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Learning Objectives
• To introduce the concept of electron spin and demonstrated it is quantized
The process of describing each atom’s electronic structure consists, essentially, of beginning with hydrogen and adding one proton and one electron at a time to create the next heavier element in the table. All stable nuclei other than hydrogen also contain one or more neutrons. Because neutrons have no electrical charge, however, they can be ignored in the following discussion. Before demonstrating how to do this, however, we must introduce the concept of electron spin.
The Stern-Gerlach Experiment
The quantum numbers \(n, \ l, \ m\) are not sufficient to fully characterize the physical state of the electrons in an atom. In 1926, Otto Stern and Walther Gerlach carried out an experiment that could not be explained in terms of the three quantum numbers \(n, \ l, \ m\) and showed that there is, in fact, another quantum-mechanical degree of freedom that needs to be included in the theory. The experiment is illustrated in the figure 8.9.1. A beam of atoms (e.g. hydrogen or silver atoms) is sent through a spatially inhomogeneous magnetic field with a definite field gradient toward one of the poles. It is observed that the beam splits into two beams as it passes through the field region.
The fact that the beam splits into 2 beams suggests that the electrons in the atoms have a degree of freedom capable of coupling to the magnetic field. That is, an electron has an intrinsic magnetic moment \(M\) arising from a degree of freedom that has no classical analog. The magnetic moment must take on only 2 values according to the Stern-Gerlach experiment. The intrinsic property that gives rise to the magnetic moment must have some analog to a spin, \(S\); unlike position and momentum, which have clear classical analogs, spin does not.
The implication of the Stern-Gerlach experiment is that we need to include a fourth quantum number, \(m_s\) in our description of the physical state of the electron. That is, in addition to give its principle, angular, and magnetic quantum numbers, we also need to say if it is a spin-up electron or a spin-down electron.
Electron Spin
When scientists analyzed the emission and absorption spectra of the elements more closely, they saw that for elements having more than one electron, nearly all the lines in the spectra were actually pairs of very closely spaced lines. Because each line represents an energy level available to electrons in the atom, there are twice as many energy levels available as would be predicted solely based on the quantum numbers n, l, and ml. Scientists also discovered that applying a magnetic field caused the lines in the pairs to split farther apart. In 1925, two graduate students in physics in the Netherlands, George Uhlenbeck (1900–1988) and Samuel Goudsmit (1902–1978), proposed that the splittings were caused by an electron spinning about its axis, much as Earth spins about its axis. When an electrically charged object spins, it produces a magnetic moment parallel to the axis of rotation, making it behave like a magnet. Although the electron cannot be viewed solely as a particle, spinning or otherwise, it is indisputable that it does have a magnetic moment. This magnetic moment is called electron spin.
In an external magnetic field, the electron has two possible orientations (Figure \(2\)). These are described by a fourth quantum number (ms), which for any electron can have only two possible values, designated +½ (up) and −½ (down) to indicate that the two orientations are opposites; the subscript s is for spin. An electron behaves like a magnet that has one of two possible orientations, aligned either with the magnetic field or against it.
The implications of electron spin for chemistry were recognized almost immediately by an Austrian physicist, Wolfgang Pauli (1900–1958; Nobel Prize in Physics, 1945), who determined that each orbital can contain no more than two electrons. He developed the Pauli exclusion principle: No two electrons in an atom can have the same values of all four quantum numbers (n, l, ml, ms).
By giving the values of n, l, and ml, we also specify a particular orbital (e.g., 1s with n = 1, l = 0, ml = 0). Because ms has only two possible values (+½ or −½), two electrons, and only two electrons, can occupy any given orbital, one with spin up and one with spin down. With this information, we can proceed to construct the entire periodic table, which was originally based on the physical and chemical properties of the known elements.
Example \(2\)
List all the allowed combinations of the four quantum numbers (n, l, ml, ms) for electrons in a 2p orbital and predict the maximum number of electrons the 2p subshell can accommodate.
Given: orbital
Asked for: allowed quantum numbers and maximum number of electrons in orbital
Strategy:
1. List the quantum numbers (n, l, ml) that correspond to an n = 2p orbital. List all allowed combinations of (n, l, ml).
2. Build on these combinations to list all the allowed combinations of (n, l, ml, ms).
3. Add together the number of combinations to predict the maximum number of electrons the 2p subshell can accommodate.
Solution:
A For a 2p orbital, we know that n = 2, l = n − 1 = 1, and ml = −l, (−l +1),…, (l − 1), l. There are only three possible combinations of (n, l, ml): (2, 1, 1), (2, 1, 0), and (2, 1, −1).
B Because ms is independent of the other quantum numbers and can have values of only +½ and −½, there are six possible combinations of (n, l, ml, ms): (2, 1, 1, +½), (2, 1, 1, −½), (2, 1, 0, +½), (2, 1, 0, −½), (2, 1, −1, +½), and (2, 1, −1, −½).
C Hence the 2p subshell, which consists of three 2p orbitals (2px, 2py, and 2pz), can contain a total of six electrons, two in each orbital.
Exercise \(2\)
List all the allowed combinations of the four quantum numbers (n, l, ml, ms) for a 6s orbital, and predict the total number of electrons it can contain.
Answer: (6, 0, 0, +½), (6, 0, 0, −½); two electrons
Summary
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. The concept of electron spin has important consequences for chemistry because the Pauli exclusion principle implies that no orbital can contain more than two electrons (with opposite spin). | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/08%3A_Electrons_in_Atoms/8.09%3A_Electron_Spin%3A_A_Fourth_Quantum_Number.txt |
Learning Objectives
• To understand the basics of adding electrons to atomic orbitals
• To understand the basics of the Aufbau principle
The electron configuration of an element is the arrangement of its electrons in its atomic orbitals. By knowing the electron configuration of an element, we can predict and explain a great deal of its chemistry.
The Aufbau Principle
We construct the periodic table by following the aufbau principle (from German, meaning “building up”). First we determine the number of electrons in the atom; then we add electrons one at a time to the lowest-energy orbital available without violating the Pauli principle. We use the orbital energy diagram of Figure 6.29, recognizing that each orbital can hold two electrons, one with spin up ↑, corresponding to ms = +½, which is arbitrarily written first, and one with spin down ↓, corresponding to ms = −½. A filled orbital is indicated by ↑↓, in which the electron spins are said to be paired. Here is a schematic orbital diagram for a hydrogen atom in its ground state:
A neutral helium atom, with an atomic number of 2 (Z = 2), has two electrons. We place one electron in the orbital that is lowest in energy, the 1s orbital. From the Pauli exclusion principle, we know that an orbital can contain two electrons with opposite spin, so we place the second electron in the same orbital as the first but pointing down, so that the electrons are paired. The orbital diagram for the helium atom is therefore
written as 1s2, where the superscript 2 implies the pairing of spins. Otherwise, our configuration would violate the Pauli principle.
The next element is lithium, with Z = 3 and three electrons in the neutral atom. We know that the 1s orbital can hold two of the electrons with their spins paired. Figure 6.29 tells us that the next lowest energy orbital is 2s, so the orbital diagram for lithium is
This electron configuration is written as 1s22s1.
The next element is beryllium, with Z = 4 and four electrons. We fill both the 1s and 2s orbitals to achieve a 1s22s2 electron configuration:
When we reach boron, with Z = 5 and five electrons, we must place the fifth electron in one of the 2p orbitals. Because all three 2p orbitals are degenerate, it doesn’t matter which one we select. The electron configuration of boron is 1s22s22p1:
At carbon, with Z = 6 and six electrons, we are faced with a choice. Should the sixth electron be placed in the same 2p orbital that already has an electron, or should it go in one of the empty 2p orbitals? If it goes in an empty 2p orbital, will the sixth electron have its spin aligned with or be opposite to the spin of the fifth? In short, which of the following three orbital diagrams is correct for carbon, remembering that the 2p orbitals are degenerate?
Because of electron-electron repulsions, it is more favorable energetically for an electron to be in an unoccupied orbital than in one that is already occupied; hence we can eliminate choice a. Similarly, experiments have shown that choice b is slightly higher in energy (less stable) than choice c because electrons in degenerate orbitals prefer to line up with their spins parallel; thus, we can eliminate choice b. Choice c illustrates Hund’s rule (named after the German physicist Friedrich H. Hund, 1896–1997), which today says that the lowest-energy electron configuration for an atom is the one that has the maximum number of electrons with parallel spins in degenerate orbitals. By Hund’s rule, the electron configuration of carbon, which is 1s22s22p2, is understood to correspond to the orbital diagram shown in c. Experimentally, it is found that the ground state of a neutral carbon atom does indeed contain two unpaired electrons.
When we get to nitrogen (Z = 7, with seven electrons), Hund’s rule tells us that the lowest-energy arrangement is
with three unpaired electrons. The electron configuration of nitrogen is thus 1s22s22p3.
At oxygen, with Z = 8 and eight electrons, we have no choice. One electron must be paired with another in one of the 2p orbitals, which gives us two unpaired electrons and a 1s22s22p4 electron configuration. Because all the 2p orbitals are degenerate, it doesn’t matter which one has the pair of electrons.
Similarly, fluorine has the electron configuration 1s22s22p5:
When we reach neon, with Z = 10, we have filled the 2p subshell, giving a 1s22s22p6 electron configuration:
Notice that for neon, as for helium, all the orbitals through the 2p level are completely filled. This fact is very important in dictating both the chemical reactivity and the bonding of helium and neon, as you will see.
Electron Configuration of Atoms: https://youtu.be/LlY-O3-bfnk
Valence Electrons
As we continue through the periodic table in this way, writing the electron configurations of larger and larger atoms, it becomes tedious to keep copying the configurations of the filled inner subshells. In practice, chemists simplify the notation by using a bracketed noble gas symbol to represent the configuration of the noble gas from the preceding row because all the orbitals in a noble gas are filled. For example, [Ne] represents the 1s22s22p6 electron configuration of neon (Z = 10), so the electron configuration of sodium, with Z = 11, which is 1s22s22p63s1, is written as [Ne]3s1:
Neon Z = 10 1s22s22p6
Sodium Z = 11 1s22s22p63s1 = [Ne]3s1
Because electrons in filled inner orbitals are closer to the nucleus and more tightly bound to it, they are rarely involved in chemical reactions. This means that the chemistry of an atom depends mostly on the electrons in its outermost shell, which are called the valence electrons. The simplified notation allows us to see the valence-electron configuration more easily. Using this notation to compare the electron configurations of sodium and lithium, we have:
Sodium 1s22s22p63s1 = [Ne]3s1
Lithium 1s22s1 = [He]2s1
It is readily apparent that both sodium and lithium have one s electron in their valence shell. We would therefore predict that sodium and lithium have very similar chemistry, which is indeed the case.
As we continue to build the eight elements of period 3, the 3s and 3p orbitals are filled, one electron at a time. This row concludes with the noble gas argon, which has the electron configuration [Ne]3s23p6, corresponding to a filled valence shell.
Example \(2\)
Draw an orbital diagram and use it to derive the electron configuration of phosphorus, Z = 15. What is its valence electron configuration?
Given: atomic number
Asked for: orbital diagram and valence electron configuration for phosphorus
Strategy:
1. Locate the nearest noble gas preceding phosphorus in the periodic table. Then subtract its number of electrons from those in phosphorus to obtain the number of valence electrons in phosphorus.
2. Referring to Figure 6.29, draw an orbital diagram to represent those valence orbitals. Following Hund’s rule, place the valence electrons in the available orbitals, beginning with the orbital that is lowest in energy. Write the electron configuration from your orbital diagram.
3. Ignore the inner orbitals (those that correspond to the electron configuration of the nearest noble gas) and write the valence electron configuration for phosphorus.
Solution:
A Because phosphorus is in the third row of the periodic table, we know that it has a [Ne] closed shell with 10 electrons. We begin by subtracting 10 electrons from the 15 in phosphorus.
B The additional five electrons are placed in the next available orbitals, which Figure 6.29 tells us are the 3s and 3p orbitals:
Because the 3s orbital is lower in energy than the 3p orbitals, we fill it first:
Hund’s rule tells us that the remaining three electrons will occupy the degenerate 3p orbitals separately but with their spins aligned:
The electron configuration is [Ne]3s23p3.
C We obtain the valence electron configuration by ignoring the inner orbitals, which for phosphorus means that we ignore the [Ne] closed shell. This gives a valence-electron configuration of 3s23p3.
Exercise \(2\)
Draw an orbital diagram and use it to derive the electron configuration of chlorine, Z = 17. What is its valence electron configuration?
Answer: [Ne]3s23p5; 3s23p5
Definition of Valence Electrons: https://youtu.be/_ldxOYwM2VM
The general order in which orbitals are filled is depicted in Figure \(1\). Subshells corresponding to each value of n are written from left to right on successive horizontal lines, where each row represents a row in the periodic table. The order in which the orbitals are filled is indicated by the diagonal lines running from the upper right to the lower left. Accordingly, the 4s orbital is filled prior to the 3d orbital because of shielding and penetration effects. Consequently, the electron configuration of potassium, which begins the fourth period, is [Ar]4s1, and the configuration of calcium is [Ar]4s2. Five 3d orbitals are filled by the next 10 elements, the transition metals, followed by three 4p orbitals. Notice that the last member of this row is the noble gas krypton (Z = 36), [Ar]4s23d104p6 = [Kr], which has filled 4s, 3d, and 4p orbitals. The fifth row of the periodic table is essentially the same as the fourth, except that the 5s, 4d, and 5p orbitals are filled sequentially.
The sixth row of the periodic table will be different from the preceding two because the 4f orbitals, which can hold 14 electrons, are filled between the 6s and the 5d orbitals. The elements that contain 4f orbitals in their valence shell are the lanthanides. When the 6p orbitals are finally filled, we have reached the next (and last known) noble gas, radon (Z = 86), [Xe]6s24f145d106p6 = [Rn]. In the last row, the 5f orbitals are filled between the 7s and the 6d orbitals, which gives the 14 actinide elements. Because the large number of protons makes their nuclei unstable, all the actinides are radioactive.
Example \(3\)
Write the electron configuration of mercury (Z = 80), showing all the inner orbitals.
Given: atomic number
Asked for: complete electron configuration
Strategy:
Using the orbital diagram in Figure \(1\) and the periodic table as a guide, fill the orbitals until all 80 electrons have been placed.
Solution:
By placing the electrons in orbitals following the order shown in Figure \(1\) and using the periodic table as a guide, we obtain
1s2 row 1 2 electrons
2s22p6 row 2 8 electrons
3s23p6 row 3 8 electrons
4s23d104p6 row 4 18 electrons
5s24d105p6 row 5 18 electrons
row 1–5 54 electrons
After filling the first five rows, we still have 80 − 54 = 26 more electrons to accommodate. According to Figure \(2\), we need to fill the 6s (2 electrons), 4f (14 electrons), and 5d (10 electrons) orbitals. The result is mercury’s electron configuration:
1s22s22p63s23p64s23d104p65s24d105p66s24f145d10 = Hg = [Xe]6s24f145d10
with a filled 5d subshell, a 6s24f145d10 valence shell configuration, and a total of 80 electrons. (You should always check to be sure that the total number of electrons equals the atomic number.)
Exercise \(3\)
Although element 114 is not stable enough to occur in nature, two isotopes of element 114 were created for the first time in a nuclear reactor in 1999 by a team of Russian and American scientists. Write the complete electron configuration for element 114.
Answer: 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p67s25f146d107p2
The electron configurations of the elements are presented in Figure \(3\), which lists the orbitals in the order in which they are filled. In several cases, the ground state electron configurations are different from those predicted by Figure \(1\). Some of these anomalies occur as the 3d orbitals are filled. For example, the observed ground state electron configuration of chromium is [Ar]4s13d5 rather than the predicted [Ar]4s23d4. Similarly, the observed electron configuration of copper is [Ar]4s13d10 instead of [Ar]s23d9. The actual electron configuration may be rationalized in terms of an added stability associated with a half-filled (ns1, np3, nd5, nf7) or filled (ns2, np6, nd10, nf14) subshell. Given the small differences between higher energy levels, this added stability is enough to shift an electron from one orbital to another. In heavier elements, other more complex effects can also be important, leading to some of the additional anomalies indicated in Figure \(3\). For example, cerium has an electron configuration of [Xe]6s24f15d1, which is impossible to rationalize in simple terms. In most cases, however, these apparent anomalies do not have important chemical consequences.
Note
Additional stability is associated with half-filled or filled subshells.
Summary
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 their spins parallel. For chemical purposes, the most important electrons are those in the outermost principal shell, the valence electrons. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/08%3A_Electrons_in_Atoms/8.10%3A_Multielectron_Atoms.txt |
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 $1$ ). (The noble gases were still unknown.) Newlands therefore suggested that the elements could be classified into octaves A group of seven elements, corresponding to the horizontal rows in the main group elements (not counting the noble gases, which were unknown at the time)., corresponding to the horizontal rows in the main group elements. 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.
Note: 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.
Note: 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.
Mendeleev’s Periodic Table
Mendeleev, who first published his periodic table in 1869 (Figure $3$ ), is usually credited with the origin of the modern periodic table. The key difference between his arrangement of the elements and that of Meyer and others is that Mendeleev did not assume that all the elements had been discovered (actually, only about two-thirds of the naturally occurring elements were known at the time). Instead, he deliberately left blanks in his table at atomic masses 44, 68, 72, and 100, in the expectation that elements with those atomic masses would be discovered. Those blanks correspond to the elements we now know as scandium, gallium, germanium, and technetium.
The groups in Mendeleev's table are determined by how many oxygen or hydrogen atoms are needed to form compounds with each element. For example, in Group I, two atoms of hydrogen, lithium, Li, sodium, Na, and potassium form compounds with one atom of oxygen. In Group VII, one atom of fluorine, F, chlorine, Cl, and bromine, Br, react with one atom of hydrogen. Notice how this approach has trouble with the transition metals. Until roughly 1960, a rectangular table developed from Mendeleev's table and based on reactivity was standard at the front of chemistry lecture halls.
The most convincing evidence in support of Mendeleev’s arrangement of the elements was the discovery of two previously unknown elements whose properties closely corresponded with his predictions (Table $1$). Two of the blanks Mendeleev had left in his original table were below aluminum and silicon, awaiting the discovery of two as-yet-unknown elements, eka-aluminum and eka-silicon (from the Sanskrit eka, meaning “one,” as in “one beyond aluminum”). The observed properties of gallium and germanium matched those of eka-aluminum and eka-silicon so well that once they were discovered, Mendeleev’s periodic table rapidly gained acceptance.
Video $1$: The genius of Mendeleev's periodic table.
When the chemical properties of an element suggested that it might have been assigned the wrong place in earlier tables, Mendeleev carefully reexamined its atomic mass. He discovered, for example, that the atomic masses previously reported for beryllium, indium, and uranium were incorrect. The atomic mass of indium had originally been reported as 75.6, based on an assumed stoichiometry of InO for its oxide. If this atomic mass were correct, then indium would have to be placed in the middle of the nonmetals, between arsenic (atomic mass 75) and selenium (atomic mass 78). Because elemental indium is a silvery-white metal, however, Mendeleev postulated that the stoichiometry of its oxide was really In2O3 rather than InO. This would mean that indium’s atomic mass was actually 113, placing the element between two other metals, cadmium and tin.
Table $1$: Comparison of the Properties Predicted by Mendeleev in 1869 for eka-Aluminum and eka-Silicon with the Properties of Gallium (Discovered in 1875) and Germanium (Discovered in 1886)
Property eka-Aluminum (predicted) Gallium (observed) eka-Silicon (predicted) Germanium (observed)
atomic mass 68 69.723 72 72.64
element metal metal dirty-gray metal gray-white metal
low mp* mp = 29.8°C high mp mp = 938°C
$\rho$ = 5.9 g/cm3 $\rho$ = 5.91 g/cm3 $\rho$ = 5.5 g/cm3 $\rho$ = 5.323 g/cm3
oxide E2O3 Ga2O3 EO2 GeO2
$\rho$ = 5.5 g/cm3 $\rho$ = 6.0 g/cm3 $\rho$ = 4.7 g/cm3 $\rho$ = 4.25 g/cm3
chloride ECl3 GaCl3 ECl4 GeCl4
volatile
mp = 78°C
bp* = 201°C
bp < 100°C bp = 87°C
*mp = melting point; bp = boiling point.
One group of elements that absent from Mendeleev’s table is the noble gases, all of which were discovered more than 20 years later, between 1894 and 1898, by Sir William Ramsay (1852–1916; Nobel Prize in Chemistry 1904). Initially, Ramsay did not know where to place these elements in the periodic table. Argon, the first to be discovered, had an atomic mass of 40. This was greater than chlorine’s and comparable to that of potassium, so Ramsay, using the same kind of reasoning as Mendeleev, decided to place the noble gases between the halogens and the alkali metals.
The Role of the Atomic Number in the Periodic Table
Despite its usefulness, Mendeleev’s periodic table was based entirely on empirical observation supported by very little understanding. It was not until 1913, when a young British physicist, H. G. J. Moseley (1887–1915), while analyzing the frequencies of x-rays emitted by the elements, discovered that the underlying foundation of the order of the elements was by the atomic number, not the atomic mass. Moseley hypothesized that the placement of each element in his series corresponded to its atomic number Z, which is the number of positive charges (protons) in its nucleus. Argon, for example, although having an atomic mass greater than that of potassium (39.9 amu versus 39.1 amu, respectively), was placed before potassium in the periodic table. While analyzing the frequencies of the emitted x-rays, Moseley noticed that the atomic number of argon is 18, whereas that of potassium is 19, which indicated that they were indeed placed correctly. Moseley also noticed three gaps in his table of x-ray frequencies, so he predicted the existence of three unknown elements: technetium (Z = 43), discovered in 1937; promethium (Z = 61), discovered in 1945; and rhenium (Z = 75), discovered in 1925.
Note: H. G. J. Moseley (1887–1915)
Moseley left his research work at the University of Oxford to join the British army as a telecommunications officer during World War I. He was killed during the Battle of Gallipoli in Turkey.
Example $1$
Before its discovery in 1999, some theoreticians believed that an element with a Z of 114 existed in nature. Use Mendeleev’s reasoning to name element 114 as eka-______; then identify the known element whose chemistry you predict would be most similar to that of element 114.
Given: atomic number
Asked for: name using prefix eka-
Strategy:
1. Using the periodic table locate the n = 7 row. Identify the location of the unknown element with Z = 114; then identify the known element that is directly above this location.
2. Name the unknown element by using the prefix eka- before the name of the known element.
Solution:
A The n = 7 row can be filled in by assuming the existence of elements with atomic numbers greater than 112, which is underneath mercury (Hg). Counting three boxes to the right gives element 114, which lies directly below lead (Pb). B If Mendeleev were alive today, he would call element 114 eka-lead.
Exercise $1$
Use Mendeleev’s reasoning to name element 112 as eka-______; then identify the known element whose chemistry you predict would be most similar to that of element 112.
Answer: eka-mercury
Summary
• The elements in the periodic table are arranged according to their properties, and the periodic table serves as an aid in predicting chemical behavior.
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; early models using limited data noted the existence of triads and octaves of elements with similar properties. The periodic table achieved its current form through the work of Dimitri Mendeleev and Julius Lothar Meyer, who both focused on the relationship between atomic mass and chemical properties. Meyer arranged the elements by their atomic volume, which today is equivalent to the molar volume, defined as molar mass divided by molar density. The correlation with the electronic structure of atoms was made when H. G. J. Moseley showed that the periodic arrangement of the elements was determined by atomic number, not atomic mass.
Contributors and Attributions
Modified by Joshua Halpern (Howard University) | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/09%3A_The_Periodic_Table_and_Some_Atomic_Properties/9.1%3A_Classifying_the_Elements%3A_The_Periodic_Law_and_the_Periodic_Table.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
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_General_Chemistry_(Petrucci_et_al.)/09%3A_The_Periodic_Table_and_Some_Atomic_Properties/9.2%3A_Metals_and_Nonmetals_and_their_Ions.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. 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 (rcov), 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 (part (a) in Figure \(2\)). 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 (rmet) is defined as half the distance between the nuclei of two adjacent metal atoms (part (b) in Figure \(2\)). 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 (rvdW), which is half the internuclear distance between two nonbonded atoms in the solid (part (c) in Figure \(2\)). 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 \(Cl_2\) molecule) and a van der Waals radius (the distance between two Cl atoms in different molecules in, for example, \(Cl_{2(s)}\) at low temperatures). These radii are generally not the same (part (d) in Figure \(2\)).
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 \(4\)).
Note
Atomic radii decrease from left to right across a row and increase from top to bottom down a column.
Trends in atomic size result from differences in the effective nuclear charges (Zeff) 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.
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.
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.
Note
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: https://youtu.be/ZYKB8SNrGVY
Ionic Radii and Isoelectronic Series
An ion is formed when either one or more electrons are removed from a neutral atom (cations) to form a positive ion or when additional electrons attach themselves to neutral atoms (anions) to form a negative one. 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 \(7\)) 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 (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 (compare F at 42 pm with F at 133 pm).
Note
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 7.3
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.
Table 9.3.3 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.
Ion Radius (pm) Atomic Number
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.
Contributors and Attributions
Modified by Joshua Halpern (Howard University) | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/09%3A_The_Periodic_Table_and_Some_Atomic_Properties/9.3%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. 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
$E_{(g)} \rightarrow E^+_{(g)} +e^- \;\;\ \text{energy required=I } \label{9.4.1}$
Because an input of energy is required, the ionization energy is always positive ($I > 0$) for the reaction as written in Equation 9.4.1. 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):
$1\; eV/atom = 96.49\; kJ/mol$
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 (I1), a second ionization energy (I2), and in general an nth ionization energy (In) according to the following reactions:
$E_{(g)} \rightarrow E^+_{(g)} +e^- \;\;\ I_1=\text{1st ionization energy} \label{9.4.2}$
$E_{(g)} \rightarrow E^+_{(g)} +e^- \;\;\ I_2=\text{2nd ionization energy} \label{9.4.3}$
$E^+_{(g)} \rightarrow E^{2+}_{(g)} +e^- \;\;\ I_3=\text{3rd ionization energy} \label{9.4.4}$
Values for the ionization energies of $Li$ and $Be$ listed in Table $1$ show that successive ionization energies for an element increase steadily; 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.
Note
Successive ionization energies for an element increase steadily.
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 $I$
$Li_{(g)}\rightarrow Li^+_{(g)} + e^-$ $1s^22s^1 \rightarrow 1s^2$ I1 = 520.2 $Be_{(g)} \rightarrow Be^+_{(g)}+e^-$ $1s^22s^2 \rightarrow 1s^22s^1$ I1 = 899.5
$Li^+_{(g)} \rightarrow Li^{2+}_{(g)} +e^-$ $1s^2 \rightarrow 1s^1$ I2 = 7298.2 $Be^+_{(g)} \rightarrow Be^{2+}_{(g)} +e^-$ $1s^22s^1 \rightarrow 1s^2$ I2 = 1757.1
$Li^{2+}_{(g)} \rightarrow Li^{3+}_{(g)} + e^-$ $1s^1 \rightarrow 1s^0$ I3 = 11,815.0 $Be^{2+}_{(g)} \rightarrow Be^{3+}_{(g)}+e^-$ $1s^2 \rightarrow 1s^1$ I3 = 14,848.8
$Be^{3+}_{(g)} \rightarrow Be^{4+}_{(g)}+e^-$ $1s^1 \rightarrow 1s^0$ I4 = 21,006.6
The most important consequence of the values listed in Table $1$ is that the chemistry of $Li$ is dominated by the $Li^+$ ion, while the chemistry of $Be$ is dominated by the +2 oxidation state. The energy required to remove the second electron from $Li$
$Li^+_{(g)} \rightarrow Li^{2+}_{(g)} + e^− \label{9.4.5}$
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 $Be$
$Be^{2+}_{(g)} \rightarrow Be^{3+}_{(g)} + e^− \label{9.4.6}$
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 $Li^+$ and $Be^{2+}$ 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.
Note
The energy required to remove electrons from a filled core is prohibitively large under normal reaction conditions.
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 579.4.4 1816.7 2744.8 11,579.4.4
Si 786.5 1577.1 3231.6 4355.5 16,090.6
P 1011.8 1909.4.4 2914.1 4963.6 6274.0 21,269.4.3
S 999.6 2251.8 3357 4556.2 7004.3 8495.8 27,109.4.3
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:
$E^{3+}_{(g)} \rightarrow E^{4+}_{(g)} + e^-$
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: 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 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. 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.
Figure $2$: First Ionization Energies of the s-, p-, d-, and f-Block 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) electron configurations.
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 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).
Note
Generally, $I_1$ increases diagonally from the lower left of the periodic table to the upper right.
The darkness of the shading inside the cells of the table indicates the relative magnitudes of the ionization energies. Elements in gray have undetermined first ionization energies. Source: Data from CRC Handbook of Chemistry and Physics (2004).
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, screening 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: As
Ionization Energy: https://youtu.be/k7j-u02ifzo
Summary
• 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 tendency of an element to lose 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_General_Chemistry_(Petrucci_et_al.)/09%3A_The_Periodic_Table_and_Some_Atomic_Properties/9.4%3A_Ionization_Energy.txt |
Learning Objectives
• To master the concept of electron affinity as a measure of the energy required to adding 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:
$E_{(g)}+e^- \rightarrow E^-_{(g)} \;\;\; \text{energy change=}EA \label{7.5.1}$
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.
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:
$Cl_{(g)}+e^- \rightarrow Cl^-_{(g)} \;\;\; EA=-346\; kJ/mol \label{7.5.2}$
In contrast, beryllium does not form a stable anion, so its effective electron affinity is
$Be_{(g)}+e^- \rightarrow Be^-_{(g)} \;\;\; EA \ge 0 \label{7.5.3}$
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:
$N_{(g)}+e^- \rightarrow N^-_{(g)} \;\;\; EA \approx 0 \label{7.5.4}$
Note
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.
There are many more exceptions to the trends across rows and down columns than with first ionization energies. Elements that do not form stable ions, such as the noble gases, are assigned an effective electron affinity that is greater than or equal to zero. Elements for which no data are available are shown in gray. Source: Data from Journal of Physical and Chemical Reference Data 28, no. 6 (1999).
Note
In general, electron affinities become more negative across a row and less negative down a column.
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.
Note
While first electron affinities can be negative, positive, or zero, second electron affinities are always positive.
If energy is required to form both positively charged ions 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
Electron Affinity: https://youtu.be/M5MsRM2Xtds
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_General_Chemistry_(Petrucci_et_al.)/09%3A_The_Periodic_Table_and_Some_Atomic_Properties/9.5%3A_Electron_Affinity.txt |
Learning Objectives
• To predict the magnetic properties of atoms and molecules based on their electronic configurations.
The magnetic moment of a system measures the strength and the direction of its magnetism. The term itself usually refers to the magnetic dipole moment. Anything that is magnetic, like a bar magnet or a loop of electric current, has a magnetic moment. A magnetic moment is a vector quantity, with a magnitude and a direction. An electron has an electron magnetic dipole moment, generated by the electron's intrinsic spin property, making it an electric charge in motion. There are many different magnetic forms: including paramagnetism, and diamagnetism, ferromagnetism, and anti-ferromagnetism. Only paramagnetism, and diamagnetism are discussed here.
Paramagnetism
Paramagnetism refers to the magnetic state of an atom with one or more unpaired electrons. The unpaired electrons are attracted by a magnetic field due to the electrons' magnetic dipole moments. Hund's Rule states that electrons must occupy every orbital singly before any orbital is doubly occupied. This may leave the atom with many unpaired electrons. Because unpaired electrons can orient in either direction, they exhibit magnetic moments that can align with a magnet. This capability allows paramagnetic atoms to be attracted to magnetic fields. Diatomic oxygen, \(O_2\) is a good example of paramagnetism (that is best understood with molecular orbital theory). The following video shows liquid oxygen attracted into a magnetic field created by a strong magnet:
Video \(1\): A chemical demonstration of the paramagnetism of molecular oxygen, as shown by the attraction of liquid oxygen to magnets.
As shown in Video \(1\), since molecular oxygen (\(O_2\) is paramagnetic, it is attracted to the magnet. In contrast, molecular nitrogen, \(N_2\), has no unpaired electrons and it is diamagnetic (discussed below); it is therefore unaffected by the magnet.
Note
Paramagnetism is a form of magnetism whereby materials are attracted by an externally applied magnetic field.
There are some exceptions to the paramagnetism rule; these concern some transition metals, in which the unpaired electron is not in a d-orbital. Examples of these metals include \(Sc^{3+}\), \(Ti^{4+}\), \(Zn^{2+}\), and \(Cu^+\). These metals are the not defined as paramagnetic: they are considered diamagnetic because all d-electrons are paired. Paramagnetic compounds sometimes display bulk magnetic properties due to the clustering of the metal atoms. This phenomenon is known as ferromagnetism, but this property is not discussed here.
Diamagnetism
Diamagnetic substances are characterized by paired electrons—except in the previously-discussed case of transition metals, there are no unpaired electrons. According to the Pauli Exclusion Principle which states that no two identical electrons may take up the same quantum state at the same time, the electron spins are oriented in opposite directions. This causes the magnetic fields of the electrons to cancel out; thus there is no net magnetic moment, and the atom cannot be attracted into a magnetic field. In fact, diamagnetic substances are weakly repelled by a magnetic field as demonstrated with the pyrolytic carbon sheet in Figure \(1\).
Note
Diamagnetic materials are repelled by the applied magnetic field.
Diamagnetism, to a greater or lesser degree, is a property of all materials and always makes a weak contribution to the material's response to a magnetic field. For materials that show some other form of magnetism (such paramagntism), the diamagnetic contribution becomes negligible.
How to tell if a substance is paramagnetic or diamagnetic
The magnetic form of a substance can be determined by examining its electron configuration: if it shows unpaired electrons, then the substance is paramagnetic; if all electrons are paired, the substance is diamagnetic. This process can be broken into four steps:
1. Find the electron configuration
2. Draw the valence orbitals
3. Look for unpaired electrons
4. Determine whether the substance is paramagnetic (one or more electrons unpaired) or diamagnetic (all electrons paired)
Determining Magnetic Properties from Orbital Diagrams: https://youtu.be/lun_w5VKD8k
Example \(1\): Chlorine atoms
Are chlorine atoms paramagnetic or diamagnetic?
Solution
Follow the four steps outlines above.
Step 1: Find the electron configuration
For Cl atoms, the electron configuration is 3s23p5
Step 2: Draw the valence orbitals
Ignore the core electrons and focus on the valence electrons only.
Step 3: Look for unpaired electrons
There is one unpaired electron.
Step 4: Determine whether the substance is paramagnetic or diamagnetic
Since there is an unpaired electron, \(Cl\) atoms are paramagnetic (but weakly since only one electron is unpaired).
Exercise \(1\)
Indicate whether boron atoms are paramagnetic or diamagnetic.
Answer:
The B atom has 2s22p1 as the electron configuration. Because it has one unpaired electron, it is paramagnetic.
Example \(2\): Zinc Atoms
Are zinc atoms paramagnetic or diamagnetic?
Solution
Step 1: Find the electron configuration
For Zn atoms, the electron configuration is 4s23d10
Step 2: Draw the valence orbitals
Step 3: Look for unpaired electrons
There are no unpaired electrons.
Step 4: Determine whether the substance is paramagnetic or diamagnetic
Because there are no unpaired electrons, Zn atoms are diamagnetic.
Exercise \(2\)
Indicate whether F- ions are paramagnetic or diamagnetic.
Answer
The F- ion has 2s22p6 has the electron configuration. Because it has no unpaired electrons, it is diamagnetic.
Problems
1. How many unpaired electrons are found in oxygen atoms ?
2. How many unpaired electrons are found in bromine atoms?
3. Indicate whether Fe2+ ions are paramagnetic or diamagnetic.
Answers
1. The O atom has 2s22p4 as the electron configuration. Therefore, O has 2 unpaired electrons.
2. The Br atom has 4s23d104p5 as the electron configuration. Therefore, Br has 1 unpaired electron.
3. The Fe2+ ion has 3d6 has the electron configuration. Because it has 4 unpaired electrons, it is paramagnetic. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/09%3A_The_Periodic_Table_and_Some_Atomic_Properties/9.6%3A_Magnetic_Properties.txt |
We defined a chemical bond as the force that holds atoms together in a chemical compound. We also introduced two idealized types of bonding: covalent bonding, in which electrons are shared between atoms in a molecule or polyatomic ion, and ionic bonding, in which positively and negatively charged ions are held together by electrostatic forces. The concepts of covalent and ionic bonding were developed to explain the properties of different kinds of chemical substances.
10: Chemical Bonding I: Basic Concepts
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.
Note
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 \(2\). 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.
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., \(H_2\)
2. For most atoms there will be a maximum of eight electrons in the valence shell (octet structure), e.g., \(CH_4\)
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.
Note
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.
Note
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.
Lewis Theory of Bonding: https://youtu.be/TZ6C5_k-SPs
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. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/10%3A_Chemical_Bonding_I%3A_Basic_Concepts/10.1%3A_Lewis_Theory%3A_An_Overview.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
Note
A chemical bond formed by sharing a pair of electrons is called a covalent bond
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 (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 Figure $2$ and Figure 8.1, 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.
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 (the octet rule). 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:
Element
Group
Valence electrons
Bonds needed to form valence octet
F
17
7
1
O
16
6
2
N
15
5
3
C
14
4
4
Lewis Structure of Molecules: https://youtu.be/xWiFCqA9Ur0
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); http://cccbdb.nist.gov.
Note
As a general rule, the distance between bonded atoms decreases as the number of shared electron pairs increases
Bond Lengths: https://youtu.be/9xn04FNkq9I
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 $2$: 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 $1$).
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 $1$. 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$\ce{Cl\bond{-}Cl > Br\bond{-}Br > F\bond{-}F > I–I}$ Similar effects are also seen for the O–O versus S–S and for N–N versus P–P single bonds.
Note
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.
Note
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 $3$); 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) 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 $2$ to estimate the $ΔH_{rxn}$ per mole of RDX.
Given: chemical reaction, structure of reactant, and Table $2$.
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)
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
$ΔH_{rxn} \approx \sum{\text{(bond energies of bonds broken)}}−\sum{\text{(bond energies of bonds formed)}}$
$= 7962 \; kJ/mol − 10,374 \; kJ/mol$
$=−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 Bond Energies: https://youtu.be/Prc6fbLXi5M Summary • The strength of a covalent bond depends on the overlap between the valence orbitals of the bonded atoms. Formal charge on an atom: \[ ΔH_{rxn} \approx \sum{\text{(bond energies of bonds broken)}}−\sum{\text{(bond energies of bonds formed)}} \label{8.3.1}$
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_General_Chemistry_(Petrucci_et_al.)/10%3A_Chemical_Bonding_I%3A_Basic_Concepts/10.2%3A_Covalent_Bonding%3A_An_Introduction.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 shared electron pairs 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. We 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 (compare Figure $2$ and Figure $2$). 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.
Note
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 that is will be highly electronegative has:
• a large (negative) electron affinity
• a high ionization energy (always endothermic, or positive for neutral atoms)
and will
• attract electrons from other atoms
• 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 Figure $1$ and Figure $2$. 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)
Pauling won two Nobel Prizes, one for chemistry in 1954 and one for peace in 1962. 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 denied a high school degree because of his refusal to take a civics class.
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 electronegativity scale).
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) 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 semimetals (or metalloids), 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.
Note
Electronegativity values increase from lower left to upper right in the periodic table.
The rules for assigning oxidation states 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 $1$
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 semimetal.
Given: four elements
Asked for: order by increasing electronegativity and classification
<p">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 semimetal according to its location about the diagonal belt of semimetals running from B to At.
Solution:
A Electronegativity increases from lower left to upper right in the periodic table (Figure $2$). 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 semimetals 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 semimetal, 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 semimetal.
Answer: Rb < Zr < Ge < N < O; metals (Rb, Zr); semimetal (Ge); nonmetal (N, O)
Electronegativity: https://youtu.be/3Pe0iShCdhM
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:
$\begin{matrix} _{less\; electronegative}& & _{more\; electronegative}\ A\; \; &-& B\; \; \; \; \ ^{\delta ^{+}} & & ^{\delta ^{-}} \end{matrix} \label{10.3.1}$
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: 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 (see Figure $2$). 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
$\begin{matrix} _{\delta ^{+}}& & _{\delta ^{-}}\ H\; \; &-& Cl \end{matrix}$
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.
Note
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 as a Measure of Bond Polarity
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:
$\mu=Qr \label{10.3.2}$
where Q is measured in coulombs (C) and r in meters. The unit for dipole moments is the debye (D):
$1\; D = 3.3356\times 10^{-30}\; C\cdot ·m \label{10.3.3}$
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 HCl using Equation 10.3.2 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
$Q=\dfrac{\mu }{r} =1.109\;\cancel{D}\left ( \dfrac{3.3356\times 10^{-30}\; C\cdot \cancel{m}}{1\; \cancel{D}} \right )\left ( \dfrac{1}{127.8\; \cancel{pm}} \right )\left ( \dfrac{1\; \cancel{pm}}{10^{-12\;} \cancel{m}} \right )=2.901\times 10^{-20}\;C \label{10.3.4}$
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.
$\dfrac{2.901\times 10^{-20}\; \cancel{C}}{1.6022\times 10^{-19}\; \cancel{C}}=0.1811\;e^{-} \label{10.3.5}$
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
$\begin{matrix} _{\delta ^{+}}& & _{\delta ^{-}}\ H\; \; &-& Cl \end{matrix}$
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 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 11.
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 10.3.2.
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
$Q=\dfrac{\mu }{r} =9.001\;\cancel{D}\left ( \dfrac{3.3356\times 10^{-30}\; C\cdot \cancel{m}}{1\; \cancel{D}} \right )\left ( \dfrac{1}{236.1\; \cancel{pm}} \right )\left ( \dfrac{1\; \cancel{pm}}{10^{-12\;} \cancel{m}} \right )=1.272\times 10^{-19}\;C$
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):
$\% \; ionic\; character=\left ( \dfrac{1.272\times 10^{-19}\; \cancel{C}}{1.6022\times 10^{-19}\; \cancel{C}} \right )\left ( 100 \right )=79.39\%\simeq 79\%$
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%
Dipole Intermolecular Force: https://youtu.be/ACq_95SIBck
Electrostatic Potential Maps
Electrostatic potential maps convey information about the charge distribution of a molecule because of the properties of the nucleus and nature of electrostatic potential energy. A region of higher than average electrostatic potential energy indicates the presence of a stronger positive charge or a weaker negative charger. Given the positive charge of the nuclei, the higher potential energy value indicates the absence of negative charges (less screening of the nuclei), which would mean that there are fewer electrons in this region. The converse is also true with a low electrostatic potential indicateing an abundance of electrons. This property of electrostatic potentials can be extrapolated to molecules as well.
Note: Constructing a Electrostatic Potential Map
The first step involved in creating an electrostatic potential map is collecting a very specific type of data: electrostatic potential energy. An advanced computer program calculates the electrostatic potential energy at a set distance from the nuclei of the molecule. Electrostatic potential energy is fundamentally a measure of the strength of the nearby charges, nuclei and electrons, at a particular position. To accurately analyze the charge distribution of a molecule, a very large quantity of electrostatic potential energy values must be calculated. The best way to convey this data is to visually represent it, as in an electrostatic potential map. A computer program then imposes the calculated data onto an electron density model of the molecule. To make the electrostatic potential energy data easy to interpret, a color spectrum, with red as the lowest electrostatic potential energy value and blue as the highest, is employed to convey the varying intensities of the electrostatic potential energy values.
The most important thing to consider when analyzing an electrostatic potential map is the charge distribution. The relative distributions of electrons will allow you to deduce everything you need to know from these maps. Recall the relationship between electrostatic potential and charge distribution. Areas of low potential, red, are characterized by an abundance of electrons. Areas of high potential, blue, are characterized by a relative absence of electrons. Oxygen has a higher electronegativity value than sulfur (Table A2), hence. oxygen atoms would have a higher electron density around them than sulfur atoms. Thus the spherical region that corresponds to an oxygen atom would have a red portion on it. Now note that there are two oxygen atoms in sulfur dioxide (Figure $7$). There are two sphere shaped objects that have red regions. These areas correspond to the location of the oxygen atoms. The blue tainted sphere at the top corresponds to the location of the sulfur atom.
Note
A high electrostatic potential indicates the relative absence of electrons and a low electrostatic potential indicates an abundance of electrons
Electrostatic potential maps can also be used to determine the nature of the molecules chemical bond. Consider $SO_2$ in Figure $7$, there is a great deal of intermediary potential energy, the non red or blue regions, in this diagram. This indicates that the electronegativity difference is not very great. In a molecule with a great electronegativity difference, charge is very polarized, and there are significant differences in electron density in different regions of the molecule. This great electronegativity difference leads to regions that are almost entirely red and almost entirely blue. Greater regions of intermediary potential, yellow and green, and smaller or no regions of extreme potential, red and blue, are key indicators of a smaller electronegativity difference. Note that the electronegativity difference is a key determinant in the nature of a chemical bond.
Example $3$
The following electrostatic potential map of phosphoric acid $\H_3PO_4$. What regions correspond to atoms of oxygen, hydrogen, and phosphorous respectively?
Solution
You do not need to know the molecular structure to answer this question. You do need to know the relative electronegative values of these atoms (Table A4).
• Oxygen has the greatest electronegative value,
• Phosphorous the second most, and
• Hydrogen has the smallest electronegative value.
Simply by knowing this, you can deduce that oxygen would be affiliated with the red region or redish regions of the diagram, and hydrogen would be affiliated with the blue region. Phosphorous would fall in between these two extremes, in the green region.
Here is the molecular diagram of phosphoric acid:
Summary
• Bond polarity and ionic character increase with an increasing difference in electronegativity.
Dipole moment
$\mu = Qr \label{10.3.2}$
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_General_Chemistry_(Petrucci_et_al.)/10%3A_Chemical_Bonding_I%3A_Basic_Concepts/10.3%3A_Polar_Covalent_Bonds_and_Electrostatic_Potential_Maps.txt |
Learning Objectives
• To use Lewis dot symbols to explain the stoichiometry of a compound
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.
Note
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.
Answer
Lewis Structure of Molecules: https://youtu.be/xWiFCqA9Ur0
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.
Note
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.
Figure $3$: The Three Allotropes of Phosphorus: White, Red, and Black. ll three forms contain only phosphorus atoms, but they differ in the arrangement and connectivity of their atoms. White phosphorus contains P4 tetrahedra, red phosphorus is a network of linked P8 and P9 units, and black phosphorus forms sheets of six-membered rings. As a result, their physical and chemical properties differ dramatically.
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 8.5.2, we obtain
$formal\; charge\left ( N \right )=5\; valence\; e^{-}-\left ( 2\; non-bonding\; e^{-} +\frac{6\; bonding\; e^{-}}{2} \right )=0 \label{8.5.2}$
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 8.5.2 to calculate the formal charge on hydrogen, we obtain
$formal\; charge\left ( H \right )=1\; valence\; e^{-}-\left ( 0\; non-bonding\; e^{-} +\frac{2\; bonding\; e^{-}}{2} \right )=0 \label{8.5.3}$
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.
Note
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.
Lewis Structure of Charged Molecules: https://youtu.be/pTkziPtvMYU
Example $2$
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 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 8.5.1, the formal charge on the nitrogen atom is therefore
$formal\; charge\left ( N \right )=5-\left ( 0+\frac{8}{2} \right )=0$
Each hydrogen atom in has one bonding pair. The formal charge on each hydrogen atom is therefore
$formal\; charge\left ( H \right )=1-\left ( 0+\frac{2}{2} \right )=0$
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.
Using Formal Charges to Distinguish Viable Lewis Structures
As an example of how formal charges can be used to determine the most stable Lewis structure for a substance, we can compare two possible structures for CO2. Both structures conform to the rules for Lewis electron structures.
CO2
1. C is less electronegative than O, so it is the central atom.
2. C has 4 valence electrons and each O has 6 valence electrons, for a total of 16 valence electrons.
3. Placing one electron pair between the C and each O gives O–C–O, with 12 electrons left over.
4. Dividing the remaining electrons between the O atoms gives three lone pairs on each atom:
This structure has an octet of electrons around each O atom but only 4 electrons around the C atom.
1. No electrons are left for the central atom.
2. To give the carbon atom an octet of electrons, we can convert two of the lone pairs on the oxygen atoms to bonding electron pairs. There are, however, two ways to do this. We can either take one electron pair from each oxygen to form a symmetrical structure or take both electron pairs from a single oxygen atom to give an asymmetrical structure:
Both Lewis electron structures give all three atoms an octet. How do we decide between these two possibilities? The formal charges for the two Lewis electron structures of CO2 are as follows:
Both Lewis structures have a net formal charge of zero, but the structure on the right has a +1 charge on the more electronegative atom (O). Thus the symmetrical Lewis structure on the left is predicted to be more stable, and it is, in fact, the structure observed experimentally. Remember, though, that formal charges do not represent the actual charges on atoms in a molecule or ion. They are used simply as a bookkeeping method for predicting the most stable Lewis structure for a compound.
Note
The Lewis structure with the set of formal charges closest to zero is usually the most stable.
Example $3$: The Thiocyanate Ion
The thiocyanate ion (SCN), which is used in printing and as a corrosion inhibitor against acidic gases, has at least two possible Lewis electron structures. Draw two possible structures, assign formal charges on all atoms in both, and decide which is the preferred arrangement of electrons.
Given: chemical species
Asked for: Lewis electron structures, formal charges, and preferred arrangement
Strategy:
1. Use the step-by-step procedure to write two plausible Lewis electron structures for SCN.
2. Calculate the formal charge on each atom using Equation 8.5.1.
3. Predict which structure is preferred based on the formal charge on each atom and its electronegativity relative to the other atoms present.
Solution:
A Possible Lewis structures for the SCN ion are as follows:
B We must calculate the formal charges on each atom to identify the more stable structure. If we begin with carbon, we notice that the carbon atom in each of these structures shares four bonding pairs, the number of bonds typical for carbon, so it has a formal charge of zero. Continuing with sulfur, we observe that in (a) the sulfur atom shares one bonding pair and has three lone pairs and has a total of six valence electrons. The formal charge on the sulfur atom is therefore $6-\left ( 6+\frac{2}{2} \right )=-1.5-\left ( 4+\frac{4}{2} \right )=-1$ In (c), nitrogen has a formal charge of −2.
C Which structure is preferred? Structure (b) is preferred because the negative charge is on the more electronegative atom (N), and it has lower formal charges on each atom as compared to structure (c): 0, −1 versus +1, −2.
Exercise $3$: The Fulminate Ion
Salts containing the fulminate ion (CNO) are used in explosive detonators. Draw three Lewis electron structures for CNO and use formal charges to predict which is more stable. (Note: N is the central atom.)
Answer
The second structure is predicted to be more stable.
Summary
• 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. In Lewis electron structures, we encounter bonding pairs, which are shared by two atoms, and lone pairs, which are not shared between atoms. If both electrons in a covalent bond come from the same atom, the bond is called a coordinate covalent bond. Lewis structures are an attempt to rationalize why certain stoichiometries are commonly observed for the elements of particular families. Neutral compounds of group 14 elements typically contain four bonds around each atom (a double bond counts as two, a triple bond as three), whereas neutral compounds of group 15 elements typically contain three bonds. In cases where it is possible to write more than one Lewis electron structure with octets around all the nonhydrogen atoms of a compound, the formal charge on each atom in alternative structures must be considered to decide which of the valid structures can be excluded and which is the most reasonable. The formal charge is the difference between the number of valence electrons of the free atom and the number of electrons assigned to it in the compound, where bonding electrons are divided equally between the bonded atoms. The Lewis structure with the lowest formal charges on the atoms is almost always the most stable one.
Contributors and Attributions
Modified by Joshua Halpern (Howard University) | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/10%3A_Chemical_Bonding_I%3A_Basic_Concepts/10.4%3A_Writing_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 a single polyatomic species including fractional bonds and fractional charges. Resonance structure are capable of describing delocalized electrons that cannot be expressed by a single Lewis formula with an integer number of covalent bonds.
When 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 (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.
Note
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 \(NO_3^-\) 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 (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 \(1\): Nitrate 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.
Resonance Structures: https://youtu.be/aSP0D72MKe4
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_General_Chemistry_(Petrucci_et_al.)/10%3A_Chemical_Bonding_I%3A_Basic_Concepts/10.5%3A_Resonance.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 (NO ,refer to figure one). 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 nitrogen monoxide. Nitrogen monoxide has 11 valence electrons. If you need more information about formal charges, see Lewis Structures. If we were to imagine nitrogen monoxide 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 nitrogen monoxide (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, 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 was to make a Lewis structure for BH3 following the basic strategies for drawing Lewis structures, one would probably come up with this structure (Figure $3$):
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:
Note
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.
Expanded Valence Shell Bonding: https://youtu.be/Y4fBdOJBSHI | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/10%3A_Chemical_Bonding_I%3A_Basic_Concepts/10.6%3A_Exceptions_to_the_Octet_Rule.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_General_Chemistry_(Petrucci_et_al.)/10%3A_Chemical_Bonding_I%3A_Basic_Concepts/10.7%3A_Shapes_of_Molecules.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_General_Chemistry_(Petrucci_et_al.)/10%3A_Chemical_Bonding_I%3A_Basic_Concepts/10.9%3A_Bond_Energies.txt |
• 11.1: What a Bonding Theory Should Do
• 11.2: Introduction to the Valence-Bond Method
Although both Lewis and VSEPR structures contain localized electron-pair bonds, but do not use atomic orbitals 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: The strength of a covalent bond is proportional to the amount of overlap between atomic orbitals and an atom can use different combinations of atomic orbitals to maximize the overlap of orbitals used by bonded atoms.
• 11.3: Hybridization of Atomic 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.
• 11.4: Multiple Covalent Bonds
So far in our molecular orbital descriptions we have not dealt with polyatomic systems with multiple bonds. To do so, we can use an approach in which we describe σσ bonding using localized electron-pair bonds formed by hybrid atomic orbitals, and ππ bonding using molecular orbitals formed by unhybridized np atomic orbitals.
• 11.5: Molecular Orbital Theory
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.
• 11.6: Delocalized Electrons: Bonding in the Benzene Molecule
• 11.7: Bonding in Metals
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.
• 11.8: Some Unresolved Issues
• 11.E: Chemical Bonding II: Additional Aspects (Exercises)
11: Chemical Bonding II: Additional Aspects
As we have talked about using Lewis structures to depict the bonding in organic 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?
In this chapter, the valence bond theory is introduced to describe bonding in organic 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. In order to explain the bonding in organic molecules, however, we will need to introduce the concept of hybrid orbitals.
Valence bond theory is adequate for describing many aspects of organic structure. In some cases, however, chemists need to use a different model, called molecular orbital (MO) theory, to talk about covalent bonds in which electrons are shared not just between two atoms, but between several, or even over an entire molecule.
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.
These two electrons are now attracted to the positive charge of both of the hydrogen nuclei, with the result that they serve as a sort of ‘chemical glue’ holding the two nuclei together.
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.
But 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.
It is not accurate, however, 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.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/11%3A_Chemical_Bonding_II%3A_Additional_Aspects/11.1%3A_What_a_Bonding_Theory_Should_Do.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 H2O and 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 H2O and 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 SrF2 and 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.
Valence Bond Theory: A Localized Bonding Approach
Valence Bond Theory has its roots in Gilbert Newton Lewis’s paper The Atom and The Molecule. Possibly unaware that Lewis’s model existed, Walter Heitler and Fritz London came up with the idea that resonance and wavefunctions contributed to chemical bonds, in which they used dihydrogen as an example. Their theory was equivalent to Lewis’s theory, with the difference of quantum mechanics being developed. Nonetheless, Heitler and London's theory proved to be successful, providing Linus Pauling and John C. Slater with an opportunity to assemble a general chemical theory containing all of these ideas. Valence Bond Theory was the result, which included the ideas of resonance, covalent-ionic superposition, atomic orbital overlap, and hybridization to describe chemical bonds.
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. 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 both 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. Maximum overlap occurs between orbitals with the same spatial orientation and similar energies.
Note
An important aspect of Valence Bond theory is the concept of maximum overlap to form the strongest possible covalent bonds.
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 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 BeH2 form, scientists developed the concept of hybridization.
Reliability of Valence Bond Theory and Uses
As one can see, Valence Bond Theory can help describe how bonds are formed. However, there are some notable failures when it comes to Valence Bond Theory. One such failure is dioxygen. Valence Bond Theory fails to predict dioxygen's paramagnitism; it predicts that oxygen is diamagnetic. A species is paramagnetic if electrons are not spin paired and diamagnetic if the electrons are spin paired. Since Valence Bond theory begins with the basis that atomic orbitals overlap to create bonds and through that reasoning, one can see that electrons are spin paired when bonds overlap, dioxygen is indeed predicted to be diamagnetic if Valence Bond Theory is used. In reality, that is not the case. Also, sp2d and sp3 both have a coordinate number of four. Thus, Valence Bond Theory cannot predict whether the molecule is a square planar or the other shape (3). One must correctly draw the Lewis structure and use VSEPR to determine the shape.
Contributors and Attributions
• Tony Chhom (UC Davis) | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/11%3A_Chemical_Bonding_II%3A_Additional_Aspects/11.2%3A_Introduction_to_the_Valence-Bond_Method.txt |
Learning Objectives
• To describe the hybridication in the bonding of simple compounds via valence bond theory.
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.
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 $0$.
Because the difference A − B can also be written as A + (−B), in Figure $1$ 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 11.3.1, where the value $\frac{1}{\sqrt{2}}$ is needed mathematically to indicate that the 2s and 2p orbitals contribute equally to each hybrid orbital.
$sp = \dfrac{1}{\sqrt{2}} (2s + 2p_z) \label{11.3.1a}$
and
$sp = \dfrac{1}{\sqrt{2}} (2s - 2p_z) \label{11.3.1b}$
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 $3$). 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:
The large lobes of the hybridized orbitals are oriented toward the vertices of a tetrahedron, with 109.5° angles between them (Figure $5$). Like all the hybridized orbitals discussed earlier, the sp3 hybrid atomic orbitals are predicted to be equal in energy.
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 form in only 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 CH4 and NH3, the hybridized orbitals are not entirely equivalent in energy, making this bonding model an active area of research.
Valence Bond Method & sp3 Hybridization: https://youtu.be/2hxKLGWQ5EQ
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 VPSER 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
1. 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.
2. 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 $6$). 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 $6$). 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
1. sp3 with four P–Cl bonds
2. sp3d with three Br–F bonds and two lone pairs
3. 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 $3$: $OF_4$
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 sp3d2 hybridization; no
Expanded Octet Hybridization: https://youtu.be/1WpxXcKl_Io
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_General_Chemistry_(Petrucci_et_al.)/11%3A_Chemical_Bonding_II%3A_Additional_Aspects/11.3%3A_Hybridization_of_Atomic_Orbitals.txt |
Learning Objectives
• To explain resonance structures using molecular orbitals.
So far in our molecular 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.
Non-singular 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. Thus each carbon forms a set of three $\sigma$ bonds: two C–H (sp2 + s) and one C–C (sp2 + sp2) (part (a) in Figure $1$). The sp2 hybridization can be represented as follows:
After hybridization, each carbon still has one unhybridized 2pz orbital that is perpendicular to the hybridized lobes and contains a single electron (part (b) in Figure $1$). The two singly occupied 2pz orbitals can overlap to form a $\pi$ bonding orbital and a $\pi$* antibonding orbital, 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. The $\pi$* orbital lies outside the internuclear region and has a nodal plane perpendicular to the internuclear axis. 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.
sp2 Hybridization: https://youtu.be/EepTvePnfBA
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 $3$).
Figure $3$: Bonding in Acetylene (a) In the formation of the $\sigma$-bonded framework, two sets of singly occupied carbon sp hybrid orbitals and two singly occupied hydrogen 1s orbitals overlap. (b) In the formation of two carbon–carbon $\pi$ bonds in acetylene, two singly occupied unhybridized 2px,y orbitals on each carbon atom overlap. With one $\sigma$ bond plus two $\pi$ bonds, the carbon–carbon bond order in acetylene is 3.
Note
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
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.
sp Hybridization: https://youtu.be/epQXzG9WDRw
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. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/11%3A_Chemical_Bonding_II%3A_Additional_Aspects/11.4%3A_Multiple_Covalent_Bonds.txt |
Learning Objectives
• To use molecular orbital theory to predict bond order
• To apply Molecular Orbital Theory to the diatomic homonuclear molecule from the elements in the second period.
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.
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 discussed previously, 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:
$\begin{matrix} MO(1)= & AO(atom\; A) & +& AO(atomB) \ MO(1)= & AO(atom\; A) & -&AO(atomB) \end{matrix} \label{Eq1}$
The molecular orbitals created from Equation $\ref{Eq1}$ 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.
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, 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.
$\sigma _{1s} \approx 1s\left ( A \right) + 1s\left ( B \right) \label{Eq2}$
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:
$\sigma _{1s}^{\star } \approx 1s\left ( A \right) - 1s\left ( B \right) \label{Eq3}$
A molecule must have as many molecular orbitals as there are atomic orbitals.
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:
$bond\; order=\dfrac{number\; of \; bonding\; electrons-number\; of \; antibonding\; electrons}{2} \label{Eq4}$
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
$\dfrac{2-0}{2}=1 \label{Eq5}$
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.
Part (a) in Figure $3$ 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 (1-0)/2=1/2 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.
Part (b) in Figure $3$ 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 (2-1)/2=1/2. 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. Part (c) in Figure $3$ 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 $\left (\sigma _{1s} \right)^{2}\left (\sigma _{1s}^{\star } \right)^{1}$ 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.
Molecular Orbital Theory: https://youtu.be/XgtOG0ezw78
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 $3$, 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 orbitals. Draw the molecular orbital energy-level diagram for the system.
2. 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.
3. 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
$\dfrac{2-0}{2} =1$
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 $2$. 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 $3$ 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.
$\sigma _{np_{z}}=np_{z}\left ( A \right)-np_{z}\left ( B \right) \label{Eq6}$
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):
$\sigma _{np_{z}}=np_{z}\left ( A \right)-np_{z}\left ( B \right) \label{Eq6a}$
The other possible combination of the two npz orbitals is the mathematical sum:
$\sigma _{np_{z}}=np_{z}\left ( A \right)+np_{z}\left ( B \right) \label{Eq7}$
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..
$\pi _{np_{x}}=np_{x}\left ( A \right)+np_{x}\left ( B \right) \label{Eq8}$
$\pi ^{\star }_{np_{x}}=np_{x}\left ( A \right)-np_{x}\left ( B \right) \label{Eq9}$
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.
Second Row Diatomic Molecules
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. When we draw a molecular orbital diagram for a molecule, there are four key points to remember:
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.
The number of molecular orbitals is always equal to the total number of atomic orbitals we started with.
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 $9$; 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 $9$. 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 $10$, 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 $11$ Experimentally, it is found that the energy gap between the ns and np atomic orbitals increases as the nuclear charge increases (Figure $11$). 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 $3$
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 $9$ and Figure $11$, 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 $11$), 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 $3$
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 $12$. 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 $\left ( \sigma _{2s} \right)^{2}\left ( \sigma ^{\star }_{2s} \right)^{2}\left ( \pi _{2p_{x,y}} \right)^{4} \left ( \sigma _{2p_{z}} \right)^{2} \left ( \pi _{2p ^{\star }_{x,y}} \right)^{2}$ 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.
Molecular Orbital Bonding for Second Row Elements: https://youtu.be/A_5Xa3sK_YE
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 $14$ 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 $4$
Use a “skewed” molecular orbital energy-level diagram like the one in Figure $12$ 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 $12$ 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 $12$ fills the five lowest-energy orbitals, as shown here:
Because χN > χC, 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 $4$
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 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.
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.
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.
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_General_Chemistry_(Petrucci_et_al.)/11%3A_Chemical_Bonding_II%3A_Additional_Aspects/11.5%3A_Molecular_Orbital_Theory.txt |
Learning Objectives
• To be able to explain how mixing atomic orbitals make molecule orbitals with delocalized bonding
The advantage of MO theory becomes more apparent when we think about $\pi$ bonds, especially in those situations where two or more $\pi$ bonds are able to interact with one another. Let’s first consider the $\pi$ bond in ethene from an MO theory standpoint (in this example we will be disregarding the various sigma bonds, and thinking only about the $\pi$ bond). According to MO theory, the two atomic 2pz orbitals combine to form two $\pi$ molecular orbitals, one a low-energy π bonding orbital and one a high-energy π-star (π*) antibonding molecular orbital. These are sometimes denoted in molecular diagrams with the Greek letter psi (Ψ) instead of π (Figure $1$).
In the bonding Ψ1 orbital, the two shaded lobes of the 2pz orbitals interact constructively with each other, as do the two unshaded lobes (remember, the shading choice represents mathematical (+) and (-) signs for the wavefunction). Therefore, there is increased electron density between the nuclei in the molecular orbital – this is why it is a bonding orbital. In the higher-energy antibonding Ψ2* orbital, the shaded lobe of one 2pz orbital interacts destructively with the unshaded lobe of the second 2pz orbital, leading to a node between the two nuclei and overall repulsion.
By the aufbau principle, the two electrons from the two atomic orbitals will be paired in the lower-energy Ψ1 orbital when the molecule is in the ground state (Figure $1$).
Now, consider the 1,3-butadiene molecule. From valence orbital theory we might expect that the C2-C3 bond in this molecule, because it is a $\sigma$ bond that would rotate freely.
However, it is experimentally found that there are significant barriers to rotation about this bond (as well as about the C1-C2 and C3-C4 double bonds), and that the entire molecule is planar. It is also observed that the C2-C3 bond, while longer than the C1-C2 and C3-C4 double bonds, is significantly shorter than a typical carbon-carbon single bond.
Molecular orbital theory accounts for these observations with the concept of delocalized π bonds. In this picture, the four 2pz orbitals are all parallel to each other (and perpendicular to the plane of the $\sigma$ bonds), and thus there is $\pi$-overlap not just between C1 and C2 and C3 and C4, but between C2 and C3 as well. The four atomic (2pz) orbitals have combined to form four $\pi$ molecular orbitals.
The lowest energy molecular orbital, Ψ1, has zero nodes, and is a bonding MO. Slightly higher in energy, but still lower than the isolated p orbitals, is the Ψ2 orbital. This orbital has one node between C2 and C3, but is still a bonding orbital due to the two constructive interactions between C1-C2 and C3-C4. The two higher-energy MO’s are denoted Ψ3* and Ψ4*, and are antibonding. Notice that Ψ3* has two nodes and one constructive interaction, while Ψ4* has three nodes and zero constructive interactions. The energy of both of these antibonding molecular orbitals is higher than that of the 2pz atomic orbitals of which they are composed.
By the aufbau principle, the four electrons from the isolated 2pz atomic orbitals are placed in the bonding Ψ1and Ψ2 MO’s. Because Ψ1includes constructive interaction between C2 and C3, there is a degree, in the 1,3-butadiene molecule, of π-bonding interaction between these two carbons, which accounts for the shorter length and the barrier to rotation. The simple Lewis structure picture of 1,3-butadiene shows the two π bonds as being isolated from one another, with each pair of π electrons ‘stuck’ in its own π bond. However, molecular orbital theory predicts (accurately) that the four π electrons are to some extent delocalized, or ‘spread out’, over the whole π system.
1,3-butadiene is the simplest example of a system of ‘conjugated’ π bonds. To be considered conjugated, two or more π bonds must be separated by only one single bond – in other words, there cannot be an intervening sp3-hybridized carbon, because this would break up the overlapping system of parallel 2pz orbitals.
Benzene
Molecular orbital theory is especially helpful in explaining the unique properties of a class of compounds called aromatics. Benzene, a common organic solvent, is the simplest example of an aromatic compound. Although benzene is most often drawn with three double bonds and three single bonds (Figure $4$), it is known that all of the carbon-carbon bonds in benzene are exactly the same length - 1.38 Å. This is shorter than a typical carbon-carbon single bond (about 1.54 Å), and slightly longer than a typical carbon-carbon double bond (about 1.34 Å).
Benzene is also a cyclic molecule in which all of the ring atoms are sp2-hybridized that allows the π electrons to be delocalized in molecular orbitals that extend all the way around the ring, above and below the plane of the ring. For this to happen, of course, the ring must be planar – otherwise the 2pz orbitals could not overlap properly. Benzene was experimentally confirmed to be flat molecule by Dame Kathleen Londsale with X-ray crystallography.
As shown in Figure $5$, the cyclic array of six \2P_z\)-orbitals (one on each carbon) overlap to generate six molecular orbitals, three bonding and three antibonding. The plus and minus signs shown in the diagram do not represent electrostatic charge, but refer to phase signs in the equations that describe these orbitals (in the diagram the phases are also color coded). When the phases correspond, the orbitals overlap to generate a common region of like phase, with those orbitals having the greatest overlap (e.g. π1) being lowest in energy. The remaining carbon valence electrons then occupy these molecular orbitals in pairs, resulting in a fully occupied (6 electrons) set of bonding molecular orbitals.
Aromaticity
It is this completely filled set of bonding orbitals, or closed shell, that gives the benzene ring its thermodynamic and chemical stability, just as a filled valence shell octet confers stability on the inert gases. This added stability is call aromaticity. and is discuss in more detail in organic chemistry courses.
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). We showed that 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 $6$. 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 $7$. 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; it 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 $7$, 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 approach, however, 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. 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 $1$
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 $1$
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 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 $8$. 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 $9$). 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.
Delocalization
Delocalization is central feature of molecular orbital theory where rather than the lone pair of electrons contained in localize bonds (as in the valence bond theory), electrons can exist in molecular orbitals that are spread over the entire molecule.
Summary
The $\pi$ 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_General_Chemistry_(Petrucci_et_al.)/11%3A_Chemical_Bonding_II%3A_Additional_Aspects/11.6%3A_Delocalized_Electrons%3A_Bonding_in_the_Benzene_Molecule.txt |
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.
11.8: Some Unresolved Issues
under construction
SF6 etc.
11.E: Chemical Bonding II: Additional Aspects (Exercises)
11.7 Bonding in Metals
Conceptual Problems
1. Can band theory be applied to metals with two electrons in their valence s orbitals? with no electrons in their valence s orbitals? Why or why not?
2. Given a sample of a metal with 1020 atoms, how does the width of the band arising from p orbital interactions compare with the width of the band arising from s orbital interactions? from d orbital interactions?
3. Diamond has one of the lowest electrical conductivities known. Based on this fact, do you expect diamond to be colored? Why? How do you account for the fact that some diamonds are colored (such as “pink” diamond or “green” diamond)?
4. Why do silver halides, used in the photographic industry, have band gaps typical of semiconducting materials, whereas alkali metal halides have very large band gaps?
5. As the ionic character of a compound increases, does its band gap increase or decrease? Why?
6. Why is silicon, rather than carbon or germanium, used in the semiconductor industry?
7. Carbon is an insulator, and silicon and germanium are semiconductors. Explain the relationship between the valence electron configuration of each element and their band structures. Which will have the higher electrical conductivity at room temperature—silicon or germanium?
8. How does doping affect the electrical conductivity of a semiconductor? Draw the effect of doping on the energy levels of the valence band and the conduction band for both an n-type and a p-type semiconductor.
Answers
1. The low electrical conductivity of diamond implies a very large band gap, corresponding to the energy of a photon of ultraviolet light rather than visible light. Consequently, diamond should be colorless. Pink or green diamonds contain small amounts of highly colored impurities that are responsible for their color.
2. -
3. As the ionic character of a compound increases, the band gap will also increase due to a decrease in orbital overlap. Remember that overlap is greatest for orbitals of the same energy, and that the difference in energy between orbitals on adjacent atoms increases as the difference in electronegativity between the atoms increases. Thus, large differences in electronegativity increase the ionic character, decrease the orbital overlap, and increase the band gap.
Numerical Problems
1. Of Ca, N, B, and Ge, which will convert pure silicon into a p-type semiconductor when doping? Explain your reasoning.
2. Of Ga, Si, Br, and P, which will convert pure germanium into an n-type semiconductor when doping? Explain your reasoning. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/11%3A_Chemical_Bonding_II%3A_Additional_Aspects/11.7%3A_Bonding_in_Metals.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_General_Chemistry_(Petrucci_et_al.)/12%3A_Intermolecular_Forces%3A_Liquids_And_Solids/12.1%3A_Intermolecular_Forces.txt |
Learning Objectives
• To describe the unique properties of liquids.
• 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.
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:
1. surface tension,
2. capillary action, and
3. viscosity.
Surface Tension
If liquids tend to adopt the shapes of their containers, then, 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 (electrostatic) bonds has as surface tension that is 15-times lower: 4.6 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 (see the chapter opener photo). 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 better describe in the zero gravity condictions of space as Figure $2$ indicates (and more so in the video link).
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.
Surface Tension, Viscosity, & Melting Point: https://youtu.be/OgKDGrdTRRM
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 put into a dish of water, water is drawn up into the tube. 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.
When a glass capillary is placed in liquid water, water rises up into the capillary. The smaller the diameter of the capillary, the higher the water rises. The height of the water does not depend on the angle at which the capillary is tilted.
Note
• Cohesive forces bind molecules of the same type together
• Adhesive forces bind a substance to a surface
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
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 (part (a) in 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 (part (b) in Figure $4$).
Note
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 $1$ 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.
Note
Viscosity increases as intermolecular interactions or molecular size increases.
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.
Oil being drained from a car
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. 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.
Vapor Pressure
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.
Vapor Pressure & Boiling Point: https://youtu.be/4QtcdpfRO1M
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 $5$), just as we did for a gas (Figure 10.19). 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 (E0) 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 E0. 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 E0 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 $6$. When a liquid is introduced into an evacuated chamber (part (a) in Figure $6$), 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 > E0 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 $6$). 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 $7$.
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 (Table 11.4) 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 (Table 11.4). 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 $8$. 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.
Note
Volatile substances have low boiling points and relatively weak intermolecular interactions; nonvolatile substances have high boiling points and relatively strong intermolecular interactions.
The exponential rise in vapor pressure with increasing temperature in Figure $8$ allows us to use natural logarithms to express the nonlinear relationship as a linear one.
$\ln\left ( P \right)=\dfrac{-\Delta H_{vap}}{R}\left ( \dfrac{1}{T} \right) + C \label{Eq1}$
where
• $\ln P$ is the natural logarithm of the vapor pressure,
• ΔHvap 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.
A plot of \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 ΔHvap of a liquid from its measured vapor pressure at two or more temperatures. The simplest way to determine ΔHvap 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 $\ref{Eq1}$, which is derived from the Clausius–Clapeyron equation:
$\ln\left ( \dfrac{P_{1}}{P_{2}} \right)=\dfrac{-\Delta H_{vap}}{R}\left ( \dfrac{1}{T_{2}}-\dfrac{1}{T_{1}} \right) \label{11.5.2}$
Conversely, if we know ΔHvap and the vapor pressure P1 at any temperature T1, we can use Equation $\ref{Eq1}$ to calculate the vapor pressure P2 at any other temperature T2, as shown in Example $2$.
Example $2$: Vapor Pressure of Mercury
The experimentally measured vapor pressures of liquid Hg at four temperatures are listed in the following table:
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 $\ref{Eq1}$ 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 $\ref{Eq1}$ 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 $\ref{Eq1}$ 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 kelvins because the equation requires absolute temperatures. Substituting the values measured at 80.0°C (T1) and 120.0°C (T2) into Equation $\ref{Eq1}$ gives
$\ln\left ( \dfrac{0.7457 \; \cancel{Torr}}{0.0888 \; \cancel{Torr}} \right)=\dfrac{-\Delta H_{vap}}{8.314 \; J/mol\cdot K}\left ( \dfrac{1}{\left ( 120+273 \right)K}-\dfrac{1}{\left ( 80.0+273 \right)K} \right)$
$\ln\left ( 8.398 \right)=\dfrac{-\Delta H_{vap}}{8.314 \; J/mol\cdot \cancel{K}}\left ( -2.88\times 10^{-4} \; \cancel{K^{-1}} \right)$
$2.13=-\Delta H_{vap} \left ( -3.46 \times 10^{-4} \right) J^{-1}\cdot mol$
$\Delta H_{vap} =61,400 \; J/mol = 61.4 \; kJ/mol$
B We can now use this value of ΔHvap to calculate the vapor pressure of the liquid (P2) at 160.0°C (T2):
$\ln\left ( \dfrac{P_{2} }{0.0888 \; torr} \right)=\dfrac{-61,400 \; \cancel{J/mol}}{8.314 \; \cancel{J/mol} \; K^{-1}}\left ( \dfrac{1}{\left ( 160+273 \right)K}-\dfrac{1}{\left ( 80.0+273 \right) K} \right)$
Using the relationship e\ln x = x, we have
$\ln\left ( \dfrac{P_{2} }{0.0888 \; Torr} \right)=3.86$
$\dfrac{P_{2} }{0.0888 \; Torr} =e^{3.86} = 47.5$
$P_{2} = 4.21 Torr$
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 $2$: 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 $8$ 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 $2$: 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 $2$ 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.
Note
As pressure increases, the boiling point of a liquid increases and vice versa.
Example $3$: Boiling Mercury
Use Figure $8$ 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
Mercury boils at 356 °C at room pressure. To see video go to https://www.youtube.com/watch?v=0iizsbXWYoo
Given: data in Figure $8$, 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 $8$ 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 $3$: 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 $8$ 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
1. 200°C
2. 450 mmHg
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.
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_General_Chemistry_(Petrucci_et_al.)/12%3A_Intermolecular_Forces%3A_Liquids_And_Solids/12.2%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 $1$.
Previously, we defined the enthalpy changes associated with various chemical and physical processes. The melting points and molar enthalpies of fusion (ΔHfus), 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 (ΔHvap) of selected compounds are listed in Table $1$.
Table $1$: Melting and Boiling Points and Enthalpies of Fusion and Vaporization for Selected Substances
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).
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.
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.
$ΔH_{sub} =ΔH_{fus} +ΔH_{vap} \label{Eq1}$
Less energy is needed to allow molecules to move past each other than to separate them totally
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, as demonstrated later in Example 8.
The Thermodynamics of Phase Changes: https://youtu.be/Uf2mAuP1BZY
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.
Example $1$: Cooling Hot 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 values given into the general equation relating heat gained to heat lost (Equation 5.39) 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
$q=mC_sΔT$
where q is heat, m is mass, Cs 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.
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_General_Chemistry_(Petrucci_et_al.)/12%3A_Intermolecular_Forces%3A_Liquids_And_Solids/12.3%3A_Some_Properties_of_Solids.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_General_Chemistry_(Petrucci_et_al.)/12%3A_Intermolecular_Forces%3A_Liquids_And_Solids/12.4%3A_Phase_Diagrams.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_General_Chemistry_(Petrucci_et_al.)/12%3A_Intermolecular_Forces%3A_Liquids_And_Solids/12.5%3A_Network_Covalent_Solids_and_Ionic_Solids.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_General_Chemistry_(Petrucci_et_al.)/12%3A_Intermolecular_Forces%3A_Liquids_And_Solids/12.6%3A_Crystal_Structures.txt |
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 ($\ce{CO2}$ gas in water)
liquid liquid liquid alcoholic beverage (ethanol in water), gasoline
liquid solid liquid tea, salt water
solid gas solid $\ce{H2}$ in $\ce{Pd}$ (used for $\ce{H2}$ storage)
solid solid liquid mercury in silver or gold (amalgam often used in dentistry)
The formation of a solution from a solute and a solvent is a physical process, not a chemical one. That is, both solute and solvent can be recovered in chemically unchanged forms using appropriate separation methods. For example, solid zinc nitrate dissolves in water to form an aqueous solution of zinc nitrate:
$\ce{Zn(NO3)2 (s) + H2O(l) \rightarrow Zn^{2+} (aq) +2NO^{-}3 (aq)} \nonumber$
Because $\ce{Zn(NO3)2}$ can be recovered easily by evaporating the water, this is a physical process. In contrast, metallic zinc appears to dissolve in aqueous hydrochloric acid. In fact, the two substances undergo a chemical reaction to form an aqueous solution of zinc chloride with evolution of hydrogen gas:
$\ce{ Zn (s) + 2H^{+}(aq) + 2Cl^{-} (aq) \rightarrow Zn^{2+} (aq) + 2Cl^{-} (aq) + H_2 (g) } \nonumber$
Dissolution of a solute in a solvent to form a solution typically does not involve a chemical transformation and largely consist of breaking intermolecular forces rather than covalent bonds.
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.
Mercury Amalgams: Examples of solid Solutions
An amalgam is an alloy of mercury with another metal. It may be a liquid, a soft paste or a solid, depending upon the proportion of mercury. These alloys are formed through metallic bonding, with the electrostatic attractive force of the conduction electrons working to bind all the positively charged metal ions together into a crystal lattice structure. Almost all metals can form amalgams with mercury including aluminum (Video $1$), the notable exceptions being iron, platinum, tungsten, and tantalum. Silver-mercury amalgams are important in dentistry, and gold-mercury amalgam is used in the extraction of gold from ore.
Substances that form a single homogeneous phase in all proportions are said to be completely 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. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/13%3A_Solutions_and_their_Physical_Properties/13.01%3A_Types_of_Solutions%3A_Some_Terminology.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_General_Chemistry_(Petrucci_et_al.)/13%3A_Solutions_and_their_Physical_Properties/13.02%3A_Solution_Concentration.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.
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.
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 $1$. The overall enthalpy change in the formation of the solution ($\Delta H_{soln}$) is the sum of the enthalpy changes in the three steps:
$\Delta H_{soln} = \Delta H_1 + \Delta H_2 + \Delta H_3 \label{13.3.2}$
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 $\ce{MgO}$ (−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 $2$).
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. 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 $3$), 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
2. aniline ($C_6H_5NH_2$) in dichloromethane ($CH_2Cl_2$)
1. iodine in water
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 $I2$ are both nonpolar molecules. The only possible attractive forces are London dispersion forces.
2. Aniline is a polar molecule with an $\ce{–NH_2}$ group, which can act as a hydrogen bond donor. Dichloromethane is also polar, but it has no obvious hydrogen bond acceptor. Therefore, the most important interactions between aniline and $CH_2Cl_2$ are likely to be London 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 $I_2$ 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:
1. hydrogen bonding
2. London interactions
3. 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, 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
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 interaction of water with Na+ and Cl− ions in an aqueous solution of $\ce{NaCl}$ was illustrated in Figure 4.3. The ion–dipole interactions between Li+ ions and acetone molecules in a solution of LiCl in acetone are shown in Figure $4$. The energetically favorable Li+–acetone interactions make $ΔH_3$ in Figure 13.1 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 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 Li+ and Cl− ions, so 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.
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. 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_General_Chemistry_(Petrucci_et_al.)/13%3A_Solutions_and_their_Physical_Properties/13.03%3A_Intermolecular_Forces_and_the_Solution_Process.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,
When a solute dissolves, its individual atoms, molecules, or ions interact with the solvent, become solvated, and are able to diffuse independently throughout the solution (part (a) in Figure $1$). This is not, however, a unidirectional process. If the molecule or ion happens to collide with the surface of a particle of the undissolved solute, it may adhere to the particle in a process called crystallization. Dissolution and crystallization continue as long as excess solid is present, resulting in a dynamic equilibrium analogous to the equilibrium that maintains the vapor pressure of a liquid. We can represent these opposing processes as follows:
$\text{solute} + \text{solvent} \ce{<=>[\ce{crystallization}][\ce{dissolution}]} \text{solution} \label{13.4.1}$
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.
Solubility as a Function of Temperature
Figure $2$ 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 $CH_3CO_2Na$) exhibit a dramatic increase in solubility with increasing temperature. Others (such as $NaCl$ and $K_2SO_4$) exhibit little variation, and still others (such as $Li_2SO_4$) become less soluble with increasing temperature.
Notice in particular the curves for $NH_4NO_3$ and $CaCl_2$. 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 $2$ 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.
Fractional Crystallization
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 (Figure $3$). For example, if we have a mixture of 150 g of sodium acetate ($CH_3CO_2Na$) and 50 g of $KBr$, 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 $2$, both compounds dissolve in water at 80°C, and all 50 g of $KBr$ remains in solution at 0°C. Only about 36 g of $CH_3CO_2Na$ are soluble in 100 g of water at 0°C, however, so approximately 114 g (150 g − 36 g) of $CH_3CO_2Na$ crystallizes out on cooling. The crystals can then be separated by filtration. Thus fractional crystallization allows us to recover about 75% of the original $CH_3CO_2Na$ in essentially pure form in only one step.
Fractional crystallization is a common technique for purifying compounds as diverse as those shown in Figure $2$ and from antibiotics to enzymes to drugs (Figure $4$). 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 $KBr$ in this example) and preferably present in relatively small amounts.
Summary
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 most substances depends strongly on the temperature. 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. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/13%3A_Solutions_and_their_Physical_Properties/13.04%3A_Solution_Formation_and_Equilibrium.txt |
The chemical structures of the solute and solvent dictate the types of forces possible and, consequently, are important factors in determining solubility. For example, under similar conditions, the water solubility of oxygen is approximately three times greater than that of helium, but 100 times less than the solubility of chloromethane, CHCl3. Considering the role of the solvent’s chemical structure, note that the solubility of oxygen in the liquid hydrocarbon hexane, C6H14, is approximately 20 times greater than it is in water.
Other factors also affect the solubility of a given substance in a given solvent. Temperature is one such factor, with gas solubility typically decreasing as temperature increases (Figure $1$). This is one of the major impacts resulting from the thermal pollution of natural bodies of water.
When the temperature of a river, lake, or stream is raised abnormally high, usually due to the discharge of hot water from some industrial process, the solubility of oxygen in the water is decreased. Decreased levels of dissolved oxygen may have serious consequences for the health of the water’s ecosystems and, in severe cases, can result in large-scale fish kills (Figure $2$).
The solubility of a gaseous solute is also affected by the partial pressure of solute in the gas to which the solution is exposed. Gas solubility increases as the pressure of the gas increases. Carbonated beverages provide a nice illustration of this relationship. The carbonation process involves exposing the beverage to a relatively high pressure of carbon dioxide gas and then sealing the beverage container, thus saturating the beverage with CO2 at this pressure. When the beverage container is opened, a familiar hiss is heard as the carbon dioxide gas pressure is released, and some of the dissolved carbon dioxide is typically seen leaving solution in the form of small bubbles (Figure $3$). At this point, the beverage is supersaturated with carbon dioxide and, with time, the dissolved carbon dioxide concentration will decrease to its equilibrium value and the beverage will become “flat.”
For many gaseous solutes, the relation between solubility, Cg, and partial pressure, Pg, is a proportional one:
$C_\ce{g}=kP_\ce{g}$
where k is a proportionality constant that depends on the identities of the gaseous solute and solvent, and on the solution temperature. This is a mathematical statement of Henry’s law: The quantity of an ideal gas that dissolves in a definite volume of liquid is directly proportional to the pressure of the gas.
Example $1$: Application of Henry’s Law
At 20 °C, the concentration of dissolved oxygen in water exposed to gaseous oxygen at a partial pressure of 101.3 kPa (760 torr) is 1.38 × 10−3 mol L−1. Use Henry’s law to determine the solubility of oxygen when its partial pressure is 20.7 kPa (155 torr), the approximate pressure of oxygen in earth’s atmosphere.
Solution
According to Henry’s law, for an ideal solution the solubility, Cg, of a gas (1.38 × 10−3 mol L−1, in this case) is directly proportional to the pressure, Pg, of the undissolved gas above the solution (101.3 kPa, or 760 torr, in this case). Because we know both Cg and Pg, we can rearrange this expression to solve for k.
\begin{align*} C_\ce{g}&=kP_\ce{g}\[4pt] k&=\dfrac{C_\ce{g}}{P_\ce{g}}\[4pt] &=\mathrm{\dfrac{1.38×10^{−3}\:mol\:L^{−1}}{101.3\:kPa}}\[4pt] &=\mathrm{1.36×10^{−5}\:mol\:L^{−1}\:kPa^{−1}}\[4pt] &\hspace{15px}\mathrm{(1.82×10^{−6}\:mol\:L^{−1}\:torr^{−1})} \end{align*}
Now we can use k to find the solubility at the lower pressure.
$C_\ce{g}=kP_\ce{g}$
$\mathrm{1.36×10^{−5}\:mol\:L^{−1}\:kPa^{−1}×20.7\:kPa\[4pt] (or\:1.82×10^{−6}\:mol\:L^{−1}\:torr^{−1}×155\:torr)\[4pt] =2.82×10^{−4}\:mol\:L^{−1}}$
Note that various units may be used to express the quantities involved in these sorts of computations. Any combination of units that yield to the constraints of dimensional analysis are acceptable.
Exercise $1$
Exposing a 100.0 mL sample of water at 0 °C to an atmosphere containing a gaseous solute at 20.26 kPa (152 torr) resulted in the dissolution of 1.45 × 10−3 g of the solute. Use Henry’s law to determine the solubility of this gaseous solute when its pressure is 101.3 kPa (760 torr).
Answer
7.25 × 10−3 g in 100.0 mL or 0.0725 g/L
Case Study: Decompression Sickness (“The Bends”)
Decompression sickness (DCS), or “the bends,” is an effect of the increased pressure of the air inhaled by scuba divers when swimming underwater at considerable depths. In addition to the pressure exerted by the atmosphere, divers are subjected to additional pressure due to the water above them, experiencing an increase of approximately 1 atm for each 10 m of depth. Therefore, the air inhaled by a diver while submerged contains gases at the corresponding higher ambient pressure, and the concentrations of the gases dissolved in the diver’s blood are proportionally higher per Henry’s law.
As the diver ascends to the surface of the water, the ambient pressure decreases and the dissolved gases becomes less soluble. If the ascent is too rapid, the gases escaping from the diver’s blood may form bubbles that can cause a variety of symptoms ranging from rashes and joint pain to paralysis and death. To avoid DCS, divers must ascend from depths at relatively slow speeds (10 or 20 m/min) or otherwise make several decompression stops, pausing for several minutes at given depths during the ascent. When these preventive measures are unsuccessful, divers with DCS are often provided hyperbaric oxygen therapy in pressurized vessels called decompression (or recompression) chambers (Figure $4$).
Deviations from Henry’s law are observed when a chemical reaction takes place between the gaseous solute and the solvent. Thus, for example, the solubility of ammonia in water does not increase as rapidly with increasing pressure as predicted by the law because ammonia, being a base, reacts to some extent with water to form ammonium ions and hydroxide ions.
Gases can form supersaturated solutions. If a solution of a gas in a liquid is prepared either at low temperature or under pressure (or both), then as the solution warms or as the gas pressure is reduced, the solution may become supersaturated. In 1986, more than 1700 people in Cameroon were killed when a cloud of gas, almost certainly carbon dioxide, bubbled from Lake Nyos (Figure $5$), a deep lake in a volcanic crater. The water at the bottom of Lake Nyos is saturated with carbon dioxide by volcanic activity beneath the lake. It is believed that the lake underwent a turnover due to gradual heating from below the lake, and the warmer, less-dense water saturated with carbon dioxide reached the surface. Consequently, tremendous quantities of dissolved CO2 were released, and the colorless gas, which is denser than air, flowed down the valley below the lake and suffocated humans and animals living in the valley.
Henry's Law (The Solubility of Gases in Solvents): https://youtu.be/fiJZCGpArJI | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/13%3A_Solutions_and_their_Physical_Properties/13.05%3A_Solubilities_of_Gases.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.
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
$\color{red} P_A=X_AP^0_A \label{13.6.1}$
where $P_A$ is the vapor pressure of component A of the solution (in this case the solvent), XA is the mole fraction of A in solution, and $P^0_A$ is the vapor pressure of pure A. Equation \ref{13.6.1} is known as Raoult’s law, after the French chemist who developed it. If the solution contains only a single nonvolatile solute (B), then $X_A + X_B = 1$, and we can substitute $X_A = 1 − X_B$ to obtain
\begin{align} P_A &=(1−X_B)P^0_A \[4pt] &=P^0_A−X_BP^0_A \label{13.6.2} \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:
$P^0_A−P_A=ΔP_A=X_BP^0_A \label{13.6.3}$
We can solve vapor pressure problems in either of two ways: by using Equation \ref{13.6.1} to calculate the actual vapor pressure above a solution of a nonvolatile solute, or by using Equation \ref{13.6.3} to calculate the decrease in vapor pressure caused by a specified amount of a nonvolatile solute.
Example $1$
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:
$moles \; EG=(302 \;\cancel{g}) \left( \dfrac{1\; mol}{62.07\; \cancel{g}} \right)=4.87\; mol\; EG \nonumber$
$moles \; \ce{H2O}=(698 \;\cancel{g}) \left( \dfrac{1\; mol}{18.02\; \cancel{g}} \right)=38.7\; mol\; H_2O \nonumber$
The mole fraction of water is thus
$X_{\ce{H2O}}=\dfrac{38.7\; \cancel{mol} \; H_2O}{38.7\; \cancel{mol}\; H_2O +4.87 \cancel{mol}\; EG} =0.888 \nonumber$
B From Raoult’s law (Equation \ref{13.6.1}), the vapor pressure of the solution is
$P_{\ce{H2O}}=(X_{H2_O})(P^0_{H2_O)}=(0.888)(760\; mmHg) =675 \;mmHg \nonumber$
Alternatively, we could solve this problem by calculating the mole fraction of ethylene glycol and then using Equation \ref{13.6.3} to calculate the resulting decrease in vapor pressure:
$X_{EG}=\dfrac{4.87\; mol\; EG}{4.87\; mol\; EG+38.7\; mol\; H_2O}=0.112 \nonumber$
$ΔP_{\ce{H2O}}=(X_{EG})(P^0_{H_2O})=(0.112)(760\; mmHg)=85.1\; mmHg \nonumber$
$P_{\ce{H2O}}=P^0H_2O−ΔP_{H_2O}=760\; mmHg−85.1\; mmHg=675\; mmHg \nonumber$
The same result is obtained using either method.
Exercise $1$
Seawater is an approximately 3.0% aqueous solution of $NaCl$ 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 $NaCl$, remembering that 1 mol of $NaCl$ 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.
Finding the Vapor Pressure of a Solution (Ionic-Nonvolatile Solute):
https://youtu.be/sRBaRXsql9s
Volatile Solutes
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 ($P_T$) is the sum of the vapor pressures of the components:
\begin{align} P_T &=P_A+P_B \[4pt] &=X_AP^0_A+X_BP^0_B \label{13.6.4} \end{align}
Because $X_B = 1 − X_A$ for a two-component system,
$P_T=X_AP^0_A+(1−X_A)P^0_B \label{13.6.5}$
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
$P_{C_6H_6}=X_{C_6H_6}P^0_{C_6H_6} \label{13.6.6}$
and the vapor pressure of toluene in the solution is
$P_{C_6H_5CH_3}=X_{C_6H_5CH_3}P^0_{C_6H_5CH3} \label{13.6.7}$
Equations \ref{13.6.6} and \ref{13.6.7} 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 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 $CCl_4$ and methanol, for example, the nonpolar $CCl_4$ molecules interrupt the extensive hydrogen bonding network in methanol, and the lighter methanol molecules have weaker London dispersion forces than the heavier $CCl_4$ molecules. Consequently, solutions of $CCl_4$ 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
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)
Finding Vapor Pressure of a Solution (Nonionic-Volatile Solute):
https://youtu.be/s06fzZZtLl0
Summary
• Henry’s law: $C = kP \nonumber$
• Raoult’s law: $P_A=X_AP^0_A \nonumber$
• vapor pressure lowering: $P^0_A−P_A=ΔP_A=X_BP^0_A \nonumber$
• vapor pressure of a system containing two volatile components: $P_T=X_AP^0_A+(1−X_A)P^0_B \nonumber$
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. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/13%3A_Solutions_and_their_Physical_Properties/13.06%3A_Vapor_Pressures_of_Solutions.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.
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 $1$, 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.
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:
$\Pi=\dfrac{nRT}{V}=MRT \label{eq1}$
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 $1$, osmotic pressures tend to be quite high, even for rather dilute solutions.
Example $1$: Yeast Cells
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% $NaCl$ by mass; the solution density is 1.02 g/mL at 25°C.
Calculate the osmotic pressure of a 4.0% aqueous $NaCl$ solution at 25°C.
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 $NaCl$ solution; internal osmotic pressure of cell
Asked for: osmotic pressure of $NaCl$ solution and concentration of glycerol needed
Strategy:
1. Calculate the molarity of the $NaCl$ 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{eq1} 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{eq1} to calculate the molarity of glycerol needed to create this osmotic pressure.
Solution:
A The solution contains 4.0 g of $NaCl$ per 100 g of solution. Using the formula mass of $NaCl$ (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}} \nonumber \[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) \nonumber \[4pt] &= 0.70\; M\; NaCl \nonumber \end{align} \nonumber
Because 1 mol of $NaCl$ 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{eq1} 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*}
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{eq1} 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 \;glycerol \end{align*}
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 $NaCl$ 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 $1$
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 $2$, 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 $3$, 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.
Osmotic Pressure: https://youtu.be/uYQxI4mi3DA
Summary
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. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/13%3A_Solutions_and_their_Physical_Properties/13.07%3A_Osmotic_Pressure.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.
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, $NaCl$, 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 Na+ ions and 0.01 M $Cl^−$ ions, for a total particle concentration of 0.02 M. Similarly, the $\ce{CaCl_2}$ solution contains 0.01 M $Ca^{2+}$ ions and 0.02 M $Cl^−$ 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 $Mg^{2+}$ or $Al^{3+}$), 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.
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 $1$). 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.
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 $2$). We can define the boiling point elevation ($ΔT_b$) as the difference between the boiling points of the solution and the pure solvent:
$ΔT_b=T_b−T^0_b \label{eq1}$
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
$ΔT_b = mK_b \label{eq2}$
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 Kb 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{eq2} 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
$\text{molality of ethylene glycol}= \left(\dfrac{4.87 \;mol}{698 \; \cancel{g} \;H_2O} \right) \left(\dfrac{1000\; \cancel{g}}{1 \;kg} \right)=6.98 m$
From Equation \ref{eq2}, the increase in boiling point is therefore
$ΔT_b=mK_b=(6.98 \cancel{m})(0.51°C/\cancel{m})=3.6°C$
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{eq2} 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.)
Freezing Point Depression
The phase diagram in Figure $1$ 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.
This phenomenon is exploited in “de-icing” schemes that use salt (Figure $3$), calcium chloride, or urea to melt ice on roads and sidewalks, and in the use of ethylene glycol as an “antifreeze” in automobile radiators. Seawater freezes at a lower temperature than fresh water, and so the Arctic and Antarctic oceans remain unfrozen even at temperatures below 0 °C (as do the body fluids of fish and other cold-blooded sea animals that live in these oceans).
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:
$ΔT_f=T^0_f−T_f \label{eq3}$
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{eq1} 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{eq2}:
$ΔT_f = mK_f \label{eq4}$
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 $K_f$ (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, 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. Heated glycols are often sprayed onto the surface of airplanes prior to takeoff in inclement weather in the winter to remove ice that has already formed and prevent the formation of more ice, which would be particularly dangerous if formed on the control surfaces of the aircraft (Video $1$).
Video $1$: Freezing point depression is exploited to remove ice from the control surfaces of aircraft.
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 $4$
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 Figure 13.9 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 Figure 13.9. 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 $4$ to calculate the freezing point depressions of the solutions.
Solution:
A From Figure 13.9, 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
$m_{\ce{NaCl}}=\left(\dfrac{36 \; \cancel{g \;NaCl}}{100 \;\cancel{g} \;H_2O}\right)\left(\dfrac{1\; mol\; NaCl}{58.44\; \cancel{ g\; NaCl}}\right)\left(\dfrac{1000\; \cancel{g}}{1\; kg}\right)=6.2\; m$
$m_{\ce{CaCl_2}}=\left(\dfrac{60\; \cancel{g\; CaCl_2}}{100\;\cancel{g}\; H_2O}\right)\left(\dfrac{1\; mol\; CaCl_2}{110.98\; \cancel{g\; CaCl_2}}\right)\left(\dfrac{1000 \;\cancel{g}}{1 kg}\right)=5.4\; m$
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 $\ce{CaCl_2}$.
B Because these salts are ionic compounds that dissociate in water to yield two and three ions per formula unit of $\ce{NaCl}$ and $\ce{CaCl_2}$, 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 $\ce{CaCl_2}$. The resulting freezing point depressions can be calculated using Equation $4$:
$\ce{NaCl}: ΔT_f=mK_f=(12\; \cancel{m})(1.86°C/\cancel{m})=22°C$
$\ce{CaCl2}: ΔT_f=mK_f=(16\;\cancel{m})(1.86°C/\cancel{m})=30°C$
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 $\ce{CaCl_2}$ is substantially more effective at lowering the freezing point of water because its solutions contain three ions per formula unit. In fact, $\ce{CaCl_2}$ 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 Example $6$.8 and Example $6$.10.
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. $KCl$, $SrCl_2$, and $HCl$ 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: $KCl$ and $HCl$, 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 > $HCl$ = $KCl$ > $SrCl_2$.
Exercise $5$
Arrange these aqueous solutions in order of increasing freezing points: 0.2 m $NaCl$, 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)
Boiling Point Elevation and Freezing Point Depression: https://youtu.be/0MZm1Ay6LhU
Determination of Molar Masses
Osmotic pressure and changes in freezing point, boiling point, and vapor pressure are directly proportional to the concentration of solute present. Consequently, we can use a measurement of one of these properties to determine the molar mass of the solute from the measurements.
Determining Molar Mass from Freezing Point Depression
A solution of 4.00 g of a nonelectrolyte dissolved in 55.0 g of benzene is found to freeze at 2.32 °C. What is the molar mass of this compound?
Solution
We can solve this problem using the following steps.
1. Determine the change in freezing point from the observed freezing point and the freezing point of pure benzene (Table $1$).
$ΔT_\ce{f}=\mathrm{5.5\:°C−2.32\:°C=3.2\:°C}$
• Determine the molal concentration from Kf, the freezing point depression constant for benzene (Table $1$), and ΔTf.
$ΔT_\ce{f}=K_\ce{f}m$
$m=\dfrac{ΔT_\ce{f}}{K_\ce{f}}=\dfrac{3.2\:°\ce C}{5.12\:°\ce C m^{−1}}=0.63\:m$
• Determine the number of moles of compound in the solution from the molal concentration and the mass of solvent used to make the solution.
$\mathrm{Moles\: of\: solute=\dfrac{0.62\:mol\: solute}{1.00\cancel{kg\: solvent}}×0.0550\cancel{kg\: solvent}=0.035\:mol}$
• Determine the molar mass from the mass of the solute and the number of moles in that mass.
$\mathrm{Molar\: mass=\dfrac{4.00\:g}{0.034\:mol}=1.2×10^2\:g/mol}$
Exercise $6$
A solution of 35.7 g of a nonelectrolyte in 220.0 g of chloroform has a boiling point of 64.5 °C. What is the molar mass of this compound?
Answer
1.8 × 102 g/mol
Determination of a Molar Mass from Osmotic Pressure
A 0.500 L sample of an aqueous solution containing 10.0 g of hemoglobin has an osmotic pressure of 5.9 torr at 22 °C. What is the molar mass of hemoglobin?
Solution
Here is one set of steps that can be used to solve the problem:
1. Convert the osmotic pressure to atmospheres, then determine the molar concentration from the osmotic pressure.
$\Pi=\mathrm{\dfrac{5.9\:torr×1\:atm}{760\:torr}=7.8×10^{−3}\:atm}$
$\Pi=MRT$
$M=\dfrac{Π}{RT}=\mathrm{\dfrac{7.8×10^{−3}\:atm}{(0.08206\:L\: atm/mol\: K)(295\:K)}=3.2×10^{−4}\:M}$
• Determine the number of moles of hemoglobin in the solution from the concentration and the volume of the solution.
$\mathrm{moles\: of\: hemoglobin=\dfrac{3.2×10^{−4}\:mol}{1\cancel{L\: solution}}×0.500\cancel{L\: solution}=1.6×10^{−4}\:mol}$
• Determine the molar mass from the mass of hemoglobin and the number of moles in that mass.
$\mathrm{molar\: mass=\dfrac{10.0\:g}{1.6×10^{−4}\:mol}=6.2×10^4\:g/mol}$
Exercise $7$
What is the molar mass of a protein if a solution of 0.02 g of the protein in 25.0 mL of solution has an osmotic pressure of 0.56 torr at 25 °C?
Answer
2.7 × 104 g/mol
Finding the Molecular Weight of an Unknown using Colligative Properties:
https://youtu.be/faSk2REYy74
Summary
• vapor pressure lowering: $P^0_A−P_A=ΔP_A=X_BP^0_A$
• vapor pressure of a system containing two volatile components: $P_T=X_AP^0_A+(1−X_A)P^0_B$
• boiling point elevation: $ΔT_b = mK_b$
• freezing point depression: $ΔT_f = mK_f$
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. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/13%3A_Solutions_and_their_Physical_Properties/13.08%3A_Freezing-Point_Depression_and_Boiling-Point_Elevation_of_Nonelectrolyte_Solutions.txt |
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 $NaCl$ 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 $NaCl$ and 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: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.
$i=\dfrac{\text{apparent number of particles in solution}}{\text{ number of moles of solute dissolved}} \label{13.9.1}$
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 $1$ 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 $1$: 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 $1$). Each of these temporary units behaves like a single dissolved particle until it dissociates. Highly charged ions such as $Mg^{2+}$, $Al^{3+}$, $\ce{SO4^{2−}}$, and $\ce{PO4^{3−}}$ 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 $1$: 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 13.9.12 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 13.9.1 to calculate the van’t Hoff factor.
Solution:
A If $FeCl_3$ 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
$\Pi=MRT=(0.200 \;mol/L) \left[0.0821\;(L⋅atm)/(K⋅mol) \right] (298\; K)=4.89\; atm$
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 $FeCl_3$ dissolved. Alternatively, we can calculate the observed particle concentration from the osmotic pressure of 4.15 atm:
$4.15\; atm=M \left[ 0.0821 \;(L⋅atm)/(K⋅mol)\right] (298 \;K)$
$0.170 mol/L=M$
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 $FeCl_3$ dissolved. From Equation \ref{13.9.1}, the van’t Hoff factor for the solution is
$i=\dfrac{\text{3.40 particles observed}}{\text{1 formula unit}\; FeCl_3}=3.40$
Exercise $1$: 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
Key Concepts and Summary
Ionic compounds may not completely dissociate in solution due to activity effects, in which case observed colligative effects may be less than predicted. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/13%3A_Solutions_and_their_Physical_Properties/13.09%3A_Solutions_of_Electrolytes.txt |
• 14.1: The Rate of a Chemical Reaction
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.
• 14.2: Measuring Reaction Rates
The method for determining a reaction rate is relatively straightforward. Since a reaction rate is based on change over time, it must be determined from tabulated values or found experimentally. With the obtained data, it is possible to calculate the reaction rate either algebraically or graphically.
• 14.3: Effect of Concentration on Reaction Rates: The Rate Law
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.
• 14.4: Zero-Order Reactions
The rates of zero-order reactions is apparently independent of reactant concentrations. This means that the rate of the reaction is equal to the rate constant, k, of that reaction, but clearly a zero-order process cannot continue after a reactant has been exhausted. Just before this point is reached, the reaction will revert to another rate law instead of falling directly to zero Situations which are apparently zero order occur when a reaction is catalyzed by attachment to a solid surface (hete
• 14.5: 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.
• 14.6: 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 2A → products. 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.
• 14.7: Reaction Kinetics: A Summary
For a zeroth-order reaction, a plot of the concentration of any reactant versus time is a straight line with a slope of −k. For a first-order reaction, a plot of the natural logarithm of the concentration of a reactant versus time is a straight line with a slope of −k. For a second-order reaction, a plot of the inverse of the concentration of a reactant versus time is a straight line with a slope of k.
• 14.8: Theoretical Models for Chemical Kinetics
• 14.9: The Effect of Temperature on Reaction Rates
• 14.10: Reaction Mechanisms
• 14.11: Catalysis
• 14.E: Exercises
14: Chemical Kinetics
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_General_Chemistry_(Petrucci_et_al.)/14%3A_Chemical_Kinetics/14.01%3A_The_Rate_of_a_Chemical_Reaction.txt |
The method for determining a reaction rate is relatively straightforward. Since a reaction rate is based on change over time, it must be determined from tabulated values or found experimentally. With the obtained data, it is possible to calculate the reaction rate either algebraically or graphically. What follows is general guidance and examples of measuring the rates of a reaction. Measuring time change is easy; a stopwatch or any other time device is sufficient. However, determining the change in concentration of the reactants or products involves more complicated processes. The change of concentration in a system can generally be acquired in two ways:
1. By monitoring the depletion of reactant over time, or
2. By monitoring the formation of product over time
It does not matter whether an experimenter monitors the reagents or products because there is no effect on the overall reaction. However, since reagents decrease during reaction, and products increase, there is a sign difference between the two rates. Reagent concentration decreases as the reaction proceeds, giving a negative number for the change in concentration. The products, on the other hand, increase concentration with time, giving a positive number. Since the convention is to express the rate of reaction as a positive number, to solve a problem, set the overall rate of the reaction equal to the negative of a reagent's disappearing rate. The overall rate also depends on stoichiometric coefficients.
It is worth noting that the process of measuring the concentration can be greatly simplified by taking advantage of the different physical or chemical properties (ie: phase difference, reduction potential, etc.) of the reagents or products involved in the reaction by using the above methods. We have emphasized the importance of taking the sign of the reaction into account to get a positive reaction rate. Now, we will turn our attention to the importance of stoichiometric coefficients.
A reaction rate can be reported quite differently depending on which product or reagent selected to be monitored.
Given a reaction:
$aA+bB \rightarrow cC + dD$
the general rate for this reaction is defined as
$rate = - \dfrac{1}{a}\dfrac{ \Delta [A]}{ \Delta t} = - \dfrac{1}{b} \dfrac{\Delta [B]}{\Delta t} = \dfrac{1}{c}\dfrac{ \Delta [C]}{\Delta t} = \dfrac{1}{d}\dfrac{ \Delta [D]}{\Delta t} \label{rate1}$
Equation $\ref{rate1}$ can also be written as:
rate of reaction = $- \dfrac{1}{a}$ (rate of disappearance of A)
= $- \dfrac{1}{b}$ (rate of disappearance of B)
= $\dfrac{1}{c}$ (rate of formation of C)
= $\dfrac{1}{d}$ (rate of formation of D)
Even though the concentrations of A, B, C and D may all change at different rates, there is only one average rate of reaction. To get this unique rate, choose any one rate and divide it by the stoichiometric coefficient. When the reaction has the formula:
$C_{R1}R_1 + \dots + C_{Rn}R_n \rightarrow C_{P1}P_1 + \dots + C_{Pn}P_n$
The general case of the unique average rate of reaction has the form:
rate of reaction = $- \dfrac{1}{C_{R1}}\dfrac{\Delta [R_1]}{\Delta t} = \dots = - \dfrac{1}{C_{Rn}}\dfrac{\Delta [R_n]}{\Delta t} = \dfrac{1}{C_{P1}}\dfrac{\Delta [P_1]}{\Delta t} = \dots = \dfrac{1}{C_{Pn}}\dfrac{\Delta [P_n]}{\Delta t}$
Average Reaction Rates: https://youtu.be/jc6jntB7GHk
Following the Course of a Reaction
Rather than performing a whole set of initial rate experiments, one can gather information about orders of reaction by following a particular reaction from start to finish. There are two different ways this can be accomplished.
1. Samples of the mixture can be collected at intervals and titrated to determine how the concentration of one of the reagents is changing.
2. A physical property of the reaction which changes as the reaction continues can be measured: for example, the volume of gas produced.
These approaches must be considered separately.
Consider that bromoethane reacts with sodium hydroxide solution as follows:
$CH_3CH_2Br + OH^- \rightarrow CH_3CH_2OH + Br^-$
During the course of the reaction, both bromoethane and sodium hydroxide are consumed. However, it is relatively easy to measure the concentration of sodium hydroxide at any one time by performing a titration with a standard acid: for example, with hydrochloric acid of a known concentration.
The process starts with known concentrations of sodium hydroxide and bromoethane, and it is often convenient for them to be equal. Because the reaction is 1:1, if the concentrations are equal at the start, they remain equal throughout the reaction. Samples are taken with a pipette at regular intervals during the reaction, and titrated with standard hydrochloric acid in the presence of a suitable indicator. The problem with this approach is that the reaction is still proceeding in the time required for the titration. In addition, only one titration attempt is possible, because by the time another sample is taken, the concentrations have changed.
There are two ways around this problem:
1. The reaction can be slowed by diluting it, adding the sample to a larger volume of cold water before the titration. Then the titration is performed as quickly as possible. This is most effective if the reaction is carried out above room temperature. Cooling it as well as diluting it slows it down even more.
2. If possible (and it is possible in this case) it is better to stop the reaction completely before titrating. In this case, this can be accomplished by adding the sample to a known, excess volume of standard hydrochloric acid. This consumes all the sodium hydroxide in the mixture, stopping the reaction.
At this point the resulting solution is titrated with standard sodium hydroxide solution to determine how much hydrochloric acid is left over in the mixture. This allows one to calculate how much acid was used, and thus how much sodium hydroxide must have been present in the original reaction mixture. This technique is known as a back titration.
This process generates a set of values for concentration of (in this example) sodium hydroxide over time. The concentrations of bromoethane are, of course, the same as those obtained if the same concentrations of each reagent were used. These values are plotted to give a concentration-time graph, such as that below:
The rates of reaction at a number of points on the graph must be calculated; this is done by drawing tangents to the graph and measuring their slopes.
These values are then tabulated. The quickest way to proceed from here is to plot a log graph as described further up the page. All rates are converted to log(rate), and all the concentrations to log(concentration). Then, log(rate) is plotted against log(concentration). The slope of the graph is equal to the order of reaction.
In the example of the reaction between bromoethane and sodium hydroxide solution, the order is calculated to be 2. Notice that this is the overall order of the reaction, not just the order with respect to the reagent whose concentration was measured. The rate of reaction decreases because the concentrations of both of the reactants decrease.
Example $1$: The course of the reaction
A familiar example is the catalytic decomposition of hydrogen peroxide (used above as an example of an initial rate experiment).
This time, measure the oxygen given off using a gas syringe, recording the volume of oxygen collected at regular intervals.
The practical side of this experiment is straightforward, but the calculation is not. The problem is that the volume of the product is measured, whereas the concentration of the reactants is used to find the reaction order. This means that the concentration of hydrogen peroxide remaining in the solution must be determined for each volume of oxygen recorded. This requires ideal gas law and stoichiometric calculations.
The table of concentrations and times is processed as described above.
Example $2$: The catalytic decomposition of hydrogen peroxide
This is an example of measuring the initial rate of a reaction producing a gas.
A simple set-up for this process is given below:
The reason for the weighing bottle containing the catalyst is to avoid introducing errors at the beginning of the experiment. The catalyst must be added to the hydrogen peroxide solution without changing the volume of gas collected. If it is added to the flask using a spatula before replacing the bung, some gas might leak out before the bung is replaced. Alternatively, air might be forced into the measuring cylinder. Either would render results meaningless.
To start the reaction, the flask is shaken until the weighing bottle falls over, and then shaken further to make sure the catalyst mixes evenly with the solution. Alternatively, a special flask with a divided bottom could be used, with the catalyst in one side and the hydrogen peroxide solution in the other. The two are easily mixed by tipping the flask. Using a 10 cm3 measuring cylinder, initially full of water, the time taken to collect a small fixed volume of gas can be accurately recorded. A small gas syringe could also be used.
To study the effect of the concentration of hydrogen peroxide on the rate, the concentration of hydrogen peroxide must be changed and everything else held constant—the temperature, the total volume of the solution, and the mass of manganese(IV) oxide. The manganese(IV) oxide must also always come from the same bottle so that its state of division is always the same.
The same apparatus can be used to determine the effects of varying the temperature, catalyst mass, or state of division due to the catalyst
Example $3$: The thiosulphate-acid reaction
Mixing dilute hydrochloric acid with sodium thiosulphate solution causes the slow formation of a pale yellow precipitate of sulfur.
$Na_2S_2O_{2(aq)} + 2HCl_{(aq)} \rightarrow 2NaCl_{(aq)} + H_2O_{(l)} + S_{(s)} + SO_{2(g)}$
A very simple, but very effective, way of measuring the time taken for a small fixed amount of precipitate to form is to stand the flask on a piece of paper with a cross drawn on it, and then look down through the solution until the cross disappears.
A known volume of sodium thiosulphate solution is placed in a flask. Then a small known volume of dilute hydrochloric acid is added, a timer is started, the flask is swirled to mix the reagents, and the flask is placed on the paper with the cross. The timer is used to determine the time for the cross to disappear. The process is repeated using a smaller volume of sodium thiosulphate, but topped up to the same original volume with water. Everything else is exactly as before.
The actual concentration of the sodium thiosulphate does not need to be known. In each case the relative concentration could be recorded. The solution with 40 cm3 of sodium thiosulphate solution plus 10 cm3 of water has a concentration which is 80% of the original, for example. The one with 10 cm3 of sodium thiosulphate solution plus 40 cm3 of water has a concentration 20% of the original.
The quantity 1/t can again be plotted as a measure of the rate, and the volume of sodium thiosulphate solution as a measure of concentration. Alternatively, relative concentrations could be plotted. In either case, the shape of the graph is the same.
The effect of temperature on this reaction can be measured by warming the sodium thiosulphate solution before adding the acid. The temperature must be measured after adding the acid, because the cold acid cools the solution slightly.This time, the temperature is changed between experiments, keeping everything else constant. To get reasonable times, a diluted version of the sodium thiosulphate solution must be used. Using the full strength, hot solution produces enough precipitate to hide the cross almost instantly.
Example $4$: The Iodine Clock Reactions
There are several reactions bearing the name "iodine clock." Each produces iodine as one of the products. This is the simplest of them, because it involves the most familiar reagents.
The reaction below is the oxidation of iodide ions by hydrogen peroxide under acidic conditions:
$H_2O_{2(aq)} + 2I_{(aq)}^- + 2H^+ \rightarrow I_{2(aq)} + 2H_2O_{(l)}$
The iodine is formed first as a pale yellow solution, darkening to orange and then dark red before dark gray solid iodine is precipitated.
Iodine reacts with starch solution to give a deep blue solution. If starch solution is added to the reaction above, as soon as the first trace of iodine is formed, the solution turns blue. This gives no useful information. However, iodine also reacts with sodium thiosulphate solution:
$2S_2O^{2-}_{3(aq)} + I_{2(aq)} \rightarrow S_2O_{6(aq)}^{2-} + 2I^-_{(aq)}$
If a very small amount of sodium thiosulphate solution is added to the reaction mixture (including the starch solution), it reacts with the iodine that is initially produced, so the iodine does not affect the starch, and there is no blue color. However, when that small amount of sodium thiosulphate is consumed, nothing inhibits further iodine produced from reacting with the starch. The mixture turns blue.
Average vs. Instantaneous Reaction Rates
Reaction rates have the general form of (change of concentration / change of time). There are two types of reaction rates. One is called the average rate of reaction, often denoted by (Δ[conc.] / Δt), while the other is referred to as the instantaneous rate of reaction, denoted as either:
$\lim_{\Delta t \rightarrow 0} \dfrac{\Delta [concentration]}{\Delta t}$
or
$\dfrac{d [concentration]}{dt}$
The average rate of reaction, as the name suggests, is an average rate, obtained by taking the change in concentration over a time period, for example: -0.3 M / 15 minutes. This is an approximation of the reaction rate in the interval; it does not necessarily mean that the reaction has this specific rate throughout the time interval or even at any instant during that time. The instantaneous rate of reaction, on the other hand, depicts a more accurate value. The instantaneous rate of reaction is defined as the change in concentration of an infinitely small time interval, expressed as the limit or derivative expression above. Instantaneous rate can be obtained from the experimental data by first graphing the concentration of a system as function of time, and then finding the slope of the tangent line at a specific point which corresponds to a time of interest. Alternatively, experimenters can measure the change in concentration over a very small time period two or more times to get an average rate close to that of the instantaneous rate. The reaction rate for that time is determined from the slope of the tangent lines.
Initial Rate of Reaction
The initial rate of reaction is the rate at which the reagents are first brought together. Like the instantaneous rate mentioned above, the initial rate can be obtained either experimentally or graphically. To experimentally determine the initial rate, an experimenter must bring the reagents together and measure the reaction rate as quickly as possible. If this is not possible, the experimenter can find the initial rate graphically. To do this, he must simply find the slope of the line tangent to the reaction curve when t=0.
The simplest initial rate experiments involve measuring the time taken for some recognizable event to happen early in a reaction. This could be the time required for 5 cm3 of gas to be produced, for a small, measurable amount of precipitate to form, or for a dramatic color change to occur. Examples of these three indicators are discussed below.
The concentration of one of the components of the reaction could be changed, holding everything else constant: the concentrations of other reactants, the total volume of the solution and the temperature. The time required for the event to occur is then measured. This process is repeated for a range of concentrations of the substance of interest. A reasonably wide range of concentrations must be measured.This process could be repeated by altering a different property.
Consider a simple example of an initial rate experiment in which a gas is produced. This might be a reaction between a metal and an acid, for example, or the catalytic decomposition of hydrogen peroxide. If volume of gas evolved is plotted against time, the first graph below results.
A measure of the rate of the reaction at any point is found by measuring the slope of the graph. The steeper the slope, the faster the rate. Because the initial rate is important, the slope at the beginning is used. In the second graph, an enlarged image of the very beginning of the first curve, the curve is approximately straight. This is only a reasonable approximation when considering an early stage in the reaction. As the reaction progresses, the curvature of the graph increases. Suppose the experiment is repeated with a different (lower) concentration of the reagent. Again, the time it takes for the same volume of gas to evolve is measured, and the initial stage of the reaction is studied.
Instantaneous Rates: https://youtu.be/GGOdoIzxvAo
Example $2$
Determine the initial rate of the reaction using the table below.
Time [A]
100 1.55
200 0.99
300 0.67
400 0.45
500 0.34
600 0.24
Solution
initial rate of reaction = $\dfrac{-(0-2.5) M}{(195-0) sec}$ = 0.0125 M per sec
Use the points [A]=2.43 M, t= 0 and [A]=1.55, t=100
initial rate of reaction = $- \dfrac{\Delta [A]}{\Delta t} = \dfrac{-(1.55-2.43) M }{\ (100-0) sec}$ = 0.0088 M per sec
Contributors and Attributions
• Jessica Lin, Brenda Mai, Elizabeth Sproat, Nyssa Spector, Joslyn Wood
• Jim Clark (Chemguide.co.uk) | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/14%3A_Chemical_Kinetics/14.02%3A_Measuring_Reaction_Rates.txt |
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.
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.
Note
Under a given set of conditions, the value of the rate constant does not change as the reaction progresses.
Although differential rate laws are generally used to describe what is occurring on a molecular level during a reaction, integrated rate laws are used to determine the reaction order and the value of the rate constant from experimental measurements. (Click the link for a presentation of the general forms for integrated rate laws.)
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
$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}$
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. For methyl bromide, in contrast, the differential rate law becomes rate =k″[CH3Br][OH], with an overall reaction order of 2. Although the two reactions proceed similarly in neutral solution, they proceed very differently in the presence of a base, providing clues as to how the reactions differ on a molecular level.
Note
Differential rate laws are generally used to describe what is occurring on a molecular level during a reaction, whereas integrated rate laws are used for determining the reaction order and the value of the rate constant from experimental measurements.
Example $1$
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. $\mathrm{2HI(g)}\xrightarrow{\textrm{Pt}}\mathrm{H_2(g)}+\mathrm{I_2(g)} \ \textrm{rate}=-\frac{1}{2}\left (\frac{\Delta[\mathrm{HI}]}{\Delta t} \right )=k[\textrm{HI}]^2$
2. $\mathrm{2N_2O(g)}\xrightarrow{\Delta}\mathrm{2N_2(g)}+\mathrm{O_2(g)} \ \textrm{rate}=-\frac{1}{2}\left (\frac{\Delta[\mathrm{N_2O}]}{\Delta t} \right )=k$
3. $\mathrm{cyclopropane(g)}\rightarrow\mathrm{propane(g)} \ \textrm{rate}=-\frac{\Delta[\mathrm{cyclopropane}]}{\Delta t}=k[\mathrm{cyclopropane}]$
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)]:
$k\textrm M^2=\dfrac{\textrm M}{\textrm s}k=\dfrac{\textrm{M/s}}{\textrm M^2}=\dfrac{1}{\mathrm{M\cdot s}}=\mathrm{M^{-1}\cdot s^{-1}}$
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 $1$
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.
a.
$\textrm{CH}_3\textrm N\textrm{=NCH}_3\textrm{(g)}\rightarrow\mathrm{C_2H_6(g)}+\mathrm{N_2(g)}\hspace{5mm}$
with
\begin{align} \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}
b.
$\mathrm{2NO_2(g)}+\mathrm{F_2(g)}\rightarrow\mathrm{2NO_2F(g)}\hspace{5mm}$
with
\begin{align} \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
1. s−1; first order in CH3N=NCH3; first order overall; doubling [CH3N=NCH3] will double the reaction rate.
2. M−1·s−1; first order in NO2, first order in F2; second order overall; doubling [NO2] will double the reaction rate.
Methods of Initial Rates
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 $1$.
$\dfrac{\mathrm{rate_1}}{\mathrm{rate_3}}=\dfrac{k[\textrm A_1]^m[\textrm B_1]^n}{k[\textrm A_3]^m[\textrm B_3]^n}$
Inserting the appropriate values from Table $1$,
$\dfrac{8.5\times10^{-3}\textrm{ M/min}}{34\times10^{-3}\textrm{ M/min}}=\dfrac{k[\textrm{0.50 M}]^m[\textrm{0.50 M}]^n}{k[\textrm{1.00 M}]^m[\textrm{0.50 M}]^n}$
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 $1$,
$\dfrac{8.5\times10^{-3}\textrm{ M/min}}{8.5\times10^{-3}\textrm{ M/min}}=\dfrac{k[\textrm{0.50 M}]^m[\textrm{0.50 M}]^n}{k[\textrm{0.50 M}]^m[\textrm{1.00 M}]^n}$
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 $1$ 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 $2$
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:
$2NO(g) + O_2(g) \rightarrow 2NO_2(g)$
Determine the 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
$k=\dfrac{\textrm{rate}}{[\mathrm{NO}]^2[\mathrm{O_2}]}=\dfrac{7.98\times10^{-3}\textrm{ M/s}}{(0.0235\textrm{ M})^2(0.0125\textrm{ M})}=1.16\times10^3\;\mathrm{ M^{-2}\cdot s^{-1}}$
Alternatively, using Experiment 2 gives
$k=\dfrac{\textrm{rate}}{[\mathrm{NO}]^2[\mathrm{O_2}]}=\dfrac{15.9\times10^{-3}\textrm{ M/s}}{(0.0235\textrm{ M})^2(0.0250\textrm{ M})}=1.15\times10^3\;\mathrm{ M^{-2}\cdot s^{-1}}$
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 $2$
The peroxydisulfate ion (S2O82) is a potent oxidizing agent that reacts rapidly with iodide ion in water:
$S_2O^{2−}_{8(aq)} + 3I^−_{(aq)} \rightarrow 2SO^{2−}_{4(aq)} + I^−_{3(aq)}$
The following table lists kinetics data for this reaction at 25°C. Determine the rate law and calculate the rate constant.
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
Initial Rates and Rate Law Expressions: https://youtu.be/VZl5dipsCEQ
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_General_Chemistry_(Petrucci_et_al.)/14%3A_Chemical_Kinetics/14.03%3A_Effect_of_Concentration_on_Reaction_Rates%3A_The_Rate_Law.txt |
A zeroth-order reaction is one whose rate is independent of concentration; its differential rate law is rate = k. 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, with y = [A], mx = −kt, and b = [A]0.) 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 $1$, 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 NAD+ (nicotinamide adenine dinucleotide) and 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 (part (a) in Figure $3$). 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.
Zero-Order Reactions: https://youtu.be/64i7uYsVsSs | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/14%3A_Chemical_Kinetics/14.04%3A_Zero-Order_Reactions.txt |
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, 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. We have already encountered two examples of first-order reactions: the hydrolysis of aspirin and the reaction of t-butyl bromide with water to give t-butanol. Another 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 $1$ 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 $1$ have been studied extensively to find ways of maximizing the concentration of the active species.
Note
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. 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
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
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:
$\mathrm{CH_3CH_2Cl(g)}\xrightarrow{\Delta}\mathrm{HCl(g)}+\mathrm{C_2H_4(g)} \nonumber$
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.
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:
SO2Cl2(g) → SO2(g) + Cl2(g)
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.
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. Part (a) in Figure $3$ 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 $3$. 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 $3$ 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}} \-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}} \k&=1.5\times10^{-3}\;\mathrm{min^{-1}}\end{align}
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 $3$ 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.
The First-Order Integrated Rate Law Equation: https://youtu.be/_JskhfxBAMI
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} \&=\textrm{0.0200 M}(e^{\large{-(1.6\times10^{-6}\textrm{ s}^{-1})[(10\textrm{ h})(60\textrm{ min/h})(60\textrm{ s/min})]}}) \&=0.0189\textrm{ M}\end{align}
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 \ &=\ln 0.0200-(1.6\times10^{-6}\textrm{ s}^{-1})[(\textrm{10 h})(\textrm{60 min/h})(\textrm{60 s/min})] \ &=-3.912-0.0576=-3.970 \ [\mathrm{CH_3CH_2Cl}]_{\textrm{10 h}}&=e^{-3.970}\textrm{ M} \ &=0.0189\textrm{ M}\end{align}
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 \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} \ 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) \ &=\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}\end{align}
Exercise $2$
In the exercise above, you found that the decomposition of sulfuryl chloride (SO2Cl2) is first order, and you calculated the rate constant at 320°C. Use the form(s) of the integrated rate law to find the amount of SO2Cl2 that remains after 20 h if a sample with an original concentration of 0.123 M is heated at 320°C. How long would it take for 90% of the SO2Cl2 to decompose?
Answer 0.0252 M; 29 h
Example Using the First-Order Integrated Rate Law Equation:
https://youtu.be/fLY6MtNl9-g | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/14%3A_Chemical_Kinetics/14.05%3A_First-Order_Reactions.txt |
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 2A → products. 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.
Note
Second-order reactions generally have the form 2A → products or A + B → products.
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:
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 $1$: 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 $1$. 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:
$\dfrac{5.0\times10^{-5}\textrm{ M/min}}{1.8\times10^{-5}\textrm{ M/min}}=2.8\hspace{5mm}\textrm{and}\hspace{5mm}\dfrac{3.4\times10^{-3}\textrm{ M}}{2.0\times10^{-3} \textrm{ M}}=1.7$
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 $1$, 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}
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 part (a) in Figure $2$. 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 part (b) in Figure $7$. 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 $1$
At high temperatures, nitrogen dioxide decomposes to nitric oxide and oxygen.
$\mathrm{2NO_2(g)}\xrightarrow{\Delta}\mathrm{2NO(g)}+\mathrm{O_2(g)}$
Experimental data for the reaction at 300°C and four initial concentrations of NO2 are listed in the following table:
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}
Exercise $1$
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:
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
Note
If a plot of reactant concentration versus time is not linear, but a plot of 1/reaction concentration versus time is linear, then the reaction is second order.
Example $2$
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 14.4.9,
\begin{align}\dfrac{1}{[\mathrm{NO_2}]_{3600}}&=\dfrac{1}{[\mathrm{NO_2}]_0}+kt=\dfrac{1}{0.056\textrm{ M}}+[(0.54 \mathrm{\;M^{-1}\cdot s^{-1}})(3600\textrm{ s})] \&=2.0\times10^3\textrm{ M}^{-1}\end{align}
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.
$t=\dfrac{(1/[\mathrm{NO_2}])-(1/[\mathrm{NO_2}]_0)}{k}=\dfrac{(1/0.0056 \textrm{ M})-(1/0.056\textrm{ M})}{0.54 \;\mathrm{M^{-1}\cdot s^{-1}}}=3.0\times10^2\textrm{ s}=5.0\textrm{ min} \nonumber$
NO2 decomposes very rapidly; under these conditions, the reaction is 90% complete in only 5.0 min.
Exercise $2$
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:
$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=-\dfrac{\Delta[\textrm B]}{\Delta t}=k[\textrm A][\textrm B] \label{14.4.10}$
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.
Second-Order Integrated Rate Law Equation: https://youtu.be/hMSgk2Rm2xA | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/14%3A_Chemical_Kinetics/14.06%3A_Second-Order_Reactions.txt |
Learning Objectives
• To use graphs to analyze the kinetics of a reaction.
You learned that the integrated rate law for each common type of reaction (zeroth, first, or second order in a single reactant) can be plotted as a straight line. Using these plots offers an alternative to the methods described for showing how reactant concentration changes with time and determining reaction order.
We will illustrate the use of these graphs by considering the thermal decomposition of NO2 gas at elevated temperatures, which occurs according to the following reaction:
$(\mathrm{2NO_2(g)}\xrightarrow{\Delta}\mathrm{2NO(g)}+\mathrm{O_2(g)} \label{14.26}$
Experimental data for this reaction at 330°C are listed in Table $1$; they are provided as [NO2], ln[NO2], and 1/[NO2] versus time to correspond to the integrated rate laws for zeroth-, first-, and second-order reactions, respectively. The actual concentrations of NO2 are plotted versus time in part (a) in Figure $1$.
Because the plot of [NO2] versus t is not a straight line, we know the reaction is not zeroth order in NO2. A plot of ln[NO2] versus t (part (b) in Figure $1$) shows us that the reaction is not first order in NO2 because a first-order reaction would give a straight line. Having eliminated zeroth-order and first-order behavior, we construct a plot of 1/[NO2] versus t (part (c) in Figure $1$). This plot is a straight line, indicating that the reaction is second order in NO2.
Table $1$: Concentration of $NO_2$ as a Function of Time at 330°C
Time (s) [NO2] (M) ln[NO2] 1/[NO2] (M−1)
0 1.00 × 10−2 −4.605 100
60 6.83 × 10−3 −4.986 146
120 5.18 × 10−3 −5.263 193
180 4.18 × 10−3 −5.477 239
240 3.50 × 10−3 −5.655 286
300 3.01 × 10−3 −5.806 332
360 2.64 × 10−3 −5.937 379
We have just determined the reaction order using data from a single experiment by plotting the concentration of the reactant as a function of time. Because of the characteristic shapes of the lines shown in Figure $2$, the graphs can be used to determine the reaction order of an unknown reaction. In contrast, the method of initial rates required multiple experiments at different NO2 concentrations as well as accurate initial rates of reaction, which can be difficult to obtain for rapid reactions.
Example $1$
Dinitrogen pentoxide (N2O5) decomposes to NO2 and O2 at relatively low temperatures in the following reaction:
2N2O5(soln) → 4NO2(soln) + O2(g)
This reaction is carried out in a CCl4 solution at 45°C. The concentrations of N2O5 as a function of time are listed in the following table, together with the natural logarithms and reciprocal N2O5 concentrations. Plot a graph of the concentration versus t, ln concentration versus t, and 1/concentration versus t and then determine the rate law and calculate the rate constant.
Time (s) [N2O5] (M) ln[N2O5] 1/[N2O5] (M−1)
0 0.0365 −3.310 27.4
600 0.0274 −3.597 36.5
1200 0.0206 −3.882 48.5
1800 0.0157 −4.154 63.7
2400 0.0117 −4.448 85.5
3000 0.00860 −4.756 116
3600 0.00640 −5.051 156
Given: balanced chemical equation, reaction times, and concentrations
Asked for: graph of data, rate law, and rate constant
Strategy:
A Use the data in the table to separately plot concentration, the natural logarithm of the concentration, and the reciprocal of the concentration (the vertical axis) versus time (the horizontal axis). Compare the graphs with those in Figure $1$ to determine the reaction order.
B Write the rate law for the reaction. Using the appropriate data from the table and the linear graph corresponding to the rate law for the reaction, calculate the slope of the plotted line to obtain the rate constant for the reaction.
Solution
A Here are plots of [N2O5] versus t, ln[N2O5] versus t, and 1/[N2O5] versus t:
The plot of ln[N2O5] versus t gives a straight line, whereas the plots of [N2O5] versus t and 1/[N2O5] versus t do not. This means that the decomposition of N2O5 is first order in [N2O5].
B The rate law for the reaction is therefore
rate = k[N2O5]
Calculating the rate constant is straightforward because we know that the slope of the plot of ln[A] versus t for a first-order reaction is −k. We can calculate the slope using any two points that lie on the line in the plot of ln[N2O5] versus t. Using the points for t = 0 and 3000 s,
$\textrm{slope}=\dfrac{\ln[\mathrm{N_2O_5}]_{3000}-\ln[\mathrm{N_2O_5}]_0}{3000\textrm{ s}-0\textrm{ s}}=\dfrac{(-4.756)-(-3.310)}{3000\textrm{ s}}=-4.820\times10^{-4}\textrm{ s}^{-1}$
Thus k = 4.820 × 10−4 s−1.
Exercise $1$
1,3-Butadiene (CH2=CH—CH=CH2; C4H6) is a volatile and reactive organic molecule used in the production of rubber. Above room temperature, it reacts slowly to form products. Concentrations of C4H6 as a function of time at 326°C are listed in the following table along with ln[C4H6] and the reciprocal concentrations. Graph the data as concentration versus t, ln concentration versus t, and 1/concentration versus t. Then determine the reaction order in C4H6, the rate law, and the rate constant for the reaction.
Time (s) [C4H6] (M) ln[C4H6] 1/[C4H6] (M−1)
0 1.72 × 10−2 −4.063 58.1
900 1.43 × 10−2 −4.247 69.9
1800 1.23 × 10−2 −4.398 81.3
3600 9.52 × 10−3 −4.654 105
6000 7.30 × 10−3 −4.920 137
Answer
second order in C4H6; rate = k[C4H6]2; k = 1.3 × 10−2 M−1·s−1
Summary
For a zeroth-order reaction, a plot of the concentration of any reactant versus time is a straight line with a slope of −k. For a first-order reaction, a plot of the natural logarithm of the concentration of a reactant versus time is a straight line with a slope of −k. For a second-order reaction, a plot of the inverse of the concentration of a reactant versus time is a straight line with a slope of k.
Key Takeaway
• Plotting the concentration of a reactant as a function of time produces a graph with a characteristic shape that can be used to identify the reaction order in that reactant.
• Anonymous | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/14%3A_Chemical_Kinetics/14.07%3A_Reaction_Kinetics%3A_A_Summary.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_General_Chemistry_(Petrucci_et_al.)/14%3A_Chemical_Kinetics/14.09%3A_The_Effect_of_Temperature_on_Reaction_Rates.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_General_Chemistry_(Petrucci_et_al.)/14%3A_Chemical_Kinetics/14.10%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.
14.E: Exercises
14.7: Reaction Kinetics: A Summary
Conceptual Problems
1. Compare first-order differential and integrated rate laws with respect to the following. Is there any information that can be obtained from the integrated rate law that cannot be obtained from the differential rate law?
1. the magnitude of the rate constant
2. the information needed to determine the order
3. the shape of the graphs
1. In the single-step, second-order reaction 2A → products, how would a graph of [A] versus time compare to a plot of 1/[A] versus time? Which of these would be the most similar to the same set of graphs for A during the single-step, second-order reaction A + B → products? Explain.
1. For reactions of the same order, what is the relationship between the magnitude of the rate constant and the reaction rate? If you were comparing reactions with different orders, could the same arguments be made? Why?
Answers
1. For a given reaction under particular conditions, the magnitude of the first-order rate constant does not depend on whether a differential rate law or an integrated rate law is used.
2. The differential rate law requires multiple experiments to determine reactant order; the integrated rate law needs only one experiment.
3. Using the differential rate law, a graph of concentration versus time is a curve with a slope that becomes less negative with time, whereas for the integrated rate law, a graph of ln[reactant] versus time gives a straight line with slope = −k. The integrated rate law allows you to calculate the concentration of a reactant at any time during the reaction; the differential rate law does not.
1. The reaction rate increases as the rate constant increases. We cannot directly compare reaction rates and rate constants for reactions of different orders because they are not mathematically equivalent.
Numerical Problems
1. One method of using graphs to determine reaction order is to use relative rate information. Plotting the log of the relative rate versus log of relative concentration provides information about the reaction. Here is an example of data from a zeroth-order reaction:
Relative [A] (M) Relative Rate (M/s)
1 1
2 1
3 1
Varying [A] does not alter the reaction rate. Using the relative rates in the table, generate plots of log(rate) versus log(concentration) for zeroth-, first- and second-order reactions. What does the slope of each line represent?
1. The table below follows the decomposition of N2O5 gas by examining the partial pressure of the gas as a function of time at 45°C. What is the reaction order? What is the rate constant? How long would it take for the pressure to reach 105 mmHg at 45°C?
Time (s) Pressure (mmHg)
0 348
400 276
1600 156
3200 69
4800 33 | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/14%3A_Chemical_Kinetics/14.11%3A_Catalysis.txt |
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_General_Chemistry_(Petrucci_et_al.)/15%3A_Principles_of_Chemical_Equilibrium/15.1%3A_Dynamic_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 system described in Equation \ref{15.2.1}, the decomposition of $N_2O_4$ to $NO_2$. Both the forward and reverse reactions for this system consist of a single elementary reaction, so the reaction rates are as follows:
$\text{forward rate} = k_f[N_2O_4] \label{15.2.1}$
and
$\text{reverse rate} = k_r[NO_2]^2 \label{15.2.2}$
At equilibrium, the forward rate equals the reverse rate (definition of equilibrium):
$k_f[N_2O_4] = k_r[NO_2]^2 \label{15.2.3}$
so
$\dfrac{k_f}{k_r}=\dfrac{[NO_2]^2}{[N_2O_4]} \label{15.2.4}$
The ratio of the rate constants gives us a new constant, the equilibrium constant ($K$), which is defined as follows:
$K=\dfrac{k_f}{k_r} \label{15.2.5}$
Hence there is a fundamental relationship between chemical kinetics and chemical equilibrium: under a given set of conditions, the composition of the equilibrium mixture is determined by the magnitudes of the rate constants for the forward and the reverse reactions.
The equilibrium constant is equal to the rate constant for the forward reaction divided by the rate constant for the reverse reaction.
Table $1$ lists the initial and equilibrium concentrations from five different experiments using the reaction system described by Equation \ref{15.1}. At equilibrium the magnitude of the quantity $[NO_2]^2/[N_2O_4]$ is essentially the same for all five experiments. In fact, no matter what the initial concentrations of $NO_2$ and $N_2O_4$ are, at equilibrium the quantity $[NO_2]^2/[N_2O_4]$ 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 $NO_2:N_2O_4$ Mixtures at 25°C
Initial Concentrations Concentrations at Equilibrium
Experiment [$N_2O_4$] (M) [$NO_2$] (M) [$N_2O_4$] (M) [$NO_2$] (M) $K = [NO_2]^2/[N_2O_4]$
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 $6.50 \times 10^{−3}$
Introduction to Dynamic Equilibrium: https://youtu.be/4AJbFuzW2cs
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
$aA+bB \rightleftharpoons cC+dD \label{15.2.6}$
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 and can be stated as follows:
$K=\dfrac{[C]^c[D]^d}{[A]^a[B]^b} \label{15.2.7}$
where $K$ is the equilibrium constant for the reaction. Equation \ref{15.2.6} is called the equilibrium equation, and the right side of Equation \ref{15.2.7} is called the equilibrium constant expression. The relationship shown in Equation \ref{15.2.7} 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.
$\ce{S(s) + O2(g) \rightleftharpoons SO2(g)}$ 300 $4.4 \times 10^{53}$
$\ce{2H2(g) + O2(g) \rightleftharpoons 2H2O (g)}$ 500 $2.4 \times 10^{47}$
$\ce{H2(g) + Cl2(g) \rightleftharpoons 2 HCl(g)}$ 300 $1.6 \times 10^{33}$
$\ce{H2(g) + Br2(g) \rightleftharpoons 2HBr(g)}$ 300 $4.1 \times 10^{18}$
$\ce{2NO(g) + O2(g) \rightleftharpoons 2NO2(g)}$ 300 $4.2 \times 10^{13}$
$\ce{3H2(g) + N2(g) \rightleftharpoons 2NH3(g)}$ 300 $2.7 \times 10^{8}$
$\ce{H2(g) + D2(g) \rightleftharpoons 2HD(g)}$ 100 $1.92$
$\ce{H2(g) + I2(g) \rightleftharpoons 2HI(g)}$ 300 $2.9 \times 10^{−1}$
$\ce{I2(g) \rightleftharpoons 2I(g)}$ 800 $4.6 \times^{ 10−7}$
$\ce{Br2(g) \rightleftharpoons 2Br(g)}$ 1000 $4.0 \times 10^{−7}$
$\ce{Cl2(g) \rightleftharpoons 2Cl (g)}$ 1000 $1.8 \times 10^{−9}$
$\ce{F2(g) \rightleftharpoons 2F(g)}$ 500 $7.4 \times 10^{−13}$
You will also notice in Table $2$ that equilibrium constants have no units, even though Equation \ref{15.2.7} 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{15.2.8}, the units of concentration cancel, which makes $K$ unitless as well:
$\dfrac{[A]_{measured}}{[A]_{standard\; state}}=\dfrac{\cancel{M}}{\cancel{M}} = \dfrac{\cancel{\frac{mol}{L}}}{\cancel{\frac{mol}{L}}} \label{15.2.8}$
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 HD:
$H_{2(g)}+D_{2(g)} \rightleftharpoons 2HD_{(g)} \label{15.2.9}$
The equilibrium constant expression for this reaction is
$K= \dfrac{[HD]^2}{[H_2][D_2]}$
with $K$ varying between 1.9 and 4 over a wide temperature range (100–1000 K). Thus an equilibrium mixture of $H_2$, $D_2$, 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
$\text{reactants} \rightleftharpoons \text{products}.$
Because there is a direct relationship between the kinetics of a reaction and the equilibrium concentrations of products and reactants (Equations \ref{15.2.8} and \ref{15.2.7}), 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 $1$
Write the equilibrium constant expression for each reaction.
• $N_{2(g)}+3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
• $CO_{(g)}+\frac{1}{2}O_{2(g)} \rightleftharpoons CO_{2(g)}$
• $2CO_{2(g)} \rightleftharpoons 2CO_{(g)}+O_{2(g)}$
Given: balanced chemical equations
Asked for: equilibrium constant expressions
Strategy:
Refer to Equation \ref{15.2.7}. 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, $N_2$ has a coefficient of 1 and H2 has a coefficient of 3. The equilibrium constant expression is as follows:
$\dfrac{[NH_3]^2}{[N_2][H_2]^3}$
The only product is carbon dioxide, which has a coefficient of 1. The reactants are $CO$, with a coefficient of 1, and $O_2$, with a coefficient of $\frac{1}{2}$. Thus the equilibrium constant expression is as follows:
$\dfrac{[CO_2]}{[CO][O_2]^{1/2}}$
This reaction is the reverse of the reaction in part b, with all coefficients multiplied by 2 to remove the fractional coefficient for $O_2$. The equilibrium constant expression is therefore the inverse of the expression in part b, with all exponents multiplied by 2:
$\dfrac{[CO]^2[O_2]}{[CO_2]^2}$
Exercise $1$
Write the equilibrium constant expression for each reaction.
1. $N_2O_{(g)} \rightleftharpoons N_{2(g)}+\frac{1}{2}O_{2(g)}$
2. $2C_8H_{18(g)}+25O_{2(g)} \rightleftharpoons 16CO_{2(g)}+18H_2O_{(g)}$
3. $H_{2(g)}+I_{2(g)} \rightleftharpoons 2HI_{(g)}$
Answer:
1. $K=\dfrac{[N_2][O_2]^{1/2}}{[N_2O]}$
2. $K=\dfrac{[CO_2]^{16}[H_2O]^{18}}{[C_8H_{18}]^2[O_2]^{25}}$
3. $K=\dfrac{[HI]^2}{[H_2][I_2]}$
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:
$3H_{2(g)}+N_{2(g)} \rightleftharpoons 2NH_{3(g)}$
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
• $K_{327°C} = 4.1$.
At which temperature would you expect to find the highest proportion of $H_2$ and $N_2$ in the equilibrium mixture?
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:
1. 327°C, where $K$ is smallest
2. 25°C
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 15.2.6 in reverse, we obtain the following:
$cC+dD \rightleftharpoons aA+bB \label{15.2.10}$
The corresponding equilibrium constant $K′$ is as follows:
$K'=\dfrac{[A]^a[B]^b}{[C]^c[D]^d} \label{15.2.11}$
This expression is the inverse of the expression for the original equilibrium constant, so $K′ = 1/K$. That is, when we write a reaction in the reverse direction, the equilibrium constant expression is inverted. For instance, the equilibrium constant for the reaction $N_2O_4$ \rightleftharpoons 2NO_2\) is as follows:
$K=\dfrac{[NO_2]^2}{[N_2O_4]} \label{15.2.12}$
but for the opposite reaction, $2 NO_2 \rightleftharpoons N_2O_4$, the equilibrium constant K′ is given by the inverse expression:
$K'=\dfrac{[N_2O_4]}{[NO_2]^2} \label{15.2.13}$
Consider another example, the formation of water: $2H_{2(g)}+O_{2(g)} \rightleftharpoons 2H_2O_{(g)}$. Because $H_2$ is a good reductant and $O_2$ is a good oxidant, this reaction has a very large equilibrium constant ($K = 2.4 \times 10^{47}$ at 500 K). Consequently, the equilibrium constant for the reverse reaction, the decomposition of water to form $O_2$ and $H_2$, is very small: $K′ = 1/K = 1/(2.4 \times 10^{47}) = 4.2 \times 10^{−48}$. As suggested by the very small equilibrium constant, and fortunately for life as we know it, a substantial amount of energy is indeed needed to dissociate water into $H_2$ and $O_2$.
The equilibrium constant for a reaction written in reverse is the 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
$2NO_2 \rightleftharpoons N_2O_4$
as
$NO_2 \rightleftharpoons \frac{1}{2}N_2O_4$
with the equilibrium constant K″ is as follows:
$K′′=\dfrac{[N_2O_4]^{1/2}}{[NO_2]} \label{15.2.14}$
The values for K′ (Equation \ref{15.2.13}) and K″ are related as follows:
$K′′=(K')^{1/2}=\sqrt{K'} \label{15.2.15}$
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.
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. $2NH_{3(g)} \rightleftharpoons N2(g)+3H_{2(g)}$
2. $\frac{1}{2}N_{2(g)}+\frac{3}{2}H_{2(g)} \rightleftharpoons NH_{3(g)}$
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:
$K=\dfrac{[NH_3]^2}{[N_2][H_2]^3}=0.118$
This reaction is the reverse of the one given, so its equilibrium constant expression is as follows:
$K'=\dfrac{1}{K}=\dfrac{[N_2][H_2]^3}{[NH_3]^2}=\dfrac{1}{0.118}=8.47$
In this reaction, the stoichiometric coefficients of the given reaction are divided by 2, so the equilibrium constant is calculated as follows:
$K′′=\dfrac{[NH_3]}{[N_2]^{1/2}[H_2]^{3/2}}=K^{1/2}=\sqrt{K}=\sqrt{0.118} = 0.344$
Exercise $3$
At 527°C, the equilibrium constant for the reaction
$2SO_{2(g)}+O_{2(g)} \rightleftharpoons 2SO_{3(g)}$
is $7.9 \times 10^4$. Calculate the equilibrium constant for the following reaction at the same temperature:
$SO_{3(g)} \rightleftharpoons SO_{2(g)}+\frac{1}{2}O_{2(g)}$
Answer: $3.6 \times 10^{−3}$
Determining the Equilibrium Expression: https://youtu.be/ZK9cMIWFerY
Summary
• 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 $K$: $K_p = K(RT)^{Δn} \nonumber$
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. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/15%3A_Principles_of_Chemical_Equilibrium/15.2%3A_The_Equilibrium_Constant_Expression.txt |
Learning Objectives
• To know the relationship between the equilibrium constant and the rate constants for the forward and reverse reactions.
• To understand how different phases affect equilibria.
Relating Stoichiometry and Equilibrium Constants
It is important to remember that an equilibrium constant is always tied to a specific chemical equation, and if we write the equation in reverse or multiply its coefficients by a common factor, the value of $K$ will change. Fortunately, the rules are very simple:
• Writing the equation in reverse will invert the equilibrium expression and
• Multiplying the coefficients by a common factor will raise K to the corresponding power.
Here are some of the possibilities for the reaction involving the equilibrium between gaseous water and its elements:
$2 H_2 + O_2\rightleftharpoons 2 H_2O$ $10 H_2 + 5 O_2 \rightleftharpoons 10 H_2O$ $H_2 + ½ O_2 \rightleftharpoons H_2O$ $H_2O \rightleftharpoons H_2 + ½ O_2$
$K_p = \dfrac{P_{H_2O}^2}{P_{H_2}^2P_{O_2}}$
$K_p = \dfrac{P_{H_2O}^{10}}{P_{H_2}^{10}P_{O_2}^5}$
$= \left(\dfrac{P_{H_2O}^2}{P_{H_2}^2P_{O_2}}\right)^{5}$
$K_p = \dfrac{P_{H_2O}}{P_{H_2}P_{O_2}^{1/2}}$
$= \left(\dfrac{P_{H_2O}^2}{P_{H_2}^2P_{O_2}}\right)^{1/2}$
$K_p = \dfrac{P_{H_2}P_{O_2}^{1/2}}{P_{H_2O}}$
$= \left(\dfrac{P_{H_2O}^2}{P_{H_2}^2P_{O_2}}\right)^{-1/2}$
Many chemical changes can be regarded as the sum or difference of two or more other reactions. If we know the equilibrium constants of the individual processes, we can easily calculate that for the overall reaction according to the following rule:
The Multi-step Equilibrium Rule
The equilibrium constant for the sum of two or more reactions is the product of the equilibrium constants for each of the steps.
Example $1$
Calculate the value of $K$ for the reaction
$CaCO_{3(s)} + H^+_{(aq)} \rightarrow Ca^{2+}_{(aq)} + HCO^–_{3(aq)}$
given the following equilibrium constants:
$CaCO_{3(s)} \rightleftharpoons Ca^{2+}_{(aq)} + CO^{2–}_{3(aq)}$
$K_1 = 10^{–6.3}$
$HCO^–_{3(aq)} \rightleftharpoons H^+_{(aq)} + CO^{2–}_{3(aq)}$
$K_2 = 10^{–10.3}$
Solution
The net reaction is the sum of reaction 1 and the reverse of reaction 2:
$CaCO_{3(s)} \rightleftharpoons Ca^{2+}_{(aq)} + CO^{2–}_{3(aq)}$
$K_1 = 10^{–6.3}$
$H^+_{(aq)} + CO^{2–}_{3(aq)} \rightleftharpoons HCO^–_{3(aq)}$
$K_{–2} = 10^{–(–10.3)}$
$CaCO_{3(s)} + H^+_{(aq)} \rightarrow Ca^{2+}_{(aq)} + HCO^–_{3(aq)}$
$K = \dfrac{K_1}{K_2} = 10^{(-6.3+10.3)} =10^{+4.0}$
Comment:
This net reaction describes the dissolution of limestone by acid; it is responsible for the eroding effect of acid rain on buildings and statues. This an example of a reaction that has practically no tendency to take place by itself (small K1) being "driven" by a second reaction having a large equilibrium constant (K–2). From the standpoint of the Le Chatelier principle, the first reaction is "pulled to the right" by the removal of carbonate by hydrogen ion. Coupled reactions of this type are widely encountered in all areas of chemistry, and especially in biochemistry, in which a dozen or so reactions may be linked.
Example $2$
The synthesis of HBr from hydrogen and liquid bromine has an equilibrium constant $K_p = 4.5 \times 10^{15}$ at 25°C. Given that the vapor pressure of liquid bromine is 0.28 atm, find $K_p$ for the homogeneous gas-phase reaction at the same temperature.
Solution
The net reaction we seek is the sum of the heterogeneous synthesis of HBr and the reverse of the vaporization of liquid bromine:
$H_{2(g)} + Br_{2(l)} \rightleftharpoons 2 HBr_{(g)}$ $K_p = 4.5\times 10^{15}$
$Br_{2(g)} \rightleftharpoons Br_{2(l)}$ $K_p = (0.28)^{–1}$
$H_{2(g)} + Br_{2(g)} \rightleftharpoons 2 HBr_{(g)}$ $K_p = 1.6 \times 10^{19}$
Relationships Involving Equilibrium Constants: https://youtu.be/2vZDpXX1zr0
Equilibria Involving 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):
$K_p=\dfrac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b} \label{15.3.1}$
Thus $K_p$ for the decomposition of $N_2O_4$ is as follows:
$K_p=\dfrac{(P_{NO_2})^2}{P_{N_2O_4}} \label{15.3.2}$
Like K, $K_p$ is a unitless quantity because the quantity that is actually used to calculate it is an “effective pressure,” the ratio of the measured pressure to a standard state of 1 bar (approximately 1 atm), which produces a unitless quantity.The “effective pressure” is called the 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 ($T$):
$\color{red} {K_p = K_c(RT)^{Δn} \label{15.3.3}}$
where $K$ is the equilibrium constant expressed in units of concentration and $Δn$ is the difference between the numbers of moles of gaseous products and gaseous reactants ($n_p − n_r$). The temperature is expressed as the absolute temperature in Kelvin. According to Equation 15.3.3, $K_p = K$ only if the moles of gaseous products and gaseous reactants are the same (i.e., $Δn = 0$). For the decomposition of $N_2O_4$, there are 2 mol of gaseous product and 1 mol of gaseous reactant, so $Δn = 1$. Thus, for this reaction,
$K_p = K(RT)^1 = K_cRT \label{15.3.4}$
Example $4$: The Haber Process
The equilibrium constant for the reaction of nitrogen and hydrogen to give ammonia is 0.118 at 745 K. The balanced equilibrium equation is as follows:
$N_{2(g)}+3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
What is $K_p$ for this reaction at the same temperature?
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 15.3.3 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 15.2.15, we have the following:
$K_p=K(RT)^{−2}=\dfrac{K}{(RT)^2}=\dfrac{0.118}{\{ [0.08206(L \cdot atm)/(mol \cdot K)][745\; K]\}^2}=3.16 \times 10^{−5}$
Because $K_p$ is a unitless quantity, the answer is $K_p = 3.16 \times 10^{−5}$.
Exercise $4$
Calculate $K_p$ for the reaction
$2SO_{2(g)}+O_{2(g)} \rightleftharpoons 2SO_{3(g)}$
at 527°C, if $K = 7.9 \times 10^4$ at this temperature.
Answer: $K_p = 1.2 \times 10^3$
Converting Kc to Kp: https://youtu.be/_2WVnlqXrV4
Combining Equilibrium Constant Expressions
Chemists frequently need to know the equilibrium constant for a reaction that has not been previously studied. In such cases, the desired reaction can often be written as the sum of other reactions for which the equilibrium constants are known. The equilibrium constant for the unknown reaction can then be calculated from the tabulated values for the other reactions.
To illustrate this procedure, let’s consider the reaction of $N_2$ with $O_2$ to give $NO_2$. This reaction is an important source of the $NO_2$ that gives urban smog its typical brown color. The reaction normally occurs in two distinct steps. In the first reaction (1), $N_2$ reacts with $O_2$ at the high temperatures inside an internal combustion engine to give $NO$. The released $NO$ then reacts with additional $O_2$ to give $NO_2$ (2). The equilibrium constant for each reaction at 100°C is also given.
1. $N_{2(g)}+O_{2(g)} \rightleftharpoons 2NO_{(g)}\;\; K_1=2.0 \times 10^{−25}$
2. $2NO_{(g)}+O_{2(g)} \rightleftharpoons 2NO_{2(g)}\;\;\;K_2=6.4 \times 10^9$
Summing reactions (1) and (2) gives the overall reaction of $N_2$ with $O_2$:
1. $N_{2(g)}+2O_{2(g)} \rightleftharpoons 2NO_{2(g)} \;\;\;K_3=?$
The equilibrium constant expressions for the reactions are as follows:
$K_1=\dfrac{[NO]^2}{[N_2][O_2]}\;\;\; K_2=\dfrac{[NO_2]^2}{[NO]^2[O_2]}\;\;\; K_3=\dfrac{[NO_2]^2}{[N_2][O_2]^2} \label{15.3.5}$
What is the relationship between $K_1$, $K_2$, and $K_3$, all at 100°C? The expression for $K_1$ has $[NO]^2$ in the numerator, the expression for $K_2$ has $[NO]^2$ in the denominator, and $[NO]^2$ does not appear in the expression for $K_3$. Multiplying $K_1$ by $K_2$ and canceling the $[NO]^2$ terms,
$K_1K_2=\dfrac{\cancel{[NO]^2}}{[N_2][O_2]} \times \dfrac{[NO_2]^2}{\cancel{[NO]^2}[O_2]}=\dfrac{[NO_2]^2}{[N_2][O_2]^2}=K_3 \label{15.3.6}$
Thus the product of the equilibrium constant expressions for $K_1$ and $K_2$ is the same as the equilibrium constant expression for $K_3$:
$K_3 = K_1K_2 = (2.0 \times 10^{−25})(6.4 \times 10^9) = 1.3 \times 10^{−15} \label{15.3.7}$
The equilibrium constant for a reaction that is the sum of two or more reactions is equal to the product of the equilibrium constants for the individual reactions. In contrast, recall that according to 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 $2$
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:
$CO_{(g)}+ \cancel{3H_{2(g)}} \rightleftharpoons \cancel{CH_{4(g)}} + H_2O_{(g)}$
$\cancel{CH_{4(g)}} +2H_2S_{(g)} \rightleftharpoons CS_{2(g)} + \cancel{3H_{2(g)}} + H_{2(g)}$
$CO_{(g)} + 3H_{2(g)} \rightleftharpoons CS_{2(g)}+H_2O_{(g)}+H_{2(g)}$
The values for $K_1$ and $K_2$ are given, so it is straightforward to calculate $K_3$:
$K_3 = K_1K_2 = (9.17 \times 10^{−2})(3.3 \times 10^4) = 3.03 \times 10^3$
Exercise $2$
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: $K_3 = 1.1 \times 10^{66}$
Heterogeneous 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.
Because the molar concentrations of pure liquids and solids normally do not vary greatly with temperature, their concentrations are treated as constants, which allows us to simplify equilibrium constant expressions that involve pure solids or liquids.The reference states for pure solids and liquids are those forms stable at 1 bar (approximately 1 atm), which are assigned an activity of 1. (Recall that the density of water, and thus its volume, changes by only a few percentage points between 0°C and 100°C.)
Consider the following reaction, which is used in the final firing of some types of pottery to produce brilliant metallic glazes:
$CO_{2(g)}+C_{(s)} \rightleftharpoons 2CO_{(g)} \label{15.3.8}$
The glaze is created when metal oxides are reduced to metals by the product, carbon monoxide. The equilibrium constant expression for this reaction is as follows:
$K=\dfrac{[CO]^2}{[CO_2][C]} \label{15.3.9}$
Because graphite is a solid, however, its molar concentration, determined from its density and molar mass, is essentially constant and has the following value:
$[C] =\dfrac{2.26 \cancel{g}/{\cancel{cm^3}}}{12.01\; \cancel{g}/mol} \times 1000 \; \cancel{cm^3}/L = 188 \; mol/L = 188\;M \label{15.3.10}$
We can rearrange Equation 15.3.8 so that the constant terms are on one side:
$K[C]=K(188)=\dfrac{[CO]^2}{[CO_2]} \label{15.3.11}$
Incorporating the constant value of $[C]$ into the equilibrium equation for the reaction in Equation 15.3.7,
$K'=\dfrac{[CO]^2}{[CO_2]} \label{15.3.12}$
The equilibrium constant for this reaction can also be written in terms of the partial pressures of the gases:
$K_p=\dfrac{(P_{CO})^2}{P_{CO_2}} \label{15.3.13}$
Incorporating all the constant values into $K′$ or $K_p$ allows us to focus on the substances whose concentrations change during the reaction.
Although the concentrations of pure liquids or solids are not written explicitly in the equilibrium constant expression, these substances must be present in the reaction mixture for chemical equilibrium to occur. Whatever the concentrations of $CO$ and $CO_2$, the system described in Equation 15.3.7 will reach chemical equilibrium only if a stoichiometric amount of solid carbon or excess solid carbon has been added so that some is still present once the system has reached equilibrium. As shown in Figure $1$, 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 $3$
Write each expression for $K$, incorporating all constants, and $K_p$ for the following equilibrium reactions.
1. $PCl_{3(l)}+Cl_{2(g)} \rightleftharpoons PCl_{5(s)}$
2. $Fe_3O_{4(s)}+4H_{2(g)} \rightleftharpoons 3Fe_{(s)}+4H_2O_{(g)}$
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 concentrations do not appear in the equilibrium constant expression because they do not change significantly. So
$K=\dfrac{1}{[Cl_2]}$
and
$K_p=\dfrac{1}{P_{Cl_2}}$
This reaction contains two pure solids ($Fe_3O_4$ and $Fe$), which do not appear in the equilibrium constant expressions. The two gases do, however, appear in the expressions:
$K=\dfrac{[H_2O]^4}{[H_2]^4}$
and
$K_p=\dfrac{(P_{H_2O})^4}{(P_{H_2})^4}$
Exercise $3$
Write the expressions for $K$ and $K_p$ for the following reactions.
1. $CaCO_{3(s)} \rightleftharpoons CaO_{(s)}+CO_{2(g)}$
2. $\underset{glucose}{C_6H_{12}O_{6(s)}} + 6O_{2(g)} \rightleftharpoons 6CO_{2(g)}+6H_2O_{(g)}$
Answer:
1. $K = [CO_2]$ and $K_p = P_{CO_2}$
2. $K=\dfrac{[CO_2]^6[H_2O]^6}{[O_2]^6}$ and $K_p=\dfrac{(P_{CO_2})^6(P_{H_2O})^6}{(P_{O_2})^6}$
For reactions carried out in solution, the concentration of the solvent is omitted from the equilibrium constant expression even when the solvent appears in the balanced chemical equation for the reaction. The concentration of the solvent is also typically much greater than the concentration of the reactants or products (recall that pure water is about 55.5 M, and pure ethanol is about 17 M). Consequently, the solvent concentration is essentially constant during chemical reactions, and the solvent is therefore treated as a pure liquid. The equilibrium constant expression for a reaction contains only those species whose concentrations could change significantly during the reaction.
Note
The concentrations of pure solids, pure liquids, and solvents are omitted from equilibrium constant expressions because they do not change significantly during reactions when enough is present to reach equilibrium.
Summary
• 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: $K=\dfrac{k_f}{k_r}$
• Equilibrium constant expression (law of mass action): $K=\dfrac{[C]^c[D]^d}{[A]^a[B]^b}$
• Equilibrium constant expression for reactions involving gases using partial pressures: $K_p=\dfrac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b} \label{15.3.1}$
• Relationship between $K_p$ and $K$: $K_p = K(RT)^{Δn}$
An equilibrated system that contains products and reactants in a single phase is a homogeneous equilibrium; a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibrium. For gases, the equilibrium constant expression can be written as the ratio of the partial pressures of the products to the partial pressures of the reactants, each raised to a power matching its coefficient in the chemical equation. An equilibrium constant calculated from partial pressures ($K_p$) is related to $K$ by the ideal gas constant ($R$), the temperature ($T$), and the change in the number of moles of gas during the reaction. An equilibrium system that contains products and reactants in a single phase is a homogeneous equilibrium; a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibrium. When a reaction can be expressed as the sum of two or more reactions, its equilibrium constant is equal to the product of the equilibrium constants for the individual reactions. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/15%3A_Principles_of_Chemical_Equilibrium/15.3%3A_Relationships_Involving_Equilibrium_Constants.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_General_Chemistry_(Petrucci_et_al.)/15%3A_Principles_of_Chemical_Equilibrium/15.4%3A_The_Magnitude_of_an_Equilibrium_Constant.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_General_Chemistry_(Petrucci_et_al.)/15%3A_Principles_of_Chemical_Equilibrium/15.5%3A_The_Reaction_Quotient_Q_-_Predicting_The_Direction_of_Net_Change.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_General_Chemistry_(Petrucci_et_al.)/15%3A_Principles_of_Chemical_Equilibrium/15.6%3A_Altering_Equilibrium_Conditions_-_Le_Chateliers_Principle.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 previously the equilibrium constant for the decomposition of $\ce{CaCO3(s)}$ to $\ce{CaO(s)}$ and $\ce{CO2(g)}$ is $K = [\ce{CO2}]$. At 800°C, the concentration of $\ce{CO2}$ in equilibrium with solid $\ce{CaCO3}$ and $\ce{CaO}$ 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
$K = \dfrac{[\text{isobutane}]}{[\text{n-butane}]}.$
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,
\begin{align} K &=\dfrac{[isobutane]}{[n-butane]} \[4pt] &=\dfrac{0.041\; M}{0.016\,M} \[4pt] &= 2.6 \label{Eq2} \end{align}
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:
$\ce{2SO2(g) + O2(g) <=> 2SO3(g)} \nonumber$
A mixture of $\ce{SO_2}$ and $\ce{O_2}$ was maintained at 800 K until the system reached equilibrium. The equilibrium mixture contained
• $5.0 \times 10^{-2}\; M\; \ce{SO_3}$,
• $3.5 \times 10^{-3}\; M\; \ce{O_2}$, and
• $3.0 \times 10^{-3}\; M\; \ce{SO_2}$.
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,
\begin{align*} K &=\dfrac{[SO_3]^2}{[SO_2]^2[O_2]} \[4pt] &=\dfrac{(5.0 \times 10^{-2})^2}{(3.0 \times 10^{-3})^2(3.5 \times 10^{-3})} \[4pt] &=7.9 \times 10^4 \end{align*}
To solve for $K_p$, we use the relationship between $K$ and $K_p$, where $Δn = 2 − 3 = −1$:
\begin{align*} K_p &=K(RT)^{Δn} \[4pt] &=7.9 \times 10^4 [(0.08206\; L⋅atm/mol⋅K)(800 K)]^{−1} \[4pt] &=1.2 \times 10^3 \end{align*}
Exercise $1$
Hydrogen gas and iodine react to form hydrogen iodide via the reaction
$\ce{H2(g) + I2(g) <=> 2HI(g)} \nonumber$
A mixture of $\ce{H_2}$ and $\ce{I_2}$ 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:
• $K = 48.8$
• $K_p = 48.8$
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:
$\ce{2 NOCl(g) <=> 2NO(g) + Cl2(g)} \nonumber$
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 $\ce{NOCl}$, $\ce{NO}$, and $\ce{Cl_2}$ 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.
$\ce{2 NOCl(g) <=> 2NO(g) + Cl2(g)} \nonumber$
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:
$\ce{2 NOCl(g) <=> 2NO(g) + Cl2(g)} \nonumber$
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:
$Δ[NO]=\left(\dfrac{0.028\; \cancel{mol \;Cl_2}}{ L}\right)\left(\dfrac{2\; mol\; NO}{1 \cancel{\;mol \;Cl_2}}\right)=0.056\; M$
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:
$Δ[NOCl]= \left(\dfrac{0.028\; \cancel{mol\; Cl_2}}{L}\right) \left(\dfrac{−2\; mol \;NOCl}{1\; \cancel{mol\; Cl_2}} \right) = -0.056 \;M$
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 $\ce{NO}$ and $\ce{NOCl}$:
$[NO]_f = 0.000\; M + 0.056 \;M = 0.056\; M$
$[NOCl]_f = 0.500\; M + (−0.056\; M) = 0.444 M$
We can now complete the table:
$2 NOCl_{(g)} \rightleftharpoons 2NO_{(g)}+Cl_{2(g)}$
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:
$K=\dfrac{[NO]^2[Cl_2]}{[NOCl]^2}=\dfrac{(0.056)^2(0.028)}{(0.444)^2}=4.5 \times 10^{−4}$
Exercise $2$
The German chemist Fritz Haber (1868–1934; Nobel Prize in Chemistry 1918) was able to synthesize ammonia ($\ce{NH3}$) by reacting $0.1248\; M \; \ce{H_2}$ and $0.0416\; M \;\ce{N_2}$ at about 500°C. At equilibrium, the mixture contained 0.00272 M $\ce{NH_3}$. What is $K$ for the reaction
$\ce{N2 + 3H2 <=> 2NH_3}$
at this temperature? What is $K_p$?
Answer:
• $K = 0.105$
• $K_p = 2.61 \times 10^{-5}$
Using Q to Find Equilibrium Concentrations: https://youtu.be/-sbeII65Z7w
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 $2$.
$\text{n-butane}(g) \rightleftharpoons \text{isobutane}(g)$
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.
$\text{n-butane}(g) \rightleftharpoons \text{isobutane}(g)$
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,
$K=\dfrac{[\text{isobutane}]}{[\text{n-butane}]}=\dfrac{x}{1.00−x}=2.6$
Rearranging and solving for $x$,
\begin{align*} x &=2.6(1.00−x) \[4pt] x &= 2.6−2.6x \[4pt] x+2.6x &= 2.6 \[4pt] x &=0.72 \end{align*}
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:
\begin{align*}[\text{n-butane}]_f &= (1.00 − x) M \[4pt] &= (1.00 − 0.72) M \[4pt] &= 0.28\; M \[5pt] [\text{isobutane}]_f &= (0.00 + x) M \[4pt] &= (0.00 + 0.72) M \[4pt] &= 0.72\; M \end{align*}
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:
\begin{align*} K &=\dfrac{[\text{isobutane}]}{[\text{n-butane}]} \[4pt] &=\left(\dfrac{0.72\; \cancel{M}}{0.28\;\cancel{M}}\right) \[4pt] &=2.6 \end{align*}
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 $3$: water–gas shift
The water–gas shift reaction is important in several chemical processes, such as the production of $\ce{H2}$ for fuel cells. This reaction can be written as follows:
$\ce{H2(g) + CO2(g) <=> H2O(g) + CO(g)} \nonumber$
$K = 0.106$ at 700 K. If a mixture of gases that initially contains 0.0150 M $\ce{H_2}$ and 0.0150 M $\ce{CO_2}$ 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 $[\ce{H_2}]_i = [\ce{CO_2}]_i = 0.0150\; M$. 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 $\ce{H_2O}$ as $x$, then $Δ[\ce{H_2O}] = +x$. 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 $\ce{CO}$ is produced for every 1 mol of $\ce{H_2O}$, so the change in the $\ce{CO}$ concentration can be expressed as $Δ[\ce{CO}] = +x$. Similarly, for every 1 mol of $\ce{H_2O}$ produced, 1 mol each of $\ce{H_2}$ and $\ce{CO_2}$ are consumed, so the change in the concentration of the reactants is $Δ[\ce{H_2}] = Δ[\ce{CO_2}] = −x$. We enter the values in the following table and calculate the final concentrations.
$H_{2(g)}+CO_{2(g)} \rightleftharpoons H_2O_{(g)}+CO_{(g)}$
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 $x$:
$K=\dfrac{[H_2O][CO]}{[H_2][CO_2]}=\dfrac{(x)(x)}{(0.0150−x)(0.0150−x}=\dfrac{x^2}{(0.0150−x)^2}=0.106$
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,
$\dfrac{x^2}{(0.0150−x)^2}=\left(\dfrac{x}{0.0150−x}\right)^2=0.106$
Taking the square root of the middle and right terms,
\begin{align*} \dfrac{x}{(0.0150−x)} &=(0.106)^{1/2} \[4pt] &=0.326 \[4pt] x &=(0.326)(0.0150)−0.326x \[4pt] 1.326x &=0.00489 \[4pt] x &=0.00369=3.69 \times 10^{−3} \end{align*}
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:
\begin{align*} K &=\dfrac{[H_2O][CO]}{[H_2][CO_2]} \[4pt] &=\dfrac{(0.00369)^2}{(0.0113)^2} \[4pt] &=0.107 \end{align*}
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:
$\ce{H2(g) + I2(g) <=> 2HI(g)}$
$K = 54$ 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 $\ce{CO2}$ and 0.570 M $\ce{H_2}$ 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 $[\ce{CO_2}]_i = 0.632\; M$ and $[\ce{H_2}]_i = 0.570\; M$. Again, x is defined as the change in the concentration of $\ce{H_2O}$: $Δ[\ce{H_2O}] = +x$. Because 1 mol of $CO$ is produced for every 1 mol of $H_2O$, the change in the concentration of $\ce{CO}$ is the same as the change in the concentration of $\ce{H2O}$, so Δ[\ce{CO}] = +x. Similarly, because 1 mol each of $\ce{H_2}$ and $\ce{CO_2}$ 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.
$H_{2(g)}+CO_{2(g)} \rightleftharpoons H_2O_{(g)}+CO_{(g)}$
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 $x$:
\begin{align*} K &=\dfrac{[\ce{H_2O}][\ce{CO}]}{[\ce{H_2}][\ce{CO_2}]} \[4pt] &=\dfrac{x^2}{(0.570−x)(0.632−x)} \[4pt] &=0.106 \end{align*}
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:
$x^2 = 0.106(0.360 − 1.20x + x^2) \nonumber$
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:
\begin{align*} x &= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} \[4pt] &= \dfrac{−0.127 \pm \sqrt{(0.127)^2−4(0.894)(−0.0382)}}{2(0.894)} \end{align*}
so
$x =0.148 \text{ and } −0.290 \nonumber$
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:
\begin{align*} K &=\dfrac{[H_2O][CO]}{[H_2][CO_2]} \[4pt] &=\dfrac{(0.148)^2}{(0.422)(0.484)} \[4pt] &=0.107 \end{align*}
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:
$N_{2(g)}+O_{2(g)} \rightleftharpoons 2NO_{(g)}$
with $K_p = 2.0 \times 10^{−31}$ at 25°C.
What is the partial pressure of $\ce{NO}$ in equilibrium with $\ce{N_2}$ and $\ce{O_2}$ in the atmosphere (at 1 atm, $P_{\ce{N_2}} = 0.78\; atm$ and $P_{\ce{O_2}} = 0.21\; atm$?
Given: balanced equilibrium equation and values of $K_p$, $P_{\ce{O_2}}$, and $P_{\ce{N_2}}$
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 $\ce{NO}$. Check your answer by substituting values into the equilibrium equation and solving for $K$.
Solution
A Because we are given $K_p$ and partial pressures are reported in atmospheres, we will use partial pressures. The initial partial pressure of $\ce{O_2}$ is 0.21 atm and that of $\ce{N_2}$ is 0.78 atm. If we define the change in the partial pressure of $\ce{NO}$ as 2x, then the change in the partial pressure of $\ce{O_2}$ and of $\ce{N_2}$ is −x because 1 mol each of $\ce{N_2}$ and of $\ce{O_2}$ is consumed for every 2 mol of $\ce{NO}$ produced. Each substance has a final partial pressure equal to the sum of the initial pressure and the change in that pressure at equilibrium.
$\ce{N2(g) + O2(g) <=> 2NO(g)}$
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,
\begin{align*} \dfrac{(2x)^2}{(0.78)(0.21)} &= 2.0 \times 10^{−31} \[4pt] \dfrac{4x^2}{0.16} &=2.0 \times10^{−31} \[4pt] x^2 &=\dfrac{0.33 \times 10^{−31}}{4} \[4pt] x&=9.1 \times 10^{−17} \end{align*}
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:
$\ce{3O2(g) <=> 2O3(g)} \nonumber$
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
$4.8 \times 10^{−31} \;atm$
Finding Equilibrium Concentrations for Reactions with a Small K Value:
https://youtu.be/EyCZTyJvrj8
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 ≥ 103). 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:
$H_{2(g)}+C_2H_{4(g)} \overset{Ni}{\rightleftharpoons} C_2H_{6(g)}$
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,
$K=\dfrac{[C_2H_6]}{[H_2][C_2H_4]}=\dfrac{0.155−x}{(0.045+x)x}=9.6 \times 10^{18}$
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:
$K=\dfrac{0.155}{0.045x} = 9.6 \times 10^{18}$
$x=3.6 \times 10^{−19}$
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:
$K=\dfrac{[C_2H_6]}{[H_2][C_2H_4]}=\dfrac{0.155}{(0.045)(3.6 \times 10^{−19})}=9.6 \times 10^{18}$
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:
$\ce{ H2(g) + Cl 2(g) <=> 2HCl(g)} \nonumber$
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:
• $[H_2]_f = 4.8 \times 10^{−32}\; M$
• $[Cl_2]_f = 0.135\; M$
• $[HCl]_f = 0.514\; M$
Summary
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.
Key Takeaway
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.
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?
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 $2A \rightleftharpoons B+C$?
2. The following table shows the reported values of the equilibrium $P_{O_2}$ at three temperatures for the reaction $Ag2O_{(s)} \rightleftharpoons 2Ag_{(s)}+12O_{2(g)}$, 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 $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$, 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 2HI_{(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)}+2H_{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 $A(s) \rightleftharpoons 2B_{(g)}+C_{(g)}$, what is $K_p$?
5. The decomposition of ammonium carbamate to $NH_3$ and $CO_2$ at 40°C is written as $NH_4CO_2NH_{2(s)} \rightleftharpoons 2NH_{3(g)}+CO_{2(g)}$. 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 $SO_2Cl_{2(g)} \rightleftharpoons SO_{2(g)}+Cl2_{(g)}$ 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 $aA \rightleftharpoons bB$, show that $K_p = K(RT)^{Δn}$ assuming ideal gas behavior.
8. For the gas-phase reaction $I_2$ \rightleftharpoons 2I, show that the total pressure is related to the equilibrium pressure by the following equation: $P_T=\sqrt{K_pP_{I_2}} + P_{I_2}$
9. Experimental data on the system $Br_{2(l)} \rightleftharpoons Br_{2(aq)}$ are given in the following table. Graph $[Br_2]$ 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 ($NH_4CO_2NH_2$) dissociates completely to ammonia and carbon dioxide when it vaporizes: $NH_4CO_2NH_{2(s)} \rightleftharpoons 2NH_{3(g)}+CO_{2(g)}$ 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 $COCl_{2(g)} \rightleftharpoons CO_{(g)}+Cl2_{(g)}$ 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: $IO^−_{4(aq)}+2H_2O_{(l)} \rightleftharpoons H_4IO^−_{6(aq)}$ 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: $I_{2(g)}+Br_{2(g)} \rightleftharpoons 2IBr_{(g)}$ 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 $C_{(s)} + 12N_{2(g)}+\frac{5}{2}H_{2(g)} \rightleftharpoons CH3NH2(g)$ 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? | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/15%3A_Principles_of_Chemical_Equilibrium/15.7%3A_Equilibrium_Calculations_-_Some_Illustrative_Examples.txt |
• 16.1: Arrhenius Theory: A Brief Review
• 16.2: Brønsted-Lowry Theory of Acids and Bases
• 16.3: Self-Ionization of Water and the pH Scale
• 16.4: Strong Acids and Strong Bases
Acids and bases that are completely ionized when dissolved in water are called strong acids and strong bases There are only a few strong acids and bases, and everyone should know their names and properties. These acids are often used in industry and everyday life. The concentrations of acids and bases are often expressed in terms of pH, and as an educated person, you should have the skill to convert concentrations into pH and pOH. The pH is an indication of the hydrogen ion concentration, [H+].
• 16.5: Weak Acids and Weak Bases
• 16.6: Polyprotic Acids
An acid that contains more than one ionizable proton is a polyprotic acid. The protons of these acids ionize in steps. The differences in the acid ionization constants for the successive ionizations of the protons in a polyprotic acid usually vary by roughly five orders of magnitude. As long as the difference between the successive values of Ka of the acid is greater than about a factor of 20, it is appropriate to break down the calculations of the concentrations sequentially.
• 16.7: Ions as Acids and Bases
• 16.8: Molecular Structure and Acid-Base Behavior
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.
• 16.9: Lewis Acids and Bases
16: Acids and Bases
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_General_Chemistry_(Petrucci_et_al.)/16%3A_Acids_and_Bases/16.1%3A_Arrhenius_Theory%3A_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
The Arrhenius definition of an acid as a compound that dissolves in water to yield hydronium ions (H3O+) 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.
Brønsted-Lowry Defintions
• A compound that donates a proton to another compound is called a Brønsted-Lowry acid.
• A compound that accepts a proton is called a Brønsted-Lowry base.
Definitions of Acids and Bases: https://youtu.be/r8reN0CSIHw
Acids may be compounds such as HCl or H2SO4, organic acids like acetic acid ($\ce{CH_3COOH}$) or ascorbic acid (vitamin C), or H2O. Anions (such as $\ce{HSO_4^-}$, $\ce{H_2PO_4^-}$, $\ce{HS^-}$, and $\ce{HCO_3^-}$) and cations (such as $\ce{H_3O^+}$, $\ce{NH_4^+}$, and $\ce{[Al(H_2O)_6]^{3+}}$) may also act as acids. Bases fall into the same three categories. 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:
$\ce{H^+ + OH^- \rightarrow H_2O} \label{14.2.1}$
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):
$\text{acid} \rightleftharpoons \text{proton} + \text{conjugate base}\label{14.2.2a}$
$\ce{HF \rightleftharpoons H^+ + F^-} \label{14.2.2b} \nonumber$
$\ce{H_2SO_4 \rightleftharpoons H^+ + HSO_4^{−}}\label{14.2.2c} \nonumber$
$\ce{H_2O \rightleftharpoons H^+ + OH^-}\label{14.2.2d} \nonumber$
$\ce{HSO_4^- \rightleftharpoons H^+ + SO_4^{2−}}\label{14.2.2e} \nonumber$
$\ce{NH_4^+ \rightleftharpoons H^+ + NH_3} \label{14.2.2f} \nonumber$
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):
$\text{base} + \text{proton} \rightleftharpoons \text{conjugate acid} \label{14.2.3a}$
$\ce{OH^- +H^+ \rightleftharpoons H2O}\label{14.2.3b} \nonumber$
$\ce{H_2O + H^+ \rightleftharpoons H3O+}\label{14.2.3c} \nonumber$
$\ce{NH_3 +H^+ \rightleftharpoons NH4+}\label{14.2.3d} \nonumber$
$\ce{S^{2-} +H^+ \rightleftharpoons HS-}\label{14.2.3e} \nonumber$
$\ce{CO_3^{2-} +H^+ \rightleftharpoons HCO3-}\label{14.2.3f} \nonumber$
$\ce{F^- +H^+ \rightleftharpoons HF} \label{14.2.3g} \nonumber$
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, $\ce{NH4+}$:
Similarly, in the reaction of acetic acid with water, acetic acid donates a proton to water, which acts as the base. In the reverse reaction, $H_3O^+$ is the acid that donates a proton to the acetate ion, which acts as the base. Once again, we have two conjugate acid–base pairs: the parent acid and its conjugate base ($CH_3CO_2H/CH_3CO_2^−$) and the parent base and its conjugate acid ($H_3O^+/H_2O$).
In the reaction of ammonia with water to give ammonium ions and hydroxide ions, ammonia acts as a base by accepting a proton from a water molecule, which in this case means that water is acting as an acid. In the reverse reaction, an ammonium ion acts as an acid by donating a proton to a hydroxide ion, and the hydroxide ion acts as a base. The conjugate acid–base pairs for this reaction are $NH_4^+/NH_3$ and $H_2O/OH^−$.
Conjugate Acid-Base Pairs: https://youtu.be/pPrp3xEQef4
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:
$\ce{HCO^{-}3(aq) + H_2O(l) <=> CO^{2-}3(aq) + H_3O^{+}(aq)} \nonumber$
$ \ce{HCO^{-}3(aq) + H_2O(l) <=> H2CO3(aq) + OH^{-}(aq)} \nonumber$
Water is the most important amphiprotic species. It can form both the hydronium ion, $\ce{H3O^{+}}$, and the hydroxide ion, $\ce{OH^-}$ when it undergoes autoionization:
$\ce{2 H_2O}_{(l)} \rightleftharpoons \ce{H_3O^+}(aq)+\ce{OH^-} (aq) \nonumber$
Example $1$: 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 $\ce{HI}$
Solution
1. $\ce{HSO3^{-}(aq) + OH^{-}(aq) <=> SO3^{2-}(aq) + H2O(l)}$
2. $\ce{HSO^{-}3(aq) + HI(aq) <=> H2SO3(aq) + I^{-}(aq)}$
Exercise $1$
Write separate equations representing the reaction of $\ce{H2PO4-}$
1. as a base with $\ce{HBr}$
2. as an acid with $\ce{OH^-}$
Answer a
$\ce{H2PO4- (aq) + HBr(aq) <=> H3PO4(aq) + Br-(aq)} \nonumber$
Answer b
$\ce{H2PO4-}(aq)+\ce{OH^-} (aq)\rightleftharpoons \ce{HPO4^2-}(aq)+ \ce{H_2O}_{(l)} \nonumber$
Key Concepts and 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:
$\ce{2 H_2O}_{(l)} \rightleftharpoons \ce{H_3O^+}(aq)+\ce{OH^-} (aq) \nonumber$
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_General_Chemistry_(Petrucci_et_al.)/16%3A_Acids_and_Bases/16.2%3A_Brnsted-Lowry_Theory_of_Acids_and_Bases.txt |
Learning Objectives
• To understand the autoionization reaction of liquid water.
• To know the relationship among pH, pOH, and $pK_w$.
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 ($H_3O^+$):
$\underset{aicd}{\ce{HCl(aq)}} + \underset{base}{\ce{H2O(l)}} \rightarrow \underset{acid}{\ce{H3O^{+}(aq)}} + \underset{base}{\ce{Cl^{-}(aq)}} \label{16.3.1a}$
In this reaction, $\ce{HCl}$ is the acid, and water acts as a base by accepting an $\ce{H^{+}}$ ion. The reaction in Equation $\ref{16.3.1a}$ is often written in a simpler form by removing $\ce{H2O}$ from each side:
$\ce {HCl (aq) \rightarrow H^{+} (aq) + Cl^{-} (aq)} \label{16.3.1b}$
In Equation $\ref{16.3.1b}$, the hydronium ion is represented by $\ce{H^{+}(aq)}$, although free $\ce{H^{+}}$ ions do not exist in liquid water as this reaction demonstrates:
$\ce{H^{+}(aq) + H2O(l) \rightarrow H3O^{+}(aq)}$
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:
$\underset{aicd}{\ce{H2O(aq)}} + \underset{base}{\ce{NH3(aq)}} \rightleftharpoons \underset{acid}{\ce{NH^{+}4 (aq)}} + \underset{base}{\ce{OH^{-}(aq)}} \label{16.3.2}$
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 Pure 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:
$\ce{2H2O(l) <=> H3O^{+}(aq) + OH^{−}(aq)} \label{16.3.3}$
In pure water, a very small fraction of water molecules donate protons to other water molecules to form hydronium ions and hydroxide ions:
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 liquid 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
$K_w=(1.003 \times 10^{−7})(1.003 \times 10^{−7})=1.006 \times 10^{−14} \label{16.3.7}$
The slight ionization of pure water is reflected in the small value of the equilibrium constant; at 25 °C.Thus, to three significant figures, $K_w = 1.01 \times 10^{−14}\; M$ at room temperature. 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.
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:
\begin{align*} K_\ce{w} &=\ce{[H_3O^+][OH^- ]} \[4pt]&=\ce{[H_3O^+]^2} \[4pt] &=\ce{[OH^- ]^2} \[4pt] &=1.0 \times 10^{−14} \end{align*}
So:
$\ce{[H_3O^+]}=\ce{[OH^- ]}=\sqrt{1.0 \times 10^{−14}} =1.0 \times 10^{−7}\; M \nonumber$
The hydronium ion concentration and the hydroxide ion concentration are the same, and we find that both equal $1.0 \times 10^{−7}\; M$.
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. Example $2$ demonstrates the quantitative aspects of this relation between hydronium and hydroxide ion concentrations.
Example $2$: The Inverse Proportionality of Hydronium and Hydroxide Concentrations
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:
$\ce{2 H_2O}_{(l)} \rightleftharpoons \ce{H_3O^+}_{(aq)} + \ce{OH^-}_{(aq)} \nonumber$
and
$K_\ce{w}=\ce{[H3O+][OH^- ]}=1.0 \times 10^{−14} \nonumber$
Thus, we can calculate the missing equilibrium concentration.
Rearrangement of the $k_w$ 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
$\ce{[H3O+]} = 1 \times 10^{−11} M \nonumber$
Self-Ionization of Water (Kw): https://youtu.be/RMpO0rqUnFg
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:
\begin{align} pH &=−\log_{10}[\ce{H^{+}}] \label{16.3.8} \[4pt] [\ce{H^{+}}] &=10^{−pH} \label{16.3.9} \end{align}
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 $[\ce{H3O^{+}}] > 1.0 \times 10^{−7}$) and basic solutions have pH > 7.00 (corresponding to $[\ce{H3O^{+}}] < 1.0 \times 10^{−7}$).
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 $[\ce{OH^{−}}]$ are related as follows:
\begin{align} pOH &=−\log_{10}[\ce{OH^{−}}] \label{16.3.10} \[4pt] [\ce{OH^{−}}] &=10^{−pOH} \label{16.3.11} \end{align}
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 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 $K_w = 1.0 \times 10^{−14}$ and $pH + pOH = 14.00$. 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.
Example $3$
The $k_w$ 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 pKw by taking the negative logarithm of $K_w$.
2. For a neutral aqueous solution, $[H_3O^+] = [OH^−]$. Use this relationship and Equation \ref{16.3.6b} 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 &= [\ce{H3O^{+}}][\ce{OH^{−}}] \[4pt] &= (x)(x) \[4pt] &=x^2 \[4pt] x &=\sqrt{K_w} \[4pt] &=\sqrt{4.99 \times 10^{−13}} \[4pt] &=7.06 \times 10^{−7}\; M \end{align*}
Because $x$ is equal to both $[\ce{H3O^{+}}]$ and $[\ce{OH^{−}}]$,
\begin{align*} pH = pOH &= −\log(7.06 \times 10^{−7}) \[4pt] &= 6.15 \, \text{(to two decimal places)} \end{align*}
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:
$pK_w = pH + pOH = y + y = 2y \nonumber$
$y=\dfrac{pK_w}{2}=\dfrac{12.302}{2}=6.15=pH=pOH \nonumber$
Exercise $3$
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
$pK_w = 13.45$ and $pH = pOH = 6.73$
Introduction to pH: https://youtu.be/pQOa3bb5YEE
Summary
• For any neutral solution, $pH + pOH = 14.00$ (at 25°C) and $pH = \ce{1/2} pK_w$.
• Ion-product constant of liquid water: $K_w = [H_3O^+][OH^−] \nonumber$
• Definition of $pH$: $pH = −\log_{10}[H^+] \nonumber$ or $[H^+] = 10^{−pH} \nonumber$
• Definition of $pOH$: $pOH = −\log_{10}[OH^+] \nonumber$ or $[OH^−] = 10^{−pOH} \nonumber$
• Relationship among $pH$, $pOH$, and $pK_w$: $pK_w= pH + pOH \nonumber$
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 ($K_w$) and is defined as $K_w = [H_3O^+][OH^−]$. At 25°C, $K_w$ is $1.01 \times 10^{−14}$; hence $pH + pOH = pK_w = 14.00$. 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.
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
autoionization
reaction between identical species yielding ionic products; for water, this reaction involves transfer of protons to yield hydronium and hydroxide ions
ion-product constant for water (Kw)
equilibrium constant for the autoionization of water | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/16%3A_Acids_and_Bases/16.3%3A_Self-Ionization_of_Water_and_the_pH_Scale.txt |
Learning Objectives
• Give the names and formulas of some strong acids and bases.
• Explain the pH scale, and convert pH and concentration of hydronium ions.
• Evaluate solution pH and pOH of strong acids or bases.
Acids and bases that are completely ionized when dissolved in water are called strong acids and strong bases There are only a few strong acids and bases, and everyone should know their names and properties. These acids are often used in industry and everyday life. The concentrations of acids and bases are often expressed in terms of pH, and as an educated person, you should have the skill to convert concentrations into pH and pOH. The pH is an indication of the hydrogen ion concentration, $\ce{[H+]}$.
Strong Acids
Strong acids are acids that are completely or nearly 100% ionized in their solutions; Table $1$ includes some common strong acids. Hence, the ionization in Equation $\ref{gen ion}$ for a strong acid HA can be represented with a single arrow:
$\ce{HA(aq) + H2O(l) \rightarrow H3O^{+}(aq) + A^{-}(aq)} \label{gen ion}$
Water is the base that reacts with the acid $\ce{HA}$, $\ce{A^{−}}$ is the conjugate base of the acid HA, and the hydronium ion is the conjugate acid of water. By definition, a strong acid yields 100% of $\ce{H3O+}$ and $\ce{A^{−}}$ when the acid ionizes in water. Table $1$ lists several strong acids.
Table $1$: Some of the common strong acids and bases are listed here.
Strong Acids Strong Bases
perchloric acid ($\ce{HClO4}$) lithium hydroxide ($\ce{LiOH}$)
hydrochloric acid ($\ce{HCl}$) sodium hydroxide ($\ce{NaOH}$)
hydrobromic acid ($\ce{HBr}$) potassium hydroxide ($\ce{KOH}$)
hydroiodic acid ($\ce{Hl}$) calcium hydroxide ($\ce{Ca(OH)2}$)
nitric acid ($\ce{HNO3}$) strontium hydroxide ($\ce{Sr(OH)2}$)
sulfuric acid ($\ce{H2SO4}$) barium hydroxide ($\ce{Ba(OH)2}$)
For a strong acid, $\ce{[H+]}$ = $\ce{[A^{-}]}$ = concentration of acid if the concentration is much higher than $1 \times 10^{-7}\, M$. However, for a very dilute strong acid solution with concentration less than $1 \times 10^{-7}\, M$, the pH is dominated by the autoionization of water
$\ce{H2O \rightleftharpoons H+ + OH-}$
Example $1$
Calculate the pH of a solution with $1.2345 \times 10^{-4}\; M \ce{HCl}$, a strong acid.
Solution
The solution of a strong acid is completely ionized. That is, this equation goes to completion
$\ce{HCl(aq) -> H(aq) + Cl^{-}(aq)} \nonumber$
Thus, $\ce{[H+]} = 1.2345 \times 10^{-4}$.
$\ce{pH} = -\log(1.2345 \times 10^{-4}) = 3.90851 \nonumber$
Exercise $1$
What is the pH for a solution containing 1.234 M $\ce{[HCl]}$?
Answer
pH = -0.0913
Example $2$
Calculate the pH of a stock $\ce{HCl}$ solution that is 32% by mass $\ce{HCl}$.
Solution
The density of such a solution is needed before we can calculate the pH. Since the density is not on the label, we need to find it from the Material Safety Data Sheet, which gives the specific gravity of 1.150. Thus, the amount of acid in 1.0 L is 1150 g.
\begin{align*} \textrm{The amount of HCl} &= 1000\times1.150\times0.32\ &= \mathrm{368\: g\: \left(\dfrac{1\: mol}{36.5\: g} \leftarrow molar\: mass\: of\: HCl\right)}\ &= \mathrm{10.08\: M}\ &= \ce{[H+]} \end{align*}
$\ce{pH} = -\log(10.08) = -1.003 \nonumber$
Discussion
Yes, pH have negative values if $\ce{[H+]} > 1.0$
Exercise $2$
Check out the information on nitric acid, a strong acid, and calculate the pH of a stock nitric acid solution.
Example $3$
Calculate the pH of a solution containing $1.00 \times 10^{-7}\; M$ of the strong acid $\ce{HCl}$.
Solution
$\ce{[H+]} = 1.0 \times 10^{-7}\; M$ from the strong acid, and if x is the amount from the ionization of water, then we have the equilibrium due to the autoionization of water. We can model this with an ICE table.
$H_2O \rightleftharpoons H^+_{(aq)} + OH^-_{(aq)} \nonumber$
ICE Table $H_2O_{(l)}$ $H^+_{(aq)}$ $OH^-_{(aq)}$
Initial - $1\times 10^{-7}$ 0
Change - +x +x
Equilibrium - $1\times 10^{-7} + x$ x
Recall that $K_{w} = \ce{[H+] [OH- ]} = 1 \times 10^{-14}$, due to the ionization equilibrium of water in the solution:
\begin{align*} (1.00 \times 10^{-7} +x) x &= 1 \times 10^{-14} \[4pt] x^2 + 1.00 \times 10^{-7}x - 1.00 \times 10^{-14} &= 0 \end{align*}
Solving this equation for $x$ from the quadratic equation results in
\begin{align*} x &= \dfrac{-1.00 \times 10^{-7} \pm \sqrt{1.00 \times 10^{-14} + (4)(1) (1.00 \times 10^{-14})}}{2}\ &= 0.61 \times 10^{-7} \end{align*}
only the additive root is physical (positive concentration); therefore
\begin{align*} \ce{[H+]} &= (1.00 + 0.61) \times 10^{-7}\; M \[4pt] \ce{pH} &= -\log(1.61 \times 10^{-7}) \[4pt] &= 6.79 \end{align*}
Discussion
If you require only 1 significant figure, the pH is about 7.
Strong Bases
Strong bases are completely ionized in solution. Table $1$ includes some common strong bases. For example, $\ce{KOH}$ dissolves in water in the reaction
$\ce{KOH \rightarrow K+ + OH-} \nonumber$
Relative to the number of strong acids, there are fewer number of strong bases and most are alkali hydroxides. Calcium hydroxide is considered a strong base, because it is completely, almost completely, ionized. However, the solubility of calcium hydroxide is very low. When $\ce{Ca(OH)2}$ dissolves in water, the ionization reaction is as follows:
$\ce{Ca(OH)2 \rightarrow Ca^2+ + 2 OH-} \nonumber$
Because of the stoichiometry of calcium hydroxide, upon dissociation, the concentration of $\ce{OH-}$ will be twice the concentration of $\ce{Ca^2+}$:
$\mathrm{[OH^-] = 2 [Ca^{2+}]} \nonumber$
Example $4$
Calculate the pOH of a solution containing $1.2345 \times 10^{-4}\; M \; \ce{Ca(OH)2}$.
Solution
Based on complete ionization of
$\ce{Ca(OH)2 \rightarrow Ca^{+2} + 2OH-} \nonumber$
\begin{align*} \ce{[OH^{-}]} &= 2 \times 1.2345 \times 10^{-4} \[4pt] &= 2.4690 \times 10^{-4}\; M \[4pt] \ce{pOH} &= -\log( 2.4690 \times 10^{-4})\[4pt] &= 3.6074 \end{align*}
Exercise $4$
The molar solubility of calcium hydroxide is 0.013 M $\ce{Ca(OH)2}$. Calculate the pOH.
Answer
pOH = 1.58
Calculating pH in Strong Acid or Strong Base Solutions: https://youtu.be/NNTptn7hV2s
Questions
1. What is the pH of a solution containing 0.01 M $\ce{HNO3}$?
2. What is the pH of a solution containing 0.0220 M $\ce{Ba(OH)2}$? Give 3 significant figures.
3. Exactly 1.00 L solution was made by dissolving 0.80 g of $\ce{NaOH}$ in water. What is $\ce{[H+]}$? (Atomic mass: $\ce{Na}$, 23.0; $\ce{O}$, 16.0; $\ce{H}$, 1.0)
4. What is the pH for a solution which is 0.050 M $\ce{HCl}$?
5. Which of the following is usually referred to as strong acid in water solution?
$\ce{HF}$, $\ce{HNO2}$, $\ce{H2CO3}$, $\ce{H2S}$, $\ce{HSO4-}$, $\ce{Cl-}$, $\ce{HNO3}$, $\ce{HCN}$
Solutions
1. Answer 2
Hint...
You do not need a calculator to evaluate $-\log (0.01) = 2$
2. Answer 12.64
Hint...
$\ce{Ba(OH)2 \rightarrow Ba^2+ + 2 OH-}$
3. Answer $5.0\times 10^{-13}$
Hint...
$\mathrm{[OH^-] = \dfrac{0.80}{40} = 0.020\: M}$; $[H^+] = \dfrac{1.0 \times 10^{-14}}{0.020} = 5\times 10^{-13} M$. The pH is 12.30.
4. Answer 1.3
Hint...
This solution contains 1.83 g of $\ce{HCl}$ per liter. $\mathrm{[H^+] = 0.050}$.
5. Answer $\ce{HNO3}$
Consider...
All others are weak acids | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/16%3A_Acids_and_Bases/16.4%3A_Strong_Acids_and_Strong_Bases.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, the concentration of water is essentially constant for all reactions in aqueous solution, so $[H_2O]$ in Equation \ref{16.5.2} can be incorporated into a new quantity, the acid ionization constant ($K_a$), also called the acid dissociation constant:
$K_a=K[H_2O]=\dfrac{[H_3O^+][A^−]}{[HA]} \label{16.5.3}$
Thus the numerical values of K and $K_a$ differ by the concentration of water (55.3 M). Again, for simplicity, $H_3O^+$ can be written as $H^+$ in Equation $\ref{16.5.3}$. 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 $H^3O^+$. 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:
$\ce{B(aq) + H2O(l) <=> BH^{+}(aq) + OH^{−} (aq)} \label{16.5.4}$
The equilibrium constant for this reaction is the base ionization constant ($K_b$), also called the base dissociation constant:
$K_b=K[H_2O]=\dfrac{[BH^+][OH^−]}{[B]} \label{16.5.5}$
Once again, the concentration of water is constant, so it does not appear in the equilibrium constant expression; instead, it is included in the $K_b$. 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 (recall that the equilibrium constant for the sum of two reactions is the product of the equilibrium constants for the individual reactions):
$\cancel{HCN_{(aq)}} \rightleftharpoons H^+_{(aq)}+\cancel{CN^−_{(aq)}} \;\;\; K_a=[H^+]\cancel{[CN^−]}/\cancel{[HCN]}$
$\cancel{CN^−_{(aq)}}+H_2O_{(l)} \rightleftharpoons OH^−_{(aq)}+\cancel{HCN_{(aq)}} \;\;\; K_b=[OH^−]\cancel{[HCN]}/\cancel{[CN^−]}$
$H_2O_{(l)} \rightleftharpoons H^+_{(aq)}+OH^−_{(aq)} \;\;\; K=K_a \times K_b=[H^+][OH^−]$
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 $pK_w$, 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 Table $1$ and Table $2$, 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 1000 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 16.5. 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:
$HCl_{(aq)} + H_2O_{(l)} \rightarrow \rightarrow H_3O^+_{(aq)}+Cl^−_{(aq)} \label{16.5.17}$
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 ($\ce{CH_3CH_2CH_2CO_2^{−}}$). 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 Equation \ref{16.5.15} and Equation \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 $pK_b$,
\begin{align*} 4.83 + pK_b &=14.00 \[4pt] pK_b &=14.00−4.83 \[4pt] &=9.17 \end{align*}
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 $K_a$,
\begin{align*} K_a(5.4 \times 10^{−4}) &=1.01 \times 10^{−14} \[4pt] K_a &=1.9 \times 10^{−11} \end{align*}
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:
\begin{align*} pK_b &=−\log(5.4 \times 10^{−4}) \[4pt] &=3.27 \[10pt] pKa + pK_b &=14.00 \[4pt] pK_a &=10.73 \ K_a &=10^{−pK_a} \[4pt] &=10^{−10.73} \[4pt] &=1.9 \times 10^{−11} \end{align*}
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
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 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.
Note: Leveling Effect
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 $\ce{OH^{−}}$ is leveled to the strength of $\ce{OH^{−}}$ because $\ce{OH^{−}}$ is the strongest base that can exist in equilibrium with water. Salts such as $\ce{K_2O}$, $\ce{NaOCH3}$ (sodium methoxide), and $\ce{NaNH2}$ (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 $\ce{OH^{−}}$ and the corresponding cation:
$\ce{K2O(s) + H2O(l) -> 2OH^{−}(aq) + 2K^{+} (aq)} \nonumber$
$\ce{NaOCH3(s) + H2O(l) -> OH^{−}(aq) + Na^{+} (aq) + CH3OH(aq)} \nonumber$
$\ce{NaNH2(s) + H2O(l) -> OH^{−}(aq) + Na^{+} (aq) + NH3(aq)} \nonumber$
Other examples that you may encounter are potassium hydride ($KH$) and organometallic compounds such as methyl lithium ($\ce{CH3Li}$).
Calculating the pH of Weak Acids and Weak Bases: https://youtu.be/zr1V1THJ5P0
Summary
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 (Ka). Similarly, the equilibrium constant for the reaction of a weak base with water is the base ionization constant (Kb). 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.
Key Takeaways
• 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.
Key Equations
• Acid ionization constant: $K_a=K[H_2O]=\dfrac{[H_3O^+][A^−]}{[HA]} \nonumber$
• Base ionization constant: $K_b=K[H_2O]=\dfrac{[BH^+][OH^−]}{[B]} \nonumber$
• 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_General_Chemistry_(Petrucci_et_al.)/16%3A_Acids_and_Bases/16.5%3A_Weak_Acids_and_Weak_Bases.txt |
Learning Objectives
• Extend previously introduced equilibrium concepts to acids and bases that may donate or accept more than one proton
We can classify acids by the number of protons per molecule that they can give up in a reaction. Acids such as $\ce{HCl}$, $\ce{HNO3}$, and $\ce{HCN}$ that contain one ionizable hydrogen atom in each molecule are called monoprotic acids. Their reactions with water are:
$\ce{HCl}(aq)+\ce{H2O}(l)⟶\ce{H3O+}(aq)+\ce{Cl-}(aq) \nonumber$
$\ce{HNO3}(aq)+\ce{H2O}(l)⟶\ce{H3O+}(aq)+\ce{NO3-}(aq) \nonumber$
$\ce{HCN}(aq)+\ce{H2O}(l)⟶\ce{H3O+}(aq)+\ce{CN-}(aq) \nonumber$
Even though it contains four hydrogen atoms, acetic acid, $\ce{CH3CO2H}$, is also monoprotic because only the hydrogen atom from the carboxyl group ($\ce{-COOH}$) reacts with bases:
Similarly, monoprotic bases are bases that will accept a single proton.
Diprotic Acids
Diprotic acids contain two ionizable hydrogen atoms per molecule; ionization of such acids occurs in two steps. The first ionization always takes place to a greater extent than the second ionization. For example, sulfuric acid, a strong acid, ionizes as follows:
• The first ionization is
$\ce{H2SO4}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HSO4-}(aq) \nonumber$
with $K_{\ce a1} > 10^2;\: {complete\: dissociation}$.
• The second ionization is
$\ce{HSO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{SO4^{2−}}(aq) \nonumber$
with $K_{\ce a2}=1.2×10^{−2}$.
This stepwise ionization process occurs for all polyprotic acids. When we make a solution of a weak diprotic acid, we get a solution that contains a mixture of acids. Carbonic acid, $\ce{H2CO3}$, is an example of a weak diprotic acid. The first ionization of carbonic acid yields hydronium ions and bicarbonate ions in small amounts.
• First Ionization
$\ce{H2CO3}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO3-}(aq) \nonumber$
with
$K_{\ce{H2CO3}}=\ce{\dfrac{[H3O+][HCO3- ]}{[H2CO3]}}=4.3×10^{−7} \nonumber$
The bicarbonate ion can also act as an acid. It ionizes and forms hydronium ions and carbonate ions in even smaller quantities.
• Second Ionization
$\ce{HCO3-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CO3^2-}(aq) \nonumber$
with
$K_{\ce{HCO3-}}=\ce{\dfrac{[H3O+][CO3^2- ]}{[HCO3- ]}}=4.7×10^{−11} \nonumber$
$K_{\ce{H2CO3}}$ is larger than $K_{\ce{HCO3-}}$ by a factor of 104, so H2CO3 is the dominant producer of hydronium ion in the solution. This means that little of the $\ce{HCO3-}$ formed by the ionization of H2CO3 ionizes to give hydronium ions (and carbonate ions), and the concentrations of H3O+ and $\ce{HCO3-}$ are practically equal in a pure aqueous solution of H2CO3.
If the first ionization constant of a weak diprotic acid is larger than the second by a factor of at least 20, it is appropriate to treat the first ionization separately and calculate concentrations resulting from it before calculating concentrations of species resulting from subsequent ionization. This can simplify our work considerably because we can determine the concentration of H3O+ and the conjugate base from the first ionization, then determine the concentration of the conjugate base of the second ionization in a solution with concentrations determined by the first ionization.
Example $1$: Ionization of a Diprotic Acid
When we buy soda water (carbonated water), we are buying a solution of carbon dioxide in water. The solution is acidic because CO2 reacts with water to form carbonic acid, H2CO3. What are $\ce{[H3O+]}$, $\ce{[HCO3- ]}$, and $\ce{[CO3^2- ]}$ in a saturated solution of CO2 with an initial [H2CO3] = 0.033 M?
$\ce{H2CO3}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO3-}(aq) \hspace{20px} K_{\ce a1}=4.3×10^{−7} \label{step1} \tag{equilibrium step 1}$
$\ce{HCO3-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CO3^2-}(aq) \hspace{20px} K_{\ce a2}=4.7×10^{−11} \label{step2} \tag{equilibrium step 2}$
Solution
As indicated by the ionization constants, H2CO3 is a much stronger acid than $\ce{HCO3-}$, so $\ce{H2CO3}$ is the dominant producer of hydronium ion in solution. Thus there are two parts in the solution of this problem:
1. Using the customary four steps, we determine the concentration of H3O+ and $\ce{HCO3-}$ produced by ionization of H2CO3.
2. Then we determine the concentration of $\ce{CO3^2-}$ in a solution with the concentration of H3O+ and $\ce{HCO3-}$ determined in (1).
To summarize:
1. First Ionization: Determine the concentrations of $\ce{H3O+}$ and $\ce{HCO3-}$.
Since \ref{step1} is has a much bigger $K_{a1}=4.3×10^{−7}$ than $K_{a2}=4.7×10^{−11}$ for \ref{step2}, we can safely ignore the second ionization step and focus only on the first step (but address it in next part of problem).
$\ce{H2CO3}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO3-}(aq) \hspace{20px} K_{\ce a1}=4.3×10^{−7} \nonumber$
As for the ionization of any other weak acid:
An abbreviated table of changes and concentrations shows:
Abbreviated table of changes and concentrations
ICE Table $\ce{H2CO3}(aq)$ $\ce{H2O}(l)$ $\ce{H3O+}(aq)$ $\ce{HCO3-}(aq)$
Initial (M) $0.033 \:M$ - $0$ $0$
Change (M) $- x$ - $+x$ $+x$
Equilibrium (M) $0.033 \:M - x$ - $x$ $x$
Substituting the equilibrium concentrations into the equilibrium constant gives us:
$K_{\ce{H2CO3}}=\ce{\dfrac{[H3O+][HCO3- ]}{[H2CO3]}}=\dfrac{(x)(x)}{0.033−x}=4.3×10^{−7} \nonumber$
Solving the preceding equation making our standard assumptions gives:
$x=1.2×10^{−4} \nonumber$
Thus:
$\ce{[H2CO3]}=0.033\:M \nonumber$
$\ce{[H3O+]}=\ce{[HCO3- ]}=1.2×10^{−4}\:M \nonumber$
2. Second Ionization: Determine the concentration of $CO_3^{2-}$ in a solution at equilibrium.
Since the \ref{step1} is has a much bigger $K_a$ than \ref{step2}, we can the equilibrium conditions calculated from first part of example as the initial conditions for an ICER Table for the \ref{step2}:
$\ce{HCO3-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CO3^2-}(aq) \nonumber$
ICER Table for the \ref{step2}:
ICE Table $\ce{HCO3-}(aq)$ $\ce{H2O}(l)$ $\ce{H3O+}(aq)$ $\ce{CO3^2-}(aq)$
Initial (M) $1.2×10^{−4}\:M$ - $1.2×10^{−4}\:M$ $0$
Change (M) $- y$ - $+y$ $+y$
Equilibrium (M) $1.2×10^{−4}\:M - y$ - $1.2×10^{−4}\:M + y$ $y$
\begin{align*} K_{\ce{HCO3-}}&=\ce{\dfrac{[H3O+][CO3^2- ]}{[HCO3- ]}} \[4pt] &=\dfrac{(1.2×10^{−4}\:M + y) (y)}{(1.2×10^{−4}\:M - y)} \end{align*} \nonumber
To avoid solving a quadratic equation, we can assume $y \ll 1.2×10^{−4}\:M$ so
$K_{\ce{HCO3-}} = 4.7×10^{−11} \approx \dfrac{(1.2×10^{−4}\:M ) (y)}{(1.2×10^{−4}\:M)} \nonumber$
Rearranging to solve for $y$
$y \approx \dfrac{ (4.7×10^{−11})(1.2×10^{−4}\:M )}{ 1.2×10^{−4}\:M} \nonumber$
$[\ce{CO3^2-}]=y \approx 4.7×10^{−11} \nonumber$
To summarize:
In part 1 of this example, we found that the $\ce{H2CO3}$ in a 0.033-M solution ionizes slightly and at equilibrium $[\ce{H2CO3}] = 0.033\, M$, $[\ce{H3O^{+}}] = 1.2 × 10^{−4}$, and $\ce{[HCO3- ]}=1.2×10^{−4}\:M$. In part 2, we determined that $\ce{[CO3^2- ]}=5.6×10^{−11}\:M$.
Exercise $2$: Hydrogen Sulfide
The concentration of $H_2S$ in a saturated aqueous solution at room temperature is approximately 0.1 M. Calculate $\ce{[H3O+]}$, $\ce{[HS^{−}]}$, and $\ce{[S^{2−}]}$ in the solution:
$\ce{H2S}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HS-}(aq) \hspace{20px} K_{\ce a1}=8.9×10^{−8} \nonumber$
$\ce{HS-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{S^2-}(aq) \hspace{20px} K_{\ce a2}=1.0×10^{−19} \nonumber$
Answer
$[\ce{H2S}] = 0.1 M$, $\ce{[H3O+]} = [HS^{−}] = 0.0001\, M$, $[S^{2−}] = 1 × 10^{−19}\, M$
We note that the concentration of the sulfide ion is the same as Ka2. This is due to the fact that each subsequent dissociation occurs to a lesser degree (as acid gets weaker).
Triprotic Acids
A triprotic acid is an acid that has three dissociable protons that undergo stepwise ionization: Phosphoric acid is a typical example:
• The first ionization is
$\ce{H3PO4}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{H2PO4-}(aq) \nonumber$
with $K_{\ce a1}=7.5×10^{−3}$.
• The second ionization is
$\ce{H2PO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HPO4^2-}(aq) \nonumber$
with $K_{\ce a2}=6.2×10^{−8}$.
• The third ionization is
$\ce{HPO4^2-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{PO4^3-}(aq) \nonumber$
with $K_{\ce a3}=4.2×10^{−13}$.
As with the diprotic acids, the differences in the ionization constants of these reactions tell us that in each successive step the degree of ionization is significantly weaker. This is a general characteristic of polyprotic acids and successive ionization constants often differ by a factor of about 105 to 106. This set of three dissociation reactions may appear to make calculations of equilibrium concentrations in a solution of H3PO4 complicated. However, because the successive ionization constants differ by a factor of 105 to 106, the calculations can be broken down into a series of parts similar to those for diprotic acids.
Polyprotic bases can accept more than one hydrogen ion in solution. The carbonate ion is an example of a diprotic base, since it can accept up to two protons. Solutions of alkali metal carbonates are quite alkaline, due to the reactions:
$\ce{H2O}(l)+\ce{CO3^2-}(aq)⇌\ce{HCO3-}(aq)+\ce{OH-}(aq) \nonumber$
and
$\ce{H2O}(l)+\ce{HCO3-}(aq)⇌\ce{H2CO3}(aq)+\ce{OH-}(aq) \nonumber$
Summary
An acid that contains more than one ionizable proton is a polyprotic acid. The protons of these acids ionize in steps. The differences in the acid ionization constants for the successive ionizations of the protons in a polyprotic acid usually vary by roughly five orders of magnitude. As long as the difference between the successive values of Ka of the acid is greater than about a factor of 20, it is appropriate to break down the calculations of the concentrations of the ions in solution into a series of steps.
Glossary
diprotic acid
acid containing two ionizable hydrogen atoms per molecule. A diprotic acid ionizes in two steps
diprotic base
base capable of accepting two protons. The protons are accepted in two steps
monoprotic acid
acid containing one ionizable hydrogen atom per molecule
stepwise ionization
process in which an acid is ionized by losing protons sequentially
triprotic acid
acid that contains three ionizable hydrogen atoms per molecule; ionization of triprotic acids occurs in three steps | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/16%3A_Acids_and_Bases/16.6%3A_Polyprotic_Acids.txt |
Learning Objectives
• Extend previously introduced equilibrium concepts to acids and bases that may donate or accept more than one proton
We can also use the relative strengths of conjugate acid–base pairs to understand the acid–base properties of solutions of salts. A neutralization reaction can be defined as the reaction of an acid and a base to produce a salt and water. That is, another cation, such as $Na^+$, replaces the proton on the acid. An example is the reaction of $\ce{CH3CO2H}$, a weak acid, with $\ce{NaOH}$, a strong base:
$\underset{acid}{\ce{CH3CO2H(l)}} +\underset{base}{\ce{NaOH(s)}} \overset{\ce{H2O}}{\longrightarrow} \underset{salt}{\ce{H2OCH3CO2Na(aq)} }+\underset{water}{\ce{H2O(l)}} \label{16.35}$
Depending on the acid–base properties of its component ions, however, a salt can dissolve in water to produce a neutral solution, a basic solution, or an acidic solution.
When a salt such as $NaCl$ dissolves in water, it produces $Na^+_{(aq)}$ and $Cl^−_{(aq)}$ ions. Using a Lewis approach, the $Na^+$ ion can be viewed as an acid because it is an electron pair acceptor, although its low charge and relatively large radius make it a very weak acid. The $Cl^−$ ion is the conjugate base of the strong acid $HCl$, so it has essentially no basic character. Consequently, dissolving $NaCl$ in water has no effect on the $pH$ of a solution, and the solution remains neutral.
Now let's compare this behavior to the behavior of aqueous solutions of potassium cyanide and sodium acetate. Again, the cations ($K^+$ and $Na^+$) have essentially no acidic character, but the anions ($CN^−$ and $CH_3CO_2^−$) are weak bases that can react with water because they are the conjugate bases of the weak acids $HCN$ and acetic acid, respectively.
$\ce{CN^{-}(aq) + H2O(l) <<=> HCN(aq) + OH^{-}(aq)}$
$\ce{CH3CO^{-}2(aq) + H2O(l) <<=> CH3CO2H(aq) + OH^{-}(aq)}$
Neither reaction proceeds very far to the right as written because the formation of the weaker acid–base pair is favored. Both $HCN$ and acetic acid are stronger acids than water, and hydroxide is a stronger base than either acetate or cyanide, so in both cases, the equilibrium lies to the left. Nonetheless, each of these reactions generates enough hydroxide ions to produce a basic solution. For example, the $pH$ of a 0.1 M solution of sodium acetate or potassium cyanide at 25°C is 8.8 or 11.1, respectively. From Table E1 and E2, we can see that $CN^−$ is a stronger base ($pK_b = 4.79$) than acetate ($pK_b = 9.24$), which is consistent with $KCN$ producing a more basic solution than sodium acetate at the same concentration.
In contrast, the conjugate acid of a weak base should be a weak acid. For example, ammonium chloride and pyridinium chloride are salts produced by reacting ammonia and pyridine, respectively, with $HCl$. As you already know, the chloride ion is such a weak base that it does not react with water. In contrast, the cations of the two salts are weak acids that react with water as follows:
$\ce{NH^{+}4(aq) + H2O(l) <<=> NH3(aq) + H3O^{+}(aq)}$
$\ce{C5H5NH^{+}(aq) + H2O(l) <<=> C5H5NH(aq) + H3O^{+}(aq)}$
Figure $2$ shows that $H_3O^+$ is a stronger acid than either $NH_4^+$ or $C_5H_5NH^+$, and conversely, ammonia and pyridine are both stronger bases than water. The equilibrium will therefore lie far to the left in both cases, favoring the weaker acid–base pair. The $H_3O^+$ concentration produced by the reactions is great enough, however, to decrease the $pH$ of the solution significantly: the $pH$ of a 0.10 M solution of ammonium chloride or pyridinium chloride at 25°C is 5.13 or 3.12, respectively.
What happens with aqueous solutions of a salt such as ammonium acetate, where both the cation and the anion can react separately with water to produce an acid and a base, respectively? According to Figure 16.10, the ammonium ion will lower the $pH$, while according to Figure 16.9, the acetate ion will raise the $pH$. This particular case is unusual, in that the cation is as strong an acid as the anion is a base (pKa ≈ pKb). Consequently, the two effects cancel, and the solution remains neutral. With salts in which the cation is a stronger acid than the anion is a base, the final solution has a $pH$ < 7.00. Conversely, if the cation is a weaker acid than the anion is a base, the final solution has a $pH$ > 7.00.
Ions as Acids and Bases: https://youtu.be/XYAGNonPSow
Acidic Metal Ions
Solutions of simple salts of metal ions can also be acidic, even though a metal ion cannot donate a proton directly to water to produce $H_3O^+$. Instead, a metal ion can act as a Lewis acid and interact with water, a Lewis base, by coordinating to a lone pair of electrons on the oxygen atom to form a hydrated metal ion (Figure $\PageIndex{1a}$). A water molecule coordinated to a metal ion is more acidic than a free water molecule for two reasons. First, repulsive electrostatic interactions between the positively charged metal ion and the partially positively charged hydrogen atoms of the coordinated water molecule make it easier for the coordinated water to lose a proton.
Figure $1$
Second, the positive charge on the $Al^{3+}$ ion attracts electron density from the oxygen atoms of the water molecules, which decreases the electron density in the $\ce{O–H}$ bonds, as shown in Figure $\PageIndex{1b}$. With less electron density between the $O$ atoms and the H atoms, the $\ce{O–H}$ bonds are weaker than in a free $H_2O$ molecule, making it easier to lose a $H^+$ ion.
The magnitude of this effect depends on the following two factors (Figure $3$):
1. The charge on the metal ion. A divalent ion ($M^{2+}$) has approximately twice as strong an effect on the electron density in a coordinated water molecule as a monovalent ion ($M^+$) of the same radius.
2. The radius of the metal ion. For metal ions with the same charge, the smaller the ion, the shorter the internuclear distance to the oxygen atom of the water molecule and the greater the effect of the metal on the electron density distribution in the water molecule.
Thus aqueous solutions of small, highly charged metal ions, such as $Al^{3+}$ and $Fe^{3+}$, are acidic:
$[Al(H_2O)_6]^{3+}_{(aq)} \rightleftharpoons [Al(H_2O)_5(OH)]^{2+}_{(aq)}+H^+_{(aq)} \label{16.36}$
The $[Al(H_2O)_6]^{3+}$ ion has a $pK_a$ of 5.0, making it almost as strong an acid as acetic acid. Because of the two factors described previously, the most important parameter for predicting the effect of a metal ion on the acidity of coordinated water molecules is the charge-to-radius ratio of the metal ion. A number of pairs of metal ions that lie on a diagonal line in the periodic table, such as $Li^+$ and $Mg^{2+}$ or $Ca^{2+}$ and $Y^{3+}$, have different sizes and charges, but similar charge-to-radius ratios. As a result, these pairs of metal ions have similar effects on the acidity of coordinated water molecules, and they often exhibit other significant similarities in chemistry as well.
Acidic Metal Ions
Solutions of small, highly charged metal ions in water are acidic.
Reactions such as those discussed in this section, in which a salt reacts with water to give an acidic or basic solution, are often called hydrolysis reactions. Using a separate name for this type of reaction is unfortunate because it suggests that they are somehow different. In fact, hydrolysis reactions are just acid–base reactions in which the acid is a cation or the base is an anion; they obey the same principles and rules as all other acid–base reactions.
Hydrolysis
A hydrolysis reaction is an acid–base reaction.
Example $1$
Predict whether aqueous solutions of these compounds are acidic, basic, or neutral.
1. $KNO_3$
2. $CrBr_3 \cdot H_2O$
3. $Na_2SO_4$
Given: compound
Asked for: acidity or basicity of aqueous solution
Strategy:
1. Assess the acid–base properties of the cation and the anion. If the cation is a weak Lewis acid, it will not affect the $pH$ of the solution. If the cation is the conjugate acid of a weak base or a relatively highly charged metal cation, however, it will react with water to produce an acidic solution.
2. f the anion is the conjugate base of a strong acid, it will not affect the $pH$ of the solution. If, however, the anion is the conjugate base of a weak acid, the solution will be basic.
Solution:
a
1. The $K^+$ cation has a small positive charge (+1) and a relatively large radius (because it is in the fourth row of the periodic table), so it is a very weak Lewis acid.
2. The $NO_3−$ anion is the conjugate base of a strong acid, so it has essentially no basic character (Table 16.1). Hence neither the cation nor the anion will react with water to produce $H^+$ or $OH^−$, and the solution will be neutral.
b.
1. The $Cr^{3+}$ ion is a relatively highly charged metal cation that should behave similarly to the $Al^{3+}$ ion and form the [Cr(H2O)6]3+ complex, which will behave as a weak acid: $Cr(H_2O)_6]^{3+}_{(aq)} \ce{ <=>>} Cr(H_2O)_5(OH)]^{2+}_{(aq)} + H^+_({aq)}$
2. The $Br^−$ anion is a very weak base (it is the conjugate base of the strong acid $HBr$), so it does not affect the $pH$ of the solution. Hence the solution will be acidic.
c.
1. The $Na^+$ ion, like the $K^+$, is a very weak acid, so it should not affect the acidity of the solution.
2. In contrast, $SO_4^{2−}$ is the conjugate base of $HSO_4^−$, which is a weak acid. Hence the $SO_4^{2−}$ ion will react with water as shown in Figure 16.6 to give a slightly basic solution.
Exercise $1$
Predict whether aqueous solutions of the following are acidic, basic, or neutral.
1. $KI$
2. $Mg(ClO_4)_2$
3. $NaHS$
Answer a
neutral
Answer b
acidic
Answer c
basic (due to the reaction of $HS^−$ with water to form $H_2S$ and $OH^−$)
Summary
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 ($A^−$), 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. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/16%3A_Acids_and_Bases/16.7%3A_Ions_as_Acids_and_Bases.txt |
Learning Objectives
• To understand how molecular structure affects the strength of an acid or base.
We have seen that the strengths of acids and bases vary over many orders of magnitude. In this section, we explore some of the structural and electronic factors that control the acidity or basicity of a molecule.
Bond Strengths
In general, the stronger the $\ce{A–H}$ or $\ce{B–H^+}$ bond, the less likely the bond is to break to form $H^+$ ions and thus the less acidic the substance. This effect can be illustrated using the hydrogen halides:
Relative Acid Strength HF HCl HBr HI
H–X Bond Energy (kJ/mol) 570 432 366 298
pKa 3.20 −6.1 −8.9 −9.3
The trend in bond energies is due to a steady decrease in overlap between the 1s orbital of hydrogen and the valence orbital of the halogen atom as the size of the halogen increases. The larger the atom to which H is bonded, the weaker the bond. Thus the bond between H and a large atom in a given family, such as I or Te, is weaker than the bond between H and a smaller atom in the same family, such as F or O. As a result, acid strengths of binary hydrides increase as we go down a column of the periodic table. For example, the order of acidity for the binary hydrides of Group 16 elements is as follows, with $pK_a$ values in parentheses:
$H_2O (14.00 = pK_w) < H_2S (7.05) < H_2Se (3.89) < H_2Te (2.6) \label{1}$
Stability of the Conjugate Base
Whether we write an acid–base reaction as $AH \rightleftharpoons A^−+H^+$ or as $BH^+ \rightleftharpoons B+H^+$, the conjugate base ($A^−$ or $B$) contains one more lone pair of electrons than the parent acid ($AH$ or $BH^+$). Any factor that stabilizes the lone pair on the conjugate base favors dissociation of $H^+$ and makes the parent acid a stronger acid. Let’s see how this explains the relative acidity of the binary hydrides of the elements in the second row of the periodic table. The observed order of increasing acidity is the following, with pKa values in parentheses:
$CH_4 (~50) \ll NH_3 (~36) < H_2O (14.00) < HF (3.20) \label{2}$
Consider, for example, the compounds at both ends of this series: methane and hydrogen fluoride. The conjugate base of $CH_4$ is $CH_3^−$, and the conjugate base of $HF$ is $F^−$. Because fluorine is much more electronegative than carbon, fluorine can better stabilize the negative charge in the $F^−$ ion than carbon can stabilize the negative charge in the CH3− ion. Consequently, $\ce{HF}$ has a greater tendency to dissociate to form $H^+$ and $F^−$ than does methane to form $H^+$ and $CH_3^−$, making HF a much stronger acid than $CH_4$.
The same trend is predicted by analyzing the properties of the conjugate acids. For a series of compounds of the general formula $HE$, as the electronegativity of E increases, the E–H bond becomes more polar, favoring dissociation to form $E^−$ and $H^+$. Due to both the increasing stability of the conjugate base and the increasing polarization of the E–H bond in the conjugate acid, acid strengths of binary hydrides increase as we go from left to right across a row of the periodic table.
Acid strengths of binary hydrides increase as we go down a column or from left to right across a row of the periodic table.
the strongest acid Known: The hydrohelium Cation
The stornger acid, the weaker the covalent bond to a hydrogen atom. So the strongest acid possible is the molecule with the weakest bond. That is the hydrohelium (1+) cation, $\ce{HeH^{+}}$, which is a positively charged ion formed by the reaction of a proton with a helium atom in the gas phase. It was first produced in the laboratory in 1925 and is isoelectronic with molecular hydrogen (\ce{H2}}). It is the strongest known acid, with a proton affinity of 177.8 kJ/mol.
Ball and stick model of the hydrohelium ion. (CC BY-SA 3.0; CCoil).
$\ce{HeH^{+}}$ cannot be prepared in a condensed phase, as it would protonate any anion, molecule or atom with which it were associated. However it is possible to estimate a hypothetical aqueous acidity using Hess's law:
HHe+(g) H+(g) + He(g) +178 kJ/mol
HHe+(aq) HHe+(g) +973 kJ/mol
H+(g) H+(aq) −1530 kJ/mol
He(g) He(aq) +19 kJ/mol
HHe+(aq) H+(aq) + He(aq) −360 kJ/mol
A free energy change of dissociation of −360 kJ/mol is equivalent to a pKa of −63.
It has been suggested that $\ce{HeH^{+}}$ should occur naturally in the interstellar medium, but it has not yet been detected.
Inductive Effects
Atoms or groups of atoms in a molecule other than those to which H is bonded can induce a change in the distribution of electrons within the molecule. This is called an inductive effect, and, much like the coordination of water to a metal ion, it can have a major effect on the acidity or basicity of the molecule. For example, the hypohalous acids (general formula HOX, with X representing a halogen) all have a hydrogen atom bonded to an oxygen atom. In aqueous solution, they all produce the following equilibrium:
$HOX_{(aq)} \rightleftharpoons H^+_{(aq)} + OX^−{(aq)} \label{3}$
The acidities of these acids vary by about three orders of magnitude, however, due to the difference in electronegativity of the halogen atoms:
HOX Electronegativity of X pKa
HOCl 3.0 7.40
HOBr 2.8 8.55
HOI 2.5 10.5
As the electronegativity of $X$ increases, the distribution of electron density within the molecule changes: the electrons are drawn more strongly toward the halogen atom and, in turn, away from the H in the O–H bond, thus weakening the O–H bond and allowing dissociation of hydrogen as $H^+$.
The acidity of oxoacids, with the general formula $HOXO_n$ (with $n$ = 0−3), depends strongly on the number of terminal oxygen atoms attached to the central atom $X$. As shown in Figure $1$, the $K_a$ values of the oxoacids of chlorine increase by a factor of about $10^4$ to $10^6$ with each oxygen as successive oxygen atoms are added. The increase in acid strength with increasing number of terminal oxygen atoms is due to both an inductive effect and increased stabilization of the conjugate base.
Any inductive effect that withdraws electron density from an O–H bond increases the acidity of the compound.
Because oxygen is the second most electronegative element, adding terminal oxygen atoms causes electrons to be drawn away from the O–H bond, making it weaker and thereby increasing the strength of the acid. The colors in Figure $1$ show how the electrostatic potential, a measure of the strength of the interaction of a point charge at any place on the surface of the molecule, changes as the number of terminal oxygen atoms increases. In Figure $1$ and Figure $2$, blue corresponds to low electron densities, while red corresponds to high electron densities. The oxygen atom in the O–H unit becomes steadily less red from $HClO$ to $HClO_4$ (also written as $HOClO_3$, while the H atom becomes steadily bluer, indicating that the electron density on the O–H unit decreases as the number of terminal oxygen atoms increases. The decrease in electron density in the O–H bond weakens it, making it easier to lose hydrogen as $H^+$ ions, thereby increasing the strength of the acid.
At least as important, however, is the effect of delocalization of the negative charge in the conjugate base. As shown in Figure $2$, the number of resonance structures that can be written for the oxoanions of chlorine increases as the number of terminal oxygen atoms increases, allowing the single negative charge to be delocalized over successively more oxygen atoms.
Electron delocalization in the conjugate base increases acid strength.
The electrostatic potential plots in Figure $2$ demonstrate that the electron density on the terminal oxygen atoms decreases steadily as their number increases. The oxygen atom in ClO− is red, indicating that it is electron rich, and the color of oxygen progressively changes to green in $ClO_4^+$, indicating that the oxygen atoms are becoming steadily less electron rich through the series. For example, in the perchlorate ion ($ClO_4^−$), the single negative charge is delocalized over all four oxygen atoms, whereas in the hypochlorite ion ($OCl^−$), the negative charge is largely localized on a single oxygen atom (Figure $2$). As a result, the perchlorate ion has no localized negative charge to which a proton can bind. Consequently, the perchlorate anion has a much lower affinity for a proton than does the hypochlorite ion, and perchloric acid is one of the strongest acids known.
As the number of terminal oxygen atoms increases, the number of resonance structures that can be written for the oxoanions of chlorine also increases, and the single negative charge is delocalized over more oxygen atoms. As these electrostatic potential plots demonstrate, the electron density on the terminal oxygen atoms decreases steadily as their number increases. As the electron density on the oxygen atoms decreases, so does their affinity for a proton, making the anion less basic. As a result, the parent oxoacid is more acidic.
Similar inductive effects are also responsible for the trend in the acidities of oxoacids that have the same number of oxygen atoms as we go across a row of the periodic table from left to right. For example, $H_3PO_4$ is a weak acid, $H_2SO_4$ is a strong acid, and $HClO_4$ is one of the strongest acids known. The number of terminal oxygen atoms increases steadily across the row, consistent with the observed increase in acidity. In addition, the electronegativity of the central atom increases steadily from P to S to $Cl$, which causes electrons to be drawn from oxygen to the central atom, weakening the $\ce{O–H}$ bond and increasing the strength of the oxoacid.
Careful inspection of the data in Table $1$ shows two apparent anomalies: carbonic acid and phosphorous acid. If carbonic acid $(H_2CO_3$) were a discrete molecule with the structure $\ce{(HO)_2C=O}$, it would have a single terminal oxygen atom and should be comparable in acid strength to phosphoric acid ($H_3PO_4$), for which pKa1 = 2.16. Instead, the tabulated value of $pK_{a1}$ for carbonic acid is 6.35, making it about 10,000 times weaker than expected. As we shall see, however, $H_2CO_3$ is only a minor component of the aqueous solutions of $CO_2$ that are referred to as carbonic acid. Similarly, if phosphorous acid ($H_3PO_3$) actually had the structure $(HO)_3P$, it would have no terminal oxygen atoms attached to phosphorous. It would therefore be expected to be about as strong an acid as $HOCl$ (pKa = 7.40). In fact, the $pK_{a1}$ for phosphorous acid is 1.30, and the structure of phosphorous acid is $\ce{(HO)_2P(=O)H}$ with one H atom directly bonded to P and one $\ce{P=O}$ bond. Thus the pKa1 for phosphorous acid is similar to that of other oxoacids with one terminal oxygen atom, such as $H_3PO_4$. Fortunately, phosphorous acid is the only common oxoacid in which a hydrogen atom is bonded to the central atom rather than oxygen.
Table $1$: Values of pKa for Selected Polyprotic Acids and Bases
*$H_2CO_3$ and $H_2SO_3$ are at best minor components of aqueous solutions of $CO_{2(g)}$ and $SO_{2(g)}$, respectively, but such solutions are commonly referred to as containing carbonic acid and sulfurous acid, respectively.
Polyprotic Acids Formula $pK_{a1}$ $pK_{a2}$ $pK_{a3}$
carbonic acid* “$H_2CO_3$” 6.35 10.33
citric acid $HO_2CCH-2C(OH)(CO_2H)CH_2CO_2H$ 3.13 4.76 6.40
malonic acid $HO-2CCH_2CO_2H$ 2.85 5.70
oxalic acid $HO_2CCO_2H$ 1.25 3.81
phosphoric acid $H_3PO_4$ 2.16 7.21 12.32
phosphorous acid $H_3PO_3$ 1.3 6.70
succinic acid $HO_2CCH_2CH_2CO_2H$ 4.21 5.64
sulfuric acid $H_2SO_4$ −2.0 1.99
sulfurous acid* “$H_2SO_3$” 1.85 7.21
Polyprotic Bases Formula $pK_{b1}$ $pK_{b2}$
ethylenediamine $H_2N(CH_2)_2NH_2$ 4.08 7.14
piperazine $HN(CH_2CH_2)_2NH$ 4.27 8.67
propylenediamine $H_2N(CH_2)_3NH_2$ 3.45 5.12
Inductive effects are also observed in organic molecules that contain electronegative substituents. The magnitude of the electron-withdrawing effect depends on both the nature and the number of halogen substituents, as shown by the pKa values for several acetic acid derivatives:
$pK_a CH_3CO_2H 4.76< CH_2ClCO_2H 2.87<CHCl_2CO_2H 1.35<CCl_3CO_2H 0.66<CF_3CO_2H 0.52 \nonumber$
As you might expect, fluorine, which is more electronegative than chlorine, causes a larger effect than chlorine, and the effect of three halogens is greater than the effect of two or one. Notice from these data that inductive effects can be quite large. For instance, replacing the $\ce{–CH_3}$ group of acetic acid by a $\ce{–CF_3}$ group results in about a 10,000-fold increase in acidity!
Example $1$
Arrange the compounds of each series in order of increasing acid or base strength.
1. sulfuric acid [$H_2SO_4$, or $(HO)_2SO_2$], fluorosulfonic acid ($FSO_3H$, or $FSO_2OH$), and sulfurous acid [$H_2SO_3$, or $(HO)_2SO$]
2. ammonia ($NH_3$), trifluoramine ($NF_3$), and hydroxylamine ($NH_2OH$)
The structures are shown here.
Given: series of compounds
Asked for: relative acid or base strengths
Strategy:
Use relative bond strengths, the stability of the conjugate base, and inductive effects to arrange the compounds in order of increasing tendency to ionize in aqueous solution.
Solution:
Although both sulfuric acid and sulfurous acid have two –OH groups, the sulfur atom in sulfuric acid is bonded to two terminal oxygen atoms versus one in sulfurous acid. Because oxygen is highly electronegative, sulfuric acid is the stronger acid because the negative charge on the anion is stabilized by the additional oxygen atom. In comparing sulfuric acid and fluorosulfonic acid, we note that fluorine is more electronegative than oxygen. Thus replacing an –OH by –F will remove more electron density from the central S atom, which will, in turn, remove electron density from the S–OH bond and the O–H bond. Because its O–H bond is weaker, $FSO_3H$ is a stronger acid than sulfuric acid. The predicted order of acid strengths given here is confirmed by the measured pKa values for these acids:
$pKa H_2SO_3 1.85<H_2SO_4^{−2} < FSO_3H−10 \nonumber$
The structures of both trifluoramine and hydroxylamine are similar to that of ammonia. In trifluoramine, all of the hydrogen atoms in NH3 are replaced by fluorine atoms, whereas in hydroxylamine, one hydrogen atom is replaced by OH. Replacing the three hydrogen atoms by fluorine will withdraw electron density from N, making the lone electron pair on N less available to bond to an $H^+$ ion. Thus $NF_3$ is predicted to be a much weaker base than $NH_3$. Similarly, because oxygen is more electronegative than hydrogen, replacing one hydrogen atom in $NH_3$ by $OH$ will make the amine less basic. Because oxygen is less electronegative than fluorine and only one hydrogen atom is replaced, however, the effect will be smaller. The predicted order of increasing base strength shown here is confirmed by the measured $pK_b$ values:
$pK_bNF_3—<<NH_2OH 8.06<NH_3 4.75 \nonumber$
Trifluoramine is such a weak base that it does not react with aqueous solutions of strong acids. Hence its base ionization constant has never been measured.
Exercise $1$
Arrange the compounds of each series in order of
1. decreasing acid strength: $H_3PO_4$, $CH_3PO_3H_2$, and $HClO_3$.
2. increasing base strength: $CH_3S^−$, $OH^−$, and $CF_3S^−$.
Answer a
$HClO-3 > CH_3PO_3H_2 > H_3PO_4$
Answer a
$CF_3S^− < CH_3S^− < OH^−$
Summary
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. Atoms or groups of atoms elsewhere in a molecule can also be important in determining acid or base strength through an inductive effect, which can weaken an $\ce{O–H}$ bond and allow hydrogen to be more easily lost as $H^+$ ions.
• Anonymous | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/16%3A_Acids_and_Bases/16.8%3A_Molecular_Structure_and_Acid-Base_Behavior.txt |
Learning Objectives
Make sure you thoroughly understand the following essential ideas which have been presented. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic.
• Write the equation for the proton transfer reaction involving a Brønsted-Lowry acid or base, and show how it can be interpreted as an electron-pair transfer reaction, clearly identifying the donor and acceptor.
• Give an example of a Lewis acid-base reaction that does not involve protons.
• Write equations illustrating the behavior of a given non-aqueous acid-base system.
The Brønsted-Lowry proton donor-acceptor concept has been one of the most successful theories of Chemistry. But as with any such theory, it is fair to ask if this is not just a special case of a more general theory that could encompass an even broader range of chemical science. In 1916, G.N. Lewis of the University of California proposed that the electron pair is the dominant actor in acid-base chemistry. The Lewis theory did not become very well known until about 1923 (the same year that Brønsted and Lowry published their work), but since then it has been recognized as a very powerful tool for describing chemical reactions of widely different kinds and is widely used in organic and inorganic chemistry. The Brønsted–Lowry concept of acids and bases defines a base as any species that can accept a proton, and an acid as any substance that can donate a proton. Lewis proposed an alternative definition that focuses on pairs of electrons instead. According to Lewis:
• An 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 base is a substance that donates an unshared pair of electrons to a recipient species with which the electrons can be shared.
Note
In modern chemistry, electron donors are often referred to as nucleophiles, while acceptors are electrophiles.
Lewis Acid-Base Neutralization Involving Electron-Pair Transfer
Just as any Arrhenius acid is also a Brønsted acid, any Brønsted acid is also a Lewis acid, so the various acid-base concepts are all "upward compatible". Although we do not really need to think about electron-pair transfers when we deal with ordinary aqueous-solution acid-base reactions, it is important to understand that it is the opportunity for electron-pair sharing that enables proton transfer to take place.
This equation for a simple acid-base neutralization shows how the Brønsted and Lewis definitions are really just different views of the same process. Take special note of the following points:
• The arrow shows the movement of a proton from the hydronium ion to the hydroxide ion.
• Note that the electron-pairs themselves do not move; they remain attached to their central atoms. The electron pair on the base is "donated" to the acceptor (the proton) only in the sense that it ends up being shared with the acceptor, rather than being the exclusive property of the oxygen atom in the hydroxide ion.
• Although the hydronium ion is the nominal Lewis acid here, it does not itself accept an electron pair, but acts merely as the source of the proton that coordinates with the Lewis base.
The point about the electron-pair remaining on the donor species is especially important to bear in mind. For one thing, it distinguishes a Lewis acid-base reaction from an oxidation-reduction reaction, in which a physical transfer of one or more electrons from donor to acceptor does occur. The product of a Lewis acid-base reaction is known formally as an "adduct" or "complex", although we do not ordinarily use these terms for simple proton-transfer reactions such as the one in the above example. Here, the proton combines with the hydroxide ion to form the "adduct" H2O. The following examples illustrate these points for some other proton-transfer reactions that you should already be familiar with.
Another example, showing the autoprotolysis of water. Note that the conjugate base is also the adduct.
Ammonia is both a Brønsted and a Lewis base, owing to the unshared electron pair on the nitrogen. The reverse of this reaction represents the hydrolysis of the ammonium ion.
Because HF is a weak acid, fluoride salts behave as bases in aqueous solution. As a Lewis base, F accepts a proton from water, which is transformed into a hydroxide ion.
The bisulfite ion is amphiprotic and can act as an electron donor or acceptor.
Note
All Brønsted–Lowry bases (proton acceptors), such as OH, H2O, and NH3, are also electron-pair donors. Thus the Lewis definition of acids and bases does not contradict the Brønsted–Lowry definition. Rather, it expands the definition of acids to include substances other than the H+ ion.
Lewis Acid-Base Neutralization without Transferring Protons
Electron-deficient molecules, such as BCl3, contain less than an octet of electrons around one atom and have a strong tendency to gain an additional pair of electrons by reacting with substances that possess a lone pair of electrons. Lewis’s definition, which is less restrictive than either the Brønsted–Lowry or the Arrhenius definition, grew out of his observation of this tendency. A general Brønsted–Lowry acid–base reaction can be depicted in Lewis electron symbols as follows:
The proton (H+), which has no valence electrons, is a Lewis acid because it accepts a lone pair of electrons on the base to form a bond. The proton, however, is just one of many electron-deficient species that are known to react with bases. For example, neutral compounds of boron, aluminum, and the other Group 13 elements, which possess only six valence electrons, have a very strong tendency to gain an additional electron pair. Such compounds are therefore potent Lewis acids that react with an electron-pair donor such as ammonia to form an acid–base adduct, a new covalent bond, as shown here for boron trifluoride (BF3):
The bond formed between a Lewis acid and a Lewis base is a coordinate covalent bond because both electrons are provided by only one of the atoms (N, in the case of F3B:NH3). After it is formed, however, a coordinate covalent bond behaves like any other covalent single bond.
Species that are very weak Brønsted–Lowry bases can be relatively strong Lewis bases. For example, many of the group 13 trihalides are highly soluble in ethers (R–O–R′) because the oxygen atom in the ether contains two lone pairs of electrons, just as in H2O. Hence the predominant species in solutions of electron-deficient trihalides in ether solvents is a Lewis acid–base adduct. A reaction of this type is shown in Figure $1$ for boron trichloride and diethyl ether:
$\textbf{Figure } 1$: Lewis Acid/Base reaction of boron trichloride and diethyl ether reaction
Note
• Electron-deficient molecules (those with less than an octet of electrons) are Lewis acids.
• The acid-base behavior of many compounds can be explained by their Lewis electron structures.
Many molecules with multiple bonds can act as Lewis acids. In these cases, the Lewis base typically donates a pair of electrons to form a bond to the central atom of the molecule, while a pair of electrons displaced from the multiple bond becomes a lone pair on a terminal atom.
$\textbf{Figure } 2$: Lewis Acid/Base reaction of the hydroxide ion with carbon dioxide
$\textbf{Example } 1$
Identify the acid and the base in each Lewis acid–base reaction.
1. BH3 + (CH3)2S → H3B:S(CH3)2
2. CaO + CO2 → CaCO3
3. BeCl2 + 2 Cl → BeCl42−
Given: reactants and products
Asked for: identity of Lewis acid and Lewis base
Strategy:
In each equation, identify the reactant that is electron deficient and the reactant that is an electron-pair donor. The electron-deficient compound is the Lewis acid, whereas the other is the Lewis base.
Solution:
1. In BH3, boron has only six valence electrons. It is therefore electron deficient and can accept a lone pair. Like oxygen, the sulfur atom in (CH3)2S has two lone pairs. Thus (CH3)2S donates an electron pair on sulfur to the boron atom of BH3. The Lewis base is (CH3)2S, and the Lewis acid is BH3.
2. As in the reaction shown in Equation 8.21, CO2 accepts a pair of electrons from the O2− ion in CaO to form the carbonate ion. The oxygen in CaO is an electron-pair donor, so CaO is the Lewis base. Carbon accepts a pair of electrons, so CO2 is the Lewis acid.
3. The chloride ion contains four lone pairs. In this reaction, each chloride ion donates one lone pair to BeCl2, which has only four electrons around Be. Thus the chloride ions are Lewis bases, and BeCl2 is the Lewis acid.
$\textbf{Exercise } 1$
Identify the acid and the base in each Lewis acid–base reaction.
1. (CH3)2O + BF3 → (CH3)2O:BF3
2. H2O + SO3 → H2SO4
Answer
1. Lewis base: (CH3)2O; Lewis acid: BF3
2. Lewis base: H2O; Lewis acid: SO3
$\textbf{Exercise } 2$
Here are several more examples of Lewis acid-base reactions that cannot be accommodated within the Brønsted or Arrhenius models. Identify the Lewis acid and Lewis base in each reaction.
1. $Al(OH)_3 + OH^– \rightarrow Al(OH)_4^–$
2. $SnS_2 + S^{2–} \rightarrow SnS_3^{2–}$
3. $Cd(CN)_2 + 2 CN^– \rightarrow Cd(CN)_4^{2+}$
4. $AgCl + 2 NH_3 \rightarrow Ag(NH_3)_2^+ + Cl^–$
5. $Fe^{2+} + NO \rightarrow Fe(NO)^{2+}$
6. $[Ni^{2+} + 6 NH_3 \rightarrow Ni(NH_3)_5^{2+}$
Non-Aqueous Acid/Base Systems
We ordinarily think of Brønsted-Lowry acid-base reactions as taking place in aqueous solutions, but this need not always be the case. A more general view encompasses a variety of acid-base solvent systems, of which the water system is only one (Table $1$). Each of these has as its basis an amphiprotic solvent (one capable of undergoing autoprotolysis), in parallel with the familiar case of water.
The limiting acid in a given solvent is the solvonium ion, such as H3O+ (hydronium) ion in water. An acid which has more of a tendency to donate a hydrogen ion than the limiting acid will be a strong acid in the solvent considered, and will exist mostly or entirely in its dissociated form. Likewise, the limiting base in a given solvent is the solvate ion, such as OH (hydroxide) ion, in water. A base which has more affinity for protons than the limiting base cannot exist in solution, as it will react with the solvent.
The ammonia system is one of the most common non-aqueous system in Chemistry. Liquid ammonia boils at –33° C, and can conveniently be maintained as a liquid by cooling with dry ice (–77° C). It is a good solvent for substances that also dissolve in water, such as ionic salts and organic compounds since it is capable of forming hydrogen bonds. Bases can exist in solution in liquid ammonia which cannot exist in aqueous solution: this is the case for any base which is stronger than the hydroxide ion, but weaker than the amide ion $NH_2^-$. The limiting acid in liquid ammonia is the ammonium ion, which has a pKa value in water of 9.25. The limiting base is the amide ion, NH2. This is a stronger base than the hydroxide ion and so cannot exist in aqueous solution. The pKa value of ammonia is estimated to be approximately 33. Any acid which is a stronger acid than the ammonium ion will be a strong acid in liquid ammonia. This is the case for acetic acid, which is completely dissociated in liquid ammonia solution. The addition of pure acetic acid and the addition of ammonium acetate have exactly the same effect on a liquid ammonia solution: the increase in its acidity: in practice, the latter is preferred for safety reasons.
Note
As with $OH^-$ and $H_3O^+$ in water, the strongest acid and base in $NH_3$ is dictated by the corresponding autoprotolysis reaction of the solvent:
$2 NH_3 \rightleftharpoons NH_4^+ + NH_2^– \nonumber$
One use of non-aqueous acid-base systems is to examine the relative strengths of the strong acids and bases, whose strengths are "leveled" by the fact that they are all totally converted into H3O+ or OH ions in water. By studying them in appropriate non-aqueous solvents which are poorer acceptors or donors of protons, their relative strengths can be determined. Many familiar substances can serve as the basis of protonic solvent systems (Table $1$).
$\textbf{Table } 1$: Popular Solvent systems
solvent
autoprotolysis reaction
pKa
water $2 H_2O \rightleftharpoons H_3O^+ + OH^–$ 14
ammonia $2 NH_3 \rightleftharpoons NH_4^+ + NH_2^–$ 33
acetic acid $2 CH_3COOH \rightleftharpoons CH_3COOH_2^+ + CH_3COO^–$ 13
ethanol $2 C_2H_5OH \rightleftharpoons C_2H_5OH_2^+ + C_2H_5O^–$ 19
hydrogen peroxide $2 HO-OH \rightleftharpoons HO-OH_2^+ + HO-O^–$ 13
hydrofluoric acid $2 HF \rightleftharpoons H^2F^+ + F^–$ 10
sulfuric acid $2 H_2SO_4 \rightleftharpoons H3SO_4^+ + HSO_4^–$ 3.5
The extreme case is a superacid, a medium in which the hydrogen ion is only very weakly solvated. The classic example is a mixture of antimony pentafluoride and liquid hydrogen fluoride:
$SbF_5 + HF \rightleftharpoons H^+ + SbF_6^−$
The limiting base, the hexfluoroantimonate anion $SbF_6^−$, is so weakly attracted to the hydrogen ion that virtually any other base will bind more strongly: hence, this mixture can be used to protonate organic molecules which would not be considered bases in other solvents. It should noted that pH is undefined in aprotic solvents, which assumes presence of hydronium ions. In other solvents, the concentration of the respective solvonium/solvate ions should be used (e.g., $[NH_4^+]$ and $[NH_2^–]$ in $NH_{3(l)}$. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/16%3A_Acids_and_Bases/16.9%3A_Lewis_Acids_and_Bases.txt |
Learning Objectives
• Recognize common ions from various salts, acids, and bases.
• Calculate concentrations involving common ions.
• Calculate ion concentrations involving chemical equilibrium.
The common-ion effect is used to describe the effect on an equilibrium involving a substance that adds an ion that is a part of the equilibrium.
Introduction
The solubility products Ksp's are equilibrium constants in heterogeneous equilibria (i.e., between two different phases). If several salts are present in a system, they all ionize in the solution. If the salts contain a common cation or anion, these salts contribute to the concentration of the common ion. Contributions from all salts must be included in the calculation of concentration of the common ion. For example, a solution containing sodium chloride and potassium chloride will have the following relationship:
$\mathrm{[Na^+] + [K^+] = [Cl^-]} \label{1}$
Consideration of charge balance or mass balance or both leads to the same conclusion.
Common Ions
When $\ce{NaCl}$ and $\ce{KCl}$ are dissolved in the same solution, the $\mathrm{ {\color{Green} Cl^-}}$ ions are common to both salts. In a system containing $\ce{NaCl}$ and $\ce{KCl}$, the $\mathrm{ {\color{Green} Cl^-}}$ ions are common ions.
$\mathrm{NaCl \rightleftharpoons Na^+ + {\color{Green} Cl^-}} \nonumber$
$\mathrm{KCl \rightleftharpoons K^+ + {\color{Green} Cl^-}} \nonumber$
$\mathrm{CaCl_2 \rightleftharpoons Ca^{2+} + {\color{Green} 2 Cl^-}} \nonumber$
$\mathrm{AlCl_3 \rightleftharpoons Al^{3+} + {\color{Green} 3 Cl^-}} \nonumber$
$\mathrm{AgCl \rightleftharpoons Ag^+ + {\color{Green} Cl^-}} \nonumber$
For example, when $\ce{AgCl}$ is dissolved into a solution already containing $\ce{NaCl}$ (actually $\ce{Na+}$ and $\ce{Cl-}$ ions), the $\ce{Cl-}$ ions come from the ionization of both $\ce{AgCl}$ and $\ce{NaCl}$. Thus, $\ce{[Cl- ]}$ differs from $\ce{[Ag+]}$. The following examples show how the concentration of the common ion is calculated.
Example $1$
What are $\ce{[Na+]}$, $\ce{[Cl- ]}$, $\ce{[Ca^2+]}$, and $\ce{[H+]}$ in a solution containing 0.10 M each of $\ce{NaCl}$, $\ce{CaCl2}$, and $\ce{HCl}$?
Solution
Due to the conservation of ions, we have
$\mathrm{[Na^+] = [Ca^{2+}] = [H^+] = 0.10\: \ce M} \nonumber$
but
\begin{alignat}{3} \nonumber &\ce{[Cl- ]} &&= && && \:\textrm{0.10 (due to NaCl)}\ \nonumber & && && + &&\mathrm{\:0.20\: (due\: to\: CaCl_2)}\ \nonumber & && && + &&\mathrm{\:0.10\: (due\: to\: HCl)}\ \nonumber & &&= && &&\mathrm{\:0.40\: M} \nonumber \end{alignat}
Exercise $1$
John poured 10.0 mL of 0.10 M $\ce{NaCl}$, 10.0 mL of 0.10 M $\ce{KOH}$, and 5.0 mL of 0.20 M $\ce{HCl}$ solutions together and then he made the total volume to be 100.0 mL. What is $\ce{[Cl- ]}$ in the final solution?
Solution
$\mathrm{[Cl^-] = \dfrac{0.1\: M\times 10\: mL+0.2\: M\times 5.0\: mL}{100.0\: mL} = 0.020\: M} \nonumber$
Le Châtelier's Principle states that if an equilibrium becomes unbalanced, the reaction will shift to restore the balance. If a common ion is added to a weak acid or weak base equilibrium, then the equilibrium will shift towards the reactants, in this case the weak acid or base.
Example $2$: Solubility of Lead Chloride
Consider the lead(II) ion concentration in this saturated solution of PbCl2. The balanced reaction is
$PbCl_{2 (s)} \rightleftharpoons Pb^{2+} _{(aq)} + 2Cl^-_{(aq)} \nonumber$
Defining $s$ as the concentration of dissolved lead(II) chloride, then:
$[Pb^{2+}] = s \nonumber$
$[Cl^- ] = 2s \nonumber$
These values can be substituted into the solubility product expression, which can be solved for $s$:
$\begin{eqnarray} K_{sp} &=& [Pb^{2+}] [Cl^-]^2 \ &=& s \times (2s)^2 \ 1.7 \times 10^{-5} &=& 4s^3 \ s^3 &=& \frac{1.7 \times 10^{-5}}{4} \ &=& 4.25 \times 10^{-6} \ s &=& \sqrt[3]{4.25 \times 10^{-6}} \ &=& 1.62 \times 10^{-2}\ mol\ dm^{-3} \end{eqnarray} \nonumber$The concentration of lead(II) ions in the solution is 1.62 x 10-2 M. Consider what happens if sodium chloride is added to this saturated solution. Sodium chloride shares an ion with lead(II) chloride. The chloride ion is common to both of them; this is the origin of the term "common ion effect".
Look at the original equilibrium expression again:
$PbCl_2 \; (s) \rightleftharpoons Pb^{2+} \; (aq) + 2Cl^- \; (aq) \nonumber$
What happens to that equilibrium if extra chloride ions are added? According to Le Châtelier, the position of equilibrium will shift to counter the change, in this case, by removing the chloride ions by making extra solid lead(II) chloride.
Of course, the concentration of lead(II) ions in the solution is so small that only a tiny proportion of the extra chloride ions can be converted into solid lead(II) chloride. The lead(II) chloride becomes even less soluble, and the concentration of lead(II) ions in the solution decreases. This type of response occurs with any sparingly soluble substance: it is less soluble in a solution which contains any ion which it has in common. This is the common ion effect.
Example $3$
If an attempt is made to dissolve some lead(II) chloride in some 0.100 M sodium chloride solution instead of in water, what is the equilibrium concentration of the lead(II) ions this time? As before, define s to be the concentration of the lead(II) ions.
$[Pb^{2+}] = s \label{2}$
The calculations are different from before. This time the concentration of the chloride ions is governed by the concentration of the sodium chloride solution. The number of ions coming from the lead(II) chloride is going to be tiny compared with the 0.100 M coming from the sodium chloride solution.
In calculations like this, it can be assumed that the concentration of the common ion is entirely due to the other solution. This simplifies the calculation.
So we assume:
$[Cl^- ] = 0.100\; M \label{3}$
The rest of the mathematics looks like this:
$\begin{split} K_{sp}& = [Pb^{2+}][Cl^-]^2 \ & = s \times (0.100)^2 \ 1.7 \times 10^{-5} & = s \times 0.00100 \end{split} \nonumber$
therefore:
$\begin{split} s & = \dfrac{1.7 \times 10^{-5}}{0.0100} \ & = 1.7 \times 10^{-3} \, \text{M} \end{split} \label{4}$
Finally, compare that value with the simple saturated solution:
Original solution:
$[Pb^{2+}] = 0.0162 \, M \label{5}$
Solution in 0.100 M NaCl solution:
$[Pb^{2+}] = 0.0017 \, M \label{6}$
The concentration of the lead(II) ions has decreased by a factor of about 10. If more concentrated solutions of sodium chloride are used, the solubility decreases further.
A Video Discussing Finding the Solubility of a Salt: Finding the Solubility of a Salt(opens in new window) [youtu.be]
Common Ion Effect with Weak Acids and Bases
Adding a common ion prevents the weak acid or weak base from ionizing as much as it would without the added common ion. The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium.
Adding a common ion to a system at equilibrium affects the equilibrium composition, but not the ionization constant.
The common ion effect of H3O+ on the ionization of acetic acid
The common ion effect suppresses the ionization of a weak base by adding more of an ion that is a product of this equilibrium. Now consider the common ion effect of $\ce{OH^{-}}$ on the ionization of ammonia
Adding the common ion of hydroxide shifts the reaction towards the left to decrease the stress (in accordance with Le Chatelier's Principle), forming more reactants. This decreases the reaction quotient, because the reaction is being pushed towards the left to reach equilibrium. The equilibrium constant, $K_b=1.8 \times 10^{-5}$, does not change. The reaction is put out of balance, or equilibrium.
$Q_a = \frac{\ce{[NH_4^{+}][OH^{-}]}}{\ce{[NH3]}} \nonumber$
At first, when more hydroxide is added, the quotient is greater than the equilibrium constant. The reaction then shifts right, causing the denominator to increase, decreasing the reaction quotient and pulling towards equilibrium and causing Q to decrease towards K.
Common Ion Effect on Solubility
When a slightly soluble ionic compound is added to water, some of it dissolves to form a solution, establishing an equilibrium between the pure solid and a solution of its ions. For the dissolution of calcium phosphate, one of the two main components of kidney stones, the equilibrium can be written as follows, with the solid salt on the left:
$\ce{Ca3(PO4)2(s) <=> 3Ca^{2+}(aq) + 2PO^{3−}4(aq)} \label{17.4.1}$
As you will discover in more advanced chemistry courses, basic anions, such as S2−, PO43, and CO32, react with water to produce OH and the corresponding protonated anion. Consequently, their calculated molarities, assuming no protonation in aqueous solution, are only approximate. The equilibrium constant for the dissolution of a sparingly soluble salt is the solubility product (Ksp) of the salt. Because the concentration of a pure solid such as Ca3(PO4)2 is a constant, it does not appear explicitly in the equilibrium constant expression. The equilibrium constant expression for the dissolution of calcium phosphate is therefore
$K=\dfrac{[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2}{[\mathrm{Ca_3(PO_4)_2}]} \label{17.4.2a}$
$[\mathrm{Ca_3(PO_4)_2}]K=K_{\textrm{sp}}=[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2 \label{17.4.2b}$
At 25°C and pH 7.00, $K_{sp}$ for calcium phosphate is 2.07 × 10−33, indicating that the concentrations of Ca2+ and PO43 ions in solution that are in equilibrium with solid calcium phosphate are very low. The values of Ksp for some common salts vary dramatically for different compounds (Table E3). Although $K_{sp}$ is not a function of pH in Equation $\ref{17.4.2a}$, changes in pH can affect the solubility of a compound.
The solubility product expression tells us that the equilibrium concentrations of the cation and the anion are inversely related. That is, as the concentration of the anion increases, the maximum concentration of the cation needed for precipitation to occur decreases—and vice versa—so that $K_{sp}$ is constant. Consequently, the solubility of an ionic compound depends on the concentrations of other salts that contain the same ions. This dependency is another example of the common ion effect where adding a common cation or anion shifts a solubility equilibrium in the direction predicted by Le Chatelier’s principle. As a result, the solubility of any sparingly soluble salt is almost always decreased by the presence of a soluble salt that contains a common ion.
Consider, for example, the effect of adding a soluble salt, such as CaCl2, to a saturated solution of calcium phosphate [Ca3(PO4)2]. We have seen that the solubility of Ca3(PO4)2 in water at 25°C is 1.14 × 10−7 M (Ksp = 2.07 × 10−33). Thus a saturated solution of Ca3(PO4)2 in water contains
• $3 \times (1.14 \times 10^{−7}\; M) = 3.42 \times 10^{−7} M \; of \; Ca^{2+}$
• $2 \times (1.14 \times 10^{−7} M) = 2.28 \times 10^{−7} M \; of \; PO_4^{3−}$
according to the stoichiometry shown in Equation $\ref{17.4.2a}$ (neglecting hydrolysis to form HPO42). If CaCl2 is added to a saturated solution of Ca3(PO4)2, the Ca2+ ion concentration will increase such that [Ca2+] > 3.42 × 10−7 M, making Q > Ksp. The only way the system can return to equilibrium is for the reaction in Equation $\ref{17.4.2a}$ to proceed to the left, resulting in precipitation of Ca3(PO4)2. This will decrease the concentration of both Ca2+ and PO43 until Q = Ksp.
Note
Adding a common ion decreases solubility, as the reaction shifts toward the left to relieve the stress of the excess product. Adding a common ion to a dissociation reaction causes the equilibrium to shift left, toward the reactants, causing precipitation.
Example $5$
Consider the reaction:
$PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq) \nonumber$
What happens to the solubility of PbCl2(s) when 0.1 M NaCl is added?
Solution
$K_{sp}=1.7 \times 10^{-5} \nonumber$
$Q_{sp}= 1.8 \times 10^{-5}\nonumber$
Identify the common ion: Cl-
Notice: Qsp > Ksp The addition of NaCl has caused the reaction to shift out of equilibrium because there are more dissociated ions. Typically, solving for the molarities requires the assumption that the solubility of PbCl2 is equivalent to the concentration of Pb2+ produced because they are in a 1:1 ratio.
Because Ksp for the reaction is 1.7×10-5, the overall reaction would be (s)(2s)2= 1.7×10-5. Solving the equation for s gives s= 1.62×10-2 M. The coefficient on Cl- is 2, so it is assumed that twice as much Cl- is produced as Pb2+, hence the '2s.' The solubility equilibrium constant can be used to solve for the molarities of the ions at equilibrium.
The molarity of Cl- added would be 0.1 M because Na+ and Cl- are in a 1:1 ration in the ionic salt, NaCl. Therefore, the overall molarity of Cl- would be 2s + 0.1, with 2s referring to the contribution of the chloride ion from the dissociation of lead chloride.
$\begin{eqnarray} Q_{sp} &=& [Pb^{2+}][Cl^-]^2 \ 1.8 \times 10^{-5} &=& (s)(2s + 0.1)^2 \ s &=& [Pb^{2+}] \ &=& 1.8 \times 10^{-3} M \ 2s &=& [Cl^-] \ &\approx & 0.1 M \end{eqnarray} \nonumber$
Notice that the molarity of Pb2+ is lower when NaCl is added. The equilibrium constant remains the same because of the increased concentration of the chloride ion. To simplify the reaction, it can be assumed that [Cl-] is approximately 0.1M since the formation of the chloride ion from the dissociation of lead chloride is so small. The reaction quotient for PbCl2 is greater than the equilibrium constant because of the added Cl-. This therefore shift the reaction left towards equilibrium, causing precipitation and lowering the current solubility of the reaction. Overall, the solubility of the reaction decreases with the added sodium chloride.
Exercise $5$
Calculate the solubility of silver carbonate in a 0.25 M solution of sodium carbonate. The solubility of silver carbonate in pure water is 8.45 × 10−12 at 25°C.
Answer
2.9 × 10−6 M (versus 1.3 × 10−4 M in pure water)
A Video Discussing the Common Ion Effect in Solubility Products: The Common Ion Effect in Solubility Products (opens in new window) [youtu.be]
Contributors and Attributions
• Emmellin Tung, Mahtab Danai (UCD)
• Jim Clark (ChemGuide)
• Chung (Peter) Chieh (Professor Emeritus, Chemistry @ University of Waterloo) | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/17%3A_Additional_Aspects_of_Acid-Base_Equilibria/17.1%3A_Common-Ion_Effect_in_Acid-Base_Equilibria.txt |
Learning Objectives
• To understand how adding a common ion affects the position of an acid–base equilibrium.
• To know how to use the Henderson-Hasselbalch approximation to calculate the pH of a buffer.
Buffers are solutions that maintain a relatively constant pH when an acid or a base is added. They therefore protect, or “buffer,” other molecules in solution from the effects of the added acid or base. Buffers contain either a weak acid ($HA$) and its conjugate base $(A^−$) or a weak base ($B$) and its conjugate acid ($BH^+$), and they are critically important for the proper functioning of biological systems. In fact, every biological fluid is buffered to maintain its physiological pH.
The Common Ion Effect: Weak Acids Combined with Conjugate Bases
To understand how buffers work, let’s look first at how the ionization equilibrium of a weak acid is affected by adding either the conjugate base of the acid or a strong acid (a source of $\ce{H^{+}}$). Le Chatelier’s principle can be used to predict the effect on the equilibrium position of the solution. A typical buffer used in biochemistry laboratories contains acetic acid and a salt such as sodium acetate. The dissociation reaction of acetic acid is as follows:
$\ce{CH3COOH (aq) <=> CH3COO^{−} (aq) + H^{+} (aq)} \label{Eq1}$
and the equilibrium constant expression is as follows:
$K_a=\dfrac{[\ce{H^{+}}][\ce{CH3COO^{-}}]}{[\ce{CH3CO2H}]} \label{Eq2}$
Sodium acetate ($\ce{CH_3CO_2Na}$) is a strong electrolyte that ionizes completely in aqueous solution to produce $\ce{Na^{+}}$ and $\ce{CH3CO2^{−}}$ ions. If sodium acetate is added to a solution of acetic acid, Le Chatelier’s principle predicts that the equilibrium in Equation \ref{Eq1} will shift to the left, consuming some of the added $\ce{CH_3COO^{−}}$ and some of the $\ce{H^{+}}$ ions originally present in solution.
Because $\ce{Na^{+}}$ is a spectator ion, it has no effect on the position of the equilibrium and can be ignored. The addition of sodium acetate produces a new equilibrium composition, in which $[\ce{H^{+}}]$ is less than the initial value. Because $[\ce{H^{+}}]$ has decreased, the pH will be higher. Thus adding a salt of the conjugate base to a solution of a weak acid increases the pH. This makes sense because sodium acetate is a base, and adding any base to a solution of a weak acid should increase the pH.
If we instead add a strong acid such as $\ce{HCl}$ to the system, $[\ce{H^{+}}]$ increases. Once again the equilibrium is temporarily disturbed, but the excess $\ce{H^{+}}$ ions react with the conjugate base ($\ce{CH_3CO_2^{−}}$), whether from the parent acid or sodium acetate, to drive the equilibrium to the left. The net result is a new equilibrium composition that has a lower [$\ce{CH_3CO_2^{−}}$] than before. In both cases, only the equilibrium composition has changed; the ionization constant $K_a$ for acetic acid remains the same. Adding a strong electrolyte that contains one ion in common with a reaction system that is at equilibrium, in this case $\ce{CH3CO2^{−}}$, will therefore shift the equilibrium in the direction that reduces the concentration of the common ion. The shift in equilibrium is via the common ion effect.
Adding a common ion to a system at equilibrium affects the equilibrium composition, but not the ionization constant.
Example $1$
A 0.150 M solution of formic acid at 25°C (pKa = 3.75) has a pH of 2.28 and is 3.5% ionized.
1. Is there a change to the pH of the solution if enough solid sodium formate is added to make the final formate concentration 0.100 M (assume that the formic acid concentration does not change)?
2. What percentage of the formic acid is ionized if 0.200 M HCl is added to the system?
Given: solution concentration and pH, $pK_a$, and percent ionization of acid; final concentration of conjugate base or strong acid added
Asked for: pH and percent ionization of formic acid
Strategy:
1. Write a balanced equilibrium equation for the ionization equilibrium of formic acid. Tabulate the initial concentrations, the changes, and the final concentrations.
2. Substitute the expressions for the final concentrations into the expression for Ka. Calculate $[\ce{H^{+}}]$ and the pH of the solution.
3. Construct a table of concentrations for the dissociation of formic acid. To determine the percent ionization, determine the anion concentration, divide it by the initial concentration of formic acid, and multiply the result by 100.
Solution:
A Because sodium formate is a strong electrolyte, it ionizes completely in solution to give formate and sodium ions. The $\ce{Na^{+}}$ ions are spectator ions, so they can be ignored in the equilibrium equation. Because water is both a much weaker acid than formic acid and a much weaker base than formate, the acid–base properties of the solution are determined solely by the formic acid ionization equilibrium:
$\ce{HCO2H (aq) <=> HCO^{−}2 (aq) + H^{+} (aq)} \nonumber$
The initial concentrations, the changes in concentration that occur as equilibrium is reached, and the final concentrations can be tabulated.
Final Concentration
ICE $[HCO_2H (aq) ]$ $[H^+ (aq) ]$ $[HCO^−_2 (aq) ]$
Initial 0.150 $1.00 \times 10^{−7}$ 0.100
Change −x +x +x
Equilibrium (0.150 − x) x (0.100 + x)
B We substitute the expressions for the final concentrations into the equilibrium constant expression and make our usual simplifying assumptions, so
\begin{align*} K_a=\dfrac{[H^+][HCO_2^−]}{[HCO_2H]} &=\dfrac{(x)(0.100+x)}{0.150−x} \[4pt] &\approx \dfrac{x(0.100)}{0.150} \[4pt] &\approx 10^{−3.75} \[4pt] &\approx 1.8 \times 10^{−4} \end{align*} \nonumber
Rearranging and solving for $x$,
\begin{align*} x &=(1.8 \times 10^{−4}) \times \dfrac{0.150 \;M}{ 0.100 \;M} \[4pt] &=2.7 \times 10^{−4}\[4pt] &=[H^+] \end{align*} \nonumber
The value of $x$ is small compared with 0.150 or 0.100 M, so our assumption about the extent of ionization is justified. Moreover,
$K_aC_{HA} = (1.8 \times 10^{−4})(0.150) = 2.7 \times 10^{−5} \nonumber$
which is greater than $1.0 \times 10^{−6}$, so again, our assumption is justified. The final pH is:
$pH= −\log(2.7 \times 10^{−4}) = 3.57 \nonumber$
compared with the initial value of 2.29. Thus adding a salt containing the conjugate base of the acid has increased the pH of the solution, as we expect based on Le Chatelier’s principle; the stress on the system has been relieved by the consumption of $\ce{H^{+}}$ ions, driving the equilibrium to the left.
C Because $HCl$ is a strong acid, it ionizes completely, and chloride is a spectator ion that can be neglected. Thus the only relevant acid–base equilibrium is again the dissociation of formic acid, and initially the concentration of formate is zero. We can construct a table of initial concentrations, changes in concentration, and final concentrations.
$HCO_2H (aq) \leftrightharpoons H^+ (aq) +HCO^−_2 (aq) \nonumber$
initial concentrations, changes in concentration, and final concentrations
$[HCO_2H (aq) ]$ $[H^+ (aq) ]$ $[HCO^−_2 (aq) ]$
initial 0.150 0.200 0
change −x +x +x
final (0.150 − x) (0.200 + x) x
To calculate the percentage of formic acid that is ionized under these conditions, we have to determine the final $[\ce{HCO2^{-}}]$. We substitute final concentrations into the equilibrium constant expression and make the usual simplifying assumptions, so
$K_a=\dfrac{[H^+][HCO_2^−]}{[HCO_2H]}=\dfrac{(0.200+x)(x)}{0.150−x} \approx \dfrac{x(0.200)}{0.150}=1.80 \times 10^{−4} \nonumber$
Rearranging and solving for $x$,
\begin{align*} x &=(1.80 \times 10^{−4}) \times \dfrac{ 0.150\; M}{ 0.200\; M} \[4pt] &=1.35 \times 10^{−4}=[HCO_2^−] \end{align*} \nonumber
Once again, our simplifying assumptions are justified. The percent ionization of formic acid is as follows:
$\text{percent ionization}=\dfrac{1.35 \times 10^{−4} \;M} {0.150\; M} \times 100\%=0.0900\% \nonumber$
Adding the strong acid to the solution, as shown in the table, decreased the percent ionization of formic acid by a factor of approximately 38 (3.45%/0.0900%). Again, this is consistent with Le Chatelier’s principle: adding $\ce{H^{+}}$ ions drives the dissociation equilibrium to the left.
Exercise $1$
A 0.225 M solution of ethylamine ($\ce{CH3CH2NH2}$ with $pK_b = 3.19$) has a pH of 12.08 and a percent ionization of 5.4% at 20°C. Calculate the following:
1. the pH of the solution if enough solid ethylamine hydrochloride ($\ce{EtNH3Cl}$) is added to make the solution 0.100 M in $\ce{EtNH3^{+}}$
2. the percentage of ethylamine that is ionized if enough solid $\ce{NaOH}$ is added to the original solution to give a final concentration of 0.050 M $\ce{NaOH}$
Answer a
11.16
Answer b
1.3%
A Video Discussing the Common Ion Effect: The Common Ion Effecr(opens in new window) [youtu.be]
The Common Ion Effect: Weak Bases Combined with Conjugate Acids
Now let’s suppose we have a buffer solution that contains equimolar concentrations of a weak base ($B$) and its conjugate acid ($BH^+$). The general equation for the ionization of a weak base is as follows:
$B (aq) +H_2O (l) \leftrightharpoons BH^+ (aq) +OH^− (aq) \label{Eq3}$
If the equilibrium constant for the reaction as written in Equation $\ref{Eq3}$ is small, for example $K_b = 10^{−5}$, then the equilibrium constant for the reverse reaction is very large: $K = \dfrac{1}{K_b} = 10^5$. Adding a strong base such as $OH^-$ to the solution therefore causes the equilibrium in Equation $\ref{Eq3}$ to shift to the left, consuming the added $OH^-$. As a result, the $OH^-$ ion concentration in solution remains relatively constant, and the pH of the solution changes very little. Le Chatelier’s principle predicts the same outcome: when the system is stressed by an increase in the $OH^-$ ion concentration, the reaction will proceed to the left to counteract the stress.
If the $pK_b$ of the base is 5.0, the $pK_a$ of its conjugate acid is
$pK_a = pK_w − pK_b = 14.0 – 5.0 = 9.0. \nonumber$
Thus the equilibrium constant for ionization of the conjugate acid is even smaller than that for ionization of the base. The ionization reaction for the conjugate acid of a weak base is written as follows:
$BH^+ (aq) +H_2O (l) \leftrightharpoons B (aq) +H_3O^+ (aq) \label{Eq4}$
Again, the equilibrium constant for the reverse of this reaction is very large: K = 1/Ka = 109. If a strong acid is added, it is neutralized by reaction with the base as the reaction in Equation $\ref{Eq4}$ shifts to the left. As a result, the $H^+$ ion concentration does not increase very much, and the pH changes only slightly. In effect, a buffer solution behaves somewhat like a sponge that can absorb $H^+$ and $OH^-$ ions, thereby preventing large changes in pH when appreciable amounts of strong acid or base are added to a solution.
Buffers are characterized by the pH range over which they can maintain a more or less constant pH and by their buffer capacity, the amount of strong acid or base that can be absorbed before the pH changes significantly. Although the useful pH range of a buffer depends strongly on the chemical properties of the weak acid and weak base used to prepare the buffer (i.e., on $K$), its buffer capacity depends solely on the concentrations of the species in the buffered solution. The more concentrated the buffer solution, the greater its buffer capacity. As illustrated in Figure $1$, when $NaOH$ is added to solutions that contain different concentrations of an acetic acid/sodium acetate buffer, the observed change in the pH of the buffer is inversely proportional to the concentration of the buffer. If the buffer capacity is 10 times larger, then the buffer solution can absorb 10 times more strong acid or base before undergoing a significant change in pH.
A buffer maintains a relatively constant pH when acid or base is added to a solution. The addition of even tiny volumes of 0.10 M $NaOH$ to 100.0 mL of distilled water results in a very large change in pH. As the concentration of a 50:50 mixture of sodium acetate/acetic acid buffer in the solution is increased from 0.010 M to 1.00 M, the change in the pH produced by the addition of the same volume of $NaOH$ solution decreases steadily. For buffer concentrations of at least 0.500 M, the addition of even 25 mL of the $NaOH$ solution results in only a relatively small change in pH.
Calculating the pH of a Buffer
The pH of a buffer can be calculated from the concentrations of the weak acid and the weak base used to prepare it, the concentration of the conjugate base and conjugate acid, and the $pK_a$ or $pK_b$ of the weak acid or weak base. The procedure is analogous to that used in Example $1$ to calculate the pH of a solution containing known concentrations of formic acid and formate.
An alternative method frequently used to calculate the pH of a buffer solution is based on a rearrangement of the equilibrium equation for the dissociation of a weak acid. The simplified ionization reaction is $HA \leftrightharpoons H^+ + A^−$, for which the equilibrium constant expression is as follows:
$K_a=\dfrac{[H^+][A^-]}{[HA]} \label{Eq5}$
This equation can be rearranged as follows:
$[H^+]=K_a\dfrac{[HA]}{[A^−]} \label{Eq6}$
Taking the logarithm of both sides and multiplying both sides by −1,
\begin{align} −\log[H^+] &=−\log K_a−\log\left(\dfrac{[HA]}{[A^−]}\right) \[4pt] &=−\log{K_a}+\log\left(\dfrac{[A^−]}{[HA]}\right) \label{Eq7} \end{align}
Replacing the negative logarithms in Equation $\ref{Eq7}$,
$pH=pK_a+\log \left( \dfrac{[A^−]}{[HA]} \right) \label{Eq8}$
or, more generally,
$pH=pK_a+\log\left(\dfrac{[base]}{[acid]}\right) \label{Eq9}$
Equation $\ref{Eq8}$ and Equation $\ref{Eq9}$ are both forms of the Henderson-Hasselbalch approximation, named after the two early 20th-century chemists who first noticed that this rearranged version of the equilibrium constant expression provides an easy way to calculate the pH of a buffer solution. In general, the validity of the Henderson-Hasselbalch approximation may be limited to solutions whose concentrations are at least 100 times greater than their $K_a$ values.
There are three special cases where the Henderson-Hasselbalch approximation is easily interpreted without the need for calculations:
• $[base] = [acid]$: Under these conditions, $\dfrac{[base]}{[acid]} = 1 \nonumber$ in Equation \ref{Eq9}. Because $\log 1 = 0$, $pH = pK_a \nonumber$ regardless of the actual concentrations of the acid and base. Recall that this corresponds to the midpoint in the titration of a weak acid or a weak base.
• $[base]/[acid] = 10$: In Equation $\ref{Eq9}$, because $\log 10 = 1$, $pH = pK_a + 1. \nonumber$
• $[base]/[acid] = 100$: In Equation $\ref{Eq9}$, because $\log 100 = 2$, $pH = pK_a + 2. \nonumber$
Each time we increase the [base]/[acid] ratio by 10, the pH of the solution increases by 1 pH unit. Conversely, if the [base]/[acid] ratio is 0.1, then pH = $pK_a$ − 1. Each additional factor-of-10 decrease in the [base]/[acid] ratio causes the pH to decrease by 1 pH unit.
If [base] = [acid] for a buffer, then pH = $pK_a$. Changing this ratio by a factor of 10 either way changes the pH by ±1 unit.
Example $2$
What is the pH of a solution that contains
1. 0.135 M $\ce{HCO2H}$ and 0.215 M $\ce{HCO2Na}$? (The $pK_a$ of formic acid is 3.75.)
2. 0.0135 M $\ce{HCO2H}$ and 0.0215 M $\ce{HCO2Na}$?
3. 0.119 M pyridine and 0.234 M pyridine hydrochloride? (The $pK_b$ of pyridine is 8.77.)
Given: concentration of acid, conjugate base, and $pK_a$; concentration of base, conjugate acid, and $pK_b$
Asked for: pH
Strategy:
Substitute values into either form of the Henderson-Hasselbalch approximation (Equations \ref{Eq8} or \ref{Eq9}) to calculate the pH.
Solution:
According to the Henderson-Hasselbalch approximation (Equation \ref{Eq8}), the pH of a solution that contains both a weak acid and its conjugate base is
$pH = pK_a + \log([A−]/[HA]). \nonumber$
A
Inserting the given values into the equation,
\begin{align*} pH &=3.75+\log\left(\dfrac{0.215}{0.135}\right) \[4pt] &=3.75+\log 1.593 \[4pt] &=3.95 \end{align*} \nonumber
This result makes sense because the $[A^−]/[HA]$ ratio is between 1 and 10, so the pH of the buffer must be between the $pK_a$ (3.75) and $pK_a + 1$, or 4.75.
B
This is identical to part (a), except for the concentrations of the acid and the conjugate base, which are 10 times lower. Inserting the concentrations into the Henderson-Hasselbalch approximation,
\begin{align*} pH &=3.75+\log\left(\dfrac{0.0215}{0.0135}\right) \[4pt] &=3.75+\log 1.593 \[4pt] &=3.95 \end{align*} \nonumber
This result is identical to the result in part (a), which emphasizes the point that the pH of a buffer depends only on the ratio of the concentrations of the conjugate base and the acid, not on the magnitude of the concentrations. Because the [A]/[HA] ratio is the same as in part (a), the pH of the buffer must also be the same (3.95).
C
In this case, we have a weak base, pyridine (Py), and its conjugate acid, the pyridinium ion ($HPy^+$). We will therefore use Equation $\ref{Eq9}$, the more general form of the Henderson-Hasselbalch approximation, in which “base” and “acid” refer to the appropriate species of the conjugate acid–base pair. We are given [base] = [Py] = 0.119 M and $[acid] = [HPy^{+}] = 0.234\, M$. We also are given $pK_b = 8.77$ for pyridine, but we need $pK_a$ for the pyridinium ion. Recall from Equation 16.23 that the $pK_b$ of a weak base and the $pK_a$ of its conjugate acid are related:
$pK_a + pK_b = pK_w. \nonumber$
Thus $pK_a$ for the pyridinium ion is $pK_w − pK_b = 14.00 − 8.77 = 5.23$. Substituting this $pK_a$ value into the Henderson-Hasselbalch approximation,
\begin{align*} pH=pK_a+\log \left(\dfrac{[base]}{[acid]}\right) \[4pt] &=5.23+\log\left(\dfrac{0.119}{0.234}\right) \[4pt] & =5.23 −0.294 \[4pt] &=4.94 \end{align*} \nonumber
Once again, this result makes sense: the $[B]/[BH^+]$ ratio is about 1/2, which is between 1 and 0.1, so the final pH must be between the $pK_a$ (5.23) and $pK_a − 1$, or 4.23.
Exercise $2$
What is the pH of a solution that contains
1. 0.333 M benzoic acid and 0.252 M sodium benzoate?
2. 0.050 M trimethylamine and 0.066 M trimethylamine hydrochloride?
The $pK_a$ of benzoic acid is 4.20, and the $pK_b$ of trimethylamine is also 4.20.
Answer a
4.08
Answer b
9.68
A Video Discussing Using the Henderson Hasselbalch Equation: Using the Henderson Hasselbalch Equation(opens in new window) [youtu.be] (opens in new window)
The Henderson-Hasselbalch approximation ((Equation $\ref{Eq8}$) can also be used to calculate the pH of a buffer solution after adding a given amount of strong acid or strong base, as demonstrated in Example $3$.
Example $3$
The buffer solution in Example $2$ contained 0.135 M $\ce{HCO2H}$ and 0.215 M $\ce{HCO2Na}$ and had a pH of 3.95.
1. What is the final pH if 5.00 mL of 1.00 M $HCl$ are added to 100 mL of this solution?
2. What is the final pH if 5.00 mL of 1.00 M $NaOH$ are added?
Given: composition and pH of buffer; concentration and volume of added acid or base
Asked for: final pH
Strategy:
1. Calculate the amounts of formic acid and formate present in the buffer solution using the procedure from Example $1$. Then calculate the amount of acid or base added.
2. Construct a table showing the amounts of all species after the neutralization reaction. Use the final volume of the solution to calculate the concentrations of all species. Finally, substitute the appropriate values into the Henderson-Hasselbalch approximation (Equation \ref{Eq9}) to obtain the pH.
Solution:
The added $\ce{HCl}$ (a strong acid) or $\ce{NaOH}$ (a strong base) will react completely with formate (a weak base) or formic acid (a weak acid), respectively, to give formic acid or formate and water. We must therefore calculate the amounts of formic acid and formate present after the neutralization reaction.
A We begin by calculating the millimoles of formic acid and formate present in 100 mL of the initial pH 3.95 buffer:
$100 \, \cancel{mL} \left( \dfrac{0.135 \, mmol\; \ce{HCO2H}}{\cancel{mL}} \right) = 13.5\, mmol\, \ce{HCO2H} \nonumber$
$100\, \cancel{mL } \left( \dfrac{0.215 \, mmol\; \ce{HCO2^{-}}}{\cancel{mL}} \right) = 21.5\, mmol\, \ce{HCO2^{-}} \nonumber$
The millimoles of $\ce{H^{+}}$ in 5.00 mL of 1.00 M $\ce{HCl}$ is as follows:
$5.00 \, \cancel{mL } \left( \dfrac{1.00 \,mmol\; \ce{H^{+}}}{\cancel{mL}} \right) = 5\, mmol\, \ce{H^{+}} \nonumber$
B Next, we construct a table of initial amounts, changes in amounts, and final amounts:
$\ce{HCO^{2−}(aq) + H^{+} (aq) <=> HCO2H (aq)} \nonumber$
initial amounts, changes in amounts, and final amounts:
$HCO^{2−} (aq)$ $H^+ (aq)$ $HCO_2H (aq)$
Initial 21.5 mmol 5.00 mmol 13.5 mmol
Change −5.00 mmol −5.00 mmol +5.00 mmol
Final 16.5 mmol ∼0 mmol 18.5 mmol
The final amount of $H^+$ in solution is given as “∼0 mmol.” For the purposes of the stoichiometry calculation, this is essentially true, but remember that the point of the problem is to calculate the final $[H^+]$ and thus the pH. We now have all the information we need to calculate the pH. We can use either the lengthy procedure of Example $1$ or the Henderson–Hasselbach approximation. Because we have performed many equilibrium calculations in this chapter, we’ll take the latter approach. The Henderson-Hasselbalch approximation requires the concentrations of $HCO_2^−$ and $HCO_2H$, which can be calculated using the number of millimoles ($n$) of each and the total volume ($VT$). Substituting these values into the Henderson-Hasselbalch approximation (Equation $\ref{Eq9}$):
\begin{align*} pH &=pK_a+\log \left( \dfrac{[HCO_2^−]}{[HCO_2H]} \right) \[4pt] &=pK_a+\log\left(\dfrac{n_{HCO_2^−}/V_f}{n_{HCO_2H}/V_f}\right) \[4pt] &=pK_a+\log \left(\dfrac{n_{HCO_2^−}}{n_{HCO_2H}}\right) \end{align*} \nonumber
Because the total volume appears in both the numerator and denominator, it cancels. We therefore need to use only the ratio of the number of millimoles of the conjugate base to the number of millimoles of the weak acid. So
\begin{align*} pH &=pK_a+\log\left(\dfrac{n_{HCO_2^−}}{n_{HCO_2H}}\right) \[4pt] &=3.75+\log\left(\dfrac{16.5\; mmol}{18.5\; mmol}\right) \[4pt] &=3.75 −0.050=3.70 \end{align*} \nonumber
Once again, this result makes sense on two levels. First, the addition of $HCl$has decreased the pH from 3.95, as expected. Second, the ratio of $HCO_2^−$ to $HCO_2H$ is slightly less than 1, so the pH should be between the $pK_a$ and $pK_a$ − 1.
A The procedure for solving this part of the problem is exactly the same as that used in part (a). We have already calculated the numbers of millimoles of formic acid and formate in 100 mL of the initial pH 3.95 buffer: 13.5 mmol of $HCO_2H$ and 21.5 mmol of $HCO_2^−$. The number of millimoles of $OH^-$ in 5.00 mL of 1.00 M $NaOH$ is as follows:
B With this information, we can construct a table of initial amounts, changes in amounts, and final amounts.
$\ce{HCO2H (aq) + OH^{−} (aq) <=> HCO^{−}2 (aq) + H2O (l)} \nonumber$
initial amounts, changes in amounts, and final amounts
$HCO_2H (aq)$ $OH^−$ $HCO^−_2 (aq)$
Initial 13.5 mmol 5.00 mmol 21.5 mmol
Change −5.00 mmol −5.00 mmol +5.00 mmol
Final 8.5 mmol ∼0 mmol 26.5 mmol
The final amount of $OH^-$ in solution is not actually zero; this is only approximately true based on the stoichiometric calculation. We can calculate the final pH by inserting the numbers of millimoles of both $HCO_2^−$ and $HCO_2H$ into the simplified Henderson-Hasselbalch expression used in part (a) because the volume cancels:
\begin{align*} pH &=pK_a+\log \left(\dfrac{n_{HCO_2^−}}{n_{HCO_2H}}\right) \[4pt] &=3.75+\log \left(\dfrac{26.5\; mmol}{8.5\; mmol} \right) \[4pt] &=3.75+0.494 =4.24 \end{align*} \nonumber
Once again, this result makes chemical sense: the pH has increased, as would be expected after adding a strong base, and the final pH is between the $pK_a$ and $pK_a$ + 1, as expected for a solution with a $HCO_2^−/HCO_2H$ ratio between 1 and 10.
Exercise $3$
The buffer solution from Example $2$ contained 0.119 M pyridine and 0.234 M pyridine hydrochloride and had a pH of 4.94.
1. What is the final pH if 12.0 mL of 1.5 M $\ce{NaOH}$ are added to 250 mL of this solution?
2. What is the final pH if 12.0 mL of 1.5 M $\ce{HCl}$ are added?
Answer a
5.30
Answer b
4.42
Only the amounts (in moles or millimoles) of the acidic and basic components of the buffer are needed to use the Henderson-Hasselbalch approximation, not their concentrations.
A Video Discussing the Change in pH with the Addition of a Strong Acid to a Buffer: The Change in pH with the Addition of a Strong Acid to a Buffer(opens in new window) [youtu.be]
The Change in pH with the Addition of a Strong Base to a Buffer:
The results obtained in Example $3$ and its corresponding exercise demonstrate how little the pH of a well-chosen buffer solution changes despite the addition of a significant quantity of strong acid or strong base. Suppose we had added the same amount of $HCl$ or $NaOH$ solution to 100 mL of an unbuffered solution at pH 3.95 (corresponding to $1.1 \times 10^{−4}$ M HCl). In this case, adding 5.00 mL of 1.00 M $HCl$ would lower the final pH to 1.32 instead of 3.70, whereas adding 5.00 mL of 1.00 M $NaOH$ would raise the final pH to 12.68 rather than 4.24. (Try verifying these values by doing the calculations yourself.) Thus the presence of a buffer significantly increases the ability of a solution to maintain an almost constant pH.
The most effective buffers contain equal concentrations of an acid and its conjugate base.
A buffer that contains approximately equal amounts of a weak acid and its conjugate base in solution is equally effective at neutralizing either added base or added acid. This is shown in Figure $2$ for an acetic acid/sodium acetate buffer. Adding a given amount of strong acid shifts the system along the horizontal axis to the left, whereas adding the same amount of strong base shifts the system the same distance to the right. In either case, the change in the ratio of $CH_3CO_2^−$ to $CH_3CO_2H$ from 1:1 reduces the buffer capacity of the solution.
A Video Discussing The Buffer Region: The Buffer Region (opens in new window) [youtu.be]
The Relationship between Titrations and Buffers
There is a strong correlation between the effectiveness of a buffer solution and the titration curves discussed in Section 16.5. Consider the schematic titration curve of a weak acid with a strong base shown in Figure $3$. As indicated by the labels, the region around $pK_a$ corresponds to the midpoint of the titration, when approximately half the weak acid has been neutralized. This portion of the titration curve corresponds to a buffer: it exhibits the smallest change in pH per increment of added strong base, as shown by the nearly horizontal nature of the curve in this region. The nearly flat portion of the curve extends only from approximately a pH value of 1 unit less than the $pK_a$ to approximately a pH value of 1 unit greater than the $pK_a$, which is why buffer solutions usually have a pH that is within ±1 pH units of the $pK_a$ of the acid component of the buffer.
This schematic plot of pH for the titration of a weak acid with a strong base shows the nearly flat region of the titration curve around the midpoint, which corresponds to the formation of a buffer. At the lower left, the pH of the solution is determined by the equilibrium for dissociation of the weak acid; at the upper right, the pH is determined by the equilibrium for reaction of the conjugate base with water.
In the region of the titration curve at the lower left, before the midpoint, the acid–base properties of the solution are dominated by the equilibrium for dissociation of the weak acid, corresponding to $K_a$. In the region of the titration curve at the upper right, after the midpoint, the acid–base properties of the solution are dominated by the equilibrium for reaction of the conjugate base of the weak acid with water, corresponding to $K_b$. However, we can calculate either $K_a$ or $K_b$ from the other because they are related by $K_w$.
Blood: A Most Important Buffer
Metabolic processes produce large amounts of acids and bases, yet organisms are able to maintain an almost constant internal pH because their fluids contain buffers. This is not to say that the pH is uniform throughout all cells and tissues of a mammal. The internal pH of a red blood cell is about 7.2, but the pH of most other kinds of cells is lower, around 7.0. Even within a single cell, different compartments can have very different pH values. For example, one intracellular compartment in white blood cells has a pH of around 5.0.
Because no single buffer system can effectively maintain a constant pH value over the entire physiological range of approximately pH 5.0 to 7.4, biochemical systems use a set of buffers with overlapping ranges. The most important of these is the $\ce{CO2}/\ce{HCO3^{−}}$ system, which dominates the buffering action of blood plasma.
The acid–base equilibrium in the $\ce{CO2}/\ce{HCO3^{−}}$ buffer system is usually written as follows:
$\ce{H2CO3 (aq) <=> H^{+} (aq) + HCO^{-}3 (aq)} \label{Eq10}$
with $K_a = 4.5 \times 10^{−7}$ and $pK_a = 6.35$ at 25°C. In fact, Equation $\ref{Eq10}$ is a grossly oversimplified version of the $\ce{CO2}/\ce{HCO3^{-}}$ system because a solution of $\ce{CO2}$ in water contains only rather small amounts of $H_2CO_3$. Thus Equation $\ref{Eq10}$ does not allow us to understand how blood is actually buffered, particularly at a physiological temperature of 37°C.
As shown in Equation $\ref{Eq11}$, $\ce{CO2}$ is in equilibrium with $\ce{H2CO3}$, but the equilibrium lies far to the left, with an $\ce{H2CO3}/\ce{CO2}$ ratio less than 0.01 under most conditions:
$\ce{CO2 (aq) + H2O (l) <=> H2CO3 (aq)} \label{Eq11}$
with $K′ = 4.0 \times 10^{−3}$ at 37°C. The true $pK_a$ of carbonic acid at 37°C is therefore 3.70, not 6.35, corresponding to a $K_a$ of $2.0 \times 10^{−4}$, which makes it a much stronger acid than Equation \ref{Eq10} suggests. Adding Equation \ref{Eq10} and Equation \ref{Eq11} and canceling $\ce{H2CO3}$ from both sides give the following overall equation for the reaction of $\ce{CO2}$ with water to give a proton and the bicarbonate ion:
$\ce{CO2 (aq) + H2O (l) <=> H2CO3 (aq)} \label{16.65a}$
with $K'=4.0 \times 10^{−3} (37°C)$
$\ce{H2CO3 (aq) <=> H^{+} (aq) + HCO^{-}3 (aq)} \label{16.65b}$
with $K_a=2.0 \times 10^{−4} (37°C)$
$\ce{CO2 (aq) + H2O (l) <=> H^{+} (aq) + HCO^{-}3 (aq)} \label{16.65c}$
with $K=8.0 \times 10^{−7} (37°C)$
The $K$ value for the reaction in Equation \ref{16.65c} is the product of the true ionization constant for carbonic acid ($K_a$) and the equilibrium constant (K) for the reaction of $\ce{CO2 (aq)}$ with water to give carbonic acid. The equilibrium equation for the reaction of $\ce{CO2}$ with water to give bicarbonate and a proton is therefore
$K=\dfrac{[\ce{H^{+}}][\ce{HCO3^{-}}]}{[\ce{CO2}]}=8.0 \times 10^{−7} \label{eq13}$
The presence of a gas in the equilibrium constant expression for a buffer is unusual. According to Henry’s law,
$[\ce{CO2}]=k P_{\ce{CO2}} \nonumber$
where $k$ is the Henry’s law constant for $\ce{CO2}$, which is $3.0 \times 10^{−5} \;M/mmHg$ at 37°C. Substituting this expression for $[\ce{CO2}]$ in Equation \ref{eq13},
$K=\dfrac{[\ce{H^{+}}][\ce{HCO3^{-}}]}{(3.0 \times 10^{−5}\; M/mmHg)(P_{\ce{CO2}})} \nonumber$
where $P_{\ce{CO2}}$ is in mmHg. Taking the negative logarithm of both sides and rearranging,
$pH=6.10+\log \left( \dfrac{ [\ce{HCO3^{−}}]}{(3.0 \times 10^{−5} M/mm \;Hg)\; (P_{\ce{CO2}}) } \right) \label{Eq15}$
Thus the pH of the solution depends on both the $\ce{CO2}$ pressure over the solution and $[\ce{HCO3^{−}}]$. Figure $4$ plots the relationship between pH and $[\ce{HCO3^{−}}]$ under physiological conditions for several different values of $P_{\ce{CO2}}$, with normal pH and $[\ce{HCO3^{−}}]$ values indicated by the dashed lines.
According to Equation \ref{Eq15}, adding a strong acid to the $\ce{CO2}/\ce{HCO3^{−}}$ system causes $[\ce{HCO3^{−}}]$ to decrease as $\ce{HCO3^{−}}$ is converted to $\ce{CO2}$. Excess $\ce{CO2}$ is released in the lungs and exhaled into the atmosphere, however, so there is essentially no change in $P_{\ce{CO2}}$. Because the change in $[\ce{HCO3^{−}}]/P_{CO_2}$ is small, Equation \ref{Eq15} predicts that the change in pH will also be rather small. Conversely, if a strong base is added, the $\ce{OH^{-}}$ reacts with $\ce{CO2}$ to form $\ce{HCO3^{−}}$, but $\ce{CO2}$ is replenished by the body, again limiting the change in both $[\ce{HCO3^{−}}]/P_{\ce{CO2}}$ and pH. The $\ce{CO2}/\ce{HCO3^{−}}$ buffer system is an example of an open system, in which the total concentration of the components of the buffer change to keep the pH at a nearly constant value.
If a passenger steps out of an airplane in Denver, Colorado, for example, the lower $P_{\ce{CO2}}$ at higher elevations (typically 31 mmHg at an elevation of 2000 m versus 40 mmHg at sea level) causes a shift to a new pH and $[\ce{HCO3^{-}}]$. The increase in pH and decrease in $[\ce{HCO3^{−}}]$ in response to the decrease in $P_{\ce{CO2}}$ are responsible for the general malaise that many people experience at high altitudes. If their blood pH does not adjust rapidly, the condition can develop into the life-threatening phenomenon known as altitude sickness.
A Video Summary of the pH Curve for a Strong Acid/Strong Base Titration:
Summary
Buffers are solutions that resist a change in pH after adding an acid or a base. Buffers contain a weak acid ($HA$) and its conjugate weak base ($A^−$). Adding a strong electrolyte that contains one ion in common with a reaction system that is at equilibrium shifts the equilibrium in such a way as to reduce the concentration of the common ion. The shift in equilibrium is called the common ion effect. Buffers are characterized by their pH range and buffer capacity. The useful pH range of a buffer depends strongly on the chemical properties of the conjugate weak acid–base pair used to prepare the buffer (the $K_a$ or $K_b$), whereas its buffer capacity depends solely on the concentrations of the species in the solution. The pH of a buffer can be calculated using the Henderson-Hasselbalch approximation, which is valid for solutions whose concentrations are at least 100 times greater than their $K_a$ values. Because no single buffer system can effectively maintain a constant pH value over the physiological range of approximately 5 to 7.4, biochemical systems use a set of buffers with overlapping ranges. The most important of these is the $CO_2/HCO_3^−$ system, which dominates the buffering action of blood plasma. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/17%3A_Additional_Aspects_of_Acid-Base_Equilibria/17.2%3A_Buffer_Solutions.txt |
Learning Objectives
• Explain the function and color changes of acid-base indicators
• Demonstrate how to select the proper indicator for a titration experiment
• Determine the acidic dissociation constants Ka or Kai of indicators.
Certain organic substances change color in dilute solution when the hydronium ion concentration reaches a particular value. For example, phenolphthalein is a colorless substance in any aqueous solution with a hydronium ion concentration greater than 5.0 × 10−9 M (pH < 8.3). In more basic solutions where the hydronium ion concentration is less than 5.0 × 10−9 M (pH > 8.3), it is red or pink. Substances such as phenolphthalein, which can be used to determine the pH of a solution, are called acid-base indicators. Acid-base indicators are either weak organic acids or weak organic bases.
The equilibrium in a solution of the acid-base indicator methyl orange, a weak acid, can be represented by an equation in which we use HIn as a simple representation for the complex methyl orange molecule:
$\underbrace{\ce{HIn}_{(aq)}}_{\ce{red}}+\ce{H2O}_{(l)}⇌\ce{H3O+}_{(aq)}+\underbrace{\ce{In-}_{(aq)}}_{\ce{yellow}}$
$K_\ce{a}=\ce{\dfrac{[H3O+][In- ]}{[HIn]}}=4.0×10^{−4}$
The anion of methyl orange, In, is yellow, and the nonionized form, HIn, is red. When we add acid to a solution of methyl orange, the increased hydronium ion concentration shifts the equilibrium toward the nonionized red form, in accordance with Le Chatelier’s principle. If we add base, we shift the equilibrium towards the yellow form. This behavior is completely analogous to the action of buffers.
An indicator’s color is the visible result of the ratio of the concentrations of the two species In and HIn. If most of the indicator (typically about 60−90% or more) is present as In, then we see the color of the In ion, which would be yellow for methyl orange. If most is present as HIn, then we see the color of the HIn molecule: red for methyl orange. For methyl orange, we can rearrange the equation for Ka and write:
$\mathrm{\dfrac{[In^-]}{[HIn]}=\dfrac{[substance\: with\: yellow\: color]}{[substance\: with\: red\: color]}=\dfrac{\mathit{K}_a}{[H_3O^+]}}$
This shows us how the ratio of $\ce{\dfrac{[In- ]}{[HIn]}}$ varies with the concentration of hydronium ion. The above expression describing the indicator equilibrium can be rearranged:
$\mathrm{\dfrac{[H_3O^+]}{\mathit{K}_a}=\dfrac{[HIn]}{[In^- ]}}$
$\mathrm{log\left(\dfrac{[H_3O^+]}{\mathit{K}_a}\right)=log\left(\dfrac{[HIn]}{[In^- ]}\right)}$
$\mathrm{log([H_3O^+])-log(\mathit{K}_a)=-log\left(\dfrac{[In^-]}{[HIn]}\right)}$
$\mathrm{-pH+p\mathit{K}_a=-log\left(\dfrac{[In^-]}{[HIn]}\right)}$
$\mathrm{pH=p\mathit{K}_a+log\left(\dfrac{[In^-]}{[HIn]}\right)\:or\:pH=p\mathit{K}_a+log\left(\dfrac{[base]}{[acid]}\right)}$
The last formula is the same as the Henderson-Hasselbalch equation, which can be used to describe the equilibrium of indicators.
When [H3O+] has the same numerical value as Ka, the ratio of [In] to [HIn] is equal to 1, meaning that 50% of the indicator is present in the red form (HIn) and 50% is in the yellow ionic form (In), and the solution appears orange in color. When the hydronium ion concentration increases to 8 × 10−4 M (a pH of 3.1), the solution turns red. No change in color is visible for any further increase in the hydronium ion concentration (decrease in pH). At a hydronium ion concentration of 4 × 10−5 M (a pH of 4.4), most of the indicator is in the yellow ionic form, and a further decrease in the hydronium ion concentration (increase in pH) does not produce a visible color change. The pH range between 3.1 (red) and 4.4 (yellow) is the color-change interval of methyl orange; the pronounced color change takes place between these pH values.
Many different substances can be used as indicators, depending on the particular reaction to be monitored. For example, red cabbage juice contains a mixture of colored substances that change from deep red at low pH to light blue at intermediate pH to yellow at high pH (Figure $1$). In all cases, though, a good indicator must have the following properties:
• The color change must be easily detected.
• The color change must be rapid.
• The indicator molecule must not react with the substance being titrated.
• To minimize errors, the indicator should have a pKin that is within one pH unit of the expected pH at the equivalence point of the titration.
Red cabbage juice contains a mixture of substances whose color depends on the pH. Each test tube contains a solution of red cabbage juice in water, but the pH of the solutions varies from pH = 2.0 (far left) to pH = 11.0 (far right). At pH = 7.0, the solution is blue.
Synthetic indicators have been developed that meet these criteria and cover virtually the entire pH range. Figure $2$ shows the approximate pH range over which some common indicators change color and their change in color. In addition, some indicators (such as thymol blue) are polyprotic acids or bases, which change color twice at widely separated pH values.
It is important to be aware that an indicator does not change color abruptly at a particular pH value; instead, it actually undergoes a pH titration just like any other acid or base. As the concentration of HIn decreases and the concentration of In− increases, the color of the solution slowly changes from the characteristic color of HIn to that of In−. As we will see in Section 16, the [In−]/[HIn] ratio changes from 0.1 at a pH one unit below pKin to 10 at a pH one unit above pKin. Thus most indicators change color over a pH range of about two pH units.
We have stated that a good indicator should have a pKin value that is close to the expected pH at the equivalence point. For a strong acid–strong base titration, the choice of the indicator is not especially critical due to the very large change in pH that occurs around the equivalence point. In contrast, using the wrong indicator for a titration of a weak acid or a weak base can result in relatively large errors, as illustrated in Figure $3$. This figure shows plots of pH versus volume of base added for the titration of 50.0 mL of a 0.100 M solution of a strong acid (HCl) and a weak acid (acetic acid) with 0.100 M $NaOH$. The pH ranges over which two common indicators (methyl red, $pK_{in} = 5.0$, and phenolphthalein, $pK_{in} = 9.5$) change color are also shown. The horizontal bars indicate the pH ranges over which both indicators change color cross the HCl titration curve, where it is almost vertical. Hence both indicators change color when essentially the same volume of $NaOH$ has been added (about 50 mL), which corresponds to the equivalence point. In contrast, the titration of acetic acid will give very different results depending on whether methyl red or phenolphthalein is used as the indicator. Although the pH range over which phenolphthalein changes color is slightly greater than the pH at the equivalence point of the strong acid titration, the error will be negligible due to the slope of this portion of the titration curve. Just as with the HCl titration, the phenolphthalein indicator will turn pink when about 50 mL of $NaOH$ has been added to the acetic acid solution. In contrast, methyl red begins to change from red to yellow around pH 5, which is near the midpoint of the acetic acid titration, not the equivalence point. Adding only about 25–30 mL of $NaOH$ will therefore cause the methyl red indicator to change color, resulting in a huge error.
The graph shows the results obtained using two indicators (methyl red and phenolphthalein) for the titration of 0.100 M solutions of a strong acid (HCl) and a weak acid (acetic acid) with 0.100 M $NaOH$. Due to the steepness of the titration curve of a strong acid around the equivalence point, either indicator will rapidly change color at the equivalence point for the titration of the strong acid. In contrast, the pKin for methyl red (5.0) is very close to the pKa of acetic acid (4.76); the midpoint of the color change for methyl red occurs near the midpoint of the titration, rather than at the equivalence point.
In general, for titrations of strong acids with strong bases (and vice versa), any indicator with a pK in between about 4.0 and 10.0 will do. For the titration of a weak acid, however, the pH at the equivalence point is greater than 7.0, so an indicator such as phenolphthalein or thymol blue, with pKin > 7.0, should be used. Conversely, for the titration of a weak base, where the pH at the equivalence point is less than 7.0, an indicator such as methyl red or bromocresol blue, with pKin < 7.0, should be used.
Example $1$
In the titration of a weak acid with a strong base, which indicator would be the best choice?
1. Methyl Orange
2. Bromocresol Green
3. Phenolphtalein
Solution
The correct answer is C. In the titration of a weak acid with a strong base, the conjugate base of the weak acid will make the pH at the equivalence point greater than 7. Therefore, you would want an indicator to change in that pH range. Both methyl orange and bromocresol green change color in an acidic pH range, while phenolphtalein changes in a basic pH.
The existence of many different indicators with different colors and pKin values also provides a convenient way to estimate the pH of a solution without using an expensive electronic pH meter and a fragile pH electrode. Paper or plastic strips impregnated with combinations of indicators are used as “pH paper,” which allows you to estimate the pH of a solution by simply dipping a piece of pH paper into it and comparing the resulting color with the standards printed on the container (Figure $4$).
pH Indicators: https://youtu.be/1IqzUa5lABs
Summary
Acid–base indicators are compounds that change color at a particular pH. They are typically weak acids or bases whose changes in color correspond to deprotonation or protonation of the indicator itself. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/17%3A_Additional_Aspects_of_Acid-Base_Equilibria/17.3%3A_Acid-Base_Indicators.txt |
Learning Objectives
• To calculate the pH at any point in an acid–base titration.
In an acid–base titration, a buret is used to deliver measured volumes of an acid or a base solution of known concentration (the titrant) to a flask that contains a solution of a base or an acid, respectively, of unknown concentration (the unknown). If the concentration of the titrant is known, then the concentration of the unknown can be determined. The following discussion focuses on the pH changes that occur during an acid–base titration. Plotting the pH of the solution in the flask against the amount of acid or base added produces a titration curve. The shape of the curve provides important information about what is occurring in solution during the titration.
Titrations of Strong Acids and Bases
Figure $\PageIndex{1a}$ shows a plot of the pH as 0.20 M $\ce{HCl}$ is gradually added to 50.00 mL of pure water. The pH of the sample in the flask is initially 7.00 (as expected for pure water), but it drops very rapidly as $\ce{HCl}$ is added. Eventually the pH becomes constant at 0.70—a point well beyond its value of 1.00 with the addition of 50.0 mL of $\ce{HCl}$ (0.70 is the pH of 0.20 M HCl). In contrast, when 0.20 M $\ce{NaOH}$ is added to 50.00 mL of distilled water, the pH (initially 7.00) climbs very rapidly at first but then more gradually, eventually approaching a limit of 13.30 (the pH of 0.20 M NaOH), again well beyond its value of 13.00 with the addition of 50.0 mL of $\ce{NaOH}$ as shown in Figure $\PageIndex{1b}$. As you can see from these plots, the titration curve for adding a base is the mirror image of the curve for adding an acid.
Suppose that we now add 0.20 M $\ce{NaOH}$ to 50.0 mL of a 0.10 M solution of $\ce{HCl}$. Because $\ce{HCl}$ is a strong acid that is completely ionized in water, the initial $[H^+]$ is 0.10 M, and the initial pH is 1.00. Adding $\ce{NaOH}$ decreases the concentration of H+ because of the neutralization reaction (Figure $\PageIndex{2a}$):
$\ce{OH^{−} + H^{+} <=> H_2O}. \nonumber$
Thus the pH of the solution increases gradually. Near the equivalence point, however, the point at which the number of moles of base (or acid) added equals the number of moles of acid (or base) originally present in the solution, the pH increases much more rapidly because most of the $\ce{H^{+}}$ ions originally present have been consumed. For the titration of a monoprotic strong acid ($\ce{HCl}$) with a monobasic strong base ($\ce{NaOH}$), we can calculate the volume of base needed to reach the equivalence point from the following relationship:
$moles\;of \;base=(volume)_b(molarity)_bV_bM_b= moles \;of \;acid=(volume)_a(molarity)_a=V_aM_a \label{Eq1}$
If 0.20 M $\ce{NaOH}$ is added to 50.0 mL of a 0.10 M solution of $\ce{HCl}$, we solve for $V_b$:
$V_b(0.20 Me)=0.025 L=25 mL \nonumber$
At the equivalence point (when 25.0 mL of $\ce{NaOH}$ solution has been added), the neutralization is complete: only a salt remains in solution (NaCl), and the pH of the solution is 7.00. Adding more $\ce{NaOH}$ produces a rapid increase in pH, but eventually the pH levels off at a value of about 13.30, the pH of 0.20 M $NaOH$.
As shown in Figure $\PageIndex{2b}$, the titration of 50.0 mL of a 0.10 M solution of $\ce{NaOH}$ with 0.20 M $\ce{HCl}$ produces a titration curve that is nearly the mirror image of the titration curve in Figure $\PageIndex{2a}$. The pH is initially 13.00, and it slowly decreases as $\ce{HCl}$ is added. As the equivalence point is approached, the pH drops rapidly before leveling off at a value of about 0.70, the pH of 0.20 M $\ce{HCl}$.
The titration of either a strong acid with a strong base or a strong base with a strong acid produces an S-shaped curve. The curve is somewhat asymmetrical because the steady increase in the volume of the solution during the titration causes the solution to become more dilute. Due to the leveling effect, the shape of the curve for a titration involving a strong acid and a strong base depends on only the concentrations of the acid and base, not their identities.
The shape of the titration curve involving a strong acid and a strong base depends only on their concentrations, not their identities.
Example $1$: Hydrochloric Acid
Calculate the pH of the solution after 24.90 mL of 0.200 M $\ce{NaOH}$ has been added to 50.00 mL of 0.100 M $\ce{HCl}$.
Given: volumes and concentrations of strong base and acid
Asked for: pH
Strategy:
1. Calculate the number of millimoles of $\ce{H^{+}}$ and $\ce{OH^{-}}$ to determine which, if either, is in excess after the neutralization reaction has occurred. If one species is in excess, calculate the amount that remains after the neutralization reaction.
2. Determine the final volume of the solution. Calculate the concentration of the species in excess and convert this value to pH.
Solution
A Because 0.100 mol/L is equivalent to 0.100 mmol/mL, the number of millimoles of $\ce{H^{+}}$ in 50.00 mL of 0.100 M $\ce{HCl}$ can be calculated as follows:
$50.00 \cancel{mL} \left ( \dfrac{0.100 \;mmol \;HCl}{\cancel{mL}} \right )= 5.00 \;mmol \;HCl=5.00 \;mmol \;H^{+} \nonumber$
The number of millimoles of $\ce{NaOH}$ added is as follows:
$24.90 \cancel{mL} \left ( \dfrac{0.200 \;mmol \;NaOH}{\cancel{mL}} \right )= 4.98 \;mmol \;NaOH=4.98 \;mmol \;OH^{-} \nonumber$
Thus $\ce{H^{+}}$ is in excess. To completely neutralize the acid requires the addition of 5.00 mmol of $\ce{OH^{-}}$ to the $\ce{HCl}$ solution. Because only 4.98 mmol of $OH^-$ has been added, the amount of excess $\ce{H^{+}}$ is 5.00 mmol − 4.98 mmol = 0.02 mmol of $H^+$.
B The final volume of the solution is 50.00 mL + 24.90 mL = 74.90 mL, so the final concentration of $\ce{H^{+}}$ is as follows:
$\left [ H^{+} \right ]= \dfrac{0.02 \;mmol \;H^{+}}{74.90 \; mL}=3 \times 10^{-4} \; M \nonumber$
Hence,
$pH \approx −\log[\ce{H^{+}}] = −\log(3 \times 10^{-4}) = 3.5 \nonumber$
This is significantly less than the pH of 7.00 for a neutral solution.
Exercise $1$
Calculate the pH of a solution prepared by adding $40.00\; mL$ of $0.237\; M$ $HCl$ to $75.00\; mL$ of a $0.133 M$ solution of $NaOH$.
Answer
11.6
pH after the addition of 10 ml of Strong Base to a Strong Acid:
https://youtu.be/_cM1_-kdJ20 (opens in new window)
pH at the Equivalence Point in a Strong Acid/Strong Base Titration:
https://youtu.be/7POGDA5Ql2M
Titrations of Weak Acids and Bases
In contrast to strong acids and bases, the shape of the titration curve for a weak acid or a weak base depends dramatically on the identity of the acid or the base and the corresponding $K_a$ or $K_b$. As we shall see, the pH also changes much more gradually around the equivalence point in the titration of a weak acid or a weak base. As you learned previously, $[\ce{H^{+}}]$ of a solution of a weak acid (HA) is not equal to the concentration of the acid but depends on both its $pK_a$ and its concentration. Because only a fraction of a weak acid dissociates, $[\(\ce{H^{+}}]$ is less than $[\ce{HA}]$. Thus the pH of a solution of a weak acid is greater than the pH of a solution of a strong acid of the same concentration.
Figure $\PageIndex{3a}$ shows the titration curve for 50.0 mL of a 0.100 M solution of acetic acid with 0.200 M $\ce{NaOH}$ superimposed on the curve for the titration of 0.100 M $\ce{HCl}$ shown in part (a) in Figure $2$. Below the equivalence point, the two curves are very different. Before any base is added, the pH of the acetic acid solution is greater than the pH of the $\ce{HCl}$ solution, and the pH changes more rapidly during the first part of the titration. Note also that the pH of the acetic acid solution at the equivalence point is greater than 7.00. That is, at the equivalence point, the solution is basic. In addition, the change in pH around the equivalence point is only about half as large as for the $\ce{HCl}$ titration; the magnitude of the pH change at the equivalence point depends on the $pK_a$ of the acid being titrated. Above the equivalence point, however, the two curves are identical. Once the acid has been neutralized, the pH of the solution is controlled only by the amount of excess $\ce{NaOH}$ present, regardless of whether the acid is weak or strong.
The shape of the titration curve of a weak acid or weak base depends heavily on their identities and the $K_a$ or $K_b$.
The titration curve in Figure $\PageIndex{3a}$ was created by calculating the starting pH of the acetic acid solution before any $\ce{NaOH}$ is added and then calculating the pH of the solution after adding increasing volumes of $NaOH$. The procedure is illustrated in the following subsection and Example $2$ for three points on the titration curve, using the $pK_a$ of acetic acid (4.76 at 25°C; $K_a = 1.7 \times 10^{-5}$.
Calculating the pH of a Solution of a Weak Acid or a Weak Base
As explained discussed, if we know $K_a$ or $K_b$ and the initial concentration of a weak acid or a weak base, we can calculate the pH of a solution of a weak acid or a weak base by setting up a ICE table (i.e, initial concentrations, changes in concentrations, and final concentrations). In this situation, the initial concentration of acetic acid is 0.100 M. If we define $x$ as $[\ce{H^{+}}]$ due to the dissociation of the acid, then the table of concentrations for the ionization of 0.100 M acetic acid is as follows:
$\ce{CH3CO2H(aq) <=> H^{+}(aq) + CH3CO2^{−}} \nonumber$
table of concentrations for the ionization of 0.100 M acetic acid
ICE $[CH_3CO_2H]$ $[H^+]$ $[CH_3CO_2^−]$
initial 0.100 $1.00 \times 10^{−7}$ 0
change −x +x +x
final 0.100 − x x x
In this and all subsequent examples, we will ignore $[H^+]$ and $[OH^-]$ due to the autoionization of water when calculating the final concentration. However, you should use Equation 16.45 and Equation 16.46 to check that this assumption is justified.
Inserting the expressions for the final concentrations into the equilibrium equation (and using approximations),
\begin{align*} K_a &=\dfrac{[H^+][CH_3CO_2^-]}{[CH_3CO_2H]} \[4pt] &=\dfrac{(x)(x)}{0.100 - x} \[4pt] &\approx \dfrac{x^2}{0.100} \[4pt] &\approx 1.74 \times 10^{-5} \end{align*} \nonumber
Solving this equation gives $x = [H^+] = 1.32 \times 10^{-3}\; M$. Thus the pH of a 0.100 M solution of acetic acid is as follows:
$pH = −\log(1.32 \times 10^{-3}) = 2.879 \nonumber$
pH at the Start of a Weak Acid/Strong Base Titration: https://youtu.be/AtdBKfrfJNg
Calculating the pH during the Titration of a Weak Acid or a Weak Base
Now consider what happens when we add 5.00 mL of 0.200 M $\ce{NaOH}$ to 50.00 mL of 0.100 M $CH_3CO_2H$ (part (a) in Figure $3$). Because the neutralization reaction proceeds to completion, all of the $OH^-$ ions added will react with the acetic acid to generate acetate ion and water:
$CH_3CO_2H_{(aq)} + OH^-_{(aq)} \rightarrow CH_3CO^-_{2\;(aq)} + H_2O_{(l)} \label{Eq2}$
All problems of this type must be solved in two steps: a stoichiometric calculation followed by an equilibrium calculation. In the first step, we use the stoichiometry of the neutralization reaction to calculate the amounts of acid and conjugate base present in solution after the neutralization reaction has occurred. In the second step, we use the equilibrium equation to determine $[\ce{H^{+}}]$ of the resulting solution.
Step 1
To determine the amount of acid and conjugate base in solution after the neutralization reaction, we calculate the amount of $\ce{CH_3CO_2H}$ in the original solution and the amount of $\ce{OH^{-}}$ in the $\ce{NaOH}$ solution that was added. The acetic acid solution contained
$50.00 \; \cancel{mL} (0.100 \;mmol (\ce{CH_3CO_2H})/\cancel{mL} )=5.00\; mmol (\ce{CH_3CO_2H}) \nonumber$
The $\ce{NaOH}$ solution contained
5.00 mL=1.00 mmol $NaOH$
Comparing the amounts shows that $CH_3CO_2H$ is in excess. Because $OH^-$ reacts with $CH_3CO_2H$ in a 1:1 stoichiometry, the amount of excess $CH_3CO_2H$ is as follows:
5.00 mmol $CH_3CO_2H$ − 1.00 mmol $OH^-$ = 4.00 mmol $CH_3CO_2H$
Each 1 mmol of $OH^-$ reacts to produce 1 mmol of acetate ion, so the final amount of $CH_3CO_2^−$ is 1.00 mmol.
The stoichiometry of the reaction is summarized in the following ICE table, which shows the numbers of moles of the various species, not their concentrations.
$\ce{CH3CO2H(aq) + OH^{−} (aq) <=> CH3CO2^{-}(aq) + H2O(l)} \nonumber$
ICE table
ICE $[\ce{CH_3CO_2H}]$ $[\ce{OH^{−}}]$ $[\ce{CH_3CO_2^{−}}]$
initial 5.00 mmol 1.00 mmol 0 mmol
change −1.00 mmol −1.00 mmol +1.00 mmol
final 4.00 mmol 0 mmol 1.00 mmol
This ICE table gives the initial amount of acetate and the final amount of $OH^-$ ions as 0. Because an aqueous solution of acetic acid always contains at least a small amount of acetate ion in equilibrium with acetic acid, however, the initial acetate concentration is not actually 0. The value can be ignored in this calculation because the amount of $CH_3CO_2^−$ in equilibrium is insignificant compared to the amount of $OH^-$ added. Moreover, due to the autoionization of water, no aqueous solution can contain 0 mmol of $OH^-$, but the amount of $OH^-$ due to the autoionization of water is insignificant compared to the amount of $OH^-$ added. We use the initial amounts of the reactants to determine the stoichiometry of the reaction and defer a consideration of the equilibrium until the second half of the problem.
Step 2
To calculate $[\ce{H^{+}}]$ at equilibrium following the addition of $NaOH$, we must first calculate [$\ce{CH_3CO_2H}$] and $[\ce{CH3CO2^{−}}]$ using the number of millimoles of each and the total volume of the solution at this point in the titration:
$final \;volume=50.00 \;mL+5.00 \;mL=55.00 \;mL \nonumber$ $\left [ CH_{3}CO_{2}H \right ] = \dfrac{4.00 \; mmol \; CH_{3}CO_{2}H }{55.00 \; mL} =7.27 \times 10^{-2} \;M \nonumber$ $\left [ CH_{3}CO_{2}^{-} \right ] = \dfrac{1.00 \; mmol \; CH_{3}CO_{2}^{-} }{55.00 \; mL} =1.82 \times 10^{-2} \;M \nonumber$
Knowing the concentrations of acetic acid and acetate ion at equilibrium and $K_a$ for acetic acid ($1.74 \times 10^{-5}$), we can calculate $[H^+]$ at equilibrium:
$K_{a}=\dfrac{\left [ CH_{3}CO_{2}^{-} \right ]\left [ H^{+} \right ]}{\left [ CH_{3}CO_{2}H \right ]} \nonumber$
$\left [ H^{+} \right ]=\dfrac{K_{a}\left [ CH_{3}CO_{2}H \right ]}{\left [ CH_{3}CO_{2}^{-} \right ]} = \dfrac{\left ( 1.72 \times 10^{-5} \right )\left ( 7.27 \times 10^{-2} \;M\right )}{\left ( 1.82 \times 10^{-2} \right )}= 6.95 \times 10^{-5} \;M \nonumber$
Calculating $−\log[\ce{H^{+}}]$ gives
$pH = −\log(6.95 \times 10^{−5}) = 4.158. \nonumber$
Comparing the titration curves for $\ce{HCl}$ and acetic acid in Figure $\PageIndex{3a}$, we see that adding the same amount (5.00 mL) of 0.200 M $\ce{NaOH}$ to 50 mL of a 0.100 M solution of both acids causes a much smaller pH change for $\ce{HCl}$ (from 1.00 to 1.14) than for acetic acid (2.88 to 4.16). This is consistent with the qualitative description of the shapes of the titration curves at the beginning of this section. In Example $2$, we calculate another point for constructing the titration curve of acetic acid.
pH Before the Equivalence Point of a Weak Acid/Strong Base Titration:
https://youtu.be/znpwGCsefXc
Example $2$
What is the pH of the solution after 25.00 mL of 0.200 M $\ce{NaOH}$ is added to 50.00 mL of 0.100 M acetic acid?
Given: volume and molarity of base and acid
Asked for: pH
Strategy:
1. Write the balanced chemical equation for the reaction. Then calculate the initial numbers of millimoles of $OH^-$ and $CH_3CO_2H$. Determine which species, if either, is present in excess.
2. Tabulate the results showing initial numbers, changes, and final numbers of millimoles.
3. If excess acetate is present after the reaction with $\ce{OH^{-}}$, write the equation for the reaction of acetate with water. Use a tabular format to obtain the concentrations of all the species present.
4. Calculate $K_b$ using the relationship $K_w = K_aK_b$. Calculate [OH−] and use this to calculate the pH of the solution.
Solution
A Ignoring the spectator ion ($Na^+$), the equation for this reaction is as follows:
$CH_3CO_2H_{ (aq)} + OH^-(aq) \rightarrow CH_3CO_2^-(aq) + H_2O(l) \nonumber$
The initial numbers of millimoles of $OH^-$ and $CH_3CO_2H$ are as follows:
25.00 mL(0.200 mmol OH−mL=5.00 mmol $OH-$
$50.00\; mL (0.100 CH_3CO_2 HL=5.00 mmol \; CH_3CO_2H \nonumber$
The number of millimoles of $OH^-$ equals the number of millimoles of $CH_3CO_2H$, so neither species is present in excess.
B Because the number of millimoles of $OH^-$ added corresponds to the number of millimoles of acetic acid in solution, this is the equivalence point. The results of the neutralization reaction can be summarized in tabular form.
$CH_3CO_2H_{(aq)}+OH^-_{(aq)} \rightleftharpoons CH_3CO_2^{-}(aq)+H_2O(l) \nonumber$
results of the neutralization reaction
ICE $[\ce{CH3CO2H}]$ $[\ce{OH^{−}}]$ $[\ce{CH3CO2^{−}}]$
initial 5.00 mmol 5.00 mmol 0 mmol
change −5.00 mmol −5.00 mmol +5.00 mmol
final 0 mmol 0 mmol 5.00 mmol
C Because the product of the neutralization reaction is a weak base, we must consider the reaction of the weak base with water to calculate [H+] at equilibrium and thus the final pH of the solution. The initial concentration of acetate is obtained from the neutralization reaction:
$[\ce{CH_3CO_2}]=\dfrac{5.00 \;mmol \; CH_3CO_2^{-}}{(50.00+25.00) \; mL}=6.67\times 10^{-2} \; M \nonumber$
The equilibrium reaction of acetate with water is as follows:
$\ce{CH_3CO^{-}2(aq) + H2O(l) <=> CH3CO2H(aq) + OH^{-} (aq)} \nonumber$
The equilibrium constant for this reaction is
$K_b = \dfrac{K_w}{K_a} \label{16.18}$
where $K_a$ is the acid ionization constant of acetic acid. We therefore define x as $[\ce{OH^{−}}]$ produced by the reaction of acetate with water. Here is the completed table of concentrations:
$H_2O_{(l)}+CH_3CO^−_{2(aq)} \rightleftharpoons CH_3CO_2H_{(aq)} +OH^−_{(aq)} \nonumber$
completed table of concentrations
$[\ce{CH3CO2^{−}}]$ $[\ce{CH3CO2H}]$ $[\ce{OH^{−}}]$
initial 0.0667 0 1.00 × 10−7
change −x +x +x
final (0.0667 − x) x x
D We can obtain $K_b$ by substituting the known values into Equation \ref{16.18}:
$K_{b}= \dfrac{K_w}{K_a} =\dfrac{1.01 \times 10^{-14}}{1.74 \times 10^{-5}} = 5.80 \times 10^{-10} \label{16.23}$
Substituting the expressions for the final values from the ICE table into Equation \ref{16.23} and solving for $x$:
\begin{align*} \dfrac{x^{2}}{0.0667} &= 5.80 \times 10^{-10} \[4pt] x &= \sqrt{(5.80 \times 10^{-10})(0.0667)} \[4pt] &= 6.22 \times 10^{-6}\end{align*} \nonumber
Thus $[OH^{−}] = 6.22 \times 10^{−6}\, M$ and the pH of the final solution is 8.794 (Figure $\PageIndex{3a}$). As expected for the titration of a weak acid, the pH at the equivalence point is greater than 7.00 because the product of the titration is a base, the acetate ion, which then reacts with water to produce $\ce{OH^{-}}$.
Exercise $2$
Calculate the pH of a solution prepared by adding 45.0 mL of a 0.213 M $\ce{HCl}$ solution to 125.0 mL of a 0.150 M solution of ammonia. The $pK_b$ of ammonia is 4.75 at 25°C.
Answer
9.23
As shown in part (b) in Figure $3$, the titration curve for NH3, a weak base, is the reverse of the titration curve for acetic acid. In particular, the pH at the equivalence point in the titration of a weak base is less than 7.00 because the titration produces an acid.
The identity of the weak acid or weak base being titrated strongly affects the shape of the titration curve. Figure $4$ illustrates the shape of titration curves as a function of the $pK_a$ or the $pK_b$. As the acid or the base being titrated becomes weaker (its $pK_a$ or $pK_b$ becomes larger), the pH change around the equivalence point decreases significantly. With very dilute solutions, the curve becomes so shallow that it can no longer be used to determine the equivalence point.
One point in the titration of a weak acid or a weak base is particularly important: the midpoint of a titration is defined as the point at which exactly enough acid (or base) has been added to neutralize one-half of the acid (or the base) originally present and occurs halfway to the equivalence point. The midpoint is indicated in Figures $\PageIndex{4a}$ and $\PageIndex{4b}$ for the two shallowest curves. By definition, at the midpoint of the titration of an acid, [HA] = [A−]. Recall that the ionization constant for a weak acid is as follows:
$K_a=\dfrac{[H_3O^+][A^−]}{[HA]} \nonumber$
If $[HA] = [A^−]$, this reduces to $K_a = [H_3O^+]$. Taking the negative logarithm of both sides,
$−\log K_a = −\log[H_3O+] \nonumber$
From the definitions of $pK_a$ and pH, we see that this is identical to
$pK_a = pH \label{16.52}$
Thus the pH at the midpoint of the titration of a weak acid is equal to the $pK_a$ of the weak acid, as indicated in part (a) in Figure $4$ for the weakest acid where we see that the midpoint for $pK_a$ = 10 occurs at pH = 10. Titration methods can therefore be used to determine both the concentration and the $pK_a$ (or the $pK_b$) of a weak acid (or a weak base).
The pH at the midpoint of the titration of a weak acid is equal to the $pK_a$ of the weak acid.
Titrations of Polyprotic Acids or Bases
When a strong base is added to a solution of a polyprotic acid, the neutralization reaction occurs in stages. The most acidic group is titrated first, followed by the next most acidic, and so forth. If the $pK_a$ values are separated by at least three $pK_a$ units, then the overall titration curve shows well-resolved “steps” corresponding to the titration of each proton. A titration of the triprotic acid $H_3PO_4$ with $\ce{NaOH}$ is illustrated in Figure $5$ and shows two well-defined steps: the first midpoint corresponds to $pK_a$1, and the second midpoint corresponds to $pK_a$2. Because HPO42 is such a weak acid, $pK_a$3 has such a high value that the third step cannot be resolved using 0.100 M $\ce{NaOH}$ as the titrant.
The titration curve for the reaction of a polyprotic base with a strong acid is the mirror image of the curve shown in Figure $5$. The initial pH is high, but as acid is added, the pH decreases in steps if the successive $pK_b$ values are well separated. Table E1 lists the ionization constants and $pK_a$ values for some common polyprotic acids and bases.
Example $3$
Calculate the pH of a solution prepared by adding 55.0 mL of a 0.120 M $\ce{NaOH}$ solution to 100.0 mL of a 0.0510 M solution of oxalic acid ($\ce{HO_2CCO_2H}$), a diprotic acid (abbreviated as $\ce{H2ox}$). Oxalic acid, the simplest dicarboxylic acid, is found in rhubarb and many other plants. Rhubarb leaves are toxic because they contain the calcium salt of the fully deprotonated form of oxalic acid, the oxalate ion ($\ce{O2CCO2^{2−}}$, abbreviated $\ce{ox^{2-}}$).Oxalate salts are toxic for two reasons. First, oxalate salts of divalent cations such as $\ce{Ca^{2+}}$ are insoluble at neutral pH but soluble at low pH. As a result, calcium oxalate dissolves in the dilute acid of the stomach, allowing oxalate to be absorbed and transported into cells, where it can react with calcium to form tiny calcium oxalate crystals that damage tissues. Second, oxalate forms stable complexes with metal ions, which can alter the distribution of metal ions in biological fluids.
Given: volume and concentration of acid and base
Asked for: pH
Strategy:
1. Calculate the initial millimoles of the acid and the base. Use a tabular format to determine the amounts of all the species in solution.
2. Calculate the concentrations of all the species in the final solution. Determine $\ce{[H{+}]}$ and convert this value to pH.
Solution:
A Table E5 gives the $pK_a$ values of oxalic acid as 1.25 and 3.81. Again we proceed by determining the millimoles of acid and base initially present:
$100.00 \cancel{mL} \left ( \dfrac{0.510 \;mmol \;H_{2}ox}{\cancel{mL}} \right )= 5.10 \;mmol \;H_{2}ox \nonumber$
$55.00 \cancel{mL} \left ( \dfrac{0.120 \;mmol \;NaOH}{\cancel{mL}} \right )= 6.60 \;mmol \;NaOH \nonumber$
The strongest acid ($H_2ox$) reacts with the base first. This leaves (6.60 − 5.10) = 1.50 mmol of $OH^-$ to react with Hox−, forming ox2 and H2O. The reactions can be written as follows:
$\underset{5.10\;mmol}{H_{2}ox}+\underset{6.60\;mmol}{OH^{-}} \rightarrow \underset{5.10\;mmol}{Hox^{-}}+ \underset{5.10\;mmol}{H_{2}O} \nonumber$
$\underset{5.10\;mmol}{Hox^{-}}+\underset{1.50\;mmol}{OH^{-}} \rightarrow \underset{1.50\;mmol}{ox^{2-}}+ \underset{1.50\;mmol}{H_{2}O} \nonumber$
In tabular form,
Solutions to Example 17.3.3
$\ce{H2ox}$ $\ce{OH^{-}}$ $\ce{Hox^{−}}$ $\ce{ox^{2−}}$
initial 5.10 mmol 6.60 mmol 0 mmol 0 mmol
change (step 1) −5.10 mmol −5.10 mmol +5.10 mmol 0 mmol
final (step 1) 0 mmol 1.50 mmol 5.10 mmol 0 mmol
change (step 2) −1.50 mmol −1.50 mmol +1.50 mmol
final 0 mmol 0 mmol 3.60 mmol 1.50 mmol
B The equilibrium between the weak acid ($\ce{Hox^{-}}$) and its conjugate base ($\ce{ox^{2-}}$) in the final solution is determined by the magnitude of the second ionization constant, $K_{a2} = 10^{−3.81} = 1.6 \times 10^{−4}$. To calculate the pH of the solution, we need to know $\ce{[H^{+}]}$, which is determined using exactly the same method as in the acetic acid titration in Example $2$:
$\text{final volume of solution} = 100.0\, mL + 55.0\, mL = 155.0 \,mL \nonumber$
Thus the concentrations of $\ce{Hox^{-}}$ and $\ce{ox^{2-}}$ are as follows:
$\left [ Hox^{-} \right ] = \dfrac{3.60 \; mmol \; Hox^{-}}{155.0 \; mL} = 2.32 \times 10^{-2} \;M \nonumber$
$\left [ ox^{2-} \right ] = \dfrac{1.50 \; mmol \; ox^{2-}}{155.0 \; mL} = 9.68 \times 10^{-3} \;M \nonumber$
We can now calculate [H+] at equilibrium using the following equation:
$K_{a2} =\dfrac{\left [ ox^{2-} \right ]\left [ H^{+} \right ] }{\left [ Hox^{-} \right ]} \nonumber$
Rearranging this equation and substituting the values for the concentrations of $\ce{Hox^{−}}$ and $\ce{ox^{2−}}$,
$\left [ H^{+} \right ] =\dfrac{K_{a2}\left [ Hox^{-} \right ]}{\left [ ox^{2-} \right ]} = \dfrac{\left ( 1.6\times 10^{-4} \right ) \left ( 2.32\times 10^{-2} \right )}{\left ( 9.68\times 10^{-3} \right )}=3.7\times 10^{-4} \; M \nonumber$
So
$pH = -\log\left [ H^{+} \right ]= -\log\left ( 3.7 \times 10^{-4} \right )= 3.43 \nonumber$
This answer makes chemical sense because the pH is between the first and second $pK_a$ values of oxalic acid, as it must be. We added enough hydroxide ion to completely titrate the first, more acidic proton (which should give us a pH greater than $pK_{a1}$), but we added only enough to titrate less than half of the second, less acidic proton, with $pK_{a2}$. If we had added exactly enough hydroxide to completely titrate the first proton plus half of the second, we would be at the midpoint of the second step in the titration, and the pH would be 3.81, equal to $pK_{a2}$.
Exercise $3$: Piperazine
Piperazine is a diprotic base used to control intestinal parasites (“worms”) in pets and humans. A dog is given 500 mg (5.80 mmol) of piperazine ($pK_{b1}$ = 4.27, $pK_{b2}$ = 8.67). If the dog’s stomach initially contains 100 mL of 0.10 M $\ce{HCl}$ (pH = 1.00), calculate the pH of the stomach contents after ingestion of the piperazine.
Answer
pH=4.9
Indicators
In practice, most acid–base titrations are not monitored by recording the pH as a function of the amount of the strong acid or base solution used as the titrant. Instead, an acid–base indicator is often used that, if carefully selected, undergoes a dramatic color change at the pH corresponding to the equivalence point of the titration. Indicators are weak acids or bases that exhibit intense colors that vary with pH. The conjugate acid and conjugate base of a good indicator have very different colors so that they can be distinguished easily. Some indicators are colorless in the conjugate acid form but intensely colored when deprotonated (phenolphthalein, for example), which makes them particularly useful.
We can describe the chemistry of indicators by the following general equation:
$\ce{ HIn (aq) <=> H^{+}(aq) + In^{-}(aq)} \nonumber$
where the protonated form is designated by $\ce{HIn}$ and the conjugate base by $\ce{In^{−}}$. The ionization constant for the deprotonation of indicator $\ce{HIn}$ is as follows:
$K_{In} =\dfrac{ [\ce{H^{+}} ][ \ce{In^{-}}]}{[\ce{HIn}]} \label{Eq3}$
The $pK_{in}$ (its $pK_a$) determines the pH at which the indicator changes color.
Many different substances can be used as indicators, depending on the particular reaction to be monitored. For example, red cabbage juice contains a mixture of colored substances that change from deep red at low pH to light blue at intermediate pH to yellow at high pH. Similarly, Hydrangea macrophylla flowers can be blue, red, pink, light purple, or dark purple depending on the soil pH (Figure $6$). Acidic soils will produce blue flowers, whereas alkaline soils will produce pinkish flowers.
Irrespective of the origins, a good indicator must have the following properties:
• The color change must be easily detected.
• The color change must be rapid.
• The indicator molecule must not react with the substance being titrated.
• To minimize errors, the indicator should have a $pK_{in}$ that is within one pH unit of the expected pH at the equivalence point of the titration.
Synthetic indicators have been developed that meet these criteria and cover virtually the entire pH range. Figure $7$ shows the approximate pH range over which some common indicators change color and their change in color. In addition, some indicators (such as thymol blue) are polyprotic acids or bases, which change color twice at widely separated pH values.
It is important to be aware that an indicator does not change color abruptly at a particular pH value; instead, it actually undergoes a pH titration just like any other acid or base. As the concentration of HIn decreases and the concentration of In− increases, the color of the solution slowly changes from the characteristic color of HIn to that of In−. As we will see later, the [In−]/[HIn] ratio changes from 0.1 at a pH one unit below pKin to 10 at a pH one unit above pKin. Thus most indicators change color over a pH range of about two pH units.
We have stated that a good indicator should have a pKin value that is close to the expected pH at the equivalence point. For a strong acid–strong base titration, the choice of the indicator is not especially critical due to the very large change in pH that occurs around the equivalence point. In contrast, using the wrong indicator for a titration of a weak acid or a weak base can result in relatively large errors, as illustrated in Figure $8$. This figure shows plots of pH versus volume of base added for the titration of 50.0 mL of a 0.100 M solution of a strong acid (HCl) and a weak acid (acetic acid) with 0.100 M $NaOH$. The pH ranges over which two common indicators (methyl red, $pK_{in} = 5.0$, and phenolphthalein, $pK_{in} = 9.5$) change color are also shown. The horizontal bars indicate the pH ranges over which both indicators change color cross the $\ce{HCl}$ titration curve, where it is almost vertical. Hence both indicators change color when essentially the same volume of $\ce{NaOH}$ has been added (about 50 mL), which corresponds to the equivalence point. In contrast, the titration of acetic acid will give very different results depending on whether methyl red or phenolphthalein is used as the indicator. Although the pH range over which phenolphthalein changes color is slightly greater than the pH at the equivalence point of the strong acid titration, the error will be negligible due to the slope of this portion of the titration curve. Just as with the $\ce{HCl}$ titration, the phenolphthalein indicator will turn pink when about 50 mL of $\ce{NaOH}$ has been added to the acetic acid solution. In contrast, methyl red begins to change from red to yellow around pH 5, which is near the midpoint of the acetic acid titration, not the equivalence point. Adding only about 25–30 mL of $\ce{NaOH}$ will therefore cause the methyl red indicator to change color, resulting in a huge error.
The graph shows the results obtained using two indicators (methyl red and phenolphthalein) for the titration of 0.100 M solutions of a strong acid (HCl) and a weak acid (acetic acid) with 0.100 M $NaOH$. Due to the steepness of the titration curve of a strong acid around the equivalence point, either indicator will rapidly change color at the equivalence point for the titration of the strong acid. In contrast, the pKin for methyl red (5.0) is very close to the $pK_a$ of acetic acid (4.76); the midpoint of the color change for methyl red occurs near the midpoint of the titration, rather than at the equivalence point.
In general, for titrations of strong acids with strong bases (and vice versa), any indicator with a pKin between about 4.0 and 10.0 will do. For the titration of a weak acid, however, the pH at the equivalence point is greater than 7.0, so an indicator such as phenolphthalein or thymol blue, with pKin > 7.0, should be used. Conversely, for the titration of a weak base, where the pH at the equivalence point is less than 7.0, an indicator such as methyl red or bromocresol blue, with pKin < 7.0, should be used.
The existence of many different indicators with different colors and pKin values also provides a convenient way to estimate the pH of a solution without using an expensive electronic pH meter and a fragile pH electrode. Paper or plastic strips impregnated with combinations of indicators are used as “pH paper,” which allows you to estimate the pH of a solution by simply dipping a piece of pH paper into it and comparing the resulting color with the standards printed on the container (Figure $9$).
pH Indicators: pH Indicators(opens in new window) [youtu.be]
Summary and Takeaway
Plots of acid–base titrations generate titration curves that can be used to calculate the pH, the pOH, the $pK_a$, and the $pK_b$ of the system. The shape of a titration curve, a plot of pH versus the amount of acid or base added, provides important information about what is occurring in solution during a titration. The shapes of titration curves for weak acids and bases depend dramatically on the identity of the compound. The equivalence point of an acid–base titration is the point at which exactly enough acid or base has been added to react completely with the other component. The equivalence point in the titration of a strong acid or a strong base occurs at pH 7.0. In titrations of weak acids or weak bases, however, the pH at the equivalence point is greater or less than 7.0, respectively. The pH tends to change more slowly before the equivalence point is reached in titrations of weak acids and weak bases than in titrations of strong acids and strong bases. The pH at the midpoint, the point halfway on the titration curve to the equivalence point, is equal to the $pK_a$ of the weak acid or the $pK_b$ of the weak base. Thus titration methods can be used to determine both the concentration and the $pK_a$ (or the $pK_b$) of a weak acid (or a weak base). Acid–base indicators are compounds that change color at a particular pH. They are typically weak acids or bases whose changes in color correspond to deprotonation or protonation of the indicator itself. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/17%3A_Additional_Aspects_of_Acid-Base_Equilibria/17.4%3A_Neutralization_Reactions_and_Titration_Curves.txt |
Learning Objectives
• Predict and calculate the pH of a solution of salts of polyprotic acids
Salts can be thought of as being derived from the neutralization of an acid and a base. A salt formed from a strong acid and a strong base will not hydrolyze (e.g., extract a proton from water). When placed in water, these salts dissociate completely, and their ions remain uncombined in solution (e.g., the NaCl salt is formed from a strong acid (HCl) and a strong base (NaOH).
In discussing the titration of acids and bases, the acid or basic properties of the corresponding salts after a neutralization reaction must be addressed to calculate the final pH of a solution. To calculate the pH of a salt solution one needs to know the concentration of the salt solution, whether the salt is an acidic, basic, or neutral salt, the equation for the interaction of the ion with the water, the equilibrium expression for this interaction and the Ka or Kb value.
Note: The Hydrolysis of Salts
As a quick review, the general rules for the hydrolysis of monoprotic salts are:
1. If neither the cation/anion can affect the pH, the solution is neutral
2. If only the cation of the salt is acidic, the solution is acidic
3. If only the anion of the salt is basic, the solution will be basic
4. If a salt has a cation that is acidic and an anion that is basic, the pH of the solution is determined by the relative strengths of the acid and base
• if $K_a = K_b$, then no net effect on pH
• if $K_a > K_b$, then the solution is slightly acidic
• if $K_b> K_b$, then the solution is slightly basic
Salts of Polyprotic Acids
Do not be intimidated by the salts of polyprotic acids. Yes they're bigger and "badder" then most other salts, but they can be handled the exact same way as other salts, just with a bit more math.
• All of the rules from above still apply, e.g., the pH of a salt of polyprotic acid will always be greater than 7.
• The same way that polyprotic acids lose H+ stepwise, salts of polyprotic acids gain H+ in the same manner, but in reverse order of the polyprotic acid.
Take for example dissociation of carbonic acid (H_2CO_3):
$H_2CO_{3(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + HCO^-_{3(aq)}$
with $K_{a1} = 2.5 \times 10^{-4}$
$HCO^-_{3(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + CO^{2-}_{3(aq)}$
with $K_{a2} = 5.61 \times 10^{-11}$. This means that when calculating the values for Kb of CO32-, The Kb of the first hydrolysis reaction will be $K_{b1} = \dfrac{K_w}{K_{a2}}$ since it will go in the reverse order.
Example $1$: $NaHSO_4$
What is the pH of a 0.100 M $\ce{NaHSO_4}$ solution? Sulfuric acid is a strong acid, and the $\ce pK_{\large\textrm a_{\Large 2}}$ of $\ce{HSO4-}$ is 1.92.
Solution
First review the ions generates immediate upon dissociation of the salt.
• $Na^+$ is a non-acidic "spectator ion" that does not affect the acidity or basicity of a solution and can be ignored.
• The $HSO_4^{-}$ anion will affect the pH
$HSO_4^{-}$ is part of the dissociation chain of diprotic sulfuric acid ($H_2SO_4$) with the following steps
$H_2SO_4 + H_2O → H_3O^+ + HSO_4^- \label{step1SO3}$
Because $H_2SO_4$ is a strong acid, this step goes to completion and no $H_2SO_4$ exists in the solution. The $HSO_4^-$ ion is a weak acid and dissociates to a small extent:
$HSO_4^- + H_2O \rightleftharpoons H_3O^+ + SO_4^{2-} \label{step2SO3}$
with $K_{a2} = 1.0 \times 10^{-2}$ from Table E1. The equilibrium favors the reactants.
The $NaHSO4$ salt is completely ionized in its solution.
$\ce{NaHSO4_{(s)} \rightarrow Na+_{(sq)} + HSO4-_{(sq)}}$
The anion then further ionizes (Equation $\ref{step2SO3}$). We can construct an ICE table to solve for the final value of this ionization and hence $[H_3O^+]$
ICE Table $HSO_4^-$ $H_2O$ $H_3O^+$ $SO_4^{2-}$
Initial 0.1 M - 0 0
Change -x - +x +x
Equilibrium 0.1 M -x - x x
$K_{a2} = \dfrac{x^2}{0.100-x} = 0.0100 \nonumber$
\begin{align} \ce{[H_3O+]}&= x \nonumber \ &= \dfrac{-0.120 + \sqrt{0.012^2 + 4\times0.00120} }{2} \nonumber\ &= \textrm{0.0292 M} \nonumber \end{align}
Thus,
$\mathrm{pH = - \log \;0.0292 = 1.54} \nonumber$
Note that while the dissociation is weak, the quadratic equation cannot be avoided since 0.0292 cannot be ignore when compared to 0.1 M.
Exercise $1$
Which of the following solutions are acidic, basic, or neutral?
1. $\ce{Na2SO4}$
2. $\ce{NaH2PO4}$
3. $\ce{Na2HPO4}$
4. $\ce{Na3PO4}$
5. $\ce{NaNO3}$
Example $2$: Qualitatively Estimating the pH of a Polyprotic Salt Solution
Predict whether the $Na_2HPO_4$ salt will form an acidic or basic solution when dissolved in water. You will need values from Table E1 to address this.
Solution
First review the ions generates immediate upon dissociation of the salt.
• $Na^+$ is a non-acidic "spectator ion" that does not affect the acidity or basicity of a solution and can be ignored.
• The $HPO_4^{2-}$ anion will affect the pH and does so via competing hydrolysis steps as discussed below.
From Table E1, the three equilibria values for the three acid/base reactions of the phosphate ion are:
$H_3PO_{4(sq)} \rightleftharpoons H^+_{(aq)} + H_2PO^{2-}_{4(aq)} \label{1}$
with $K_{a1} = 6.9 \times 10^{-3}$
$H_2PO_{4(sq)}^{-} \rightleftharpoons H^+_{(aq)} + HPO^{2-}_{4(aq)} \label{2}$
with $K_{a2} = 6.2 \times 10^{ -8}$
$HPO_{4(sq)}^{2-} \rightleftharpoons H^+_{(aq)} + PO^{3-}_{4(aq)} \label{3}$
with $K_{a3} = 4.8 \times 10^{-11}$
The low $K_a$ value for the deprotonation of $HPO_{4(sq)}^{2-}$ means that it is a poor acid and better functions as a (Brønsted-Lowry) base:
$HPO^{2-}_{4 (aq)} + H_2O_{(l)} \rightleftharpoons H_2PO^-_{4 (aq)} + OH^-_{ (aq)} \label{3b}$
The two possible reactions that $HPO^{2-}_{4(aq)}$ undergoes are the acid reaction in Equation $\ref{2}$ and the basic reaction in Equation $\ref{3b}$. Whether this ion will cause the solution to be acidic or basic depends on whether the $K_a$ for Equation $\ref{2}$ is bigger than the $K_b$ for Equation $\ref{3b}$.
The $K_b$ of this reaction is related to the $K_{a2}$ by
$K_b = \dfrac{K_w}{ K_a} \nonumber$
$K_b = \dfrac{1 \times 10 ^{-14}}{ 6.2 \times 10^{-8}} = 1.6 \times 10^{-7} \text{ for } HPO_4^{2-} \nonumber$
Since $K_b > K_a$, the solution will be basic. To determine just how basic requires the use of an ICE table as discussed in Example $3$.
Solving the final pH for a $Na_2HPO_4$ salt solution is more complex than for the $NaHSO_4$, since the latter involves a single equilibrium. Two equilibria must be simultaneously solved for $Na_2HPO_4$. However, often time (but not always), they can be easily approximated given the relative $pk_a$ describing the reversible reaction.
Example $3$: Quantitatively Estimating the pH of a Polyprotic Salt Solution
Calculate the pH of the solution containing 3.875 g of $Na_2HPO_4$ that has been dissolved in a 250 mL of water.
Solution
First it is best to identify the molarity of $Na_2HPO_4$ that was initially generated (i.e., before any hydrolysis reactions happens).
$c = \dfrac{ 3.875 \;g / \text{molecular mass}}{\text{volume}} = \dfrac{3.875\;g / 141.98\; g/mol}{250 \; mL} = 0.11\;M\nonumber$
As discussed in Example $1$, the low $K_a$ value for the deprotonation of $HPO_{4(sq)}^{2-}$ means that it is a poor acid and instead functions as a base instead. From comparing the ionization constant in Example $1$, the $Na_2HPO_4$ is a far better base than acid. If we ignore the acid properties of the ion and focus on the basic we get the following reaction
$HPO^{2-}_{4 (aq)} + H_2O_{(l)} \rightleftharpoons H_2PO^-_{4 (aq)} + OH^-_{ (aq)}$
with a $K_b$ of $1.6 \times 10^{-7}$. Now do an ICE table to get the final $[H^+]$ concentration.
ICE Table $HPO ^{2-}_{4 (aq)}$ $H_2O_{(l)}$ $H_2PO^-_{4 (aq)}$ $OH^-_{ (aq)}$
Initial 0.11 M - - 0
Change -x - +x +x
Equilibrium 0.11 M - x - x x
$K_b = \dfrac{[H_2PO^-_{4 (aq)}][OH^-_{ (aq)} ]}{[HPO ^{2-}_{4 (aq)}]} = 1.6 \times 10^{-7}$
$\dfrac{x^2}{0.11\; M -x} = 1.6 \times 10^{-7}$
if $x << 0.11\;M$ then $0.11\;M -x \approx 0.11 \; M$ then
$x =1.3 \times 10^{-4} = [OH^-]$
$pOH = -\log_{10} [OH^-] = 3.8$
$pH = pK_w - pOH] Assuming that the reaction is at 25°, then $pK_w = 14$ and \[pH = 14- 3.8 = 10.2$
As expected, the solution is basic. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/17%3A_Additional_Aspects_of_Acid-Base_Equilibria/17.5%3A_Solutions_of_Salts_of_Polyprotic_Acids.txt |
17.1: Common-Ion Effect in Acid-Base Equilibria
Questions
1. The solubility product Ksp for bismuth sulfide $\ce{Bi2S3}$ is $1.6 \times 10^{-72}$ at 25 °C. What is the molar solubility of bismuth sulfide in a solution that is 0.0010 M in sodium sulfide $\ce{Na2S}$?
2. John poured 1.0 mL of 0.10 M $\ce{NaCl}$, 1.0 mL of 0.10 M $\ce{KOH}$, and 1.0 mL 0.20 $\ce{HCl}$ solutions together and then he made the total volume to be 100.0 mL. What is the $\ce{[Cl- ]}$ in the final solution?
3. The Ksp for $\ce{AgCl}$ is $1.0 \times 10^{-10}$. From which of the following solutions would silver chloride precipitate?
A: A solution 0.10 M in $\ce{Ag+}$ and 1.00 M in $\ce{Cl-}$
B: A solution $1.0\times 10^{-5}\; M$ in $\ce{Ag+}$ and 0.20 M in $\ce{Cl-}$
C: A solution $1.0 \times 10^{-7}\; M$ in $\ce{Ag+}$ and 1.0E-7 M in $\ce{Cl-}$
1. A only
2. B only
3. C only
4. A and B
5. A, B, and C
Substance Ksp
magnesium hydroxide $1.2\times 10^{-11}$
magnesium carbonate $1.6\times 10^{-5}$
magnesium fluoride $6.4\times 10^{-9}$
4. Addition of which of the following substances will cause the precipitation of a salt from one liter of a $1\times 10^{-4}\; M$ $\ce{Mg^2+}$ solution?
1. $1\times 10^{-4}$ mole NaOH
2. $1\times 10^{-1}$ mole nitric acid
3. $1\times 10^{-5}$ mole potassium acetate
4. $1\times 10^{-4}$ mole ammonium nitrate
5. $1\times 10^{-2}$ mole sodium fluoride
5. The Ksp for strontium chromate is $3.6 \times 10^{-5}$ and Ksp for barium chromate is $1.2\times 10^{-10}$. What concentration of potassium chromate will precipitate the maximum amount of either the barium or the strontium chromate from an equimolar 0.10 M solution of barium and strontium ions without precipitating the other?
6. Iron(II) hydroxide is only sparingly soluble in water at 25 °C; its Ksp is equal to $7.9\times 10^{-16}$. Calculate the solubility of iron(II) hydroxide in a solution of pH 6.0.
Solutions
1. Answer 1.08e-13
Consider...
x = molar solubility; (2x)2 (0.0010 + 3x)3 = Ksp; x = ?
2. Answer 0.0030 M
Consider...
Evaluate $\ce{[Na+]}$, $\ce{[K+]}$, $\ce{[H+]}$ and $\ce{[OH- ]}$ for fun!
3. Answer d.
Consider...
Calculate the $\ce{[Ag+] [Cl- ]}$ for A, B, and C, and compare their values with Ksp.
4. Answer e.
Consider...
A lot of calculations to figure out, but it's a common ion problem.
5. Answer 3.6e-4
Consider...
This problem requires some thinking.
6. Answer 7.9 M
Consider...
$\ce{[OH- ]}$ = 10(-14+6) = 1e-8. x * (1e-8 + x)2 = Ksp; x = ?
17.3: Acid-Base Indicators
Q17.3.1
There are numerous natural indicators present in plants. The dye in red cabbage, the purple color of grapes, even the color of some flowers are some examples. What is the cause for some fruits to change color when they ripen?
S17.3.1
$\ce{[H+]}$ of the juice changes. The changes in pH or $\ce{[H+]}$ cause the dye to change color if their conjugate acid-base pairs have different colors. There may be other reasons too. Do colors indicate how good or bad they taste?
Q17.3.2
Choose the true statement:
1. All weak acids are indicators.
2. All weak bases are indicators.
3. Weak acids and bases are indicators.
4. All indicators are weak acids.
5. An acid-base conjugate pair has different colors.
6. Any indicator changes color when the pH of its solution is 7.
S17.3.2
d. Color change is a requirement for indicators.
Q17.3.3
Do all indicators change color at pH 7 (y/n)?
S17.3.3
No! Phenolphthalein changes color at pH ~9. Bromothymol blue has a pKn value of 7.1. At pH 7, its color changes from yellow to blue. Some indicators change color at pH other than 7. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/17%3A_Additional_Aspects_of_Acid-Base_Equilibria/17.E%3A_Exercises.txt |
• 18.1: Solubility Product Constant, Ksp
The equilibrium constant for a dissolution reaction, called the solubility product (Ksp), is a measure of the solubility of a compound. Whereas solubility is usually expressed in terms of mass of solute per 100 mL of solvent, Ksp is defined in terms of the molar concentrations of the component ions. In contrast, the ion product (Q) describes concentrations that are not necessarily equilibrium concentrations. Comparing Q and Ksp enables us to determine whether a precipitate will form.
• 18.2: Relationship Between Solubility and Ksp
Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium. Equilibrium is the state at which the concentrations of products and reactant are constant after the reaction has taken place. The solubility product constant describes the equilibrium between a solid and its constituent ions in a solution. The value of the constant identifies the degree to which the compound can dissociate in water. The higher the, the more soluble the compound is.
• 18.3: Common-Ion Effect in Solubility Equilibria
Adding a common cation or common anion to a solution of a sparingly soluble salt shifts the solubility equilibrium in the direction predicted by Le Châtelier’s principle. The solubility of the salt is almost always decreased by the presence of a common ion.
• 18.4: Limitations of the Kₛₚ Concept
The solubility constant, as with all equilibrium constants, are properly defined in terms of effective concentrations (activities) and the use of true concentrations to approximate activities in equilibrium constants can often fail. As with all approximations, it is important to understand the conditions that it will fail. three examples are discussed.
• 18.5: Criteria for Precipitation and its Completeness
A slightly soluble electrolyte begins to precipitate when the magnitude of the reaction quotient for the dissolution reaction exceeds the magnitude of the solubility product. Precipitation continues until the reaction quotient equals the solubility product. A reagent can be added to a solution of ions to allow one ion to selectively precipitate out of solution. The common ion effect can also play a role in precipitation reactions.
• 18.6: Fractional Precipitation
Formation of crystals from a saturated solution is a heterogeneous equilibrium phenomenon, and it can be applied to separate various chemicals or ions in a solution. When solubilities of two metal salts are very different, they can be separated by precipitation.
• 18.7: Solubility and pH
The anion in many sparingly soluble salts is the conjugate base of a weak acid. At low pH, protonation of the anion can dramatically increase the solubility of the salt. Oxides can be classified as acidic oxides or basic oxides. Acidic oxides either react with water to give an acidic solution or dissolve in strong base; most acidic oxides are nonmetal oxides or oxides of metals in high oxidation states.
• 18.8: Equilibria Involving Complex Ions
A complex ion is a species formed between a central metal ion and one or more surrounding ligands, molecules or ions that contain at least one lone pair of electrons. Small, highly charged metal ions have the greatest tendency to act as Lewis acids and form complex ions. The equilibrium constant for the formation of the complex ion is the formation constant (Kf). The formation of a complex ion by adding a complexing agent increases the solubility of a compound.
• 18.9: Qualitative Cation Analysis
In qualitative analysis, the identity, not the amount, of metal ions present in a mixture is determined. The technique consists of selectively precipitating only a few kinds of metal ions at a time under given sets of conditions. Consecutive precipitation steps become progressively less selective until almost all the metal ions are precipitated. Other additional steps are needed to separate metal ions that precipitate together.
18: Solubility and Complex-Ion Equilibria
Learning Objectives
• To calculate the solubility of an ionic compound from its Ksp
We begin our discussion of solubility and complexation equilibria—those associated with the formation of complex ions—by developing quantitative methods for describing dissolution and precipitation reactions of ionic compounds in aqueous solution. Just as with acid–base equilibria, we can describe the concentrations of ions in equilibrium with an ionic solid using an equilibrium constant expression.
The Solubility Product
When a slightly soluble ionic compound is added to water, some of it dissolves to form a solution, establishing an equilibrium between the pure solid and a solution of its ions. For the dissolution of calcium phosphate, one of the two main components of kidney stones, the equilibrium can be written as follows, with the solid salt on the left:
$Ca_3(PO_4)_{2(s)} \rightleftharpoons 3Ca^{2+}_{(aq)} + 2PO^{3−}_{4(aq)} \label{Eq1}$
As you will discover in Section 17.4 and in more advanced chemistry courses, basic anions, such as S2−, PO43, and CO32, react with water to produce OH and the corresponding protonated anion. Consequently, their calculated molarities, assuming no protonation in aqueous solution, are only approximate.
The equilibrium constant for the dissolution of a sparingly soluble salt is the solubility product (Ksp) of the salt. Because the concentration of a pure solid such as Ca3(PO4)2 is a constant, it does not appear explicitly in the equilibrium constant expression. The equilibrium constant expression for the dissolution of calcium phosphate is therefore
$K=\dfrac{[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2}{[\mathrm{Ca_3(PO_4)_2}]} \label{Eq2a}$
$[\mathrm{Ca_3(PO_4)_2}]K=K_{\textrm{sp}}=[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2 \label{Eq2b}$
At 25°C and pH 7.00, Ksp for calcium phosphate is 2.07 × 10−33, indicating that the concentrations of Ca2+ and PO43 ions in solution that are in equilibrium with solid calcium phosphate are very low. The values of Ksp for some common salts are listed in Table $1$, which shows that the magnitude of Ksp varies dramatically for different compounds. Although Ksp is not a function of pH in Equations $\ref{Eq2a}$ and $\ref{Eq2b}$, changes in pH can affect the solubility of a compound as discussed later.
Note
As with K, the concentration of a pure solid does not appear explicitly in Ksp.
Table $1$: Solubility Products for Selected Ionic Substances at 25°C
Solid Color Ksp Solid Color Ksp
*These contain the Hg22+ ion.
Acetates Iodides
Ca(O2CCH3)2·3H2O white 4 × 10−3 Hg2I2* yellow 5.2 × 10−29
Bromides PbI2 yellow 9.8 × 10−9
AgBr off-white 5.35 × 10−13 Oxalates
Hg2Br2* yellow 6.40 × 10−23 Ag2C2O4 white 5.40 × 10−12
Carbonates MgC2O4·2H2O white 4.83 × 10−6
CaCO3 white 3.36 × 10−9 PbC2O4 white 4.8 × 10−10
PbCO3 white 7.40 × 10−14 Phosphates
Chlorides Ag3PO4 white 8.89 × 10−17
AgCl white 1.77 × 10−10 Sr3(PO4)2 white 4.0 × 10−28
Hg2Cl2* white 1.43 × 10−18 FePO4·2H2O pink 9.91 × 10−16
PbCl2 white 1.70 × 10−5 Sulfates
Chromates Ag2SO4 white 1.20 × 10−5
CaCrO4 yellow 7.1 × 10−4 BaSO4 white 1.08 × 10−10
PbCrO4 yellow 2.8 × 10−13 PbSO4 white 2.53 × 10−8
Fluorides Sulfides
BaF2 white 1.84 × 10−7 Ag2S black 6.3 × 10−50
PbF2 white 3.3 × 10−8 CdS yellow 8.0 × 10−27
Hydroxides PbS black 8.0 × 10−28
Ca(OH)2 white 5.02 × 10−6 ZnS white 1.6 × 10−24
Cu(OH)2 pale blue 1 × 10−14
Mn(OH)2 light pink 1.9 × 10−13
Cr(OH)3 gray-green 6.3 × 10−31
Fe(OH)3 rust red 2.79 × 10−39
Solubility products are determined experimentally by directly measuring either the concentration of one of the component ions or the solubility of the compound in a given amount of water. However, whereas solubility is usually expressed in terms of mass of solute per 100 mL of solvent, $K_{sp}$, like $K$, is defined in terms of the molar concentrations of the component ions.
A color photograph of a kidney stone, 8 mm in length. Kidney stones form from sparingly soluble calcium salts and are largely composed of Ca(O2CCO2)·H2O and Ca3(PO4)2. from Wikipedia.
Definition of a Solubility Product: https://youtu.be/VzxSmH_iwHE
Example $1$
Calcium oxalate monohydrate [Ca(O2CCO2)·H2O, also written as CaC2O4·H2O] is a sparingly soluble salt that is the other major component of kidney stones [along with Ca3(PO4)2]. Its solubility in water at 25°C is 7.36 × 10−4 g/100 mL. Calculate its Ksp.
Given: solubility in g/100 mL
Asked for: Ksp
Strategy:
1. Write the balanced dissolution equilibrium and the corresponding solubility product expression.
2. Convert the solubility of the salt to moles per liter. From the balanced dissolution equilibrium, determine the equilibrium concentrations of the dissolved solute ions. Substitute these values into the solubility product expression to calculate Ksp.
Solution
A We need to write the solubility product expression in terms of the concentrations of the component ions. For calcium oxalate monohydrate, the balanced dissolution equilibrium and the solubility product expression (abbreviating oxalate as ox2) are as follows:
$\mathrm{Ca(O_2CCO_2)}\cdot\mathrm{H_2O(s)}\rightleftharpoons \mathrm{Ca^{2+}(aq)}+\mathrm{^-O_2CCO_2^-(aq)}+\mathrm{H_2O(l)}\hspace{5mm}K_{\textrm{sp}}=[\mathrm{Ca^{2+}}][\mathrm{ox^{2-}}]$
Neither solid calcium oxalate monohydrate nor water appears in the solubility product expression because their concentrations are essentially constant.
B Next we need to determine [Ca2+] and [ox2−] at equilibrium. We can use the mass of calcium oxalate monohydrate that dissolves in 100 mL of water to calculate the number of moles that dissolve in 100 mL of water. From this we can determine the number of moles that dissolve in 1.00 L of water. For dilute solutions, the density of the solution is nearly the same as that of water, so dissolving the salt in 1.00 L of water gives essentially 1.00 L of solution. Because each 1 mol of dissolved calcium oxalate monohydrate dissociates to produce 1 mol of calcium ions and 1 mol of oxalate ions, we can obtain the equilibrium concentrations that must be inserted into the solubility product expression. The number of moles of calcium oxalate monohydrate that dissolve in 100 mL of water is as follows:
$\dfrac{7.36\times10^{-4}\textrm{ g}}{146.1\textrm{ g/mol}}=5.04\times10^{-6}\textrm{ mol }\mathrm{Ca(O_2CCO_2)\cdot H_2O}$
The number of moles of calcium oxalate monohydrate that dissolve in 1.00 L of the saturated solution is as follows:
$\left(\dfrac{5.04\times10^{-6}\textrm{ mol }\mathrm{Ca(O_2CCO_2\cdot)H_2O}}{\textrm{100 mL}}\right)\left(\dfrac{\textrm{1000 mL}}{\textrm{1.00 L}}\right)=5.04\times10^{-5}\textrm{ mol/L}=5.04\times10^{-5}\textrm{ M}$
Because of the stoichiometry of the reaction, the concentration of Ca2+ and ox2− ions are both 5.04 × 10−5 M. Inserting these values into the solubility product expression,
$K_{sp} = [Ca^{2+}][ox^{2−}] = (5.04 \times 10^{−5})(5.04 \times10^{−5}) = 2.54 \times 10^{−9}$
In our calculation, we have ignored the reaction of the weakly basic anion with water, which tends to make the actual solubility of many salts greater than the calculated value.
Exercise $1$
One crystalline form of calcium carbonate (CaCO3) is the mineral sold as “calcite” in mineral and gem shops. The solubility of calcite in water is 0.67 mg/100 mL. Calculate its Ksp.
Answer 4.5 × 10−9
The reaction of weakly basic anions with H2O tends to make the actual solubility of many salts higher than predicted.
A crystal of calcite (CaCO3), illustrating the phenomenon of double refraction. When a transparent crystal of calcite is placed over a page, we see two images of the letters. Image used with permisison from Wikipedia
Calcite, a structural material for many organisms, is found in the teeth of sea urchins. The urchins create depressions in limestone that they can settle in by grinding the rock with their teeth. Limestone, however, also consists of calcite, so how can the urchins grind the rock without also grinding their teeth? Researchers have discovered that the teeth are shaped like needles and plates and contain magnesium. The concentration of magnesium increases toward the tip, which contributes to the hardness. Moreover, each tooth is composed of two blocks of the polycrystalline calcite matrix that are interleaved near the tip. This creates a corrugated surface that presumably increases grinding efficiency. Toolmakers are particularly interested in this approach to grinding.
Tabulated values of Ksp can also be used to estimate the solubility of a salt with a procedure that is essentially the reverse of the one used in Example $1$. In this case, we treat the problem as a typical equilibrium problem and set up a table of initial concentrations, changes in concentration, and final concentrations (ICE Tables), remembering that the concentration of the pure solid is essentially constant.
Example $2$
We saw that the Ksp for Ca3(PO4)2 is 2.07 × 10−33 at 25°C. Calculate the aqueous solubility of Ca3(PO4)2 in terms of the following:
1. the molarity of ions produced in solution
2. the mass of salt that dissolves in 100 mL of water at 25°C
Given: Ksp
Asked for: molar concentration and mass of salt that dissolves in 100 mL of water
Strategy:
1. Write the balanced equilibrium equation for the dissolution reaction and construct a table showing the concentrations of the species produced in solution. Insert the appropriate values into the solubility product expression and calculate the molar solubility at 25°C.
2. Calculate the mass of solute in 100 mL of solution from the molar solubility of the salt. Assume that the volume of the solution is the same as the volume of the solvent.
Solution:
1. A The dissolution equilibrium for Ca3(PO4)2 (Equation $\ref{Eq2a}$) is shown in the following ICE table. Because we are starting with distilled water, the initial concentration of both calcium and phosphate ions is zero. For every 1 mol of Ca3(PO4)2 that dissolves, 3 mol of Ca2+ and 2 mol of PO43− ions are produced in solution. If we let x equal the solubility of Ca3(PO4)2 in moles per liter, then the change in [Ca2+] will be +3x, and the change in [PO43−] will be +2x. We can insert these values into the table.
Ca3(PO4)2(s) ⇌ 3Ca2+(aq) + 2PO43−(aq)
Ca3(PO4)2 [Ca2+] [PO43−]
initial pure solid 0 0
change +3x +2x
final pure solid 3x 2x
Although the amount of solid Ca3(PO4)2 changes as some of it dissolves, its molar concentration does not change. We now insert the expressions for the equilibrium concentrations of the ions into the solubility product expression (Equation 17.2):
\begin{align}K_{\textrm{sp}}=[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2&=(3x)^3(2x)^2 \2.07\times10^{-33}&=108x^5 \1.92\times10^{-35}&=x^5 \1.14\times10^{-7}\textrm{ M}&=x\end{align}
This is the molar solubility of calcium phosphate at 25°C. However, the molarity of the ions is 2x and 3x, which means that [PO43−] = 2.28 × 10−7 and [Ca2+] = 3.42 × 10−7.
1. B To find the mass of solute in 100 mL of solution, we assume that the density of this dilute solution is the same as the density of water because of the low solubility of the salt, so that 100 mL of water gives 100 mL of solution. We can then determine the amount of salt that dissolves in 100 mL of water:
$\left(\dfrac{1.14\times10^{-7}\textrm{ mol}}{\textrm{1 L}}\right)\textrm{100 mL}\left(\dfrac{\textrm{1 L}}{\textrm{1000 mL}} \right )\left(\dfrac{310.18 \textrm{ g }\mathrm{Ca_3(PO_4)_2}}{\textrm{1 mol}}\right)=3.54\times10^{-6}\textrm{ g }\mathrm{Ca_3(PO_4)_2}$
Exercise $2$
The solubility product of silver carbonate (Ag2CO3) is 8.46 × 10−12 at 25°C. Calculate the following:
1. the molarity of a saturated solution
2. the mass of silver carbonate that will dissolve in 100 mL of water at this temperature
Answer
1. 1.28 × 10−4 M
2. 3.54 mg
The Ion Product
The ion product (Q) of a salt is the product of the concentrations of the ions in solution raised to the same powers as in the solubility product expression. It is analogous to the reaction quotient (Q) discussed for gaseous equilibria. Whereas Ksp describes equilibrium concentrations, the ion product describes concentrations that are not necessarily equilibrium concentrations.
The ion product Q is analogous to the reaction quotient Q for gaseous equilibria.
As summarized in Figure $1$ "The Relationship between ", there are three possible conditions for an aqueous solution of an ionic solid:
• Q < Ksp. The solution is unsaturated, and more of the ionic solid, if available, will dissolve.
• Q = Ksp. The solution is saturated and at equilibrium.
• Q > Ksp. The solution is supersaturated, and ionic solid will precipitate.
The process of calculating the value of the ion product and comparing it with the magnitude of the solubility product is a straightforward way to determine whether a solution is unsaturated, saturated, or supersaturated. More important, the ion product tells chemists whether a precipitate will form when solutions of two soluble salts are mixed.
Example $3$
We mentioned that barium sulfate is used in medical imaging of the gastrointestinal tract. Its solubility product is 1.08 × 10−10 at 25°C, so it is ideally suited for this purpose because of its low solubility when a “barium milkshake” is consumed by a patient. The pathway of the sparingly soluble salt can be easily monitored by x-rays. Will barium sulfate precipitate if 10.0 mL of 0.0020 M Na2SO4 is added to 100 mL of 3.2 × 10−4 M BaCl2? Recall that NaCl is highly soluble in water.
Given: Ksp and volumes and concentrations of reactants
Asked for: whether precipitate will form
Strategy:
1. Write the balanced equilibrium equation for the precipitation reaction and the expression for Ksp.
2. Determine the concentrations of all ions in solution when the solutions are mixed and use them to calculate the ion product (Q).
3. Compare the values of Q and Ksp to decide whether a precipitate will form.
Solution
A The only slightly soluble salt that can be formed when these two solutions are mixed is BaSO4 because NaCl is highly soluble. The equation for the precipitation of BaSO4 is as follows:
$BaSO_{4(s)} \rightleftharpoons Ba^{2+}_{(aq)} + SO^{2−}_{4(aq)}$
The solubility product expression is as follows:
Ksp = [Ba2+][SO42] = 1.08×10−10
B To solve this problem, we must first calculate the ion product—Q = [Ba2+][SO42]—using the concentrations of the ions that are present after the solutions are mixed and before any reaction occurs. The concentration of Ba2+ when the solutions are mixed is the total number of moles of Ba2+ in the original 100 mL of BaCl2 solution divided by the final volume (100 mL + 10.0 mL = 110 mL):
$\textrm{moles Ba}^{2+}=\textrm{100 mL}\left(\dfrac{\textrm{1 L}}{\textrm{1000 mL}}\right)\left(\dfrac{3.2\times10^{-4}\textrm{ mol}}{\textrm{1 L}} \right )=3.2\times10^{-5}\textrm{ mol Ba}^{2+}$
$[\mathrm{Ba^{2+}}]=\left(\dfrac{3.2\times10^{-5}\textrm{ mol Ba}^{2+}}{\textrm{110 mL}}\right)\left(\dfrac{\textrm{1000 mL}}{\textrm{1 L}}\right)=2.9\times10^{-4}\textrm{ M Ba}^{2+}$
Similarly, the concentration of SO42 after mixing is the total number of moles of SO42 in the original 10.0 mL of Na2SO4 solution divided by the final volume (110 mL):
$\textrm{moles SO}_4^{2-}=\textrm{10.0 mL}\left(\dfrac{\textrm{1 L}}{\textrm{1000 mL}}\right)\left(\dfrac{\textrm{0.0020 mol}}{\textrm{1 L}}\right)=2.0\times10^{-5}\textrm{ mol SO}_4^{2-}$
$[\mathrm{SO_4^{2-}}]=\left(\dfrac{2.0\times10^{-5}\textrm{ mol SO}_4^{2-}}{\textrm{110 mL}} \right )\left(\dfrac{\textrm{1000 mL}}{\textrm{1 L}}\right)=1.8\times10^{-4}\textrm{ M SO}_4^{2-}$
We can now calculate Q:
Q = [Ba2+][SO42] = (2.9×10−4)(1.8×10−4) = 5.2×10−8
C We now compare Q with the Ksp. If Q > Ksp, then BaSO4 will precipitate, but if Q < Ksp, it will not. Because Q > Ksp, we predict that BaSO4 will precipitate when the two solutions are mixed. In fact, BaSO4 will continue to precipitate until the system reaches equilibrium, which occurs when [Ba2+][SO42] = Ksp = 1.08 × 10−10.
Exercise $3$
The solubility product of calcium fluoride (CaF2) is 3.45 × 10−11. If 2.0 mL of a 0.10 M solution of NaF is added to 128 mL of a 2.0 × 10−5M solution of Ca(NO3)2, will CaF2 precipitate?
Answer yes (Q = 4.7 × 10−11 > Ksp)
Summary
The solubility product (Ksp) is used to calculate equilibrium concentrations of the ions in solution, whereas the ion product (Q) describes concentrations that are not necessarily at equilibrium. The equilibrium constant for a dissolution reaction, called the solubility product (Ksp), is a measure of the solubility of a compound. Whereas solubility is usually expressed in terms of mass of solute per 100 mL of solvent, Ksp is defined in terms of the molar concentrations of the component ions. In contrast, the ion product (Q) describes concentrations that are not necessarily equilibrium concentrations. Comparing Q and Ksp enables us to determine whether a precipitate will form when solutions of two soluble salts are mixed. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/18%3A_Solubility_and_Complex-Ion_Equilibria/18.1%3A_Solubility_Product_Constant_Ksp.txt |
Learning Objectives
• Quantitatively related $K_{sp}$ to solubility
Considering the relation between solubility and $K_{sp}$ is important when describing the solubility of slightly ionic compounds. However, this article discusses ionic compounds that are difficult to dissolve; they are considered "slightly soluble" or "almost insoluble." Solubility product constants ($K_{sq}$) are given to those solutes, and these constants can be used to find the molar solubility of the compounds that make the solute. This relationship also facilitates finding the $K_{sq}$ of a slightly soluble solute from its solubility.
Introduction
Recall that the definition of solubility is the maximum possible concentration of a solute in a solution at a given temperature and pressure. We can determine the solubility product of a slightly soluble solid from that measure of its solubility at a given temperature and pressure, provided that the only significant reaction that occurs when the solid dissolves is its dissociation into solvated ions, that is, the only equilibrium involved is:
$\ce{M}_p\ce{X}_q(s)⇌p\mathrm{M^{m+}}(aq)+q\mathrm{X^{n−}}(aq)$
In this case, we calculate the solubility product by taking the solid’s solubility expressed in units of moles per liter (mol/L), known as its molar solubility.
Calculation of Ksp from Equilibrium Concentrations
We began the chapter with an informal discussion of how the mineral fluorite is formed. Fluorite, $\ce{CaF2}$, is a slightly soluble solid that dissolves according to the equation:
$\ce{CaF2}(s)⇌\ce{Ca^2+}(aq)+\ce{2F-}(aq)\nonumber$
The concentration of Ca2+ in a saturated solution of CaF2 is 2.1 × 10–4 M; therefore, that of F is 4.2 × 10–4 M, that is, twice the concentration of $\ce{Ca^{2+}}$. What is the solubility product of fluorite?
Solution
First, write out the Ksp expression, then substitute in concentrations and solve for Ksp:
$\ce{CaF2(s) <=> Ca^{2+}(aq) + 2F^{-}(aq)} \nonumber$
A saturated solution is a solution at equilibrium with the solid. Thus:
\begin{align*} K_\ce{sp} &= \ce{[Ca^{2+}][F^{-}]^2} \[4pt] &=(2.1×10^{−4})(4.2×10^{−4})^2 \[4pt] &=3.7×10^{−11}\end{align*}
As with other equilibrium constants, we do not include units with Ksp.
Exercise $1$
In a saturated solution that is in contact with solid Mg(OH)2, the concentration of Mg2+ is 3.7 × 10–5 M. What is the solubility product for Mg(OH)2?
$\ce{Mg(OH)2}(s)⇌\ce{Mg^2+}(aq)+\ce{2OH-}(aq)\nonumber$
Answer
2.0 × 10–13
Determination of Molar Solubility from Ksp
The Ksp of copper(I) bromide, $\ce{CuBr}$, is 6.3 × 10–9. Calculate the molar solubility of copper bromide.
Solution
The solubility product constant of copper(I) bromide is 6.3 × 10–9.
The reaction is:
$\ce{CuBr}(s)⇌\ce{Cu+}(aq)+\ce{Br-}(aq)\nonumber$
First, write out the solubility product equilibrium constant expression:
$K_\ce{sp}=\ce{[Cu+][Br- ]}\nonumber$
Create an ICE table (as introduced in the chapter on fundamental equilibrium concepts), leaving the $\ce{CuBr}$ column empty as it is a solid and does not contribute to the Ksp:
At equilibrium:
\begin{align*} K_\ce{sp} &=\ce{[Cu+][Br- ]} \[4pt] 6.3×10^{−9} &=(x)(x)=x^2 \[4pt] x&=\sqrt{(6.3×10^{−9})}=7.9×10^{−5} \end{align*}
Therefore, the molar solubility of $\ce{CuBr}$ is 7.9 × 10–5 M.
Finding the Solubility of a Salt: https://youtu.be/98BuldrICXM
Summary
Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium. Equilibrium is the state at which the concentrations of products and reactant are constant after the reaction has taken place. The solubility product constant ($K_{sp}$) describes the equilibrium between a solid and its constituent ions in a solution. The value of the constant identifies the degree to which the compound can dissociate in water. The higher the $K_{sp}$, the more soluble the compound is. $K_{sq}$ is defined in terms of activity rather than concentration because it is a measure of a concentration that depends on certain conditions such as temperature, pressure, and composition. It is influenced by surroundings. $K_{sp}$ is used to describe the saturated solution of ionic compounds. (A saturated solution is in a state of equilibrium between the dissolved, dissociated, undissolved solid, and the ionic compound.) | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/18%3A_Solubility_and_Complex-Ion_Equilibria/18.2%3A_Relationship_Between_Solubility_and_Ksp.txt |
Learning Objectives
• Recognize common ions from various salts, acids, and bases.
• Calculate concentrations involving common ions.
• Calculate ion concentrations involving chemical equilibrium.
The common-ion effect is used to describe the effect on an equilibrium involving a substance that adds an ion that is a part of the equilibrium.
Introduction
The solubility products Ksp's are equilibrium constants in hetergeneous equilibria (i.e., between two different phases). If several salts are present in a system, they all ionize in the solution. If the salts contain a common cation or anion, these salts contribute to the concentration of the common ion. Contributions from all salts must be included in the calculation of concentration of the common ion. For example, a solution containing sodium chloride and potassium chloride will have the following relationship:
$\mathrm{[Na^+] + [K^+] = [Cl^-]} \label{1}$
Consideration of charge balance or mass balance or both leads to the same conclusion. The solubility product expression tells us that the equilibrium concentrations of the cation and the anion are inversely related. That is, as the concentration of the anion increases, the maximum concentration of the cation needed for precipitation to occur decreases—and vice versa—so that Ksp is constant. Consequently, the solubility of an ionic compound depends on the concentrations of other salts that contain the same ions. Adding a common cation or anion shifts a solubility equilibrium in the direction predicted by Le Chatelier’s principle. As a result, the solubility of any sparingly soluble salt is almost always decreased by the presence of a soluble salt that contains a common ion. The exceptions generally involve the formation of complex ions, which is discussed later.
Common Ions
When $\ce{NaCl}$ and $\ce{KCl}$ are dissolved in the same solution, the $\mathrm{ {\color{Green} Cl^-}}$ ions are common to both salts. In a system containing $\ce{NaCl}$ and $\ce{KCl}$, the $\mathrm{ {\color{Green} Cl^-}}$ ions are common ions.
$\mathrm{NaCl \rightleftharpoons Na^+ + {\color{Green} Cl^-}}$
$\mathrm{KCl \rightleftharpoons K^+ + {\color{Green} Cl^-}}$
$\mathrm{CaCl_2 \rightleftharpoons Ca^{2+} + {\color{Green} 2 Cl^-}}$
$\mathrm{AlCl_3 \rightleftharpoons Al^{3+} + {\color{Green} 3 Cl^-}}$
$\mathrm{AgCl \rightleftharpoons Ag^+ + {\color{Green} Cl^-}}$
For example, when $\ce{AgCl}$ is dissolved into a solution already containing $\ce{NaCl}$ (actually $\ce{Na+}$ and $\ce{Cl-}$ ions), the $\ce{Cl-}$ ions come from the ionization of both $\ce{AgCl}$ and $\ce{NaCl}$. Thus, $\ce{[Cl- ]}$ differs from $\ce{[Ag+]}$. The following examples show how the concentration of the common ion is calculated.
Example $1$
What are $\ce{[Na+]}$, $\ce{[Cl- ]}$, $\ce{[Ca^2+]}$, and $\ce{[H+]}$ in a solution containing 0.10 M each of $\ce{NaCl}$, $\ce{CaCl2}$, and $\ce{HCl}$?
Solution
Due to the conservation of ions, we have
$\mathrm{[Na^+] = [Ca^{2+}] = [H^+] = 0.10\: \ce M}\nonumber.$
but
\begin{alignat}{3} &\ce{[Cl- ]} &&= && && \:\textrm{0.10 (due to NaCl)}\nonumber \ & && && + &&\mathrm{\:0.20\: (due\: to\: CaCl_2)}\nonumber\ & && && + &&\mathrm{\:0.10\: (due\: to\: HCl)}\nonumber\ & &&= && &&\mathrm{\:0.40\: M}\nonumber \end{alignat}
Exercise $1$
John poured 10.0 mL of 0.10 M $\ce{NaCl}$, 10.0 mL of 0.10 M $\ce{KOH}$, and 5.0 mL of 0.20 M $\ce{HCl}$ solutions together and then he made the total volume to be 100.0 mL. What is $\ce{[Cl- ]}$ in the final solution?
$\mathrm{[Cl^-] = \dfrac{0.1\: M\times 10\: mL+0.2\: M\times 5.0\: mL}{100.0\: mL} = 0.020\: M}\nonumber$
Le Châtelier's Principle states that if an equilibrium becomes unbalanced, the reaction will shift to restore the balance. If a common ion is added to a weak acid or weak base equilibrium, then the equilibrium will shift towards the reactants, in this case the weak acid or base.
Example $2$
Consider the lead(II) ion concentration in this saturated solution of PbCl2. The balanced reaction is
$PbCl_{2 (s)} \rightleftharpoons Pb^{2+} _{(aq)} + 2Cl^-_{(aq)}\nonumber$
Defining $s$ as the concentration of dissolved lead(II) chloride, then:
$[Pb^{2+}] = s\nonumber$
$[Cl^- ] = 2s\nonumber$
These values can be substituted into the solubility product expression, which can be solved for $s$:
\begin{align*} K_{sp} &= [Pb^{2+}] [Cl^-]^2 \[4pt] &= s \times (2s)^2 \[4pt] 1.7 \times 10^{-5} &= 4s^3 \[4pt] s^3 &= \frac{1.7 \times 10^{-5}}{4} \[4pt] &= 4.25 \times 10^{-6} \[4pt] s &= \sqrt[3]{4.25 \times 10^{-6}} \[4pt] &= 1.62 \times 10^{-2}\, mol\ dm^{-3} \end{align*}
The concentration of lead(II) ions in the solution is 1.62 x 10-2 M. Consider what happens if sodium chloride is added to this saturated solution. Sodium chloride shares an ion with lead(II) chloride. The chloride ion is common to both of them; this is the origin of the term "common ion effect".
Look at the original equilibrium expression again:
$PbCl_2 \; (s) \rightleftharpoons Pb^{2+} \; (aq) + 2Cl^- \; (aq)\nonumber$
What happens to that equilibrium if extra chloride ions are added? According to Le Châtelier, the position of equilibrium will shift to counter the change, in this case, by removing the chloride ions by making extra solid lead(II) chloride.
Of course, the concentration of lead(II) ions in the solution is so small that only a tiny proportion of the extra chloride ions can be converted into solid lead(II) chloride. The lead(II) chloride becomes even less soluble, and the concentration of lead(II) ions in the solution decreases. This type of response occurs with any sparingly soluble substance: it is less soluble in a solution which contains any ion which it has in common. This is the common ion effect.
A Simple Example
If an attempt is made to dissolve some lead(II) chloride in some 0.100 M sodium chloride solution instead of in water, what is the equilibrium concentration of the lead(II) ions this time? As before, define s to be the concentration of the lead(II) ions.
$[Pb^{2+}] = s \label{2}$
The calculations are different from before. This time the concentration of the chloride ions is governed by the concentration of the sodium chloride solution. The number of ions coming from the lead(II) chloride is going to be tiny compared with the 0.100 M coming from the sodium chloride solution. In calculations like this, it can be assumed that the concentration of the common ion is entirely due to the other solution. This simplifies the calculation.
So we assume:
$[Cl^- ] = 0.100\; M \label{3}$
The rest of the mathematics looks like this:
$\begin{split} K_{sp}& = [Pb^{2+}][Cl^-]^2 \ & = s \times (0.100)^2 \ 1.7 \times 10^{-5} & = s \times 0.00100 \end{split}$
therefore:
$\begin{split} s & = \dfrac{1.7 \times 10^{-5}}{0.0100} \ & = 1.7 \times 10^{-3} \, \text{M} \end{split} \label{4}$
Finally, compare that value with the simple saturated solution:
Original solution:
$[Pb^{2+}] = 0.0162 \, M \label{5}$
Solution in 0.100 M NaCl solution:
$[Pb^{2+}] = 0.0017 \, M \label{6}$
The concentration of the lead(II) ions has decreased by a factor of about 10. If more concentrated solutions of sodium chloride are used, the solubility decreases further.
Adding a common ion to a system at equilibrium affects the equilibrium composition, but not the ionization constant.
Common Ion Effect with Weak Acids and Bases
Adding a common ion prevents the weak acid or weak base from ionizing as much as it would without the added common ion. The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium.
Example $3$
The common ion effect of H3O+ on the ionization of acetic acid
The common ion effect suppresses the ionization of a weak base by adding more of an ion that is a product of this equilibrium.
Example $4$
Consider the common ion effect of OH- on the ionization of ammonia
Adding the common ion of hydroxide shifts the reaction towards the left to decrease the stress (in accordance with Le Châtelier's Principle), forming more reactants. This decreases the reaction quotient, because the reaction is being pushed towards the left to reach equilibrium. The equilibrium constant, $K_b=1.8 \times 10^{-5}$, does not change. The reaction is put out of balance, or equilibrium.
$Q_a = \dfrac{[NH_4^+][OH^-]}{[NH_3]}\nonumber$
At first, when more hydroxide is added, the quotient is greater than the equilibrium constant. The reaction then shifts right, causing the denominator to increase, decreasing the reaction quotient and pulling towards equilibrium and causing $Q$ to decrease towards $K$.
Common Ion Effect on Solubility
Consider, for example, the effect of adding a soluble salt, such as CaCl2, to a saturated solution of calcium phosphate [Ca3(PO4)2].
$\ce{Ca3(PO4)2(s) <=> 3Ca^{2+}(aq) + 2PO^{3−}4(aq)} \label{Eq1}$
We have seen that the solubility of Ca3(PO4)2 in water at 25°C is 1.14 × 10−7 M (Ksp = 2.07 × 10−33). Thus a saturated solution of Ca3(PO4)2 in water contains
$3 × (1.14 × 10^{−7}\, M) = 3.42 × 10^{−7}\, M\, \ce{Ca^{2+}}$
and
$2 × (1.14 × 10^{−7}\, M) = 2.28 × 10^{−7}\, M\, \ce{PO4^{3−}}$
according to the stoichiometry shown in Equation $\ref{Eq1}$ (neglecting hydrolysis to form HPO42). If CaCl2 is added to a saturated solution of Ca3(PO4)2, the Ca2+ ion concentration will increase such that [Ca2+] > 3.42 × 10−7 M, making Q > Ksp. The only way the system can return to equilibrium is for the reaction in Equation $\ref{Eq1}$ to proceed to the left, resulting in precipitation of $\ce{Ca3(PO4)2}$. This will decrease the concentration of both Ca2+ and PO43 until Q = Ksp.
Adding a common ion decreases solubility, as the reaction shifts toward the left to relieve the stress of the excess product. Adding a common ion to a dissociation reaction causes the equilibrium to shift left, toward the reactants, causing precipitation.
Example $5$
Consider the reaction:
$PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq)\nonumber$
What happens to the solubility of PbCl2(s) when 0.1 M NaCl is added?
Solution
$K_{sp}=1.7 \times 10^{-5}\nonumber$
$Q_{sp}= 1.8 \times 10^{-5}\nonumber$
Identify the common ion: Cl-
Notice: Qsp > Ksp The addition of NaCl has caused the reaction to shift out of equilibrium because there are more dissociated ions. Typically, solving for the molarities requires the assumption that the solubility of PbCl2 is equivalent to the concentration of Pb2+ produced because they are in a 1:1 ratio.
Because Ksp for the reaction is 1.7×10-5, the overall reaction would be (s)(2s)2= 1.7×10-5. Solving the equation for s gives s= 1.62×10-2 M. The coefficient on Cl- is 2, so it is assumed that twice as much Cl- is produced as Pb2+, hence the '2s.' The solubility equilibrium constant can be used to solve for the molarities of the ions at equilibrium.
The molarity of Cl- added would be 0.1 M because Na+ and Cl- are in a 1:1 ration in the ionic salt, NaCl. Therefore, the overall molarity of Cl- would be 2s + 0.1, with 2s referring to the contribution of the chloride ion from the dissociation of lead chloride.
$\begin{eqnarray} Q_{sp} &=& [Pb^{2+}][Cl^-]^2\nonumber \ 1.8 \times 10^{-5} &=& (s)(2s + 0.1)^2 \ s &=& [Pb^{2+}]\nonumber \ &=& 1.8 \times 10^{-3} M\nonumber\ 2s &=& [Cl^-]\nonumber\ &\approx & 0.1 M \end{eqnarray}$
Notice that the molarity of Pb2+ is lower when NaCl is added. The equilibrium constant remains the same because of the increased concentration of the chloride ion. To simplify the reaction, it can be assumed that [Cl-] is approximately 0.1M since the formation of the chloride ion from the dissociation of lead chloride is so small. The reaction quotient for PbCl2 is greater than the equilibrium constant because of the added Cl-. This therefore shift the reaction left towards equilibrium, causing precipitation and lowering the current solubility of the reaction. Overall, the solubility of the reaction decreases with the added sodium chloride.
The common ion effect usually decreases the solubility of a sparingly soluble salt.
Example $6$
Calculate the solubility of calcium phosphate [Ca3(PO4)2] in 0.20 M CaCl2.
Given: concentration of CaCl2 solution
Asked for: solubility of Ca3(PO4)2 in CaCl2 solution
Strategy:
1. Write the balanced equilibrium equation for the dissolution of Ca3(PO4)2. Tabulate the concentrations of all species produced in solution.
2. Substitute the appropriate values into the expression for the solubility product and calculate the solubility of Ca3(PO4)2.
Solution
A The balanced equilibrium equation is given in the following table. If we let x equal the solubility of Ca3(PO4)2 in moles per liter, then the change in [Ca2+] is once again +3x, and the change in [PO43] is +2x. We can insert these values into the ICE table.
$Ca_3(PO_4)_{2(s)} \rightleftharpoons 3Ca^{2+}_{(aq)} + 2PO^{3−}_{4(aq)}$
Ca3(PO4)2 [Ca2+] [PO43]
initial pure solid 0.20 0
change +3x +2x
final pure solid 0.20 + 3x 2x
B The Ksp expression is as follows:
Ksp = [Ca2+]3[PO43]2 = (0.20 + 3x)3(2x)2 = 2.07×10−33
Because Ca3(PO4)2 is a sparingly soluble salt, we can reasonably expect that x << 0.20. Thus (0.20 + 3x) M is approximately 0.20 M, which simplifies the Ksp expression as follows:
\begin{align*}K_{\textrm{sp}}=(0.20)^3(2x)^2&=2.07\times10^{-33} \[4pt] x^2&=6.5\times10^{-32} \[4pt] x&=2.5\times10^{-16}\textrm{ M}\end{align*}
This value is the solubility of Ca3(PO4)2 in 0.20 M CaCl2 at 25°C. It is approximately nine orders of magnitude less than its solubility in pure water, as we would expect based on Le Chatelier’s principle. With one exception, this example is identical to Example $2$—here the initial [Ca2+] was 0.20 M rather than 0.
Exercise $4$
Calculate the solubility of silver carbonate in a 0.25 M solution of sodium carbonate. The solubility of silver carbonate in pure water is 8.45 × 10−12 at 25°C.
Answer
2.9 × 10−6 M (versus 1.3 × 10−4 M in pure water)
The Common Ion Effect in Solubility Products: https://youtu.be/_P3wozLs0Tc
Summary
Adding a common cation or common anion to a solution of a sparingly soluble salt shifts the solubility equilibrium in the direction predicted by Le Chatelier’s principle. The solubility of the salt is almost always decreased by the presence of a common ion. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/18%3A_Solubility_and_Complex-Ion_Equilibria/18.3%3A_Common-Ion_Effect_in_Solubility_Equilibria.txt |
Solubility equilibria defined by a single, simple Ks expression are probably the exception rather than the rule. Such equilibria are often in competition with other reactions with such species as H+or OH, complexing agents, oxidation-reduction, formation of other sparingly soluble species or, in the case of carbonates and sulfites, of gaseous products. The exact treatments of these systems can be extremely complicated, involving the solution of large sets of simultaneous equations. For most practical purposes it is sufficient to recognize the general trends, and to carry out approximate calculations.
The solubility product of an ionic compound describes the concentrations of ions in equilibrium with a solid, but what happens if some of the cations become associated with anions rather than being completely surrounded by solvent? Then predictions of the total solubility of the compound based on the assumption that the solute exists solely as discrete ions would differ substantially from the actual solubility, as would predictions of ionic concentrations. In general, four situations explain why the solubility of a compound may be other than expected: ion pair formation, the incomplete dissociation of molecular solutes, the formation of complex ions, and changes in pH.
The Diverse Ion Effect (The Salt Effect)
A sparingly-soluble salt will be more soluble in a solution that contains non-participating ions. This is just the opposite of the common ion effect, and it might at first seem rather counter-intuitive: why would adding more ions of any kind make a salt more likely to dissolve?
Although the concentrations of ions in equilibrium with a sparingly soluble solid are so low that they are essentially the same as the activities, the presence of other ions at concentrations of about 0.001 M or greater can materially reduce the activities of the dissolution products, permitting the solubilities to be greater than what simple equilibrium calculations would predict.
The Diverse Ion Effect typically has the opposite effect on solubility than the common ion effect.
The higher the charge of the foreign ion, the more pronounced is the salt effect. This tells us that inter-ionic (and thus electrostatic) interactions must play a role. The details are rather complicated, but the general idea is that all ions in solution, besides possessing tightly-held waters of hydration, tend to attract oppositely-charged ions ("counter-ions") around them. This "atmosphere" of counterions is always rather diffuse, but much less so (and more tightly bound) when one or both kinds of ions have greater charges. From a distance, these ion-counterion pairs (or ion pairs) bodies appear to be almost electrically neutral, which keeps them from interacting with each other (as to form a precipitate).
Example $1$
How does the solubility of $AgCl$ change in absence and presence of $NaNO_3$, given that the molar solubility product of $AgCl$ is $1.76 \times 10^{−10}$.
Solution
The Uncommon Ion Effect is a "real" effect and requires the activity definition of $K_{sp}$ to address. The exact definition of the solubility constant is:
$K_{sp}=a\{Ag^+\} \cdot a\{Cl^−\}=[Ag+] \gamma(Ag^+) [Cl^−] \gamma (Cl^−) \nonumber$
Where $a\{Ag^+\}$ and $a\{Cl^−\}$ are the activities of these ions in the presence of the $NaNO_3$. The $\gamma$ values are the are the corresponding activity coefficients for the ions.
• In dilute solutions (i.e. absence of $NaNO_3$\) $\gamma (Ag^+)= \gamma (Cl^−)=1 \nonumber$ So, the solubility of $AgCl$ is then $[Ag^+]=\sqrt{K_{sp}} =1.33 \times 10^{−5}$
• In the presence of $NaNO_3$, the activity coefficients deviate from unity. So $[Ag^+] = \dfrac{\sqrt{K_{sp}}}{ \sqrt{\gamma(Ag^+)\gamma (Cl^−) }}$. The denominator has to evaluated from an activity table and is 0.769 so $[Ag^+] = \dfrac{\sqrt{K_{sp}}}{0.769} = 1.73 \times 10^{-5}\]. Hence, in the presence of the \(NaNO_3$ soluble salt, the molar solubility of the slightly soluble $AgCl$ salt increases. An opposite effect is expected if there were a common ions between the two salts.
Incomplete Dissociation of Solutes into Ions
An ion pair consists of a cation and an anion that are in intimate contact in solution, rather than separated by solvent (Figure $1$). The ions in an ion pair are held together by the same attractive electrostatic force in ionic solids. As a result, the ions in an ion pair migrate as a single unit, whose net charge is the sum of the charges on the ions. In many ways, we can view an ion pair as a species intermediate between the ionic solid (in which each ion participates in many cation–anion interactions that hold the ions in a rigid array) and the completely dissociated ions in solution (where each is fully surrounded by water molecules and free to migrate independently).
The overall effect is to reduce the concentrations of the less-shielded ions that are available to combine to form a precipitate. We say that the thermodynamically-effective concentrations of these ions are less than their "analytical" concentrations. Chemists refer to these effective concentrations as ionic activities, and they denote them by curly brackets {Ag+} as opposed to square brackets [Ag+] which refer to the nominal or analytical concentrations.
As illustrated for calcium sulfate in the following equation, a second equilibrium must be included to describe the solubility of salts that form ion pairs:
$\mathrm{CaSO_4(s)}\rightleftharpoons\mathrm{Ca^{2+}}\cdot\underset{\textrm{ion pair}}{\mathrm{SO_4^{2-}(aq)}}\rightleftharpoons \mathrm{Ca^{2+}(aq)}+\mathrm{SO_4^{2-}(aq)} \label{17.5.3.1}$
The ion pair is represented by the symbols of the individual ions separated by a dot, which indicates that they are associated in solution. The formation of an ion pair is a dynamic process, just like any other equilibrium, so a particular ion pair may exist only briefly before dissociating into the free ions, each of which may later associate briefly with other ions.
Ion-pair formation can have a major effect on the measured solubility of a salt. For example, the measured Ksp for calcium sulfate is 4.93 × 10−5 at 25°C. The solubility of CaSO4 should be 7.02 × 10−3 M if the only equilibrium involved were as follows:
$CaSO_{4(s)} \rightleftharpoons Ca^{2+}_{(aq)} + SO^{2−}_{4(aq)} \label{17.5.2}$
In fact, the experimentally measured solubility of calcium sulfate at 25°C is 1.6 × 10−2 M, almost twice the value predicted from its Ksp. The reason for the discrepancy is that the concentration of ion pairs in a saturated CaSO4 solution is almost as high as the concentration of the hydrated ions. Recall that the magnitude of attractive electrostatic interactions is greatest for small, highly charged ions. Hence ion pair formation is most important for salts that contain M2+ and M3+ ions, such as Ca2+ and La3+, and is relatively unimportant for salts that contain monopositive cations, except for the smallest, Li+. We therefore expect a saturated solution of CaSO4 to contain a high concentration of ion pairs and its solubility to be greater than predicted from its Ksp.
Simultaneous Equilibria (Side Reactions)
A molecular solute may also be more soluble than predicted by the measured concentrations of ions in solution due to incomplete dissociation. This is particularly common with weak organic acids. Although strong acids (HA) dissociate completely into their constituent ions (H+ and A) in water, weak acids such as carboxylic acids do not (Ka = 1.5 × 10−5). However, the molecular (undissociated) form of a weak acid (HA) is often quite soluble in water; for example, acetic acid (CH3CO2H) is completely miscible with water. Many carboxylic acids, however, have only limited solubility in water, such as benzoic acid (C6H5CO2H), with Ka = 6.25 × 10−5. Just as with calcium sulfate, we need to include an additional equilibrium to describe the solubility of benzoic acid:
$C_6H_5CO_2H_{(s)} \rightleftharpoons C_6H_5CO_2H_{(aq)} \rightleftharpoons C_6H_5CO^−_{2(aq)} + H^+_{(aq)} \label{17.5.3}$
In a case like this, measuring only the concentration of the ions grossly underestimates the total concentration of the organic acid in solution. In the case of benzoic acid, for example, the pH of a saturated solution at 25°C is 2.85, corresponding to [H+] = [C6H5CO2] = 1.4 × 10−3 M. The total concentration of benzoic acid in the solution, however, is 2.8 × 10−2 M. Thus approximately 95% of the benzoic acid in solution is in the form of hydrated neutral molecules—$C_6H_5CO_2H_{(aq)}$—and only about 5% is present as the dissociated ions (Figure $3$).
Incomplete dissociation of a molecular solute that is miscible with water can increase the solubility of the solute.
Although ion pairs, such as Ca2+•SO42−, and undissociated electrolytes, such as C6H5CO2H, are both electrically neutral, there is a major difference in the forces responsible for their formation. Simple electrostatic attractive forces between the cation and the anion hold the ion pair together, whereas a polar covalent O−H bond holds together the undissociated electrolyte.
Example $2$: $CdI_2$ Solubility
The dissolution of cadmium iodide is water is commonly represented as
$CdI_{2(s)} \rightleftharpoons Cd^{2+} + 2 I^–$
What you were likely taught about the dissociation of salts in water is wrong! To most students (and to most of their teachers!), this would imply that a 0.1M solution of this salt would contain 0.1M of Cd2+(aq) — and this would be seriously in error because it fails to take into account that the two ions react with each other.
Firstly, they combine to form neutral, largely-covalent molecular species:
$Cd^{2+}_{(aq)} + 2 I^–_{(aq)} → CdI_{2(aq)}$
This non-ionic form accounts for 78% of the Cd present in the solution! In addition, they form a molecular ion $CdI^–_{(aq)}$ according to the following scheme:
Table $1$: CdI2 species in solution
$CdI_{2(s)} \rightleftharpoons Cd^{2+} + 2 I^–$ $K_1 = 10^{–3.9}$
$Cd^{2+} + I^– \rightleftharpoons CdI^+$ $K_2= 10^{+2.3}$
$CdI2_{(s)} \rightleftharpoons CdI^++ I^–$ $K = 10^{–1.6} = 0.023$
As a consequence, the concentration of "free" Cd2+(aq) in an aqueous cadmium iodide solution is only about 2% of the value you would calculate by taking K1 as the solubility product. The principal component of such as solution is actually [covalently-bound] CdI2(aq). It turns out that
Many salts, especially those of metals beyond Group 2, are similarly only partially ionized in aqueous solution:
salt molarity % cation other species
Table $2$: Incomplete Dissociation of Select Salts
KCl 0.52 95 KCl(aq) 5%
MgSO4 0.04 58 MgSO4(aq) 42%
CaCl2 0.44 70 CaCl+(aq) 30%
CuSO4 0.045 56 CuSO4(aq) 44%
CdI2 0.50 2 CdI2(aq) 76%, CdI(aq) 22%
FeCl3 0.1 10 FeCl2+(aq) 42%, FeCl2(aq) 40%, FeOH2+(aq) 6%, Fe(OH)2+(aq) 2%
The data shown Tables $1$ and $2$ are taken from the article Salts are Mostly NOT Ionized by Stephen Hawkes: 1996 J Chem Educ. 73(5) 421-423. This fact was stated by Arrhenius in 1887, but has been largely ignored and is rarely mentioned in standard textbooks.
Summary
The solubility constant, as with all equilibrium constants, are properly defined in terms of effective concentrations (activities) and the use of true concentrations to approximate activities in equilibrium constants can often fail. As with all approximations, it is important to understand the conditions that it will fail. three examples are discussed. The salt effect refers to the fact that the presence of a salt which has no ion in common with the solute, has an effect on the ionic strength of the solution and hence on activity coefficients, so that the equilibrium constant, expressed as a concentration quotient, changes. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/18%3A_Solubility_and_Complex-Ion_Equilibria/18.4%3A_Limitations_of_the_Ksp_Concept.txt |
The equation that describes the equilibrium between solid calcium carbonate and its solvated ions is:
$\ce{CaCO3}(s)⇌\ce{Ca^2+}(aq)+\ce{CO3^2-}(aq)$
We can establish this equilibrium either by adding solid calcium carbonate to water or by mixing a solution that contains calcium ions with a solution that contains carbonate ions. If we add calcium carbonate to water, the solid will dissolve until the concentrations are such that the value of the reaction quotient $\ce{(Q=[Ca^2+][CO3^2- ])}$ is equal to the solubility product (Ksp = 4.8 × 10–9). If we mix a solution of calcium nitrate, which contains Ca2+ ions, with a solution of sodium carbonate, which contains $\ce{CO3^2-}$ ions, the slightly soluble ionic solid CaCO3 will precipitate, provided that the concentrations of Ca2+ and $\ce{CO3^2-}$ ions are such that Q is greater than Ksp for the mixture. The reaction shifts to the left and the concentrations of the ions are reduced by formation of the solid until the value of Q equals Ksp. A saturated solution in equilibrium with the undissolved solid will result. If the concentrations are such that Q is less than Ksp, then the solution is not saturated and no precipitate will form.
We can compare numerical values of Q with Ksp to predict whether precipitation will occur, as Example $7$ shows. (Note: Since all forms of equilibrium constants are temperature dependent, we will assume a room temperature environment going forward in this chapter unless a different temperature value is explicitly specified.)
The first step in the preparation of magnesium metal is the precipitation of Mg(OH)2 from sea water by the addition of lime, Ca(OH)2, a readily available inexpensive source of OH ion:
$\ce{Mg(OH)2}(s)⇌\ce{Mg^2+}(aq)+\ce{2OH-}(aq) \hspace{20px} K_\ce{sp}=2.1×10^{−13}$
The concentration of Mg2+(aq) in sea water is 0.0537 M. Will Mg(OH)2 precipitate when enough Ca(OH)2 is added to give a [OH] of 0.0010 M?
Solution
This problem asks whether the reaction:
$\ce{Mg(OH)2}(s)⇌\ce{Mg^2+}(aq)+\ce{2OH-}(aq)$
shifts to the left and forms solid Mg(OH)2 when [Mg2+] = 0.0537 M and [OH] = 0.0010 M. The reaction shifts to the left if Q is greater than Ksp. Calculation of the reaction quotient under these conditions is shown here:
$\mathrm{Q=[Mg^{2+}][OH^-]^2=(0.0537)(0.0010)^2=5.4×10^{−8}}$
Because Q is greater than Ksp (Q = 5.4 × 10–8 is larger than Ksp = 2.1 × 10–13), we can expect the reaction to shift to the left and form solid magnesium hydroxide. Mg(OH)2(s) forms until the concentrations of magnesium ion and hydroxide ion are reduced sufficiently so that the value of Q is equal to Ksp.
Exercise $1$
Use the solubility products in Appendix J to determine whether CaHPO4 will precipitate from a solution with [Ca2+] = 0.0001 M and $\ce{[HPO4^2- ]}$ = 0.001 M.
Answer
No precipitation of CaHPO4; Q = 1 × 10–7, which is less than Ksp
Precipitation of AgCl upon Mixing Solutions
Does silver chloride precipitate when equal volumes of a 2.0 × 10–4-M solution of AgNO3 and a 2.0 × 10–4-M solution of NaCl are mixed?
(Note: The solution also contains Na+ and $\ce{NO3-}$ ions, but when referring to solubility rules, one can see that sodium nitrate is very soluble and cannot form a precipitate.)
Solution
The equation for the equilibrium between solid silver chloride, silver ion, and chloride ion is:
$\ce{AgCl}(s)⇌\ce{Ag+}(aq)+\ce{Cl-}(aq)$
The solubility product is 1.8 × 10–10 (see Appendix J).
AgCl will precipitate if the reaction quotient calculated from the concentrations in the mixture of AgNO3 and NaCl is greater than Ksp. The volume doubles when we mix equal volumes of AgNO3 and NaCl solutions, so each concentration is reduced to half its initial value. Consequently, immediately upon mixing, [Ag+] and [Cl] are both equal to:
$\dfrac{1}{2}(2.0×10^{−4})\:M=1.0×10^{−4}\:M$
The reaction quotient, Q, is momentarily greater than Ksp for AgCl, so a supersaturated solution is formed:
$Q=\ce{[Ag+][Cl- ]}=(1.0×10^{−4})(1.0×10^{−4})=1.0×10^{−8}>K_\ce{sp}$
Since supersaturated solutions are unstable, AgCl will precipitate from the mixture until the solution returns to equilibrium, with Q equal to Ksp.
Exercise $2$
Will KClO4 precipitate when 20 mL of a 0.050-M solution of K+ is added to 80 mL of a 0.50-M solution of $\ce{ClO4-}$? (Remember to calculate the new concentration of each ion after mixing the solutions before plugging into the reaction quotient expression.)
Answer
No, Q = 4.0 × 10–3, which is less than Ksp = 1.07 × 10–2
In the previous two examples, we have seen that Mg(OH)2 or AgCl precipitate when Q is greater than Ksp. In general, when a solution of a soluble salt of the Mm+ ion is mixed with a solution of a soluble salt of the Xn– ion, the solid, MpXq precipitates if the value of Q for the mixture of Mm+ and Xn– is greater than Ksp for MpXq. Thus, if we know the concentration of one of the ions of a slightly soluble ionic solid and the value for the solubility product of the solid, then we can calculate the concentration that the other ion must exceed for precipitation to begin. To simplify the calculation, we will assume that precipitation begins when the reaction quotient becomes equal to the solubility product constant.
Precipitation of Calcium Oxalate
Blood will not clot if calcium ions are removed from its plasma. Some blood collection tubes contain salts of the oxalate ion, $\ce{C2O4^2-}$, for this purpose (Figure $4$). At sufficiently high concentrations, the calcium and oxalate ions form solid, CaC2O4•H2O (which also contains water bound in the solid). The concentration of Ca2+ in a sample of blood serum is 2.2 × 10–3 M. What concentration of $\ce{C2O4^2-}$ ion must be established before CaC2O4•H2O begins to precipitate?
Solution
The equilibrium expression is:
$\ce{CaC2O4}(s)⇌\ce{Ca^2+}(aq)+\ce{C2O4^2-}(aq)$
For this reaction:
$K_\ce{sp}=\ce{[Ca^2+][C2O4^2- ]}=2.27×10^{−9}$
(see Appendix J)
CaC2O4 does not appear in this expression because it is a solid. Water does not appear because it is the solvent.
Solid CaC2O4 does not begin to form until Q equals Ksp. Because we know Ksp and [Ca2+], we can solve for the concentration of $\ce{C2O4^2-}$ that is necessary to produce the first trace of solid:
$Q=K_\ce{sp}=\ce{[Ca^2+][C2O4^2- ]}=2.27×10^{−9}$
$(2.2×10^{−3})\ce{[C2O4^2- ]}=2.27×10^{−9}$
$\ce{[C2O4^2- ]}=\dfrac{2.27×10^{−9}}{2.2×10^{−3}}=1.0×10^{−6}$
A concentration of $\ce{[C2O4^2- ]}$ = 1.0 × 10–6 M is necessary to initiate the precipitation of CaC2O4 under these conditions.
Exercise $3$
If a solution contains 0.0020 mol of $\ce{CrO4^2-}$ per liter, what concentration of Ag+ ion must be reached by adding solid AgNO3 before Ag2CrO4 begins to precipitate? Neglect any increase in volume upon adding the solid silver nitrate.
Answer:
7.0 × 10–5 M
It is sometimes useful to know the concentration of an ion that remains in solution after precipitation. We can use the solubility product for this calculation too: If we know the value of Ksp and the concentration of one ion in solution, we can calculate the concentration of the second ion remaining in solution. The calculation is of the same type as that in Example—calculation of the concentration of a species in an equilibrium mixture from the concentrations of the other species and the equilibrium constant. However, the concentrations are different; we are calculating concentrations after precipitation is complete, rather than at the start of precipitation.
Concentrations Following Precipitation
Clothing washed in water that has a manganese [Mn2+(aq)] concentration exceeding 0.1 mg/L (1.8 × 10–6 M) may be stained by the manganese upon oxidation, but the amount of Mn2+ in the water can be reduced by adding a base. If a person doing laundry wishes to add a buffer to keep the pH high enough to precipitate the manganese as the hydroxide, Mn(OH)2, what pH is required to keep [Mn2+] equal to 1.8 × 10–6 M?
Solution The dissolution of Mn(OH)2 is described by the equation:
$\ce{Mn(OH)2}(s)⇌\ce{Mn^2+}(aq)+\ce{2OH-}(aq) \hspace{20px} K_\ce{sp}=4.5×10^{−14}$
We need to calculate the concentration of OH when the concentration of Mn2+ is 1.8 × 10–6 M. From that, we calculate the pH. At equilibrium:
$K_\ce{sp}=\ce{[Mn^2+][OH- ]^2}$
or
$(1.8×10^{−6})\ce{[OH- ]^2}=4.5×10^{−14}$
so
$\ce{[OH- ]}=1.6×10^{−4}\:M$
Now we calculate the pH from the pOH:
$\mathrm{pOH=−\log[OH^-]=−\log(1.6×10^{−4})=3.80}$
$\mathrm{pH=14.00−pOH=14.00−3.80=10.20}$
If the person doing laundry adds a base, such as the sodium silicate (Na4SiO4) in some detergents, to the wash water until the pH is raised to 10.20, the manganese ion will be reduced to a concentration of 1.8 × 10–6 M; at that concentration or less, the ion will not stain clothing.
Exercise $4$
The first step in the preparation of magnesium metal is the precipitation of Mg(OH)2 from sea water by the addition of Ca(OH)2. The concentration of Mg2+(aq) in sea water is 5.37 × 10–2 M. Calculate the pH at which [Mg2+] is diminished to 1.0 × 10–5 M by the addition of Ca(OH)2.
Answer
11.09
Due to their light sensitivity, mixtures of silver halides are used in fiber optics for medical lasers, in photochromic eyeglass lenses (glass lenses that automatically darken when exposed to sunlight), and—before the advent of digital photography—in photographic film. Even though AgCl (Ksp = 1.6 × 10–10), AgBr (Ksp = 7.7 × 10–13), and AgI (Ksp = 8.3 × 10–17) are each quite insoluble, we cannot prepare a homogeneous solid mixture of them by adding Ag+ to a solution of Cl, Br, and I; essentially all of the AgI will precipitate before any of the other solid halides form because of its smaller value for Ksp. However, we can prepare a homogeneous mixture of the solids by slowly adding a solution of Cl, Br, and I to a solution of Ag+.
When two anions form slightly soluble compounds with the same cation, or when two cations form slightly soluble compounds with the same anion, the less soluble compound (usually, the compound with the smaller Ksp) generally precipitates first when we add a precipitating agent to a solution containing both anions (or both cations). When the Ksp values of the two compounds differ by two orders of magnitude or more (e.g., 10–2 vs. 10–4), almost all of the less soluble compound precipitates before any of the more soluble one does. This is an example of selective precipitation, where a reagent is added to a solution of dissolved ions causing one of the ions to precipitate out before the rest.
Determining if a Precipitate forms (The Ion Product): https://youtu.be/Naf7PoHPz8Y
Summary
A slightly soluble electrolyte begins to precipitate when the magnitude of the reaction quotient for the dissolution reaction exceeds the magnitude of the solubility product. Precipitation continues until the reaction quotient equals the solubility product. A reagent can be added to a solution of ions to allow one ion to selectively precipitate out of solution. The common ion effect can also play a role in precipitation reactions. In the presence of an ion in common with one of the ions in the solution, Le Chatelier’s principle applies and more precipitate comes out of solution so that the molar solubility is reduced.
Glossary
common ion effect
effect on equilibrium when a substance with an ion in common with the dissolved species is added to the solution; causes a decrease in the solubility of an ionic species, or a decrease in the ionization of a weak acid or base
molar solubility
solubility of a compound expressed in units of moles per liter (mol/L)
selective precipitation
process in which ions are separated using differences in their solubility with a given precipitating reagent
solubility product (Ksp)
equilibrium constant for the dissolution of a slightly soluble electrolyte | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/18%3A_Solubility_and_Complex-Ion_Equilibria/18.5%3A_Criteria_for_Precipitation_and_its_Completeness.txt |
Learning Objectives
• Calculate ion concentrations to maintain a heterogeneous equilibrium.
• Calculate pH required to precipitate a metal hydroxide.
• Design experiments to separate metal ions in a solution of mixtures of metals.
A mixture of metal ions in a solution can be separated by precipitation with anions such as $\ce{Cl-}$, $\ce{Br-}$, $\ce{SO4^2-}$, $\ce{CO3^2-}$, $\ce{S^2-}$, $\ce{Cr2O4^2-}$, $\ce{PO4^2-}$, $\ce{OH-}$ etc. When a metal ion or a group of metal ions form insoluble salts with a particular anion, they can be separated from others by precipitation. We can also separate the anions by precipitating them with appropriate metal ions. There are no definite dividing lines between insoluble salts, sparingly soluble, and soluble salts, but concentrations of their saturated solutions are small, medium, and large. Solubility products are usually listed for insoluble and sparingly soluble salts, but they are not given for soluble salts. Solubility products for soluble salts are very large.
What type of salts are usually soluble, sparingly soluble and insoluble? The following are some general guidelines, but these are not precise laws.
• All nitrates are soluble. The singly charged large $\ce{NO3-}$ ions form salts with high solubilities. So do $\ce{ClO4-}$, $\ce{ClO3-}$, $\ce{NO2-}$, $\ce{HCOO-}$, and $\ce{CH3COO-}$.
• All chlorides, bromides, and iodides are soluble except those of $\ce{Ag+}$, $\ce{Hg2^2+}$, and $\ce{Pb^2+}$. $\ce{CaF2}$, $\ce{BaF2}$, and $\ce{PbF2}$ are also insoluble.
• All sulfates are soluble, except those of $\ce{Ba^2+}$, $\ce{Sr^2+}$, and $\ce{Pb^2+}$. The doubly charged sulfates are usually less soluble than halides and nitrates.
• Most singly charge cations $\ce{K+}$, $\ce{Na+}$, $\ce{NH4+}$ form soluble salts. However, $\ce{K3Co(NO2)6}$ and $\ce{(NH4)3Co(NO2)6}$ are insoluble.
These are handy rules for us to have if we deal with salts often. On the other hand, solubility is an important physical property of a substance, and these properties are listed in handbooks.
Chemical Separation of Metal Ions
Formation of crystals from a saturated solution is a heterogeneous equilibrium phenomenon, and it can be applied to separate various chemicals or ions in a solution. When solubilities of two metal salts are very different, they can be separated by precipitation. The Ksp values for various salts are valuable information, and some data are given in Table E3. In the first two examples, we show how barium and strontium can be separated as chromate.
Example $1$
The Ksp for strontium chromate is $3.6 \times 10^{-5}$ and the Ksp for barium chromate is $1.2 \times10^{-10}$. What concentration of potassium chromate will precipitate the maximum amount of either the barium or the strontium chromate from an equimolar 0.30 M solution of barium and strontium ions without precipitating the other?
Solution
Since the Ksp for barium chromate is smaller, we know that $\ce{BaCrO4}$ will form a precipitate first as $\ce{[CrO4^2- ]}$ increases so that Qsp for $\ce{BaCrO4}$ also increases from zero to Ksp of $\ce{BaCrO4}$, at which point, $\ce{BaCrO4}$ precipitates. As $\ce{[CrO4^2- ]}$ increases, $\ce{[Ba^2+]}$ decreases. Further increase of $\ce{[CrO4^2- ]}$ till Qsp for $\ce{SrCrO4}$ increases to Ksp of $\ce{SrCrO4}$; it then precipitates.
Let us write the equilibrium equations and data down to help us think. Let $x$ be the concentration of chromate to precipitate $\ce{Sr^2+}$, and $y$ be that to precipitate $\ce{Ba^2+}$:
$\ce{SrCrO4(s) \rightarrow Sr^{2+}(aq) + CrO4^{2-}(aq)} \nonumber$
According to the definition of Ksp we have we have $K_{\ce{sp}} = (0.30)(x) = 3.6 \times 10^{-5}$. Solving for $x$ gives
$x = \dfrac{3.6 \times 10^{-5}}{0.30} = 1.2 \times 10^{-4} M \nonumber$
Further, let $y$ be the concentration of chromate to precipitate precipitate $\ce{Ba^2+}$:
$\ce{BaCrO4(s) \rightarrow Ba^{2+}(aq) + CrO4^{2-}(aq)} \nonumber$
with $K_{\ce{sp}} = (0.30)(y) = 1.2 \times 10^{-10}$. Solving for $y$ gives
$y = \dfrac{1.2 \times 10^{-10}}{0.30} = 4.0 \times 10^{-10} \;M \nonumber$
The Ksp's for the two salts indicate $\ce{BaCrO4}$ to be much less soluble, and it will precipitate before any $\ce{SrCrO4}$ precipitates. If chromate concentration is maintained less than $1.2 \times 10^{-4} M$, then all $\ce{Sr^2+}$ ions will remain in the solution.
Discussion
In reality, controling the increase of $\ce{[CrO4^2- ]}$ is very difficult.
Example $2$
The Ksp for strontium chromate is $3.6\times 10^{-5}$ and the Ksp for barium chromate is $1.2\times 10^{-10}$. Potassium chromate is added a small amount at a time to first precipitate $\ce{BaCrO4}$. Calculate $\ce{[Ba^2+]}$ when the first trace of $\ce{SrCrO4}$ precipitate starts to form in a solution that contains 0.30 M each of $\ce{Ba^2+}$ and $\ce{Sr^2+}$ ions.
Solution
From the solution given in Example $1$, $\ce{[CrO4^2- ]} = 3.6\times 10^{-4}\; M$ when $\ce{SrCrO4}$ starts to form. At this concentration, the $\ce{[Ba^2+]}$ is estimated at $3.6 \times 10^{-4} = 1.2\times 10^{-10}$.
The Ksp of $\ce{BaCrO4}$.
Thus,
$\ce{[Ba^2+]} = 3.33 \times 10^{-7}\, M \nonumber$
Very small indeed, compared to 0.30. In the fresh precipitate of $\ce{SrCrO4}$, the molar ratio of $\ce{SrCrO4}$ to $\ce{BaCrO4}$ is
$\dfrac{0.30}{3.33 \times 10^{-7}} = 9.0 \times 10^{5}. \nonumber$
Hence, the amount of $\ce{Ba^2+}$ ion in the solid is only $1 \times 10^{-6}$ (i.e., 1 ppm) of all metal ions, providing that all the solid was removed when
$\ce{[CrO4^{2-}]} = 3.6 \times 10^{-4} M. \nonumber$
Discussion
The calculation shown here indicates that the separation of $\ce{Sr}$ and $\ce{Ba}$ is pretty good. In practice, an impurity level of 1 ppm is a very small value.
Example $3$
What reagent should you use to separate silver and lead ions that are present in a solution? What data or information will be required for this task?
Solution
The Ksp's for salts of silver and lead are required. We list the Ksp's for chlorides and sulfates in a table here. These value are found in the Handbook Menu of our website as Salts Ksp.
Solutions to Example 17.6.3
Salt Ksp Salt Ksp
$\ce{AgCl}$ $1.8 \times 10^{-10}$ $\ce{Ag2SO4}$ $1.4\times 10^{-5}$
$\ce{Hg2Cl2}$ $1.3\times 10^{-18}$ $\ce{BaSO4}$ $1.1\times 10^{-10}$
$\ce{PbCl2}$ $1.7 \times 10^{-5}$ $\ce{CaSO4}$ $2.4\times 10^{-5}$
$\ce{PbSO4}$ $6.3\times 10^{-7}$
$\ce{SrSO4}$ $3.2\times 10^{-7}$
Because the Ksp's $\ce{AgCl}$ and $\ce{PbCl2}$ are very different, chloride, $\ce{Cl-}$, apppears a good choice of negative ions for their separation.
The literature also indicates that $\ce{PbCl2}$ is rather soluble in warm water, and by heating the solution to 350 K (80oC), you can keep $\ce{Pb^2+}$ ions in solution and precipitate $\ce{AgCl}$ as a solid. The solubility of $\ce{AgCl}$ is very small even at high temperatures.
Discussion
Find more detailed information about the solubility of lead chloride as a function of temperature.
Can sulfate be used to separate silver and lead ions? Which one will form a precipitate first as the sulfate ion concentration increases? What is the $\ce{[Pb^2+]}$ when $\ce{Ag2SO4}$ begins to precipitate in a solution that contains 0.10 M $\ce{Ag+}$?
The Separation of Two Ions by a Difference in Solubility: The Separation of Two Ions by a Difference in Solubility(opens in new window) [youtu.be] | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/18%3A_Solubility_and_Complex-Ion_Equilibria/18.6%3A_Fractional_Precipitation.txt |
Learning Objectives
• To understand why the solubility of many compounds depends on pH.
The solubility of many compounds depends strongly on the pH of the solution. For example, the anion in many sparingly soluble salts is the conjugate base of a weak acid that may become protonated in solution. In addition, the solubility of simple binary compounds such as oxides and sulfides, both strong bases, is often dependent on pH. In this section, we discuss the relationship between the solubility of these classes of compounds and pH.
The Effect of Acid–Base Equilibria the Solubility of Salts
We begin our discussion by examining the effect of pH on the solubility of a representative salt, $\ce{M^{+}A^{−}}$, where $\ce{A^{−}}$ is the conjugate base of the weak acid $\ce{HA}$. When the salt dissolves in water, the following reaction occurs:
$\ce{MA (s) \rightleftharpoons M^{+} (aq) + A^{-} (aq)} \label{17.13a} \nonumber$
with
$K_{sp} = [\ce{M^{+}}][\ce{A^{−}}] \label{17.13b} \nonumber$
The anion can also react with water in a hydrolysis reaction:
$\ce{A^{-} (aq) + H2O (l) \rightleftharpoons OH^{-} (aq) + HA (aq)} \label{17.14}$
Because of the reaction described in Equation $\ref{17.14}$, the predicted solubility of a sparingly soluble salt that has a basic anion such as S2−, PO43, or CO32 is increased. If instead a strong acid is added to the solution, the added H+ will react essentially completely with A to form HA. This reaction decreases [A], which decreases the magnitude of the ion product
$Q = [\ce{M^{+}}][\ce{A^{-}}]$
According to Le Chatelier’s principle, more MA will dissolve until $Q = K_{sp}$. Hence an acidic pH dramatically increases the solubility of virtually all sparingly soluble salts whose anion is the conjugate base of a weak acid. In contrast, pH has little to no effect on the solubility of salts whose anion is the conjugate base of a stronger weak acid or a strong acid, respectively (e.g., chlorides, bromides, iodides, and sulfates). For example, the hydroxide salt Mg(OH)2 is relatively insoluble in water:
$Mg(OH)_{2(s)} \rightleftharpoons Mg^{2+} (aq) + 2OH^− (aq) \label{17.15a}$
with
$K_{sp} = 5.61 \times 10^{−12} \label{17.15b}$
When acid is added to a saturated solution that contains excess solid Mg(OH)2, the following reaction occurs, removing OH from solution:
$H^+ (aq) + OH^− (aq) \rightarrow H_2O (l) \label{17.16}$
The overall equation for the reaction of Mg(OH)2 with acid is thus
$Mg(OH)_{2(s)} + 2H^+ (aq) \rightleftharpoons Mg^{2+} (aq) + 2H_2O (l) \label{17.17}$
As more acid is added to a suspension of Mg(OH)2, the equilibrium shown in Equation $\ref{17.17}$ is driven to the right, so more Mg(OH)2 dissolves.
Such pH-dependent solubility is not restricted to salts that contain anions derived from water. For example, CaF2 is a sparingly soluble salt:
$CaF_{2(s)} \rightleftharpoons Ca^{2+} (aq) + 2F^− (aq) \label{17.18a}$
with
$K_{sp} = 3.45 \times 10^{−11} \label{17.18b}$
When strong acid is added to a saturated solution of CaF2, the following reaction occurs:
$H^+ (aq) + F^− (aq) \rightleftharpoons HF (aq) \label{17.19}$
Because the forward reaction decreases the fluoride ion concentration, more CaF2 dissolves to relieve the stress on the system. The net reaction of CaF2 with strong acid is thus
$CaF_{2(s)} + 2H^+ (aq) \rightarrow Ca^{2+} (aq) + 2HF (aq) \label{17.20}$
Example $1$ shows how to calculate the solubility effect of adding a strong acid to a solution of a sparingly soluble salt.
Sparingly soluble salts derived from weak acids tend to be more soluble in an acidic solution.
Example $1$
Lead oxalate (PbC2O4), lead iodide (PbI2), and lead sulfate (PbSO4) are all rather insoluble, with Ksp values of 4.8 × 10−10, 9.8 × 10−9, and 2.53 × 10−8, respectively. What effect does adding a strong acid, such as perchloric acid, have on their relative solubilities?
Given: Ksp values for three compounds
Asked for: relative solubilities in acid solution
Strategy:
Write the balanced chemical equation for the dissolution of each salt. Because the strongest conjugate base will be most affected by the addition of strong acid, determine the relative solubilities from the relative basicity of the anions.
Solution
The solubility Equilibria for the three salts are as follows:
$PbC_2O_{4(s)} \rightleftharpoons Pb^{2+} (aq) + C_2O^{2−}_{4(aq)} \nonumber$
$PbI_{2(s)} \rightleftharpoons Pb^{2+} (aq) + 2I^− (aq) \nonumber$
$PbSO_{4(s)} \rightleftharpoons Pb^{2+} (aq) + SO^{2−}_{4(aq)} \nonumber$
The addition of a strong acid will have the greatest effect on the solubility of a salt that contains the conjugate base of a weak acid as the anion. Because HI is a strong acid, we predict that adding a strong acid to a saturated solution of PbI2 will not greatly affect its solubility; the acid will simply dissociate to form H+(aq) and the corresponding anion. In contrast, oxalate is the fully deprotonated form of oxalic acid (HO2CCO2H), which is a weak diprotic acid (pKa1 = 1.23 and pKa2 = 4.19). Consequently, the oxalate ion has a significant affinity for one proton and a lower affinity for a second proton. Adding a strong acid to a saturated solution of lead oxalate will result in the following reactions:
$C_2O^{2−}_{4(aq)} + H^+ (aq) \rightarrow HO_2CCO^−_{2(aq)} \nonumber$
$HO_2CCO^−_{2(aq)} + H^+ (aq) \rightarrow HO_2CCO_2H (aq) \nonumber$
These reactions will decrease [C2O42], causing more lead oxalate to dissolve to relieve the stress on the system. The pKa of HSO4 (1.99) is similar in magnitude to the pKa1 of oxalic acid, so adding a strong acid to a saturated solution of PbSO4 will result in the following reaction:
$SO^{2-}_{4(aq)} + H^+ (aq) \rightleftharpoons HSO^-_{4(aq)} \nonumber$
Because HSO4 has a pKa of 1.99, this reaction will lie largely to the left as written. Consequently, we predict that the effect of added strong acid on the solubility of PbSO4 will be significantly less than for PbC2O4.
Exercise $1$
Which of the following insoluble salts—AgCl, Ag2CO3, Ag3PO4, and/or AgBr—will be substantially more soluble in 1.0 M HNO3 than in pure water?
Answer
Ag2CO3 and Ag3PO4
Solubility Products and pH: https://youtu.be/XJ0s5SATZgQ
Caves and their associated pinnacles and spires of stone provide one of the most impressive examples of pH-dependent solubility Equilbria(part (a) in Figure $1$:). Perhaps the most familiar caves are formed from limestone, such as Carlsbad Caverns in New Mexico, Mammoth Cave in Kentucky, and Luray Caverns in Virginia. The primary reactions that are responsible for the formation of limestone caves are as follows:
$\ce{CO2(aq) + H2O (l) \rightleftharpoons H^{+} (aq) + HCO^{−}3(aq)} \label{17.21}$
$\ce{HCO^{−}3(aq) \rightleftharpoons H^{+} (aq) + CO^{2-}3(aq)} \label{17.22}$
$\ce{Ca^{2+} (aq) + CO^{2−}3(aq) \rightleftharpoons CaCO3(s)} \label{17.23}$
Limestone deposits that form caves consist primarily of CaCO3 from the remains of living creatures such as clams and corals, which used it for making structures such as shells. When a saturated solution of CaCO3 in CO2-rich water rises toward Earth’s surface or is otherwise heated, CO2 gas is released as the water warms. CaCO3 then precipitates from the solution according to the following equation (part (b) in Figure $1$:):
$Ca^{2+} (aq) + 2HCO^−_{3(aq)} \rightleftharpoons CaCO_{3(s)} + CO_{2(g)} + H_2O (l) \label{17.24}$
The forward direction is the same reaction that produces the solid called scale in teapots, coffee makers, water heaters, boilers, and other places where hard water is repeatedly heated.
When groundwater-containing atmospheric CO2 (Equations $\ref{17.21}$ and $\ref{17.22}$) finds its way into microscopic cracks in the limestone deposits, CaCO3 dissolves in the acidic solution in the reverse direction of Equation $\ref{17.24}$. The cracks gradually enlarge from 10–50 µm to 5–10 mm, a process that can take as long as 10,000 yr. Eventually, after about another 10,000 yr, a cave forms. Groundwater from the surface seeps into the cave and clings to the ceiling, where the water evaporates and causes the equilibrium in Equation $\ref{17.24}$ to shift to the right. A circular layer of solid CaCO3 is deposited, which eventually produces a long, hollow spire of limestone called a stalactite that grows down from the ceiling. Below, where the droplets land when they fall from the ceiling, a similar process causes another spire, called a stalagmite, to grow up. The same processes that carve out hollows below ground are also at work above ground, in some cases producing fantastically convoluted landscapes like that of Yunnan Province in China (Figure $2$).
Acidic, Basic, and Amphoteric Oxides and Hydroxides
One of the earliest classifications of substances was based on their solubility in acidic versus basic solution, which led to the classification of oxides and hydroxides as being either basic or acidic. Basic oxides and hydroxides either react with water to produce a basic solution or dissolve readily in aqueous acid. Acidic oxides or hydroxides either react with water to produce an acidic solution or are soluble in aqueous base. There is a clear correlation between the acidic or the basic character of an oxide and the position of the element combined with oxygen in the periodic table. Oxides of metallic elements are generally basic oxides, and oxides of nonmetallic elements are acidic oxides. Compare, for example, the reactions of a typical metal oxide, cesium oxide, and a typical nonmetal oxide, sulfur trioxide, with water:
$Cs_2O (s) + H_2O (l) \rightarrow 2Cs^+ (aq) + 2OH^− (aq) \label{17.25}$
$SO_{3(g)} + H_2O (l) \rightarrow H_2SO_{4(aq)} \label{17.26}$
Cesium oxide reacts with water to produce a basic solution of cesium hydroxide, whereas sulfur trioxide reacts with water to produce a solution of sulfuric acid—very different behaviors indeed
Metal oxides generally react with water to produce basic solutions, whereas nonmetal oxides produce acidic solutions.
The difference in reactivity is due to the difference in bonding in the two kinds of oxides. Because of the low electronegativity of the metals at the far left in the periodic table, their oxides are best viewed as containing discrete Mn+ cations and O2− anions. At the other end of the spectrum are nonmetal oxides; due to their higher electronegativities, nonmetals form oxides with covalent bonds to oxygen. Because of the high electronegativity of oxygen, however, the covalent bond between oxygen and the other atom, E, is usually polarized: Eδ+–Oδ−. The atom E in these oxides acts as a Lewis acid that reacts with the oxygen atom of water to produce an oxoacid. Oxides of metals in high oxidation states also tend to be acidic oxides for the same reason: they contain covalent bonds to oxygen. An example of an acidic metal oxide is MoO3, which is insoluble in both water and acid but dissolves in strong base to give solutions of the molybdate ion (MoO42):
$MoO_{3(s)} + 2OH^− (aq) \rightarrow MoO^{2−}_{4(aq)} + H_2O (l) \label{17.27}$
As shown in Figure $3$, there is a gradual transition from basic metal oxides to acidic nonmetal oxides as we go from the lower left to the upper right in the periodic table, with a broad diagonal band of oxides of intermediate character separating the two extremes. Many of the oxides of the elements in this diagonal region of the periodic table are soluble in both acidic and basic solutions; consequently, they are called amphoteric oxides (from the Greek ampho, meaning “both,” as in amphiprotic). Amphoteric oxides either dissolve in acid to produce water or dissolve in base to produce a soluble complex. As shown in Video $1$, for example, mixing the amphoteric oxide Cr(OH)3 (also written as Cr2O3•3H2O) with water gives a muddy, purple-brown suspension. Adding acid causes the Cr(OH)3 to dissolve to give a bright violet solution of Cr3+(aq), which contains the [Cr(H2O)6]3+ ion, whereas adding strong base gives a green solution of the [Cr(OH)4] ion. The chemical equations for the reactions are as follows:
$\mathrm{Cr(OH)_3(s)}+\mathrm{3H^+(aq)}\rightarrow\underset{\textrm{violet}}{\mathrm{Cr^{3+}(aq)}}+\mathrm{3H_2O(l)} \label{17.28}$
$\mathrm{Cr(OH)_3(s)}+\mathrm{OH^-(aq)}\rightarrow\underset{\textrm{green}}{\mathrm{[Cr(OH)_4]^-}}\mathrm{(aq)}\label{17.29}$
Video $1$: Chromium(III) Hydroxide [Cr(OH)3 or Cr2O3•3H2O] is an Example of an Amphoteric Oxide. All three beakers originally contained a suspension of brownish purple Cr(OH)3(s) (center). When concentrated acid (6 M H2SO4) was added to the beaker on the left, Cr(OH)3 dissolved to produce violet [Cr(H2O)6]3+ ions and water. The addition of concentrated base (6 M NaOH) to the beaker on the right caused Cr(OH)3 to dissolve, producing green [Cr(OH)4]ions. For a more complete description, see https://www.youtube.com/watch?v=IQNcLH6OZK0
Example $2$
Aluminum hydroxide, written as either Al(OH)3 or Al2O3•3H2O, is amphoteric. Write chemical equations to describe the dissolution of aluminum hydroxide in (a) acid and (b) base.
Given: amphoteric compound
Asked for: dissolution reactions in acid and base
Strategy:
Using Equations $\ref{17.28}$ and $\ref{17.29}$ as a guide, write the dissolution reactions in acid and base solutions.
Solution
1. An acid donates protons to hydroxide to give water and the hydrated metal ion, so aluminum hydroxide, which contains three OH ions per Al, needs three H+ ions:
$Al(OH)_{3(s)} + 3H^+ (aq) \rightarrow Al^{3+} (aq) + 3H_2O (l) \nonumber$
In aqueous solution, Al3+ forms the complex ion [Al(H2O)6]3+.
1. In basic solution, OH is added to the compound to produce a soluble and stable poly(hydroxo) complex:
$Al(OH)_{3(s)} + OH^− (aq) \rightarrow [Al(OH)_4]^− (aq) \nonumber$
Exercise $2$
Copper(II) hydroxide, written as either Cu(OH)2 or CuO•H2O, is amphoteric. Write chemical equations that describe the dissolution of cupric hydroxide both in an acid and in a base.
Answer
$Cu(OH)_{2(s)} + 2H^+ (aq) \rightarrow Cu^{2+} (aq) + 2H_2O (l) \nonumber$
$Cu(OH)_{2(s)} + 2OH^− (aq) \rightarrow [Cu(OH)_4]^2_{−(aq)} \nonumber$
Selective Precipitation Using pH
Many dissolved metal ions can be separated by the selective precipitation of the cations from solution under specific conditions. In this technique, pH is often used to control the concentration of the anion in solution, which controls which cations precipitate.
The concentration of anions in solution can often be controlled by adjusting the pH, thereby allowing the selective precipitation of cations.
Suppose, for example, we have a solution that contains 1.0 mM Zn2+ and 1.0 mM Cd2+ and want to separate the two metals by selective precipitation as the insoluble sulfide salts, ZnS and CdS. The relevant solubility equilbria can be written as follows:
$ZnS (s) \rightleftharpoons Zn^{2+} (aq) + S^{2−} (aq) \label{17.30a}$
with
$K_{sp}= 1.6 \times 10^{−24} \label{17.30b}$
and
$CdS (s) \rightleftharpoons Cd^{2+} (aq) + S^{2−} (aq) \label{17.31a}$
with
$K_{sp} = 8.0 \times 10^{−27} \label{17.31b}$
Because the S2− ion is quite basic and reacts extensively with water to give HS and OH, the solubility equilbria are more accurately written as $MS (s) \rightleftharpoons M^{2+} (aq) + HS^− (aq) + OH^−$ rather than $MS (s) \rightleftharpoons M^{2+} (aq) + S^{2−} (aq)$. Here we use the simpler form involving S2−, which is justified because we take the reaction of S2− with water into account later in the solution, arriving at the same answer using either equilibrium equation.
The sulfide concentrations needed to cause $ZnS$ and $CdS$ to precipitate are as follows:
$K_{sp} = [Zn^{2+}][S^{2−}] \label{17.32a}$
$1.6 \times 10^{−24} = (0.0010\; M)[S^{2−}]\label{17.32b}$
$1.6 \times 10^{−21}\; M = [S^{2−}]\label{17.32c}$
and
$K_{sp} = [Cd^{2+}][S^{2−}] \label{17.33a}$
$8.0 \times 10^{−27} = (0.0010\; M)[S^{2−}]\label{17.33b}$
$8.0 \times 10^{−24}\; M = [S^{2−}] \label{17.33c}$
Thus sulfide concentrations between 1.6 × 10−21 M and 8.0 × 10−24 M will precipitate CdS from solution but not ZnS. How do we obtain such low concentrations of sulfide? A saturated aqueous solution of H2S contains 0.10 M H2S at 20°C. The pKa1for H2S is 6.97, and pKa2 corresponding to the formation of [S2−] is 12.90. The equations for these reactions are as follows:
$H_2S (aq) \rightleftharpoons H^+ (aq) + HS^− (aq) \label{17.34a}$
with
$pK_{a1} = 6.97 \; \text{and hence} \; K_{a1} = 1.1 \times 10^{−7} \label{17.34b}$
$HS^− (aq) \rightleftharpoons H^+ (aq) + S^{2−} (aq) \label{17.34c}$
with
$pK_{a2} = 12.90 \; \text{and hence} \; K_{a2} = 1.3 \times 10^{−13} \label{17.34d}$
We can show that the concentration of S2− is 1.3 × 10−13 by comparing Ka1 and Ka2 and recognizing that the contribution to [H+] from the dissociation of HS is negligible compared with [H+] from the dissociation of H2S. Thus substituting 0.10 M in the equation for Ka1 for the concentration of H2S, which is essentially constant regardless of the pH, gives the following:
$K_{\textrm{a1}}=1.1\times10^{-7}=\dfrac{[\mathrm{H^+}][\mathrm{HS^-}]}{[\mathrm{H_2S}]}=\dfrac{x^2}{0.10\textrm{ M}} \x=1.1\times10^{-4}\textrm{ M}=[\mathrm{H^+}]=[\mathrm{HS^-}] \label{17.35}$
Substituting this value for [H+] and [HS] into the equation for Ka2,
$K_{\textrm{a2}}=1.3\times10^{-13}=\dfrac{[\mathrm{H^+}][\mathrm{S^{2-}}]}{[\mathrm{HS^-}]}=\dfrac{(1.1\times10^{-4}\textrm{ M})x}{1.1\times10^{-4}\textrm{ M}}=x=[\mathrm{S^{2-}}]$
Although [S2−] in an H2S solution is very low (1.3 × 10−13 M), bubbling H2S through the solution until it is saturated would precipitate both metal ions because the concentration of S2− would then be much greater than 1.6 × 10−21 M. Thus we must adjust [S2−] to stay within the desired range. The most direct way to do this is to adjust [H+] by adding acid to the H2S solution (recall Le Chatelier's principle), thereby driving the equilibrium in Equation $\ref{17.34d}$ to the left. The overall equation for the dissociation of H2S is as follows:
$H_2S (aq) \rightleftharpoons 2H^+ (aq) + S^{2−} (aq) \label{17.36}$
Now we can use the equilibrium constant K for the overall reaction, which is the product of Ka1 and Ka2, and the concentration of H2S in a saturated solution to calculate the H+ concentration needed to produce [S2−] of 1.6 × 10−21 M:
$K=K_{\textrm{a1}}K_{\textrm{a2}}=(1.1\times10^{-7})(1.3\times10^{-13})=1.4\times10^{-20}=\dfrac{[\mathrm{H^+}]^2[\mathrm{S^{2-}}]}{[\mathrm{H_2S}]} \label{17.37}$
\begin{align}\mathrm{[H^+]^2}=\dfrac{K[\mathrm{H_2S}]}{[\mathrm{S^{2-}}]}=\dfrac{(1.4\times10^{-20})(\textrm{0.10 M})}{1.6\times10^{-21}\textrm{ M}}&=0.88 \ [\mathrm{H^+}]&=0.94\end{align} \label{17.38}
Thus adding a strong acid such as HCl to make the solution 0.94 M in H+ will prevent the more soluble ZnS from precipitating while ensuring that the less soluble CdS will precipitate when the solution is saturated with H2S.
Example $3$
A solution contains 0.010 M Ca2+ and 0.010 M La3+. What concentration of HCl is needed to precipitate La2(C2O4)3•9H2O but not Ca(C2O4)•H2O if the concentration of oxalic acid is 1.0 M? Ksp values are 2.32 × 10−9 for Ca(C2O4) and 2.5 × 10−27 for La2(C2O4)3; pKa1 = 1.25 and pKa2 = 3.81 for oxalic acid.
Given: concentrations of cations, Ksp values, and concentration and pKa values for oxalic acid
Asked for: concentration of HCl needed for selective precipitation of La2(C2O4)3
Strategy:
1. Write each solubility product expression and calculate the oxalate concentration needed for precipitation to occur. Determine the concentration range needed for selective precipitation of La2(C2O4)3•9H2O.
2. Add the equations for the first and second dissociations of oxalic acid to get an overall equation for the dissociation of oxalic acid to oxalate. Substitute the [ox2] needed to precipitate La2(C2O4)3•9H2O into the overall equation for the dissociation of oxalic acid to calculate the required [H+].
Solution
A Because the salts have different stoichiometries, we cannot directly compare the magnitudes of the solubility products. Instead, we must use the equilibrium constant expression for each solubility product to calculate the concentration of oxalate needed for precipitation to occur. Using ox2 for oxalate, we write the solubility product expression for calcium oxalate as follows:
$K_{sp} = [Ca^{2+}][ox^{2−}] = (0.010)[ox^{2−}] = 2.32 \times 10^{−9} \nonumber$
$[ox^{2−}] = 2.32 \times 10^{−7}\; M \nonumber$
The expression for lanthanum oxalate is as follows:
$K_{sp} = [La^{3+}]^2[ox^{2−}]^3 = (0.010)^2[ox^{2−}]^3 = 2.5 \times 10^{−27} \nonumber$
$[ox^{2−}] = 2.9 \times 10^{−8}\; M \nonumber$
Thus lanthanum oxalate is less soluble and will selectively precipitate when the oxalate concentration is between $2.9 \times 10^{−8} M$ and $2.32 \times 10^{−7} M$.
B To prevent Ca2+ from precipitating as calcium oxalate, we must add enough H+ to give a maximum oxalate concentration of 2.32 × 10−7 M. We can calculate the required [H+] by using the overall equation for the dissociation of oxalic acid to oxalate:
$HO_2CCO_2H (aq) \rightleftharpoons 2H^+ (aq) + C_2O^{2−}_{4(aq)}$
K = Ka1Ka2 = (10−1.25)(10−3.81) = 10−5.06 = 8.7×10−6
Substituting the desired oxalate concentration into the equilibrium constant expression,
\begin{align}8.7\times10^{-6}=\dfrac{[\mathrm{H^+}]^2[\mathrm{ox^{2-}}]}{[\mathrm{HO_2CCO_2H}]} &= \dfrac{[\mathrm{H^+}]^2(2.32\times10^{-7})}{1.0} \ [\mathrm{H^+}] &=\textrm{6.1 M}\end{align} \nonumber
Thus adding enough HCl to give [H+] = 6.1 M will cause only La2(C2O4)3•9H2O to precipitate from the solution.
Exercise $3$
A solution contains 0.015 M Fe2+ and 0.015 M Pb2+. What concentration of acid is needed to ensure that Pb2+ precipitates as PbS in a saturated solution of H2S, but Fe2+ does not precipitate as FeS? Ksp values are 6.3 × 10−18 for FeS and 8.0 × 10−28 for PbS.
Answer
0.018 M H+
Summary
The anion in sparingly soluble salts is often the conjugate base of a weak acid that may become protonated in solution, so the solubility of simple oxides and sulfides, both strong bases, often depends on pH. The anion in many sparingly soluble salts is the conjugate base of a weak acid. At low pH, protonation of the anion can dramatically increase the solubility of the salt. Oxides can be classified as acidic oxides or basic oxides. Acidic oxides either react with water to give an acidic solution or dissolve in strong base; most acidic oxides are nonmetal oxides or oxides of metals in high oxidation states. Basic oxides either react with water to give a basic solution or dissolve in strong acid; most basic oxides are oxides of metallic elements. Oxides or hydroxides that are soluble in both acidic and basic solutions are called amphoteric oxides. Most elements whose oxides exhibit amphoteric behavior are located along the diagonal line separating metals and nonmetals in the periodic table. In solutions that contain mixtures of dissolved metal ions, the pH can be used to control the anion concentration needed to selectively precipitate the desired cation.
• Anonymous | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/18%3A_Solubility_and_Complex-Ion_Equilibria/18.7%3A_Solubility_and_pH.txt |
Learning Objectives
• To be introduced to complex ions, including ligands.
Previously, you learned that metal ions in aqueous solution are hydrated—that is, surrounded by a shell of usually four or six water molecules. A hydrated ion is one kind of a complex ion (or, simply, complex), a species formed between a central metal ion and one or more surrounding ligands, molecules or ions that contain at least one lone pair of electrons, such as the [Al(H2O)6]3+ ion.
A complex ion forms from a metal ion and a ligand because of a Lewis acid–base interaction. The positively charged metal ion acts as a Lewis acid, and the ligand, with one or more lone pairs of electrons, acts as a Lewis base. Small, highly charged metal ions, such as Cu2+ or Ru3+, have the greatest tendency to act as Lewis acids, and consequently, they have the greatest tendency to form complex ions.
As an example of the formation of complex ions, consider the addition of ammonia to an aqueous solution of the hydrated Cu2+ ion {[Cu(H2O)6]2+}. Because it is a stronger base than H2O, ammonia replaces the water molecules in the hydrated ion to form the [Cu(NH3)4(H2O)2]2+ ion. Formation of the [Cu(NH3)4(H2O)2]2+ complex is accompanied by a dramatic color change, as shown in Figure $1$. The solution changes from the light blue of [Cu(H2O)6]2+ to the blue-violet characteristic of the [Cu(NH3)4(H2O)2]2+ ion.
The Formation Constant
The replacement of water molecules from [Cu(H2O)6]2+ by ammonia occurs in sequential steps. Omitting the water molecules bound to Cu2+ for simplicity, we can write the equilibrium reactions as follows:
\begin{align}\mathrm{Cu^{2+}(aq)}+\mathrm{NH_{3(aq)}}&\rightleftharpoons\mathrm{[Cu(NH_3)]^{2+}_{(aq)}}\hspace{5mm}K_1 \ \mathrm{[Cu(NH_3)]^{2+}_{(aq)}}+\mathrm{NH_{3(aq)}}&\rightleftharpoons\mathrm{[Cu(NH_3)_2]^{2+}_{(aq)}}\hspace{3mm}K_2 \ \mathrm{[Cu(NH_3)_2]^{2+}_{(aq)}}+\mathrm{NH_{3(aq)}}&\rightleftharpoons\mathrm{[Cu(NH_3)_3]^{2+}_{(aq)}}\hspace{3mm}K_3 \ \mathrm{[Cu(NH_3)_3]^{2+}_{(aq)}}+\mathrm{NH_{3(aq)}}&\rightleftharpoons \mathrm{[Cu(NH_3)_4]^{2+}_{(aq)}}\hspace{3mm}K_4 \end{align} \label{Eq1}
The sum of the stepwise reactions is the overall equation for the formation of the complex ion: The hydrated Cu2+ ion contains six H2O ligands, but the complex ion that is produced contains only four $NH_3$ ligands, not six.
$Cu^{2+}_{(aq)} + 4NH_{3(aq)} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)} \label{Eq2}$
The equilibrium constant for the formation of the complex ion from the hydrated ion is called the formation constant (Kf). The equilibrium constant expression for Kf has the same general form as any other equilibrium constant expression. In this case, the expression is as follows:
$K_\textrm f=\dfrac{\left[[\mathrm{Cu(NH_3)_4}]^{2+}\right]}{[\mathrm{Cu^{2+}}][\mathrm{NH_3}]^4}=2.1\times10^{13}=K_1K_2K_3K_4\label{Eq3}$
The formation constant (Kf) has the same general form as any other equilibrium constant expression.
Water, a pure liquid, does not appear explicitly in the equilibrium constant expression, and the hydrated Cu2+(aq) ion is represented as Cu2+ for simplicity. As for any equilibrium, the larger the value of the equilibrium constant (in this case, Kf), the more stable the product. With Kf = 2.1 × 1013, the [Cu(NH3)4(H2O)2]2+ complex ion is very stable. The formation constants for some common complex ions are listed in Table $1$.
Table $1$: Formation Constants for Selected Complex Ions in Aqueous Solution*
Complex Ion Equilibrium Equation Kf
*Reported values are overall formation constants. Source: Data from Lange’s Handbook of Chemistry, 15th ed. (1999).
Ammonia Complexes [Ag(NH3)2]+ Ag+ + 2NH3 ⇌ [Ag(NH3)2]+ 1.1 × 107
[Cu(NH3)4]2+ Cu2+ + 4NH3 ⇌ [Cu(NH3)4]2+ 2.1 × 1013
[Ni(NH3)6]2+ Ni2+ + 6NH3 ⇌ [Ni(NH3)6]2+ 5.5 × 108
Cyanide Complexes [Ag(CN)2] Ag+ + 2CN ⇌ [Ag(CN)2] 1.1 × 1018
[Ni(CN)4]2− Ni2+ + 4CN ⇌ [Ni(CN)4]2− 2.2 × 1031
[Fe(CN)6]3− Fe3+ + 6CN ⇌ [Fe(CN)6]3− 1 × 1042
Hydroxide Complexes [Zn(OH)4]2− Zn2+ + 4OH ⇌ [Zn(OH)4]2− 4.6 × 1017
[Cr(OH)4] Cr3+ + 4OH ⇌ [Cr(OH)4] 8.0 × 1029
Halide Complexes [HgCl4]2− Hg2+ + 4Cl ⇌ [HgCl4]2− 1.2 × 1015
[CdI4]2− Cd2+ + 4I ⇌ [CdI4]2− 2.6 × 105
[AlF6]3− Al3+ + 6F ⇌ [AlF6]3− 6.9 × 1019
Other Complexes [Ag(S2O3)2]3− Ag+ + 2S2O32− ⇌ [Ag(S2O3)2]3− 2.9 × 1013
[Fe(C2O4)3]3− Fe3+ + 3C2O42− ⇌ [Fe(C2O4)3]3− 2.0 × 1020
Example $1$
If 12.5 g of Cu(NO3)2•6H2O is added to 500 mL of 1.00 M aqueous ammonia, what is the equilibrium concentration of Cu2+(aq)?
Given: mass of Cu2+ salt and volume and concentration of ammonia solution
Asked for: equilibrium concentration of Cu2+(aq)
Strategy:
1. Calculate the initial concentration of Cu2+ due to the addition of copper(II) nitrate hexahydrate. Use the stoichiometry of the reaction shown in Equation $\ref{Eq2}$ to construct a table showing the initial concentrations, the changes in concentrations, and the final concentrations of all species in solution.
2. Substitute the final concentrations into the expression for the formation constant (Equation $\ref{Eq3}$) to calculate the equilibrium concentration of Cu2+(aq).
Solution
Adding an ionic compound that contains Cu2+ to an aqueous ammonia solution will result in the formation of [Cu(NH3)4]2+(aq), as shown in Equation $\ref{Eq2}$. We assume that the volume change caused by adding solid copper(II) nitrate to aqueous ammonia is negligible.
A The initial concentration of Cu2+ from the amount of added copper nitrate prior to any reaction is as follows:
$12.5\mathrm{\;\cancel{g}\;Cu(NO_3)_2}\cdot\mathrm{6H_2O}\left(\dfrac{\textrm{1 mol}}{\textrm{295.65} \cancel{g}} \right )\left(\dfrac{1}{\textrm{500}\; \cancel{mL}} \right )\left(\dfrac{\textrm{1000}\; \cancel{mL}}{\textrm{1 L}} \right )=\textrm{0.0846 M}$
Because the stoichiometry of the reaction is four NH3 to one Cu2+, the amount of NH3 required to react completely with the Cu2+ is 4(0.0846) = 0.338 M. The concentration of ammonia after complete reaction is 1.00 M − 0.338 M = 0.66 M. These results are summarized in the first two lines of the following table. Because the equilibrium constant for the reaction is large (2.1 × 1013), the equilibrium will lie far to the right. Thus we will assume that the formation of [Cu(NH3)4]2+ in the first step is complete and allow some of it to dissociate into Cu2+ and NH3 until equilibrium has been reached. If we define x as the amount of Cu2+ produced by the dissociation reaction, then the stoichiometry of the reaction tells us that the change in the concentration of [Cu(NH3)4]2+ is −x, and the change in the concentration of ammonia is +4x, as indicated in the table. The final concentrations of all species (in the bottom row of the table) are the sums of the concentrations after complete reaction and the changes in concentrations.
Cu2+ + 4NH3 ⇌ [Cu(NH3)4]2+
[Cu2+] [NH3] [[Cu(NH3)4]2+]
initial 0.0846 1.00 0
after complete reaction 0 0.66 0.0846
change +x +4x x
final x 0.66 + 4x 0.0846 − x
B Substituting the final concentrations into the expression for the formation constant (Equation $\ref{Eq3}$) and assuming that x << 0.0846, which allows us to remove x from the sum and difference,
\begin{align}K_\textrm f&=\dfrac{\left[[\mathrm{Cu(NH_3)_4}]^{2+}\right]}{[\mathrm{Cu^{2+}}][\mathrm{NH_3}]^4}=\dfrac{0.0846-x}{x(0.66+4x)^4}\approx\dfrac{0.0846}{x(0.66)^4}=2.1\times10^{13} \x&=2.1\times10^{-14}\end{align}
The value of x indicates that our assumption was justified. The equilibrium concentration of Cu2+(aq) in a 1.00 M ammonia solution is therefore 2.1 × 10−14 M.
Exercise $1$
The ferrocyanide ion {[Fe(CN)6]4−} is very stable, with a Kf of 1 × 1035. Calculate the concentration of cyanide ion in equilibrium with a 0.65 M solution of K4[Fe(CN)6].
Answer 2 × 10−6 M
The Effect of the Formation of Complex Ions on Solubility
What happens to the solubility of a sparingly soluble salt if a ligand that forms a stable complex ion is added to the solution? One such example occurs in conventional black-and-white photography. Recall that black-and-white photographic film contains light-sensitive microcrystals of AgBr, or mixtures of AgBr and other silver halides. AgBr is a sparingly soluble salt, with a Ksp of 5.35 × 10−13 at 25°C. When the shutter of the camera opens, the light from the object being photographed strikes some of the crystals on the film and initiates a photochemical reaction that converts AgBr to black Ag metal. Well-formed, stable negative images appear in tones of gray, corresponding to the number of grains of AgBr converted, with the areas exposed to the most light being darkest. To fix the image and prevent more AgBr crystals from being converted to Ag metal during processing of the film, the unreacted AgBr on the film is removed using a complexation reaction to dissolve the sparingly soluble salt.
The reaction for the dissolution of silver bromide is as follows:
$AgBr_{(s)} \rightleftharpoons Ag^+_{(aq)} + Br^{−}_{(aq)} \label{Eq4a}$
with
$K_{sp} = 5.35 \times 10^{−13} \text{ at 25°C} \label{Eq4b}$
The equilibrium lies far to the left, and the equilibrium concentrations of Ag+ and Br ions are very low (7.31 × 10−7 M). As a result, removing unreacted AgBr from even a single roll of film using pure water would require tens of thousands of liters of water and a great deal of time. Le Chatelier’s principle tells us, however, that we can drive the reaction to the right by removing one of the products, which will cause more AgBr to dissolve. Bromide ion is difficult to remove chemically, but silver ion forms a variety of stable two-coordinate complexes with neutral ligands, such as ammonia, or with anionic ligands, such as cyanide or thiosulfate (S2O32). In photographic processing, excess AgBr is dissolved using a concentrated solution of sodium thiosulfate.
The reaction of Ag+ with thiosulfate is as follows:
$Ag^+_{(aq)} + 2S_2O^{2−}_{3(aq)} \rightleftharpoons [Ag(S_2O_3)_2]^{3−}_{(aq)} \label{Eq5a}$
with
$K_f = 2.9 \times 10^{13} \label{Eq5b}$
The magnitude of the equilibrium constant indicates that almost all Ag+ ions in solution will be immediately complexed by thiosulfate to form [Ag(S2O3)2]3−. We can see the effect of thiosulfate on the solubility of AgBr by writing the appropriate reactions and adding them together:
\begin{align}\mathrm{AgBr(s)}\rightleftharpoons\mathrm{Ag^+(aq)}+\mathrm{Br^-(aq)}\hspace{3mm}K_{\textrm{sp}}&=5.35\times10^{-13} \ \mathrm{Ag^+(aq)}+\mathrm{2S_2O_3^{2-}(aq)}\rightleftharpoons\mathrm{[Ag(S_2O_3)_2]^{3-}(aq)}\hspace{3mm}K_\textrm f&=2.9\times10^{13} \ \mathrm{AgBr(s)}+\mathrm{2S_2O_3^{2-}(aq)}\rightleftharpoons\mathrm{[Ag(S_2O_3)_2]^{3-}(aq)}+\mathrm{Br^-(aq)}\hspace{3mm}K&=K_{\textrm{sp}}K_{\textrm f}=15\end{align} \label{Eq6}
Comparing K with Ksp shows that the formation of the complex ion increases the solubility of AgBr by approximately 3 × 1013. The dramatic increase in solubility combined with the low cost and the low toxicity explains why sodium thiosulfate is almost universally used for developing black-and-white film. If desired, the silver can be recovered from the thiosulfate solution using any of several methods and recycled.
If a complex ion has a large Kf, the formation of a complex ion can dramatically increase the solubility of sparingly soluble salts.
Example $2$
Due to the common ion effect, we might expect a salt such as AgCl to be much less soluble in a concentrated solution of KCl than in water. Such an assumption would be incorrect, however, because it ignores the fact that silver ion tends to form a two-coordinate complex with chloride ions (AgCl2). Calculate the solubility of AgCl in each situation:
1. in pure water
2. in 1.0 M KCl solution, ignoring the formation of any complex ions
3. the same solution as in part (b) except taking the formation of complex ions into account, assuming that AgCl2 is the only Ag+ complex that forms in significant concentrations
At 25°C, Ksp = 1.77 × 10−10 for AgCl and Kf = 1.1 × 105 for AgCl2.
Given: Ksp of AgCl, Kf of AgCl2, and KCl concentration
Asked for: solubility of AgCl in water and in KCl solution with and without the formation of complex ions
Strategy:
1. Write the solubility product expression for AgCl and calculate the concentration of Ag+ and Cl in water.
2. Calculate the concentration of Ag+ in the KCl solution.
3. Write balanced chemical equations for the dissolution of AgCl and for the formation of the AgCl2 complex. Add the two equations and calculate the equilibrium constant for the overall equilibrium.
4. Write the equilibrium constant expression for the overall reaction. Solve for the concentration of the complex ion.
Solution
1. A If we let x equal the solubility of AgCl, then at equilibrium [Ag+] = [Cl] = x M. Substituting this value into the solubility product expression,
Ksp = [Ag+][Cl] = (x)(x) = x2 =1.77×10−10
x = 1.33×10−5
Thus the solubility of AgCl in pure water at 25°C is 1.33 × 10−5 M.
1. B If x equals the solubility of AgCl in the KCl solution, then at equilibrium [Ag+] = x M and [Cl] = (1.0 + x) M. Substituting these values into the solubility product expression and assuming that x << 1.0,
Ksp = [Ag+][Cl] = (x)(1.0 + x) ≈ x(1.0) = 1.77×10−10 = x
If the common ion effect were the only important factor, we would predict that AgCl is approximately five orders of magnitude less soluble in a 1.0 M KCl solution than in water.
1. C To account for the effects of the formation of complex ions, we must first write the equilibrium equations for both the dissolution and the formation of complex ions. Adding the equations corresponding to Ksp and Kf gives us an equation that describes the dissolution of AgCl in a KCl solution. The equilibrium constant for the reaction is therefore the product of Ksp and Kf:
\begin{align}\mathrm{AgCl(s)}\rightleftharpoons\mathrm{Ag^+(aq)}+\mathrm{Cl^-(aq)}\hspace{3mm}K_{\textrm{sp}}&=1.77\times10^{-10} \ \mathrm{Ag^+(aq)}+\mathrm{2Cl^{-}}\rightleftharpoons\mathrm{[AgCl_2]^{-}}\hspace{3mm}K_\textrm f&=1.1\times10^{5} \ \mathrm{AgCl(s)}+\mathrm{Cl^{-}}\rightleftharpoons\mathrm{[AgCl_2]^{-}}\hspace{3mm}K&=K_{\textrm{sp}}K_{\textrm f}=1.9\times10^{-5}\end{align}
D If we let x equal the solubility of AgCl in the KCl solution, then at equilibrium [AgCl2] = x and [Cl] = 1.0 − x. Substituting these quantities into the equilibrium constant expression for the net reaction and assuming that x << 1.0,
$K=\dfrac{[\mathrm{AgCl_2^-}]}{[\mathrm{Cl^-}]}=\dfrac{x}{1.0-x}\approx1.9\times10^{-5}=x$
That is, AgCl dissolves in 1.0 M KCl to produce a 1.9 × 10−5 M solution of the AgCl2 complex ion. Thus we predict that AgCl has approximately the same solubility in a 1.0 M KCl solution as it does in pure water, which is 105 times greater than that predicted based on the common ion effect. (In fact, the measured solubility of AgCl in 1.0 M KCl is almost a factor of 10 greater than that in pure water, largely due to the formation of other chloride-containing complexes.)
Exercise $2$
Calculate the solubility of mercury(II) iodide (HgI2) in each situation:
1. pure water
2. a 3.0 M solution of NaI, assuming [HgI4]2− is the only Hg-containing species present in significant amounts
Ksp = 2.9 × 10−29 for HgI2 and Kf = 6.8 × 1029 for [HgI4]2−.
Answer
1. 1.9 × 10−10 M
2. 1.4 M
Solubility of Complex Ions: https://youtu.be/f4pkKDg2XTA
Complexing agents, molecules or ions that increase the solubility of metal salts by forming soluble metal complexes, are common components of laundry detergents. Long-chain carboxylic acids, the major components of soaps, form insoluble salts with Ca2+ and Mg2+, which are present in high concentrations in “hard” water. The precipitation of these salts produces a bathtub ring and gives a gray tinge to clothing. Adding a complexing agent such as pyrophosphate (O3POPO34, or P2O74) or triphosphate (P3O105) to detergents prevents the magnesium and calcium salts from precipitating because the equilibrium constant for complex-ion formation is large:
$Ca^{2+}_{(aq)} + O_3POPO^{4−}_{4(aq)} \rightleftharpoons [Ca(O_3POPO_3)]^{2−}_{(aq)} \label{Eq7a}$
with
$K_f = 4\times 10^4\label{Eq7b}$
However, phosphates can cause environmental damage by promoting eutrophication, the growth of excessive amounts of algae in a body of water, which can eventually lead to large decreases in levels of dissolved oxygen that kill fish and other aquatic organisms. Consequently, many states in the United States have banned the use of phosphate-containing detergents, and France has banned their use beginning in 2007. “Phosphate-free” detergents contain different kinds of complexing agents, such as derivatives of acetic acid or other carboxylic acids. The development of phosphate substitutes is an area of intense research.
Commercial water softeners also use a complexing agent to treat hard water by passing the water over ion-exchange resins, which are complex sodium salts. When water flows over the resin, sodium ion is dissolved, and insoluble salts precipitate onto the resin surface. Water treated in this way has a saltier taste due to the presence of Na+, but it contains fewer dissolved minerals.
Another application of complexing agents is found in medicine. Unlike x-rays, magnetic resonance imaging (MRI) can give relatively good images of soft tissues such as internal organs. MRI is based on the magnetic properties of the 1H nucleus of hydrogen atoms in water, which is a major component of soft tissues. Because the properties of water do not depend very much on whether it is inside a cell or in the blood, it is hard to get detailed images of these tissues that have good contrast. To solve this problem, scientists have developed a class of metal complexes known as “MRI contrast agents.” Injecting an MRI contrast agent into a patient selectively affects the magnetic properties of water in cells of normal tissues, in tumors, or in blood vessels and allows doctors to “see” each of these separately (Figure $2$). One of the most important metal ions for this application is Gd3+, which with seven unpaired electrons is highly paramagnetic. Because Gd3+(aq) is quite toxic, it must be administered as a very stable complex that does not dissociate in the body and can be excreted intact by the kidneys. The complexing agents used for gadolinium are ligands such as DTPA5 (diethylene triamine pentaacetic acid), whose fully protonated form is shown here.
Figure $2$: An MRI Image of the Heart, Arteries, and Veins. When a patient is injected with a paramagnetic metal cation in the form of a stable complex known as an MRI contrast agent, the magnetic properties of water in cells are altered. Because the different environments in different types of cells respond differently, a physician can obtain detailed images of soft tissues.
Summary
The formation of complex ions can substantially increase the solubility of sparingly soluble salts if the complex ion has a large Kf. A complex ion is a species formed between a central metal ion and one or more surrounding ligands, molecules or ions that contain at least one lone pair of electrons. Small, highly charged metal ions have the greatest tendency to act as Lewis acids and form complex ions. The equilibrium constant for the formation of the complex ion is the formation constant (Kf). The formation of a complex ion by adding a complexing agent increases the solubility of a compound. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/18%3A_Solubility_and_Complex-Ion_Equilibria/18.8%3A_Equilibria_Involving_Complex_Ions.txt |
Learning Objectives
• To know how to separate metal ions by selective precipitation.
• To understand how several common metal cations can be identified in a solution using selective precipitation.
The composition of relatively complex mixtures of metal ions can be determined using qualitative analysis, a procedure for discovering the identity of metal ions present in the mixture (rather than quantitative information about their amounts). The procedure used to separate and identify more than 20 common metal cations from a single solution consists of selectively precipitating only a few kinds of metal ions at a time under given sets of conditions. Consecutive precipitation steps become progressively less selective until almost all of the metal ions are precipitated, as illustrated in Figure $1$.
Group 1: Insoluble Chlorides
Most metal chloride salts are soluble in water; only $\ce{Ag^{+}}$, $\ce{Pb^{2+}}$, and $\ce{Hg2^{2+}}$ form chlorides that precipitate from water. Thus the first step in a qualitative analysis is to add about 6 M $\ce{HCl}$, thereby causing $\ce{AgCl}$, $\ce{PbCl2}$, and/or $\ce{Hg2Cl2}$ to precipitate. If no precipitate forms, then these cations are not present in significant amounts. The precipitate can be collected by filtration or centrifugation.
Group 2: Acid-Insoluble Sulfides
Next, the acidic solution is saturated with $\ce{H2S}$ gas. Only those metal ions that form very insoluble sulfides, such as $\ce{As^{3+}}$, $\ce{Bi^{3+}}$, $\ce{Cd^{2+}}$, $\ce{Cu^{2+}}$, $\ce{Hg^{2+}}$, $\ce{Sb^{3+}}$, and $\ce{Sn^{2+}}$, precipitate as their sulfide salts under these acidic conditions. All others, such as $\ce{Fe^{2+}}$ and $\ce{Zn^{2+}}$, remain in solution. Once again, the precipitates are collected by filtration or centrifugation.
Group 3: Base-Insoluble Sulfides (and Hydroxides)
Ammonia or $\ce{NaOH}$ is now added to the solution until it is basic, and then $\ce{(NH4)2S}$ is added. This treatment removes any remaining cations that form insoluble hydroxides or sulfides. The divalent metal ions $\ce{Co^{2+}}$, $\ce{Fe^{2+}}$, $\ce{Mn^{2+}}$, $\ce{Ni^{2+}}$, and $\ce{Zn^{2+}}$ precipitate as their sulfides, and the trivalent metal ions $\ce{Al^{3+}}$ and $\ce{Cr^{3+}}$ precipitate as their hydroxides: $\ce{Al(OH)3}$ and $\ce{Cr(OH)3}$. If the mixture contains $\ce{Fe^{3+}}$, sulfide reduces the cation to $\ce{Fe^{2+}}$, which precipitates as $\ce{FeS}$.
Group 4: Insoluble Carbonates or Phosphates
The next metal ions to be removed from solution are those that form insoluble carbonates and phosphates. When $\ce{Na2CO3}$ is added to the basic solution that remains after the precipitated metal ions are removed, insoluble carbonates precipitate and are collected. Alternatively, adding $\ce{(NH4)2HPO4}$ causes the same metal ions to precipitate as insoluble phosphates.
Group 5: Alkali Metals
At this point, we have removed all the metal ions that form water-insoluble chlorides, sulfides, carbonates, or phosphates. The only common ions that might remain are any alkali metals ($\ce{Li^{+}}$, $\ce{Na^{+}}$, $\ce{K^{+}}$, $\ce{Rb^{+}}$, and $\ce{Cs^{+}}$) and ammonium ($\ce{NH4^{+}}$). We now take a second sample from the original solution and add a small amount of $\ce{NaOH}$ to neutralize the ammonium ion and produce $\ce{NH3}$. (We cannot use the same sample we used for the first four groups because we added ammonium to that sample in earlier steps.) Any ammonia produced can be detected by either its odor or a litmus paper test. A flame test on another original sample is used to detect sodium, which produces a characteristic bright yellow color. The other alkali metal ions also give characteristic colors in flame tests, which allows them to be identified if only one is present.
Metal ions that precipitate together are separated by various additional techniques, such as forming complex ions, changing the pH of the solution, or increasing the temperature to redissolve some of the solids. For example, the precipitated metal chlorides of group 1 cations, containing $\ce{Ag^{+}}$, $\ce{Pb^{2+}}$, and $\ce{Hg2^{2+}}$, are all quite insoluble in water. Because $\ce{PbCl2}$ is much more soluble in hot water than are the other two chloride salts, however, adding water to the precipitate and heating the resulting slurry will dissolve any $\ce{PbCl2}$ present. Isolating the solution and adding a small amount of $\ce{Na2CrO4}$ solution to it will produce a bright yellow precipitate of $\ce{PbCrO4}$ if $\ce{Pb^{2+}}$ were in the original sample (Figure $2$).
As another example, treating the precipitates from group 1 cations with aqueous ammonia will dissolve any $\ce{AgCl}$ because $\ce{Ag^{+}}$ forms a stable complex with ammonia: $\ce{[Ag(NH3)2]^{+}}$. In addition, $\ce{Hg2Cl2}$ disproportionates in ammonia.
$\ce{2Hg2^{2+} \rightarrow Hg + Hg^{2+}} \nonumber$
to form a black solid that is a mixture of finely divided metallic mercury and an insoluble mercury(II) compound, which is separated from solution:
$\ce{Hg2Cl2(s) + 2NH3(aq) \rightarrow Hg(l) + Hg(NH_2)Cl(s) + NH^{+}4(aq) + Cl^{−}(aq)} \nonumber$
Any silver ion in the solution is then detected by adding $\ce{HCl}$, which reverses the reaction and gives a precipitate of white $\ce{AgCl}$ that slowly darkens when exposed to light:
$\ce{[Ag(NH3)2]^{+} (aq) + 2H^{+}(aq) + Cl^{−}(aq) \rightarrow AgCl(s) + 2NH^{+}4(aq)} \nonumber$
Similar but slightly more complex reactions are also used to separate and identify the individual components of the other groups.
Summary
In qualitative analysis, the identity, not the amount, of metal ions present in a mixture is determined. The technique consists of selectively precipitating only a few kinds of metal ions at a time under given sets of conditions. Consecutive precipitation steps become progressively less selective until almost all the metal ions are precipitated. Other additional steps are needed to separate metal ions that precipitate together. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/18%3A_Solubility_and_Complex-Ion_Equilibria/18.9%3A_Qualitative_Cation_Analysis.txt |
Learning Objectives
• Distinguish between spontaneous and nonspontaneous processes
• Describe the dispersal of matter and energy that accompanies certain spontaneous processes
In this section, consider the differences between two types of changes in a system: Those that occur spontaneously and those that occur only with the continuous input of energy. In doing so, we’ll gain an understanding as to why some systems are naturally inclined to change in one direction under certain conditions. We’ll also gain insight into how the spontaneity of a process affects the distribution of energy and matter within the system.
Spontaneous and Nonspontaneous Processes
Processes have a natural tendency to occur in one direction under a given set of conditions. Water will naturally flow downhill, but uphill flow requires outside intervention such as the use of a pump. A spontaneous process is one that occurs naturally under certain conditions. A nonspontaneous process, on the other hand, will not take place unless it is “driven” by the continual input of energy from an external source. A process that is spontaneous in one direction under a particular set of conditions is nonspontaneous in the reverse direction. At room temperature and typical atmospheric pressure, for example, ice will spontaneously melt, but water will not spontaneously freeze.
The spontaneity of a process is not correlated to the speed of the process. A spontaneous change may be so rapid that it is essentially instantaneous or so slow that it cannot be observed over any practical period of time. To illustrate this concept, consider the decay of radioactive isotopes, a topic more thoroughly treated in the chapter on nuclear chemistry. Radioactive decay is by definition a spontaneous process in which the nuclei of unstable isotopes emit radiation as they are converted to more stable nuclei. All the decay processes occur spontaneously, but the rates at which different isotopes decay vary widely. Technetium-99m is a popular radioisotope for medical imaging studies that undergoes relatively rapid decay and exhibits a half-life of about six hours. Uranium-238 is the most abundant isotope of uranium, and its decay occurs much more slowly, exhibiting a half-life of more than four billion years (Figure $1$).
As another example, consider the conversion of diamond into graphite (Figure $2$).
$\ce{C(s, diamond)}⟶\ce{C(s, graphite)} \label{Eq1}$
The phase diagram for carbon indicates that graphite is the stable form of this element under ambient atmospheric pressure, while diamond is the stable allotrope at very high pressures, such as those present during its geologic formation. Thermodynamic calculations of the sort described in the last section of this chapter indicate that the conversion of diamond to graphite at ambient pressure occurs spontaneously, yet diamonds are observed to exist, and persist, under these conditions. Though the process is spontaneous under typical ambient conditions, its rate is extremely slow, and so for all practical purposes diamonds are indeed “forever.” Situations such as these emphasize the important distinction between the thermodynamic and the kinetic aspects of a process. In this particular case, diamonds are said to be thermodynamically unstable but kinetically stable under ambient conditions.
Dispersal of Matter and Energy
As we extend our discussion of thermodynamic concepts toward the objective of predicting spontaneity, consider now an isolated system consisting of two flasks connected with a closed valve. Initially there is an ideal gas on the left and a vacuum on the right (Figure $3$). When the valve is opened, the gas spontaneously expands to fill both flasks. Recalling the definition of pressure-volume work from the chapter on thermochemistry, note that no work has been done because the pressure in a vacuum is zero.
\begin{align} w&=−PΔV \[4pt]&=0 \,\,\, \mathrm{(P=0\: in\: a\: vaccum)} \label{Eq2} \end{align}
Note as well that since the system is isolated, no heat has been exchanged with the surroundings (q = 0). The first law of thermodynamics confirms that there has been no change in the system’s internal energy as a result of this process.
\begin{align} ΔU&=q+w \tag{First Law of Thermodynamics} \[4pt] &=0+0=0 \label{Eq3}\end{align}
The spontaneity of this process is therefore not a consequence of any change in energy that accompanies the process. Instead, the movement of the gas appears to be related to the greater, more uniform dispersal of matter that results when the gas is allowed to expand. Initially, the system was comprised of one flask containing matter and another flask containing nothing. After the spontaneous process took place, the matter was distributed both more widely (occupying twice its original volume) and more uniformly (present in equal amounts in each flask).
Now consider two objects at different temperatures: object X at temperature TX and object Y at temperature TY, with TX > TY (Figure $4$). When these objects come into contact, heat spontaneously flows from the hotter object (X) to the colder one (Y). This corresponds to a loss of thermal energy by X and a gain of thermal energy by Y.
$q_\ce{X}<0 \hspace{20px} \ce{and} \hspace{20px} q_\ce{Y}=−q_\ce{X}>0 \label{Eq4}$
From the perspective of this two-object system, there was no net gain or loss of thermal energy, rather the available thermal energy was redistributed among the two objects. This spontaneous process resulted in a more uniform dispersal of energy.
As illustrated by the two processes described, an important factor in determining the spontaneity of a process is the extent to which it changes the dispersal or distribution of matter and/or energy. In each case, a spontaneous process took place that resulted in a more uniform distribution of matter or energy.
Example $1$: Redistribution of Matter during a Spontaneous Process
Describe how matter and energy are redistributed when the following spontaneous processes take place:
1. A solid sublimes.
2. A gas condenses.
3. A drop of food coloring added to a glass of water forms a solution with uniform color.
Solution
1. Sublimation is the conversion of a solid (relatively high density) to a gas (much lesser density). This process yields a much greater dispersal of matter, since the molecules will occupy a much greater volume after the solid-to-gas transition. However, an input of energy from the surroundings ss required for the molecules to leave the solid phase and enter the gas phase.
2. Condensation is the conversion of a gas (relatively low density) to a liquid (much greater density). This process yields a much lesser dispersal of matter, since the molecules will occupy a much lesser volume after the gas-to-liquid transition. As the gas molecules move together to form the droplets of liquid, they form intermolecular forces and thus release energy to the surroundings.
3. The process in question is dilution. The food dye molecules initially occupy a much smaller volume (the drop of dye solution) than they occupy once the process is complete (in the full glass of water). The process therefore entails a greater dispersal of matter. The process may also yield a more uniform dispersal of matter, since the initial state of the system involves two regions of different dye concentrations (high in the drop, zero in the water), and the final state of the system contains a single dye concentration throughout. This process can occur with out a change in energy because the molecules have kinetic energy relative to the temperature of the water, and so will be constantly in motion.
Exercise $1$
Describe how matter and energy are redistributed when you empty a canister of compressed air into a room.
Answer
This process entails both a greater and more uniform dispersal of matter as the compressed air in the canister is permitted to expand into the lower-pressure air of the room. The process also requires an input of energy to disrupt the intermolecular forces between the closely-spaced gas molecules that are originally compressed into the container. If you were to touch the nozzle of the canister, you would notice that it is cold because the exiting molecules are taking energy away from their surroundings, and the canister is part of the surroundings.
Summary
Chemical and physical processes have a natural tendency to occur in one direction under certain conditions. A spontaneous process occurs without the need for a continual input of energy from some external source, while a nonspontaneous process requires such. Systems undergoing a spontaneous process may or may not experience a gain or loss of energy, but they will experience a change in the way matter and/or energy is distributed within the system. In this section we have only discussed nuclear decay, physical changes of pure substances, and macroscopic events such as water flowing downhill. In the following sections we will discuss mixtures and chemical reactions, situations in which the description of sponteneity becomes more challenging.
Glossary
nonspontaneous process
process that requires continual input of energy from an external source
spontaneous change
process that takes place without a continuous input of energy from an external source | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/19%3A_Spontaneous_Change%3A_Entropy_and_Gibbs_Energy/19.1%3A_Spontaneity%3A_The_Meaning_of_Spontaneous_Change.txt |
Learning Objectives
• Define entropy
• Explain the relationship between entropy and the number of microstates
• Predict the sign of the entropy change for chemical and physical processes
The first law of thermodynamics governs changes in the state function we have called internal energy (U). Changes in the internal energy (ΔU) are closely related to changes in the enthalpy (ΔH), which is a measure of the heat flow between a system and its surroundings at constant pressure. You also learned previously that the enthalpy change for a chemical reaction can be calculated using tabulated values of enthalpies of formation. This information, however, does not tell us whether a particular process or reaction will occur spontaneously.
Let’s consider a familiar example of spontaneous change. If a hot frying pan that has just been removed from the stove is allowed to come into contact with a cooler object, such as cold water in a sink, heat will flow from the hotter object to the cooler one, in this case usually releasing steam. Eventually both objects will reach the same temperature, at a value between the initial temperatures of the two objects. This transfer of heat from a hot object to a cooler one obeys the first law of thermodynamics: energy is conserved.
Now consider the same process in reverse. Suppose that a hot frying pan in a sink of cold water were to become hotter while the water became cooler. As long as the same amount of thermal energy was gained by the frying pan and lost by the water, the first law of thermodynamics would be satisfied. Yet we all know that such a process cannot occur: heat always flows from a hot object to a cold one, never in the reverse direction. That is, by itself the magnitude of the heat flow associated with a process does not predict whether the process will occur spontaneously.
For many years, chemists and physicists tried to identify a single measurable quantity that would enable them to predict whether a particular process or reaction would occur spontaneously. Initially, many of them focused on enthalpy changes and hypothesized that an exothermic process would always be spontaneous. But although it is true that many, if not most, spontaneous processes are exothermic, there are also many spontaneous processes that are not exothermic. For example, at a pressure of 1 atm, ice melts spontaneously at temperatures greater than 0°C, yet this is an endothermic process because heat is absorbed. Similarly, many salts (such as NH4NO3, NaCl, and KBr) dissolve spontaneously in water even though they absorb heat from the surroundings as they dissolve (i.e., ΔHsoln > 0). Reactions can also be both spontaneous and highly endothermic, like the reaction of barium hydroxide with ammonium thiocyanate shown in Figure $1$.
Thus enthalpy is not the only factor that determines whether a process is spontaneous. For example, after a cube of sugar has dissolved in a glass of water so that the sucrose molecules are uniformly dispersed in a dilute solution, they never spontaneously come back together in solution to form a sugar cube. Moreover, the molecules of a gas remain evenly distributed throughout the entire volume of a glass bulb and never spontaneously assemble in only one portion of the available volume. To help explain why these phenomena proceed spontaneously in only one direction requires an additional state function called entropy (S), a thermodynamic property of all substances that is proportional to their degree of disorder. In Chapter 13, we introduced the concept of entropy in relation to solution formation. Here we further explore the nature of this state function and define it mathematically.
Entropy and Microstates
In 1824, at the age of 28, Nicolas Léonard Sadi Carnot (Figure $2$) published the results of an extensive study regarding the efficiency of steam heat engines. In a later review of Carnot’s findings, Rudolf Clausius introduced a new thermodynamic property that relates the spontaneous heat flow accompanying a process to the temperature at which the process takes place. This new property was expressed as the ratio of the reversible heat (qrev) and the kelvin temperature (T). The term reversible process refers to a process that takes place at such a slow rate that it is always at equilibrium and its direction can be changed (it can be “reversed”) by an infinitesimally small change is some condition. Note that the idea of a reversible process is a formalism required to support the development of various thermodynamic concepts; no real processes are truly reversible, rather they are classified as irreversible.
Similar to other thermodynamic properties, this new quantity is a state function, and so its change depends only upon the initial and final states of a system. In 1865, Clausius named this property entropy (S) and defined its change for any process as the following:
$\color{red} ΔS=\dfrac{q_\ce{rev}}{T} \label{Eq1}$
The entropy change for a real, irreversible process is then equal to that for the theoretical reversible process that involves the same initial and final states.
Following the work of Carnot and Clausius, Ludwig Boltzmann developed a molecular-scale statistical model that related the entropy of a system to the number of microstates possible for the system. A microstate (W) is a specific configuration of the locations and energies of the atoms or molecules that comprise a system like the following:
$S=k \ln W \label{Eq2}$
Here k is the Boltzmann constant and has a value of 1.38 × 10−23 J/K.
As for other state functions, the change in entropy for a process is the difference between its final (Sf) and initial (Si) values:
$ΔS=S_\ce{f}−S_\ce{i}=k \ln W_\ce{f} − k \ln W_\ce{i}=k \ln\dfrac{W_\ce{f}}{W_\ce{i}} \label{Eq2a}$
For processes involving an increase in the number of microstates, Wf > Wi, the entropy of the system increases, ΔS > 0. Conversely, processes that reduce the number of microstates, Wf < Wi, yield a decrease in system entropy, ΔS < 0. This molecular-scale interpretation of entropy provides a link to the probability that a process will occur as illustrated in the next paragraphs.
Consider the general case of a system comprised of N particles distributed among n boxes. The number of microstates possible for such a system is nN. For example, distributing four particles among two boxes will result in 24 = 16 different microstates as illustrated in Figure $3$. Microstates with equivalent particle arrangements (not considering individual particle identities) are grouped together and are called distributions. The probability that a system will exist with its components in a given distribution is proportional to the number of microstates within the distribution. Since entropy increases logarithmically with the number of microstates, the most probable distribution is therefore the one of greatest entropy.
For this system, the most probable configuration is one of the six microstates associated with distribution (c) where the particles are evenly distributed between the boxes, that is, a configuration of two particles in each box. The probability of finding the system in this configuration is
$\dfrac{6}{16}\) or $\dfrac{3}{8}$ The least probable configuration of the system is one in which all four particles are in one box, corresponding to distributions (a) and (d), each with a probability of $\dfrac{1}{16}$ The probability of finding all particles in only one box (either the left box or right box) is then $\left(\dfrac{1}{16}+\dfrac{1}{16}\right)=\dfrac{2}{16}$ or $\dfrac{1}{8}$ As you add more particles to the system, the number of possible microstates increases exponentially (2N). A macroscopic (laboratory-sized) system would typically consist of moles of particles (N ~ 1023), and the corresponding number of microstates would be staggeringly huge. Regardless of the number of particles in the system, however, the distributions in which roughly equal numbers of particles are found in each box are always the most probable configurations. The previous description of an ideal gas expanding into a vacuum is a macroscopic example of this particle-in-a-box model. For this system, the most probable distribution is confirmed to be the one in which the matter is most uniformly dispersed or distributed between the two flasks. The spontaneous process whereby the gas contained initially in one flask expands to fill both flasks equally therefore yields an increase in entropy for the system. A similar approach may be used to describe the spontaneous flow of heat. Consider a system consisting of two objects, each containing two particles, and two units of energy (represented as “*”) in Figure \(4$. The hot object is comprised of particles A and B and initially contains both energy units. The cold object is comprised of particles C and D, which initially has no energy units. Distribution (a) shows the three microstates possible for the initial state of the system, with both units of energy contained within the hot object. If one of the two energy units is transferred, the result is distribution (b) consisting of four microstates. If both energy units are transferred, the result is distribution (c) consisting of three microstates. And so, we may describe this system by a total of ten microstates. The probability that the heat does not flow when the two objects are brought into contact, that is, that the system remains in distribution (a), is $\dfrac{3}{10}$. More likely is the flow of heat to yield one of the other two distribution, the combined probability being $\dfrac{7}{10}$. The most likely result is the flow of heat to yield the uniform dispersal of energy represented by distribution (b), the probability of this configuration being $\dfrac{4}{10}$. As for the previous example of matter dispersal, extrapolating this treatment to macroscopic collections of particles dramatically increases the probability of the uniform distribution relative to the other distributions. This supports the common observation that placing hot and cold objects in contact results in spontaneous heat flow that ultimately equalizes the objects’ temperatures. And, again, this spontaneous process is also characterized by an increase in system entropy.
Consider the system shown here. What is the change in entropy for a process that converts the system from distribution (a) to (c)?
Solution
We are interested in the following change:
The initial number of microstates is one, the final six:
$ΔS=k\ln\dfrac{W_\ce{c}}{W_\ce{a}}=\mathrm{1.38×10^{−23}\:J/K×\ln\dfrac{6}{1}=2.47×10^{−23}\:J/K}$
The sign of this result is consistent with expectation; since there are more microstates possible for the final state than for the initial state, the change in entropy should be positive.
Exercise $1$
Consider the system shown in Figure $3$. What is the change in entropy for the process where all the energy is transferred from the hot object (AB) to the cold object (CD)?
Answer: 0 J/K
Entropy: https://youtu.be/dkanY87VsjY
Predicting the Sign of ΔS
The relationships between entropy, microstates, and matter/energy dispersal described previously allow us to make generalizations regarding the relative entropies of substances and to predict the sign of entropy changes for chemical and physical processes. Consider the phase changes illustrated in Figure $5$. In the solid phase, the atoms or molecules are restricted to nearly fixed positions with respect to each other and are capable of only modest oscillations about these positions. With essentially fixed locations for the system’s component particles, the number of microstates is relatively small. In the liquid phase, the atoms or molecules are free to move over and around each other, though they remain in relatively close proximity to one another. This increased freedom of motion results in a greater variation in possible particle locations, so the number of microstates is correspondingly greater than for the solid. As a result, Sliquid > Ssolid and the process of converting a substance from solid to liquid (melting) is characterized by an increase in entropy, ΔS > 0. By the same logic, the reciprocal process (freezing) exhibits a decrease in entropy, ΔS < 0.
Now consider the vapor or gas phase. The atoms or molecules occupy a much greater volume than in the liquid phase; therefore each atom or molecule can be found in many more locations than in the liquid (or solid) phase. Consequently, for any substance, Sgas > Sliquid > Ssolid, and the processes of vaporization and sublimation likewise involve increases in entropy, ΔS > 0. Likewise, the reciprocal phase transitions, condensation and deposition, involve decreases in entropy, ΔS < 0.
According to kinetic-molecular theory, the temperature of a substance is proportional to the average kinetic energy of its particles. Raising the temperature of a substance will result in more extensive vibrations of the particles in solids and more rapid translations of the particles in liquids and gases. At higher temperatures, the distribution of kinetic energies among the atoms or molecules of the substance is also broader (more dispersed) than at lower temperatures. Thus, the entropy for any substance increases with temperature (Figure $6$ ).
The entropy of a substance is influenced by structure of the particles (atoms or molecules) that comprise the substance. With regard to atomic substances, heavier atoms possess greater entropy at a given temperature than lighter atoms, which is a consequence of the relation between a particle’s mass and the spacing of quantized translational energy levels (which is a topic beyond the scope of our treatment). For molecules, greater numbers of atoms (regardless of their masses) increase the ways in which the molecules can vibrate and thus the number of possible microstates and the system entropy.
Finally, variations in the types of particles affects the entropy of a system. Compared to a pure substance, in which all particles are identical, the entropy of a mixture of two or more different particle types is greater. This is because of the additional orientations and interactions that are possible in a system comprised of nonidentical components. For example, when a solid dissolves in a liquid, the particles of the solid experience both a greater freedom of motion and additional interactions with the solvent particles. This corresponds to a more uniform dispersal of matter and energy and a greater number of microstates. The process of dissolution therefore involves an increase in entropy, ΔS > 0.
Considering the various factors that affect entropy allows us to make informed predictions of the sign of ΔS for various chemical and physical processes as illustrated in Example .
Predict the sign of the entropy change for the following processes. Indicate the reason for each of your predictions.
1. One mole liquid water at room temperature $⟶$ one mole liquid water at 50 °C
2. $\ce{Ag+}(aq)+\ce{Cl-}(aq)⟶\ce{AgCl}(s)$
3. $\ce{C6H6}(l)+\dfrac{15}{2}\ce{O2}(g)⟶\ce{6CO2}(g)+\ce{3H2O}(l)$
4. $\ce{NH3}(s)⟶\ce{NH3}(l)$
Solution
1. positive, temperature increases
2. negative, reduction in the number of ions (particles) in solution, decreased dispersal of matter
3. negative, net decrease in the amount of gaseous species
4. positive, phase transition from solid to liquid, net increase in dispersal of matter
Exercise $2$
Predict the sign of the enthalpy change for the following processes. Give a reason for your prediction.
1. $\ce{NaNO3}(s)⟶\ce{Na+}(aq)+\ce{NO3-}(aq)$
2. the freezing of liquid water
3. $\ce{CO2}(s)⟶\ce{CO2}(g)$
4. $\ce{CaCO}(s)⟶\ce{CaO}(s)+\ce{CO2}(g)$
Answer:
(a) Positive; The solid dissolves to give an increase of mobile ions in solution. (b) Negative; The liquid becomes a more ordered solid. (c) Positive; The relatively ordered solid becomes a gas. (d) Positive; There is a net production of one mole of gas.
Note
Entropy (S) is a thermodynamic property of all substances. The greater the number of possible microstates for a system, the greater the disorder and the higher the entropy.
Experiments show that the magnitude of ΔSvap is 80–90 J/(mol•K) for a wide variety of liquids with different boiling points. However, liquids that have highly ordered structures due to hydrogen bonding or other intermolecular interactions tend to have significantly higher values of ΔSvap. For instance, ΔSvap for water is 102 J/(mol•K). Another process that is accompanied by entropy changes is the formation of a solution. As illustrated in Figure $4$, the formation of a liquid solution from a crystalline solid (the solute) and a liquid solvent is expected to result in an increase in the number of available microstates of the system and hence its entropy. Indeed, dissolving a substance such as NaCl in water disrupts both the ordered crystal lattice of NaCl and the ordered hydrogen-bonded structure of water, leading to an increase in the entropy of the system. At the same time, however, each dissolved Na+ ion becomes hydrated by an ordered arrangement of at least six water molecules, and the Cl ions also cause the water to adopt a particular local structure. Both of these effects increase the order of the system, leading to a decrease in entropy. The overall entropy change for the formation of a solution therefore depends on the relative magnitudes of these opposing factors. In the case of an NaCl solution, disruption of the crystalline NaCl structure and the hydrogen-bonded interactions in water is quantitatively more important, so ΔSsoln > 0.
Dissolving NaCl in water results in an increase in the entropy of the system. Each hydrated ion, however, forms an ordered arrangement with water molecules, which decreases the entropy of the system. The magnitude of the increase is greater than the magnitude of the decrease, so the overall entropy change for the formation of an NaCl solution is positive.
Example $3$
Predict which substance in each pair has the higher entropy and justify your answer.
1. 1 mol of NH3(g) or 1 mol of He(g), both at 25°C
2. 1 mol of Pb(s) at 25°C or 1 mol of Pb(l) at 800°C
Given: amounts of substances and temperature
Asked for: higher entropy
Strategy:
From the number of atoms present and the phase of each substance, predict which has the greater number of available microstates and hence the higher entropy.
Solution:
1. Both substances are gases at 25°C, but one consists of He atoms and the other consists of NH3 molecules. With four atoms instead of one, the NH3 molecules have more motions available, leading to a greater number of microstates. Hence we predict that the NH3 sample will have the higher entropy.
2. The nature of the atomic species is the same in both cases, but the phase is different: one sample is a solid, and one is a liquid. Based on the greater freedom of motion available to atoms in a liquid, we predict that the liquid sample will have the higher entropy.
Exercise $3$
Predict which substance in each pair has the higher entropy and justify your answer.
1. 1 mol of He(g) at 10 K and 1 atm pressure or 1 mol of He(g) at 250°C and 0.2 atm
2. a mixture of 3 mol of H2(g) and 1 mol of N2(g) at 25°C and 1 atm or a sample of 2 mol of NH3(g) at 25°C and 1 atm
Answer
1. 1 mol of He(g) at 250°C and 0.2 atm (higher temperature and lower pressure indicate greater volume and more microstates)
2. a mixture of 3 mol of H2(g) and 1 mol of N2(g) at 25°C and 1 atm (more molecules of gas are present)
Video Solution
Reversible and Irreversible Changes
Changes in entropy (ΔS), together with changes in enthalpy (ΔH), enable us to predict in which direction a chemical or physical change will occur spontaneously. Before discussing how to do so, however, we must understand the difference between a reversible process and an irreversible one. In a reversible process, every intermediate state between the extremes is an equilibrium state, regardless of the direction of the change. In contrast, an irreversible process is one in which the intermediate states are not equilibrium states, so change occurs spontaneously in only one direction. As a result, a reversible process can change direction at any time, whereas an irreversible process cannot. When a gas expands reversibly against an external pressure such as a piston, for example, the expansion can be reversed at any time by reversing the motion of the piston; once the gas is compressed, it can be allowed to expand again, and the process can continue indefinitely. In contrast, the expansion of a gas into a vacuum (Pext = 0) is irreversible because the external pressure is measurably less than the internal pressure of the gas. No equilibrium states exist, and the gas expands irreversibly. When gas escapes from a microscopic hole in a balloon into a vacuum, for example, the process is irreversible; the direction of airflow cannot change.
Because work done during the expansion of a gas depends on the opposing external pressure (w = PextΔV), work done in a reversible process is always equal to or greater than work done in a corresponding irreversible process: wrev ≥ wirrev. Whether a process is reversible or irreversible, ΔU = q + w. Because U is a state function, the magnitude of ΔU does not depend on reversibility and is independent of the path taken. So
$ΔU = q_{rev} + w_{rev} = q_{irrev} + w_{irrev} \label{Eq1a}$
Note
Work done in a reversible process is always equal to or greater than work done in a corresponding irreversible process: wrev ≥ wirrev.
In other words, ΔU for a process is the same whether that process is carried out in a reversible manner or an irreversible one. We now return to our earlier definition of entropy, using the magnitude of the heat flow for a reversible process (qrev) to define entropy quantitatively.
Quantum states and Energy Spreading
At the atomic and molecular level, all energy is quantized; each particle possesses discrete states of kinetic energy and is able to accept thermal energy only in packets whose values correspond to the energies of one or more of these states. Polyatomic molecules can store energy in rotational and vibrational motions, and all molecules (even monatomic ones) will possess translational kinetic energy (thermal energy) at all temperatures above absolute zero. The energy difference between adjacent translational states is so minute that translational kinetic energy can be regarded as continuous (non-quantized) for most practical purposes.
The number of ways in which thermal energy can be distributed amongst the allowed states within a collection of molecules is easily calculated from simple statistics, but we will confine ourselves to an example here. Suppose that we have a system consisting of three molecules and three quanta of energy to share among them. We can give all the kinetic energy to any one molecule, leaving the others with none, we can give two units to one molecule and one unit to another, or we can share out the energy equally and give one unit to each molecule. All told, there are ten possible ways of distributing three units of energy among three identical molecules as shown here:
Each of these ten possibilities represents a distinct microstate that will describe the system at any instant in time. Those microstates that possess identical distributions of energy among the accessible quantum levels (and differ only in which particular molecules occupy the levels) are known as configurations. Because all microstates are equally probable, the probability of any one configuration is proportional to the number of microstates that can produce it. Thus in the system shown above, the configuration labeled ii will be observed 60% of the time, while iii will occur only 10% of the time.
As the number of molecules and the number of quanta increases, the number of accessible microstates grows explosively; if 1000 quanta of energy are shared by 1000 molecules, the number of available microstates will be around 10600— a number that greatly exceeds the number of atoms in the observable universe! The number of possible configurations (as defined above) also increases, but in such a way as to greatly reduce the probability of all but the most probable configurations. Thus for a sample of a gas large enough to be observable under normal conditions, only a single configuration (energy distribution amongst the quantum states) need be considered; even the second-most-probable configuration can be neglected.
The bottom line: any collection of molecules large enough in numbers to have chemical significance will have its therrmal energy distributed over an unimaginably large number of microstates. The number of microstates increases exponentially as more energy states ("configurations" as defined above) become accessible owing to
• Addition of energy quanta (higher temperature),
• Increase in the number of molecules (resulting from dissociation, for example).
• the volume of the system increases (which decreases the spacing between energy states, allowing more of them to be populated at a given temperature.)
Key Concepts and Summary
• For a given system, the greater the number of microstates, the higher the entropy.
• During a spontaneous process, the entropy of the universe increases.
$\Delta S=\frac{q_{\textrm{rev}}}{T}$
Entropy (S) is a state function whose value increases with an increase in the number of available microstates. A reversible process is one for which all intermediate states between extremes are equilibrium states; it can change direction at any time. In contrast, an irreversible process occurs in one direction only. The change in entropy of the system or the surroundings is the quantity of heat transferred divided by the temperature. Entropy (S) may be interpreted as a measure of the dispersal or distribution of matter and/or energy in a system, and it is often described as representing the “disorder” of the system.
For a given substance, Ssolid < Sliquid < Sgas in a given physical state at a given temperature, entropy is typically greater for heavier atoms or more complex molecules. Entropy increases when a system is heated and when solutions form. Using these guidelines, the sign of entropy changes for some chemical reactions may be reliably predicted.
Key Equations
• $ΔS=\dfrac{q_\ce{rev}}{T}$
• $S = k \ln W$
• $ΔS=k\ln\dfrac{W_\ce{f}}{W_\ce{i}}$
Glossary
entropy (S)
state function that is a measure of the matter and/or energy dispersal within a system, determined by the number of system microstates often described as a measure of the disorder of the system
microstate (W)
possible configuration or arrangement of matter and energy within a system
reversible process
process that takes place so slowly as to be capable of reversing direction in response to an infinitesimally small change in conditions; hypothetical construct that can only be approximated by real processes removed | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/19%3A_Spontaneous_Change%3A_Entropy_and_Gibbs_Energy/19.2%3A_The_Concept_of_Entropy.txt |
Learning Objectives
• To use thermodynamic cycles to calculate changes in entropy.
The atoms, molecules, or ions that compose a chemical system can undergo several types of molecular motion, including translation, rotation, and vibration (Figure $1$). The greater the molecular motion of a system, the greater the number of possible microstates and the higher the entropy. A perfectly ordered system with only a single microstate available to it would have an entropy of zero. The only system that meets this criterion is a perfect crystal at a temperature of absolute zero (0 K), in which each component atom, molecule, or ion is fixed in place within a crystal lattice and exhibits no motion (ignoring quantum effects). Such a state of perfect order (or, conversely, zero disorder) corresponds to zero entropy. In practice, absolute zero is an ideal temperature that is unobtainable, and a perfect single crystal is also an ideal that cannot be achieved. Nonetheless, the combination of these two ideals constitutes the basis for the third law of thermodynamics: the entropy of any perfectly ordered, crystalline substance at absolute zero is zero.
Third Law of Thermodynamics
The entropy of any perfectly ordered, crystalline substance at absolute zero is zero.
The third law of thermodynamics has two important consequences: it defines the sign of the entropy of any substance at temperatures above absolute zero as positive, and it provides a fixed reference point that allows us to measure the absolute entropy of any substance at any temperature.In practice, chemists determine the absolute entropy of a substance by measuring the molar heat capacity (Cp) as a function of temperature and then plotting the quantity Cp/T versus T. The area under the curve between 0 K and any temperature T is the absolute entropy of the substance at T. In contrast, other thermodynamic properties, such as internal energy and enthalpy, can be evaluated in only relative terms, not absolute terms. In this section, we examine two different ways to calculate ΔS for a reaction or a physical change. The first, based on the definition of absolute entropy provided by the third law of thermodynamics, uses tabulated values of absolute entropies of substances. The second, based on the fact that entropy is a state function, uses a thermodynamic cycle similar to those discussed previously.
Calculating ΔS from Standard Molar Entropy Values
One way of calculating ΔS for a reaction is to use tabulated values of the standard molar entropy (S°), which is the entropy of 1 mol of a substance at a standard temperature of 298 K; the units of S° are J/(mol•K). Unlike enthalpy or internal energy, it is possible to obtain absolute entropy values by measuring the entropy change that occurs between the reference point of 0 K [corresponding to S = 0 J/(mol•K)] and 298 K.
As shown in Table $1$, for substances with approximately the same molar mass and number of atoms, S° values fall in the order S°(gas) > S°(liquid) > S°(solid). For instance, S° for liquid water is 70.0 J/(mol•K), whereas S° for water vapor is 188.8 J/(mol•K). Likewise, S° is 260.7 J/(mol•K) for gaseous I2 and 116.1 J/(mol•K) for solid I2. This order makes qualitative sense based on the kinds and extents of motion available to atoms and molecules in the three phases. The correlation between physical state and absolute entropy is illustrated in Figure $2$, which is a generalized plot of the entropy of a substance versus temperature.
Table $1$: Standard Molar Entropy Values of Selected Substances at 25°C
Gases Liquids Solids
Substance S° [J/(mol•K)] Substance S° [J/(mol•K)] Substance S° [J/(mol•K)]
He 126.2 H2O 70.0 C (diamond) 2.4
H2 130.7 CH3OH 126.8 C (graphite) 5.7
Ne 146.3 Br2 152.2 LiF 35.7
Ar 154.8 CH3CH2OH 160.7 SiO2 (quartz) 41.5
Kr 164.1 C6H6 173.4 Ca 41.6
Xe 169.7 CH3COCl 200.8 Na 51.3
H2O 188.8 C6H12 (cyclohexane) 204.4 MgF2 57.2
N2 191.6 C8H18 (isooctane) 329.3 K 64.7
O2 205.2 NaCl 72.1
CO2 213.8 KCl 82.6
I2 260.7 I2 116.1
Note
Entropy increases with softer, less rigid solids, solids that contain larger atoms, and solids with complex molecular structures.
A closer examination of Table $1$ also reveals that substances with similar molecular structures tend to have similar S° values. Among crystalline materials, those with the lowest entropies tend to be rigid crystals composed of small atoms linked by strong, highly directional bonds, such as diamond [S° = 2.4 J/(mol•K)]. In contrast, graphite, the softer, less rigid allotrope of carbon, has a higher S° [5.7 J/(mol•K)] due to more disorder in the crystal. Soft crystalline substances and those with larger atoms tend to have higher entropies because of increased molecular motion and disorder. Similarly, the absolute entropy of a substance tends to increase with increasing molecular complexity because the number of available microstates increases with molecular complexity. For example, compare the S° values for CH3OH(l) and CH3CH2OH(l). Finally, substances with strong hydrogen bonds have lower values of S°, which reflects a more ordered structure.
Note
ΔS° for a reaction can be calculated from absolute entropy values using the same “products minus reactants” rule used to calculate ΔH°.
To calculate ΔS° for a chemical reaction from standard molar entropies, we use the familiar “products minus reactants” rule, in which the absolute entropy of each reactant and product is multiplied by its stoichiometric coefficient in the balanced chemical equation. Example $1$ illustrates this procedure for the combustion of the liquid hydrocarbon isooctane (C8H18; 2,2,4-trimethylpentane).
Example $1$
Use the data in Table $1$ to calculate ΔS° for the reaction of liquid isooctane with O2(g) to give CO2(g) and H2O(g) at 298 K.
Given: standard molar entropies, reactants, and products
Asked for: ΔS°
Strategy:
Write the balanced chemical equation for the reaction and identify the appropriate quantities in Table $1$. Subtract the sum of the absolute entropies of the reactants from the sum of the absolute entropies of the products, each multiplied by their appropriate stoichiometric coefficients, to obtain ΔS° for the reaction.
Solution:
The balanced chemical equation for the complete combustion of isooctane (C8H18) is as follows:
$\mathrm{C_8H_{18}(l)}+\dfrac{25}{2}\mathrm{O_2(g)}\rightarrow\mathrm{8CO_2(g)}+\mathrm{9H_2O(g)}$
We calculate ΔS° for the reaction using the “products minus reactants” rule, where m and n are the stoichiometric coefficients of each product and each reactant:
\begin{align}\Delta S^\circ_{\textrm{rxn}}&=\sum mS^\circ(\textrm{products})-\sum nS^\circ(\textrm{reactants})
\ &=[8S^\circ(\mathrm{CO_2})+9S^\circ(\mathrm{H_2O})]-[S^\circ(\mathrm{C_8H_{18}})+\dfrac{25}{2}S^\circ(\mathrm{O_2})]
\ &=\left \{ [8\textrm{ mol }\mathrm{CO_2}\times213.8\;\mathrm{J/(mol\cdot K)}]+[9\textrm{ mol }\mathrm{H_2O}\times188.8\;\mathrm{J/(mol\cdot K)}] \right \}
\ &-\left \{[1\textrm{ mol }\mathrm{C_8H_{18}}\times329.3\;\mathrm{J/(mol\cdot K)}]+\left [\dfrac{25}{2}\textrm{ mol }\mathrm{O_2}\times205.2\textrm{ J}/(\mathrm{mol\cdot K})\right ] \right \}
\ &=515.3\;\mathrm{J/K}\end{align}
ΔS° is positive, as expected for a combustion reaction in which one large hydrocarbon molecule is converted to many molecules of gaseous products.
Exercise $1$
Use the data in Table $1$ to calculate ΔS° for the reaction of H2(g) with liquid benzene (C6H6) to give cyclohexane (C6H12).
Answer: −361.1 J/K
Calculating ΔS from Thermodynamic Cycles
We can also calculate a change in entropy using a thermodynamic cycle. As you learned previously, the molar heat capacity (Cp) is the amount of heat needed to raise the temperature of 1 mol of a substance by 1°C at constant pressure. Similarly, Cv is the amount of heat needed to raise the temperature of 1 mol of a substance by 1°C at constant volume. The increase in entropy with increasing temperature in Figure $2$ is approximately proportional to the heat capacity of the substance.
Recall that the entropy change (ΔS) is related to heat flow (qrev) by ΔS = qrev/T. Because qrev = nCpΔT at constant pressure or nCvΔT at constant volume, where n is the number of moles of substance present, the change in entropy for a substance whose temperature changes from T1 to T2 is as follows:
$\Delta S=\dfrac{q_{\textrm{rev}}}{T}=nC_\textrm p\dfrac{\Delta T}{T}\hspace{4mm}(\textrm{constant pressure})$
As you will discover in more advanced math courses than is required here, it can be shown that this is equal to the following:For a review of natural logarithms, see Essential Skills 6 in Chapter 11.
$\Delta S=nC_\textrm p\ln\dfrac{T_2}{T_1}\hspace{4mm}(\textrm{constant pressure}) \label{18.20}$
Similarly,
$\Delta S=nC_{\textrm v}\ln\dfrac{T_2}{T_1}\hspace{4mm}(\textrm{constant volume}) \label{18.21}$
Thus we can use a combination of heat capacity measurements (Equation 18.20 or Equation 18.21) and experimentally measured values of enthalpies of fusion or vaporization if a phase change is involved (Equation 18.18) to calculate the entropy change corresponding to a change in the temperature of a sample.
We can use a thermodynamic cycle to calculate the entropy change when the phase change for a substance such as sulfur cannot be measured directly. As noted in the exercise in Example 6, elemental sulfur exists in two forms (part (a) in Figure $3$): an orthorhombic form with a highly ordered structure (Sα) and a less-ordered monoclinic form (Sβ). The orthorhombic (α) form is more stable at room temperature but undergoes a phase transition to the monoclinic (β) form at temperatures greater than 95.3°C (368.5 K). The transition from Sα to Sβ can be described by the thermodynamic cycle shown in part (b) in Figure $3$, in which liquid sulfur is an intermediate. The change in entropy that accompanies the conversion of liquid sulfur to Sβ (−ΔSfus(β) = ΔS3 in the cycle) cannot be measured directly. Because entropy is a state function, however, ΔS3 can be calculated from the overall entropy change (ΔSt) for the Sα–Sβ transition, which equals the sum of the ΔS values for the steps in the thermodynamic cycle, using Equation 18.20 and tabulated thermodynamic parameters (the heat capacities of Sα and Sβ, ΔHfus(α), and the melting point of Sα.)
If we know the melting point of Sα (Tm = 115.2°C = 388.4 K) and ΔSt for the overall phase transition [calculated to be 1.09 J/(mol•K) in the exercise in Example 6], we can calculate ΔS3 from the values given in part (b) in Figure $3$ where Cp(α) = 22.70 J/mol•K and Cp(β) = 24.77 J/mol•K (subscripts on ΔS refer to steps in the cycle):
\begin{align}\Delta S_{\textrm t}&=\Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4 \ 1.09\;\mathrm{J/(mol\cdot K)}&=C_{\textrm p({\alpha})}\ln\left(\dfrac{T_2}{T_1}\right)+\dfrac{\Delta H_{\textrm{fus}}}{T_{\textrm m}}+\Delta S_3+C_{\textrm p(\beta)}\ln\left(\dfrac{T_4}{T_3}\right) \ &=22.70\;\mathrm{J/(mol\cdot K)}\ln\left(\dfrac{388.4}{368.5}\right)+\left(\dfrac{1.722\;\mathrm{kJ/mol}}{\textrm{388.4 K}}\times1000\textrm{ J/kJ}\right) \ &+\Delta S_3+24.77\;\mathrm{J/(mol\cdot K)}\ln\left(\dfrac{368.5}{388.4}\right) \ &=[1.194\;\mathrm{J/(mol\cdot K)}]+[4.434\;\mathrm{J/(mol\cdot K)}]+\Delta S_3+[-1.303\;\mathrm{J/(mol\cdot K)}]\end{align}
Solving for ΔS3 gives a value of −3.24 J/(mol•K). As expected for the conversion of a less ordered state (a liquid) to a more ordered one (a crystal), ΔS3 is negative.
How are Entropies Measured
The absolute entropy of a substance at any temperature above 0 K must be determined by calculating the increments of heat q required to bring the substance from 0 K to the temperature of interest, and then summing the ratios q/T. Two kinds of experimental measurements are needed:
1. The enthalpies associated with any phase changes the substance may undergo within the temperature range of interest. Melting of a solid and vaporization of a liquid correspond to sizeable increases in the number of microstates available to accept thermal energy, so as these processes occur, energy will flow into a system, filling these new microstates to the extent required to maintain a constant temperature (the freezing or boiling point); these inflows of thermal energy correspond to the heats of fusion and vaporization. The entropy increase associated with melting, for example, is just ΔHfusion/Tm.
2. The heat capacity C of a phase expresses the quantity of heat required to change the temperature by a small amount ΔT , or more precisely, by an infinitesimal amount dT . Thus the entropy increase brought about by warming a substance over a range of temperatures that does not encompass a phase transition is given by the sum of the quantities C dT/T for each increment of temperature dT . This is of course just the integral
$S_{0 \rightarrow T^o} = \int _{0}^{T^o} \dfrac{C_p}{T} dt$
Because the heat capacity is itself slightly temperature dependent, the most precise determinations of absolute entropies require that the functional dependence of C on T be used in the above integral in place of a constant C.
$S_{0 \rightarrow T^o} = \int _{0}^{T^o} \dfrac{C_p(T)}{T} dt$
When this is not known, one can take a series of heat capacity measurements over narrow temperature increments ΔT and measure the area under each section of the curve.
The area under each section of the plot represents the entropy change associated with heating the substance through an interval ΔT. To this must be added the enthalpies of melting, vaporization, and of any solid-solid phase changes. Values of Cp for temperatures near zero are not measured directly, but can be estimated from quantum theory.
The cumulative areas from 0 K to any given temperature (taken from the experimental plot on the left) are then plotted as a function of T, and any phase-change entropies such as Svap = Hvap / Tb are added to obtain the absolute entropy at temperature T.
Summary
• Entropy changes can be calculated using the “products minus reactants” rule or from a combination of heat capacity measurements and measured values of enthalpies of fusion or vaporization.
The third law of thermodynamics states that the entropy of any perfectly ordered, crystalline substance at absolute zero is zero. At temperatures greater than absolute zero, entropy has a positive value, which allows us to measure the absolute entropy of a substance. Measurements of the heat capacity of a substance and the enthalpies of fusion or vaporization can be used to calculate the changes in entropy that accompany a physical change. The entropy of 1 mol of a substance at a standard temperature of 298 K is its standard molar entropy (S°). We can use the “products minus reactants” rule to calculate the standard entropy change (ΔS°) for a reaction using tabulated values of S° for the reactants and the products. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/19%3A_Spontaneous_Change%3A_Entropy_and_Gibbs_Energy/19.3%3A_Evaluating_Entropy_and_Entropy_Changes.txt |
Learning Objectives
• State and explain the second and third laws of thermodynamics
• Calculate entropy changes for phase transitions and chemical reactions under standard conditions
In the quest to identify a property that may reliably predict the spontaneity of a process, we have identified a very promising candidate: entropy. Processes that involve an increase in entropy of the systemS > 0) are very often spontaneous; however, examples to the contrary are plentiful. By expanding consideration of entropy changes to include the surroundings, we may reach a significant conclusion regarding the relation between this property and spontaneity. In thermodynamic models, the system and surroundings comprise everything, that is, the universe, and so the following is true:
$ΔS_\ce{univ}=ΔS_\ce{sys}+ΔS_\ce{surr} \label{$1$}$
To illustrate this relation, consider again the process of heat flow between two objects, one identified as the system and the other as the surroundings. There are three possibilities for such a process:
1. The objects are at different temperatures, and heat flows from the hotter to the cooler object. This is always observed to occur spontaneously. Designating the hotter object as the system and invoking the definition of entropy yields the following:
$ΔS_\ce{sys}=\dfrac{−q_\ce{rev}}{T_\ce{sys}}\hspace{20px}\ce{and}\hspace{20px}ΔS_\ce{surr}=\dfrac{q_\ce{rev}}{T_\ce{surr}} \label{$2$}$
The arithmetic signs of qrev denote the loss of heat by the system and the gain of heat by the surroundings. Since Tsys > Tsurr in this scenario, the magnitude of the entropy change for the surroundings will be greater than that for the system, and so the sum of ΔSsys and ΔSsurr will yield a positive value for ΔSuniv. This process involves an increase in the entropy of the universe.
2. The objects are at different temperatures, and heat flows from the cooler to the hotter object. This is never observed to occur spontaneously. Again designating the hotter object as the system and invoking the definition of entropy yields the following:
$ΔS_\ce{sys}=\dfrac{q_\ce{rev}}{T_\ce{sys}}\hspace{20px}\ce{and}\hspace{20px}ΔS_\ce{surr}=\dfrac{−q_\ce{rev}}{T_\ce{surr}} \label{$3$}$
The arithmetic signs of qrev denote the gain of heat by the system and the loss of heat by the surroundings. The magnitude of the entropy change for the surroundings will again be greater than that for the system, but in this case, the signs of the heat changes will yield a negative value for ΔSuniv. This process involves a decrease in the entropy of the universe.
3. The temperature difference between the objects is infinitesimally small, TsysTsurr, and so the heat flow is thermodynamically reversible. See the previous section’s discussion). In this case, the system and surroundings experience entropy changes that are equal in magnitude and therefore sum to yield a value of zero for ΔSuniv. This process involves no change in the entropy of the universe.
These results lead to a profound statement regarding the relation between entropy and spontaneity known as the second law of thermodynamics: all spontaneous changes cause an increase in the entropy of the universe. A summary of these three relations is provided in Table $1$.
Table $1$: The Second Law of Thermodynamics
ΔSuniv > 0 spontaneous
ΔSuniv < 0 nonspontaneous (spontaneous in opposite direction)
ΔSuniv = 0 reversible (system is at equilibrium)
Definition: The Second Law of Thermodynamics
All spontaneous changes cause an increase in the entropy of the universe.
For many realistic applications, the surroundings are vast in comparison to the system. In such cases, the heat gained or lost by the surroundings as a result of some process represents a very small, nearly infinitesimal, fraction of its total thermal energy. For example, combustion of a fuel in air involves transfer of heat from a system (the fuel and oxygen molecules undergoing reaction) to surroundings that are infinitely more massive (the earth’s atmosphere). As a result, qsurr is a good approximation of qrev, and the second law may be stated as the following:
$ΔS_\ce{univ}=ΔS_\ce{sys}+ΔS_\ce{surr}=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T} \label{$4$}$
We may use this equation to predict the spontaneity of a process as illustrated in Example $1$.
Will Ice Spontaneously Melt?
The entropy change for the process
$\ce{H2O}(s)⟶\ce{H2O}(l) \nonumber$
is 22.1 J/K and requires that the surroundings transfer 6.00 kJ of heat to the system. Is the process spontaneous at −10.00 °C? Is it spontaneous at +10.00 °C?
Solution
We can assess the spontaneity of the process by calculating the entropy change of the universe. If ΔSuniv is positive, then the process is spontaneous. At both temperatures, ΔSsys = 22.1 J/K and qsurr = −6.00 kJ.
At −10.00 °C (263.15 K), the following is true:
\begin{align} ΔS_\ce{univ}&=ΔS_\ce{sys}+ΔS_\ce{surr}=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T} \nonumber\ &=\mathrm{22.1\: J/K+\dfrac{−6.00×10^3\:J}{263.15\: K}=−0.7\:J/K} \nonumber \end{align} \nonumber
Suniv < 0, so melting is nonspontaneous (not spontaneous) at −10.0 °C.
At 10.00 °C (283.15 K), the following is true:
$ΔS_\ce{univ}=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T} \nonumber$
$\mathrm{=22.1\:J/K+\dfrac{−6.00×10^3\:J}{283.15\: K}=+0.9\: J/K} \nonumber$
Suniv > 0, so melting is spontaneous at 10.00 °C.
Exercise $1$
Using this information, determine if liquid water will spontaneously freeze at the same temperatures. What can you say about the values of Suniv?
Answer:
Entropy is a state function, and freezing is the opposite of melting. At −10.00 °C spontaneous, +0.7 J/K; at +10.00 °C nonspontaneous, −0.9 J/K.
Gibbs Energy and Changes of Gibbs Energy
One of the challenges of using the second law of thermodynamics to determine if a process is spontaneous is that we must determine the entropy change for the system and the entropy change for the surroundings. An alternative approach involving a new thermodynamic property defined in terms of system properties only was introduced in the late nineteenth century by American mathematician Josiah Willard Gibbs. This new property is called the Gibbs free energy change (G) (or simply the free energy), and it is defined in terms of a system’s enthalpy and entropy as the following:
$G=H−TS \label{2}$
Free energy is a state function, and at constant temperature and pressure, the standard free energy change (ΔG°) may be expressed as the following:
$ΔG=ΔH−TΔS \label{3}$
(For simplicity’s sake, the subscript “sys” will be omitted henceforth.) We can understand the relationship between this system property and the spontaneity of a process by recalling the previously derived second law expression:
$ΔS_\ce{univ}=ΔS+\dfrac{q_\ce{surr}}{T} \label{4}$
The first law requires that qsurr = −qsys, and at constant pressure qsys = ΔH, and so this expression may be rewritten as the following:
$ΔS_\ce{univ}=ΔS−\dfrac{ΔH}{T} \label{$5$}$
ΔH is the enthalpy change of the system. Multiplying both sides of this equation by −T, and rearranging yields the following:
$−TΔS_\ce{univ}=ΔH−TΔS \label{$6$}$
Comparing this equation to the previous one for free energy change shows the following relation:
$ΔG=−TΔS_\ce{univ} \label{$7$}$
The free energy change is therefore a reliable indicator of the spontaneity of a process, being directly related to the previously identified spontaneity indicator, $ΔS_{univ}$. Table $2$ expands on Table $2$ and summarizes the relation between the spontaneity of a process and the arithmetic signs of $\Delta G$ and $\Delta S$ indicators.
Table $2$: Relation between Process Spontaneity and Signs of Thermodynamic Properties
ΔSuniv > 0 ΔG < 0 spontaneous
ΔSuniv < 0 ΔG > 0 nonspontaneous
ΔSuniv = 0 ΔG = 0 reversible (at equilibrium)
. Willard Gibbs (1839–1903)
Born in Connecticut, Josiah Willard Gibbs attended Yale, as did his father, a professor of sacred literature at Yale, who was involved in the Amistad trial. In 1863, Gibbs was awarded the first engineering doctorate granted in the United States. He was appointed professor of mathematical physics at Yale in 1871, the first such professorship in the United States. His series of papers entitled “On the Equilibrium of Heterogeneous Substances” was the foundation of the field of physical chemistry and is considered one of the great achievements of the 19th century. Gibbs, whose work was translated into French by Le Chatelier, lived with his sister and brother-in-law until his death in 1903, shortly before the inauguration of the Nobel Prizes.
Gibbs energy is a state function, so its value depends only on the conditions of the initial and final states of the system that have undergone some change. A convenient and common approach to the calculation of free energy changes for physical and chemical reactions is by use of widely available compilations of standard state thermodynamic data. One method involves the use of standard enthalpies and entropies to compute standard free energy changes according to the following relation as demonstrated in Example $1$.
$ΔG°=ΔH°−TΔS° \label{$7$}$
The Definition of Gibbs Free Energy: https://youtu.be/iuWkcHUh-1o
Example $2$: Evaluation of ΔG°
Change from ΔH° and ΔS° Use standard enthalpy and entropy data from Appendix G to calculate the standard free energy change for the vaporization of water at room temperature (298 K). What does the computed value for ΔG° say about the spontaneity of this process?
Solution
The process of interest is the following:
$\ce{H2O}(l)⟶\ce{H2O}(g) \label{$8$} \nonumber$
The standard change in free energy may be calculated using the following equation:
$ΔG^\circ_{298}=ΔH°−TΔS° \label{$9$} \nonumber$
From Appendix G, here is the data:
Substance $ΔH^\circ_\ce{f}\ce{(kJ/mol)}$ $S^\circ_{298}\textrm{(J/K⋅mol)}$
H2O(l) −286.83 70.0
H2O(g) −241.82 188.8
Combining at 298 K:
$ΔH°=ΔH^\circ_{298}=ΔH^\circ_\ce{f}(\ce{H2O}(g))−ΔH^\circ_\ce{f}(\ce{H2O}(l)) \nonumber\ =\mathrm{[−241.82\: kJ−(−285.83)]\:kJ/mol=44.01\: kJ/mol} \nonumber$
$ΔS°=ΔS^\circ_{298}=S^\circ_{298}(\ce{H2O}(g))−S^\circ_{298}(\ce{H2O}(l)) \nonumber\ =\mathrm{188.8\:J/mol⋅K−70.0\:J/K=118.8\:J/mol⋅K} \nonumber$
$ΔG°=ΔH°−TΔS° \nonumber$
Converting everything into kJ and combining at 298 K:
$ΔG^\circ_{298}=ΔH°−TΔS° \nonumber$
$\mathrm{=44.01\: kJ/mol−(298\: K×118.8\:J/mol⋅K)×\dfrac{1\: kJ}{1000\: J}} \nonumber$
$\mathrm{44.01\: kJ/mol−35.4\: kJ/mol=8.6\: kJ/mol} \nonumber$
At 298 K (25 °C) $ΔG^\circ_{298}>0$, and so boiling is nonspontaneous (not spontaneous).
Exercise $2$
Use standard enthalpy and entropy data from Appendix G to calculate the standard free energy change for the reaction shown here (298 K). What does the computed value for ΔG° say about the spontaneity of this process?
$\ce{C2H6}(g)⟶\ce{H2}(g)+\ce{C2H4}(g) \nonumber$
Answer:
$ΔG^\circ_{298}=\mathrm{102.0\: kJ/mol}$; the reaction is nonspontaneous (not spontaneous) at 25 °C.
Free energy changes may also use the standard free energy of formation $(ΔG^\circ_\ce{f})$, for each of the reactants and products involved in the reaction. The standard free energy of formation is the free energy change that accompanies the formation of one mole of a substance from its elements in their standard states. Similar to the standard enthalpies of formation, $(ΔG^\circ_\ce{f})$ is by definition zero for elemental substances under standard state conditions. The approach to computing the free energy change for a reaction using this approach is the same as that demonstrated previously for enthalpy and entropy changes. For the reaction
$m\ce{A}+n\ce{B}⟶x\ce{C}+y\ce{D},$
the standard free energy change at room temperature may be calculated as
$ΔG^\circ_{298}=ΔG°=∑νΔG^\circ_{298}(\ce{products})−∑νΔG^\circ_{298}(\ce{reactants})$
$=[xΔG^\circ_\ce{f}(\ce{C})+yΔG^\circ_\ce{f}(\ce{D})]−[mΔG^\circ_\ce{f}(\ce{A})+nΔG^\circ_\ce{f}(\ce{B})].$
Example $3$: Calculation of $ΔG^\circ_{298}$
Consider the decomposition of yellow mercury(II) oxide.
$\ce{HgO}(s,\,\ce{yellow})⟶\ce{Hg}(l)+\dfrac{1}{2}\ce{O2}(g) \nonumber$
Calculate the standard free energy change at room temperature, $ΔG^\circ_{298}$, using (a) standard free energies of formation and (b) standard enthalpies of formation and standard entropies. Do the results indicate the reaction to be spontaneous or nonspontaneous under standard conditions?
Solution
The required data are available in Appendix G and are shown here.
Compound $ΔG^\circ_\ce{f}\:\mathrm{(kJ/mol)}$ $ΔH^\circ_\ce{f}\:\mathrm{(kJ/mol)}$ $S^\circ_{298}\:\textrm{(J/K⋅mol)}$
HgO (s, yellow) −58.43 −90.46 71.13
Hg(l) 0 0 75.9
O2(g) 0 0 205.2
(a) Using free energies of formation:
$ΔG^\circ_{298}=∑νGS^\circ_{298}(\ce{products})−∑νΔG^\circ_{298}(\ce{reactants}) \nonumber$
$=\left[1ΔG^\circ_{298}\ce{Hg}(l)+\dfrac{1}{2}ΔG^\circ_{298}\ce{O2}(g)\right]−1ΔG^\circ_{298}\ce{HgO}(s,\,\ce{yellow}) \nonumber$
$\mathrm{=\left[1\:mol(0\: kJ/mol)+\dfrac{1}{2}mol(0\: kJ/mol)\right]−1\: mol(−58.43\: kJ/mol)=58.43\: kJ/mol} \nonumber$
(b) Using enthalpies and entropies of formation:
$ΔH^\circ_{298}=∑νΔH^\circ_{298}(\ce{products})−∑νΔH^\circ_{298}(\ce{reactants}) \nonumber$
$=\left[1ΔH^\circ_{298}\ce{Hg}(l)+\dfrac{1}{2}ΔH^\circ_{298}\ce{O2}(g)\right]−1ΔH^\circ_{298}\ce{HgO}(s,\,\ce{yellow}) \nonumber$
$\mathrm{=[1\: mol(0\: kJ/mol)+\dfrac{1}{2}mol(0\: kJ/mol)]−1\: mol(−90.46\: kJ/mol)=90.46\: kJ/mol} \nonumber$
$ΔS^\circ_{298}=∑νΔS^\circ_{298}(\ce{products})−∑νΔS^\circ_{298}(\ce{reactants}) \nonumber$
$=\left[1ΔS^\circ_{298}\ce{Hg}(l)+\dfrac{1}{2}ΔS^\circ_{298}\ce{O2}(g)\right]−1ΔS^\circ_{298}\ce{HgO}(s,\,\ce{yellow}) \nonumber$
$\mathrm{=\left[1\: mol(75.9\: J/mol\: K)+\dfrac{1}{2}mol(205.2\: J/mol\: K)\right]−1\: mol(71.13\: J/mol\: K)=107.4\: J/mol\: K} \nonumber$
$ΔG°=ΔH°−TΔS°=\mathrm{90.46\: kJ−298.15\: K×107.4\: J/K⋅mol×\dfrac{1\: kJ}{1000\: J}} \nonumber$
$ΔG°=\mathrm{(90.46−32.01)\:kJ/mol=58.45\: kJ/mol} \nonumber$
Both ways to calculate the standard free energy change at 25 °C give the same numerical value (to three significant figures), and both predict that the process is nonspontaneous (not spontaneous) at room temperature.
Exercise $3$
Calculate ΔG° using (a) free energies of formation and (b) enthalpies of formation and entropies (Appendix G). Do the results indicate the reaction to be spontaneous or nonspontaneous at 25 °C?
$\ce{C2H4}(g)⟶\ce{H2}(g)+\ce{C2H2}(g) \nonumber$
Answer
−141.5 kJ/mol, nonspontaneous
Key Concepts and Summary
The second law of thermodynamics states that a spontaneous process increases the entropy of the universe, Suniv > 0. If ΔSuniv < 0, the process is nonspontaneous, and if ΔSuniv = 0, the system is at equilibrium. Gibbs free energy (G) is a state function defined with regard to system quantities only and may be used to predict the spontaneity of a process. A negative value for ΔG indicates a spontaneous process; a positive ΔG indicates a nonspontaneous process; and a ΔG of zero indicates that the system is at equilibrium. A number of approaches to the computation of free energy changes are possible.
Key Equations
• $ΔS^\circ=ΔS^\circ_{298}=∑νS^\circ_{298}(\ce{products})−∑νS^\circ_{298}(\ce{reactants})$
• $ΔS=\dfrac{q_\ce{rev}}{T}$
• ΔSuniv = ΔSsys + ΔSsurr
• $ΔS_\ce{univ}=ΔS_\ce{sys}+ΔS_\ce{surr}=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T}$
• ΔG = ΔHTΔS
• ΔG = ΔG° + RT ln Q
• ΔG° = −RT ln K
Glossary
Gibbs free energy change (G)
thermodynamic property defined in terms of system enthalpy and entropy; all spontaneous processes involve a decrease in G
standard free energy change (ΔG°)
change in free energy for a process occurring under standard conditions (1 bar pressure for gases, 1 M concentration for solutions)
standard free energy of formation $(ΔG^\circ_\ce{f})$
change in free energy accompanying the formation of one mole of substance from its elements in their standard states
second law of thermodynamics
entropy of the universe increases for a spontaneous process
standard entropy (S°)
entropy for a substance at 1 bar pressure; tabulated values are usually determined at 298.15 K and denoted $S^\circ_{298}$
standard entropy change (ΔS°)
change in entropy for a reaction calculated using the standard entropies, usually at room temperature and denoted $ΔS^\circ_{298}$ | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/19%3A_Spontaneous_Change%3A_Entropy_and_Gibbs_Energy/19.4%3A_Criteria_for_Spontaneous_Change%3A_The_Second_Law_of_Thermodynamics.txt |
Learning Objectives
• Define Gibbs free energy, and describe its relation to spontaneity
• Calculate free energy change for a process using free energies of formation for its reactants and products
• Calculate free energy change for a process using enthalpies of formation and the entropies for its reactants and products
The Gibbs free energy ($G$), often called simply free energy, was named in honor of J. Willard Gibbs (1838–1903), an American physicist who first developed the concept. It is defined in terms of three other state functions with which you are already familiar: enthalpy, temperature, and entropy:
$G = H − TS \label{Eq1}$
Because it is a combination of state functions, $G$ is also a state function.
The criterion for predicting spontaneity is based on ΔG, the change in G, at constant temperature and pressure. Although very few chemical reactions actually occur under conditions of constant temperature and pressure, most systems can be brought back to the initial temperature and pressure without significantly affecting the value of thermodynamic state functions such as G. At constant temperature and pressure,
$\Delta G=\Delta H-T\Delta S \label{18.5.2}$
where all thermodynamic quantities are those of the system. Under standad conditions Equation $\ref{18.5.2}$ is then expressed at
$\Delta G^o=\Delta H^o-T\Delta S^o \label{18.5.3}$
Since $G$ is a state function, $\Delta G^o$ can be obtained from the standard free-energy of formation values in Table T1 (or T2) via the similar relationship used to calculate other state functions like $\Delta H^o$ and $\Delta S^o$:
$\Delta G^o = \sum n \Delta G_f^o\;(products) - \sum m \Delta G_f^o\; (reactants) \label{19.7}$
Example $1$: Calculation of $ΔG^\circ_{298}$
Consider the decomposition of yellow mercury(II) oxide.
$\ce{HgO}(s,\,\ce{yellow})⟶\ce{Hg}(l)+\dfrac{1}{2}\ce{O2}(g) \nonumber$
Calculate the standard free energy change at room temperature, $ΔG^\circ_{298}$, using (a) standard free energies of formation and (b) standard enthalpies of formation and standard entropies. Do the results indicate the reaction to be spontaneous or nonspontaneous under standard conditions? The required data are available in Table T1 .
Solution
The required data are available in Table T1 and are shown here.
Compound $ΔG^\circ_\ce{f}\:\mathrm{(kJ/mol)}$ $ΔH^\circ_\ce{f}\:\mathrm{(kJ/mol)}$ $S^\circ_{298}\:\textrm{(J/K⋅mol)}$
HgO (s, yellow) −58.43 −90.46 71.13
Hg(l) 0 0 75.9
O2(g) 0 0 205.2
(a) Using free energies of formation:
$ΔG^\circ_{298}=∑νGS^\circ_{298}(\ce{products})−∑νΔG^\circ_{298}(\ce{reactants}) \nonumber$
$=\left[1ΔG^\circ_{298}\ce{Hg}(l)+\dfrac{1}{2}ΔG^\circ_{298}\ce{O2}(g)\right]−1ΔG^\circ_{298}\ce{HgO}(s,\,\ce{yellow}) \nonumber$
$\mathrm{=\left[1\:mol(0\: kJ/mol)+\dfrac{1}{2}mol(0\: kJ/mol)\right]−1\: mol(−58.43\: kJ/mol)=58.43\: kJ/mol} \nonumber$
(b) Using enthalpies and entropies of formation:
$ΔH^\circ_{298}=∑νΔH^\circ_{298}(\ce{products})−∑νΔH^\circ_{298}(\ce{reactants}) \nonumber$
$=\left[1ΔH^\circ_{298}\ce{Hg}(l)+\dfrac{1}{2}ΔH^\circ_{298}\ce{O2}(g)\right]−1ΔH^\circ_{298}\ce{HgO}(s,\,\ce{yellow}) \nonumber$
$\mathrm{=[1\: mol(0\: kJ/mol)+\dfrac{1}{2}mol(0\: kJ/mol)]−1\: mol(−90.46\: kJ/mol)=90.46\: kJ/mol} \nonumber$
$ΔS^\circ_{298}=∑νΔS^\circ_{298}(\ce{products})−∑νΔS^\circ_{298}(\ce{reactants}) \nonumber$
$=\left[1ΔS^\circ_{298}\ce{Hg}(l)+\dfrac{1}{2}ΔS^\circ_{298}\ce{O2}(g)\right]−1ΔS^\circ_{298}\ce{HgO}(s,\,\ce{yellow}) \nonumber$
$\mathrm{=\left[1\: mol(75.9\: J/mol\: K)+\dfrac{1}{2}mol(205.2\: J/mol\: K)\right]−1\: mol(71.13\: J/mol\: K)=107.4\: J/mol\: K} \nonumber$
Now use these values in Equation $\ref{18.5.3}$ to get $ΔG^o$:
$ΔG°=ΔH°−TΔS°=\mathrm{90.46\: kJ−298.15\: K×107.4\: J/K⋅mol×\dfrac{1\: kJ}{1000\: J}} \nonumber$
$ΔG°=\mathrm{(90.46−32.01)\:kJ/mol=58.45\: kJ/mol} \nonumber$
Both ways to calculate the standard free energy change at 25 °C give the same numerical value (to three significant figures), and both predict that the process is nonspontaneous (not spontaneous) at room temperature (since $ΔG^o > 0$.
Exercise $1$
Calculate ΔG° using (a) free energies of formation and (b) enthalpies of formation and entropies (Appendix G). Do the results indicate the reaction to be spontaneous or nonspontaneous at 25 °C?
$\ce{C2H4}(g)⟶\ce{H2}(g)+\ce{C2H2}(g) \nonumber$
Answer
−141.5 kJ/mol, nonspontaneous
Calculating Gibbs Free Energy (Grxn) for a Reaction: https://youtu.be/wmreE6zeFQo | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/19%3A_Spontaneous_Change%3A_Entropy_and_Gibbs_Energy/19.5%3A_Standard_Gibbs_Energy_Change_G.txt |
Learning Objectives
• To know the relationship between free energy and the equilibrium constant.
• The sign of the standard free energy change ΔG° of a chemical reaction determines whether the reaction will tend to proceed in the forward or reverse direction.
• Similarly, the relative signs of ΔG° and ΔS° determine whether the spontaniety of a chemical reaction will be affected by the temperature, and if so, in what way.
ΔG is meaningful only for changes in which the temperature and pressure remain constant. These are the conditions under which most reactions are carried out in the laboratory; the system is usually open to the atmosphere (constant pressure) and we begin and end the process at room temperature (after any heat we have added or which is liberated by the reaction has dissipated.) The importance of the Gibbs function can hardly be over-stated: it serves as the single master variable that determines whether a given chemical change is thermodynamically possible. Thus if the free energy of the reactants is greater than that of the products, the entropy of the world will increase when the reaction takes place as written, and so the reaction will tend to take place spontaneously. Conversely, if the free energy of the products exceeds that of the reactants, then the reaction will not take place in the direction written, but it will tend to proceed in the reverse direction.
Temperature Dependence to ΔG
In a spontaneous change, Gibbs energy always decreases and never increases. This of course reflects the fact that the entropy of the world behaves in the exact opposite way (owing to the negative sign in the TΔS term).
$H_2O_{(l)} \rightarrow H_2O_{(s)} \label{23.5.6}$
water below its freezing point undergoes a decrease in its entropy, but the heat released into the surroundings more than compensates for this, so the entropy of the world increases, the free energy of the H2O diminishes, and the process proceeds spontaneously.
Note
In a spontaneous change, Gibbs energy always decreases and never increases.
An important consequence of the one-way downward path of the free energy is that once it reaches its minimum possible value, all net change comes to a halt. This, of course, represents the state of chemical equilibrium. These relations are nicely summarized as follows:
• ΔG < 0: reaction can spontaneously proceed to the right: $A \rightarrow B$
• ΔG > 0: reaction can spontaneously proceed to the left: $A \leftarrow B$
• ΔG = 0: the reaction is at equilibrium; the quantities of [A] and [B] will not change
Recall the condition for spontaneous change
$ΔG = ΔH – TΔS < 0 \label{Master}$
it is apparent that the temperature dependence of ΔG depends almost entirely on the entropy change associated with the process. (We say "almost" because the values of ΔH and ΔS are themselves slightly temperature dependent; both gradually increase with temperature). In particular, notice that in the above equation the sign of the entropy change determines whether the reaction becomes more or less spontaneous as the temperature is raised. For any given reaction, the sign of ΔH can also be positive or negative. This means that there are four possibilities for the influence that temperature can have on the spontaneity of a process.
The following cases generalizes these relations for the four sign-combinations of ΔH and ΔS. (Note that use of the standard ΔH° and ΔS° values in the example reactions is not strictly correct here, and can yield misleading results when used generally.)
> 0
Under these conditions, both the ΔH and TΔS terms will be negative, so ΔG will be negative regardless of the temperature. An exothermic reaction whose entropy increases will be spontaneous at all temperatures.
Example Reaction
$C_{(graphite)} + O_{2(g)} \rightleftharpoons CO_{2(g)}$
• ΔH° = –393 kJ
• ΔS° = +2.9 J K–1
• ΔG° = –394 kJ at 298 K
The positive entropy change is due mainly to the greater mass of CO2 molecules compared to those of O2.
< 0
If the reaction is sufficiently exothermic it can force ΔG negative only at temperatures below which |TΔS| < |ΔH|. This means that there is a temperature T = ΔH / ΔS at which the reaction is at equilibrium; the reaction will only proceed spontaneously below this temperature. The freezing of a liquid or the condensation of a gas are the most common examples of this condition.
Example reaction:
$3 H_2 + N_2 \rightleftharpoons 2 NH_{3(g)}$
• ΔH° = –46.2 kJ
• ΔS° = –389 J K–1
• ΔG° = –16.4 kJ at 298 K
The decrease in moles of gas in the Haber ammonia synthesis drives the entropy change negative, making the reaction spontaneous only at low temperatures. Thus higher T, which speeds up the reaction, also reduces its extent.
> 0
This is the reverse of the previous case; the entropy increase must overcome the handicap of an endothermic process so that TΔS > ΔH. Since the effect of the temperature is to "magnify" the influence of a positive ΔS, the process will be spontaneous at temperatures above T = ΔH / ΔS. (Think of melting and boiling.)
Example reaction:
$N_2O_{4(g)} \rightleftharpoons 2 NO_{2(g)}$
• ΔH° = 55.3 kJ
• ΔS° = +176 J K–1
• ΔG° = +2.8 kJ at 298 K
Dissociation reactions are typically endothermic with positive entropy change, and are therefore spontaneous at high temperatures.Ultimately, all molecules decompose to their atoms at sufficiently high temperatures.
< 0
With both ΔH and ΔS working against it, this kind of process will not proceed spontaneously at any temperature. Substance A always has a greater number of accessible energy states, and is therefore always the preferred form.
Example reaction:
$½ N_2 + O_2 \rightleftharpoons NO_{2(g)}$
• ΔH° = 33.2 kJ
• ΔS° = –249 J K1
• ΔG° = +51.3 kJ at 298 K
This reaction is not spontaneous at any temperature, meaning that its reverse is always spontaneous. But because the reverse reaction is kinetically inhibited, NO2 can exist indefinitely at ordinary temperatures even though it is thermodynamically unstable.
The above cases and associated plots are the important ones; do not try to memorize them, but make sure you understand and can explain or reproduce them for a given set of ΔH and ΔS.
• Their most important differentiating features are the position of the ΔH line (above or below the is TΔS line), and the slope of the latter, which of course depends on the sign of ΔS.
• The reaction A → B will occur spontaneously only when ΔG is negative (blue arrows pointing down.)
• Owing to the slight temperature dependence of ΔS, the TΔS plots are not quite straight lines as shown here. Similarly, the lines representing ΔH are even more curved.
The other two plots on each diagram are only for the chemistry-committed.
• Each pair of energy-level diagrams depicts the relative spacing of the microscopic energy levels in the reactants and products as reflected by the value of ΔS°. (The greater the entropy, the more closely-spaced are the quantized microstates.)
• The red shading indicates the range of energy levels that are accessible to the system at each temperature. The spontaneous direction of the reaction will always be in the direction in which the red shading overlaps the greater number of energy levels, resulting in the maximum dispersal of thermal energy.
• Note that the vertical offsets correspond to ΔH° for the reaction.
• Never forget that it is the ability of thermal energy to spread into as many of these states as possible that determines the tendency of the process to take place. None of this is to scale, of course!
Liquid-Vapor Equilibrium
To further understand how the various components of ΔG dictate whether a process occurs spontaneously, we now look at a simple and familiar physical change: the conversion of liquid water to water vapor. If this process is carried out at 1 atm and the normal boiling point of 100.00°C (373.15 K), we can calculate ΔG from the experimentally measured value of ΔHvap (40.657 kJ/mol). For vaporizing 1 mol of water, $ΔH = 40,657; J$, so the process is highly endothermic. From the definition of ΔS, we know that for 1 mol of water,
$\Delta S_{\textrm{vap}}=\dfrac{\Delta H_{\textrm{vap}}}{T_\textrm b}=\dfrac{\textrm{40,657 J}}{\textrm{373.15 K}}=\textrm{108.96 J/K}$
Hence there is an increase in the disorder of the system. At the normal boiling point of water,
$\Delta G_{100^\circ\textrm C}=\Delta H_{100^\circ\textrm C}-T\Delta S_{100^\circ\textrm C}=\textrm{40,657 J}-[(\textrm{373.15 K})(\textrm{108.96 J/K})]=\textrm{0 J}$
The energy required for vaporization offsets the increase in disorder of the system. Thus ΔG = 0, and the liquid and vapor are in equilibrium, as is true of any liquid at its boiling point under standard conditions.
Now suppose we were to superheat 1 mol of liquid water to 110°C. The value of ΔG for the vaporization of 1 mol of water at 110°C, assuming that ΔH and ΔS do not change significantly with temperature, becomes
$\Delta G_{110^\circ\textrm C}=\Delta H-T\Delta S=\textrm{40,657 J}-[(\textrm{383.15 K})(\textrm{108.96 J/K})]=-\textrm{1091 J}$
At 110°C, ΔG < 0, and vaporization is predicted to occur spontaneously and irreversibly.
We can also calculate ΔG for the vaporization of 1 mol of water at a temperature below its normal boiling point—for example, 90°C—making the same assumptions:
$\Delta G_{90^\circ\textrm C}=\Delta H-T\Delta S=\textrm{40,657 J}-[(\textrm{363.15 K})(\textrm{108.96 J/K})]=\textrm{1088 J}$
At 90°C, ΔG > 0, and water does not spontaneously convert to water vapor. When using all the digits in the calculator display in carrying out our calculations, ΔG110°C = 1090 J = −ΔG90°C, as we would predict.
We can also calculate the temperature at which liquid water is in equilibrium with water vapor. Inserting the values of ΔH and ΔS into the definition of ΔG (Equation $\ref{Master}$), setting ΔG = 0, and solving for T,
0 J=40,657 J−T(108.96 J/K)
T=373.15 K
Thus ΔG = 0 at T = 373.15 K and 1 atm, which indicates that liquid water and water vapor are in equilibrium; this temperature is called the normal boiling point of water. At temperatures greater than 373.15 K, ΔG is negative, and water evaporates spontaneously and irreversibly. Below 373.15 K, ΔG is positive, and water does not evaporate spontaneously. Instead, water vapor at a temperature less than 373.15 K and 1 atm will spontaneously and irreversibly condense to liquid water. Figure $1$ shows how the ΔH and TΔS terms vary with temperature for the vaporization of water. When the two lines cross, ΔG = 0, and ΔH = TΔS.
A similar situation arises in the conversion of liquid egg white to a solid when an egg is boiled. The major component of egg white is a protein called albumin, which is held in a compact, ordered structure by a large number of hydrogen bonds. Breaking them requires an input of energy (ΔH > 0), which converts the albumin to a highly disordered structure in which the molecules aggregate as a disorganized solid (ΔS > 0). At temperatures greater than 373 K, the TΔS term dominates, and ΔG < 0, so the conversion of a raw egg to a hard-boiled egg is an irreversible and spontaneous process above 373 K.
Free Energy and the Equilibrium Constant
ΔG is key in determining whether or not a reaction will take place in a given direction. It turns out, however, that it is almost never necessary to explicitly evaluate ΔG. It is far more convenient to work with the equilibrium constant of a reaction, within which ΔG is "hidden". This is just as well, because for most reactions (those that take place in solutions or gas mixtures) the value of ΔG depends on the proportions of the various reaction components in the mixture; it is not a simple sum of the "products minus reactants" type, as is the case with ΔH.
Because ΔH° and ΔS° determine the magnitude of ΔG° and because K is a measure of the ratio of the concentrations of products to the concentrations of reactants, we should be able to express K in terms of ΔG° and vice versa. ΔG is equal to the maximum amount of work a system can perform on its surroundings while undergoing a spontaneous change. For a reversible process that does not involve external work, we can express the change in free energy in terms of volume, pressure, entropy, and temperature, thereby eliminating ΔH from the equation for ΔG. The general relationship can be shown as follow (derivation not shown):
$\Delta G = V \Delta P − S \Delta T \label{18.29}$
If a reaction is carried out at constant temperature (ΔT = 0), then Equation $\ref{18.29}$ simplifies to
$\Delta{G} = V\Delta{P} \label{18.30}$
Under normal conditions, the pressure dependence of free energy is not important for solids and liquids because of their small molar volumes. For reactions that involve gases, however, the effect of pressure on free energy is very important.
Assuming ideal gas behavior, we can replace the $V$ in Equation $\ref{18.30}$ by nRT/P (where n is the number of moles of gas and R is the ideal gas constant) and express $\Delta{G}$ in terms of the initial and final pressures ($P_i$ and $P_f$, respectively):
$\Delta G=\left(\dfrac{nRT}{P}\right)\Delta P=nRT\dfrac{\Delta P}{P}=nRT\ln\left(\dfrac{P_\textrm f}{P_\textrm i}\right) \label{18.31}$
If the initial state is the standard state with Pi = 1 atm, then the change in free energy of a substance when going from the standard state to any other state with a pressure P can be written as follows:
$G − G^° = nRT\ln{P}$
This can be rearranged as follows:
$G = G^° + nRT\ln {P} \label{18.32}$
As you will soon discover, Equation $\ref{18.32}$ allows us to relate ΔG° and Kp. Any relationship that is true for $K_p$ must also be true for $K$ because $K_p$ and $K$ are simply different ways of expressing the equilibrium constant using different units.
Let’s consider the following hypothetical reaction, in which all the reactants and the products are ideal gases and the lowercase letters correspond to the stoichiometric coefficients for the various species:
$aA+bB \rightleftharpoons cC+dD \label{18.33}$
Because the free-energy change for a reaction is the difference between the sum of the free energies of the products and the reactants, we can write the following expression for ΔG:
$\delta{G}=\sum_m G_{products}−\sum_n G_{reactants}=(cG_C+dG_D)−(aG_A+bG_B) \label{18.34}$
Substituting Equation $\ref{18.32}$ for each term into Equation $\ref{18.34}$,
$ΔG=[(cG^o_C+cRT \ln P_C)+(dG^o_D+dRT\ln P_D)]−[(aG^o_A+aRT\ln P_A)+(bG^o_B+bRT\ln P_B)]$
Combining terms gives the following relationship between ΔG and the reaction quotient Q:
$\Delta G=\Delta G^\circ+RT \ln\left(\dfrac{P^c_\textrm CP^d_\textrm D}{P^a_\textrm AP^b_\textrm B}\right)=\Delta G^\circ+RT\ln Q \label{18.35}$
where ΔG° indicates that all reactants and products are in their standard states. For gases at equilibrium ($Q = K_p$,), and as you’ve learned in this chapter, ΔG = 0 for a system at equilibrium. Therefore, we can describe the relationship between ΔG° and Kp for gases as follows:
$0 = ΔG° + RT\ln K_p \label{18.36a}$
$\color{red} ΔG°= −RT\ln K_p \label{18.36b}$
If the products and reactants are in their standard states and ΔG° < 0, then Kp > 1, and products are favored over reactants. Conversely, if ΔG° > 0, then Kp < 1, and reactants are favored over products. If ΔG° = 0, then $K_p = 1$, and neither reactants nor products are favored: the system is at equilibrium.
Note
For a spontaneous process under standard conditions, $K_{eq}$ and $K_p$ are greater than 1.
Example $1$
We previosuly calculated that ΔG° = −32.7 kJ/mol of N2 for the reaction
$N_{2(g)}+3H_{2(g)} \rightleftharpoons 2NH_{3(g)} \nonumber$
This calculation was for the reaction under standard conditions—that is, with all gases present at a partial pressure of 1 atm and a temperature of 25°C. Calculate ΔG for the same reaction under the following nonstandard conditions:
• $P_{\textrm N_2}$ = 2.00 atm,
• $P_{\textrm H_2}$ = 7.00 atm,
• $P_{\textrm{NH}_3}$ = 0.021 atm,
• and T = 100°C.
Does the reaction favor products or reactants?
Given: balanced chemical equation, partial pressure of each species, temperature, and ΔG°
Asked for: whether products or reactants are favored
Strategy:
1. Using the values given and Equation $\ref{18.35}$, calculate Q.
2. Substitute the values of ΔG° and Q into Equation $\ref{18.35}$ to obtain ΔG for the reaction under nonstandard conditions.
Solution:
A The relationship between ΔG° and ΔG under nonstandard conditions is given in Equation $\ref{18.35}$. Substituting the partial pressures given, we can calculate Q:
$Q=\dfrac{P^2_{\textrm{NH}_3}}{P_{\textrm N_2}P^3_{\textrm H_2}}=\dfrac{(0.021)^2}{(2.00)(7.00)^3}=6.4\times10^{-7} \nonumber$
B Substituting the values of ΔG° and Q into Equation $\ref{18.35}$,
$\Delta G=\Delta G^\circ+RT\ln Q=-32.7\textrm{ kJ}+\left[(\textrm{8.314 J/K})(\textrm{373 K})\left(\dfrac{\textrm{1 kJ}}{\textrm{1000 J}}\right)\ln(6.4\times10^{-7})\right]$
$=-32.7\textrm{ kJ}+(-44\textrm{ kJ})$
$=-77\textrm{ kJ/mol of N}_2$
Because ΔG < 0 and Q < 1.0, the reaction is spontaneous to the right as written, so products are favored over reactants.
Exercise $1$
Calculate ΔG for the reaction of nitric oxide with oxygen to give nitrogen dioxide under these conditions: T = 50°C, PNO = 0.0100 atm, $P_{\mathrm{O_2}}$ = 0.200 atm, and $P_{\mathrm{NO_2}}$ = 1.00 × 10−4 atm. The value of ΔG° for this reaction is −72.5 kJ/mol of O2. Are products or reactants favored?
Answer: −92.9 kJ/mol of O2; the reaction is spontaneous to the right as written, so products are favored.
Example $2$
Calculate Kp for the reaction of H2 with N2 to give NH3 at 25°C. As calculated in Example 10, ΔG° for this reaction is −32.7 kJ/mol of N2.
Given: balanced chemical equation from Example 10, ΔG°, and temperature
Asked for: Kp
Strategy:
Substitute values for ΔG° and T (in kelvins) into Equation $\ref{18.36}$ to calculate Kp, the equilibrium constant for the formation of ammonia.
Solution
In Example 10, we used tabulated values of ΔGf to calculate ΔG° for this reaction (−32.7 kJ/mol of N2). For equilibrium conditions, rearranging Equation $\ref{18.36b}$,
\begin{align} \Delta G^\circ &=-RT\ln K_\textrm p \nonumber \ \dfrac{-\Delta G^\circ}{RT} &=\ln K_\textrm p \nonumber \end{align}
Inserting the value of ΔG° and the temperature (25°C = 298 K) into this equation,
\begin{align}\ln K_\textrm p &=-\dfrac{(-\textrm{32.7 kJ})(\textrm{1000 J/kJ})}{(\textrm{8.314 J/K})(\textrm{298 K})}=13.2 \nonumber \ K_\textrm p &=5.4\times10^5 \nonumber\end{align}
Thus the equilibrium constant for the formation of ammonia at room temperature is favorable. However, the rate at which the reaction occurs at room temperature is too slow to be useful.
Exercise $2$
Calculate Kp for the reaction of NO with O2 to give NO2 at 25°C. As calculated in the exercise in Example 10, ΔG° for this reaction is −70.5 kJ/mol of O2.
Answer: 2.2 × 1012
Although Kp is defined in terms of the partial pressures of the reactants and the products, the equilibrium constant K is defined in terms of the concentrations of the reactants and the products. We described the relationship between the numerical magnitude of Kp and K in Chapter 15 and showed that they are related:
$K_p = K(RT)^{Δn} \label{18.37}$
where Δn is the number of moles of gaseous product minus the number of moles of gaseous reactant. For reactions that involve only solutions, liquids, and solids, Δn = 0, so Kp = K. For all reactions that do not involve a change in the number of moles of gas present, the relationship in Equation $\ref{18.36b}$ can be written in a more general form:
$ΔG° = −RT \ln K \label{18.38}$
Only when a reaction results in a net production or consumption of gases is it necessary to correct Equation $\ref{18.38}$ for the difference between Kp and K. Although we typically use concentrations or pressures in our equilibrium calculations, recall that equilibrium constants are generally expressed as unitless numbers because of the use of activities or fugacities in precise thermodynamic work. Systems that contain gases at high pressures or concentrated solutions that deviate substantially from ideal behavior require the use of fugacities or activities, respectively.
Combining Equations $\ref{18.38}$ with $ΔG^o = ΔH^o − TΔS^o$ provides insight into how the components of ΔG° influence the magnitude of the equilibrium constant:
$ΔG° = ΔH° − TΔS° = −RT \ln K \label{18.39}$
Equation $\ref{18.39}$ is quite powerful and connected the nature of the system under equilibrium $K$ to the condition of the system under standard conditions $\Delta G^o$.; that is quite powerful. Notice that $K$ becomes larger as ΔS° becomes more positive, indicating that the magnitude of the equilibrium constant is directly influenced by the tendency of a system to move toward maximum disorder. Moreover, K increases as ΔH° decreases. Thus the magnitude of the equilibrium constant is also directly influenced by the tendency of a system to seek the lowest energy state possible.
Note
The magnitude of the equilibrium constant is directly influenced by the tendency of a system to move toward maximum entropy and seek the lowest energy state possible.
To further illustrate the relation between these two essential thermodynamic concepts, consider the observation that reactions spontaneously proceed in a direction that ultimately establishes equilibrium. As may be shown by plotting the free energy change versus the extent of the reaction (for example, as reflected in the value of Q), equilibrium is established when the system’s free energy is minimized (Figure $2$). If a system is present with reactants and products present in nonequilibrium amounts (QK), the reaction will proceed spontaneously in the direction necessary to establish equilibrium.
Relating Grxn and Kp: https://youtu.be/T-OYNTYN__4
ΔG° and ΔG: Predicting the Direction of Chemical Change
We have seen that there is no way to measure absolute enthalpies, although we can measure changes in enthalpy (ΔH) during a chemical reaction. Because enthalpy is one of the components of Gibbs free energy, we are consequently unable to measure absolute free energies; we can measure only changes in free energy. The standard free-energy change (ΔG°) is the change in free energy when one substance or a set of substances in their standard states is converted to one or more other substances, also in their standard states. The standard free-energy change can be calculated from the definition of free energy, if the standard enthalpy and entropy changes are known, using Equation $\ref{Eq5}$:
$ΔG° = ΔH° − TΔS° \label{Eq5}$
If ΔS° and ΔH° for a reaction have the same sign, then the sign of ΔG° depends on the relative magnitudes of the ΔH° and TΔS° terms. It is important to recognize that a positive value of ΔG° for a reaction does not mean that no products will form if the reactants in their standard states are mixed; it means only that at equilibrium the concentrations of the products will be less than the concentrations of the reactants.
Note
A positive ΔG° means that the equilibrium constant is less than 1.
Example $3$
Calculate the standard free-energy change (ΔG°) at 25°C for the reaction
$H_{2(g)}+O_{2(g)} \rightleftharpoons H_2O_{2(l)} \nonumber$
At 25°C, the standard enthalpy change (ΔH°) is −187.78 kJ/mol, and the absolute entropies of the products and reactants are:
• S°(H2O2) = 109.6 J/(mol•K),
• S°(O2) = 205.2 J/(mol•K), and
• S°(H2) = 130.7 J/(mol•K).
Is the reaction spontaneous as written?
Given: balanced chemical equation, ΔH° and S° for reactants and products
Asked for: spontaneity of reaction as written
Strategy:
1. Calculate ΔS° from the absolute molar entropy values given.
2. Use Equation $\ref{Eq5}$, the calculated value of ΔS°, and other data given to calculate ΔG° for the reaction. Use the value of ΔG° to determine whether the reaction is spontaneous as written.
Solution
A To calculate ΔG° for the reaction, we need to know ΔH°, ΔS°, and T. We are given ΔH°, and we know that T = 298.15 K. We can calculate ΔS° from the absolute molar entropy values provided using the “products minus reactants” rule:
\begin{align}\Delta S^\circ &=S^\circ(\mathrm{H_2O_2})-[S^\circ(\mathrm{O_2})+S^\circ(\mathrm{H_2})] \nonumber \ &=[\mathrm{1\;mol\;H_2O_2}\times109.6\;\mathrm{J/(mol\cdot K})] \nonumber \ &-\left \{ [\textrm{1 mol H}_2\times130.7\;\mathrm{J/(mol\cdot K)}]+[\textrm{1 mol O}_2\times205.2\;\mathrm{J/(mol\cdot K)}] \right \} \nonumber \&=-226.3\textrm{ J/K }(\textrm{per mole of }\mathrm{H_2O_2}) \end{align}
As we might expect for a reaction in which 2 mol of gas is converted to 1 mol of a much more ordered liquid, ΔS° is very negative for this reaction.
B Substituting the appropriate quantities into Equation $\ref{Eq5}$,
\begin{align}\Delta G^\circ=\Delta H^\circ -T\Delta S^\circ &=-187.78\textrm{ kJ/mol}-(\textrm{298.15 K}) [-226.3\;\mathrm{J/(mol\cdot K)}\times\textrm{1 kJ/1000 J}] \nonumber \ &=-187.78\textrm{ kJ/mol}+67.47\textrm{ kJ/mol}=-120.31\textrm{ kJ/mol} \nonumber \end{align}
The negative value of ΔG° indicates that the reaction is spontaneous as written. Because ΔS° and ΔH° for this reaction have the same sign, the sign of ΔG° depends on the relative magnitudes of the ΔH° and TΔS° terms. In this particular case, the enthalpy term dominates, indicating that the strength of the bonds formed in the product more than compensates for the unfavorable ΔS° term and for the energy needed to break bonds in the reactants.
Exercise $3$
Calculate the standard free-energy change (ΔG°) at 25°C for the reaction
$2H_2(g)+N_2(g) \rightleftharpoons N_2H_4(l) \nonumber$
. At 25°C, the standard enthalpy change (ΔH°) is 50.6 kJ/mol, and the absolute entropies of the products and reactants are S°(N2H4) = 121.2 J/(mol•K), S°(N2) = 191.6 J/(mol•K), and S°(H2) = 130.7 J/(mol•K). Is the reaction spontaneous as written?
Answer:
Video Solution
149.5 kJ/mol; no
Tabulated values of standard free energies of formation allow chemists to calculate the values of ΔG° for a wide variety of chemical reactions rather than having to measure them in the laboratory. The standard free energy of formation ($ΔG^∘_f$)of a compound is the change in free energy that occurs when 1 mol of a substance in its standard state is formed from the component elements in their standard states. By definition, the standard free energy of formation of an element in its standard state is zero at 298.15 K. One mole of Cl2 gas at 298.15 K, for example, has $ΔG^∘_f = 0$. The standard free energy of formation of a compound can be calculated from the standard enthalpy of formation (ΔHf) and the standard entropy of formation (ΔSf) using the definition of free energy:
$Δ^o_{f} =ΔH^o_{f} −TΔS^o_{f} \label{Eq6}$
Using standard free energies of formation to calculate the standard free energy of a reaction is analogous to calculating standard enthalpy changes from standard enthalpies of formation using the familiar “products minus reactants” rule:
$ΔG^o_{rxn}=\sum mΔG^o_{f} (products)− \sum nΔ^o_{f} (reactants) \label{Eq7a}$
where m and n are the stoichiometric coefficients of each product and reactant in the balanced chemical equation. A very large negative ΔG° indicates a strong tendency for products to form spontaneously from reactants; it does not, however, necessarily indicate that the reaction will occur rapidly. To make this determination, we need to evaluate the kinetics of the reaction.
Example $4$
Calculate ΔG° for the reaction of isooctane with oxygen gas to give carbon dioxide and water (described in Example 7). Use the following data:
• ΔG°f(isooctane) = −353.2 kJ/mol,
• ΔG°f(CO2) = −394.4 kJ/mol, and
• ΔG°f(H2O) = −237.1 kJ/mol. Is the reaction spontaneous as written?
Given: balanced chemical equation and values of ΔG°f for isooctane, CO2, and H2O
Asked for: spontaneity of reaction as written
Strategy:
Use the “products minus reactants” rule to obtain ΔGrxn, remembering that ΔG°f for an element in its standard state is zero. From the calculated value, determine whether the reaction is spontaneous as written.
Solution
The balanced chemical equation for the reaction is as follows:
$\mathrm{C_8H_{18}(l)}+\frac{25}{2}\mathrm{O_2(g)}\rightarrow\mathrm{8CO_2(g)}+\mathrm{9H_2O(l)} \nonumber$
We are given ΔGf values for all the products and reactants except O2(g). Because oxygen gas is an element in its standard state, ΔGf (O2) is zero. Using the “products minus reactants” rule,
\begin{align} \Delta G^\circ &=[8\Delta G^\circ_\textrm f(\mathrm{CO_2})+9\Delta G^\circ_\textrm f(\mathrm{H_2O})]-\left[1\Delta G^\circ_\textrm f(\mathrm{C_8H_{18}})+\dfrac{25}{2}\Delta G^\circ_\textrm f(\mathrm{O_2})\right] \nonumber \ &=[(\textrm{8 mol})(-394.4\textrm{ kJ/mol})+(\textrm{9 mol})(-237.1\textrm{ kJ/mol})] \nonumber\&-\left [(\textrm{1 mol})(-353.2\textrm{ kJ/mol})+\left(\dfrac{25}{2}\;\textrm{mol}\right)(0 \textrm{ kJ/mol}) \right ] \nonumber \ &=-4935.9\textrm{ kJ }(\textrm{per mol of }\mathrm{C_8H_{18}}) \nonumber \end{align}
Because ΔG° is a large negative number, there is a strong tendency for the spontaneous formation of products from reactants (though not necessarily at a rapid rate). Also notice that the magnitude of ΔG° is largely determined by the ΔGf of the stable products: water and carbon dioxide.
Exercise $4$
Calculate ΔG° for the reaction of benzene with hydrogen gas to give cyclohexane using the following data
• ΔGf(benzene) = 124.5 kJ/mol
• ΔGf (cyclohexane) = 217.3 kJ/mol.
Is the reaction spontaneous as written?
Answer:
• 92.8 kJ; no
Video Solution
Calculated values of ΔG° are extremely useful in predicting whether a reaction will occur spontaneously if the reactants and products are mixed under standard conditions. We should note, however, that very few reactions are actually carried out under standard conditions, and calculated values of ΔG° may not tell us whether a given reaction will occur spontaneously under nonstandard conditions. What determines whether a reaction will occur spontaneously is the free-energy change (ΔG) under the actual experimental conditions, which are usually different from ΔG°. If the ΔH and TΔS terms for a reaction have the same sign, for example, then it may be possible to reverse the sign of ΔG by changing the temperature, thereby converting a reaction that is not thermodynamically spontaneous, having Keq < 1, to one that is, having a Keq > 1, or vice versa. Because ΔH and ΔS usually do not vary greatly with temperature in the absence of a phase change, we can use tabulated values of ΔH° and ΔS° to calculate ΔG° at various temperatures, as long as no phase change occurs over the temperature range being considered.
Note
In the absence of a phase change, neither $ΔH$ nor $ΔS$ vary greatly with temperature.
Example $5$
Calculate (a) ΔG° and (b) ΔG300°C for the reaction N2(g)+3H2(g)⇌2NH3(g), assuming that ΔH and ΔS do not change between 25°C and 300°C. Use these data:
• S°(N2) = 191.6 J/(mol•K),
• S°(H2) = 130.7 J/(mol•K),
• S°(NH3) = 192.8 J/(mol•K), and
• ΔHf (NH3) = −45.9 kJ/mol.
Given: balanced chemical equation, temperatures, S° values, and ΔHf for NH3
Asked for: ΔG° and ΔG at 300°C
Strategy:
1. Convert each temperature to kelvins. Then calculate ΔS° for the reaction. Calculate ΔH° for the reaction, recalling that ΔHf for any element in its standard state is zero.
2. Substitute the appropriate values into Equation $\ref{Eq5}$ to obtain ΔG° for the reaction.
3. Assuming that ΔH and ΔS are independent of temperature, substitute values into Equation $\ref{Eq2}$ to obtain ΔG for the reaction at 300°C.
Solution
A To calculate ΔG° for the reaction using Equation $\ref{Eq5}$, we must know the temperature as well as the values of ΔS° and ΔH°. At standard conditions, the temperature is 25°C, or 298 K. We can calculate ΔS° for the reaction from the absolute molar entropy values given for the reactants and the products using the “products minus reactants” rule:
\begin{align}\Delta S^\circ_{\textrm{rxn}}&=2S^\circ(\mathrm{NH_3})-[S^\circ(\mathrm{N_2})+3S^\circ(\mathrm{H_2})] \nonumber\ &=[\textrm{2 mol NH}_3\times192.8\;\mathrm{J/(mol\cdot K)}] \nonumber\ &-\left \{[\textrm{1 mol N}_2\times191.6\;\mathrm{J/(mol\cdot K)}]+[\textrm{3 mol H}_2\times130.7\;\mathrm{J/(mol\cdot K)}]\right \} \nonumber\ &=-198.1\textrm{ J/K (per mole of N}_2)\end{align} \nonumber
We can also calculate ΔH° for the reaction using the “products minus reactants” rule. The value of ΔHf (NH3) is given, and ΔHf is zero for both N2 and H2:
\begin{align}\Delta H^\circ_{\textrm{rxn}}&=2\Delta H^\circ_\textrm f(\mathrm{NH_3})-[\Delta H^\circ_\textrm f(\mathrm{N_2})+3\Delta H^\circ_\textrm f(\mathrm{H_2})] \nonumber \ &=[2\times(-45.9\textrm{ kJ/mol})]-[(1\times0\textrm{ kJ/mol})+(3\times0 \textrm{ kJ/mol})] \nonumber \ &=-91.8\textrm{ kJ(per mole of N}_2) \nonumber\end{align} \nonumber
B Inserting the appropriate values into Equation $\ref{Eq5}$
$\Delta G^\circ_{\textrm{rxn}}=\Delta H^\circ-T\Delta S^\circ=(-\textrm{91.8 kJ})-(\textrm{298 K})(-\textrm{198.1 J/K})(\textrm{1 kJ/1000 J})=-\textrm{32.7 kJ (per mole of N}_2) \nonumber$
C To calculate ΔG for this reaction at 300°C, we assume that ΔH and ΔS are independent of temperature (i.e., ΔH300°C = H° and ΔS300°C = ΔS°) and insert the appropriate temperature (573 K) into Equation $\ref{Eq2}$:
\begin{align}\Delta G_{300^\circ\textrm C}&=\Delta H_{300^\circ\textrm C}-(\textrm{573 K})(\Delta S_{300^\circ\textrm C})=\Delta H^\circ -(\textrm{573 K})\Delta S^\circ \nonumber \ &=(-\textrm{91.8 kJ})-(\textrm{573 K})(-\textrm{198.1 J/K})(\textrm{1 kJ/1000 J})=21.7\textrm{ kJ (per mole of N}_2) \nonumber \end{align} \nonumber
In this example, changing the temperature has a major effect on the thermodynamic spontaneity of the reaction. Under standard conditions, the reaction of nitrogen and hydrogen gas to produce ammonia is thermodynamically spontaneous, but in practice, it is too slow to be useful industrially. Increasing the temperature in an attempt to make this reaction occur more rapidly also changes the thermodynamics by causing the −TΔS° term to dominate, and the reaction is no longer spontaneous at high temperatures; that is, its Keq is less than one. This is a classic example of the conflict encountered in real systems between thermodynamics and kinetics, which is often unavoidable.
Exercise $5$
Calculate
1. $ΔG°$ and
2. $ΔG_{750°C}$
for the following reaction
$2NO_{(g)}+O_{2\; (g)} \rightleftharpoons 2NO_{2\; (g)} \nonumber$
which is important in the formation of urban smog. Assume that $ΔH$ and $ΔS$ do not change between 25.0°C and 750°C and use these data:
• S°(NO) = 210.8 J/(mol•K),
• S°(O2) = 205.2 J/(mol•K),
• S°(NO2) = 240.1 J/(mol•K),
• ΔHf(NO2) = 33.2 kJ/mol, and
• ΔHf (NO) = 91.3 kJ/mol.
Answer
1. −72.5 kJ/mol of $O_2$
2. 33.8 kJ/mol of $O_2$
Video Solution
The effect of temperature on the spontaneity of a reaction, which is an important factor in the design of an experiment or an industrial process, depends on the sign and magnitude of both ΔH° and ΔS°. The temperature at which a given reaction is at equilibrium can be calculated by setting ΔG° = 0 in Equation $\ref{Eq5}$, as illustrated in Example $4$.
Example $6$
The reaction of nitrogen and hydrogen gas to produce ammonia is one in which ΔH° and ΔS° are both negative. Such reactions are predicted to be thermodynamically spontaneous at low temperatures but nonspontaneous at high temperatures. Use the data in Example $3$ to calculate the temperature at which this reaction changes from spontaneous to nonspontaneous, assuming that ΔH° and ΔS° are independent of temperature.
Given: ΔH° and ΔS°
Asked for: temperature at which reaction changes from spontaneous to nonspontaneous
Strategy:
Set ΔG° equal to zero in Equation $\ref{Eq5}$ and solve for T, the temperature at which the reaction becomes nonspontaneous.
Solution
In Example $3$, we calculated that ΔH° is −91.8 kJ/mol of N2 and ΔS° is −198.1 J/K per mole of N2, corresponding to ΔG° = −32.7 kJ/mol of N2 at 25°C. Thus the reaction is indeed spontaneous at low temperatures, as expected based on the signs of ΔH° and ΔS°. The temperature at which the reaction becomes nonspontaneous is found by setting ΔG° equal to zero and rearranging Equation $\ref{Eq5}$ to solve for T:
\begin{align}\Delta G^\circ &=\Delta H^\circ - T\Delta S^\circ=0 \ \Delta H^\circ &=T\Delta S^\circ \ T=\dfrac{\Delta H^\circ}{\Delta S^\circ}&=\dfrac{(-\textrm{91.8 kJ})(\textrm{1000 J/kJ})}{-\textrm{198.1 J/K}}=\textrm{463 K}\end{align}
This is a case in which a chemical engineer is severely limited by thermodynamics. Any attempt to increase the rate of reaction of nitrogen with hydrogen by increasing the temperature will cause reactants to be favored over products above 463 K.
Exercise $6$
ΔH° and ΔS° are both negative for the reaction of nitric oxide and oxygen to form nitrogen dioxide. Use those data to calculate the temperature at which this reaction changes from spontaneous to nonspontaneous.
Answer: 792.6 K
Summary
• The change in Gibbs free energy, which is based solely on changes in state functions, is the criterion for predicting the spontaneity of a reaction.
We can predict whether a reaction will occur spontaneously by combining the entropy, enthalpy, and temperature of a system in a new state function called Gibbs free energy (G). The change in free energy (ΔG) is the difference between the heat released during a process and the heat released for the same process occurring in a reversible manner. If a system is at equilibrium, ΔG = 0. If the process is spontaneous, ΔG < 0. If the process is not spontaneous as written but is spontaneous in the reverse direction, ΔG > 0. At constant temperature and pressure, ΔG is equal to the maximum amount of work a system can perform on its surroundings while undergoing a spontaneous change. The standard free-energy change (ΔG°) is the change in free energy when one substance or a set of substances in their standard states is converted to one or more other substances, also in their standard states. The standard free energy of formation (ΔGf), is the change in free energy that occurs when 1 mol of a substance in its standard state is formed from the component elements in their standard states. Tabulated values of standard free energies of formation are used to calculate ΔG° for a reaction. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/19%3A_Spontaneous_Change%3A_Entropy_and_Gibbs_Energy/19.6%3A_Gibbs_Energy_Change_and_Equilibrium.txt |
Learning Objectives
• To know the relationship between free energy and the equilibrium constant.
As was previously demonstrated, the spontaneity of a process may depend upon the temperature of the system. Phase transitions, for example, will proceed spontaneously in one direction or the other depending upon the temperature of the substance in question. Likewise, some chemical reactions can also exhibit temperature dependent spontaneities. To illustrate this concept, the equation relating free energy change to the enthalpy and entropy changes for the process is considered:
$ΔG=ΔH−TΔS$
The spontaneity of a process, as reflected in the arithmetic sign of its free energy change, is then determined by the signs of the enthalpy and entropy changes and, in some cases, the absolute temperature. Since T is the absolute (kelvin) temperature, it can only have positive values. Four possibilities therefore exist with regard to the signs of the enthalpy and entropy changes:
1. Both ΔH and ΔS are positive. This condition describes an endothermic process that involves an increase in system entropy. In this case, ΔG will be negative if the magnitude of the TΔS term is greater than ΔH. If the TΔS term is less than ΔH, the free energy change will be positive. Such a process is spontaneous at high temperatures and nonspontaneous at low temperatures.
2. Both ΔH and ΔS are negative. This condition describes an exothermic process that involves a decrease in system entropy. In this case, ΔG will be negative if the magnitude of the TΔS term is less than ΔH. If the TΔS term’s magnitude is greater than ΔH, the free energy change will be positive. Such a process is spontaneous at low temperatures and nonspontaneous at high temperatures.
3. ΔH is positive and ΔS is negative. This condition describes an endothermic process that involves a decrease in system entropy. In this case, ΔG will be positive regardless of the temperature. Such a process is nonspontaneous at all temperatures.
4. ΔH is negative and ΔS is positive. This condition describes an exothermic process that involves an increase in system entropy. In this case, ΔG will be negative regardless of the temperature. Such a process is spontaneous at all temperatures.
These four scenarios are summarized in Figure $1$.
Predicting the Temperature Dependence of Spontaneity
The incomplete combustion of carbon is described by the following equation:
$\ce{2C}(s)+\ce{O2}(g)⟶\ce{2CO}(g) \nonumber$
How does the spontaneity of this process depend upon temperature?
Solution
Combustion processes are exothermic (ΔH < 0). This particular reaction involves an increase in entropy due to the accompanying increase in the amount of gaseous species (net gain of one mole of gas, ΔS > 0). The reaction is therefore spontaneous (ΔG < 0) at all temperatures.
Exercise $3$
Popular chemical hand warmers generate heat by the air-oxidation of iron:
$\ce{4Fe}(s)+\ce{3O2}(g)⟶\ce{2Fe2O3}(s) \nonumber$
How does the spontaneity of this process depend upon temperature?
Answer
ΔH and ΔS are negative; the reaction is spontaneous at low temperatures.
When considering the conclusions drawn regarding the temperature dependence of spontaneity, it is important to keep in mind what the terms “high” and “low” mean. Since these terms are adjectives, the temperatures in question are deemed high or low relative to some reference temperature. A process that is nonspontaneous at one temperature but spontaneous at another will necessarily undergo a change in “spontaneity” (as reflected by its ΔG) as temperature varies. This is clearly illustrated by a graphical presentation of the free energy change equation, in which ΔG is plotted on the y axis versus T on the x axis:
$ΔG=ΔH−TΔS$
$y=b+mx$
Such a plot is shown in Figure $2$. A process whose enthalpy and entropy changes are of the same arithmetic sign will exhibit a temperature-dependent spontaneity as depicted by the two yellow lines in the plot. Each line crosses from one spontaneity domain (positive or negative ΔG) to the other at a temperature that is characteristic of the process in question. This temperature is represented by the x-intercept of the line, that is, the value of T for which ΔG is zero:
$ΔG=0=ΔH−TΔS$
$T=\dfrac{ΔH}{ΔS}$
And so, saying a process is spontaneous at “high” or “low” temperatures means the temperature is above or below, respectively, that temperature at which ΔG for the process is zero. As noted earlier, this condition describes a system at equilibrium.
Equilibrium Temperature for a Phase Transition
As defined in the chapter on liquids and solids, the boiling point of a liquid is the temperature at which its solid and liquid phases are in equilibrium (that is, when vaporization and condensation occur at equal rates). Use the information in Appendix G to estimate the boiling point of water.
Solution
The process of interest is the following phase change:
$\ce{H2O}(l)⟶\ce{H2O}(g) \nonumber$
When this process is at equilibrium, ΔG = 0, so the following is true:
$0=ΔH°−TΔS°\hspace{40px}\ce{or}\hspace{40px}T=\dfrac{ΔH°}{ΔS°} \nonumber$
Using the standard thermodynamic data from Appendix G,
\begin{align} ΔH°&=ΔH^\circ_\ce{f}(\ce{H2O}(g))−ΔH^\circ_\ce{f}(\ce{H2O}(l)) \nonumber\ &=\mathrm{−241.82\: kJ/mol−(−285.83\: kJ/mol)=44.01\: kJ/mol} \nonumber \end{align}
\begin{align} ΔS°&=ΔS^\circ_{298}(\ce{H2O}(g))−ΔS^\circ_{298}(\ce{H2O}(l)) \nonumber\ &=\mathrm{188.8\: J/K⋅mol−70.0\: J/K⋅mol=118.8\: J/K⋅mol} \nonumber \end{align}
$T=\dfrac{ΔH°}{ΔS°}=\mathrm{\dfrac{44.01×10^3\:J/mol}{118.8\:J/K⋅mol}=370.5\:K=97.3\:°C} \nonumber$
The accepted value for water’s normal boiling point is 373.2 K (100.0 °C), and so this calculation is in reasonable agreement. Note that the values for enthalpy and entropy changes data used were derived from standard data at 298 K (Appendix G). If desired, you could obtain more accurate results by using enthalpy and entropy changes determined at (or at least closer to) the actual boiling point.
Exercise $4$
Use the information in Appendix G to estimate the boiling point of CS2.
Answer
313 K (accepted value 319 K)
Temperature Dependence of the Equilibrium Constant
The fact that ΔG° and K are related provides us with another explanation of why equilibrium constants are temperature dependent. This relationship can be expressed as follows:
$\ln K=-\dfrac{\Delta H^\circ}{RT}+\dfrac{\Delta S^\circ}{R} \label{18.40}$
Assuming ΔH° and ΔS° are temperature independent, for an exothermic reaction (ΔH° < 0), the magnitude of K decreases with increasing temperature, whereas for an endothermic reaction (ΔH° > 0), the magnitude of K increases with increasing temperature. The quantitative relationship expressed in Equation $\ref{18.40}$ agrees with the qualitative predictions made by applying Le Chatelier’s principle. Because heat is produced in an exothermic reaction, adding heat (by increasing the temperature) will shift the equilibrium to the left, favoring the reactants and decreasing the magnitude of K. Conversely, because heat is consumed in an endothermic reaction, adding heat will shift the equilibrium to the right, favoring the products and increasing the magnitude of K. Equation $\ref{18.40}$ also shows that the magnitude of ΔH° dictates how rapidly K changes as a function of temperature. In contrast, the magnitude and sign of ΔS° affect the magnitude of K but not its temperature dependence.
If we know the value of K at a given temperature and the value of ΔH° for a reaction, we can estimate the value of K at any other temperature, even in the absence of information on ΔS°. Suppose, for example, that K1 and K2 are the equilibrium constants for a reaction at temperatures T1 and T2, respectively. Applying Equation $\ref{18.40}$ gives the following relationship at each temperature:
\begin{align}\ln K_1&=\dfrac{-\Delta H^\circ}{RT_1}+\dfrac{\Delta S^\circ}{R} \ \ln K_2 &=\dfrac{-\Delta H^\circ}{RT_2}+\dfrac{\Delta S^\circ}{R}\end{align}
Subtracting $\ln K_1$ from $\ln K_2$,
$\ln K_2-\ln K_1=\ln\dfrac{K_2}{K_1}=\dfrac{\Delta H^\circ}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right) \label{18.41}$
Thus calculating ΔH° from tabulated enthalpies of formation and measuring the equilibrium constant at one temperature (K1) allow us to calculate the value of the equilibrium constant at any other temperature (K2), assuming that ΔH° and ΔS° are independent of temperature. The linear relation between $\ln K$and the standard enthalpies and entropies in Equation $\ref{18.41}$ is known as the van’t Hoff equation. It shows that a plot of $\ln K$ vs. $1/T$ should be a line with slope $-\Delta_r{H^o}/R$ and intercept $\Delta_r{S^o}/R$.
Hence, these thermodynamic enthalpy and entropy changes for a reversible reaction can be determined from plotting $\ln K$ vs. $1/T$ data without the aid of calorimetry. Of course, the main assumption here is that $\Delta_r{H^o}$ and $\Delta_r{S^o}$ are only very weakly dependent on $T$, which is usually valid over a narrow temperature range.
Example $4$
The equilibrium constant for the formation of NH3 from H2 and N2 at 25°C is Kp = 5.4 × 105. What is Kp at 500°C? (Use the data from Example 10.)
Given: balanced chemical equation, ΔH°, initial and final T, and Kp at 25°C
Asked for: Kp at 500°C
Strategy:
Convert the initial and final temperatures to kelvins. Then substitute appropriate values into Equation $\ref{18.41}$ to obtain K2, the equilibrium constant at the final temperature.
Solution:
The value of ΔH° for the reaction obtained using Hess’s law is −91.8 kJ/mol of N2. If we set T1 = 25°C = 298.K and T2 = 500°C = 773 K, then from Equation $\ref{18.41}$ we obtain the following:
\begin{align}\ln\dfrac{K_2}{K_1}&=\dfrac{\Delta H^\circ}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right) \&=\dfrac{(-\textrm{91.8 kJ})(\textrm{1000 J/kJ})}{\textrm{8.314 J/K}}\left(\dfrac{1}{\textrm{298 K}}-\dfrac{1}{\textrm{773 K}}\right)=-22.8 \ \dfrac{K_2}{K_1}&=1.3\times10^{-10} \ K_2&=(5.4\times10^5)(1.3\times10^{-10})=7.0\times10^{-5}\end{align}
Thus at 500°C, the equilibrium strongly favors the reactants over the products.
Exercise $4$
In the exercise in Example $3$, you calculated Kp = 2.2 × 1012 for the reaction of NO with O2 to give NO2 at 25°C. Use the ΔHf values in the exercise in Example 10 to calculate Kp for this reaction at 1000°C.
Answer: 5.6 × 10−4
The Van't Hoff Equation: https://youtu.be/4vk6idAXp_A
Summary
For a reversible process that does not involve external work, we can express the change in free energy in terms of volume, pressure, entropy, and temperature. If we assume ideal gas behavior, the ideal gas law allows us to express ΔG in terms of the partial pressures of the reactants and products, which gives us a relationship between ΔG and Kp, the equilibrium constant of a reaction involving gases, or K, the equilibrium constant expressed in terms of concentrations. If ΔG° < 0, then K or Kp > 1, and products are favored over reactants. If ΔG° > 0, then K or Kp < 1, and reactants are favored over products. If ΔG° = 0, then K or Kp = 1, and the system is at equilibrium. We can use the measured equilibrium constant K at one temperature and ΔH° to estimate the equilibrium constant for a reaction at any other temperature.
• Anonymous | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/19%3A_Spontaneous_Change%3A_Entropy_and_Gibbs_Energy/19.7%3A_G_and_K_as_Functions_of_Temperature.txt |
Learning Objectives
Endergonic reactions can also be pushed by coupling them to another reaction, which is strongly exergonic, often through shared intermediates.
Many chemicals' reactions are endergonic (i.e., not spontaneous ($\Delta G > 0$)) and require energy to be externally applied to occur. However, these reaction can be coupled to a separate, exergonic (thermodynamically favorable $\Delta G <0$) reactions that 'drive' the thermodynamically unfavorable one by coupling or 'mechanistically joining' the two reactions often via a share intermediate. Since Gibbs Energy is a state function, the $\Delta G$ values for each half-reaction may be summed, to yield the combined $\Delta G$ of the coupled reaction.
One simple example of the coupling of reaction is the decomposition of calcium carbonate:
$CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)} \;\;\;\;\;\;\; \Delta G^o = 130.40 \;kJ/mol \label{1}$
The strongly positive $\Delta G$ for this reaction is reactant-favored. If the temperature is raised above 837 ºC, this reaction becomes spontaneous and favors the products. Now, let's consider a second and completely different reaction that can be coupled ot this reaction. The combustion of coal released by burning the coal $\Delta G^o = -394.36 \;kJ/mol$ is greater than the energy required to decompose calcium carbonate ($\Delta G^o = 130.40 \;kJ/mo$).
$C_{(s)} + O_2 \rightleftharpoons CO_{2(g)} \;\;\;\;\;\;\; \Delta G^o = -394.36 \;kJ/mol \label{2}$
If reactions $\ref{1}$ and $\ref{2}$ were added
$CaCO_{3(s)} + C_{(s)} + O_2 \rightleftharpoons CaO_{(s)} + 2CO_[{2(g)} \;\;\;\; \Delta G^o = -263.96 \;kJ/mol \label{3}$ and then Hess's Law were applied, the combined reaction (Equation $\ref{3}$) is product-favored with $\Delta G^o = -263.96 \;kJ/mol$. This is because the reactant-favored reaction (Equation $\ref{2}$) is linked to a strong spontaneous reaction so that both reactions yield products. Notice that the $\Delta G$ for the coupled reaction is the sum of the constituent reactions; this is a consequence of Gibbs energy being a state function:
$\Delta G^o = 130.40 \;kJ/mol+ -394.36 \;kJ/mol = -263.96 \;kJ/mol$
Coupled Reactions in Biology
This is a common feature in biological systems where some enzyme-catalyzed reactions are interpretable as two coupled half-reactions, one spontaneous and the other non-spontaneous. Organisms often the hydrolysis of ATP (adenosine triphosphate) to generate ADP (adenosine diphosphate) as the spontaneous coupling reaction (Figure $1$).
$ATP + H_2O \rightleftharpoons ADP + P_i \label{4}$
• $P_i$ is inorganic phosphate ion
The phosphoanhydride bonds formed by ejecting water between two phosphate group of ATP exhibit a large negative $-\Delta G$ of hydrolysis and are thus often termed "high energy" bonds. However, as with all bonds, energy is requires to break these bonds, but the thermodynamic Gibbs energy difference is strongly "energy releasing" when including the solvation thermodynamics of the phosphate ions; $\Delta G$ for this reaction is - 31 kJ/mol.
ATP is the major 'energy' molecule produced by metabolism, and it serves as a sort of 'energy source' in cell: ATP is dispatched to wherever a non-spontaneous reaction needs to occurs so that the two reactions are coupled so that the overall reaction is thermodynamically favored.
Example $1$: Phosphorylating Carboxylic Acids
Aldehydes $RCHO$ are organic compounds that can be oxidized to generate carboxylic acids and nicotinamide adenine dinucleotide (NAD) is a coenzyme found in all living cells and in the reduced form, $NAD^+$, it acts as an oxidizing agent that can accept electrons from other molecules.
The NAD+-linked oxidation of an aldehyde is practically irreversible with an equilibrium that strongly favors the products ($\Delta G >> 0$:
$RCHO + NAD^+ + H_2O \rightleftharpoons RCOOH + NADH + H^+ \label{Spontaneous}$
The position of equilibrium for phosphorylating carboxylic acids lies very much to the left:
$RCOOH + P_i \rightleftharpoons RC(=O)(O-P_i) + H_2O \label{Nonspontaneous}$
• (P_i\) is inorganic phosphate ion.
The non-spontaneous formation of a phosphorylated carboxylic acid can be driven by coupling it to the (spontaneous) NAD+-linked oxidation of an aldehyde?
Similarly, ATP hydrolysis can be used to combine amino acids together to generate polypeptides (and proteins) as graphically illustrated by Figure $2$. In this case, the reverse of Equation $\ref{4}$ is initially coupled to the oxidizing glucose by oxygen
$C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O \label{5}$ Reaction $\ref{5}$ is strongly spontaneous with $\Delta G = −2880 \;kJ/mol$ or close to 100x greater energy capability than the hydrolysis of ATP in Equation $\ref{4}$. Hence, the equilibrium for this reaction so strongly favors the products that a single arrow is typically used in the chemical equation as it is essential irreversible. It may not be surpising that glucose and all sugars are very energetic moleculess since they are the primary energy source for life.
Summary
Two (or more) reactions may be combined such that a spontaneous reaction may be made 'drive' an nonspontaneous one. Such reactions may be considered coupled. Changes in Gibbs energy of the coupled reactions are additive.
Contributors and Attributions
• Stackexcahnge (TomD) | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/19%3A_Spontaneous_Change%3A_Entropy_and_Gibbs_Energy/19.8%3A_Coupled_Reactions.txt |
Learning Objectives
• To distinguish between galvanic and electrolytic cells.
In any electrochemical process, electrons flow from one chemical substance to another, driven by an oxidation–reduction (redox) reaction. A redox reaction occurs when electrons are transferred from a substance that is oxidized to one that is being reduced. The reductant is the substance that loses electrons and is oxidized in the process; the oxidant is the species that gains electrons and is reduced in the process. The associated potential energy is determined by the potential difference between the valence electrons in atoms of different elements.
Because it is impossible to have a reduction without an oxidation and vice versa, a redox reaction can be described as two half-reactions, one representing the oxidation process and one the reduction process. For the reaction of zinc with bromine, the overall chemical reaction is as follows:
$Zn_{(s)} + Br_{2(aq)} \rightarrow Zn^{2+}_{(aq)} + 2Br^−_{(aq)} \label{19.1}$
The half-reactions are as follows:
reduction half-reaction:
$Br_{2(aq)} + 2e^− \rightarrow 2Br^−_{(aq)} \label{19.2}$
oxidation half-reaction:
$Zn_{(s)} \rightarrow Zn^{2+}_{(aq)} + 2e^− \label{19.3}$
Each half-reaction is written to show what is actually occurring in the system; Zn is the reductant in this reaction (it loses electrons), and Br2 is the oxidant (it gains electrons). Adding the two half-reactions gives the overall chemical reaction (Equation $\ref{19.1}$). A redox reaction is balanced when the number of electrons lost by the reductant equals the number of electrons gained by the oxidant. Like any balanced chemical equation, the overall process is electrically neutral; that is, the net charge is the same on both sides of the equation.
Note
In any redox reaction, the number of electrons lost by the reductant equals the number of electrons gained by the oxidant.
In most of our discussions of chemical reactions, we have assumed that the reactants are in intimate physical contact with one another. Acid–base reactions, for example, are usually carried out with the acid and the base dispersed in a single phase, such as a liquid solution. With redox reactions, however, it is possible to physically separate the oxidation and reduction half-reactions in space, as long as there is a complete circuit, including an external electrical connection, such as a wire, between the two half-reactions. As the reaction progresses, the electrons flow from the reductant to the oxidant over this electrical connection, producing an electric current that can be used to do work. An apparatus that is used to generate electricity from a spontaneous redox reaction or, conversely, that uses electricity to drive a nonspontaneous redox reaction is called an electrochemical cell.
There are two types of electrochemical cells: galvanic cells and electrolytic cells. Galvanic cells are named for the Italian physicist and physician Luigi Galvani (1737–1798), who observed that dissected frog leg muscles twitched when a small electric shock was applied, demonstrating the electrical nature of nerve impulses. A galvanic (voltaic) cell uses the energy released during a spontaneous redox reaction (ΔG < 0) to generate electricity. This type of electrochemical cell is often called a voltaic cell after its inventor, the Italian physicist Alessandro Volta (1745–1827). In contrast, an electrolytic cell consumes electrical energy from an external source, using it to cause a nonspontaneous redox reaction to occur (ΔG > 0). Both types contain two electrodes, which are solid metals connected to an external circuit that provides an electrical connection between the two parts of the system (Figure $1$). The oxidation half-reaction occurs at one electrode (the anode), and the reduction half-reaction occurs at the other (the cathode). When the circuit is closed, electrons flow from the anode to the cathode. The electrodes are also connected by an electrolyte, an ionic substance or solution that allows ions to transfer between the electrode compartments, thereby maintaining the system’s electrical neutrality. In this section, we focus on reactions that occur in galvanic cells.
Electrochemical Cells: https://youtu.be/nyS1BQ2ZVIg
Galvanic (Voltaic) Cells
To illustrate the basic principles of a galvanic cell, let’s consider the reaction of metallic zinc with cupric ion (Cu2+) to give copper metal and Zn2+ ion. The balanced chemical equation is as follows:
$Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)} \label{19.4}$
We can cause this reaction to occur by inserting a zinc rod into an aqueous solution of copper(II) sulfate. As the reaction proceeds, the zinc rod dissolves, and a mass of metallic copper forms. These changes occur spontaneously, but all the energy released is in the form of heat rather than in a form that can be used to do work.
This same reaction can be carried out using the galvanic cell illustrated in part (a) in Figure $3$. To assemble the cell, a copper strip is inserted into a beaker that contains a 1 M solution of Cu2+ ions, and a zinc strip is inserted into a different beaker that contains a 1 M solution of Zn2+ ions. The two metal strips, which serve as electrodes, are connected by a wire, and the compartments are connected by a salt bridge, a U-shaped tube inserted into both solutions that contains a concentrated liquid or gelled electrolyte. The ions in the salt bridge are selected so that they do not interfere with the electrochemical reaction by being oxidized or reduced themselves or by forming a precipitate or complex; commonly used cations and anions are Na+ or K+ and NO3 or SO42, respectively. (The ions in the salt bridge do not have to be the same as those in the redox couple in either compartment.) When the circuit is closed, a spontaneous reaction occurs: zinc metal is oxidized to Zn2+ ions at the zinc electrode (the anode), and Cu2+ ions are reduced to Cu metal at the copper electrode (the cathode). As the reaction progresses, the zinc strip dissolves, and the concentration of Zn2+ ions in the Zn2+ solution increases; simultaneously, the copper strip gains mass, and the concentration of Cu2+ ions in the Cu2+ solution decreases (part (b) in Figure $3$). Thus we have carried out the same reaction as we did using a single beaker, but this time the oxidative and reductive half-reactions are physically separated from each other. The electrons that are released at the anode flow through the wire, producing an electric current. Galvanic cells therefore transform chemical energy into electrical energy that can then be used to do work.
The electrolyte in the salt bridge serves two purposes: it completes the circuit by carrying electrical charge and maintains electrical neutrality in both solutions by allowing ions to migrate between them. The identity of the salt in a salt bridge is unimportant, as long as the component ions do not react or undergo a redox reaction under the operating conditions of the cell. Without such a connection, the total positive charge in the Zn2+ solution would increase as the zinc metal dissolves, and the total positive charge in the Cu2+ solution would decrease. The salt bridge allows charges to be neutralized by a flow of anions into the Zn2+ solution and a flow of cations into the Cu2+ solution. In the absence of a salt bridge or some other similar connection, the reaction would rapidly cease because electrical neutrality could not be maintained.
A voltmeter can be used to measure the difference in electrical potential between the two compartments. Opening the switch that connects the wires to the anode and the cathode prevents a current from flowing, so no chemical reaction occurs. With the switch closed, however, the external circuit is closed, and an electric current can flow from the anode to the cathode. The potential (Ecell) of the cell, measured in volts, is the difference in electrical potential between the two half-reactions and is related to the energy needed to move a charged particle in an electric field. In the cell we have described, the voltmeter indicates a potential of 1.10 V (part (a) in Figure $3$). Because electrons from the oxidation half-reaction are released at the anode, the anode in a galvanic cell is negatively charged. The cathode, which attracts electrons, is positively charged.
Note
A galvanic (voltaic) cell converts the energy released by a spontaneous chemical reaction to electrical energy. An electrolytic cell consumes electrical energy from an external source to drive a nonspontaneous chemical reaction.
Not all electrodes undergo a chemical transformation during a redox reaction. The electrode can be made from an inert, highly conducting metal such as platinum to prevent it from reacting during a redox process, where it does not appear in the overall electrochemical reaction. This phenomenon is illustrated in Example $1$.
Example $1$
A chemist has constructed a galvanic cell consisting of two beakers. One beaker contains a strip of tin immersed in aqueous sulfuric acid, and the other contains a platinum electrode immersed in aqueous nitric acid. The two solutions are connected by a salt bridge, and the electrodes are connected by a wire. Current begins to flow, and bubbles of a gas appear at the platinum electrode. The spontaneous redox reaction that occurs is described by the following balanced chemical equation:
$3Sn_{(s)} + 2NO^-_{3(aq)} + 8H^+_{(aq)} \rightarrow 3Sn^{2+}_{(aq)} + 2NO_{(g)} + 4H_2O_{(l)}$
For this galvanic cell,
1. write the half-reaction that occurs at each electrode.
2. indicate which electrode is the cathode and which is the anode.
3. indicate which electrode is the positive electrode and which is the negative electrode.
Given: galvanic cell and redox reaction
Asked for: half-reactions, identity of anode and cathode, and electrode assignment as positive or negative
Strategy:
1. Identify the oxidation half-reaction and the reduction half-reaction. Then identify the anode and cathode from the half-reaction that occurs at each electrode.
2. From the direction of electron flow, assign each electrode as either positive or negative.
Solution:
1. A In the reduction half-reaction, nitrate is reduced to nitric oxide. (The nitric oxide would then react with oxygen in the air to form NO2, with its characteristic red-brown color.) In the oxidation half-reaction, metallic tin is oxidized. The half-reactions corresponding to the actual reactions that occur in the system are as follows:
reduction: NO3(aq) + 4H+(aq) + 3e → NO(g) + 2H2O(l)
oxidation: Sn(s) → Sn2+(aq) + 2e
Thus nitrate is reduced to NO, while the tin electrode is oxidized to Sn2+.
1. Because the reduction reaction occurs at the Pt electrode, it is the cathode. Conversely, the oxidation reaction occurs at the tin electrode, so it is the anode.
2. B Electrons flow from the tin electrode through the wire to the platinum electrode, where they transfer to nitrate. The electric circuit is completed by the salt bridge, which permits the diffusion of cations toward the cathode and anions toward the anode. Because electrons flow from the tin electrode, it must be electrically negative. In contrast, electrons flow toward the Pt electrode, so that electrode must be electrically positive.
Exercise $1$
Consider a simple galvanic cell consisting of two beakers connected by a salt bridge. One beaker contains a solution of MnO4 in dilute sulfuric acid and has a Pt electrode. The other beaker contains a solution of Sn2+ in dilute sulfuric acid, also with a Pt electrode. When the two electrodes are connected by a wire, current flows and a spontaneous reaction occurs that is described by the following balanced chemical equation:
$2MnO^−_{4(aq)} + 5Sn^{2+}_{(aq)} + 16H^+_{(aq)} \rightarrow 2Mn^{2+}_{(aq)} + 5Sn^{4+}_{(aq)} + 8H_2O_{(l)}$
For this galvanic cell,
1. write the half-reaction that occurs at each electrode.
2. indicate which electrode is the cathode and which is the anode.
3. indicate which electrode is positive and which is negative.
Answer
1. MnO4(aq) + 8H+(aq) + 5e → Mn2+(aq) + 4H2O(l); Sn2+(aq) → Sn4+(aq) + 2e
2. The Pt electrode in the permanganate solution is the cathode; the one in the tin solution is the anode.
3. The cathode (electrode in beaker that contains the permanganate solution) is positive, and the anode (electrode in beaker that contains the tin solution) is negative.
Constructing Cell Diagrams
Because it is somewhat cumbersome to describe any given galvanic cell in words, a more convenient notation has been developed. In this line notation, called a cell diagram, the identity of the electrodes and the chemical contents of the compartments are indicated by their chemical formulas, with the anode written on the far left and the cathode on the far right. Phase boundaries are shown by single vertical lines, and the salt bridge, which has two phase boundaries, by a double vertical line. Thus the cell diagram for the Zn/Cu cell shown in part (a) in Figure $3$ is written as follows:
A cell diagram includes solution concentrations when they are provided.
Galvanic cells can have arrangements other than the examples we have seen so far. For example, the voltage produced by a redox reaction can be measured more accurately using two electrodes immersed in a single beaker containing an electrolyte that completes the circuit. This arrangement reduces errors caused by resistance to the flow of charge at a boundary, called the junction potential. One example of this type of galvanic cell is as follows:
$Pt_{(s)}\, | \, H_{2(g)} | HCl_{(aq)}\,|\, AgCl_{(s)} \,Ag_{(s)} \label{19.5}$
This cell diagram does not include a double vertical line representing a salt bridge because there is no salt bridge providing a junction between two dissimilar solutions. Moreover, solution concentrations have not been specified, so they are not included in the cell diagram. The half-reactions and the overall reaction for this cell are as follows:
cathode reaction:
$AgCl_{(s)} + e^− \rightarrow Ag_{(s)} + Cl^−_{(aq)} \label{19.6}$
anode reaction:
$\dfrac{1}{2}\,\mathrm{H_{2(g)}}\rightarrow\mathrm{H^+_{(aq)}}+\mathrm{e^-} \label{19.7}$
overall:
$\mathrm{AgCl_{(s)}}+\dfrac{1}{2}\mathrm{H_{2(g)}}\rightarrow\mathrm{Ag_{(s)}}+\mathrm{Cl^-_{(aq)}}+\mathrm{H^+_{(aq)}} \label{19.8}$
A single-compartment galvanic cell will initially exhibit the same voltage as a galvanic cell constructed using separate compartments, but it will discharge rapidly because of the direct reaction of the reactant at the anode with the oxidized member of the cathodic redox couple. Consequently, cells of this type are not particularly useful for producing electricity.
Example $2$
Draw a cell diagram for the galvanic cell described in Example 1. The balanced chemical reaction is as follows:
$3Sn_{(s)} + 2NO^−_{3(aq)} + 8H^+_{(aq)} \rightarrow 3Sn^{2+}_{(aq)} + 2NO_{(g)} + 4H_2O_{(l)}$
Given: galvanic cell and redox reaction
Asked for: cell diagram
Strategy:
Using the symbols described, write the cell diagram beginning with the oxidation half-reaction on the left.
Solution
The anode is the tin strip, and the cathode is the Pt electrode. Beginning on the left with the anode, we indicate the phase boundary between the electrode and the tin solution by a vertical bar. The anode compartment is thus Sn(s)∣Sn2+(aq). We could include H2SO4(aq) with the contents of the anode compartment, but the sulfate ion (as HSO4) does not participate in the overall reaction, so it does not need to be specifically indicated. The cathode compartment contains aqueous nitric acid, which does participate in the overall reaction, together with the product of the reaction (NO) and the Pt electrode. These are written as HNO3(aq)∣NO(g)∣Pt(s), with single vertical bars indicating the phase boundaries. Combining the two compartments and using a double vertical bar to indicate the salt bridge,
$Sn_{(s)}\,|\,Sn^{2+}_{(aq)}\,||\,HNO_{3(aq)}\,|\,NO_{(g)}\,|\,Pt_{(s)}$
The solution concentrations were not specified, so they are not included in this cell diagram.
Exercise $2$
Draw the cell diagram for the following reaction, assuming the concentration of Ag+ and Mg2+ are each 1 M:
$Mg_{(s)} + 2Ag^+_{(aq)} \rightarrow Mg^{2+}_{(aq)} + 2Ag_{(s)}$
Answer
$Mg(s) \,|\,Mg^{2+}(aq, 1 M \,||\,Ag^+(aq, 1 M)\,|\,Ag(s)$
Cell Diagrams: https://youtu.be/IKqOAfivem8
Summary
• A galvanic (voltaic) cell uses the energy released during a spontaneous redox reaction to generate electricity, whereas an electrolytic cell consumes electrical energy from an external source to force a reaction to occur.
Electrochemistry is the study of the relationship between electricity and chemical reactions. The oxidation–reduction reaction that occurs during an electrochemical process consists of two half-reactions, one representing the oxidation process and one the reduction process. The sum of the half-reactions gives the overall chemical reaction. The overall redox reaction is balanced when the number of electrons lost by the reductant equals the number of electrons gained by the oxidant. An electric current is produced from the flow of electrons from the reductant to the oxidant. An electrochemical cell can either generate electricity from a spontaneous redox reaction or consume electricity to drive a nonspontaneous reaction. In a galvanic (voltaic) cell, the energy from a spontaneous reaction generates electricity, whereas in an electrolytic cell, electrical energy is consumed to drive a nonspontaneous redox reaction. Both types of cells use two electrodes that provide an electrical connection between systems that are separated in space. The oxidative half-reaction occurs at the anode, and the reductive half-reaction occurs at the cathode. A salt bridge connects the separated solutions, allowing ions to migrate to either solution to ensure the system’s electrical neutrality. A voltmeter is a device that measures the flow of electric current between two half-reactions. The potential of a cell, measured in volts, is the energy needed to move a charged particle in an electric field. An electrochemical cell can be described using line notation called a cell diagram, in which vertical lines indicate phase boundaries and the location of the salt bridge. Resistance to the flow of charge at a boundary is called the junction potential. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/20%3A_Electrochemistry/20.1%3A_Electrode_Potentials_and_their_Measurement.txt |
Learning Objectives
• To use redox potentials to predict whether a reaction is spontaneous.
• To balance redox reactions using half-reactions.
In a galvanic cell, current is produced when electrons flow externally through the circuit from the anode to the cathode because of a difference in potential energy between the two electrodes in the electrochemical cell. In the Zn/Cu system, the valence electrons in zinc have a substantially higher potential energy than the valence electrons in copper because of shielding of the s electrons of zinc by the electrons in filled d orbitals. Hence electrons flow spontaneously from zinc to copper(II) ions, forming zinc(II) ions and metallic copper. Just like water flowing spontaneously downhill, which can be made to do work by forcing a waterwheel, the flow of electrons from a higher potential energy to a lower one can also be harnessed to perform work.
Because the potential energy of valence electrons differs greatly from one substance to another, the voltage of a galvanic cell depends partly on the identity of the reacting substances. If we construct a galvanic cell similar to the one in part (a) in Figure 19.3 but instead of copper use a strip of cobalt metal and 1 M Co2+ in the cathode compartment, the measured voltage is not 1.10 V but 0.51 V. Thus we can conclude that the difference in potential energy between the valence electrons of cobalt and zinc is less than the difference between the valence electrons of copper and zinc by 0.59 V.
The measured potential of a cell also depends strongly on the concentrations of the reacting species and the temperature of the system. To develop a scale of relative potentials that will allow us to predict the direction of an electrochemical reaction and the magnitude of the driving force for the reaction, the potentials for oxidations and reductions of different substances must be measured under comparable conditions. To do this, chemists use the standard cell potential (E°cell), defined as the potential of a cell measured under standard conditions—that is, with all species in their standard states (1 M for solutions,Concentrated solutions of salts (about 1 M) generally do not exhibit ideal behavior, and the actual standard state corresponds to an activity of 1 rather than a concentration of 1 M. Corrections for nonideal behavior are important for precise quantitative work but not for the more qualitative approach that we are taking here. 1 atm for gases, pure solids or pure liquids for other substances) and at a fixed temperature, usually 25°C.
Note
Measured redox potentials depend on the potential energy of valence electrons, the concentrations of the species in the reaction, and the temperature of the system.
Measuring Standard Electrode Potentials
It is physically impossible to measure the potential of a single electrode: only the difference between the potentials of two electrodes can be measured. (This is analogous to measuring absolute enthalpies or free energies. Recall that only differences in enthalpy and free energy can be measured.) We can, however, compare the standard cell potentials for two different galvanic cells that have one kind of electrode in common. This allows us to measure the potential difference between two dissimilar electrodes. For example, the measured standard cell potential (E°) for the Zn/Cu system is 1.10 V, whereas E° for the corresponding Zn/Co system is 0.51 V. This implies that the potential difference between the Co and Cu electrodes is 1.10 V − 0.51 V = 0.59 V. In fact, that is exactly the potential measured under standard conditions if a cell is constructed with the following cell diagram:
$Co_{(s)} ∣ Co^{2+}(aq, 1 M)∥Cu^{2+}(aq, 1 M) ∣ Cu (s)\;\;\; E°=0.59\; V \label{19.9}$
This cell diagram corresponds to the oxidation of a cobalt anode and the reduction of Cu2+ in solution at the copper cathode.
All tabulated values of standard electrode potentials by convention are listed for a reaction written as a reduction, not as an oxidation, to be able to compare standard potentials for different substances (Table P1). The standard cell potential (E°cell) is therefore the difference between the tabulated reduction potentials of the two half-reactions, not their sum:
$E°_{cell} = E°_{cathode} − E°_{anode} \label{19.10}$
In contrast, recall that half-reactions are written to show the reduction and oxidation reactions that actually occur in the cell, so the overall cell reaction is written as the sum of the two half-reactions. According to Equation $\ref{19.10}$, when we know the standard potential for any single half-reaction, we can obtain the value of the standard potential of many other half-reactions by measuring the standard potential of the corresponding cell.
Note
The overall cell reaction is the sum of the two half-reactions, but the cell potential is the difference between the reduction potentials:
$E°_{cell} = E°_{cathode} − E°_{anode}$
Although it is impossible to measure the potential of any electrode directly, we can choose a reference electrode whose potential is defined as 0 V under standard conditions. The standard hydrogen electrode (SHE) is universally used for this purpose and is assigned a standard potential of 0 V. It consists of a strip of platinum wire in contact with an aqueous solution containing 1 M H+. The [H+] in solution is in equilibrium with H2 gas at a pressure of 1 atm at the Pt-solution interface (Figure $2$). Protons are reduced or hydrogen molecules are oxidized at the Pt surface according to the following equation:
$2H^+_{(aq)}+2e^− \rightleftharpoons H_{2(g)} \label{19.11}$
One especially attractive feature of the SHE is that the Pt metal electrode is not consumed during the reaction.
Figure $3$ shows a galvanic cell that consists of a SHE in one beaker and a Zn strip in another beaker containing a solution of Zn2+ ions. When the circuit is closed, the voltmeter indicates a potential of 0.76 V. The zinc electrode begins to dissolve to form Zn2+, and H+ ions are reduced to H2 in the other compartment. Thus the hydrogen electrode is the cathode, and the zinc electrode is the anode. The diagram for this galvanic cell is as follows:
$Zn_{(s)}∣Zn^{2+}_{(aq)}∥H^+(aq, 1 M)∣H_2(g, 1 atm)∣Pt_{(s)} \label{19.12}$
The half-reactions that actually occur in the cell and their corresponding electrode potentials are as follows:
• cathode: $2H^+_{(aq)} + 2e^− \rightarrow H_{2(g)}\;\;\; E°_{cathode}=0 V \label{19.13}$
• anode: $Zn_{(s)} \rightarrow Zn^{2+}_{(aq)}+2e^−\;\;\; E°_{anode}=−0.76\; V \label{19.14}$
• overall: $Zn_{(s)}+2H^+_{(aq)} \rightarrow Zn^{2+}_{(aq)}+H_{2(g)} \label{19.15}$
$E°_{cell}=E°_{cathode}−E°_{anode}=0.76\; V$
Although the reaction at the anode is an oxidation, by convention its tabulated E° value is reported as a reduction potential. The potential of a half-reaction measured against the SHE under standard conditions is called the standard electrode potential for that half-reaction.In this example, the standard reduction potential for Zn2+(aq) + 2e → Zn(s) is −0.76 V, which means that the standard electrode potential for the reaction that occurs at the anode, the oxidation of Zn to Zn2+, often called the Zn/Zn2+ redox couple, or the Zn/Zn2+ couple, is −(−0.76 V) = 0.76 V. We must therefore subtract E°anode from E°cathode to obtain E°cell: 0 − (−0.76 V) = 0.76 V.
Because electrical potential is the energy needed to move a charged particle in an electric field, standard electrode potentials for half-reactions are intensive properties and do not depend on the amount of substance involved. Consequently, E° values are independent of the stoichiometric coefficients for the half-reaction, and, most important, the coefficients used to produce a balanced overall reaction do not affect the value of the cell potential.
Note
E° values do NOT depend on the stoichiometric coefficients for a half-reaction, because it is an intensive property.
Standard Electrode Potentials
To measure the potential of the Cu/Cu2+ couple, we can construct a galvanic cell analogous to the one shown in Figure $3$ but containing a Cu/Cu2+ couple in the sample compartment instead of Zn/Zn2+. When we close the circuit this time, the measured potential for the cell is negative (−0.34 V) rather than positive. The negative value of E°cell indicates that the direction of spontaneous electron flow is the opposite of that for the Zn/Zn2+ couple. Hence the reactions that occur spontaneously, indicated by a positive E°cell, are the reduction of Cu2+ to Cu at the copper electrode. The copper electrode gains mass as the reaction proceeds, and H2 is oxidized to H+ at the platinum electrode. In this cell, the copper strip is the cathode, and the hydrogen electrode is the anode. The cell diagram therefore is written with the SHE on the left and the Cu2+/Cu couple on the right:
$Pt_{(s)}∣H_2(g, 1 atm)∣H^+(aq, 1\; M)∥Cu^{2+}(aq, 1 M)∣Cu_{(s)} \label{19.16}$
The half-cell reactions and potentials of the spontaneous reaction are as follows:
• Cathode: $Cu^{2+}{(aq)} + 2e^− \rightarrow Cu_{(g)}\;\;\; E°_{cathode} = 0.34\; V \label{19.17}$
• Anode: $H_{2(g)} \rightarrow 2H^+_{(aq)} + 2e^−\;\;\; E°_{anode} = 0\; V \label{19.18}$
• Overall: $H_{2(g)} + Cu^{2+}_{(aq)} \rightarrow 2H^+_{(aq)} + Cu_{(s)} \label{19.19}$
$E°_{cell} = E°_{cathode}− E°_{anode} = 0.34\; V$
Thus the standard electrode potential for the Cu2+/Cu couple is 0.34 V.
Electrode Potentials and ECell: https://youtu.be/zeeAXleT1c0
Balancing Redox Reactions Using the Half-Reaction Method
Previously, we described a method for balancing redox reactions using oxidation numbers. Oxidation numbers were assigned to each atom in a redox reaction to identify any changes in the oxidation states. Here we present an alternative approach to balancing redox reactions, the half-reaction method, in which the overall redox reaction is divided into an oxidation half-reaction and a reduction half-reaction, each balanced for mass and charge. This method more closely reflects the events that take place in an electrochemical cell, where the two half-reactions may be physically separated from each other.
We can illustrate how to balance a redox reaction using half-reactions with the reaction that occurs when Drano, a commercial solid drain cleaner, is poured into a clogged drain. Drano contains a mixture of sodium hydroxide and powdered aluminum, which in solution reacts to produce hydrogen gas:
$Al_{(s)} + OH^−_{(aq)} \rightarrow Al(OH)^−_{4(aq)} + H_{2(g)} \label{19.20}$
In this reaction, $Al_{(s)}$ is oxidized to Al3+, and H+ in water is reduced to H2 gas, which bubbles through the solution, agitating it and breaking up the clogs.
The overall redox reaction is composed of a reduction half-reaction and an oxidation half-reaction. From the standard electrode potentials listed Table P1, we find the corresponding half-reactions that describe the reduction of H+ ions in water to H2and the oxidation of Al to Al3+ in basic solution:
• reduction: $2H_2O_{(l)} + 2e^− \rightarrow 2OH^−_{(aq)} + H_{2(g)} \label{19.21}$
• oxidation: $Al_{(s)} + 4OH^−_{(aq)} \rightarrow Al(OH)^−_{4(aq)} + 3e^− \label{19.22}$
The half-reactions chosen must exactly reflect the reaction conditions, such as the basic conditions shown here. Moreover, the physical states of the reactants and the products must be identical to those given in the overall reaction, whether gaseous, liquid, solid, or in solution.
In Equation $\ref{19.21}$, two H+ ions gain one electron each in the reduction; in Equation $\ref{19.22}$, the aluminum atom loses three electrons in the oxidation. The charges are balanced by multiplying the reduction half-reaction (Equation $\ref{19.21}$) by 3 and the oxidation half-reaction (Equation $\ref{19.22}$) by 2 to give the same number of electrons in both half-reactions:
• Adding the two half-reactions,
$6H_2O_{(l)} + 2Al_{(s)} + 8OH^−_{(aq)} \rightarrow 2Al(OH)^−{4(aq)} + 3H_{2(g)} + 6OH^−_{(aq)} \label{19.25}$
Simplifying by canceling substances that appear on both sides of the equation,
$6H_2O_{(l)} + 2Al_{(s)} + 2OH^−_{(aq)} \rightarrow 2Al(OH)^−_{4(aq)} + 3H_{2(g)} \label{19.26}$
We have a −2 charge on the left side of the equation and a −2 charge on the right side. Thus the charges are balanced, but we must also check that atoms are balanced:
$2Al + 8O + 14H = 2Al + 8O + 14H \label{19.27}$
The atoms also balance, so Equation $\ref{19.26}$ is a balanced chemical equation for the redox reaction depicted in Equation $\ref{19.20}$.
Note
The half-reaction method requires that half-reactions exactly reflect reaction conditions, and the physical states of the reactants and the products must be identical to those in the overall reaction.
We can also balance a redox reaction by first balancing the atoms in each half-reaction and then balancing the charges. With this alternative method, we do not need to use the half-reactions listed in Table P1 but instead focus on the atoms whose oxidation states change, as illustrated in the following steps:
Step 1: Write the reduction half-reaction and the oxidation half-reaction.
For the reaction shown in Equation $\ref{19.20}$, hydrogen is reduced from H+ in OH to H2, and aluminum is oxidized from Al° to Al3+:
• Step 2: Balance the atoms by balancing elements other than O and H. Then balance O atoms by adding H2O and balance H atoms by adding H+.
Elements other than O and H in the previous two equations are balanced as written, so we proceed with balancing the O atoms. We can do this by adding water to the appropriate side of each half-reaction:
• Balancing H atoms by adding H+, we obtain the following:
• We have now balanced the atoms in each half-reaction, but the charges are not balanced.
Step 3: Balance the charges in each half-reaction by adding electrons.
Two electrons are gained in the reduction of H+ ions to H2, and three electrons are lost during the oxidation of Al° to Al3+:
• Step 4: Multiply the reductive and oxidative half-reactions by appropriate integers to obtain the same number of electrons in both half-reactions.
In this case, we multiply Equation $\ref{19.34}$ (the reductive half-reaction) by 3 and Equation $\ref{19.35}$ (the oxidative half-reaction) by 2 to obtain the same number of electrons in both half-reactions:
• Step 5: Add the two half-reactions and cancel substances that appear on both sides of the equation.
Adding and, in this case, canceling 8H+, 3H2O, and 6e,
$2Al_{(s)} + 5H_2O_{(l)} + 3OH^−_{(aq)} + H^+_{(aq)} \rightarrow 2Al(OH)^−_{4(aq)} + 3H_{2(g)} \label{19.38}$
We have three OH and one H+ on the left side. Neutralizing the H+ gives us a total of 5H2O + H2O = 6H2O and leaves 2OH on the left side:
$2Al_{(s)} + 6H_2O_{(l)} + 2OH^−_{(aq)} \rightarrow 2Al(OH)^−_{4(aq)} + 3H_{2(g)} \label{19.39}$
Step 6: Check to make sure that all atoms and charges are balanced.
Equation $\ref{19.39}$ is identical to Equation $\ref{19.26}$, obtained using the first method, so the charges and numbers of atoms on each side of the equation balance.
Example $1$
In acidic solution, the redox reaction of dichromate ion ($Cr_2O_7^{2−}$) and iodide ($I^−$) can be monitored visually. The yellow dichromate solution reacts with the colorless iodide solution to produce a solution that is deep amber due to the presence of a green $Cr^{3+}_{(aq)}$ complex and brown I2(aq) ions (Figure $4$):
$Cr_2O^{2−}_{7(aq)} + I^−_{(aq)} \rightarrow Cr^{3+}_{(aq)} + I_{2(aq)}$
Balance this equation using half-reactions.
Given: redox reaction and Table P1
Asked for: balanced chemical equation using half-reactions
Strategy:
Follow the steps to balance the redox reaction using the half-reaction method.
Solution
From the standard electrode potentials listed in Table P1 we find the half-reactions corresponding to the overall reaction:
• reduction: $Cr_2O^{2−}_{7(aq)} + 14H^+_{(aq)} + 6e^− \rightarrow 2Cr^{3+}(_{(aq)} + 7H_2O_{(l)}$
• oxidation: $2I^−_{(aq)} \rightarrow I_{2(aq)} + 2e^−$
Balancing the number of electrons by multiplying the oxidation reaction by 3,
• oxidation: $6I^−_{(aq)} \rightarrow 3I_{2(aq)} + 6e^−$
Adding the two half-reactions and canceling electrons,
$Cr_2O^{2−}_{7(aq)} + 14H^+_{(aq)} + 6I^−_{(aq)} \rightarrow 2Cr^{3+}_{(aq)} + 7H_2O_{(l)} + 3I_{2(aq)}$
We must now check to make sure the charges and atoms on each side of the equation balance:
(−2) + 14 + (−6) = +6
+6 = +6
2Cr + 7O+ 14H+ 6I = 2Cr + 7O + 14H + 6I
The charges and atoms balance, so our equation is balanced.
We can also use the alternative procedure, which does not require the half-reactions listed in Table P1.
Step 1: Chromium is reduced from $Cr^{6+}$ in $Cr_2O_7^{2−}$ to $Cr^{3+}$, and $I^−$ ions are oxidized to $I_2$. Dividing the reaction into two half-reactions,
• reduction: $Cr_2O^{2−}_{7(aq)} \rightarrow Cr^{3+}_{(aq)}$
• oxidation: $I^−_{(aq)} \rightarrow I_{2(aq)}$
Step 2: Balancing the atoms other than oxygen and hydrogen,
• reduction: $Cr_2O^{2−}_{7(aq)} \rightarrow 2Cr^{3+}_{(aq)}$
• oxidation: $2I^−_{(aq)} \rightarrow I_{2(aq)}$
We now balance the O atoms by adding H2O—in this case, to the right side of the reduction half-reaction. Because the oxidation half-reaction does not contain oxygen, it can be ignored in this step.
• reduction: $Cr_2O^{2−}_{7(aq)} \rightarrow 2Cr^{3+}_{(aq)} + 7H_2O_{(l)}$
Next we balance the H atoms by adding H+ to the left side of the reduction half-reaction. Again, we can ignore the oxidation half-reaction.
• reduction: $Cr_2O^{2−}_{7(aq)} + 14H^+_{(aq)} \rightarrow 2Cr^{3+}_{(aq)} + 7H_2O_{(l)}$
Step 3: We must now add electrons to balance the charges. The reduction half-reaction (2Cr+6 to 2Cr+3) has a +12 charge on the left and a +6 charge on the right, so six electrons are needed to balance the charge. The oxidation half-reaction (2I to I2) has a −2 charge on the left side and a 0 charge on the right, so it needs two electrons to balance the charge:
• reduction: Cr2O72(aq) + 14H+(aq) + 6e → 2Cr3+(aq) + 7H2O(l)
• oxidation: 2I(aq) → I2(aq) + 2e
Step 4: To have the same number of electrons in both half-reactions, we must multiply the oxidation half-reaction by 3:
• oxidation: 6I(aq) → 3I2(s) + 6e
Step 5: Adding the two half-reactions and canceling substances that appear in both reactions,
• Cr2O72−(aq) + 14H+(aq) + 6I(aq) → 2Cr3+(aq) + 7H2O(l) + 3I2(aq)
Step 6: This is the same equation we obtained using the first method. Thus the charges and atoms on each side of the equation balance.
Exercise $1$
Copper is found as the mineral covellite ($CuS$). The first step in extracting the copper is to dissolve the mineral in nitric acid ($HNO_3$), which oxidizes sulfide to sulfate and reduces nitric acid to $NO$:
$CuS_{(s)} + HNO_{3(aq)} \rightarrow NO_{(g)} + CuSO_{4(aq)}$
Balance this equation using the half-reaction method.
Answer $3CuS_{(s)} + 8HNO{3(aq)} \rightarrow 8NO_{(g)} + 3CuSO_{4(aq)} + 4H_2O_{(l)}$
Balancing a Redox Reaction in Acidic Conditions: https://youtu.be/IB-fWLsI0lc
Calculating Standard Cell Potentials
The standard cell potential for a redox reaction (E°cell) is a measure of the tendency of reactants in their standard states to form products in their standard states; consequently, it is a measure of the driving force for the reaction, which earlier we called voltage. We can use the two standard electrode potentials we found earlier to calculate the standard potential for the Zn/Cu cell represented by the following cell diagram:
$Zn{(s)}∣Zn^{2+}(aq, 1 M)∥Cu^{2+}(aq, 1 M)∣Cu_{(s)} \label{19.40}$
We know the values of E°anode for the reduction of Zn2+ and E°cathode for the reduction of Cu2+, so we can calculate E°cell:
$E°_{cell} = E°_{cathode} − E°_{anode} = 1.10\; V$
This is the same value that is observed experimentally. If the value of E°cell is positive, the reaction will occur spontaneously as written. If the value of E°cell is negative, then the reaction is not spontaneous, and it will not occur as written under standard conditions; it will, however, proceed spontaneously in the opposite direction. As we shall see, this does not mean that the reaction cannot be made to occur at all under standard conditions. With a sufficient input of electrical energy, virtually any reaction can be forced to occur. Example 4 and its corresponding exercise illustrate how we can use measured cell potentials to calculate standard potentials for redox couples.
Note
A positive E°cell means that the reaction will occur spontaneously as written. A negative E°cell means that the reaction will proceed spontaneously in the opposite direction.
Example $2$
A galvanic cell with a measured standard cell potential of 0.27 V is constructed using two beakers connected by a salt bridge. One beaker contains a strip of gallium metal immersed in a 1 M solution of GaCl3, and the other contains a piece of nickel immersed in a 1 M solution of NiCl2. The half-reactions that occur when the compartments are connected are as follows:
cathode: Ni2+(aq) + 2e → Ni(s)
anode: Ga(s) → Ga3+(aq) + 3e
If the potential for the oxidation of Ga to Ga3+ is 0.55 V under standard conditions, what is the potential for the oxidation of Ni to Ni2+?
Given: galvanic cell, half-reactions, standard cell potential, and potential for the oxidation half-reaction under standard conditions
Asked for: standard electrode potential of reaction occurring at the cathode
Strategy:
1. Write the equation for the half-reaction that occurs at the anode along with the value of the standard electrode potential for the half-reaction.
2. Use Equation $\ref{19.10}$ to calculate the standard electrode potential for the half-reaction that occurs at the cathode. Then reverse the sign to obtain the potential for the corresponding oxidation half-reaction under standard conditions.
Solution
A We have been given the potential for the oxidation of Ga to Ga3+ under standard conditions, but to report the standard electrode potential, we must reverse the sign. For the reduction reaction Ga3+(aq) + 3e → Ga(s), E°anode = −0.55 V.
B Using the value given for E°cell and the calculated value of E°anode, we can calculate the standard potential for the reduction of Ni2+ to Ni from Equation $\ref{19.10}$:
cell = E°cathode − E°anode
0.27 V = E°cathode − (−0.55 V)
cathode = −0.28 V
This is the standard electrode potential for the reaction Ni2+(aq) + 2e → Ni(s). Because we are asked for the potential for the oxidation of Ni to Ni2+ under standard conditions, we must reverse the sign of E°cathode. Thus E° = −(−0.28 V) = 0.28 V for the oxidation. With three electrons consumed in the reduction and two produced in the oxidation, the overall reaction is not balanced. Recall, however, that standard potentials are independent of stoichiometry.
Exercise $2$
A galvanic cell is constructed with one compartment that contains a mercury electrode immersed in a 1 M aqueous solution of mercuric acetate $Hg(CH_3CO_2)_2$ and one compartment that contains a strip of magnesium immersed in a 1 M aqueous solution of $MgCl_2$. When the compartments are connected, a potential of 3.22 V is measured and the following half-reactions occur:
• cathode: Hg2+(aq) + 2e → Hg(l)
• anode: Mg(s) → Mg2+(aq) + 2e
If the potential for the oxidation of Mg to Mg2+ is 2.37 V under standard conditions, what is the standard electrode potential for the reaction that occurs at the anode?
Answer 0.85 V
We can use this procedure described to measure the standard potentials for a wide variety of chemical substances, some of which are listed in Table P2. These data allow us to compare the oxidative and reductive strengths of a variety of substances. The half-reaction for the standard hydrogen electrode (SHE) lies more than halfway down the list in Table $1$. All reactants that lie below the SHE in the table are stronger oxidants than H+, and all those that lie above the SHE are weaker. The strongest oxidant in the table is F2, with a standard electrode potential of 2.87 V. This high value is consistent with the high electronegativity of fluorine and tells us that fluorine has a stronger tendency to accept electrons (it is a stronger oxidant) than any other element.
Table $1$: Standard Potentials for Selected Reduction Half-Reactions at 25°C
Half-Reaction E° (V)
Li+(aq) + e $\rightleftharpoons$ Li(s) –3.040
Be2+(aq) + 2e $\rightleftharpoons$ Be(s) –1.99
Al3+(aq) + 3e $\rightleftharpoons$ Al(s) –1.676
Zn2+(aq) + 2e $\rightleftharpoons$ Zn(s) –0.7618
Ag2S(s) + 2e $\rightleftharpoons$ 2Ag(s) + S2−(aq) –0.71
Fe2+(aq) + 2e $\rightleftharpoons$ Fe(s) –0.44
Cr3+(aq) + e $\rightleftharpoons$ Cr2+(aq) –0.424
Cd2+(aq) + 2e $\rightleftharpoons$ Cd(s) –0.4030
PbSO4(s) + 2e $\rightleftharpoons$ Pb(s) + SO42−(aq) –0.356
Ni2+(aq) + 2e $\rightleftharpoons$ Ni(s) –0.257
2SO42−(aq) + 4H+(aq) + 2e $\rightleftharpoons$ S2O62−(aq) + 2H2O(l) –0.25
Sn2+(aq) + 2e $\rightleftharpoons$ Sn(s) −0.14
2H+(aq) + 2e $\rightleftharpoons$ H2(g) 0.00
Sn4+(aq) + 2e $\rightleftharpoons$ Sn2+(aq) 0.154
Cu2+(aq) + e $\rightleftharpoons$ Cu+(aq) 0.159
AgCl(s) + e $\rightleftharpoons$ Ag(s) + Cl(aq) 0.2223
Cu2+(aq) + 2e $\rightleftharpoons$ Cu(s) 0.3419
O2(g) + 2H2O(l) + 4e $\rightleftharpoons$ 4OH(aq) 0.401
H2SO3(aq) + 4H+(aq) + 4e $\rightleftharpoons$ S(s) + 3H2O(l) 0.45
I2(s) + 2e $\rightleftharpoons$ 2I(aq) 0.5355
MnO42−(aq) + 2H2O(l) + 2e $\rightleftharpoons$ MnO2(s) + 4OH(aq) 0.60
O2(g) + 2H+(aq) + 2e $\rightleftharpoons$ H2O2(aq) 0.695
H2SeO3(aq) + 4H+ + 4e $\rightleftharpoons$ Se(s) + 3H2O(l) 0.74
Fe3+(aq) + e $\rightleftharpoons$ Fe2+(aq) 0.771
Ag+(aq) + e $\rightleftharpoons$ Ag(s) 0.7996
NO3(aq) + 3H+(aq) + 2e $\rightleftharpoons$ HNO2(aq) + H2O(l) 0.94
Br2(aq) + 2e $\rightleftharpoons$ 2Br(aq) 1.087
MnO2(s) + 4H+(aq) + 2e $\rightleftharpoons$ Mn2+(aq) + 2H2O(l) 1.23
O2(g) + 4H+(aq) + 4e $\rightleftharpoons$ 2H2O(l) 1.229
Cr2O72−(aq) + 14H+(aq) + 6e $\rightleftharpoons$ 2Cr3+(aq) + 7H2O(l) 1.36
Cl2(g) + 2e $\rightleftharpoons$ 2Cl(aq) 1.396
$Ce^{4+}(aq) + e^− \rightleftharpoons Ce^{3+}(aq)$
1.44
PbO2(s) + HSO4(aq) + 3H+(aq) + 2e $\rightleftharpoons$ PbSO4(s) + 2H2O(l) 1.690
H2O2(aq) + 2H+(aq) + 2e $\rightleftharpoons$ 2H2O(l) 1.763
F2(g) + 2e$\rightleftharpoons$ 2F(aq) 2.87
Similarly, all species in Table $1$ that lie above H2 are stronger reductants than H2, and those that lie below H2 are weaker. The strongest reductant in the table is thus metallic lithium, with a standard electrode potential of −3.04 V. This fact might be surprising because cesium, not lithium, is the least electronegative element. The apparent anomaly can be explained by the fact that electrode potentials are measured in aqueous solution, where intermolecular interactions are important, whereas ionization potentials and electron affinities are measured in the gas phase. Due to its small size, the Li+ ion is stabilized in aqueous solution by strong electrostatic interactions with the negative dipole end of water molecules. These interactions result in a significantly greater ΔHhydration for Li+ compared with Cs+. Lithium metal is therefore the strongest reductant (most easily oxidized) of the alkali metals in aqueous solution.
Note
Species in Talbe Table $1$ (or Table P2) that lie above H2 are stronger reducing agents (more easily oxidized) than H2. Species that lie below H2 are stronger oxidizing agents.
Because the half-reactions shown in Table $1$ are arranged in order of their E° values, we can use the table to quickly predict the relative strengths of various oxidants and reductants. Any species on the left side of a half-reaction will spontaneously oxidize any species on the right side of another half-reaction that lies below it in the table. Conversely, any species on the right side of a half-reaction will spontaneously reduce any species on the left side of another half-reaction that lies above it in the table. We can use these generalizations to predict the spontaneity of a wide variety of redox reactions (E°cell > 0), as illustrated below.
Example $3$
The black tarnish that forms on silver objects is primarily Ag2S. The half-reaction for reversing the tarnishing process is as follows:
Ag2S(s)+2e→2Ag(s)+S2−(aq) E°=−0.69 V
1. Referring to Table $1$, predict which species—H2O2(aq), Zn(s), I(aq), Sn2+(aq)—can reduce Ag2S to Ag under standard conditions.
2. Of these species—H2O2(aq), Zn(s), I(aq), Sn2+(aq), identify which is the strongest reducing agent in aqueous solution and thus the best candidate for a commercial product.
3. From the data in Table $1$, suggest an alternative reducing agent that is readily available, inexpensive, and possibly more effective at removing tarnish.
Given: reduction half-reaction, standard electrode potential, and list of possible reductants
Asked for: reductants for Ag2S, strongest reductant, and potential reducing agent for removing tarnish
Strategy:
A From their positions inTable $1$, decide which species can reduce Ag2S. Determine which species is the strongest reductant.
B Use Table $1$ to identify a reductant for Ag2S that is a common household product.
Solution
We can solve the problem in one of two ways: (1) compare the relative positions of the four possible reductants with that of the Ag2S/Ag couple in Table $1$ or (2) compare E° for each species with E° for the Ag2S/Ag couple (−0.69 V).
1. A The species in Table $1$ are arranged from top to bottom in order of increasing reducing strength. Of the four species given in the problem, I(aq), Sn2+(aq), and H2O2(aq) lie above Ag2S, and one [Zn(s)] lies below it. We can therefore conclude that Zn(s) can reduce Ag2S(s) under standard conditions, whereas I(aq), Sn2+(aq), and H2O2(aq) cannot. Sn2+(aq) and H2O2(aq) appear twice in the table: on the left side (oxidant) in one half-reaction and on the right side (reductant) in another.
2. The strongest reductant is Zn(s), the species on the right side of the half-reaction that lies closer to the bottom of Table $1$ than the half-reactions involving I(aq), Sn2+(aq), and H2O2(aq). (Commercial products that use a piece of zinc are often marketed as a “miracle product” for removing tarnish from silver. All that is required is to add warm water and salt for electrical conductivity.)
3. B Of the reductants that lie below Zn(s) in Table $1$, and therefore are stronger reductants, only one is commonly available in household products: Al(s), which is sold as aluminum foil for wrapping foods.
Example $4$
Use the data in Table $1$ to determine whether each reaction is likely to occur spontaneously under standard conditions:
1. Sn(s) + Be2+(aq) → Sn2+(aq) + Be(s)
2. MnO2(s) + H2O2(aq) + 2H+(aq) → O2(g) + Mn2+(aq) + 2H2O(l)
Given: redox reaction and list of standard electrode potentials (Table P2 )
Asked for: reaction spontaneity
Strategy:
1. Identify the half-reactions in each equation. Using Table $1$, determine the standard potentials for the half-reactions in the appropriate direction.
2. Use Equation $\ref{19.10}$ to calculate the standard cell potential for the overall reaction. From this value, determine whether the overall reaction is spontaneous.
Solution
1. A Metallic tin is oxidized to Sn2+(aq), and Be2+(aq) is reduced to elemental beryllium. We can find the standard electrode potentials for the latter (reduction) half-reaction (−1.85 V) and for the former (oxidation) half-reaction (−0.14 V) directly from Table $1$.
B Adding the two half-reactions gives the overall reaction:
$\textrm{cathode:} \; \mathrm{Be^{2+}(aq)} +\mathrm{2e^-} \rightarrow \mathrm{Be(s)}$
$\textrm{anode:} \; \mathrm{Sn(s) \rightarrow \mathrm{Sn^{2+}}(s)} +\mathrm{2e^-}$
$\textrm{total:} \; \mathrm{Sn(s)+ \mathrm{Be^{2+}(aq)} \rightarrow \mathrm{Sn^{2+}}(aq)} + \mathrm{Be(s)}$
$E^\circ_{\textrm{cathode}}=\textrm{–1.99 V} \ E^\circ_{\textrm{anode}}=\textrm{-0.14 V} \ E^\circ_{\textrm{cell}}=E^\circ_{\textrm{cathode}}-E^\circ_{\textrm{anode}} \ \hspace{5mm} =-\textrm{1.85 V}$
The standard cell potential is quite negative, so the reaction will not occur spontaneously as written. That is, metallic tin cannot reduce Be2+ to beryllium metal under standard conditions. Instead, the reverse process, the reduction of stannous ions (Sn2+) by metallic beryllium, which has a positive value of E°cell, will occur spontaneously.
1. A MnO2 is the oxidant (Mn4+ is reduced to Mn2+), while H2O2 is the reductant (O2− is oxidized to O2). We can obtain the standard electrode potentials for the reduction and oxidation half-reactions directly from Table $1$.
B The two half-reactions and their corresponding potentials are as follows
\begin{align}\textrm{cathode:} & \mathrm{MnO_2(s)}+\mathrm{4H^+(aq)}+\mathrm{2e^-}\rightarrow\mathrm{Mn^{2+}(aq)}+\mathrm{2H_2O(l)} \nonumber \ \textrm{anode:} & \mathrm{H_2O_2(aq)}\rightarrow\mathrm{O_2(g)}+\mathrm{2H^+(aq)}+\mathrm{2e^-} \nonumber\ \textrm{overall:} & \mathrm{MnO_2(s)}+\mathrm{H_2O_2(aq)}+\mathrm{2H^+(aq)}\rightarrow\mathrm{O_2(g)}+\mathrm{Mn^{2+}(aq)}+\mathrm{2H_2O(l)} \nonumber\end{align}
$E^\circ_{\textrm{cathode}}=\textrm{1.22 V} \nonumber \ E^\circ_{\textrm{anode}}=\textrm{0.70 V} \nonumber \ E^\circ_{\textrm{cell}}=E^\circ_{\textrm{cathode}}-E^\circ_{\textrm{anode}} \nonumber \ \hspace{5mm} =-\textrm{0.53 V}$
The standard potential for the reaction is positive, indicating that under standard conditions, it will occur spontaneously as written. Hydrogen peroxide will reduce MnO2, and oxygen gas will evolve from the solution.
Exercise $4$
Use the data in Table $1$ to determine whether each reaction is likely to occur spontaneously under standard conditions:
1. 2Ce4+(aq) + 2Cl(aq) → 2Ce3+(aq) + Cl2(g)
2. 4MnO2(s) + 3O2(g) + 4OH(aq) → 4MnO4(aq) + 2H2O
Answer
1. spontaneous (E°cell = 0.36 V)
2. nonspontaneous (E°cell = −0.20 V)
Although the sign of E°cell tells us whether a particular redox reaction will occur spontaneously under standard conditions, it does not tell us to what extent the reaction proceeds, and it does not tell us what will happen under nonstandard conditions. To answer these questions requires a more quantitative understanding of the relationship between electrochemical cell potential and chemical thermodynamics.
Reference Electrodes and Measuring Concentrations
When using a galvanic cell to measure the concentration of a substance, we are generally interested in the potential of only one of the electrodes of the cell, the so-called indicator electrode, whose potential is related to the concentration of the substance being measured. To ensure that any change in the measured potential of the cell is due to only the substance being analyzed, the potential of the other electrode, the reference electrode, must be constant. You are already familiar with one example of a reference electrode: the SHE. The potential of a reference electrode must be unaffected by the properties of the solution, and if possible, it should be physically isolated from the solution of interest. To measure the potential of a solution, we select a reference electrode and an appropriate indicator electrode. Whether reduction or oxidation of the substance being analyzed occurs depends on the potential of the half-reaction for the substance of interest (the sample) and the potential of the reference electrode.
Note
The potential of any reference electrode should not be affected by the properties of the solution to be analyzed, and it should also be physically isolated.
There are many possible choices of reference electrode other than the SHE. The SHE requires a constant flow of highly flammable hydrogen gas, which makes it inconvenient to use. Consequently, two other electrodes are commonly chosen as reference electrodes. One is the silver–silver chloride electrode, which consists of a silver wire coated with a very thin layer of AgCl that is dipped into a chloride ion solution with a fixed concentration. The cell diagram and reduction half-reaction are as follows:
$Cl^−_{(aq)}∣AgCl_{(s)}∣Ag_{(s)} \label{19.44}$
$AgCl_{(s)}+e^− \rightarrow Ag_{(s)} + Cl^−_{(aq)}$
If a saturated solution of KCl is used as the chloride solution, the potential of the silver–silver chloride electrode is 0.197 V versus the SHE. That is, 0.197 V must be subtracted from the measured value to obtain the standard electrode potential measured against the SHE.
A second common reference electrode is the saturated calomel electrode (SCE), which has the same general form as the silver–silver chloride electrode. The SCE consists of a platinum wire inserted into a moist paste of liquid mercury (Hg2Cl2; called calomel in the old chemical literature) and KCl. This interior cell is surrounded by an aqueous KCl solution, which acts as a salt bridge between the interior cell and the exterior solution (part (a) in Figure $4$). Although it sounds and looks complex, this cell is actually easy to prepare and maintain, and its potential is highly reproducible. The SCE cell diagram and corresponding half-reaction are as follows:
$Pt_{(s)} ∣ Hg_2Cl_{2(s)}∣KCl_{(aq, sat)} \label{19.45}$
$Hg_2Cl_{2(s)} + 2e^− \rightarrow 2Hg_{(l)} + 2Cl^−{(aq)} \label{19.46}$
At 25°C, the potential of the SCE is 0.2415 V versus the SHE, which means that 0.2415 V must be subtracted from the potential versus an SCE to obtain the standard electrode potential.
One of the most common uses of electrochemistry is to measure the H+ ion concentration of a solution. A glass electrode is generally used for this purpose, in which an internal Ag/AgCl electrode is immersed in a 0.10 M HCl solution that is separated from the solution by a very thin glass membrane (part (b) in Figure $5$). The glass membrane absorbs protons, which affects the measured potential. The extent of the adsorption on the inner side is fixed because [H+] is fixed inside the electrode, but the adsorption of protons on the outer surface depends on the pH of the solution. The potential of the glass electrode depends on [H+] as follows (recall that pH = −log[H+]:
$E_{glass} = E′ + (0.0591\; V \times \log[H^+]) = E′ − 0.0591\; V \times pH \label{19.47}$
The voltage E′ is a constant that depends on the exact construction of the electrode. Although it can be measured, in practice, a glass electrode is calibrated; that is, it is inserted into a solution of known pH, and the display on the pH meter is adjusted to the known value. Once the electrode is properly calibrated, it can be placed in a solution and used to determine an unknown pH.
Ion-selective electrodes are used to measure the concentration of a particular species in solution; they are designed so that their potential depends on only the concentration of the desired species (part (c) in Figure $5$). These electrodes usually contain an internal reference electrode that is connected by a solution of an electrolyte to a crystalline inorganic material or a membrane, which acts as the sensor. For example, one type of ion-selective electrode uses a single crystal of Eu-doped $LaF_3$ as the inorganic material. When fluoride ions in solution diffuse to the surface of the solid, the potential of the electrode changes, resulting in a so-called fluoride electrode. Similar electrodes are used to measure the concentrations of other species in solution. Some of the species whose concentrations can be determined in aqueous solution using ion-selective electrodes and similar devices are listed in Table $2$.
Table $2$: Some Species Whose Aqueous Concentrations Can Be Measured Using Electrochemical Methods
Species Type of Sample
H+ laboratory samples, blood, soil, and ground and surface water
NH3/NH4+ wastewater and runoff water
K+ blood, wine, and soil
CO2/HCO3 blood and groundwater
F groundwater, drinking water, and soil
Br grains and plant extracts
I milk and pharmaceuticals
NO3 groundwater, drinking water, soil, and fertilizer
The Standard Hydrogen Electrode (SHE): https://youtu.be/GS-SE7IDDtY
Summary
The flow of electrons in an electrochemical cell depends on the identity of the reacting substances, the difference in the potential energy of their valence electrons, and their concentrations. The potential of the cell under standard conditions (1 M for solutions, 1 atm for gases, pure solids or liquids for other substances) and at a fixed temperature (25°C) is called the standard cell potential (E°cell). Only the difference between the potentials of two electrodes can be measured. By convention, all tabulated values of standard electrode potentials are listed as standard reduction potentials. The overall cell potential is the reduction potential of the reductive half-reaction minus the reduction potential of the oxidative half-reaction (E°cell = E°cathode − E°anode). The potential of the standard hydrogen electrode (SHE) is defined as 0 V under standard conditions. The potential of a half-reaction measured against the SHE under standard conditions is called its standard electrode potential. The standard cell potential is a measure of the driving force for a given redox reaction. All E° values are independent of the stoichiometric coefficients for the half-reaction. Redox reactions can be balanced using the half-reaction method, in which the overall redox reaction is divided into an oxidation half-reaction and a reduction half-reaction, each balanced for mass and charge. The half-reactions selected from tabulated lists must exactly reflect reaction conditions. In an alternative method, the atoms in each half-reaction are balanced, and then the charges are balanced. Whenever a half-reaction is reversed, the sign of E° corresponding to that reaction must also be reversed.
The oxidative and reductive strengths of a variety of substances can be compared using standard electrode potentials. Apparent anomalies can be explained by the fact that electrode potentials are measured in aqueous solution, which allows for strong intermolecular electrostatic interactions, and not in the gas phase.
If E°cell is positive, the reaction will occur spontaneously under standard conditions. If E°cell is negative, then the reaction is not spontaneous under standard conditions, although it will proceed spontaneously in the opposite direction. The potential of an indicator electrode is related to the concentration of the substance being measured, whereas the potential of the reference electrode is held constant. Whether reduction or oxidation occurs depends on the potential of the sample versus the potential of the reference electrode. In addition to the SHE, other reference electrodes are the silver–silver chloride electrode; the saturated calomel electrode (SCE); the glass electrode, which is commonly used to measure pH; and ion-selective electrodes, which depend on the concentration of a single ionic species in solution. Differences in potential between the SHE and other reference electrodes must be included when calculating values for E°.
Contributors and Attributions
• Anonymous
• cathode: $Cu^{2+}_{(aq)} + 2e^− \rightarrow Cu_{(s)} \;\;\; E°_{cathode} = 0.34\; V \label{19.41}$
• anode: $Zn_{(s)} \rightarrow Zn^{2+}(aq, 1 M) + 2e^−\;\;\; E°_{anode} = −0.76\; V \label{19.42}$
• overall: $Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)} \label{19.43}$
• Redox reactions can be balanced using the half-reaction method.
• The standard cell potential is a measure of the driving force for the reaction.
• The relative strengths of various oxidants and reductants can be predicted using E° values. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/20%3A_Electrochemistry/20.2%3A_Standard_Electrode_Potentials.txt |
Learning Objectives
• To understand the relationship between cell potential and the equilibrium constant.
• To use cell potentials to calculate solution concentrations.
Changes in reaction conditions can have a tremendous effect on the course of a redox reaction. For example, under standard conditions, the reaction of $\ce{Co(s)}$ with $\ce{Ni^{2+}(aq)}$ to form $\ce{Ni(s)}$ and $\ce{Co^{2+}(aq)}$ occurs spontaneously, but if we reduce the concentration of $\ce{Ni^{2+}}$ by a factor of 100, so that $\ce{[Ni^{2+}]}$ is 0.01 M, then the reverse reaction occurs spontaneously instead. The relationship between voltage and concentration is one of the factors that must be understood to predict whether a reaction will be spontaneous.
The Relationship between Cell Potential & Gibbs Energy
Electrochemical cells convert chemical energy to electrical energy and vice versa. The total amount of energy produced by an electrochemical cell, and thus the amount of energy available to do electrical work, depends on both the cell potential and the total number of electrons that are transferred from the reductant to the oxidant during the course of a reaction. The resulting electric current is measured in coulombs (C), an SI unit that measures the number of electrons passing a given point in 1 s. A coulomb relates energy (in joules) to electrical potential (in volts). Electric current is measured in amperes (A); 1 A is defined as the flow of 1 C/s past a given point (1 C = 1 A·s):
$\dfrac{\textrm{1 J}}{\textrm{1 V}}=\textrm{1 C}=\mathrm{A\cdot s} \label{20.5.1}$
In chemical reactions, however, we need to relate the coulomb to the charge on a mole of electrons. Multiplying the charge on the electron by Avogadro’s number gives us the charge on 1 mol of electrons, which is called the faraday (F), named after the English physicist and chemist Michael Faraday (1791–1867):
\begin{align}F &=(1.60218\times10^{-19}\textrm{ C})\left(\dfrac{6.02214 \times 10^{23}\, J}{\textrm{1 mol e}^-}\right) \[4pt] &=9.64833212\times10^4\textrm{ C/mol e}^- \[4pt] &\simeq 96,485\, J/(\mathrm{V\cdot mol\;e^-})\end{align} \label{20.5.2}
The total charge transferred from the reductant to the oxidant is therefore $nF$, where $n$ is the number of moles of electrons.
Michael Faraday (1791–1867)
Faraday was a British physicist and chemist who was arguably one of the greatest experimental scientists in history. The son of a blacksmith, Faraday was self-educated and became an apprentice bookbinder at age 14 before turning to science. His experiments in electricity and magnetism made electricity a routine tool in science and led to both the electric motor and the electric generator. He discovered the phenomenon of electrolysis and laid the foundations of electrochemistry. In fact, most of the specialized terms introduced in this chapter (electrode, anode, cathode, and so forth) are due to Faraday. In addition, he discovered benzene and invented the system of oxidation state numbers that we use today. Faraday is probably best known for “The Chemical History of a Candle,” a series of public lectures on the chemistry and physics of flames.
The maximum amount of work that can be produced by an electrochemical cell ($w_{max}$) is equal to the product of the cell potential ($E^°_{cell}$) and the total charge transferred during the reaction ($nF$):
$w_{max} = nFE_{cell} \label{20.5.3}$
Work is expressed as a negative number because work is being done by a system (an electrochemical cell with a positive potential) on its surroundings.
The change in free energy ($\Delta{G}$) is also a measure of the maximum amount of work that can be performed during a chemical process ($ΔG = w_{max}$). Consequently, there must be a relationship between the potential of an electrochemical cell and $\Delta{G}$; this relationship is as follows:
$\Delta{G} = −nFE_{cell} \label{20.5.4}$
A spontaneous redox reaction is therefore characterized by a negative value of $\Delta{G}$ and a positive value of $E^°_{cell}$, consistent with our earlier discussions. When both reactants and products are in their standard states, the relationship between ΔG° and $E^°_{cell}$ is as follows:
$\Delta{G^°} = −nFE^°_{cell} \label{20.5.5}$
A spontaneous redox reaction is characterized by a negative value of ΔG°, which corresponds to a positive value of E°cell.
Example $1$
Suppose you want to prepare elemental bromine from bromide using the dichromate ion as an oxidant. Using the data in Table P2, calculate the free-energy change (ΔG°) for this redox reaction under standard conditions. Is the reaction spontaneous?
Given: redox reaction
Asked for: $ΔG^o$ for the reaction and spontaneity
Strategy:
1. From the relevant half-reactions and the corresponding values of $E^o$, write the overall reaction and calculate $E^°_{cell}$.
2. Determine the number of electrons transferred in the overall reaction. Then use Equation \ref{20.5.5} to calculate $ΔG^o$. If $ΔG^o$ is negative, then the reaction is spontaneous.
A
As always, the first step is to write the relevant half-reactions and use them to obtain the overall reaction and the magnitude of $E^o$. From Table P2, we can find the reduction and oxidation half-reactions and corresponding $E^o$ values:
\begin{align*} & \textrm{cathode:} &\quad & \mathrm{Cr_2O_7^{2-}(aq)} + \mathrm{14H^+(aq)}+\mathrm{6e^-}\rightarrow \mathrm{2Cr^{3+}(aq)} +\mathrm{7H_2O(l)} &\quad & E^\circ_{\textrm{cathode}} =\textrm{1.36 V} \ & \textrm{anode:} &\quad & \mathrm{2Br^{-}(aq)} \rightarrow \mathrm{Br_2(aq)} +\mathrm{2e^-} &\quad & E^\circ_{\textrm{anode}} =\textrm{1.09 V} \end{align*} \nonumber
To obtain the overall balanced chemical equation, we must multiply both sides of the oxidation half-reaction by 3 to obtain the same number of electrons as in the reduction half-reaction, remembering that the magnitude of $E^o$ is not affected:
\begin{align*} & \textrm{cathode:} &\quad & \mathrm{Cr_2O_7^{2-}(aq)} + \mathrm{14H^+(aq)}+\mathrm{6e^-}\rightarrow \mathrm{2Cr^{3+}(aq)} +\mathrm{7H_2O(l)} &\quad & E^\circ_{\textrm{cathode}} =\textrm{1.36 V} \[4pt] & \textrm{anode:} &\quad & \mathrm{6Br^{-}(aq)} \rightarrow \mathrm{3Br_2(aq)} +\mathrm{6e^-} &\quad & E^\circ_{\textrm{anode}} =\textrm{1.09 V} \[4pt] \hline & \textrm{overall:} &\quad & \mathrm{Cr_2O_7^{2-}(aq)} + \mathrm{6Br^{-}(aq)} + \mathrm{14H^+(aq)} \rightarrow \mathrm{2Cr^{3+}(aq)} + \mathrm{3Br_2(aq)} +\mathrm{7H_2O(l)} &\quad & E^\circ_{\textrm{cell}} =\textrm{0.27 V} \end{align*} \nonumber
B
We can now calculate ΔG° using Equation $\ref{20.5.5}$. Because six electrons are transferred in the overall reaction, the value of $n$ is 6:
\begin{align*}\Delta G^\circ &=-(n)(F)(E^\circ_{\textrm{cell}}) \[4pt] & =-(\textrm{6 mole})[96,485\;\mathrm{J/(V\cdot mol})(\textrm{0.27 V})] \& =-15.6 \times 10^4\textrm{ J} \ & =-156\;\mathrm{kJ/mol\;Cr_2O_7^{2-}} \end{align*} \nonumber
Thus $ΔG^o$ is −168 kJ/mol for the reaction as written, and the reaction is spontaneous.
Exercise $1$
Use the data in Table P2 to calculate $ΔG^o$ for the reduction of ferric ion by iodide:
$\ce{2Fe^{3+}(aq) + 2I^{−}(aq) → 2Fe^{2+}(aq) + I2(s)}\nonumber$
Is the reaction spontaneous?
Answer
−44 kJ/mol I2; yes
Relating G and Ecell: Relating G and Ecell(opens in new window) [youtu.be]
Potentials for the Sums of Half-Reactions
Although Table P2 list several half-reactions, many more are known. When the standard potential for a half-reaction is not available, we can use relationships between standard potentials and free energy to obtain the potential of any other half-reaction that can be written as the sum of two or more half-reactions whose standard potentials are available. For example, the potential for the reduction of $\ce{Fe^{3+}(aq)}$ to $\ce{Fe(s)}$ is not listed in the table, but two related reductions are given:
$\ce{Fe^{3+}(aq) + e^{−} -> Fe^{2+}(aq)} \;\;\;E^° = +0.77 V \label{20.5.6}$
$\ce{Fe^{2+}(aq) + 2e^{−} -> Fe(s)} \;\;\;E^° = −0.45 V \label{20.5.7}$
Although the sum of these two half-reactions gives the desired half-reaction, we cannot simply add the potentials of two reductive half-reactions to obtain the potential of a third reductive half-reaction because $E^o$ is not a state function. However, because $ΔG^o$ is a state function, the sum of the $ΔG^o$ values for the individual reactions gives us $ΔG^o$ for the overall reaction, which is proportional to both the potential and the number of electrons ($n$) transferred. To obtain the value of $E^o$ for the overall half-reaction, we first must add the values of $ΔG^o (= −nFE^o)$ for each individual half-reaction to obtain $ΔG^o$ for the overall half-reaction:
\begin{align*}\ce{Fe^{3+}(aq)} + \ce{e^-} &\rightarrow \ce\mathrm{Fe^{2+}(aq)} &\quad \Delta G^\circ &=-(1)(F)(\textrm{0.77 V})\[4pt] \ce{Fe^{2+}(aq)}+\ce{2e^-} &\rightarrow\ce{Fe(s)} &\quad\Delta G^\circ &=-(2)(F)(-\textrm{0.45 V})\[4pt] \ce{Fe^{3+}(aq)}+\ce{3e^-} &\rightarrow \ce{Fe(s)} &\quad\Delta G^\circ & =[-(1)(F)(\textrm{0.77 V})]+[-(2)(F)(-\textrm{0.45 V})] \end{align*} \nonumber
Solving the last expression for ΔG° for the overall half-reaction,
$\Delta{G^°} = F[(−0.77 V) + (−2)(−0.45 V)] = F(0.13 V) \label{20.5.9}$
Three electrons ($n = 3$) are transferred in the overall reaction, so substituting into Equation $\ref{20.5.5}$ and solving for $E^o$ gives the following:
\begin{align*}\Delta G^\circ & =-nFE^\circ_{\textrm{cell}} \[4pt] F(\textrm{0.13 V}) & =-(3)(F)(E^\circ_{\textrm{cell}}) \[4pt] E^\circ & =-\dfrac{0.13\textrm{ V}}{3}=-0.043\textrm{ V}\end{align*} \nonumber
This value of $E^o$ is very different from the value that is obtained by simply adding the potentials for the two half-reactions (0.32 V) and even has the opposite sign.
Values of $E^o$ for half-reactions cannot be added to give $E^o$ for the sum of the half-reactions; only values of $ΔG^o = −nFE^°_{cell}$ for half-reactions can be added.
The Relationship between Cell Potential & the Equilibrium Constant
We can use the relationship between $\Delta{G^°}$ and the equilibrium constant $K$, to obtain a relationship between $E^°_{cell}$ and $K$. Recall that for a general reaction of the type $aA + bB \rightarrow cC + dD$, the standard free-energy change and the equilibrium constant are related by the following equation:
$\Delta{G°} = −RT \ln K \label{20.5.10}$
Given the relationship between the standard free-energy change and the standard cell potential (Equation $\ref{20.5.5}$), we can write
$−nFE^°_{cell} = −RT \ln K \label{20.5.12}$
Rearranging this equation,
$E^\circ_{\textrm{cell}}= \left( \dfrac{RT}{nF} \right) \ln K \label{20.5.12B}$
For $T = 298\, K$, Equation $\ref{20.5.12B}$ can be simplified as follows:
\begin{align} E^\circ_{\textrm{cell}} &=\left(\dfrac{RT}{nF}\right)\ln K \[4pt] &=\left[ \dfrac{[8.314\;\mathrm{J/(mol\cdot K})(\textrm{298 K})]}{n[96,485\;\mathrm{J/(V\cdot mol)}]}\right]2.303 \log K \[4pt] &=\left(\dfrac{\textrm{0.0592 V}}{n}\right)\log K \label{20.5.13} \end{align}
Thus $E^°_{cell}$ is directly proportional to the logarithm of the equilibrium constant. This means that large equilibrium constants correspond to large positive values of $E^°_{cell}$ and vice versa.
Example $2$
Use the data in Table P2 to calculate the equilibrium constant for the reaction of metallic lead with PbO2 in the presence of sulfate ions to give PbSO4 under standard conditions. (This reaction occurs when a car battery is discharged.) Report your answer to two significant figures.
Given: redox reaction
Asked for: $K$
Strategy:
1. Write the relevant half-reactions and potentials. From these, obtain the overall reaction and $E^o_{cell}$.
2. Determine the number of electrons transferred in the overall reaction. Use Equation $\ref{20.5.13}$ to solve for $\log K$ and then $K$.
Solution
A The relevant half-reactions and potentials from Table P2 are as follows:
\begin{align*} & \textrm {cathode:} & & \mathrm{PbO_2(s)}+\mathrm{SO_4^{2-}(aq)}+\mathrm{4H^+(aq)}+\mathrm{2e^-}\rightarrow\mathrm{PbSO_4(s)}+\mathrm{2H_2O(l)} & & E^\circ_\textrm{cathode}=\textrm{1.69 V} \[4pt] & \textrm{anode:} & & \mathrm{Pb(s)}+\mathrm{SO_4^{2-}(aq)}\rightarrow\mathrm{PbSO_4(s)}+\mathrm{2e^-} & & E^\circ_\textrm{anode}=-\textrm{0.36 V} \[4pt] \hline & \textrm {overall:} & & \mathrm{Pb(s)}+\mathrm{PbO_2(s)}+\mathrm{2SO_4^{2-}(aq)}+\mathrm{4H^+(aq)}\rightarrow\mathrm{2PbSO_4(s)}+\mathrm{2H_2O(l)} & & E^\circ_\textrm{cell}=\textrm{2.05 V} \end{align*} \nonumber
B Two electrons are transferred in the overall reaction, so $n = 2$. Solving Equation $\ref{20.5.13}$ for log K and inserting the values of $n$ and $E^o$,
\begin{align*}\log K & =\dfrac{nE^\circ}{\textrm{0.0591 V}}=\dfrac{2(\textrm{2.05 V})}{\textrm{0.0591 V}}=69.37 \[4pt] K & =2.3\times10^{69}\end{align*} \nonumber
Thus the equilibrium lies far to the right, favoring a discharged battery (as anyone who has ever tried unsuccessfully to start a car after letting it sit for a long time will know).
Exercise $2$
Use the data in Table P2 to calculate the equilibrium constant for the reaction of $\ce{Sn^{2+}(aq)}$ with oxygen to produce $\ce{Sn^{4+}(aq)}$ and water under standard conditions. Report your answer to two significant figures. The reaction is as follows:
$\ce{2Sn^{2+}(aq) + O2(g) + 4H^{+}(aq) <=> 2Sn^{4+}(aq) + 2H2O(l)} \nonumber$
Answer
$5.7 \times 10^{72}$
Figure $1$ summarizes the relationships that we have developed based on properties of the system—that is, based on the equilibrium constant, standard free-energy change, and standard cell potential—and the criteria for spontaneity (ΔG° < 0). Unfortunately, these criteria apply only to systems in which all reactants and products are present in their standard states, a situation that is seldom encountered in the real world. A more generally useful relationship between cell potential and reactant and product concentrations, as we are about to see, uses the relationship between $\Delta{G}$ and the reaction quotient $Q$.
Electrode Potentials and ECell: Electrode Potentials and Ecell(opens in new window) [youtu.be]
Summary
A coulomb (C) relates electrical potential, expressed in volts, and energy, expressed in joules. The current generated from a redox reaction is measured in amperes (A), where 1 A is defined as the flow of 1 C/s past a given point. The faraday (F) is Avogadro’s number multiplied by the charge on an electron and corresponds to the charge on 1 mol of electrons. The product of the cell potential and the total charge is the maximum amount of energy available to do work, which is related to the change in free energy that occurs during the chemical process. Adding together the ΔG values for the half-reactions gives ΔG for the overall reaction, which is proportional to both the potential and the number of electrons (n) transferred. Spontaneous redox reactions have a negative ΔG and therefore a positive Ecell. Because the equilibrium constant K is related to ΔG, E°cell and K are also related. Large equilibrium constants correspond to large positive values of E°. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/20%3A_Electrochemistry/20.3%3A_Ecell_G_and_K.txt |
Learning Objectives
• Relate cell potentials to Gibbs energy changes
• Use the Nernst equation to determine cell potentials at nonstandard conditions
• Perform calculations that involve converting between cell potentials, free energy changes, and equilibrium constants
The Nernst Equation enables the determination of cell potential under non-standard conditions. It relates the measured cell potential to the reaction quotient and allows the accurate determination of equilibrium constants (including solubility constants).
The Effect of Concentration on Cell Potential: The Nernst Equation
Recall that the actual free-energy change for a reaction under nonstandard conditions, $\Delta{G}$, is given as follows:
$\Delta{G} = \Delta{G°} + RT \ln Q \label{Eq1}$
We also know that $ΔG = −nFE_{cell}$ (under non-standard conditions) and $ΔG^o = −nFE^o_{cell}$ (under standard conditions). Substituting these expressions into Equation $\ref{Eq1}$, we obtain
$−nFE_{cell} = −nFE^o_{cell} + RT \ln Q \label{Eq2}$
Dividing both sides of this equation by $−nF$,
$E_\textrm{cell}=E^\circ_\textrm{cell}-\left(\dfrac{RT}{nF}\right)\ln Q \label{Eq3}$
Equation $\ref{Eq3}$ is called the Nernst equation, after the German physicist and chemist Walter Nernst (1864–1941), who first derived it. The Nernst equation is arguably the most important relationship in electrochemistry. When a redox reaction is at equilibrium ($ΔG = 0$), then Equation $\ref{Eq3}$ reduces to Equation $\ref{Eq31}$ and $\ref{Eq32}$ because $Q = K$, and there is no net transfer of electrons (i.e., Ecell = 0).
$E_\textrm{cell}=E^\circ_\textrm{cell}-\left(\dfrac{RT}{nF}\right)\ln K = 0 \label{Eq31}$
since
$E^\circ_\textrm{cell}= \left(\dfrac{RT}{nF}\right)\ln K \label{Eq32}$
Substituting the values of the constants into Equation $\ref{Eq3}$ with $T = 298\, K$ and converting to base-10 logarithms give the relationship of the actual cell potential (Ecell), the standard cell potential (E°cell), and the reactant and product concentrations at room temperature (contained in $Q$):
$E_{\textrm{cell}}=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \label{Eq4}$
The Power of the Nernst Equation
The Nernst Equation ($\ref{Eq3}$) can be used to determine the value of Ecell, and thus the direction of spontaneous reaction, for any redox reaction under any conditions.
Equation $\ref{Eq4}$ allows us to calculate the potential associated with any electrochemical cell at 298 K for any combination of reactant and product concentrations under any conditions. We can therefore determine the spontaneous direction of any redox reaction under any conditions, as long as we have tabulated values for the relevant standard electrode potentials. Notice in Equation $\ref{Eq4}$ that the cell potential changes by 0.0591/n V for each 10-fold change in the value of $Q$ because log 10 = 1.
Example $1$
The following reaction proceeds spontaneously under standard conditions because E°cell > 0 (which means that ΔG° < 0):
$\ce{2Ce^{4+}(aq) + 2Cl^{–}(aq) -> 2Ce^{3+}(aq) + Cl2(g)}\;\; E^°_{cell} = 0.25\, V \nonumber$
Calculate $E_{cell}$ for this reaction under the following nonstandard conditions and determine whether it will occur spontaneously: [Ce4+] = 0.013 M, [Ce3+] = 0.60 M, [Cl] = 0.0030 M, $P_\mathrm{Cl_2}$ = 1.0 atm, and T = 25°C.
Given: balanced redox reaction, standard cell potential, and nonstandard conditions
Asked for: cell potential
Strategy:
Determine the number of electrons transferred during the redox process. Then use the Nernst equation to find the cell potential under the nonstandard conditions.
Solution
We can use the information given and the Nernst equation to calculate Ecell. Moreover, because the temperature is 25°C (298 K), we can use Equation $\ref{Eq4}$ instead of Equation $\ref{Eq3}$. The overall reaction involves the net transfer of two electrons:
$2Ce^{4+}_{(aq)} + 2e^− \rightarrow 2Ce^{3+}_{(aq)}\nonumber$
$2Cl^−_{(aq)} \rightarrow Cl_{2(g)} + 2e^−\nonumber$
so n = 2. Substituting the concentrations given in the problem, the partial pressure of Cl2, and the value of E°cell into Equation $\ref{Eq4}$,
\begin{align*}E_\textrm{cell} & =E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \ & =\textrm{0.25 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{[\mathrm{Ce^{3+}}]^2P_\mathrm{Cl_2}}{[\mathrm{Ce^{4+}}]^2[\mathrm{Cl^-}]^2}\right) \ & =\textrm{0.25 V}-[(\textrm{0.0296 V})(8.37)]=\textrm{0.00 V}\end{align*} \nonumber
Thus the reaction will not occur spontaneously under these conditions (because E = 0 V and ΔG = 0). The composition specified is that of an equilibrium mixture
Exercise $1$
Molecular oxygen will not oxidize $MnO_2$ to permanganate via the reaction
$\ce{4MnO2(s) + 3O2(g) + 4OH^{−} (aq) -> 4MnO^{−}4(aq) + 2H2O(l)} \;\;\; E°_{cell} = −0.20\; V\nonumber$
Calculate $E_{cell}$ for the reaction under the following nonstandard conditions and decide whether the reaction will occur spontaneously: pH 10, $P_\mathrm{O_2}$= 0.20 atm, [MNO4] = 1.0 × 10−4 M, and T = 25°C.
Answer
Ecell = −0.22 V; the reaction will not occur spontaneously.
Applying the Nernst equation to a simple electrochemical cell such as the Zn/Cu cell allows us to see how the cell voltage varies as the reaction progresses and the concentrations of the dissolved ions change. Recall that the overall reaction for this cell is as follows:
$Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)\;\;\;E°cell = 1.10 V \label{Eq5}$
The reaction quotient is therefore $Q = [Zn^{2+}]/[Cu^{2+}]$. Suppose that the cell initially contains 1.0 M Cu2+ and 1.0 × 10−6 M Zn2+. The initial voltage measured when the cell is connected can then be calculated from Equation $\ref{Eq4}$:
\begin{align}E_\textrm{cell} & =E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log\dfrac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]}\ & =\textrm{1.10 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{1.0\times10^{-6}}{1.0}\right)=\textrm{1.28 V}\end{align} \label{Eq6}
Thus the initial voltage is greater than E° because $Q<1$. As the reaction proceeds, [Zn2+] in the anode compartment increases as the zinc electrode dissolves, while [Cu2+] in the cathode compartment decreases as metallic copper is deposited on the electrode. During this process, the ratio Q = [Zn2+]/[Cu2+] steadily increases, and the cell voltage therefore steadily decreases. Eventually, [Zn2+] = [Cu2+], so Q = 1 and Ecell = E°cell. Beyond this point, [Zn2+] will continue to increase in the anode compartment, and [Cu2+] will continue to decrease in the cathode compartment. Thus the value of Q will increase further, leading to a further decrease in Ecell. When the concentrations in the two compartments are the opposite of the initial concentrations (i.e., 1.0 M Zn2+ and 1.0 × 10−6 M Cu2+), Q = 1.0 × 106, and the cell potential will be reduced to 0.92 V.
The variation of Ecell with $\log{Q}$ over this range is linear with a slope of −0.0591/n, as illustrated in Figure $1$. As the reaction proceeds still further, $Q$ continues to increase, and Ecell continues to decrease. If neither of the electrodes dissolves completely, thereby breaking the electrical circuit, the cell voltage will eventually reach zero. This is the situation that occurs when a battery is “dead.” The value of $Q$ when Ecell = 0 is calculated as follows:
\begin{align}E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q=0 \ E^\circ &=\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \ \log Q &=\dfrac{E^\circ n}{\textrm{0.0591 V}}=\dfrac{(\textrm{1.10 V})(2)}{\textrm{0.0591 V}}=37.23 \ Q &=10^{37.23}=1.7\times10^{37}\end{align} \label{Eq7}
Recall that at equilibrium, $Q = K$. Thus the equilibrium constant for the reaction of Zn metal with Cu2+ to give Cu metal and Zn2+ is 1.7 × 1037 at 25°C.
The Nernst Equation: The Nernst Equation (opens in new window) [youtu.be]
Concentration Cells
A voltage can also be generated by constructing an electrochemical cell in which each compartment contains the same redox active solution but at different concentrations. The voltage is produced as the concentrations equilibrate. Suppose, for example, we have a cell with 0.010 M AgNO3 in one compartment and 1.0 M AgNO3 in the other. The cell diagram and corresponding half-reactions are as follows:
$\ce{Ag(s)\,|\,Ag^{+}}(aq, 0.010 \;M)\,||\,\ce{Ag^{+}}(aq, 1.0 \;M)\,|\,\ce{Ag(s)} \label{Eq8}$
cathode:
$\ce{Ag^{+}} (aq, 1.0\; M) + \ce{e^{−}} \rightarrow \ce{Ag(s)} \label{Eq9}$
anode:
$\ce{Ag(s)} \rightarrow \ce{Ag^{+}}(aq, 0.010\; M) + \ce{e^{−}} \label{Eq10}$
Overall
$\ce{Ag^{+}}(aq, 1.0 \;M) \rightarrow \ce{Ag^{+}}(aq, 0.010\; M) \label{Eq11}$
As the reaction progresses, the concentration of $Ag^+$ will increase in the left (oxidation) compartment as the silver electrode dissolves, while the $Ag^+$ concentration in the right (reduction) compartment decreases as the electrode in that compartment gains mass. The total mass of $Ag(s)$ in the cell will remain constant, however. We can calculate the potential of the cell using the Nernst equation, inserting 0 for E°cell because E°cathode = −E°anode:
\begin{align*} E_\textrm{cell}&=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \[4pt] &=0-\left(\dfrac{\textrm{0.0591 V}}{1}\right)\log\left(\dfrac{0.010}{1.0}\right) \[4pt] &=\textrm{0.12 V} \end{align*} \nonumber
An electrochemical cell of this type, in which the anode and cathode compartments are identical except for the concentration of a reactant, is called a concentration cell. As the reaction proceeds, the difference between the concentrations of Ag+ in the two compartments will decrease, as will Ecell. Finally, when the concentration of Ag+ is the same in both compartments, equilibrium will have been reached, and the measured potential difference between the two compartments will be zero (Ecell = 0).
Example $2$
Calculate the voltage in a galvanic cell that contains a manganese electrode immersed in a 2.0 M solution of MnCl2 as the cathode, and a manganese electrode immersed in a 5.2 × 10−2 M solution of MnSO4 as the anode (T = 25°C).
Given: galvanic cell, identities of the electrodes, and solution concentrations
Asked for: voltage
Strategy:
1. Write the overall reaction that occurs in the cell.
2. Determine the number of electrons transferred. Substitute this value into the Nernst equation to calculate the voltage.
Solution
A This is a concentration cell, in which the electrode compartments contain the same redox active substance but at different concentrations. The anions (Cl and SO42) do not participate in the reaction, so their identity is not important. The overall reaction is as follows:
$\ce{ Mn^{2+}}(aq, 2.0\, M) \rightarrow \ce{Mn^{2+}} (aq, 5.2 \times 10^{−2}\, M)\nonumber$
B For the reduction of Mn2+(aq) to Mn(s), n = 2. We substitute this value and the given Mn2+ concentrations into Equation $\ref{Eq4}$:
\begin{align*} E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \[4pt] &=\textrm{0 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{5.2\times10^{-2}}{2.0}\right) \[4pt] &=\textrm{0.047 V}\end{align*} \nonumber
Thus manganese will dissolve from the electrode in the compartment that contains the more dilute solution and will be deposited on the electrode in the compartment that contains the more concentrated solution.
Exercise $2$
Suppose we construct a galvanic cell by placing two identical platinum electrodes in two beakers that are connected by a salt bridge. One beaker contains 1.0 M HCl, and the other a 0.010 M solution of Na2SO4 at pH 7.00. Both cells are in contact with the atmosphere, with $P_\mathrm{O_2}$ = 0.20 atm. If the relevant electrochemical reaction in both compartments is the four-electron reduction of oxygen to water:
$\ce{O2(g) + 4H^{+}(aq) + 4e^{−} \rightarrow 2H2O(l)} \nonumber$
What will be the potential when the circuit is closed?
Answer
0.41 V
Using Cell Potentials to Measure Solubility Products
Because voltages are relatively easy to measure accurately using a voltmeter, electrochemical methods provide a convenient way to determine the concentrations of very dilute solutions and the solubility products ($K_{sp}$) of sparingly soluble substances. As you learned previously, solubility products can be very small, with values of less than or equal to 10−30. Equilibrium constants of this magnitude are virtually impossible to measure accurately by direct methods, so we must use alternative methods that are more sensitive, such as electrochemical methods.
To understand how an electrochemical cell is used to measure a solubility product, consider the cell shown in Figure $1$, which is designed to measure the solubility product of silver chloride:
$K_{sp} = [\ce{Ag^{+}}][\ce{Cl^{−}}]. \nonumber$
In one compartment, the cell contains a silver wire inserted into a 1.0 M solution of Ag+; the other compartment contains a silver wire inserted into a 1.0 M Cl solution saturated with AgCl. In this system, the Ag+ ion concentration in the first compartment equals Ksp. We can see this by dividing both sides of the equation for Ksp by [Cl] and substituting:
\begin{align*}[\ce{Ag^{+}}] &= \dfrac{K_{sp}}{[\ce{Cl^{−}}]} \[4pt] &= \dfrac{K_{sp}}{1.0} = K_{sp}. \end{align*} \nonumber
The overall cell reaction is as follows:
Ag+(aq, concentrated) → Ag+(aq, dilute)
Thus the voltage of the concentration cell due to the difference in [Ag+] between the two cells is as follows:
\begin{align} E_\textrm{cell} &=\textrm{0 V}-\left(\dfrac{\textrm{0.0591 V}}{1}\right)\log\left(\dfrac{[\mathrm{Ag^+}]_\textrm{dilute}}{[\mathrm{Ag^+}]_\textrm{concentrated}}\right) \nonumber \[4pt] &= -\textrm{0.0591 V } \log\left(\dfrac{K_{\textrm{sp}}}{1.0}\right) \nonumber \[4pt] &=-\textrm{0.0591 V }\log K_{\textrm{sp}} \label{Eq122} \end{align}
By closing the circuit, we can measure the potential caused by the difference in [Ag+] in the two cells. In this case, the experimentally measured voltage of the concentration cell at 25°C is 0.580 V. Solving Equation $\ref{Eq122}$ for $K_{sp}$,
\begin{align*}\log K_\textrm{sp} & =\dfrac{-E_\textrm{cell}}{\textrm{0.0591 V}}=\dfrac{-\textrm{0.580 V}}{\textrm{0.0591 V}}=-9.81 \[4pt] K_\textrm{sp} & =1.5\times10^{-10}\end{align*} \nonumber
Thus a single potential measurement can provide the information we need to determine the value of the solubility product of a sparingly soluble salt.
Example $3$: Solubility of lead(II) sulfate
To measure the solubility product of lead(II) sulfate (PbSO4) at 25°C, you construct a galvanic cell like the one shown in Figure $1$, which contains a 1.0 M solution of a very soluble Pb2+ salt [lead(II) acetate trihydrate] in one compartment that is connected by a salt bridge to a 1.0 M solution of Na2SO4 saturated with PbSO4 in the other. You then insert a Pb electrode into each compartment and close the circuit. Your voltmeter shows a voltage of 230 mV. What is Ksp for PbSO4? Report your answer to two significant figures.
Given: galvanic cell, solution concentrations, electrodes, and voltage
Asked for: Ksp
Strategy:
1. From the information given, write the equation for Ksp. Express this equation in terms of the concentration of Pb2+.
2. Determine the number of electrons transferred in the electrochemical reaction. Substitute the appropriate values into Equation $\ref{Eq12}$ and solve for Ksp.
Solution
A You have constructed a concentration cell, with one compartment containing a 1.0 M solution of $\ce{Pb^{2+}}$ and the other containing a dilute solution of Pb2+ in 1.0 M Na2SO4. As for any concentration cell, the voltage between the two compartments can be calculated using the Nernst equation. The first step is to relate the concentration of Pb2+ in the dilute solution to Ksp:
\begin{align*}[\mathrm{Pb^{2+}}][\mathrm{SO_4^{2-}}] & =K_\textrm{sp} \ [\mathrm{Pb^{2+}}] &=\dfrac{K_\textrm{sp}}{[\mathrm{SO_4^{2-}}]}=\dfrac{K_\textrm{sp}}{\textrm{1.0 M}}=K_\textrm{sp}\end{align*} \nonumber
B The reduction of Pb2+ to Pb is a two-electron process and proceeds according to the following reaction:
Pb2+(aq, concentrated) → Pb2+(aq, dilute)
so
\begin{align*}E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{0.0591}{n}\right)\log Q \ \textrm{0.230 V} & =\textrm{0 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{[\mathrm{Pb^{2+}}]_\textrm{dilute}}{[\mathrm{Pb^{2+}}]_\textrm{concentrated}}\right)=-\textrm{0.0296 V}\log\left(\dfrac{K_\textrm{sp}}{1.0}\right) \ -7.77 & =\log K_\textrm{sp} \ 1.7\times10^{-8} & =K_\textrm{sp}\end{align*} \nonumber
Exercise $3$
A concentration cell similar to the one described in Example $3$ contains a 1.0 M solution of lanthanum nitrate [La(NO3)3] in one compartment and a 1.0 M solution of sodium fluoride saturated with LaF3 in the other. A metallic La strip is inserted into each compartment, and the circuit is closed. The measured potential is 0.32 V. What is the Ksp for LaF3? Report your answer to two significant figures.
Answer
5.7 × 10−17
Using Cell Potentials to Measure Concentrations
Another use for the Nernst equation is to calculate the concentration of a species given a measured potential and the concentrations of all the other species. We saw an example of this in Example $3$, in which the experimental conditions were defined in such a way that the concentration of the metal ion was equal to Ksp. Potential measurements can be used to obtain the concentrations of dissolved species under other conditions as well, which explains the widespread use of electrochemical cells in many analytical devices. Perhaps the most common application is in the determination of [H+] using a pH meter, as illustrated below.
Example $4$: Measuring pH
Suppose a galvanic cell is constructed with a standard Zn/Zn2+ couple in one compartment and a modified hydrogen electrode in the second compartment. The pressure of hydrogen gas is 1.0 atm, but [H+] in the second compartment is unknown. The cell diagram is as follows:
$\ce{Zn(s)}|\ce{Zn^{2+}}(aq, 1.0\, M) || \ce{H^{+}} (aq, ?\, M)| \ce{H2} (g, 1.0\, atm)| Pt(s) \nonumber$
What is the pH of the solution in the second compartment if the measured potential in the cell is 0.26 V at 25°C?
Given: galvanic cell, cell diagram, and cell potential
Asked for: pH of the solution
Strategy:
1. Write the overall cell reaction.
2. Substitute appropriate values into the Nernst equation and solve for −log[H+] to obtain the pH.
Solution
A Under standard conditions, the overall reaction that occurs is the reduction of protons by zinc to give H2 (note that Zn lies below H2 in Table P2):
Zn(s) + 2H2+(aq) → Zn2+(aq) + H2(g) E°=0.76 V
B By substituting the given values into the simplified Nernst equation (Equation $\ref{Eq4}$), we can calculate [H+] under nonstandard conditions:
\begin{align*}E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}n\right)\log\left(\dfrac{[\mathrm{Zn^{2+}}]P_\mathrm{H_2}}{[\mathrm{H^+}]^2}\right) \ \textrm{0.26 V} &=\textrm{0.76 V}-\left(\dfrac{\textrm{0.0591 V}}2\right)\log\left(\dfrac{(1.0)(1.0)}{[\mathrm{H^+}]^2}\right) \ 16.9 &=\log\left(\dfrac{1}{[\mathrm{H^+}]^2}\right)=\log[\mathrm{H^+}]^{-2}=(-2)\log[\mathrm{H^+}] \ 8.46 &=-\log[\mathrm{H^+}] \ 8.5 &=\mathrm{pH}\end{align*} \nonumber
Thus the potential of a galvanic cell can be used to measure the pH of a solution.
Exercise $4$
Suppose you work for an environmental laboratory and you want to use an electrochemical method to measure the concentration of Pb2+ in groundwater. You construct a galvanic cell using a standard oxygen electrode in one compartment (E°cathode = 1.23 V). The other compartment contains a strip of lead in a sample of groundwater to which you have added sufficient acetic acid, a weak organic acid, to ensure electrical conductivity. The cell diagram is as follows:
$Pb_{(s)} ∣Pb^{2+}(aq, ? M)∥H^+(aq), 1.0 M∣O_2(g, 1.0 atm)∣Pt_{(s)}\nonumber$
When the circuit is closed, the cell has a measured potential of 1.62 V. Use Table P2 to determine the concentration of Pb2+ in the groundwater.
Answer
$1.2 \times 10^{−9}\; M$
Summary
The Nernst equation can be used to determine the direction of spontaneous reaction for any redox reaction in aqueous solution. The Nernst equation allows us to determine the spontaneous direction of any redox reaction under any reaction conditions from values of the relevant standard electrode potentials. Concentration cells consist of anode and cathode compartments that are identical except for the concentrations of the reactant. Because ΔG = 0 at equilibrium, the measured potential of a concentration cell is zero at equilibrium (the concentrations are equal). A galvanic cell can also be used to measure the solubility product of a sparingly soluble substance and calculate the concentration of a species given a measured potential and the concentrations of all the other species. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/20%3A_Electrochemistry/20.4%3A_Cell_Potential_as_a_Function_of_Concentrations.txt |
Because galvanic cells can be self-contained and portable, they can be used as batteries and fuel cells. A battery (storage cell) is a galvanic cell (or a series of galvanic cells) that contains all the reactants needed to produce electricity. In contrast, a fuel cell is a galvanic cell that requires a constant external supply of one or more reactants to generate electricity. In this section, we describe the chemistry behind some of the more common types of batteries and fuel cells.
Batteries
There are two basic kinds of batteries: disposable, or primary, batteries, in which the electrode reactions are effectively irreversible and which cannot be recharged; and rechargeable, or secondary, batteries, which form an insoluble product that adheres to the electrodes. These batteries can be recharged by applying an electrical potential in the reverse direction. The recharging process temporarily converts a rechargeable battery from a galvanic cell to an electrolytic cell.
Batteries are cleverly engineered devices that are based on the same fundamental laws as galvanic cells. The major difference between batteries and the galvanic cells we have previously described is that commercial batteries use solids or pastes rather than solutions as reactants to maximize the electrical output per unit mass. The use of highly concentrated or solid reactants has another beneficial effect: the concentrations of the reactants and the products do not change greatly as the battery is discharged; consequently, the output voltage remains remarkably constant during the discharge process. This behavior is in contrast to that of the Zn/Cu cell, whose output decreases logarithmically as the reaction proceeds (Figure $1$). When a battery consists of more than one galvanic cell, the cells are usually connected in series—that is, with the positive (+) terminal of one cell connected to the negative (−) terminal of the next, and so forth. The overall voltage of the battery is therefore the sum of the voltages of the individual cells.
The major difference between batteries and the galvanic cells is that commercial typically batteries use solids or pastes rather than solutions as reactants to maximize the electrical output per unit mass. An obvious exception is the standard car battery which used solution phase chemistry.
Leclanché Dry Cell
The dry cell, by far the most common type of battery, is used in flashlights, electronic devices such as the Walkman and Game Boy, and many other devices. Although the dry cell was patented in 1866 by the French chemist Georges Leclanché and more than 5 billion such cells are sold every year, the details of its electrode chemistry are still not completely understood. In spite of its name, the Leclanché dry cell is actually a “wet cell”: the electrolyte is an acidic water-based paste containing $MnO_2$, $NH_4Cl$, $ZnCl_2$, graphite, and starch (part (a) in Figure $1$). The half-reactions at the anode and the cathode can be summarized as follows:
• cathode (reduction):
$\ce{2MnO2(s) + 2NH^{+}4(aq) + 2e^{−} -> Mn2O3(s) + 2NH3(aq) + H2O(l)} \nonumber$
• anode (oxidation):
$\ce{Zn(s) -> Zn^{2+}(aq) + 2e^{−}} \nonumber$
The $\ce{Zn^{2+}}$ ions formed by the oxidation of $\ce{Zn(s)}$ at the anode react with $\ce{NH_3}$ formed at the cathode and $\ce{Cl^{−}}$ ions present in solution, so the overall cell reaction is as follows:
• overall reaction:
$\ce{2MnO2(s) + 2NH4Cl(aq) + Zn(s) -> Mn2O3(s) + Zn(NH3)2Cl2(s) + H2O(l)} \label{Eq3}$
The dry cell produces about 1.55 V and is inexpensive to manufacture. It is not, however, very efficient in producing electrical energy because only the relatively small fraction of the $\ce{MnO2}$ that is near the cathode is actually reduced and only a small fraction of the zinc cathode is actually consumed as the cell discharges. In addition, dry cells have a limited shelf life because the $\ce{Zn}$ anode reacts spontaneously with $\ce{NH4Cl}$ in the electrolyte, causing the case to corrode and allowing the contents to leak out.
The alkaline battery is essentially a Leclanché cell adapted to operate under alkaline, or basic, conditions. The half-reactions that occur in an alkaline battery are as follows:
• cathode (reduction)
$\ce{2MnO2(s) + H2O(l) + 2e^{−} -> Mn2O3(s) + 2OH^{−}(aq)} \nonumber$
• anode (oxidation):
$\ce{Zn(s) + 2OH^{−}(aq) -> ZnO(s) + H2O(l) + 2e^{−}} \nonumber$
• overall reaction:
$\ce{Zn(s) + 2MnO2(s) -> ZnO(s) + Mn2O3(s)} \nonumber$
This battery also produces about 1.5 V, but it has a longer shelf life and more constant output voltage as the cell is discharged than the Leclanché dry cell. Although the alkaline battery is more expensive to produce than the Leclanché dry cell, the improved performance makes this battery more cost-effective.
Button Batteries
Although some of the small button batteries used to power watches, calculators, and cameras are miniature alkaline cells, most are based on a completely different chemistry. In these "button" batteries, the anode is a zinc–mercury amalgam rather than pure zinc, and the cathode uses either $\ce{HgO}$ or $\ce{Ag2O}$ as the oxidant rather than $\ce{MnO2}$ in Figure $\PageIndex{1b}$).
The cathode, anode and overall reactions and cell output for these two types of button batteries are as follows (two half-reactions occur at the anode, but the overall oxidation half-reaction is shown):
• cathode (mercury battery): $\ce{HgO(s) + H2O(l) + 2e^{−} -> Hg(l) + 2OH^{−}(aq)} \nonumber$
• Anode (mercury battery): $\ce{Zn + 2OH^{−} -> ZnO + H2O + 2e^{−}} \nonumber$
• overall reaction (mercury battery): $\ce{Zn(s) + 2HgO(s) -> 2Hg(l) + ZnO(s)} \nonumber$ with $E_{cell} = 1.35 \,V$.
• cathode reaction (silver battery): $\ce{Ag2O(s) + H2O(l) + 2e^{−} -> 2Ag(s) + 2OH^{−}(aq)} \nonumber$
• anode (silver battery): $\ce{Zn + 2OH^{−} -> ZnO + H2O + 2e^{−}} \nonumber$
• Overall reaction (silver battery): $\ce{Zn(s) + 2Ag2O(s) -> 2Ag(s) + ZnO(s)} \nonumber$ with $E_{cell} = 1.6 \,V$.
The major advantages of the mercury and silver cells are their reliability and their high output-to-mass ratio. These factors make them ideal for applications where small size is crucial, as in cameras and hearing aids. The disadvantages are the expense and the environmental problems caused by the disposal of heavy metals, such as $\ce{Hg}$ and $\ce{Ag}$.
Lithium–Iodine Battery
None of the batteries described above is actually “dry.” They all contain small amounts of liquid water, which adds significant mass and causes potential corrosion problems. Consequently, substantial effort has been expended to develop water-free batteries. One of the few commercially successful water-free batteries is the lithium–iodine battery. The anode is lithium metal, and the cathode is a solid complex of $I_2$. Separating them is a layer of solid $LiI$, which acts as the electrolyte by allowing the diffusion of Li+ ions. The electrode reactions are as follows:
• cathode (reduction):
$I_{2(s)} + 2e^− \rightarrow {2I^-}_{(LiI)}\label{Eq11}$
• anode (oxidation):
$2Li_{(s)} \rightarrow 2Li^+_{(LiI)} + 2e^− \label{Eq12}$
• overall:
$2Li_{(s)}+ I_{2(s)} \rightarrow 2LiI_{(s)} \label{Eq12a}$
with $E_{cell} = 3.5 \, V$
As shown in part (c) in Figure $1$, a typical lithium–iodine battery consists of two cells separated by a nickel metal mesh that collects charge from the anode. Because of the high internal resistance caused by the solid electrolyte, only a low current can be drawn. Nonetheless, such batteries have proven to be long-lived (up to 10 yr) and reliable. They are therefore used in applications where frequent replacement is difficult or undesirable, such as in cardiac pacemakers and other medical implants and in computers for memory protection. These batteries are also used in security transmitters and smoke alarms. Other batteries based on lithium anodes and solid electrolytes are under development, using $TiS_2$, for example, for the cathode.
Dry cells, button batteries, and lithium–iodine batteries are disposable and cannot be recharged once they are discharged. Rechargeable batteries, in contrast, offer significant economic and environmental advantages because they can be recharged and discharged numerous times. As a result, manufacturing and disposal costs drop dramatically for a given number of hours of battery usage. Two common rechargeable batteries are the nickel–cadmium battery and the lead–acid battery, which we describe next.
Nickel–Cadmium (NiCad) Battery
The nickel–cadmium, or NiCad, battery is used in small electrical appliances and devices like drills, portable vacuum cleaners, and AM/FM digital tuners. It is a water-based cell with a cadmium anode and a highly oxidized nickel cathode that is usually described as the nickel(III) oxo-hydroxide, NiO(OH). As shown in Figure $2$, the design maximizes the surface area of the electrodes and minimizes the distance between them, which decreases internal resistance and makes a rather high discharge current possible.
The electrode reactions during the discharge of a $NiCad$ battery are as follows:
• cathode (reduction):
$2NiO(OH)_{(s)} + 2H_2O_{(l)} + 2e^− \rightarrow 2Ni(OH)_{2(s)} + 2OH^-_{(aq)} \label{Eq13}$
• anode (oxidation):
$Cd_{(s)} + 2OH^-_{(aq)} \rightarrow Cd(OH)_{2(s)} + 2e^- \label{Eq14}$
• overall:
$Cd_{(s)} + 2NiO(OH)_{(s)} + 2H_2O_{(l)} \rightarrow Cd(OH)_{2(s)} + 2Ni(OH)_{2(s)} \label{Eq15}$
$E_{cell} = 1.4 V$
Because the products of the discharge half-reactions are solids that adhere to the electrodes [Cd(OH)2 and 2Ni(OH)2], the overall reaction is readily reversed when the cell is recharged. Although NiCad cells are lightweight, rechargeable, and high capacity, they have certain disadvantages. For example, they tend to lose capacity quickly if not allowed to discharge fully before recharging, they do not store well for long periods when fully charged, and they present significant environmental and disposal problems because of the toxicity of cadmium.
A variation on the NiCad battery is the nickel–metal hydride battery (NiMH) used in hybrid automobiles, wireless communication devices, and mobile computing. The overall chemical equation for this type of battery is as follows:
$NiO(OH)_{(s)} + MH \rightarrow Ni(OH)_{2(s)} + M_{(s)} \label{Eq16}$
The NiMH battery has a 30%–40% improvement in capacity over the NiCad battery; it is more environmentally friendly so storage, transportation, and disposal are not subject to environmental control; and it is not as sensitive to recharging memory. It is, however, subject to a 50% greater self-discharge rate, a limited service life, and higher maintenance, and it is more expensive than the NiCad battery.
Directive 2006/66/EC of the European Union prohibits the placing on the market of portable batteries that contain more than 0.002% of cadmium by weight. The aim of this directive was to improve "the environmental performance of batteries and accumulators"
Lead–Acid (Lead Storage) Battery
The lead–acid battery is used to provide the starting power in virtually every automobile and marine engine on the market. Marine and car batteries typically consist of multiple cells connected in series. The total voltage generated by the battery is the potential per cell (E°cell) times the number of cells.
As shown in Figure $3$, the anode of each cell in a lead storage battery is a plate or grid of spongy lead metal, and the cathode is a similar grid containing powdered lead dioxide ($PbO_2$). The electrolyte is usually an approximately 37% solution (by mass) of sulfuric acid in water, with a density of 1.28 g/mL (about 4.5 M $H_2SO_4$). Because the redox active species are solids, there is no need to separate the electrodes. The electrode reactions in each cell during discharge are as follows:
• cathode (reduction):
$PbO_{2(s)} + HSO^−_{4(aq)} + 3H^+_{(aq)} + 2e^− \rightarrow PbSO_{4(s)} + 2H_2O_{(l)} \label{Eq17}$
with $E^°_{cathode} = 1.685 \; V$
• anode (oxidation):
$Pb_{(s)} + HSO^−_{4(aq)} \rightarrow PbSO_{4(s) }+ H^+_{(aq)} + 2e^−\label{Eq18}$
with $E^°_{anode} = −0.356 \; V$
• overall:
$Pb_{(s)} + PbO_{2(s)} + 2HSO^−_{4(aq)} + 2H^+_{(aq)} \rightarrow 2PbSO_{4(s)} + 2H_2O_{(l)} \label{Eq19}$
and $E^°_{cell} = 2.041 \; V$
As the cell is discharged, a powder of $PbSO_4$ forms on the electrodes. Moreover, sulfuric acid is consumed and water is produced, decreasing the density of the electrolyte and providing a convenient way of monitoring the status of a battery by simply measuring the density of the electrolyte. This is often done with the use of a hydrometer.
A hydrometer can be used to test the specific gravity of each cell as a measure of its state of charge (www.youtube.com/watch?v=SRcOqfL6GqQ).
When an external voltage in excess of 2.04 V per cell is applied to a lead–acid battery, the electrode reactions reverse, and $PbSO_4$ is converted back to metallic lead and $PbO_2$. If the battery is recharged too vigorously, however, electrolysis of water can occur:
$2H_2O_{(l)} \rightarrow 2H_{2(g)} +O_{2 (g)} \label{EqX}$
This results in the evolution of potentially explosive hydrogen gas. The gas bubbles formed in this way can dislodge some of the $PbSO_4$ or $PbO_2$ particles from the grids, allowing them to fall to the bottom of the cell, where they can build up and cause an internal short circuit. Thus the recharging process must be carefully monitored to optimize the life of the battery. With proper care, however, a lead–acid battery can be discharged and recharged thousands of times. In automobiles, the alternator supplies the electric current that causes the discharge reaction to reverse.
Fuel Cells
A fuel cell is a galvanic cell that requires a constant external supply of reactants because the products of the reaction are continuously removed. Unlike a battery, it does not store chemical or electrical energy; a fuel cell allows electrical energy to be extracted directly from a chemical reaction. In principle, this should be a more efficient process than, for example, burning the fuel to drive an internal combustion engine that turns a generator, which is typically less than 40% efficient, and in fact, the efficiency of a fuel cell is generally between 40% and 60%. Unfortunately, significant cost and reliability problems have hindered the wide-scale adoption of fuel cells. In practice, their use has been restricted to applications in which mass may be a significant cost factor, such as US manned space vehicles.
These space vehicles use a hydrogen/oxygen fuel cell that requires a continuous input of H2(g) and O2(g), as illustrated in Figure $4$. The electrode reactions are as follows:
• cathode (reduction):
$O_{2(g)} + 4H^+ + 4e^− \rightarrow 2H_2O_{(g)} \label{Eq20}$
• anode (oxidation):
$2H_{2(g)} \rightarrow 4H^+ + 4e^− \label{Eq21}$
• overall:
$2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(g)} \label{Eq22}$
The overall reaction represents an essentially pollution-free conversion of hydrogen and oxygen to water, which in space vehicles is then collected and used. Although this type of fuel cell should produce 1.23 V under standard conditions, in practice the device achieves only about 0.9 V. One of the major barriers to achieving greater efficiency is the fact that the four-electron reduction of $O_2 (g)$ at the cathode is intrinsically rather slow, which limits current that can be achieved. All major automobile manufacturers have major research programs involving fuel cells: one of the most important goals is the development of a better catalyst for the reduction of $O_2 (g)$.
Summary
Commercial batteries are galvanic cells that use solids or pastes as reactants to maximize the electrical output per unit mass. A battery is a contained unit that produces electricity, whereas a fuel cell is a galvanic cell that requires a constant external supply of one or more reactants to generate electricity. One type of battery is the Leclanché dry cell, which contains an electrolyte in an acidic water-based paste. This battery is called an alkaline battery when adapted to operate under alkaline conditions. Button batteries have a high output-to-mass ratio; lithium–iodine batteries consist of a solid electrolyte; the nickel–cadmium (NiCad) battery is rechargeable; and the lead–acid battery, which is also rechargeable, does not require the electrodes to be in separate compartments. A fuel cell requires an external supply of reactants as the products of the reaction are continuously removed. In a fuel cell, energy is not stored; electrical energy is provided by a chemical reaction. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/20%3A_Electrochemistry/20.5%3A_Batteries%3A_Producing_Electricity_Through_Chemical_Reactions.txt |
Learning Objectives
• To understand the process of corrosion.
Corrosion is a galvanic process by which metals deteriorate through oxidation—usually but not always to their oxides. For example, when exposed to air, iron rusts, silver tarnishes, and copper and brass acquire a bluish-green surface called a patina. Of the various metals subject to corrosion, iron is by far the most important commercially. An estimated \$100 billion per year is spent in the United States alone to replace iron-containing objects destroyed by corrosion. Consequently, the development of methods for protecting metal surfaces from corrosion constitutes a very active area of industrial research. In this section, we describe some of the chemical and electrochemical processes responsible for corrosion. We also examine the chemical basis for some common methods for preventing corrosion and treating corroded metals.
Corrosion is a REDOX process.
Under ambient conditions, the oxidation of most metals is thermodynamically spontaneous, with the notable exception of gold and platinum. Hence it is actually somewhat surprising that any metals are useful at all in Earth’s moist, oxygen-rich atmosphere. Some metals, however, are resistant to corrosion for kinetic reasons. For example, aluminum in soft-drink cans and airplanes is protected by a thin coating of metal oxide that forms on the surface of the metal and acts as an impenetrable barrier that prevents further destruction. Aluminum cans also have a thin plastic layer to prevent reaction of the oxide with acid in the soft drink. Chromium, magnesium, and nickel also form protective oxide films. Stainless steels are remarkably resistant to corrosion because they usually contain a significant proportion of chromium, nickel, or both.
In contrast to these metals, when iron corrodes, it forms a red-brown hydrated metal oxide ($\ce{Fe2O3 \cdot xH2O}$), commonly known as rust, that does not provide a tight protective film (Figure $1$). Instead, the rust continually flakes off to expose a fresh metal surface vulnerable to reaction with oxygen and water. Because both oxygen and water are required for rust to form, an iron nail immersed in deoxygenated water will not rust—even over a period of several weeks. Similarly, a nail immersed in an organic solvent such as kerosene or mineral oil will not rust because of the absence of water even if the solvent is saturated with oxygen.
In the corrosion process, iron metal acts as the anode in a galvanic cell and is oxidized to Fe2+; oxygen is reduced to water at the cathode. The relevant reactions are as follows:
• at cathode: $\ce{O2(g) + 4H^{+}(aq) + 4e^{−} -> 2H2O(l)} \nonumber$ with $E^o_{SRP}=1.23\; V$.
• at anode: $\ce{Fe(s) -> Fe^{2+}(aq) + 2e^{−}}\nonumber$ with $E^o_{SRP} = −0.45\; V$.
• overall: $\ce{2Fe(s) + O2(g) + 4H^{+}(aq) -> 2Fe^{2+}(aq) + 2H2O(l)} \label{Eq3}$ with $E^o_{cell} = 1.68\; V$.
The $\ce{Fe^{2+}}$ ions produced in the initial reaction are then oxidized by atmospheric oxygen to produce the insoluble hydrated oxide containing $\ce{Fe^{3+}}$, as represented in the following equation:
$\ce{4Fe^{2+}(aq) + O2(g) + (2 + 4x)H2O \rightarrow 2Fe2O3 \cdot xH2O + 4H^{+}(aq)} \label{Eq4}$
The sign and magnitude of $E^o_{cell}$ for the corrosion process (Equation $\ref{Eq3}$) indicate that there is a strong driving force for the oxidation of iron by O2 under standard conditions (1 M H+). Under neutral conditions, the driving force is somewhat less but still appreciable (E = 1.25 V at pH 7.0). Normally, the reaction of atmospheric CO2 with water to form H+ and HCO3 provides a low enough pH to enhance the reaction rate, as does acid rain. Automobile manufacturers spend a great deal of time and money developing paints that adhere tightly to the car’s metal surface to prevent oxygenated water, acid, and salt from coming into contact with the underlying metal. Unfortunately, even the best paint is subject to scratching or denting, and the electrochemical nature of the corrosion process means that two scratches relatively remote from each other can operate together as anode and cathode, leading to sudden mechanical failure (Figure $2$).
Prophylactic Protection
One of the most common techniques used to prevent the corrosion of iron is applying a protective coating of another metal that is more difficult to oxidize. Faucets and some external parts of automobiles, for example, are often coated with a thin layer of chromium using an electrolytic process. With the increased use of polymeric materials in cars, however, the use of chrome-plated steel has diminished in recent years. Similarly, the “tin cans” that hold soups and other foods are actually consist of steel container that is coated with a thin layer of tin. While neither chromium nor tin metals are intrinsically resistant to corrosion, they both form protective oxide coatings that hinder access of oxygen and water to the underlying steel (iron alloy).
As with a protective paint, scratching a protective metal coating will allow corrosion to occur. In this case, however, the presence of the second metal can actually increase the rate of corrosion. The values of the standard electrode potentials for $\ce{Sn^{2+}}$ (E° = −0.14 V) and Fe2+ (E° = −0.45 V) in Table P2 show that $\ce{Fe}$ is more easily oxidized than $\ce{Sn}$. As a result, the more corrosion-resistant metal (in this case, tin) accelerates the corrosion of iron by acting as the cathode and providing a large surface area for the reduction of oxygen (Figure $3$). This process is seen in some older homes where copper and iron pipes have been directly connected to each other. The less easily oxidized copper acts as the cathode, causing iron to dissolve rapidly near the connection and occasionally resulting in a catastrophic plumbing failure.
Cathodic Protection
One way to avoid these problems is to use a more easily oxidized metal to protect iron from corrosion. In this approach, called cathodic protection, a more reactive metal such as $\ce{Zn}$ (E° = −0.76 V for $\ce{Zn^{2+} + 2e^{−} -> Zn}$) becomes the anode, and iron becomes the cathode. This prevents oxidation of the iron and protects the iron object from corrosion. The reactions that occur under these conditions are as follows:
$\underbrace{O_{2(g)} + 4e^− + 4H^+_{(aq)} \rightarrow 2H_2O_{(l)} }_{\text{reduction at cathode}}\label{Eq5}$
$\underbrace{Zn_{(s)} \rightarrow Zn^{2+}_{(aq)} + 2e^−}_{\text{oxidation at anode}} \label{Eq6}$
$\underbrace{ 2Zn_{(s)} + O_{2(g)} + 4H^+_{(aq)} \rightarrow 2Zn^{2+}_{(aq)} + 2H_2O_{(l)} }_{\text{overall}}\label{Eq7}$
The more reactive metal reacts with oxygen and will eventually dissolve, “sacrificing” itself to protect the iron object. Cathodic protection is the principle underlying galvanized steel, which is steel protected by a thin layer of zinc. Galvanized steel is used in objects ranging from nails to garbage cans.
In a similar strategy, sacrificial electrodes using magnesium, for example, are used to protect underground tanks or pipes (Figure $4$). Replacing the sacrificial electrodes is more cost-effective than replacing the iron objects they are protecting.
Example $1$
Suppose an old wooden sailboat, held together with iron screws, has a bronze propeller (recall that bronze is an alloy of copper containing about 7%–10% tin).
1. If the boat is immersed in seawater, what corrosion reaction will occur? What is $E^o°_{cell}$?
2. How could you prevent this corrosion from occurring?
Given: identity of metals
Asked for: corrosion reaction, $E^o°_{cell}$, and preventive measures
Strategy:
1. Write the reactions that occur at the anode and the cathode. From these, write the overall cell reaction and calculate $E^o°_{cell}$.
2. Based on the relative redox activity of various substances, suggest possible preventive measures.
Solution
1. A According to Table P2, both copper and tin are less active metals than iron (i.e., they have higher positive values of $E^o°_{cell}$ than iron). Thus if tin or copper is brought into electrical contact by seawater with iron in the presence of oxygen, corrosion will occur. We therefore anticipate that the bronze propeller will act as the cathode at which $\ce{O2}$ is reduced, and the iron screws will act as anodes at which iron dissolves:
\begin{align*} & \textrm{cathode:} & & \mathrm{O_2(s)} + \mathrm{4H^+(aq)}+\mathrm{4e^-}\rightarrow \mathrm{2H_2O(l)} & & E^\circ_{\textrm{cathode}} =\textrm{1.23 V} \ & \textrm{anode:} & & \mathrm{Fe(s)} \rightarrow \mathrm{Fe^{2+}} +\mathrm{2e^-} & & E^\circ_{\textrm{anode}} =-\textrm{0.45 V} \ & \textrm{overall:} & & \mathrm{2Fe(s)}+\mathrm{O_2(g)}+\mathrm{4H^+(aq)} \rightarrow \mathrm{2Fe^{2+}(aq)} +\mathrm{2H_2O(l)} & & E^\circ_{\textrm{overall}} =\textrm{1.68 V} \end{align*} \nonumber
Over time, the iron screws will dissolve, and the boat will fall apart.
1. B Possible ways to prevent corrosion, in order of decreasing cost and inconvenience, are as follows: disassembling the boat and rebuilding it with bronze screws; removing the boat from the water and storing it in a dry place; or attaching an inexpensive piece of zinc metal to the propeller shaft to act as a sacrificial electrode and replacing it once or twice a year. Because zinc is a more active metal than iron, it will act as the sacrificial anode in the electrochemical cell and dissolve (Equation $\ref{Eq7}$).
Exercise $1$
Suppose the water pipes leading into your house are made of lead, while the rest of the plumbing in your house is iron. To eliminate the possibility of lead poisoning, you call a plumber to replace the lead pipes. He quotes you a very low price if he can use up his existing supply of copper pipe to do the job.
1. Do you accept his proposal?
2. What else should you have the plumber do while at your home?
Answer a
Not unless you plan to sell the house very soon because the $\ce{Cu/Fe}$ pipe joints will lead to rapid corrosion.
Answer b
Any existing $\ce{Pb/Fe}$ joints should be examined carefully for corrosion of the iron pipes due to the $\ce{Pb–Fe}$ junction; the less active $\ce{Pb}$ will have served as the cathode for the reduction of $\ce{O2}$, promoting oxidation of the more active $\ce{Fe}$ nearby.
Summary
Corrosion is a galvanic process that can be prevented using cathodic protection. The deterioration of metals through oxidation is a galvanic process called corrosion. Protective coatings consist of a second metal that is more difficult to oxidize than the metal being protected. Alternatively, a more easily oxidized metal can be applied to a metal surface, thus providing cathodic protection of the surface. A thin layer of zinc protects galvanized steel. Sacrificial electrodes can also be attached to an object to protect it. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/20%3A_Electrochemistry/20.6%3A_Corrosion%3A_Unwanted_Voltaic_Cells.txt |
Learning Objectives
• To understand electrolysis and describe it quantitatively.
In this chapter, we have described various galvanic cells in which a spontaneous chemical reaction is used to generate electrical energy. In an electrolytic cell, however, the opposite process, called electrolysis, occurs: an external voltage is applied to drive a nonspontaneous reaction. In this section, we look at how electrolytic cells are constructed and explore some of their many commercial applications.
Electrolytic Cells
If we construct an electrochemical cell in which one electrode is copper metal immersed in a 1 M Cu2+ solution and the other electrode is cadmium metal immersed in a $\,1\; M\, Cd^{2+}$ solution and then close the circuit, the potential difference between the two compartments will be 0.74 V. The cadmium electrode will begin to dissolve (Cd is oxidized to Cd2+) and is the anode, while metallic copper will be deposited on the copper electrode (Cu2+ is reduced to Cu), which is the cathode (Figure $\PageIndex{1a}$).
The overall reaction is as follows:
$\ce{Cd (s) + Cu^{2+} (aq) \rightarrow Cd^{2+} (aq) + Cu (s)} \nonumber$
with $E°_{cell} = 0.74\; V$
This reaction is thermodynamically spontaneous as written ($ΔG^o < 0$):
\begin{align*} \Delta G^\circ &=-nFE^\circ_\textrm{cell} \[4pt] &=-(\textrm{2 mol e}^-)[\mathrm{96,485\;J/(V\cdot mol)}](\mathrm{0.74\;V}) \[4pt] &=-\textrm{140 kJ (per mole Cd)} \end{align*} \nonumber
In this direction, the system is acting as a galvanic cell.
In an electrolytic cell, an external voltage is applied to drive a nonspontaneous reaction.
The reverse reaction, the reduction of Cd2+ by Cu, is thermodynamically nonspontaneous and will occur only with an input of 140 kJ. We can force the reaction to proceed in the reverse direction by applying an electrical potential greater than 0.74 V from an external power supply. The applied voltage forces electrons through the circuit in the reverse direction, converting a galvanic cell to an electrolytic cell. Thus the copper electrode is now the anode (Cu is oxidized), and the cadmium electrode is now the cathode (Cd2+ is reduced) (Figure $\PageIndex{1b}$). The signs of the cathode and the anode have switched to reflect the flow of electrons in the circuit. The half-reactions that occur at the cathode and the anode are as follows:
• half-reaction at the cathode:
$\ce{Cd^{2+}(aq) + 2e^{−} \rightarrow Cd(s)}\label{20.9.3}$
with $E^°_{cathode} = −0.40 \, V$
• half-reaction at the anode:
$\ce{Cu(s) \rightarrow Cu^{2+}(aq) + 2e^{−}} \label{20.9.4}$
with $E^°_{anode} = 0.34 \, V$
• Overall Reaction:
$\ce{Cd^{2+}(aq) + Cu(s) \rightarrow Cd(s) + Cu^{2+}(aq) } \label{20.9.5}$
with $E^°_{cell} = −0.74 \: V$
Because $E^°_{cell} < 0$, the overall reaction—the reduction of $Cd^{2+}$ by $Cu$—clearly cannot occur spontaneously and proceeds only when sufficient electrical energy is applied. The differences between galvanic and electrolytic cells are summarized in Table $1$.
Table $1$: Comparison of Galvanic and Electrolytic Cells
Property Galvanic Cell Electrolytic Cell
ΔG < 0 > 0
Ecell > 0 < 0
Electrode Process
anode oxidation oxidation
cathode reduction reduction
Sign of Electrode
anode +
cathode +
Electrolytic Reactions
At sufficiently high temperatures, ionic solids melt to form liquids that conduct electricity extremely well due to the high concentrations of ions. If two inert electrodes are inserted into molten $\ce{NaCl}$, for example, and an electrical potential is applied, $\ce{Cl^{-}}$ is oxidized at the anode, and $\ce{Na^{+}}$ is reduced at the cathode. The overall reaction is as follows:
$\ce{ 2NaCl (l) \rightarrow 2Na(l) + Cl2(g)} \label{20.9.6}$
This is the reverse of the formation of $\ce{NaCl}$ from its elements. The product of the reduction reaction is liquid sodium because the melting point of sodium metal is 97.8°C, well below that of $\ce{NaCl}$ (801°C). Approximately 20,000 tons of sodium metal are produced commercially in the United States each year by the electrolysis of molten $\ce{NaCl}$ in a Downs cell (Figure $2$). In this specialized cell, $\ce{CaCl2}$ (melting point = 772°C) is first added to the $\ce{NaCl}$ to lower the melting point of the mixture to about 600°C, thereby lowering operating costs.
Similarly, in the Hall–Heroult process used to produce aluminum commercially, a molten mixture of about 5% aluminum oxide (Al2O3; melting point = 2054°C) and 95% cryolite (Na3AlF6; melting point = 1012°C) is electrolyzed at about 1000°C, producing molten aluminum at the cathode and CO2 gas at the carbon anode. The overall reaction is as follows:
$\ce{2Al2O3(l) + 3C(s) -> 4Al(l) + 3CO2(g)} \label{20.9.7}$
Oxide ions react with oxidized carbon at the anode, producing CO2(g).
There are two important points to make about these two commercial processes and about the electrolysis of molten salts in general.
1. The electrode potentials for molten salts are likely to be very different from the standard cell potentials listed in Table P2, which are compiled for the reduction of the hydrated ions in aqueous solutions under standard conditions.
2. Using a mixed salt system means there is a possibility of competition between different electrolytic reactions. When a mixture of NaCl and CaCl2 is electrolyzed, Cl is oxidized because it is the only anion present, but either Na+ or Ca2+ can be reduced. Conversely, in the Hall–Heroult process, only one cation is present that can be reduced (Al3+), but there are three species that can be oxidized: C, O2−, and F.
In the Hall–Heroult process, C is oxidized instead of O2− or F because oxygen and fluorine are more electronegative than carbon, which means that C is a weaker oxidant than either O2 or F2. Similarly, in the Downs cell, we might expect electrolysis of a NaCl/CaCl2 mixture to produce calcium rather than sodium because Na is slightly less electronegative than Ca (χ = 0.93 versus 1.00, respectively), making Na easier to oxidize and, conversely, Na+ more difficult to reduce. In fact, the reduction of Na+ to Na is the observed reaction. In cases where the electronegativities of two species are similar, other factors, such as the formation of complex ions, become important and may determine the outcome.
Example $1$
If a molten mixture of MgCl2 and KBr is electrolyzed, what products will form at the cathode and the anode, respectively?
Given: identity of salts
Asked for: electrolysis products
Strategy:
1. List all the possible reduction and oxidation products. Based on the electronegativity values shown in Figure 7.5, determine which species will be reduced and which species will be oxidized.
2. Identify the products that will form at each electrode.
Solution
A The possible reduction products are Mg and K, and the possible oxidation products are Cl2 and Br2. Because Mg is more electronegative than K (χ = 1.31 versus 0.82), it is likely that Mg will be reduced rather than K. Because Cl is more electronegative than Br (3.16 versus 2.96), Cl2 is a stronger oxidant than Br2.
B Electrolysis will therefore produce Br2 at the anode and Mg at the cathode.
Exercise $1$
Predict the products if a molten mixture of AlBr3 and LiF is electrolyzed.
Answer
Br2 and Al
Electrolysis can also be used to drive the thermodynamically nonspontaneous decomposition of water into its constituent elements: H2 and O2. However, because pure water is a very poor electrical conductor, a small amount of an ionic solute (such as H2SO4 or Na2SO4) must first be added to increase its electrical conductivity. Inserting inert electrodes into the solution and applying a voltage between them will result in the rapid evolution of bubbles of H2 and O2 (Figure $3$).
The reactions that occur are as follows:
• cathode: $2H^+_{(aq)} + 2e^− \rightarrow H_{2(g)} \;\;\; E^°_{cathode} = 0 V \label{20.9.8}$
• anode: $2H_2O_{(l)} → O_{2(g)} + 4H^+_{(aq)} + 4e^−\;\;\; E^°_{anode} = 1.23\; V \label{20.9.9}$
• overall: $2H_2O_{(l)} → O_{2(g)} + 2H_{2(g)}\;\;\; E^°_{cell} = −1.23 \;V \label{20.9.10}$
For a system that contains an electrolyte such as Na2SO4, which has a negligible effect on the ionization equilibrium of liquid water, the pH of the solution will be 7.00 and [H+] = [OH] = 1.0 × 10−7. Assuming that $P_\mathrm{O_2}$ = $P_\mathrm{H_2}$ = 1 atm, we can use the standard potentials to calculate E for the overall reaction:
\begin{align}E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log(P_\mathrm{O_2}P^2_\mathrm{H_2}) \ &=-\textrm{1.23 V}-\left(\dfrac{\textrm{0.0591 V}}{4}\right)\log(1)=-\textrm{1.23 V}\end{align} \label{20.9.11}
Thus Ecell is −1.23 V, which is the value of E°cell if the reaction is carried out in the presence of 1 M H+ rather than at pH 7.0.
In practice, a voltage about 0.4–0.6 V greater than the calculated value is needed to electrolyze water. This added voltage, called an overvoltage, represents the additional driving force required to overcome barriers such as the large activation energy for the formation of a gas at a metal surface. Overvoltages are needed in all electrolytic processes, which explain why, for example, approximately 14 V must be applied to recharge the 12 V battery in your car.
In general, any metal that does not react readily with water to produce hydrogen can be produced by the electrolytic reduction of an aqueous solution that contains the metal cation. The p-block metals and most of the transition metals are in this category, but metals in high oxidation states, which form oxoanions, cannot be reduced to the metal by simple electrolysis. Active metals, such as aluminum and those of groups 1 and 2, react so readily with water that they can be prepared only by the electrolysis of molten salts. Similarly, any nonmetallic element that does not readily oxidize water to O2 can be prepared by the electrolytic oxidation of an aqueous solution that contains an appropriate anion. In practice, among the nonmetals, only F2 cannot be prepared using this method. Oxoanions of nonmetals in their highest oxidation states, such as NO3, SO42, PO43, are usually difficult to reduce electrochemically and usually behave like spectator ions that remain in solution during electrolysis.
In general, any metal that does not react readily with water to produce hydrogen can be produced by the electrolytic reduction of an aqueous solution that contains the metal cation.
Electroplating
In a process called electroplating, a layer of a second metal is deposited on the metal electrode that acts as the cathode during electrolysis. Electroplating is used to enhance the appearance of metal objects and protect them from corrosion. Examples of electroplating include the chromium layer found on many bathroom fixtures or (in earlier days) on the bumpers and hubcaps of cars, as well as the thin layer of precious metal that coats silver-plated dinnerware or jewelry. In all cases, the basic concept is the same. A schematic view of an apparatus for electroplating silverware and a photograph of a commercial electroplating cell are shown in Figure $4$.
The half-reactions in electroplating a fork, for example, with silver are as follows:
• cathode (fork): $\ce{Ag^{+}(aq) + e^{−} -> Ag(s)} \quad E°_{cathode} = 0.80 V\ \nonumber$
• anode (silver bar): $\ce{Ag(s) -> Ag^{+}(aq) + e^{-}} \quad E°_{anode} = 0.80 V \nonumber$
The overall reaction is the transfer of silver metal from one electrode (a silver bar acting as the anode) to another (a fork acting as the cathode). Because $E^o_{cell} = 0\, V$, it takes only a small applied voltage to drive the electroplating process. In practice, various other substances may be added to the plating solution to control its electrical conductivity and regulate the concentration of free metal ions, thus ensuring a smooth, even coating.
Quantitative Considerations
If we know the stoichiometry of an electrolysis reaction, the amount of current passed, and the length of time, we can calculate the amount of material consumed or produced in a reaction. Conversely, we can use stoichiometry to determine the combination of current and time needed to produce a given amount of material.
The quantity of material that is oxidized or reduced at an electrode during an electrochemical reaction is determined by the stoichiometry of the reaction and the amount of charge that is transferred. For example, in the reaction
$\ce{Ag^{+}(aq) + e^{−} → Ag(s)} \nonumber$
1 mol of electrons reduces 1 mol of $\ce{Ag^{+}}$ to $\ce{Ag}$ metal. In contrast, in the reaction
$\ce{Cu^{2+}(aq) + 2e^{−} → Cu(s)} \nonumber$
1 mol of electrons reduces only 0.5 mol of $\ce{Cu^{2+}}$ to $\ce{Cu}$ metal. Recall that the charge on 1 mol of electrons is 1 faraday (1 F), which is equal to 96,485 C. We can therefore calculate the number of moles of electrons transferred when a known current is passed through a cell for a given period of time. The total charge ($q$ in coulombs) transferred is the product of the current ($I$ in amperes) and the time ($t$, in seconds):
$q = I \times t \label{20.9.14}$
The stoichiometry of the reaction and the total charge transferred enable us to calculate the amount of product formed during an electrolysis reaction or the amount of metal deposited in an electroplating process.
For example, if a current of 0.60 A passes through an aqueous solution of $\ce{CuSO4}$ for 6.0 min, the total number of coulombs of charge that passes through the cell is as follows:
\begin{align*} q &= \textrm{(0.60 A)(6.0 min)(60 s/min)} \[4pt] &=\mathrm{220\;A\cdot s} \[4pt] &=\textrm{220 C} \end{align*} \nonumber
The number of moles of electrons transferred to $\ce{Cu^{2+}}$ is therefore
\begin{align*} \textrm{moles e}^- &=\dfrac{\textrm{220 C}}{\textrm{96,485 C/mol}} \[4pt] &=2.3\times10^{-3}\textrm{ mol e}^- \end{align*} \nonumber
Because two electrons are required to reduce a single Cu2+ ion, the total number of moles of Cu produced is half the number of moles of electrons transferred, or 1.2 × 10−3 mol. This corresponds to 76 mg of Cu. In commercial electrorefining processes, much higher currents (greater than or equal to 50,000 A) are used, corresponding to approximately 0.5 F/s, and reaction times are on the order of 3–4 weeks.
Example $2$
A silver-plated spoon typically contains about 2.00 g of Ag. If 12.0 h are required to achieve the desired thickness of the Ag coating, what is the average current per spoon that must flow during the electroplating process, assuming an efficiency of 100%?
Given: mass of metal, time, and efficiency
Asked for: current required
Strategy:
1. Calculate the number of moles of metal corresponding to the given mass transferred.
2. Write the reaction and determine the number of moles of electrons required for the electroplating process.
3. Use the definition of the faraday to calculate the number of coulombs required. Then convert coulombs to current in amperes.
Solution
A We must first determine the number of moles of Ag corresponding to 2.00 g of Ag:
$\textrm{moles Ag}=\dfrac{\textrm{2.00 g}}{\textrm{107.868 g/mol}}=1.85\times10^{-2}\textrm{ mol Ag}$
B The reduction reaction is Ag+(aq) + e → Ag(s), so 1 mol of electrons produces 1 mol of silver.
C Using the definition of the faraday,
coulombs = (1.85 × 102mol e)(96,485 C/mol e) = 1.78 × 103 C / mole
The current in amperes needed to deliver this amount of charge in 12.0 h is therefore
\begin{align*}\textrm{amperes} &=\dfrac{1.78\times10^3\textrm{ C}}{(\textrm{12.0 h})(\textrm{60 min/h})(\textrm{60 s/min})}\ & =4.12\times10^{-2}\textrm{ C/s}=4.12\times10^{-2}\textrm{ A}\end{align*} \nonumber
Because the electroplating process is usually much less than 100% efficient (typical values are closer to 30%), the actual current necessary is greater than 0.1 A.
Exercise $2$
A typical aluminum soft-drink can weighs about 29 g. How much time is needed to produce this amount of Al(s) in the Hall–Heroult process, using a current of 15 A to reduce a molten Al2O3/Na3AlF6 mixture?
Answer
5.8 h
Electroplating: Electroplating(opens in new window) [youtu.be]
Summary
In electrolysis, an external voltage is applied to drive a nonspontaneous reaction. The quantity of material oxidized or reduced can be calculated from the stoichiometry of the reaction and the amount of charge transferred. Relationship of charge, current and time:
$q = I \times t \nonumber$
In electrolysis, an external voltage is applied to drive a nonspontaneous reaction. Electrolysis can also be used to produce H2 and O2 from water. In practice, an additional voltage, called an overvoltage, must be applied to overcome factors such as a large activation energy and a junction potential. Electroplating is the process by which a second metal is deposited on a metal surface, thereby enhancing an object’s appearance or providing protection from corrosion. The amount of material consumed or produced in a reaction can be calculated from the stoichiometry of an electrolysis reaction, the amount of current passed, and the duration of the electrolytic reaction. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/20%3A_Electrochemistry/20.7%3A_Electrolysis%3A_Causing_Nonspontaneous_Reactions_to_Occur.txt |
Learning Objectives
• Define Electrorefining and electrosynthesis
• Explain the basics of the industrial electrolysis of salt water to generate chlorine gas.
Electrolysis reactions are the basic foundations of today's modern industry. There are various elements, chemical compounds, and organic compounds that are only produced by electrolysis including aluminum, chlorine, and NaOH. Electrolysis is the process by which an electric current spurs an otherwise non-spontaneous reaction.
Electrorefining
The electrorefining process refines metals or compounds at a high purity for a low cost. The pure metal can coat an otherwise worthless object. Let's consider the electrorefining process of copper: At the anode, there is an impure piece of copper that has other metals such as Ag, Au, Pt, Sn, Bi, Sb, As, Fe, Ni, Co, and Zn. The copper in this impure ore is oxidized to form Cu2+ at the anode, and moves through an aqueous sulfuric acid-Copper (II) sulfate solution into the cathode. When it reaches the cathode, the Cu2+ is reduced to Cu. This whole process takes place at a fairly low voltage (about .15 to .30 V), so Ag, Au, and Pt are not oxidized at the anode, as their standard oxidation electrode potentials are -.800, -1.36 and -1.20 respectively; these unoxidized impurities turn into a mixture called anode mud, a sludge at the bottom of the tank. This sludge can be recovered and used in different processes. Unlike Ag, Au and Pt, the impurities of Sb, Bi and Sn in the ore are indeed oxidized at the anode, but they are precipitated as they form hydroxides and oxides. Finally, Fe, Ni, Co and Zn are oxidized as well, but they are dissolved in water. Therefore, the only solid we are left with is the pure solid copper plate at the cathode, which has a purity level of about 99.999%. The image below gives an outline about the fate of the main components of an impure iron ore.
Electrosynthesis
Electrosynthesis is the method of producing substances through electrolysis reactions. This is useful when reaction conditions must be carefully controlled. One example of electrosynthesis is that of MnO2, Manganese dioxide. MnO2 occurs naturally in the form of the mineral pyrolusite, but this mineral is not easily used due to the nature of its size and lattice structure. However, MnO2 can be obtained a different way, through the electrolysis of MnSO4 in a sulfuric acid solution.
\begin{align} &\textrm{Oxidation: } &&\mathrm{Mn^{2+} + 2H_2O \rightarrow MnO_2 + 4H^+ + 2e^-} &&\mathrm{\hspace{12px}E^0_{MnO_2/H_2}=-1.23}\ &\textrm{Reduction: } &&\mathrm{2e^- + 2H^+ \rightarrow H_2} &&\mathrm{{-E}^0_{H^+/H_2}= -0}\ &\textrm{Overall: } &&\mathrm{Mn^{2+} + 2H_2O \rightarrow MnO_2 + 2H^+ +H_2} &&\mathrm{\hspace{12px}E^0_{MnO_2/H_2} -E^0_{H^+/H_2}= -1.23 - 0= -1.23} \end{align}
The commercial process for organic chemicals that is currently practiced on a scale comparable to that of inorganic chemicals and metals is the electrohydrodimerization of acrylonitrile to adiponitrile.
\begin{align} &\textrm{Anode: } &&\mathrm{H_2O \rightarrow 2H^+ + \dfrac{1}{2} O_2 + 2e^-}\ &\textrm{Cathode: } &&\ce{2CH2=CHCN + 2H2O + 2e- \rightarrow NC(CH2)CN + 2OH-}\ &\textrm{Overall: } &&\underset{\textrm{acrylonitrile}} {\ce{2CH2=CHCN}} + H_2O \rightarrow \dfrac{1}{2} O_2 + \underset{\textrm{adiponitrile}} {\ce{ NC(CH2)4CN}} \end{align}
The importance of adiponitrile is that it can be readily converted to other useful compounds.
The Chlor-Alkali Process
This process is the electrolysis of sodium chloride (NaCl) at an industrial level. We will begin by discussing the equation for the chlor-alkali process, followed by discussing three different types of the process: the diaphragm cell, the mercury cell and the membrane cell. We will begin the explanation of the chlor-alkali process by determining the reactions that occur during the electrolysis of NaCl. Because NaCl is in an aqueous solution, we also have to consider the electrolysis of water at both the anode and the cathode. Therefore, there are two possible reduction equations and two possible oxidation reactions.
Reduction:
\begin{align}
& \mathrm{Na^+_{\large{(aq)}} + 2e^- \rightarrow Na_{\large{(s)}}}
&& \mathrm{E^0_{Na^+/Na}= -2.71\: V \label{1}}\
& \mathrm{2 H_2O_{\large{(l)}} + 2e^- \rightarrow H_{2\large{(g)}} +2OH^-_{\large{(aq)}}}
&& \mathrm{E^0_{H_2O/H_2}= -0.83\: V \label{2}}
\end{align}
Oxidation:
\begin{align} &\mathrm{2Cl^-_{\large{(aq)}} \rightarrow Cl_2 + 2e^-} &&\mathrm{-E^0_{Cl_2/Cl^-}= -1.36\: V \label{3}}\ &\mathrm{2H_2O_{\large{(l)}} \rightarrow O_2 + 4H^+ + 4e^-} &&\mathrm{-E^0_{O_2/H_2O}= -1.23\: V \label{4}} \end{align}
As we can see, due to the very much more negative electrode potential, the reduction of sodium ions is much less likely to occur than the reduction of water, so we can assume that in the electrolysis of NaCl, the reduction that occurs is Equation $\ref{2}$. Therefore, we should try to determine what the oxidation reaction that occurs is.
Let's say we have Equation $\ref{2}$ as the reduction and Equation $\ref{3}$ as the oxidation. We would get:
\begin{align} &\textrm{Reduction: } &&\mathrm{2 H_2O_{\large{(l)}} + 2e^- \rightarrow H_{2\large{(g)}} +2OH^-_{\large{(aq)}}} &&\mathrm{\hspace{12px}E^0_{H_20/H_2}= -0.83\: V \label{2a}}\ &\textrm{Oxidation: } &&\mathrm{2Cl^-_{\large{(aq)}} \rightarrow Cl_2 + 2e^-} &&\mathrm{-E^0_{Cl_2/Cl^-}= -(1.36\: V) \label{3a}} \ &\textrm{Overall: } &&\mathrm{2 H_2O_{\large{(l)}} + 2Cl^-_{\large{(aq)}}} &&\mathrm{\hspace{12px}E^0_{H_20/H_2} - E^0_{Cl_2/Cl^-}\label{5a}}\ & &&\mathrm{\hspace{10px}\rightarrow H_{2\large{(g)}} +2OH^-_{\large{(aq)}} +Cl_2} &&\mathrm{\hspace{22px} = -0.83 + (-1.36)= -2.19} \end{align}
Alternatively, we could also have Equation $\ref{2}$ with $\ref{4}$
\begin{align} &\textrm{Reduction: } &&\mathrm{2\,[2 H_2O_{\large{(l)}} + 2e^- \rightarrow H_{2\large{(g)}} +2OH^-_{\large{(aq)}}]} &&\mathrm{\hspace{12px}E_{H_2O/H_2O}=-.83\: V \label{2b}}\ &\textrm{Oxidation: } &&\mathrm{2H_2O_{\large{(l)}} \rightarrow O_2 + 4H^+ + 4e^-} &&\mathrm{-E_{O_2/H_20}^0= -(1.23\: V) \label{4b}}\ &\textrm{Overall: } &&\mathrm{2H_2O_{\large{(l)}} \rightarrow 2H_{2\large{(g)}} + O_{2\large{(g)}}} &&\mathrm{\hspace{12px}E^0_{H_2O/H_2} - E^0_{O_2/H_20}\label{6b}}\ & && && \mathrm{\hspace{22px}= -0.83\: V -(1.23\: V)= -2.06} \end{align}
At first glance it would appear as though Equation $\ref{6b}$ would occur due to the smaller (less negative) electrode potential. However, O2 actually has a fairly large overpotential, so instead Cl2 is more likely to form, making Equation $\ref{5a}$ the most probable outcome for the electrolysis of NaCl.
Chlor-Alkali Process in a Diaphragm Cell
Depending on the method used, there can be several different products produced through the chlor-alkali process. The value of these products is what makes the chlor-alkali process so important. The name comes from the two main products of the process, chlorine and the alkali, sodium hydroxide (NaOH). Therefore, one of the main uses of the chlor-alkali process is the production of NaOH. As described earlier, the equation for the chlor-alkali process, that is, the electrolysis of NaCl, is as follows:
\begin{align} &\textrm{Reduction: } &&\mathrm{2 H_2O_{\large{(l)}} + 2e^- \rightarrow H_{2\large{(g)}} +2OH^-_{\large{(aq)}}} &&\mathrm{E_{H_2O/H_2O}= -0.83\; V}\ &\textrm{Oxidation: } &&\mathrm{2Cl^-_{\large{(aq)}} \rightarrow Cl_2 + 2e^-} &&\mathrm{E^0_{Cl_2/Cl^-}= -1.36\;V}\ &\textrm{Overall: } &&\mathrm{2Cl^- + 2H_2O_{\large{(l)}} \rightarrow 2 OH^- + H_{2\large{(g)}} +Cl_{2\large{(g)}}} &&\mathrm{E^0_{H_2O/H_2}- E^0_{Cl/Cl_2^-}}\ & && &&\hspace{12px}\mathrm{= -0.83\:V- (-1.36\: V)= -2.19\: V} \end{align}
**The chlor-alkali process often occurs in an apparatus called a diaphragm cell, which is illustrated in Figure $1$.
Figure $1$: Basic membrane cell used in the electrolysis of brine. At the anode (A), chloride (Cl) is oxidized to chlorine. The ion-selective membrane (B) allows the counterion Na+ to freely flow across, but prevents anions such as hydroxide (OH) and chloride from diffusing across. At the cathode (C), water is reduced to hydroxide and hydrogen gas. The net process is the electrolysis of an aqueous solution of NaCl into industrially useful products sodium hydroxide (NaOH) and chlorine gas. of Wikipedia (Jkwchui)
Note the following aspects of the Diagram cell:
Anode
• The sodium chloride is put into the anode compartment, in aqueous form.
• The actual physical anode is made of either graphite or titanium.
• In the anode compartment, Cl2 gas is produced as Cl- is oxidized.
Cathode
• On the cathode side, OH- (aq) and H2 gas are formed as water is reduced.
• You may wonder, why are there sodium and chloride ions of the cathode side if we put the sodium chloride into the anode compartment? To answer this question, we can consider the difference in solution levels between the anode and cathode. This causes the gradual flow of NaCl into the cathode, and prevents the backflow of NaOH into the anode. If Cl2 and NaOH come into contact, Cl2 turns into ClO-, ClO3-, Cl- ions. The water level difference prevents this contact and also encourages the flow of NaCl to the cathode side, so NaOH (aq) can form.
• Now, you may notice that the solution with NaOH in the cathode will also have aqueous NaCl mixed with it due to the flow of NaCl from the anode to the cathode. Therefore, if we want very pure NaOH for uses such as rayon manufacture, we need to somehow purify the salt out of the NaOH. In general, before purification, the solution is about 14-16% NaCl (aq) and 10-12% NaOH (aq). However, the NaOH (aq) can be concentrated and the NaCl (s) can be crystallized to form a solution with 50% NaOH (aq) and about 1% NaCl (aq).
Note
If chlorine comes into contact with hydrogen, it produces a mixture which will explode violently on exposure to sunlight or heat. Hydrogen chloride gas would be produced. Obviously, the two gases need to be kept apart.
Chlor-Alkali Process in Mercury Cell
To even further improve the purity of NaOH, a mercury cell can be used for the location of electrolysis, opposed to a diaphragm cell. In the mercury-cell process, also known as the Castner–Kellner process, a saturated brine solution floats on top of a thin layer of mercury. The mercury is the cathode, where sodium is produced and forms a sodium-mercury amalgam with the mercury. The amalgam is continuously drawn out of the cell and reacted with water which decomposes the amalgam into sodium hydroxide and mercury. The mercury is recycled into the electrolytic cell. Chlorine is produced at the anode and evaporates out of the cell. Mercury cells are being phased out due to concerns about mercury poisoning from mercury cell pollution
Anode Side
• The anodes are placed in the aqueous NaCl solution, above the liquid mercury.
• The reduction of Cl- occurs to produce chlorine gas, Cl2 (g).
Cathode Side
• A layer of Hg (l) at the bottom of the tank serves as the cathode.
• With a mercury cathode, the reaction of H2O (l) to H2 has a fairly high over potential, so the reduction of Na+ to Na occurs instead. The Na is soluble in Hg (l) and the two combine to form the Na-Hg alloy amalgam. This amalgam can be removed and then mixed with water to cause the following reaction:
• $\ce{2Na\:(in\: Hg) + 2H2O \rightarrow 2 Na+ + 2 OH- + H_{2\large{(g)}} + Hg_{\large{(l)}}}$
• The Hg (l) that forms is recycled back into the liquid at the bottom of the tank that acts as a cathode.
• H2 gas is released.
• NaOH is left in a very pure, aqueous form.
**Some of the main problems with the mercury cell are as follows:
• The reaction needs a higher voltage than the diaphragm cell: 4.5 V in the mercury cell compared to 3.5 in a diaphragm cell.
• Requires quite a bit of electrical energy, as it needs 3400 kWh/ton Cl2 opposed to 2500 in a diaphragm cell.
• Potential damage to the environment due to mercury deposits. Luckily deposits were as large as 200 g mercury per ton chlorine gas, but now, they never exceed 0.25 g per ton chlorine gas.
Membrane Cell Process
A third way to make even more pure NaOH is to use a membrane cell. It is preferred over the diaphragm cell or mercury cell method because it uses the least amount of electric energy and produces the highest quality NaOH. For instance, it can produce NaOH with a degree of chlorine ion contamination of only 50 ppm. An ion-permeable membrane is used to separate the anode and cathode.
Anode
Saturated brine is fed to the compartment. Current passed through the cell splits the sodium chloride into its constituent components, Na+ and Cl-. Chloride ions are oxidized to chlorine gas at the anode.
$2Cl^-_{(aq)} \rightarrow 2e^- + Cl_{2(g)}$
The chlorine is purified by liquifying it under pressure. The oxygen stays as a gas when it is compressed at ordinary temperatures. The membrane passes Na+ ions to the cathode compartment.
Cathode
The hydrogen is produced at the nickel cathode:
$2H^+_{(aq)} + 2e^- \rightarrow H_{2(g)}$
• H2O is reduced to form OH- and H2 gas.
• $\ce{2H2O \rightarrow H_2 + 2OH- }$
• Na+ ions that had flowed over and OH- produced through the reduction of water react to form aqueous NaOH.
• H2 gas also is produced as a byproduct.
The membrane is made from a polymer which only allows positive ions to pass through it. That means that the only the sodium ions from the sodium chloride solution can pass through the membrane - and not the chloride ions. The advantage of this is that the sodium hydroxide solution being formed in the right-hand compartment never gets contaminated with any sodium chloride solution. The sodium chloride solution being used has to be pure. If it contained any other metal ions, these would also pass through the membrane and so contaminate the sodium hydroxide solution. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/20%3A_Electrochemistry/20.8%3A_Industrial_Electrolysis_Processes.txt |
20.1: Electrode Potentials and their Measurement
Conceptual Problems
1. Is $2NaOH_{(aq)} + H_2SO_{4(aq)} \rightarrow Na_2SO_{4(aq)} + 2H_2O_{(l)}$ an oxidation–reduction reaction? Why or why not?
2. If two half-reactions are physically separated, how is it possible for a redox reaction to occur? What is the name of the apparatus in which two half-reactions are carried out simultaneously?
3. What is the difference between a galvanic cell and an electrolytic cell? Which would you use to generate electricity?
4. What is the purpose of a salt bridge in a galvanic cell? Is it always necessary to use a salt bridge in a galvanic cell?
5. One criterion for a good salt bridge is that it contains ions that have similar rates of diffusion in aqueous solution, as K+ and Cl ions do. What would happen if the diffusion rates of the anions and cations differed significantly?
6. It is often more accurate to measure the potential of a redox reaction by immersing two electrodes in a single beaker rather than in two beakers. Why?
Conceptual Answer
1. A large difference in cation/anion diffusion rates would increase resistance in the salt bridge and limit electron flow through the circuit.
1. Numerical Problems
1. Copper(I) sulfate forms a bright blue solution in water. If a piece of zinc metal is placed in a beaker of aqueous CuSO4 solution, the blue color fades with time, the zinc strip begins to erode, and a black solid forms around the zinc strip. What is happening? Write half-reactions to show the chemical changes that are occurring. What will happen if a piece of copper metal is placed in a colorless aqueous solution of $ZnCl_2$?
2. Consider the following spontaneous redox reaction: NO3(aq) + H+(aq) + SO32(aq) → SO42(aq) + HNO2(aq).
1. Write the two half-reactions for this overall reaction.
2. If the reaction is carried out in a galvanic cell using an inert electrode in each compartment, which electrode corresponds to which half-reaction?
3. Which electrode is negatively charged, and which is positively charged?
1. The reaction Pb(s) + 2VO2+(aq) + 4H+(aq) → Pb2+(aq) + 2V3+(aq) + 2H2O(l) occurs spontaneously.
1. Write the two half-reactions for this redox reaction.
2. If the reaction is carried out in a galvanic cell using an inert electrode in each compartment, which reaction occurs at the cathode and which occurs at the anode?
3. Which electrode is positively charged, and which is negatively charged?
1. Phenolphthalein is an indicator that turns pink under basic conditions. When an iron nail is placed in a gel that contains [Fe(CN)6]3−, the gel around the nail begins to turn pink. What is occurring? Write the half-reactions and then write the overall redox reaction.
1. Sulfate is reduced to HS in the presence of glucose, which is oxidized to bicarbonate. Write the two half-reactions corresponding to this process. What is the equation for the overall reaction?
1. Write the spontaneous half-reactions and the overall reaction for each proposed cell diagram. State which half-reaction occurs at the anode and which occurs at the cathode.
1. Pb(s)∣PbSO4(s)∣SO42(aq)∥Cu2+(aq)∣Cu(s)
2. Hg(l)∣Hg2Cl2(s)∣Cl(aq) ∥ Cd2+(aq)∣Cd(s)
1. For each galvanic cell represented by these cell diagrams, determine the spontaneous half-reactions and the overall reaction. Indicate which reaction occurs at the anode and which occurs at the cathode.
1. Zn(s)∣Zn2+(aq) ∥ H+(aq)∣H2(g), Pt(s)
2. Ag(s)∣AgCl(s)∣Cl(aq) ∥ H+(aq)∣H2(g)∣Pt(s)
3. Pt(s)∣H2(g)∣H+(aq) ∥ Fe2+(aq), Fe3+(aq)∣Pt(s)
1. For each redox reaction, write the half-reactions and draw the cell diagram for a galvanic cell in which the overall reaction occurs spontaneously. Identify each electrode as either positive or negative.
1. Ag(s) + Fe3+(aq) → Ag+(aq) + Fe2+(aq)
2. Fe3+(aq) + 1/2H2(g) → Fe2+(aq) + H+(aq)
1. Write the half-reactions for each overall reaction, decide whether the reaction will occur spontaneously, and construct a cell diagram for a galvanic cell in which a spontaneous reaction will occur.
1. 2Cl(aq) + Br2(l) → Cl2(g) + 2Br(aq)
2. 2NO2(g) + 2OH(aq) → NO2(aq) + NO3(aq) + H2O(l)
3. 2H2O(l) + 2Cl(aq) → H2(g) + Cl2(g) + 2OH(aq)
4. C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
1. Write the half-reactions for each overall reaction, decide whether the reaction will occur spontaneously, and construct a cell diagram for a galvanic cell in which a spontaneous reaction will occur.
1. Co(s) + Fe2+(aq) → Co2+(aq) + Fe(s)
2. O2(g) + 4H+(aq) + 4Fe2+(aq) → 2H2O(l) + 4Fe3+(aq)
3. 6Hg2+(aq) + 2NO3(aq) + 8H+ → 3Hg22+(aq) + 2NO(g) + 4H2O(l)
4. CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
Numerical Answers
reduction: SO42−(aq) + 9H+(aq) + 8e → HS(aq) + 4H2O(l)
oxidation: C6H12O6(aq) + 12H2O(l) → 6HCO3(g) + 30H+(aq) + 24e
overall: C6H12O6(aq) + 3SO42−(aq) → 6HCO3(g) + 3H+(aq) + 3HS(aq)
1. reduction: 2H+(aq) + 2e → H2(aq); cathode;
oxidation: Zn(s) → Zn2+(aq) + 2e; anode;
overall: Zn(s) + 2H+(aq) → Zn2+(aq) + H2(aq)
1. reduction: AgCl(s) + e → Ag(s) + Cl(aq); cathode;
oxidation: H2(g) → 2H+(aq) + 2e; anode;
overall: AgCl(s) + H2(g) → 2H+(aq) + Ag(s) + Cl(aq)
1. reduction: Fe3+(aq) + e → Fe2+(aq); cathode;
oxidation: H2(g) → 2H+(aq) + 2e; anode;
overall: 2Fe3+(aq) + H2(g) → 2H+(aq) + 2Fe2+(aq)
20.2: Standard Electrode Potentials
Conceptual Problems
1. Is a hydrogen electrode chemically inert? What is the major disadvantage to using a hydrogen electrode?
2. List two factors that affect the measured potential of an electrochemical cell and explain their impact on the measurements.
3. What is the relationship between electron flow and the potential energy of valence electrons? If the valence electrons of substance A have a higher potential energy than those of substance B, what is the direction of electron flow between them in a galvanic cell?
4. If the components of a galvanic cell include aluminum and bromine, what is the predicted direction of electron flow? Why?
5. Write a cell diagram representing a cell that contains the Ni/Ni2+ couple in one compartment and the SHE in the other compartment. What are the values of E°cathode, E°anode, and E°cell?
6. Explain why E° values are independent of the stoichiometric coefficients in the corresponding half-reaction.
7. Identify the oxidants and the reductants in each redox reaction.
1. Cr(s) + Ni2+(aq) → Cr2+(aq) + Ni(s)
2. Cl2(g) + Sn2+(aq) → 2Cl(aq) + Sn4+(aq)
3. H3AsO4(aq) + 8H+(aq) + 4Zn(s) → AsH3(g) + 4H2O(l) + 4Zn2+(aq)
4. 2NO2(g) + 2OH(aq) → NO2(aq) + NO3(aq) + H2O(l)
1. Identify the oxidants and the reductants in each redox reaction.
1. Br2(l) + 2I(aq) → 2Br(aq) + I2(s)
2. Cu2+(aq) + 2Ag(s) → Cu(s) + 2Ag+(aq)
3. H+(aq) + 2MnO4(aq) + 5H2SO3(aq) → 2Mn2+(aq) + 3H2O(l) + 5HSO4(aq)
4. IO3(aq) + 5I(aq) + 6H+(aq) → 3I2(s) + 3H2O(l)
1. All reference electrodes must conform to certain requirements. List the requirements and explain their significance.
1. For each application, describe the reference electrode you would use and explain why. In each case, how would the measured potential compare with the corresponding E°?
1. measuring the potential of a Cl/Cl2 couple
2. measuring the pH of a solution
3. measuring the potential of a MnO4/Mn2+ couple
Conceptual Answers
1. Ni(s)∣Ni2+(aq)∥H+(aq, 1 M)∣H2(g, 1 atm)∣Pt(s)
$E^\circ_{\textrm{anode}} \ E^\circ_{\textrm{cathode}} \ E^\circ_{\textrm{cell}}$
$\mathrm{Ni^{2+}(aq)}+\mathrm{2e^-}\rightarrow\mathrm{Ni(s)};\;-\textrm{0.257 V} \ \mathrm{2H^+(aq)}+\mathrm{2e^-}\rightarrow\mathrm{H_2(g)};\textrm{ 0.000 V} \ \mathrm{2H^+(aq)}+\mathrm{Ni(s)}\rightarrow\mathrm{H_2(g)}+\mathrm{Ni^{2+}(aq)};\textrm{ 0.257 V}$
1. oxidant: Ni2+(aq); reductant: Cr(s)
2. oxidant: Cl2(g); reductant: Sn2+(aq)
3. oxidant: H3AsO4(aq); reductant: Zn(s)
4. oxidant: NO2(g); reductant: NO2(g)
Numerical Problems
1. Draw the cell diagram for a galvanic cell with an SHE and a copper electrode that carries out this overall reaction: H2(g) + Cu2+(aq) → 2H+(aq) + Cu(s).
2. Draw the cell diagram for a galvanic cell with an SHE and a zinc electrode that carries out this overall reaction: Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g).
3. Balance each reaction and calculate the standard electrode potential for each. Be sure to include the physical state of each product and reactant.
1. Cl2(g) + H2(g) → 2Cl(aq) + 2H+(aq)
2. Br2(aq) + Fe2+(aq) → 2Br(aq) + Fe3+(aq)
3. Fe3+(aq) + Cd(s) → Fe2+(aq) + Cd2+(aq)
1. Balance each reaction and calculate the standard reduction potential for each. Be sure to include the physical state of each product and reactant.
1. Cu+(aq) + Ag+(aq) → Cu2+(aq) + Ag(s)
2. Sn(s) + Fe3+(aq) → Sn2+(aq) + Fe2+(aq)
3. Mg(s) + Br2(l) → 2Br(aq) + Mg2+(aq)
1. Write a balanced chemical equation for each redox reaction.
1. H2PO2(aq) + SbO2(aq) → HPO32−(aq) + Sb(s) in basic solution
2. HNO2(aq) + I(aq) → NO(g) + I2(s) in acidic solution
3. N2O(g) + ClO(aq) → Cl(aq) + NO2(aq) in basic solution
4. Br2(l) → Br(aq) + BrO3(aq) in basic solution
5. Cl(CH2)2OH(aq) + K2Cr2O7(aq) → ClCH2CO2H(aq) + Cr3+(aq) in acidic solution
1. Write a balanced chemical equation for each redox reaction.
1. I(aq) + HClO2(aq) → IO3(aq) + Cl2(g) in acidic solution
2. Cr2+(aq) + O2(g) → Cr3+(aq) + H2O(l) in acidic solution
3. CrO2(aq) + ClO(aq) → CrO42−(aq) + Cl(aq) in basic solution
4. S(s) + HNO2(aq) → H2SO3(aq) + N2O(g) in acidic solution
5. F(CH2)2OH(aq) + K2Cr2O7(aq) → FCH2CO2H(aq) + Cr3+(aq) in acidic solution
1. The standard cell potential for the oxidation of Pb to Pb2+ with the concomitant reduction of Cu+ to Cu is 0.39 V. You know that E° for the Pb2+/Pb couple is −0.13 V. What is E° for the Cu+/Cu couple?
1. You have built a galvanic cell similar to the one in Figure 19.7 using an iron nail, a solution of FeCl2, and an SHE. When the cell is connected, you notice that the iron nail begins to corrode. What else do you observe? Under standard conditions, what is Ecell?
1. Carbon is used to reduce iron ore to metallic iron. The overall reaction is as follows:
2Fe2O3•xH2O(s) + 3C(s) → 4Fe(l) + 3CO2(g) + 2xH2O(g)
Write the two half-reactions for this overall reaction.
1. Will each reaction occur spontaneously under standard conditions?
1. Cu(s) + 2H+(aq) → Cu2+(aq) + H2(g)
2. Zn2+(aq) + Pb(s) → Zn(s) + Pb2+(aq)
1. Each reaction takes place in acidic solution. Balance each reaction and then determine whether it occurs spontaneously as written under standard conditions.
1. Se(s) + Br2(l) → H2SeO3(aq) + Br(aq)
2. NO3(aq) + S(s) → HNO2(aq) + H2SO3(aq)
3. Fe3+(aq) + Cr3+(aq) → Fe2+(aq) + Cr2O72−(aq)
1. Calculate E°cell and ΔG° for the redox reaction represented by the cell diagram Pt(s)∣Cl2(g, 1 atm)∥ZnCl2(aq, 1 M)∣Zn(s). Will this reaction occur spontaneously?
1. If you place Zn-coated (galvanized) tacks in a glass and add an aqueous solution of iodine, the brown color of the iodine solution fades to a pale yellow. What has happened? Write the two half-reactions and the overall balanced chemical equation for this reaction. What is E°cell?
1. Your lab partner wants to recover solid silver from silver chloride by using a 1.0 M solution of HCl and 1 atm H2 under standard conditions. Will this plan work?
Numerical Answers
1. Pt(s)∣H2(g, 1 atm) | H+(aq, 1M)∥Cu2+(aq)∣Cu(s)
1. Cl2(g) + H2(g) → 2Cl(aq) + 2H+(aq); E° = 1.358 V
2. Br2(l) + 2Fe2+(aq) → 2Br(aq) + 2Fe3+(aq); E° = 0.316 V
3. 2Fe3+(aq) + Cd(s) → 2Fe2+(aq) + Cd2+(aq); E° = 1.174 V
Conceptual Problems
1. The order of electrode potentials cannot always be predicted by ionization potentials and electron affinities. Why? Do you expect sodium metal to have a higher or a lower electrode potential than predicted from its ionization potential? What is its approximate electrode potential?
2. Without referring to tabulated data, of Br2/Br, Ca2+/Ca, O2/OH, and Al3+/Al, which would you expect to have the least negative electrode potential and which the most negative? Why?
3. Because of the sulfur-containing amino acids present in egg whites, eating eggs with a silver fork will tarnish the fork. As a chemist, you have all kinds of interesting cleaning products in your cabinet, including a 1 M solution of oxalic acid (H2C2O4). Would you choose this solution to clean the fork that you have tarnished from eating scrambled eggs?
4. The electrode potential for the reaction Cu2+(aq) + 2e → Cu(s) is 0.34 V under standard conditions. Is the potential for the oxidation of 0.5 mol of Cu equal to −0.34/2 V? Explain your answer.
5. Refer to Table 19.2 to predict
1. Which species—Sn4+(aq), Cl(aq), Ag+(aq), Cr3+(aq), and/or H2O2(aq)—can oxidize MnO2(s) to MNO4 under standard conditions.
2. Which species—Sn4+(aq), Cl(aq), Ag+(aq), Cr3+(aq), and/or H2O2(aq)—is the strongest oxidizing agent in aqueous solution.
Conceptual Answer
1. No; E° = −0.691 V for Ag2S(s) + 2e → Ag(s) + S2−(aq), which is too negative for Ag2S to be spontaneously reduced by oxalic acid [E° = 0.49 V for 2CO2(g) + 2H+(aq) + 2e → H2C2O4(aq)]
5.
1. Ag+(aq); H2O2(aq)
2. H2O2(aq)
20.3: Ecell, ΔG, and K
Conceptual Problems
1. State whether you agree or disagree with this reasoning and explain your answer: Standard electrode potentials arise from the number of electrons transferred. The greater the number of electrons transferred, the greater the measured potential difference. If 1 mol of a substance produces 0.76 V when 2 mol of electrons are transferred—as in Zn(s) → Zn2+(aq) + 2e—then 0.5 mol of the substance will produce 0.76/2 V because only 1 mol of electrons is transferred.
2. What is the relationship between the measured cell potential and the total charge that passes through a cell? Which of these is dependent on concentration? Which is dependent on the identity of the oxidant or the reductant? Which is dependent on the number of electrons transferred?
3. In the equation wmax = −nFE°cell, which quantities are extensive properties and which are intensive properties?
4. For any spontaneous redox reaction, E is positive. Use thermodynamic arguments to explain why this is true.
5. State whether you agree or disagree with this statement and explain your answer: Electrochemical methods are especially useful in determining the reversibility or irreversibility of reactions that take place in a cell.
6. Although the sum of two half-reactions gives another half-reaction, the sum of the potentials of the two half-reactions cannot be used to obtain the potential of the net half-reaction. Why? When does the sum of two half-reactions correspond to the overall reaction? Why?
7. Occasionally, you will find high-quality electronic equipment that has its electronic components plated in gold. What is the advantage of this?
8. Blood analyzers, which measure pH, $P_\mathrm{CO_2}$, and $P_\mathrm{O_2}$, are frequently used in clinical emergencies. For example, blood $P_\mathrm{CO_2}$ is measured with a pH electrode covered with a plastic membrane that is permeable to CO2. Based on your knowledge of how electrodes function, explain how such an electrode might work. Hint: CO2(g) + H2O(l) → HCO3(aq) + H+(aq).
9. Concentration cells contain the same species in solution in two different compartments. Explain what produces a voltage in a concentration cell. When does V = 0 in such a cell?
10. Describe how an electrochemical cell can be used to measure the solubility of a sparingly soluble salt.
Conceptual Answers
1. extensive: wmax and n; intensive: E°cell
1. Gold is highly resistant to corrosion because of its very positive reduction potential.
Numerical Problems
1. The chemical equation for the combustion of butane is as follows:
$\mathrm{C_4H_{10}(g)+\frac{13}{2}O_2(g)\rightarrow4CO_2(g)+5H_2O(g)}$
This reaction has ΔH° = −2877 kJ/mol. Calculate E°cell and then determine ΔG°. Is this a spontaneous process? What is the change in entropy that accompanies this process at 298 K?
1. How many electrons are transferred during the reaction Pb(s) + Hg2Cl2(s) → PbCl2(aq) + 2Hg(l)? What is the standard cell potential? Is the oxidation of Pb by Hg2Cl2 spontaneous? Calculate ΔG° for this reaction.
1. For the cell represented as Al(s)∣Al3+(aq)∥Sn2+(aq), Sn4+(aq)∣Pt(s), how many electrons are transferred in the redox reaction? What is the standard cell potential? Is this a spontaneous process? What is ΔG°?
1. Explain why the sum of the potentials for the half-reactions Sn2+(aq) + 2e → Sn(s) and Sn4+(aq) + 2e → Sn2+(aq) does not equal the potential for the reaction Sn4+(aq) + 4e → Sn(s). What is the net cell potential? Compare the values of ΔG° for the sum of the potentials and the actual net cell potential.
1. Based on Table 19.2 and Chapter 29, do you agree with the proposed potentials for the following half-reactions? Why or why not?
1. Cu2+(aq) + 2e → Cu(s), E° = 0.68 V
2. Ce4+(aq) + 4e → Ce(s), E° = −0.62 V
1. For each reaction, calculate E°cell and then determine ΔG°. Indicate whether each reaction is spontaneous.
1. 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
2. K2S2O6(aq) + I2(s) → 2KI(aq) + 2K2SO4(aq)
3. Sn(s) + CuSO4(aq) → Cu(s) + SnSO4(aq)
1. What is the standard change in free energy for the reaction between Ca2+ and Na(s) to give Ca(s) and Na+? Do the sign and magnitude of ΔG° agree with what you would expect based on the positions of these elements in the periodic table? Why or why not?
1. In acidic solution, permanganate (MnO4) oxidizes Cl to chlorine gas, and MnO4 is reduced to Mn2+(aq).
1. Write the balanced chemical equation for this reaction.
2. Determine E°cell.
3. Calculate the equilibrium constant.
1. Potentiometric titrations are an efficient method for determining the endpoint of a redox titration. In such a titration, the potential of the solution is monitored as measured volumes of an oxidant or a reductant are added. Data for a typical titration, the potentiometric titration of Fe(II) with a 0.1 M solution of Ce(IV), are given in the following table. The starting potential has been arbitrarily set equal to zero because it is the change in potential with the addition of the oxidant that is important.
Titrant (mL) E (mV)
2.00 50
6.00 100
9.00 255
10.00 960
11.00 1325
12.00 1625
14.00 1875
1. Write the balanced chemical equation for the oxidation of Fe2+ by Ce4+.
2. Plot the data and then locate the endpoint.
3. How many millimoles of Fe2+ did the solution being titrated originally contain?
1. The standard electrode potential (E°) for the half-reaction Ni2+(aq) + 2e → Ni(s) is −0.257 V. What pH is needed for this reaction to take place in the presence of 1.00 atm H2(g) as the reductant if [Ni2+] is 1.00 M?
1. The reduction of Mn(VII) to Mn(s) by H2(g) proceeds in five steps that can be readily followed by changes in the color of the solution. Here is the redox chemistry:
1. MnO4−(aq) + e → MnO42−(aq); E° = +0.56 V (purple → dark green)
2. MnO42−(aq) + 2e + 4H+(aq) → MnO2(s); E° = +2.26 V (dark green → dark brown solid)
3. MnO2(s) + e + 4H+(aq) → Mn3+(aq); E° = +0.95 V (dark brown solid → red-violet)
4. Mn3+(aq) + e → Mn2+(aq); E° = +1.51 V (red-violet → pale pink)
5. Mn2+(aq) + 2e → Mn(s); E° = −1.18 V (pale pink → colorless)
1. Is the reduction of MnO4 to Mn3+(aq) by H2(g) spontaneous under standard conditions? What is E°cell?
2. Is the reduction of Mn3+(aq) to Mn(s) by H2(g) spontaneous under standard conditions? What is E°cell?
1. Mn(III) can disproportionate (both oxidize and reduce itself) by means of the following half-reactions:
Mn3+(aq) + e → Mn2+(aq) E°=1.51 V
Mn3+(aq) + 2H2O(l) → MnO2(s) + 4H+(aq) + e E°=0.95 V
1. What is E° for the disproportionation reaction?
2. Is disproportionation more or less thermodynamically favored at low pH than at pH 7.0? Explain your answer.
3. How could you prevent the disproportionation reaction from occurring?
1. For the reduction of oxygen to water, E° = 1.23 V. What is the potential for this half-reaction at pH 7.00? What is the potential in a 0.85 M solution of NaOH?
1. The biological molecule abbreviated as NADH (reduced nicotinamide adenine dinucleotide) can be formed by reduction of NAD+ (nicotinamide adenine dinucleotide) via the half-reaction NAD+ + H+ + 2e → NADH; E° = −0.32 V.
1. Would NADH be able to reduce acetate to pyruvate?
2. Would NADH be able to reduce pyruvate to lactate?
3. What potential is needed to convert acetate to lactate?
acetate + CO2 + 2H+ +2e → pyruvate +H2O E° = −0.70 V
pyruvate + 2H+ + 2e → lactate E° = −0.185 V
1. Given the following biologically relevant half-reactions, will FAD (flavin adenine dinucleotide), a molecule used to transfer electrons whose reduced form is FADH2, be an effective oxidant for the conversion of acetaldehyde to acetate at pH 4.00?
acetate + 2H+ +2e → acetaldehyde + H2O E° = −0.58 V
FAD + 2H+ +2e → FADH2 E° = −0.18 V
1. Ideally, any half-reaction with E° > 1.23 V will oxidize water as a result of the half-reaction O2(g) + 4H+(aq) + 4e → 2H2O(l).
1. Will FeO42 oxidize water if the half-reaction for the reduction of Fe(VI) → Fe(III) is FeO42−(aq) + 8H+(aq) + 3e → Fe3+(aq) + 4H2O; E° = 1.9 V?
2. What is the highest pH at which this reaction will proceed spontaneously if [Fe3+] = [FeO42−] = 1.0 M and $P_\mathrm{O_2}$= 1.0 atm?
1. Under acidic conditions, ideally any half-reaction with E° > 1.23 V will oxidize water via the reaction O2(g) + 4H+(aq) + 4e → 2H2O(l).
1. Will aqueous acidic KMnO4 evolve oxygen with the formation of MnO2?
2. At pH 14.00, what is E° for the oxidation of water by aqueous KMnO4 (1 M) with the formation of MnO2?
3. At pH 14.00, will water be oxidized if you are trying to form MnO2 from MnO42− via the reaction 2MnO42−(aq) + 2H2O(l) → 2MnO2(s) + O2(g) + 4OH(aq)?
1. Complexing agents can bind to metals and result in the net stabilization of the complexed species. What is the net thermodynamic stabilization energy that results from using CN as a complexing agent for Mn3+/Mn2+?
Mn3+(aq) + e → Mn2+(aq) E° = 1.51 V
Mn(CN)63−(aq) + e → Mn(CN)64− E° = −0.24 V
1. You have constructed a cell with zinc and lead amalgam electrodes described by the cell diagram Zn(Hg)(s)∣Zn(NO3)2(aq)∥Pb(NO3)2(aq)∣Pb(Hg)(s). If you vary the concentration of Zn(NO3)2 and measure the potential at different concentrations, you obtain the following data:
Zn(NO3)2 (M) Ecell (V)
0.0005 0.7398
0.002 0.7221
0.01 0.7014
1. Write the half-reactions that occur in this cell.
2. What is the overall redox reaction?
3. What is E°cell? What is ΔG° for the overall reaction?
4. What is the equilibrium constant for this redox reaction?
1. Hydrogen gas reduces Ni2+ according to the following reaction: Ni2+(aq) + H2(g) → Ni(s) + 2H+(aq); E°cell = −0.25 V; ΔH = 54 kJ/mol.
1. What is K for this redox reaction?
2. Is this reaction likely to occur?
3. What conditions can be changed to increase the likelihood that the reaction will occur as written?
4. Is the reaction more likely to occur at higher or lower pH?
1. The silver–silver bromide electrode has a standard potential of 0.07133 V. What is Ksp of AgBr?
Numerical Answers
1. 6e; E°cell = 1.813 V; the reaction is spontaneous; ΔG° = −525 kJ/mol Al.
1. yes; E° = 0.40 V
1. yes; E° = 0.45 V
2. 0.194 V
3. yes; E° = 0.20 V
20.4: Ecell as a Function of Concentrations
Conceptual Problems
1. What advantage is there to using an alkaline battery rather than a Leclanché dry cell?
2. Why does the density of the fluid in lead–acid batteries drop when the battery is discharged?
3. What type of battery would you use for each application and why?
1. powering an electric motor scooter
2. a backup battery for a smartphone
3. powering an iPod
1. Why are galvanic cells used as batteries and fuel cells? What is the difference between a battery and a fuel cell? What is the advantage to using highly concentrated or solid reactants in a battery?
Conceptual Answer
1. lead storage battery
2. lithium–iodine battery
3. NiCad, NiMH, or lithium ion battery (rechargeable)
Numerical Problem
1. This reaction is characteristic of a lead storage battery:
Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l)
If you have a battery with an electrolyte that has a density of 1.15 g/cm3 and contains 30.0% sulfuric acid by mass, is the potential greater than or less than that of the standard cell?
Numerica Answer
1. [H2SO4] = 3.52 M; E > E°
20.7: Electrolysis: Causing Nonspontaneous Reactions to Occur
Conceptual Problems
1. Why might an electrochemical reaction that is thermodynamically favored require an overvoltage to occur?
2. How could you use an electrolytic cell to make quantitative comparisons of the strengths of various oxidants and reductants?
3. Why are mixtures of molten salts, rather than a pure salt, generally used during electrolysis?
4. Two solutions, one containing Fe(NO3)2·6H2O and the other containing the same molar concentration of Fe(NO3)3·6H2O, were electrolyzed under identical conditions. Which solution produced the most metal? Justify your answer.
Numerical Problems
1. The electrolysis of molten salts is frequently used in industry to obtain pure metals. How many grams of metal are deposited from these salts for each mole of electrons?
1. AlCl3
2. MgCl2
3. FeCl3
1. Electrolysis is the most direct way of recovering a metal from its ores. However, the Na+(aq)/Na(s), Mg2+(aq)/Mg(s), and Al3+(aq)/Al(s) couples all have standard electrode potentials (E°) more negative than the reduction potential of water at pH 7.0 (−0.42 V), indicating that these metals can never be obtained by electrolysis of aqueous solutions of their salts. Why? What reaction would occur instead?
1. What volume of chlorine gas at standard temperature and pressure is evolved when a solution of MgCl2 is electrolyzed using a current of 12.4 A for 1.0 h?
1. What mass of copper metal is deposited if a 5.12 A current is passed through a Cu(NO3)2 solution for 1.5 h.
1. What mass of PbO2 is reduced when a current of 5.0 A is withdrawn over a period of 2.0 h from a lead storage battery?
1. Electrolysis of Cr3+(aq) produces Cr2+(aq). If you had 500 mL of a 0.15 M solution of Cr3+(aq), how long would it take to reduce the Cr3+ to Cr2+ using a 0.158 A current?
1. Predict the products obtained at each electrode when aqueous solutions of the following are electrolyzed.
1. AgNO3
2. RbI
1. Predict the products obtained at each electrode when aqueous solutions of the following are electrolyzed.
1. MgBr2
2. Hg(CH3CO2)2
3. Al2(SO4)3
Answers
1. 5.2 L
1. cathode: Ag(s); anode: O2(g);
2. cathode: H2(g); anode: I2(s)
20.8: Industrial Electrolysis Processes
True or False:
1. The main product of the chlor-alkali process is NaCl.
2. A diaphragm cell yields the purest concentration of NaOH.
3. In a diaphragm cell, the oxidation of Cl- ions occurs at the anode and the reduction of Na+ occurs at the cathode.
4. H2 gas is a byproduct of the chlor-alkali process.
5. Electrorefining for copper is effective because out of the all of the metals in an ore, only copper is oxidized at the anode at low voltages.
6. A mercury cell uses mercury liquid as the cathode.
7. The chlor-alkali process is spontaneous.
8. A mercury cell requires a higher voltage and more energy than a diaphragm cell.
9. In a membrane cell, all ions can freely be exchanged between the membrane.
10. The reaction between NaOH and Cl2 gas is necessary for the chlor-alkali process.
Check Your Understanding: Answers
1. False - The chlor-alkali process is the electrolysis of NaCl, not the electrosynthesis. NaCl is one of the main reactants of the chlor-alkali process
2. False - A diaphragm cell yields the most impure NaOH. The mercury cell yields purer NaOH, followed by the membrane cell with the purest.
3. False - Due to the high voltage required to reduce Na ions (meaning the large negative standard electrode reduction potential of sodium), the reduction of water occurs instead.
4. True - Yes, H2 gas is produced at the cathode while Cl2 gas is produced at the anode.
5. False - Though Au, Ag and Pt are not oxidized, Sb, Bi and Sn are oxidized but are precipitated as they form hydroxides and oxides. Fe, Ni, Co and Zn are oxidized as well, but they are dissolved in water. Thus, even though copper is the only pure metal solid we have left, other impurities are oxidized as well.
6. True - Liquid mercury at the bottom of the tank acts as the cathode. More mercury is constantly cycled back into the tank as amalgam is treated with water.
7. False - Electrosynthesis requires energy.
8. True - A mercury cell needs 3400 kWh/ton Cl2 opposed to 2500 in a diaphragm cell.
9. False - In a membrane cell, only cations can pass through the membrane into the cathode. This explains why the NaOH on the cathode side has so little chlorine ion impurities.
10. False - If Cl2 and NaOH come into contact, Cl2 turns into ClO-, ClO3-, Cl ions. This is not desirable, so the difference in solution levels is used to prevent backflow of NaOH into the anode side. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/20%3A_Electrochemistry/20.E%3A_Exercises.txt |
• 21.1: Periodic Trends and Charge Density
The most important unifying principle in describing the chemistry of the elements is that the systematic increase in atomic number and the orderly filling of atomic orbitals lead to periodic trends in atomic properties. The most fundamental property leading to periodic variations is the effective nuclear charge (Zeff). Because of the position of the diagonal line separating metals and nonmetals in the periodic table, the chemistry of groups 13, 14, and 15 is relatively complex.
• 21.2: Group 1: The Alkali Metals
Li, Na, K, Rb, and Cs are all group IA elements, also known as the alkali metals. The seventh member of the group, francium (Fr) is radioactive and so rare that only 20 atoms of Fr may exist on Earth at any given moment. The term alkali is derived from an Arabic word meaning “ashes.” Compounds of potassium as well as other alkali metals were obtained from wood ashes by early chemists.
• 21.3: Group 2: The Alkaline Earth Metals
Group 2 elements almost exclusively form ionic compounds containing the M2+ ion, they are more reactive toward group 15 elements, and they have a greater tendency to form complexes with Lewis bases than do the alkali metals. Pure samples of most of the alkaline earth metals can be obtained by electrolysis of the chlorides or oxides. Beryllium was first obtained by the reduction of its chloride; radium chloride, which is radioactive, was obtained through a series of reactions and separations.
• 21.4: Group 13: The Boron Family
Compounds of the group 13 elements with oxygen are thermodynamically stable. Many of the anomalous properties of the group 13 elements can be explained by the increase in Zeff moving down the group. Isolation of the group 13 elements requires a large amount of energy because compounds of the group 13 elements with oxygen are thermodynamically stable. Boron behaves chemically like a nonmetal, whereas its heavier congeners exhibit metallic behavior.
• 21.5: Group 14: The Carbon Family
The group 14 elements show the greatest diversity in chemical behavior of any group; covalent bond strengths decease with increasing atomic size, and ionization energies are greater than expected, increasing from C to Pb. The group 14 elements show the greatest range of chemical behavior of any group in the periodic table. Because the covalent bond strength decreases with increasing atomic size and greater-than-expected ionization energies due to an increase in Zeff.
• 21.E: Exercises
21: Chemistry of The Main-Group Elements I
Learning Objectives
• To know important periodic trends in several atomic properties.
As we begin our summary of periodic trends, recall that the single most important unifying principle in understanding the chemistry of the elements is the systematic increase in atomic number, accompanied by the orderly filling of atomic orbitals by electrons, which leads to periodicity in such properties as atomic and ionic size, ionization energy, electronegativity, and electron affinity. The same factors also lead to periodicity in valence electron configurations, which for each group results in similarities in oxidation states and the formation of compounds with common stoichiometries.
The most important periodic trends in atomic properties are summarized in Figure \(1\). Recall that these trends are based on periodic variations in a single fundamental property, the effective nuclear charge (\(Z_{eff}\)), which increases from left to right and from top to bottom in the periodic table.
The diagonal line in Figure \(1\) separates the metals (to the left of the line) from the nonmetals (to the right of the line). Because metals have relatively low electronegativities, they tend to lose electrons in chemical reactions to elements that have relatively high electronegativities, forming compounds in which they have positive oxidation states. Conversely, nonmetals have high electronegativities, and they therefore tend to gain electrons in chemical reactions to form compounds in which they have negative oxidation states. The semimetals lie along the diagonal line dividing metals and nonmetals. It is not surprising that they tend to exhibit properties and reactivities intermediate between those of metals and nonmetals. Because the elements of groups 13, 14, and 15 span the diagonal line separating metals and nonmetals, their chemistry is more complex than predicted based solely on their valence electron configurations.
Unique Chemistry of the Lightest Elements
The chemistry of the second-period element of each group (n = 2: Li, Be, B, C, N, O, and F) differs in many important respects from that of the heavier members, or congeners, of the group. Consequently, the elements of the third period (n = 3: Na, Mg, Al, Si, P, S, and Cl) are generally more representative of the group to which they belong. The anomalous chemistry of second-period elements results from three important characteristics: small radii, energetically unavailable d orbitals, and a tendency to form pi (π) bonds with other atoms.
Note
In contrast to the chemistry of the second-period elements, the chemistry of the third-period elements is more representative of the chemistry of the respective group.
Due to their small radii, second-period elements have electron affinities that are less negative than would be predicted from general periodic trends. When an electron is added to such a small atom, increased electron–electron repulsions tend to destabilize the anion. Moreover, the small sizes of these elements prevent them from forming compounds in which they have more than four nearest neighbors. Thus BF3 forms only the four-coordinate, tetrahedral BF4 ion, whereas under the same conditions AlF3 forms the six-coordinate, octahedral AlF63 ion. Because of the smaller atomic size, simple binary ionic compounds of second-period elements also have more covalent character than the corresponding compounds formed from their heavier congeners. The very small cations derived from second-period elements have a high charge-to-radius ratio and can therefore polarize the filled valence shell of an anion. As such, the bonding in such compounds has a significant covalent component, giving the compounds properties that can differ significantly from those expected for simple ionic compounds. As an example, LiCl, which is partially covalent in character, is much more soluble than NaCl in solvents with a relatively low dielectric constant, such as ethanol (ε = 25.3 versus 80.1 for H2O).
Because d orbitals are never occupied for principal quantum numbers less than 3, the valence electrons of second-period elements occupy 2s and 2p orbitals only. The energy of the 3d orbitals far exceeds the energy of the 2s and 2p orbitals, so using them in bonding is energetically prohibitive. Consequently, electron configurations with more than four electron pairs around a central, second-period element are simply not observed. You may recall that the role of d orbitals in bonding in main group compounds with coordination numbers of 5 or higher remains somewhat controversial. In fact, theoretical descriptions of the bonding in molecules such as SF6 have been published without mentioning the participation of d orbitals on sulfur. Arguments based on d-orbital availability and on the small size of the central atom, however, predict that coordination numbers greater than 4 are unusual for the elements of the second period, which is in agreement with experimental results.
One of the most dramatic differences between the lightest main group elements and their heavier congeners is the tendency of the second-period elements to form species that contain multiple bonds. For example, N is just above P in group 15: N2 contains an N≡N bond, but each phosphorus atom in tetrahedral P4 forms three P–P bonds. This difference in behavior reflects the fact that within the same group of the periodic table, the relative energies of the π bond and the sigma (σ) bond differ. A C=C bond, for example, is approximately 80% stronger than a C–C bond. In contrast, an Si=Si bond, with less p-orbital overlap between the valence orbitals of the bonded atoms because of the larger atomic size, is only about 40% stronger than an Si–Si bond. Consequently, compounds that contain both multiple and single C to C bonds are common for carbon, but compounds that contain only sigma Si–Si bonds are more energetically favorable for silicon and the other third-period elements.
The Inert-Pair Effect
The inert-pair effect refers to the empirical observation that the heavier elements of groups 13–17 often have oxidation states that are lower by 2 than the maximum predicted for their group. For example, although an oxidation state of +3 is common for group 13 elements, the heaviest element in group 13, thallium (Tl), is more likely to form compounds in which it has a +1 oxidation state. There appear to be two major reasons for the inert-pair effect: increasing ionization energies and decreasing bond strengths.
Note
In moving down a group in the p-block, increasing ionization energies and decreasing bond strengths result in an inert-pair effect.
The ionization energies increase because filled (n − 1)d or (n − 2)f subshells are relatively poor at shielding electrons in ns orbitals. Thus the two electrons in the ns subshell experience an unusually high effective nuclear charge, so they are strongly attracted to the nucleus, reducing their participation in bonding. It is therefore substantially more difficult than expected to remove these ns2electrons, as shown in Table \(1\) by the difference between the first ionization energies of thallium and aluminum. Because Tl is less likely than Al to lose its two ns2 electrons, its most common oxidation state is +1 rather than +3.
Table \(1\): Ionization Energies (I) and Average M–Cl Bond Energies for the Group 13 Elements
Element Electron Configuration I1 (kJ/mol) I1 + I2 + I3 (kJ/mol) Average M–Cl Bond Energy (kJ/mol)
B [He] 2s22p1 801 6828 536
Al [Ne] 3s23p1 578 5139 494
Ga [Ar] 3d104s24p1 579 5521 481
In [Kr] 4d105s2p1 558 5083 439
Tl [Xe] 4f145d106s2p1 589 5439 373
Source of data: John A. Dean, Lange’s Handbook of Chemistry, 15th ed. (New York: McGraw-Hill, 1999).
Going down a group, the atoms generally became larger, and the overlap between the valence orbitals of the bonded atoms decreases. Consequently, bond strengths tend to decrease down a column. As shown by the M–Cl bond energies listed in Table \(1\), the strength of the bond between a group 13 atom and a chlorine atom decreases by more than 30% from B to Tl. Similar decreases are observed for the atoms of groups 14 and 15.
The net effect of these two factors—increasing ionization energies and decreasing bond strengths—is that in going down a group in the p-block, the additional energy released by forming two additional bonds eventually is not great enough to compensate for the additional energy required to remove the two ns2 electrons.
Example \(1\)
Based on the positions of the group 13 elements in the periodic table and the general trends outlined in this section,
1. classify these elements as metals, semimetals, or nonmetals.
2. predict which element forms the most stable compounds in the +1 oxidation state.
3. predict which element differs the most from the others in its chemistry.
4. predict which element of another group will exhibit chemistry most similar to that of Al.
Given: positions of elements in the periodic table
Asked for: classification, oxidation-state stability, and chemical reactivity
Strategy:
From the position of the diagonal line in the periodic table separating metals and nonmetals, classify the group 13 elements. Then use the trends discussed in this section to compare their relative stabilities and chemical reactivities.
Solution
1. Group 13 spans the diagonal line separating the metals from the nonmetals. Although Al and B both lie on the diagonal line, only B is a semimetal; the heavier elements are metals.
2. All five elements in group 13 have an ns2np1 valence electron configuration, so they are expected to form ions with a +3 charge from the loss of all valence electrons. The inert-pair effect should be most important for the heaviest element (Tl), so it is most likely to form compounds in an oxidation state that is lower by 2. Thus the +1 oxidation state is predicted to be most important for thallium.
3. Among the main group elements, the lightest member of each group exhibits unique chemistry because of its small size resulting in a high concentration of charge, energetically unavailable d orbitals, and a tendency to form multiple bonds. In group 13, we predict that the chemistry of boron will be quite different from that of its heavier congeners.
4. Within the s and p blocks, similarities between elements in different groups are most marked between the lightest member of one group and the element of the next group immediately below and to the right of it. These elements exhibit similar electronegativities and charge-to-radius ratios. Because Al is the second member of group 13, we predict that its chemistry will be most similar to that of Be, the lightest member of group 2.
Exercise \(1\)
Based on the positions of the group 14 elements C, Si, Ge, Sn, and Pb in the periodic table and the general trends outlined in this section,
1. classify these elements as metals, semimetals, or nonmetals.
2. predict which element forms the most stable compounds in the +2 oxidation state.
3. predict which element differs the most from the others in its chemistry.
4. predict which element of group 14 will be chemically most similar to a group 15 element.
Answer
1. nonmetal: C; semimetals: Si and Ge; metals: Sn and Pb
2. Pb is most stable as M2+.
3. C is most different.
4. C and P are most similar in chemistry.
Summary
The most important unifying principle in describing the chemistry of the elements is that the systematic increase in atomic number and the orderly filling of atomic orbitals lead to periodic trends in atomic properties. The most fundamental property leading to periodic variations is the effective nuclear charge (Zeff). Because of the position of the diagonal line separating metals and nonmetals in the periodic table, the chemistry of groups 13, 14, and 15 is relatively complex. The second-period elements (n = 2) in each group exhibit unique chemistry compared with their heavier congeners because of their smaller radii, energetically unavailable d orbitals, and greater ability to form π bonds with other atoms. Increasing ionization energies and decreasing bond strengths lead to the inert-pair effect, which causes the heaviest elements of groups 13–17 to have a stable oxidation state that is lower by 2 than the maximum predicted for their respective groups.
Key Takeaway
• The chemistry of the third-period element in a group is most representative of the chemistry of the group because the chemistry of the second-period elements is dominated by their small radii, energetically unavailable d orbitals, and tendency to form π bonds with other atoms. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/21%3A_Chemistry_of_The_Main-Group_Elements_I/21.1%3A_Periodic_Trends_and_Charge_Density.txt |
Learning Objectives
1. To describe how the alkali metals are isolated.
2. To be familiar with the reactions, compounds, and complexes of the alkali metals.
The alkali metals are so reactive that they are never found in nature in elemental form. Although some of their ores are abundant, isolating them from their ores is somewhat difficult. For these reasons, the group 1 elements were unknown until the early 19th century, when Sir Humphry Davy first prepared sodium (Na) and potassium (K) by passing an electric current through molten alkalis. (The ashes produced by the combustion of wood are largely composed of potassium and sodium carbonate.) Lithium (Li) was discovered 10 years later when the Swedish chemist Johan Arfwedson was studying the composition of a new Brazilian mineral. Cesium (Cs) and rubidium (Rb) were not discovered until the 1860s, when Robert Bunsen conducted a systematic search for new elements. Known to chemistry students as the inventor of the Bunsen burner, Bunsen’s spectroscopic studies of ores showed sky blue and deep red emission lines that he attributed to two new elements, Cs and Rb, respectively. Francium (Fr) is found in only trace amounts in nature, so our knowledge of its chemistry is limited. All the isotopes of Fr have very short half-lives, in contrast to the other elements in group 1.
Humphry Davy (1778–1829)
Davy was born in Penzance, Cornwall, England. He was a bit of a wild man in the laboratory, often smelling and tasting the products of his experiments, which almost certainly shortened his life. He discovered the physiological effects that cause nitrous oxide to be called “laughing gas” (and became addicted to it!), and he almost lost his eyesight in an explosion of nitrogen trichloride (NCl3), which he was the first to prepare. Davy was one of the first to recognize the utility of Alessandro Volta’s “electric piles” (batteries). By connecting several “piles” in series and inserting electrodes into molten salts of the alkali metals and alkaline earth metals, he was able to isolate six previously unknown elements as pure metals: sodium, potassium, calcium, strontium, barium, and magnesium. He also discovered boron and was the first to prepare phosphine (PH3) and hydrogen telluride (H2Te), both of which are highly toxic.
Robert Wilhelm Bunsen (1811–1899)
Bunsen was born and educated in Göttingen, Germany. His early work dealt with organic arsenic compounds, whose highly toxic nature and explosive tendencies almost killed him and did cost him an eye. He designed the Bunsen burner, a reliable gas burner, and used it and emission spectra to discover cesium (named for its blue line) and rubidium (named for its red line).
Preparation of the Alkali Metals
Because the alkali metals are among the most potent reductants known, obtaining them in pure form requires a considerable input of energy. Pure lithium and sodium for example, are typically prepared by the electrolytic reduction of molten chlorides:
$\mathrm{LiCl(l)}\rightarrow\mathrm{Li(l)}+\frac{1}{2}\mathrm{Cl_2(g)} \label{21.15}$
In practice, CaCl2 is mixed with LiCl to lower the melting point of the lithium salt. The electrolysis is carried out in an argon atmosphere rather than the nitrogen atmosphere typically used for substances that are highly reactive with O2 and water because Li reacts with nitrogen gas to form lithium nitride (Li3N). Metallic sodium is produced by the electrolysis of a molten mixture of NaCl and CaCl2. In contrast, potassium is produced commercially from the reduction of KCl by Na, followed by the fractional distillation of K(g). Although rubidium and cesium can also be produced by electrolysis, they are usually obtained by reacting their hydroxide salts with a reductant such as Mg:
$2RbOH_{(s)} + Mg_{(s)} \rightarrow 2Rb_{(l)} + Mg(OH)_{2(s)} \label{21.6}$
Massive deposits of essentially pure NaCl and KCl are found in nature and are the major sources of sodium and potassium. The other alkali metals are found in low concentrations in a wide variety of minerals, but ores that contain high concentrations of these elements are relatively rare. No concentrated sources of rubidium are known, for example, even though it is the 16th most abundant element on Earth. Rubidium is obtained commercially by isolating the 2%–4% of Rb present as an impurity in micas, minerals that are composed of sheets of complex hydrated potassium–aluminum silicates.
Alkali metals are recovered from silicate ores in a multistep process that takes advantage of the pH-dependent solubility of selected salts of each metal ion. The steps in this process are leaching, which uses sulfuric acid to dissolve the desired alkali metal ion and Al3+ from the ore; basic precipitation to remove Al3+ from the mixture as Al(OH)3; selective precipitation of the insoluble alkali metal carbonate; dissolution of the salt again in hydrochloric acid; and isolation of the metal by evaporation and electrolysis. Figure $1$ illustrates the isolation of liquid lithium from a lithium silicate ore by this process.
General Properties of the Alkali Metals
Various properties of the group 1 elements are summarized in Table $1$. In keeping with overall periodic trends, the atomic and ionic radii increase smoothly from Li to Cs, and the first ionization energies decrease as the atoms become larger. As a result of their low first ionization energies, the alkali metals have an overwhelming tendency to form ionic compounds where they have a +1 charge. All the alkali metals have relatively high electron affinities because the addition of an electron produces an anion (M−) with an ns2 electron configuration. The densities of the elements generally increase from Li to Cs, reflecting another common trend: because the atomic masses of the elements increase more rapidly than the atomic volumes as you go down a group, the densest elements are near the bottom of the periodic table. An unusual trend in the group 1 elements is the smooth decrease in the melting and boiling points from Li to Cs. As a result, Cs (melting point = 28.5°C) is one of only three metals (the others are Ga and Hg) that are liquids at body temperature (37°C).
Table $1$: Selected Properties of the Group 1 Elements
Lithium Sodium Potassium Rubidium Cesium Francium
*The values cited are for four-coordinate ions except for Rb+ and Cs+, whose values are given for the six-coordinate ion.
atomic symbol Li Na K Rb Cs Fr
atomic number 3 11 19 37 55 87
atomic mass 6.94 22.99 39.10 85.47 132.91 223
valence electron configuration 2s1 3s1 4s1 5s1 6s1 7s1
melting point/boiling point (°C) 180.5/1342 97.8/883 63.5/759 39.3/688 28.5/671 27/—
density (g/cm3) at 25°C 0.534 0.97 0.89 1.53 1.93
atomic radius (pm) 167 190 243 265 298
first ionization energy (kJ/mol) 520 496 419 403 376 393
most common oxidation state +1 +1 +1 +1 +1 +1
ionic radius (pm)* 76 102 138 152 167
electron affinity (kJ/mol) −60 −53 −48 −47 −46
electronegativity 1.0 0.9 0.8 0.8 0.8 0.7
standard electrode potential (E°, V) −3.04 −2.71 −2.93 −2.98 −3.03
product of reaction with O2 Li2O Na2O2 KO2 RbO2 CsO2
type of oxide basic basic basic basic basic
product of reaction with N2 Li3N none none none none
product of reaction with X2 LiX NaX KX RbX CsX
product of reaction with H2 LiH NaH KH RbH CsH
The standard reduction potentials (E°) of the alkali metals do not follow the trend based on ionization energies. Unexpectedly, lithium is the strongest reductant, and sodium is the weakest (Table $1$). Because Li+ is much smaller than the other alkali metal cations, its hydration energy is the highest. The high hydration energy of Li+ more than compensates for its higher ionization energy, making lithium metal the strongest reductant in aqueous solution. This apparent anomaly is an example of how the physical or the chemical behaviors of the elements in a group are often determined by the subtle interplay of opposing periodic trends.
Reactions and Compounds of the Alkali Metals
All alkali metals are electropositive elements with an ns1 valence electron configuration, forming the monocation (M+) by losing the single valence electron. Because removing a second electron would require breaking into the (n − 1) closed shell, which is energetically prohibitive, the chemistry of the alkali metals is largely that of ionic compounds that contain M+ ions. However, as we discuss later, the lighter group 1 elements also form a series of organometallic compounds that contain polar covalent M–C bonds.
All the alkali metals react vigorously with the halogens (group 17) to form the corresponding ionic halides, where $X$ is a halogen:
$2M_{(s)} + X_{2(s, l, g)} \rightarrow 2M^+X^−_{(s)} \label{21.7}$
Similarly, the alkali metals react with the heavier chalcogens (sulfur, selenium, and tellurium in group 16) to produce metal chalcogenides, where Y is S, Se, or Te:
$2M_{(s)} + Y_{(s)} \rightarrow M_2Y_{(s)} \label{21.8}$
When excess chalcogen is used, however, a variety of products can be obtained that contain chains of chalcogen atoms, such as the sodium polysulfides (Na2Sn, where n = 2–6). For example, Na2S3 contains the S32− ion, which is V shaped with an S–S–S angle of about 103°. The one-electron oxidation product of the trisulfide ion (S3) is responsible for the intense blue color of the gemstones lapis lazuli and blue ultramarine (Figure $2$).
Reacting the alkali metals with oxygen, the lightest element in group 16, is more complex, and the stoichiometry of the product depends on both the metal:oxygen ratio and the size of the metal atom. For instance, when alkali metals burn in air, the observed products are Li2O (white), Na2O2 (pale yellow), KO2 (orange), RbO2 (brown), and CsO2 (orange). Only Li2O has the stoichiometry expected for a substance that contains two M+ cations and one O2− ion. In contrast, Na2O2 contains the O22− (peroxide) anion plus two Na+ cations. The other three salts, with stoichiometry MO2, contain the M+ cation and the O2 (superoxide) ion. Because O2− is the smallest of the three oxygen anions, it forms a stable ionic lattice with the smallest alkali metal cation (Li+). In contrast, the larger alkali metals—potassium, rubidium, and cesium—react with oxygen in air to give the metal superoxides. Because the Na+ cation is intermediate in size, sodium reacts with oxygen to form a compound with an intermediate stoichiometry: sodium peroxide. Under specific reaction conditions, however, it is possible to prepare the oxide, peroxide, and superoxide salts of all five alkali metals, except for lithium superoxide (LiO2).
The chemistry of the alkali metals is largely that of ionic compounds containing the M+ ions.
The alkali metal peroxides and superoxides are potent oxidants that react, often vigorously, with a wide variety of reducing agents, such as charcoal or aluminum metal. For example, Na2O2 is used industrially for bleaching paper, wood pulp, and fabrics such as linen and cotton. In submarines, Na2O2 and KO2 are used to purify and regenerate the air by removing the CO2 produced by respiration and replacing it with O2. Both compounds react with CO2 in a redox reaction in which O22− or O2 is simultaneously oxidized and reduced, producing the metal carbonate and O2:
$2Na_2O_{2(s)} + 2CO_{2(g)} \rightarrow 2Na_2CO_{3(s)} + O_{2(g)} \label{21.9}$
$4KO_{2(s)} + 2CO_{2(g)} \rightarrow 2K_2CO_{3(s)} + 3O_{2(g)} \label{21.10}$
The presence of water vapor, the other product of respiration, makes KO2 even more effective at removing CO2 because potassium bicarbonate, rather than potassium carbonate, is formed:
$4KO_{2(s)} + 4CO_{2(g)} + 2H_2O_{(g)} \rightarrow 4KHCO_{3(s)} + 3O_{2(g)} \label{21.11}$
Notice that 4 mol of CO2 are removed in this reaction, rather than 2 mol in Equation 21.10.
Lithium, the lightest alkali metal, is the only one that reacts with atmospheric nitrogen, forming lithium nitride (Li3N). Lattice energies again explain why the larger alkali metals such as potassium do not form nitrides: packing three large K+ cations around a single relatively small anion is energetically unfavorable. In contrast, all the alkali metals react with the larger group 15 elements phosphorus and arsenic to form metal phosphides and arsenides (where Z is P or As):
$12M_{(s)} + Z_{4(s)} \rightarrow 4M_3Z_{(s)} \label{21.12}$
Because of lattice energies, only lithium forms a stable oxide and nitride.
The alkali metals react with all group 14 elements, but the compositions and properties of the products vary significantly. For example, reaction with the heavier group 14 elements gives materials that contain polyatomic anions and three-dimensional cage structures, such as K4Si4 whose structure is shown here. In contrast, lithium and sodium are oxidized by carbon to produce a compound with the stoichiometry M2C2 (where M is Li or Na):
$2M_{(s)} + 2C_{(s)} \rightarrow M_2C_{2(s)} \label{21.13}$
The same compounds can be obtained by reacting the metal with acetylene (C2H2). In this reaction, the metal is again oxidized, and hydrogen is reduced:
$2M_{(s)} + C_2H_{2(g)} \rightarrow M_2C_{2(s)} + H_{2(g)} \label{21.14}$
The acetylide ion (C22−), formally derived from acetylene by the loss of both hydrogens as protons, is a very strong base. Reacting acetylide salts with water produces acetylene and MOH(aq).
The heavier alkali metals (K, Rb, and Cs) also react with carbon in the form of graphite. Instead of disrupting the hexagonal sheets of carbon atoms, however, the metals insert themselves between the sheets of carbon atoms to give new substances called graphite intercalation compounds (part (a) in Figure $3$). The stoichiometries of these compounds include MC60 and MC48, which are black/gray; MC36 and MC24, which are blue; and MC8, which is bronze (part (b) in Figure $3$). The remarkably high electrical conductivity of these compounds (about 200 times greater than graphite) is attributed to a net transfer of the valence electron of the alkali metal to the graphite layers to produce, for example, K+C8.
All the alkali metals react directly with gaseous hydrogen at elevated temperatures to produce ionic hydrides (M+H):
$2M_{(s)} + H_{2(g)} \rightarrow 2MH_{(s)} \label{21.15a}$
All are also capable of reducing water to produce hydrogen gas:
$\mathrm{M(s)}+\mathrm{H_2O(l)}\rightarrow\frac{1}{2}\mathrm{H_2(g)}+\mathrm{MOH(aq)} \label{21.16}$
Although lithium reacts rather slowly with water, sodium reacts quite vigorously (Figure $4$), and the heavier alkali metals (K, Rb, and Cs) react so vigorously that they invariably explode. This trend, which is not consistent with the relative magnitudes of the reduction potentials of the elements, serves as another example of the complex interplay of different forces and phenomena—in this case, kinetics and thermodynamics. Although the driving force for the reaction is greatest for lithium, the heavier metals have lower melting points. The heat liberated by the reaction causes them to melt, and the larger surface area of the liquid metal in contact with water greatly accelerates the reaction rate.
Alkali metal cations are found in a wide variety of ionic compounds. In general, any alkali metal salt can be prepared by reacting the alkali metal hydroxide with an acid and then evaporating the water:
$2MOH_{(aq)} + H_2SO_{4(aq)} \rightarrow M_2SO_{4(aq)} + 2H_2O_{(l)} \label{21.17}$
$MOH_{(aq)} + HNO_{3(aq)} \rightarrow MNO_{3(aq)} + H_2O_{(l)} \label{21.18}$
Hydroxides of alkali metals also can react with organic compounds that contain an acidic hydrogen to produce a salt. An example is the preparation of sodium acetate (CH3CO2Na) by reacting sodium hydroxide and acetic acid:
$CH_3CO_2H_{(aq)} + NaOH_{(s)} \rightarrow CH_3CO_2Na_{(aq)} + H_2O_{(l)} \label{21.19}$
Soap is a mixture of the sodium and potassium salts of naturally occurring carboxylic acids, such as palmitic acid [CH3(CH2)14CO2H] and stearic acid [CH3(CH2)16CO2H]. Lithium salts, such as lithium stearate [CH3(CH2)14CO2Li], are used as additives in motor oils and greases.
Complexes of the Alkali Metals
Because of their low positive charge (+1) and relatively large ionic radii, alkali metal cations have only a weak tendency to react with simple Lewis bases to form metal complexes. Complex formation is most significant for the smallest cation (Li+) and decreases with increasing radius. In aqueous solution, for example, Li+ forms the tetrahedral [Li(H2O)4]+ complex. In contrast, the larger alkali metal cations form octahedral [M(H2O)6]+ complexes. Complex formation is primarily due to the electrostatic interaction of the metal cation with polar water molecules. Because of their high affinity for water, anhydrous salts that contain Li+ and Na+ ions (such as Na2SO4) are often used as drying agents. These compounds absorb trace amounts of water from nonaqueous solutions to form hydrated salts, which are then easily removed from the solution by filtration.
Because of their low positive charge (+1) and relatively large ionic radii, alkali metal cations have only a weak tendency to form complexes with simple Lewis bases.
Electrostatic interactions also allow alkali metal ions to form complexes with certain cyclic polyethers and related compounds, such as crown ethers and cryptands. As discussed in Chapter 13, crown ethers are cyclic polyethers that contain four or more oxygen atoms separated by two or three carbon atoms. All crown ethers have a central cavity that can accommodate a metal ion coordinated to the ring of oxygen atoms, and crown ethers with rings of different sizes prefer to bind metal ions that fit into the cavity. For example, 14-crown-4, with the smallest cavity that can accommodate a metal ion, has the highest affinity for Li+, whereas 18-crown-6 forms the strongest complexes with K+.
Cryptands are more nearly spherical analogues of crown ethers and are even more powerful and selective complexing agents. Cryptands consist of three chains containing oxygen that are connected by two nitrogen atoms (part (b) in Figure 13.7). They can completely surround (encapsulate) a metal ion of the appropriate size, coordinating to the metal by a lone pair of electrons on each O atom and the two N atoms. Like crown ethers, cryptands with different cavity sizes are highly selective for metal ions of particular sizes. Crown ethers and cryptands are often used to dissolve simple inorganic salts such as KMnO4 in nonpolar organic solvents.
Liquid Ammonia Solutions
A remarkable feature of the alkali metals is their ability to dissolve reversibly in liquid ammonia. Just as in their reactions with water, reacting alkali metals with liquid ammonia eventually produces hydrogen gas and the metal salt of the conjugate base of the solvent—in this case, the amide ion (NH2) rather than hydroxide:
$\mathrm{M(s)}+\mathrm{NH_3(l)}\rightarrow\frac{1}{2}\mathrm{H_2(g)}+\mathrm{M^+(am)}+\mathrm{NH_2^-(am)} \label{21.20}$
where the (am) designation refers to an ammonia solution, analogous to (aq) used to indicate aqueous solutions. Without a catalyst, the reaction in Equation 21.20 tends to be rather slow. In many cases, the alkali metal amide salt (MNH2) is not very soluble in liquid ammonia and precipitates, but when dissolved, very concentrated solutions of the alkali metal are produced. One mole of Cs metal, for example, will dissolve in as little as 53 mL (40 g) of liquid ammonia. The pure metal is easily recovered when the ammonia evaporates.
Solutions of alkali metals in liquid ammonia are intensely colored and good conductors of electricity due to the presence of solvated electrons (e, NH3), which are not attached to single atoms. A solvated electron is loosely associated with a cavity in the ammonia solvent that is stabilized by hydrogen bonds. Alkali metal–liquid ammonia solutions of about 3 M or less are deep blue (Figure $5$) and conduct electricity about 10 times better than an aqueous NaCl solution because of the high mobility of the solvated electrons. As the concentration of the metal increases above 3 M, the color changes to metallic bronze or gold, and the conductivity increases to a value comparable with that of the pure liquid metals.
In addition to solvated electrons, solutions of alkali metals in liquid ammonia contain the metal cation (M+), the neutral metal atom (M), metal dimers (M2), and the metal anion (M). The anion is formed by adding an electron to the singly occupied ns valence orbital of the metal atom. Even in the absence of a catalyst, these solutions are not very stable and eventually decompose to the thermodynamically favored products: M+NH2 and hydrogen gas (Equation 21.20). Nonetheless, the solvated electron is a potent reductant that is often used in synthetic chemistry.
Organometallic Compounds of the Group 1 Elements
Compounds that contain a metal covalently bonded to a carbon atom of an organic species are called organometallic compounds. The properties and reactivities of organometallic compounds differ greatly from those of either the metallic or organic components. Because of its small size, lithium, for example, forms an extensive series of covalent organolithium compounds, such as methyllithium (LiCH3), which are by far the most stable and best-known group 1 organometallic compounds. These volatile, low-melting-point solids or liquids can be sublimed or distilled at relatively low temperatures and are soluble in nonpolar solvents. Like organic compounds, the molten solids do not conduct electricity to any significant degree. Organolithium compounds have a tendency to form oligomers with the formula (RLi)n, where R represents the organic component. For example, in both the solid state and solution, methyllithium exists as a tetramer with the structure shown in Figure $6$, where each triangular face of the Li4 tetrahedron is bridged by the carbon atom of a methyl group. Effectively, the carbon atom of each CH3 group is using a single pair of electrons in an sp3 hybrid lobe to bridge three lithium atoms, making this an example of two-electron, four-center bonding. Clearly, such a structure, in which each carbon atom is apparently bonded to six other atoms, cannot be explained using any of the electron-pair bonding schemes. Molecular orbital theory can explain the bonding in methyllithium, but the description is beyond the scope of this text.
• The properties and reactivities of organometallic compounds differ greatly from those of either the metallic or organic components.
• Organosodium and organopotassium compounds are more ionic than organolithium compounds. They contain discrete M+ and R ions and are insoluble or only sparingly soluble in nonpolar solvents.
Uses of the Alkali Metals
Because sodium remains liquid over a wide temperature range (97.8–883°C), it is used as a coolant in specialized high-temperature applications, such as nuclear reactors and the exhaust valves in high-performance sports car engines. Cesium, because of its low ionization energy, is used in photosensors in automatic doors, toilets, burglar alarms, and other electronic devices. In these devices, cesium is ionized by a beam of visible light, thereby producing a small electric current; blocking the light interrupts the electric current and triggers a response.
Compounds of sodium and potassium are produced on a huge scale in industry. Each year, the top 50 industrial compounds include NaOH, used in a wide variety of industrial processes; Na2CO3, used in the manufacture of glass; K2O, used in porcelain glazes; and Na4SiO4, used in detergents.
Several other alkali metal compounds are also important. For example, Li2CO3 is one of the most effective treatments available for manic depression or bipolar disorder. It appears to modulate or dampen the effect on the brain of changes in the level of neurotransmitters, which are biochemical substances responsible for transmitting nerve impulses between neurons. Consequently, patients who take “lithium” do not exhibit the extreme mood swings that characterize this disorder.
$1$
For each application, choose the more appropriate substance based on the properties and reactivities of the alkali metals and their compounds. Explain your choice in each case.
1. For a reaction that requires a strong base in a solution of tetrahydrofuran (THF), would you use LiOH or CsOH?
2. To extinguish a fire caused by burning lithium metal, would you use water, CO2, N2 gas, or sand (SiO2)?
3. Both LiNO3and CsNO3 are highly soluble in acetone (2-propanone). Which of these alkali metal salts would you use to precipitate I from an acetone solution?
Given: application and selected alkali metals
Asked for: appropriate metal for each application
Strategy:
Use the properties and reactivities discussed in this section to determine which alkali metal is most suitable for the indicated application.
Solution
1. Both LiOH and CsOH are ionic compounds that contain the hydroxide anion. Li+, however, is much smaller than Cs+, so the Li+ cation will be more effectively solvated by the oxygen of THF with its lone pairs of electrons. This difference will have two effects: (1) LiOH is likely to be much more soluble than CsOH in the nonpolar solvent, which could be a significant advantage, and (2) the solvated Li+ ions are less likely to form tight ion pairs with the OH ions in the relatively nonpolar solution, making the OH more basic and thus more reactive. Thus LiOH is the better choice.
2. Lithium is a potent reductant that reacts with water to form LiOH and H2 gas, so adding a source of hydrogen such as water to a lithium fire is likely to produce an explosion. Lithium also reacts with oxygen and nitrogen in the air to form Li2O and Li3N, respectively, so we would not expect nitrogen to extinguish a lithium fire. Because CO2 is a gaseous molecule that contains carbon in its highest accessible oxidation state (+4), adding CO2 to a strong reductant such as Li should result in a vigorous redox reaction. Thus water, N2, and CO2 are all unsuitable choices for extinguishing a lithium fire. In contrast, sand is primarily SiO2, which is a network solid that is not readily reduced. Smothering a lithium fire with sand is therefore the best choice.
3. The salt with the smaller cation has the higher lattice energy, and high lattice energies tend to decrease the solubility of a salt. However, the solvation energy of the cation is also important in determining solubility, and small cations tend to have higher solvation energies. Recall that high solvation energies tend to increase the solubility of ionic substances. Thus CsI should be the least soluble of the alkali metal iodides, and LiI the most soluble. Consequently, CsNO3 is the better choice.
$1$
Indicate which of the alternative alkali metals or their compounds given is more appropriate for each application.
1. drying agent for an organic solvent—Li2SO4 or Rb2SO4
2. removing trace amounts of N2 from highly purified Ar gas—Li, K, or Cs
3. reacting with an alkyl halide (formula RX) to prepare an organometallic compound (formula MR)—Li or K
Answer
1. Li2SO4
2. Li
3. Li
$2$
Predict the products of each reaction and then balance each chemical equation.
1. Na(s) + O2(g) →
2. Li2O(s) + H2O(l) →
3. K(s) + CH3OH(l) →
4. Li(s) + CH3Cl(l) →
5. Li3N(s) + KCl(s) →
Given: reactants
Asked for: products and balanced chemical equation
Strategy:
A Determine whether one of the reactants is an oxidant or a reductant or a strong acid or a strong base. If so, a redox reaction or an acid–base reaction is likely to occur. Identify the products of the reaction.
B If a reaction is predicted to occur, balance the chemical equation.
Solution
1. A Sodium is a reductant, and oxygen is an oxidant, so a redox reaction is most likely. We expect an electron to be transferred from Na (thus forming Na+) to O2. We now need to determine whether the reduced product is a superoxide (O2), peroxide (O22−), or oxide (O2−). Under normal reaction conditions, the product of the reaction of an alkali metal with oxygen depends on the identity of the metal. Because of differences in lattice energy, Li produces the oxide (Li2O), the heavier metals (K, Rb, Cs) produce the superoxide (MO2), and Na produces the peroxide (Na2O2).
B The balanced chemical equation is 2Na(s) + O2(g) → Na2O2(s).
1. A Li2O is an ionic salt that contains the oxide ion (O2−), which is the completely deprotonated form of water and thus is expected to be a strong base. The other reactant, water, is both a weak acid and a weak base, so we can predict that an acid–base reaction will occur.
B The balanced chemical equation is Li2O(s) + H2O(l) → 2LiOH(aq).
1. A Potassium is a reductant, whereas methanol is both a weak acid and a weak base (similar to water). A weak acid produces H+, which can act as an oxidant by accepting an electron to form $\frac{1}{2}\mathrm{H_2}$. This reaction, therefore, is an acid dissociation that is driven to completion by a reduction of the protons as they are released.
B The balanced chemical equation is as follows: $\mathrm{K(s)}+\mathrm{CH_3OH(l)}\rightarrow\frac{1}{2}\mathrm{H_2(g)}+\mathrm{CH_3OK(soln)}$.
1. A One of the reactants is an alkali metal, a potent reductant, and the other is an alkyl halide. Any compound that contains a carbon–halogen bond can, in principle, be reduced, releasing a halide ion and forming an organometallic compound. That outcome seems likely in this case because organolithium compounds are among the most stable organometallic compounds known.
B Two moles of lithium are required to balance the equation: 2Li(s) + CH3Cl(l) → LiCl(s) + CH3Li(soln).
1. A Lithium nitride and potassium chloride are largely ionic compounds. The nitride ion (N3−) is a very strong base because it is the fully deprotonated form of ammonia, a weak acid. An acid–base reaction requires an acid as well as a base, however, and KCl is not acidic. What about a redox reaction? Both substances contain ions that have closed-shell valence electron configurations. The nitride ion could act as a reductant by donating electrons to an oxidant and forming N2. KCl is not an oxidant, however, and a redox reaction requires an oxidant as well as a reductant.
B We conclude that the two substances will not react with each other.
$2$
Predict the products of each reaction and balance each chemical equation.
1. K(s) + N2(g) →
2. Li3N(s) + H2O(l) →
3. Na(s) + (CH3)2NH(soln) →
4. C6H5Li(soln) + D2O(l) → C6H5D(l) + LiOD(soln)
5. CH3CH2Cl(soln) + 2Li →
Answer
1. no reaction
2. Li3N(s) + 3H2O(l) → NH3(aq) + 3LiOH(aq)
3. $\mathrm{Na(s)}+\mathrm{(CH_3)_2NH(soln)}\rightarrow\frac{1}{2}\mathrm{H_2(g)}+\mathrm{Na[(CH_3)_2N](soln)}$
4. C6H5Li(soln) + D2O(l) → C6H5D(l) + LiOD(soln)
5. CH3CH2Cl(soln) + 2Li → CH3CH2Li(soln) + LiCl(soln)
Summary
The alkali metals are potent reductants whose chemistry is largely that of ionic compounds containing the M+ ion. Alkali metals have only a weak tendency to form complexes with simple Lewis bases. The first alkali metals to be isolated (Na and K) were obtained by passing an electric current through molten potassium and sodium carbonates. The alkali metals are among the most potent reductants known; most can be isolated by electrolysis of their molten salts or, in the case of rubidium and cesium, by reacting their hydroxide salts with a reductant. They can also be recovered from their silicate ores using a multistep process. Lithium, the strongest reductant, and sodium, the weakest, are examples of the physical and chemical effects of opposing periodic trends. The alkali metals react with halogens (group 17) to form ionic halides; the heavier chalcogens (group 16) to produce metal chalcogenides; and oxygen to form compounds, whose stoichiometry depends on the size of the metal atom. The peroxides and superoxides are potent oxidants. The only alkali metal to react with atmospheric nitrogen is lithium. Heavier alkali metals react with graphite to form graphite intercalation compounds, substances in which metal atoms are inserted between the sheets of carbon atoms. With heavier group 14 elements, alkali metals react to give polyatomic anions with three-dimensional cage structures. All alkali metals react with hydrogen at high temperatures to produce the corresponding hydrides, and all reduce water to produce hydrogen gas. Alkali metal salts are prepared by reacting a metal hydroxide with an acid, followed by evaporation of the water. Both Li and Na salts are used as drying agents, compounds that are used to absorb water. Complexing agents such as crown ethers and cryptands can accommodate alkali metal ions of the appropriate size. Alkali metals can also react with liquid ammonia to form solutions that slowly decompose to give hydrogen gas and the metal salt of the amide ion (NH2). These solutions, which contain unstable solvated electrons loosely associated with a cavity in the solvent, are intensely colored, good conductors of electricity, and excellent reductants. Alkali metals can react with organic compounds that contain an acidic proton to produce salts. They can also form organometallic compounds, which have properties that differ from those of their metallic and organic components. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/21%3A_Chemistry_of_The_Main-Group_Elements_I/21.2%3A_Group_1%3A_The_Alkali_Metals.txt |
Learning Objectives
• To describe how to isolate the alkaline earth metals.
• To be familiar with the reactions, compounds, and complexes of the alkaline earth metals.
Like the alkali metals, the alkaline earth metals are so reactive that they are never found in elemental form in nature. Because they form +2 ions that have very negative reduction potentials, large amounts of energy are needed to isolate them from their ores. Four of the six group 2 elements—magnesium (Mg), calcium (Ca), strontium (Sr), and barium (Ba)—were first isolated in the early 19th century by Sir Humphry Davy, using a technique similar to the one he used to obtain the first alkali metals. In contrast to the alkali metals, however, compounds of the alkaline earth metals had been recognized as unique for many centuries. In fact, the name alkali comes from the Arabic al-qili, meaning “ashes,” which were known to neutralize acids. Medieval alchemists found that a portion of the ashes would melt on heating, and these substances were later identified as the carbonates of sodium and potassium ($M_2CO_3$). The ashes that did not melt (but did dissolve in acid), originally called alkaline earths, were subsequently identified as the alkaline earth oxides (MO). In 1808, Davy was able to obtain pure samples of Mg, Ca, Sr, and Ba by electrolysis of their chlorides or oxides.
Beryllium (Be), the lightest alkaline earth metal, was first obtained in 1828 by Friedrich Wöhler in Germany and simultaneously by Antoine Bussy in France. The method used by both men was reduction of the chloride by the potent “new” reductant, potassium:
$\mathrm{BeCl_2(s)}+\mathrm{2K(s)}\xrightarrow\Delta\mathrm{Be(s)}+\mathrm{2KCl(s)} \label{Eq1}$
Radium was discovered in 1898 by Pierre and Marie Curie, who processed tons of residue from uranium mines to obtain about 120 mg of almost pure $RaCl_2$. Marie Curie was awarded the Nobel Prize in Chemistry in 1911 for its discovery. Because of its low abundance and high radioactivity however, radium has few uses.
Preparation of the Alkaline Earth Metals
The alkaline earth metals are produced for industrial use by electrolytic reduction of their molten chlorides, as indicated in this equation for calcium:
$CaCl_{2\;(l)} \rightarrow Ca_{(l)} + Cl_{2\;(g)} \label{Eq2}$
The group 2 metal chlorides are obtained from a variety of sources. For example, $BeCl_2$ is produced by reacting $HCl$ with beryllia ($BeO$), which is obtained from the semiprecious stone beryl $[Be_3Al_2(SiO_3)_6]$.
Chemical reductants can also be used to obtain the group 2 elements. For example, magnesium is produced on a large scale by heating a form of limestone called dolomite (CaCO3·MgCO3) with an inexpensive iron/silicon alloy at 1150°C. Initially $CO_2$ is released, leaving behind a mixture of $CaO$ and MgO; Mg2+ is then reduced:
$2CaO·MgO_{(s)} + Fe/Si_{(s)} \rightarrow 2Mg(l) + Ca_2SiO_{4\;(s)} + Fe(s) \label{Eq3}$
An early source of magnesium was an ore called magnesite ($MgCO_3$) from the district of northern Greece called Magnesia. Strontium was obtained from strontianite ($SrCO_3$) found in a lead mine in the town of Strontian in Scotland. The alkaline earth metals are somewhat easier to isolate from their ores, as compared to the alkali metals, because their carbonate and some sulfate and hydroxide salts are insoluble.
General Properties of the Alkaline Earth Metals
Several important properties of the alkaline earth metals are summarized in Table $1$. Although many of these properties are similar to those of the alkali metals (Table $1$), certain key differences are attributable to the differences in the valence electron configurations of the two groups (ns2 for the alkaline earth metals versus ns1 for the alkali metals).
Table $1$: Selected Properties of the Group 2 Elements
Beryllium Magnesium Calcium Strontium Barium Radium
*The values cited are for six-coordinate ions except for Be2+, for which the value for the four-coordinate ion is given.
atomic symbol Be Mg Ca Sr Ba Ra
atomic number 4 12 20 38 56 88
atomic mass 9.01 24.31 40.08 87.62 137.33 226
valence electron configuration 2s2 3s2 4s2 5s2 6s2 7s2
melting point/boiling point (°C) 1287/2471 650/1090 842/1484 777/1382 727/1897 700/—
density (g/cm3) at 25°C 1.85 1.74 1.54 2.64 3.62 ~5
atomic radius (pm) 112 145 194 219 253
first ionization energy (kJ/mol) 900 738 590 549 503
most common oxidation state +2 +2 +2 +2 +2 +2
ionic radius (pm)* 45 72 100 118 135
electron affinity (kJ/mol) ≥ 0 ≥ 0 −2 −5 −14
electronegativity 1.6 1.3 1.0 1.0 0.9 0.9
standard electrode potential (E°, V) −1.85 −2.37 −2.87 −2.90 −2.91 −2.8
product of reaction with O2 BeO MgO CaO SrO BaO2
type of oxide amphoteric weakly basic basic basic basic
product of reaction with N2 none Mg3N2 Ca3N2 Sr3N2 Ba3N2
product of reaction with X2 BeX2 MgX2 CaX2 SrX2 BaX2
product of reaction with H2 none MgH2 CaH2 SrH2 BaH2
As with the alkali metals, the atomic and ionic radii of the alkaline earth metals increase smoothly from Be to Ba, and the ionization energies decrease. As we would expect, the first ionization energy of an alkaline earth metal, with an ns2 valence electron configuration, is always significantly greater than that of the alkali metal immediately preceding it. The group 2 elements do exhibit some anomalies, however. For example, the density of Ca is less than that of Be and Mg, the two lightest members of the group, and Mg has the lowest melting and boiling points. In contrast to the alkali metals, the heaviest alkaline earth metal (Ba) is the strongest reductant, and the lightest (Be) is the weakest. The standard electrode potentials of Ca and Sr are not very different from that of Ba, indicating that the opposing trends in ionization energies and hydration energies are of roughly equal importance.
One major difference between the group 1 and group 2 elements is their electron affinities. With their half-filled ns orbitals, the alkali metals have a significant affinity for an additional electron. In contrast, the alkaline earth metals generally have little or no tendency to accept an additional electron because their ns valence orbitals are already full; an added electron would have to occupy one of the vacant np orbitals, which are much higher in energy.
Reactions and Compounds of the Alkaline Earth Metals
With their low first and second ionization energies, the group 2 elements almost exclusively form ionic compounds that contain M2+ ions. As expected, however, the lightest element (Be), with its higher ionization energy and small size, forms compounds that are largely covalent. Some compounds of Mg2+ also have significant covalent character. Hence organometallic compounds like those discussed for Li in group 1 are also important for Be and Mg in group 2.
The group 2 elements almost exclusively form ionic compounds containing M2+ ions.
All alkaline earth metals react vigorously with the halogens (group 17) to form the corresponding halides (MX2). Except for the beryllium halides, these compounds are all primarily ionic in nature, containing the M2+ cation and two X anions. The beryllium halides, with properties more typical of covalent compounds, have a polymeric halide-bridged structure in the solid state, as shown for BeCl2. These compounds are volatile, producing vapors that contain the linear X–Be–X molecules predicted by the valence-shell electron-pair repulsion (VSEPR) model. As expected for compounds with only four valence electrons around the central atom, the beryllium halides are potent Lewis acids. They react readily with Lewis bases, such as ethers, to form tetrahedral adducts in which the central beryllium is surrounded by an octet of electrons:
$BeCl_{2(s)} + 2(CH_3CH_2)_2O_{(l)} \rightarrow BeCl_2[O(CH_2CH_3)_2]_{2(soln)} \label{Eq4}$
Because of their higher ionization energy and small size, both Be and Mg form organometallic compounds.
The reactions of the alkaline earth metals with oxygen are less complex than those of the alkali metals. All group 2 elements except barium react directly with oxygen to form the simple oxide MO. Barium forms barium peroxide (BaO2) because the larger O22− ion is better able to separate the large Ba2+ ions in the crystal lattice. In practice, only BeO is prepared by direct reaction with oxygen, and this reaction requires finely divided Be and high temperatures because Be is relatively inert. The other alkaline earth oxides are usually prepared by the thermal decomposition of carbonate salts:
$\mathrm{MCO_3(s)}\xrightarrow\Delta\mathrm{MO(s)}+\mathrm{CO_2(g)} \label{Eq5}$
The reactions of the alkaline earth metals with the heavier chalcogens (Y) are similar to those of the alkali metals. When the reactants are present in a 1:1 ratio, the binary chalcogenides (MY) are formed; at lower M:Y ratios, salts containing polychalcogenide ions (Yn2−) are formed.
In the reverse of Equation $\ref{Eq5}$, the oxides of Ca, Sr, and Ba react with CO2 to regenerate the carbonate. Except for BeO, which has significant covalent character and is therefore amphoteric, all the alkaline earth oxides are basic. Thus they react with water to form the hydroxides—M(OH)2:
$MO_{(s)} + H_2O_{(l)} \rightarrow M^{2+}_{(aq)} + 2OH^−_{(aq)} \label{Eq6}$
and they dissolve in aqueous acid. Hydroxides of the lighter alkaline earth metals are insoluble in water, but their solubility increases as the atomic number of the metal increases. Because BeO and MgO are much more inert than the other group 2 oxides, they are used as refractory materials in applications involving high temperatures and mechanical stress. For example, MgO (melting point = 2825°C) is used to coat the heating elements in electric ranges.
The carbonates of the alkaline earth metals also react with aqueous acid to give CO2 and H2O:
$MCO_{3(s)} + 2H^+_{(aq)} \rightarrow M^{2+}_{(aq)} + CO_{2(g)} + H_2O_{(l)} \label{Eq7}$
The reaction in Equation $\ref{Eq7}$ is the basis of antacids that contain MCO3, which is used to neutralize excess stomach acid.
The trend in the reactivities of the alkaline earth metals with nitrogen is the opposite of that observed for the alkali metals. Only the lightest element (Be) does not react readily with N2 to form the nitride (M3N2), although finely divided Be will react at high temperatures. The higher lattice energy due to the highly charged M2+ and N3− ions is apparently sufficient to overcome the chemical inertness of the N2 molecule, with its N≡N bond. Similarly, all the alkaline earth metals react with the heavier group 15 elements to form binary compounds such as phosphides and arsenides with the general formula M3Z2.
Higher lattice energies cause the alkaline earth metals to be more reactive than the alkali metals toward group 15 elements.
When heated, all alkaline earth metals, except for beryllium, react directly with carbon to form ionic carbides with the general formula MC2. The most important alkaline earth carbide is calcium carbide (CaC2), which reacts readily with water to produce acetylene. For many years, this reaction was the primary source of acetylene for welding and lamps on miners’ helmets. In contrast, beryllium reacts with elemental carbon to form Be2C, which formally contains the C4− ion (although the compound is covalent). Consistent with this formulation, reaction of Be2C with water or aqueous acid produces methane:
$Be_2C_{(s)} + 4H_2O_{(l)} \rightarrow 2Be(OH)_{2(s)} + CH_{4(g)} \label{Eq8}$
Beryllium does not react with hydrogen except at high temperatures (1500°C), although BeH2 can be prepared at lower temperatures by an indirect route. All the heavier alkaline earth metals (Mg through Ba) react directly with hydrogen to produce the binary hydrides (MH2). The hydrides of the heavier alkaline earth metals are ionic, but both BeH2 and MgH2 have polymeric structures that reflect significant covalent character. All alkaline earth hydrides are good reducing agents that react rapidly with water or aqueous acid to produce hydrogen gas:
$CaH_{2(s)} + 2H_2O_{(l)} \rightarrow Ca(OH)_{2(s)} + 2H_{2(g)} \label{Eq9}$
Like the alkali metals, the heavier alkaline earth metals are sufficiently electropositive to dissolve in liquid ammonia. In this case, however, two solvated electrons are formed per metal atom, and no equilibriums involving metal dimers or metal anions are known. Also, like the alkali metals, the alkaline earth metals form a wide variety of simple ionic salts with oxoanions, such as carbonate, sulfate, and nitrate. The nitrate salts tend to be soluble, but the carbonates and sulfates of the heavier alkaline earth metals are quite insoluble because of the higher lattice energy due to the doubly charged cation and anion. The solubility of the carbonates and the sulfates decreases rapidly down the group because hydration energies decrease with increasing cation size.
The solubility of alkaline earth carbonate and sulfates decrease down the group because the hydration energies decrease.
Complexes of the Alkaline Earth Metals
Because of their higher positive charge (+2) and smaller ionic radii, the alkaline earth metals have a much greater tendency to form complexes with Lewis bases than do the alkali metals. This tendency is most important for the lightest cation (Be2+) and decreases rapidly with the increasing radius of the metal ion.
The alkaline earth metals have a substantially greater tendency to form complexes with Lewis bases than do the alkali metals.
The chemistry of Be2+ is dominated by its behavior as a Lewis acid, forming complexes with Lewis bases that produce an octet of electrons around beryllium. For example, Be2+ salts dissolve in water to form acidic solutions that contain the tetrahedral [Be(H2O)4]2+ ion. Because of its high charge-to-radius ratio, the Be2+ ion polarizes coordinated water molecules, thereby increasing their acidity:
$[Be(H_2O)_4]^{2+}_{(aq)} \rightarrow [Be(H_2O)_3(OH)]^+_{(aq)} + H^+_{(aq)} \label{Eq10}$
Similarly, in the presence of a strong base, beryllium and its salts form the tetrahedral hydroxo complex: [Be(OH)4]2−. Hence beryllium oxide is amphoteric. Beryllium also forms a very stable tetrahedral fluoride complex: [BeF4]2−. Recall that beryllium halides behave like Lewis acids by forming adducts with Lewis bases (Equation $\ref{Eq4}$).
The heavier alkaline earth metals also form complexes, but usually with a coordination number of 6 or higher. Complex formation is most important for the smaller cations (Mg2+ and Ca2+). Thus aqueous solutions of Mg2+ contain the octahedral [Mg(H2O)6]2+ ion. Like the alkali metals, the alkaline earth metals form complexes with neutral cyclic ligands like the crown ethers and cryptands discussed in Section 21.3.
Organometallic Compounds Containing Group 2 Elements
Like the alkali metals, the lightest alkaline earth metals (Be and Mg) form the most covalent-like bonds with carbon, and they form the most stable organometallic compounds. Organometallic compounds of magnesium with the formula RMgX, where R is an alkyl or aryl group and X is a halogen, are universally called Grignard reagents, after Victor Grignard (1871–1935), the French chemist who discovered them. Grignard reagents can be used to synthesize various organic compounds, such as alcohols, aldehydes, ketones, carboxylic acids, esters, thiols, and amines.
Uses of the Alkaline Earth Metals
Elemental magnesium is the only alkaline earth metal that is produced on a large scale (about 5 × 105 tn per year). Its low density (1.74 g/cm3 compared with 7.87 g/cm3 for iron and 2.70 g/cm3 for aluminum) makes it an important component of the lightweight metal alloys used in aircraft frames and aircraft and automobile engine parts (Figure $1$). Most commercial aluminum actually contains about 5% magnesium to improve its corrosion resistance and mechanical properties. Elemental magnesium also serves as an inexpensive and powerful reductant for the production of a number of metals, including titanium, zirconium, uranium, and even beryllium, as shown in the following equation:
$TiCl_{4\;(l)} + 2Mg(s) \rightarrow Ti_{(s)} + 2MgCl_{2\;(s)} \label{11}$
The only other alkaline earth that is widely used as the metal is beryllium, which is extremely toxic. Ingestion of beryllium or exposure to beryllium-containing dust causes a syndrome called berylliosis, characterized by severe inflammation of the respiratory tract or other tissues. A small percentage of beryllium dramatically increases the strength of copper or nickel alloys, which are used in nonmagnetic, nonsparking tools (such as wrenches and screwdrivers), camera springs, and electrical contacts. The low atomic number of beryllium gives it a very low tendency to absorb x-rays and makes it uniquely suited for applications involving radioactivity. Both elemental Be and BeO, which is a high-temperature ceramic, are used in nuclear reactors, and the windows on all x-ray tubes and sources are made of beryllium foil.
Millions of tons of calcium compounds are used every year. As discussed in earlier chapters, CaCl2 is used as “road salt” to lower the freezing point of water on roads in cold temperatures. In addition, CaCO3 is a major component of cement and an ingredient in many commercial antacids. “Quicklime” (CaO), produced by heating CaCO3 (Equation $\ref{Eq5}$), is used in the steel industry to remove oxide impurities, make many kinds of glass, and neutralize acidic soil. Other applications of group 2 compounds described in earlier chapters include the medical use of BaSO4 in “barium milkshakes” for identifying digestive problems by x-rays and the use of various alkaline earth compounds to produce the brilliant colors seen in fireworks.
Example $1$
For each application, choose the most appropriate substance based on the properties and reactivities of the alkaline earth metals and their compounds. Explain your choice in each case. Use any tables you need in making your decision, such as Ksp values (Table 17.1), lattice energies (Table 8.1), and band-gap energies.
1. To neutralize excess stomach acid that causes indigestion, would you use BeCO3, CaCO3, or BaCO3?
2. To remove CO2 from the atmosphere in a space capsule, would you use MgO, CaO, or BaO?
3. As a component of the alloy in an automotive spark plug electrode, would you use Be, Ca, or Ba?
Given: application and selected alkaline earth metals
Asked for: most appropriate substance for each application
Strategy:
Based on the discussion in this section and any relevant information elsewhere in this book, determine which substance is most appropriate for the indicated use.
Solution
1. All the alkaline earth carbonates will neutralize an acidic solution by Equation $\ref{Eq7}$. Because beryllium and its salts are toxic, however, BeCO3 cannot be used as an antacid. Of the remaining choices, CaCO3 is somewhat more soluble than BaCO3 (according to the Ksp values in Table 17.1), suggesting that it will act more rapidly. Moreover, the formula mass of CaCO3 is 100.1 amu, whereas that of BaCO3 is almost twice as large. Therefore, neutralizing a given amount of acid would require twice the mass of BaCO3 compared with CaCO3. Furthermore, reaction of BaCO3 with acid produces a solution containing Ba2+ ions, which are toxic. (Ba2+ is a stimulant that can cause ventricular fibrillation of the heart.) Finally, CaCO3 is produced on a vast scale, so CaCO3 is likely to be significantly less expensive than any barium compound. Consequently, CaCO3 is the best choice for an antacid.
2. This application involves reacting CO2 with an alkaline earth oxide to form the carbonate, which is the reverse of the thermal decomposition reaction in which MCO3 decomposes to CO2 and the metal oxide MO (Equation $\ref{Eq5}$). Owing to their higher lattice energies, the smallest alkaline earth metals should form the most stable oxides. Hence their carbonates should decompose at the lowest temperatures, as is observed (BeCO3 decomposes at 100°C; BaCO3 at 1360°C). If the carbonate with the smallest alkaline earth metal decomposes most readily, we would expect the reverse reaction (formation of a carbonate) to occur most readily with the largest metal cation (Ba2+). Hence BaO is the best choice.
3. The alloy in a spark plug electrode must release electrons and promote their flow across the gap between the electrodes at high temperatures. Of the three metals listed, Ba has the lowest ionization energy and thus releases electrons most readily. Heating a barium-containing alloy to high temperatures will cause some ionization to occur, providing the initial step in forming a spark.
Exercise $1$
Which of the indicated alkaline earth metals or their compounds is most appropriate for each application?
1. drying agent for removing water from the atmosphere—CaCl2, MgSO4, or BaF2
2. removal of scale deposits (largely CaCO3) in water pipes—HCl(aq) or H2SO4(aq)
3. removal of traces of N2 from purified argon gas—Be, Ca, or Ba
Answer
1. MgSO4
2. HCl
3. Ba
Example $2$
Predict the products of each reaction and then balance each chemical equation.
1. CaO(s) + HCl(g) →
2. MgO(s) + excess OH(aq) →
3. $\mathrm{CaH_2(s)}+\mathrm{TiO_2(s)}\xrightarrow\Delta$
Given: reactants
Asked for: products and balanced chemical equation
Strategy:
Follow the procedure given in Example 3 to predict the products of each reaction and then balance each chemical equation.
Solution
1. A Gaseous HCl is an acid, and CaO is a basic oxide that contains the O2− ion. This is therefore an acid–base reaction that produces CaCl2 and H2O.
B The balanced chemical equation is $CaO_{(s)} + 2HCl_{(g)} → CaCl_{2(aq)} + H_2O_{(l)}$
1. A Magnesium oxide is a basic oxide, so it can either react with water to give a basic solution or dissolve in an acidic solution. Hydroxide ion is also a base. Because we have two bases but no acid, an acid–base reaction is impossible. A redox reaction is not likely because MgO is neither a good oxidant nor a good reductant.
B We conclude that no reaction occurs.
1. A Because CaH2 contains the hydride ion (H), it is a good reductant. It is also a strong base because H ions can react with H+ ions to form H2. Titanium oxide (TiO2) is a metal oxide that contains the metal in its highest oxidation state (+4 for a group 4 metal); it can act as an oxidant by accepting electrons. We therefore predict that a redox reaction will occur, in which H is oxidized and Ti4+ is reduced. The most probable reduction product is metallic titanium, but what is the oxidation product? Oxygen must appear in the products, and both CaO and H2O are stable compounds. The +1 oxidation state of hydrogen in H2O is a sign that an oxidation has occurred (2H → 2H+ + 4e).
B The balanced chemical equation is
$\mathrm{CaH_2(s)}+\mathrm{TiO_2(s)}\xrightarrow\Delta\mathrm{Ti(s)}+\mathrm{CaO(s)}+\mathrm{H_2O(l)}$
We could also write the products as Ti(s) + Ca(OH)2(s).
Exercise $2$
Predict the products of each reaction and then balance each chemical equation.
1. $\mathrm{BeCl_2(s)}+\mathrm{Mg(s)}\xrightarrow\Delta$
2. BaCl2(aq) + Na2SO4(aq) →
3. BeO(s) + OH(aq) + H2O(l) →
Answer
1. $\mathrm{BeCl_2(s)}+\mathrm{Mg(s)}\xrightarrow\Delta\mathrm{Be(s)}+ \mathrm{MgCl_2(s)}$
2. BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq)
3. BeO(s) + 2OH(aq) + H2O(l) → [Be(OH)4]2−(aq)
Summary
Group 2 elements almost exclusively form ionic compounds containing the M2+ ion, they are more reactive toward group 15 elements, and they have a greater tendency to form complexes with Lewis bases than do the alkali metals. Pure samples of most of the alkaline earth metals can be obtained by electrolysis of the chlorides or oxides. Beryllium was first obtained by the reduction of its chloride; radium chloride, which is radioactive, was obtained through a series of reactions and separations. In contrast to the alkali metals, the alkaline earth metals generally have little or no affinity for an added electron. All alkaline earth metals react with the halogens to produce the corresponding halides, with oxygen to form the oxide (except for barium, which forms the peroxide), and with the heavier chalcogens to form chalcogenides or polychalcogenide ions. All oxides except BeO react with CO2 to form carbonates, which in turn react with acid to produce CO2 and H2O. Except for Be, all the alkaline earth metals react with N2 to form nitrides, and all react with carbon and hydrogen to form carbides and hydrides. Alkaline earth metals dissolve in liquid ammonia to give solutions that contain two solvated electrons per metal atom. The alkaline earth metals have a greater tendency than the alkali metals to form complexes with crown ethers, cryptands, and other Lewis bases. The most important alkaline earth organometallic compounds are Grignard reagents (RMgX), which are used to synthesize organic compounds. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/21%3A_Chemistry_of_The_Main-Group_Elements_I/21.3%3A_Group_2%3A_The_Alkaline_Earth_Metals.txt |
Learning Objectives
• To understand the trends in properties and the reactivity of the group 13 elements.
Group 13 is the first group to span the dividing line between metals and nonmetals, so its chemistry is more diverse than that of groups 1 and 2, which include only metallic elements. Except for the lightest element (boron), the group 13 elements are all relatively electropositive; that is, they tend to lose electrons in chemical reactions rather than gain them. Although group 13 includes aluminum, the most abundant metal on Earth, none of these elements was known until the early 19th century because they are never found in nature in their free state. Elemental boron and aluminum, which were first prepared by reducing B2O3 and AlCl3, respectively, with potassium, could not be prepared until potassium had been isolated and shown to be a potent reductant. Indium (In) and thallium (Tl) were discovered in the 1860s by using spectroscopic techniques, long before methods were available for isolating them. Indium, named for its indigo (deep blue-violet) emission line, was first observed in the spectrum of zinc ores, while thallium (from the Greek thallos, meaning “a young, green shoot of a plant”) was named for its brilliant green emission line. Gallium (Ga; Mendeleev’s eka-aluminum) was discovered in 1875 by the French chemist Paul Émile Lecoq de Boisbaudran during a systematic search for Mendeleev’s “missing” element in group 13.
Group 13 elements are never found in nature in their free state.
Preparation and General Properties of the Group 13 Elements
As reductants, the group 13 elements are less powerful than the alkali metals and alkaline earth metals. Nevertheless, their compounds with oxygen are thermodynamically stable, and large amounts of energy are needed to isolate even the two most accessible elements—boron and aluminum—from their oxide ores.
Although boron is relatively rare (it is about 10,000 times less abundant than aluminum), concentrated deposits of borax [Na2B4O5(OH)4·8H2O] are found in ancient lake beds (Figure $1$) and were used in ancient times for making glass and glazing pottery. Boron is produced on a large scale by reacting borax with acid to produce boric acid [B(OH)3], which is then dehydrated to the oxide (B2O3). Reduction of the oxide with magnesium or sodium gives amorphous boron that is only about 95% pure:
$\mathrm{Na_2B_4O_5(OH)_4\cdot8H_2O(s)}\xrightarrow{\textrm{acid}}\mathrm{B(OH)_3(s)}\xrightarrow{\Delta}\mathrm{B_2O_3(s)} \label{Eq1}$
$\mathrm{B_2O_3(s)}+\mathrm{3Mg(s)}\xrightarrow{\Delta}\mathrm{2B(s)}+\mathrm{3MgO(s)} \label{Eq2}$
Pure, crystalline boron, however, is extremely difficult to obtain because of its high melting point (2300°C) and the highly corrosive nature of liquid boron. It is usually prepared by reducing pure BCl3 with hydrogen gas at high temperatures or by the thermal decomposition of boron hydrides such as diborane (B2H6):
$\mathrm{BCl_3(g)}+\frac{3}{2}\mathrm{H_2(g)}\rightarrow\mathrm{B(s)}+\mathrm{3HCl(g)} \label{Eq3}$
$B_2H_{6(g)} \rightarrow 2B_{(s)} + 3H_{2(g)} \label{Eq4}$
The reaction shown in Equation $\ref{Eq3}$ is used to prepare boron fibers, which are stiff and light. Hence they are used as structural reinforcing materials in objects as diverse as the US space shuttle and the frames of lightweight bicycles that are used in races such as the Tour de France. Boron is also an important component of many ceramics and heat-resistant borosilicate glasses, such as Pyrex, which is used for ovenware and laboratory glassware.
In contrast to boron, deposits of aluminum ores such as bauxite, a hydrated form of Al2O3, are abundant. With an electrical conductivity about twice that of copper on a weight for weight basis, aluminum is used in more than 90% of the overhead electric power lines in the United States. However, because aluminum–oxygen compounds are stable, obtaining aluminum metal from bauxite is an expensive process. Aluminum is extracted from oxide ores by treatment with a strong base, which produces the soluble hydroxide complex [Al(OH)4]. Neutralization of the resulting solution with gaseous CO2 results in the precipitation of Al(OH)3:
$2[Al(OH)_4]^−_{(aq)} + CO_{2(g)} \rightarrow 2Al(OH)_{3(s)} + CO^{2−}_{3(aq)} + H_2O_{(l)} \label{Eq5}$
Thermal dehydration of Al(OH)3 produces Al2O3, and metallic aluminum is obtained by the electrolytic reduction of Al2O3 using the Hall–Heroult process. Of the group 13 elements, only aluminum is used on a large scale: for example, each Boeing 777 airplane is about 50% aluminum by mass.
The other members of group 13 are rather rare: gallium is approximately 5000 times less abundant than aluminum, and indium and thallium are even scarcer. Consequently, these metals are usually obtained as by-products in the processing of other metals. The extremely low melting point of gallium (29.6°C), however, makes it easy to separate from aluminum. Due to its low melting point and high boiling point, gallium is used as a liquid in thermometers that have a temperature range of almost 2200°C. Indium and thallium, the heavier group 13 elements, are found as trace impurities in sulfide ores of zinc and lead. Indium is used as a crushable seal for high-vacuum cryogenic devices, and its alloys are used as low-melting solders in electronic circuit boards. Thallium, on the other hand, is so toxic that the metal and its compounds have few uses. Both indium and thallium oxides are released in flue dust when sulfide ores are converted to metal oxides and SO2. Until relatively recently, these and other toxic elements were allowed to disperse in the air, creating large “dead zones” downwind of a smelter. The flue dusts are now trapped and serve as a relatively rich source of elements such as In and Tl (as well as Ge, Cd, Te, and As).
Table $1$ summarizes some important properties of the group 13 elements. Notice the large differences between boron and aluminum in size, ionization energy, electronegativity, and standard reduction potential, which is consistent with the observation that boron behaves chemically like a nonmetal and aluminum like a metal. All group 13 elements have ns2np1 valence electron configurations, and all tend to lose their three valence electrons to form compounds in the +3 oxidation state. The heavier elements in the group can also form compounds in the +1 oxidation state formed by the formal loss of the single np valence electron. Because the group 13 elements generally contain only six valence electrons in their neutral compounds, these compounds are all moderately strong Lewis acids.
Table $1$: Selected Properties of the Group 13 Elements
Property Boron Aluminum* Gallium Indium Thallium
*This is the name used in the United States; the rest of the world inserts an extra i and calls it aluminium.
The configuration shown does not include filled d and f subshells.
The values cited are for six-coordinate ions in the most common oxidation state, except for Al3+, for which the value for the four-coordinate ion is given. The B3+ ion is not a known species; the radius cited is an estimated four-coordinate value.
§X is Cl, Br, or I. Reaction with F2 gives the trifluorides (MF3) for all group 13 elements.
atomic symbol B Al Ga In Tl
atomic number 5 13 31 49 81
atomic mass (amu) 10.81 26.98 69.72 114.82 204.38
valence electron configuration 2s22p1 3s23p1 4s24p1 5s25p1 6s26p1
melting point/boiling point (°C) 2075/4000 660/2519 29.7/2204 156.6/2072 304/1473
density (g/cm3) at 25°C 2.34 2.70 5.91 7.31 11.8
atomic radius (pm) 87 118 136 156 156
first ionization energy (kJ/mol) 801 578 579 558 589
most common oxidation state +3 +3 +3 +3 +1
ionic radius (pm) −25 54 62 80 162
electron affinity (kJ/mol) −27 −42 −40 −39 −37
electronegativity 2.0 1.6 1.8 1.8 1.8
standard reduction potential (E°, V) −0.87 −1.66 −0.55 −0.34 +0.741 of M3+(aq)
product of reaction with O2 B2O3 Al2O3 Ga2O3 In2O3 Tl2O
type of oxide acidic amphoteric amphoteric amphoteric basic
product of reaction with N2 BN AlN GaN InN none
product of reaction with X2§ BX3 Al2X6 Ga2X6 In2X6 TlX
Neutral compounds of the group 13 elements are electron deficient, so they are generally moderately strong Lewis acids.
In contrast to groups 1 and 2, the group 13 elements show no consistent trends in ionization energies, electron affinities, and reduction potentials, whereas electronegativities actually increase from aluminum to thallium. Some of these anomalies, especially for the series Ga, In, Tl, can be explained by the increase in the effective nuclear charge (Zeff) that results from poor shielding of the nuclear charge by the filled (n − 1)d10 and (n − 2)f14 subshells. Consequently, although the actual nuclear charge increases by 32 as we go from indium to thallium, screening by the filled 5d and 4f subshells is so poor that Zeff increases significantly from indium to thallium. Thus the first ionization energy of thallium is actually greater than that of indium.
Anomalies in periodic trends among Ga, In, and Tl can be explained by the increase in the effective nuclear charge due to poor shielding.
Reactions and Compounds of Boron
Elemental boron is a semimetal that is remarkably unreactive; in contrast, the other group 13 elements all exhibit metallic properties and reactivity. We therefore consider the reactions and compounds of boron separately from those of other elements in the group. All group 13 elements have fewer valence electrons than valence orbitals, which generally results in delocalized, metallic bonding. With its high ionization energy, low electron affinity, low electronegativity, and small size, however, boron does not form a metallic lattice with delocalized valence electrons. Instead, boron forms unique and intricate structures that contain multicenter bonds, in which a pair of electrons holds together three or more atoms.
Elemental boron forms multicenter bonds, whereas the other group 13 elements exhibit metallic bonding.
The basic building block of elemental boron is not the individual boron atom, as would be the case in a metal, but rather the B12 icosahedron. Because these icosahedra do not pack together very well, the structure of solid boron contains voids, resulting in its low density (Figure $3$). Elemental boron can be induced to react with many nonmetallic elements to give binary compounds that have a variety of applications. For example, plates of boron carbide (B4C) can stop a 30-caliber, armor-piercing bullet, yet they weigh 10%–30% less than conventional armor. Other important compounds of boron with nonmetals include boron nitride (BN), which is produced by heating boron with excess nitrogen (Equation $\ref{Eq22.6}$); boron oxide (B2O3), which is formed when boron is heated with excess oxygen (Equation $\ref{Eq22.7}$); and the boron trihalides (BX3), which are formed by heating boron with excess halogen (Equation $\ref{Eq22.8}$).
$\mathrm{2B(s)}+\mathrm{N_2(g)}\xrightarrow{\Delta}\mathrm{2BN(s)} \label{Eq22.6}$
$\mathrm{4B(s)} + \mathrm{3O_2(g)}\xrightarrow{\Delta}\mathrm{2B_2O_3(s)}\label{Eq22.7}$
$\mathrm{2B(s)} +\mathrm{3X_2(g)}\xrightarrow{\Delta}\mathrm{2BX_3(g)}\label{Eq22.8}$
As is typical of elements lying near the dividing line between metals and nonmetals, many compounds of boron are amphoteric, dissolving in either acid or base.
Boron nitride is similar in many ways to elemental carbon. With eight electrons, the B–N unit is isoelectronic with the C–C unit, and B and N have the same average size and electronegativity as C. The most stable form of BN is similar to graphite, containing six-membered B3N3 rings arranged in layers. At high temperature and pressure, hexagonal BN converts to a cubic structure similar to diamond, which is one of the hardest substances known. Boron oxide (B2O3) contains layers of trigonal planar BO3 groups (analogous to BX3) in which the oxygen atoms bridge two boron atoms. It dissolves many metal and nonmetal oxides, including SiO2, to give a wide range of commercially important borosilicate glasses. A small amount of CoO gives the deep blue color characteristic of “cobalt blue” glass.
At high temperatures, boron also reacts with virtually all metals to give metal borides that contain regular three-dimensional networks, or clusters, of boron atoms. The structures of two metal borides—ScB12 and CaB6—are shown in Figure $4$. Because metal-rich borides such as ZrB2 and TiB2 are hard and corrosion resistant even at high temperatures, they are used in applications such as turbine blades and rocket nozzles.
Boron hydrides were not discovered until the early 20th century, when the German chemist Alfred Stock undertook a systematic investigation of the binary compounds of boron and hydrogen, although binary hydrides of carbon, nitrogen, oxygen, and fluorine have been known since the 18th century. Between 1912 and 1936, Stock oversaw the preparation of a series of boron–hydrogen compounds with unprecedented structures that could not be explained with simple bonding theories. All these compounds contain multicenter bonds. The simplest example is diborane (B2H6), which contains two bridging hydrogen atoms (part (a) in Figure $5$. An extraordinary variety of polyhedral boron–hydrogen clusters is now known; one example is the B12H122− ion, which has a polyhedral structure similar to the icosahedral B12 unit of elemental boron, with a single hydrogen atom bonded to each boron atom.
A related class of polyhedral clusters, the carboranes, contain both CH and BH units; an example is shown here. Replacing the hydrogen atoms bonded to carbon with organic groups produces substances with novel properties, some of which are currently being investigated for their use as liquid crystals and in cancer chemotherapy.
The enthalpy of combustion of diborane (B2H6) is −2165 kJ/mol, one of the highest values known:
$B_2H_{6(g)} + 3O_{2(g)} \rightarrow B_2O_{3(s)} + 3H_2O(l)\;\;\; ΔH_{comb} = −2165\; kJ/mol \label{Eq 22.9}$
Consequently, the US military explored using boron hydrides as rocket fuels in the 1950s and 1960s. This effort was eventually abandoned because boron hydrides are unstable, costly, and toxic, and, most important, B2O3 proved to be highly abrasive to rocket nozzles. Reactions carried out during this investigation, however, showed that boron hydrides exhibit unusual reactivity.
Because boron and hydrogen have almost identical electronegativities, the reactions of boron hydrides are dictated by minor differences in the distribution of electron density in a given compound. In general, two distinct types of reaction are observed: electron-rich species such as the BH4 ion are reductants, whereas electron-deficient species such as B2H6 act as oxidants.
Example $1$
For each reaction, explain why the given products form.
1. B2H6(g) + 3O2(g) → B2O3(s) + 3H2O(l)
2. BCl3(l) + 3H2O(l) → B(OH)3(aq) + 3HCl(aq)
3. $\mathrm{2BI_3(s)}+\mathrm{3H_2(g)}\xrightarrow{\Delta}\frac{1}{6}\mathrm{B_{12}(s)}+\mathrm{6HI(g)}$
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the reaction products form.
Solution
1. Molecular oxygen is an oxidant. If the other reactant is a potential reductant, we expect that a redox reaction will occur. Although B2H6 contains boron in its highest oxidation state (+3), it also contains hydrogen in the −1 oxidation state (the hydride ion). Because hydride is a strong reductant, a redox reaction will probably occur. We expect that H will be oxidized to H+ and O2 will be reduced to O2−, but what are the actual products? A reasonable guess is B2O3 and H2O, both stable compounds.
2. Neither BCl3 nor water is a powerful oxidant or reductant, so a redox reaction is unlikely; a hydrolysis reaction is more probable. Nonmetal halides are acidic and react with water to form a solution of the hydrohalic acid and a nonmetal oxide or hydroxide. In this case, the most probable boron-containing product is boric acid [B(OH)3].
3. We normally expect a boron trihalide to behave like a Lewis acid. In this case, however, the other reactant is elemental hydrogen, which usually acts as a reductant. The iodine atoms in BI3 are in the lowest accessible oxidation state (−1), and boron is in the +3 oxidation state. Consequently, we can write a redox reaction in which hydrogen is oxidized and boron is reduced. Because compounds of boron in lower oxidation states are rare, we expect that boron will be reduced to elemental boron. The other product of the reaction must therefore be HI.
Exercise $1$
Predict the products of the reactions and write a balanced chemical equation for each reaction.
1. $\mathrm{B_2H_6(g)}+\mathrm{H_2O(l)}\xrightarrow{\Delta}$
2. $\mathrm{BBr_3(l)}+\mathrm{O_2(g)}\rightarrow$
3. $\mathrm{B_2O_3(s)}+\mathrm{Ca(s)}\xrightarrow{\Delta}$
Answer
1. $\mathrm{B_2H_6(g)}+\mathrm{H_2O(l)}\xrightarrow{\Delta}\mathrm{2B(OH)_3(s)}+\mathrm{6H_2(g)}$
2. $\mathrm{BBr_3(l)}+\mathrm{O_2(g)}\rightarrow\textrm{no reaction}$
3. $\mathrm{6B_2O_3(s)}+18\mathrm{Ca(s)}\xrightarrow{\Delta}\mathrm{B_{12}(s)}+\mathrm{18CaO(s)}$
Reactions and Compounds of the Heavier Group 13 Elements
All four of the heavier group 13 elements (Al, Ga, In, and Tl) react readily with the halogens to form compounds with a 1:3 stoichiometry:
$2M_{(s)} + 3X_{2(s,l,g)} \rightarrow 2MX_{3(s)} \text{ or } M_2X_6 \label{Eq10}$
The reaction of Tl with iodine is an exception: although the product has the stoichiometry TlI3, it is not thallium(III) iodide, but rather a thallium(I) compound, the Tl+ salt of the triiodide ion (I3). This compound forms because iodine is not a powerful enough oxidant to oxidize thallium to the +3 oxidation state.
Of the halides, only the fluorides exhibit behavior typical of an ionic compound: they have high melting points (>950°C) and low solubility in nonpolar solvents. In contrast, the trichorides, tribromides, and triiodides of aluminum, gallium, and indium, as well as TlCl3 and TlBr3, are more covalent in character and form halogen-bridged dimers (part (b) in Figure $4$). Although the structure of these dimers is similar to that of diborane (B2H6), the bonding can be described in terms of electron-pair bonds rather than the delocalized electron-deficient bonding found in diborane. Bridging halides are poor electron-pair donors, so the group 13 trihalides are potent Lewis acids that react readily with Lewis bases, such as amines, to form a Lewis acid–base adduct:
$Al_2Cl_{6(soln)} + 2(CH_3)_3N_{(soln)} \rightarrow 2(CH_3)_3N:AlCl_{3(soln)} \label{Eq11}$
In water, the halides of the group 13 metals hydrolyze to produce the metal hydroxide ($M(OH)_3\)): \[MX_{3(s)} + 3H_2O_{(l)} \rightarrow M(OH)_{3(s)} + 3HX_{(aq)} \label{Eq12}$
In a related reaction, Al2(SO4)3 is used to clarify drinking water by the precipitation of hydrated Al(OH)3, which traps particulates. The halides of the heavier metals (In and Tl) are less reactive with water because of their lower charge-to-radius ratio. Instead of forming hydroxides, they dissolve to form the hydrated metal complex ions: [M(H2O)6]3+.
Of the group 13 halides, only the fluorides behave as typical ionic compounds.
Like boron (Equation $\ref{Eq22.7}$), all the heavier group 13 elements react with excess oxygen at elevated temperatures to give the trivalent oxide (M2O3), although Tl2O3 is unstable:
$\mathrm{4M(s)}+\mathrm{3O_2(g)}\xrightarrow{\Delta}\mathrm{2M_2O_3(s)} \label{Eq13}$
Aluminum oxide (Al2O3), also known as alumina, is a hard, high-melting-point, chemically inert insulator used as a ceramic and as an abrasive in sandpaper and toothpaste. Replacing a small number of Al3+ ions in crystalline alumina with Cr3+ ions forms the gemstone ruby, whereas replacing Al3+ with a mixture of Fe2+, Fe3+, and Ti4+ produces blue sapphires. The gallium oxide compound MgGa2O4 gives the brilliant green light familiar to anyone who has ever operated a xerographic copy machine. All the oxides dissolve in dilute acid, but Al2O3 and Ga2O3 are amphoteric, which is consistent with their location along the diagonal line of the periodic table, also dissolving in concentrated aqueous base to form solutions that contain M(OH)4 ions.
Group 13 trihalides are potent Lewis acids that react with Lewis bases to form a Lewis acid–base adduct.
Aluminum, gallium, and indium also react with the other group 16 elements (chalcogens) to form chalcogenides with the stoichiometry M2Y3. However, because Tl(III) is too strong an oxidant to form a stable compound with electron-rich anions such as S2−, Se2−, and Te2−, thallium forms only the thallium(I) chalcogenides with the stoichiometry Tl2Y. Only aluminum, like boron, reacts directly with N2 (at very high temperatures) to give AlN, which is used in transistors and microwave devices as a nontoxic heat sink because of its thermal stability; GaN and InN can be prepared using other methods. All the metals, again except Tl, also react with the heavier group 15 elements (pnicogens) to form the so-called III–V compounds, such as GaAs. These are semiconductors, whose electronic properties, such as their band gaps, differ from those that can be achieved using either pure or doped group 14 elements. For example, nitrogen- and phosphorus-doped gallium arsenide (GaAs1−x−yPxNy) is used in the displays of calculators and digital watches.
All group 13 oxides dissolve in dilute acid, but Al2O3 and Ga2O3 are amphoteric.
Unlike boron, the heavier group 13 elements do not react directly with hydrogen. Only the aluminum and gallium hydrides are known, but they must be prepared indirectly; AlH3 is an insoluble, polymeric solid that is rapidly decomposed by water, whereas GaH3 is unstable at room temperature.
Complexes of Group 13 Elements
Boron has a relatively limited tendency to form complexes, but aluminum, gallium, indium, and, to some extent, thallium form many complexes. Some of the simplest are the hydrated metal ions [M(H2O)63+], which are relatively strong Brønsted–Lowry acids that can lose a proton to form the M(H2O)5(OH)2+ ion:
$[M(H_2O)_6]^{3+}_{(aq)} \rightarrow M(H_2O)_5(OH)^{2+}_{(aq)} + H^+_{(aq)} \label{Eq14}$
Group 13 metal ions also form stable complexes with species that contain two or more negatively charged groups, such as the oxalate ion. The stability of such complexes increases as the number of coordinating groups provided by the ligand increases.
Example $2$
For each reaction, explain why the given products form.
1. $\mathrm{2Al(s)} + \mathrm{Fe_2O_3(s)}\xrightarrow{\Delta}\mathrm{2Fe(l)} + \mathrm{Al_2O_3(s)}$
2. $\mathrm{2Ga(s)} + \mathrm{6H_2O(l)}+ \mathrm{2OH^-(aq)}\xrightarrow{\Delta}\mathrm{3H_2(g)} + \mathrm{2Ga(OH)^-_4(aq)}$
3. $\mathrm{In_2Cl_6(s)}\xrightarrow{\mathrm{H_2O(l)}}\mathrm{2In^{3+}(aq)}+\mathrm{6Cl^-(aq)}$
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the reaction products form.
Solution
1. Aluminum is an active metal and a powerful reductant, and Fe2O3 contains Fe(III), a potential oxidant. Hence a redox reaction is probable, producing metallic Fe and Al2O3. Because Al is a main group element that lies above Fe, which is a transition element, it should be a more active metal than Fe. Thus the reaction should proceed to the right. In fact, this is the thermite reaction, which is so vigorous that it produces molten Fe and can be used for welding.
2. Gallium lies immediately below aluminum in the periodic table and is amphoteric, so it will dissolve in either acid or base to produce hydrogen gas. Because gallium is similar to aluminum in many of its properties, we predict that gallium will dissolve in the strong base.
3. The metallic character of the group 13 elements increases with increasing atomic number. Indium trichloride should therefore behave like a typical metal halide, dissolving in water to form the hydrated cation.
Exercise $2$
Predict the products of the reactions and write a balanced chemical equation for each reaction.
1. LiH(s) + Al2Cl6(soln)→
2. Al2O3(s) + OH(aq)→
3. Al(s) + N2(g) $\xrightarrow{\Delta}$
4. Ga2Cl6(soln) + Cl(soln)→
Answer
1. 8LiH(s) + Al2Cl6(soln)→2LiAlH4(soln) + 6LiCl(s)
2. Al2O3(s) + 2OH(aq) + 3H2O(l) → 2Al(OH)4(aq)
3. 2Al(s) + N2(g) $\xrightarrow{\Delta}$ 2AlN(s)
4. Ga2Cl6(soln) + 2Cl(soln) → 2GaCl4(soln)
Summary
Compounds of the group 13 elements with oxygen are thermodynamically stable. Many of the anomalous properties of the group 13 elements can be explained by the increase in Zeff moving down the group. Isolation of the group 13 elements requires a large amount of energy because compounds of the group 13 elements with oxygen are thermodynamically stable. Boron behaves chemically like a nonmetal, whereas its heavier congeners exhibit metallic behavior. Many of the inconsistencies observed in the properties of the group 13 elements can be explained by the increase in Zeff that arises from poor shielding of the nuclear charge by the filled (n − 1)d10 and (n − 2)f14 subshells. Instead of forming a metallic lattice with delocalized valence electrons, boron forms unique aggregates that contain multicenter bonds, including metal borides, in which boron is bonded to other boron atoms to form three-dimensional networks or clusters with regular geometric structures. All neutral compounds of the group 13 elements are electron deficient and behave like Lewis acids. The trivalent halides of the heavier elements form halogen-bridged dimers that contain electron-pair bonds, rather than the delocalized electron-deficient bonds characteristic of diborane. Their oxides dissolve in dilute acid, although the oxides of aluminum and gallium are amphoteric. None of the group 13 elements reacts directly with hydrogen, and the stability of the hydrides prepared by other routes decreases as we go down the group. In contrast to boron, the heavier group 13 elements form a large number of complexes in the +3 oxidation state. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/21%3A_Chemistry_of_The_Main-Group_Elements_I/21.4%3A_Group_13%3A_The_Boron_Family.txt |
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