$ $x$ $x$
B We can now use the equilibrium equation and the given $K$ to solve for $x$:
$K=\dfrac{[H_2O][CO]}{[H_2][CO_2]}=\dfrac{(x)(x)}{(0.0150−x)(0.0150−x}=\dfrac{x^2}{(0.0150−x)^2}=0.106\nonumber$
We could solve this equation with the quadratic formula, but it is far easier to solve for $x$ by recognizing that the left side of the equation is a perfect square; that is,
$\dfrac{x^2}{(0.0150−x)^2}=\left(\dfrac{x}{0.0150−x}\right)^2=0.106\nonumber$
Taking the square root of the middle and right terms,
$\dfrac{x}{(0.0150−x)} =(0.106)^{1/2}=0.326\nonumber$
$x =(0.326)(0.0150)−0.326x\nonumber$
$1.326x=0.00489\nonumber$
$x =0.00369=3.69 \times 10^{−3}\nonumber$
C The final concentrations of all species in the reaction mixture are as follows:
• $[H_2]_f=[H_2]_i+Δ[H_2]=(0.0150−0.00369) \;M=0.0113\; M$
• $[CO_2]_f =[CO_2]_i+Δ[CO_2]=(0.0150−0.00369)\; M=0.0113\; M$
• $[H_2O]_f=[H_2O]_i+Δ[H_2O]=(0+0.00369) \;M=0.00369\; M$
• $[CO]_f=[CO]_i+Δ[CO]=(0+0.00369)\; M=0.00369 \;M$
We can check our work by inserting the calculated values back into the equilibrium constant expression:
$K=\dfrac{[H_2O][CO]}{[H_2][CO_2]}=\dfrac{(0.00369)^2}{(0.0113)^2}=0.107\nonumber$
To two significant figures, this K is the same as the value given in the problem, so our answer is confirmed.
Exercise $3$
Hydrogen gas reacts with iodine vapor to give hydrogen iodide according to the following chemical equation:
$H_{2(g)}+I_{2(g)} \rightleftharpoons 2HI_{(g)}\nonumber$
$K = 54$ at 425°C. If 0.172 M $H_2$ and $I_2$ are injected into a reactor and maintained at 425°C until the system equilibrates, what is the final concentration of each substance in the reaction mixture?
Answer
• $[HI]_f = 0.270 \;M$
• $[H_2]_f = [I_2]_f = 0.037\; M$
In Example $3$, the initial concentrations of the reactants were the same, which gave us an equation that was a perfect square and simplified our calculations. Often, however, the initial concentrations of the reactants are not the same, and/or one or more of the products may be present when the reaction starts. Under these conditions, there is usually no way to simplify the problem, and we must determine the equilibrium concentrations with other means. Such a case is described in Example $4$.
Example $4$
In the water–gas shift reaction shown in Example $3$, a sample containing 0.632 M CO2 and 0.570 M $H_2$ is allowed to equilibrate at 700 K. At this temperature, $K = 0.106$. What is the composition of the reaction mixture at equilibrium?
Given: balanced equilibrium equation, concentrations of reactants, and $K$
Asked for: composition of reaction mixture at equilibrium
Strategy:
1. Write the equilibrium equation. Construct a table showing the initial concentrations of all substances in the mixture. Complete the table showing the changes in the concentrations ($x) and the final concentrations. 2. Write the equilibrium constant expression for the reaction. Substitute the known K value and the final concentrations to solve for \(x$.
3. Calculate the final concentration of each substance in the reaction mixture. Check your answers by substituting these values into the equilibrium constant expression to obtain $K$.
Solution
A $[CO_2]_i = 0.632\; M$ and $[H_2]_i = 0.570\; M$. Again, $x$ is defined as the change in the concentration of $H_2O$: $Δ[H_2O] = +x$. Because 1 mol of $CO$ is produced for every 1 mol of $H_2O$, the change in the concentration of $CO$ is the same as the change in the concentration of H2O, so Δ[CO] = +x. Similarly, because 1 mol each of $H_2$ and $CO_2$ are consumed for every 1 mol of $H_2O$ produced, $Δ[H_2] = Δ[CO_2] = −x$. The final concentrations are the sums of the initial concentrations and the changes in concentrations at equilibrium.
$H_{2(g)}+CO_{2(g)} \rightleftharpoons H_2O_{(g)}+CO_{(g)}\nonumber$
ICE $H_{2(g)}$ $CO_{2(g)}$ $H_2O_{(g)}$ $CO_{(g)}$
Initial 0.570 0.632 0 0
Change $−x$ $−x$ $+x$ $+x$
Final $(0.570 − x)$ $(0.632 − x)$ $x$ $x$
B We can now use the equilibrium equation and the known $K$ value to solve for $x$:
$K=\dfrac{[H_2O][CO]}{[H_2][CO_2]}=\dfrac{x^2}{(0.570−x)(0.632−x)}=0.106\nonumber$
In contrast to Example $3$, however, there is no obvious way to simplify this expression. Thus we must expand the expression and multiply both sides by the denominator:
$x^2 = 0.106(0.360 − 1.202x + x^2)\nonumber$
Collecting terms on one side of the equation,
$0.894x^2 + 0.127x − 0.0382 = 0\nonumber$
This equation can be solved using the quadratic formula:
$x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} = \dfrac{−0.127 \pm \sqrt{(0.127)^2−4(0.894)(−0.0382)}}{2(0.894)}\nonumber$
$x =0.148 \text{ and } −0.290\nonumber$
Only the answer with the positive value has any physical significance, so $Δ[H_2O] = Δ[CO] = +0.148 M$, and $Δ[H_2] = Δ[CO_2] = −0.148\; M$.
C The final concentrations of all species in the reaction mixture are as follows:
• $[H_2]_f[ = [H_2]_i+Δ[H_2]=0.570 \;M −0.148\; M=0.422 M$
• $[CO_2]_f =[CO_2]_i+Δ[CO_2]=0.632 \;M−0.148 \;M=0.484 M$
• $[H_2O]_f =[H_2O]_i+Δ[H_2O]=0\; M+0.148\; M =0.148\; M$
• $[CO]_f=[CO]_i+Δ[CO]=0 M+0.148\;M=0.148 M$
We can check our work by substituting these values into the equilibrium constant expression:
$K=\dfrac{[H_2O][CO]}{[H_2][CO_2]}=\dfrac{(0.148)^2}{(0.422)(0.484)}=0.107\nonumber$
Because $K$ is essentially the same as the value given in the problem, our calculations are confirmed.
Exercise $4$
The exercise in Example $1$ showed the reaction of hydrogen and iodine vapor to form hydrogen iodide, for which $K = 54$ at 425°C. If a sample containing 0.200 M $H_2$ and 0.0450 M $I_2$ is allowed to equilibrate at 425°C, what is the final concentration of each substance in the reaction mixture?
Answer
• $[H_I]_f = 0.0882\; M$
• $[H_2]_f = 0.156\; M$
• $[I_2]_f = 9.2 \times 10^{−4} M$
In many situations it is not necessary to solve a quadratic (or higher-order) equation. Most of these cases involve reactions for which the equilibrium constant is either very small ($K ≤ 10^{−3}$) or very large ($K ≥ 10^3$), which means that the change in the concentration (defined as $x$) is essentially negligible compared with the initial concentration of a substance. Knowing this simplifies the calculations dramatically, as illustrated in Example $5$.
Example $5$
Atmospheric nitrogen and oxygen react to form nitric oxide:
$N_{2(g)}+O_{2(g)} \rightleftharpoons 2NO_{(g)}\nonumber$
with $K_p = 2.0 \times 10^{−31}$ at 25°C.
What is the partial pressure of NO in equilibrium with $N_2$ and $O_2$ in the atmosphere (at 1 atm, $P_{N_2} = 0.78\; atm$ and $P_{O_2} = 0.21\; atm$?
Given: balanced equilibrium equation and values of $K_p$, $P_{O_2}$, and $P_{N_2}$
Asked for: partial pressure of NO
Strategy:
1. Construct a table and enter the initial partial pressures, the changes in the partial pressures that occur during the course of the reaction, and the final partial pressures of all substances.
2. Write the equilibrium equation for the reaction. Then substitute values from the table to solve for the change in concentration ($x). 3. Calculate the partial pressure of \(NO$. Check your answer by substituting values into the equilibrium equation and solving for $K$.
Solution
A Because we are given Kp and partial pressures are reported in atmospheres, we will use partial pressures. The initial partial pressure of $O_2$ is 0.21 atm and that of $N_2$ is 0.78 atm. If we define the change in the partial pressure of $NO$ as $2x$, then the change in the partial pressure of $O_2$ and of $N_2$ is $−x$ because 1 mol each of $N_2$ and of $O_2$ is consumed for every 2 mol of NO produced. Each substance has a final partial pressure equal to the sum of the initial pressure and the change in that pressure at equilibrium.
$N_{2(g)}+O_{2(g)} \rightleftharpoons 2NO_{(g)}\nonumber$
ICE $P_{N_2}$ $P_{O_2}$ $P_{NO}$
Initial 0.78 0.21 0
Change $−x$ $−x$ $+2x$
Final $(0.78 − x)$ $(0.21 − x)$ $2x$
B Substituting these values into the equation for the equilibrium constant,
$K_p=\dfrac{(P_{NO})^2}{(P_{N_2})(P_{O_2})}=\dfrac{(2x)^2}{(0.78−x)(0.21−x)}=2.0 \times 10^{−31}\nonumber$
In principle, we could multiply out the terms in the denominator, rearrange, and solve the resulting quadratic equation. In practice, it is far easier to recognize that an equilibrium constant of this magnitude means that the extent of the reaction will be very small; therefore, the $x$ value will be negligible compared with the initial concentrations. If this assumption is correct, then to two significant figures, $(0.78 − x) = 0.78$ and $(0.21 − x) = 0.21$. Substituting these expressions into our original equation,
$\dfrac{(2x)^2}{(0.78)(0.21)} = 2.0 \times 10^{−31\nonumber} \nonumber$
$\dfrac{4x^2}{0.16} =2.0 \times10^{−31}\nonumber$
$x^2=\dfrac{0.33 \times 10^{−31}}{4}\nonumber$
$x^=9.1 \times 10^{−17}\nonumber$
C Substituting this value of $x$ into our expressions for the final partial pressures of the substances,
• $P_{NO}=2x \; atm=1.8 \times 10^{−16} \;atm$
• $P_{N_2}=(0.78−x) \;atm=0.78 \;atm$
• $P_{O_2}=(0.21−x) \;atm=0.21\; atm$
From these calculations, we see that our initial assumption regarding $x$ was correct: given two significant figures, $2.0 \times 10^{−16}$ is certainly negligible compared with 0.78 and 0.21. When can we make such an assumption? As a general rule, if $x$ is less than about 5% of the total, or $10^{−3} > K > 10^3$, then the assumption is justified. Otherwise, we must use the quadratic formula or some other approach. The results we have obtained agree with the general observation that toxic $NO$, an ingredient of smog, does not form from atmospheric concentrations of $N_2$ and $O_2$ to a substantial degree at 25°C. We can verify our results by substituting them into the original equilibrium equation:
$K_p=\dfrac{(P_{NO})^2}{(P_{N_2})(P_{O_2})}=\dfrac{(1.8 \times 10^{−16})^2}{(0.78)(0.21)}=2.0 \times 10^{−31}\nonumber$
The final $K_p$ agrees with the value given at the beginning of this example.
Exercise $5$
Under certain conditions, oxygen will react to form ozone, as shown in the following equation:
$3O_{2(g)} \rightleftharpoons 2O_{3(g)}\nonumber$
with $K_p = 2.5 \times 10^{−59}$ at 25°C. What ozone partial pressure is in equilibrium with oxygen in the atmosphere ($P_{O_2}=0.21\; atm$)?
Answer
$4.8 \times 10^{−31} \;atm$
Another type of problem that can be simplified by assuming that changes in concentration are negligible is one in which the equilibrium constant is very large ($K \geq 10^3$). A large equilibrium constant implies that the reactants are converted almost entirely to products, so we can assume that the reaction proceeds 100% to completion. When we solve this type of problem, we view the system as equilibrating from the products side of the reaction rather than the reactants side. This approach is illustrated in Example $6$.
Example $6$
The chemical equation for the reaction of hydrogen with ethylene ($C_2H_4$) to give ethane ($C_2H_6$) is as follows:
$H_{2(g)}+C_2H_{4(g)} \overset{Ni}{\rightleftharpoons} C_2H_{6(g)}\nonumber$
with $K = 9.6 \times 10^{18}$ at 25°C. If a mixture of 0.200 M $H_2$ and 0.155 M $C_2H_4$ is maintained at 25°C in the presence of a powdered nickel catalyst, what is the equilibrium concentration of each substance in the mixture?
Given: balanced chemical equation, $K$, and initial concentrations of reactants
Asked for: equilibrium concentrations
Strategy:
1. Construct a table showing initial concentrations, concentrations that would be present if the reaction were to go to completion, changes in concentrations, and final concentrations.
2. Write the equilibrium constant expression for the reaction. Then substitute values from the table into the expression to solve for $x$ (the change in concentration).
3. Calculate the equilibrium concentrations. Check your answers by substituting these values into the equilibrium equation.
Solution:
A From the magnitude of the equilibrium constant, we see that the reaction goes essentially to completion. Because the initial concentration of ethylene (0.155 M) is less than the concentration of hydrogen (0.200 M), ethylene is the limiting reactant; that is, no more than 0.155 M ethane can be formed from 0.155 M ethylene. If the reaction were to go to completion, the concentration of ethane would be 0.155 M and the concentration of ethylene would be 0 M. Because the concentration of hydrogen is greater than what is needed for complete reaction, the concentration of unreacted hydrogen in the reaction mixture would be 0.200 M − 0.155 M = 0.045 M. The equilibrium constant for the forward reaction is very large, so the equilibrium constant for the reverse reaction must be very small. The problem then is identical to that in Example $5$. If we define $−x$ as the change in the ethane concentration for the reverse reaction, then the change in the ethylene and hydrogen concentrations is $+x$. The final equilibrium concentrations are the sums of the concentrations for the forward and reverse reactions.
$H_{2(g)}+C_2H_{4(g)} \overset{Ni}{\rightleftharpoons} C_2H_{6(g)}\nonumber$
IACE $[H_{2(g)}]$ $[C_2H_{4(g)}]$ $[C_2H_{6(g)}]$
Initial 0.200 0.155 0
Assuming 100% reaction 0.045 0 0.155
Change $+x$ $+x$ $−x$
Final $(0.045 + x)$ $(0 + x)$ $(0.155 − x)$
B Substituting values into the equilibrium constant expression,
$K=\dfrac{[C_2H_6]}{[H_2][C_2H_4]}=\dfrac{0.155−x}{(0.045+x)x}=9.6 \times 10^{18}\nonumber$
Once again, the magnitude of the equilibrium constant tells us that the equilibrium will lie far to the right as written, so the reverse reaction is negligible. Thus $x$ is likely to be very small compared with either 0.155 M or 0.045 M, and the equation can be simplified ($(0.045 + x)$ = 0.045 and $(0.155 − x) = 0.155$) as follows:
$K=\dfrac{0.155}{0.045x} = 9.6 \times 10^{18}\nonumber$
$x=3.6 \times 10^{−19}\nonumber$
C The small $x$ value indicates that our assumption concerning the reverse reaction is correct, and we can therefore calculate the final concentrations by evaluating the expressions from the last line of the table:
• $[C_2H_6]_f = (0.155 − x)\; M = 0.155 \; M$
• $[C_2H_4]_f = x\; M = 3.6 \times 10^{−19} M$
• $[H_2]_f = (0.045 + x) \;M = 0.045 \;M$
We can verify our calculations by substituting the final concentrations into the equilibrium constant expression:
$K=\dfrac{[C_2H_6]}{[H_2][C_2H_4]}=\dfrac{0.155}{(0.045)(3.6 \times 10^{−19})}=9.6 \times 10^{18}\nonumber$
This $K$ value agrees with our initial value at the beginning of the example.
Exercise $6$
Hydrogen reacts with chlorine gas to form hydrogen chloride:
$H_{2(g)}+Cl_{2(g)} \rightleftharpoons 2HCl_{(g)}\nonumber$
with $K_p = 4.0 \times 10^{31}$ at 47°C. If a mixture of 0.257 M $H_2$ and 0.392 M $Cl_2$ is allowed to equilibrate at 47°C, what is the equilibrium composition of the mixture?
Answer
$[H_2]_f = 4.8 \times 10^{−32}\; M$ $[Cl_2]_f = 0.135\; M$ $[HCl]_f = 0.514\; M$
A Video Discussing Using ICE Tables to find Eq. Concentrations & Kc: Using ICE Tables to find Eq. Concentrations & Kc(opens in new window) [youtu.be]
Summary
Various methods can be used to solve the two fundamental types of equilibrium problems: (1) those in which we calculate the concentrations of reactants and products at equilibrium and (2) those in which we use the equilibrium constant and the initial concentrations of reactants to determine the composition of the equilibrium mixture. When an equilibrium constant is calculated from equilibrium concentrations, molar concentrations or partial pressures are substituted into the equilibrium constant expression for the reaction. Equilibrium constants can be used to calculate the equilibrium concentrations of reactants and products by using the quantities or concentrations of the reactants, the stoichiometry of the balanced chemical equation for the reaction, and a tabular format to obtain the final concentrations of all species at equilibrium. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/15%3A_Chemical_Equilibrium/15.05%3A_Calculating_Equilibrium_Constants.txt |
Learning Objectives
• To predict in which direction a reaction will proceed.
We previously saw that knowing the magnitude of the equilibrium constant under a given set of conditions allows chemists to predict the extent of a reaction. Often, however, chemists must decide whether a system has reached equilibrium or if the composition of the mixture will continue to change with time. In this section, we describe how to quantitatively analyze the composition of a reaction mixture to make this determination.
The Reaction Quotient
To determine whether a system has reached equilibrium, chemists use a Quantity called the reaction Quotient ($Q$). The expression for the reaction Quotient has precisely the same form as the equilibrium constant expression, except that $Q$ may be derived from a set of values measured at any time during the reaction of any mixture of the reactants and the products, regardless of whether the system is at equilibrium. Therefore, for the following general reaction:
$aA+bB \rightleftharpoons cC+dD \nonumber$
the reaction quotient is defined as follows:
$Q=\dfrac{[C]^c[D]^d}{[A]^a[B]^b} \label{15.6.1}$
To understand how information is obtained using a reaction Quotient, consider the dissociation of dinitrogen tetroxide to nitrogen dioxide,
$\ce{N2O4(g) <=> 2NO2(g)} \nonumber$
for which $K = 4.65 \times 10^{−3}$ at 298 K. We can write $Q$ for this reaction as follows:
$Q=\dfrac{[\ce{NO2}]^2}{[\ce{N2O4}]} \label{15.6.2}$
The following table lists data from three experiments in which samples of the reaction mixture were obtained and analyzed at equivalent time intervals, and the corresponding values of $Q$ were calculated for each. Each experiment begins with different proportions of product and reactant:
Table $1$: Equilibrium Experiment data
Experiment $[\ce{NO2}]\; (M)$ $[\ce{N2O4}]\; (M)$ $Q = \dfrac{[\ce{NO2}]^2}{[\ce{N2O4}]}$
1 0 0.0400 $\dfrac{0^2}{0.0400}=0$
2 0.0600 0 $\dfrac{(0.0600)^2}{0}=\text{undefined}$
3 0.0200 0.0600 $\dfrac{(0.0200)^2}{0.0600}=6.67 \times 10^{−3}$
As these calculations demonstrate, $Q$ can have any numerical value between 0 and infinity (undefined); that is, $Q$ can be greater than, less than, or equal to $K$.
Comparing the magnitudes of $Q$ and $K$ enables us to determine whether a reaction mixture is already at equilibrium and, if it is not, predict how its composition will change with time to reach equilibrium (i.e., whether the reaction will proceed to the right or to the left as written). All you need to remember is that the composition of a system not at equilibrium will change in a way that makes $Q$ approach $K$:
• If $Q = K$, for example, then the system is already at equilibrium, and no further change in the composition of the system will occur unless the conditions are changed.
• If $Q < K$, then the ratio of the concentrations of products to the concentrations of reactants is less than the ratio at equilibrium. Therefore, the reaction will proceed to the right as written, forming products at the expense of reactants.
• If $Q > K$, then the ratio of the concentrations of products to the concentrations of reactants is greater than at equilibrium, so the reaction will proceed to the left as written, forming reactants at the expense of products.
These points are illustrated graphically in Figure $1$.
If $Q < K$, the reaction will proceed to the right as written. If $Q > K$, the reaction will proceed to the left as written. If $Q = K$, then the system is at equilibrium.
A Video Discussing Using the Reaction Quotient (Q): Using the Reaction Quotient (Q) (opens in new window) [youtu.be]
Example $1$
At elevated temperatures, methane ($CH_4$) reacts with water to produce hydrogen and carbon monoxide in what is known as a steam-reforming reaction:
$\ce{CH4(g) + H2O(g) <=> CO(g) + 3H2(g)} \nonumber$
$K = 2.4 \times 10^{−4}$ at 900 K. Huge amounts of hydrogen are produced from natural gas in this way and are then used for the industrial synthesis of ammonia. If $1.2 \times 10^{−2}$ mol of $CH_4$, 8.0 × 10−3 mol of $H_2O$, $1.6 \times 10^{−2}$ mol of $CO$, and $6.0 \times 10^{−3}$ mol of $H_2$ are placed in a 2.0 L steel reactor and heated to 900 K, will the reaction be at equilibrium or will it proceed to the right to produce $\ce{CO}$ and $\ce{H_2}$ or to the left to form $\ce{CH_4}$ and $\ce{H_2O}$?
Given: balanced chemical equation, $K$, amounts of reactants and products, and volume
Asked for: direction of reaction
Strategy:
1. Calculate the molar concentrations of the reactants and the products.
2. Use Equation $\ref{15.6.1}$ to determine $Q$. Compare $Q$ and $K$ to determine in which direction the reaction will proceed.
Solution:
A We must first find the initial concentrations of the substances present. For example, we have $1.2 \times 10^{−2} mol$ of $\ce{CH_4}$ in a 2.0 L container, so
$[\ce{CH4}]=\dfrac{1.2\times 10^{−2} \, \text{mol}}{2.0\; \text{L}}=6.0 \times 10^{−3} M \nonumber$
We can calculate the other concentrations in a similar way:
• $[\ce{H2O}] = 4.0 \times 10^{−3} M$,
• $[\ce{CO}] = 8.0 \times 10^{−3} M$, and
• $[\ce{H_2}] = 3.0 \times 10^{−3} M$.
B We now compute $Q$ and compare it with $K$:
\begin{align*} Q&=\dfrac{[\ce{CO}][\ce{H_2}]^3}{[\ce{CH_4}][\ce{H_2O}]} \[4pt] &=\dfrac{(8.0 \times 10^{−3})(3.0 \times 10^{−3})^3}{(6.0\times 10^{−3})(4.0 \times 10^{−3})} \[4pt] &=9.0 \times 10^{−6} \end{align*} \nonumber
Because $K = 2.4 \times 10^{−4}$, we see that $Q < K$. Thus the ratio of the concentrations of products to the concentrations of reactants is less than the ratio for an equilibrium mixture. The reaction will therefore proceed to the right as written, forming $\ce{H2}$ and $\ce{CO}$ at the expense of $\ce{H_2O}$ and $\ce{CH4}$.
Exercise $2$
In the water–gas shift reaction introduced in Example $1$, carbon monoxide produced by steam-reforming reaction of methane reacts with steam at elevated temperatures to produce more hydrogen:
$\ce{CO(g) + H_2O(g) <=> CO2(g) + H2(g)} \nonumber$
$K = 0.64$ at 900 K. If 0.010 mol of both $\ce{CO}$ and $\ce{H_2O}$, 0.0080 mol of $\ce{CO_2}$, and 0.012 mol of $\ce{H_2}$ are injected into a 4.0 L reactor and heated to 900 K, will the reaction proceed to the left or to the right as written?
Answer
$Q = 0.96$. Since (Q > K), so the reaction will proceed to the left, and $CO$ and $H_2O$ will form.
Predicting the Direction of a Reaction with a Graph
By graphing a few equilibrium concentrations for a system at a given temperature and pressure, we can readily see the range of reactant and product concentrations that correspond to equilibrium conditions, for which $Q = K$. Such a graph allows us to predict what will happen to a reaction when conditions change so that $Q$ no longer equals $K$, such as when a reactant concentration or a product concentration is increased or decreased.
Reaction 1
Lead carbonate decomposes to lead oxide and carbon dioxide according to the following equation:
$\ce{PbCO3(s) <=> PbO(s) + CO2(g)} \label{15.6.3}$
Because $\ce{PbCO_3}$ and $\ce{PbO}$ are solids, the equilibrium constant is simply
$K = [\ce{CO_2}]. \nonumber$
At a given temperature, therefore, any system that contains solid $\ce{PbCO_3}$ and solid $\ce{PbO}$ will have exactly the same concentration of $\ce{CO_2}$ at equilibrium, regardless of the ratio or the amounts of the solids present. This situation is represented in Figure $3$, which shows a plot of $[\ce{CO_2}]$ versus the amount of $\ce{PbCO_3}$ added. Initially, the added $\ce{PbCO_3}$ decomposes completely to $\ce{CO_2}$ because the amount of $\ce{PbCO_3}$ is not sufficient to give a $\ce{CO_2}$ concentration equal to $K$. Thus the left portion of the graph represents a system that is not at equilibrium because it contains only $\ce{CO2(g)}$ and $\ce{PbO(s)}$. In contrast, when just enough $\ce{PbCO_3}$ has been added to give $[CO_2] = K$, the system has reached equilibrium, and adding more $\ce{PbCO_3}$ has no effect on the $\ce{CO_2}$ concentration: the graph is a horizontal line.
Thus any $\ce{CO_2}$ concentration that is not on the horizontal line represents a nonequilibrium state, and the system will adjust its composition to achieve equilibrium, provided enough $\ce{PbCO_3}$ and $\ce{PbO}$ are present. For example, the point labeled A in Figure $2$ lies above the horizontal line, so it corresponds to a $[\ce{CO_2}]$ that is greater than the equilibrium concentration of $\ce{CO_2}$ (i.e., $Q > K$). To reach equilibrium, the system must decrease $[\ce{CO_2}]$, which it can do only by reacting $\ce{CO_2}$ with solid $\ce{PbO}$ to form solid $\ce{PbCO_3}$. Thus the reaction in Equation $\ref{15.6.3}$ will proceed to the left as written, until $[\ce{CO_2}] = K$. Conversely, the point labeled B in Figure $2$ lies below the horizontal line, so it corresponds to a $[\ce{CO_2}]$ that is less than the equilibrium concentration of $\ce{CO_2}$ (i.e., $Q < K$). To reach equilibrium, the system must increase $[\ce{CO_2}]$, which it can do only by decomposing solid $\ce{PbCO_3}$ to form $\ce{CO_2}$ and solid $\ce{PbO}$. The reaction in Equation \ref{15.6.3} will therefore proceed to the right as written, until $[\ce{CO_2}] = K$.
Reaction 2
In contrast, the reduction of cadmium oxide by hydrogen gives metallic cadmium and water vapor:
$\ce{CdO(s) + H2(g) <=> Cd(s) + H_2O(g)} \label{15.6.4}$
and the equilibrium constant is
$K = \dfrac{[\ce{H_2O}]}{[\ce{H_2}]}. \nonumber$
If $[\ce{H_2O}]$ is doubled at equilibrium, then $[\ce{H2}]$ must also be doubled for the system to remain at equilibrium. A plot of $[\ce{H_2O}]$ versus $[\ce{H_2}]$ at equilibrium is a straight line with a slope of $K$ (Figure $3$). Again, only those pairs of concentrations of $\ce{H_2O}$ and $\ce{H_2}$ that lie on the line correspond to equilibrium states. Any point representing a pair of concentrations that does not lie on the line corresponds to a nonequilibrium state. In such cases, the reaction in Equation $\ref{15.6.4}$ will proceed in whichever direction causes the composition of the system to move toward the equilibrium line. For example, point A in Figure $3$ lies below the line, indicating that the $[\ce{H_2O}]/[\ce{H_2}]$ ratio is less than the ratio of an equilibrium mixture (i.e., $Q < K$). Thus the reaction in Equation \ref{15.6.4} will proceed to the right as written, consuming $\ce{H_2}$ and producing $\ce{H_2O}$, which causes the concentration ratio to move up and to the left toward the equilibrium line. Conversely, point B in Figure $3$ lies above the line, indicating that the $[\ce{H_2O}]/[\ce{H_2}]$ ratio is greater than the ratio of an equilibrium mixture ($Q > K$). Thus the reaction in Equation $\ref{15.6.4}$ will proceed to the left as written, consuming $\ce{H_2O}$ and producing $\ce{H_2}$, which causes the concentration ratio to move down and to the right toward the equilibrium line.
Reaction 3
In another example, solid ammonium iodide dissociates to gaseous ammonia and hydrogen iodide at elevated temperatures:
$\ce{ NH4I(s) <=> NH3(g) + HI(g)} \label{15.6.5}$
For this system, $K$ is equal to the product of the concentrations of the two products:
$K = [\ce{NH_3}][\ce{HI}]. \nonumber$
If we double the concentration of $\ce{NH3}$, the concentration of $\ce{HI}$ must decrease by approximately a factor of 2 to maintain equilibrium, as shown in Figure $4$. As a result, for a given concentration of either $\ce{HI}$ or $\ce{NH_3}$, only a single equilibrium composition that contains equal concentrations of both $\ce{NH_3}$ and $\ce{HI}$ is possible, for which
$[\ce{NH_3}] = [\ce{HI}] = \sqrt{K}. \nonumber$
Any point that lies below and to the left of the equilibrium curve (such as point A in Figure $4$) corresponds to $Q < K$, and the reaction in Equation $\ref{15.6.5}$ will therefore proceed to the right as written, causing the composition of the system to move toward the equilibrium line. Conversely, any point that lies above and to the right of the equilibrium curve (such as point B in Figure $\ref{15.6.5}$) corresponds to $Q > K$, and the reaction in Equation $\ref{15.6.5}$ will therefore proceed to the left as written, again causing the composition of the system to move toward the equilibrium line. By graphing equilibrium concentrations for a given system at a given temperature and pressure, we can predict the direction of reaction of that mixture when the system is not at equilibrium.
Summary
The reaction Quotient ($Q$) is used to determine whether a system is at equilibrium and if it is not, to predict the direction of reaction. The reaction Quotient ($Q$ or $Q_p$) has the same form as the equilibrium constant expression, but it is derived from concentrations obtained at any time. When a reaction system is at equilibrium, $Q = K$. Graphs derived by plotting a few equilibrium concentrations for a system at a given temperature and pressure can be used to predict the direction in which a reaction will proceed. Points that do not lie on the line or curve represent nonequilibrium states, and the system will adjust, if it can, to achieve equilibrium. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/15%3A_Chemical_Equilibrium/15.06%3A_Applications_of_Equilibrium_Constants.txt |
Learning Objectives
• Describe the ways in which an equilibrium system can be stressed
• Predict the response of a stressed equilibrium using Le Chatelier’s principle
As we saw in the previous section, reactions proceed in both directions (reactants go to products and products go to reactants). We can tell a reaction is at equilibrium if the reaction quotient ($Q$) is equal to the equilibrium constant ($K$). We next address what happens when a system at equilibrium is disturbed so that $Q$ is no longer equal to $K$. If a system at equilibrium is subjected to a perturbance or stress (such as a change in concentration) the position of equilibrium changes. Since this stress affects the concentrations of the reactants and the products, the value of $Q$ will no longer equal the value of $K$. To re-establish equilibrium, the system will either shift toward the products (if $(Q \leq K)$ or the reactants (if $(Q \geq K)$ until $Q$ returns to the same value as $K$. This process is described by Le Chatelier's principle.
Le Chatelier's principle
When a chemical system at equilibrium is disturbed, it returns to equilibrium by counteracting the disturbance. As described in the previous paragraph, the disturbance causes a change in $Q$; the reaction will shift to re-establish $Q = K$.
Predicting the Direction of a Reversible Reaction
Le Chatelier's principle can be used to predict changes in equilibrium concentrations when a system that is at equilibrium is subjected to a stress. However, if we have a mixture of reactants and products that have not yet reached equilibrium, the changes necessary to reach equilibrium may not be so obvious. In such a case, we can compare the values of $Q$ and $K$ for the system to predict the changes.
A chemical system at equilibrium can be temporarily shifted out of equilibrium by adding or removing one or more of the reactants or products. The concentrations of both reactants and products then undergo additional changes to return the system to equilibrium.
The stress on the system in Figure $1$ is the reduction of the equilibrium concentration of SCN (lowering the concentration of one of the reactants would cause $Q$ to be larger than K). As a consequence, Le Chatelier's principle leads us to predict that the concentration of Fe(SCN)2+ should decrease, increasing the concentration of SCN part way back to its original concentration, and increasing the concentration of Fe3+ above its initial equilibrium concentration.
The effect of a change in concentration on a system at equilibrium is illustrated further by the equilibrium of this chemical reaction:
$\ce{H2(g) + I2(g) \rightleftharpoons 2HI(g)} \label{15.7.1a}$
$K_c=\mathrm{50.0 \; at\; 400°C} \label{15.7.1b}$
The numeric values for this example have been determined experimentally. A mixture of gases at 400 °C with $\mathrm{[H_2] = [I_2]} = 0.221\; M$ and $\ce{[HI]} = 1.563 \;M$ is at equilibrium; for this mixture, $Q_c = K_c = 50.0$. If $\ce{H_2}$ is introduced into the system so quickly that its concentration doubles before it begins to react (new $\ce{[H_2]} = 0.442\; M$), the reaction will shift so that a new equilibrium is reached, at which
• $\ce{[H_2]} = 0.374\; M$,
• $\ce{[I_2]} = 0.153\; M$, and
• $\ce{[HI]} = 1.692\; M$.
This gives:
\begin{align*} Q_c &=\mathrm{\dfrac{[HI]^2}{[H_2][I_2]}} \[4pt] &=\dfrac{(1.692)^2}{(0.374)(0.153)} \[4pt] &= 50.0 =K_c \label{15.7.2} \end{align*}
We have stressed this system by introducing additional $\ce{H_2}$. The stress is relieved when the reaction shifts to the right, using up some (but not all) of the excess $\ce{H_2}$, reducing the amount of uncombined $\ce{I_2}$, and forming additional $\ce{HI}$. Le Chatelier’s Principle (Changing Concentrations):
A Video Discussing Le Chatelier’s Principle (Changing Concentrations): Le Chatelier’s Principle (Changing Concentrations)(opens in new window) [youtu.be] (opens in new window)
Effect of Change in Pressure on Equilibrium
Sometimes we can change the position of equilibrium by changing the pressure of a system. However, changes in pressure have a measurable effect only in systems in which gases are involved, and then only when the chemical reaction produces a change in the total number of gas molecules in the system. An easy way to recognize such a system is to look for different numbers of moles of gas on the reactant and product sides of the equilibrium. While evaluating pressure (as well as related factors like volume), it is important to remember that equilibrium constants are defined with regard to concentration (for Kc) or partial pressure (for KP). Some changes to total pressure, like adding an inert gas that is not part of the equilibrium, will change the total pressure but not the partial pressures of the gases in the equilibrium constant expression. Thus, addition of a gas not involved in the equilibrium will not perturb the equilibrium.
As we increase the pressure of a gaseous system at equilibrium, either by decreasing the volume of the system or by adding more of one of the components of the equilibrium mixture, we introduce a stress by increasing the partial pressures of one or more of the components. In accordance with Le Chatelier's principle, a shift in the equilibrium that reduces the total number of molecules per unit of volume will be favored because this relieves the stress. The reverse reaction would be favored by a decrease in pressure.
Consider what happens when we increase the pressure on a system in which $\ce{NO}$, $\ce{O_2}$, and $\ce{NO_2}$ are at equilibrium:
$\ce{2NO(g) + O2(g) \rightleftharpoons 2NO2(g)} \label{15.7.3}$
The formation of additional amounts of $\ce{NO2}$ decreases the total number of molecules in the system because each time two molecules of $\ce{NO_2}$ form, a total of three molecules of $\ce{NO}$ and $\ce{O_2}$ are consumed. This reduces the total pressure exerted by the system and reduces, but does not completely relieve, the stress of the increased pressure. On the other hand, a decrease in the pressure on the system favors decomposition of $\ce{NO_2}$ into $\ce{NO}$ and $\ce{O_2}$, which tends to restore the pressure.
Now consider this reaction:
$\ce{N2(g) + O2(g) \rightleftharpoons 2NO(g)} \label{15.7.4}$
Because there is no change in the total number of molecules in the system during reaction, a change in pressure does not favor either formation or decomposition of gaseous nitrogen monoxide.
Le Chatelier’s Principle (Changes in Pressure or Volume):
Effect of Change in Temperature on Equilibrium
Changing concentration or pressure perturbs an equilibrium because the reaction quotient is shifted away from the equilibrium value. Changing the temperature of a system at equilibrium has a different effect: A change in temperature actually changes the value of the equilibrium constant. However, we can qualitatively predict the effect of the temperature change by treating it as a stress on the system and applying Le Chatelier's principle.
When hydrogen reacts with gaseous iodine, heat is evolved.
$\ce{H2(g) + I2(g) \rightleftharpoons 2HI(g)}\;\;\ ΔH=\mathrm{−9.4\;kJ\;(exothermic)} \label{15.7.5}$
Because this reaction is exothermic, we can write it with heat as a product.
$\ce{H2(g) + I2(g) \rightleftharpoons 2HI(g)} + \text{heat} \label{15.7.6}$
Increasing the temperature of the reaction increases the internal energy of the system. Thus, increasing the temperature has the effect of increasing the amount of one of the products of this reaction. The reaction shifts to the left to relieve the stress, and there is an increase in the concentration of H2 and I2 and a reduction in the concentration of HI. Lowering the temperature of this system reduces the amount of energy present, favors the production of heat, and favors the formation of hydrogen iodide.
When we change the temperature of a system at equilibrium, the equilibrium constant for the reaction changes. Lowering the temperature in the HI system increases the equilibrium constant: At the new equilibrium the concentration of HI has increased and the concentrations of H2 and I2 decreased. Raising the temperature decreases the value of the equilibrium constant, from 67.5 at 357 °C to 50.0 at 400 °C.
Temperature affects the equilibrium between $\ce{NO_2}$ and $\ce{N_2O_4}$ in this reaction
$\ce{N2O4(g) \rightleftharpoons 2NO2(g)}\;\;\; ΔH=\mathrm{57.20\; kJ} \label{15.7.7}$
The positive ΔH value tells us that the reaction is endothermic and could be written
$\ce{heat}+\ce{N_2O4(g) \rightleftharpoons 2NO2(g)} \label{15.7.8}$
At higher temperatures, the gas mixture has a deep brown color, indicative of a significant amount of brown $\ce{NO_2}$ molecules. If, however, we put a stress on the system by cooling the mixture (withdrawing energy), the equilibrium shifts to the left to supply some of the energy lost by cooling. The concentration of colorless $\ce{N_2O_4}$ increases, and the concentration of brown $\ce{NO_2}$ decreases, causing the brown color to fade.
The overview of how different disturbances affect the reaction equilibrium properties is tabulated in Table $1$.
Table $1$: Effects of Disturbances of Equilibrium and $K$
Disturbance Observed Change as Equilibrium is Restored Direction of Shift Effect on K
reactant added added reactant is partially consumed toward products none
product added added product is partially consumed toward reactants none
decrease in volume/increase in gas pressure pressure decreases toward side with fewer moles of gas none
increase in volume/decrease in gas pressure pressure increases toward side with more moles of gas none
temperature increase heat is absorbed toward products for endothermic, toward reactants for exothermic changes
temperature decrease heat is given off toward reactants for endothermic, toward products for exothermic changes
Example $1$
Write an equilibrium constant expression for each reaction and use this expression to predict what will happen to the concentration of the substance in bold when the indicated change is made if the system is to maintain equilibrium.
1. $2HgO_{(s)} \rightleftharpoons 2Hg_{(l)} + \mathbf{O}_{2(g)}$: the amount of HgO is doubled.
2. $NH_4HS_{(s)} \rightleftharpoons \mathbf{NH}_{3(g)} + H_2S_{(g)}$: the concentration of $H_2S$ is tripled.
3. $\textbf{n-butane}_{(g)} \rightleftharpoons isobutane_{(g)}$: the concentration of isobutane is halved.
Given: equilibrium systems and changes
Asked for: equilibrium constant expressions and effects of changes
Strategy:
Write the equilibrium constant expression, remembering that pure liquids and solids do not appear in the expression. From this expression, predict the change that must occur to maintain equilibrium when the indicated changes are made.
Solution:
Because $HgO_{(s)}$ and $Hg_{(l)}$ are pure substances, they do not appear in the equilibrium constant expression. Thus, for this reaction, $K = [O_2]$. The equilibrium concentration of $O_2$ is a constant and does not depend on the amount of $HgO$ present. Hence adding more $HgO$ will not affect the equilibrium concentration of $O_2$, so no compensatory change is necessary.
$NH_4HS$ does not appear in the equilibrium constant expression because it is a solid. Thus $K = [NH_3][H_2S]$, which means that the concentrations of the products are inversely proportional. If adding $H_2S$ triples the $H_2S$ concentration, for example, then the $NH_3$ concentration must decrease by about a factor of 3 for the system to remain at equilibrium so that the product of the concentrations equals $K$.
For this reaction, $K = \frac{[isobutane]}{[\textit{n-butane}]}$, so halving the concentration of isobutane means that the n-butane concentration must also decrease by about half if the system is to maintain equilibrium.
Exercise $1$
Write an equilibrium constant expression for each reaction. What must happen to the concentration of the substance in bold when the indicated change occurs if the system is to maintain equilibrium?
1. $\ce{HBr (g) + NaH (s) \rightleftharpoons NaBr (s)} + \mathbf{H_2(g)}$: the concentration of $\ce{HBr}$ is decreased by a factor of 3.
2. $\ce{6Li (s)} + \mathbf{N_2(g)} \ce{ \rightleftharpoons 2Li3N(s)}$: the amount of $\ce{Li}$ is tripled.
3. $\mathbf{SO_2(g)} + \ce{ Cl2(g) \rightleftharpoons SO2Cl2(l)}$: the concentration of $\ce{Cl2}$ is doubled.
Answer a
$K = \dfrac{[H_2]}{[HBr]}$; $[H_2]$ must decrease by about a factor of 3.
Answer b
$K = \dfrac{1}{[N_2]}$; solid lithium does not appear in the equilibrium constant expression, so no compensatory change is necessary.
Answer c
$K = \dfrac{1}{[SO_2][Cl_2]}$; $[SO_2]$ must decrease by about half.
Le Chatelier’s Principle (Changes in Temperature):
Catalysts Do Not Affect Equilibrium
As we learned during our study of kinetics, a catalyst can speed up the rate of a reaction. Though this increase in reaction rate may cause a system to reach equilibrium more quickly (by speeding up the forward and reverse reactions), a catalyst has no effect on the value of an equilibrium constant nor on equilibrium concentrations. The interplay of changes in concentration or pressure, temperature, and the lack of an influence of a catalyst on a chemical equilibrium is illustrated in the industrial synthesis of ammonia from nitrogen and hydrogen according to the equation
$\ce{N2(g) + 3H2(g) \rightleftharpoons 2NH3(g)} \label{15.7.9}$
A large quantity of ammonia is manufactured by this reaction. Each year, ammonia is among the top 10 chemicals, by mass, manufactured in the world. About 2 billion pounds are manufactured in the United States each year. Ammonia plays a vital role in our global economy. It is used in the production of fertilizers and is, itself, an important fertilizer for the growth of corn, cotton, and other crops. Large quantities of ammonia are converted to nitric acid, which plays an important role in the production of fertilizers, explosives, plastics, dyes, and fibers, and is also used in the steel industry.
Fritz Haber
Haber was born in Breslau, Prussia (presently Wroclaw, Poland) in December 1868. He went on to study chemistry and, while at the University of Karlsruhe, he developed what would later be known as the Haber process: the catalytic formation of ammonia from hydrogen and atmospheric nitrogen under high temperatures and pressures. For this work, Haber was awarded the 1918 Nobel Prize in Chemistry for synthesis of ammonia from its elements (Equation \ref{15.7.9}). The Haber process was a boon to agriculture, as it allowed the production of fertilizers to no longer be dependent on mined feed stocks such as sodium nitrate.
Currently, the annual production of synthetic nitrogen fertilizers exceeds 100 million tons and synthetic fertilizer production has increased the number of humans that arable land can support from 1.9 persons per hectare in 1908 to 4.3 in 2008. The availability of nitrogen is a strong limiting factor to the growth of plants. Despite accounting for 78% of air, diatomic nitrogen ($\ce{N_2}$) is nutritionally unavailable to a majority of plants due the tremendous stability of the nitrogen-nitrogen triple bond. Therefore, the nitrogen must be converted to a more bioavailable form (this conversion is called nitrogen fixation). Legumes achieve this conversion at ambient temperature by exploiting bacteria equipped with suitable enzymes.
In addition to his work in ammonia production, Haber is also remembered by history as one of the fathers of chemical warfare. During World War I, he played a major role in the development of poisonous gases used for trench warfare. Regarding his role in these developments, Haber said, “During peace time a scientist belongs to the World, but during war time he belongs to his country.”1 Haber defended the use of gas warfare against accusations that it was inhumane, saying that death was death, by whatever means it was inflicted. He stands as an example of the ethical dilemmas that face scientists in times of war and the double-edged nature of the sword of science.
Like Haber, the products made from ammonia can be multifaceted. In addition to their value for agriculture, nitrogen compounds can also be used to achieve destructive ends. Ammonium nitrate has also been used in explosives, including improvised explosive devices. Ammonium nitrate was one of the components of the bomb used in the attack on the Alfred P. Murrah Federal Building in downtown Oklahoma City on April 19, 1995.
Summary
Systems at equilibrium can be disturbed by changes to temperature, concentration, and, in some cases, volume and pressure; volume and pressure changes will disturb equilibrium if the number of moles of gas is different on the reactant and product sides of the reaction. The system's response to these disturbances is described by Le Chatelier's principle: The system will respond in a way that counteracts the disturbance. Not all changes to the system result in a disturbance of the equilibrium. Adding a catalyst affects the rates of the reactions but does not alter the equilibrium, and changing pressure or volume will not significantly disturb systems with no gases or with equal numbers of moles of gas on the reactant and product side.
Footnotes
1. 1 Herrlich, P. “The Responsibility of the Scientist: What Can History Teach Us About How Scientists Should Handle Research That Has the Potential to Create Harm?” EMBO Reports 14 (2013): 759–764.
Glossary
Le Chatelier's principle
when a chemical system at equilibrium is disturbed, it returns to equilibrium by counteracting the disturbance
position of equilibrium
concentrations or partial pressures of components of a reaction at equilibrium (commonly used to describe conditions before a disturbance)
stress
change to a reaction's conditions that may cause a shift in the equilibrium | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/15%3A_Chemical_Equilibrium/15.07%3A_Le_Chatelier%27s_Principle.txt |
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
15.1: The Concept of Equilibrium
Conceptual Problems
1. What is meant when a reaction is described as “having reached equilibrium”? What does this statement mean regarding the forward and reverse reaction rates? What does this statement mean regarding the amounts or concentrations of the reactants and the products?
2. Is it correct to say that the reaction has “stopped” when it has reached equilibrium? Explain your answer and support it with a specific example.
3. Why is chemical equilibrium described as a dynamic process? Describe this process in the context of a saturated solution of $NaCl$ in water. What is occurring on a microscopic level? What is happening on a macroscopic level?
4. Which of these systems exists in a state of chemical equilibrium?
1. oxygen and hemoglobin in the human circulatory system
2. iodine crystals in an open beaker
3. the combustion of wood
4. the amount of $\ce{^{14}C}$ in a decomposing organism
Conceptual Answer
1. When a reaction is described as "having reached equilibrium" this means that the forward reaction rate is now equal to the reverse reaction rate. In regards to the amounts or concentrations of the reactants and the products, there is no change due to the forward reaction rate being equal to the reverse reaction rate.
2. It is not correct to say that the reaction has "stopped" when it has reached equilibrium because it is not necessarily a static process where it can be assumed that the reaction rates cancel each other out to equal zero or be "stopped" but rather a dynamic process in which reactants are converted to products at the same rate products are converted to reactants. For example, a soda has carbon dioxide dissolved in the liquid and carbon dioxide between the liquid and the cap that is constantly being exchanged with each other. The system is in equilibrium and the reaction taking place is: $CO_{2}\,(g)+2\,H_{2}O\,(l)\rightleftharpoons H_{2}CO_3\,(aq)$.
3. Chemical equilibrium is described as a dynamic process because there is a movement in which the forward and reverse reactions occur at the same rate to reach a point where the amounts or concentrations of the reactants and products are unchanging with time. Chemical equilibrium can be described in a saturated solution of $NaCl$ as on the microscopic level $Na^+$ and $Cl^−$ ions continuously leave the surface of an $NaCl$ crystal to enter the solution, while at the same time $Na^+$ and $Cl^−$ ions in solution precipitate on the surface of the crystal. At the macroscopic level, the salt can be seen to dissolve or not dissolve depending whether chemical equilibrium was established.
4.
a. Exists in a state of equilibrium as the chemical reaction that occurs in the body is: $Hb\,(aq)+4\,H_{2}O\,(l)\rightleftharpoons Hb(O_{2})_{4}\,(aq)$.
b. Exists in a state of equilibrium as the chemical reaction occurs is: $H_{2}\,(g)+I_{2}\,(g) \rightleftharpoons 2\,HI\,(g)$
c. Does not exist in a state of equilibrium as it is not a reversible process. The chemical reaction that takes place is: $6\,C_{10}H_{15}O_{7}\,(s)+heat\rightarrow C_{50}H_{10}O\,(s)+10\,CH_{2}O\,(g)$.
d. Does not exist in a state of chemical equilibrium as it is not a reversible process. The chemical reaction that takes place is: $CH_{2}O+O_{2}\rightarrow H_{2}O\,(l)+CO_{2}\,(g)+nutrients$.
15.2: The Equilibrium Constant
Conceptual Problems
1. For an equilibrium reaction, what effect does reversing the reactants and products have on the value of the equilibrium constant?
2. Which of the following equilibriums are homogeneous and which are heterogeneous?
1. $2\,HF\,(g)\rightleftharpoons H_{2}\,(g)+F_{2}\,(g)$
2. $C\,(s) + 2\,H_{2}\,(g) \rightleftharpoons CH_{4}\,(g)$
3. $H_{2}C=CH_{2}\,(g) + H_{2}\,(g) \rightleftharpoons C_{2}H_{6}\,(g)$
4. $2\,Hg\,(l) + O_{2}\,(g) \rightleftharpoons 2\,HgO\,(s)$
3. Classify each equilibrium system as either homogeneous or heterogeneous.
1. $NH_{4}CO_{2}NH_{2}\,(s) \rightleftharpoons 2NH_{3}\,(g)+CO_{2}\,(g)$
2. $C\,(s) + O_{2}\,(g) \rightleftharpoons CO_{2}\,(g)$
3. $2\,Mg\,(s) + O_{2}\,(g) \rightleftharpoons 2\,MgO\,(s)$
4. $AgCl\,(s) \rightleftharpoons Ag^+\,(aq)+Cl^−\,(aq)$
4. If an equilibrium reaction is endothermic, what happens to the equilibrium constant if the temperature of the reaction is increased? if the temperature is decreased?
5. Industrial production of $NO$ by the reaction $N_{2}\,(g)+O_{2}\,(g) \rightleftharpoons 2\,NO_\,(g)$ is carried out at elevated temperatures to drive the reaction toward the formation of the product. After sufficient product has formed, the reaction mixture is quickly cooled. Why?
6. How would you differentiate between a system that has reached chemical equilibrium and one that is reacting so slowly that changes in concentration are difficult to observe?
7. What is the relationship between the equilibrium constant, the concentration of each component of the system, and the rate constants for the forward and reverse reactions?
8. Write the equilibrium constant expressions for $K$ and $K_p$ for each reaction.
1. $CO\,(g) + H_2O\,(g) \rightleftharpoons CO_{2}\,(g)+H_{2}\,(g)$
2. $PCl_{3}\,(g)+Cl_{2}\,(g) \rightleftharpoons PCl_{5}\,(g)$
3. $2\,O_{3}\,(g) \rightleftharpoons 3\,O_{2}\,(g)$
9. Write the equilibrium constant expressions for $K$ and $K_p$ as appropriate for each reaction.
1. $2\,NO\,(g)+O_{2}\,(g) \rightleftharpoons 2\,NO_{2}\,(g)$
2. $\dfrac{1}{2}\,H_2\,(g)+12\,I_{2}\,(g) \rightleftharpoons HI\,(g)$
3. $cis-stilbene\,(soln) \rightleftharpoons trans-silbene\,(soln)$
10. Why is it incorrect to state that pure liquids, pure solids, and solvents are not part of an equilibrium constant expression?
11. Write the equilibrium constant expressions for $K$ and $K_p$ for each equilibrium reaction.
1. $2\,S\,(s)+3\,O_{2}\,(g) \rightleftharpoons 2\,SO_{3}\,(g)$
2. $C\,(s) + CO_{2}\,(g) \rightleftharpoons 2\,CO\,(g)$
3. $2\,ZnS\,(s)+3\,O_{2}\,(g) \rightleftharpoons 2\,ZnO\,(s)+2\,SO_{2}\,(g)$
12. Write the equilibrium constant expressions for $K$ and $K_p$ for each equilibrium reaction.
1. $2\,HgO\,(s) \rightleftharpoons 2\,Hg\,(l)+O_{2}\,(g)$
2. $H_{2}\,(g)+I_{2}\,(s) \rightleftharpoons 2\,HI\,(g)$
3. $NH_4CO_2NH_{2}\,(s) \rightleftharpoons 2\,NH_{3}\,(g)+CO_{2}\,(g)$
13. At room temperature, the equilibrium constant for the reaction $2\,A\,(g) \rightleftharpoons B\,(g)$ is 1. What does this indicate about the concentrations of $A$ and $B$ at equilibrium? Would you expect $K$ and $K_p$ to vary significantly from each other? If so, how would their difference be affected by temperature?
14. For a certain series of reactions, if $\dfrac{[OH^−][HCO_3^−]}{[CO_3^{2−}]} = K_1$ and $\dfrac{[OH^−][H_2CO_3]}{[HCO_3^−]} = K_2$, what is the equilibrium constant expression for the overall reaction? Write the overall equilibrium equation.
15. In the equation for an enzymatic reaction, $ES$ represents the complex formed between the substrate $S$ and the enzyme protein $E$. In the final step of the following oxidation reaction, the product $P$ dissociates from the $ESO_2$ complex, which regenerates the active enzyme:
$E + S \rightleftharpoons ES\,\,\,K_1$
$ES + O_2 \rightleftharpoons ESO_2\,\,\,K_2$
$ESO_2 \rightleftharpoons E+P\,\,\,K_3$
Give the overall reaction equation and show that $K = K_1 \times K_2 \times K_3$.
Conceptual Answers
1. By reversing the reactants and products for an equilibrium reaction, the equilibrium constant becomes: $K′ = \dfrac{1}{K}$.
2.
a. This equilibrium is homogenous as all substances are in the same state.
b. This equilibrium is heterogeneous as not all substances are in the same state.
c. This equilibrium is homogeneous as all substances are in the same state.
d. This equilibrium is heterogeneous as not all substances are in the same state.
3.
a. This equilibrium is heterogeneous as not all substances are in the same state.
b. This equilibrium is heterogeneous as not all substances are in the same state.
c. This equilibrium is heterogeneous as not all substances are in the same state.
d. This equilibrium is heterogeneous as not all substances are in the same state.
4. According to Le Chatelier’s principle, equilibrium will shift in the direction to counteract the effect of a constraint (such as concentration of a reactant, pressure, and temperature). Thus, in an endothermic reaction, the equilibrium shifts to the right-hand side when the temperature is increased which increases the equilibrium constant and the equilibrium shifts to the left-hand side when the temperature is decreased which decreases the equilibrium constant.
5. After sufficient industrial production of $NO$ by the reaction of $N_{2}\,(g)+O_{2}\,(g) \rightleftharpoons 2\,NO_\,(g)$ at elevated temperatures to drive the reaction toward the formation of the product, the reaction mixture is cooled quickly because it quenches the reaction and prevents the system from reverting to the low-temperature equilibrium composition that favors the reactants.
6. To differentiate between a system that has reached equilibrium and one that is reacting slowly that changes in concentrations are difficult to observe we can use Le Chatelier’s principle to observe any shifts in the reaction upon addition of a constraint (such as concentration, pressure, or temperature).
7. The relationship between the equilibrium constant, the concentration of each component of a system, and the rate constants for the forward and reverse reactions considering a reaction of a general form: $a\,A+b\,B \rightleftharpoons c\,C+d\,D$ is $K=\dfrac{[C]^c[D]^d}{[A]^a[B]^b}=\dfrac{k_f}{k_r}$
8.
a.
$K=\dfrac{[CO_2][H_2]}{[CO][H_2O]}$
$K_p=\dfrac{(P_{CO_2})(P_{H_2})}{(P_{CO})(P_{H_2O})}$
b.
$K=\dfrac{[PCl_5]}{[PCl_3][Cl_2]}$
$K_p=\dfrac{(P_{PCl_5})}{(P_{Cl_3})(P_{Cl_2})}$
c.
$K=\dfrac{[O_2]^3}{[O_3]^2}$
$K_p=\dfrac{(P_{O_2})^3}{(P_{O_3})^2}$
9.
a.
$K=\dfrac{[NO_2]^2}{[NO]^2[O_2]}$
$K_p=\dfrac{(P_{NO_{2}})^2}{(P_{NO})^2(P_{O_2})}$
b.
$K=\dfrac{[HI]}{[H_2]^{\dfrac{1}{2}}[O_2]}$
$K_p=\dfrac{(P_{HI})}{(P_{H_{2}})^{\frac{1}{2}}(P_{O_{2}})}$
c.
$K=\dfrac{trans-stilbene}{cis-stilbene}$
$K_p=\dfrac{(P_{trans-stilbene})}{(P_{cis-stilbene})}$
10. It is incorrect to state that pure liquids, pure solids, and solvents are not part of an equilibrium constant expression because they are not reactive enough or cause a change in the concentrations of the ions or the species that exist in the gas phase.
11.
1.
$K=\dfrac{[SO_3]^2}{[O_2]^3}$
$K_p=\dfrac{(P_{SO_3})^2}{(P_{O_2})^3}$
b.
$K=\dfrac{[CO]^2}{[CO_2]}$
$K_p=\dfrac{(P_{CO})^2}{(P_{CO_2})}$
c.
$K=\dfrac{[SO_2]^2}{[O_2]^3}$
$K_p=\dfrac{(P_{SO_2})^2}{(P_{O_2})^3}$
12.
a.
$K=[O_2]$
$K_p=(P_{O_{2}})$
b.
$K=\dfrac{[HI]^2}{[H_2]}$
$K_p=\dfrac{(P_{HI})^2}{(P_{H_2})}$
c.
$K=[NH_3]^2[CO_2]$
$K_p=(P_{NH_3})^2(P_{CO_2})$
13.
$K=\dfrac{[B]}{[A]^2} \rightarrow 1=\frac{[B]}{[A]^2} \rightarrow [A]^2=[B] \rightarrow [A]=\sqrt{B}$
$K$ and $K_p$ vary by $RT$, but it largely depends on $T$ as $R$ is constant. A raise or decrease in temperature would cause a difference.
$K_p=K(RT)^{Δn}=K(RT)^{-1}=\dfrac{K}{RT}$
$Δn=(total\,moles\,of\,gas\,on\,the\,product\,side)-(total\,of\,moles\,on\,the\,reactant\,side)=1-2=−1$
14.
$K=\dfrac{K_1}{K_2}={\dfrac{\dfrac{[OH^−][HCO_3^−]}{[CO_3^{2−}]}}{\dfrac{[OH^−][H_2CO_3]}{[HCO_3^−]}}=\dfrac{[HCO_3^−]^2}{[CO_3^{2-}][H_2CO_3]}}$
$CO_3^{2-}\,(g)+H_2CO_{3}\,(g) \rightleftharpoons 2\,HCO_3^{-}\,(g)$
15.
$K = K_1 \times K_2 \times K_3=\frac{[ES]}{[E][S]}\times\frac{[ESO_2]}{[ES][O_2]}\times\frac{[E][P]}{[ESO_2]}=\frac{[P]}{[S][O_2]}$
$S+O_2 \rightleftharpoons P$
Numerical Problems
1. Explain what each of the following values for $K$ tells you about the relative concentrations of the reactants versus the products in a given equilibrium reaction: $K = 0.892$; $K = 3.25 \times 10^8$; $K = 5.26 \times 10^{−11}$. Are products or reactants favored at equilibrium?
2. Write the equilibrium constant expression for each reaction. Are these equilibrium constant expressions equivalent? Explain.
1. $N_2O_{4}\,(g) \rightleftharpoons 2\,NO_{2}\,(g)$
2. $\frac{1}{2}\,N_2O_{4}\,(g) \rightleftharpoons NO_{2}\,(g)$
3. Write the equilibrium constant expression for each reaction.
1. $\frac{1}{2}N_{2}\,(g)+\frac{3}{2}H_{2}\,(g) \rightleftharpoons NH_{3}\,(g)$
2. $\frac{1}{3}N_{2}\,(g)+H_{2}\,(g) \rightleftharpoons \frac{2}{3}NH_{3}\,(g)$
How are these two expressions mathematically related to the equilibrium constant expression for
$N_{2}\,(g)+3\,H_{2}\,(g) \rightleftharpoons 2\,NH_{3}\,(g) ?$
1. Write an equilibrium constant expression for each reaction.
1. $C\,(s) + 2\,H_2O\,(g) \rightleftharpoons CO_{2}\,(g)+2\,H_{2}\,(g)$
2. $SbCl_{3}\,(g)+Cl_{2}\,(g) \rightleftharpoons SbCl_{5}\,(g)$
3. $2\,O_{3}\,(g) \rightleftharpoons 3\,O_{2}\,(g)$
2. Give an equilibrium constant expression for each reaction.
a. $2\,NO\,(g) + O_{2}\,(g) \rightleftharpoons 2\,NO_{2}\,(g)$
b. $\frac{1}{2}H_{2}\,(g)+\frac{1}{2}I_{2}\,(g) \rightleftharpoons HI\,(g)$
c. $CaCO_{3}\,(s) + 2\,HOCl\,(aq) \rightleftharpoons Ca^{2+}\,(aq) + 2\,OCl^−\,(aq) + H_2O\,(l) + CO_{2}\,(g)$
6. Calculate $K$ and $K_p$ for each reaction.
1. $2\,NOBr\,(g) \rightleftharpoons 2\,NO\,(g)+Br_2\,(g)$: at 727°C, the equilibrium concentration of $NO$ is 1.29 M, $Br_2$ is 10.52 M, and $NOBr$ is 0.423 M.
2. $C\,(s) + CO_{2}\,(g) \rightleftharpoons 2\,CO\,(g)$: at 1,200 K, a 2.00 L vessel at equilibrium has partial pressures of 93.5 atm $CO_2$ and 76.8 atm $CO$, and the vessel contains 3.55 g of carbon.
7. Calculate $K$ and $K_p$ for each reaction.
1. $N_2O_4\,(g) \rightleftharpoons 2\,NO_{2}\,(g)$: at the equilibrium temperature of −40°C, a 0.150 M sample of $N_2O_4$ undergoes a decomposition of 0.456%.
2. $CO\,(g)+2\,H_{2}\,(g) \rightleftharpoons CH_3OH\,(g)$: an equilibrium is reached at 227°C in a 15.5 L reaction vessel with a total pressure of $6.71 \times 10^2$ atm. It is found to contain 37.8 g of hydrogen gas, 457.7 g of carbon monoxide, and 7,193 g of methanol.
8. Determine $K$ and $K_p$ (where applicable) for each reaction.
1. $2\,H_2S\,(g) \rightleftharpoons 2\,H_{2}\,(g)+S_{2}\,(g)$: at 1065°C, an equilibrium mixture consists of $1.00 \times 10^{−3}$ M $H_2$, $1.20 \times 10^{−3}$ M $S_2$, and $3.32 \times 10^{−3}$ M $H_2S$.
2. $Ba(OH)_{2}\,(s) \rightleftharpoons 2\,OH^−\,(aq)+Ba^{2+}\,(aq)$: at 25°C, a 250 mL beaker contains 0.330 mol of barium hydroxide in equilibrium with 0.0267 mol of barium ions and 0.0534 mol of hydroxide ions.
9. Determine $K$ and $K_p$ for each reaction.
1. $2\,NOCl\,(g) \rightleftharpoons 2\,NO\,(g)+Cl_{2}\,(g)$: at 500 K, a 24.3 mM sample of $NOCl$ has decomposed, leaving an equilibrium mixture that contains 72.7% of the original amount of $NOCl$.
2. $Cl_{2}\,(g)+PCl_{3}\,(g) \rightleftharpoons PCl_{5}\,(g)$: at 250°C, a 500 mL reaction vessel contains 16.9 g of $Cl_2$ gas, 0.500 g of $PCl_3$, and 10.2 g of $PCl_5$ at equilibrium.
10. The equilibrium constant expression for a reaction is $\dfrac{[CO_2]^2}{[SO_2]^2[O_2]}$. What is the balanced chemical equation for the overall reaction if one of the reactants is $Na_2CO_{3}\,(s)$?
11. The equilibrium constant expression for a reaction is $\dfrac{[NO][H_{2}O]^{\dfrac{3}{2}}}{[NH_3][O_2]^{\dfrac{5}{4}}}$. What is the balanced chemical equation for the overall reaction?
12. Given $K =\dfrac{k_f}{k_r}$, what happens to the magnitude of the equilibrium constant if the reaction rate of the forward reaction is doubled? What happens if the reaction rate of the reverse reaction for the overall reaction is decreased by a factor of 3?
13. The value of the equilibrium constant for $2\,H_{2}\,(g)+S_{2}\,(g) \rightleftharpoons 2\,H_2S\,(g)$ is $1.08 \times 10^7$ at 700°C. What is the value of the equilibrium constant for the following related reactions
1. $H_{2}\,(g)+12\,S_{2}\,(g) \rightleftharpoons H_2S\,(g)$
2. $4\,H_{2}\,(g)+2\,S_{2}\,(g) \rightleftharpoons 4\,H_2S\,(g)$
3. $H_2S\,(g) \rightleftharpoons H_{2}\,(g)+12\,S_{2}\,(g)$
Numerical Answers
1. In the given equilibrium reaction where $K = 0.892\approx1$ has a concentration of the reactants that is approximately equal to the concentration of the products so neither formation of the reactants or products is favored. In the given equilibrium reaction where $K = 3.25 \times 10^8>1$ has a concentration of the products that is relatively small compared to the concentration of the reactants so the formation of the products is favored. In the given equilibrium reaction where $K = 5.26 \times 10^{−11}<1$ has a concentration of products that is relatively large compared to that of the concentration of the reactants so the formation of the reactants is favored.
2.
a. $K=\dfrac{[NO_2]^{2}}{[N_2O_4]}$
b. $K=\dfrac{[NO_2]}{[N_2O_4]^{\dfrac{1}{2}}}$
Although the equilibrium constant expressions have a 2:1 ratio of concentration for the products to the concentration of the reactants for the same species involved to get the $K$ value for a. We would need to square it to get the $K$ value for b.
3.
1. $K’=\dfrac{[NH_3]}{[N_2]^{\dfrac{1}{2}}[H_2]^{\dfrac{3}{2}}}$
2. $K’’=\dfrac{[NH_3]^{\dfrac{2}{3}}}{[N_2]^{\dfrac{1}{2}}[H_2]}$
$K=\dfrac{[NH_3]^2}{[N_2][H_2]^3}$
$K'=K^{\dfrac{1}{2}}$
$K''=K^{\dfrac{1}{3}}$
4.
a. $K=\dfrac{[H_2]^2[CO_2]}{[H_{2}O]^2}$
b. $K=\dfrac{[SbCl_5]}{[SbCl_3][Cl_2]}$
c. $K=\dfrac{[O_2]^3}{[O_3]^2}$
5.
1. $K=\dfrac{[NO_2]^2}{[NO]^2[O_2]}$
2. $K=\dfrac{[HI]}{[H_2]^{\dfrac{1}{2}}[I_2]^{\dfrac{1}{2}}}$
3. $K=\dfrac{[Ca^{2+}][OCl^−]^2[CO_2]}{[HOCl]^2}$
6.
a.
$K=\dfrac{[NO]^2[Br]}{[NOBr]^2}=\frac{[1.29\,M]^2[10.52\,M]}{[0.423\,M]^2}=97.8$
$K_p=K(RT)^{Δn}=(97.8)((0.08206\frac{L\cdot atm}{mol\cdot K})((727+273.15)K))^{3-2}=8.03x10^{4}$
b.
$K_p=K(RT)^{Δn}=K(RT)^{2-1}=K(RT) \rightarrow K=\dfrac{K_p}{RT}=\frac{63.1}{(0.08206\frac{L\cdot atm}{mol\cdot K})(1,200\,K)}=6.41$
$K=\dfrac{(P_{CO})^2}{P_{CO_2}}=\frac{(76.8\,atm)^2}{93.5\,atm}=63.1$
7.
1.
$K=\dfrac{[NO_{2}]^2}{[N_{2}O_{4}]}=\frac{[0.001368\,M]}{[0.149316\,M]}=1.25x10^{-5}$
$[NO_2]=(2)(0.150\,M)(0.00456)=0.001368\,M$
$[N_2O_4]=(0.150\,M)(1-0.00456)=0.149316\,M$
$K_p=K(RT)^{Δn}=(1.25x10^{-5})((0.08206\frac{L\cdot atm}{mol\cdot K})((-40+273.15)K))^{2-1}=2.39x10^{-4}$
b.
$K=\dfrac{[CH_{3}OH]}{[CO][H_{2}]^2}=\frac{[14.5\,M]}{[1.05\,M][1.21\,M]^2}=9.47$
$[CH_{3}OH]=7,193\,g\,CH_{3}OH \times \frac{1\,mol\,CH_{3}OH}{32.04\,g\,CH_{3}OH} \times \frac{1}{15.5\,L}=14.5\,M$
$[CO]=457.7\,g\,CO \times \frac{1\,mol\,CO}{28.01\,g\,CO} \times \frac{1}{15.5\,L}=1.05\,M$
$[H_{2}]=37.8\,g\,H_{2} \times \frac{1\,mol\,H_{2}}{2.02\,g\,H_{2}} \times \frac{1}{15.5\,L}=1.21\,M$
$K_p=K(RT)^{Δn}=(9.47)((0.08206\frac{L\cdot atm}{mol\cdot K})((227+273.15)K))^{1-3}=5.62x10^{-2}$
8.
a.
$K=\dfrac{[H_{2}]^2[S_{2}]}{[H_{2}S]^2}=\frac{[1.00 \times 10^{-3}\,M]^2[1.20 \times 10^{-3}\,M]}{[3.32 \times\ 10^{-3}\,M]^2}=1.09 \times 10^{-4}$
$K_p=K(RT)^{Δn}=(1.09 \times 10^{-4})((0.08206\frac{L\cdot atm}{mol\cdot K})((1065+273.15)K))^{3-2}=1.20 \times 10^{-2}$
b.
$K=[OH^{-}]^2[Ba^{2+}]=[0.2136\,M]^2[0.1068\,M]=4.87 \times 10^{-3}$
$[OH^{-}]=0.0534\,mol\,OH^{-} \times \frac{1}{0.25\,L}=0.2136\,M$
$[Ba^{2+}]=0.0267\,mol\,Ba^{2+} \times \frac{1}{0.25\,L}=0.1068\,M$
$K_p=K(RT)^{Δn}=(4.87 \times 10^{-3})((0.08206\frac{L\cdot atm}{mol\cdot K})((25+273.15)K))^{3-1}=2.92$
9.
a.
$K=\dfrac{[NO]^2[Cl_{2}]}{[NOCl]^2}=4.59 \times 10^{-4}$
$K_p=K(RT)^{Δn}=(4.59 \times 10^{-4})((0.08206\frac{L\cdot atm}{mol\cdot K})(500K))^{3-2}=1.88 \times 10^{-2}$
b.
$K=\dfrac{[PCl_5]}{[Cl_{2}][PCl_{3}]}=\frac{[9.80 \times 10^{-2}\,M]}{[4.77 \times 10^{-1}\,M][7.28 \times 10^{-3}\,M]}=28.2$
$[PCl_{5}]=10.2\,g\,PCl_{5} \times \frac{1\,mol\,PCl_{5}}{208.2388\,g\,PCl_{5}} \times \frac{1}{0.5\,L}=9.80 \times 10^{-2}\,M$
$[Cl_{2}]=16.9\,g\,Cl_{2} \times \frac{1\,mol\,Cl_{2}}{70.9\,g\,Cl_{2}} \times \frac{1}{0.5\,L}=4.77 \times 10^{-1}\,M$
$[PCl_{3}]=0.500\,g\,PCl_{3} \times \frac{1\,mol\,PCl_{3}}{137.33\,g\,PCl_{3}} \times \frac{1}{0.5\,L}=7.28 \times 10^{-3}\,M$
$K_p=K(RT)^{Δn}=(28.2)((0.08206\frac{L\cdot atm}{mol\cdot K})(250+273.15)K)^{1-2}=6.57 \times 10^{-1}$
10. $2\,SO_{2}\,(g)+O_{2}\,(g)+2\,Na_{2}CO_{3}\,(s) \rightleftharpoons 2\,CO_{2}\,(g)+2\,Na_{2}SO_{4}\,(s)$
11. $NH_{3}\,(g) + \frac{5}{4}\,O_{2}\,(g)⇌NO \,(g)+\frac{3}{2}\,H_{2}O\,(g)$
12. If the reaction rate of the forward reaction is doubled the magnitude of the equilibrium constant is doubled. If the reaction rate of the reverse reaction for the overall reaction is described by a factor of 3 the magnitude of the equilibrium constant is increased by a factor of 3.
13. $K’= \dfrac{[H_{2}S]^2}{[H_{2}]^2[S_{2}]}= 1.08 \times 10^{7}$
a. $K= \dfrac{[H_{2}S]}{[H_{2}][S_{2}]^\frac{1}{2}}=K’^{\frac{1}{2}}=(1.08 \times 10^{7})^{\frac{1}{2}}=3.29 \times 10^{3}$ b. $K= \dfrac{[H_{2}S]^4}{[H_{2}]^4[S_{2}]^2}=K’^{2}=(1.08 \times 10^{7})^{2}=1.17 \times 10^{14}$ c. $K= \dfrac{[H_{2}][S_2]^\frac{1}{2}}{[H_{2}S]}=K’^{-\frac{1}{2}}= (1.08 \times 10^{7})^{-\frac{1}{2}}=3.04 \times 10^{-4}$
15.3: Interpreting & Working with Equilibrium Constants
Conceptual Problems
1. Describe how to determine the magnitude of the equilibrium constant for a reaction when not all concentrations of the substances are known.
2. Calculations involving systems with very small or very large equilibrium constants can be dramatically simplified by making certain assumptions about the concentrations of products and reactants. What are these assumptions when $K$ is (a) very large and (b) very small? Illustrate this technique using the system $A+2B \rightleftharpoons C$ for which you are to calculate the concentration of the product at equilibrium starting with only A and B. Under what circumstances should simplifying assumptions not be used?
Conceptual Answers
1. The magnitude of the equilibrium constant for a reaction depends on the form in which the chemical reaction is written. For example, writing a chemical reaction in different but chemically equivalent forms causes the magnitude of the equilibrium constant to be different but can be related by comparing their respective magnitudes.
2.
a. When $K$ is very large the reactants are converted almost entirely to products, so we can assume that the reaction proceeds 100% to completion.
$K= \dfrac{[C]}{[A][B]^2}=\frac{[C]}{very\,small}=\frac{1}{0}= \infty \rightarrow [C]= \infty$
b. When $K$ is very small the reactants do not tend to form products readily, and the equilibrium lies to the left as written, favoring the formation of the reactants.
$K=\dfrac{[C]}{[A][B]^2}=\frac{very\,small}{[A][B]^2}=\frac{0}{1}=0 \rightarrow [C]=0$
Simplifying assumptions should not be used if the equilibrium constant is not known to be very large or very small.
Numerical Problems
Please be sure you are familiar with the quadratic formula before proceeding to the Numerical Problems.
1. In the equilibrium reaction $A+B \rightleftharpoons C$, what happens to $K$ if the concentrations of the reactants are doubled? tripled? Can the same be said about the equilibrium reaction $2\,A \rightleftharpoons B+C$?
2. The following table shows the reported values of the equilibrium $P_{O_2}$ at three temperatures for the reaction $Ag_{2}O\,(s) \rightleftharpoons 2\,Ag\,(s)+ \frac{1}{2}\,O_{2}\,(g)$, for which ΔH° = 31 kJ/mol. Are these data consistent with what you would expect to occur? Why or why not?
T (°C) $P_{O_2}\;(mmHg)$
150 182
184 143
191 126
1. Given the equilibrium system $N_2O_{4}\,(g) \rightleftharpoons 2\,NO_{2}\,(g)$, what happens to $K_p$ if the initial pressure of $N_2O_4$ is doubled? If $K_p$ is $1.7 \times 10^{−1}$ at 2300°C, and the system initially contains 100% $N_2O_4$ at a pressure of $2.6 \times 10^2$ atm, what is the equilibrium pressure of each component?
2. At 430°C, 4.20 mol of $HI$ in a 9.60 L reaction vessel reaches equilibrium according to the following equation: $H_{2}\,(g)+I_{2}\,(g) \rightleftharpoons 2\,HI\,(g)$ At equilibrium, $[H_2] = 0.047\;M$ and $[HI] = 0.345\;M$ What are $K$ and $K_p$ for this reaction?
3. Methanol, a liquid used as an automobile fuel additive, is commercially produced from carbon monoxide and hydrogen at 300°C according to the following reaction: $CO\,(g)+2\,H_{2}\,(g) \rightleftharpoons CH_3OH\,(g)$ with $K_p = 1.3 \times 10^{−4}$. If 56.0 g of $CO$ is mixed with excess hydrogen in a 250 mL flask at this temperature, and the hydrogen pressure is continuously maintained at 100 atm, what would be the maximum percent yield of methanol? What pressure of hydrogen would be required to obtain a minimum yield of methanol of 95% under these conditions?
4. Starting with pure A, if the total equilibrium pressure is 0.969 atm for the reaction $A\,(s) \rightleftharpoons 2\,B\,(g)+C\,(g)$, what is $K_p$?
5. The decomposition of ammonium carbamate to $NH_3$ and $CO_2$ at 40°C is written as $NH_4CO_2NH_{2}\,(s) \rightleftharpoons 2\,NH_{3}\,(g)+CO_{2}\,(g)$. If the partial pressure of $NH_3$ at equilibrium is 0.242 atm, what is the equilibrium partial pressure of $CO_2$? What is the total gas pressure of the system? What is $K_p$?
6. At 375 $K$, $K_p$ for the reaction $SO_{2}Cl_{2}\,(g) \rightleftharpoons SO_{2}\,(g)+Cl_{2}\,(g)$ is 2.4, with pressures expressed in atmospheres. At 303 $K$, $K_p$ is $2.9 \times 10^{−2}$.
1. What is $K$ for the reaction at each temperature?
2. If a sample at 375 K has 0.100 M $Cl_2$ and 0.200 M $SO_2$ at equilibrium, what is the concentration of $SO_2Cl_2$?
3. If the sample given in part b is cooled to 303 K, what is the pressure inside the bulb?
7. For the gas-phase reaction $a\,A \rightleftharpoons b\,B$, show that $K_p = K(RT)^{Δn}$ assuming ideal gas behavior.
8. For the gas-phase reaction $I_2 \rightleftharpoons 2\,I$, show that the total pressure is related to the equilibrium pressure by the following equation: $P_T=\sqrt{K_{p}P_{I_{2}}} + P_{I_{2}}$
9. Experimental data on the system $Br_{2}\,(l) \rightleftharpoons Br_{2}\,(aq)$ are given in the following table. Graph $Br_{2}\,(aq)$ versus moles of $Br_{2}\,(l)$ present; then write the equilibrium constant expression and determine K.
Grams $Br_{2}$ in 100 mL Water $Br_{2}$ (M)
1.0 0.0626
2.5 0.156
3.0 0.188
4.0 0.219
4.5 0.219
1. Data accumulated for the reaction (\n-butane(g) \rightleftharpoons isobutane(g)\) at equilibrium are shown in the following table. What is the equilibrium constant for this conversion? If 1 mol of n-butane is allowed to equilibrate under the same reaction conditions, what is the final number of moles of n-butane and isobutane?
Moles n-butane Moles Isobutane
0.5 1.25
1.0 2.5
1.50 3.75
1. Solid ammonium carbamate ($NH_{4}CO_{2}NH_{2}$) dissociates completely to ammonia and carbon dioxide when it vaporizes: $NH_{4}CO_{2}NH_{2}\,(s) \rightleftharpoons 2\,NH_{3}\,(g)+CO_{2}\,(g)$ At 25°C, the total pressure of the gases in equilibrium with the solid is 0.116 atm. What is the equilibrium partial pressure of each gas? What is $K_p$? If the concentration of $CO_2$ is doubled and then equilibrates to its initial equilibrium partial pressure +x atm, what change in the $NH_{3}$ concentration is necessary for the system to restore equilibrium?
2. The equilibrium constant for the reaction $COCl_{2}\,(g) \rightleftharpoons CO\,(g)+Cl_{2}\,(g)$ is $K_p = 2.2 \times 10^{−10}$ at 100°C. If the initial concentration of $COCl_{2}$ is $3.05 \times 10^{−3}\; M$, what is the partial pressure of each gas at equilibrium at 100°C? What assumption can be made to simplify your calculations?
3. Aqueous dilution of $IO_{4}^{−}$ results in the following reaction: $IO^−_{4}\,(aq)+2\,H_{2}O_(l)\, \rightleftharpoons H_4IO^−_{6}\,(aq)$ with $K = 3.5 \times 10^{−2}$. If you begin with 50 mL of a 0.896 M solution of $IO_4^−$ that is diluted to 250 mL with water, how many moles of $H_4IO_6^−$ are formed at equilibrium?
4. Iodine and bromine react to form $IBr$, which then sublimes. At 184.4°C, the overall reaction proceeds according to the following equation: $I_{2}\,(g)+Br_{2}\,(g) \rightleftharpoons 2\,IBr\,(g)$ with $K_p = 1.2 \times 10^2$. If you begin the reaction with 7.4 g of $I_2$ vapor and 6.3 g of $Br_2$ vapor in a 1.00 L container, what is the concentration of $IBr\,(g)$ at equilibrium? What is the partial pressure of each gas at equilibrium? What is the total pressure of the system?
5. For the reaction $2\,C\,(s) + \,N_{2}\,(g)+5\,H_{2}\, \rightleftharpoons 2\,CH_{3}NH_{2}\,(g)$ with $K = 1.8 \times 10^{−6}$. If you begin the reaction with 1.0 mol of $N_2$, 2.0 mol of $H_2$, and sufficient $C\,(s)$ in a 2.00 L container, what are the concentrations of $N_2$ and $CH_3NH_2$ at equilibrium? What happens to $K$ if the concentration of $H_2$ is doubled?
Numerical Answers
1. In both cases, the equilibrium constant will remain the same as it does not depend on the concentrations.
2. No, the data is not consistent with what I would expect to occur because enthalpy is positive indicating that the reaction is endothermic thus heat is on the left side of the reaction. As the temperature is raised $P_{O_2}$ would be expected to increase to counteract the constraint.
3.
If the initial pressure of $N_2O_4$ was doubled then $K_p$ is one half of the original value.
$K_p=\dfrac{(P_{NO_{2}})^2}{(P_{N_{2}O_{4}})} \rightarrow 1.7 \times 10^{-1} = \frac{(2x)^2}{2.6 \times 10^2 -x} \rightarrow 44.2-0.17x=4x^2 \rightarrow x \approx 3.303\,atm$
$P_{N_{2}O_{4}}$ $P_{NO_{2}}$
I $2.6 \times 10^{2}$ 0
C -x +2x
E $2.6 \times 10^2-x$ 2x
$P_{N_{2}O_{4}}=2.6 \times 10^2-x=260-3.303=2.6 \times 10^2\,atm$
$P_{NO_{2}}=2x=(2)(3.303)=6.6\,atm$
4.
$K=\frac{[HI]^2}{[H_2][I_2]}=\frac{0.345\,M}{(0.047\,M)(0.047\,M)}=157$
$K_p=K(RT)^{Δn}=(157)((0.08206\frac{L\cdot atm}{mol\cdot K})(430+273.15)K)^{2-2}=157$
5.
$Maximum\;Percent\;Yield=\frac{Actual}{Theoretical} \times 100\%=\frac{212.593}{376.127} \times 100\%=56.52\% \approx 57\%$
$PV=nRT \rightarrow P=\frac{(1.999\;mol)(0.08206\frac{L\cdot atm}{mol\cdot K})(300+273.15)K}{0.250\;L}=376.127\;atm$
$[CO]=56.0\;g\;CO \times \frac{1\;mol\;CO}{28.01\,g\,CO}=1.999\,mol\,CO$
$K_p=\frac{P_{CH_{3}OH}}{(P_{CO})(P_{H_2})^{2}} \rightarrow 1.3 \times 10^{-4}= \frac{x}{(376.02-x)(100^2)} \rightarrow 1.3= \frac{x}{376.07-x} \rightarrow 488.965-1.3x=x \rightarrow 488.965=2.3x \rightarrow x=212.593\,atm$
$K_p=\frac{(P_{CH_{3}OH})}{(P_{CO})(P_{H_2})^2} \rightarrow 1.3 \times 10^{-4}= \frac{357.320}{(376.127-357.320)(P_{H_{2}})^2} \rightarrow 1.3 \times 10^{-4}= \frac{357.320}{(18.80635)(P_{H_{2}})^2} \rightarrow 0.002444(P_{H_{2}})^2=357.320 \rightarrow (P_{H_{2}})=382.300 \approx 3.8 \times 10^2\,atm$
$Minimum\,Percent\,Yield=\frac{Actual}{Theoretical} \times 100 \% \rightarrow 95\%=\frac{x}{376.127} \times 100\% \rightarrow x=357.320\,atm$
$CO$ $2\,H_2$ $CH_{3}OH$
I 376.127 100 0
C -x -2x +x
E 376.127-x 100 (maintained) x
6. $K_p=\frac{(P_B)^2(P_C)}{P_A}=\frac{[2x]^2[x]}{[0.969-x]}=\frac{4x^3}{0.969-x}$
$A$ $2\,B$ $C$
I 0.969 0 0
C -x +2x +x
E 0.969-x 2x x
7.
$P_{CO_{2}}=P_{tot}-P_{NH_{3}}=P_{tot}-0.242\,atm$
$P_{tot}=P_{NH_3}+P_{CO_2}=0.242\,atm+P_{CO_2}$
$K_p=(P_{NH_3})^2(P_{CO_2})=(0.242\,atm)^2(P_{CO_2})$
8.
a.
$At\,375\,K:K_p=K(RT)^{Δn} \rightarrow K=\frac{K_p}{(RT)^{Δn}}=\frac{2.9 \times 10^{-2}}{((0.08206\frac{L\cdot atm}{mol\cdot K})(375\,K))^{2-1}}=7.80 \times 10^{-2}$
$At\,303\,K:K_p=K(RT)^{Δn} \rightarrow K=\frac{K_p}{(RT)^{Δn}}=\frac{2.9 \times 10^{-2}}{((0.08206\frac{L\cdot atm}{mol\cdot K})(303\,K))^{2-1}}=1.17 \times 10^{-3}$
b. $K=\frac{[SO_{2}][Cl_{2}]}{[SO_{2}Cl_{2}]} \rightarrow [SO_{2}Cl_{2}]=\frac{[0.200\,M][0.100\,M]}{7.80 \times 10^{-2}}=2.56 \times 10^{-1}\,M$
c. If the sample given in part b is cooled to 303 $K$, the pressure inside the bulb would decrease.
9.
$K_p=\frac{(P_B)^b}{(P_A)^a}=\frac{((\frac{n_B}{V})(RT))^b}{((\frac{n_A}{V})(RT))^a}=\frac{[B]^{b}(RT)^b}{[A]^a(RT)^a}=K(RT)^{b-a}=K(RT)^{Δn}$
$PV=nRT \rightarrow P=\frac{n}{V}RT$
$K=\frac{[B]^{b}}{[A]^{a}}$
$Δn=b-a$
10.
$P_T=P_I+P_{I_2}=\sqrt{K_pP_{I_2}}+P_{I_2}$
$K_p=\frac{(P_I)^2}{P_{I_2}} \rightarrow (P_I)^2=K_p(P_{I_2}) \rightarrow P_I=\sqrt{K_pP_{I_2}}$
11.
The graph should be a positive linear correlation.
$Br_2\,(l)\,(M)\,(x-axis)$ $Br_2\,(aq)\,(M)\,(y-axis)\,(same\,value\,for\,K)$
$6.26 \times 10^-2$ 0.0626
$1.56 \times 10^-1$ 0.156
$1.88 \times 10^-2$ 0.188
$2.50 \times 10^-2$ 0.219
$2.82 \times 10^-2$ 0.219
$[Br_2\,(l)]=1.0\,g\,Br_2 \times \frac{1\,mol\,Br_2}{159.808\,g\,Br_2} \times \frac{1}{0.1\,L}=6.26 \times 10^{-2}$
$K=\frac{[Br_2\,(aq)]}{[Br_2\,(l)]}=\frac{[Br_2\,(aq)]}{1}=[Br_2\,(aq)]$
12. $K=\frac{[isobutane]}{[n-butane]}=\frac{x}{1-x}$
$n-butane$ $isobutane$
I 1 0
C -x +x
E 1-x x
13.
$P_{NH_3}=2x=2(0.0387)=7.73 \times 10^{-2}\,atm$
$P_{CO_2}=x=3.87 \times 10^{-2}\,atm$
$K_p=(P_{NH_3})^2(P_{CO_2})=(2x)^2(x)=4x^3=4(0.0387)^3=2.32 \times 10^{-4}$
$P_{tot}=P_{NH_3}+P_{CO_2} \rightarrow 0.116=2x+x \rightarrow 0.116=3x \rightarrow x=0.0387$
If the concentration of $CO_{2}$ is doubled and then equilibrates to its initial equilibrium partial
pressure +x atm, the concentration of $NH_{3}$ should also be doubled for the system to restore
equilibrium.
$NH_{3}$ $CO_{2}$
I 0 0
C 2x x
E 2x x
14.
$P_{COCl_{2}}=9.34 \times 10^{-2}-x=9.34 \times 10^{-2}-9.34 \times 10^{-22}=9.34 \times 10^{-2}\,atm$
$P_{CO}=x=9.34 \times 10^{-22}\,atm$
$P_{Cl_{2}}=x=9.34 \times 10^{-22}\,atm$
Assume that the equilibrium mainly lies on the reactants side because the $K_p$ value is less than 1.
$K_p=\frac{(P_{CO})(P_{Cl_{2}})}{(P_{COCl_{2}})} \rightarrow 2.2 \times 10^{-10} =\frac{x^{2}}{9.34 \times 10^{-2}-x} \rightarrow 2.0548 \times 10^{-11}-2.2 \times 10^{-10}x=x^{2} \rightarrow x^{2}+2.2 \times 10^{-10}x-2.0548 \times 10^{-11}=0 \rightarrow x=9.34 \times 10^{-22}$
$PV=nRT \rightarrow P=\frac{nRT}{V}=(3.05 \times 10^{-3})(0.08206\frac{L\cdot atm}{mol\cdot K})(100+273.15)K=9.34 \times 10^{-2}$
$COCl_{2}$ $CO$ $Cl_{2}$
I $9.34 \times 10^{-2}$ 0 0
C -x +x +x
E $9.34 \times 10^{-2}-x$ x x
15. $[H_{4}IO_{6}^{-}]=x=1.6 \times 10^{-3}\,mol$
$K=\frac{[H_{4}IO_{6}^{-}]}{[IO_{4}^{-}]} \rightarrow 3.5 \times 10^{-2} =\frac{x}{(0.0448-x)} \rightarrow x=1.568 \times 10^{-3}$
$IO_{4}^{-}:50\,mL\,IO_{4}^{-} \times \frac{1\,L\,IO_{4}^{-}}{1,000\,mL\,IO_{4}^{-}} \times \frac{0.896\,mol\,IO_{4}^{-}}{1\,L\,IO_{4}^{-}}=0.0448\,mol$
$IO_{4}^{-}$ $H_{4}IO_{6}^{-}$
I 0.0448 0
C -x +x
E 0.0448-x x
16. $PV=nRT \rightarrow \frac{P}{RT}=\frac{n}{V} \rightarrow \frac{12.9468}{(0.08206\frac{L\cdot atm}{mol\cdot K})(184.4+273.15)K}=3.5 \times 10^{-1}\,M$
$K_p=\frac{(P_{IBr})^2}{(P_{I_{2}})(P_{Br_{2}})} \rightarrow 1.2 \times 10^{-2} = \frac{2x}{(1.096-x)(1.479-x)}=\frac{2x}{x^2-2.575x+1.62098} \rightarrow 0.012x^2-0.0309x+0.0194518=2x \rightarrow 0.012x^2-2.0309+0.0194518=0 \rightarrow x=12.9468$
$I_{2}$ $Br_{2}$ $IBr$
I 1.096 1.479 0
C -x -x 2x
E 1.096-x 1.479-x 2x
$PV=nRT \rightarrow P=\frac{nRT}{V}=(2.92 \times 10^{-2}\,M)(0.08206\frac{L\cdot atm}{mol\cdot K})(184.4+273.15)K=1.096\,atm$
$[I_{2}]=7.4\,g\,I_{2} \times \frac{1\,mol\,I_{2}}{253\,g\,I_{2}} \times \frac{1}{1.00\,L}=2.92 \times 10^{-2}\,M$
$PV=nRT \rightarrow P={nRT}{V}=(3.94 \times 10^{-2}\,M)(0.08206\frac{L\cdot atm}{mol\cdot K})(184.4+273.15)K=1.479\,atm$
$[Br_{2}]=6.3\,g\,Br_{2} \times \frac{1\,mol\,Br_{2}}{159.808\,g\,Br_{2}}=3.94 \times 10^{-2}\,M$
17.
$[N_{2}]=0.5-12x=0.5-12(0.000471330)=0.494\,M$
$[CH_{3}NH_{2}]=2x=2(0.0004713300=9.43 \times 10^{-4}\,M$
$If\,the\,concentration\,of\,H_{2}\,is\,doubled\,,then\,K=\frac{[CH_{3}NH_{2}]^{2}}{[N_2][H_2]^5}=\frac{(9.43 \times 10^{-4})^{2}}{(0.494)(1.998)^5}=5.65 \times 10^{-8}$
$2 \times [H_{2}]=2(1.0-2.5x)=2(1.0-2.5(0.000471330))=1.998\,M$
$K=\frac{[CH_{3}NH_{2}]^2}{[N_2][H_2]^5} \rightarrow 1.8 \times 10^{-6}=\frac{(2x)^2}{(0.5-x)(1.0-5x)^5} \rightarrow x=0.000471330\,M$
$C$ $N_2$
$H_2$
$CH_{3}NH_{2}$
I - 0.5 1.0 0
C - -x -5x 2x
E - 0.5-x 1.0-5x 2x
$[N_2]=1.00\,mol\,N_2 \times \frac{1}{2.00\,L}=0.5\,M$
$[H_2]=2.00\,mol\,H_2 \times \frac{1}{2.00\,L}=1.0\,M$
15.6: Applications of Equilibrium Constants
Conceptual Problems
1. During a set of experiments, graphs were drawn of [reactants] versus [products] at equilibrium. Using Figure 15.8 and Figure 15.9 as your guides, sketch the shape of each graph using appropriate labels.
1. $H_2O\,(l) \rightleftharpoons H_2O\,(g)$
2. $2\,MgO\,(s) \rightleftharpoons 2\,Mg\,(s)+O_{2}\,(g)$
3. $2\,O_{3}\,(g) \rightleftharpoons 3\,O_{2}\,(g)$
4. $2\,PbS\,(g)+3\,O_{2}\,(g) \rightleftharpoons 2\,PbO\,(s)+2\,SO_{2}\,(g)$
2. Write an equilibrium constant expression for each reaction system. Given the indicated changes, how must the concentration of the species in bold change if the system is to maintain equilibrium?
1. $2\,NaHCO_{3}\,(s) \rightleftharpoons Na_2CO_{3}\,(s) + CO_{2}\,(g)+ H_2O\,(g)$: $[CO_2]$ is doubled.
2. $N_2F_{4}\,(g) \rightleftharpoons 2\,NF_{2}\,(g)$: $[NF_{2}]$ is decreased by a factor of 2.
3. $H_{2}\,(g) + I_{2}\,(g) \rightleftharpoons 2\,HI\,(g)$: $[I_2]$ is doubled.
3. Write an equilibrium constant expression for each reaction system. Given the indicated changes, how must the concentration of the species in bold change if the system is to maintain equilibrium?
1. $CS_{2}\,(g) + 4\,H_{2}\,(g) \rightleftharpoons CH_{4}\,(g) + 2\,H_2S\,(g)$: $[CS_2]$ is doubled.
2. $PCl_{5}\,(g) \rightleftharpoons PCl_{3}\,(g) + Cl_{2}\,(g)$: $[Cl_2]$ is decreased by a factor of 2.
3. $4\,NH_{3}\,(g) + 5\,O_{2}\,(g) \rightleftharpoons 4\,NO\,(g) + 6\,H_2O\,(g)$: $[NO]$ is doubled.
Conceptual Answer
1.
a. According to Figure 15.8, we could obtain a graph with the x-axis labeled $[H_2O]\,(l)\,(M)$ and y-axis labeled $[H_2O]\,(g)\,(M)$. The graph should have a positive linear correlation. For any equilibrium concentration of $H_2O\,(g)$, there is only one equilibrium $H_2O\;(l)$. Because the magnitudes of the two concentrations are directly proportional, a large $[H_2O]\,(g)$ at equilibrium requires a large $[H_2O]\,(l)$ and vice versa. In this case, the slope of the line is equal to $K$.
b. According to Figure 15.9, we could obtain a graph with the x-axis labeled $[O2]\,(M)$ and y-axis labeled $[MgO]\,(M)$. Because $O_2\,(g)$ is the only one in gaseous form, the graph would depend on the concentration of $O_2$.
c. According to Figure 15.8, we could obtain a graph with the x-axis labeled $[O_3]\,(M)$ and y-axis labeled $[O2]\,(M)$. The graph should have a positive linear correlation. For every $3\,O_2\,(g)$ there is $2\,O_3\,(g)$. Because the magnitudes of the two concentrations are directly proportional, a large $[O3]\,(g)$ at equilibrium requires a large $[O_2]\,(g)$ and vice versa. In this case, the slope of the line is equal to $K$.
d. According to figure 15.8, we could obtain a graph with the x-axis labeled $[O2]\,(M)$ and y-axis labeled $[SO2]\,(M)$. The graph should have a positive linear correlation. For every $3\,O_{2}\,(g)$ there is $2\,SO_2\,(g)$. Because the magnitudes of the two concentrations are directly proportional, a large $O_2\,(g)$ at equilibrium requires a large $SO_2\,(g)$ and vice versa. In this case, the slope of the line is equal to $K$.
2.
a.
$K=[Na_{2}CO_{3}][CO_2][H_{2}O]$
If $[CO_2]$ is doubled, $[H_2O]$ should be halved if the system is to maintain equilibrium.
b.
$K=\frac{[NF_2]^2}{[N_{2}F_{4}]}$
If $[NF_2]$ is decreased by a factor of 2, then $[N_{2}F_{4}]$ must also be decreased by a factor of 2 if the system is to maintain equilibrium.
c.
$K=\frac{[HI]^2}{[H_2][I_{2}]}$
If $[I_{2}]$ is doubled then $[HI]$ must also be doubled if the system is to maintain equilibrium.
3.
1.
$K=\dfrac{[CH_4][H_2S]^2}{[CS_2][H_2]^4}$
If $[CS_2]$ is doubled then $[H_2]$ must be decreased by a factor of 2√4≅ 1.189 if the system is to maintain equilibrium.
b.
$K=\dfrac{[PCl_3]}{[Cl_2][PCl_5]}$
If $[Cl_2]$ is halved then $[PCl_5]$ must also be halved if the system is to maintain equilibrium.
c.
$K=\dfrac{[NO]^4[H_2O]^6}{[NH_3][O_2]^5}$
If $[NO]$ is doubled then $[H_2O]$ must also be multiplied by 22/3≅1.587 if the system is to maintain equilibrium.
Numerical Problems
1. The data in the following table were collected at 450°C for the reaction $N_{2}\,(g)+3\,H_{2}\,(g) \rightleftharpoons 2\,NH_{3}\,(g)$:
Equilibrium Partial Pressure (atm)
P (atm) $NH_3$ $N_2$ $H_2$
30 (equilibrium) 1.740 6.588 21.58
100 15.20 19.17 65.13
600 321.6 56.74 220.8
The reaction equilibrates at a pressure of 30 atm. The pressure on the system is first increased to 100 atm and then to 600 atm. Is the system at equilibrium at each of these higher pressures? If not, in which direction will the reaction proceed to reach equilibrium?
1. For the reaction $A \rightleftharpoons B+C$, $K$ at 200°C is 2.0. A 6.00 L flask was used to carry out the reaction at this temperature. Given the experimental data in the following table, all at 200°C, when the data for each experiment were collected, was the reaction at equilibrium? If it was not at equilibrium, in which direction will the reaction proceed?
Experiment A B C
1 2.50 M 2.50 M 2.50 M
2 1.30 atm 1.75 atm 14.15 atm
3 12.61 mol 18.72 mol 6.51 mol
1. The following two reactions are carried out at 823 K:
$CoO\,(s)+H_{2}\,(g) \rightleftharpoons Co\,(s)+H_2O\,(g) \text{ with } K=67$
$CoO\,(s)+CO\,(g) \rightleftharpoons Co\,(s)+CO_{2}\,(g) \text{ with } K=490$
1. Write the equilibrium expression for each reaction.
2. Calculate the partial pressure of both gaseous components at equilibrium in each reaction if a 1.00 L reaction vessel initially contains 0.316 mol of $H_2$ or $CO$ plus 0.500 mol $CoO$.
3. Using the information provided, calculate Kp for the following reaction: $H_{2}\,(g)+CO_{2}\,(g) \rightleftharpoons CO\,(g)+H_2O\,(g)$
4. Describe the shape of the graphs of [reactants] versus [products] as the amount of $CoO$ changes.
1. Hydrogen iodide (HI) is synthesized via $H_{2}\,(g)+I_{2}\,(g) \rightleftharpoons 2\,HI\,(g)$, for which $K_p = 54.5$ at 425°C. Given a 2.0 L vessel containing $1.12 \times 10^{−2}\,mol$ of $H_2$ and $1.8 \times 10^{−3}\,mol$ of $I_2$ at equilibrium, what is the concentration of $HI$? Excess hydrogen is added to the vessel so that the vessel now contains $3.64 \times 10^{−1}\,mol$ of $H_2$. Calculate $Q$ and then predict the direction in which the reaction will proceed. What are the new equilibrium concentrations?
Answers
1. The system is not at equilibrium at each of these higher pressures. To reach equilibrium, the reaction will proceed to the right to decrease the pressure because the equilibrium partial pressure is less than the total pressure.
$K_p=\frac{[NH_{3}]^2}{[N_{2}][H_{2}]^3}=\frac{[15.20]^2}{[19.17][65.13]^3}=4.4 \times 10^{-5}$
$K_p=\frac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^3}=\frac{[321.6]^2}{[56.74][220.8]^3}= 1.7 \times 10^{-4}$
2.
1. $K=\frac{[B][C]}{[A]}=\frac{[2.50][2.50]}{[2.50]}=2.50$
2. $K_p=K(RT)^{Δn} \rightarrow K=\frac{K_p}{(RT)^{Δn}}=\frac{19.0}{((0.08206\frac{L\cdot atm}{mol\cdot K})(200+273.15)K)^{2-1})}=0.49$
$K_p=\frac{(P_B)(P_C)}{(P_A)}=\frac{(1.75)(14.15)}{1.30}=19.0$
3. $K=\frac{[B][C]}{[A]^{2}}=\frac{(18.72)(6.51)}{12.61}=9.7$
Experiment 1 is about the same as the given $K$ value and thus considered to be about equilibrium. The second experiment has a $K$ value that is about 1 so neither the formation of the reactants or products is favored. The third experiment has a $K$ value that is larger than 1 so the formation of the products is favored.
3.
a.
$K=\frac{[H_{2}O]}{[H_{2}]}$
$K=\frac{[CO_{2}]}{[Co]}$
b.
$[H_2]=[CO]=0.316\,mol\,H_{2} \times \frac{1}{1.00\,L}=0.316\,M$
$[CoO]=0.5\,mol\,CoO \times \frac{1}{1.00\,L}=0.5\,M$
Reaction 1:
$PV=nRT \rightarrow P=\frac{nRT}{V}=(4.65 \times 10^{-3})(0.08206\frac{L\cdot atm}{mol\cdot K})(823\,K)=0.314$
$[H_{2}]=0.316-x=0.316-0.311=4.65x10^{-3}$
$PV=nRT \rightarrow P=\frac{nRT}{V}=(0.311)(0.08206\frac{L\cdot atm}{mol\cdot K})(823\,K)=21.0$
$[H_{2}O]=x=0.311$
$H_{2}$ $H_{2}O$
I 0.316 0
C -x +x
E 0.316-x x
$K=\frac{[H_{2}O]}{[H_{2}]} \rightarrow 67=\frac{x}{0.316-x} \rightarrow x=0.311$
Reaction 2:
$PV=nRT \rightarrow P=\frac{nRT}{V}=(0.001)(0.08206\frac{L\cdot atm}{mol\cdot K})(823\,K)=6.75 \times 10^{-2}\,atm$
$[CO]=0.316-x=0.316-0.315=0.001\,M$
$PV=nRT \rightarrow P=\frac{nRT}{V}=(0.315)(0.08206\frac{L\cdot atm}{mol\cdot K})(823\,K)=21.3\,atm$
$[CO_{2}]=x=0.315\,M$
$CO$ $CO_{2}$
I 0.316 0
C -x +x
E 0.316-x x
$K=\frac{[CO_{2}]}{[Co]} \rightarrow 490=\frac{x}{0.316-x} \rightarrow x=0.315$
c.
$H_{2}\,(g)+CO_{2}\,(g) \rightleftharpoons CO\,(g)+H_{2}O\,(g)$
$K_p=\frac{(P_{CO})(P_{H_{2}O})}{(P_{H_{2}})(P_{CO_{2}})}=\frac{(6.75 \times 10^{-2})(21)}{(0.314)(21.3)}=0.21$
d. The shape of the graphs [reactants] versus [products] does not change as the amount of $CoO$ changes because it is a solid.
4.
$PV=nRT \rightarrow \frac{n}{V}=\frac{P}{RT} \rightarrow \frac{n}{V}=\frac{0.101798}{(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)}=1.5 \times 10^{-4}\,M\,HI$
$[HI]=2x=2(0.050899)=0.101798\,atm$
$K_p=\frac{(P_{HI})^2}{(P_{H_{2}})(P_{I_{2}})} \rightarrow 54.5=\frac{(2x)^2}{(3.2 \times 10^{-1}-x)(5.16 \times 10^{-2}-x)} \rightarrow x=0.050899\,atm$
$H_{2}$ $I_{2}$ $HI$
I $3.2 \times 10^{-1}$ $5.16 \times 10^{-2}$ 0
C -x -x +2x
E $3.2 \times 10^{-1}-x$ $5.16 \times 10^{-2}-x$ 2x
$PV=nRT \rightarrow P=\frac{nRT}{V}=(5.6 \times 10^{-3})(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)=3.2 \times 10^{-1}\,atm$
$[H_{2}]=1.12 \times 10^{-2}\,mol\,H_{2} \times \frac{1}{2.0\,L}=5.6 \times 10^{-3}\,M$
$PV=nRT \rightarrow P=\frac{nRT}{V}=(9.0 \times 10^{-4})(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)=5.16 \times 10^{-2}\,atm$
$[I_2]= 1.8 \times 10^{-3}\,mol\,I_{2} \times \frac{1}{2.0\,L}=9.0 \times 10^{-4}\,M$
For excess hydrogen:
$Q=\frac{[HI]}{[H_{2}][I_{2}]}=\frac{1.8 \times 10^{-3}}{(594.410)(1.09 \times 10^{-3})}=2.8 \times 10^{-3}$
The reaction will proceed to the right to reach equilibrium.
$PV=nRT \rightarrow \frac{n}{V}=\frac{P}{RT} \rightarrow \frac{0.103162}{(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)}=1.8 \times 10^{-3}\,M$
$[HI]=2x=2(0.051581)=0.103162\,atm$
$PV=nRT \rightarrow \frac{n}{V}=\frac{P}{RT} \rightarrow \frac{10.375}{(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)}=594.410\,M$
$[H_{2}]=10.427-x=10.427-0.051581=10.375\,atm$
$PV=nRT \rightarrow \frac{n}{V}=\frac{P}{RT} \rightarrow \frac{1.9 \times 10^{-5}}{(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)}=1.09 \times 10^{-3}\,M$
$[I_{2}]=5.16 \times 10^{-2}-x=5.16 \times 10^{-2}-0.051581=1.9 \times 10^{-5}\,atm$
$K_p=\frac{(P_{HI})^2}{(P_{H_{2}})(P_{I_{2}})} \rightarrow 54.5=\frac{(2x)^2}{(10.427-x)(5.16 \times 10^{-2}-x)} \rightarrow x=0.051581\,atm$
$H_{2}$ $I_{2}$ $HI$
I 10.427 $5.16 \times 10^{-2}$ 0
C -x -x +2x
E 10.427-x $5.16 \times 10^{-2}-x$ 2x
$PV=nRT \rightarrow P=\frac{nRT}{V}=(0.182\,M)(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)=10.427\,atm$
$[H_{2}]=3.64 \times 10^{-1}\,mol\,H_{2} \times \frac{1}{2.0\,L}=0.182\,M$ | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/15%3A_Chemical_Equilibrium/15.E%3A_Exercises.txt |
chemical equilibrium – condition where the concentration of products and reactants do not change with time
15.1: The Concept of Equilibrium
at equilibrium
$k_f[A] = k_r[B] \nonumber$
there for the ratio
$\displaystyle\frac{[B]}{[A]} = \frac{k_f}{k_r} = \textit{constant} \nonumber$
15.2: The Equilibrium Constant
equilibrium condition can be reached from either forward or reverse direction
Cato Maximillian Galdberg (1836-1902), and Peter Wauge (1833-1900)
• Law of mass action – relationship between concentrations of reactants and products at equilibrium
If $aA + bB\rightleftharpoons pP+qQ$
then an equilibrium expression can be constructed
$\displaystyle K_c=\frac{[P]^p[Q]^q}{[A]^a[B]^b} \nonumber$
• equilibrium expression depends only on stoichiometry of reaction and not mechanisms
• equilibrium constant:
• does not depend on initial concentrations
• does not matter if other substances present as long as they do not react with reactants or products
• varies with temperatures
• no units
15.2.1 Expressing Equilibrium Constants in Terms of Pressure, $K_p$
$\displaystyle K_p=\frac{(P_P)^p(P_Q)^q}{(P_A)^a(P_B)^b} \nonumber$
15.2.2 The Magnitude of Equilibrium Constants
• $K\gg 1$; equilibrium lies to the right; products favored
• $K \ll 1$; equilibrium lies to the left; reactants favored
15.2.3 The Direction of the Chemical Equation and $K$
• equilibrium expression written in one direction is the reciprocal of the one in the other direction
15.4: Heterogeneous Equilibria
• homogeneous equilibria – substances in the same phase
• heterogeneous equilibria – substances in different phases
• a pure solid, a pure liquid, and a solvent in dilute solutions all appear in equilibrium laws, but they are all assigned activities that are equal to 1
• by convention the actvities of the pure solid, pure liquid, or solute are not explicitly written as part of the equilibrium law
15.5: Calculating Equilibrium Constants
determining unknown equilibrium concentrations
• tabulate known initial and equilibrium concentrations
• calculate change in concentration that occurs as system reaches equilibrium
• use stoichiometry to determine change in concentration of unknown species
• from initial concentrations and changes in concentrations, calculate equilibrium concentrations
15.5.1 Relating Kc and Kp
$PV = nRT \nonumber$
$P = (n/V)RT = MRT \nonumber$
$PA = [A](RT) \nonumber$
$K_p=K_c(RT)D^n \nonumber$
• D n = change in moles from reactants to products
15.6: Applications of Equilibrium Constants
• equilibrium constant:
1. product direction reaction mixture will proceed
2. calculate concentrations of reactants and products once equilibrium is reached
15.6.1 Predicting the Direction of Reaction
• reaction quotient
• at equilibrium Q=K
• Q>K; reaction moves right to left
• Q<K; reaction moves left to right
15.7: Le Chatelier's Principle
If system at equilibrium is disturbed by change in temperature, pressure or concentration then system will shift equilibrium position
15.7.1 Change in Reactant or Product Concentration
• addition of substance will result in consummation of part of added substance
• if substance removed, reaction will move to produce more of the substance
15.7.2 Effects of Volume and Pressure Changes
• reducing volume, reaction shifts to reduce number of gas molecules
• increase volume, reaction shifts to produce more gas molecules
• increase pressure, decrease volume reduces total number of moles
• pressure volume changes do not affect K as long as temperature is constant
• changes concentrations of gaseous substances
15.7.3 Effect on Temperature Change
• endothermic: reactants + heat « products
• exothermic: reactants « products + heat
• increase temperature, equilibrium shifts in direction that absorbs heat
• endothermic: increase T, increase K
• exothermic: increase T, decrease K
• cooling shifts equilibrium to produce heat
15.7.4 The Effect of Catalysts
• catalysts increase rate at which equilibrium is obtained
• does not change composition of equilibrium mixture | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/15%3A_Chemical_Equilibrium/15.S%3A_Chemical_Equilibrium_%28Summary%29.txt |
Acids and bases have been defined differently by three sets of theories. One is the Arrhenius definition, 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. On the other hand, the Bronsted-Lowry definition defines acids as substances that donate protons (H+) whereas bases are substances that accept protons. Also, the Lewis theory of acids and bases states that acids are electron pair acceptors while bases are electron pair donors. Acids and bases can be defined by their physical and chemical observations.
• 16.1: Acids and Bases - A Brief Review
In chemistry, acids and bases have been defined differently by three sets of theories: 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 alter in the chapter and include the Brønsted-Lowry definition and the Lewis theory.
• 16.2: Brønsted–Lowry Acids and Bases
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.
• 16.3: The Autoionization of Water
Water is amphiprotic: it can act as an acid by donating a proton to a base to form the hydroxide ion, or as a base by accepting a proton from an acid to form the hydronium ion ($H_3O^+$). The autoionization of liquid water produces $OH^−$ and $H_3O^+$ ions. The equilibrium constant for this reaction is called the ion-product constant of liquid water (Kw) and is defined as $K_w = [H_3O^+][OH^−]$. At 25°C, $K_w$ is $1.01 \times 10^{−14}$; hence $pH + pOH = pK_w = 14.00$.
• 16.4: The pH Scale
The concentration of hydronium ion in a solution of an acid in water is greater than $1.0 \times 10^{-7}\; M$ at 25 °C. The concentration of hydroxide ion in a solution of a base in water is greater than $1.0 \times 10^{-7}\; M$ at 25 °C. The concentration of H3O+ in a solution can be expressed as the pH of the solution; $\ce{pH} = -\log \ce{H3O+}$. The concentration of OH− can be expressed as the pOH of the solution: $\ce{pOH} = -\log[\ce{OH-}]$. In pure water, pH = 7 and pOH = 7.
• 16.5: Strong Acids and Bases
Acid–base reactions always contain two conjugate acid–base pairs. Each acid and each base has an associated ionization constant that corresponds to its acid or base strength. Two species that differ by only a proton constitute a conjugate acid–base pair. The magnitude of the equilibrium constant for an ionization reaction can be used to determine the relative strengths of acids and bases.
• 16.6: Weak Acids
The strengths of Brønsted-Lowry acids and bases in aqueous solutions can be determined by their acid or base ionization constants. Stronger acids form weaker conjugate bases, and weaker acids form stronger conjugate bases. Thus strong acids are completely ionized in aqueous solution because their conjugate bases are weaker bases than water. Weak acids are only partially ionized because their conjugate bases are strong enough to compete successfully with water for possession of protons.
• 16.7: Weak Bases
The pH of a solution of a weak base can be calculated in a way which is very similar to that used for a weak acid. Instead of an acid constant Ka, a base constant Kb must be used.
• 16.8: Relationship Between Ka and Kb
• 16.9: Acid-Base Properties of Salt Solutions
A salt can dissolve in water to produce a neutral, a basic, or an acidic solution, depending on whether it contains the conjugate base of a weak acid as the anion ( A−A− ), the conjugate acid of a weak base as the cation ( BH+ ), or both. Salts that contain small, highly charged metal ions produce acidic solutions in water. The reaction of a salt with water to produce an acidic or a basic solution is called a hydrolysis reaction.
• 16.10: Acid-Base Behavior and Chemical Structure
Inductive effects and charge delocalization significantly influence the acidity or basicity of a compound. The acid–base strength of a molecule depends strongly on its structure. The weaker the A–H or B–H+ bond, the more likely it is to dissociate to form an $H^+$ ion. In addition, any factor that stabilizes the lone pair on the conjugate base favors the dissociation of $H^+$, making the conjugate acid a stronger acid.
• 16.11: Lewis Acids and Bases
Lewis proposed that the electron pair is the dominant actor in acid-base chemistry. An Lewis acid is a substance that accepts a pair of electrons, and in doing so, forms a covalent bond with the entity that supplies the electrons. A Lewis base is a substance that donates an unshared pair of electrons to a recipient species with which the electrons can be shared. Lewis acis/base theory is a powerful tool for describing many chemical reactions used in organic and inorganic chemistry.
• 16.E: Acid–Base Equilibria (Exercises)
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al.
• 16.S: Acid–Base Equilibria (Summary)
16: AcidBase Equilibria
Acids and bases have been known for a long time. When Robert Boyle 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 CO2), and interact with alkalis to form neutral substances. In 1815, Humphry Davy 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 Svante Arrhenius 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 H2O, which will be discussed later in a subsection. Acids and bases can be defined by their physical and chemical observations (Table $1$).
ACIDS BASES
Table $1$: General Properties of Acids and Bases
produce a piercing pain in a wound. give a slippery feel.
taste sour. taste bitter.
are colorless when placed in phenolphthalein (an indicator). are pink when placed in phenolphthalein (an indicator).
are red on blue litmus paper (a pH indicator). are blue on red litmus paper (a pH indicator).
have a pH<7. have a pH>7.
produce hydrogen gas when reacted with metals.
produce carbon dioxide when reacted with carbonates.
Common examples: Lemons, oranges, vinegar, urine, sulfuric acid, hydrochloric acid 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 electrolytes. Strong acids and bases will be strong electrolytes. Weak acids and bases will be weak electrolytes. This affects the amount of conductivity.
The Arrhenius Definition of Acids and Bases
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 Arrhenius acid is a compound that increases the concentration of $H^+$ ions that are present when added to water. These $H^+$ ions form the hydronium ion ($H_3O^+$) when they combine with water molecules. This process is represented in a chemical equation by adding H2O to the reactants side.
$HCl_{(aq)} \rightarrow H^+_{(aq)} + Cl^-_{(aq)} \nonumber$
In this reaction, hydrochloric acid ($HCl$) dissociates into hydrogen ($H^+$) and chlorine ($Cl^-$) ions when dissolved in water, thereby releasing H+ ions into solution. Formation of the hydronium ion equation:
$HCl_{(aq)} + H_2O_{(l)} \rightarrow H_3O^+_{(aq)} + Cl^-_{(aq)} \nonumber$
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 $H_2SO_4$ or $HCl$ 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 ($NaNH_2$) in liquid ammonia is not alkaline, despite the fact that the amide ion ($NH^−_2$) will readily deprotonate ammonia. Thus, the Arrhenius definition can only describe acids and bases in an aqueous environment.
Limitation of the Arrhenius Definition of Acids and Bases
The Arrhenius definition can only describe acids and bases in an aqueous environment.
In chemistry, acids and bases have been defined differently by three sets of theories: 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 alter in the chapter and include the Brønsted-Lowry definition the defines acids as substances that donate protons ($H^+$) whereas bases are substances that accept protons and the Lewis theory of acids and bases states that acids are electron pair acceptors while bases are electron pair donors.
• Anonymous | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/16%3A_AcidBase_Equilibria/16.01%3A_Acids_and_Bases_-_A_Brief_Review.txt |
Learning Objectives
• Identify acids, bases, and conjugate acid-base pairs according to the Brønsted-Lowry definition
• Write equations for acid and base ionization reactions
• Use the ion-product constant for water to calculate hydronium and hydroxide ion concentrations
• Describe the acid-base behavior of amphiprotic substances
Acids and bases have been known for a long time. When Robert Boyle 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 CO2), and interact with alkalis to form neutral substances. In 1815, Humphry Davy 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 Svante Arrhenius 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.
Previously, we defined acids and bases as Arrhenius did: An acid is a compound that dissolves in water to yield hydronium ions ($H_3O^+$) and a base as a compound that dissolves in water to yield hydroxide ions ($\ce{OH-}$). This definition is not wrong; it is simply 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 Brønsted-Lowry acid, and a compound that accepts a proton is called a Brønsted-Lowry base. An acid-base reaction is the transfer of a proton from a proton donor (acid) to a proton acceptor (base). In a subsequent chapter of this text we will introduce the most general model of acid-base behavior introduced by the American chemist G. N. Lewis.
Acids may be compounds such as $HCl$ or $H_2SO_4$, organic acids like acetic acid ($\ce{CH_3COOH}$) or ascorbic acid (vitamin C), or $H_2O$. 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. Bases 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:
$\ce{H^+ + OH^- \rightarrow H_2O} \label{16.2.1}$
We call the product that remains after an acid donates a proton the conjugate base of the acid. This species is a base because it can accept a proton (to re-form the acid):
$\text{acid} \rightleftharpoons \text{proton} + \text{conjugate base}\label{16.2.2a}$
$\ce{HF \rightleftharpoons H^+ + F^-} \label{16.2.2b}$
$\ce{H_2SO_4 \rightleftharpoons H^+ + HSO_4^{−}}\label{16.2.2c}$
$\ce{H_2O \rightleftharpoons H^+ + OH^-}\label{16.2.2d}$
$\ce{HSO_4^- \rightleftharpoons H^+ + SO_4^{2−}}\label{16.2.2e}$
$\ce{NH_4^+ \rightleftharpoons H^+ + NH_3} \label{16.2.2f}$
We call the product that results when a base accepts a proton the base’s conjugate acid. This species is an acid because it can give up a proton (and thus re-form the base):
$\text{base} + \text{proton} \rightleftharpoons \text{conjugate acid} \label{16.2.3a}$
$\ce{OH^- +H^+ \rightleftharpoons H2O}\label{16.2.3b}$
$\ce{H_2O + H^+ \rightleftharpoons H3O+}\label{16.2.3c}$
$\ce{NH_3 +H^+ \rightleftharpoons NH4+}\label{16.2.3d}$
$\ce{S^{2-} +H^+ \rightleftharpoons HS-}\label{16.2.3e}$
$\ce{CO_3^{2-} +H^+ \rightleftharpoons HCO3-}\label{16.2.3f}$
$\ce{F^- +H^+ \rightleftharpoons HF} \label{16.2.3g}$
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 acid ionization. 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 base ionization reaction occurs in which protons are transferred from water molecules to base molecules. For example, adding pyridine to water yields hydroxide ions and pyridinium ions:
" height="167" width="612" src="/@api/deki/files/56740/CNX_Chem_14_01_NH3_img.jpg">
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 autoionization. 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 (Kw):
$\ce{H_2O}_{(l)}+\ce{H_2O}_{(l)} \rightleftharpoons \ce{H_3O^+}_{(aq)}+H\ce{O^-}_{(aq)}\;\;\; K_\ce{w}=\ce{[H_3O^+][OH^- ]} \label{16.2.4}$
The slight ionization of pure water is reflected in the small value of the equilibrium constant; at 25 °C, Kw 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.
A Video Discussing Conjugate Acid-Base Pairs: Conjugate Acid-Base Pairs [youtu.be]
Example $1$: Ion Concentrations in Pure Water
What are the hydronium ion concentration and the hydroxide ion concentration in pure water at 25 °C?
Solution
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:
$K_\ce{w}=\ce{[H_3O^+][OH^- ]}=\ce{[H_3O^+]^2+}=\ce{[OH^- ]^2+}=1.0 \times 10^{−14} \nonumber$
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 $1.0 \times 10^{−7}\; M$.
Exercise $1$
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?
Answer
$\ce{[H_3O^+]} = \ce{[OH^- ]} = 4.9 \times 10^{−7}\; M$
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 determined by the autoionization reaction but it does shift the relative concentrations of $\ce{[OH^-]}$ and $\ce{[H_3O^+]}$. Example 16.2.2 demonstrates the quantitative aspects of this relation between hydronium and hydroxide ion concentrations.
A Video Describing the Self-Ionization of Water (Kw): Self-Ionization of Water (Kw) [youtu.be]
Example $2$: The Inverse Proportionality of $\ce{[H_3O^+]}$ and $\ce{[OH^- ]}$
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?
Solution
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 Kw expression yields that $[\ce{OH^- }]$ is directly proportional to the inverse of [H3O+]:
$[\ce{OH^- }]=\dfrac{K_{\ce w}}{[\ce{H_3O^+}]}=\dfrac{1.0 \times 10^{−14}}{2.0 \times 10^{−6}}=5.0 \times 10^{−9} \nonumber$
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:
$K_\ce{w}=\ce{[H_3O^+][OH^- ]}=(2.0 \times 10^{−6})(5.0 \times 10^{−9})=1.0 \times 10^{−14} \nonumber$
Exercise $2$
What is the hydronium ion concentration in an aqueous solution with a hydroxide ion concentration of 0.001 M at 25 °C?
Answer
$\ce{[H3O+]} = 1 \times 10^{−11} M \nonumber$
Amphiprotic Species
Like water, many molecules and ions may either gain or lose a proton under the appropriate conditions. Such species are said to be amphiprotic. Another term used to describe such species is amphoteric, 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:
$\ce{HCO^-}_{3(aq)} + \ce{H_2O}_{(l)} \rightleftharpoons \ce{CO^{2-}}_{3(aq)} + \ce{H_3O^+}_{(aq)} \label{16.2.5a}$
$ \ce{HCO^-}_{3(aq)} + \ce{H_2O}_{(l)} \rightleftharpoons \ce{H_2CO}_{3(aq)} + \ce{OH^-}_{(aq)} \label{16.2.5b}$
Example $3$: The Acid-Base Behavior of an Amphoteric Substance
Write separate equations representing the reaction of $\ce{HSO3-}$
1. as an acid with $\ce{OH^-}$
2. as a base with HI
Solution
1. $HSO_{3(aq)}^- + OH_{(aq)}^- \rightleftharpoons SO_{3(aq)}^{2-} + H_2O_{(l)}$
2. $HSO_{3(aq)}^- + HI_{(aq)} \rightleftharpoons H_2SO_{3(aq)}+ I_{(aq)}^-$
Example $4$
Write separate equations representing the reaction of $\ce{H2PO4-}$
1. as a base with HBr
2. as an acid with $\ce{OH^-}$
Answer
1. $H_2PO_{4(aq)}^- + HBr_{(aq)} \rightleftharpoons H_3PO_{4(aq)} + Br^-_{(aq)}$
2. $H_2PO_{4(aq)}^- + OH^-_{(aq)} \rightleftharpoons HPO_{4(aq)}^{2-} + H_2O_{(l)}$
Summary
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, H3O+, and the hydroxide ion, $\ce{OH^-}$ when it undergoes autoionization:
$\ce{2 H_2O}_{(l)} \rightleftharpoons H_3O^+_{(aq)} + OH^-_{(aq)} \nonumber$
The ion product of water, Kw is the equilibrium constant for the autoionization reaction:
$K_\ce{w}=\mathrm{[H_2O^+][OH^- ]=1.0 \times 10^{−14} \; at\; 25°C} \nonumber$
Key Equations
• $K_{\ce w} = \ce{[H3O+][OH^- ]} = 1.0 \times 10^{−14}\textrm{ (at 25 °C)} \nonumber$
Glossary
acid ionization
reaction involving the transfer of a proton from an acid to water, yielding hydronium ions and the conjugate base of the acid
amphiprotic
species that may either gain or lose a proton in a reaction
amphoteric
species that can act as either an acid or a base
autoionization
reaction between identical species yielding ionic products; for water, this reaction involves transfer of protons to yield hydronium and hydroxide ions
base ionization
reaction involving the transfer of a proton from water to a base, yielding hydroxide ions and the conjugate acid of the base
Brønsted-Lowry acid
proton donor
Brønsted-Lowry base
proton acceptor
conjugate acid
substance formed when a base gains a proton
conjugate base
substance formed when an acid loses a proton
ion-product constant for water (Kw)
equilibrium constant for the autoionization of water | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/16%3A_AcidBase_Equilibria/16.02%3A_BrnstedLowry_Acids_and_Bases.txt |
Learning Objectives
• To understand the autoionization reaction of liquid water.
• To know the relationship among pH, pOH, and $pK_w$.
As you learned previously acids and bases can be defined in several different ways (Table $1$). Recall that the Arrhenius definition of an acid is a substance that dissociates in water to produce $H^+$ ions (protons), and an Arrhenius base is a substance that dissociates in water to produce $OH^−$ (hydroxide) ions. According to this view, an acid–base reaction involves the reaction of a proton with a hydroxide ion to form water. Although Brønsted and Lowry defined an acid similarly to Arrhenius by describing an acid as any substance that can donate a proton, the Brønsted–Lowry definition of a base is much more general than the Arrhenius definition. In Brønsted–Lowry terms, a base is any substance that can accept a proton, so a base is not limited to just a hydroxide ion. This means that for every Brønsted–Lowry acid, there exists a corresponding conjugate base with one fewer proton. Consequently, all Brønsted–Lowry acid–base reactions actually involve two conjugate acid–base pairs and the transfer of a proton from one substance (the acid) to another (the base). In contrast, the Lewis definition of acids and bases, focuses on accepting or donating pairs of electrons rather than protons. A Lewis base is an electron-pair donor, and a Lewis acid is an electron-pair acceptor.
Table $1$: Definitions of Acids and Bases
Definition Acids Bases
Arrhenius $H^+$ donor $OH^−$ donor
Brønsted–Lowry $H^+$ donor $H^+$ acceptor
Lewis electron-pair acceptor electron-pair donor
Because this chapter deals with acid–base equilibria in aqueous solution, our discussion will use primarily the Brønsted–Lowry definitions and nomenclature. Remember, however, that all three definitions are just different ways of looking at the same kind of reaction: a proton is an acid, and the hydroxide ion is a base—no matter which definition you use. In practice, chemists tend to use whichever definition is most helpful to make a particular point or understand a given system. If, for example, we refer to a base as having one or more lone pairs of electrons that can accept a proton, we are simply combining the Lewis and Brønsted–Lowry definitions to emphasize the characteristic properties of a base.
Acid–Base Properties of Water
Recall that because of its highly polar structure, liquid water can act as either an acid (by donating a proton to a base) or a base (by using a lone pair of electrons to accept a proton). For example, when a strong acid such as HCl dissolves in water, it dissociates into chloride ions ($Cl^−$) and protons ($H^+$). The proton, in turn, reacts with a water molecule to form the hydronium ion ($H_3O^+$):
$\underset{acid}{HCl_{(aq)}} + \underset{base}{H_2O_{(l)}} \rightarrow \underset{acid}{H_3O^+_{(aq)}} + \underset{base}{Cl^-_{(aq)}} \label{16.3.1a}$
In this reaction, $HCl$ is the acid, and water acts as a base by accepting an $H^+$ ion. The reaction in Equation \ref{16.3.1a} is often written in a simpler form by removing $H_2O$ from each side:
$HCl_{(aq)} \rightarrow H^+_{(aq)} + Cl^-_{(aq)} \label{16.3.1b}$
In Equation \ref{16.3.1b}, the hydronium ion is represented by $H^+$, although free $H^+$ ions do not exist in liquid water as this reaction demonstrates:
$H^+_{(aq)} + H_2O_{(l)} \rightarrow H_3O^+_{(aq)} \nonumber$
Water can also act as an acid, as shown in Equation \ref{16.3.2}. In this equilibrium reaction, $H_2O$ donates a proton to $NH_3$, which acts as a base:
$\underset{acid}{H_2O_{(l)}} + \underset{base}{NH_{3(aq)}} \rightleftharpoons \underset{acid}{NH^+_{4 (aq)}} + \underset{base}{OH^-_{(aq)}} \label{16.3.2}$
Water is thus termed amphiprotic, meaning that it can behave as either an acid or a base, depending on the nature of the other reactant. Notice that Equation $\ref{16.3.2}$ is an equilibrium reaction as indicated by the double arrow and hence has an equilibrium constant associated with it.
The Ion-Product Constant of Liquid Water
Because water is amphiprotic, one water molecule can react with another to form an $OH^−$ ion and an $H_3O^+$ ion in an autoionization process:
$2H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)}+OH^−_{(aq)} \label{16.3.3}$
The equilibrium constant $K$ for this reaction can be written as follows:
$K_{a}=\dfrac{a_{H_3O^+}·a_{OH^-}}{a_{H_2O}^2} \approx \frac{[H_{3}O^{+}][HO^{-}]}{(1)^{2}}=[H_{3}O^{+}][HO^{-}] \label{16.3.4}$
where $a$ is the activity of a species. Because water is the solvent, and the solution is assumed to be dilute, the activity of the water is approximated by the activity of pure water, which is defined as having a value of 1. The activity of each solute is approximated by the molarity of the solute.
Note
It is a common error to claim that the molar concentration of the solvent is in some way involved in the equilibrium law. This error is a result of a misunderstanding of solution thermodynamics. For example, it is often claimed that Ka = Keq[H2O] for aqueous solutions. This equation is incorrect because it is an erroneous interpretation of the correct equation Ka = Keq($\textit{a}_{H_2O}$). Because $\textit{a}_{H_2O}$ = 1 for a dilute solution, Ka = Keq(1), or Ka = Keq.
In this reaction, one water molecule acts as an acid and one water molecule acts as a base. Thus, this reaction actually can be designated as the $K_a$ of water and as the $K_b$ of water. It is most common, however, to designate this reaction and the associated law of mass action as the $K_w$ of water:
$K_{w}=[H_{3}O^{+}][HO^{-}] \label{16.3.5}$
When pure liquid water is in equilibrium with hydronium and hydroxide ions at 25 °C, the concentrations of the hydronium ion and the hydroxide ion are equal:
$[H_3O^+] = [OH^−] = 1.003 \times 10^{−7}\; M \label{16.3.6}$
Thus the number of dissociated water molecules is very small indeed, approximately 2 ppb.
Substituting the values for $[H_3O^+]$ and $[OH^−]$ at 25 °C into this expression
$K_w=(1.003 \times10^{−7})(1.003 \times 10^{−7})=1.006 \times 10^{−14} \label{16.3.7}$
Thus, to three significant figures, $K_w = 1.01 \times 10^{−14}$ at room temperature, and $K_w = 1.01 \times 10^{-14} = [H_3O^+][OH^-] \label{16.3.7b}$.
Like any other equilibrium constant, $K_w$ varies with temperature, ranging from $1.15 \times 10^{−15}$ at 0 °C to $4.99 \times 10^{−13}$ at 100 °C.
In pure water, the concentrations of the hydronium ion and the hydroxide ion are equal, and the solution is therefore neutral. If $[H_3O^+] > [OH^−]$, however, the solution is acidic, whereas if $[H_3O^+] < [OH^−]$, the solution is basic. For an aqueous solution, the $H_3O^+$ concentration is a quantitative measure of acidity: the higher the $H_3O^+$ concentration, the more acidic the solution. Conversely, the higher the $OH^−$ concentration, the more basic the solution. In most situations that you will encounter, the $H_3O^+$ and $OH^−$ concentrations from the dissociation of water are so small ($1.003 \times 10^{−7} M$) that they can be ignored in calculating the $H_3O^+$ or $OH^−$ concentrations of solutions of acids and bases, but this is not always the case.
A Video Describing the Self-Ionization of Water (Kw): Self-Ionization of Water (Kw): [youtu.be]
The Relationship among pH, pOH, and $pK_w$
The pH scale is a concise way of describing the $H_3O^+$ concentration and hence the acidity or basicity of a solution. Recall that pH and the $H^+$ ($H_3O^+$) concentration are related as follows:
$pH=−\log_{10}[H^+] \label{16.3.8}$
$[H^+]=10^{−pH} \label{16.3.9}$
Because the scale is logarithmic, a pH difference of 1 between two solutions corresponds to a difference of a factor of 10 in their hydronium ion concentrations. Recall also that the pH of a neutral solution is 7.00 ($[H_3O^+] = 1.0 \times 10^{−7}\; M$), whereas acidic solutions have pH < 7.00 (corresponding to $[H_3O^+] > 1.0 \times 10^{−7}$) and basic solutions have pH > 7.00 (corresponding to $[H_3O^+] < 1.0 \times 10^{−7}$).
Similar notation systems are used to describe many other chemical quantities that contain a large negative exponent. For example, chemists use an analogous pOH scale to describe the hydroxide ion concentration of a solution. The pOH and $[OH^−]$ are related as follows:
$pOH=−\log_{10}[OH^−] \label{16.3.10}$
$[OH^−]=10^{−pOH} \label{16.3.11}$
The constant $K_w$ can also be expressed using this notation, where $pK_w = −\log\; K_w$.
Because a neutral solution has $[OH^−] = 1.0 \times 10^{−7}$, the pOH of a neutral solution is 7.00. Consequently, the sum of the pH and the pOH for a neutral solution at 25 °C is 7.00 + 7.00 = 14.00. We can show that the sum of pH and pOH is equal to 14.00 for any aqueous solution at 25 °C by taking the negative logarithm of both sides of Equation \ref{16.3.6b}:
\begin{align} −\log_{10} K_w &= pK_w \[4pt] &=−\log([H_3O^{+}][OH^{−}]) \[4pt] &= (−\log[H_3O^{+}])+(−\log[OH^{−}])\[4pt] &= pH+pOH \label{16.3.12} \end{align}
Thus at any temperature, $pH + pOH = pK_w$, so at 25 °C, where $K_w = 1.0 \times 10^{−14}$, pH + pOH = 14.00. More generally, the pH of any neutral solution is half of the $pK_w$ at that temperature. The relationship among pH, pOH, and the acidity or basicity of a solution is summarized graphically in Figure $1$ over the common pH range of 0 to 14. Notice the inverse relationship between the pH and pOH scales.
For any neutral solution, pH + pOH = 14.00 (at 25 °C) with pH=pOH=7.
A Video Introduction to pH: Introduction to pH [youtu.be]
Example $1$
The Kw for water at 100 °C is $4.99 \times 10^{−13}$. Calculate $pK_w$ for water at this temperature and the pH and the pOH for a neutral aqueous solution at 100 °C. Report pH and pOH values to two decimal places.
Given: $K_w$
Asked for: $pK_w$, $pH$, and $pOH$
Strategy:
1. Calculate $pK_w$ by taking the negative logarithm of $K_w$.
2. For a neutral aqueous solution, $[H_3O^+] = [OH^−]$. Use this relationship and Equation \ref{16.3.7b} to calculate $[H_3O^+]$ and $[OH^−]$. Then determine the pH and the pOH for the solution.
Solution:
A
Because $pK_w$ is the negative logarithm of Kw, we can write
$pK_w = −\log K_w = −\log(4.99 \times 10^{−13}) = 12.302 \nonumber$
The answer is reasonable: $K_w$ is between $10^{−13}$ and $10^{−12}$, so $pK_w$ must be between 12 and 13.
B
Equation \ref{16.3.6b} shows that $K_w = [H_3O^+][OH^−]$. Because $[H_3O^+] = [OH^−]$ in a neutral solution, we can let $x = [H_3O^+] = [OH^−]$:
\begin{align*} K_w &=[H_3O^+][OH^−] \[4pt] &=(x)(x)=x^2 \[4pt] x&=\sqrt{K_w} \[4pt] &=\sqrt{4.99 \times 10^{−13}} \[4pt] &=7.06 \times 10^{−7}\; M \end{align*} \nonumber
Because $x$ is equal to both $[H_3O^+]$ and $[OH^−]$,
\begin{align*} pH &= pOH = −\log(7.06 \times 10^{−7}) \[4pt] &= 6.15 \,\,\, \text{(to two decimal places)} \end{align*} \nonumber
We could obtain the same answer more easily (without using logarithms) by using the $pK_w$. In this case, we know that $pK_w = 12.302$, and from Equation \ref{16.3.12}, we know that $pK_w = pH + pOH$. Because $pH = pOH$ in a neutral solution, we can use Equation \ref{16.3.12} directly, setting $pH = pOH = y$. Solving to two decimal places we obtain the following:
\begin{align*} pK_w &= pH + pOH \[4pt] &= y + y \[4pt] &= 2y \[4pt] y &=\dfrac{pK_w}{2} \[4pt] &=\dfrac{12.302}{2} \[4pt] &=6.15 =pH=pOH \end{align*} \nonumber
Exercise $1$
Humans maintain an internal temperature of about 37 °C. At this temperature, $K_w = 3.55 \times 10^{−14}$. Calculate $pK_w$ and the pH and the pOH of a neutral solution at 37 °C. Report pH and pOH values to two decimal places.
Answer
• $pK_w = 13.45$
• $pH = pOH = 6.73$
Summary
Water is amphiprotic: it can act as an acid by donating a proton to a base to form the hydroxide ion, or as a base by accepting a proton from an acid to form the hydronium ion ($H_3O^+$). The autoionization of liquid water produces $OH^−$ and $H_3O^+$ ions. The equilibrium constant for this reaction is called the ion-product constant of liquid water (Kw) and is defined as $K_w = [H_3O^+][OH^−]$. At 25 °C, $K_w$ is $1.01 \times 10^{−14}$; hence $pH + pOH = pK_w = 14.00$.
• For any neutral solution, $pH + pOH = 14.00$ (at 25 °C) and $pH = 1/2 pK_w$.
• Ion-product constant of liquid water: $K_w = [H_3O^+][OH^−] \nonumber$
• Definition of $pH$: $pH = −\log10[H^+] \nonumber$ or $[H^+] = 10^{−pH} \nonumber$
• Definition of $pOH$: $pOH = −\log_{10}[OH^+] \nonumber$ or $[OH^−] = 10^{−pOH} \nonumber$
• Relationship among $pH$, $pOH$, and $pK_w$: $pK_w= pH + pOH \nonumber$ | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/16%3A_AcidBase_Equilibria/16.03%3A_The_Autoionization_of_Water.txt |
Learning Objectives
• To define the pH scale as a measure of acidity of a solution
Because of its amphoteric nature (i.e., acts as both an acid or a base), water does not always remain as $H_2O$ molecules. In fact, two water molecules react to form hydronium and hydroxide ions:
$2\, H_2O \;(l) \rightleftharpoons H_3O^+ \;(aq) + OH^− \; (aq) \label{1}$
This is also called the self-ionization of water. The concentration of $H_3O^+$ and $OH^-$ are equal in pure water because of the 1:1 stoichiometric ratio of Equation 1. The molarity of H3O+ and OH- in water are also both $1.0 \times 10^{-7} \,M$ at 25° C. Therefore, a constant of water ($K_w$) is created to show the equilibrium condition for the self-ionization of water. The product of the molarity of hydronium and hydroxide ion is always $1.0 \times 10^{-14}$.
$K_w= [H_3O^+][OH^-] = 1.0 \times 10^{-14} \label{2}$
This equations also applies to all aqueous solutions. However, $K_w$ does change at different temperatures, which affects the pH range discussed below.
$H^+$ and $H_3O^+$ is often used interchangeably to represent the hydrated proton, commonly call the hydronium ion.
The equation for water equilibrium is:
$H_2O \rightleftharpoons H^+ + OH^- \label{3}$
• If an acid ($H^+$) is added to the water, the equilibrium shifts to the left and the $OH^-$ ion concentration decreases
• If base ( $OH^-$) is added to water, the equilibrium shifts to left and the $H^+$ concentration decreases.
pH and pOH
The constant of water determines the range of the pH scale. To understand what the pKw is, it is important to understand first what the "p" means in pOH, and pH. The Danish biochemist Søren Sørenson proposed the term pH to refer to the "potential of hydrogen ion." He defined the "p" as the negative of the logarithm, -log, of [H+]. Therefore the pH is the negative logarithm of the molarity of H. The pOH is the negative logarithm of the molarity of OH- and the pKw is the negative logarithm of the constant of water. These definitions give the following equations:
$pH= -\log [H^+] \label{4a}$
$pOH= -\log [OH^-] \label{4b}$
$pK_w= -\log [K_w] \label{4c}$
At room temperature (25 °C),
$K_w =1.0 \times 10^{-14} \label{4d}$
So
$pK_w=-\log [1.0 \times 10^{-14}] \label{4e}$
Using the properties of logarithms, Equation $\ref{4e}$ can be rewritten as
$10^{-pK_w}=10^{-14}. \label{4f}$
By substituting, we see that pKw is 14. The equation also shows that each increasing unit on the scale decreases by the factor of ten on the concentration of $H^+$. For example, a pH of 1 has a molarity ten times more concentrated than a solution of pH 2.
Since
$pK_w\ = 14 \label{5a}$
$pK_w= pH + pOH = 14 \label{5b}$
Definitions
• A solution with more $OH^-$ ions than $H^+$ ion is basic; for aqueous solutions at 25°C that corresponds to pH > 7.
• A solution with more $H^+$ ions than $OH^-$ ions is acidic; for aqueous solutions at 25°C that corresponds to pH < 7.
• A solution with that same concentration of $H^+$ ions than $OH^-$ ions is neutral; for aqueous solutions at 25°C that corresponds to pH = 7.
Since the autoionization constant $K_w$ is temperature dependent, these correlations between pH values and the acidic/neutral/basic adjectives will be different at temperatures other than 25 °C. For example, the hydronium molarity of pure water at 80 °C is 4.9 × 10−7 M, which corresponds to pH and pOH values of:
$\mathrm{pH=-\log[H_3O^+]=-\log(4.9\times 10^{−7})=6.31}\label{12}$
$\mathrm{pOH=-\log[OH^-]=-\log(4.9\times 10^{−7})=6.31}\label{13}$
At this temperature, then, neutral solutions exhibit pH = pOH = 6.31, acidic solutions exhibit pH less than 6.31 and pOH greater than 6.31, whereas basic solutions exhibit pH greater than 6.31 and pOH less than 6.31. This distinction can be important when studying certain processes that occur at nonstandard temperatures, such as enzyme reactions in warm-blooded organisms. Unless otherwise noted, references to pH values are presumed to be those at standard temperature (25 °C) (Table $1$).
The pH scale is logarithmic, meaning that an increase or decrease of an integer value changes the concentration by a tenfold. For example, a pH of 3 is ten times more acidic than a pH of 4. Likewise, a pH of 3 is one hundred times more acidic than a pH of 5. Similarly a pH of 11 is ten times more basic than a pH of 10.
Figure $1$ depicts the pH scale with common solutions and where they are on the scale.
The pH scale does not have upper nor lower bounds!
It is common that the pH scale is argued to range from 0-14 or perhaps 1-14, but neither is correct. The pH range does not have an upper nor lower bound, since as defined above, the pH is an indication of concentration of H+. For example, at a pH of zero the hydronium ion concentration is one molar, while at pH 14 the hydroxide ion concentration is one molar. Typically the concentrations of H+ in water in most solutions fall between a range of 1 M (pH=0) and 10-14 M (pH=14). Hence a range of 0 to 14 provides sensible (but not absolute) "bookends" for the scale. One can go somewhat below zero and somewhat above 14 in water, because the concentrations of hydronium ions or hydroxide ions can exceed one molar.
Example $1$
If the concentration of $NaOH$ in a solution is $2.5 \times 10^{-4}\; M$, what is the concentration of $H_3O^+$?
Solution
Because
$1.0 \times 10^{-14} = [H_3O^+][OH^-] \nonumber$
to find the concentration of H3O+, solve for the [H3O+].
$\dfrac{1.0 \times 10^{-14}}{[OH^-]} = [H_3O^+] \nonumber$
$\dfrac{1.0 \times 10^{-14}}{2.5 \times 10^{-4}} = [H_3O^+] = 4.0 \times 10^{-11}\; M \nonumber$
Example $2$
1. Find the pH of a solution of 0.002 M of HCl.
2. Find the pH of a solution of 0.00005 M NaOH.
Solution
1. The equation for pH is -log [H+]
$[H^+]= 2.0 \times 10^{-3}\; M \nonumber$
$pH = -\log [2.0 \times 10^{-3}] = 2.70 \nonumber$
1. The equation for pOH is -log [OH-]
$[OH^-]= 5.0 \times 10^{-5}\; M \nonumber$
$pOH = -\log [5.0 \times 10^{-5}] = 4.30 \nonumber$
$pK_w = pH + pOH \nonumber$
and
$pH = pK_w - pOH \nonumber$
then
$pH = 14 - 4.30 = 9.70 \nonumber$
Example $3$: Soil
If moist soil has a pH of 7.84, what is the H+ concentration of the soil solution?
Solution
$pH = -\log [H^+] \nonumber$
$7.84 = -\log [H^+] \nonumber$
$[H^+] = 1.45 \times 10^{-8} M \nonumber$
Hint: Place -7.84 in your calculator and take the antilog (often inverse log or 10x) = 1.45 x 10-8M
A Video Introduction to pH: Introduction to pH [youtu.be]
Summary
The concentration of hydronium ion in a solution of an acid in water is greater than $1.0 \times 10^{-7}\; M$ at 25 °C. The concentration of hydroxide ion in a solution of a base in water is greater than $1.0 \times 10^{-7}\; M$ at 25 °C. The concentration of H3O+ in a solution can be expressed as the pH of the solution; $\ce{pH} = -\log \ce{H3O+}$. The concentration of OH can be expressed as the pOH of the solution: $\ce{pOH} = -\log[\ce{OH-}]$. In pure water, pH = 7.00 and pOH = 7.00.
Contributors and Attributions
• Emmellin Tung (UCD), Sharon Tsao (UCD), Divya Singh (UCD), Patrick Gormley (Lapeer Community School District)
• Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected]). | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/16%3A_AcidBase_Equilibria/16.04%3A_The_pH_Scale.txt |
Learning Objectives
• To know the relationship between acid or base strength and the magnitude of $K_a$, $K_b$, $pK_a$, and $pK_b$.
• To understand the leveling effect.
The magnitude of the equilibrium constant for an ionization reaction can be used to determine the relative strengths of acids and bases. For example, the general equation for the ionization of a weak acid in water, where HA is the parent acid and A− is its conjugate base, is as follows:
$HA_{(aq)}+H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)}+A^−_{(aq)} \label{16.5.1}$
The equilibrium constant for this dissociation is as follows:
$K=\dfrac{[H_3O^+][A^−]}{[H_2O][HA]} \label{16.5.2}$
As we noted earlier, because water is the solvent, it has an activity equal to 1, so the $[H_2O]$ term in Equation $\ref{16.5.2}$ is actually the $\textit{a}_{H_2O}$, which is equal to 1.
Again, for simplicity, $H_3O^+$ can be written as $H^+$ in Equation $\ref{16.5.3}$.
$HA_{(aq)} \rightleftharpoons H^+_{(aq)}+A^−_{(aq)} \label{16.5.3}$
Keep in mind, though, that free $H^+$ does not exist in aqueous solutions and that a proton is transferred to $H_2O$ in all acid ionization reactions to form hydronium ions, $H_3O^+$. The larger the $K_a$, the stronger the acid and the higher the $H^+$ concentration at equilibrium. Like all equilibrium constants, acid–base ionization constants are actually measured in terms of the activities of $H^+$ or $OH^−$, thus making them unitless. The values of $K_a$ for a number of common acids are given in Table $1$.
Table $1$: Values of $K_a$, $pK_a$, $K_b$, and $pK_b$ for Selected Acids ($HA$ and Their Conjugate Bases ($A^−$)
Acid $HA$ $K_a$ $pK_a$ $A^−$ $K_b$ $pK_b$
*The number in parentheses indicates the ionization step referred to for a polyprotic acid.
hydroiodic acid $HI$ $2 \times 10^{9}$ −9.3 $I^−$ $5.5 \times 10^{−24}$ 23.26
sulfuric acid (1)* $H_2SO_4$ $1 \times 10^{2}$ −2.0 $HSO_4^−$ $1 \times 10^{−16}$ 16.0
nitric acid $HNO_3$ $2.3 \times 10^{1}$ −1.37 $NO_3^−$ $4.3 \times 10^{−16}$ 15.37
hydronium ion $H_3O^+$ $1.0$ 0.00 $H_2O$ $1.0 \times 10^{−14}$ 14.00
sulfuric acid (2)* $HSO_4^−$ $1.0 \times 10^{−2}$ 1.99 $SO_4^{2−}$ $9.8 \times 10^{−13}$ 12.01
hydrofluoric acid $HF$ $6.3 \times 10^{−4}$ 3.20 $F^−$ $1.6 \times 10^{−11}$ 10.80
nitrous acid $HNO_2$ $5.6 \times 10^{−4}$ 3.25 $NO2^−$ $1.8 \times 10^{−11}$ 10.75
formic acid $HCO_2H$ $1.78 \times 10^{−4}$ 3.750 $HCO_2−$ $5.6 \times 10^{−11}$ 10.25
benzoic acid $C_6H_5CO_2H$ $6.3 \times 10^{−5}$ 4.20 $C_6H_5CO_2^−$ $1.6 \times 10^{−10}$ 9.80
acetic acid $CH_3CO_2H$ $1.7 \times 10^{−5}$ 4.76 $CH_3CO_2^−$ $5.8 \times 10^{−10}$ 9.24
pyridinium ion $C_5H_5NH^+$ $5.9 \times 10^{−6}$ 5.23 $C_5H_5N$ $1.7 \times 10^{−9}$ 8.77
hypochlorous acid $HOCl$ $4.0 \times 10^{−8}$ 7.40 $OCl^−$ $2.5 \times 10^{−7}$ 6.60
hydrocyanic acid $HCN$ $6.2 \times 10^{−10}$ 9.21 $CN^−$ $1.6 \times 10^{−5}$ 4.79
ammonium ion $NH_4^+$ $5.6 \times 10^{−10}$ 9.25 $NH_3$ $1.8 \times 10^{−5}$ 4.75
water $H_2O$ $1.0 \times 10^{−14}$ 14.00 $OH^−$ $1.00$ 0.00
acetylene $C_2H_2$ $1 \times 10^{−26}$ 26.0 $HC_2^−$ $1 \times 10^{12}$ −12.0
ammonia $NH_3$ $1 \times 10^{−35}$ 35.0 $NH_2^−$ $1 \times 10^{21}$ −21.0
Weak bases react with water to produce the hydroxide ion, as shown in the following general equation, where B is the parent base and BH+ is its conjugate acid:
$B_{(aq)}+H_2O_{(l)} \rightleftharpoons BH^+_{(aq)}+OH^−_{(aq)} \label{16.5.4}$
The equilibrium constant for this reaction is the base ionization constant (Kb), also called the base dissociation constant:
$K_b= \frac{[BH^+][OH^−]}{[B]} \label{16.5.5}$
Once again, the activity of water has a value of 1, so water does not appear in the equilibrium constant expression. The larger the $K_b$, the stronger the base and the higher the $OH^−$ concentration at equilibrium. The values of $K_b$ for a number of common weak bases are given in Table $2$.
Table $2$: Values of $K_b$, $pK_b$, $K_a$, and $pK_a$ for Selected Weak Bases (B) and Their Conjugate Acids (BH+)
Base $B$ $K_b$ $pK_b$ $BH^+$ $K_a$ $pK_a$
*As in Table $1$.
hydroxide ion $OH^−$ $1.0$ 0.00* $H_2O$ $1.0 \times 10^{−14}$ 14.00
phosphate ion $PO_4^{3−}$ $2.1 \times 10^{−2}$ 1.68 $HPO_4^{2−}$ $4.8 \times 10^{−13}$ 12.32
dimethylamine $(CH_3)_2NH$ $5.4 \times 10^{−4}$ 3.27 $(CH_3)_2NH_2^+$ $1.9 \times 10^{−11}$ 10.73
methylamine $CH_3NH_2$ $4.6 \times 10^{−4}$ 3.34 $CH_3NH_3^+$ $2.2 \times 10^{−11}$ 10.66
trimethylamine $(CH_3)_3N$ $6.3 \times 10^{−5}$ 4.20 $(CH_3)_3NH^+$ $1.6 \times 10^{−10}$ 9.80
ammonia $NH_3$ $1.8 \times 10^{−5}$ 4.75 $NH_4^+$ $5.6 \times 10^{−10}$ 9.25
pyridine $C_5H_5N$ $1.7 \times 10^{−9}$ 8.77 $C_5H_5NH^+$ $5.9 \times 10^{−6}$ 5.23
aniline $C_6H_5NH_2$ $7.4 \times 10^{−10}$ 9.13 $C_6H_5NH_3^+$ $1.3 \times 10^{−5}$ 4.87
water $H_2O$ $1.0 \times 10^{−14}$ 14.00 $H_3O^+$ $1.0^*$ 0.00
There is a simple relationship between the magnitude of $K_a$ for an acid and $K_b$ for its conjugate base. Consider, for example, the ionization of hydrocyanic acid ($HCN$) in water to produce an acidic solution, and the reaction of $CN^−$ with water to produce a basic solution:
$HCN_{(aq)} \rightleftharpoons H^+_{(aq)}+CN^−_{(aq)} \label{16.5.6}$
$CN^−_{(aq)}+H_2O_{(l)} \rightleftharpoons OH^−_{(aq)}+HCN_{(aq)} \label{16.5.7}$
The equilibrium constant expression for the ionization of HCN is as follows:
$K_a=\dfrac{[H^+][CN^−]}{[HCN]} \label{16.5.8}$
The corresponding expression for the reaction of cyanide with water is as follows:
$K_b=\dfrac{[OH^−][HCN]}{[CN^−]} \label{16.5.9}$
If we add Equations $\ref{16.5.6}$ and $\ref{16.5.7}$, we obtain the following:
add Equations $\ref{16.5.6}$ and $\ref{16.5.7}$, we obtain
Reaction Equilibrium Constants
$\cancel{HCN_{(aq)}} \rightleftharpoons H^+_{(aq)}+\cancel{CN^−_{(aq)}}$ $K_a=[H^+]\cancel{[CN^−]}/\cancel{[HCN]}$
$\cancel{CN^−_{(aq)}}+H_2O_{(l)} \rightleftharpoons OH^−_{(aq)}+\cancel{HCN_{(aq)}}$ $K_b=[OH^−]\cancel{[HCN]}/\cancel{[CN^−]}$
$H_2O_{(l)} \rightleftharpoons H^+_{(aq)}+OH^−_{(aq)}$ $K=K_a \times K_b=[H^+][OH^−]$
In this case, the sum of the reactions described by $K_a$ and $K_b$ is the equation for the autoionization of water, and the product of the two equilibrium constants is $K_w$:
$K_aK_b = K_w \label{16.5.10}$
Thus if we know either $K_a$ for an acid or $K_b$ for its conjugate base, we can calculate the other equilibrium constant for any conjugate acid–base pair.
Just as with $pH$, $pOH$, and pKw, we can use negative logarithms to avoid exponential notation in writing acid and base ionization constants, by defining $pK_a$ as follows:
$pKa = −\log_{10}K_a \label{16.5.11}$
$K_a=10^{−pK_a} \label{16.5.12}$
and $pK_b$ as
$pK_b = −\log_{10}K_b \label{16.5.13}$
$K_b=10^{−pK_b} \label{16.5.14}$
Similarly, Equation $\ref{16.5.10}$, which expresses the relationship between $K_a$ and $K_b$, can be written in logarithmic form as follows:
$pK_a + pK_b = pK_w \label{16.5.15}$
At 25 °C, this becomes
$pK_a + pK_b = 14.00 \label{16.5.16}$
The values of $pK_a$ and $pK_b$ are given for several common acids and bases in Tables $1$ and $2$, respectively, and a more extensive set of data is provided in Tables E1 and E2. Because of the use of negative logarithms, smaller values of $pK_a$ correspond to larger acid ionization constants and hence stronger acids. For example, nitrous acid ($HNO_2$), with a $pK_a$ of 3.25, is about a million times stronger acid than hydrocyanic acid (HCN), with a $pK_a$ of 9.21. Conversely, smaller values of $pK_b$ correspond to larger base ionization constants and hence stronger bases.
The relative strengths of some common acids and their conjugate bases are shown graphically in Figure $1$. The conjugate acid–base pairs are listed in order (from top to bottom) of increasing acid strength, which corresponds to decreasing values of $pK_a$. This order corresponds to decreasing strength of the conjugate base or increasing values of $pK_b$. At the bottom left of Figure $2$ are the common strong acids; at the top right are the most common strong bases. Notice the inverse relationship between the strength of the parent acid and the strength of the conjugate base. Thus the conjugate base of a strong acid is a very weak base, and the conjugate base of a very weak acid is a strong base.
The conjugate base of a strong acid is a weak base and vice versa.
We can use the relative strengths of acids and bases to predict the direction of an acid–base reaction by following a single rule: an acid–base equilibrium always favors the side with the weaker acid and base, as indicated by these arrows:
$\text{stronger acid + stronger base} \ce{ <=>>} \text{weaker acid + weaker base} \nonumber$
In an acid–base reaction, the proton always reacts with the stronger base.
For example, hydrochloric acid is a strong acid that ionizes essentially completely in dilute aqueous solution to produce $H_3O^+$ and $Cl^−$; only negligible amounts of $HCl$ molecules remain undissociated. Hence the ionization equilibrium lies virtually all the way to the right, as represented by a single arrow:
$HCl_{(aq)} + H_2O_{(l)} \rightarrow H_3O^+_{(aq)}+Cl^−_{(aq)} \label{16.5.17}$
In contrast, acetic acid is a weak acid, and water is a weak base. Consequently, aqueous solutions of acetic acid contain mostly acetic acid molecules in equilibrium with a small concentration of $H_3O^+$ and acetate ions, and the ionization equilibrium lies far to the left, as represented by these arrows:
$\ce{ CH_3CO_2H_{(aq)} + H_2O_{(l)} <<=> H_3O^+_{(aq)} + CH_3CO_{2(aq)}^- } \nonumber$
Similarly, in the reaction of ammonia with water, the hydroxide ion is a strong base, and ammonia is a weak base, whereas the ammonium ion is a stronger acid than water. Hence this equilibrium also lies to the left:
$H_2O_{(l)} + NH_{3(aq)} \ce{ <<=>} NH^+_{4(aq)} + OH^-_{(aq)} \nonumber$
All acid–base equilibria favor the side with the weaker acid and base. Thus the proton is bound to the stronger base.
Example $1$: Butyrate and Dimethylammonium Ions
1. Calculate $K_b$ and $pK_b$ of the butyrate ion ($CH_3CH_2CH_2CO_2^−$). The $pK_a$ of butyric acid at 25°C is 4.83. Butyric acid is responsible for the foul smell of rancid butter.
2. Calculate $K_a$ and $pK_a$ of the dimethylammonium ion ($(CH_3)_2NH_2^+$). The base ionization constant $K_b$ of dimethylamine ($(CH_3)_2NH$) is $5.4 \times 10^{−4}$ at 25°C.
Given: $pK_a$ and $K_b$
Asked for: corresponding $K_b$ and $pK_b$, $K_a$ and $pK_a$
Strategy:
The constants $K_a$ and $K_b$ are related as shown in Equation $\ref{16.5.10}$. The $pK_a$ and $pK_b$ for an acid and its conjugate base are related as shown in Equations $\ref{16.5.15}$ and $\ref{16.5.16}$. Use the relationships pK = −log K and K = 10−pK (Equations $\ref{16.5.11}$ and $\ref{16.5.13}$) to convert between $K_a$ and $pK_a$ or $K_b$ and $pK_b$.
Solution:
We are given the $pK_a$ for butyric acid and asked to calculate the $K_b$ and the $pK_b$ for its conjugate base, the butyrate ion. Because the $pK_a$ value cited is for a temperature of 25°C, we can use Equation $\ref{16.5.16}$: $pK_a$ + $pK_b$ = pKw = 14.00. Substituting the $pK_a$ and solving for the $pK_b$,
$4.83+pK_b=14.00 \nonumber$
$pK_b=14.00−4.83=9.17 \nonumber$
Because $pK_b = −\log K_b$, $K_b$ is $10^{−9.17} = 6.8 \times 10^{−10}$.
In this case, we are given $K_b$ for a base (dimethylamine) and asked to calculate $K_a$ and $pK_a$ for its conjugate acid, the dimethylammonium ion. Because the initial quantity given is $K_b$ rather than $pK_b$, we can use Equation $\ref{16.5.10}$: $K_aK_b = K_w$. Substituting the values of $K_b$ and $K_w$ at 25°C and solving for $K_a$,
$K_a(5.4 \times 10^{−4})=1.01 \times 10^{−14} \nonumber$
$K_a=1.9 \times 10^{−11} \nonumber$
Because $pK_a$ = −log $K_a$, we have $pK_a = −\log(1.9 \times 10^{−11}) = 10.72$. We could also have converted $K_b$ to $pK_b$ to obtain the same answer:
$pK_b=−\log(5.4 \times 10^{−4})=3.27 \nonumber$
$pKa+pK_b=14.00 \nonumber$
$pK_a=10.73 \nonumber$
$K_a=10^{−pK_a}=10^{−10.73}=1.9 \times 10^{−11} \nonumber$
If we are given any one of these four quantities for an acid or a base ($K_a$, $pK_a$, $K_b$, or $pK_b$), we can calculate the other three.
Exercise $1$: Lactic Acid
Lactic acid ($CH_3CH(OH)CO_2H$) is responsible for the pungent taste and smell of sour milk; it is also thought to produce soreness in fatigued muscles. Its $pK_a$ is 3.86 at 25°C. Calculate $K_a$ for lactic acid and $pK_b$ and $K_b$ for the lactate ion.
Answer
• $K_a = 1.4 \times 10^{−4}$ for lactic acid;
• $pK_b$ = 10.14 and
• $K_b = 7.2 \times 10^{−11}$ for the lactate ion
A Video Calculating pH in Strong Acid or Strong Base Solutions: Calculating pH in Strong Acid or Strong Base Solutions [youtu.be]
Solutions of Strong Acids and Bases: The Leveling Effect
You will notice in Table $1$ that acids like $H_2SO_4$ and $HNO_3$ lie above the hydronium ion, meaning that they have $pK_a$ values less than zero and are stronger acids than the $H_3O^+$ ion. Recall from Chapter 4 that the acidic proton in virtually all oxoacids is bonded to one of the oxygen atoms of the oxoanion. Thus nitric acid should properly be written as $HONO_2$. Unfortunately, however, the formulas of oxoacids are almost always written with hydrogen on the left and oxygen on the right, giving $HNO_3$ instead. In fact, all six of the common strong acids that we first encountered in Chapter 4 have $pK_a$ values less than zero, which means that they have a greater tendency to lose a proton than does the $H_3O^+$ ion. Conversely, the conjugate bases of these strong acids are weaker bases than water. Consequently, the proton-transfer equilibria for these strong acids lie far to the right, and adding any of the common strong acids to water results in an essentially stoichiometric reaction of the acid with water to form a solution of the $H_3O^+$ ion and the conjugate base of the acid.
Although $K_a$ for $HI$ is about 108 greater than $K_a$ for $HNO_3$, the reaction of either $HI$ or $HNO_3$ with water gives an essentially stoichiometric solution of $H_3O^+$ and I− or $NO_3^−$. In fact, a 0.1 M aqueous solution of any strong acid actually contains 0.1 M $H_3O^+$, regardless of the identity of the strong acid. This phenomenon is called the leveling effect: any species that is a stronger acid than the conjugate acid of water ($H_3O^+$) is leveled to the strength of $H_3O^+$ in aqueous solution because $H_3O^+$ is the strongest acid that can exist in equilibrium with water. Consequently, it is impossible to distinguish between the strengths of acids such as HI and HNO3 in aqueous solution, and an alternative approach must be used to determine their relative acid strengths.
One method is to use a solvent such as anhydrous acetic acid. Because acetic acid is a stronger acid than water, it must also be a weaker base, with a lesser tendency to accept a proton than $H_2O$. Measurements of the conductivity of 0.1 M solutions of both HI and $HNO_3$ in acetic acid show that HI is completely dissociated, but $HNO_3$ is only partially dissociated and behaves like a weak acid in this solvent. This result clearly tells us that HI is a stronger acid than $HNO_3$. The relative order of acid strengths and approximate $K_a$ and $pK_a$ values for the strong acids at the top of Table $1$ were determined using measurements like this and different nonaqueous solvents.
In aqueous solutions, $H_3O^+$ is the strongest acid and $OH^−$ is the strongest base that can exist in equilibrium with $H_2O$.
The leveling effect applies to solutions of strong bases as well: In aqueous solution, any base stronger than OH− is leveled to the strength of OH− because OH− is the strongest base that can exist in equilibrium with water. Salts such as $K_2O$, $NaOCH_3$ (sodium methoxide), and $NaNH_2$ (sodamide, or sodium amide), whose anions are the conjugate bases of species that would lie below water in Table $2$, are all strong bases that react essentially completely (and often violently) with water, accepting a proton to give a solution of $OH^−$ and the corresponding cation:
$K_2O_{(s)}+H_2O_{(l)} \rightarrow 2OH^−_{(aq)}+2K^+_{(aq)} \label{16.5.18}$
$NaOCH_{3(s)}+H_2O_{(l)} \rightarrow OH^−_{(aq)}+Na^+_{(aq)}+CH_3OH_{(aq)} \label{16.5.19}$
$NaNH_{2(s)}+H_2O_{(l)} \rightarrow OH^−_{(aq)}+Na^+_{(aq)}+NH_{3(aq)} \label{16.5.20}$
Other examples that you may encounter are potassium hydride ($KH$) and organometallic compounds such as methyl lithium ($CH_3Li$).
Polyprotic Acids and Bases
As you learned, polyprotic acids such as $H_2SO_4$, $H_3PO_4$, and $H_2CO_3$ contain more than one ionizable proton, and the protons are lost in a stepwise manner. The fully protonated species is always the strongest acid because it is easier to remove a proton from a neutral molecule than from a negatively charged ion. Thus acid strength decreases with the loss of subsequent protons, and, correspondingly, the $pK_a$ increases. Consider $H_2SO_4$, for example:
$HSO^−_{4 (aq)} \ce{ <=>>} SO^{2−}_{4(aq)}+H^+_{(aq)} \;\;\; pK_a=-2 \nonumber$
The equilibrium in the first reaction lies far to the right, consistent with $H_2SO_4$ being a strong acid. In contrast, in the second reaction, appreciable quantities of both $HSO_4^−$ and $SO_4^{2−}$ are present at equilibrium.
For a polyprotic acid, acid strength decreases and the $pK_a$ increases with the sequential loss of each proton.
The hydrogen sulfate ion ($HSO_4^−$) is both the conjugate base of $H_2SO_4$ and the conjugate acid of $SO_4^{2−}$. Just like water, HSO4− can therefore act as either an acid or a base, depending on whether the other reactant is a stronger acid or a stronger base. Conversely, the sulfate ion ($SO_4^{2−}$) is a polyprotic base that is capable of accepting two protons in a stepwise manner:
$SO^{2−}_{4 (aq)} + H_2O_{(aq)} \ce{ <=>>} HSO^{−}_{4(aq)}+OH_{(aq)}^- \nonumber$
$HSO^{−}_{4 (aq)} + H_2O_{(aq)} \ce{ <=>>} H_2SO_{4(aq)}+OH_{(aq)}^- \label{16.6}$
Like any other conjugate acid–base pair, the strengths of the conjugate acids and bases are related by $pK_a$ + $pK_b$ = pKw. Consider, for example, the $HSO_4^−/ SO_4^{2−}$ conjugate acid–base pair. From Table $1$, we see that the $pK_a$ of $HSO_4^−$ is 1.99. Hence the $pK_b$ of $SO_4^{2−}$ is 14.00 − 1.99 = 12.01. Thus sulfate is a rather weak base, whereas $OH^−$ is a strong base, so the equilibrium shown in Equation $\ref{16.6}$ lies to the left. The $HSO_4^−$ ion is also a very weak base ($pK_a$ of $H_2SO_4$ = 2.0, $pK_b$ of $HSO_4^− = 14 − (−2.0) = 16$), which is consistent with what we expect for the conjugate base of a strong acid.
Example $2$
Predict whether the equilibrium for each reaction lies to the left or the right as written.
• $NH^+_{4(aq)}+PO^{3−}_{4(aq)} \rightleftharpoons NH_{3(aq)}+HPO^{2−}_{4(aq)}$
• $CH_3CH_2CO_2H_{(aq)}+CN^−_{(aq)} \rightleftharpoons CH_3CH_2CO^−_{2(aq)}+HCN_{(aq)}$
Given: balanced chemical equation
Asked for: equilibrium position
Strategy:
Identify the conjugate acid–base pairs in each reaction. Then refer to Tables $1$and$2$ and Figure $2$ to determine which is the stronger acid and base. Equilibrium always favors the formation of the weaker acid–base pair.
Solution:
The conjugate acid–base pairs are $NH_4^+/NH_3$ and $HPO_4^{2−}/PO_4^{3−}$. According to Tables $1$ and $2$, $NH_4^+$ is a stronger acid ($pK_a = 9.25$) than $HPO_4^{2−}$ (pKa = 12.32), and $PO_4^{3−}$ is a stronger base ($pK_b = 1.68$) than $NH_3$ ($pK_b = 4.75$). The equilibrium will therefore lie to the right, favoring the formation of the weaker acid–base pair:
$\underset{\text{stronger acid}}{NH^+_{4(aq)}} + \underset{\text{stronger base}}{PO^{3-}_{4(aq)}} \ce{<=>>} \underset{\text{weaker base}}{NH_{3(aq)}} +\underset{\text{weaker acid}} {HPO^{2-}_{4(aq)}} \nonumber$
The conjugate acid–base pairs are $CH_3CH_2CO_2H/CH_3CH_2CO_2^−$ and $HCN/CN^−$. According to Table $1$, HCN is a weak acid (pKa = 9.21) and $CN^−$ is a moderately weak base (pKb = 4.79). Propionic acid ($CH_3CH_2CO_2H$) is not listed in Table $1$, however. In a situation like this, the best approach is to look for a similar compound whose acid–base properties are listed. For example, propionic acid and acetic acid are identical except for the groups attached to the carbon atom of the carboxylic acid ($\ce{−CH_2CH_3}$ versus $\ce{−CH_3}$), so we might expect the two compounds to have similar acid–base properties. In particular, we would expect the $pK_a$ of propionic acid to be similar in magnitude to the $pK_a$ of acetic acid. (In fact, the $pK_a$ of propionic acid is 4.87, compared to 4.76 for acetic acid, which makes propionic acid a slightly weaker acid than acetic acid.) Thus propionic acid should be a significantly stronger acid than $HCN$. Because the stronger acid forms the weaker conjugate base, we predict that cyanide will be a stronger base than propionate. The equilibrium will therefore lie to the right, favoring the formation of the weaker acid–base pair:
$\underset{\text{stronger acid}}{CH_3CH_2CO_2H_{(aq)}} + \underset{\text{stronger base}}{CN^-_{(aq)}} \ce{<=>>} \underset{\text{weaker base}}{CH_3CH_2CO^-_{2(aq)}} +\underset{\text{weaker acid}} {HCN_{(aq)}} \nonumber$
Exercise $1$
Predict whether the equilibrium for each reaction lies to the left or the right as written.
1. $H_2O_{(l)}+HS^−_{(aq)} \rightleftharpoons OH^−_{(aq)}+H_2S_{(aq)}$
2. $HCO^−_{2(aq)}+HSO^−_{4(aq)} \rightleftharpoons HCO_2H_{(aq)}+SO^{2−}_{4(aq)}$
Answer a
left
Answer b
left
A Video Discussing Polyprotic Acids: Polyprotic Acids [youtu.be]
Summary
Acid–base reactions always contain two conjugate acid–base pairs. Each acid and each base has an associated ionization constant that corresponds to its acid or base strength. Two species that differ by only a proton constitute a conjugate acid–base pair. The magnitude of the equilibrium constant for an ionization reaction can be used to determine the relative strengths of acids and bases. For an aqueous solution of a weak acid, the dissociation constant is called the acid ionization constant ($K_a$). Similarly, the equilibrium constant for the reaction of a weak base with water is the base ionization constant ($K_b$). For any conjugate acid–base pair, $K_aK_b = K_w$. Smaller values of $pK_a$ correspond to larger acid ionization constants and hence stronger acids. Conversely, smaller values of $pK_b$ correspond to larger base ionization constants and hence stronger bases. At 25°C, $pK_a + pK_b = 14.00$. Acid–base reactions always proceed in the direction that produces the weaker acid–base pair. No acid stronger than $H_3O^+$ and no base stronger than $OH^−$ can exist in aqueous solution, leading to the phenomenon known as the leveling effect. Polyprotic acids (and bases) lose (and gain) protons in a stepwise manner, with the fully protonated species being the strongest acid and the fully deprotonated species the strongest base.
Key Equations
• Acid ionization constant: $K_a=\dfrac{[H_3O^+][A^−]}{[HA]} \nonumber$
• Base ionization constant: $K_b= \dfrac{[BH^+][OH^−]}{[B]} \nonumber$
• Relationship between $K_a$ and $K_b$ of a conjugate acid–base pair: $K_aK_b = K_w \nonumber$
• Definition of $pK_a$: $pKa = −\log_{10}K_a \nonumber$ $K_a=10^{−pK_a} \nonumber$
• Definition of $pK_b$: $pK_b = −\log_{10}K_b \nonumber$ $K_b=10^{−pK_b} \nonumber$
• Relationship between $pK_a$ and $pK_b$ of a conjugate acid–base pair: $pK_a + pK_b = pK_w \nonumber$ $pK_a + pK_b = 14.00 \; \text{at 25°C} \nonumber$ | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/16%3A_AcidBase_Equilibria/16.05%3A_Strong_Acids_and_Bases.txt |
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