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Learning Objectives
By the end of this section, you will be able to:
• Derive the predicted ground-state electron configurations of atoms
• Identify and explain exceptions to predicted electron configurations for atoms and ions
• Relate electron configurations to element classifications in the periodic table
Having introduced the basics of atomic structure and quantum mechanics, we can use our understanding of quantum numbers to determine how atomic orbitals relate to one another. This allows us to determine which orbitals are occupied by electrons in each atom. The specific arrangement of electrons in orbitals of an atom determines many of the chemical properties of that atom.
Orbital Energies and Atomic Structure
The energy of atomic orbitals increases as the principal quantum number, n, increases. In any atom with two or more electrons, the repulsion between the electrons makes energies of subshells with different values of l differ so that the energy of the orbitals increases within a shell in the order s < p < d < f. Figure $1$: depicts how these two trends in increasing energy relate. The 1s orbital at the bottom of the diagram is the orbital with electrons of lowest energy. The energy increases as we move up to the 2s and then 2p, 3s, and 3p orbitals, showing that the increasing n value has more influence on energy than the increasing l value for small atoms. However, this pattern does not hold for larger atoms. The 3d orbital is higher in energy than the 4s orbital. Such overlaps continue to occur frequently as we move up the chart.
Electrons in successive atoms on the periodic table tend to fill low-energy orbitals first. Thus, many students find it confusing that, for example, the 5p orbitals fill immediately after the 4d, and immediately before the 6s. The filling order is based on observed experimental results, and has been confirmed by theoretical calculations. As the principal quantum number, n, increases, the size of the orbital increases and the electrons spend more time farther from the nucleus. Thus, the attraction to the nucleus is weaker and the energy associated with the orbital is higher (less stabilized). But this is not the only effect we have to take into account. Within each shell, as the value of l increases, the electrons are less penetrating (meaning there is less electron density found close to the nucleus), in the order s > p > d > f. Electrons that are closer to the nucleus slightly repel electrons that are farther out, offsetting the more dominant electron–nucleus attractions slightly (recall that all electrons have −1 charges, but nuclei have +Z charges). This phenomenon is called shielding and will be discussed in more detail in the next section. Electrons in orbitals that experience more shielding are less stabilized and thus higher in energy. For small orbitals (1s through 3p), the increase in energy due to n is more significant than the increase due to l; however, for larger orbitals the two trends are comparable and cannot be simply predicted. We will discuss methods for remembering the observed order.
The arrangement of electrons in the orbitals of an atom is called the electron configuration of the atom. We describe an electron configuration with a symbol that contains three pieces of information (Figure $2$):
1. The number of the principal quantum shell, $n$,
2. The letter that designates the orbital type (the subshell, $l$), and
3. A superscript number that designates the number of electrons in that particular subshell.
For example, the notation 2p4 (read "two–p–four") indicates four electrons in a p subshell ($l = 1$) with a principal quantum number ($n$) of 2. The notation 3d8 (read "three–d–eight") indicates eight electrons in the d subshell (i.e., $l = 2$) of the principal shell for which $n = 3$.
The Aufbau Principle
To determine the electron configuration for any particular atom, we can “build” the structures in the order of atomic numbers. Beginning with hydrogen, and continuing across the periods of the periodic table, we add one proton at a time to the nucleus and one electron to the proper subshell until we have described the electron configurations of all the elements. This procedure is called the Aufbau principle, from the German word Aufbau (“to build up”). Each added electron occupies the subshell of lowest energy available (in the order shown in Figure $1$), subject to the limitations imposed by the allowed quantum numbers according to the Pauli exclusion principle. Electrons enter higher-energy subshells only after lower-energy subshells have been filled to capacity. Figure $3$ illustrates the traditional way to remember the filling order for atomic orbitals.
Since the arrangement of the periodic table is based on the electron configurations, Figure $4$ provides an alternative method for determining the electron configuration. The filling order simply begins at hydrogen and includes each subshell as you proceed in increasing Z order. For example, after filling the 3p block up to Ar, we see the orbital will be 4s (K, Ca), followed by the 3d orbitals.
We will now construct the ground-state electron configuration and orbital diagram for a selection of atoms in the first and second periods of the periodic table. Orbital diagrams are pictorial representations of the electron configuration, showing the individual orbitals and the pairing arrangement of electrons. We start with a single hydrogen atom (atomic number 1), which consists of one proton and one electron. Referring to Figure $3$ or Figure $4$, we would expect to find the electron in the 1s orbital. By convention, the $m_s=+\frac{1}{2}$ value is usually filled first. The electron configuration and the orbital diagram are:
Following hydrogen is the noble gas helium, which has an atomic number of 2. The helium atom contains two protons and two electrons. The first electron has the same four quantum numbers as the hydrogen atom electron ($n = 1$, $l = 0$, $m_l = 0$, $m_s=+\frac{1}{2}$. The second electron also goes into the 1s orbital and fills that orbital. The second electron has the same n, l, and ml quantum numbers, but must have the opposite spin quantum number, $m_s=-\frac{1}{2}$. This is in accord with the Pauli exclusion principle: No two electrons in the same atom can have the same set of four quantum numbers. For orbital diagrams, this means two arrows go in each box (representing two electrons in each orbital) and the arrows must point in opposite directions (representing paired spins). The electron configuration and orbital diagram of helium are:
The $n = 1$ shell is completely filled in a helium atom.
The next atom is the alkali metal lithium with an atomic number of 3. The first two electrons in lithium fill the 1s orbital and have the same sets of four quantum numbers as the two electrons in helium. The remaining electron must occupy the orbital of next lowest energy, the 2s orbital (Figure $3$ or Figure $4$). Thus, the electron configuration and orbital diagram of lithium are:
An atom of the alkaline earth metal beryllium, with an atomic number of 4, contains four protons in the nucleus and four electrons surrounding the nucleus. The fourth electron fills the remaining space in the 2s orbital.
An atom of boron (atomic number 5) contains five electrons. The n = 1 shell is filled with two electrons and three electrons will occupy the n = 2 shell. Because any s subshell can contain only two electrons, the fifth electron must occupy the next energy level, which will be a 2p orbital. There are three degenerate 2p orbitals (ml = −1, 0, +1) and the electron can occupy any one of these p orbitals. When drawing orbital diagrams, we include empty boxes to depict any empty orbitals in the same subshell that we are filling.
Carbon (atomic number 6) has six electrons. Four of them fill the 1s and 2s orbitals. The remaining two electrons occupy the 2p subshell. We now have a choice of filling one of the 2p orbitals and pairing the electrons or of leaving the electrons unpaired in two different, but degenerate, p orbitals. The orbitals are filled as described by Hund’s rule: the lowest-energy configuration for an atom with electrons within a set of degenerate orbitals is that having the maximum number of unpaired electrons. Thus, the two electrons in the carbon 2p orbitals have identical $n$, $l$, and $m_s$ quantum numbers and differ in their $m_l$ quantum number (in accord with the Pauli exclusion principle). The electron configuration and orbital diagram for carbon are:
Nitrogen (atomic number 7) fills the 1s and 2s subshells and has one electron in each of the three 2p orbitals, in accordance with Hund’s rule. These three electrons have unpaired spins. Oxygen (atomic number 8) has a pair of electrons in any one of the 2p orbitals (the electrons have opposite spins) and a single electron in each of the other two. Fluorine (atomic number 9) has only one 2p orbital containing an unpaired electron. All of the electrons in the noble gas neon (atomic number 10) are paired, and all of the orbitals in the n = 1 and the n = 2 shells are filled. The electron configurations and orbital diagrams of these four elements are:
The alkali metal sodium (atomic number 11) has one more electron than the neon atom. This electron must go into the lowest-energy subshell available, the 3s orbital, giving a 1s22s22p63s1 configuration. The electrons occupying the outermost shell orbital(s) (highest value of n) are called valence electrons, and those occupying the inner shell orbitals are called core electrons (Figure $5$). Since the core electron shells correspond to noble gas electron configurations, we can abbreviate electron configurations by writing the noble gas that matches the core electron configuration, along with the valence electrons in a condensed format. For our sodium example, the symbol [Ne] represents core electrons, (1s22s22p6) and our abbreviated or condensed configuration is [Ne]3s1.
Similarly, the abbreviated configuration of lithium can be represented as [He]2s1, where [He] represents the configuration of the helium atom, which is identical to that of the filled inner shell of lithium. Writing the configurations in this way emphasizes the similarity of the configurations of lithium and sodium. Both atoms, which are in the alkali metal family, have only one electron in a valence s subshell outside a filled set of inner shells.
\begin{aligned} & Li :[ He ] 2 s^1 \[4pt] & Na :[ Ne ] 3 s^1 \end{aligned} \nonumber
The alkaline earth metal magnesium (atomic number 12), with its 12 electrons in a [Ne]3s2 configuration, is analogous to its family member beryllium, [He]2s2. Both atoms have a filled s subshell outside their filled inner shells. Aluminum (atomic number 13), with 13 electrons and the electron configuration [Ne]3s23p1, is analogous to its family member boron, [He]2s22p1.
The electron configurations of silicon (14 electrons), phosphorus (15 electrons), sulfur (16 electrons), chlorine (17 electrons), and argon (18 electrons) are analogous in the electron configurations of their outer shells to their corresponding family members carbon, nitrogen, oxygen, fluorine, and neon, respectively, except that the principal quantum number of the outer shell of the heavier elements has increased by one to n = 3. Figure $6$: shows the lowest energy, or ground-state, electron configuration for these elements as well as that for atoms of each of the known elements.
When we come to the next element in the periodic table, the alkali metal potassium (atomic number 19), we might expect that we would begin to add electrons to the 3d subshell. However, all available chemical and physical evidence indicates that potassium is like lithium and sodium, and that the next electron is not added to the 3d level but is, instead, added to the 4s level (Figure $6$). As discussed previously, the 3d orbital with no radial nodes is higher in energy because it is less penetrating and more shielded from the nucleus than the 4s, which has three radial nodes. Thus, potassium has an electron configuration of [Ar]4s1. Hence, potassium corresponds to Li and Na in its valence shell configuration. The next electron is added to complete the 4s subshell and calcium has an electron configuration of [Ar]4s2. This gives calcium an outer-shell electron configuration corresponding to that of beryllium and magnesium.
Beginning with the transition metal scandium (atomic number 21), additional electrons are added successively to the 3d subshell. This subshell is filled to its capacity with 10 electrons (remember that for l = 2 [d orbitals], there are 2l + 1 = 5 values of ml, meaning that there are five d orbitals that have a combined capacity of 10 electrons). The 4p subshell fills next. Note that for three series of elements, scandium (Sc) through copper (Cu), yttrium (Y) through silver (Ag), and lutetium (Lu) through gold (Au), a total of 10 d electrons are successively added to the (n – 1) shell next to the n shell to bring that (n – 1) shell from 8 to 18 electrons. For two series, lanthanum (La) through lutetium (Lu) and actinium (Ac) through lawrencium (Lr), 14 f electrons (l = 3, 2l + 1 = 7 ml values; thus, seven orbitals with a combined capacity of 14 electrons) are successively added to the (n – 2) shell to bring that shell from 18 electrons to a total of 32 electrons.
Example $1$: Quantum Numbers and Electron Configurations
What is the electron configuration and orbital diagram for a phosphorus atom? What are the four quantum numbers for the last electron added?
Solution
The atomic number of phosphorus is 15. Thus, a phosphorus atom contains 15 electrons. The order of filling of the energy levels is 1s, 2s, 2p, 3s, 3p, 4s, . . . The 15 electrons of the phosphorus atom will fill up to the 3p orbital, which will contain three electrons:
The last electron added is a 3p electron. Therefore, n = 3 and, for a p-type orbital, l = 1. The ml value could be –1, 0, or +1. The three p orbitals are degenerate, so any of these ml values is correct. For unpaired electrons, convention assigns the value of $+ \frac{1}{2}$ for the spin quantum number; thus, $m_s=+ \frac{1}{2}$
Exercise $1$
Identify the atoms from the electron configurations given:
1. [Ar]4s23d5
2. [Kr]5s24d105p6
Answer
1. Mn
2. Xe
The periodic table can be a powerful tool in predicting the electron configuration of an element. However, we do find exceptions to the order of filling of orbitals that are shown in Figure $3$: or Figure $4$. For instance, the electron configurations (shown in Figure $6$) of the transition metals chromium (Cr; atomic number 24) and copper (Cu; atomic number 29), among others, are not those we would expect. In general, such exceptions involve subshells with very similar energy, and small effects can lead to changes in the order of filling.
In the case of Cr and Cu, we find that half-filled and completely filled subshells apparently represent conditions of preferred stability. This stability is such that an electron shifts from the 4s into the 3d orbital to gain the extra stability of a half-filled 3d subshell (in Cr) or a filled 3d subshell (in Cu). Other exceptions also occur. For example, niobium (Nb, atomic number 41) is predicted to have the electron configuration [Kr]5s24d3. Experimentally, we observe that its ground-state electron configuration is actually [Kr]5s14d4. We can rationalize this observation by saying that the electron–electron repulsions experienced by pairing the electrons in the 5s orbital are larger than the gap in energy between the 5s and 4d orbitals. There is no simple method to predict the exceptions for atoms where the magnitude of the repulsions between electrons is greater than the small differences in energy between subshells.
Electron Configurations and the Periodic Table
As described earlier, the periodic table arranges atoms based on increasing atomic number so that elements with the same chemical properties recur periodically. When their electron configurations are added to the table (Figure $6$), we also see a periodic recurrence of similar electron configurations in the outer shells of these elements. Because they are in the outer shells of an atom, valence electrons play the most important role in chemical reactions. The outer electrons have the highest energy of the electrons in an atom and are more easily lost or shared than the core electrons. Valence electrons are also the determining factor in some physical properties of the elements.
Elements in any one group (or column) have the same number of valence electrons; the alkali metals lithium and sodium each have only one valence electron, the alkaline earth metals beryllium and magnesium each have two, and the halogens fluorine and chlorine each have seven valence electrons. The similarity in chemical properties among elements of the same group occurs because they have the same number of valence electrons. It is the loss, gain, or sharing of valence electrons that defines how elements react.
It is important to remember that the periodic table was developed on the basis of the chemical behavior of the elements, well before any idea of their atomic structure was available. Now we can understand why the periodic table has the arrangement it has—the arrangement puts elements whose atoms have the same number of valence electrons in the same group. This arrangement is emphasized in Figure $6$: , which shows in periodic-table form the electron configuration of the last subshell to be filled by the Aufbau principle. The colored sections of Figure $6$: show the three categories of elements classified by the orbitals being filled: main group, transition, and inner transition elements. These classifications determine which orbitals are counted in the valence shell, or highest energy level orbitals of an atom.
1. Main group elements (sometimes called representative elements) are those in which the last electron added enters an s or a p orbital in the outermost shell, shown in blue and red in Figure $6$. This category includes all the nonmetallic elements, as well as many metals and the metalloids. The valence electrons for main group elements are those with the highest n level. For example, gallium (Ga, atomic number 31) has the electron configuration [Ar]4s23d104p1, which contains three valence electrons (underlined). The completely filled d orbitals count as core, not valence, electrons.
2. Transition elements or transition metals. These are metallic elements in which the last electron added enters a d orbital. The valence electrons (those added after the last noble gas configuration) in these elements include the ns and (n – 1) d electrons. The official IUPAC definition of transition elements specifies those with partially filled d orbitals. Thus, the elements with completely filled orbitals (Zn, Cd, Hg, as well as Cu, Ag, and Au in Figure $6$) are not technically transition elements. However, the term is frequently used to refer to the entire d block (colored yellow in Figure $6$), and we will adopt this usage in this textbook.
3. Inner transition elements are metallic elements in which the last electron added occupies an f orbital. They are shown in green in Figure $6$. The valence shells of the inner transition elements consist of the (n – 2)f, the (n – 1)d, and the ns subshells. There are two inner transition series:
1. The lanthanide series: lanthanum (La) through lutetium (Lu)
2. The actinide series: actinium (Ac) through lawrencium (Lr)
Lanthanum and actinium, because of their similarities to the other members of the series, are included and used to name the series, even though they are transition metals with no f electrons.
Electron Configurations of Ions
Ions are formed when atoms gain or lose electrons. A cation (positively charged ion) forms when one or more electrons are removed from a parent atom. For main group elements, the electrons that were added last are the first electrons removed. For transition metals and inner transition metals, however, electrons in the s orbital are easier to remove than the d or f electrons, and so the highest ns electrons are lost, and then the (n – 1)d or (n – 2)f electrons are removed. An anion (negatively charged ion) forms when one or more electrons are added to a parent atom. The added electrons fill in the order predicted by the Aufbau principle.
Example $2$: Predicting Electron Configurations of Ions
What is the electron configuration of:
1. Na+
2. P3–
3. Al2+
4. Fe2+
5. Sm3+
Solution
First, write out the electron configuration for each parent atom. We have chosen to show the full, unabbreviated configurations to provide more practice for students who want it, but listing the core-abbreviated electron configurations is also acceptable.
Next, determine whether an electron is gained or lost. Remember electrons are negatively charged, so ions with a positive charge have lost an electron. For main group elements, the last orbital gains or loses the electron. For transition metals, the last s orbital loses an electron before the d orbitals.
1. Na: 1s22s22p63s1. Sodium cation loses one electron, so Na+: 1s22s22p63s1 = Na+: 1s22s22p6.
2. P: 1s22s22p63s23p3. Phosphorus trianion gains three electrons, so P3−: 1s22s22p63s23p6.
3. Al: 1s22s22p63s23p1. Aluminum dication loses two electrons Al2+: 1s22s22p63s23p1 = Al2+: 1s22s22p63s1.
4. Fe: 1s22s22p63s23p64s23d6. Iron(II) loses two electrons and, since it is a transition metal, they are removed from the 4s orbital Fe2+: 1s22s22p63s23p64s23d6 = 1s22s22p63s23p63d6.
5. Sm: 1s22s22p63s23p64s23d104p65s24d105p66s24f6. Samarium trication loses three electrons. The first two will be lost from the 6s orbital, and the final one is removed from the 4f orbital. Sm3+: 1s22s22p63s23p64s23d104p65s24d105p66s24f6 = 1s22s22p63s23p64s23d104p65s24d105p64f5.
Exercise $21$
Which ion with a +2 charge has the electron configuration 1s22s22p63s23p64s23d104p64d5? Which ion with a +3 charge has this configuration?
Answer
Tc2+, Ru3+ | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/06%3A_Electronic_Structure_and_Periodic_Properties/6.04%3A_Electronic_Structure_of_Atoms_%28Electron_Configurations%29.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe and explain the observed trends in atomic size, ionization energy, and electron affinity of the elements
The elements in groups (vertical columns) of the periodic table exhibit similar chemical behavior. This similarity occurs because the members of a group have the same number and distribution of electrons in their valence shells. However, there are also other patterns in chemical properties on the periodic table. For example, as we move down a group, the metallic character of the atoms increases. Oxygen, at the top of group 16 (6A), is a colorless gas; in the middle of the group, selenium is a semiconducting solid; and, toward the bottom, polonium is a silver-grey solid that conducts electricity.
As we go across a period from left to right, we add a proton to the nucleus and an electron to the valence shell with each successive element. As we go down the elements in a group, the number of electrons in the valence shell remains constant, but the principal quantum number increases by one each time. An understanding of the electronic structure of the elements allows us to examine some of the properties that govern their chemical behavior. These properties vary periodically as the electronic structure of the elements changes. They are (1) size (radius) of atoms and ions, (2) ionization energies, and (3) electron affinities.
Link to Learning
Explore visualizations of the periodic trends discussed in this section (and many more trends). With just a few clicks, you can create three-dimensional versions of the periodic table showing atomic size or graphs of ionization energies from all measured elements.
Variation in Covalent Radius
The quantum mechanical picture makes it difficult to establish a definite size of an atom. However, there are several practical ways to define the radius of atoms and, thus, to determine their relative sizes that give roughly similar values. We will use the covalent radius (Figure $1$), which is defined as one-half the distance between the nuclei of two identical atoms when they are joined by a covalent bond (this measurement is possible because atoms within molecules still retain much of their atomic identity). We know that as we scan down a group, the principal quantum number, n, increases by one for each element. Thus, the electrons are being added to a region of space that is increasingly distant from the nucleus. Consequently, the size of the atom (and its covalent radius) must increase as we increase the distance of the outermost electrons from the nucleus. This trend is illustrated for the covalent radii of the halogens in Table $1$ and Figure $1$. The trends for the entire periodic table can be seen in Figure $2$.
Table $1$: Covalent Radii of the Halogen Group Elements
Atom Covalent radius (pm) Nuclear charge
F 64 +9
Cl 99 +17
Br 114 +35
I 133 +53
At 148 +85
As shown in Figure $2$, as we move across a period from left to right, we generally find that each element has a smaller covalent radius than the element preceding it. This might seem counterintuitive because it implies that atoms with more electrons have a smaller atomic radius. This can be explained with the concept of effective nuclear charge, $Z_{eff}$. This is the pull exerted on a specific electron by the nucleus, taking into account any electron–electron repulsions. For hydrogen, there is only one electron and so the nuclear charge ($Z$) and the effective nuclear charge ($Z_{eff}$) are equal. For all other atoms, the inner electrons partially shield the outer electrons from the pull of the nucleus, and thus:
$Z_{\text {eff }}=Z-\text { shielding } \nonumber$
Shielding is determined by the probability of another electron being between the electron of interest and the nucleus, as well as by the electron–electron repulsions the electron of interest encounters. Core electrons are adept at shielding, while electrons in the same valence shell do not block the nuclear attraction experienced by each other as efficiently. Thus, each time we move from one element to the next across a period, Z increases by one, but the shielding increases only slightly. Thus, Zeff increases as we move from left to right across a period. The stronger pull (higher effective nuclear charge) experienced by electrons on the right side of the periodic table draws them closer to the nucleus, making the covalent radii smaller.
Thus, as we would expect, the outermost or valence electrons are easiest to remove because they have the highest energies, are shielded more, and are farthest from the nucleus. As a general rule, when the representative elements form cations, they do so by the loss of the ns or np electrons that were added last in the Aufbau process. The transition elements, on the other hand, lose the ns electrons before they begin to lose the (n – 1)d electrons, even though the ns electrons are added first, according to the Aufbau principle.
Example $1$: Sorting Atomic Radii
Predict the order of increasing covalent radius for Ge, Fl, Br, Kr.
Solution
Radius increases as we move down a group, so Ge < Fl (Note: Fl is the symbol for flerovium, element 114, NOT fluorine). Radius decreases as we move across a period, so Kr < Br < Ge. Putting the trends together, we obtain Kr < Br < Ge < Fl.
Exercise $1$
Give an example of an atom whose size is smaller than fluorine.
Answer
Ne or He
Variation in Ionic Radii
Ionic radius is the measure used to describe the size of an ion. A cation always has fewer electrons and the same number of protons as the parent atom; it is smaller than the atom from which it is derived (Figure $3$). For example, the covalent radius of an aluminum atom (1s22s22p63s23p1) is 118 pm, whereas the ionic radius of an Al3+ (1s22s22p6) is 68 pm. As electrons are removed from the outer valence shell, the remaining core electrons occupying smaller shells experience a greater effective nuclear charge Zeff (as discussed) and are drawn even closer to the nucleus.
Cations with larger charges are smaller than cations with smaller charges (e.g., V2+ has an ionic radius of 79 pm, while that of V3+ is 64 pm). Proceeding down the groups of the periodic table, we find that cations of successive elements with the same charge generally have larger radii, corresponding to an increase in the principal quantum number, n.
An anion (negative ion) is formed by the addition of one or more electrons to the valence shell of an atom. This results in a greater repulsion among the electrons and a decrease in $Z_{eff}$ per electron. Both effects (the increased number of electrons and the decreased $Z_{eff}$) cause the radius of an anion to be larger than that of the parent atom (Figure $3$). For example, a sulfur atom ([Ne]3s23p4) has a covalent radius of 104 pm, whereas the ionic radius of the sulfide anion ([Ne]3s23p6) is 170 pm. For consecutive elements proceeding down any group, anions have larger principal quantum numbers and, thus, larger radii.
Atoms and ions that have the same electron configuration are said to be isoelectronic. Examples of isoelectronic species are N3–, O2–, F, Ne, Na+, Mg2+, and Al3+ (1s22s22p6). Another isoelectronic series is P3–, S2–, Cl, Ar, K+, Ca2+, and Sc3+ ([Ne]3s23p6). For atoms or ions that are isoelectronic, the number of protons determines the size. The greater the nuclear charge, the smaller the radius in a series of isoelectronic ions and atoms.
Variation in Ionization Energies
The amount of energy required to remove the most loosely bound electron from a gaseous atom in its ground state is called its first ionization energy ($IE_1$). The first ionization energy for an element, $\ce{X}$, is the energy required to form a cation with +1 charge:
$\ce{X (g) -> X^{+}(g) + e^{-}} \quad\quad IE_1 \nonumber$
The energy required to remove the second most loosely bound electron is called the second ionization energy (IE2).
$\ce{X^{+} (g) -> X^{2+}(g) + e^{-}} \quad\quad IE_2 \nonumber$
The energy required to remove the third electron is the third ionization energy, and so on. Energy is always required to remove electrons from atoms or ions, so ionization processes are endothermic and IE values are always positive. For larger atoms, the most loosely bound electron is located farther from the nucleus and so is easier to remove. Thus, as size (atomic radius) increases, the ionization energy should decrease. Relating this logic to what we have just learned about radii, we would expect first ionization energies to decrease down a group and to increase across a period.
Figure $4$ graphs the relationship between the first ionization energy and the atomic number of several elements. The values of first ionization energy for the elements are given in Figure $5$. Within a period, the IE1 generally increases with increasing Z. Down a group, the IE1 value generally decreases with increasing Z. There are some systematic deviations from this trend, however. Note that the ionization energy of boron (atomic number 5) is less than that of beryllium (atomic number 4) even though the nuclear charge of boron is greater by one proton. This can be explained because the energy of the subshells increases as l increases, due to penetration and shielding (as discussed previously in this chapter). Within any one shell, the s electrons are lower in energy than the p electrons. This means that an s electron is harder to remove from an atom than a p electron in the same shell. The electron removed during the ionization of beryllium ([He]2s2) is an s electron, whereas the electron removed during the ionization of boron ([He]2s22p1) is a p electron; this results in a lower first ionization energy for boron, even though its nuclear charge is greater by one proton. Thus, we see a small deviation from the predicted trend occurring each time a new subshell begins.
Another deviation occurs as orbitals become more than one-half filled. The first ionization energy for oxygen is slightly less than that for nitrogen, despite the trend in increasing IE1 values across a period. Looking at the orbital diagram of oxygen, we can see that removing one electron will eliminate the electron–electron repulsion caused by pairing the electrons in the 2p orbital and will result in a half-filled orbital (which is energetically favorable). Analogous changes occur in succeeding periods (note the dip for sulfur after phosphorus in Figure $5$).
Removing an electron from a cation is more difficult than removing an electron from a neutral atom because of the greater electrostatic attraction to the cation. Likewise, removing an electron from a cation with a higher positive charge is more difficult than removing an electron from an ion with a lower charge. Thus, successive ionization energies for one element always increase. As seen in Table $2$, there is a large increase in the ionization energies for each element. This jump corresponds to removal of the core electrons, which are harder to remove than the valence electrons. For example, Sc and Ga both have three valence electrons, so the rapid increase in ionization energy occurs after the third ionization.
Table $2$: Successive Ionization Energies for Selected Elements (kJ/mol)
Element IE1 IE2 IE3 IE4 IE5 IE6 IE7
K 418.8 3051.8 4419.6 5876.9 7975.5 9590.6 11343
Ca 589.8 1145.4 4912.4 6490.6 8153.0 10495.7 12272.9
Sc 633.1 1235.0 2388.7 7090.6 8842.9 10679.0 13315.0
Ga 578.8 1979.4 2964.6 6180 8298.7 10873.9 13594.8
Ge 762.2 1537.5 3302.1 4410.6 9021.4 Not available Not available
As 944.5 1793.6 2735.5 4836.8 6042.9 12311.5 Not available
Example $2$: Ranking Ionization Energies
Predict the order of increasing energy for the following processes: IE1 for Al, IE1 for Tl, IE2 for Na, IE3 for Al.
Solution
Removing the 6p1 electron from Tl is easier than removing the 3p1 electron from Al because the higher n orbital is farther from the nucleus, so IE1(Tl) < IE1(Al). Ionizing the third electron from Al ($\ce{Al^{2+} -> Al^{3+} + e^{-}}$) requires more energy because the cation Al2+ exerts a stronger pull on the electron than the neutral Al atom, so IE1(Al) < IE3(Al). The second ionization energy for sodium removes a core electron, which is a much higher energy process than removing valence electrons. Putting this all together, we obtain: IE1(Tl) < IE1(Al) < IE3(Al) < IE2(Na).
Exercise $2$
Which has the lowest value for IE1: O, Po, Pb, or Ba?
Answer
Ba
Variation in Electron Affinities
The electron affinity (EA) is the energy change for the process of adding an electron to a gaseous atom to form an anion (negative ion).
$\ce{X (g) + e^{-} -> X^{-}(g)} \quad EA_1 \nonumber$
This process can be either endothermic or exothermic, depending on the element. The EA of some of the elements is given in Figure $6$. You can see that many of these elements have negative values of EA, which means that energy is released when the gaseous atom accepts an electron. However, for some elements, energy is required for the atom to become negatively charged and the value of their EA is positive. Just as with ionization energy, subsequent EA values are associated with forming ions with more charge. The second EA is the energy associated with adding an electron to an anion to form a –2 ion, and so on.
As we might predict, it becomes easier to add an electron across a series of atoms as the effective nuclear charge of the atoms increases. We find, as we go from left to right across a period, EAs tend to become more negative. The exceptions found among the elements of group 2 (2A), group 15 (5A), and group 18 (8A) can be understood based on the electronic structure of these groups. The noble gases, group 18 (8A), have a completely filled shell and the incoming electron must be added to a higher n level, which is more difficult to do. Group 2 (2A) has a filled ns subshell, and so the next electron added goes into the higher energy np, so, again, the observed EA value is not as the trend would predict. Finally, group 15 (5A) has a half-filled np subshell and the next electron must be paired with an existing np electron. In all of these cases, the initial relative stability of the electron configuration disrupts the trend in EA.
We also might expect the atom at the top of each group to have the most negative EA; their first ionization potentials suggest that these atoms have the largest effective nuclear charges. However, as we move down a group, we see that the second element in the group most often has the most negative EA. This can be attributed to the small size of the n = 2 shell and the resulting large electron–electron repulsions. For example, chlorine, with an EA value of –348 kJ/mol, has the highest value of any element in the periodic table. The EA of fluorine is –322 kJ/mol. When we add an electron to a fluorine atom to form a fluoride anion (F), we add an electron to the n = 2 shell. The electron is attracted to the nucleus, but there is also significant repulsion from the other electrons already present in this small valence shell. The chlorine atom has the same electron configuration in the valence shell, but because the entering electron is going into the n = 3 shell, it occupies a considerably larger region of space and the electron–electron repulsions are reduced. The entering electron does not experience as much repulsion and the chlorine atom accepts an additional electron more readily, resulting in a more negative EA.
The properties discussed in this section (size of atoms and ions, effective nuclear charge, ionization energies, and electron affinities) are central to understanding chemical reactivity. For example, because fluorine has an energetically favorable EA and a large energy barrier to ionization (IE), it is much easier to form fluorine anions than cations. Metallic properties including conductivity and malleability (the ability to be formed into sheets) depend on having electrons that can be removed easily. Thus, metallic character increases as we move down a group and decreases across a period in the same trend observed for atomic size because it is easier to remove an electron that is farther away from the nucleus. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/06%3A_Electronic_Structure_and_Periodic_Properties/6.05%3A_Periodic_Variations_in_Element_Properties.txt |
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Glossary Entries
Word(s)DefinitionImageCaptionLinkSource
amplitudeextent of the displacement caused by a wave
atomic orbitalmathematical function that describes the behavior of an electron in an atom (also called the wavefunction)
Aufbau principleprocedure in which the electron configuration of the elements is determined by “building” them in order of atomic numbers, adding one proton to the nucleus and one electron to the proper subshell at a time
blackbodyidealized perfect absorber of all incident electromagnetic radiation; such bodies emit electromagnetic radiation in characteristic continuous spectra called blackbody radiation
Bohr’s model of the hydrogen atomstructural model in which an electron moves around the nucleus only in circular orbits, each with a specific allowed radius
continuous spectrumelectromagnetic radiation given off in an unbroken series of wavelengths (e.g., white light from the sun)
core electronelectron in an atom that occupies the orbitals of the inner shells
covalent radiusone-half the distance between the nuclei of two identical atoms when they are joined by a covalent bond
d orbitalregion of space with high electron density that is either four lobed or contains a dumbbell and torus shape; describes orbitals with l = 2.
degenerate orbitalsorbitals that have the same energy
effective nuclear chargecharge that leads to the Coulomb force exerted by the nucleus on an electron, calculated as the nuclear charge minus shielding
electromagnetic radiationenergy transmitted by waves that have an electric-field component and a magnetic-field component
electromagnetic spectrumrange of energies that electromagnetic radiation can comprise, including radio, microwaves, infrared, visible, ultraviolet, X-rays, and gamma rays
electron affinityenergy change associated with addition of an electron to a gaseous atom or ion
electron configurationlisting that identifies the electron occupancy of an atom’s shells and subshells
electron densitya measure of the probability of locating an electron in a particular region of space, it is equal to the squared absolute value of the wave function ψ
excited statestate having an energy greater than the ground-state energy
f orbitalmultilobed region of space with high electron density, describes orbitals with l = 3
frequency (ν)number of wave cycles (peaks or troughs) that pass a specified point in space per unit time
ground statestate in which the electrons in an atom, ion, or molecule have the lowest energy possible
Heisenberg uncertainty principlerule stating that it is impossible to exactly determine both certain conjugate dynamical properties such as the momentum and the position of a particle at the same time. The uncertainty principle is a consequence of quantum particles exhibiting wave–particle duality
hertz (Hz)the unit of frequency, which is the number of cycles per second, s−1
Hund’s ruleevery orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin
intensityproperty of wave-propagated energy related to the amplitude of the wave, such as brightness of light or loudness of sound
interference patternpattern typically consisting of alternating bright and dark fringes; it results from constructive and destructive interference of waves
ionization energyenergy required to remove an electron from a gaseous atom or ion
isoelectronicgroup of ions or atoms that have identical electron configurations
line spectrumelectromagnetic radiation emitted at discrete wavelengths by a specific atom (or atoms) in an excited state
magnetic quantum number (ml)quantum number signifying the orientation of an atomic orbital around the nucleus
nodeany point of a standing wave with zero amplitude
orbital diagrampictorial representation of the electron configuration showing each orbital as a box and each electron as an arrow
p orbitaldumbbell-shaped region of space with high electron density, describes orbitals with l = 1
Pauli exclusion principlespecifies that no two electrons in an atom can have the same value for all four quantum numbers
photonsmallest possible packet of electromagnetic radiation, a particle of light
principal quantum number (n)quantum number specifying the shell an electron occupies in an atom
quantizationlimitation of some property to specific discrete values, not continuous
quantum mechanicsfield of study that includes quantization of energy, wave-particle duality, and the Heisenberg uncertainty principle to describe matter
quantum numbernumber having only specific allowed values and used to characterize the arrangement of electrons in an atom
s orbitalspherical region of space with high electron density, describes orbitals with l = 0
secondary (angular momentum) quantum number (l)quantum number distinguishing the different shapes of orbitals; it is also a measure of the orbital angular momentum
shellatomic orbitals with the same principal quantum number, n
spin quantum number (ms)number specifying the electron spin direction, either or
standing wave(also, stationary wave) localized wave phenomenon characterized by discrete wavelengths determined by the boundary conditions used to generate the waves; standing waves are inherently quantized
subshellatomic orbitals with the same values of n and l
valence electronselectrons in the high energy outer shell(s) of an atom
valence shellhigh energy outer shell(s) of an atom
waveoscillation of a property over time or space; can transport energy from one point to another
wave-particle dualityobservation that elementary particles can exhibit both wave-like and particle-like properties
wavefunction (ψ)mathematical description of an atomic orbital that describes the shape of the orbital; it can be used to calculate the probability of finding the electron at any given location in the orbital, as well as dynamical variables such as the energy and the angular momentum
wavelength (λ)distance between two consecutive peaks or troughs in a wave
6.07: Key Equations
c = λν
where h = 6.626 10−34 J s | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/06%3A_Electronic_Structure_and_Periodic_Properties/6.06%3A_Key_Terms.txt |
Light and other forms of electromagnetic radiation move through a vacuum with a constant speed, c, of 2.998 108 m s−1. This radiation shows wavelike behavior, which can be characterized by a frequency, ν, and a wavelength, λ, such that c = λν. Light is an example of a travelling wave. Other important wave phenomena include standing waves, periodic oscillations, and vibrations. Standing waves exhibit quantization, since their wavelengths are limited to discrete integer multiples of some characteristic lengths. Electromagnetic radiation that passes through two closely spaced narrow slits having dimensions roughly similar to the wavelength will show an interference pattern that is a result of constructive and destructive interference of the waves. Electromagnetic radiation also demonstrates properties of particles called photons. The energy of a photon is related to the frequency (or alternatively, the wavelength) of the radiation as E = (or ), where h is Planck's constant. That light demonstrates both wavelike and particle-like behavior is known as wave-particle duality. All forms of electromagnetic radiation share these properties, although various forms including X-rays, visible light, microwaves, and radio waves interact differently with matter and have very different practical applications. Electromagnetic radiation can be generated by exciting matter to higher energies, such as by heating it. The emitted light can be either continuous (incandescent sources like the sun) or discrete (from specific types of excited atoms). Continuous spectra often have distributions that can be approximated as blackbody radiation at some appropriate temperature. The line spectrum of hydrogen can be obtained by passing the light from an electrified tube of hydrogen gas through a prism. This line spectrum was simple enough that an empirical formula called the Rydberg formula could be derived from the spectrum. Three historically important paradoxes from the late 19th and early 20th centuries that could not be explained within the existing framework of classical mechanics and classical electromagnetism were the blackbody problem, the photoelectric effect, and the discrete spectra of atoms. The resolution of these paradoxes ultimately led to quantum theories that superseded the classical theories.
Bohr incorporated Planck’s and Einstein’s quantization ideas into a model of the hydrogen atom that resolved the paradox of atom stability and discrete spectra. The Bohr model of the hydrogen atom explains the connection between the quantization of photons and the quantized emission from atoms. Bohr described the hydrogen atom in terms of an electron moving in a circular orbit about a nucleus. He postulated that the electron was restricted to certain orbits characterized by discrete energies. Transitions between these allowed orbits result in the absorption or emission of photons. When an electron moves from a higher-energy orbit to a more stable one, energy is emitted in the form of a photon. To move an electron from a stable orbit to a more excited one, a photon of energy must be absorbed. Using the Bohr model, we can calculate the energy of an electron and the radius of its orbit in any one-electron system.
Macroscopic objects act as particles. Microscopic objects (such as electrons) have properties of both a particle and a wave. Their exact trajectories cannot be determined. The quantum mechanical model of atoms describes the three-dimensional position of the electron in a probabilistic manner according to a mathematical function called a wavefunction, often denoted as ψ. Atomic wavefunctions are also called orbitals. The squared magnitude of the wavefunction describes the distribution of the probability of finding the electron in a particular region in space. Therefore, atomic orbitals describe the areas in an atom where electrons are most likely to be found.
An atomic orbital is characterized by three quantum numbers. The principal quantum number, n, can be any positive integer. The general region for value of energy of the orbital and the average distance of an electron from the nucleus are related to n. Orbitals having the same value of n are said to be in the same shell. The secondary (angular momentum) quantum number, l, can have any integer value from 0 to n – 1. This quantum number describes the shape or type of the orbital. Orbitals with the same principal quantum number and the same l value belong to the same subshell. The magnetic quantum number, ml, with 2l + 1 values ranging from –l to +l, describes the orientation of the orbital in space. In addition, each electron has a spin quantum number, ms, that can be equal to No two electrons in the same atom can have the same set of values for all the four quantum numbers.
The relative energy of the subshells determine the order in which atomic orbitals are filled (1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, and so on). Electron configurations and orbital diagrams can be determined by applying the Pauli exclusion principle (no two electrons can have the same set of four quantum numbers) and Hund’s rule (whenever possible, electrons retain unpaired spins in degenerate orbitals).
Electrons in the outermost orbitals, called valence electrons, are responsible for most of the chemical behavior of elements. In the periodic table, elements with analogous valence electron configurations usually occur within the same group. There are some exceptions to the predicted filling order, particularly when half-filled or completely filled orbitals can be formed. The periodic table can be divided into three categories based on the orbital in which the last electron to be added is placed: main group elements (s and p orbitals), transition elements (d orbitals), and inner transition elements (f orbitals).
Electron configurations allow us to understand many periodic trends. Covalent radius increases as we move down a group because the n level (orbital size) increases. Covalent radius mostly decreases as we move left to right across a period because the effective nuclear charge experienced by the electrons increases, and the electrons are pulled in tighter to the nucleus. Anionic radii are larger than the parent atom, while cationic radii are smaller, because the number of valence electrons has changed while the nuclear charge has remained constant. Ionization energy (the energy associated with forming a cation) decreases down a group and mostly increases across a period because it is easier to remove an electron from a larger, higher energy orbital. Electron affinity (the energy associated with forming an anion) is more favorable (exothermic) when electrons are placed into lower energy orbitals, closer to the nucleus. Therefore, electron affinity becomes increasingly negative as we move left to right across the periodic table and decreases as we move down a group. For both IE and electron affinity data, there are exceptions to the trends when dealing with completely filled or half-filled subshells. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/06%3A_Electronic_Structure_and_Periodic_Properties/6.08%3A_Summary.txt |
1.
The light produced by a red neon sign is due to the emission of light by excited neon atoms. Qualitatively describe the spectrum produced by passing light from a neon lamp through a prism.
2.
An FM radio station found at 103.1 on the FM dial broadcasts at a frequency of 1.031 108 s−1 (103.1 MHz). What is the wavelength of these radio waves in meters?
3.
FM-95, an FM radio station, broadcasts at a frequency of 9.51 107 s−1 (95.1 MHz). What is the wavelength of these radio waves in meters?
4.
A bright violet line occurs at 435.8 nm in the emission spectrum of mercury vapor. What amount of energy, in joules, must be released by an electron in a mercury atom to produce a photon of this light?
5.
Light with a wavelength of 614.5 nm looks orange. What is the energy, in joules, per photon of this orange light? What is the energy in eV (1 eV = 1.602 10−19 J)?
6.
Heated lithium atoms emit photons of light with an energy of 2.961 10−19 J. Calculate the frequency and wavelength of one of these photons. What is the total energy in 1 mole of these photons? What is the color of the emitted light?
7.
A photon of light produced by a surgical laser has an energy of 3.027 10−19 J. Calculate the frequency and wavelength of the photon. What is the total energy in 1 mole of photons? What is the color of the emitted light?
8.
When rubidium ions are heated to a high temperature, two lines are observed in its line spectrum at wavelengths (a) 7.9 10−7 m and (b) 4.2 10−7 m. What are the frequencies of the two lines? What color do we see when we heat a rubidium compound?
9.
The emission spectrum of cesium contains two lines whose frequencies are (a) 3.45 1014 Hz and (b) 6.53 1014 Hz. What are the wavelengths and energies per photon of the two lines? What color are the lines?
10.
Photons of infrared radiation are responsible for much of the warmth we feel when holding our hands before a fire. These photons will also warm other objects. How many infrared photons with a wavelength of 1.5 10−6 m must be absorbed by the water to warm a cup of water (175 g) from 25.0 °C to 40 °C?
11.
One of the radiographic devices used in a dentist's office emits an X-ray of wavelength 2.090 10−11 m. What is the energy, in joules, and frequency of this X-ray?
12.
The eyes of certain reptiles pass a single visual signal to the brain when the visual receptors are struck by photons of a wavelength of 850 nm. If a total energy of 3.15 10−14 J is required to trip the signal, what is the minimum number of photons that must strike the receptor?
13.
RGB color television and computer displays use cathode ray tubes that produce colors by mixing red, green, and blue light. If we look at the screen with a magnifying glass, we can see individual dots turn on and off as the colors change. Using a spectrum of visible light, determine the approximate wavelength of each of these colors. What is the frequency and energy of a photon of each of these colors?
14.
Answer the following questions about a Blu-ray laser:
1. The laser on a Blu-ray player has a wavelength of 405 nm. In what region of the electromagnetic spectrum is this radiation? What is its frequency?
2. A Blu-ray laser has a power of 5 milliwatts (1 watt = 1 J s−1). How many photons of light are produced by the laser in 1 hour?
3. The ideal resolution of a player using a laser (such as a Blu-ray player), which determines how close together data can be stored on a compact disk, is determined using the following formula: Resolution = 0.60(λ/NA), where λ is the wavelength of the laser and NA is the numerical aperture. Numerical aperture is a measure of the size of the spot of light on the disk; the larger the NA, the smaller the spot. In a typical Blu-ray system, NA = 0.95. If the 405-nm laser is used in a Blu-ray player, what is the closest that information can be stored on a Blu-ray disk?
4. The data density of a Blu-ray disk using a 405-nm laser is 1.5 107 bits mm−2. Disks have an outside diameter of 120 mm and a hole of 15-mm diameter. How many data bits can be contained on the disk? If a Blu-ray disk can hold 9,400,000 pages of text, how many data bits are needed for a typed page? (Hint: Determine the area of the disk that is available to hold data. The area inside a circle is given by A = πr2, where the radius r is one-half of the diameter.)
15.
What is the threshold frequency for sodium metal if a photon with frequency 6.66 1014 s−1 ejects an electron with 7.74 10−20 J kinetic energy? Will the photoelectric effect be observed if sodium is exposed to orange light?
16.
Why is the electron in a Bohr hydrogen atom bound less tightly when it has a quantum number of 3 than when it has a quantum number of 1?
17.
What does it mean to say that the energy of the electrons in an atom is quantized?
18.
Using the Bohr model, determine the energy, in joules, necessary to ionize a ground-state hydrogen atom. Show your calculations.
19.
The electron volt (eV) is a convenient unit of energy for expressing atomic-scale energies. It is the amount of energy that an electron gains when subjected to a potential of 1 volt; 1 eV = 1.602 10–19 J. Using the Bohr model, determine the energy, in electron volts, of the photon produced when an electron in a hydrogen atom moves from the orbit with n = 5 to the orbit with n = 2. Show your calculations.
20.
Using the Bohr model, determine the lowest possible energy, in joules, for the electron in the Li2+ ion.
21.
Using the Bohr model, determine the lowest possible energy for the electron in the He+ ion.
22.
Using the Bohr model, determine the energy of an electron with n = 6 in a hydrogen atom.
23.
Using the Bohr model, determine the energy of an electron with n = 8 in a hydrogen atom.
24.
How far from the nucleus in angstroms (1 angstrom = 1 10–10 m) is the electron in a hydrogen atom if it has an energy of –8.72 10–20 J?
25.
What is the radius, in angstroms, of the orbital of an electron with n = 8 in a hydrogen atom?
26.
Using the Bohr model, determine the energy in joules of the photon produced when an electron in a He+ ion moves from the orbit with n = 5 to the orbit with n = 2.
27.
Using the Bohr model, determine the energy in joules of the photon produced when an electron in a Li2+ ion moves from the orbit with n = 2 to the orbit with n = 1.
28.
Consider a large number of hydrogen atoms with electrons randomly distributed in the n = 1, 2, 3, and 4 orbits.
1. How many different wavelengths of light are emitted by these atoms as the electrons fall into lower-energy orbits?
2. Calculate the lowest and highest energies of light produced by the transitions described in part (a).
3. Calculate the frequencies and wavelengths of the light produced by the transitions described in part (b).
29.
How are the Bohr model and the Rutherford model of the atom similar? How are they different?
30.
The spectra of hydrogen and of calcium are shown here.
What causes the lines in these spectra? Why are the colors of the lines different? Suggest a reason for the observation that the spectrum of calcium is more complicated than the spectrum of hydrogen.
31.
How are the Bohr model and the quantum mechanical model of the hydrogen atom similar? How are they different?
32.
What are the allowed values for each of the four quantum numbers: n, l, ml, and ms?
33.
Describe the properties of an electron associated with each of the following four quantum numbers: n, l, ml, and ms.
34.
Answer the following questions:
1. Without using quantum numbers, describe the differences between the shells, subshells, and orbitals of an atom.
2. How do the quantum numbers of the shells, subshells, and orbitals of an atom differ?
35.
Identify the subshell in which electrons with the following quantum numbers are found:
1. n = 2, l = 1
2. n = 4, l = 2
3. n = 6, l = 0
36.
Which of the subshells described in the previous question contain degenerate orbitals? How many degenerate orbitals are in each?
37.
Identify the subshell in which electrons with the following quantum numbers are found:
1. n = 3, l = 2
2. n = 1, l = 0
3. n = 4, l = 3
38.
Which of the subshells described in the previous question contain degenerate orbitals? How many degenerate orbitals are in each?
39.
Sketch the boundary surface of a and a py orbital. Be sure to show and label the axes.
40.
Sketch the px and dxz orbitals. Be sure to show and label the coordinates.
41.
Consider the orbitals shown here in outline.
1. What is the maximum number of electrons contained in an orbital of type (x)? Of type (y)? Of type (z)?
2. How many orbitals of type (x) are found in a shell with n = 2? How many of type (y)? How many of type (z)?
3. Write a set of quantum numbers for an electron in an orbital of type (x) in a shell with n = 4. Of an orbital of type (y) in a shell with n = 2. Of an orbital of type (z) in a shell with n = 3.
4. What is the smallest possible n value for an orbital of type (x)? Of type (y)? Of type (z)?
5. What are the possible l and ml values for an orbital of type (x)? Of type (y)? Of type (z)?
42.
State the Heisenberg uncertainty principle. Describe briefly what the principle implies.
43.
How many electrons could be held in the second shell of an atom if the spin quantum number ms could have three values instead of just two? (Hint: Consider the Pauli exclusion principle.)
44.
Which of the following equations describe particle-like behavior? Which describe wavelike behavior? Do any involve both types of behavior? Describe the reasons for your choices.
1. c = λν
2. E = hν
45.
Write a set of quantum numbers for each of the electrons with an n of 4 in a Se atom.
46.
Read the labels of several commercial products and identify monatomic ions of at least four transition elements contained in the products. Write the complete electron configurations of these cations.
47.
Read the labels of several commercial products and identify monatomic ions of at least six main group elements contained in the products. Write the complete electron configurations of these cations and anions.
48.
Using complete subshell notation (not abbreviations, 1s22s22p6, and so forth), predict the electron configuration of each of the following atoms:
1. C
2. P
3. V
4. Sb
5. Sm
49.
Using complete subshell notation (1s22s22p6, and so forth), predict the electron configuration of each of the following atoms:
1. N
2. Si
3. Fe
4. Te
5. Tb
50.
Is 1s22s22p6 the symbol for a macroscopic property or a microscopic property of an element? Explain your answer.
51.
What additional information do we need to answer the question “Which ion has the electron configuration 1s22s22p63s23p6”?
52.
Draw the orbital diagram for the valence shell of each of the following atoms:
1. C
2. P
3. V
4. Sb
5. Ru
53.
Use an orbital diagram to describe the electron configuration of the valence shell of each of the following atoms:
1. N
2. Si
3. Fe
4. Te
5. Mo
54.
Using complete subshell notation (1s22s22p6, and so forth), predict the electron configurations of the following ions.
1. (f) Gd3+
55.
Which atom has the electron configuration 1s22s22p63s23p64s23d104p65s24d2?
56.
Which atom has the electron configuration 1s22s22p63s23p63d74s2?
57.
Which ion with a +1 charge has the electron configuration 1s22s22p63s23p63d104s24p6? Which ion with a –2 charge has this configuration?
58.
Which of the following atoms contains only three valence electrons: Li, B, N, F, Ne?
59.
Which of the following has two unpaired electrons?
1. Mg
2. Si
3. S
4. Both Mg and S
5. Both Si and S.
60.
Which atom would be expected to have a half-filled 6p subshell?
61.
Which atom would be expected to have a half-filled 4s subshell?
62.
In one area of Australia, the cattle did not thrive despite the presence of suitable forage. An investigation showed the cause to be the absence of sufficient cobalt in the soil. Cobalt forms cations in two oxidation states, Co2+ and Co3+. Write the electron structure of the two cations.
63.
Thallium was used as a poison in the Agatha Christie mystery story “The Pale Horse.” Thallium has two possible cationic forms, +1 and +3. The +1 compounds are the more stable. Write the electron structure of the +1 cation of thallium.
64.
Write the electron configurations for the following atoms or ions:
1. B3+
2. O
3. Cl3+
4. Ca2+
5. Ti
65.
Cobalt–60 and iodine–131 are radioactive isotopes commonly used in nuclear medicine. How many protons, neutrons, and electrons are in atoms of these isotopes? Write the complete electron configuration for each isotope.
66.
Write a set of quantum numbers for each of the electrons with an n of 3 in a Sc atom.
67.
Based on their positions in the periodic table, predict which has the smallest atomic radius: Mg, Sr, Si, Cl, I.
68.
Based on their positions in the periodic table, predict which has the largest atomic radius: Li, Rb, N, F, I.
69.
Based on their positions in the periodic table, predict which has the largest first ionization energy: Mg, Ba, B, O, Te.
70.
Based on their positions in the periodic table, predict which has the smallest first ionization energy: Li, Cs, N, F, I.
71.
Based on their positions in the periodic table, rank the following atoms in order of increasing first ionization energy: F, Li, N, Rb
72.
Based on their positions in the periodic table, rank the following atoms in order of increasing first ionization energy: Mg, O, S, Si
73.
Atoms of which group in the periodic table have a valence shell electron configuration of ns2np3?
74.
Atoms of which group in the periodic table have a valence shell electron configuration of ns2?
75.
Based on their positions in the periodic table, list the following atoms in order of increasing radius: Mg, Ca, Rb, Cs.
76.
Based on their positions in the periodic table, list the following atoms in order of increasing radius: Sr, Ca, Si, Cl.
77.
Based on their positions in the periodic table, list the following ions in order of increasing radius: K+, Ca2+, Al3+, Si4+.
78.
List the following ions in order of increasing radius: Li+, Mg2+, Br, Te2–.
79.
Which atom and/or ion is (are) isoelectronic with Br+: Se2+, Se, As, Kr, Ga3+, Cl?
80.
Which of the following atoms and ions is (are) isoelectronic with S2+: Si4+, Cl3+, Ar, As3+, Si, Al3+?
81.
Compare both the numbers of protons and electrons present in each to rank the following ions in order of increasing radius: As3–, Br, K+, Mg2+.
82.
Of the five elements Al, Cl, I, Na, Rb, which has the most exothermic reaction? (E represents an atom.) What name is given to the energy for the reaction? Hint: Note the process depicted does not correspond to electron affinity.)
83.
Of the five elements Sn, Si, Sb, O, Te, which has the most endothermic reaction? (E represents an atom.) What name is given to the energy for the reaction?
84.
The ionic radii of the ions S2–, Cl, and K+ are 184, 181, 138 pm respectively. Explain why these ions have different sizes even though they contain the same number of electrons.
85.
Which main group atom would be expected to have the lowest second ionization energy?
86.
Explain why Al is a member of group 13 rather than group 3? | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/06%3A_Electronic_Structure_and_Periodic_Properties/6.09%3A_Exercises.txt |
A chemical bond is an attraction between atoms that allows the formation of chemical substances that contain two or more atoms. The bond is caused by the electrostatic force of attraction between opposite charges, either between electrons and nuclei, or as the result of a dipole attraction. All bonds can be explained by quantum theory, but, in practice, simplification rules allow chemists to predict the strength, directionality, and polarity of bonds. The octet rule and VSEPR theory are two examples. More sophisticated theories are valence bond theory which includes orbital hybridization and resonance, and the linear combination of atomic orbitals molecular orbital method. Electrostatics are used to describe bond polarities and the effects they have on chemical substances.
• 7.0: Introduction
It has long been known that pure carbon occurs in different forms (allotropes) including graphite and diamonds. But it was not until 1985 that a new form of carbon was recognized: buckminsterfullerene, commonly known as a “buckyball.” Experimental evidence revealed the formula, C60, and then scientists determined how 60 carbon atoms could form one symmetric, stable molecule. They were guided by bonding theory—the topic of this chapter—which explains how individual atoms connect to form more comp
• 7.1: Ionic Bonding
Atoms gain or lose electrons to form ions with particularly stable electron configurations. The charges of cations formed by the representative metals may be determined readily because, with few exceptions, the electronic structures of these ions have either a noble gas configuration or a completely filled electron shell. The charges of anions formed by the nonmetals may also be readily determined because these ions form when nonmetal atoms gain enough electrons to fill their valence shells.
• 7.2: Covalent Bonding
Covalent bonds form when electrons are shared between atoms and are attracted by the nuclei of both atoms. In pure covalent bonds, the electrons are shared equally. In polar covalent bonds, the electrons are shared unequally, as one atom exerts a stronger force of attraction on the electrons than the other. The ability of an atom to attract a pair of electrons in a chemical bond is called its electronegativity.
• 7.3: Lewis Symbols and Structures
Valence electronic structures can be visualized by drawing Lewis symbols (for atoms and monatomic ions) and Lewis structures (for molecules and polyatomic ions). Lone pairs, unpaired electrons, and single, double, or triple bonds are used to indicate where the valence electrons are located around each atom in a Lewis structure. Most structures—especially those containing second row elements—obey the octet rule, in which every atom (except H) is surrounded by eight electrons.
• 7.4: Formal Charges and Resonance
In a Lewis structure, formal charges can be assigned to each atom by treating each bond as if one-half of the electrons are assigned to each atom. These hypothetical formal charges are a guide to determining the most appropriate Lewis structure. A structure in which the formal charges are as close to zero as possible is preferred. Resonance occurs in cases where two or more Lewis structures with identical arrangements of atoms but different distributions of electrons can be written.
• 7.5: Strengths of Ionic and Covalent Bonds
The strength of a covalent bond is measured by its bond dissociation energy, that is, the amount of energy required to break that particular bond in a mole of molecules. Multiple bonds are stronger than single bonds between the same atoms. Reaction enthalpies can be estimated based on the energy input required to break bonds and the energy released when new bonds are formed. For ionic bonds, the lattice energy is the energy required to separate one mole of a compound into its gas phase ions.
• 7.6: Molecular Structure and Polarity
VSEPR theory predicts the three-dimensional arrangement of atoms in a molecule. It states that valence electrons will assume an electron-pair geometry that minimizes repulsions between areas of high electron density (bonds and/or lone pairs). Molecular structure, which refers only to the placement of atoms in a molecule and not the electrons, is equivalent to electron-pair geometry only when there are no lone electron pairs around the central atom.
• 7.7: Key Terms
• 7.8: Key Equations
• 7.9: Summary
• 7.10: Exercises
These are homework exercises to accompany the Textmap created for "Chemistry" by OpenStax. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
Thumbnail: Covalently bonded hydrogen and carbon in a w:molecule of methane. (CC BY-SA 2.5; DynaBlast via Wikipedia)
07: Chemical Bonding and Molecular Geometry
It has long been known that pure carbon occurs in different forms (allotropes) including graphite and diamonds. But it was not until 1985 that a new form of carbon was recognized: buckminsterfullerene. This molecule was named after the architect and inventor R. Buckminster Fuller (1895–1983), whose signature architectural design was the geodesic dome, characterized by a lattice shell structure supporting a spherical surface. Experimental evidence revealed the formula, C60, and then scientists determined how 60 carbon atoms could form one symmetric, stable molecule. They were guided by bonding theory—the topic of this chapter—which explains how individual atoms connect to form more complex structures. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/07%3A_Chemical_Bonding_and_Molecular_Geometry/7.00%3A_Introduction.txt |
Learning Objectives
By the end of this section, you will be able to:
• Explain the formation of cations, anions, and ionic compounds
• Predict the charge of common metallic and nonmetallic elements, and write their electron configurations
As you have learned, ions are atoms or molecules bearing an electrical charge. A cation (a positive ion) forms when a neutral atom loses one or more electrons from its valence shell, and an anion (a negative ion) forms when a neutral atom gains one or more electrons in its valence shell.
Compounds composed of ions are called ionic compounds (or salts), and their constituent ions are held together by ionic bonds: electrostatic forces of attraction between oppositely charged cations and anions. The properties of ionic compounds shed some light on the nature of ionic bonds. Ionic solids exhibit a crystalline structure and tend to be rigid and brittle; they also tend to have high melting and boiling points, which suggests that ionic bonds are very strong. Ionic solids are also poor conductors of electricity for the same reason—the strength of ionic bonds prevents ions from moving freely in the solid state. Most ionic solids, however, dissolve readily in water. Once dissolved or melted, ionic compounds are excellent conductors of electricity and heat because the ions can move about freely.
Neutral atoms and their associated ions have very different physical and chemical properties. Sodium atoms form sodium metal, a soft, silvery-white metal that burns vigorously in air and reacts explosively with water. Chlorine atoms form chlorine gas, Cl2, a yellow-green gas that is extremely corrosive to most metals and very poisonous to animals and plants. The vigorous reaction between the elements sodium and chlorine forms the white, crystalline compound sodium chloride, common table salt, which contains sodium cations and chloride anions (Figure $1$). The compound composed of these ions exhibits properties entirely different from the properties of the elements sodium and chlorine. Chlorine is poisonous, but sodium chloride is essential to life; sodium atoms react vigorously with water, but sodium chloride simply dissolves in water.
The Formation of Ionic Compounds
Binary ionic compounds are composed of just two elements: a metal (which forms the cations) and a nonmetal (which forms the anions). For example, NaCl is a binary ionic compound. We can think about the formation of such compounds in terms of the periodic properties of the elements. Many metallic elements have relatively low ionization potentials and lose electrons easily. These elements lie to the left in a period or near the bottom of a group on the periodic table. Nonmetal atoms have relatively high electron affinities and thus readily gain electrons lost by metal atoms, thereby filling their valence shells. Nonmetallic elements are found in the upper-right corner of the periodic table.
As all substances must be electrically neutral, the total number of positive charges on the cations of an ionic compound must equal the total number of negative charges on its anions. The formula of an ionic compound represents the simplest ratio of the numbers of ions necessary to give identical numbers of positive and negative charges. For example, the formula for aluminum oxide, Al2O3, indicates that this ionic compound contains two aluminum cations, Al3+, for every three oxide anions, O2− [thus, (2 +3) + (3 –2) = 0].
It is important to note, however, that the formula for an ionic compound does not represent the physical arrangement of its ions. It is incorrect to refer to a sodium chloride (NaCl) “molecule” because there is not a single ionic bond, per se, between any specific pair of sodium and chloride ions. The attractive forces between ions are isotropic—the same in all directions—meaning that any particular ion is equally attracted to all of the nearby ions of opposite charge. This results in the ions arranging themselves into a tightly bound, three-dimensional lattice structure. Sodium chloride, for example, consists of a regular arrangement of equal numbers of Na+ cations and Cl anions (Figure $2$).
The strong electrostatic attraction between Na+ and Cl ions holds them tightly together in solid NaCl. It requires 769 kJ of energy to dissociate one mole of solid NaCl into separate gaseous Na+ and Cl ions:
$\ce{NaCl(s) -> Na^{+}(g) + Cl^{-}(g)} \quad \quad \Delta H=769 \,\text{kJ} \nonumber$
Electronic Structures of Cations
When forming a cation, an atom of a main group element tends to lose all of its valence electrons, thus assuming the electronic structure of the noble gas that precedes it in the periodic table. For groups 1 (the alkali metals) and 2 (the alkaline earth metals), the group numbers are equal to the numbers of valence shell electrons and, consequently, to the charges of the cations formed from atoms of these elements when all valence shell electrons are removed. For example, calcium is a group 2 element whose neutral atoms have 20 electrons and a ground state electron configuration of 1s22s22p63s23p64s2. When a Ca atom loses both of its valence electrons, the result is a cation with 18 electrons, a 2+ charge, and an electron configuration of 1s22s22p63s23p6. The Ca2+ ion is therefore isoelectronic with the noble gas Ar.
For groups 13–17, the group numbers exceed the number of valence electrons by 10 (accounting for the possibility of full d subshells in atoms of elements in the fourth and greater periods). Thus, the charge of a cation formed by the loss of all valence electrons is equal to the group number minus 10. For example, aluminum (in group 13) forms 3+ ions (Al3+).
Exceptions to the expected behavior involve elements toward the bottom of the groups. In addition to the expected ions Tl3+, Sn4+, Pb4+, and Bi5+, a partial loss of these atoms’ valence shell electrons can also lead to the formation of Tl+, Sn2+, Pb2+, and Bi3+ ions. The formation of these 1+, 2+, and 3+ cations is ascribed to the inert pair effect, which reflects the relatively low energy of the valence s-electron pair for atoms of the heavy elements of groups 13, 14, and 15. Mercury (group 12) also exhibits an unexpected behavior: it forms a diatomic ion, (an ion formed from two mercury atoms, with an Hg-Hg bond), in addition to the expected monatomic ion Hg2+ (formed from only one mercury atom).
Transition and inner transition metal elements behave differently than main group elements. Most transition metal cations have 2+ or 3+ charges that result from the loss of their outermost s electron(s) first, sometimes followed by the loss of one or two d electrons from the next-to-outermost shell. For example, iron (1s22s22p63s23p63d64s2) forms the ion Fe2+ (1s22s22p63s23p63d6) by the loss of the 4s electrons and the ion Fe3+ (1s22s22p63s23p63d5) by the loss of the 4s electrons and one of the 3d electrons. Although the d orbitals of the transition elements are—according to the Aufbau principle—the last to fill when building up electron configurations, the outermost s electrons are the first to be lost when these atoms ionize. When the inner transition metals form ions, they usually have a 3+ charge, resulting from the loss of their outermost s electrons and a d or f electron.
Example $1$: Determining the Electronic Structures of Cations
There are at least 14 elements categorized as “essential trace elements” for the human body. They are called “essential” because they are required for healthy bodily functions, “trace” because they are required only in small amounts, and “elements” in spite of the fact that they are really ions. Two of these essential trace elements, chromium and zinc, are required as Cr3+ and Zn2+. Write the electron configurations of these cations.
Solution
First, write the electron configuration for the neutral atoms:
Zn: [Ar]3d104s2
Cr: [Ar]3d54s1
Next, remove electrons from the highest energy orbital. For the transition metals, electrons are removed from the s orbital first and then from the d orbital. For the p-block elements, electrons are removed from the p orbitals and then from the s orbital. Zinc is a member of group 12, so it should have a charge of 2+, and thus loses only the two electrons in its s orbital. Chromium is a transition element and should lose its s electrons and then its d electrons when forming a cation. Thus, we find the following electron configurations of the ions:
Zn2+: [Ar]3d10
Cr3+: [Ar]3d3
Exercise $1$
Potassium and magnesium are required in our diet. Write the electron configurations of the ions expected from these elements.
Answer
K+: [Ar], Mg2+: [Ne]
Electronic Structures of Anions
Most monatomic anions form when a neutral nonmetal atom gains enough electrons to completely fill its outer s and p orbitals, thereby reaching the electron configuration of the next noble gas. Thus, it is simple to determine the charge on such a negative ion: The charge is equal to the number of electrons that must be gained to fill the s and p orbitals of the parent atom. Oxygen, for example, has the electron configuration 1s22s22p4, whereas the oxygen anion has the electron configuration of the noble gas neon (Ne), 1s22s22p6. The two additional electrons required to fill the valence orbitals give the oxide ion the charge of 2– (O2–).
Example $1$: Determining the Electronic Structure of Anions
Selenium and iodine are two essential trace elements that form anions. Write the electron configurations of the anions.
Solution
Se2: [Ar]3d104s24p6
I: [Kr]4d105s25p6
Exercise $2$
Write the electron configurations of a phosphorus atom and its negative ion. Give the charge on the anion.
Answer
P: [Ne]3s23p3; P3–: [Ne]3s23p6 | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/07%3A_Chemical_Bonding_and_Molecular_Geometry/7.01%3A_Ionic_Bonding.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the formation of covalent bonds
• Define electronegativity and assess the polarity of covalent bonds
Ionic bonding results from the electrostatic attraction of oppositely charged ions that are typically produced by the transfer of electrons between metallic and nonmetallic atoms. A different type of bonding results from the mutual attraction of atoms for a “shared” pair of electrons. Such bonds are called covalent bonds. Covalent bonds are formed between two atoms when both have similar tendencies to attract electrons to themselves (i.e., when both atoms have identical or fairly similar ionization energies and electron affinities). For example, two hydrogen atoms bond covalently to form an H2 molecule; each hydrogen atom in the H2 molecule has two electrons stabilizing it, giving each atom the same number of valence electrons as the noble gas He.
Compounds that contain covalent bonds exhibit different physical properties than ionic compounds. Because the attraction between molecules, which are electrically neutral, is weaker than that between electrically charged ions, covalent compounds generally have much lower melting and boiling points than ionic compounds. In fact, many covalent compounds are liquids or gases at room temperature, and, in their solid states, they are typically much softer than ionic solids. Furthermore, whereas ionic compounds are good conductors of electricity when dissolved in water, most covalent compounds are insoluble in water; since they are electrically neutral, they are poor conductors of electricity in any state.
Formation of Covalent Bonds
Nonmetal atoms frequently form covalent bonds with other nonmetal atoms. For example, the hydrogen molecule, H2, contains a covalent bond between its two hydrogen atoms. Figure $1$ illustrates why this bond is formed. Starting on the far right, we have two separate hydrogen atoms with a particular potential energy, indicated by the red line. Along the x-axis is the distance between the two atoms. As the two atoms approach each other (moving left along the x-axis), their valence orbitals (1s) begin to overlap. The single electrons on each hydrogen atom then interact with both atomic nuclei, occupying the space around both atoms. The strong attraction of each shared electron to both nuclei stabilizes the system, and the potential energy decreases as the bond distance decreases. If the atoms continue to approach each other, the positive charges in the two nuclei begin to repel each other, and the potential energy increases. The bond length is determined by the distance at which the lowest potential energy is achieved.
It is essential to remember that energy must be added to break chemical bonds (an endothermic process), whereas forming chemical bonds releases energy (an exothermic process). In the case of H2, the covalent bond is very strong; a large amount of energy, 436 kJ, must be added to break the bonds in one mole of hydrogen molecules and cause the atoms to separate:
$\ce{H2(g) -> 2 H(g)} \quad \Delta H=436 \,\text{kJ} \nonumber$
Conversely, the same amount of energy is released when one mole of H2 molecules forms from two moles of H atoms:
$\ce{2 H(g) -> H2(g)} \quad \Delta H=-436\,\text{kJ} \nonumber$
Pure vs. Polar Covalent Bonds
If the atoms that form a covalent bond are identical, as in H2, Cl2, and other diatomic molecules, then the electrons in the bond must be shared equally. We refer to this as a pure covalent bond. Electrons shared in pure covalent bonds have an equal probability of being near each nucleus.
In the case of Cl2, each atom starts off with seven valence electrons, and each Cl shares one electron with the other, forming one covalent bond:
$\ce{Cl + Cl -> Cl2} \nonumber$
The total number of electrons around each individual atom consists of six nonbonding electrons and two shared (i.e., bonding) electrons for eight total electrons, matching the number of valence electrons in the noble gas argon. Since the bonding atoms are identical, Cl2 also features a pure covalent bond.
When the atoms linked by a covalent bond are different, the bonding electrons are shared, but no longer equally. Instead, the bonding electrons are more attracted to one atom than the other, giving rise to a shift of electron density toward that atom. This unequal distribution of electrons is known as a polar covalent bond, characterized by a partial positive charge on one atom and a partial negative charge on the other. The atom that attracts the electrons more strongly acquires the partial negative charge and vice versa. For example, the electrons in the H–Cl bond of a hydrogen chloride molecule spend more time near the chlorine atom than near the hydrogen atom. Thus, in an HCl molecule, the chlorine atom carries a partial negative charge and the hydrogen atom has a partial positive charge. Figure $2$ shows the distribution of electrons in the H–Cl bond. Note that the shaded area around Cl is much larger than it is around H. Compare this to Figure $1$, which shows the even distribution of electrons in the H2 nonpolar bond.
We sometimes designate the positive and negative atoms in a polar covalent bond using a lowercase Greek letter “delta,” δ, with a plus sign or minus sign to indicate whether the atom has a partial positive charge (δ+) or a partial negative charge (δ–). This symbolism is shown for the H–Cl molecule in Figure $2$.
Electronegativity
Whether a bond is nonpolar or polar covalent is determined by a property of the bonding atoms called electronegativity. Electronegativity is a measure of the tendency of an atom to attract electrons (or electron density) towards itself. It determines how the shared electrons are distributed between the two atoms in a bond. The more strongly an atom attracts the electrons in its bonds, the larger its electronegativity. Electrons in a polar covalent bond are shifted toward the more electronegative atom; thus, the more electronegative atom is the one with the partial negative charge. The greater the difference in electronegativity, the more polarized the electron distribution and the larger the partial charges of the atoms.
Figure $3$ shows the electronegativity values of the elements as proposed by one of the most famous chemists of the twentieth century: Linus Pauling (Figure $4$). In general, electronegativity increases from left to right across a period in the periodic table and decreases down a group. Thus, the nonmetals, which lie in the upper right, tend to have the highest electronegativities, with fluorine the most electronegative element of all (EN = 4.0). Metals tend to be less electronegative elements, and the group 1 metals have the lowest electronegativities. Note that noble gases are excluded from this figure because these atoms usually do not share electrons with others atoms since they have a full valence shell. (While noble gas compounds such as XeO2 do exist, they can only be formed under extreme conditions, and thus they do not fit neatly into the general model of electronegativity.)
Electronegativity versus Electron Affinity
We must be careful not to confuse electronegativity and electron affinity. The electron affinity of an element is a measurable physical quantity, namely, the energy released or absorbed when an isolated gas-phase atom acquires an electron, measured in kJ/mol. Electronegativity, on the other hand, describes how tightly an atom attracts electrons in a bond. It is a dimensionless quantity that is calculated, not measured. Pauling derived the first electronegativity values by comparing the amounts of energy required to break different types of bonds. He chose an arbitrary relative scale ranging from 0 to 4.
Portrait of a Chemist: Linus Pauling
Linus Pauling, shown in Figure $4$, is the only person to have received two unshared (individual) Nobel Prizes: one for chemistry in 1954 for his work on the nature of chemical bonds and one for peace in 1962 for his opposition to weapons of mass destruction. He developed many of the theories and concepts that are foundational to our current understanding of chemistry, including electronegativity and resonance structures.
Pauling also contributed to many other fields besides chemistry. His research on sickle cell anemia revealed the cause of the disease—the presence of a genetically inherited abnormal protein in the blood—and paved the way for the field of molecular genetics. His work was also pivotal in curbing the testing of nuclear weapons; he proved that radioactive fallout from nuclear testing posed a public health risk.
Electronegativity and Bond Type
The absolute value of the difference in electronegativity (ΔEN) of two bonded atoms provides a rough measure of the polarity to be expected in the bond and, thus, the bond type. When the difference is very small or zero, the bond is covalent and nonpolar. When it is large, the bond is polar covalent or ionic. The absolute values of the electronegativity differences between the atoms in the bonds H–H, H–Cl, and Na–Cl are 0 (nonpolar), 0.9 (polar covalent), and 2.1 (ionic), respectively. The degree to which electrons are shared between atoms varies from completely equal (pure covalent bonding) to not at all (ionic bonding). Figure $5$ shows the relationship between electronegativity difference and bond type.
A rough approximation of the electronegativity differences associated with covalent, polar covalent, and ionic bonds is shown in Figure $5$. This table is just a general guide, however, with many exceptions. For example, the H and F atoms in HF have an electronegativity difference of 1.9, and the N and H atoms in NH3 a difference of 0.9, yet both of these compounds form bonds that are considered polar covalent. Likewise, the Na and Cl atoms in NaCl have an electronegativity difference of 2.1, and the Mn and I atoms in MnI2 have a difference of 1.0, yet both of these substances form ionic compounds.
The best guide to the covalent or ionic character of a bond is to consider the types of atoms involved and their relative positions in the periodic table. Bonds between two nonmetals are generally covalent; bonding between a metal and a nonmetal is often ionic.
Some compounds contain both covalent and ionic bonds. The atoms in polyatomic ions, such as OH, and are held together by polar covalent bonds. However, these polyatomic ions form ionic compounds by combining with ions of opposite charge. For example, potassium nitrate, KNO3, contains the K+ cation and the polyatomic anion. Thus, bonding in potassium nitrate is ionic, resulting from the electrostatic attraction between the ions K+ and as well as covalent between the nitrogen and oxygen atoms in
Example $1$: Electronegativity and Bond Polarity
Bond polarities play an important role in determining the structure of proteins. Using the electronegativity values in Figure $3$, arrange the following covalent bonds—all commonly found in amino acids—in order of increasing polarity. Then designate the positive and negative atoms using the symbols δ+ and δ–:
C–H, C–N, C–O, N–H, O–H, S–H
Solution
The polarity of these bonds increases as the absolute value of the electronegativity difference increases. The atom with the δ– designation is the more electronegative of the two. Table $1$ shows these bonds in order of increasing polarity.
Table $1$: Bond Polarity and Electronegativity Difference
Bond ΔEN Polarity
C–H 0.4
S–H 0.4
C–N 0.5
N–H 0.9
C–O 1.0
O–H 1.4
Exercise $1$
Silicones are polymeric compounds containing, among others, the following types of covalent bonds: Si–O, Si–C, C–H, and C–C. Using the electronegativity values in Figure $3$, arrange the bonds in order of increasing polarity and designate the positive and negative atoms using the symbols δ+ and δ–.
Answer
Bond Electronegativity Difference Polarity
C–C 0.0 nonpolar
C–H 0.4
Si–C 0.7
Si–O 1.7 | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/07%3A_Chemical_Bonding_and_Molecular_Geometry/7.02%3A_Covalent_Bonding.txt |
Learning Objectives
By the end of this section, you will be able to:
• Write Lewis symbols for neutral atoms and ions
• Draw Lewis structures depicting the bonding in simple molecules
Thus far in this chapter, we have discussed the various types of bonds that form between atoms and/or ions. In all cases, these bonds involve the sharing or transfer of valence shell electrons between atoms. In this section, we will explore the typical method for depicting valence shell electrons and chemical bonds, namely Lewis symbols and Lewis structures.
Lewis Symbols
We use Lewis symbols to describe valence electron configurations of atoms and monatomic ions. A Lewis symbol consists of an elemental symbol surrounded by one dot for each of its valence electrons:
Figure $1$: shows the Lewis symbols for the elements of the third period of the periodic table.
Lewis symbols can also be used to illustrate the formation of cations from atoms, as shown here for sodium and calcium:
Likewise, they can be used to show the formation of anions from atoms, as shown here for chlorine and sulfur:
Figure $2$ demonstrates the use of Lewis symbols to show the transfer of electrons during the formation of ionic compounds.
Lewis Structures
We also use Lewis symbols to indicate the formation of covalent bonds, which are shown in Lewis structures, drawings that describe the bonding in molecules and polyatomic ions. For example, when two chlorine atoms form a chlorine molecule, they share one pair of electrons:
The Lewis structure indicates that each Cl atom has three pairs of electrons that are not used in bonding (called lone pairs) and one shared pair of electrons (written between the atoms). A dash (or line) is sometimes used to indicate a shared pair of electrons:
A single shared pair of electrons is called a single bond. Each Cl atom interacts with eight valence electrons: the six in the lone pairs and the two in the single bond.
The Octet Rule
The other halogen molecules (F2, Br2, I2, and At2) form bonds like those in the chlorine molecule: one single bond between atoms and three lone pairs of electrons per atom. This allows each halogen atom to have a noble gas electron configuration. The tendency of main group atoms to form enough bonds to obtain eight valence electrons is known as the octet rule.
The number of bonds that an atom can form can often be predicted from the number of electrons needed to reach an octet (eight valence electrons); this is especially true of the nonmetals of the second period of the periodic table (C, N, O, and F). For example, each atom of a group 14 element has four electrons in its outermost shell and therefore requires four more electrons to reach an octet. These four electrons can be gained by forming four covalent bonds, as illustrated here for carbon in CCl4 (carbon tetrachloride) and silicon in SiH4 (silane). Because hydrogen only needs two electrons to fill its valence shell, it is an exception to the octet rule. The transition elements and inner transition elements also do not follow the octet rule:
Group 15 elements such as nitrogen have five valence electrons in the atomic Lewis symbol: one lone pair and three unpaired electrons. To obtain an octet, these atoms form three covalent bonds, as in NH3 (ammonia). Oxygen and other atoms in group 16 obtain an octet by forming two covalent bonds:
Double and Triple Bonds
As previously mentioned, when a pair of atoms shares one pair of electrons, we call this a single bond. However, a pair of atoms may need to share more than one pair of electrons in order to achieve the requisite octet. A double bond forms when two pairs of electrons are shared between a pair of atoms, as between the carbon and oxygen atoms in CH2O (formaldehyde) and between the two carbon atoms in C2H4 (ethylene):
A triple bond forms when three electron pairs are shared by a pair of atoms, as in carbon monoxide (CO) and the cyanide ion (CN):
Writing Lewis Structures with the Octet Rule
For very simple molecules and molecular ions, we can write the Lewis structures by merely pairing up the unpaired electrons on the constituent atoms. See these examples:
For more complicated molecules and molecular ions, it is helpful to follow the step-by-step procedure outlined here.
Writing Lewis Structures
1. Determine the total number of valence (outer shell) electrons. For cations, subtract one electron for each positive charge. For anions, add one electron for each negative charge.
2. Draw a skeleton structure of the molecule or ion, arranging the atoms around a central atom. (Generally, the least electronegative element should be placed in the center.) Connect each atom to the central atom with a single bond (one electron pair).
3. Distribute the remaining electrons as lone pairs on the terminal atoms (except hydrogen), completing an octet around each atom.
4. Place all remaining electrons on the central atom.
5. Rearrange the electrons of the outer atoms to make multiple bonds with the central atom in order to obtain octets wherever possible.
Let us determine the Lewis structures of $\ce{SiH4}$, $\ce{CHO2^{-}}$, $\ce{NO^{+}}$, and $\ce{OF2}$ as examples in following this procedure:
1. Determine the total number of valence (outer shell) electrons in the molecule or ion.
• For a molecule, we add the number of valence electrons on each atom in the molecule:
\begin{align*} \ce{SiH4}& \[4pt] \text{Si: 4 valence electrons/atom} \times \text{ 1 atom } &= 4 \[4pt] + \quad \text{H : 1 valence electron/atom} \times \text{ 4 atoms} &=4 \[4pt] \hline \text{total valence electrons} &= 8 \end{align*} \nonumber
• For a negative ion, such as $\ce{CHO2^{−}}$, we add the number of valence electrons on the atoms to the number of negative charges on the ion (one electron is gained for each single negative charge):
\begin{align*} \ce{CHO2^{-}} & \[4pt] \text{C: 4 valence electrons/atom} \times \text{ 1 atom } &= 4 \[4pt] \text{H: 1 valence electron/atom} \times \text{ 1 atoms} &=1 \[4pt] \text{O: 6 valence electron/atom} \times \text{ 2 atoms} &=12 \[4pt] + \quad \quad \quad\quad \quad \quad \quad \quad \quad \quad \quad \quad \text{1 electron} &= 1 \[4pt] \hline \text{total valence electrons} &= 18 \end{align*} \nonumber
• For a positive ion, such as $\ce{NO^{+}}$, we add the number of valence electrons on the atoms in the ion and then subtract the number of positive charges on the ion (one electron is lost for each single positive charge) from the total number of valence electrons:
\begin{align*} \ce{NO^{+}}& \[4pt] \text{N: 5 valence electrons/atom} \times \text{ 1 atom } &= 5 \[4pt] \text{O: 6 valence electron/atom} \times \text{ 1 atoms} &=6 \[4pt] + \quad \text{e: -1 electron (positive charge)} \times \text{ 1 electron} &=-1 \[4pt] \hline \text{total valence electrons} &= 10 \end{align*} \nonumber
• Since $\ce{OF2}$ is a neutral molecule, we simply add the number of valence electrons:
\begin{align*} \ce{OF2}& \[4pt] \text{O: 6 valence electrons/atom} \times \text{ 1 atom } &= 6 \[4pt] + \quad \text{F: 7 valence electron/atom} \times \text{ 2 atoms} &=14 \[4pt] \hline \text{total valence electrons} &= 20 \end{align*} \nonumber
2. Draw a skeleton structure of the molecule or ion, arranging the atoms around a central atom and connecting each atom to the central atom with a single (one electron pair) bond. (Note that we denote ions with brackets around the structure, indicating the charge outside the brackets:)
When several arrangements of atoms are possible, as for $\ce{CHO2^{-}}$, we must use experimental evidence to choose the correct one. In general, the less electronegative elements are more likely to be central atoms. In $\ce{CHO2^{−}}$, the less electronegative carbon atom occupies the central position with the oxygen and hydrogen atoms surrounding it. Other examples include $\ce{P}$ in $\ce{POCl3}$, $\ce{S}$ in $\ce{SO2}$, and $\ce{Cl}$ in $\ce{ClO4^{-}}$. An exception is that hydrogen is almost never a central atom. As the most electronegative element, fluorine also cannot be a central atom.
3. Distribute the remaining electrons as lone pairs on the terminal atoms (except hydrogen) to complete their valence shells with an octet of electrons.
• There are no remaining electrons on $\ce{SiH4}$, so it is unchanged:
4. Place all remaining electrons on the central atom.
• For $\ce{SiH4}$, $\ce{CHO2^{−}}$, and $\ce{NO^{+}}$, there are no remaining electrons; we already placed all of the electrons determined in Step 1.
• For $\ce{OF2}$, we had 16 electrons remaining in Step 3, and we placed 12, leaving 4 to be placed on the central atom:
5. Rearrange the electrons of the outer atoms to make multiple bonds with the central atom in order to obtain octets wherever possible.
• $\ce{SiH4}$: Si already has an octet, so nothing needs to be done.
• $\ce{ CHO 2^{ −}}$: We have distributed the valence electrons as lone pairs on the oxygen atoms, but the carbon atom lacks an octet:
• $\ce{NO^{+}}$: For this ion, we added eight valence electrons, but neither atom has an octet. We cannot add any more electrons since we have already used the total that we found in Step 1, so we must move electrons to form a multiple bond:
This still does not produce an octet, so we must move another pair, forming a triple bond:
• In $\ce{OF2}$, each atom has an octet as drawn, so nothing changes.
Example $1$: Writing Lewis Structures
NASA’s Cassini-Huygens mission detected a large cloud of toxic hydrogen cyanide (HCN) on Titan, one of Saturn’s moons. Titan also contains ethane (H3CCH3), acetylene (HCCH), and ammonia (NH3). What are the Lewis structures of these molecules?
Solution
Step 1. Calculate the number of valence electrons.
HCN: (1 × 1) + (4 × 1) + (5 × 1) = 10
H3CCH3: (1 × 3) + (2 × 4) + (1 × 3) = 14
HCCH: (1 × 1) + (2 × 4) + (1 × 1) = 10
NH3: (5 × 1) + (3 × 1) = 8
Step 2 Draw a skeleton and connect the atoms with single bonds. Remember that H is never a central atom:
Step 3. Where needed, distribute electrons to the terminal atoms:
HCN: six electrons placed on N
H3CCH3: no electrons remain
HCCH: no terminal atoms capable of accepting electrons
NH3: no terminal atoms capable of accepting electrons
Step 4. Where needed, place remaining electrons on the central atom:
HCN: no electrons remain
H3CCH3: no electrons remain
HCCH: four electrons placed on carbon
NH3: two electrons placed on nitrogen
Step 5. Where needed, rearrange electrons to form multiple bonds in order to obtain an octet on each atom:
HCN: form two more C–N bonds
H3CCH3: all atoms have the correct number of electrons
HCCH: form a triple bond between the two carbon atoms
NH3: all atoms have the correct number of electrons
Exercise $1$
Both carbon monoxide, CO, and carbon dioxide, CO2, are products of the combustion of fossil fuels. Both of these gases also cause problems: CO is toxic and CO2 has been implicated in global climate change. What are the Lewis structures of these two molecules?
Answer
How Sciences Interconnect: Fullerene Chemistry
Carbon, in various forms and compounds, has been known since prehistoric times, . Soot has been used as a pigment (often called carbon black) for thousands of years. Charcoal, high in carbon content, has likewise been critical to human development. Carbon is the key additive to iron in the steelmaking process, and diamonds have a unique place in both culture and industry. With all this usage came significant study, particularly with the emergence of organic chemistry. And even with all the known forms and functions of the element, scientists began to uncover the potential for even more varied and extensive carbon structures.
As early as the 1960s, chemists began to observe complex carbon structures, but they had little evidence to support their concepts, or their work did not make it into the mainstream. Eiji Osawa predicted a spherical form based on observations of a similar structure, but his work was not widely known outside Japan. In a similar manner, the most comprehensive advance was likely computational chemist Elena Galpern's, who in 1973 predicted a highly stable, 60-carbon molecule; her work was also isolated to her native Russia. Still later, Harold Kroto, working with Canadian radio astronomers, sought to uncover the nature of long carbon chains that had been discovered in interstellar space.
Kroto sought to use a machine developed by Richard Smalley's team at Rice University to learn more about these structures. Together with Robert Curl, who had introduced them, and three graduate students—James Heath, Sean O’Brien, and Yuan Liu—they performed an intensive series of experiments that led to a major discovery.
In 1996, the Nobel Prize in Chemistry was awarded to Richard Smalley (Figure $3$), Robert Curl, and Harold Kroto for their work in discovering a new form of carbon, the C60 buckminsterfullerene molecule. An entire class of compounds, including spheres and tubes of various shapes, were discovered based on C60. This type of molecule, called a fullerene, shows promise in a variety of applications. Because of their size and shape, fullerenes can encapsulate other molecules, so they have shown potential in various applications from hydrogen storage to targeted drug delivery systems. They also possess unique electronic and optical properties that have been put to good use in solar powered devices and chemical sensors.
Exceptions to the Octet Rule
Many covalent molecules have central atoms that do not have eight electrons in their Lewis structures. These molecules fall into three categories:
• Odd-electron molecules have an odd number of valence electrons, and therefore have an unpaired electron.
• Electron-deficient molecules have a central atom that has fewer electrons than needed for a noble gas configuration.
• Hypervalent molecules have a central atom that has more electrons than needed for a noble gas configuration.
Odd-electron Molecules
We call molecules that contain an odd number of electrons free radicals. Nitric oxide, NO, is an example of an odd-electron molecule; it is produced in internal combustion engines when oxygen and nitrogen react at high temperatures.
To draw the Lewis structure for an odd-electron molecule like NO, we follow the same five steps we would for other molecules, but with a few minor changes:
1. Determine the total number of valence (outer shell) electrons. The sum of the valence electrons is 5 (from N) + 6 (from O) = 11. The odd number immediately tells us that we have a free radical, so we know that not every atom can have eight electrons in its valence shell.
2. Draw a skeleton structure of the molecule. We can easily draw a skeleton with an N–O single bond:
N–O
3. Distribute the remaining electrons as lone pairs on the terminal atoms. In this case, there is no central atom, so we distribute the electrons around both atoms. We give eight electrons to the more electronegative atom in these situations; thus oxygen has the filled valence shell:
4. Place all remaining electrons on the central atom. Since there are no remaining electrons, this step does not apply.
5. Rearrange the electrons to make multiple bonds with the central atom in order to obtain octets wherever possible. We know that an odd-electron molecule cannot have an octet for every atom, but we want to get each atom as close to an octet as possible. In this case, nitrogen has only five electrons around it. To move closer to an octet for nitrogen, we take one of the lone pairs from oxygen and use it to form a NO double bond. (We cannot take another lone pair of electrons on oxygen and form a triple bond because nitrogen would then have nine electrons:)
Electron-deficient Molecules
We will also encounter a few molecules that contain central atoms that do not have a filled valence shell. Generally, these are molecules with central atoms from groups 2 and 13, outer atoms that are hydrogen, or other atoms that do not form multiple bonds. For example, in the Lewis structures of beryllium dihydride, BeH2, and boron trifluoride, BF3, the beryllium and boron atoms each have only four and six electrons, respectively. It is possible to draw a structure with a double bond between a boron atom and a fluorine atom in BF3, satisfying the octet rule, but experimental evidence indicates the bond lengths are closer to that expected for B–F single bonds. This suggests the best Lewis structure has three B–F single bonds and an electron deficient boron. The reactivity of the compound is also consistent with an electron deficient boron. However, the B–F bonds are slightly shorter than what is actually expected for B–F single bonds, indicating that some double bond character is found in the actual molecule.
An atom like the boron atom in BF3, which does not have eight electrons, is very reactive. It readily combines with a molecule containing an atom with a lone pair of electrons. For example, NH3 reacts with BF3 because the lone pair on nitrogen can be shared with the boron atom:
Hypervalent Molecules
Elements in the second period of the periodic table (n = 2) can accommodate only eight electrons in their valence shell orbitals because they have only four valence orbitals (one 2s and three 2p orbitals). Elements in the third and higher periods (n ≥ 3) have more than four valence orbitals and can share more than four pairs of electrons with other atoms because they have empty d orbitals in the same shell. Molecules formed from these elements are sometimes called hypervalent molecules. Figure $4$: shows the Lewis structures for two hypervalent molecules, PCl5 and SF6.
In some hypervalent molecules, such as IF5 and XeF4, some of the electrons in the outer shell of the central atom are lone pairs:
When we write the Lewis structures for these molecules, we find that we have electrons left over after filling the valence shells of the outer atoms with eight electrons. These additional electrons must be assigned to the central atom.
Example $2$: Writing Lewis Structures - Octet Rule Violations
Xenon is a noble gas, but it forms a number of stable compounds. We examined XeF4 earlier. What are the Lewis structures of XeF2 and XeF6?
Solution
We can draw the Lewis structure of any covalent molecule by following the six steps discussed earlier. In this case, we can condense the last few steps, since not all of them apply.
Step 1. Calculate the number of valence electrons:
XeF2: 8 + (2 × 7) = 22
XeF6: 8 + (6 × 7) = 50
Step 2. Draw a skeleton joining the atoms by single bonds. Xenon will be the central atom because fluorine cannot be a central atom:
Step 3. Distribute the remaining electrons.
XeF2: We place three lone pairs of electrons around each F atom, accounting for 12 electrons and giving each F atom 8 electrons. Thus, six electrons (three lone pairs) remain. These lone pairs must be placed on the Xe atom. This is acceptable because Xe atoms have empty valence shell d orbitals and can accommodate more than eight electrons. The Lewis structure of XeF2 shows two bonding pairs and three lone pairs of electrons around the Xe atom:
XeF6: We place three lone pairs of electrons around each F atom, accounting for 36 electrons. Two electrons remain, and this lone pair is placed on the Xe atom:
Exercise $2$: Interhalogens
The halogens form a class of compounds called the interhalogens, in which halogen atoms covalently bond to each other. Write the Lewis structures for the interhalogens $\ce{BrCl3}$ and $\ce{ICl4^{-}}$.
Answer | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/07%3A_Chemical_Bonding_and_Molecular_Geometry/7.03%3A_Lewis_Symbols_and_Structures.txt |
Learning Objectives
By the end of this section, you will be able to:
• Compute formal charges for atoms in any Lewis structure
• Use formal charges to identify the most reasonable Lewis structure for a given molecule
• Explain the concept of resonance and draw Lewis structures representing resonance forms for a given molecule
In the previous section, we discussed how to write Lewis structures for molecules and polyatomic ions. As we have seen, however, in some cases, there is seemingly more than one valid structure for a molecule. We can use the concept of formal charges to help us predict the most appropriate Lewis structure when more than one is reasonable.
Calculating Formal Charge
The formal charge of an atom in a molecule is the hypothetical charge the atom would have if we could redistribute the electrons in the bonds evenly between the atoms. Another way of saying this is that formal charge results when we take the number of valence electrons of a neutral atom, subtract the nonbonding electrons, and then subtract the number of bonds connected to that atom in the Lewis structure.
Thus, we calculate formal charge as follows:
$\text { formal charge }=\# \text { valence shell electrons (free atom) }-\# \text { lone pair electrons }-\frac{1}{2} \# \text { bonding electrons } \nonumber$
We can double-check formal charge calculations by determining the sum of the formal charges for the whole structure. The sum of the formal charges of all atoms in a molecule must be zero; the sum of the formal charges in an ion should equal the charge of the ion.
We must remember that the formal charge calculated for an atom is not the actual charge of the atom in the molecule. Formal charge is only a useful bookkeeping procedure; it does not indicate the presence of actual charges.
Example $1$: Calculating Formal Charge from Lewis Structures
Assign formal charges to each atom in the interhalogen ion
Solution
Step 1. We divide the bonding electron pairs equally for all I–Cl bonds:
Step 2. We assign lone pairs of electrons to their atoms. Each Cl atom now has seven electrons assigned to it, and the I atom has eight.
Step 3. Subtract this number from the number of valence electrons for the neutral atom:
I: 7 – 8 = –1
Cl: 7 – 7 = 0
The sum of the formal charges of all the atoms equals –1, which is identical to the charge of the ion (–1).
Exercise $1$
Calculate the formal charge for each atom in the carbon monoxide molecule:
Answer
C −1, O +1
Example $2$: Calculating Formal Charge from Lewis Structures
Assign formal charges to each atom in the interhalogen molecule BrCl3.
Solution
Step 1. Assign one of the electrons in each Br–Cl bond to the Br atom and one to the Cl atom in that bond:
Step 2. Assign the lone pairs to their atom. Now each Cl atom has seven electrons and the Br atom has seven electrons.
Step 3. Subtract this number from the number of valence electrons for the neutral atom. This gives the formal charge:
Br: 7 – 7 = 0
Cl: 7 – 7 = 0
All atoms in BrCl3 have a formal charge of zero, and the sum of the formal charges totals zero, as it must in a neutral molecule.
Exercise $2$
Determine the formal charge for each atom in NCl3.
Answer
N: 0; all three Cl atoms: 0
Using Formal Charge to Predict Molecular Structure
The arrangement of atoms in a molecule or ion is called its molecular structure. In many cases, following the steps for writing Lewis structures may lead to more than one possible molecular structure—different multiple bond and lone-pair electron placements or different arrangements of atoms, for instance. A few guidelines involving formal charge can be helpful in deciding which of the possible structures is most likely for a particular molecule or ion:
1. A molecular structure in which all formal charges are zero is preferable to one in which some formal charges are not zero.
2. If the Lewis structure must have nonzero formal charges, the arrangement with the smallest nonzero formal charges is preferable.
3. Lewis structures are preferable when adjacent formal charges are zero or of the opposite sign.
4. When we must choose among several Lewis structures with similar distributions of formal charges, the structure with the negative formal charges on the more electronegative atoms is preferable.
To see how these guidelines apply, let us consider some possible structures for carbon dioxide, CO2. We know from our previous discussion that the less electronegative atom typically occupies the central position, but formal charges allow us to understand why this occurs. We can draw three possibilities for the structure: carbon in the center and double bonds, carbon in the center with a single and triple bond, and oxygen in the center with double bonds:
Comparing the three formal charges, we can definitively identify the structure on the left as preferable because it has only formal charges of zero (Guideline 1).
As another example, the thiocyanate ion, an ion formed from a carbon atom, a nitrogen atom, and a sulfur atom, could have three different molecular structures: NCS, CNS, or CSN. The formal charges present in each of these molecular structures can help us pick the most likely arrangement of atoms. Possible Lewis structures and the formal charges for each of the three possible structures for the thiocyanate ion are shown here:
Note that the sum of the formal charges in each case is equal to the charge of the ion (–1). However, the first arrangement of atoms is preferred because it has the lowest number of atoms with nonzero formal charges (Guideline 2). Also, it places the least electronegative atom in the center, and the negative charge on the more electronegative element (Guideline 4).
Example $3$: Using Formal Charge to Determine Molecular Structure
Nitrous oxide, N2O, commonly known as laughing gas, is used as an anesthetic in minor surgeries, such as the routine extraction of wisdom teeth. Which is the more likely structure for nitrous oxide?
Solution
Determining formal charge yields the following:
The structure with a terminal oxygen atom best satisfies the criteria for the most stable distribution of formal charge:
The number of atoms with formal charges are minimized (Guideline 2), there is no formal charge with a magnitude greater than one (Guideline 2), the negative formal charge is on the more electronegative element (Guideline 4), and the less electronegative atom is in the center position.
Exercise $3$
Which is the most likely molecular structure for the nitrite ion?
Answer
ONO
Resonance
Notice that the more likely structure for the nitrite anion in Example $3$ may actually be drawn in two different ways, distinguished by the locations of the N-O and N=O bonds:
If nitrite ions do indeed contain a single and a double bond, we would expect for the two bond lengths to be different. A double bond between two atoms is shorter (and stronger) than a single bond between the same two atoms. Experiments show, however, that both N–O bonds in have the same strength and length, and are identical in all other properties.
It is not possible to write a single Lewis structure for in which nitrogen has an octet and both bonds are equivalent. Instead, we use the concept of resonance: if two or more Lewis structures with the same arrangement of atoms can be written for a molecule or ion, the actual distribution of electrons is an average of that shown by the various Lewis structures. The actual distribution of electrons in each of the nitrogen-oxygen bonds in is the average of a double bond and a single bond. We call the individual Lewis structures resonance forms. The actual electronic structure of the molecule (the average of the resonance forms) is called a resonance hybrid of the individual resonance forms. A double-headed arrow between Lewis structures indicates that they are resonance forms.
We should remember that a molecule described as a resonance hybrid never possesses an electronic structure described by either resonance form. It does not fluctuate between resonance forms; rather, the actual electronic structure is always the average of that shown by all resonance forms. George Wheland, one of the pioneers of resonance theory, used a historical analogy to describe the relationship between resonance forms and resonance hybrids. A medieval traveler, having never before seen a rhinoceros, described it as a hybrid of a dragon and a unicorn because it had many properties in common with both. Just as a rhinoceros is neither a dragon sometimes nor a unicorn at other times, a resonance hybrid is neither of its resonance forms at any given time. Like a rhinoceros, it is a real entity that experimental evidence has shown to exist. It has some characteristics in common with its resonance forms, but the resonance forms themselves are convenient, imaginary images (like the unicorn and the dragon).
The carbonate anion, provides a second example of resonance:
One oxygen atom must have a double bond to carbon to complete the octet on the central atom. All oxygen atoms, however, are equivalent, and the double bond could form from any one of the three atoms. This gives rise to three resonance forms of the carbonate ion. Because we can write three identical resonance structures, we know that the actual arrangement of electrons in the carbonate ion is the average of the three structures. Again, experiments show that all three C–O bonds are exactly the same.
Link to Learning
Use this online quiz to practice your skills in drawing resonance structures and estimating formal charges. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/07%3A_Chemical_Bonding_and_Molecular_Geometry/7.04%3A_Formal_Charges_and_Resonance.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the energetics of covalent and ionic bond formation and breakage
• Use the Born-Haber cycle to compute lattice energies for ionic compounds
• Use average covalent bond energies to estimate enthalpies of reaction
A bond’s strength describes how strongly each atom is joined to another atom, and therefore how much energy is required to break the bond between the two atoms. In this section, you will learn about the bond strength of covalent bonds, and then compare that to the strength of ionic bonds, which is related to the lattice energy of a compound.
Bond Strength: Covalent Bonds
Stable molecules exist because covalent bonds hold the atoms together. We measure the strength of a covalent bond by the energy required to break it, that is, the energy necessary to separate the bonded atoms. Separating any pair of bonded atoms requires energy. The stronger a bond, the greater the energy required to break it.
The energy required to break a specific covalent bond in one mole of gaseous molecules is called the bond energy or the bond dissociation energy. The bond energy for a diatomic molecule, $D_{\ce{X–Y}}$, is defined as the standard enthalpy change for the endothermic reaction:
$XY (g) \longrightarrow X (g)+ Y (g) \quad D_{\ce{X-Y}}=\Delta H^{\circ} \nonumber$
For example, the bond energy of the pure covalent H–H bond, DH–H, is 436 kJ per mole of H–H bonds broken:
$H_2(g) \longrightarrow 2 H (g) \quad D_{\ce{H-H}}=\Delta H^{\circ}=436\,\text{kJ} \nonumber$
Molecules with three or more atoms have two or more bonds. The sum of all bond energies in such a molecule is equal to the standard enthalpy change for the endothermic reaction that breaks all the bonds in the molecule. For example, the sum of the four C–H bond energies in CH4, 1660 kJ, is equal to the standard enthalpy change of the reaction:
The average C–H bond energy, DC–H, is 1660/4 = 415 kJ/mol because there are four moles of C–H bonds broken per mole of the reaction. Although the four C–H bonds are equivalent in the original molecule, they do not each require the same energy to break; once the first bond is broken (which requires 439 kJ/mol), the remaining bonds are easier to break. The 415 kJ/mol value is the average, not the exact value required to break any one bond.
The strength of a bond between two atoms increases as the number of electron pairs in the bond increases. Generally, as the bond strength increases, the bond length decreases. Thus, we find that triple bonds are stronger and shorter than double bonds between the same two atoms; likewise, double bonds are stronger and shorter than single bonds between the same two atoms. Average bond energies for some common bonds appear in Table $1$, and a comparison of bond lengths and bond strengths for some common bonds appears in Table $2$. When one atom bonds to various atoms in a group, the bond strength typically decreases as we move down the group. For example, C–F is 439 kJ/mol, C–Cl is 330 kJ/mol, and C–Br is 275 kJ/mol.
Table $1$: Bond Energies (kJ/mol)
Bond Bond Energy Bond Bond Energy Bond Bond Energy
H–H 436 C–S 260 F–Cl 255
H–C 415 C–Cl 330 F–Br 235
H–N 390 C–Br 275 Si–Si 230
H–O 464 C–I 240 Si–P 215
H–F 569 N–N 160 Si–S 225
H–Si 395 418 Si–Cl 359
H–P 320 946 Si–Br 290
H–S 340 N–O 200 Si–I 215
H–Cl 432 N–F 270 P–P 215
H–Br 370 N–P 210 P–S 230
H–I 295 N–Cl 200 P–Cl 330
C–C 345 N–Br 245 P–Br 270
611 O–O 140 P–I 215
837 498 S–S 215
C–N 290 O–F 160 S–Cl 250
615 O–Si 370 S–Br 215
891 O–P 350 Cl–Cl 243
C–O 350 O–Cl 205 Cl–Br 220
741 O–I 200 Cl–I 210
1080 F–F 160 Br–Br 190
C–F 439 F–Si 540 Br–I 180
C–Si 360 F–P 489 I–I 150
C–P 265 F–S 285
Table $2$: Average Bond Lengths and Bond Energies for Some Common Bonds
Bond Bond Length (Å) Bond Energy (kJ/mol)
C–C 1.54 345
1.34 611
1.20 837
C–N 1.43 290
1.38 615
1.16 891
C–O 1.43 350
1.23 741
1.13 1080
We can use bond energies to calculate approximate enthalpy changes for reactions where enthalpies of formation are not available. Calculations of this type will also tell us whether a reaction is exothermic or endothermic. An exothermic reaction (ΔH negative, heat produced) results when the bonds in the products are stronger than the bonds in the reactants. An endothermic reaction (ΔH positive, heat absorbed) results when the bonds in the products are weaker than those in the reactants.
The enthalpy change, ΔH, for a chemical reaction is approximately equal to the sum of the energy required to break all bonds in the reactants (energy “in”, positive sign) plus the energy released when all bonds are formed in the products (energy “out,” negative sign). This can be expressed mathematically in the following way:
$\Delta H=\sum D_{\text {bonds broken }}-\sum D_{\text {bonds formed }} \nonumber$
In this expression, the symbol Ʃ means “the sum of” and D represents the bond energy in kilojoules per mole, which is always a positive number. The bond energy is obtained from a table (like Table $2$) and will depend on whether the particular bond is a single, double, or triple bond. Thus, in calculating enthalpies in this manner, it is important that we consider the bonding in all reactants and products. Because D values are typically averages for one type of bond in many different molecules, this calculation provides a rough estimate, not an exact value, for the enthalpy of reaction.
Consider the following reaction:
$\ce{H2(g) + Cl2(g) -> 2 HCl(g)} \nonumber$
or
$\ce{H-H(g) + Cl-Cl(g) -> 2 H-Cl(g)} \nonumber$
To form two moles of HCl, one mole of H–H bonds and one mole of Cl–Cl bonds must be broken. The energy required to break these bonds is the sum of the bond energy of the H–H bond (436 kJ/mol) and the Cl–Cl bond (243 kJ/mol). During the reaction, two moles of H–Cl bonds are formed (bond energy = 432 kJ/mol), releasing 2 432 kJ; or 864 kJ. Because the bonds in the products are stronger than those in the reactants, the reaction releases more energy than it consumes:
\begin{aligned} \Delta H & =\sum D_{\text {bonds broken }}-\sum D_{\text {bonds formed }} \[4pt] & =\left[ D_{\ce{H-H}} + D_{\ce{Cl-Cl|} \right] - 2 D_{\ce{H-Cl}} \[4pt] & =[436+243]-2(432)=-185 kJ \end{aligned} \nonumber
This excess energy is released as heat, so the reaction is exothermic. Appendix G gives a value for the standard molar enthalpy of formation of HCl(g), $\Delta H^o_f$ of –92.307 kJ/mol. Twice that value is –184.6 kJ, which agrees well with the answer obtained earlier for the formation of two moles of HCl.
Example $1$: Using Bond Energies to Calculate Approximate Enthalpy Changes
Methanol, CH3OH, may be an excellent alternative fuel. The high-temperature reaction of steam and carbon produces a mixture of the gases carbon monoxide, CO, and hydrogen, H2, from which methanol can be produced. Using the bond energies in Table $2$, calculate the approximate enthalpy change, ΔH, for the reaction here:
$\ce{CO(g) + 2 H2(g) -> CH3OH (g)} \nonumber$
Solution
First, we need to write the Lewis structures of the reactants and the products:
From this, we see that ΔH for this reaction involves the energy required to break a C–O triple bond and two H–H single bonds, as well as the energy produced by the formation of three C–H single bonds, a C–O single bond, and an O–H single bond. We can express this as follows:
\begin{aligned} \Delta H & =\Sigma D_{\text {bonds broken }}-\Sigma D_{\text {bonds formed }} \[4pt] & =\left[ D_{ C \equiv O }+2\left( D_{ H - H }\right)\right]-\left[3\left( D_{ C - H }\right)+ D_{ C - O }+ D_{ O - H }\right] \end{aligned} \nonumber
Using the bond energy values in Table $2$, we obtain:
\begin{aligned} \Delta H & =[1080+2(436)]-[3(415)+350+464] \[4pt] & =-107 kJ \end{aligned} \nonumber
We can compare this value to the value calculated based on data from Appendix G:
\begin{aligned} \Delta H & =\left[\Delta H_{ f }^{\circ} CH_3 OH (g)\right]-\left[\Delta H_{ f }^{\circ} CO (g)+2 \times \Delta H_{ f }^{\circ} H_2\right] \[4pt] & =[-201.0]-[-110.52+2 \times 0] \[4pt] & =-90.5 kJ \end{aligned} \nonumber
Note that there is a fairly significant gap between the values calculated using the two different methods. This occurs because D values are the average of different bond strengths; therefore, they often give only rough agreement with other data.
Exercise $1$
Ethyl alcohol, CH3CH2OH, was one of the first organic chemicals deliberately synthesized by humans. It has many uses in industry, and it is the alcohol contained in alcoholic beverages. It can be obtained by the fermentation of sugar or synthesized by the hydration of ethylene in the following reaction:
Using the bond energies in Table $2$, calculate an approximate enthalpy change, ΔH, for this reaction.
Answer
–35 kJ
Ionic Bond Strength and Lattice Energy
An ionic compound is stable because of the electrostatic attraction between its positive and negative ions. The lattice energy of a compound is a measure of the strength of this attraction. The lattice energy ($ΔH_{lattice}$) of an ionic compound is defined as the energy required to separate one mole of the solid into its component gaseous ions. For the ionic solid MX, the lattice energy is the enthalpy change of the process:
$MX (s) \longrightarrow M^{n+}(g)+ X^{n-}(g) \quad \Delta H_{\text {lattice }} \nonumber$
Note that we are using the convention where the ionic solid is separated into ions, so our lattice energies will be endothermic (positive values). Some texts use the equivalent but opposite convention, defining lattice energy as the energy released when separate ions combine to form a lattice and giving negative (exothermic) values. Thus, if you are looking up lattice energies in another reference, be certain to check which definition is being used. In both cases, a larger magnitude for lattice energy indicates a more stable ionic compound. For sodium chloride, ΔHlattice = 769 kJ. Thus, it requires 769 kJ to separate one mole of solid NaCl into gaseous Na+ and Cl ions. When one mole each of gaseous Na+ and Cl ions form solid NaCl, 769 kJ of heat is released.
The lattice energy ΔHlattice of an ionic crystal can be expressed by the following equation (derived from Coulomb’s law, governing the forces between electric charges):
$\Delta H_{\text {lattice }}=\frac{ C \left( Z^{+}\right)\left( Z^{-}\right)}{ R_0} \nonumber$
in which C is a constant that depends on the type of crystal structure; Z+ and Z are the charges on the ions; and Ro is the interionic distance (the sum of the radii of the positive and negative ions). Thus, the lattice energy of an ionic crystal increases rapidly as the charges of the ions increase and the sizes of the ions decrease. When all other parameters are kept constant, doubling the charge of both the cation and anion quadruples the lattice energy. For example, the lattice energy of LiF (Z+ and Z = 1) is 1023 kJ/mol, whereas that of MgO (Z+ and Z = 2) is 3900 kJ/mol (Ro is nearly the same—about 200 pm for both compounds).
Different interatomic distances produce different lattice energies. For example, we can compare the lattice energy of MgF2 (2957 kJ/mol) to that of MgI2 (2327 kJ/mol) to observe the effect on lattice energy of the smaller ionic size of F as compared to I.
Example $2$: Lattice Energy Comparisons
The precious gem ruby is aluminum oxide, Al2O3, containing traces of Cr3+. The compound Al2Se3 is used in the fabrication of some semiconductor devices. Which has the larger lattice energy, Al2O3 or Al2Se3?
Solution
In these two ionic compounds, the charges Z+ and Z are the same, so the difference in lattice energy will depend upon Ro. The O2– ion is smaller than the Se2 ion. Thus, Al2O3 would have a shorter interionic distance than Al2Se3, and Al2O3 would have the larger lattice energy.
Exercise $1$
Zinc oxide, ZnO, is a very effective sunscreen. How would the lattice energy of ZnO compare to that of NaCl?
Answer
ZnO would have the larger lattice energy because the Z values of both the cation and the anion in ZnO are greater, and the interionic distance of ZnO is smaller than that of NaCl.
The Born-Haber Cycle
It is not possible to measure lattice energies directly. However, the lattice energy can be calculated using the equation given in the previous section or by using a thermochemical cycle. The Born-Haber cycle is an application of Hess’s law that breaks down the formation of an ionic solid into a series of individual steps:
• $\Delta H_{ f }^{\circ}$ the standard enthalpy of formation of the compound
• IE, the ionization energy of the metal
• EA, the electron affinity of the nonmetal
• $\Delta H_{s}^{\circ}$ the enthalpy of sublimation of the metal
• $D$, the bond dissociation energy of the nonmetal
• $ΔH_{lattice}$, the lattice energy of the compound
Figure $1$: diagrams the Born-Haber cycle for the formation of solid cesium fluoride.
We begin with the elements in their most common states, $\ce{Cs(s)}$ and $\ce{F2(g)}$. The represents the conversion of solid cesium into a gas, and then the ionization energy converts the gaseous cesium atoms into cations. In the next step, we account for the energy required to break the F–F bond to produce fluorine atoms. Converting one mole of fluorine atoms into fluoride ions is an exothermic process, so this step gives off energy (the electron affinity) and is shown as decreasing along the y-axis. We now have one mole of Cs cations and one mole of F anions. These ions combine to produce solid cesium fluoride. The enthalpy change in this step is the negative of the lattice energy, so it is also an exothermic quantity. The total energy involved in this conversion is equal to the experimentally determined enthalpy of formation, $\Delta H_{ f }^{\circ}$ of the compound from its elements. In this case, the overall change is exothermic.
Hess’s law can also be used to show the relationship between the enthalpies of the individual steps and the enthalpy of formation. Table $3$ shows this for fluoride, CsF.
Table $3$
Enthalpy of sublimation of $\ce{Cs(s)}$ $\ce{Cs(s) -> Cs (g)}$ $\Delta H=\Delta H_s^{\circ}=76.5\,\text{kJ} / \text{mol}$
One-half of the bond energy of $\ce{F2}$ $\ce{ 1/2 F2(g) -> F(g)}$ $\Delta H=\frac{1}{2} D=79.4\, \text{kJ} / \text{mol}$
Ionization energy of $\ce{Cs(g)}$ $\ce{Cs(g) -> Cs^{+}(g) + e^{-}}$ $\Delta H=IE=375.7\, \text{kJ} / \text{mol}$
Electron affinity of $\ce{F}$ $\ce{F(g) + e^{-} -> F^{-}(g)}$ $\Delta H=EA=-328.2\, \text{kJ} / \text{mol}$
Negative of the lattice energy of $\ce{CsF(s)}$ $\ce{Cs^{+}(g) + F^{-}(g) -> CsF(s)}$ $\Delta H=-\Delta H_{\text {lattice }}=?$
Enthalpy of formation of $\ce{CsF(s)}$, add steps 1–5
$\Delta H=\Delta H_f^{\circ}=\Delta H_s^{\circ}+\frac{1}{2} D+I E+(E A)+\left(-\Delta H_{\text {lattice }}\right)$
$\ce{Cs(s) + 1/2 F2(g) -> CsF(s)}$
$\Delta H=-553.5 \, \text{kJ} / \text{mol}$
Thus, the lattice energy can be calculated from other values. For cesium fluoride, using this data, the lattice energy is:
$\Delta H_{\text {lattice }}=76.5+79.4+375.7+(-328.2)-(-553.5)=756.9\, \text{kJ} / \text{mol} \nonumber$
The Born-Haber cycle may also be used to calculate any one of the other quantities in the equation for lattice energy, provided that the remainder is known. For example, if the relevant enthalpy of sublimation $\Delta H_s^{\circ}$, ionization energy (IE), bond dissociation enthalpy (D), lattice energy ΔHlattice, and standard enthalpy of formation are known, the Born-Haber cycle can be used to determine the electron affinity of an atom.
Lattice energies calculated for ionic compounds are typically much higher than bond dissociation energies measured for covalent bonds. Whereas lattice energies typically fall in the range of 600–4000 kJ/mol (some even higher), covalent bond dissociation energies are typically between 150–400 kJ/mol for single bonds. Keep in mind, however, that these are not directly comparable values. For ionic compounds, lattice energies are associated with many interactions, as cations and anions pack together in an extended lattice. For covalent bonds, the bond dissociation energy is associated with the interaction of just two atoms. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/07%3A_Chemical_Bonding_and_Molecular_Geometry/7.05%3A_Strengths_of_Ionic_and_Covalent_Bonds.txt |
Learning Objectives
By the end of this section, you will be able to:
• Predict the structures of small molecules using valence shell electron pair repulsion (VSEPR) theory
• Explain the concepts of polar covalent bonds and molecular polarity
• Assess the polarity of a molecule based on its bonding and structure
Thus far, we have used two-dimensional Lewis structures to represent molecules. However, molecular structure is actually three-dimensional, and it is important to be able to describe molecular bonds in terms of their distances, angles, and relative arrangements in space (Figure $1$). A bond angle is the angle between any two bonds that include a common atom, usually measured in degrees. A bond distance (or bond length) is the distance between the nuclei of two bonded atoms along the straight line joining the nuclei. Bond distances are measured in Ångstroms (1 Å = 10–10 m) or picometers (1 pm = 10–12 m, 100 pm = 1 Å).
VSEPR Theory
Valence shell electron-pair repulsion theory (VSEPR theory) enables us to predict the molecular structure, including approximate bond angles around a central atom, of a molecule from an examination of the number of bonds and lone electron pairs in its Lewis structure. The VSEPR model assumes that electron pairs in the valence shell of a central atom will adopt an arrangement that minimizes repulsions between these electron pairs by maximizing the distance between them. The electrons in the valence shell of a central atom form either bonding pairs of electrons, located primarily between bonded atoms, or lone pairs. The electrostatic repulsion of these electrons is reduced when the various regions of high electron density assume positions as far from each other as possible.
VSEPR theory predicts the arrangement of electron pairs around each central atom and, usually, the correct arrangement of atoms in a molecule. We should understand, however, that the theory only considers electron-pair repulsions. Other interactions, such as nuclear-nuclear repulsions and nuclear-electron attractions, are also involved in the final arrangement that atoms adopt in a particular molecular structure.
As a simple example of VSEPR theory, let us predict the structure of a gaseous BeF2 molecule. The Lewis structure of BeF2 (Figure $2$) shows only two electron pairs around the central beryllium atom. With two bonds and no lone pairs of electrons on the central atom, the bonds are as far apart as possible, and the electrostatic repulsion between these regions of high electron density is reduced to a minimum when they are on opposite sides of the central atom. The bond angle is 180° (Figure $2$).
Figure $3$ illustrates this and other electron-pair geometries that minimize the repulsions among regions of high electron density (bonds and/or lone pairs). Two regions of electron density around a central atom in a molecule form a linear geometry; three regions form a trigonal planar geometry; four regions form a tetrahedral geometry; five regions form a trigonal bipyramidal geometry; and six regions form an octahedral geometry.
Electron-pair Geometry versus Molecular Structure
It is important to note that electron-pair geometry around a central atom is not the same thing as its molecular structure. The electron-pair geometries shown in Figure $3$ describe all regions where electrons are located, bonds as well as lone pairs. Molecular structure describes the location of the atoms, not the electrons.
We differentiate between these two situations by naming the geometry that includes all electron pairs the electron-pair geometry. The structure that includes only the placement of the atoms in the molecule is called the molecular structure. The electron-pair geometries will be the same as the molecular structures when there are no lone electron pairs around the central atom, but they will be different when there are lone pairs present on the central atom.
For example, the methane molecule, CH4, which is the major component of natural gas, has four bonding pairs of electrons around the central carbon atom; the electron-pair geometry is tetrahedral, as is the molecular structure (Figure $4$). On the other hand, the ammonia molecule, NH3, also has four electron pairs associated with the nitrogen atom, and thus has a tetrahedral electron-pair geometry. One of these regions, however, is a lone pair, which is not included in the molecular structure, and this lone pair influences the shape of the molecule (Figure $5$).
As seen in Figure $5$, small distortions from the ideal angles in Figure $3$ can result from differences in repulsion between various regions of electron density. VSEPR theory predicts these distortions by establishing an order of repulsions and an order of the amount of space occupied by different kinds of electron pairs. The order of electron-pair repulsions from greatest to least repulsion is:
$\text { lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair } \nonumber$
This order of repulsions determines the amount of space occupied by different regions of electrons. A lone pair of electrons occupies a larger region of space than the electrons in a triple bond; in turn, electrons in a triple bond occupy more space than those in a double bond, and so on. The order of sizes from largest to smallest is:
$\text { lone pair }>\text { triple bond }>\text { double bond }>\text { single bond } \nonumber$
Consider formaldehyde, H2CO, which is used as a preservative for biological and anatomical specimens (Figure $1$). This molecule has regions of high electron density that consist of two single bonds and one double bond. The basic geometry is trigonal planar with 120° bond angles, but we see that the double bond causes slightly larger angles (121°), and the angle between the single bonds is slightly smaller (118°).
In the ammonia molecule, the three hydrogen atoms attached to the central nitrogen are not arranged in a flat, trigonal planar molecular structure, but rather in a three-dimensional trigonal pyramid (Figure $5$) with the nitrogen atom at the apex and the three hydrogen atoms forming the base. The ideal bond angles in a trigonal pyramid are based on the tetrahedral electron pair geometry. Again, there are slight deviations from the ideal because lone pairs occupy larger regions of space than do bonding electrons. The H–N–H bond angles in NH3 are slightly smaller than the 109.5° angle in a regular tetrahedron (Figure $3$) because the lone pair-bonding pair repulsion is greater than the bonding pair-bonding pair repulsion (Figure $5$). Figure $6$ illustrates the ideal molecular structures, which are predicted based on the electron-pair geometries for various combinations of lone pairs and bonding pairs.
According to VSEPR theory, the terminal atom locations (Xs in Figure $6$) are equivalent within the linear, trigonal planar, and tetrahedral electron-pair geometries (the first three rows of the table). It does not matter which X is replaced with a lone pair because the molecules can be rotated to convert positions. For trigonal bipyramidal electron-pair geometries, however, there are two distinct X positions, as shown in Figure $7$ an axial position (if we hold a model of a trigonal bipyramid by the two axial positions, we have an axis around which we can rotate the model) and an equatorial position (three positions form an equator around the middle of the molecule). As shown in Figure $6$, the axial position is surrounded by bond angles of 90°, whereas the equatorial position has more space available because of the 120° bond angles. In a trigonal bipyramidal electron-pair geometry, lone pairs always occupy equatorial positions because these more spacious positions can more easily accommodate the larger lone pairs.
Theoretically, we can come up with three possible arrangements for the three bonds and two lone pairs for the ClF3 molecule (Figure $7$). The stable structure is the one that puts the lone pairs in equatorial locations, giving a T-shaped molecular structure.
When a central atom has two lone electron pairs and four bonding regions, we have an octahedral electron-pair geometry. The two lone pairs are on opposite sides of the octahedron (180° apart), giving a square planar molecular structure that minimizes lone pair-lone pair repulsions (Figure $6$).
Predicting Electron Pair Geometry and Molecular Structure
The following procedure uses VSEPR theory to determine the electron pair geometries and the molecular structures:
1. Write the Lewis structure of the molecule or polyatomic ion.
2. Count the number of regions of electron density (lone pairs and bonds) around the central atom. A single, double, or triple bond counts as one region of electron density.
3. Identify the electron-pair geometry based on the number of regions of electron density: linear, trigonal planar, tetrahedral, trigonal bipyramidal, or octahedral (Figure $6$, first column).
4. Use the number of lone pairs to determine the molecular structure (Figure $6$). If more than one arrangement of lone pairs and chemical bonds is possible, choose the one that will minimize repulsions, remembering that lone pairs occupy more space than multiple bonds, which occupy more space than single bonds. In trigonal bipyramidal arrangements, repulsion is minimized when every lone pair is in an equatorial position. In an octahedral arrangement with two lone pairs, repulsion is minimized when the lone pairs are on opposite sides of the central atom.
The following examples illustrate the use of VSEPR theory to predict the molecular structure of molecules or ions that have no lone pairs of electrons. In this case, the molecular structure is identical to the electron pair geometry.
Example $1$: Predicting Electron-pair Geometry and Molecular Structure: CO2 and BCl3
Predict the electron-pair geometry and molecular structure for each of the following:
1. carbon dioxide, CO2, a molecule produced by the combustion of fossil fuels
2. boron trichloride, BCl3, an important industrial chemical
Solution
(a) We write the Lewis structure of CO2 as:
This shows us two regions of high electron density around the carbon atom—each double bond counts as one region, and there are no lone pairs on the carbon atom. Using VSEPR theory, we predict that the two regions of electron density arrange themselves on opposite sides of the central atom with a bond angle of 180°. The electron-pair geometry and molecular structure are identical, and CO2 molecules are linear.
(b) We write the Lewis structure of BCl3 as:
Thus we see that BCl3 contains three bonds, and there are no lone pairs of electrons on boron. The arrangement of three regions of high electron density gives a trigonal planar electron-pair geometry. The B–Cl bonds lie in a plane with 120° angles between them. BCl3 also has a trigonal planar molecular structure (Figure $8$).
The electron-pair geometry and molecular structure of BCl3 are both trigonal planar. Note that the VSEPR geometry indicates the correct bond angles (120°), unlike the Lewis structure shown above.
Exercise $1$
Carbonate, is a common polyatomic ion found in various materials from eggshells to antacids. What are the electron-pair geometry and molecular structure of this polyatomic ion?
Answer
The electron-pair geometry is trigonal planar and the molecular structure is trigonal planar. Due to resonance, all three C–O bonds are identical. Whether they are single, double, or an average of the two, each bond counts as one region of electron density.
Example $2$: Predicting Electron-pair Geometry and Molecular Structure: Ammonium
Two of the top 50 chemicals produced in the United States, ammonium nitrate and ammonium sulfate, both used as fertilizers, contain the ammonium ion. Predict the electron-pair geometry and molecular structure of the cation.
Solution
We write the Lewis structure of as:
We can see that $\ce{NH4^{+}}$ contains four bonds from the nitrogen atom to hydrogen atoms and no lone pairs. We expect the four regions of high electron density to arrange themselves so that they point to the corners of a tetrahedron with the central nitrogen atom in the middle (Figure $6$). Therefore, the electron pair geometry of Figure $9$.
Exercise $2$
Identify a molecule with trigonal bipyramidal molecular structure.
Answer
Any molecule with five electron pairs around the central atoms including no lone pairs will be trigonal bipyramidal. PF5 is a common example.
The next several examples illustrate the effect of lone pairs of electrons on molecular structure.
Example $3$: Predicting Electron-pair Geometry and Molecular Structure: Lone Pairs on the Central Atom
Predict the electron-pair geometry and molecular structure of a water molecule.
Solution
The Lewis structure of H2O indicates that there are four regions of high electron density around the oxygen atom: two lone pairs and two chemical bonds:
We predict that these four regions are arranged in a tetrahedral fashion (Figure $10$), as indicated in Figure $6$. Thus, the electron-pair geometry is tetrahedral and the molecular structure is bent with an angle slightly less than 109.5°. In fact, the bond angle is 104.5°.
Exercise $3$
The hydronium ion, H3O+, forms when acids are dissolved in water. Predict the electron-pair geometry and molecular structure of this cation.
Answer
electron pair geometry: tetrahedral; molecular structure: trigonal pyramidal
Example $4$: Predicting Electron-pair Geometry and Molecular Structure: SF4
Sulfur tetrafluoride, SF4, is extremely valuable for the preparation of fluorine-containing compounds used as herbicides (i.e., SF4 is used as a fluorinating agent). Predict the electron-pair geometry and molecular structure of a SF4 molecule.
Solution
The Lewis structure of SF4 indicates five regions of electron density around the sulfur atom: one lone pair and four bonding pairs:
We expect these five regions to adopt a trigonal bipyramidal electron-pair geometry. To minimize lone pair repulsions, the lone pair occupies one of the equatorial positions. The molecular structure (Figure $11$) is that of a seesaw (Figure $6$).
Exercise $4$
Predict the electron pair geometry and molecular structure for molecules of XeF2.
Answer
The electron-pair geometry is trigonal bipyramidal. The molecular structure is linear.
Example $5$: Predicting Electron-pair Geometry and Molecular Structure: XeF4
Of all the noble gases, xenon is the most reactive, frequently reacting with elements such as oxygen and fluorine. Predict the electron-pair geometry and molecular structure of the XeF4 molecule.
Solution
The Lewis structure of XeF4 indicates six regions of high electron density around the xenon atom: two lone pairs and four bonds:
These six regions adopt an octahedral arrangement (Figure $6$), which is the electron-pair geometry. To minimize repulsions, the lone pairs should be on opposite sides of the central atom (Figure $12$). The five atoms are all in the same plane and have a square planar molecular structure.
Exercise $5$
In a certain molecule, the central atom has three lone pairs and two bonds. What will the electron pair geometry and molecular structure be?
Answer
electron pair geometry: trigonal bipyramidal; molecular structure: linear
Molecular Structure for Multicenter Molecules
When a molecule or polyatomic ion has only one central atom, the molecular structure completely describes the shape of the molecule. Larger molecules do not have a single central atom, but are connected by a chain of interior atoms that each possess a “local” geometry. The way these local structures are oriented with respect to each other also influences the molecular shape, but such considerations are largely beyond the scope of this introductory discussion. For our purposes, we will only focus on determining the local structures.
Example $6$: Predicting Structure in Multicenter Molecules
The Lewis structure for the simplest amino acid, glycine, H2NCH2CO2H, is shown here. Predict the local geometry for the nitrogen atom, the two carbon atoms, and the oxygen atom with a hydrogen atom attached:
Solution
Consider each central atom independently. The electron-pair geometries:
• nitrogen––four regions of electron density; tetrahedral
• carbon (CH2)––four regions of electron density; tetrahedral
• carbon (CO2)—three regions of electron density; trigonal planar
• oxygen (OH)—four regions of electron density; tetrahedral
The local structures:
• nitrogen––three bonds, one lone pair; trigonal pyramidal
• carbon (CH2)—four bonds, no lone pairs; tetrahedral
• carbon (CO2)—three bonds (double bond counts as one bond), no lone pairs; trigonal planar
• oxygen (OH)—two bonds, two lone pairs; bent (109°)
Exercise $6$
Another amino acid is alanine, which has the Lewis structure shown here. Predict the electron-pair geometry and local structure of the nitrogen atom, the three carbon atoms, and the oxygen atom with hydrogen attached:
Answer
electron-pair geometries: nitrogen––tetrahedral; carbon (CH)—tetrahedral; carbon (CH3)—tetrahedral; carbon (CO2)—trigonal planar; oxygen (OH)—tetrahedral; local structures: nitrogen—trigonal pyramidal; carbon (CH)—tetrahedral; carbon (CH3)—tetrahedral; carbon (CO2)—trigonal planar; oxygen (OH)—bent (109°)
Link to Learning
The molecular shape simulator lets you build various molecules and practice naming their electron-pair geometries and molecular structures.
Example $7$: Molecular Simulation
Using molecular shape simulator allows us to control whether bond angles and/or lone pairs are displayed by checking or unchecking the boxes under “Options” on the right. We can also use the “Name” checkboxes at bottom-left to display or hide the electron pair geometry (called “electron geometry” in the simulator) and/or molecular structure (called “molecular shape” in the simulator).
Build the molecule HCN in the simulator based on the following Lewis structure:
$\ce{H-C#N} \nonumber$
Click on each bond type or lone pair at right to add that group to the central atom. Once you have the complete molecule, rotate it to examine the predicted molecular structure. What molecular structure is this?
Solution
The molecular structure is linear.
Exercise $7$
Build a more complex molecule in the simulator. Identify the electron-group geometry, molecular structure, and bond angles. Then try to find a chemical formula that would match the structure you have drawn.
Answer
Answers will vary. For example, an atom with four single bonds, a double bond, and a lone pair has an octahedral electron-group geometry and a square pyramidal molecular structure. XeOF4 is a molecule that adopts this structure.
Molecular Polarity and Dipole Moment
As discussed previously, polar covalent bonds connect two atoms with differing electronegativities, leaving one atom with a partial positive charge (δ+) and the other atom with a partial negative charge (δ–), as the electrons are pulled toward the more electronegative atom. This separation of charge gives rise to a bond dipole moment. The magnitude of a bond dipole moment is represented by the Greek letter mu (µ) and is given by the formula shown here, where Q is the magnitude of the partial charges (determined by the electronegativity difference) and r is the distance between the charges:
$\mu= Qr \nonumber$
This bond moment can be represented as a vector, a quantity having both direction and magnitude (Figure $13$). Dipole vectors are shown as arrows pointing along the bond from the less electronegative atom toward the more electronegative atom. A small plus sign is drawn on the less electronegative end to indicate the partially positive end of the bond. The length of the arrow is proportional to the magnitude of the electronegativity difference between the two atoms.
A whole molecule may also have a separation of charge, depending on its molecular structure and the polarity of each of its bonds. If such a charge separation exists, the molecule is said to be a polar molecule (or dipole); otherwise the molecule is said to be nonpolar. The dipole moment measures the extent of net charge separation in the molecule as a whole. We determine the dipole moment by adding the bond moments in three-dimensional space, taking into account the molecular structure.
For diatomic molecules, there is only one bond, so its bond dipole moment determines the molecular polarity. Homonuclear diatomic molecules such as Br2 and N2 have no difference in electronegativity, so their dipole moment is zero. For heteronuclear molecules such as CO, there is a small dipole moment. For HF, there is a larger dipole moment because there is a larger difference in electronegativity.
When a molecule contains more than one bond, the geometry must be taken into account. If the bonds in a molecule are arranged such that their bond moments cancel (vector sum equals zero), then the molecule is nonpolar. This is the situation in CO2 (Figure $14$). Each of the bonds is polar, but the molecule as a whole is nonpolar. From the Lewis structure, and using VSEPR theory, we determine that the CO2 molecule is linear with polar C=O bonds on opposite sides of the carbon atom. The bond moments cancel because they are pointed in opposite directions. In the case of the water molecule (Figure $14$), the Lewis structure again shows that there are two bonds to a central atom, and the electronegativity difference again shows that each of these bonds has a nonzero bond moment. In this case, however, the molecular structure is bent because of the lone pairs on O, and the two bond moments do not cancel. Therefore, water does have a net dipole moment and is a polar molecule (dipole).
The OCS molecule has a structure similar to CO2, but a sulfur atom has replaced one of the oxygen atoms. To determine if this molecule is polar, we draw the molecular structure. VSEPR theory predicts a linear molecule:
The C-O bond is considerably polar. Although C and S have very similar electronegativity values, S is slightly more electronegative than C, and so the C-S bond is just slightly polar. Because oxygen is more electronegative than sulfur, the oxygen end of the molecule is the negative end.
Chloromethane, CH3Cl, is a tetrahedral molecule with three slightly polar C-H bonds and a more polar C-Cl bond. The relative electronegativities of the bonded atoms is H < C < Cl, and so the bond moments all point toward the Cl end of the molecule and sum to yield a considerable dipole moment (the molecules are relatively polar).
For molecules of high symmetry such as BF3 (trigonal planar), CH4 (tetrahedral), PF5 (trigonal bipyramidal), and SF6 (octahedral), all the bonds are of identical polarity (same bond moment) and they are oriented in geometries that yield nonpolar molecules (dipole moment is zero). Molecules of less geometric symmetry, however, may be polar even when all bond moments are identical. For these molecules, the directions of the equal bond moments are such that they sum to give a nonzero dipole moment and a polar molecule. Examples of such molecules include hydrogen sulfide, H2S (nonlinear), and ammonia, NH3 (trigonal pyramidal).
To summarize, to be polar, a molecule must:
1. Contain at least one polar covalent bond.
2. Have a molecular structure such that the sum of the vectors of each bond dipole moment does not cancel.
Properties of Polar Molecules
Polar molecules tend to align when placed in an electric field with the positive end of the molecule oriented toward the negative plate and the negative end toward the positive plate (Figure $15$). We can use an electrically charged object to attract polar molecules, but nonpolar molecules are not attracted. Also, polar solvents are better at dissolving polar substances, and nonpolar solvents are better at dissolving nonpolar substances.
Link to Learning
The molecule polarity simulation provides many ways to explore dipole moments of bonds and molecules.
Example $8$: Polarity Simulations
Open the molecule polarity simulation and select the “Three Atoms” tab at the top. This should display a molecule ABC with three electronegativity adjustors. You can display or hide the bond moments, molecular dipoles, and partial charges at the right. Turning on the Electric Field will show whether the molecule moves when exposed to a field, similar to Figure $15$.
Use the electronegativity controls to determine how the molecular dipole will look for the starting bent molecule if:
1. A and C are very electronegative and B is in the middle of the range.
2. A is very electronegative, and B and C are not.
Solution
1. Molecular dipole moment points immediately between A and C.
2. Molecular dipole moment points along the A–B bond, toward A.
Exercise $8$
Determine the partial charges that will give the largest possible bond dipoles.
Answer
The largest bond moments will occur with the largest partial charges. The two solutions above represent how unevenly the electrons are shared in the bond. The bond moments will be maximized when the electronegativity difference is greatest. The controls for A and C should be set to one extreme, and B should be set to the opposite extreme. Although the magnitude of the bond moment will not change based on whether B is the most electronegative or the least, the direction of the bond moment will. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/07%3A_Chemical_Bonding_and_Molecular_Geometry/7.06%3A_Molecular_Structure_and_Polarity.txt |
Example and Directions
Words (or words that have the same definition)The definition is case sensitive(Optional) Image to display with the definition [Not displayed in Glossary, only in pop-up on pages](Optional) Caption for Image(Optional) External or Internal Link(Optional) Source for Definition
(Eg. "Genetic, Hereditary, DNA ...")(Eg. "Relating to genes or heredity")The infamous double helix https://bio.libretexts.org/CC-BY-SA; Delmar Larsen
Glossary Entries
Word(s)DefinitionImageCaptionLinkSource
axial positionlocation in a trigonal bipyramidal geometry in which there is another atom at a 180° angle and the equatorial positions are at a 90° angle
bond angleangle between any two covalent bonds that share a common atom
bond dipole momentseparation of charge in a bond that depends on the difference in electronegativity and the bond distance represented by partial charges or a vector
bond distance(also, bond length) distance between the nuclei of two bonded atoms
bond energy(also, bond dissociation energy) energy required to break a covalent bond in a gaseous substance
bond lengthdistance between the nuclei of two bonded atoms at which the lowest potential energy is achieved
Born-Haber cyclethermochemical cycle relating the various energetic steps involved in the formation of an ionic solid from the relevant elements
covalent bondbond formed when electrons are shared between atoms
dipole momentproperty of a molecule that describes the separation of charge determined by the sum of the individual bond moments based on the molecular structure
double bondcovalent bond in which two pairs of electrons are shared between two atoms
electron-pair geometryarrangement around a central atom of all regions of electron density (bonds, lone pairs, or unpaired electrons)
electronegativitytendency of an atom to attract electrons in a bond to itself
equatorial positionone of the three positions in a trigonal bipyramidal geometry with 120° angles between them; the axial positions are located at a 90° angle
formal chargecharge that would result on an atom by taking the number of valence electrons on the neutral atom and subtracting the nonbonding electrons and the number of bonds (one-half of the bonding electrons)
free radicalmolecule that contains an odd number of electrons
hypervalent moleculemolecule containing at least one main group element that has more than eight electrons in its valence shell
inert pair effecttendency of heavy atoms to form ions in which their valence s electrons are not lost
ionic bondstrong electrostatic force of attraction between cations and anions in an ionic compound
lattice energy (ΔHlattice)energy required to separate one mole of an ionic solid into its component gaseous ions
Lewis structurediagram showing lone pairs and bonding pairs of electrons in a molecule or an ion
Lewis symbolsymbol for an element or monatomic ion that uses a dot to represent each valence electron in the element or ion
linearshape in which two outside groups are placed on opposite sides of a central atom
lone pairtwo (a pair of) valence electrons that are not used to form a covalent bond
molecular structurearrangement of atoms in a molecule or ion
molecular structurestructure that includes only the placement of the atoms in the molecule
octahedralshape in which six outside groups are placed around a central atom such that a three-dimensional shape is generated with four groups forming a square and the other two forming the apex of two pyramids, one above and one below the square plane
octet ruleguideline that states main group atoms will form structures in which eight valence electrons interact with each nucleus, counting bonding electrons as interacting with both atoms connected by the bond
polar covalent bondcovalent bond between atoms of different electronegativities; a covalent bond with a positive end and a negative end
polar molecule(also, dipole) molecule with an overall dipole moment
pure covalent bond(also, nonpolar covalent bond) covalent bond between atoms of identical electronegativities
resonancesituation in which one Lewis structure is insufficient to describe the bonding in a molecule and the average of multiple structures is observed
resonance formstwo or more Lewis structures that have the same arrangement of atoms but different arrangements of electrons
resonance hybridaverage of the resonance forms shown by the individual Lewis structures
single bondbond in which a single pair of electrons is shared between two atoms
tetrahedralshape in which four outside groups are placed around a central atom such that a three-dimensional shape is generated with four corners and 109.5° angles between each pair and the central atom
trigonal bipyramidalshape in which five outside groups are placed around a central atom such that three form a flat triangle with 120° angles between each pair and the central atom, and the other two form the apex of two pyramids, one above and one below the triangular plane
trigonal planarshape in which three outside groups are placed in a flat triangle around a central atom with 120° angles between each pair and the central atom
triple bondbond in which three pairs of electrons are shared between two atoms
valence shell electron-pair repulsion theory (VSEPR)theory used to predict the bond angles in a molecule based on positioning regions of high electron density as far apart as possible to minimize electrostatic repulsion
vectorquantity having magnitude and direction | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/07%3A_Chemical_Bonding_and_Molecular_Geometry/7.07%3A_Key_Terms.txt |
Bond energy for a diatomic molecule:
Enthalpy change: ΔH = ƩDbonds broken – ƩDbonds formed
Lattice energy for a solid MX:
Lattice energy for an ionic crystal:
7.09: Summary
Atoms gain or lose electrons to form ions with particularly stable electron configurations. The charges of cations formed by the representative metals may be determined readily because, with few exceptions, the electronic structures of these ions have either a noble gas configuration or a completely filled electron shell. The charges of anions formed by the nonmetals may also be readily determined because these ions form when nonmetal atoms gain enough electrons to fill their valence shells.
Covalent bonds form when electrons are shared between atoms and are attracted by the nuclei of both atoms. In pure covalent bonds, the electrons are shared equally. In polar covalent bonds, the electrons are shared unequally, as one atom exerts a stronger force of attraction on the electrons than the other. The ability of an atom to attract a pair of electrons in a chemical bond is called its electronegativity. The difference in electronegativity between two atoms determines how polar a bond will be. In a diatomic molecule with two identical atoms, there is no difference in electronegativity, so the bond is nonpolar or pure covalent. When the electronegativity difference is very large, as is the case between metals and nonmetals, the bonding is characterized as ionic.
Valence electronic structures can be visualized by drawing Lewis symbols (for atoms and monatomic ions) and Lewis structures (for molecules and polyatomic ions). Lone pairs, unpaired electrons, and single, double, or triple bonds are used to indicate where the valence electrons are located around each atom in a Lewis structure. Most structures—especially those containing second row elements—obey the octet rule, in which every atom (except H) is surrounded by eight electrons. Exceptions to the octet rule occur for odd-electron molecules (free radicals), electron-deficient molecules, and hypervalent molecules.
In a Lewis structure, formal charges can be assigned to each atom by treating each bond as if one-half of the electrons are assigned to each atom. These hypothetical formal charges are a guide to determining the most appropriate Lewis structure. A structure in which the formal charges are as close to zero as possible is preferred. Resonance occurs in cases where two or more Lewis structures with identical arrangements of atoms but different distributions of electrons can be written. The actual distribution of electrons (the resonance hybrid) is an average of the distribution indicated by the individual Lewis structures (the resonance forms).
The strength of a covalent bond is measured by its bond dissociation energy, that is, the amount of energy required to break that particular bond in a mole of molecules. Multiple bonds are stronger than single bonds between the same atoms. The enthalpy of a reaction can be estimated based on the energy input required to break bonds and the energy released when new bonds are formed. For ionic bonds, the lattice energy is the energy required to separate one mole of a compound into its gas phase ions. Lattice energy increases for ions with higher charges and shorter distances between ions. Lattice energies are often calculated using the Born-Haber cycle, a thermochemical cycle including all of the energetic steps involved in converting elements into an ionic compound.
VSEPR theory predicts the three-dimensional arrangement of atoms in a molecule. It states that valence electrons will assume an electron-pair geometry that minimizes repulsions between areas of high electron density (bonds and/or lone pairs). Molecular structure, which refers only to the placement of atoms in a molecule and not the electrons, is equivalent to electron-pair geometry only when there are no lone electron pairs around the central atom. A dipole moment measures a separation of charge. For one bond, the bond dipole moment is determined by the difference in electronegativity between the two atoms. For a molecule, the overall dipole moment is determined by both the individual bond moments and how these dipoles are arranged in the molecular structure. Polar molecules (those with an appreciable dipole moment) interact with electric fields, whereas nonpolar molecules do not. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/07%3A_Chemical_Bonding_and_Molecular_Geometry/7.08%3A_Key_Equations.txt |
1.
Does a cation gain protons to form a positive charge or does it lose electrons?
2.
Iron(III) sulfate [Fe2(SO4)3] is composed of Fe3+ and ions. Explain why a sample of iron(III) sulfate is uncharged.
3.
Which of the following atoms would be expected to form negative ions in binary ionic compounds and which would be expected to form positive ions: P, I, Mg, Cl, In, Cs, O, Pb, Co?
4.
Which of the following atoms would be expected to form negative ions in binary ionic compounds and which would be expected to form positive ions: Br, Ca, Na, N, F, Al, Sn, S, Cd?
5.
Predict the charge on the monatomic ions formed from the following atoms in binary ionic compounds:
1. (f) Cs
6.
Predict the charge on the monatomic ions formed from the following atoms in binary ionic compounds:
1. (f) In
7.
Write the electron configuration for each of the following ions:
1. (k) Fe2+
8.
Write the electron configuration for the monatomic ions formed from the following elements (which form the greatest concentration of monatomic ions in seawater):
1. Cl
2. Na
3. Mg
4. Ca
5. K
6. Br
7. Sr
8. F
9.
Write out the full electron configuration for each of the following atoms and for the monatomic ion found in binary ionic compounds containing the element:
1. (f) S
10.
From the labels of several commercial products, prepare a list of six ionic compounds in the products. For each compound, write the formula. (You may need to look up some formulas in a suitable reference.)
11.
Why is it incorrect to speak of a molecule of solid NaCl?
12.
What information can you use to predict whether a bond between two atoms is covalent or ionic?
13.
Predict which of the following compounds are ionic and which are covalent, based on the location of their constituent atoms in the periodic table:
1. (i) CaO
(j) IBr
(k) CO2
14.
Explain the difference between a nonpolar covalent bond, a polar covalent bond, and an ionic bond.
15.
From its position in the periodic table, determine which atom in each pair is more electronegative:
1. (f) Ba or P
(g) N or K
16.
From its position in the periodic table, determine which atom in each pair is more electronegative:
1. (f) Se or P
(g) C or Si
17.
From their positions in the periodic table, arrange the atoms in each of the following series in order of increasing electronegativity:
1. C, F, H, N, O
2. Br, Cl, F, H, I
3. F, H, O, P, S
4. Al, H, Na, O, P
5. Ba, H, N, O, As
18.
From their positions in the periodic table, arrange the atoms in each of the following series in order of increasing electronegativity:
1. As, H, N, P, Sb
2. Cl, H, P, S, Si
3. Br, Cl, Ge, H, Sr
4. Ca, H, K, N, Si
5. Cl, Cs, Ge, H, Sr
19.
Which atoms can bond to sulfur so as to produce a positive partial charge on the sulfur atom?
20.
Which is the most polar bond?
1. C–C
2. C–H
3. N–H
4. O–H
5. Se–H
21.
Identify the more polar bond in each of the following pairs of bonds:
1. (f) SO or PO
(g) CN or NN
22.
Which of the following molecules or ions contain polar bonds?
1. (f) H2S
(g)
23.
Write the Lewis symbols for each of the following ions:
1. (f) Li+
24.
Many monatomic ions are found in seawater, including the ions formed from the following list of elements. Write the Lewis symbols for the monatomic ions formed from the following elements:
1. Cl
2. Na
3. Mg
4. Ca
5. K
6. Br
7. Sr
8. F
25.
Write the Lewis symbols of the ions in each of the following ionic compounds and the Lewis symbols of the atom from which they are formed:
1. (f) KF
26.
In the Lewis structures listed here, M and X represent various elements in the third period of the periodic table. Write the formula of each compound using the chemical symbols of each element:
(a)
(b)
(c)
(d)
27.
Write the Lewis structure for the diatomic molecule P2, an unstable form of phosphorus found in high-temperature phosphorus vapor.
28.
Write Lewis structures for the following:
1. (k) CN
29.
Write Lewis structures for the following:
1. (i) HCCH
(j) ClCN
(k)
30.
Write Lewis structures for the following:
1. ClF3
2. PCl5
3. BF3
31.
Write Lewis structures for the following:
1. SeF6
2. XeF4
3. Cl2BBCl2 (contains a B–B bond)
32.
Write Lewis structures for:
1. HONO
33.
Correct the following statement: “The bonds in solid PbCl2 are ionic; the bond in a HCl molecule is covalent. Thus, all of the valence electrons in PbCl2 are located on the Cl ions, and all of the valence electrons in a HCl molecule are shared between the H and Cl atoms.”
34.
Write Lewis structures for the following molecules or ions:
1. SbH3
2. XeF2
3. Se8 (a cyclic molecule with a ring of eight Se atoms)
35.
Methanol, H3COH, is used as the fuel in some race cars. Ethanol, C2H5OH, is used extensively as motor fuel in Brazil. Both methanol and ethanol produce CO2 and H2O when they burn. Write the chemical equations for these combustion reactions using Lewis structures instead of chemical formulas.
36.
Many planets in our solar system contain organic chemicals including methane (CH4) and traces of ethylene (C2H4), ethane (C2H6), propyne (H3CCCH), and diacetylene (HCCCCH). Write the Lewis structures for each of these molecules.
37.
Carbon tetrachloride was formerly used in fire extinguishers for electrical fires. It is no longer used for this purpose because of the formation of the toxic gas phosgene, Cl2CO. Write the Lewis structures for carbon tetrachloride and phosgene.
38.
Identify the atoms that correspond to each of the following electron configurations. Then, write the Lewis symbol for the common ion formed from each atom:
1. 1s22s22p5
2. 1s22s22p63s2
3. 1s22s22p63s23p64s23d10
4. 1s22s22p63s23p64s23d104p4
5. 1s22s22p63s23p64s23d104p1
39.
The arrangement of atoms in several biologically important molecules is given here. Complete the Lewis structures of these molecules by adding multiple bonds and lone pairs. Do not add any more atoms.
(a) the amino acid serine:
(b) urea:
(c) pyruvic acid:
(d) uracil:
(e) carbonic acid:
40.
A compound with a molar mass of about 28 g/mol contains 85.7% carbon and 14.3% hydrogen by mass. Write the Lewis structure for a molecule of the compound.
41.
A compound with a molar mass of about 42 g/mol contains 85.7% carbon and 14.3% hydrogen by mass. Write the Lewis structure for a molecule of the compound.
42.
Two arrangements of atoms are possible for a compound with a molar mass of about 45 g/mol that contains 52.2% C, 13.1% H, and 34.7% O by mass. Write the Lewis structures for the two molecules.
43.
How are single, double, and triple bonds similar? How do they differ?
44.
Write resonance forms that describe the distribution of electrons in each of these molecules or ions.
1. selenium dioxide, OSeO
2. nitrate ion,
3. nitric acid, HNO3 (N is bonded to an OH group and two O atoms)
4. benzene, C6H6:
(e) the formate ion:
45.
Write resonance forms that describe the distribution of electrons in each of these molecules or ions.
1. sulfur dioxide, SO2
2. carbonate ion,
3. hydrogen carbonate ion, (C is bonded to an OH group and two O atoms)
4. pyridine:
(e) the allyl ion:
46.
Write the resonance forms of ozone, O3, the component of the upper atmosphere that protects the Earth from ultraviolet radiation.
47.
Sodium nitrite, which has been used to preserve bacon and other meats, is an ionic compound. Write the resonance forms of the nitrite ion,
48.
In terms of the bonds present, explain why acetic acid, CH3CO2H, contains two distinct types of carbon-oxygen bonds, whereas the acetate ion, formed by loss of a hydrogen ion from acetic acid, only contains one type of carbon-oxygen bond. The skeleton structures of these species are shown:
49.
Write the Lewis structures for the following, and include resonance structures where appropriate. Indicate which has the strongest carbon-oxygen bond.
1. CO2
2. CO
50.
Toothpastes containing sodium hydrogen carbonate (sodium bicarbonate) and hydrogen peroxide are widely used. Write Lewis structures for the hydrogen carbonate ion and hydrogen peroxide molecule, with resonance forms where appropriate.
51.
Determine the formal charge of each element in the following:
1. HCl
2. CF4
3. PCl3
4. PF5
52.
Determine the formal charge of each element in the following:
1. H3O+
2. NH3
3. H2O2
53.
Calculate the formal charge of chlorine in the molecules Cl2, BeCl2, and ClF5.
54.
Calculate the formal charge of each element in the following compounds and ions:
1. F2CO
2. NO
3. H2CCH2
4. ClF3
5. SeF6
55.
Draw all possible resonance structures for each of these compounds. Determine the formal charge on each atom in each of the resonance structures:
1. O3
2. SO2
56.
Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in nitrosyl chloride: ClNO or ClON?
57.
Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in hypochlorous acid: HOCl or OClH?
58.
Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in sulfur dioxide: OSO or SOO?
59.
Draw the structure of hydroxylamine, H3NO, and assign formal charges; look up the structure. Is the actual structure consistent with the formal charges?
60.
Iodine forms a series of fluorides (listed here). Write Lewis structures for each of the four compounds and determine the formal charge of the iodine atom in each molecule:
1. IF
2. IF3
3. IF5
4. IF7
61.
Write the Lewis structure and chemical formula of the compound with a molar mass of about 70 g/mol that contains 19.7% nitrogen and 80.3% fluorine by mass, and determine the formal charge of the atoms in this compound.
62.
Which of the following structures would we expect for nitrous acid? Determine the formal charges:
63.
Sulfuric acid is the industrial chemical produced in greatest quantity worldwide. About 90 billion pounds are produced each year in the United States alone. Write the Lewis structure for sulfuric acid, H2SO4, which has two oxygen atoms and two OH groups bonded to the sulfur.
64.
Which bond in each of the following pairs of bonds is the strongest?
1. (e) C–H or O–H
(f) C–N or C–O
65.
Using the bond energies in Table 7.2, determine the approximate enthalpy change for each of the following reactions:
66.
Using the bond energies in Table 7.2, determine the approximate enthalpy change for each of the following reactions:
67.
When a molecule can form two different structures, the structure with the stronger bonds is usually the more stable form. Use bond energies to predict the correct structure of the hydroxylamine molecule:
68.
How does the bond energy of HCl(g) differ from the standard enthalpy of formation of HCl(g)?
69.
Using the standard enthalpy of formation data in Appendix G, show how the standard enthalpy of formation of HCl(g) can be used to determine the bond energy.
70.
Using the standard enthalpy of formation data in Appendix G, calculate the bond energy of the carbon-sulfur double bond in CS2.
71.
Using the standard enthalpy of formation data in Appendix G, determine which bond is stronger: the S–F bond in SF4(g) or in SF6(g)?
72.
Using the standard enthalpy of formation data in Appendix G, determine which bond is stronger: the P–Cl bond in PCl3(g) or in PCl5(g)?
73.
Complete the following Lewis structure by adding bonds (not atoms), and then indicate the longest bond:
74.
Use the bond energy to calculate an approximate value of ΔH for the following reaction. Which is the more stable form of FNO2?
75.
Use principles of atomic structure to answer each of the following:1
1. The radius of the Ca atom is 197 pm; the radius of the Ca2+ ion is 99 pm. Account for the difference.
2. The lattice energy of CaO(s) is –3460 kJ/mol; the lattice energy of K2O is –2240 kJ/mol. Account for the difference.
3. Given these ionization values, explain the difference between Ca and K with regard to their first and second ionization energies.
Element First Ionization Energy (kJ/mol) Second Ionization Energy (kJ/mol)
K 419 3050
Ca 590 1140
(d) The first ionization energy of Mg is 738 kJ/mol and that of Al is 578 kJ/mol. Account for this difference.
76.
The lattice energy of LiF is 1023 kJ/mol, and the Li–F distance is 200.8 pm. NaF crystallizes in the same structure as LiF but with a Na–F distance of 231 pm. Which of the following values most closely approximates the lattice energy of NaF: 510, 890, 1023, 1175, or 4090 kJ/mol? Explain your choice.
77.
For which of the following substances is the least energy required to convert one mole of the solid into separate ions?
1. MgO
2. SrO
3. KF
4. CsF
5. MgF2
78.
The reaction of a metal, M, with a halogen, X2, proceeds by an exothermic reaction as indicated by this equation: For each of the following, indicate which option will make the reaction more exothermic. Explain your answers.
1. a large radius vs. a small radius for M+2
2. a high ionization energy vs. a low ionization energy for M
3. an increasing bond energy for the halogen
4. a decreasing electron affinity for the halogen
5. an increasing size of the anion formed by the halogen
79.
The lattice energy of LiF is 1023 kJ/mol, and the Li–F distance is 201 pm. MgO crystallizes in the same structure as LiF but with a Mg–O distance of 205 pm. Which of the following values most closely approximates the lattice energy of MgO: 256 kJ/mol, 512 kJ/mol, 1023 kJ/mol, 2046 kJ/mol, or 4008 kJ/mol? Explain your choice.
80.
Which compound in each of the following pairs has the larger lattice energy? Note: Mg2+ and Li+ have similar radii; O2– and F have similar radii. Explain your choices.
1. MgO or MgSe
2. LiF or MgO
3. Li2O or LiCl
4. Li2Se or MgO
81.
Which compound in each of the following pairs has the larger lattice energy? Note: Ba2+ and
K+ have similar radii; S2– and Cl have similar radii. Explain your choices.
1. K2O or Na2O
2. K2S or BaS
3. KCl or BaS
4. BaS or BaCl2
82.
Which of the following compounds requires the most energy to convert one mole of the solid into separate ions?
1. MgO
2. SrO
3. KF
4. CsF
5. MgF2
83.
Which of the following compounds requires the most energy to convert one mole of the solid into separate ions?
1. K2S
2. K2O
3. CaS
4. Cs2S
5. CaO
84.
The lattice energy of KF is 794 kJ/mol, and the interionic distance is 269 pm. The Na–F
distance in NaF, which has the same structure as KF, is 231 pm. Which of the following values is the closest approximation of the lattice energy of NaF: 682 kJ/mol, 794 kJ/mol, 924 kJ/mol, 1588 kJ/mol, or 3175 kJ/mol? Explain your answer.
85.
Explain why the HOH molecule is bent, whereas the HBeH molecule is linear.
86.
What feature of a Lewis structure can be used to tell if a molecule’s (or ion’s) electron-pair geometry and molecular structure will be identical?
87.
Explain the difference between electron-pair geometry and molecular structure.
88.
Why is the H–N–H angle in NH3 smaller than the H–C–H bond angle in CH4? Why is the H–N–H angle in identical to the H–C–H bond angle in CH4?
89.
Explain how a molecule that contains polar bonds can be nonpolar.
90.
As a general rule, MXn molecules (where M represents a central atom and X represents terminal atoms; n = 2 – 5) are polar if there is one or more lone pairs of electrons on M. NH3 (M = N, X = H, n = 3) is an example. There are two molecular structures with lone pairs that are exceptions to this rule. What are they?
91.
Predict the electron pair geometry and the molecular structure of each of the following molecules or ions:
1. SF6
2. PCl5
3. BeH2
92.
Identify the electron pair geometry and the molecular structure of each of the following molecules or ions:
1. CF4
2. BF3
3. BeCl2
93.
What are the electron-pair geometry and the molecular structure of each of the following molecules or ions?
1. (f)
94.
Predict the electron pair geometry and the molecular structure of each of the following ions:
1. (f) XeF4
95.
Identify the electron pair geometry and the molecular structure of each of the following molecules:
1. (f) XeO2F2 (Xe is the central atom)
(g) (Cl is the central atom)
96.
Predict the electron pair geometry and the molecular structure of each of the following:
1. (f)
97.
Which of the following molecules and ions contain polar bonds? Which of these molecules and ions have dipole moments?
1. (f)
98.
Which of these molecules and ions contain polar bonds? Which of these molecules and ions have dipole moments?
1. (f) XeF4
99.
Which of the following molecules have dipole moments?
1. CS2
2. SeS2
3. CCl2F2
4. PCl3 (P is the central atom)
5. ClNO (N is the central atom)
100.
Identify the molecules with a dipole moment:
1. SF4
2. CF4
3. Cl2CCBr2
4. CH3Cl
5. H2CO
101.
The molecule XF3 has a dipole moment. Is X boron or phosphorus?
102.
The molecule XCl2 has a dipole moment. Is X beryllium or sulfur?
103.
Is the Cl2BBCl2 molecule polar or nonpolar?
104.
There are three possible structures for PCl2F3 with phosphorus as the central atom. Draw them and discuss how measurements of dipole moments could help distinguish among them.
105.
Describe the molecular structure around the indicated atom or atoms:
1. (i) each of the carbon atoms in allene, H2CCCH2
106.
Draw the Lewis structures and predict the shape of each compound or ion:
1. CO2
2. SO3
107.
A molecule with the formula AB2, in which A and B represent different atoms, could have one of three different shapes. Sketch and name the three different shapes that this molecule might have. Give an example of a molecule or ion for each shape.
108.
A molecule with the formula AB3, in which A and B represent different atoms, could have one of three different shapes. Sketch and name the three different shapes that this molecule might have. Give an example of a molecule or ion that has each shape.
109.
Draw the Lewis electron dot structures for these molecules, including resonance structures where appropriate:
1. CS2
2. CS
3. predict the molecular shapes for and CS2 and explain how you arrived at your predictions
110.
What is the molecular structure of the stable form of FNO2? (N is the central atom.)
111.
A compound with a molar mass of about 42 g/mol contains 85.7% carbon and 14.3% hydrogen. What is its molecular structure?
112.
Use the simulation to perform the following exercises for a two-atom molecule:
1. Adjust the electronegativity value so the bond dipole is pointing toward B. Then determine what the electronegativity values must be to switch the dipole so that it points toward A.
2. With a partial positive charge on A, turn on the electric field and describe what happens.
3. With a small partial negative charge on A, turn on the electric field and describe what happens.
4. Reset all, and then with a large partial negative charge on A, turn on the electric field and describe what happens.
113.
Use the simulation to perform the following exercises for a real molecule. You may need to rotate the molecules in three dimensions to see certain dipoles.
1. Sketch the bond dipoles and molecular dipole (if any) for O3. Explain your observations.
2. Look at the bond dipoles for NH3. Use these dipoles to predict whether N or H is more electronegative.
3. Predict whether there should be a molecular dipole for NH3 and, if so, in which direction it will point. Check the molecular dipole box to test your hypothesis.
114.
Use the Molecule Shape simulator to build a molecule. Starting with the central atom, click on the double bond to add one double bond. Then add one single bond and one lone pair. Rotate the molecule to observe the complete geometry. Name the electron group geometry and molecular structure and predict the bond angle. Then click the check boxes at the bottom and right of the simulator to check your answers.
115.
Use the Molecule Shape simulator to explore real molecules. On the Real Molecules tab, select H2O. Switch between the “real” and “model” modes. Explain the difference observed.
116.
Use the Molecule Shape simulator to explore real molecules. On the Real Molecules tab, select “model” mode and S2O. What is the model bond angle? Explain whether the “real” bond angle should be larger or smaller than the ideal model angle.
Footnotes
• 1This question is taken from the Chemistry Advanced Placement Examination and is used with the permission of the Educational Testing Service. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/07%3A_Chemical_Bonding_and_Molecular_Geometry/7.10%3A_Exercises.txt |
We have examined the basic ideas of bonding, showing that atoms share electrons to form molecules with stable Lewis structures and that we can predict the shapes of those molecules by valence shell electron pair repulsion (VSEPR) theory. These ideas provide an important starting point for understanding chemical bonding. But these models sometimes fall short in their abilities to predict the behavior of real substances. How can we reconcile the geometries of s, p, and d atomic orbitals with molecular shapes that show angles like 120° and 109.5°? Furthermore, we know that electrons and magnetic behavior are related through electromagnetic fields.
• 8.0: Introduction
Oxygen demonstrates very different magnetic behavior than nitrogen. We can pour liquid nitrogen through a magnetic field with no visible interactions, while liquid oxygen is attracted to the magnet and floats in the magnetic field. We need to understand the additional concepts of valence bond theory, orbital hybridization, and molecular orbital theory to understand these observations.
• 8.1: Valence Bond Theory
Valence bond theory describes bonding as a consequence of the overlap of two separate atomic orbitals on different atoms that creates a region with one pair of electrons shared between the two atoms. When the orbitals overlap along an axis containing the nuclei, they form a σ bond. When they overlap in a fashion that creates a node along this axis, they form a π bond.
• 8.2: Hybrid Atomic Orbitals
We can use hybrid orbitals, which are mathematical combinations of some or all of the valence atomic orbitals, to describe the electron density around covalently bonded atoms. These hybrid orbitals either form sigma (σ) bonds directed toward other atoms of the molecule or contain lone pairs of electrons. We can determine the type of hybridization around a central atom from the geometry of the regions of electron density about it.
• 8.3: Multiple Bonds
Multiple bonds consist of a σ bond located along the axis between two atoms and one or two π bonds. The σ bonds are usually formed by the overlap of hybridized atomic orbitals, while the π bonds are formed by the side-by-side overlap of unhybridized orbitals. Resonance occurs when there are multiple unhybridized orbitals with the appropriate alignment to overlap, so the placement of π bonds can vary.
• 8.4: Molecular Orbital Theory
Molecular orbital (MO) theory describes the behavior of electrons in a molecule in terms of combinations of the atomic wavefunctions. The resulting molecular orbitals may extend over all the atoms in the molecule. Bonding molecular orbitals are formed by in-phase combinations of atomic wavefunctions, and electrons in these orbitals stabilize a molecule. Antibonding molecular orbitals result from out-of-phase combinations and electrons in these orbitals make a molecule less stable.
• 8.5: Key Terms
• 8.6: Key Equations
• 8.7: Summary
• 8.8: Exercises
These are homework exercises to accompany the Textmap created for "Chemistry" by OpenStax. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
08: Advanced Theories of Covalent Bonding
We have examined the basic ideas of bonding, showing that atoms share electrons to form molecules with stable Lewis structures and that we can predict the shapes of those molecules by valence shell electron pair repulsion (VSEPR) theory. These ideas provide an important starting point for understanding chemical bonding. But these models sometimes fall short in their abilities to predict the behavior of real substances. How can we reconcile the geometries of s, p, and d atomic orbitals with molecular shapes that show angles like 120° and 109.5°? Furthermore, we know that electrons and magnetic behavior are related through electromagnetic fields. Both N2 and O2 have fairly similar Lewis structures that contain lone pairs of electrons.
Yet oxygen demonstrates very different magnetic behavior than nitrogen. We can pour liquid nitrogen through a magnetic field with no visible interactions, while liquid oxygen (shown in Figure \(1\)) is attracted to the magnet and floats in the magnetic field. We need to understand the additional concepts of valence bond theory, orbital hybridization, and molecular orbital theory to understand these observations. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/08%3A_Advanced_Theories_of_Covalent_Bonding/8.00%3A_Introduction.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the formation of covalent bonds in terms of atomic orbital overlap
• Define and give examples of σ and π bonds
As we know, a scientific theory is a strongly supported explanation for observed natural laws or large bodies of experimental data. For a theory to be accepted, it must explain experimental data and be able to predict behavior. For example, VSEPR theory has gained widespread acceptance because it predicts three-dimensional molecular shapes that are consistent with experimental data collected for thousands of different molecules. However, VSEPR theory does not provide an explanation of chemical bonding.
There are successful theories that describe the electronic structure of atoms. We can use quantum mechanics to predict the specific regions around an atom where electrons are likely to be located: A spherical shape for an s orbital, a dumbbell shape for a p orbital, and so forth. However, these predictions only describe the orbitals around free atoms. When atoms bond to form molecules, atomic orbitals are not sufficient to describe the regions where electrons will be located in the molecule. A more complete understanding of electron distributions requires a model that can account for the electronic structure of molecules. One popular theory holds that a covalent bond forms when a pair of electrons is shared by two atoms and is simultaneously attracted by the nuclei of both atoms. In the following sections, we will discuss how such bonds are described by valence bond theory and hybridization.
Valence bond theory describes a covalent bond as the overlap of half-filled atomic orbitals (each containing a single electron) that yield a pair of electrons shared between the two bonded atoms. We say that orbitals on two different atoms overlap when a portion of one orbital and a portion of a second orbital occupy the same region of space. According to valence bond theory, a covalent bond results when two conditions are met: (1) an orbital on one atom overlaps an orbital on a second atom and (2) the single electrons in each orbital combine to form an electron pair. The mutual attraction between this negatively charged electron pair and the two atoms’ positively charged nuclei serves to physically link the two atoms through a force we define as a covalent bond. The strength of a covalent bond depends on the extent of overlap of the orbitals involved. Orbitals that overlap extensively form bonds that are stronger than those that have less overlap.
The energy of the system depends on how much the orbitals overlap. Figure $1$ illustrates how the sum of the energies of two hydrogen atoms (the colored curve) changes as they approach each other. When the atoms are far apart there is no overlap, and by convention we set the sum of the energies at zero. As the atoms move together, their orbitals begin to overlap. Each electron begins to feel the attraction of the nucleus in the other atom. In addition, the electrons begin to repel each other, as do the nuclei. While the atoms are still widely separated, the attractions are slightly stronger than the repulsions, and the energy of the system decreases. (A bond begins to form.) As the atoms move closer together, the overlap increases, so the attraction of the nuclei for the electrons continues to increase (as do the repulsions among electrons and between the nuclei). At some specific distance between the atoms, which varies depending on the atoms involved, the energy reaches its lowest (most stable) value. This optimum distance between the two bonded nuclei is the bond distance between the two atoms. The bond is stable because at this point, the attractive and repulsive forces combine to create the lowest possible energy configuration. If the distance between the nuclei were to decrease further, the repulsions between nuclei and the repulsions as electrons are confined in closer proximity to each other would become stronger than the attractive forces. The energy of the system would then rise (making the system destabilized), as shown at the far left of Figure $1$.
The bond energy is the difference between the energy minimum (which occurs at the bond distance) and the energy of the two separated atoms. This is the quantity of energy released when the bond is formed. Conversely, the same amount of energy is required to break the bond. For the H2 molecule shown in Figure $1$, at the bond distance of 74 pm the system is $7.24 \times 10^{−19}~\text{J}$ lower in energy than the two separated hydrogen atoms. This may seem like a small number. However, we know from our earlier description of thermochemistry that bond energies are often discussed on a per-mole basis. For example, it requires $7.24 \times 10^{−19}~\text{J}$ to break one H–H bond, but it takes $4.36 \times 10^5~\text{J}$ to break 1 mole of H–H bonds. A comparison of some bond lengths and energies is shown in Table $1$. We can find many of these bonds in a variety of molecules, and this table provides average values. For example, breaking the first C–H bond in CH4 requires 439.3 kJ/mol, while breaking the first C–H bond in H–CH2C6H5 (a common paint thinner) requires 375.5 kJ/mol.
Table $1$: Representative Bond Energies and Lengths
Bond Length (pm) Energy (kJ/mol) Bond Length (pm) Energy (kJ/mol)
H–H 74 436 C–O 140.1 358
H–C 106.8 413 119.7 745
H–N 101.5 391 113.7 1072
H–O 97.5 467 H–Cl 127.5 431
C–C 150.6 347 H–Br 141.4 366
133.5 614 H–I 160.9 298
120.8 839 O–O 148 146
C–N 142.1 305 120.8 498
130.0 615 F–F 141.2 159
116.1 891 Cl–Cl 198.8 243
In addition to the distance between two orbitals, the orientation of orbitals also affects their overlap (other than for two s orbitals, which are spherically symmetric). Greater overlap is possible when orbitals are oriented such that they overlap on a direct line between the two nuclei. Figure $2$ illustrates this for two p orbitals from different atoms; the overlap is greater when the orbitals overlap end to end rather than at an angle.
The overlap of two s orbitals (as in H2), the overlap of an s orbital and a p orbital (as in HCl), and the end-to-end overlap of two p orbitals (as in Cl2) all produce sigma bonds (σ bonds), as illustrated in Figure $3$. A σ bond is a covalent bond in which the electron density is concentrated in the region along the internuclear axis; that is, a line between the nuclei would pass through the center of the overlap region. Single bonds in Lewis structures are described as σ bonds in valence bond theory.
A pi bond (π bond) is a type of covalent bond that results from the side-by-side overlap of two p orbitals, as illustrated in Figure $4$. In a π bond, the regions of orbital overlap lie on opposite sides of the internuclear axis. Along the axis itself, there is a node, that is, a plane with no probability of finding an electron.
While all single bonds are σ bonds, multiple bonds consist of both σ and π bonds. As the Lewis structures below suggest, O2 contains a double bond, and N2 contains a triple bond. The double bond consists of one σ bond and one π bond, and the triple bond consists of one σ bond and two π bonds. Between any two atoms, the first bond formed will always be a σ bond, but there can only be one σ bond in any one location. In any multiple bond, there will be one σ bond, and the remaining one or two bonds will be π bonds. These bonds are described in more detail later in this chapter.
As seen in Figure $1$, an average carbon-carbon single bond is 347 kJ/mol, while in a carbon-carbon double bond, the π bond increases the bond strength by 267 kJ/mol. Adding an additional π bond causes a further increase of 225 kJ/mol. We can see a similar pattern when we compare other σ and π bonds. Thus, each individual π bond is generally weaker than a corresponding σ bond between the same two atoms. In a σ bond, there is a greater degree of orbital overlap than in a π bond.
Example $1$: Counting σ and π Bonds
Butadiene, C4H6, is used to make synthetic rubber. Identify the number of σ and π bonds contained in this molecule.
Solution
There are six σ C–H bonds and one σ C–C bond, for a total of seven from the single bonds. There are two double bonds that each have a π bond in addition to the σ bond. This gives a total nine σ and two π bonds overall.
Exercise $1$
Identify each illustration as depicting a σ or π bond:
1. side-by-side overlap of a 4p and a 2p orbital
2. end-to-end overlap of a 4p and 4p orbital
3. end-to-end overlap of a 4p and a 2p orbital
Answer
(a) is a π bond with a node along the axis connecting the nuclei while (b) and (c) are σ bonds that overlap along the axis. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/08%3A_Advanced_Theories_of_Covalent_Bonding/8.01%3A_Valence_Bond_Theory.txt |
Learning Objectives
By the end of this section, you will be able to:
• Explain the concept of atomic orbital hybridization
• Determine the hybrid orbitals associated with various molecular geometries
Thinking in terms of overlapping atomic orbitals is one way for us to explain how chemical bonds form in diatomic molecules. However, to understand how molecules with more than two atoms form stable bonds, we require a more detailed model. As an example, let us consider the water molecule, in which we have one oxygen atom bonding to two hydrogen atoms. Oxygen has the electron configuration 1s22s22p4, with two unpaired electrons (one in each of the two 2p orbitals). Valence bond theory would predict that the two O–H bonds form from the overlap of these two 2p orbitals with the 1s orbitals of the hydrogen atoms. If this were the case, the bond angle would be 90°, as shown in Figure \(1\), because p orbitals are perpendicular to each other. Experimental evidence shows that the bond angle is 104.5°, not 90°. The prediction of the valence bond theory model does not match the real-world observations of a water molecule; a different model is needed.
Quantum-mechanical calculations suggest why the observed bond angles in H2O differ from those predicted by the overlap of the 1s orbital of the hydrogen atoms with the 2p orbitals of the oxygen atom. The mathematical expression known as the wave function, ψ, contains information about each orbital and the wavelike properties of electrons in an isolated atom. When atoms are bound together in a molecule, the wave functions combine to produce new mathematical descriptions that have different shapes. This process of combining the wave functions for atomic orbitals is called hybridization and is mathematically accomplished by the linear combination of atomic orbitals, LCAO, (a technique that we will encounter again later). The new orbitals that result are called hybrid orbitals. The valence orbitals in an isolated oxygen atom are a 2s orbital and three 2p orbitals. The valence orbitals in an oxygen atom in a water molecule differ; they consist of four equivalent hybrid orbitals that point approximately toward the corners of a tetrahedron (Figure \(2\)). Consequently, the overlap of the O and H orbitals should result in a tetrahedral bond angle (109.5°). The observed angle of 104.5° is experimental evidence for which quantum-mechanical calculations give a useful explanation: Valence bond theory must include a hybridization component to give accurate predictions.
The following ideas are important in understanding hybridization:
1. Hybrid orbitals do not exist in isolated atoms. They are formed only in covalently bonded atoms.
2. Hybrid orbitals have shapes and orientations that are very different from those of the atomic orbitals in isolated atoms.
3. A set of hybrid orbitals is generated by combining atomic orbitals. The number of hybrid orbitals in a set is equal to the number of atomic orbitals that were combined to produce the set.
4. All orbitals in a set of hybrid orbitals are equivalent in shape and energy.
5. The type of hybrid orbitals formed in a bonded atom depends on its electron-pair geometry as predicted by the VSEPR theory.
6. Hybrid orbitals overlap to form σ bonds. Unhybridized orbitals overlap to form π bonds.
In the following sections, we shall discuss the common types of hybrid orbitals.
sp Hybridization
The beryllium atom in a gaseous BeCl2 molecule is an example of a central atom with no lone pairs of electrons in a linear arrangement of three atoms. There are two regions of valence electron density in the BeCl2 molecule that correspond to the two covalent Be–Cl bonds. To accommodate these two electron domains, two of the Be atom’s four valence orbitals will mix to yield two hybrid orbitals. This hybridization process involves mixing of the valence s orbital with one of the valence p orbitals to yield two equivalent sp hybrid orbitals that are oriented in a linear geometry (Figure \(3\)). In this figure, the set of sp orbitals appears similar in shape to the original p orbital, but there is an important difference. The number of atomic orbitals combined always equals the number of hybrid orbitals formed. The p orbital is one orbital that can hold up to two electrons. The sp set is two equivalent orbitals that point 180° from each other. The two electrons that were originally in the s orbital are now distributed to the two sp orbitals, which are half filled. In gaseous BeCl2, these half-filled hybrid orbitals will overlap with orbitals from the chlorine atoms to form two identical σ bonds.
We illustrate the electronic differences in an isolated Be atom and in the bonded Be atom in the orbital energy-level diagram in Figure \(4\). These diagrams represent each orbital by a horizontal line (indicating its energy) and each electron by an arrow. Energy increases toward the top of the diagram. We use one upward arrow to indicate one electron in an orbital and two arrows (up and down) to indicate two electrons of opposite spin.
When atomic orbitals hybridize, the valence electrons occupy the newly created orbitals. The Be atom had two valence electrons, so each of the sp orbitals gets one of these electrons. Each of these electrons pairs up with the unpaired electron on a chlorine atom when a hybrid orbital and a chlorine orbital overlap during the formation of the Be–Cl bonds.
Any central atom surrounded by just two regions of valence electron density in a molecule will exhibit sp hybridization. Other examples include the mercury atom in the linear HgCl2 molecule, the zinc atom in Zn(CH3)2, which contains a linear C–Zn–C arrangement, and the carbon atoms in HCCH and CO2.
Link to Learning
Check out the University of Wisconsin-Oshkosh website to learn about visualizing hybrid orbitals in three dimensions.
sp2 Hybridization
The valence orbitals of a central atom surrounded by three regions of electron density consist of a set of three sp2 hybrid orbitals and one unhybridized p orbital. This arrangement results from sp2 hybridization, the mixing of one s orbital and two p orbitals to produce three identical hybrid orbitals oriented in a trigonal planar geometry (Figure \(5\)).
Although quantum mechanics yields the “plump” orbital lobes as depicted in Figure \(5\), sometimes for clarity these orbitals are drawn thinner and without the minor lobes, as in Figure \(6\), to avoid obscuring other features of a given illustration. We will use these “thinner” representations whenever the true view is too crowded to easily visualize.
The observed structure of the borane molecule, BH3, suggests sp2 hybridization for boron in this compound. The molecule is trigonal planar, and the boron atom is involved in three bonds to hydrogen atoms (Figure \(7\)). We can illustrate the comparison of orbitals and electron distribution in an isolated boron atom and in the bonded atom in BH3 as shown in the orbital energy level diagram in Figure \(8\). We redistribute the three valence electrons of the boron atom in the three sp2 hybrid orbitals, and each boron electron pairs with a hydrogen electron when B–H bonds form.
Any central atom surrounded by three regions of electron density will exhibit sp2 hybridization. This includes molecules with a lone pair on the central atom, such as ClNO (Figure \(9\)), or molecules with two single bonds and a double bond connected to the central atom, as in formaldehyde, CH2O, and ethene, H2CCH2.
sp3 Hybridization
The valence orbitals of an atom surrounded by a tetrahedral arrangement of bonding pairs and lone pairs consist of a set of four sp3 hybrid orbitals. The hybrids result from the mixing of one s orbital and all three p orbitals that produces four identical sp3 hybrid orbitals (Figure \(10\)). Each of these hybrid orbitals points toward a different corner of a tetrahedron.
A molecule of methane, CH4, consists of a carbon atom surrounded by four hydrogen atoms at the corners of a tetrahedron. The carbon atom in methane exhibits sp3 hybridization. We illustrate the orbitals and electron distribution in an isolated carbon atom and in the bonded atom in CH4 in Figure \(11\). The four valence electrons of the carbon atom are distributed equally in the hybrid orbitals, and each carbon electron pairs with a hydrogen electron when the C–H bonds form.
In a methane molecule, the 1s orbital of each of the four hydrogen atoms overlaps with one of the four sp3 orbitals of the carbon atom to form a sigma (σ) bond. This results in the formation of four strong, equivalent covalent bonds between the carbon atom and each of the hydrogen atoms to produce the methane molecule, CH4.
The structure of ethane, C2H6, is similar to that of methane in that each carbon in ethane has four neighboring atoms arranged at the corners of a tetrahedron—three hydrogen atoms and one carbon atom (Figure \(12\)). However, in ethane an sp3 orbital of one carbon atom overlaps end to end with an sp3 orbital of a second carbon atom to form a σ bond between the two carbon atoms. Each of the remaining sp3 hybrid orbitals overlaps with an s orbital of a hydrogen atom to form carbon–hydrogen σ bonds. The structure and overall outline of the bonding orbitals of ethane are shown in Figure \(12\). The orientation of the two CH3 groups is not fixed relative to each other. Experimental evidence shows that rotation around σ bonds occurs easily.
An sp3 hybrid orbital can also hold a lone pair of electrons. For example, the nitrogen atom in ammonia is surrounded by three bonding pairs and a lone pair of electrons directed to the four corners of a tetrahedron. The nitrogen atom is sp3 hybridized with one hybrid orbital occupied by the lone pair.
The molecular structure of water is consistent with a tetrahedral arrangement of two lone pairs and two bonding pairs of electrons. Thus we say that the oxygen atom is sp3 hybridized, with two of the hybrid orbitals occupied by lone pairs and two by bonding pairs. Since lone pairs occupy more space than bonding pairs, structures that contain lone pairs have bond angles slightly distorted from the ideal. Perfect tetrahedra have angles of 109.5°, but the observed angles in ammonia (107.3°) and water (104.5°) are slightly smaller. Other examples of sp3 hybridization include CCl4, PCl3, and NCl3.
sp3d and sp3d2 Hybridization
To describe the five bonding orbitals in a trigonal bipyramidal arrangement, we must use five of the valence shell atomic orbitals (the s orbital, the three p orbitals, and one of the d orbitals), which gives five sp3d hybrid orbitals. With an octahedral arrangement of six hybrid orbitals, we must use six valence shell atomic orbitals (the s orbital, the three p orbitals, and two of the d orbitals in its valence shell), which gives six sp3d2 hybrid orbitals. These hybridizations are only possible for atoms that have d orbitals in their valence subshells (that is, not those in the first or second period).
In a molecule of phosphorus pentachloride, PCl5, there are five P–Cl bonds (thus five pairs of valence electrons around the phosphorus atom) directed toward the corners of a trigonal bipyramid. We use the 3s orbital, the three 3p orbitals, and one of the 3d orbitals to form the set of five sp3d hybrid orbitals (Figure \(14\)) that are involved in the P–Cl bonds. Other atoms that exhibit sp3d hybridization include the sulfur atom in SF4 and the chlorine atoms in ClF3 and in (The electrons on fluorine atoms are omitted for clarity.)
The sulfur atom in sulfur hexafluoride, SF6, exhibits sp3d2 hybridization. A molecule of sulfur hexafluoride has six bonding pairs of electrons connecting six fluorine atoms to a single sulfur atom. There are no lone pairs of electrons on the central atom. To bond six fluorine atoms, the 3s orbital, the three 3p orbitals, and two of the 3d orbitals form six equivalent sp3d2 hybrid orbitals, each directed toward a different corner of an octahedron. Other atoms that exhibit sp3d2 hybridization include the phosphorus atom in the iodine atom in the interhalogens IF5, and the xenon atom in XeF4.
Assignment of Hybrid Orbitals to Central Atoms
The hybridization of an atom is determined based on the number of regions of electron density that surround it. The geometrical arrangements characteristic of the various sets of hybrid orbitals are shown in Figure \(16\). These arrangements are identical to those of the electron-pair geometries predicted by VSEPR theory. VSEPR theory predicts the shapes of molecules, and hybrid orbital theory provides an explanation for how those shapes are formed. To find the hybridization of a central atom, we can use the following guidelines:
1. Determine the Lewis structure of the molecule.
2. Determine the number of regions of electron density around an atom using VSEPR theory, in which single bonds, multiple bonds, radicals, and lone pairs each count as one region.
3. Assign the set of hybridized orbitals from Figure \(16\) that corresponds to this geometry.
It is important to remember that hybridization was devised to rationalize experimentally observed molecular geometries. The model works well for molecules containing small central atoms, in which the valence electron pairs are close together in space. However, for larger central atoms, the valence-shell electron pairs are farther from the nucleus, and there are fewer repulsions. Their compounds exhibit structures that are often not consistent with VSEPR theory, and hybridized orbitals are not necessary to explain the observed data. For example, we have discussed the H–O–H bond angle in H2O, 104.5°, which is more consistent with sp3 hybrid orbitals (109.5°) on the central atom than with 2p orbitals (90°). Sulfur is in the same group as oxygen, and H2S has a similar Lewis structure. However, it has a much smaller bond angle (92.1°), which indicates much less hybridization on sulfur than oxygen. Continuing down the group, tellurium is even larger than sulfur, and for H2Te, the observed bond angle (90°) is consistent with overlap of the 5p orbitals, without invoking hybridization. We invoke hybridization where it is necessary to explain the observed structures.
Example \(1\): Assigning Hybridization
Ammonium sulfate is important as a fertilizer. What is the hybridization of the sulfur atom in the sulfate ion,
Solution
The Lewis structure of sulfate shows there are four regions of electron density. The hybridization is sp3.
Exercise \(1\)
What is the hybridization of the selenium atom in SeF4?
Answer
The selenium atom is sp3d hybridized.
Example \(2\): Assigning Hybridization
Urea, NH2C(O)NH2, is sometimes used as a source of nitrogen in fertilizers. What is the hybridization of the carbon atom in urea?
Solution
The Lewis structure of urea is
The carbon atom is surrounded by three regions of electron density, positioned in a trigonal planar arrangement. The hybridization in a trigonal planar electron pair geometry is sp2 (Figure \(16\)), which is the hybridization of the carbon atom in urea.
Exercise \(1\)
Acetic acid, H3CC(O)OH, is the molecule that gives vinegar its odor and sour taste. What is the hybridization of the two carbon atoms in acetic acid?
Answer
H3C, sp3; C(O)OH, sp2
Footnotes
• 1Note that orbitals may sometimes be drawn in an elongated “balloon” shape rather than in a more realistic “plump” shape in order to make the geometry easier to visualize. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/08%3A_Advanced_Theories_of_Covalent_Bonding/8.02%3A_Hybrid_Atomic_Orbitals.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe multiple covalent bonding in terms of atomic orbital overlap
• Relate the concept of resonance to π-bonding and electron delocalization
The hybrid orbital model appears to account well for the geometry of molecules involving single covalent bonds. Is it also capable of describing molecules containing double and triple bonds? We have already discussed that multiple bonds consist of σ and π bonds. Next we can consider how we visualize these components and how they relate to hybrid orbitals. The Lewis structure of ethene, C2H4, shows us that each carbon atom is surrounded by one other carbon atom and two hydrogen atoms.
The three bonding regions form a trigonal planar electron-pair geometry. Thus we expect the σ bonds from each carbon atom are formed using a set of sp2 hybrid orbitals that result from hybridization of two of the 2p orbitals and the 2s orbital (Figure \(1\)).
These orbitals form the C–H single bonds and the σ bond in the Figure \(2\). The π bond in the C=C double bond results from the overlap of the third (remaining) 2p orbital on each carbon atom that is not involved in hybridization. This unhybridized p orbital (lobes shown in red and blue in Figure \(2\)) is perpendicular to the plane of the sp2 hybrid orbitals. Thus the unhybridized 2p orbitals overlap in a side-by-side fashion, above and below the internuclear axis (Figure \(2\)) and form a π bond.
In an ethene molecule, the four hydrogen atoms and the two carbon atoms are all in the same plane. If the two planes of sp2 hybrid orbitals tilted relative to each other, the p orbitals would not be oriented to overlap efficiently to create the π bond. The planar configuration for the ethene molecule occurs because it is the most stable bonding arrangement. This is a significant difference between σ and π bonds; rotation around single (σ) bonds occurs easily because the end-to-end orbital overlap does not depend on the relative orientation of the orbitals on each atom in the bond. In other words, rotation around the internuclear axis does not change the extent to which the σ bonding orbitals overlap because the bonding electron density is symmetric about the axis. Rotation about the internuclear axis is much more difficult for multiple bonds; however, this would drastically alter the off-axis overlap of the π bonding orbitals, essentially breaking the π bond.
In molecules with sp hybrid orbitals, two unhybridized p orbitals remain on the atom (Figure \(3\)). We find this situation in acetylene, \(\ce{H−C≡C−H}\) (Figure \(4\)), The remaining sp orbitals form σ bonds with hydrogen atoms. The two unhybridized p orbitals per carbon are positioned such that they overlap side by side and, hence, form two π bonds. The two carbon atoms of acetylene are thus bound together by one σ bond and two π bonds, giving a triple bond.
Hybridization involves only σ bonds, lone pairs of electrons, and single unpaired electrons (radicals). Structures that account for these features describe the correct hybridization of the atoms. However, many structures also include resonance forms. Remember that resonance forms occur when various arrangements of π bonds are possible. Since the arrangement of π bonds involves only the unhybridized orbitals, resonance does not influence the assignment of hybridization.
For example, molecule benzene has two resonance forms (Figure \(5\)). We can use either of these forms to determine that each of the carbon atoms is bonded to three other atoms with no lone pairs, so the correct hybridization is sp2. The electrons in the unhybridized p orbitals form π bonds. Neither resonance structure completely describes the electrons in the π bonds. They are not located in one position or the other, but in reality are delocalized throughout the ring. Valence bond theory does not easily address delocalization. Bonding in molecules with resonance forms is better described by molecular orbital theory. (See the next module.)
Example \(1\): Assignment of Hybridization Involving Resonance
Some acid rain results from the reaction of sulfur dioxide with atmospheric water vapor, followed by the formation of sulfuric acid. Sulfur dioxide, SO2, is a major component of volcanic gases as well as a product of the combustion of sulfur-containing coal. What is the hybridization of the S atom in SO2?
Solution
The resonance structures of SO2 are
The sulfur atom is surrounded by two bonds and one lone pair of electrons in either resonance structure. Therefore, the electron-pair geometry is trigonal planar, and the hybridization of the sulfur atom is sp2.
Exercise \(1\)
Another acid in acid rain is nitric acid, HNO3, which is produced by the reaction of nitrogen dioxide, NO2, with atmospheric water vapor. What is the hybridization of the nitrogen atom in NO2? (Note: the lone electron on nitrogen occupies a hybridized orbital just as a lone pair would.)
Answer
sp2 | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/08%3A_Advanced_Theories_of_Covalent_Bonding/8.03%3A_Multiple_Bonds.txt |
Learning Objectives
By the end of this section, you will be able to:
• Outline the basic quantum-mechanical approach to deriving molecular orbitals from atomic orbitals
• Describe traits of bonding and antibonding molecular orbitals
• Calculate bond orders based on molecular electron configurations
• Write molecular electron configurations for first- and second-row diatomic molecules
• Relate these electron configurations to the molecules’ stabilities and magnetic properties
For almost every covalent molecule that exists, we can now draw the Lewis structure, predict the electron-pair geometry, predict the molecular geometry, and come close to predicting bond angles. However, one of the most important molecules we know, the oxygen molecule O2, presents a problem with respect to its Lewis structure. We would write the following Lewis structure for O2:
This electronic structure adheres to all the rules governing Lewis theory. There is an O=O double bond, and each oxygen atom has eight electrons around it. However, this picture is at odds with the magnetic behavior of oxygen. By itself, O2 is not magnetic, but it is attracted to magnetic fields. Thus, when we pour liquid oxygen past a strong magnet, it collects between the poles of the magnet and defies gravity. Such attraction to a magnetic field is called paramagnetism, and it arises in molecules that have unpaired electrons. And yet, the Lewis structure of O2 indicates that all electrons are paired. How do we account for this discrepancy?
Magnetic susceptibility measures the force experienced by a substance in a magnetic field. When we compare the weight of a sample to the weight measured in a magnetic field (Figure $1$), paramagnetic samples that are attracted to the magnet will appear heavier because of the force exerted by the magnetic field. We can calculate the number of unpaired electrons based on the increase in weight.
Experiments show that each O2 molecule has two unpaired electrons. The Lewis-structure model does not predict the presence of these two unpaired electrons. Unlike oxygen, the apparent weight of most molecules decreases slightly in the presence of an inhomogeneous magnetic field. Materials in which all of the electrons are paired are diamagnetic and weakly repel a magnetic field. Paramagnetic and diamagnetic materials do not act as permanent magnets. Only in the presence of an applied magnetic field do they demonstrate attraction or repulsion.
Link to Learning
Water, like most molecules, contains all paired electrons. Living things contain a large percentage of water, so they demonstrate diamagnetic behavior. If you place a frog near a sufficiently large magnet, it will levitate. You can see videos of diamagnetic floating frogs, strawberries, and more.
Molecular orbital theory (MO theory) provides an explanation of chemical bonding that accounts for the paramagnetism of the oxygen molecule. It also explains the bonding in a number of other molecules, such as violations of the octet rule and more molecules with more complicated bonding (beyond the scope of this text) that are difficult to describe with Lewis structures. Additionally, it provides a model for describing the energies of electrons in a molecule and the probable location of these electrons. Unlike valence bond theory, which uses hybrid orbitals that are assigned to one specific atom, MO theory uses the combination of atomic orbitals to yield molecular orbitals that are delocalized over the entire molecule rather than being localized on its constituent atoms. MO theory also helps us understand why some substances are electrical conductors, others are semiconductors, and still others are insulators. Table $1$ summarizes the main points of the two complementary bonding theories. Both theories provide different, useful ways of describing molecular structure.
Table $1$: Comparison of Bonding Theories
Valence Bond Theory Molecular Orbital Theory
considers bonds as localized between one pair of atoms considers electrons delocalized throughout the entire molecule
creates bonds from overlap of atomic orbitals (s, p, d…) and hybrid orbitals (sp, sp2, sp3…) combines atomic orbitals to form molecular orbitals (σ, σ*, π, π*)
forms σ or π bonds creates bonding and antibonding interactions based on which orbitals are filled
predicts molecular shape based on the number of regions of electron density predicts the arrangement of electrons in molecules
needs multiple structures to describe resonance
Molecular orbital theory describes the distribution of electrons in molecules in much the same way that the distribution of electrons in atoms is described using atomic orbitals. Using quantum mechanics, the behavior of an electron in a molecule is still described by a wave function, Ψ, analogous to the behavior in an atom. Just like electrons around isolated atoms, electrons around atoms in molecules are limited to discrete (quantized) energies. The region of space in which a valence electron in a molecule is likely to be found is called a molecular orbital (Ψ2). Like an atomic orbital, a molecular orbital is full when it contains two electrons with opposite spin.
We will consider the molecular orbitals in molecules composed of two identical atoms (H2 or Cl2, for example). Such molecules are called homonuclear diatomic molecules. In these diatomic molecules, several types of molecular orbitals occur.
The mathematical process of combining atomic orbitals to generate molecular orbitals is called the linear combination of atomic orbitals (LCAO). The wave function describes the wavelike properties of an electron. Molecular orbitals are combinations of atomic orbital wave functions. Combining waves can lead to constructive interference, in which peaks line up with peaks, or destructive interference, in which peaks line up with troughs (Figure $2$). In orbitals, the waves are three dimensional, and they combine with in-phase waves producing regions with a higher probability of electron density and out-of-phase waves producing nodes, or regions of no electron density.
There are two types of molecular orbitals that can form from the overlap of two atomic s orbitals on adjacent atoms. The two types are illustrated in Figure $3$. The in-phase combination produces a lower energy σs molecular orbital (read as "sigma-s") in which most of the electron density is directly between the nuclei. The out-of-phase addition (which can also be thought of as subtracting the wave functions) produces a higher energy molecular orbital (read as "sigma-s-star") molecular orbital in which there is a node between the nuclei. The asterisk signifies that the orbital is an antibonding orbital. Electrons in a σs orbital are attracted by both nuclei at the same time and are more stable (of lower energy) than they would be in the isolated atoms. Adding electrons to these orbitals creates a force that holds the two nuclei together, so we call these orbitals bonding orbitals. Electrons in the orbitals are located well away from the region between the two nuclei. The attractive force between the nuclei and these electrons pulls the two nuclei apart. Hence, these orbitals are called antibonding orbitals. Electrons fill the lower-energy bonding orbital before the higher-energy antibonding orbital, just as they fill lower-energy atomic orbitals before they fill higher-energy atomic orbitals.
Link to Learning
You can watch animations visualizing the calculated atomic orbitals combining to form various molecular orbitals at the Orbitron website.
In p orbitals, the wave function gives rise to two lobes with opposite phases, analogous to how a two-dimensional wave has both parts above and below the average. We indicate the phases by shading the orbital lobes different colors. When orbital lobes of the same phase overlap, constructive wave interference increases the electron density. When regions of opposite phase overlap, the destructive wave interference decreases electron density and creates nodes. When p orbitals overlap end to end, they create σ and σ* orbitals (Figure $4$). If two atoms are located along the x-axis in a Cartesian coordinate system, the two px orbitals overlap end to end and form σpx (bonding) and (antibonding) (read as "sigma-p-x" and "sigma-p-x star," respectively). Just as with s-orbital overlap, the asterisk indicates the orbital with a node between the nuclei, which is a higher-energy, antibonding orbital.
The side-by-side overlap of two p orbitals gives rise to a pi (π) bonding molecular orbital and a π* antibonding molecular orbital, as shown in Figure $5$. In valence bond theory, we describe π bonds as containing a nodal plane containing the internuclear axis and perpendicular to the lobes of the p orbitals, with electron density on either side of the node. In molecular orbital theory, we describe the π orbital by this same shape, and a π bond exists when this orbital contains electrons. Electrons in this orbital interact with both nuclei and help hold the two atoms together, making it a bonding orbital. For the out-of-phase combination, there are two nodal planes created, one along the internuclear axis and a perpendicular one between the nuclei.
In the molecular orbitals of diatomic molecules, each atom also has two sets of p orbitals oriented side by side (py and pz), so these four atomic orbitals combine pairwise to create two π orbitals and two π* orbitals. The $π_{py}$ and orbitals are oriented at right angles to the $π_{pz}$ and $π_{pz}^{*}$ orbitals. Except for their orientation, the $π_{py}$ and $π_{pz}$ orbitals are identical and have the same energy; they are degenerate orbitals. The $π_{py}^{*}$ and $π_{pz}^{*}$ antibonding orbitals are also degenerate and identical except for their orientation. A total of six molecular orbitals results from the combination of the six atomic p orbitals in two atoms: $σ_{px}$ and $σ_{px}^{*}$, $π_{py}$ and $π_{py}^{*}$, $π_{pz}$ and $π_{pz}^{*}$.
Example $1$: Molecular Orbitals
Predict what type (if any) of molecular orbital would result from adding the wave functions so each pair of orbitals shown overlap. The orbitals are all similar in energy.
Solution
1. is an in-phase combination, resulting in a σ3p orbital
2. will not result in a new orbital because the in-phase component (bottom) and out-of-phase component (top) cancel out. Only orbitals with the correct alignment can combine.
3. is an out-of-phase combination, resulting in a $\pi_{3 p}^*) orbital. Exercise \(1$
Label the molecular orbital shown as σ or π, bonding or antibonding and indicate where the node occurs.
Answer
The orbital is located along the internuclear axis, so it is a σ orbital. There is a node bisecting the internuclear axis, so it is an antibonding orbital.
Portrait of a Chemist: Walter Kohn: Nobel Laureate
Walter Kohn (Figure $6$) is a theoretical physicist who studies the electronic structure of solids. His work combines the principles of quantum mechanics with advanced mathematical techniques. This technique, called density functional theory, makes it possible to compute properties of molecular orbitals, including their shape and energies. Kohn and mathematician John Pople were awarded the Nobel Prize in Chemistry in 1998 for their contributions to our understanding of electronic structure. Kohn also made significant contributions to the physics of semiconductors.
Kohn’s biography has been remarkable outside the realm of physical chemistry as well. He was born in Austria, and during World War II he was part of the Kindertransport program that rescued 10,000 children from the Nazi regime. His summer jobs included discovering gold deposits in Canada and helping Polaroid explain how its instant film worked. Dr. Kohn passed away in 2016 at the age of 93.
How Sciences Interconnect: Computational Chemistry in Drug Design
While the descriptions of bonding described in this chapter involve many theoretical concepts, they also have many practical, real-world applications. For example, drug design is an important field that uses our understanding of chemical bonding to develop pharmaceuticals. This interdisciplinary area of study uses biology (understanding diseases and how they operate) to identify specific targets, such as a binding site that is involved in a disease pathway. By modeling the structures of the binding site and potential drugs, computational chemists can predict which structures can fit together and how effectively they will bind (see Figure $7$). Thousands of potential candidates can be narrowed down to a few of the most promising candidates. These candidate molecules are then carefully tested to determine side effects, how effectively they can be transported through the body, and other factors. Dozens of important new pharmaceuticals have been discovered with the aid of computational chemistry, and new research projects are underway.
Molecular Orbital Energy Diagrams
The relative energy levels of atomic and molecular orbitals are typically shown in a molecular orbital diagram (Figure $8$). For a diatomic molecule, the atomic orbitals of one atom are shown on the left, and those of the other atom are shown on the right. Each horizontal line represents one orbital that can hold two electrons. The molecular orbitals formed by the combination of the atomic orbitals are shown in the center. Dashed lines show which of the atomic orbitals combine to form the molecular orbitals. For each pair of atomic orbitals that combine, one lower-energy (bonding) molecular orbital and one higher-energy (antibonding) orbital result. Thus we can see that combining the six 2p atomic orbitals results in three bonding orbitals (one σ and two π) and three antibonding orbitals (one σ* and two π*).
We predict the distribution of electrons in these molecular orbitals by filling the orbitals in the same way that we fill atomic orbitals, by the Aufbau principle. Lower-energy orbitals fill first, electrons spread out among degenerate orbitals before pairing, and each orbital can hold a maximum of two electrons with opposite spins (Figure $8$). Just as we write electron configurations for atoms, we can write the molecular electronic configuration by listing the orbitals with superscripts indicating the number of electrons present. For clarity, we place parentheses around molecular orbitals with the same energy. In this case, each orbital is at a different energy, so parentheses separate each orbital. Thus we would expect a diatomic molecule or ion containing seven electrons (such as $\ce{Be_2^{+}}$) would have the molecular electron configuration $\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^1$. It is common to omit the core electrons from molecular orbital diagrams and configurations and include only the valence electrons.
Bond Order
The filled molecular orbital diagram shows the number of electrons in both bonding and antibonding molecular orbitals. The net contribution of the electrons to the bond strength of a molecule is identified by determining the bond order that results from the filling of the molecular orbitals by electrons.
When using Lewis structures to describe the distribution of electrons in molecules, we define bond order as the number of bonding pairs of electrons between two atoms. Thus a single bond has a bond order of 1, a double bond has a bond order of 2, and a triple bond has a bond order of 3. We define bond order differently when we use the molecular orbital description of the distribution of electrons, but the resulting bond order is usually the same. The MO technique is more accurate and can handle cases when the Lewis structure method fails, but both methods describe the same phenomenon.
In the molecular orbital model, an electron contributes to a bonding interaction if it occupies a bonding orbital and it contributes to an antibonding interaction if it occupies an antibonding orbital. The bond order is calculated by subtracting the destabilizing (antibonding) electrons from the stabilizing (bonding) electrons. Since a bond consists of two electrons, we divide by two to get the bond order. We can determine bond order with the following equation:
$\text { bond order }=\frac{\text { (number of bonding electrons })-(\text { number of antibonding electrons })}{2} \nonumber$
The order of a covalent bond is a guide to its strength; a bond between two given atoms becomes stronger as the bond order increases (Table 8.1). If the distribution of electrons in the molecular orbitals between two atoms is such that the resulting bond would have a bond order of zero, a stable bond does not form. We next look at some specific examples of MO diagrams and bond orders.
Bonding in Diatomic Molecules
A dihydrogen molecule (H2) forms from two hydrogen atoms. When the atomic orbitals of the two atoms combine, the electrons occupy the molecular orbital of lowest energy, the σ1s bonding orbital. A dihydrogen molecule, H2, readily forms because the energy of a H2 molecule is lower than that of two H atoms. The σ1s orbital that contains both electrons is lower in energy than either of the two 1s atomic orbitals.
A molecular orbital can hold two electrons, so both electrons in the $\ce{H2}$ molecule are in the σ1s bonding orbital; the electron configuration is $\left(\sigma_{1 s}\right)^2$. We represent this configuration by a molecular orbital energy diagram (Figure $9$) in which a single upward arrow indicates one electron in an orbital, and two (upward and downward) arrows indicate two electrons of opposite spin.
A dihydrogen molecule contains two bonding electrons and no antibonding electrons so we have
$\text { bond order in } H_2=\frac{(2-0)}{2}=1 \nonumber$
Because the bond order for the H–H bond is equal to 1, the bond is a single bond.
A helium atom has two electrons, both of which are in its 1s orbital. Two helium atoms do not combine to form a dihelium molecule, He2, with four electrons, because the stabilizing effect of the two electrons in the lower-energy bonding orbital would be offset by the destabilizing effect of the two electrons in the higher-energy antibonding molecular orbital. We would write the hypothetical electron configuration of He2 as
$\text { bond order in } He_2=\frac{(2-2)}{2}=0 \nonumber$
A bond order of zero indicates that no bond is formed between two atoms.
The Diatomic Molecules of the Second Period
Eight possible homonuclear diatomic molecules might be formed by the atoms of the second period of the periodic table: Li2, Be2, B2, C2, N2, O2, F2, and Ne2. However, we can predict that the Be2 molecule and the Ne2 molecule would not be stable. We can see this by a consideration of the molecular electron configurations (Table $2$).
We predict valence molecular orbital electron configurations just as we predict electron configurations of atoms. Valence electrons are assigned to valence molecular orbitals with the lowest possible energies. Consistent with Hund’s rule, whenever there are two or more degenerate molecular orbitals, electrons fill each orbital of that type singly before any pairing of electrons takes place.
As we saw in valence bond theory, σ bonds are generally more stable than π bonds formed from degenerate atomic orbitals. Similarly, in molecular orbital theory, σ orbitals are usually more stable than π orbitals. However, this is not always the case. The MOs for the valence orbitals of the second period are shown in Figure $11$. Looking at Ne2 molecular orbitals, we see that the order is consistent with the generic diagram shown in the previous section. However, for atoms with three or fewer electrons in the p orbitals (Li through N) we observe a different pattern, in which the σp orbital is higher in energy than the πp set. Obtain the molecular orbital diagram for a homonuclear diatomic ion by adding or subtracting electrons from the diagram for the neutral molecule.
This switch in orbital ordering occurs because of a phenomenon called s-p mixing. s-p mixing does not create new orbitals; it merely influences the energies of the existing molecular orbitals. The σs wavefunction mathematically combines with the σp wavefunction, with the result that the σs orbital becomes more stable, and the σp orbital becomes less stable (Figure $12$). Similarly, the antibonding orbitals also undergo s-p mixing, with the σs* becoming more stable and the σp* becoming less stable.
s-p mixing occurs when the s and p orbitals have similar energies. The energy difference between 2s and 2p orbitals in O, F, and Ne is greater than that in Li, Be, B, C, and N. Because of this, O2, F2, and Ne2 exhibit negligible s-p mixing (not sufficient to change the energy ordering), and their MO diagrams follow the normal pattern, as shown in Figure $11$. All of the other period 2 diatomic molecules do have s-p mixing, which leads to the pattern where the σp orbital is raised above the πp set.
Using the MO diagrams shown in Figure $11$: , we can add in the electrons and determine the molecular electron configuration and bond order for each of the diatomic molecules. As shown in Table $2$, Be2 and Ne2 molecules would have a bond order of 0, and these molecules do not exist.
Table $2$: Electron Configuration and Bond Order for Molecular Orbitals in Homonuclear Diatomic Molecules of Period Two Elements
Molecule Electron Configuration Bond Order
Li2 $\left(\sigma_{2 s}\right)^2$ 1
Be2 (unstable) $\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2$ 0
B2 $\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\pi_{2 p y}, \pi_{2 p z}\right)^2$ 1
C2 $\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\pi_{2 p y}, \pi_{2 p z}\right)^4$ 2
N2 $\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\pi_{2 p y}, \pi_{2 p z}\right)^4\left(\sigma_{2 p x}\right)^2$ 3
O2 $\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p x}\right)^2\left(\pi_{2 p y}, \pi_{2 p z}\right)^4\left(\pi_{2 p y}^*, \pi_{2 p z}^*\right)^2$ 2
F2 $\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p x}\right)^2\left(\pi_{2 p y}, \pi_{2 p z}\right)^4\left(\pi_{2 p y}^*, \pi_{2 p z}^*\right)^4$ 1
Ne2 (unstable) $\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p x}\right)^2\left(\pi_{2 p y}, \pi_{2 p z}\right)^4\left(\pi_{2 p y}^*, \pi_{2 p z}^*\right)^4\left(\sigma_{2 p x}^*\right)^2$ 0
The combination of two lithium atoms to form a lithium molecule, Li2, is analogous to the formation of H2, but the atomic orbitals involved are the valence 2s orbitals. Each of the two lithium atoms has one valence electron. Hence, we have two valence electrons available for the σ2s bonding molecular orbital. Because both valence electrons would be in a bonding orbital, we would predict the Li2 molecule to be stable. The molecule is, in fact, present in appreciable concentration in lithium vapor at temperatures near the boiling point of the element. All of the other molecules in Table $2$ with a bond order greater than zero are also known.
The O2 molecule has enough electrons to half fill the $\left(\pi_{2 p y}^*, \pi_{2 p z}^*\right)$ level. We expect the two electrons that occupy these two degenerate orbitals to be unpaired, and this molecular electronic configuration for O2 is in accord with the fact that the oxygen molecule has two unpaired electrons (Figure 8.40). The presence of two unpaired electrons has proved to be difficult to explain using Lewis structures, but the molecular orbital theory explains it quite well. In fact, the unpaired electrons of the oxygen molecule provide a strong piece of support for the molecular orbital theory.
How Sciences Interconnect: Band Theory
When two identical atomic orbitals on different atoms combine, two molecular orbitals result (see Figure $3$). The bonding orbital is lower in energy than the original atomic orbitals because the atomic orbitals are in-phase in the molecular orbital. The antibonding orbital is higher in energy than the original atomic orbitals because the atomic orbitals are out-of-phase.
In a solid, similar things happen, but on a much larger scale. Remember that even in a small sample there are a huge number of atoms (typically > 1023 atoms), and therefore a huge number of atomic orbitals that may be combined into molecular orbitals. When N valence atomic orbitals, all of the same energy and each containing one (1) electron, are combined, N/2 (filled) bonding orbitals and N/2 (empty) antibonding orbitals will result. Each bonding orbital will show an energy lowering as the atomic orbitals are mostly in-phase, but each of the bonding orbitals will be a little different and have slightly different energies. The antibonding orbitals will show an increase in energy as the atomic orbitals are mostly out-of-phase, but each of the antibonding orbitals will also be a little different and have slightly different energies. The allowed energy levels for all the bonding orbitals are so close together that they form a band, called the valence band. Likewise, all the antibonding orbitals are very close together and form a band, called the conduction band. Figure $13$ shows the bands for three important classes of materials: insulators, semiconductors, and conductors.
In order to conduct electricity, electrons must move from the filled valence band to the empty conduction band where they can move throughout the solid. The size of the band gap, or the energy difference between the top of the valence band and the bottom of the conduction band, determines how easy it is to move electrons between the bands. Only a small amount of energy is required in a conductor because the band gap is very small. This small energy difference is “easy” to overcome, so they are good conductors of electricity. In an insulator, the band gap is so “large” that very few electrons move into the conduction band; as a result, insulators are poor conductors of electricity. Semiconductors conduct electricity when “moderate” amounts of energy are provided to move electrons out of the valence band and into the conduction band. Semiconductors, such as silicon, are found in many electronics.
Semiconductors are used in devices such as computers, smartphones, and solar cells. Solar cells produce electricity when light provides the energy to move electrons out of the valence band. The electricity that is generated may then be used to power a light or tool, or it can be stored for later use by charging a battery. As of December 2014, up to 46% of the energy in sunlight could be converted into electricity using solar cells.
Example $2$: Molecular Orbital Diagrams, Bond Order, and Number of Unpaired Electrons
Draw the molecular orbital diagram for the oxygen molecule, O2. From this diagram, calculate the bond order for O2. How does this diagram account for the paramagnetism of O2?
Solution
We draw a molecular orbital energy diagram similar to that shown in Figure $11$. Each oxygen atom contributes six electrons, so the diagram appears as shown in Figure $14$.
We calculate the bond order as
$O_2=\frac{(8-4)}{2}=2 \nonumber$
Oxygen's paramagnetism is explained by the presence of two unpaired electrons in the (π2py, π2pz)* molecular orbitals.
Exercise $2$
The main component of air is N2. From the molecular orbital diagram of N2, predict its bond order and whether it is diamagnetic or paramagnetic.
Answer
N2 has a bond order of 3 and is diamagnetic.
Example $3$: Ion Predictions with MO Diagrams
Give the molecular orbital configuration for the valence electrons in $\ce{C_2^{2-}}$. Will this ion be stable?
Solution
Looking at the appropriate MO diagram, we see that the π orbitals are lower in energy than the σp orbital. The valence electron configuration for C2 is $\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\pi_{2 p y}, \pi_{2 p z}\right)^4$. Adding two more electrons to generate the $\ce{C_2^{2-}}$ anion will give a valence electron configuration of $\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\pi_{2 p y}, \pi_{2 p z}\right)^4\left(\sigma_{2 p x}\right)^2$. Since this has six more bonding electrons than antibonding, the bond order will be 3, and the ion should be stable.
Exercise $3$
How many unpaired electrons would be present on a $\ce{Be_2^{2-}}$ ion? Would it be paramagnetic or diamagnetic?
Answer
two, paramagnetic
Link to Learning
Creating molecular orbital diagrams for molecules with more than two atoms relies on the same basic ideas as the diatomic examples presented here. However, with more atoms, computers are required to calculate how the atomic orbitals combine. See three-dimensional drawings of the molecular orbitals for $\ce{C6H6}$. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/08%3A_Advanced_Theories_of_Covalent_Bonding/8.04%3A_Molecular_Orbital_Theory.txt |
Example and Directions
Words (or words that have the same definition)The definition is case sensitive(Optional) Image to display with the definition [Not displayed in Glossary, only in pop-up on pages](Optional) Caption for Image(Optional) External or Internal Link(Optional) Source for Definition
(Eg. "Genetic, Hereditary, DNA ...")(Eg. "Relating to genes or heredity")The infamous double helix https://bio.libretexts.org/CC-BY-SA; Delmar Larsen
Glossary Entries
Word(s)DefinitionImageCaptionLinkSource
antibonding orbitalmolecular orbital located outside of the region between two nuclei; electrons in an antibonding orbital destabilize the molecule
bond ordernumber of pairs of electrons between two atoms; it can be found by the number of bonds in a Lewis structure or by the difference between the number of bonding and antibonding electrons divided by two
bonding orbitalmolecular orbital located between two nuclei; electrons in a bonding orbital stabilize a molecule
degenerate orbitalsorbitals that have the same energy
diamagnetismphenomenon in which a material is not magnetic itself but is repelled by a magnetic field; it occurs when there are only paired electrons present
homonuclear diatomic moleculemolecule consisting of two identical atoms
hybrid orbitalorbital created by combining atomic orbitals on a central atom
hybridizationmodel that describes the changes in the atomic orbitals of an atom when it forms a covalent compound
linear combination of atomic orbitalstechnique for combining atomic orbitals to create molecular orbitals
molecular orbitalregion of space in which an electron has a high probability of being found in a molecule
molecular orbital diagramvisual representation of the relative energy levels of molecular orbitals
molecular orbital theorymodel that describes the behavior of electrons delocalized throughout a molecule in terms of the combination of atomic wave functions
nodeplane separating different lobes of orbitals, where the probability of finding an electron is zero
overlapcoexistence of orbitals from two different atoms sharing the same region of space, leading to the formation of a covalent bond
paramagnetismphenomenon in which a material is not magnetic itself but is attracted to a magnetic field; it occurs when there are unpaired electrons present
pi bond (π bond)covalent bond formed by side-by-side overlap of atomic orbitals; the electron density is found on opposite sides of the internuclear axis
s-p mixingchange that causes σp orbitals to be less stable than πp orbitals due to the mixing of s and p-based molecular orbitals of similar energies.
sigma bond (σ bond)covalent bond formed by overlap of atomic orbitals along the internuclear axis
sp hybrid orbitalone of a set of two orbitals with a linear arrangement that results from combining one s and one p orbital
sp2 hybrid orbitalone of a set of three orbitals with a trigonal planar arrangement that results from combining one s and two p orbitals
sp3 hybrid orbitalone of a set of four orbitals with a tetrahedral arrangement that results from combining one s and three p orbitals
sp3d hybrid orbitalone of a set of five orbitals with a trigonal bipyramidal arrangement that results from combining one s, three p, and one d orbital
sp3d2 hybrid orbitalone of a set of six orbitals with an octahedral arrangement that results from combining one s, three p, and two d orbitals
valence bond theorydescription of bonding that involves atomic orbitals overlapping to form σ or π bonds, within which pairs of electrons are shared
π bonding orbitalmolecular orbital formed by side-by-side overlap of atomic orbitals, in which the electron density is found on opposite sides of the internuclear axis
π* bonding orbitalantibonding molecular orbital formed by out of phase side-by-side overlap of atomic orbitals, in which the electron density is found on both sides of the internuclear axis, and there is a node between the nuclei
σ bonding orbitalmolecular orbital in which the electron density is found along the axis of the bond
σ* bonding orbitalantibonding molecular orbital formed by out-of-phase overlap of atomic orbital along the axis of the bond, generating a node between the nuclei
8.07: Summary
Valence bond theory describes bonding as a consequence of the overlap of two separate atomic orbitals on different atoms that creates a region with one pair of electrons shared between the two atoms. When the orbitals overlap along an axis containing the nuclei, they form a σ bond. When they overlap in a fashion that creates a node along this axis, they form a π bond.
We can use hybrid orbitals, which are mathematical combinations of some or all of the valence atomic orbitals, to describe the electron density around covalently bonded atoms. These hybrid orbitals either form sigma (σ) bonds directed toward other atoms of the molecule or contain lone pairs of electrons. We can determine the type of hybridization around a central atom from the geometry of the regions of electron density about it. Two such regions imply sp hybridization; three, sp2 hybridization; four, sp3 hybridization; five, sp3d hybridization; and six, sp3d2 hybridization. Pi (π) bonds are formed from unhybridized atomic orbitals (p or d orbitals).
Multiple bonds consist of a σ bond located along the axis between two atoms and one or two π bonds. The σ bonds are usually formed by the overlap of hybridized atomic orbitals, while the π bonds are formed by the side-by-side overlap of unhybridized orbitals. Resonance occurs when there are multiple unhybridized orbitals with the appropriate alignment to overlap, so the placement of π bonds can vary.
Molecular orbital (MO) theory describes the behavior of electrons in a molecule in terms of combinations of the atomic wave functions. The resulting molecular orbitals may extend over all the atoms in the molecule. Bonding molecular orbitals are formed by in-phase combinations of atomic wave functions, and electrons in these orbitals stabilize a molecule. Antibonding molecular orbitals result from out-of-phase combinations of atomic wave functions and electrons in these orbitals make a molecule less stable. Molecular orbitals located along an internuclear axis are called σ MOs. They can be formed from s orbitals or from p orbitals oriented in an end-to-end fashion. Molecular orbitals formed from p orbitals oriented in a side-by-side fashion have electron density on opposite sides of the internuclear axis and are called π orbitals.
We can describe the electronic structure of diatomic molecules by applying molecular orbital theory to the valence electrons of the atoms. Electrons fill molecular orbitals following the same rules that apply to filling atomic orbitals; Hund’s rule and the Aufbau principle tell us that lower-energy orbitals will fill first, electrons will spread out before they pair up, and each orbital can hold a maximum of two electrons with opposite spins. Materials with unpaired electrons are paramagnetic and attracted to a magnetic field, while those with all-paired electrons are diamagnetic and repelled by a magnetic field. Correctly predicting the magnetic properties of molecules is in advantage of molecular orbital theory over Lewis structures and valence bond theory. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/08%3A_Advanced_Theories_of_Covalent_Bonding/8.05%3A_Key_Terms.txt |
1.
Explain how σ and π bonds are similar and how they are different.
2.
Draw a curve that describes the energy of a system with H and Cl atoms at varying distances. Then, find the minimum energy of this curve two ways.
1. Use the bond energy found in Table 8.1 to calculate the energy for one single HCl bond (Hint: How many bonds are in a mole?)
2. Use the enthalpy of reaction and the bond energies for H2 and Cl2 to solve for the energy of one mole of HCl bonds.
3.
Explain why bonds occur at specific average bond distances instead of the atoms approaching each other infinitely close.
4.
Use valence bond theory to explain the bonding in F2, HF, and ClBr. Sketch the overlap of the atomic orbitals involved in the bonds.
5.
Use valence bond theory to explain the bonding in O2. Sketch the overlap of the atomic orbitals involved in the bonds in O2.
6.
How many σ and π bonds are present in the molecule HCN?
7.
A friend tells you N2 has three π bonds due to overlap of the three p-orbitals on each N atom. Do you agree?
8.
Draw the Lewis structures for CO2 and CO, and predict the number of σ and π bonds for each molecule.
1. CO2
2. CO
9.
Why is the concept of hybridization required in valence bond theory?
10.
Give the shape that describes each hybrid orbital set:
1. sp2
2. sp3d
3. sp
4. sp3d2
11.
Explain why a carbon atom cannot form five bonds using sp3d hybrid orbitals.
12.
What is the hybridization of the central atom in each of the following?
1. BeH2
2. SF6
3. PCl5
13.
A molecule with the formula AB3 could have one of four different shapes. Give the shape and the hybridization of the central A atom for each.
14.
Methionine, CH3SCH2CH2CH(NH2)CO2H, is an amino acid found in proteins. The Lewis structure of this compound is shown below. What is the hybridization type of each carbon, oxygen, the nitrogen, and the sulfur?
15.
Sulfuric acid is manufactured by a series of reactions represented by the following equations:
Draw a Lewis structure, predict the molecular geometry by VSEPR, and determine the hybridization of sulfur for the following:
1. circular S8 molecule
2. SO2 molecule
3. SO3 molecule
4. H2SO4 molecule (the hydrogen atoms are bonded to oxygen atoms)
16.
Two important industrial chemicals, ethene, C2H4, and propene, C3H6, are produced by the steam (or thermal) cracking process:
For each of the four carbon compounds, do the following:
1. Draw a Lewis structure.
2. Predict the geometry about the carbon atom.
3. Determine the hybridization of each type of carbon atom.
17.
Analysis of a compound indicates that it contains 77.55% Xe and 22.45% F by mass.
1. What is the empirical formula for this compound? (Assume this is also the molecular formula in responding to the remaining parts of this exercise).
2. Write a Lewis structure for the compound.
3. Predict the shape of the molecules of the compound.
4. What hybridization is consistent with the shape you predicted?
18.
Consider nitrous acid, HNO2 (HONO).
1. Write a Lewis structure.
2. What are the electron pair and molecular geometries of the internal oxygen and nitrogen atoms in the HNO2 molecule?
3. What is the hybridization on the internal oxygen and nitrogen atoms in HNO2?
19.
Strike-anywhere matches contain a layer of KClO3 and a layer of P4S3. The heat produced by the friction of striking the match causes these two compounds to react vigorously, which sets fire to the wooden stem of the match. KClO3 contains the ion. P4S3 is an unusual molecule with the skeletal structure.
1. Write Lewis structures for P4S3 and the ion.
2. Describe the geometry about the P atoms, the S atom, and the Cl atom in these species.
3. Assign a hybridization to the P atoms, the S atom, and the Cl atom in these species.
4. Determine the oxidation states and formal charge of the atoms in P4S3 and the ion.
20.
Identify the hybridization of each carbon atom in the following molecule. (The arrangement of atoms is given; you need to determine how many bonds connect each pair of atoms.)
21.
Write Lewis structures for NF3 and PF5. On the basis of hybrid orbitals, explain the fact that NF3, PF3, and PF5 are stable molecules, but NF5 does not exist.
22.
In addition to NF3, two other fluoro derivatives of nitrogen are known: N2F4 and N2F2. What shapes do you predict for these two molecules? What is the hybridization for the nitrogen in each molecule?
23.
The bond energy of a C–C single bond averages 347 kJ mol−1; that of a triple bond averages 839 kJ mol−1. Explain why the triple bond is not three times as strong as a single bond.
24.
For the carbonate ion, draw all of the resonance structures. Identify which orbitals overlap to create each bond.
25.
A useful solvent that will dissolve salts as well as organic compounds is the compound acetonitrile, H3CCN. It is present in paint strippers.
1. Write the Lewis structure for acetonitrile, and indicate the direction of the dipole moment in the molecule.
2. Identify the hybrid orbitals used by the carbon atoms in the molecule to form σ bonds.
3. Describe the atomic orbitals that form the π bonds in the molecule. Note that it is not necessary to hybridize the nitrogen atom.
26.
For the molecule allene, give the hybridization of each carbon atom. Will the hydrogen atoms be in the same plane or perpendicular planes?
27.
Identify the hybridization of the central atom in each of the following molecules and ions that contain multiple bonds:
1. (f) XeO2F2 (Xe is the central atom)
(g) (Cl is the central atom)
28.
Describe the molecular geometry and hybridization of the N, P, or S atoms in each of the following compounds.
1. H3PO4, phosphoric acid, used in cola soft drinks
2. NH4NO3, ammonium nitrate, a fertilizer and explosive
3. S2Cl2, disulfur dichloride, used in vulcanizing rubber
4. K4[O3POPO3], potassium pyrophosphate, an ingredient in some toothpastes
29.
For each of the following molecules, indicate the hybridization requested and whether or not the electrons will be delocalized:
1. ozone (O3) central O hybridization
2. carbon dioxide (CO2) central C hybridization
3. nitrogen dioxide (NO2) central N hybridization
4. phosphate ion central P hybridization
30.
For each of the following structures, determine the hybridization requested and whether the electrons will be delocalized:
(a) Hybridization of each carbon
(b) Hybridization of sulfur
(c) All atoms
31.
Draw the orbital diagram for carbon in CO2 showing how many carbon atom electrons are in each orbital.
32.
Sketch the distribution of electron density in the bonding and antibonding molecular orbitals formed from two s orbitals and from two p orbitals.
33.
How are the following similar, and how do they differ?
1. σ molecular orbitals and π molecular orbitals
2. ψ for an atomic orbital and ψ for a molecular orbital
3. bonding orbitals and antibonding orbitals
34.
If molecular orbitals are created by combining five atomic orbitals from atom A and five atomic orbitals from atom B combine, how many molecular orbitals will result?
35.
Can a molecule with an odd number of electrons ever be diamagnetic? Explain why or why not.
36.
Can a molecule with an even number of electrons ever be paramagnetic? Explain why or why not.
37.
Why are bonding molecular orbitals lower in energy than the parent atomic orbitals?
38.
Calculate the bond order for an ion with this configuration:
39.
Explain why an electron in the bonding molecular orbital in the H2 molecule has a lower energy than an electron in the 1s atomic orbital of either of the separated hydrogen atoms.
40.
Predict the valence electron molecular orbital configurations for the following, and state whether they will be stable or unstable ions.
41.
Determine the bond order of each member of the following groups, and determine which member of each group is predicted by the molecular orbital model to have the strongest bond.
1. H2,
2. O2,
3. Li2, Be2
4. F2,
5. N2,
42.
For the first ionization energy for an N2 molecule, what molecular orbital is the electron removed from?
43.
Compare the atomic and molecular orbital diagrams to identify the member of each of the following pairs that has the highest first ionization energy (the most tightly bound electron) in the gas phase:
1. H and H2
2. N and N2
3. O and O2
4. C and C2
5. B and B2
44.
Which of the period 2 homonuclear diatomic molecules are predicted to be paramagnetic?
45.
A friend tells you that the 2s orbital for fluorine starts off at a much lower energy than the 2s orbital for lithium, so the resulting σ2s molecular orbital in F2 is more stable than in Li2. Do you agree?
46.
True or false: Boron contains 2s22p1 valence electrons, so only one p orbital is needed to form molecular orbitals.
47.
What charge would be needed on F2 to generate an ion with a bond order of 2?
48.
Predict whether the MO diagram for S2 would show s-p mixing or not.
49.
Explain why is diamagnetic, while which has the same number of valence electrons, is paramagnetic.
50.
Using the MO diagrams, predict the bond order for the stronger bond in each pair:
1. B2 or
2. F2 or
3. O2 or
4. or | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/08%3A_Advanced_Theories_of_Covalent_Bonding/8.08%3A_Exercises.txt |
In this chapter, we examine the relationships between gas temperature, pressure, amount, and volume. We will study a simple theoretical model and use it to analyze the experimental behavior of gases. The results of these analyses will show us the limitations of the theory and how to improve on it.
• 9.0: Introduction
Gases have played an important part in the development of chemistry. In the seventeenth and eighteenth centuries, many scientists investigated gas behavior, providing the first mathematical descriptions of the behavior of matter.
• 9.1: Gas Pressure
Gases exert pressure, which is force per unit area. The pressure of a gas may be expressed in the SI unit of pascal or kilopascal, as well as in many other units including torr, atmosphere, and bar. Atmospheric pressure is measured using a barometer; other gas pressures can be measured using one of several types of manometers.
• 9.2: Relating Pressure, Volume, Amount, and Temperature - The Ideal Gas Law
The behavior of gases can be described by several laws based on experimental observations of their properties. including Amontons’s law, Charles’s law, Boyle’s lawand Avogadro’s law. These laws can be extracted directly from the ideal gas law.
• 9.3: Stoichiometry of Gaseous Substances, Mixtures, and Reactions
The ideal gas law can be used to derive a number of convenient equations relating directly measured quantities to properties of interest for gaseous substances and mixtures. Appropriate rearrangement of the ideal gas equation may be made to permit the calculation of gas densities and molar masses. Dalton’s law of partial pressures may be used to relate measured gas pressures for gaseous mixtures to their compositions.
• 9.4: Effusion and Diffusion of Gases
Gaseous atoms and molecules move freely and randomly through space. Diffusion is the process whereby gaseous atoms and molecules are transferred from regions of relatively high concentration to regions of relatively low concentration. Effusion is a similar process with gaseous species passing from a container to vacuum through a small orifices. The rates of effusion of gases are inversely proportional to the square roots of their densities or to the square roots of their atoms/molecules’ masses.
• 9.5: The Kinetic-Molecular Theory
The kinetic molecular theory is a simple but very effective model that effectively explains ideal gas behavior. The theory assumes that gases consist of widely separated molecules of negligible volume that are in constant motion, colliding elastically with one another and the walls of their container with average velocities determined by their absolute temperatures. The individual molecules of a gas exhibit a range of velocities.
• 9.6: Non-Ideal Gas Behavior
Gas molecules possess a finite volume and experience forces of attraction for one another. Consequently, gas behavior is not necessarily described well by the ideal gas law. Under conditions of low pressure and high temperature, these factors are negligible, the ideal gas equation is an accurate description of gas behavior, and the gas is said to exhibit ideal behavior. The van der Waals equation is a modified version of the ideal gas law that can be used to account for the non-ideal behavior.
• 9.7: Key Terms
• 9.8: Key Equations
• 9.9: Summary
• 9.10: Exercises
These are homework exercises to accompany the Textmap created for "Chemistry" by OpenStax. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
Thumbnail: As long as black-body radiation (not shown) doesn't escape a system, atoms in thermal agitation undergo essentially elastic collisions. On average, two atoms rebound from each other with the same kinetic energy as before a collision. Five atoms are colored red so their paths of motion are easier to see. (Public Domain; Greg L via Wikipedia)
09: Gases
We are surrounded by an ocean of gas—the atmosphere—and many of the properties of gases are familiar to us from our daily activities. Heated gases expand, which can make a hot air balloon rise (Figure \(1\)) or cause a blowout in a bicycle tire left in the sun on a hot day. Gases have played an important part in the development of chemistry. In the seventeenth and eighteenth centuries, many scientists investigated gas behavior, providing the first mathematical descriptions of the behavior of matter.
In this chapter, we will examine the relationships between gas temperature, pressure, amount, and volume. We will study a simple theoretical model and use it to analyze the experimental behavior of gases. The results of these analyses will show us the limitations of the theory and how to improve on it. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/09%3A_Gases/9.00%3A_Introduction.txt |
Learning Objectives
By the end of this section, you will be able to:
• Define the property of pressure
• Define and convert among the units of pressure measurements
• Describe the operation of common tools for measuring gas pressure
The earth’s atmosphere exerts a pressure, as does any other gas. Although we do not normally notice atmospheric pressure, we are sensitive to pressure changes—for example, when your ears “pop” during take-off and landing while flying, or when you dive underwater. Gas pressure is caused by the force exerted by gas molecules colliding with the surfaces of objects (Figure $1$). Although the force of each collision is very small, any surface of appreciable area experiences a large number of collisions in a short time, which can result in a high pressure. In fact, normal air pressure is strong enough to crush a metal container when not balanced by equal pressure from inside the container.
Link to Learning
A dramatic illustration of atmospheric pressure is provided in this brief video, which shows a railway tanker car imploding when its internal pressure is decreased.
A smaller scale demonstration of this phenomenon is briefly explained.
Atmospheric pressure is caused by the weight of the column of air molecules in the atmosphere above an object, such as the tanker car. At sea level, this pressure is roughly the same as that exerted by a full-grown African elephant standing on a doormat, or a typical bowling ball resting on your thumbnail. These may seem like huge amounts, and they are, but life on earth has evolved under such atmospheric pressure. If you actually perch a bowling ball on your thumbnail, the pressure experienced is twice the usual pressure, and the sensation is unpleasant.
In general, pressure is defined as the force exerted on a given area: $P=\frac{F}{A}). Note that pressure is directly proportional to force and inversely proportional to area. Thus, pressure can be increased either by increasing the amount of force or by decreasing the area over which it is applied; pressure can be decreased by decreasing the force or increasing the area. Let’s apply this concept to determine which exerts a greater pressure in Figure \(2$ —the elephant or the figure skater? A large African elephant can weigh 7 tons, supported on four feet, each with a diameter of about 1.5 ft (footprint area of 250 in2), so the pressure exerted by each foot is about 14 lb/in2:
$\text { pressure per elephant foot }=14,000 \frac{ lb }{\text { elephant }} \times \frac{1 \text { elephant }}{4 \text { feet }} \times \frac{1 \text { foot }}{250 in^2}=14 lb / in^2 \nonumber$
The figure skater weighs about 120 lbs, supported on two skate blades, each with an area of about 2 in2, so the pressure exerted by each blade is about 30 lb/in2:
$\text { pressure per skate blade }=120 \frac{ lb }{\text { skater }} \times \frac{1 \text { skater }}{2 \text { blades }} \times \frac{1 \text { blade }}{2 \text { in }^2}=30 lb / in^2 \nonumber$
Even though the elephant is more than one hundred-times heavier than the skater, it exerts less than one-half of the pressure. On the other hand, if the skater removes their skates and stands with bare feet (or regular footwear) on the ice, the larger area over which their weight is applied greatly reduces the pressure exerted:
$\text { pressure per human foot }=120 \frac{ lb }{\text { skater }} \times \frac{1 \text { skater }}{2 \text { feet }} \times \frac{1 \text { foot }}{30 in^2}=2 lb / in^2 \nonumber$
The SI unit of pressure is the pascal (Pa), with 1 Pa = 1 N/m2, where N is the newton, a unit of force defined as 1 kg m/s2. One pascal is a small pressure; in many cases, it is more convenient to use units of kilopascal (1 kPa = 1000 Pa) or bar (1 bar = 100,000 Pa). In the United States, pressure is often measured in pounds of force on an area of one square inch—pounds per square inch (psi)—for example, in car tires. Pressure can also be measured using the unit atmosphere (atm), which originally represented the average sea level air pressure at the approximate latitude of Paris (45°). Table $1$ provides some information on these and a few other common units for pressure measurements
Table $1$: Pressure Units
Unit Name and Abbreviation Definition or Relation to Other Unit
pascal (Pa) 1 Pa = 1 N/m2
recommended IUPAC unit
kilopascal (kPa) 1 kPa = 1000 Pa
pounds per square inch (psi) air pressure at sea level is ~14.7 psi
atmosphere (atm) 1 atm = 101,325 Pa = 760 torr
air pressure at sea level is ~1 atm
bar (bar, or b) 1 bar = 100,000 Pa (exactly)
commonly used in meteorology
millibar (mbar, or mb) 1000 mbar = 1 bar
inches of mercury (in. Hg) 1 in. Hg = 3386 Pa
used by aviation industry, also some weather reports
torr
named after Evangelista Torricelli, inventor of the barometer
millimeters of mercury (mm Hg) 1 mm Hg ~1 torr
Example $1$: Conversion of Pressure Units
The United States National Weather Service reports pressure in both inches of Hg and millibars. Convert a pressure of 29.2 in. Hg into:
1. torr
2. atm
3. kPa
4. mbar
Solution
This is a unit conversion problem. The relationships between the various pressure units are given in Table $1$.
1. $29.2 \text { in } Hg_{ g } \times \frac{25.4 mm }{1 im } \times \frac{1 \text { torr }}{1 mm Hg_{ g }}=742 \text { torr } \nonumber$
2. $742 \text { torr } \times \frac{1 atm }{760 \text { torr }}=0.976 atm \nonumber$
3. $742 \text { torr } \times \frac{101.325 kPa }{760 \text { torr }}=98.9 kPa \nonumber$
4. $98.9 kPa \times \frac{1000 Pa }{1 kPa } \times \frac{1 bar }{100,000 Pa } \times \frac{1000 mbar }{1 bar }=989 mbar \nonumber$
Exercise $1$
A typical barometric pressure in Kansas City is 740 torr. What is this pressure in atmospheres, in millimeters of mercury, in kilopascals, and in bar?
Answer
0.974 atm; 740 mm Hg; 98.7 kPa; 0.987 bar
We can measure atmospheric pressure, the force exerted by the atmosphere on the earth’s surface, with a barometer (Figure $3$). A barometer is a glass tube that is closed at one end, filled with a nonvolatile liquid such as mercury, and then inverted and immersed in a container of that liquid. The atmosphere exerts pressure on the liquid outside the tube, the column of liquid exerts pressure inside the tube, and the pressure at the liquid surface is the same inside and outside the tube. The height of the liquid in the tube is therefore proportional to the pressure exerted by the atmosphere.
If the liquid is water, normal atmospheric pressure will support a column of water over 10 meters high, which is rather inconvenient for making (and reading) a barometer. Because mercury (Hg) is about 13.6-times denser than water, a mercury barometer only needs to be $\frac{1}{13.6}}$ as tall as a water barometer—a more suitable size. Standard atmospheric pressure of 1 atm at sea level (101,325 Pa) corresponds to a column of mercury that is about 760 mm (29.92 in.) high. The torr was originally intended to be a unit equal to one millimeter of mercury, but it no longer corresponds exactly. The pressure exerted by a fluid due to gravity is known as hydrostatic pressure, $p$:
$p=h \rho g \nonumber$
where h is the height of the fluid, ρ (lowercase Greek letter rho) is the density of the fluid, and g is acceleration due to gravity.
Example $2$: Calculation of Barometric Pressure
Show the calculation supporting the claim that atmospheric pressure near sea level corresponds to the pressure exerted by a column of mercury that is about 760 mm high. The density of mercury = 13.6 g/cm3.
Solution
The hydrostatic pressure is given by p = hρg, with h = 760 mm, ρ = 13.6 g/cm3, and g = 9.81 m/s2. Plugging these values into the equation and doing the necessary unit conversions will give us the value we seek. (Note: We are expecting to find a pressure of ~101,325 Pa.)
$101,325 N / m^2=101,325 \frac{ kg \cdot m / s^2}{ m^2}=101,325 \frac{ kg }{ m \cdot s^2} \nonumber$
$\begin{gathered} p=\left(760 mm \times \frac{1 m }{1000 mm }\right) \times\left(\frac{13.6 g }{1 cm^3} \times \frac{1 kg }{1000 g } \times \frac{(100 cm )^3}{(1 m )^3}\right) \times\left(\frac{9.81 m }{1 s^2}\right) \[4pt] =(0.760 m )\left(13,600 kg / m^3\right)\left(9.81 m / s^2\right)=1.01 \times 10^5 kg / ms^2=1.01 \times 10^5 N / m^2 \[4pt] =1.01 \times 10^5 Pa \end{gathered} \nonumber$
Exercise $2$
Calculate the height of a column of water at 25 °C that corresponds to normal atmospheric pressure. The density of water at this temperature is 1.0 g/cm3.
Answer
10.3 m
A manometer is a device similar to a barometer that can be used to measure the pressure of a gas trapped in a container. A closed-end manometer is a U-shaped tube with one closed arm, one arm that connects to the gas to be measured, and a nonvolatile liquid (usually mercury) in between. As with a barometer, the distance between the liquid levels in the two arms of the tube (h in the diagram) is proportional to the pressure of the gas in the container. An open-end manometer (Figure $4$) is the same as a closed-end manometer, but one of its arms is open to the atmosphere. In this case, the distance between the liquid levels corresponds to the difference in pressure between the gas in the container and the atmosphere.
Example $3$: Calculation of Pressure Using a Closed-End Manometer
The pressure of a sample of gas is measured with a closed-end manometer, as shown to the right. The liquid in the manometer is mercury. Determine the pressure of the gas in:
1. torr
2. Pa
3. bar
Solution
The pressure of the gas is equal to a column of mercury of height 26.4 cm. (The pressure at the bottom horizontal line is equal on both sides of the tube. The pressure on the left is due to the gas and the pressure on the right is due to 26.4 cm Hg, or mercury.) We could use the equation $p = hρg$ as in Example $2$, but it is simpler to just convert between units using Table $1$.
1. $26.4 em Hg_{ g } \times \frac{10 mm Hg_{ g }}{1 emHg } \times \frac{1 \text { torr }}{1 mm \Pi Hg }=264 \text { torr } \nonumber$
2. $264 \text { torr } \times \frac{1 \text { atm }}{760 \text { torr }} \times \frac{101,325 Pa }{1 \text { atm }}=35,200 Pa \nonumber$
3. $35,200 Pa \times \frac{1 bar }{100,000 Pa }=0.352 bar \nonumber$
Exercise $3$
The pressure of a sample of gas is measured with a closed-end manometer. The liquid in the manometer is mercury. Determine the pressure of the gas in:
1. torr
2. Pa
3. bar
Answer
(a) ~150 torr; (b) ~20,000 Pa; (c) ~0.20 bar
Example $4$: Calculation of Pressure Using an Open-End Manometer
The pressure of a sample of gas is measured at sea level with an open-end Hg (mercury) manometer, as shown to the right. Determine the pressure of the gas in:
1. mm Hg
2. atm
3. kPa
Solution
The pressure of the gas equals the hydrostatic pressure due to a column of mercury of height 13.7 cm plus the pressure of the atmosphere at sea level. (The pressure at the bottom horizontal line is equal on both sides of the tube. The pressure on the left is due to the gas and the pressure on the right is due to 13.7 cm of Hg plus atmospheric pressure.)
1. In mm Hg, this is: 137 mm Hg + 760 mm Hg = 897 mm Hg
2. $897 mm Hg_{ g } \times \frac{1 atm }{760 mm Hg_{ g }}=1.18 atm \nonumber$
3. $1.18 \text { atm } \times \frac{101.325 kPa }{1-20 \times 10^2 kPa }=1.20 \nonumber$
Exercise $4$
The pressure of a sample of gas is measured at sea level with an open-end Hg manometer, as shown to the right. Determine the pressure of the gas in:
1. mm Hg
2. atm
3. kPa
Answer
(a) 642 mm Hg; (b) 0.845 atm; (c) 85.6 kPa
Chemistry in Everyday Life: Measuring Blood Pressure
Blood pressure is measured using a device called a sphygmomanometer (Greek sphygmos = “pulse”). It consists of an inflatable cuff to restrict blood flow, a manometer to measure the pressure, and a method of determining when blood flow begins and when it becomes impeded (Figure $5$). Since its invention in 1881, it has been an essential medical device. There are many types of sphygmomanometers: manual ones that require a stethoscope and are used by medical professionals; mercury ones, used when the most accuracy is required; less accurate mechanical ones; and digital ones that can be used with little training but that have limitations. When using a sphygmomanometer, the cuff is placed around the upper arm and inflated until blood flow is completely blocked, then slowly released. As the heart beats, blood forced through the arteries causes a rise in pressure. This rise in pressure at which blood flow begins is the systolic pressure—the peak pressure in the cardiac cycle. When the cuff’s pressure equals the arterial systolic pressure, blood flows past the cuff, creating audible sounds that can be heard using a stethoscope. This is followed by a decrease in pressure as the heart’s ventricles prepare for another beat. As cuff pressure continues to decrease, eventually sound is no longer heard; this is the diastolic pressure—the lowest pressure (resting phase) in the cardiac cycle. Blood pressure units from a sphygmomanometer are in terms of millimeters of mercury (mm Hg).
How Sciences Interconnect: Meteorology, Climatology, and Atmospheric Science
Throughout the ages, people have observed clouds, winds, and precipitation, trying to discern patterns and make predictions: when it is best to plant and harvest; whether it is safe to set out on a sea voyage; and much more. We now face complex weather and atmosphere-related challenges that will have a major impact on our civilization and the ecosystem. Several different scientific disciplines use chemical principles to help us better understand weather, the atmosphere, and climate. These are meteorology, climatology, and atmospheric science. Meteorology is the study of the atmosphere, atmospheric phenomena, and atmospheric effects on earth’s weather. Meteorologists seek to understand and predict the weather in the short term, which can save lives and benefit the economy. Weather forecasts (Figure $6$) are the result of thousands of measurements of air pressure, temperature, and the like, which are compiled, modeled, and analyzed in weather centers worldwide.
In terms of weather, low-pressure systems occur when the earth’s surface atmospheric pressure is lower than the surrounding environment: Moist air rises and condenses, producing clouds. Movement of moisture and air within various weather fronts instigates most weather events.
The atmosphere is the gaseous layer that surrounds a planet. Earth’s atmosphere, which is roughly 100–125 km thick, consists of roughly 78.1% nitrogen and 21.0% oxygen, and can be subdivided further into the regions shown in Figure $7$ the exosphere (furthest from earth, > 700 km above sea level), the thermosphere (80–700 km), the mesosphere (50–80 km), the stratosphere (second lowest level of our atmosphere, 12–50 km above sea level), and the troposphere (up to 12 km above sea level, roughly 80% of the earth’s atmosphere by mass and the layer where most weather events originate). As you go higher in the troposphere, air density and temperature both decrease.
Climatology is the study of the climate, averaged weather conditions over long time periods, using atmospheric data. However, climatologists study patterns and effects that occur over decades, centuries, and millennia, rather than shorter time frames of hours, days, and weeks like meteorologists. Atmospheric science is an even broader field, combining meteorology, climatology, and other scientific disciplines that study the atmosphere. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/09%3A_Gases/9.01%3A_Gas_Pressure.txt |
Learning Objectives
By the end of this section, you will be able to:
• Identify the mathematical relationships between the various properties of gases
• Use the ideal gas law, and related gas laws, to compute the values of various gas properties under specified conditions
During the seventeenth and especially eighteenth centuries, driven both by a desire to understand nature and a quest to make balloons in which they could fly (Figure $1$), a number of scientists established the relationships between the macroscopic physical properties of gases, that is, pressure, volume, temperature, and amount of gas. Although their measurements were not precise by today’s standards, they were able to determine the mathematical relationships between pairs of these variables (e.g., pressure and temperature, pressure and volume) that hold for an ideal gas—a hypothetical construct that real gases approximate under certain conditions. Eventually, these individual laws were combined into a single equation—the ideal gas law—that relates gas quantities for gases and is quite accurate for low pressures and moderate temperatures. We will consider the key developments in individual relationships (for pedagogical reasons not quite in historical order), then put them together in the ideal gas law.
Pressure and Temperature: Amontons’s Law
Imagine filling a rigid container attached to a pressure gauge with gas and then sealing the container so that no gas may escape. If the container is cooled, the gas inside likewise gets colder and its pressure is observed to decrease. Since the container is rigid and tightly sealed, both the volume and number of moles of gas remain constant. If we heat the sphere, the gas inside gets hotter (Figure $2$) and the pressure increases.
This relationship between temperature and pressure is observed for any sample of gas confined to a constant volume. An example of experimental pressure-temperature data is shown for a sample of air under these conditions in Figure $3$. We find that temperature and pressure are linearly related, and if the temperature is on the kelvin scale, then P and T are directly proportional (again, when volume and moles of gas are held constant); if the temperature on the kelvin scale increases by a certain factor, the gas pressure increases by the same factor.
Guillaume Amontons was the first to empirically establish the relationship between the pressure and the temperature of a gas (~1700), and Joseph Louis Gay-Lussac determined the relationship more precisely (~1800). Because of this, the P-T relationship for gases is known as either Amontons’s law or Gay-Lussac’s law. Under either name, it states that the pressure of a given amount of gas is directly proportional to its temperature on the kelvin scale when the volume is held constant. Mathematically, this can be written:
$P \propto T \text { or } P=\text { constant } \times T \text { or } P=k \times T \nonumber$
where $∝$ means “is proportional to,” and $k$ is a proportionality constant that depends on the identity, amount, and volume of the gas.
For a confined, constant volume of gas, the $\frac{P}{T}$ ratio is therefore constant (i.e., $\frac{P}{T}=k$). If the gas is initially in “Condition 1” (with P = P1 and T = T1), and then changes to “Condition 2” (with P = P2 and T = T2), we have that $\frac{P_1}{T_1}=k$ and $\frac{P_2}{T_2}=k$ which reduces to
$\frac{P_1}{T_1} = \frac{P_2}{T_2}. \nonumber$
This equation is useful for pressure-temperature calculations for a confined gas at constant volume. Note that temperatures must be on the kelvin scale for any gas law calculations (0 on the kelvin scale and the lowest possible temperature is called absolute zero). (Also note that there are at least three ways we can describe how the pressure of a gas changes as its temperature changes: We can use a table of values, a graph, or a mathematical equation.)
Example $1$: Predicting Change in Pressure with Temperature
A can of hair spray is used until it is empty except for the propellant, isobutane gas.
1. On the can is the warning “Store only at temperatures below 120 °F (48.8 °C). Do not incinerate.” Why?
2. The gas in the can is initially at 24 °C and 360 kPa, and the can has a volume of 350 mL. If the can is left in a car that reaches 50 °C on a hot day, what is the new pressure in the can?
Solution
(a) The can contains an amount of isobutane gas at a constant volume, so if the temperature is increased by heating, the pressure will increase proportionately. High temperature could lead to high pressure, causing the can to burst. (Also, isobutane is combustible, so incineration could cause the can to explode.)
(b) We are looking for a pressure change due to a temperature change at constant volume, so we will use Amontons’s/Gay-Lussac’s law. Taking P1 and T1 as the initial values, T2 as the temperature where the pressure is unknown and P2 as the unknown pressure, and converting °C to K, we have:
$\frac{P_1}{T_1}=\frac{P_2}{T_2} \text { which means that } \frac{360 ~\text{kPa} }{297 ~\text{K} }=\dfrac{P_2}{323 ~\text{K} } \nonumber$
Rearranging and solving gives:
$P_2=\frac{360 ~\text{kPa} \times 323 \cancel{\text{K}}}{297\, \cancel{\text{K}}}=390~\text{kPa} \nonumber$
Exercise $1$
A sample of nitrogen, N2, occupies 45.0 mL at 27 °C and 600 torr. What pressure will it have if cooled to –73 °C while the volume remains constant?
Answer
400 torr
Volume and Temperature: Charles’s Law
If we fill a balloon with air and seal it, the balloon contains a specific amount of air at atmospheric pressure, let’s say 1 atm. If we put the balloon in a refrigerator, the gas inside gets cold and the balloon shrinks (although both the amount of gas and its pressure remain constant). If we make the balloon very cold, it will shrink a great deal, and it expands again when it warms up.
Link to Learning
This video shows how cooling and heating a gas causes its volume to decrease or increase, respectively.
These examples of the effect of temperature on the volume of a given amount of a confined gas at constant pressure are true in general: The volume increases as the temperature increases, and decreases as the temperature decreases. Volume-temperature data for a 1-mole sample of methane gas at 1 atm are listed and graphed in Figure 9.12.
The relationship between the volume and temperature of a given amount of gas at constant pressure is known as Charles’s law in recognition of the French scientist and balloon flight pioneer Jacques Alexandre César Charles. Charles’s law states that the volume of a given amount of gas is directly proportional to its temperature on the kelvin scale when the pressure is held constant.
Mathematically, this can be written as:
$V \alpha T \text { or } V=\text { constant } \cdot T \text { or } V=k \cdot T \text { or } V_1 / T_1=V_2 / T_2 \nonumber$
with $k$ being a proportionality constant that depends on the amount and pressure of the gas.
For a confined, constant pressure gas sample, $\frac{V}{T}$ is constant (i.e., the ratio = k), and as seen with the P-T relationship, this leads to another form of Charles’s law:
$\frac{V_1}{T_1}=\frac{V_2}{T_2}. \label{charles}$
Example $2$: Predicting Change in Volume with Temperature
A sample of carbon dioxide, CO2, occupies 0.300 L at 10 °C and 750 torr. What volume will the gas have at 30 °C and 750 torr?
Solution
Because we are looking for the volume change caused by a temperature change at constant pressure, this is a job for Charles’s law. Taking V1 and T1 as the initial values, T2 as the temperature at which the volume is unknown and V2 as the unknown volume, and converting °C into K we have:
$\frac{V_1}{T_1}=\frac{V_2}{T_2} \text { which means that } \frac{0.300 L }{283 K }=\frac{V_2}{303 K } \nonumber$
Rearranging and solving gives:
$V_2=\dfrac{0.300 L \times 303~\cancel{\text{K}} }{283~ \cancel{\text{K}} }=0.321~\text{L} \nonumber$
This answer supports our expectation from Charles’s law, namely, that raising the gas temperature (from 283 K to 303 K) at a constant pressure will yield an increase in its volume (from 0.300 L to 0.321 L).
Exercise $2$
A sample of oxygen, O2, occupies 32.2 mL at 30 °C and 452 torr. What volume will it occupy at –70 °C and the same pressure?
Answer
21.6 mL
Example $3$: Measuring Temperature with a Volume Change
Temperature is sometimes measured with a gas thermometer by observing the change in the volume of the gas as the temperature changes at constant pressure. The hydrogen in a particular hydrogen gas thermometer has a volume of 150.0 cm3 when immersed in a mixture of ice and water (0.00 °C). When immersed in boiling liquid ammonia, the volume of the hydrogen, at the same pressure, is 131.7 cm3. Find the temperature of boiling ammonia on the kelvin and Celsius scales.
Solution
A volume change caused by a temperature change at constant pressure means we should use Charles’s law. Taking V1 and T1 as the initial values, T2 as the temperature at which the volume is unknown and V2 as the unknown volume, and converting °C into K we have:
$\frac{V_1}{T_1}=\frac{V_2}{T_2} \text { which means that } \frac{150.0 cm^3}{273.15 K }=\frac{131.7 cm^3}{T_2} \nonumber$
Rearrangement gives
$T_2=\dfrac{131.7 ~\cancel{\text{cm}^3} \times 273.15 ~\text{K} }{150.0 ~\cancel{\text{cm}^3}}=239.8 ~\text{K} \nonumber$
Subtracting 273.15 from 239.8 K, we find that the temperature of the boiling ammonia on the Celsius scale is –33.4 °C.
Exercise $3$
What is the volume of a sample of ethane at 467 K and 1.1 atm if it occupies 405 mL at 298 K and 1.1 atm?
Answer
635 mL
Volume and Pressure: Boyle’s Law
If we partially fill an airtight syringe with air, the syringe contains a specific amount of air at constant temperature, say 25 °C. If we slowly push in the plunger while keeping temperature constant, the gas in the syringe is compressed into a smaller volume and its pressure increases; if we pull out the plunger, the volume increases and the pressure decreases. This example of the effect of volume on the pressure of a given amount of a confined gas is true in general. Decreasing the volume of a contained gas will increase its pressure, and increasing its volume will decrease its pressure. In fact, if the volume increases by a certain factor, the pressure decreases by the same factor, and vice versa. Volume-pressure data for an air sample at room temperature are graphed in Figure $4$.
Unlike the P-T and V-T relationships, pressure and volume are not directly proportional to each other. Instead, P and V exhibit inverse proportionality: Increasing the pressure results in a decrease of the volume of the gas. Mathematically this can be written:
$P \propto \dfrac{1}{V} \label{Boyle1}$
or
$P=k \cdot \dfrac{1}{V} \label{Boyle2}$
or
$P \cdot V=k \label{Boyle3}$
or
$P_1 V_1=P_2 V_2 \label{Boyle4}$
with k being a constant. Graphically, this relationship is shown by the straight line that results when plotting the inverse of the pressure
The relationship between the volume and pressure of a given amount of gas at constant temperature was first published by the English natural philosopher Robert Boyle over 300 years ago. It is summarized in the statement now known as Boyle’s law: The volume of a given amount of gas held at constant temperature is inversely proportional to the pressure under which it is measured.
Example $4$: Volume of a Gas Sample
The sample of gas in Figure $4$: has a volume of 15.0 mL at a pressure of 13.0 psi. Determine the pressure of the gas at a volume of 7.5 mL, using:
1. the P-V graph in Figure $5$
2. the $\frac{1}{P}$ vs. $V$ graph in Figure $5$
3. the Boyle’s law equations (Equations \ref{Boyle1} - \ref{Boyle4})
Comment on the likely accuracy of each method.
Solution
(a) Estimating from the P-V graph gives a value for P somewhere around 27 psi.
(b) Estimating from the \frac{1}{P}\) versus $V$ graph give a value of about 26 psi.
(c) From Boyle’s law, we know that the product of pressure and volume (PV) for a given sample of gas at a constant temperature is always equal to the same value. Therefore we have P1V1 = k and P2V2 = k which means that P1V1 = P2V2.
Using P1 and V1 as the known values 13.0 psi and 15.0 mL, P2 as the pressure at which the volume is unknown, and V2 as the unknown volume, we have:
\begin{align*} P_1 ~ V_1 &=P_2 ~V_2 \[4pt] (13.0 ~\text{psi}) \times (15.0~\text{mL} &= P_2 \times (7.5~\text{mL}) \end{align*}
Solving:
$P_2=\frac{13.0 ~\text{psi} \times 15.0 ~\cancel{\text{mL}}}{7.5 ~\cancel{\text{mL}} }=26 ~\text{psi} \nonumber$
It was more difficult to estimate well from the P-V graph, so (a) is likely more inaccurate than (b) or (c). The calculation will be as accurate as the equation and measurements allow.
Exercise $4$
The sample of gas in Figure $4$ has a volume of 30.0 mL at a pressure of 6.5 psi. Determine the volume of the gas at a pressure of 11.0 psi, using:
1. the P-V graph in Figure $5$
2. the $\frac{1}{P}$ vs. $V$ graph in Figure $5$
3. the Boyle’s law equations (\ref{Boyle1} - \ref{Boyle4})
Comment on the likely accuracy of each method.
Answer
(a) about 17–18 mL; (b) ~18 mL; (c) 17.7 mL; it was more difficult to estimate well from the P-V graph, so (a) is likely more inaccurate than (b); the calculation will be as accurate as the equation and measurements allow
Chemistry in Everyday Life: Breathing and Boyle’s Law
What do you do about 20 times per minute for your whole life, without break, and often without even being aware of it? The answer, of course, is respiration, or breathing. How does it work? It turns out that the gas laws apply here. Your lungs take in gas that your body needs (oxygen) and get rid of waste gas (carbon dioxide). Lungs are made of spongy, stretchy tissue that expands and contracts while you breathe. When you inhale, your diaphragm and intercostal muscles (the muscles between your ribs) contract, expanding your chest cavity and making your lung volume larger. The increase in volume leads to a decrease in pressure (Boyle’s law). This causes air to flow into the lungs (from high pressure to low pressure). When you exhale, the process reverses: Your diaphragm and rib muscles relax, your chest cavity contracts, and your lung volume decreases, causing the pressure to increase (Boyle’s law again), and air flows out of the lungs (from high pressure to low pressure). You then breathe in and out again, and again, repeating this Boyle’s law cycle for the rest of your life (Figure $6$).
Moles of Gas and Volume: Avogadro’s Law
The Italian scientist Amedeo Avogadro advanced a hypothesis in 1811 to account for the behavior of gases, stating that equal volumes of all gases, measured under the same conditions of temperature and pressure, contain the same number of molecules. Over time, this relationship was supported by many experimental observations as expressed by Avogadro’s law: For a confined gas, the volume (V) and number of moles (n) are directly proportional if the pressure and temperature both remain constant.
In equation form, this is written as:
$V \propto n \label{Avogadro1}$
or
$V=k \times n \label{Avogadro2}$
or
$\dfrac{V_1}{n_1}=\dfrac{V_2}{n_2} \label{Avogadro3}$
Mathematical relationships can also be determined for the other variable pairs, such as P versus n, and n versus T.
Link to Learning
Visit this interactive PhET simulation to investigate the relationships between pressure, volume, temperature, and amount of gas. Use the simulation to examine the effect of changing one parameter on another while holding the other parameters constant (as described in the preceding sections on the various gas laws).
The Ideal Gas Law
To this point, four separate laws have been discussed that relate pressure, volume, temperature, and the number of moles of the gas:
• Boyle’s law: $PV$ = constant at constant T and n
• Amontons’s law: $\frac{P}{T}$ = constant at constant V and n
• Charles’s law: $\frac{V}{T}$ = constant at constant P and n
• Avogadro’s law: $\frac{V}{n}$ = constant at constant P and T
Combining these four laws yields the ideal gas law, a relation between the pressure, volume, temperature, and number of moles of a gas:
$P V=n R T \nonumber$
where P is the pressure of a gas, V is its volume, n is the number of moles of the gas, T is its temperature on the kelvin scale, and R is a constant called the ideal gas constant or the universal gas constant. The units used to express pressure, volume, and temperature will determine the proper form of the gas constant as required by dimensional analysis, the most commonly encountered values being 0.08206 L atm mol–1 K–1 and 8.314 kPa L mol–1 K–1.
Gases whose properties of P, V, and T are accurately described by the ideal gas law (or the other gas laws) are said to exhibit ideal behavior or to approximate the traits of an ideal gas. An ideal gas is a hypothetical construct that may be used along with kinetic molecular theory to effectively explain the gas laws as will be described in a later module of this chapter. Although all the calculations presented in this module assume ideal behavior, this assumption is only reasonable for gases under conditions of relatively low pressure and high temperature. In the final module of this chapter, a modified gas law will be introduced that accounts for the non-ideal behavior observed for many gases at relatively high pressures and low temperatures.
The ideal gas equation contains five terms, the gas constant R and the variable properties P, V, n, and T. Specifying any four of these terms will permit use of the ideal gas law to calculate the fifth term as demonstrated in the following example exercises.
Example $5$: Using the Ideal Gas Law
Methane, CH4, is being considered for use as an alternative automotive fuel to replace gasoline. One gallon of gasoline could be replaced by 655 g of CH4. What is the volume of this much methane at 25 °C and 745 torr?
Solution
We must rearrange PV = nRT to solve for V:
$V=\frac{n R T}{P} \nonumber$
If we choose to use R = 0.08206 L atm mol–1 K–1, then the amount must be in moles, temperature must be in kelvin, and pressure must be in atm.
Converting into the “right” units:
$n=655 ~ \cancel{\text{g} ~ \ce{CH4}} \times \frac{1 ~\text{mol} }{16.043 \cancel{\text{g}~ \ce{CH4}}}=40.8 ~\text{mol} \nonumber$
$T=25^{\circ} C +273=298 ~\text{K} \nonumber$
$P=745 \text { torr } \times \frac{1 ~\text{atm} }{760 \text { torr }}=0.980 ~\text{atm} \nonumber$
$V=\frac{n R T}{P}=\frac{(40.8 \cancel{\text{mol}} )\left(0.08206 ~L ~ \cancel{\text{atm mol}^{-1} \text{K}^{-1}} \right)(298 ~\cancel{\text{K}} )}{0.980~ \cancel{\text{atm}} }=1.02 \times 10^3 ~\text{L} \nonumber$
It would require 1020 L (269 gal) of gaseous methane at about 1 atm of pressure to replace 1 gal of gasoline. It requires a large container to hold enough methane at 1 atm to replace several gallons of gasoline.
Exercise $5$
Calculate the pressure in bar of 2520 moles of hydrogen gas stored at 27 °C in the 180-L storage tank of a modern hydrogen-powered car.
Answer
350 bar
If the number of moles of an ideal gas are kept constant under two different sets of conditions, a useful mathematical relationship called the combined gas law is obtained:
$\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2} \nonumber$
using units of atm, L, and K. Both sets of conditions are equal to the product of n R (where n = the number of moles of the gas and R is the ideal gas law constant).
Example $6$: Using the Combined Gas Law
When filled with air, a typical scuba tank with a volume of 13.2 L has a pressure of 153 atm (Figure $7$). If the water temperature is 27 °C, how many liters of air will such a tank provide to a diver’s lungs at a depth of approximately 70 feet in the ocean where the pressure is 3.13 atm?
Letting 1 represent the air in the scuba tank and 2 represent the air in the lungs, and noting that body temperature (the temperature the air will be in the lungs) is 37 °C, we have:
$\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2} \rightarrow \frac{(153 atm )(13.2 L )}{(300 K )}=\frac{(3.13 atm )\left(V_2\right)}{(310 K )} \nonumber$
Solving for V2:
$V_2=\frac{(153 \text { atm })(13.2 L )(310 K )}{(300 K )(3.13 \text { atm })}=667 L \nonumber$
(Note: Be advised that this particular example is one in which the assumption of ideal gas behavior is not very reasonable, since it involves gases at relatively high pressures and low temperatures. Despite this limitation, the calculated volume can be viewed as a good “ballpark” estimate.)
Exercise $6$
A sample of ammonia is found to occupy 0.250 L under laboratory conditions of 27 °C and 0.850 atm. Find the volume of this sample at 0 °C and 1.00 atm.
Answer
0.193 L
Chemistry in Everyday Life: The Interdependence between Ocean Depth and Pressure in Scuba Diving
Whether scuba diving at the Great Barrier Reef in Australia (shown in Figure $8$) or in the Caribbean, divers must understand how pressure affects a number of issues related to their comfort and safety.
Pressure increases with ocean depth, and the pressure changes most rapidly as divers reach the surface. The pressure a diver experiences is the sum of all pressures above the diver (from the water and the air). Most pressure measurements are given in units of atmospheres, expressed as “atmospheres absolute” or ATA in the diving community: Every 33 feet of salt water represents 1 ATA of pressure in addition to 1 ATA of pressure from the atmosphere at sea level. As a diver descends, the increase in pressure causes the body’s air pockets in the ears and lungs to compress; on the ascent, the decrease in pressure causes these air pockets to expand, potentially rupturing eardrums or bursting the lungs. Divers must therefore undergo equalization by adding air to body airspaces on the descent by breathing normally and adding air to the mask by breathing out of the nose or adding air to the ears and sinuses by equalization techniques; the corollary is also true on ascent, divers must release air from the body to maintain equalization. Buoyancy, or the ability to control whether a diver sinks or floats, is controlled by the buoyancy compensator (BCD). If a diver is ascending, the air in their BCD expands because of lower pressure according to Boyle’s law (decreasing the pressure of gases increases the volume). The expanding air increases the buoyancy of the diver, and they begin to ascend. The diver must vent air from the BCD or risk an uncontrolled ascent that could rupture the lungs. In descending, the increased pressure causes the air in the BCD to compress and the diver sinks much more quickly; the diver must add air to the BCD or risk an uncontrolled descent, facing much higher pressures near the ocean floor. The pressure also impacts how long a diver can stay underwater before ascending. The deeper a diver dives, the more compressed the air that is breathed because of increased pressure: If a diver dives 33 feet, the pressure is 2 ATA and the air would be compressed to one-half of its original volume. The diver uses up available air twice as fast as at the surface.
Standard Conditions of Temperature and Pressure
We have seen that the volume of a given quantity of gas and the number of molecules (moles) in a given volume of gas vary with changes in pressure and temperature. Chemists sometimes make comparisons against a standard temperature and pressure (STP) for reporting properties of gases: 273.15 K and 1 atm (101.325 kPa).1 At STP, one mole of an ideal gas has a volume of about 22.4 L—this is referred to as the standard molar volume (Figure $9$).
Footnotes
• 1The IUPAC definition of standard pressure was changed from 1 atm to 1 bar (100 kPa) in 1982, but the prior definition remains in use by many literature resources and will be used in this text. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/09%3A_Gases/9.02%3A_Relating_Pressure_Volume_Amount_and_Temperature_-_The_Ideal_Gas_Law.txt |
Learning Objectives
By the end of this section, you will be able to:
• Use the ideal gas law to compute gas densities and molar masses
• Perform stoichiometric calculations involving gaseous substances
• State Dalton’s law of partial pressures and use it in calculations involving gaseous mixtures
The study of the chemical behavior of gases was part of the basis of perhaps the most fundamental chemical revolution in history. French nobleman Antoine Lavoisier, widely regarded as the “father of modern chemistry,” changed chemistry from a qualitative to a quantitative science through his work with gases. He discovered the law of conservation of matter, discovered the role of oxygen in combustion reactions, determined the composition of air, explained respiration in terms of chemical reactions, and more. He was a casualty of the French Revolution, guillotined in 1794. Of his death, mathematician and astronomer Joseph-Louis Lagrange said, “It took the mob only a moment to remove his head; a century will not suffice to reproduce it.”2 Much of the knowledge we do have about Lavoisier's contributions is due to his wife, Marie-Anne Paulze Lavoisier, who worked with him in his lab. A trained artist fluent in several languages, she created detailed illustrations of the equipment in his lab, and translated texts from foreign scientists to complement his knowledge. After his execution, she was instrumental in publishing Lavoisier's major treatise, which unified many concepts of chemistry and laid the groundwork for significant further study.
As described in an earlier chapter of this text, we can turn to chemical stoichiometry for answers to many of the questions that ask “How much?” The essential property involved in such use of stoichiometry is the amount of substance, typically measured in moles (n). For gases, molar amount can be derived from convenient experimental measurements of pressure, temperature, and volume. Therefore, these measurements are useful in assessing the stoichiometry of pure gases, gas mixtures, and chemical reactions involving gases. This section will not introduce any new material or ideas, but will provide examples of applications and ways to integrate concepts already discussed.
Gas Density and Molar Mass
The ideal gas law described previously in this chapter relates the properties of pressure P, volume V, temperature T, and molar amount n. This law is universal, relating these properties in identical fashion regardless of the chemical identity of the gas:
$P V=n R T \nonumber$
The density d of a gas, on the other hand, is determined by its identity. As described in another chapter of this text, the density of a substance is a characteristic property that may be used to identify the substance.
$d=\frac{m}{V} \nonumber$
Rearranging the ideal gas equation to isolate V and substituting into the density equation yields
$d=\frac{m P}{n R T}=\left(\frac{m}{n}\right) \frac{P}{R T} \nonumber$
The ratio m/n is the definition of molar mass, ℳ:
$ℳ =\frac{m}{n} \nonumber$
The density equation can then be written
$d=\frac{ ℳ P}{R T} \nonumber$
This relation may be used for calculating the densities of gases of known identities at specified values of pressure and temperature as demonstrated in Example $1$.
Example $1$: Measuring Gas Density
What is the density of molecular nitrogen gas at STP?
Solution
The molar mass of molecular nitrogen, N2, is 28.01 g/mol. Substituting this value along with standard temperature and pressure into the gas density equation yields
$d=\frac{ ℳ P }{R T}=\frac{(28.01~\text{g/mol })(1.00 ~\text{atm} )}{\left(0.0821 ~\text{L} \cdot \text{atm} \cdot \text{mol}^{-1}\text{K}^{-1}\right)(273~\text{K} )}=1.25 ~\text{g / L} \nonumber$
Exercise $1$
What is the density of molecular hydrogen gas at 17.0 °C and a pressure of 760 torr?
Answer
d = 0.0847 g/L
When the identity of a gas is unknown, measurements of the mass, pressure, volume, and temperature of a sample can be used to calculate the molar mass of the gas (a useful property for identification purposes). Combining the ideal gas equation
$P V=n R T \nonumber$
and the definition of molar mass
$ℳ =\frac{m}{n} \nonumber$
yields the following equation:
$ℳ =\frac{m R T}{P V} \nonumber$
Determining the molar mass of a gas via this approach is demonstrated in Example $2$.
Example $2$: Determining the Molecular Formula of a Gas from its Molar Mass and Empirical Formula
Cyclopropane, a gas once used with oxygen as a general anesthetic, is composed of 85.7% carbon and 14.3% hydrogen by mass. Find the empirical formula. If 1.56 g of cyclopropane occupies a volume of 1.00 L at 0.984 atm and 50 °C, what is the molecular formula for cyclopropane?
Solution
First determine the empirical formula of the gas. Assume 100 g and convert the percentage of each element into grams. Determine the number of moles of carbon and hydrogen in the 100-g sample of cyclopropane. Divide by the smallest number of moles to relate the number of moles of carbon to the number of moles of hydrogen.
\begin{align*} 85.7 ~\text{g}~ C \times \frac{1 ~\text{mol} ~C }{12.01 ~\text{g}~ C }=7.136 ~\text{mol}~ C \[4pt] 14.3 ~\text{g}~ H \times \frac{1 ~\text{mol} ~H }{1.01 ~\text{g}~ H }=14.158 ~\text{mol} ~ H \end{align*} \nonumber
In the last step, realize that the smallest whole number ratio is the empirical formula:
\begin{align*} \dfrac{7.136}{7.136}&=1.00 ~\text{mol}~ C \[4pt] \dfrac{14.158}{7.136}&=1.98 ~\text{mol} ~ H \end{align*} \nonumber
Empirical formula is $\ce{CH2}$ [empirical mass (EM) of 14.03 g/empirical unit].
Next, use the provided values for mass, pressure, temperature and volume to compute the molar mass of the gas:
$ℳ =\frac{m R T}{P V}=\frac{(1.56 ~\text{g} )\left(0.0821 ~\text{L} \cdot \text{atm} \cdot \text{mol}^{-1} \text{K}^{-1}\right)(323 ~\text{K} )}{(0.984 ~\text{atm} )(1.00~\text{L} )}=42.0 ~\text{g/mol} \nonumber$
Comparing the molar mass to the empirical formula mass shows how many empirical formula units make up a molecule:
$\frac{ ℳ }{E M}=\frac{42.0~\text{g/mol} }{14.0 ~ \text{g/mol} }=3 \nonumber$
The molecular formula is thus derived from the empirical formula by multiplying each of its subscripts by three:
$\ce{(CH2)3= C3H6} \nonumber$
Exercise $2$
Acetylene, a fuel used welding torches, is composed of 92.3% C and 7.7% H by mass. Find the empirical formula. If 1.10 g of acetylene occupies of volume of 1.00 L at 1.15 atm and 59.5 °C, what is the molecular formula for acetylene?
Answer
Empirical formula, CH; Molecular formula, C2H2
Example $3$: Determining the Molar Mass of a Volatile Liquid
The approximate molar mass of a volatile liquid can be determined by:
1. Heating a sample of the liquid in a flask with a tiny hole at the top, which converts the liquid into gas that may escape through the hole
2. Removing the flask from heat at the instant when the last bit of liquid becomes gas, at which time the flask will be filled with only gaseous sample at ambient pressure
3. Sealing the flask and permitting the gaseous sample to condense to liquid, and then weighing the flask to determine the sample’s mass (see Figure $1$)
Using this procedure, a sample of chloroform gas weighing 0.494 g is collected in a flask with a volume of 129 cm3 at 99.6 °C when the atmospheric pressure is 742.1 mm Hg. What is the approximate molar mass of chloroform?
Solution
Since $ℳ =\frac{m}{n}$ and $n=\frac{P V}{R T}$ substituting and rearranging gives $ℳ =\frac{m R T}{P V}$,
then
$ℳ =\frac{m R T}{P V}=\frac{(0.494~\text{g} ) \times 0.08206 ~\text{L} \cdot \text{atm/mol K } \times 372.8 ~\text{K} }{0.976 ~\text{atm} \times 0.129~ \text{L} }=120 ~ \text{g/mol} \nonumber$
Exercise $3$
A sample of phosphorus that weighs $3.243 \times 10^{−2}~\text{g}$ exerts a pressure of 31.89 kPa in a 56.0-mL bulb at 550 °C. What are the molar mass and molecular formula of phosphorus vapor?
Answer
124 g/mol P4
The Pressure of a Mixture of Gases: Dalton’s Law
Unless they chemically react with each other, the individual gases in a mixture of gases do not affect each other’s pressure. Each individual gas in a mixture exerts the same pressure that it would exert if it were present alone in the container (Figure $2$). The pressure exerted by each individual gas in a mixture is called its partial pressure. This observation is summarized by Dalton’s law of partial pressures: The total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases:
$P_{\text {Total }}=P_A+P_B+P_C+\ldots=\sum_{ i } P_{ i } \nonumber$
In the equation PTotal is the total pressure of a mixture of gases, PA is the partial pressure of gas A; PB is the partial pressure of gas B; PC is the partial pressure of gas C; and so on.
The partial pressure of gas A is related to the total pressure of the gas mixture via its mole fraction ($X$), a unit of concentration defined as the number of moles of a component of a solution divided by the total number of moles of all components:
$P_A=X_A \times P_{\text {Total }} \nonumber$
where
$X_A=\frac{n_A}{n_{\text {Total }}} \nonumber$
where PA, XA, and nA are the partial pressure, mole fraction, and number of moles of gas A, respectively, and nTotal is the number of moles of all components in the mixture.
Example $4$: The Pressure of a Mixture of Gases
A 10.0-L vessel contains 2.50 10−3 mol of H2, 1.00 10−3 mol of He, and 3.00 10−4 mol of Ne at 35 °C.
1. What are the partial pressures of each of the gases?
2. What is the total pressure in atmospheres?
Solution
The gases behave independently, so the partial pressure of each gas can be determined from the ideal gas equation, using $P=\frac{n R T}{V}$:
\begin{align*} P_{ H_2} &=\frac{\left(2.50 \times 10^{-3} ~\cancel{\text{mol}} \right)\left(0.08206~ \cancel{\text{L}}~\text{atm} \cancel{~\text{mol}^{-1} ~ \text{K}^{-1}}\right)(308 ~ \cancel{\text{K}} )}{10.0 ~\cancel{\text{L}} }=6.32 \times 10^{-3} ~\text{atm} \[4pt] &=\frac{\left(1.00 \times 10^{-3} ~\cancel{\text{mol}} \right)\left(0.08206 ~ \cancel{\text{L}}~\text{atm} \cancel{~\text{mol}^{-1} ~ \text{K}^{-1}} \right)(308 ~ \cancel{\text{K}} )}{10.0 ~\cancel{\text{L}} }=2.53 \times 10^{-3} ~\text{atm} \[4pt] &=\frac{\left(3.00 \times 10^{-4} ~\cancel{\text{mol}} \right)\left(0.08206 ~ \cancel{\text{L}}~\text{atm} \cancel{~\text{mol}^{-1} ~ \text{K}^{-1}}\right)(308 ~ \cancel{\text{K}} )}{10.0 ~\cancel{\text{L}}}=7.58 \times 10^{-4} ~\text{atm} \end{align*} \nonumber
The total pressure is given by the sum of the partial pressures:
$P_{ T } = P_{ H_2} + P_{ He } + P_{ Ne }=(0.00632+0.00253+0.00076) ~\text{atm}=9.61 \times 10^{-3} ~\text{atm}\nonumber$
Exercise $4$
A 5.73-L flask at 25 °C contains 0.0388 mol of N2, 0.147 mol of CO, and 0.0803 mol of H2. What is the total pressure in the flask in atmospheres?
Answer
1.137 atm
Here is another example of this concept, but dealing with mole fraction calculations.
Example $5$: The Pressure of a Mixture of Gases
A gas mixture used for anesthesia contains 2.83 mol oxygen, $\ce{O2}$, and 8.41 mol nitrous oxide, $\ce{N2O}$. The total pressure of the mixture is 192 kPa.
1. What are the mole fractions of O2 and N2O?
2. What are the partial pressures of O2 and N2O?
Solution
The mole fraction is given by $X_A=\frac{n_A}{n_{\text {Total }}}$ and the partial pressure is $P_A = X_A P_{Total}$.
For $\ce{O2}$,
$X_{O_2}=\frac{n_{O_2}}{n_{\text {Total }}}=\frac{2.83 mol }{(2.83+8.41) mol }=0.252 \nonumber$
and $P_{O_2}=X_{O_2} \times P_{\text {Total }}=0.252 \times 192 kPa =48.4 kPa$
For $\ce{N2O}$,
$X_{ N_2O }=\frac{n_{ N_2O }}{n_{\text {Total }}}=\frac{8.41 mol }{(2.83+8.41) mol }=0.748 \nonumber$
and
$P_{N_2O }=X_{N_2O } \times P_{\text {Total }}=0.748 \times 192 kPa =144 kPa \nonumber$
Exercise $5$
What is the pressure of a mixture of 0.200 g of H2, 1.00 g of N2, and 0.820 g of Ar in a container with a volume of 2.00 L at 20 °C?
Answer
1.87 atm
Collection of Gases over Water
A simple way to collect gases that do not react with water is to capture them in a bottle that has been filled with water and inverted into a dish filled with water. The pressure of the gas inside the bottle can be made equal to the air pressure outside by raising or lowering the bottle. When the water level is the same both inside and outside the bottle (Figure $3$), the pressure of the gas is equal to the atmospheric pressure, which can be measured with a barometer.
However, there is another factor we must consider when we measure the pressure of the gas by this method. Water evaporates and there is always gaseous water (water vapor) above a sample of liquid water. As a gas is collected over water, it becomes saturated with water vapor and the total pressure of the mixture equals the partial pressure of the gas plus the partial pressure of the water vapor. The pressure of the pure gas is therefore equal to the total pressure minus the pressure of the water vapor—this is referred to as the “dry” gas pressure, that is, the pressure of the gas only, without water vapor. The vapor pressure of water, which is the pressure exerted by water vapor in equilibrium with liquid water in a closed container, depends on the temperature (Figure $4$); more detailed information on the temperature dependence of water vapor can be found in Table $1$, and vapor pressure will be discussed in more detail in the next chapter on liquids.
Table $1$: Vapor Pressure of Ice and Water in Various Temperatures at Sea Level
Temperature (°C) Pressure (torr) Temperature (°C) Pressure (torr) Temperature (°C) Pressure (torr)
–10 1.95 18 15.5 30 31.8
–5 3.0 19 16.5 35 42.2
–2 3.9 20 17.5 40 55.3
0 4.6 21 18.7 50 92.5
2 5.3 22 19.8 60 149.4
4 6.1 23 21.1 70 233.7
6 7.0 24 22.4 80 355.1
8 8.0 25 23.8 90 525.8
10 9.2 26 25.2 95 633.9
12 10.5 27 26.7 99 733.2
14 12.0 28 28.3 100.0 760.0
16 13.6 29 30.0 101.0 787.6
Example $6$: Pressure of a Gas Collected Over Water
If 0.200 L of argon is collected over water at a temperature of 26 °C and a pressure of 750 torr in a system like that shown in Figure $3$, what is the partial pressure of argon?
Solution
According to Dalton’s law, the total pressure in the bottle (750 torr) is the sum of the partial pressure of argon and the partial pressure of gaseous water:
$P_{ T }=P_{ Ar }+P_{ H_2O } \nonumber$
Rearranging this equation to solve for the pressure of argon gives:
$P_{ Ar }=P_{ T }-P_{ H_2O } \nonumber$
The pressure of water vapor above a sample of liquid water at 26 °C is 25.2 torr (Appendix E), so:
$P_{ Ar }=750 \text { torr }-25.2 \text { torr }=725 \text { torr } \nonumber$
Exercise $6$
A sample of oxygen collected over water at a temperature of 29.0 °C and a pressure of 764 torr has a volume of 0.560 L. What volume would the dry oxygen from this sample have under the same conditions of temperature and pressure?
Answer
0.537 L
Chemical Stoichiometry and Gases
Chemical stoichiometry describes the quantitative relationships between reactants and products in chemical reactions.
We have previously measured quantities of reactants and products using masses for solids and volumes in conjunction with the molarity for solutions; now we can also use gas volumes to indicate quantities. If we know the volume, pressure, and temperature of a gas, we can use the ideal gas equation to calculate how many moles of the gas are present. If we know how many moles of a gas are involved, we can calculate the volume of a gas at any temperature and pressure.
Avogadro’s Law Revisited
Sometimes we can take advantage of a simplifying feature of the stoichiometry of gases that solids and solutions do not exhibit: All gases that show ideal behavior contain the same number of molecules in the same volume (at the same temperature and pressure). Thus, the ratios of volumes of gases involved in a chemical reaction are given by the coefficients in the equation for the reaction, provided that the gas volumes are measured at the same temperature and pressure.
We can extend Avogadro’s law (that the volume of a gas is directly proportional to the number of moles of the gas) to chemical reactions with gases: Gases combine, or react, in definite and simple proportions by volume, provided that all gas volumes are measured at the same temperature and pressure. For example, since nitrogen and hydrogen gases react to produce ammonia gas according to
$\ce{N2(g) +3 H2(g) -> 2 NH3(g)} \nonumber$
a given volume of nitrogen gas reacts with three times that volume of hydrogen gas to produce two times that volume of ammonia gas, if pressure and temperature remain constant.
The explanation for this is illustrated in Figure $5$. According to Avogadro’s law, equal volumes of gaseous N2, H2, and NH3, at the same temperature and pressure, contain the same number of molecules. Because one molecule of N2 reacts with three molecules of H2 to produce two molecules of NH3, the volume of H2 required is three times the volume of N2, and the volume of NH3 produced is two times the volume of N2.
Example $7$: Reaction of Gases
Propane, $\ce{C3H8(g)}$, is used in gas grills to provide the heat for cooking. What volume of O2(g) measured at 25 °C and 760 torr is required to react with 2.7 L of propane measured under the same conditions of temperature and pressure? Assume that the propane undergoes complete combustion.
Solution
The ratio of the volumes of C3H8 and O2 will be equal to the ratio of their coefficients in the balanced equation for the reaction:
$\underset{\ce{C3H8(g) + 5 O2(g)}}{\text{1 volume + 5 volumes }} \ce{ -> } \underset{\ce{3 CO2(g) + 4 H2O(l)}}{\text{3 volumes + 4 volumes}} \nonumber$
From the equation, we see that one volume of $\ce{C3H8}$ will react with five volumes of $\ce{O2}$:
$2.7~\cancel{L} ~\ce{C3H8} \times \dfrac{5~\text{L}~\ce{O2}}{1 ~ \cancel{\text{L} \ce{C3H8}}}=13.5 ~ \text{L}~\ce{O2} \nonumber$
A volume of 13.5 L of O2 will be required to react with 2.7 L of C3H8.
Exercise $7$
An acetylene tank for an oxyacetylene welding torch provides 9340 L of acetylene gas, C2H2, at 0 °C and 1 atm. How many tanks of oxygen, each providing 7.00 103 L of O2 at 0 °C and 1 atm, will be required to burn the acetylene?
$2 C_2 H_2+5 O_2 \longrightarrow 4 CO_2+2 H_2O \nonumber$
Answer
3.34 tanks (2.34 104 L)
Example $8$: Volumes of Reacting Gases
Ammonia is an important fertilizer and industrial chemical. Suppose that a volume of 683 billion cubic feet of gaseous ammonia, measured at 25 °C and 1 atm, was manufactured. What volume of H2(g), measured under the same conditions, was required to prepare this amount of ammonia by reaction with N2?
$\ce{N2(g) + 3 H2(g) -> 2 NH3(g)} \nonumber$
Solution
Because equal volumes of H2 and NH3 contain equal numbers of molecules and each three molecules of H2 that react produce two molecules of NH3, the ratio of the volumes of H2 and NH3 will be equal to 3:2. Two volumes of NH3, in this case in units of billion ft3, will be formed from three volumes of H2:
$683 \text { billion } ft^3 NH_3 \times \frac{3 \text { billion } ft^3 H_2}{2 \text { billion } ft^3 NH_3}=1.02 \times 10^3 \text { billion } ft^3 H_2 \nonumber$
The manufacture of 683 billion ft3 of NH3 required 1020 billion ft3 of H2. (At 25 °C and 1 atm, this is the volume of a cube with an edge length of approximately 1.9 miles.)
Exercise $8$
What volume of O2(g) measured at 25 °C and 760 torr is required to react with 17.0 L of ethylene, C2H4(g), measured under the same conditions of temperature and pressure? The products are CO2 and water vapor.
Answer
51.0 L
Example $9$: Volume of Gaseous Product
What volume of hydrogen at 27 °C and 723 torr may be prepared by the reaction of 8.88 g of gallium with an excess of hydrochloric acid?
$\ce{2 Ga(s) + 6 HCl(aq) -> 2 GaCl3(aq) + 3 H2(g)} \nonumber$
Solution
Convert the provided mass of the limiting reactant, Ga, to moles of hydrogen produced:
$8.88 ~ \cancel{\text{g}~\ce{Ga}} \times \frac{1 ~ \cancel{\text{mol} ~\ce{Ga}}}{69.723~ \cancel{\text{g}~\ce{Ga}} } \times \frac{3 ~ \text{mol} ~\ce{H_2}}{2~ \cancel{\text{mol}~\ce{Ga}} }=0.191 ~\text{mol}~\ce{H_2} \nonumber$
Convert the provided temperature and pressure values to appropriate units (K and atm, respectively), and then use the molar amount of hydrogen gas and the ideal gas equation to calculate the volume of gas:
$V=\left(\frac{n R T}{P}\right)=\frac{0.191 ~ \cancel{\text{mol}} \times 0.08206 ~\text{L} \cancel{\text{atm mol}}^{-1}\text{K}^{-1} \times 300 ~\text{K} }{0.951 ~ \cancel{\text{atm}} }=4.94 ~\text{L} \nonumber$
Exercise $9$
Sulfur dioxide is an intermediate in the preparation of sulfuric acid. What volume of SO2 at 343 °C and 1.21 atm is produced by burning l.00 kg of sulfur in excess oxygen?
Answer
1.30 103 L
How Sciences Interconnect: Greenhouse Gases and Climate Change
The thin skin of our atmosphere keeps the earth from being an ice planet and makes it habitable. In fact, this is due to less than 0.5% of the air molecules. Of the energy from the sun that reaches the earth, almost $\frac{1}{3}$ is reflected back into space, with the rest absorbed by the atmosphere and the surface of the earth. Some of the energy that the earth absorbs is re-emitted as infrared (IR) radiation, a portion of which passes back out through the atmosphere into space. Most if this IR radiation, however, is absorbed by certain atmospheric gases, effectively trapping heat within the atmosphere in a phenomenon known as the greenhouse effect. This effect maintains global temperatures within the range needed to sustain life on earth. Without our atmosphere, the earth's average temperature would be lower by more than 30 °C (nearly 60 °F). The major greenhouse gases (GHGs) are water vapor, carbon dioxide, methane, and ozone. Since the Industrial Revolution, human activity has been increasing the concentrations of GHGs, which have changed the energy balance and are significantly altering the earth’s climate (Figure $6$).
There is strong evidence from multiple sources that higher atmospheric levels of CO2 are caused by human activity, with fossil fuel burning accounting for about $\frac{3}{4}$ of the recent increase in CO2. Reliable data from ice cores reveals that CO2 concentration in the atmosphere is at the highest level in the past 800,000 years; other evidence indicates that it may be at its highest level in 20 million years. In recent years, the CO2 concentration has increased preindustrial levels of ~280 ppm to more than 400 ppm today (Figure $7$).
Link to Learning
Click here to see a 2-minute video explaining greenhouse gases and global warming.
Portrait of a Chemist: Susan Solomon
Atmospheric and climate scientist Susan Solomon (Figure $8$) is the author of one of The New York Times books of the year (The Coldest March, 2001), one of Time magazine’s 100 most influential people in the world (2008), and a working group leader of the Intergovernmental Panel on Climate Change (IPCC), which was the recipient of the 2007 Nobel Peace Prize. She helped determine and explain the cause of the formation of the ozone hole over Antarctica, and has authored many important papers on climate change. She has been awarded the top scientific honors in the US and France (the National Medal of Science and the Grande Medaille, respectively), and is a member of the National Academy of Sciences, the Royal Society, the French Academy of Sciences, and the European Academy of Sciences. Formerly a professor at the University of Colorado, she is now at MIT, and continues to work at NOAA.
For more information, watch this video about Susan Solomon.
Footnotes
• 2“Quotations by Joseph-Louis Lagrange,” last modified February 2006, accessed February 10, 2015, www-history.mcs.st-andrews.ac.../Lagrange.html | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/09%3A_Gases/9.03%3A_Stoichiometry_of_Gaseous_Substances_Mixtures_and_Reactions.txt |
Learning Objectives
By the end of this section, you will be able to:
• Define and explain effusion and diffusion
• State Graham’s law and use it to compute relevant gas properties
If you have ever been in a room when a piping hot pizza was delivered, you have been made aware of the fact that gaseous molecules can quickly spread throughout a room, as evidenced by the pleasant aroma that soon reaches your nose. Although gaseous molecules travel at tremendous speeds (hundreds of meters per second), they collide with other gaseous molecules and travel in many different directions before reaching the desired target. At room temperature, a gaseous molecule will experience billions of collisions per second. The mean free path is the average distance a molecule travels between collisions. The mean free path increases with decreasing pressure; in general, the mean free path for a gaseous molecule will be hundreds of times the diameter of the molecule
In general, we know that when a sample of gas is introduced to one part of a closed container, its molecules very quickly disperse throughout the container; this process by which molecules disperse in space in response to differences in concentration is called diffusion (shown in Figure $1$). The gaseous atoms or molecules are, of course, unaware of any concentration gradient, they simply move randomly—regions of higher concentration have more particles than regions of lower concentrations, and so a net movement of species from high to low concentration areas takes place. In a closed environment, diffusion will ultimately result in equal concentrations of gas throughout, as depicted in Figure $1$. The gaseous atoms and molecules continue to move, but since their concentrations are the same in both bulbs, the rates of transfer between the bulbs are equal (no net transfer of molecules occurs).
We are often interested in the rate of diffusion, the amount of gas passing through some area per unit time:
$\text { rate of diffusion }=\frac{\text { amount of gas passing through an area }}{\text { unit of time }} \nonumber$
The diffusion rate depends on several factors: temperature; the mass of the atoms or molecules; the concentration gradient (the increase or decrease in concentration from one point to another); the amount of surface area available for diffusion; and the distance the gas particles must travel. Note also that the time required for diffusion to occur is inversely proportional to the rate of diffusion, as shown in the rate of diffusion equation.
A process involving movement of gaseous species similar to diffusion is effusion, the escape of gas molecules through a tiny hole such as a pinhole in a balloon into a vacuum (Figure $2$). Although diffusion and effusion rates both depend on the molar mass of the gas involved, their rates are not equal; however, the ratios of their rates are the same.
If a mixture of gases is placed in a container with porous walls, the gases effuse through the small openings in the walls. The lighter gases pass through the small openings more rapidly (at a higher rate) than the heavier ones (Figure $3$). In 1832, Thomas Graham studied the rates of effusion of different gases and formulated Graham’s law of effusion: The rate of effusion of a gas is inversely proportional to the square root of the mass of its particles:
$\text { rate of effusion } \propto \frac{1}{\sqrt{ M }} \nonumber$
This means that if two gases A and B are at the same temperature and pressure, the ratio of their effusion rates is inversely proportional to the ratio of the square roots of the masses of their particles:
$\frac{\text { rate of effusion of } A }{\text { rate of effusion of } B }=\frac{\sqrt{ M_{ B }}}{\sqrt{ M_{ A }}} \nonumber$
Example $1$: Applying Graham’s Law to Rates of Effusion
Calculate the ratio of the rate of effusion of hydrogen to the rate of effusion of oxygen.
Solution
From Graham’s law, we have:
$\frac{\text { rate of effusion of hydrogen }}{\text { rate of effusion of oxygen }}=\frac{\sqrt{32 gmol^{-1}}}{\sqrt{2 gmol^{-1}}}=\frac{\sqrt{16}}{\sqrt{1}}=\frac{4}{1} \nonumber$
Hydrogen effuses four times as rapidly as oxygen.
Exercise $1$
At a particular pressure and temperature, nitrogen gas effuses at the rate of 79 mL/s. Under the same conditions, at what rate will sulfur dioxide effuse?
Answer
52 mL/s
Example $2$: Effusion Time Calculations
It takes 243 s for 4.46 10−5 mol Xe to effuse through a tiny hole. Under the same conditions, how long will it take 4.46 10−5 mol Ne to effuse?
Solution
It is important to resist the temptation to use the times directly, and to remember how rate relates to time as well as how it relates to mass. Recall the definition of rate of effusion:
$\text { rate of effusion }=\frac{\text { amount of gas transferred }}{\text { time }} \nonumber$
and combine it with Graham’s law:
$\frac{\text { rate of effusion of gas Xe }}{\text { rate of effusion of gas Ne}}=\frac{\sqrt{ M_{ Ne }}}{\sqrt{ M_{ Xe }}} \nonumber$
To get:
$\frac{\frac{\text { amount of Xe transferred }}{\text { time for Xe }}}{\frac{\text { amount of Ne transferred }}{\text { time for Ne }}}=\frac{\sqrt{ M_{ Ne }}}{\sqrt{ M_{ Xe }}} \nonumber$
Noting that amount of A = amount of B, and solving for time for Ne:
$\frac{\frac{\text { amount of } Xe }{\text { time for Xe }}}{\frac{\text { amount of } Ne }{\text { time for } Ne }}=\frac{\text { time for } Ne }{\text { time for Xe }}=\frac{\sqrt{ M_{ Ne }}}{\sqrt{ M_{ Xe }}}=\frac{\sqrt{ M_{ Ne }}}{\sqrt{ M_{ Xe }}} \nonumber$
and substitute values:
$\frac{\text { time for Ne }}{243 s }=\sqrt{\frac{20.2~ \cancel{\text{g mol}} }{131.3 ~ \cancel{\text{g mol}} }}=0.392 \nonumber$
Finally, solve for the desired quantity:
$\text { time for } Ne =0.392 \times 243~\text{s} =95.3~\text{s} \nonumber$
Note that this answer is reasonable: Since Ne is lighter than Xe, the effusion rate for Ne will be larger than that for Xe, which means the time of effusion for Ne will be smaller than that for Xe.
Exercise $2$
A party balloon filled with helium deflates to $\frac{2}{3}$ of its original volume in 8.0 hours. How long will it take an identical balloon filled with the same number of moles of air (ℳ = 28.2 g/mol) to deflate to$\frac{1}{2}$ of its original volume?
Answer
32 h
Example $3$: Determining Molar Mass Using Graham’s Law
An unknown gas effuses 1.66 times more rapidly than CO2. What is the molar mass of the unknown gas? Can you make a reasonable guess as to its identity?
Solution
From Graham’s law, we have:
$\frac{\text { rate of effusion of Unknown }}{\text { rate of effusion of } CO_2}=\frac{\sqrt{ M_{ CO_2}}}{\sqrt{ M_{\text {Unknown }}}} \nonumber$
Plug in known data:
$\frac{1.66}{1}=\frac{\sqrt{44.0 g / mol }}{\sqrt{ M_{\text {Unknown }}}} \nonumber$
Solve:
$M_{\text {Unknown }}=\frac{44.0 g / mol }{(1.66)^2}=16.0 g / mol \nonumber$
The gas could well be CH4, the only gas with this molar mass.
Exercise $3$
Hydrogen gas effuses through a porous container 8.97-times faster than an unknown gas. Estimate the molar mass of the unknown gas.
Answer
163 g/mol
How Sciences Interconnect: Use of Diffusion for Nuclear Energy Applications: Uranium Enrichment
Gaseous diffusion has been used to produce enriched uranium for use in nuclear power plants and weapons. Naturally occurring uranium contains only 0.72% of 235U, the kind of uranium that is “fissile,” that is, capable of sustaining a nuclear fission chain reaction. Nuclear reactors require fuel that is 2–5% 235U, and nuclear bombs need even higher concentrations. One way to enrich uranium to the desired levels is to take advantage of Graham’s law. In a gaseous diffusion enrichment plant, uranium hexafluoride (UF6, the only uranium compound that is volatile enough to work) is slowly pumped through large cylindrical vessels called diffusers, which contain porous barriers with microscopic openings. The process is one of diffusion because the other side of the barrier is not evacuated. The 235UF6 molecules have a higher average speed and diffuse through the barrier a little faster than the heavier 238UF6 molecules. The gas that has passed through the barrier is slightly enriched in 235UF6 and the residual gas is slightly depleted. The small difference in molecular weights between 235UF6 and 238UF6 only about 0.4% enrichment, is achieved in one diffuser (Figure $4$). But by connecting many diffusers in a sequence of stages (called a cascade), the desired level of enrichment can be attained.
The large scale separation of gaseous 235UF6 from 238UF6 was first done during the World War II, at the atomic energy installation in Oak Ridge, Tennessee, as part of the Manhattan Project (the development of the first atomic bomb). Although the theory is simple, this required surmounting many daunting technical challenges to make it work in practice. The barrier must have tiny, uniform holes (about 10–6 cm in diameter) and be porous enough to produce high flow rates. All materials (the barrier, tubing, surface coatings, lubricants, and gaskets) need to be able to contain, but not react with, the highly reactive and corrosive UF6.
Because gaseous diffusion plants require very large amounts of energy (to compress the gas to the high pressures required and drive it through the diffuser cascade, to remove the heat produced during compression, and so on), it is now being replaced by gas centrifuge technology, which requires far less energy. A current hot political issue is how to deny this technology to Iran, to prevent it from producing enough enriched uranium for them to use to make nuclear weapons. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/09%3A_Gases/9.04%3A_Effusion_and_Diffusion_of_Gases.txt |
Learning Objectives
• State the postulates of the kinetic-molecular theory
• Use this theory’s postulates to explain the gas laws
The gas laws that we have seen to this point, as well as the ideal gas equation, are empirical, that is, they have been derived from experimental observations. The mathematical forms of these laws closely describe the macroscopic behavior of most gases at pressures less than about 1 or 2 atm. Although the gas laws describe relationships that have been verified by many experiments, they do not tell us why gases follow these relationships.
The kinetic molecular theory (KMT) is a simple microscopic model that effectively explains the gas laws described in previous modules of this chapter. This theory is based on the following five postulates described here. (Note: The term “molecule” will be used to refer to the individual chemical species that compose the gas, although some gases are composed of atomic species, for example, the noble gases.)
1. Gases are composed of molecules that are in continuous motion, travelling in straight lines and changing direction only when they collide with other molecules or with the walls of a container.
2. The molecules composing the gas are negligibly small compared to the distances between them.
3. The pressure exerted by a gas in a container results from collisions between the gas molecules and the container walls.
4. Gas molecules exert no attractive or repulsive forces on each other or the container walls; therefore, their collisions are elastic (do not involve a loss of energy).
5. The average kinetic energy of the gas molecules is proportional to the kelvin temperature of the gas.
The test of the KMT and its postulates is its ability to explain and describe the behavior of a gas. The various gas laws can be derived from the assumptions of the KMT, which have led chemists to believe that the assumptions of the theory accurately represent the properties of gas molecules. We will first look at the individual gas laws (Boyle’s, Charles’s, Amontons’s, Avogadro’s, and Dalton’s laws) conceptually to see how the KMT explains them. Then, we will more carefully consider the relationships between molecular masses, speeds, and kinetic energies with temperature, and explain Graham’s law.
The Kinetic-Molecular Theory Explains the Behavior of Gases, Part I
Recalling that gas pressure is exerted by rapidly moving gas molecules and depends directly on the number of molecules hitting a unit area of the wall per unit of time, we see that the KMT conceptually explains the behavior of a gas as follows:
• Amontons’s law. If the temperature is increased, the average speed and kinetic energy of the gas molecules increase. If the volume is held constant, the increased speed of the gas molecules results in more frequent and more forceful collisions with the walls of the container, therefore increasing the pressure (Figure $1$).
• Charles’s law. If the temperature of a gas is increased, a constant pressure may be maintained only if the volume occupied by the gas increases. This will result in greater average distances traveled by the molecules to reach the container walls, as well as increased wall surface area. These conditions will decrease the both the frequency of molecule-wall collisions and the number of collisions per unit area, the combined effects of which balance the effect of increased collision forces due to the greater kinetic energy at the higher temperature.
• Boyle’s law. If the gas volume volume of a given amount of gas at a given temperature is decreased (that is, if the gas is compressed), the molecules will be exposed to a decreased container wall area. Collisions with the container wall will therefore occur more frequently and the pressure exerted by the gas will increase (Figure $1$).
• Avogadro’s law. At constant pressure and temperature, the frequency and force of molecule-wall collisions are constant. Under such conditions, increasing the number of gaseous molecules will require a proportional increase in the container volume in order to yield a decrease in the number of collisions per unit area to compensate for the increased frequency of collisions (Figure $1$).
• Dalton’s Law. Because of the large distances between them, the molecules of one gas in a mixture bombard the container walls with the same frequency whether other gases are present or not, and the total pressure of a gas mixture equals the sum of the (partial) pressures of the individual gases.
molecular speeds and Kinetic Energy
The previous discussion showed that the KMT qualitatively explains the behaviors described by the various gas laws. The postulates of this theory may be applied in a more quantitative fashion to derive these individual laws. To do this, we must first look at speeds and kinetic energies of gas molecules, and the temperature of a gas sample.
In a gas sample, individual molecules have widely varying speeds; however, because of the vast number of molecules and collisions involved, the molecular speed distribution and average speed are constant. This molecular speed distribution is known as a Maxwell-Boltzmann distribution, and it depicts the relative numbers of molecules in a bulk sample of gas that possesses a given speed (Figure $2$).
The kinetic energy (KE) of a particle of mass ($m$) and speed ($u$) is given by:
$KE =\frac{1}{2} m u^2 \nonumber$
Expressing mass in kilograms and speed in meters per second will yield energy values in units of joules (J = kg m2 s–2). To deal with a large number of gas molecules, we use averages for both speed and kinetic energy. In the KMT, the root mean square speed of a particle, urms, is defined as the square root of the average of the squares of the speeds with n = the number of particles:
$u_{ rms }=\sqrt{\overline{\overline{u^2}}}=\sqrt{\frac{u_1^2+u_2^2+u_3^2+u_4^2+\ldots}{n}} \nonumber$
The average kinetic energy for a mole of particles, KEavg, is then equal to:
$KE_{ avg }=\frac{1}{2} M u_{ rms }^2 \nonumber$
where M is the molar mass expressed in units of kg/mol. The KEavg of a mole of gas molecules is also directly proportional to the temperature of the gas and may be described by the equation:
$KE_{ avg }=\frac{3}{2} R T \nonumber$
where R is the gas constant and T is the kelvin temperature. When used in this equation, the appropriate form of the gas constant is 8.314 J/mol⋅K (8.314 kg m2s2mol–1K–1). These two separate equations for KEavg may be combined and rearranged to yield a relation between molecular speed and temperature:
$\frac{1}{2} M u_{ rms }^2=\frac{3}{2} R T \nonumber$
with
$u_{ rms }=\sqrt{\frac{3 R T}{M}} \nonumber$
Example $1$: Calculation of urms
Calculate the root-mean-square speed for a nitrogen molecule at 30 °C.
Solution
Convert the temperature into Kelvin:
$30^{\circ} C +273=303 K \nonumber$
Determine the molar mass of nitrogen in kilograms:
$\frac{28.0 ~ \cancel{\text{g}}} }{1 ~\text{mol} } \times \frac{1~\text{kg} }{1000 ~ \cancel{\text{g}}}=0.028~ \text{kg/mol} \nonumber$
Replace the variables and constants in the root-mean-square speed equation, replacing Joules with the equivalent kg m2s–2:
$u_{ rms }=\sqrt{\frac{3 R T}{M}} \nonumber$
$u_{ rms }=\sqrt{\frac{3(8.314 J / mol K )(303 K )}{(0.028 kg / mol )}}=\sqrt{2.70 \times 10^5 m^2 s^{-2}}=519 m / s \nonumber$
Exercise $1$
Calculate the root-mean-square speed for a mole of oxygen molecules at –23 °C.
Answer
441 m/s
If the temperature of a gas increases, its KEavg increases, more molecules have higher speeds and fewer molecules have lower speeds, and the distribution shifts toward higher speeds overall, that is, to the right. If temperature decreases, KEavg decreases, more molecules have lower speeds and fewer molecules have higher speeds, and the distribution shifts toward lower speeds overall, that is, to the left. This behavior is illustrated for nitrogen gas in Figure $3$.
At a given temperature, all gases have the same KEavg for their molecules. Gases composed of lighter molecules have more high-speed particles and a higher urms, with a speed distribution that peaks at relatively higher speeds. Gases consisting of heavier molecules have more low-speed particles, a lower urms, and a speed distribution that peaks at relatively lower speeds. This trend is demonstrated by the data for a series of noble gases shown in Figure $4$.
Link to Learning
The gas simulator may be used to examine the effect of temperature on molecular speeds. Examine the simulator’s “energy histograms” (molecular speed distributions) and “species information” (which gives average speed values) for molecules of different masses at various temperatures.
The Kinetic-Molecular Theory Explains the Behavior of Gases, Part II
According to Graham’s law, the molecules of a gas are in rapid motion and the molecules themselves are small. The average distance between the molecules of a gas is large compared to the size of the molecules. As a consequence, gas molecules can move past each other easily and diffuse at relatively fast rates.
The rate of effusion of a gas depends directly on the (average) speed of its molecules:
$\text { effusion rate } \propto u_{\text {rms }} \nonumber$
Using this relation, and the equation relating molecular speed to mass, Graham’s law may be easily derived as shown here:
$u_{ rms }=\sqrt{\frac{3 R T}{M}} \nonumber$
$M=\frac{3 R T}{u_{ rms }^2}=\frac{3 R T}{\bar{u}^2} \nonumber$
$\frac{\text { effusion rate } A }{\text { effusion rate } B }=\frac{u_{ rms A }}{u_{ rms B }}=\frac{\sqrt{\frac{3 R T}{M_{ A }}}}{\sqrt{\frac{3 R T}{M_{ B }}}}=\sqrt{\frac{M_{ B }}{M_{ A }}} \nonumber$
The ratio of the rates of effusion is thus derived to be inversely proportional to the ratio of the square roots of their masses. This is the same relation observed experimentally and expressed as Graham’s law. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/09%3A_Gases/9.05%3A_The_Kinetic-Molecular_Theory.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the physical factors that lead to deviations from ideal gas behavior
• Explain how these factors are represented in the van der Waals equation
• Define compressibility (Z) and describe how its variation with pressure reflects non-ideal behavior
• Quantify non-ideal behavior by comparing computations of gas properties using the ideal gas law and the van der Waals equation
Thus far, the ideal gas law, $PV = nRT$, has been applied to a variety of different types of problems, ranging from reaction stoichiometry and empirical and molecular formula problems to determining the density and molar mass of a gas. As mentioned in the previous modules of this chapter, however, the behavior of a gas is often non-ideal, meaning that the observed relationships between its pressure, volume, and temperature are not accurately described by the gas laws. In this section, the reasons for these deviations from ideal gas behavior are considered.
One way in which the accuracy of PV = nRT can be judged is by comparing the actual volume of 1 mole of gas (its molar volume, Vm) to the molar volume of an ideal gas at the same temperature and pressure. This ratio is called the compressibility factor (Z) with:
$Z =\frac{\text { molar volume of gas at same } T \text { and } P}{\text { molar volume of ideal gas at same } T \text { and } P}=\left(\frac{P V_m}{R T}\right)_{\text {measured }} \nonumber$
Ideal gas behavior is therefore indicated when this ratio is equal to 1, and any deviation from 1 is an indication of non-ideal behavior. Figure $1$ shows plots of Z over a large pressure range for several common gases.
As is apparent from Figure $1$, the ideal gas law does not describe gas behavior well at relatively high pressures. To determine why this is, consider the differences between real gas properties and what is expected of a hypothetical ideal gas.
Particles of a hypothetical ideal gas have no significant volume and do not attract or repel each other. In general, real gases approximate this behavior at relatively low pressures and high temperatures. However, at high pressures, the molecules of a gas are crowded closer together, and the amount of empty space between the molecules is reduced. At these higher pressures, the volume of the gas molecules themselves becomes appreciable relative to the total volume occupied by the gas. The gas therefore becomes less compressible at these high pressures, and although its volume continues to decrease with increasing pressure, this decrease is not proportional as predicted by Boyle’s law.
At relatively low pressures, gas molecules have practically no attraction for one another because they are (on average) so far apart, and they behave almost like particles of an ideal gas. At higher pressures, however, the force of attraction is also no longer insignificant. This force pulls the molecules a little closer together, slightly decreasing the pressure (if the volume is constant) or decreasing the volume (at constant pressure) (Figure $2$). This change is more pronounced at low temperatures because the molecules have lower KE relative to the attractive forces, and so they are less effective in overcoming these attractions after colliding with one another.
There are several different equations that better approximate gas behavior than does the ideal gas law. The first, and simplest, of these was developed by the Dutch scientist Johannes van der Waals in 1879. The van der Waals equation improves upon the ideal gas law by adding two terms: one to account for the volume of the gas molecules and another for the attractive forces between them.
The constant a corresponds to the strength of the attraction between molecules of a particular gas, and the constant b corresponds to the size of the molecules of a particular gas. The “correction” to the pressure term in the ideal gas law is
Table $1$: Values of van der Waals Constants for Some Common Gases
Gas a (L2 atm/mol2) b (L/mol)
N2 1.39 0.0391
O2 1.36 0.0318
CO2 3.59 0.0427
H2O 5.46 0.0305
He 0.0342 0.0237
CCl4 20.4 0.1383
At low pressures, the correction for intermolecular attraction, a, is more important than the one for molecular volume, b. At high pressures and small volumes, the correction for the volume of the molecules becomes important because the molecules themselves are incompressible and constitute an appreciable fraction of the total volume. At some intermediate pressure, the two corrections have opposing influences and the gas appears to follow the relationship given by PV = nRT over a small range of pressures. This behavior is reflected by the “dips” in several of the compressibility curves shown in Figure $1$. The attractive force between molecules initially makes the gas more compressible than an ideal gas, as pressure is raised (Z decreases with increasing P). At very high pressures, the gas becomes less compressible (Z increases with P), as the gas molecules begin to occupy an increasingly significant fraction of the total gas volume.
Strictly speaking, the ideal gas equation functions well when intermolecular attractions between gas molecules are negligible and the gas molecules themselves do not occupy an appreciable part of the whole volume. These criteria are satisfied under conditions of low pressure and high temperature. Under such conditions, the gas is said to behave ideally, and deviations from the gas laws are small enough that they may be disregarded—this is, however, very often not the case.
Example $1$: Comparison of Ideal Gas Law and van der Waals Equation
A 4.25-L flask contains 3.46 mol CO2 at 229 °C. Calculate the pressure of this sample of CO2:
1. from the ideal gas law
2. from the van der Waals equation
3. Explain the reason(s) for the difference.
Solution
(a) From the ideal gas law:
$P=\frac{n R T}{V}=\frac{3.46 ~\text{mol} \times 0.08206 E atm mol^{-1} K^{-1} \times 502 K }{4.25 I }=33.5 atm \nonumber$
(b) From the van der Waals equation:
$\left(P+\frac{n^2 a}{V^2}\right) \times(V-n b)=n R T \longrightarrow P=\frac{n R T}{(V-n b)}-\frac{n^2 a}{V^2} \nonumber$
$P=\frac{3.46 ~\text{mol} \times 0.08206 ~\text{L atm mol}^{-1} ~ \text{K}^{-1} \times 502 ~\text{K} }{\left(4.25 ~\text{L} - 3.46 ~\text{mol} \times 0.0427 ~\text{L mol}^{-1}\right)} - \frac{(3.46 ~\text{mol})^2 \times 3.59 ~\text{L}^2 ~\text{atm mol}^2}{(4.24 ~\text{L})^2} \nonumber$
This finally yields P = 32.4 atm.
(c) This is not very different from the value from the ideal gas law because the pressure is not very high and the temperature is not very low. The value is somewhat different because CO2 molecules do have some volume and attractions between molecules, and the ideal gas law assumes they do not have volume or attractions.
Exercise $1$
A 560-mL flask contains 21.3 g N2 at 145 °C. Calculate the pressure of N2:
1. from the ideal gas law
2. from the van der Waals equation
3. Explain the reason(s) for the difference.
Answer
(a) 46.562 atm; (b) 46.594 atm; (c) The van der Waals equation takes into account the volume of the gas molecules themselves as well as intermolecular attractions. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/09%3A_Gases/9.06%3A_Non-Ideal_Gas_Behavior.txt |
Example and Directions
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Glossary Entries
Word(s)DefinitionImageCaptionLinkSource
absolute zerotemperature at which the volume of a gas would be zero according to Charles’s law.
Amontons’s law(also, Gay-Lussac’s law) pressure of a given number of moles of gas is directly proportional to its kelvin temperature when the volume is held constant
atmosphere (atm)unit of pressure; 1 atm = 101,325 Pa
Avogadro’s lawvolume of a gas at constant temperature and pressure is proportional to the number of gas molecules
bar(bar or b) unit of pressure; 1 bar = 100,000 Pa
barometerdevice used to measure atmospheric pressure
Boyle’s lawvolume of a given number of moles of gas held at constant temperature is inversely proportional to the pressure under which it is measured
Charles’s lawvolume of a given number of moles of gas is directly proportional to its kelvin temperature when the pressure is held constant
compressibility factor (Z)ratio of the experimentally measured molar volume for a gas to its molar volume as computed from the ideal gas equation
Dalton’s law of partial pressurestotal pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases
diffusionmovement of an atom or molecule from a region of relatively high concentration to one of relatively low concentration (discussed in this chapter with regard to gaseous species, but applicable to species in any phase)
effusiontransfer of gaseous atoms or molecules from a container to a vacuum through very small openings
Graham’s law of effusionrates of diffusion and effusion of gases are inversely proportional to the square roots of their molecular masses
hydrostatic pressurepressure exerted by a fluid due to gravity
ideal gashypothetical gas whose physical properties are perfectly described by the gas laws
ideal gas constant (R)constant derived from the ideal gas equation R = 0.08206 L atm mol–1 K–1 or 8.314 L kPa mol–1 K–1
ideal gas lawrelation between the pressure, volume, amount, and temperature of a gas under conditions derived by combination of the simple gas laws
kinetic molecular theorytheory based on simple principles and assumptions that effectively explains ideal gas behavior
manometerdevice used to measure the pressure of a gas trapped in a container
mean free pathaverage distance a molecule travels between collisions
mole fraction (X)concentration unit defined as the ratio of the molar amount of a mixture component to the total number of moles of all mixture components
partial pressurepressure exerted by an individual gas in a mixture
pascal (Pa)SI unit of pressure; 1 Pa = 1 N/m2
pounds per square inch (psi)unit of pressure common in the US
pressureforce exerted per unit area
rate of diffusionamount of gas diffusing through a given area over a given time
root mean square speed (urms)measure of average speed for a group of particles calculated as the square root of the average squared speed
standard conditions of temperature and pressure (STP)273.15 K (0 °C) and 1 atm (101.325 kPa)
standard molar volumevolume of 1 mole of gas at STP, approximately 22.4 L for gases behaving ideally
torrunit of pressure;
van der Waals equationmodified version of the ideal gas equation containing additional terms to account for non-ideal gas behavior
vapor pressure of waterpressure exerted by water vapor in equilibrium with liquid water in a closed container at a specific temperature | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/09%3A_Gases/9.07%3A_Key_Terms.txt |
p = hρg
PV = nRT
PTotal = PA + PB + PC + … = ƩiPi
PA = XA PTotal
9.09: Summary
Gases exert pressure, which is force per unit area. The pressure of a gas may be expressed in the SI unit of pascal or kilopascal, as well as in many other units including torr, atmosphere, and bar. Atmospheric pressure is measured using a barometer; other gas pressures can be measured using one of several types of manometers.
The behavior of gases can be described by several laws based on experimental observations of their properties. The pressure of a given amount of gas is directly proportional to its absolute temperature, provided that the volume does not change (Amontons’s law). The volume of a given gas sample is directly proportional to its absolute temperature at constant pressure (Charles’s law). The volume of a given amount of gas is inversely proportional to its pressure when temperature is held constant (Boyle’s law). Under the same conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules (Avogadro’s law).
The equations describing these laws are special cases of the ideal gas law, PV = nRT, where P is the pressure of the gas, V is its volume, n is the number of moles of the gas, T is its kelvin temperature, and R is the ideal (universal) gas constant.
The ideal gas law can be used to derive a number of convenient equations relating directly measured quantities to properties of interest for gaseous substances and mixtures. Appropriate rearrangement of the ideal gas equation may be made to permit the calculation of gas densities and molar masses. Dalton’s law of partial pressures may be used to relate measured gas pressures for gaseous mixtures to their compositions. Avogadro’s law may be used in stoichiometric computations for chemical reactions involving gaseous reactants or products.
Gaseous atoms and molecules move freely and randomly through space. Diffusion is the process whereby gaseous atoms and molecules are transferred from regions of relatively high concentration to regions of relatively low concentration. Effusion is a similar process in which gaseous species pass from a container to a vacuum through very small orifices. The rates of effusion of gases are inversely proportional to the square roots of their densities or to the square roots of their atoms/molecules’ masses (Graham’s law).
The kinetic molecular theory is a simple but very effective model that effectively explains ideal gas behavior. The theory assumes that gases consist of widely separated molecules of negligible volume that are in constant motion, colliding elastically with one another and the walls of their container with average speeds determined by their absolute temperatures. The individual molecules of a gas exhibit a range of speeds, the distribution of these speeds being dependent on the temperature of the gas and the mass of its molecules.
Gas molecules possess a finite volume and experience forces of attraction for one another. Consequently, gas behavior is not necessarily described well by the ideal gas law. Under conditions of low pressure and high temperature, these factors are negligible, the ideal gas equation is an accurate description of gas behavior, and the gas is said to exhibit ideal behavior. However, at lower temperatures and higher pressures, corrections for molecular volume and molecular attractions are required to account for finite molecular size and attractive forces. The van der Waals equation is a modified version of the ideal gas law that can be used to account for the non-ideal behavior of gases under these conditions. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/09%3A_Gases/9.08%3A_Key_Equations.txt |
1.
Why are sharp knives more effective than dull knives? (Hint: Think about the definition of pressure.)
2.
Why do some small bridges have weight limits that depend on how many wheels or axles the crossing vehicle has?
3.
Why should you roll or belly-crawl rather than walk across a thinly-frozen pond?
4.
A typical barometric pressure in Redding, California, is about 750 mm Hg. Calculate this pressure in atm and kPa.
5.
A typical barometric pressure in Denver, Colorado, is 615 mm Hg. What is this pressure in atmospheres and kilopascals?
6.
A typical barometric pressure in Kansas City is 740 torr. What is this pressure in atmospheres, in millimeters of mercury, and in kilopascals?
7.
Canadian tire pressure gauges are marked in units of kilopascals. What reading on such a gauge corresponds to 32 psi?
8.
During the Viking landings on Mars, the atmospheric pressure was determined to be on the average about 6.50 millibars (1 bar = 0.987 atm). What is that pressure in torr and kPa?
9.
The pressure of the atmosphere on the surface of the planet Venus is about 88.8 atm. Compare that pressure in psi to the normal pressure on earth at sea level in psi.
10.
A medical laboratory catalog describes the pressure in a cylinder of a gas as 14.82 MPa. What is the pressure of this gas in atmospheres and torr?
11.
Consider this scenario and answer the following questions: On a mid-August day in the northeastern United States, the following information appeared in the local newspaper: atmospheric pressure at sea level 29.97 in. Hg, 1013.9 mbar.
1. What was the pressure in kPa?
2. The pressure near the seacoast in the northeastern United States is usually reported near 30.0 in. Hg. During a hurricane, the pressure may fall to near 28.0 in. Hg. Calculate the drop in pressure in torr.
12.
Why is it necessary to use a nonvolatile liquid in a barometer or manometer?
13.
The pressure of a sample of gas is measured at sea level with a closed-end manometer. The liquid in the manometer is mercury. Determine the pressure of the gas in:
1. torr
2. Pa
3. bar
14.
The pressure of a sample of gas is measured with an open-end manometer, partially shown to the right. The liquid in the manometer is mercury. Assuming atmospheric pressure is 29.92 in. Hg, determine the pressure of the gas in:
1. torr
2. Pa
3. bar
15.
The pressure of a sample of gas is measured at sea level with an open-end mercury manometer. Assuming atmospheric pressure is 760.0 mm Hg, determine the pressure of the gas in:
1. mm Hg
2. atm
3. kPa
16.
The pressure of a sample of gas is measured at sea level with an open-end mercury manometer. Assuming atmospheric pressure is 760 mm Hg, determine the pressure of the gas in:
1. mm Hg
2. atm
3. kPa
17.
How would the use of a volatile liquid affect the measurement of a gas using open-ended manometers vs. closed-end manometers?
18.
Sometimes leaving a bicycle in the sun on a hot day will cause a blowout. Why?
19.
Explain how the volume of the bubbles exhausted by a scuba diver (Figure 9.16) change as they rise to the surface, assuming that they remain intact.
20.
One way to state Boyle’s law is “All other things being equal, the pressure of a gas is inversely proportional to its volume.” (a) What is the meaning of the term “inversely proportional?” (b) What are the “other things” that must be equal?
21.
An alternate way to state Avogadro’s law is “All other things being equal, the number of molecules in a gas is directly proportional to the volume of the gas.” (a) What is the meaning of the term “directly proportional?” (b) What are the “other things” that must be equal?
22.
How would the graph in Figure 9.12 change if the number of moles of gas in the sample used to determine the curve were doubled?
23.
How would the graph in Figure 9.13 change if the number of moles of gas in the sample used to determine the curve were doubled?
24.
In addition to the data found in Figure 9.13, what other information do we need to find the mass of the sample of air used to determine the graph?
25.
Determine the volume of 1 mol of CH4 gas at 150 K and 1 atm, using Figure 9.12.
26.
Determine the pressure of the gas in the syringe shown in Figure 9.13 when its volume is 12.5 mL, using:
1. the appropriate graph
2. Boyle’s law
27.
A spray can is used until it is empty except for the propellant gas, which has a pressure of 1344 torr at 23 °C. If the can is thrown into a fire (T = 475 °C), what will be the pressure in the hot can?
28.
What is the temperature of an 11.2-L sample of carbon monoxide, CO, at 744 torr if it occupies 13.3 L at 55 °C and 744 torr?
29.
A 2.50-L volume of hydrogen measured at –196 °C is warmed to 100 °C. Calculate the volume of the gas at the higher temperature, assuming no change in pressure.
30.
A balloon inflated with three breaths of air has a volume of 1.7 L. At the same temperature and pressure, what is the volume of the balloon if five more same-sized breaths are added to the balloon?
31.
A weather balloon contains 8.80 moles of helium at a pressure of 0.992 atm and a temperature of 25 °C at ground level. What is the volume of the balloon under these conditions?
32.
The volume of an automobile air bag was 66.8 L when inflated at 25 °C with 77.8 g of nitrogen gas. What was the pressure in the bag in kPa?
33.
How many moles of gaseous boron trifluoride, BF3, are contained in a 4.3410-L bulb at 788.0 K if the pressure is 1.220 atm? How many grams of BF3?
34.
Iodine, I2, is a solid at room temperature but sublimes (converts from a solid into a gas) when warmed. What is the temperature in a 73.3-mL bulb that contains 0.292 g of I2 vapor at a pressure of 0.462 atm?
35.
How many grams of gas are present in each of the following cases?
1. 0.100 L of CO2 at 307 torr and 26 °C
2. 8.75 L of C2H4, at 378.3 kPa and 483 K
3. 221 mL of Ar at 0.23 torr and –54 °C
36.
A high altitude balloon is filled with 1.41 104 L of hydrogen at a temperature of 21 °C and a pressure of 745 torr. What is the volume of the balloon at a height of 20 km, where the temperature is –48 °C and the pressure is 63.1 torr?
37.
A cylinder of medical oxygen has a volume of 35.4 L, and contains O2 at a pressure of 151 atm and a temperature of 25 °C. What volume of O2 does this correspond to at normal body conditions, that is, 1 atm and 37 °C?
38.
A large scuba tank (Figure 9.16) with a volume of 18 L is rated for a pressure of 220 bar. The tank is filled at 20 °C and contains enough air to supply 1860 L of air to a diver at a pressure of 2.37 atm (a depth of 45 feet). Was the tank filled to capacity at 20 °C?
39.
A 20.0-L cylinder containing 11.34 kg of butane, C4H10, was opened to the atmosphere. Calculate the mass of the gas remaining in the cylinder if it were opened and the gas escaped until the pressure in the cylinder was equal to the atmospheric pressure, 0.983 atm, and a temperature of 27 °C.
40.
While resting, the average 70-kg human male consumes 14 L of pure O2 per hour at 25 °C and 100 kPa. How many moles of O2 are consumed by a 70 kg man while resting for 1.0 h?
41.
For a given amount of gas showing ideal behavior, draw labeled graphs of:
1. the variation of P with V
2. the variation of V with T
3. the variation of P with T
4. the variation of with V
42.
A liter of methane gas, CH4, at STP contains more atoms of hydrogen than does a liter of pure hydrogen gas, H2, at STP. Using Avogadro’s law as a starting point, explain why.
43.
The effect of chlorofluorocarbons (such as CCl2F2) on the depletion of the ozone layer is well known. The use of substitutes, such as CH3CH2F(g), for the chlorofluorocarbons, has largely corrected the problem. Calculate the volume occupied by 10.0 g of each of these compounds at STP:
1. CCl2F2(g)
2. CH3CH2F(g)
44.
As 1 g of the radioactive element radium decays over 1 year, it produces 1.16 1018 alpha particles (helium nuclei). Each alpha particle becomes an atom of helium gas. What is the pressure in pascal of the helium gas produced if it occupies a volume of 125 mL at a temperature of 25 °C?
45.
A balloon with a volume of 100.21 L at 21 °C and 0.981 atm is released and just barely clears the top of Mount Crumpit in British Columbia. If the final volume of the balloon is 144.53 L at a temperature of 5.24 °C, what is the pressure experienced by the balloon as it clears Mount Crumpet?
46.
If the temperature of a fixed amount of a gas is doubled at constant volume, what happens to the pressure?
47.
If the volume of a fixed amount of a gas is tripled at constant temperature, what happens to the pressure?
48.
What is the density of laughing gas, dinitrogen monoxide, N2O, at a temperature of 325 K and a pressure of 113.0 kPa?
49.
Calculate the density of Freon 12, CF2Cl2, at 30.0 °C and 0.954 atm.
50.
Which is denser at the same temperature and pressure, dry air or air saturated with water vapor? Explain.
51.
A cylinder of O2(g) used in breathing by patients with emphysema has a volume of 3.00 L at a pressure of 10.0 atm. If the temperature of the cylinder is 28.0 °C, what mass of oxygen is in the cylinder?
52.
What is the molar mass of a gas if 0.0494 g of the gas occupies a volume of 0.100 L at a temperature 26 °C and a pressure of 307 torr?
53.
What is the molar mass of a gas if 0.281 g of the gas occupies a volume of 125 mL at a temperature 126 °C and a pressure of 777 torr?
54.
How could you show experimentally that the molecular formula of propene is C3H6, not CH2?
55.
The density of a certain gaseous fluoride of phosphorus is 3.93 g/L at STP. Calculate the molar mass of this fluoride and determine its molecular formula.
56.
Consider this question: What is the molecular formula of a compound that contains 39% C, 45% N, and 16% H if 0.157 g of the compound occupies l25 mL with a pressure of 99.5 kPa at 22 °C?
1. Outline the steps necessary to answer the question.
2. Answer the question.
57.
A 36.0–L cylinder of a gas used for calibration of blood gas analyzers in medical laboratories contains 350 g CO2, 805 g O2, and 4,880 g N2. At 25 degrees C, what is the pressure in the cylinder in atmospheres?
58.
A cylinder of a gas mixture used for calibration of blood gas analyzers in medical laboratories contains 5.0% CO2, 12.0% O2, and the remainder N2 at a total pressure of 146 atm. What is the partial pressure of each component of this gas? (The percentages given indicate the percent of the total pressure that is due to each component.)
59.
A sample of gas isolated from unrefined petroleum contains 90.0% CH4, 8.9% C2H6, and 1.1% C3H8 at a total pressure of 307.2 kPa. What is the partial pressure of each component of this gas? (The percentages given indicate the percent of the total pressure that is due to each component.)
60.
A mixture of 0.200 g of H2, 1.00 g of N2, and 0.820 g of Ar is stored in a closed container at STP. Find the volume of the container, assuming that the gases exhibit ideal behavior.
61.
Most mixtures of hydrogen gas with oxygen gas are explosive. However, a mixture that contains less than 3.0 % O2 is not. If enough O2 is added to a cylinder of H2 at 33.2 atm to bring the total pressure to 34.5 atm, is the mixture explosive?
62.
A commercial mercury vapor analyzer can detect, in air, concentrations of gaseous Hg atoms (which are poisonous) as low as 2 10−6 mg/L of air. At this concentration, what is the partial pressure of gaseous mercury if the atmospheric pressure is 733 torr at 26 °C?
63.
A sample of carbon monoxide was collected over water at a total pressure of 756 torr and a temperature of 18 °C. What is the pressure of the carbon monoxide? (See Table 9.2 for the vapor pressure of water.)
64.
In an experiment in a general chemistry laboratory, a student collected a sample of a gas over water. The volume of the gas was 265 mL at a pressure of 753 torr and a temperature of 27 °C. The mass of the gas was 0.472 g. What was the molar mass of the gas?
65.
Joseph Priestley first prepared pure oxygen by heating mercuric oxide, HgO:
1. Outline the steps necessary to answer the following question: What volume of O2 at 23 °C and 0.975 atm is produced by the decomposition of 5.36 g of HgO?
2. Answer the question.
66.
Cavendish prepared hydrogen in 1766 by the novel method of passing steam through a red-hot gun barrel:
1. Outline the steps necessary to answer the following question: What volume of H2 at a pressure of 745 torr and a temperature of 20 °C can be prepared from the reaction of 15.O g of H2O?
2. Answer the question.
67.
The chlorofluorocarbon CCl2F2 can be recycled into a different compound by reaction with hydrogen to produce CH2F2(g), a compound useful in chemical manufacturing:
1. Outline the steps necessary to answer the following question: What volume of hydrogen at 225 atm and 35.5 °C would be required to react with 1 ton (1.000 103 kg) of CCl2F2?
2. Answer the question.
68.
Automobile air bags are inflated with nitrogen gas, which is formed by the decomposition of solid sodium azide (NaN3). The other product is sodium metal. Calculate the volume of nitrogen gas at 27 °C and 756 torr formed by the decomposition of 125 g of sodium azide.
69.
Lime, CaO, is produced by heating calcium carbonate, CaCO3; carbon dioxide is the other product.
1. Outline the steps necessary to answer the following question: What volume of carbon dioxide at 875 K and 0.966 atm is produced by the decomposition of 1 ton (1.000 103 kg) of calcium carbonate?
2. Answer the question.
70.
Before small batteries were available, carbide lamps were used for bicycle lights. Acetylene gas, C2H2, and solid calcium hydroxide were formed by the reaction of calcium carbide, CaC2, with water. The ignition of the acetylene gas provided the light. Currently, the same lamps are used by some cavers, and calcium carbide is used to produce acetylene for carbide cannons.
1. Outline the steps necessary to answer the following question: What volume of C2H2 at 1.005 atm and 12.2 °C is formed by the reaction of 15.48 g of CaC2 with water?
2. Answer the question.
71.
Calculate the volume of oxygen required to burn 12.00 L of ethane gas, C2H6, to produce carbon dioxide and water, if the volumes of C2H6 and O2 are measured under the same conditions of temperature and pressure.
72.
What volume of O2 at STP is required to oxidize 8.0 L of NO at STP to NO2? What volume of NO2 is produced at STP?
73.
Consider the following questions:
1. What is the total volume of the CO2(g) and H2O(g) at 600 °C and 0.888 atm produced by the combustion of 1.00 L of C2H6(g) measured at STP?
2. What is the partial pressure of H2O in the product gases?
74.
Methanol, CH3OH, is produced industrially by the following reaction:
Assuming that the gases behave as ideal gases, find the ratio of the total volume of the reactants to the final volume.
75.
What volume of oxygen at 423.0 K and a pressure of 127.4 kPa is produced by the decomposition of 129.7 g of BaO2 to BaO and O2?
76.
A 2.50-L sample of a colorless gas at STP decomposed to give 2.50 L of N2 and 1.25 L of O2 at STP. What is the colorless gas?
77.
Ethanol, C2H5OH, is produced industrially from ethylene, C2H4, by the following sequence of reactions:
What volume of ethylene at STP is required to produce 1.000 metric ton (1000 kg) of ethanol if the overall yield of ethanol is 90.1%?
78.
One molecule of hemoglobin will combine with four molecules of oxygen. If 1.0 g of hemoglobin combines with 1.53 mL of oxygen at body temperature (37 °C) and a pressure of 743 torr, what is the molar mass of hemoglobin?
79.
A sample of a compound of xenon and fluorine was confined in a bulb with a pressure of 18 torr. Hydrogen was added to the bulb until the pressure was 72 torr. Passage of an electric spark through the mixture produced Xe and HF. After the HF was removed by reaction with solid KOH, the final pressure of xenon and unreacted hydrogen in the bulb was 36 torr. What is the empirical formula of the xenon fluoride in the original sample? (Note: Xenon fluorides contain only one xenon atom per molecule.)
80.
One method of analyzing amino acids is the van Slyke method. The characteristic amino groups (−NH2) in protein material are allowed to react with nitrous acid, HNO2, to form N2 gas. From the volume of the gas, the amount of amino acid can be determined. A 0.0604-g sample of a biological sample containing glycine, CH2(NH2)COOH, was analyzed by the van Slyke method and yielded 3.70 mL of N2 collected over water at a pressure of 735 torr and 29 °C. What was the percentage of glycine in the sample?
81.
A balloon filled with helium gas takes 6 hours to deflate to 50% of its original volume. How long will it take for an identical balloon filled with the same volume of hydrogen gas (instead of helium) to decrease its volume by 50%?
82.
Explain why the numbers of molecules are not identical in the left- and right-hand bulbs shown in the center illustration of Figure 9.27.
83.
Starting with the definition of rate of effusion and Graham’s finding relating rate and molar mass, show how to derive the Graham’s law equation, relating the relative rates of effusion for two gases to their molecular masses.
84.
Heavy water, D2O (molar mass = 20.03 g mol–1), can be separated from ordinary water, H2O (molar mass = 18.01), as a result of the difference in the relative rates of diffusion of the molecules in the gas phase. Calculate the relative rates of diffusion of H2O and D2O.
85.
Which of the following gases diffuse more slowly than oxygen? F2, Ne, N2O, C2H2, NO, Cl2, H2S
86.
During the discussion of gaseous diffusion for enriching uranium, it was claimed that 235UF6 diffuses 0.4% faster than 238UF6. Show the calculation that supports this value. The molar mass of 235UF6 = 235.043930 + 6 18.998403 = 349.034348 g/mol, and the molar mass of 238UF6 = 238.050788 + 6 18.998403 = 352.041206 g/mol.
87.
Calculate the relative rate of diffusion of 1H2 (molar mass 2.0 g/mol) compared with 2H2 (molar mass 4.0 g/mol) and the relative rate of diffusion of O2 (molar mass 32 g/mol) compared with O3 (molar mass 48 g/mol).
88.
A gas of unknown identity diffuses at a rate of 83.3 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate of 102 mL/s. Calculate the molecular mass of the unknown gas.
89.
When two cotton plugs, one moistened with ammonia and the other with hydrochloric acid, are simultaneously inserted into opposite ends of a glass tube that is 87.0 cm long, a white ring of NH4Cl forms where gaseous NH3 and gaseous HCl first come into contact. At approximately what distance from the ammonia moistened plug does this occur? (Hint: Calculate the rates of diffusion for both NH3 and HCl, and find out how much faster NH3 diffuses than HCl.)
90.
Using the postulates of the kinetic molecular theory, explain why a gas uniformly fills a container of any shape.
91.
Can the speed of a given molecule in a gas double at constant temperature? Explain your answer.
92.
Describe what happens to the average kinetic energy of ideal gas molecules when the conditions are changed as follows:
1. The pressure of the gas is increased by reducing the volume at constant temperature.
2. The pressure of the gas is increased by increasing the temperature at constant volume.
3. The average speed of the molecules is increased by a factor of 2.
93.
The distribution of molecular speeds in a sample of helium is shown in Figure 9.34. If the sample is cooled, will the distribution of speeds look more like that of H2 or of H2O? Explain your answer.
94.
What is the ratio of the average kinetic energy of a SO2 molecule to that of an O2 molecule in a mixture of two gases? What is the ratio of the root mean square speeds, urms, of the two gases?
95.
A 1-L sample of CO initially at STP is heated to 546 K, and its volume is increased to 2 L.
1. What effect do these changes have on the number of collisions of the molecules of the gas per unit area of the container wall?
2. What is the effect on the average kinetic energy of the molecules?
3. What is the effect on the root mean square speed of the molecules?
96.
The root mean square speed of H2 molecules at 25 °C is about 1.6 km/s. What is the root mean square speed of a N2 molecule at 25 °C?
97.
Answer the following questions:
1. Is the pressure of the gas in the hot-air balloon shown at the opening of this chapter greater than, less than, or equal to that of the atmosphere outside the balloon?
2. Is the density of the gas in the hot-air balloon shown at the opening of this chapter greater than, less than, or equal to that of the atmosphere outside the balloon?
3. At a pressure of 1 atm and a temperature of 20 °C, dry air has a density of 1.2256 g/L. What is the (average) molar mass of dry air?
4. The average temperature of the gas in a hot-air balloon is 1.30 102 °F. Calculate its density, assuming the molar mass equals that of dry air.
5. The lifting capacity of a hot-air balloon is equal to the difference in the mass of the cool air displaced by the balloon and the mass of the gas in the balloon. What is the difference in the mass of 1.00 L of the cool air in part (c) and the hot air in part (d)?
6. An average balloon has a diameter of 60 feet and a volume of 1.1 105 ft3. What is the lifting power of such a balloon? If the weight of the balloon and its rigging is 500 pounds, what is its capacity for carrying passengers and cargo?
7. A balloon carries 40.0 gallons of liquid propane (density 0.5005 g/L). What volume of CO2 and H2O gas is produced by the combustion of this propane?
8. A balloon flight can last about 90 minutes. If all of the fuel is burned during this time, what is the approximate rate of heat loss (in kJ/min) from the hot air in the bag during the flight?
98.
Show that the ratio of the rate of diffusion of Gas 1 to the rate of diffusion of Gas 2, is the same at 0 °C and 100 °C.
99.
Graphs showing the behavior of several different gases follow. Which of these gases exhibit behavior significantly different from that expected for ideal gases?
100.
Explain why the plot of PV for CO2 differs from that of an ideal gas.
101.
Under which of the following sets of conditions does a real gas behave most like an ideal gas, and for which conditions is a real gas expected to deviate from ideal behavior? Explain.
1. high pressure, small volume
2. high temperature, low pressure
3. low temperature, high pressure
102.
Describe the factors responsible for the deviation of the behavior of real gases from that of an
ideal gas.
103.
For which of the following gases should the correction for the molecular volume be largest:
CO, CO2, H2, He, NH3, SF6?
104.
A 0.245-L flask contains 0.467 mol CO2 at 159 °C. Calculate the pressure:
1. using the ideal gas law
2. using the van der Waals equation
3. Explain the reason for the difference.
4. Identify which correction (that for P or V) is dominant and why.
105.
Answer the following questions:
1. If XX behaved as an ideal gas, what would its graph of Z vs. P look like?
2. For most of this chapter, we performed calculations treating gases as ideal. Was this justified?
3. What is the effect of the volume of gas molecules on Z? Under what conditions is this effect small? When is it large? Explain using an appropriate diagram.
4. What is the effect of intermolecular attractions on the value of Z? Under what conditions is this effect small? When is it large? Explain using an appropriate diagram.
5. In general, under what temperature conditions would you expect Z to have the largest deviations from the Z for an ideal gas? | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/09%3A_Gases/9.10%3A_Exercises.txt |
The great distances between atoms and molecules in a gaseous phase, and the corresponding absence of any significant interactions between them, allows for simple descriptions of many physical properties that are the same for all gases, regardless of their chemical identities. As described in the final module of the chapter on gases, this situation changes at high pressures and low temperatures—conditions that permit the atoms and molecules to interact to a much greater extent. In the liquid and solid states, these interactions are of considerable strength and play an important role in determining a number of physical properties that do depend on the chemical identity of the substance. In this chapter, the nature of these interactions and their effects on various physical properties of liquid and solid phases will be examined.
• 10.0: Introduction
In the liquid and solid states, these interactions are of considerable strength and play an important role in determining a number of physical properties that do depend on the chemical identity of the substance. In this chapter, the nature of these interactions and their effects on various physical properties of liquid and solid phases will be examined.
• 10.1: Intermolecular Forces
The physical properties of condensed matter (liquids and solids) can be explained in terms of the kinetic molecular theory. In a liquid, intermolecular attractive forces hold the molecules in contact, although they still have sufficient kinetic energy to move past each other. Intermolecular attractive forces, collectively referred to as van der Waals forces, are responsible for the behavior of liquids and solids and are electrostatic in nature.
• 10.2: Properties of Liquids
The intermolecular forces between molecules in the liquid state vary depending upon their chemical identities and result in corresponding variations in various physical properties. Cohesive forces between like molecules are responsible for a liquid’s viscosity (resistance to flow) and surface tension. Adhesive forces between the molecules of a liquid and different molecules composing a surface in contact with the liquid are responsible for surface wetting and capillary rise.
• 10.3: Phase Transitions
Phase transitions are processes that convert matter from one physical state into another. There are six phase transitions between the three phases of matter. Melting, vaporization, and sublimation are all endothermic processes, requiring an input of heat to overcome intermolecular attractions. The reciprocal transitions of freezing, condensation, and deposition are all exothermic processes, involving heat as intermolecular attractive forces are established or strengthened.
• 10.4: Phase Diagrams
The temperature and pressure conditions at which a substance exists in solid, liquid, and gaseous states are summarized in a phase diagram for that substance. Phase diagrams are combined plots of three pressure-temperature equilibrium curves: solid-liquid, liquid-gas, and solid-gas. These curves represent the relationships between phase-transition temperatures and pressures. The intersection of all three curves represents the substance’s triple point at which all three phases coexist.
• 10.5: The Solid State of Matter
Some substances form crystalline solids consisting of particles in a very organized structure; others form amorphous (noncrystalline) solids with an internal structure that is not ordered. The main types of crystalline solids are ionic solids, metallic solids, covalent network solids, and molecular solids. The properties of the different kinds of crystalline solids are due to the types of particles of which they consist, the arrangements of the particles, and the strengths of the attractions bet
• 10.6: Lattice Structures in Crystalline Solids
The structures of crystalline metals and simple ionic compounds can be described in terms of packing of spheres. Metal atoms can pack in hexagonal closest-packed structures, cubic closest-packed structures, body-centered structures, and simple cubic structures. The anions in simple ionic structures commonly adopt one of these structures, and the cations occupy the spaces remaining between the anions.
• 10.7: Key Terms
• 10.8: Key Equations
• 10.9: Summary
• 10.10: Exercises
These are homework exercises to accompany the Textmap created for "Chemistry" by OpenStax.
10: Liquids and Solids
Leprosy has been a devastating disease throughout much of human history. Aside from the symptoms and complications of the illness, its social stigma led sufferers to be cast out of communities and isolated in colonies; in some regions this practice lasted well into the twentieth century. At that time, the best potential treatment for leprosy was oil from the chaulmoogra tree, but the oil was extremely thick, causing blisters and making usage painful and ineffective. Healthcare professionals seeking a better application contacted Alice Ball, a young chemist at the University of Hawaii, who had focused her masters thesis on a similar plant. Ball initiated a sequence of procedures (repeated acidification and purification to change the characteristics of the oil and isolate the active substances (esters, discussed later in this text). The "Ball Method" as it later came to be called, became the standard treatment for leprosy for decades.
In the liquid and solid states, atomic and molecular interactions are of considerable strength and play an important role in determining a number of physical properties of the substance. For example, the thickness, or viscosity, of the chaulmoogra oil was due to its intermolecular forces. In this chapter, the nature of these interactions and their effects on various physical properties of liquid and solid phases will be examined. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/10%3A_Liquids_and_Solids/10.00%3A_Introduction.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the types of intermolecular forces possible between atoms or molecules in condensed phases (dispersion forces, dipole-dipole attractions, and hydrogen bonding)
• Identify the types of intermolecular forces experienced by specific molecules based on their structures
• Explain the relation between the intermolecular forces present within a substance and the temperatures associated with changes in its physical state
As was the case for gaseous substances, the kinetic molecular theory may be used to explain the behavior of solids and liquids. In the following description, the term particle will be used to refer to an atom, molecule, or ion. Note that we will use the popular phrase “intermolecular attraction” to refer to attractive forces between the particles of a substance, regardless of whether these particles are molecules, atoms, or ions.
Consider these two aspects of the molecular-level environments in solid, liquid, and gaseous matter:
• Particles in a solid are tightly packed together and often arranged in a regular pattern; in a liquid, they are close together with no regular arrangement; in a gas, they are far apart with no regular arrangement.
• Particles in a solid vibrate about fixed positions and do not generally move in relation to one another; in a liquid, they move past each other but remain in essentially constant contact; in a gas, they move independently of one another except when they collide.
The differences in the properties of a solid, liquid, or gas reflect the strengths of the attractive forces between the atoms, molecules, or ions that make up each phase. The phase in which a substance exists depends on the relative extents of its intermolecular forces (IMFs) and the kinetic energies (KE) of its molecules. IMFs are the various forces of attraction that may exist between the atoms and molecules of a substance due to electrostatic phenomena, as will be detailed in this module. These forces serve to hold particles close together, whereas the particles’ KE provides the energy required to overcome the attractive forces and thus increase the distance between particles. Figure \(1\) illustrates how changes in physical state may be induced by changing the temperature, hence, the average KE, of a given substance.
As an example of the processes depicted in this figure, consider a sample of water. When gaseous water is cooled sufficiently, the attractions between H2O molecules will be capable of holding them together when they come into contact with each other; the gas condenses, forming liquid H2O. For example, liquid water forms on the outside of a cold glass as the water vapor in the air is cooled by the cold glass, as seen in Figure \(2\).
We can also liquefy many gases by compressing them, if the temperature is not too high. The increased pressure brings the molecules of a gas closer together, such that the attractions between the molecules become strong relative to their KE. Consequently, they form liquids. Butane, C4H10, is the fuel used in disposable lighters and is a gas at standard temperature and pressure. Inside the lighter’s fuel compartment, the butane is compressed to a pressure that results in its condensation to the liquid state, as shown in Figure \(3\).
Finally, if the temperature of a liquid becomes sufficiently low, or the pressure on the liquid becomes sufficiently high, the molecules of the liquid no longer have enough KE to overcome the IMF between them, and a solid forms. A more thorough discussion of these and other changes of state, or phase transitions, is provided in a later module of this chapter.
Link to Learning
Access this interactive simulation on states of matter, phase transitions, and intermolecular forces. This simulation is useful for visualizing concepts introduced throughout this chapter.
Forces between Molecules
Under appropriate conditions, the attractions between all gas molecules will cause them to form liquids or solids. This is due to intermolecular forces, not intramolecular forces. Intramolecular forces are those within the molecule that keep the molecule together, for example, the bonds between the atoms. Intermolecular forces are the attractions between molecules, which determine many of the physical properties of a substance. Figure \(4\): illustrates these different molecular forces. The strengths of these attractive forces vary widely, though usually the IMFs between small molecules are weak compared to the intramolecular forces that bond atoms together within a molecule. For example, to overcome the IMFs in one mole of liquid HCl and convert it into gaseous HCl requires only about 17 kilojoules. However, to break the covalent bonds between the hydrogen and chlorine atoms in one mole of HCl requires about 25 times more energy—430 kilojoules.
All of the attractive forces between neutral atoms and molecules are known as van der Waals forces, although they are usually referred to more informally as intermolecular attraction. We will consider the various types of IMFs in the next three sections of this module.
Dispersion Forces
One of the three van der Waals forces is present in all condensed phases, regardless of the nature of the atoms or molecules composing the substance. This attractive force is called the London dispersion force in honor of German-born American physicist Fritz London who, in 1928, first explained it. This force is often referred to as simply the dispersion force. Because the electrons of an atom or molecule are in constant motion (or, alternatively, the electron’s location is subject to quantum-mechanical variability), at any moment in time, an atom or molecule can develop a temporary, instantaneous dipole if its electrons are distributed asymmetrically. The presence of this dipole can, in turn, distort the electrons of a neighboring atom or molecule, producing an induced dipole. These two rapidly fluctuating, temporary dipoles thus result in a relatively weak electrostatic attraction between the species—a so-called dispersion force like that illustrated in Figure \(5\).
Dispersion forces that develop between atoms in different molecules can attract the two molecules to each other. The forces are relatively weak, however, and become significant only when the molecules are very close. Larger and heavier atoms and molecules exhibit stronger dispersion forces than do smaller and lighter atoms and molecules. F2 and Cl2 are gases at room temperature (reflecting weaker attractive forces); Br2 is a liquid, and I2 is a solid (reflecting stronger attractive forces). Trends in observed melting and boiling points for the halogens clearly demonstrate this effect, as seen in Table \(1\).
Table \(1\): Melting and Boiling Points of the Halogens
Halogen Molar Mass Atomic Radius Melting Point Boiling Point
fluorine, F2 38 g/mol 72 pm 53 K 85 K
chlorine, Cl2 71 g/mol 99 pm 172 K 238 K
bromine, Br2 160 g/mol 114 pm 266 K 332 K
iodine, I2 254 g/mol 133 pm 387 K 457 K
astatine, At2 420 g/mol 150 pm 575 K 610 K
The increase in melting and boiling points with increasing atomic/molecular size may be rationalized by considering how the strength of dispersion forces is affected by the electronic structure of the atoms or molecules in the substance. In a larger atom, the valence electrons are, on average, farther from the nuclei than in a smaller atom. Thus, they are less tightly held and can more easily form the temporary dipoles that produce the attraction. The measure of how easy or difficult it is for another electrostatic charge (for example, a nearby ion or polar molecule) to distort a molecule’s charge distribution (its electron cloud) is known as polarizability. A molecule that has a charge cloud that is easily distorted is said to be very polarizable and will have large dispersion forces; one with a charge cloud that is difficult to distort is not very polarizable and will have small dispersion forces.
Example \(1\): London Forces and Their Effects
Order the following compounds of a group 14 element and hydrogen from lowest to highest boiling point: CH4, SiH4, GeH4, and SnH4. Explain your reasoning.
Solution
Applying the skills acquired in the chapter on chemical bonding and molecular geometry, all of these compounds are predicted to be nonpolar, so they may experience only dispersion forces: the smaller the molecule, the less polarizable and the weaker the dispersion forces; the larger the molecule, the larger the dispersion forces. The molar masses of CH4, SiH4, GeH4, and SnH4 are approximately 16 g/mol, 32 g/mol, 77 g/mol, and 123 g/mol, respectively. Therefore, CH4 is expected to have the lowest boiling point and SnH4 the highest boiling point. The ordering from lowest to highest boiling point is expected to be CH4 < SiH4 < GeH4 < SnH4.
A graph of the actual boiling points of these compounds versus the period of the group 14 element shows this prediction to be correct:
Exercise \(1\)
Order the following hydrocarbons from lowest to highest boiling point: C2H6, C3H8, and C4H10.
Answer
C2H6 < C3H8 < C4H10. All of these compounds are nonpolar and only have London dispersion forces: the larger the molecule, the larger the dispersion forces and the higher the boiling point. The ordering from lowest to highest boiling point is therefore C2H6 < C3H8 < C4H10.
The shapes of molecules also affect the magnitudes of the dispersion forces between them. For example, boiling points for the isomers n-pentane, isopentane, and neopentane (shown in Figure \(6\)) are 36 °C, 27 °C, and 9.5 °C, respectively. Even though these compounds are composed of molecules with the same chemical formula, C5H12, the difference in boiling points suggests that dispersion forces in the liquid phase are different, being greatest for n-pentane and least for neopentane. The elongated shape of n-pentane provides a greater surface area available for contact between molecules, resulting in correspondingly stronger dispersion forces. The more compact shape of isopentane offers a smaller surface area available for intermolecular contact and, therefore, weaker dispersion forces. Neopentane molecules are the most compact of the three, offering the least available surface area for intermolecular contact and, hence, the weakest dispersion forces. This behavior is analogous to the connections that may be formed between strips of VELCRO brand fasteners: the greater the area of the strip’s contact, the stronger the connection.
Chemistry in Everyday Life: Geckos and Intermolecular Forces
Geckos have an amazing ability to adhere to most surfaces. They can quickly run up smooth walls and across ceilings that have no toe-holds, and they do this without having suction cups or a sticky substance on their toes. And while a gecko can lift its feet easily as it walks along a surface, if you attempt to pick it up, it sticks to the surface. How are geckos (as well as spiders and some other insects) able to do this? Although this phenomenon has been investigated for hundreds of years, scientists only recently uncovered the details of the process that allows geckos’ feet to behave this way.
Geckos’ toes are covered with hundreds of thousands of tiny hairs known as setae, with each seta, in turn, branching into hundreds of tiny, flat, triangular tips called spatulae. The huge numbers of spatulae on its setae provide a gecko, shown in Figure \(7\), with a large total surface area for sticking to a surface. In 2000, Kellar Autumn, who leads a multi-institutional gecko research team, found that geckos adhered equally well to both polar silicon dioxide and nonpolar gallium arsenide. This proved that geckos stick to surfaces because of dispersion forces—weak intermolecular attractions arising from temporary, synchronized charge distributions between adjacent molecules. Although dispersion forces are very weak, the total attraction over millions of spatulae is large enough to support many times the gecko’s weight.
In 2014, two scientists developed a model to explain how geckos can rapidly transition from “sticky” to “non-sticky.” Alex Greaney and Congcong Hu at Oregon State University described how geckos can achieve this by changing the angle between their spatulae and the surface. Geckos’ feet, which are normally nonsticky, become sticky when a small shear force is applied. By curling and uncurling their toes, geckos can alternate between sticking and unsticking from a surface, and thus easily move across it. Later research led by Alyssa Stark at University of Akron showed that geckos can maintain their hold on hydrophobic surfaces (similar to the leaves in their habitats) equally well whether the surfaces were wet or dry. Stark's experiment used a ribbon to gently pull the geckos until they slipped, so that the researchers could determine the geckos' ability to hold various surfaces under wet and dry conditions. Further investigations may eventually lead to the development of better adhesives and other applications.
Link to Learning
Watch this video to learn more about Kellar Autumn’s research that determined that van der Waals forces are responsible for a gecko’s ability to cling and climb.
Dipole-Dipole Attractions
Recall from the chapter on chemical bonding and molecular geometry that polar molecules have a partial positive charge on one side and a partial negative charge on the other side of the molecule—a separation of charge called a dipole. Consider a polar molecule such as hydrogen chloride, HCl. In the HCl molecule, the more electronegative Cl atom bears the partial negative charge, whereas the less electronegative H atom bears the partial positive charge. An attractive force between HCl molecules results from the attraction between the positive end of one HCl molecule and the negative end of another. This attractive force is called a dipole-dipole attraction—the electrostatic force between the partially positive end of one polar molecule and the partially negative end of another, as illustrated in Figure \(8\).
The effect of a dipole-dipole attraction is apparent when we compare the properties of HCl molecules to nonpolar F2 molecules. Both HCl and F2 consist of the same number of atoms and have approximately the same molecular mass. At a temperature of 150 K, molecules of both substances would have the same average KE. However, the dipole-dipole attractions between HCl molecules are sufficient to cause them to “stick together” to form a liquid, whereas the relatively weaker dispersion forces between nonpolar F2 molecules are not, and so this substance is gaseous at this temperature. The higher normal boiling point of HCl (188 K) compared to F2 (85 K) is a reflection of the greater strength of dipole-dipole attractions between HCl molecules, compared to the attractions between nonpolar F2 molecules. We will often use values such as boiling or freezing points, or enthalpies of vaporization or fusion, as indicators of the relative strengths of IMFs of attraction present within different substances.
Example \(2\): Dipole-Dipole Forces and Their Effects
Predict which will have the higher boiling point: N2 or CO. Explain your reasoning.
Solution
CO and N2 are both diatomic molecules with masses of about 28 amu, so they experience similar London dispersion forces. Because CO is a polar molecule, it experiences dipole-dipole attractions. Because N2 is nonpolar, its molecules cannot exhibit dipole-dipole attractions. The dipole-dipole attractions between CO molecules are comparably stronger than the dispersion forces between nonpolar N2 molecules, so CO is expected to have the higher boiling point.
Exercise \(2\)
Predict which will have the higher boiling point: ICl or Br2. Explain your reasoning.
Answer
ICl. ICl and Br2 have similar masses (~160 amu) and therefore experience similar London dispersion forces. ICl is polar and thus also exhibits dipole-dipole attractions; Br2 is nonpolar and does not. The relatively stronger dipole-dipole attractions require more energy to overcome, so ICl will have the higher boiling point.
Hydrogen Bonding
Nitrosyl fluoride (ONF, molecular mass 49 amu) is a gas at room temperature. Water (H2O, molecular mass 18 amu) is a liquid, even though it has a lower molecular mass. We clearly cannot attribute this difference between the two compounds to dispersion forces. Both molecules have about the same shape and ONF is the heavier and larger molecule. It is, therefore, expected to experience more significant dispersion forces. Additionally, we cannot attribute this difference in boiling points to differences in the dipole moments of the molecules. Both molecules are polar and exhibit comparable dipole moments. The large difference between the boiling points is due to a particularly strong dipole-dipole attraction that may occur when a molecule contains a hydrogen atom bonded to a fluorine, oxygen, or nitrogen atom (the three most electronegative elements). The very large difference in electronegativity between the H atom (2.1) and the atom to which it is bonded (4.0 for an F atom, 3.5 for an O atom, or 3.0 for a N atom), combined with the very small size of a H atom and the relatively small sizes of F, O, or N atoms, leads to highly concentrated partial charges with these atoms. Molecules with F-H, O-H, or N-H moieties are very strongly attracted to similar moieties in nearby molecules, a particularly strong type of dipole-dipole attraction called hydrogen bonding. Examples of hydrogen bonds include HF⋯HF, H2O⋯HOH, and H3N⋯HNH2, in which the hydrogen bonds are denoted by dots. Figure \(9\) illustrates hydrogen bonding between water molecules.
Despite use of the word “bond,” keep in mind that hydrogen bonds are intermolecular attractive forces, not intramolecular attractive forces (covalent bonds). Hydrogen bonds are much weaker than covalent bonds, only about 5 to 10% as strong, but are generally much stronger than other dipole-dipole attractions and dispersion forces.
Hydrogen bonds have a pronounced effect on the properties of condensed phases (liquids and solids). For example, consider the trends in boiling points for the binary hydrides of group 15 (NH3, PH3, AsH3, and SbH3), group 16 hydrides (H2O, H2S, H2Se, and H2Te), and group 17 hydrides (HF, HCl, HBr, and HI). The boiling points of the heaviest three hydrides for each group are plotted in Figure \(10\). As we progress down any of these groups, the polarities of the molecules decrease slightly, whereas the sizes of the molecules increase substantially. The effect of increasingly stronger dispersion forces dominates that of increasingly weaker dipole-dipole attractions, and the boiling points are observed to increase steadily.
If we use this trend to predict the boiling points for the lightest hydride for each group, we would expect NH3 to boil at about −120 °C, H2O to boil at about −80 °C, and HF to boil at about −110 °C. However, when we measure the boiling points for these compounds, we find that they are dramatically higher than the trends would predict, as shown in Figure \(11\). The stark contrast between our naïve predictions and reality provides compelling evidence for the strength of hydrogen bonding.
Example \(3\): Effect of Hydrogen Bonding on Boiling Points
Consider the compounds dimethylether (CH3OCH3), ethanol (CH3CH2OH), and propane (CH3CH2CH3). Their boiling points, not necessarily in order, are −42.1 °C, −24.8 °C, and 78.4 °C. Match each compound with its boiling point. Explain your reasoning.
Solution
The VSEPR-predicted shapes of CH3OCH3, CH3CH2OH, and CH3CH2CH3 are similar, as are their molar masses (46 g/mol, 46 g/mol, and 44 g/mol, respectively), so they will exhibit similar dispersion forces. Since CH3CH2CH3 is nonpolar, it may exhibit only dispersion forces. Because CH3OCH3 is polar, it will also experience dipole-dipole attractions. Finally, CH3CH2OH has an −OH group, and so it will experience the uniquely strong dipole-dipole attraction known as hydrogen bonding. So the ordering in terms of strength of IMFs, and thus boiling points, is CH3CH2CH3 < CH3OCH3 < CH3CH2OH. The boiling point of propane is −42.1 °C, the boiling point of dimethylether is −24.8 °C, and the boiling point of ethanol is 78.5 °C.
Exercise \(3\)
Ethane (CH3CH3) has a melting point of −183 °C and a boiling point of −89 °C. Predict the melting and boiling points for methylamine (CH3NH2). Explain your reasoning.
Answer
The melting point and boiling point for methylamine are predicted to be significantly greater than those of ethane. CH3CH3 and CH3NH2 are similar in size and mass, but methylamine possesses an −NH group and therefore may exhibit hydrogen bonding. This greatly increases its IMFs, and therefore its melting and boiling points. It is difficult to predict values, but the known values are a melting point of −93 °C and a boiling point of −6 °C.
How Sciences Interconnect: Hydrogen Bonding and DNA
Deoxyribonucleic acid (DNA) is found in every living organism and contains the genetic information that determines the organism’s characteristics, provides the blueprint for making the proteins necessary for life, and serves as a template to pass this information on to the organism’s offspring. A DNA molecule consists of two (anti-)parallel chains of repeating nucleotides, which form its well-known double helical structure, as shown in Figure \(12\).
Each nucleotide contains a (deoxyribose) sugar bound to a phosphate group on one side, and one of four nitrogenous bases on the other. Two of the bases, cytosine (C) and thymine (T), are single-ringed structures known as pyrimidines. The other two, adenine (A) and guanine (G), are double-ringed structures called purines. These bases form complementary base pairs consisting of one purine and one pyrimidine, with adenine pairing with thymine, and cytosine with guanine. Each base pair is held together by hydrogen bonding. A and T share two hydrogen bonds, C and G share three, and both pairings have a similar shape and structure Figure \(13\).
The cumulative effect of millions of hydrogen bonds effectively holds the two strands of DNA together. Importantly, the two strands of DNA can relatively easily “unzip” down the middle since hydrogen bonds are relatively weak compared to the covalent bonds that hold the atoms of the individual DNA molecules together. This allows both strands to function as a template for replication. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/10%3A_Liquids_and_Solids/10.01%3A_Intermolecular_Forces.txt |
Learning Objectives
By the end of this section, you will be able to:
• Distinguish between adhesive and cohesive forces
• Define viscosity, surface tension, and capillary rise
• Describe the roles of intermolecular attractive forces in each of these properties/phenomena
When you pour a glass of water, or fill a car with gasoline, you observe that water and gasoline flow freely. But when you pour syrup on pancakes or add oil to a car engine, you note that syrup and motor oil do not flow as readily. The viscosity of a liquid is a measure of its resistance to flow. Water, gasoline, and other liquids that flow freely have a low viscosity. Honey, syrup, motor oil, and other liquids that do not flow freely, like those shown in Figure $1$, have higher viscosities. We can measure viscosity by measuring the rate at which a metal ball falls through a liquid (the ball falls more slowly through a more viscous liquid) or by measuring the rate at which a liquid flows through a narrow tube (more viscous liquids flow more slowly).
The IMFs between the molecules of a liquid, the size and shape of the molecules, and the temperature determine how easily a liquid flows. As Table $1$ shows, the more structurally complex are the molecules in a liquid and the stronger the IMFs between them, the more difficult it is for them to move past each other and the greater is the viscosity of the liquid. As the temperature increases, the molecules move more rapidly and their kinetic energies are better able to overcome the forces that hold them together; thus, the viscosity of the liquid decreases.
Table $1$: Viscosities of Common Substances at 25 °C
Substance Formula Viscosity (mPa·s)
water H2O 0.890
mercury Hg 1.526
ethanol C2H5OH 1.074
octane C8H18 0.508
ethylene glycol CH2(OH)CH2(OH) 16.1
honey variable ~2,000–10,000
motor oil variable ~50–500
The various IMFs between identical molecules of a substance are examples of cohesive forces. The molecules within a liquid are surrounded by other molecules and are attracted equally in all directions by the cohesive forces within the liquid. However, the molecules on the surface of a liquid are attracted only by about one-half as many molecules. Because of the unbalanced molecular attractions on the surface molecules, liquids contract to form a shape that minimizes the number of molecules on the surface—that is, the shape with the minimum surface area. A small drop of liquid tends to assume a spherical shape, as shown in Figure $2$, because in a sphere, the ratio of surface area to volume is at a minimum. Larger drops are more greatly affected by gravity, air resistance, surface interactions, and so on, and as a result, are less spherical.
Surface tension is defined as the energy required to increase the surface area of a liquid, or the force required to increase the length of a liquid surface by a given amount. This property results from the cohesive forces between molecules at the surface of a liquid, and it causes the surface of a liquid to behave like a stretched rubber membrane. Surface tensions of several liquids are presented in Table $2$. Among common liquids, water exhibits a distinctly high surface tension due to strong hydrogen bonding between its molecules. As a result of this high surface tension, the surface of water represents a relatively “tough skin” that can withstand considerable force without breaking. A steel needle carefully placed on water will float. Some insects, like the one shown in Figure $3$, even though they are denser than water, move on its surface because they are supported by the surface tension.
Table $2$: Surface Tensions of Common Substances at 25 °C
Substance Formula Surface Tension (mN/m)
water H2O 71.99
mercury Hg 458.48
ethanol C2H5OH 21.97
octane C8H18 21.14
ethylene glycol CH2(OH)CH2(OH) 47.99
Surface tension is affected by a variety of variables, including the introduction of additional substances on the surface. In the late 1800s, Agnes Pockels, who was initially blocked from pursuing a scientific career but studied on her own, began investigating the impact and characteristics of soapy and greasy films in water. Using homemade materials, she developed an instrument known as a trough for measuring surface contaminants and their effects. With the support of renowned scientist Lord Rayleigh, her 1891 paper showed that surface contamination significantly reduces surface tension, and also that changing the characteristics of the surface (compressing or expanding it) also affects surface tension. Decades later, Irving Langmuir and Katharine Blodgett built on Pockels' work in their own trough and important advances in surface chemistry. Langmuir pioneered methods for producing single-molecule layers of film; Blodgett applied these to the development of non-reflective glass (critical for film-making and other applications), and also studied methods related to cleaning surfaces, which are important in semiconductor fabrication.
The IMFs of attraction between two different molecules are called adhesive forces. Consider what happens when water comes into contact with some surface. If the adhesive forces between water molecules and the molecules of the surface are weak compared to the cohesive forces between the water molecules, the water does not “wet” the surface. For example, water does not wet waxed surfaces or many plastics such as polyethylene. Water forms drops on these surfaces because the cohesive forces within the drops are greater than the adhesive forces between the water and the plastic. Water spreads out on glass because the adhesive force between water and glass is greater than the cohesive forces within the water. When water is confined in a glass tube, its meniscus (surface) has a concave shape because the water wets the glass and creeps up the side of the tube. On the other hand, the cohesive forces between mercury atoms are much greater than the adhesive forces between mercury and glass. Mercury therefore does not wet glass, and it forms a convex meniscus when confined in a tube because the cohesive forces within the mercury tend to draw it into a drop (Figure $4$).
If you place one end of a paper towel in spilled wine, as shown in Figure $5$, the liquid wicks up the paper towel. A similar process occurs in a cloth towel when you use it to dry off after a shower. These are examples of capillary action—when a liquid flows within a porous material due to the attraction of the liquid molecules to the surface of the material and to other liquid molecules. The adhesive forces between the liquid and the porous material, combined with the cohesive forces within the liquid, may be strong enough to move the liquid upward against gravity.
Towels soak up liquids like water because the fibers of a towel are made of molecules that are attracted to water molecules. Most cloth towels are made of cotton, and paper towels are generally made from paper pulp. Both consist of long molecules of cellulose that contain many −OH groups. Water molecules are attracted to these −OH groups and form hydrogen bonds with them, which draws the H2O molecules up the cellulose molecules. The water molecules are also attracted to each other, so large amounts of water are drawn up the cellulose fibers.
Capillary action can also occur when one end of a small diameter tube is immersed in a liquid, as illustrated in Figure $6$. If the liquid molecules are strongly attracted to the tube molecules, the liquid creeps up the inside of the tube until the weight of the liquid and the adhesive forces are in balance. The smaller the diameter of the tube is, the higher the liquid climbs. It is partly by capillary action occurring in plant cells called xylem that water and dissolved nutrients are brought from the soil up through the roots and into a plant. Capillary action is the basis for thin layer chromatography, a laboratory technique commonly used to separate small quantities of mixtures. You depend on a constant supply of tears to keep your eyes lubricated and on capillary action to pump tear fluid away.
The height to which a liquid will rise in a capillary tube is determined by several factors as shown in the following equation:
$h=\dfrac{2 T \cos \theta}{r \rho g} \label{capillary}$
In this equation, $h$ is the height of the liquid inside the capillary tube relative to the surface of the liquid outside the tube, $T$ is the surface tension of the liquid, $θ$ is the contact angle between the liquid and the tube, $r$ is the radius of the tube, $ρ$ is the density of the liquid, and $g$ is the acceleration due to gravity, 9.8 m/s2. When the tube is made of a material to which the liquid molecules are strongly attracted, they will spread out completely on the surface, which corresponds to a contact angle of 0°. This is the situation for water rising in a glass tube.
Example $1$: Capillary Rise
At 25 °C, how high will water rise in a glass capillary tube with an inner diameter of 0.25 mm?
For water, T = 71.99 mN/m and ρ = 1.0 g/cm3.
Solution
The liquid will rise to a height h given by Equation \ref{capillary}:
$h=\dfrac{2 T \cos \theta}{r \rho g} \nonumber$
The Newton is defined as a kg m/s2, and so the provided surface tension is equivalent to 0.07199 kg/s2. The provided density must be converted into units that will cancel appropriately: ρ = 1000 kg/m3. The diameter of the tube in meters is 0.00025 m, so the radius is 0.000125 m. For a glass tube immersed in water, the contact angle is θ = 0°, so cos θ = 1. Finally, acceleration due to gravity on the earth is g = 9.8 m/s2. Substituting these values into the equation, and cancelling units, we have:
$h=\frac{2\left(0.07199 kg / s^2\right)}{(0.000125 m )\left(1000 kg / m^3\right)\left(9.8 m / s^2\right)}=0.12 m =12 cm \nonumber$
Exercise $1$
Water rises in a glass capillary tube to a height of 8.4 cm. What is the diameter of the capillary tube?
Answer
diameter = 0.36 mm
Chemistry in Everyday Life: Biomedical Applications of Capillary Action
Many medical tests require drawing a small amount of blood, for example to determine the amount of glucose in someone with diabetes or the hematocrit level in an athlete. This procedure can be easily done because of capillary action, the ability of a liquid to flow up a small tube against gravity, as shown in Figure $7$. When your finger is pricked, a drop of blood forms and holds together due to surface tension—the unbalanced intermolecular attractions at the surface of the drop. Then, when the open end of a narrow-diameter glass tube touches the drop of blood, the adhesive forces between the molecules in the blood and those at the glass surface draw the blood up the tube. How far the blood goes up the tube depends on the diameter of the tube (and the type of fluid). A small tube has a relatively large surface area for a given volume of blood, which results in larger (relative) attractive forces, allowing the blood to be drawn farther up the tube. The liquid itself is held together by its own cohesive forces. When the weight of the liquid in the tube generates a downward force equal to the upward force associated with capillary action, the liquid stops rising. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/10%3A_Liquids_and_Solids/10.02%3A_Properties_of_Liquids.txt |
Learning Objectives
By the end of this section, you will be able to:
• Define phase transitions and phase transition temperatures
• Explain the relation between phase transition temperatures and intermolecular attractive forces
• Describe the processes represented by typical heating and cooling curves, and compute heat flows and enthalpy changes accompanying these processes
We witness and utilize changes of physical state, or phase transitions, in a great number of ways. As one example of global significance, consider the evaporation, condensation, freezing, and melting of water. These changes of state are essential aspects of our earth’s water cycle as well as many other natural phenomena and technological processes of central importance to our lives. In this module, the essential aspects of phase transitions are explored.
Vaporization and Condensation
When a liquid vaporizes in a closed container, gas molecules cannot escape. As these gas phase molecules move randomly about, they will occasionally collide with the surface of the condensed phase, and in some cases, these collisions will result in the molecules re-entering the condensed phase. The change from the gas phase to the liquid is called condensation. When the rate of condensation becomes equal to the rate of vaporization, neither the amount of the liquid nor the amount of the vapor in the container changes. The vapor in the container is then said to be in equilibrium with the liquid. Keep in mind that this is not a static situation, as molecules are continually exchanged between the condensed and gaseous phases. Such is an example of a dynamic equilibrium, the status of a system in which reciprocal processes (for example, vaporization and condensation) occur at equal rates. The pressure exerted by the vapor in equilibrium with a liquid in a closed container at a given temperature is called the liquid’s vapor pressure (or equilibrium vapor pressure). The area of the surface of the liquid in contact with a vapor and the size of the vessel have no effect on the vapor pressure, although they do affect the time required for the equilibrium to be reached. We can measure the vapor pressure of a liquid by placing a sample in a closed container, like that illustrated in Figure $1$, and using a manometer to measure the increase in pressure that is due to the vapor in equilibrium with the condensed phase.
The chemical identities of the molecules in a liquid determine the types (and strengths) of intermolecular attractions possible; consequently, different substances will exhibit different equilibrium vapor pressures. Relatively strong intermolecular attractive forces will serve to impede vaporization as well as favoring “recapture” of gas-phase molecules when they collide with the liquid surface, resulting in a relatively low vapor pressure. Weak intermolecular attractions present less of a barrier to vaporization, and a reduced likelihood of gas recapture, yielding relatively high vapor pressures. The following example illustrates this dependence of vapor pressure on intermolecular attractive forces.
Example $1$: Explaining Vapor Pressure in Terms of IMFs
Given the shown structural formulas for these four compounds, explain their relative vapor pressures in terms of types and extents of IMFs:
Solution
Diethyl ether has a very small dipole and most of its intermolecular attractions are London forces. Although this molecule is the largest of the four under consideration, its IMFs are the weakest and, as a result, its molecules most readily escape from the liquid. It also has the highest vapor pressure. Due to its smaller size, ethanol exhibits weaker dispersion forces than diethyl ether. However, ethanol is capable of hydrogen bonding and, therefore, exhibits stronger overall IMFs, which means that fewer molecules escape from the liquid at any given temperature, and so ethanol has a lower vapor pressure than diethyl ether. Water is much smaller than either of the previous substances and exhibits weaker dispersion forces, but its extensive hydrogen bonding provides stronger intermolecular attractions, fewer molecules escaping the liquid, and a lower vapor pressure than for either diethyl ether or ethanol. Ethylene glycol has two −OH groups, so, like water, it exhibits extensive hydrogen bonding. It is much larger than water and thus experiences larger London forces. Its overall IMFs are the largest of these four substances, which means its vaporization rate will be the slowest and, consequently, its vapor pressure the lowest.
Exercise $1$
At 20 °C, the vapor pressures of several alcohols are given in this table. Explain these vapor pressures in terms of types and extents of IMFs for these alcohols:
Compound methanol CH3OH ethanol C2H5OH propanol C3H7OH butanol C4H9OH
Vapor Pressure at 20 °C 11.9 kPa 5.95 kPa 2.67 kPa 0.56 kPa
Answer
All these compounds exhibit hydrogen bonding; these strong IMFs are difficult for the molecules to overcome, so the vapor pressures are relatively low. As the size of molecule increases from methanol to butanol, dispersion forces increase, which means that the vapor pressures decrease as observed:
Pmethanol > Pethanol > Ppropanol > Pbutanol.
As temperature increases, the vapor pressure of a liquid also increases due to the increased average KE of its molecules. Recall that at any given temperature, the molecules of a substance experience a range of kinetic energies, with a certain fraction of molecules having a sufficient energy to overcome IMF and escape the liquid (vaporize). At a higher temperature, a greater fraction of molecules have enough energy to escape from the liquid, as shown in Figure $2$. The escape of more molecules per unit of time and the greater average speed of the molecules that escape both contribute to the higher vapor pressure.
Boiling Points
When the vapor pressure increases enough to equal the external atmospheric pressure, the liquid reaches its boiling point. The boiling point of a liquid is the temperature at which its equilibrium vapor pressure is equal to the pressure exerted on the liquid by its gaseous surroundings. For liquids in open containers, this pressure is that due to the earth’s atmosphere. The normal boiling point of a liquid is defined as its boiling point when surrounding pressure is equal to 1 atm (101.3 kPa). Figure $3$ shows the variation in vapor pressure with temperature for several different substances. Considering the definition of boiling point, these curves may be seen as depicting the dependence of a liquid’s boiling point on surrounding pressure.
Example $2$: A Boiling Point at Reduced Pressure
A typical atmospheric pressure in Leadville, Colorado (elevation 10,200 feet) is 68 kPa. Use the graph in Figure $3$ to determine the boiling point of water at this elevation.
Solution
The graph of the vapor pressure of water versus temperature in Figure $3$ indicates that the vapor pressure of water is 68 kPa at about 90 °C. Thus, at about 90 °C, the vapor pressure of water will equal the atmospheric pressure in Leadville, and water will boil.
Exercise $2$
The boiling point of ethyl ether was measured to be 10 °C at a base camp on the slopes of Mount Everest. Use Figure $3$ to determine the approximate atmospheric pressure at the camp.
Answer
Approximately 40 kPa (0.4 atm)
The quantitative relation between a substance’s vapor pressure and its temperature is described by the Clausius-Clapeyron equation:
$P=A e^{-\Delta H_{\text {vap }} / R T} \nonumber$
where ΔHvap is the enthalpy of vaporization for the liquid, R is the gas constant, and A is a constant whose value depends on the chemical identity of the substance. Temperature T must be in Kelvin in this equation. This equation is often rearranged into logarithmic form to yield the linear equation:
$\ln P=-\frac{\Delta H_{ vap }}{R T}+\ln A \nonumber$
This linear equation may be expressed in a two-point format that is convenient for use in various computations, as demonstrated in the example exercises that follow. If at temperature T1, the vapor pressure is P1, and at temperature T2, the vapor pressure is P2, the corresponding linear equations are:
$\ln P_1=-\frac{\Delta H_{ vap }}{R T_1}+\ln A \nonumber$
and
$\ln P_2=-\frac{\Delta H_{ vap }}{R T_2}+\ln A \nonumber$
Since the constant, A, is the same, these two equations may be rearranged to isolate ln A and then set them equal to one another:
$\ln P_1+\frac{\Delta H_{\text {vap }}}{R T_1}=\ln P_2+\frac{\Delta H_{\text {vap }}}{R T_2} \nonumber$
which can be combined into:
$\ln \left(\frac{P_2}{P_1}\right)=\frac{\Delta H_{\text {vap }}}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \nonumber$
Example $3$: Estimating Enthalpy of Vaporization
Isooctane (2,2,4-trimethylpentane) has an octane rating of 100. It is used as one of the standards for the octane-rating system for gasoline. At 34.0 °C, the vapor pressure of isooctane is 10.0 kPa, and at 98.8 °C, its vapor pressure is 100.0 kPa. Use this information to estimate the enthalpy of vaporization for isooctane.
Solution
The enthalpy of vaporization, ΔHvap, can be determined by using the Clausius-Clapeyron equation:
$\ln \left(\frac{P_2}{P_1}\right)=\frac{\Delta H_{\text {vap }}}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \nonumber$
Since we have two vapor pressure-temperature values (T1 = 34.0 °C = 307.2 K, P1 = 10.0 kPa and T2 = 98.8 °C = 372.0 K, P2 = 100 kPa), we can substitute them into this equation and solve for ΔHvap. Rearranging the Clausius-Clapeyron equation and solving for ΔHvap yields:
$\Delta H_{\text {vap }}=\frac{R \cdot \ln \left(\frac{P_2}{P_1}\right)}{\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}=\frac{(8.3145 J / mol \cdot K ) \cdot \ln \left(\frac{100 kPa }{10.0 kPa }\right)}{\left(\frac{1}{307.2 K }-\frac{1}{372.0 K }\right)}=33,800 J / mol =33.8 kJ / mol \nonumber$
Note that the pressure can be in any units, so long as they agree for both P values, but the temperature must be in kelvin for the Clausius-Clapeyron equation to be valid.
Exercise $3$
At 20.0 °C, the vapor pressure of ethanol is 5.95 kPa, and at 63.5 °C, its vapor pressure is 53.3 kPa. Use this information to estimate the enthalpy of vaporization for ethanol.
Answer
41,360 J/mol or 41.4 kJ/mol
Example $4$: Estimating Temperature (or Vapor Pressure)
For benzene (C6H6), the normal boiling point is 80.1 °C and the enthalpy of vaporization is 30.8 kJ/mol. What is the boiling point of benzene in Denver, where atmospheric pressure = 83.4 kPa?
Solution
If the temperature and vapor pressure are known at one point, along with the enthalpy of vaporization, ΔHvap, then the temperature that corresponds to a different vapor pressure (or the vapor pressure that corresponds to a different temperature) can be determined by using the Clausius-Clapeyron equation:
$\ln \left(\frac{P_2}{P_1}\right)=\frac{\Delta H_{\text {vap }}}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \nonumber$
Since the normal boiling point is the temperature at which the vapor pressure equals atmospheric pressure at sea level, we know one vapor pressure-temperature value (T1 = 80.1 °C = 353.3 K, P1 = 101.3 kPa, ΔHvap = 30.8 kJ/mol) and want to find the temperature (T2) that corresponds to vapor pressure P2 = 83.4 kPa. We can substitute these values into the Clausius-Clapeyron equation and then solve for T2. Rearranging the Clausius-Clapeyron equation and solving for T2 yields:
$T_2=\left(\frac{-R \cdot \ln \left(\frac{P_2}{P_1}\right)}{\Delta H_{\text {vap }}}+\frac{1}{T_1}\right)^{-1}=\left(\frac{-(8.3145 J / mol \cdot K ) \cdot \ln \left(\frac{83.4 kPa }{101.3 kPa }\right)}{30,800 J / mol }+\frac{1}{353.3 K }\right)^{-1}=346.9 \nonumber$
Exercise $4$
For acetone (CH3)2CO, the normal boiling point is 56.5 °C and the enthalpy of vaporization is 31.3 kJ/mol. What is the vapor pressure of acetone at 25.0 °C?
Answer
30.1 kPa
Enthalpy of Vaporization
Vaporization is an endothermic process. The cooling effect can be evident when you leave a swimming pool or a shower. When the water on your skin evaporates, it removes heat from your skin and causes you to feel cold. The energy change associated with the vaporization process is the enthalpy of vaporization, ΔHvap. For example, the vaporization of water at standard temperature is represented by:
$H_2 O (l) \longrightarrow H_2 O (g) \quad \quad \Delta H_{ vap }=44.01 kJ / mol \nonumber$
As described in the chapter on thermochemistry, the reverse of an endothermic process is exothermic. And so, the condensation of a gas releases heat:
$H_2 O (g) \longrightarrow H_2 O (l) \quad \quad \Delta H_{\text {con }}=-\Delta H_{\text {vap }}=-44.01 kJ / mol \nonumber$
Example $5$: Using Enthalpy of Vaporization
One way our body is cooled is by evaporation of the water in sweat (Figure $4$). In very hot climates, we can lose as much as 1.5 L of sweat per day. Although sweat is not pure water, we can get an approximate value of the amount of heat removed by evaporation by assuming that it is. How much heat is required to evaporate 1.5 L of water (1.5 kg) at T = 37 °C (normal body temperature); ΔHvap = 43.46 kJ/mol at 37 °C.
Solution
We start with the known volume of sweat (approximated as just water) and use the given information to convert to the amount of heat needed:
$1.5 \, \cancel{\text{L}} \times \dfrac{1000\, \cancel{\text{g}}}{1 \, \cancel{\text{L}}} \times \frac{1\, \cancel{\text{mol}}}{18\,\cancel{\text{g}}} \times \frac{43.46\, \text{kJ} }{1\, \cancel{\text{mol}}} = 3.6 \times 10^3\,\text{kJ} \nonumber$
Thus, 3600 kJ of heat are removed by the evaporation of 1.5 L of water.
Exercise $1$
How much heat is required to evaporate 100.0 g of liquid ammonia, NH3, at its boiling point if its enthalpy of vaporization is 4.8 kJ/mol?
Answer
28 kJ
Melting and Freezing
When we heat a crystalline solid, we increase the average energy of its atoms, molecules, or ions and the solid gets hotter. At some point, the added energy becomes large enough to partially overcome the forces holding the molecules or ions of the solid in their fixed positions, and the solid begins the process of transitioning to the liquid state, or melting. At this point, the temperature of the solid stops rising, despite the continual input of heat, and it remains constant until all of the solid is melted. Only after all of the solid has melted will continued heating increase the temperature of the liquid (Figure $5$).
If we stop heating during melting and place the mixture of solid and liquid in a perfectly insulated container so no heat can enter or escape, the solid and liquid phases remain in equilibrium. This is almost the situation with a mixture of ice and water in a very good thermos bottle; almost no heat gets in or out, and the mixture of solid ice and liquid water remains for hours. In a mixture of solid and liquid at equilibrium, the reciprocal processes of melting and freezing occur at equal rates, and the quantities of solid and liquid therefore remain constant. The temperature at which the solid and liquid phases of a given substance are in equilibrium is called the melting point of the solid or the freezing point of the liquid. Use of one term or the other is normally dictated by the direction of the phase transition being considered, for example, solid to liquid (melting) or liquid to solid (freezing).
The enthalpy of fusion and the melting point of a crystalline solid depend on the strength of the attractive forces between the units present in the crystal. Molecules with weak attractive forces form crystals with low melting points. Crystals consisting of particles with stronger attractive forces melt at higher temperatures.
The amount of heat required to change one mole of a substance from the solid state to the liquid state is the enthalpy of fusion, ΔHfus of the substance. The enthalpy of fusion of ice is 6.0 kJ/mol at 0 °C. Fusion (melting) is an endothermic process:
$H_2 O (s) \longrightarrow H_2 O (l) \quad \Delta H_{\text {fus }}=6.01 kJ / mol \nonumber$
The reciprocal process, freezing, is an exothermic process whose enthalpy change is −6.0 kJ/mol at 0 °C:
$H_2 O (l) \longrightarrow H_2 O (s) \quad \Delta H_{\text {frz }}=-\Delta H_{\text {fus }}=-6.01 kJ / mol \nonumber$
Sublimation and Deposition
Some solids can transition directly into the gaseous state, bypassing the liquid state, via a process known as sublimation. At room temperature and standard pressure, a piece of dry ice (solid CO2) sublimes, appearing to gradually disappear without ever forming any liquid. Snow and ice sublime at temperatures below the melting point of water, a slow process that may be accelerated by winds and the reduced atmospheric pressures at high altitudes. When solid iodine is warmed, the solid sublimes and a vivid purple vapor forms (Figure $6$). The reverse of sublimation is called deposition, a process in which gaseous substances condense directly into the solid state, bypassing the liquid state. The formation of frost is an example of deposition.
Like vaporization, the process of sublimation requires an input of energy to overcome intermolecular attractions. The enthalpy of sublimation, ΔHsub, is the energy required to convert one mole of a substance from the solid to the gaseous state. For example, the sublimation of carbon dioxide is represented by:
$CO_2(s) \longrightarrow CO_2(g) \quad \quad \Delta H_{\text {sub }}=26.1 kJ / mol \nonumber$
Likewise, the enthalpy change for the reverse process of deposition is equal in magnitude but opposite in sign to that for sublimation:
$CO_2(g) \longrightarrow CO_2(s) \quad \quad \Delta H_{\text {dep }}=-\Delta H_{\text {sub }}=-26.1 kJ / mol \nonumber$
Consider the extent to which intermolecular attractions must be overcome to achieve a given phase transition. Converting a solid into a liquid requires that these attractions be only partially overcome; transition to the gaseous state requires that they be completely overcome. As a result, the enthalpy of fusion for a substance is less than its enthalpy of vaporization. This same logic can be used to derive an approximate relation between the enthalpies of all phase changes for a given substance. Though not an entirely accurate description, sublimation may be conveniently modeled as a sequential two-step process of melting followed by vaporization in order to apply Hess’s Law. Viewed in this manner, the enthalpy of sublimation for a substance may be estimated as the sum of its enthalpies of fusion and vaporization, as illustrated in Figure $7$. For example:
$\begin{array}{lc} \text { solid } \longrightarrow \text { liquid } & \Delta H_{\text {fus }} \ \text { liquid } \longrightarrow \text { gas } & \Delta H_{\text {vap }} \ \hline \text { solid } \longrightarrow \text { gas } & \Delta H_{\text {sub }}=\Delta H_{\text {fus }}+\Delta H_{\text {vap }} \end{array} \nonumber$
Heating and Cooling Curves
In the chapter on thermochemistry, the relation between the amount of heat absorbed or released by a substance, $q$, and its accompanying temperature change, $ΔT$, was introduced:
$q=m c \Delta T \nonumber$
where $m$ is the mass of the substance and $c$ is its specific heat. The relation applies to matter being heated or cooled, but not undergoing a change in state. When a substance being heated or cooled reaches a temperature corresponding to one of its phase transitions, further gain or loss of heat is a result of diminishing or enhancing intermolecular attractions, instead of increasing or decreasing molecular kinetic energies. While a substance is undergoing a change in state, its temperature remains constant. Figure $8$ shows a typical heating curve.
Consider the example of heating a pot of water to boiling. A stove burner will supply heat at a roughly constant rate; initially, this heat serves to increase the water’s temperature. When the water reaches its boiling point, the temperature remains constant despite the continued input of heat from the stove burner. This same temperature is maintained by the water as long as it is boiling. If the burner setting is increased to provide heat at a greater rate, the water temperature does not rise, but instead the boiling becomes more vigorous (rapid). This behavior is observed for other phase transitions as well: For example, temperature remains constant while the change of state is in progress.
Example $6$: Total Heat Needed to Change Temperature and Phase for a Substance
How much heat is required to convert 135 g of ice at −15 °C into water vapor at 120 °C?
Solution
The transition described involves the following steps:
1. Heat ice from −15 °C to 0 °C
2. Melt ice
3. Heat water from 0 °C to 100 °C
4. Boil water
5. Heat steam from 100 °C to 120 °C
The heat needed to change the temperature of a given substance (with no change in phase) is: $q = m c ΔT$. The heat needed to induce a given change in phase is given by $q = n ΔH$.
Using these equations with the appropriate values for specific heat of ice, water, and steam, and enthalpies of fusion and vaporization, we have:
\begin{aligned} q_{\text {total }} = &(m \cdot c \cdot \Delta T)_{\text {ice }}+n \cdot \Delta H_{\text {fus }}+(m \cdot c \cdot \Delta T)_{\text {water }}+n \cdot \Delta H_{\text {vap }}+(m \cdot c \cdot \Delta T)_{\text {steam }}\[4pt] =&\left(135 g \cdot 2.09 J / g \cdot{ }^{\circ} C \cdot 15^{\circ} C \right)+\left(135 \cdot \frac{1 mol }{18.02 g } \cdot 6.01 kJ / mol \right) \[4pt] & +\left(135 g \cdot 4.18 J / g \cdot{ }^{\circ} C \cdot 100^{\circ} C \right)+\left(135 g \cdot \frac{1 mol }{18.02 g } \cdot 40.67 kJ / mol \right) \[4pt] & +\left(135 g \cdot 1.86 J / g \cdot{ }^{\circ} C \cdot 20^{\circ} C \right) \[4pt] =& 4230 J +45.0 kJ +56,500 J +305 kJ +5022 J \end{aligned} \nonumber
Converting the quantities in J to kJ permits them to be summed, yielding the total heat required:
$q_{\text {total }} =4.23 kJ +45.0 kJ +56.5 kJ +305 kJ +5.02 kJ =416 kJ \nonumber$
NOTE: The value of ΔHvap at the boiling point of water (40.67 kJ/mol) is used here instead of the value at standard temperature (44.01 kJ/mol).
Exercise $6$
What is the total amount of heat released when 94.0 g water at 80.0 °C cools to form ice at −30.0 °C?
Answer
68.7 kJ | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/10%3A_Liquids_and_Solids/10.03%3A_Phase_Transitions.txt |
Learning Objectives
By the end of this section, you will be able to:
• Explain the construction and use of a typical phase diagram
• Use phase diagrams to identify stable phases at given temperatures and pressures, and to describe phase transitions resulting from changes in these properties
• Describe the supercritical fluid phase of matter
In the previous module, the variation of a liquid’s equilibrium vapor pressure with temperature was described. Considering the definition of boiling point, plots of vapor pressure versus temperature represent how the boiling point of the liquid varies with pressure. Also described was the use of heating and cooling curves to determine a substance’s melting (or freezing) point. Making such measurements over a wide range of pressures yields data that may be presented graphically as a phase diagram. A phase diagram combines plots of pressure versus temperature for the liquid-gas, solid-liquid, and solid-gas phase-transition equilibria of a substance. These diagrams indicate the physical states that exist under specific conditions of pressure and temperature, and also provide the pressure dependence of the phase-transition temperatures (melting points, sublimation points, boiling points). A typical phase diagram for a pure substance is shown in Figure \(1\).
To illustrate the utility of these plots, consider the phase diagram for water shown in Figure \(2\).
We can use the phase diagram to identify the physical state of a sample of water under specified conditions of pressure and temperature. For example, a pressure of 50 kPa and a temperature of −10 °C correspond to the region of the diagram labeled “ice.” Under these conditions, water exists only as a solid (ice). A pressure of 50 kPa and a temperature of 50 °C correspond to the “water” region—here, water exists only as a liquid. At 25 kPa and 200 °C, water exists only in the gaseous state. Note that on the H2O phase diagram, the pressure and temperature axes are not drawn to a constant scale in order to permit the illustration of several important features as described here.
The curve BC in Figure \(2\) is the plot of vapor pressure versus temperature as described in the previous module of this chapter. This “liquid-vapor” curve separates the liquid and gaseous regions of the phase diagram and provides the boiling point for water at any pressure. For example, at 1 atm, the boiling point is 100 °C. Notice that the liquid-vapor curve terminates at a temperature of 374 °C and a pressure of 218 atm, indicating that water cannot exist as a liquid above this temperature, regardless of the pressure. The physical properties of water under these conditions are intermediate between those of its liquid and gaseous phases. This unique state of matter is called a supercritical fluid, a topic that will be described in the next section of this module.
The solid-vapor curve, labeled AB in Figure \(2\), indicates the temperatures and pressures at which ice and water vapor are in equilibrium. These temperature-pressure data pairs correspond to the sublimation, or deposition, points for water. If we could zoom in on the solid-gas line in Figure \(2\), we would see that ice has a vapor pressure of about 0.20 kPa at −10 °C. Thus, if we place a frozen sample in a vacuum with a pressure less than 0.20 kPa, ice will sublime. This is the basis for the “freeze-drying” process often used to preserve foods, such as the ice cream shown in Figure \(3\).
The solid-liquid curve labeled BD shows the temperatures and pressures at which ice and liquid water are in equilibrium, representing the melting/freezing points for water. Note that this curve exhibits a slight negative slope (greatly exaggerated for clarity), indicating that the melting point for water decreases slightly as pressure increases. Water is an unusual substance in this regard, as most substances exhibit an increase in melting point with increasing pressure. This behavior is partly responsible for the movement of glaciers, like the one shown in Figure \(4\). The bottom of a glacier experiences an immense pressure due to its weight that can melt some of the ice, forming a layer of liquid water on which the glacier may more easily slide.
The point of intersection of all three curves is labeled B in Figure \(2\). At the pressure and temperature represented by this point, three phases of water coexist in equilibrium. This temperature-pressure data pair is called the triple point. At pressures lower than the triple point, water cannot exist as a liquid, regardless of the temperature.
Example \(1\): Determining the State of Water
Using the phase diagram for water given in Figure \(2\), determine the state of water at the following temperatures and pressures:
1. −10 °C and 50 kPa
2. 25 °C and 90 kPa
3. 50 °C and 40 kPa
4. 80 °C and 5 kPa
5. −10 °C and 0.3 kPa
6. 50 °C and 0.3 kPa
Solution
Using the phase diagram for water, we can determine that the state of water at each temperature and pressure given are as follows: (a) solid; (b) liquid; (c) liquid; (d) gas; (e) solid; (f) gas.
Exercise \(1\)
What phase changes can water undergo as the temperature changes if the pressure is held at 0.3 kPa? If the pressure is held at 50 kPa?
Answer
At 0.3 kPa: at −58 °C. At 50 kPa: at 0 °C, l ⟶ g at 78 °C
Consider the phase diagram for carbon dioxide shown in Figure \(5\) as another example. The solid-liquid curve exhibits a positive slope, indicating that the melting point for CO2 increases with pressure as it does for most substances (water being a notable exception as described previously). Notice that the triple point is well above 1 atm, indicating that carbon dioxide cannot exist as a liquid under ambient pressure conditions. Instead, cooling gaseous carbon dioxide at 1 atm results in its deposition into the solid state. Likewise, solid carbon dioxide does not melt at 1 atm pressure but instead sublimes to yield gaseous CO2. Finally, notice that the critical point for carbon dioxide is observed at a relatively modest temperature and pressure in comparison to water.
Example \(2\): Determining the State of Carbon Dioxide
Using the phase diagram for carbon dioxide shown in Figure \(5\), determine the state of CO2 at the following temperatures and pressures:
1. −30 °C and 2000 kPa
2. −90 °C and 1000 kPa
3. −60 °C and 100 kPa
4. −40 °C and 1500 kPa
5. 0 °C and 100 kPa
6. 20 °C and 100 kPa
Solution
Using the phase diagram for carbon dioxide provided, we can determine that the state of CO2 at each temperature and pressure given are as follows: (a) liquid; (b) solid; (c) gas; (d) liquid; (e) gas; (f) gas.
Exercise \(2\)
Identify the phase changes that carbon dioxide will undergo as its temperature is increased from −100 °C while holding its pressure constant at 1500 kPa. At 50 kPa. At what approximate temperatures do these phase changes occur?
Answer
at 1500 kPa: at −55 °C, at −10 °C;
at 50 kPa: at −60 °C
Supercritical Fluids
If we place a sample of water in a sealed container at 25 °C, remove the air, and let the vaporization-condensation equilibrium establish itself, we are left with a mixture of liquid water and water vapor at a pressure of 0.03 atm. A distinct boundary between the more dense liquid and the less dense gas is clearly observed. As we increase the temperature, the pressure of the water vapor increases, as described by the liquid-gas curve in the phase diagram for water (Figure \(2\)), and a two-phase equilibrium of liquid and gaseous phases remains. At a temperature of 374 °C, the vapor pressure has risen to 218 atm, and any further increase in temperature results in the disappearance of the boundary between liquid and vapor phases. All of the water in the container is now present in a single phase whose physical properties are intermediate between those of the gaseous and liquid states. This phase of matter is called a supercritical fluid, and the temperature and pressure above which this phase exists is the critical point (Figure \(6\)). Above its critical temperature, a gas cannot be liquefied no matter how much pressure is applied. The pressure required to liquefy a gas at its critical temperature is called the critical pressure. The critical temperatures and critical pressures of some common substances are given in the following table.
Substance Critical Temperature (°C) Critical Pressure (kPa)
hydrogen −240.0 1300
nitrogen −147.2 3400
oxygen −118.9 5000
carbon dioxide 31.1 7400
ammonia 132.4 11,300
sulfur dioxide 157.2 7800
water 374.0 22,000
Link to Learning
Observe the liquid-to-supercritical fluid transition for carbon dioxide.
Like a gas, a supercritical fluid will expand and fill a container, but its density is much greater than typical gas densities, typically being close to those for liquids. Similar to liquids, these fluids are capable of dissolving nonvolatile solutes. They exhibit essentially no surface tension and very low viscosities, however, so they can more effectively penetrate very small openings in a solid mixture and remove soluble components. These properties make supercritical fluids extremely useful solvents for a wide range of applications. For example, supercritical carbon dioxide has become a very popular solvent in the food industry, being used to decaffeinate coffee, remove fats from potato chips, and extract flavor and fragrance compounds from citrus oils. It is nontoxic, relatively inexpensive, and not considered to be a pollutant. After use, the CO2 can be easily recovered by reducing the pressure and collecting the resulting gas.
Example \(3\): The Critical Temperature of Carbon Dioxide
If we shake a carbon dioxide fire extinguisher on a cool day (18 °C), we can hear liquid CO2 sloshing around inside the cylinder. However, the same cylinder appears to contain no liquid on a hot summer day (35 °C). Explain these observations.
Solution
On the cool day, the temperature of the CO2 is below the critical temperature of CO2, 304 K or 31 °C, so liquid CO2 is present in the cylinder. On the hot day, the temperature of the CO2 is greater than its critical temperature of 31 °C. Above this temperature no amount of pressure can liquefy CO2 so no liquid CO2 exists in the fire extinguisher.
Exercise \(3\)
Ammonia can be liquefied by compression at room temperature; oxygen cannot be liquefied under these conditions. Why do the two gases exhibit different behavior?
Answer
The critical temperature of ammonia is 405.5 K, which is higher than room temperature. The critical temperature of oxygen is below room temperature; thus oxygen cannot be liquefied at room temperature.
Chemistry in Everyday Life: Decaffeinating Coffee Using Supercritical CO2
Coffee is the world’s second most widely traded commodity, following only petroleum. Across the globe, people love coffee’s aroma and taste. Many of us also depend on one component of coffee—caffeine—to help us get going in the morning or stay alert in the afternoon. But late in the day, coffee’s stimulant effect can keep you from sleeping, so you may choose to drink decaffeinated coffee in the evening.
Since the early 1900s, many methods have been used to decaffeinate coffee. All have advantages and disadvantages, and all depend on the physical and chemical properties of caffeine. Because caffeine is a somewhat polar molecule, it dissolves well in water, a polar liquid. However, since many of the other 400-plus compounds that contribute to coffee’s taste and aroma also dissolve in H2O, hot water decaffeination processes can also remove some of these compounds, adversely affecting the smell and taste of the decaffeinated coffee. Dichloromethane (CH2Cl2) and ethyl acetate (CH3CO2C2H5) have similar polarity to caffeine, and are therefore very effective solvents for caffeine extraction, but both also remove some flavor and aroma components, and their use requires long extraction and cleanup times. Because both of these solvents are toxic, health concerns have been raised regarding the effect of residual solvent remaining in the decaffeinated coffee.
Supercritical fluid extraction using carbon dioxide is now being widely used as a more effective and environmentally friendly decaffeination method (Figure \(7\)). At temperatures above 304.2 K and pressures above 7376 kPa, CO2 is a supercritical fluid, with properties of both gas and liquid. Like a gas, it penetrates deep into the coffee beans; like a liquid, it effectively dissolves certain substances. Supercritical carbon dioxide extraction of steamed coffee beans removes 97−99% of the caffeine, leaving coffee’s flavor and aroma compounds intact. Because CO2 is a gas under standard conditions, its removal from the extracted coffee beans is easily accomplished, as is the recovery of the caffeine from the extract. The caffeine recovered from coffee beans via this process is a valuable product that can be used subsequently as an additive to other foods or drugs. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/10%3A_Liquids_and_Solids/10.04%3A_Phase_Diagrams.txt |
Learning Objectives
By the end of this section, you will be able to:
• Define and describe the bonding and properties of ionic, molecular, metallic, and covalent network crystalline solids
• Describe the main types of crystalline solids: ionic solids, metallic solids, covalent network solids, and molecular solids
• Explain the ways in which crystal defects can occur in a solid
When most liquids are cooled, they eventually freeze and form crystalline solids, solids in which the atoms, ions, or molecules are arranged in a definite repeating pattern. It is also possible for a liquid to freeze before its molecules become arranged in an orderly pattern. The resulting materials are called amorphous solids or noncrystalline solids (or, sometimes, glasses). The particles of such solids lack an ordered internal structure and are randomly arranged (Figure \(1\)).
Metals and ionic compounds typically form ordered, crystalline solids. Substances that consist of large molecules, or a mixture of molecules whose movements are more restricted, often form amorphous solids. For examples, candle waxes are amorphous solids composed of large hydrocarbon molecules. Some substances, such as silicon dioxide (shown in Figure \(2\)), can form either crystalline or amorphous solids, depending on the conditions under which it is produced. Also, amorphous solids may undergo a transition to the crystalline state under appropriate conditions.
Crystalline solids are generally classified according to the nature of the forces that hold its particles together. These forces are primarily responsible for the physical properties exhibited by the bulk solids. The following sections provide descriptions of the major types of crystalline solids: ionic, metallic, covalent network, and molecular.
Ionic Solids
Ionic solids, such as sodium chloride and nickel oxide, are composed of positive and negative ions that are held together by electrostatic attractions, which can be quite strong (Figure \(3\)). Many ionic crystals also have high melting points. This is due to the very strong attractions between the ions—in ionic compounds, the attractions between full charges are (much) larger than those between the partial charges in polar molecular compounds. This will be looked at in more detail in a later discussion of lattice energies. Although they are hard, they also tend to be brittle, and they shatter rather than bend. Ionic solids do not conduct electricity; however, they do conduct when molten or dissolved because their ions are free to move. Many simple compounds formed by the reaction of a metallic element with a nonmetallic element are ionic.
Metallic Solids
Metallic solids such as crystals of copper, aluminum, and iron are formed by metal atoms Figure \(4\). 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. The melting points of the metals vary widely. Mercury is a liquid at room temperature, and the alkali metals melt below 200 °C. Several post-transition metals also have low melting points, whereas the transition metals melt at temperatures above 1000 °C. These differences reflect differences in strengths of metallic bonding among the metals.
Covalent Network Solids
Covalent network solids include crystals of diamond, silicon, some other nonmetals, and some covalent compounds such as silicon dioxide (sand) and silicon carbide (carborundum, the abrasive on sandpaper). Many minerals have networks of covalent bonds. The atoms in these solids are held together by a network of covalent bonds, as shown in Figure \(5\). To break or to melt a covalent network solid, covalent bonds must be broken. Because covalent bonds are relatively strong, covalent network solids are typically characterized by hardness, strength, and high melting points. For example, diamond is one of the hardest substances known and melts above 3500 °C.
Molecular Solids
Molecular solids, such as ice, sucrose (table sugar), and iodine, as shown in Figure \(6\): , are composed of neutral molecules. The strengths of the attractive forces between the units present in different crystals vary widely, as indicated by the melting points of the crystals. Small symmetrical molecules (nonpolar molecules), such as H2, N2, O2, and F2, have weak attractive forces and form molecular solids with very low melting points (below −200 °C). Substances consisting of larger, nonpolar molecules have larger attractive forces and melt at higher temperatures. Molecular solids composed of molecules with permanent dipole moments (polar molecules) melt at still higher temperatures. Examples include ice (melting point, 0 °C) and table sugar (melting point, 185 °C).
Properties of Solids
A crystalline solid, like those listed in Table \(1\), has a precise melting temperature because each atom or molecule of the same type is held in place with the same forces or energy. Thus, the attractions between the units that make up the crystal all have the same strength and all require the same amount of energy to be broken. The gradual softening of an amorphous material differs dramatically from the distinct melting of a crystalline solid. This results from the structural nonequivalence of the molecules in the amorphous solid. Some forces are weaker than others, and when an amorphous material is heated, the weakest intermolecular attractions break first. As the temperature is increased further, the stronger attractions are broken. Thus amorphous materials soften over a range of temperatures.
Table \(1\): Types of Crystalline Solids and Their Properties
Type of Solid Type of Particles Type of Attractions Properties Examples
ionic ions ionic bonds hard, brittle, conducts electricity as a liquid but not as a solid, high to very high melting points NaCl, Al2O3
metallic atoms of electropositive elements metallic bonds shiny, malleable, ductile, conducts heat and electricity well, variable hardness and melting temperature Cu, Fe, Ti, Pb, U
covalent network atoms of electronegative elements covalent bonds very hard, not conductive, very high melting points C (diamond), SiO2, SiC
molecular molecules (or atoms) IMFs variable hardness, variable brittleness, not conductive, low melting points H2O, CO2, I2, C12H22O11
How Sciences Interconnect: Graphene: Material of the Future
Carbon is an essential element in our world. The unique properties of carbon atoms allow the existence of carbon-based life forms such as ourselves. Carbon forms a huge variety of substances that we use on a daily basis, including those shown in Figure \(7\). You may be familiar with diamond and graphite, the two most common allotropes of carbon. (Allotropes are different structural forms of the same element.) Diamond is one of the hardest-known substances, whereas graphite is soft enough to be used as pencil lead. These very different properties stem from the different arrangements of the carbon atoms in the different allotropes.
You may be less familiar with a recently discovered form of carbon: graphene. Graphene was first isolated in 2004 by using tape to peel off thinner and thinner layers from graphite. It is essentially a single sheet (one atom thick) of graphite. Graphene, illustrated in Figure \(8\): , is not only strong and lightweight, but it is also an excellent conductor of electricity and heat. These properties may prove very useful in a wide range of applications, such as vastly improved computer chips and circuits, better batteries and solar cells, and stronger and lighter structural materials. The 2010 Nobel Prize in Physics was awarded to Andre Geim and Konstantin Novoselov for their pioneering work with graphene.
Crystal Defects
In a crystalline solid, the atoms, ions, or molecules are arranged in a definite repeating pattern, but occasional defects may occur in the pattern. Several types of defects are known, as illustrated in Figure \(9\). Vacancies are defects that occur when positions that should contain atoms or ions are vacant. Less commonly, some atoms or ions in a crystal may occupy positions, called interstitial sites, located between the regular positions for atoms. Other distortions are found in impure crystals, as, for example, when the cations, anions, or molecules of the impurity are too large to fit into the regular positions without distorting the structure. Trace amounts of impurities are sometimes added to a crystal (a process known as doping) in order to create defects in the structure that yield desirable changes in its properties. For example, silicon crystals are doped with varying amounts of different elements to yield suitable electrical properties for their use in the manufacture of semiconductors and computer chips. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/10%3A_Liquids_and_Solids/10.05%3A_The_Solid_State_of_Matter.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the arrangement of atoms and ions in crystalline structures
• Compute ionic radii using unit cell dimensions
• Explain the use of X-ray diffraction measurements in determining crystalline structures
Over 90% of naturally occurring and man-made solids are crystalline. Most solids form with a regular arrangement of their particles because the overall attractive interactions between particles are maximized, and the total intermolecular energy is minimized, when the particles pack in the most efficient manner. The regular arrangement at an atomic level is often reflected at a macroscopic level. In this module, we will explore some of the details about the structures of metallic and ionic crystalline solids, and learn how these structures are determined experimentally.
The Structures of Metals
We will begin our discussion of crystalline solids by considering elemental metals, which are relatively simple because each contains only one type of atom. A pure metal is a crystalline solid with metal atoms packed closely together in a repeating pattern. Some of the properties of metals in general, such as their malleability and ductility, are largely due to having identical atoms arranged in a regular pattern. The different properties of one metal compared to another partially depend on the sizes of their atoms and the specifics of their spatial arrangements. We will explore the similarities and differences of four of the most common metal crystal geometries in the sections that follow.
Unit Cells of Metals
The structure of a crystalline solid, whether a metal or not, is best described by considering its simplest repeating unit, which is referred to as its unit cell. The unit cell consists of lattice points that represent the locations of atoms or ions. The entire structure then consists of this unit cell repeating in three dimensions, as illustrated in Figure $1$.
Let us begin our investigation of crystal lattice structure and unit cells with the most straightforward structure and the most basic unit cell. To visualize this, imagine taking a large number of identical spheres, such as tennis balls, and arranging them uniformly in a container. The simplest way to do this would be to make layers in which the spheres in one layer are directly above those in the layer below, as illustrated in Figure $2$. This arrangement is called simple cubic structure, and the unit cell is called the simple cubic unit cell or primitive cubic unit cell.
In a simple cubic structure, the spheres are not packed as closely as they could be, and they only “fill” about 52% of the volume of the container. This is a relatively inefficient arrangement, and only one metal (polonium, Po) crystallizes in a simple cubic structure. As shown in Figure $3$, a solid with this type of arrangement consists of planes (or layers) in which each atom contacts only the four nearest neighbors in its layer; one atom directly above it in the layer above; and one atom directly below it in the layer below. The number of other particles that each particle in a crystalline solid contacts is known as its coordination number. For a polonium atom in a simple cubic array, the coordination number is, therefore, six.
In a simple cubic lattice, the unit cell that repeats in all directions is a cube defined by the centers of eight atoms, as shown in Figure $4$. Atoms at adjacent corners of this unit cell contact each other, so the edge length of this cell is equal to two atomic radii, or one atomic diameter. A cubic unit cell contains only the parts of these atoms that are within it. Since an atom at a corner of a simple cubic unit cell is contained by a total of eight unit cells, only one-eighth of that atom is within a specific unit cell. And since each simple cubic unit cell has one atom at each of its eight “corners,” there is $8 \times \frac{1}{8}=1$ atom within one simple cubic unit cell.
Example $1$: Calculation of Atomic Radius and Density for Metals, Part 1
The edge length of the unit cell of alpha polonium is 336 pm.
1. Determine the radius of a polonium atom.
2. Determine the density of alpha polonium.
Solution
Alpha polonium crystallizes in a simple cubic unit cell:
1. Two adjacent Po atoms contact each other, so the edge length of this cell is equal to two Po atomic radii: $l = 2r$. Therefore, the radius of Po is: $r=\frac{l}{2}=\frac{336 pm }{2}=168 pm . \nonumber$
2. Density is given by $\text{density}=\frac{\text{mass}}{\text{volume}}$. The density of polonium can be found by determining the density of its unit cell (the mass contained within a unit cell divided by the volume of the unit cell). Since a Po unit cell contains one-eighth of a Po atom at each of its eight corners, a unit cell contains one Po atom.
The mass of a Po unit cell can be found by $1 \text { Po unit cell } \times \frac{1 \text { Po atom }}{1 \text { Po unit cell }} \times \frac{1 mol Po }{6.022 \times 10^{23} \text { Po atoms }} \times \frac{208.998 g }{1 mol Po }=3.47 \times 10^{-22} g \nonumber$
The volume of a Po unit cell can be found by: $V=l^3=\left(336 \times 10^{-10} cm \right)^3=3.79 \times 10^{-23} cm^3 \nonumber$
(Note that the edge length was converted from pm to cm to get the usual volume units for density.) Therefore, the density is $\text { Po }=\frac{3.471 \times 10^{-22} g }{3.79 \times 10^{-23} cm^3}=9.16 g / cm^3 \nonumber$
Exercise $1$
The edge length of the unit cell for nickel is 0.3524 nm. The density of Ni is 8.90 g/cm3. Does nickel crystallize in a simple cubic structure? Explain.
Answer
No. If Ni was simple cubic, its density would be given by:
$1 Ni \text { atom } \times \frac{1 mol Ni }{6.022 \times 10^{23} Ni \text { atoms }} \times \frac{58.693 g }{1 mol Ni }=9.746 \times 10^{-23} g \nonumber$
$V=l^3=\left(3.524 \times 10^{-8} cm \right)^3=4.376 \times 10^{-23} cm^3 \nonumber$
Then the density of Ni would be
$=\frac{9.746 \times 10^{-23} g }{4.376 \times 10^{-23} cm^3}=2.23 g / cm^3 \nonumber$
Since the actual density of Ni is not close to this, Ni does not form a simple cubic structure.
Most metal crystals are one of the four major types of unit cells. For now, we will focus on the three cubic unit cells: simple cubic (which we have already seen), body-centered cubic unit cell, and face-centered cubic unit cell—all of which are illustrated in Figure $5$. (Note that there are actually seven different lattice systems, some of which have more than one type of lattice, for a total of 14 different types of unit cells. We leave the more complicated geometries for later in this module.)
Some metals crystallize in an arrangement that has a cubic unit cell with atoms at all of the corners and an atom in the center, as shown in Figure $6$. This is called a body-centered cubic (BCC) solid. Atoms in the corners of a BCC unit cell do not contact each other but contact the atom in the center. A BCC unit cell contains two atoms: one-eighth of an atom at each of the eight corners atom from the corners ($8 \times \frac{1}{8}=1$ atom from the corners) plus one atom from the center. Any atom in this structure touches four atoms in the layer above it and four atoms in the layer below it. Thus, an atom in a BCC structure has a coordination number of eight.
Atoms in BCC arrangements are much more efficiently packed than in a simple cubic structure, occupying about 68% of the total volume. Isomorphous metals with a BCC structure include K, Ba, Cr, Mo, W, and Fe at room temperature. (Elements or compounds that crystallize with the same structure are said to be isomorphous.)
Many other metals, such as aluminum, copper, and lead, crystallize in an arrangement that has a cubic unit cell with atoms at all of the corners and at the centers of each face, as illustrated in Figure $7$. This arrangement is called a face-centered cubic (FCC) solid. A FCC unit cell contains four atoms: one-eighth of an atom at each of the eight corners atom from the corners ($8 \times \frac{1}{8}=1$ atom from the corners) and one-half of an atom on each of the six faces atoms from the faces ($6 \times \frac{1}{2}=3$ atom from the faces). The atoms at the corners touch the atoms in the centers of the adjacent faces along the face diagonals of the cube. Because the atoms are on identical lattice points, they have identical environments.
Atoms in an FCC arrangement are packed as closely together as possible, with atoms occupying 74% of the volume. This structure is also called cubic closest packing (CCP). In CCP, there are three repeating layers of hexagonally arranged atoms. Each atom contacts six atoms in its own layer, three in the layer above, and three in the layer below. In this arrangement, each atom touches 12 near neighbors, and therefore has a coordination number of 12. The fact that FCC and CCP arrangements are equivalent may not be immediately obvious, but why they are actually the same structure is illustrated in Figure $8$.
Because closer packing maximizes the overall attractions between atoms and minimizes the total intermolecular energy, the atoms in most metals pack in this manner. We find two types of closest packing in simple metallic crystalline structures: CCP, which we have already encountered, and hexagonal closest packing (HCP) shown in Figure $9$. Both consist of repeating layers of hexagonally arranged atoms. In both types, a second layer (B) is placed on the first layer (A) so that each atom in the second layer is in contact with three atoms in the first layer. The third layer is positioned in one of two ways. In HCP, atoms in the third layer are directly above atoms in the first layer (i.e., the third layer is also type A), and the stacking consists of alternating type A and type B close-packed layers (i.e., ABABAB⋯). In CCP, atoms in the third layer are not above atoms in either of the first two layers (i.e., the third layer is type C), and the stacking consists of alternating type A, type B, and type C close-packed layers (i.e., ABCABCABC⋯). About two–thirds of all metals crystallize in closest-packed arrays with coordination numbers of 12. Metals that crystallize in an HCP structure include Cd, Co, Li, Mg, Na, and Zn, and metals that crystallize in a CCP structure include Ag, Al, Ca, Cu, Ni, Pb, and Pt.
Example $2$: Calculation of Atomic Radius and Density for Metals, Part 2
Calcium crystallizes in a face-centered cubic structure. The edge length of its unit cell is 558.8 pm.
1. What is the atomic radius of Ca in this structure?
2. Calculate the density of Ca.
Solution
(a) In an FCC structure, Ca atoms contact each other across the diagonal of the face, so the length of the diagonal is equal to four Ca atomic radii (d = 4r). Two adjacent edges and the diagonal of the face form a right triangle, with the length of each side equal to 558.8 pm and the length of the hypotenuse equal to four Ca atomic radii:
$a^2+a^2=d^2 \longrightarrow(558.8 pm )^2+(558.5 pm )^2=(4 r)^2 \nonumber$
Solving this gives
$r=\sqrt{\frac{(558.8 pm )^2+(558.5 pm )^2}{16}}=197.6 pm \text { for a Ca radius. } \nonumber$
(b) Density is given by $\text { density }=\frac{\text { mass }}{\text { volume }}$. The density of calcium can be found by determining the density of its unit cell: for example, the mass contained within a unit cell divided by the volume of the unit cell. A face-centered Ca unit cell has one-eighth of an atom at each of the eight corners ($8 \times \frac{1}{8}=1$ atom from the corners) and one-half of an atom on each of the six faces ($6 \times \frac{1}{2}=3$ atom), for a total of four atoms in the unit cell.
The mass of the unit cell can be found by:
$1 Ca \text { unit cell } \times \frac{4 Ca \text { atoms }}{1 Ca \text { unit cell }} \times \frac{1 mol Ca }{6.022 \times 10^{23} Ca \text { atoms }} \times \frac{40.078 g }{1 mol Ca }=2.662 \times 10^{-22} g \nonumber$
The volume of a Ca unit cell can be found by:
$V=a^3=\left(558.8 \times 10^{-10} cm \right)^3=1.745 \times 10^{-22} cm^3 \nonumber$
(Note that the edge length was converted from pm to cm to get the usual volume units for density.)
Then, the density of Ca is
$\frac{2.662 \times 10^{-22} g }{1.745 \times 10^{-22} cm^3}=1.53 g / cm^3 \nonumber$
Exercise $2$
Silver crystallizes in an FCC structure. The edge length of its unit cell is 409 pm.
1. What is the atomic radius of Ag in this structure?
2. Calculate the density of Ag.
Answer
1. 144 pm
2. 10.5 g/cm3
In general, a unit cell is defined by the lengths of three axes (a, b, and c) and the angles (α, β, and γ) between them, as illustrated in Figure $10$. The axes are defined as being the lengths between points in the space lattice. Consequently, unit cell axes join points with identical environments.
There are seven different lattice systems, some of which have more than one type of lattice, for a total of fourteen different unit cells, which have the shapes shown in Figure $11$.
The Structures of Ionic Crystals
Ionic crystals consist of two or more different kinds of ions that usually have different sizes. The packing of these ions into a crystal structure is more complex than the packing of metal atoms that are the same size.
Most monatomic ions behave as charged spheres, and their attraction for ions of opposite charge is the same in every direction. Consequently, stable structures for ionic compounds result (1) when ions of one charge are surrounded by as many ions as possible of the opposite charge and (2) when the cations and anions are in contact with each other. Structures are determined by two principal factors: the relative sizes of the ions and the ratio of the numbers of positive and negative ions in the compound.
In simple ionic structures, we usually find the anions, which are normally larger than the cations, arranged in a closest-packed array. (As seen previously, additional electrons attracted to the same nucleus make anions larger and fewer electrons attracted to the same nucleus make cations smaller when compared to the atoms from which they are formed.) The smaller cations commonly occupy one of two types of holes (or interstices) remaining between the anions. The smaller of the holes is found between three anions in one plane and one anion in an adjacent plane. The four anions surrounding this hole are arranged at the corners of a tetrahedron, so the hole is called a tetrahedral hole. The larger type of hole is found at the center of six anions (three in one layer and three in an adjacent layer) located at the corners of an octahedron; this is called an octahedral hole. Figure $12$ illustrates both of these types of holes.
Depending on the relative sizes of the cations and anions, the cations of an ionic compound may occupy tetrahedral or octahedral holes, as illustrated in Figure $13$. Relatively small cations occupy tetrahedral holes, and larger cations occupy octahedral holes. If the cations are too large to fit into the octahedral holes, the anions may adopt a more open structure, such as a simple cubic array. The larger cations can then occupy the larger cubic holes made possible by the more open spacing.
There are two tetrahedral holes for each anion in either an HCP or CCP array of anions. A compound that crystallizes in a closest-packed array of anions with cations in the tetrahedral holes can have a maximum cation:anion ratio of 2:1; all of the tetrahedral holes are filled at this ratio. Examples include Li2O, Na2O, Li2S, and Na2S. Compounds with a ratio of less than 2:1 may also crystallize in a closest-packed array of anions with cations in the tetrahedral holes, if the ionic sizes fit. In these compounds, however, some of the tetrahedral holes remain vacant.
Example $3$: Occupancy of Tetrahedral Holes
Zinc sulfide is an important industrial source of zinc and is also used as a white pigment in paint. Zinc sulfide crystallizes with zinc ions occupying one-half of the tetrahedral holes in a closest-packed array of sulfide ions. What is the formula of zinc sulfide?
Solution
Because there are two tetrahedral holes per anion (sulfide ion) and one-half of these holes are occupied by zinc ions, there must be $\frac{1}{2} \times 2$ or 1, zinc ion per sulfide ion. Thus, the formula is ZnS.
Exercise $3$
Lithium selenide can be described as a closest-packed array of selenide ions with lithium ions in all of the tetrahedral holes. What it the formula of lithium selenide?
Answer
Li2Se
The ratio of octahedral holes to anions in either an HCP or CCP structure is 1:1. Thus, compounds with cations in octahedral holes in a closest-packed array of anions can have a maximum cation:anion ratio of 1:1. In NiO, MnS, NaCl, and KH, for example, all of the octahedral holes are filled. Ratios of less than 1:1 are observed when some of the octahedral holes remain empty.
Example $4$: Stoichiometry of Ionic Compounds
Sapphire is aluminum oxide. Aluminum oxide crystallizes with aluminum ions in two-thirds of the octahedral holes in a closest-packed array of oxide ions. What is the formula of aluminum oxide?
Solution
Because there is one octahedral hole per anion (oxide ion) and only two-thirds of these holes are occupied, the ratio of aluminum to oxygen must be $\frac{2}{3}:1$, which would give $\ce{Al_{2/3}O}$. The simplest whole number ratio is 2:3, so the formula is $\ce{Al2O3}$.
Exercise $4$
The white pigment titanium oxide crystallizes with titanium ions in one-half of the octahedral holes in a closest-packed array of oxide ions. What is the formula of titanium oxide?
Answer
$\ce{TiO2}$
In a simple cubic array of anions, there is one cubic hole that can be occupied by a cation for each anion in the array. In CsCl, and in other compounds with the same structure, all of the cubic holes are occupied. Half of the cubic holes are occupied in SrH2, UO2, SrCl2, and CaF2.
Different types of ionic compounds often crystallize in the same structure when the relative sizes of their ions and their stoichiometries (the two principal features that determine structure) are similar.
Unit Cells of Ionic Compounds
Many ionic compounds crystallize with cubic unit cells, and we will use these compounds to describe the general features of ionic structures.
When an ionic compound is composed of cations and anions of similar size in a 1:1 ratio, it typically forms a simple cubic structure. Cesium chloride, CsCl, (illustrated in Figure $14$) is an example of this, with Cs+ and Cl having radii of 174 pm and 181 pm, respectively. We can think of this as chloride ions forming a simple cubic unit cell, with a cesium ion in the center; or as cesium ions forming a unit cell with a chloride ion in the center; or as simple cubic unit cells formed by Cs+ ions overlapping unit cells formed by Cl ions. Cesium ions and chloride ions touch along the body diagonals of the unit cells. One cesium ion and one chloride ion are present per unit cell, giving the l:l stoichiometry required by the formula for cesium chloride. Note that there is no lattice point in the center of the cell, and CsCl is not a BCC structure because a cesium ion is not identical to a chloride ion.
We have said that the location of lattice points is arbitrary. This is illustrated by an alternate description of the CsCl structure in which the lattice points are located in the centers of the cesium ions. In this description, the cesium ions are located on the lattice points at the corners of the cell, and the chloride ion is located at the center of the cell. The two unit cells are different, but they describe identical structures.
When an ionic compound is composed of a 1:1 ratio of cations and anions that differ significantly in size, it typically crystallizes with an FCC unit cell, like that shown in Figure $15$. Sodium chloride, NaCl, is an example of this, with Na+ and Cl having radii of 102 pm and 181 pm, respectively. We can think of this as chloride ions forming an FCC cell, with sodium ions located in the octahedral holes in the middle of the cell edges and in the center of the cell. The sodium and chloride ions touch each other along the cell edges. The unit cell contains four sodium ions and four chloride ions, giving the 1:1 stoichiometry required by the formula, NaCl.
The cubic form of zinc sulfide, zinc blende, also crystallizes in an FCC unit cell, as illustrated in Figure $16$. This structure contains sulfide ions on the lattice points of an FCC lattice. (The arrangement of sulfide ions is identical to the arrangement of chloride ions in sodium chloride.) The radius of a zinc ion is only about 40% of the radius of a sulfide ion, so these small Zn2+ ions are located in alternating tetrahedral holes, that is, in one half of the tetrahedral holes. There are four zinc ions and four sulfide ions in the unit cell, giving the empirical formula ZnS.
A calcium fluoride unit cell, like that shown in Figure $17$, is also an FCC unit cell, but in this case, the cations are located on the lattice points; equivalent calcium ions are located on the lattice points of an FCC lattice. All of the tetrahedral sites in the FCC array of calcium ions are occupied by fluoride ions. There are four calcium ions and eight fluoride ions in a unit cell, giving a calcium:fluorine ratio of 1:2, as required by the chemical formula, CaF2. Close examination of Figure $17$ will reveal a simple cubic array of fluoride ions with calcium ions in one half of the cubic holes. The structure cannot be described in terms of a space lattice of points on the fluoride ions because the fluoride ions do not all have identical environments. The orientation of the four calcium ions about the fluoride ions differs.
Calculation of Ionic Radii
If we know the edge length of a unit cell of an ionic compound and the position of the ions in the cell, we can calculate ionic radii for the ions in the compound if we make assumptions about individual ionic shapes and contacts.
Example $5$: Calculation of Ionic Radii
The edge length of the unit cell of LiCl (NaCl-like structure, FCC) is 0.514 nm or 5.14 Å. Assuming that the lithium ion is small enough so that the chloride ions are in contact, as in Figure $15$, calculate the ionic radius for the chloride ion. Note: The length unit angstrom, Å, is often used to represent atomic-scale dimensions and is equivalent to 10−10 m.
Solution
On the face of a LiCl unit cell, chloride ions contact each other across the diagonal of the face:
Drawing a right triangle on the face of the unit cell, we see that the length of the diagonal is equal to four chloride radii (one radius from each corner chloride and one diameter—which equals two radii—from the chloride ion in the center of the face), so d = 4r. From the Pythagorean theorem, we have:
$\nonumber$a^2+a^2=d^2 \nonumber \]
which yields:
$(0.514 nm )^2+(0.514 nm )^2=(4 r)^2=16 r^2 \nonumber$
Solving this gives:
$r=\sqrt{\frac{(0.514 nm )^2+(0.514 nm )^2}{16}}=0.182 nm (1.82 \AA) \text { for a } Cl^{-} \text {radius. } \nonumber$
Exercise $1$
The edge length of the unit cell of KCl (NaCl-like structure, FCC) is 6.28 Å. Assuming anion-cation contact along the cell edge, calculate the radius of the potassium ion. The radius of the chloride ion is 1.82 Å.
Answer
The radius of the potassium ion is 1.33 Å.
It is important to realize that values for ionic radii calculated from the edge lengths of unit cells depend on numerous assumptions, such as a perfect spherical shape for ions, which are approximations at best. Hence, such calculated values are themselves approximate and comparisons cannot be pushed too far. Nevertheless, this method has proved useful for calculating ionic radii from experimental measurements such as X-ray crystallographic determinations.
X-Ray Crystallography
The size of the unit cell and the arrangement of atoms in a crystal may be determined from measurements of the diffraction of X-rays by the crystal, termed X-ray crystallography. Diffraction is the change in the direction of travel experienced by an electromagnetic wave when it encounters a physical barrier whose dimensions are comparable to those of the wavelength of the light. X-rays are electromagnetic radiation with wavelengths about as long as the distance between neighboring atoms in crystals (on the order of a few Å).
When a beam of monochromatic X-rays strikes a crystal, its rays are scattered in all directions by the atoms within the crystal. When scattered waves traveling in the same direction encounter one another, they undergo interference, a process by which the waves combine to yield either an increase or a decrease in amplitude (intensity) depending upon the extent to which the combining waves’ maxima are separated (see Figure $18$).
When X-rays of a certain wavelength, $λ$, are scattered by atoms in adjacent crystal planes separated by a distance, $d$, they may undergo constructive interference when the difference between the distances traveled by the two waves prior to their combination is an integer factor, $n$, of the wavelength. This condition is satisfied when the angle of the diffracted beam, $θ$, is related to the wavelength and interatomic distance by the equation:
$n \lambda=2 d \sin \theta \label{Bragg}$
This relation is known as the Bragg equation in honor of W. H. Bragg, the English physicist who first explained this phenomenon. Figure $19$: illustrates two examples of diffracted waves from the same two crystal planes. The figure on the left depicts waves diffracted at the Bragg angle, resulting in constructive interference, while that on the right shows diffraction and a different angle that does not satisfy the Bragg condition, resulting in destructive interference.
Link to Learning
Visit this site for more details on the Bragg equation and a simulator that allows you to explore the effect of each variable on the intensity of the diffracted wave.
An X-ray diffractometer, such as the one illustrated in Figure $20$, may be used to measure the angles at which X-rays are diffracted when interacting with a crystal as described earlier. From such measurements, the Bragg equation may be used to compute distances between atoms as demonstrated in the following example exercise.
Example $6$: Using the Bragg Equation
In a diffractometer, X-rays with a wavelength of 0.1315 nm were used to produce a diffraction pattern for copper. The first order diffraction (n = 1) occurred at an angle θ = 25.25°. Determine the spacing between the diffracting planes in copper.
Solution
The distance between the planes is found by solving the Bragg equation (Equation \ref{Bragg}), $nλ = 2d \sin θ$, for $d$.
This gives:
$d=\frac{n \lambda}{2 \sin \theta}=\frac{1(0.1315 nm )}{2 \sin \left(25.25^{\circ}\right)}=0.154 nm \nonumber$
Exercise $6$
A crystal with spacing between planes equal to 0.394 nm diffracts X-rays with a wavelength of 0.147 nm. What is the angle for the first order diffraction?
Answer
10.8°
Portrait of a Chemist: X-ray Crystallographer Rosalind Franklin
The discovery of the structure of DNA in 1953 by Francis Crick and James Watson is one of the great achievements in the history of science. They were awarded the 1962 Nobel Prize in Physiology or Medicine, along with Maurice Wilkins, who provided experimental proof of DNA’s structure. British chemist Rosalind Franklin made invaluable contributions to this monumental achievement through her work in measuring X-ray diffraction images of DNA. Early in her career, Franklin’s research on the structure of coals proved helpful to the British war effort. After shifting her focus to biological systems in the early 1950s, Franklin and doctoral student Raymond Gosling discovered that DNA consists of two forms: a long, thin fiber formed when wet (type “B”) and a short, wide fiber formed when dried (type “A”). Her X-ray diffraction images of DNA (Figure $21$) provided the crucial information that allowed Watson and Crick to confirm that DNA forms a double helix, and to determine details of its size and structure. Franklin also conducted pioneering research on viruses and the RNA that contains their genetic information, uncovering new information that radically changed the body of knowledge in the field. After developing ovarian cancer, Franklin continued to work until her death in 1958 at age 37. Among many posthumous recognitions of her work, the Chicago Medical School of Finch University of Health Sciences changed its name to the Rosalind Franklin University of Medicine and Science in 2004, and adopted an image of her famous X-ray diffraction image of DNA as its official university logo. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/10%3A_Liquids_and_Solids/10.06%3A_Lattice_Structures_in_Crystalline_Solids.txt |
Example and Directions
Words (or words that have the same definition)The definition is case sensitive(Optional) Image to display with the definition [Not displayed in Glossary, only in pop-up on pages](Optional) Caption for Image(Optional) External or Internal Link(Optional) Source for Definition
(Eg. "Genetic, Hereditary, DNA ...")(Eg. "Relating to genes or heredity")The infamous double helix https://bio.libretexts.org/CC-BY-SA; Delmar Larsen
Glossary Entries
Word(s)DefinitionImageCaptionLinkSource
adhesive forceforce of attraction between molecules of different chemical identities
amorphous solid(also, noncrystalline solid) solid in which the particles lack an ordered internal structure
body-centered cubic (BCC) solidcrystalline structure that has a cubic unit cell with lattice points at the corners and in the center of the cell
body-centered cubic unit cellsimplest repeating unit of a body-centered cubic crystal; it is a cube containing lattice points at each corner and in the center of the cube
boiling pointtemperature at which the vapor pressure of a liquid equals the pressure of the gas above it
Bragg equationequation that relates the angles at which X-rays are diffracted by the atoms within a crystal
capillary actionflow of liquid within a porous material due to the attraction of the liquid molecules to the surface of the material and to other liquid molecules
Clausius-Clapeyron equationmathematical relationship between the temperature, vapor pressure, and enthalpy of vaporization for a substance
cohesive forceforce of attraction between identical molecules
condensationchange from a gaseous to a liquid state
coordination numbernumber of atoms closest to any given atom in a crystal or to the central metal atom in a complex
covalent network solidsolid whose particles are held together by covalent bonds
critical pointtemperature and pressure above which a gas cannot be condensed into a liquid
crystalline solidsolid in which the particles are arranged in a definite repeating pattern
cubic closest packing (CCP)crystalline structure in which planes of closely packed atoms or ions are stacked as a series of three alternating layers of different relative orientations (ABC)
depositionchange from a gaseous state directly to a solid state
diffractionredirection of electromagnetic radiation that occurs when it encounters a physical barrier of appropriate dimensions
dipole-dipole attractionintermolecular attraction between two permanent dipoles
dispersion force(also, London dispersion force) attraction between two rapidly fluctuating, temporary dipoles; significant only when particles are very close together
dynamic equilibriumstate of a system in which reciprocal processes are occurring at equal rates
face-centered cubic (FCC) solidcrystalline structure consisting of a cubic unit cell with lattice points on the corners and in the center of each face
face-centered cubic unit cellsimplest repeating unit of a face-centered cubic crystal; it is a cube containing lattice points at each corner and in the center of each face
freezingchange from a liquid state to a solid state
freezing pointtemperature at which the solid and liquid phases of a substance are in equilibrium; see also melting point
hexagonal closest packing (HCP)crystalline structure in which close packed layers of atoms or ions are stacked as a series of two alternating layers of different relative orientations (AB)
hole(also, interstice) space between atoms within a crystal
hydrogen bondingoccurs when exceptionally strong dipoles attract; bonding that exists when hydrogen is bonded to one of the three most electronegative elements: F, O, or N
induced dipoletemporary dipole formed when the electrons of an atom or molecule are distorted by the instantaneous dipole of a neighboring atom or molecule
instantaneous dipoletemporary dipole that occurs for a brief moment in time when the electrons of an atom or molecule are distributed asymmetrically
intermolecular forcenoncovalent attractive force between atoms, molecules, and/or ions
interstitial sitesspaces between the regular particle positions in any array of atoms or ions
ionic solidsolid composed of positive and negative ions held together by strong electrostatic attractions
isomorphouspossessing the same crystalline structure
meltingchange from a solid state to a liquid state
melting pointtemperature at which the solid and liquid phases of a substance are in equilibrium; see also freezing point
metallic solidsolid composed of metal atoms
molecular solidsolid composed of neutral molecules held together by intermolecular forces of attraction
normal boiling pointtemperature at which a liquid’s vapor pressure equals 1 atm (760 torr)
octahedral holeopen space in a crystal at the center of six particles located at the corners of an octahedron
phase diagrampressure-temperature graph summarizing conditions under which the phases of a substance can exist
polarizabilitymeasure of the ability of a charge to distort a molecule’s charge distribution (electron cloud)
simple cubic structurecrystalline structure with a cubic unit cell with lattice points only at the corners
simple cubic unit cell(also, primitive cubic unit cell) unit cell in the simple cubic structure
space latticeall points within a crystal that have identical environments
sublimationchange from solid state directly to gaseous state
supercritical fluidsubstance at a temperature and pressure higher than its critical point; exhibits properties intermediate between those of gaseous and liquid states
surface tensionenergy required to increase the area, or length, of a liquid surface by a given amount
tetrahedral holetetrahedral space formed by four atoms or ions in a crystal
triple pointtemperature and pressure at which three phases of a substance are in equilibrium
unit cellsmallest portion of a space lattice that is repeated in three dimensions to form the entire lattice
vacancydefect that occurs when a position that should contain an atom or ion is vacant
van der Waals forceattractive or repulsive force between molecules, including dipole-dipole, dipole-induced dipole, and London dispersion forces; does not include forces due to covalent or ionic bonding, or the attraction between ions and molecules
vapor pressure(also, equilibrium vapor pressure) pressure exerted by a vapor in equilibrium with a solid or a liquid at a given temperature
vaporizationchange from liquid state to gaseous state
viscositymeasure of a liquid’s resistance to flow
X-ray crystallographyexperimental technique for determining distances between atoms in a crystal by measuring the angles at which X-rays are diffracted when passing through the crystal | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/10%3A_Liquids_and_Solids/10.07%3A_Key_Terms.txt |
The physical properties of condensed matter (liquids and solids) can be explained in terms of the kinetic molecular theory. In a liquid, intermolecular attractive forces hold the molecules in contact, although they still have sufficient KE to move past each other.
Intermolecular attractive forces, collectively referred to as van der Waals forces, are responsible for the behavior of liquids and solids and are electrostatic in nature. Dipole-dipole attractions result from the electrostatic attraction of the partial negative end of one polar molecule for the partial positive end of another. The temporary dipole that results from the motion of the electrons in an atom can induce a dipole in an adjacent atom and give rise to the London dispersion force. London forces increase with increasing molecular size. Hydrogen bonds are a special type of dipole-dipole attraction that results when hydrogen is bonded to one of the three most electronegative elements: F, O, or N.
The intermolecular forces between molecules in the liquid state vary depending upon their chemical identities and result in corresponding variations in various physical properties. Cohesive forces between like molecules are responsible for a liquid’s viscosity (resistance to flow) and surface tension (elasticity of a liquid surface). Adhesive forces between the molecules of a liquid and different molecules composing a surface in contact with the liquid are responsible for phenomena such as surface wetting and capillary rise.
Phase transitions are processes that convert matter from one physical state into another. There are six phase transitions between the three phases of matter. Melting, vaporization, and sublimation are all endothermic processes, requiring an input of heat to overcome intermolecular attractions. The reciprocal transitions of freezing, condensation, and deposition are all exothermic processes, involving heat as intermolecular attractive forces are established or strengthened. The temperatures at which phase transitions occur are determined by the relative strengths of intermolecular attractions and are, therefore, dependent on the chemical identity of the substance.
The temperature and pressure conditions at which a substance exists in solid, liquid, and gaseous states are summarized in a phase diagram for that substance. Phase diagrams are combined plots of pressure-temperature equilibrium curves representing the relationships between phase transition temperatures and pressures. The point of intersection of any three curves in a phase diagram represents a substance’s triple point—the temperature and pressure at which three different phases are in equilibrium. At pressures below a solid-liquid-gas triple point, a substance cannot exist in the liquid state, regardless of its temperature. The terminus of the liquid-gas curve represents the substance’s critical point, the pressure and temperature above which a liquid phase cannot exist.
Some substances form crystalline solids consisting of particles in a very organized structure; others form amorphous (noncrystalline) solids with an internal structure that is not ordered. The main types of crystalline solids are ionic solids, metallic solids, covalent network solids, and molecular solids. The properties of the different kinds of crystalline solids are due to the types of particles of which they consist, the arrangements of the particles, and the strengths of the attractions between them. Because their particles experience identical attractions, crystalline solids have distinct melting temperatures; the particles in amorphous solids experience a range of interactions, so they soften gradually and melt over a range of temperatures. Some crystalline solids have defects in the definite repeating pattern of their particles. These defects (which include vacancies, atoms or ions not in the regular positions, and impurities) change physical properties such as electrical conductivity, which is exploited in the silicon crystals used to manufacture computer chips.
The structures of crystalline metals and simple ionic compounds can be described in terms of packing of spheres. Metal atoms can pack in hexagonal closest-packed structures, cubic closest-packed structures, body-centered structures, and simple cubic structures. The anions in simple ionic structures commonly adopt one of these structures, and the cations occupy the spaces remaining between the anions. Small cations usually occupy tetrahedral holes in a closest-packed array of anions. Larger cations usually occupy octahedral holes. Still larger cations can occupy cubic holes in a simple cubic array of anions. The structure of a solid can be described by indicating the size and shape of a unit cell and the contents of the cell. The type of structure and dimensions of the unit cell can be determined by X-ray diffraction measurements. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/10%3A_Liquids_and_Solids/10.09%3A_Summary.txt |
1.
In terms of their bulk properties, how do liquids and solids differ? How are they similar?
2.
In terms of the kinetic molecular theory, in what ways are liquids similar to solids? In what ways are liquids different from solids?
3.
In terms of the kinetic molecular theory, in what ways are liquids similar to gases? In what ways are liquids different from gases?
4.
Explain why liquids assume the shape of any container into which they are poured, whereas solids are rigid and retain their shape.
5.
What is the evidence that all neutral atoms and molecules exert attractive forces on each other?
6.
Open the PhET States of Matter Simulation to answer the following questions:
1. Select the Solid, Liquid, Gas tab. Explore by selecting different substances, heating and cooling the systems, and changing the state. What similarities do you notice between the four substances for each phase (solid, liquid, gas)? What differences do you notice?
2. For each substance, select each of the states and record the given temperatures. How do the given temperatures for each state correlate with the strengths of their intermolecular attractions? Explain.
3. Select the Interaction Potential tab, and use the default neon atoms. Move the Ne atom on the right and observe how the potential energy changes. Select the Total Force button, and move the Ne atom as before. When is the total force on each atom attractive and large enough to matter? Then select the Component Forces button, and move the Ne atom. When do the attractive (van der Waals) and repulsive (electron overlap) forces balance? How does this relate to the potential energy versus the distance between atoms graph? Explain.
7.
Define the following and give an example of each:
1. dispersion force
2. dipole-dipole attraction
3. hydrogen bond
8.
The types of intermolecular forces in a substance are identical whether it is a solid, a liquid, or a gas. Why then does a substance change phase from a gas to a liquid or to a solid?
9.
Why do the boiling points of the noble gases increase in the order He < Ne < Ar < Kr < Xe?
10.
Neon and HF have approximately the same molecular masses.
1. Explain why the boiling points of Neon and HF differ.
2. Compare the change in the boiling points of Ne, Ar, Kr, and Xe with the change of the boiling points of HF, HCl, HBr, and HI, and explain the difference between the changes with increasing atomic or molecular mass.
11.
Arrange each of the following sets of compounds in order of increasing boiling point temperature:
1. HCl, H2O, SiH4
2. F2, Cl2, Br2
3. CH4, C2H6, C3H8
4. O2, NO, N2
12.
The molecular mass of butanol, C4H9OH, is 74.14; that of ethylene glycol, CH2(OH)CH2OH, is 62.08, yet their boiling points are 117.2 °C and 174 °C, respectively. Explain the reason for the difference.
13.
On the basis of intermolecular attractions, explain the differences in the boiling points of n–butane (−1 °C) and chloroethane (12 °C), which have similar molar masses.
14.
On the basis of dipole moments and/or hydrogen bonding, explain in a qualitative way the differences in the boiling points of acetone (56.2 °C) and 1-propanol (97.4 °C), which have similar molar masses.
15.
The melting point of H2O(s) is 0 °C. Would you expect the melting point of H2S(s) to be −85 °C, 0 °C, or 185 °C? Explain your answer.
16.
Silane (SiH4), phosphine (PH3), and hydrogen sulfide (H2S) melt at −185 °C, −133 °C, and −85 °C, respectively. What does this suggest about the polar character and intermolecular attractions of the three compounds?
17.
Explain why a hydrogen bond between two water molecules is weaker than a hydrogen bond between two hydrogen fluoride molecules.
18.
Under certain conditions, molecules of acetic acid, CH3COOH, form “dimers,” pairs of acetic acid molecules held together by strong intermolecular attractions:
Draw a dimer of acetic acid, showing how two CH3COOH molecules are held together, and stating the type of IMF that is responsible.
19.
Proteins are chains of amino acids that can form in a variety of arrangements, one of which is a helix. What kind of IMF is responsible for holding the protein strand in this shape? On the protein image, show the locations of the IMFs that hold the protein together:
20.
The density of liquid NH3 is 0.64 g/mL; the density of gaseous NH3 at STP is 0.0007 g/mL. Explain the difference between the densities of these two phases.
21.
Identify the intermolecular forces present in the following solids:
1. CH3CH2OH
2. CH3CH2CH3
3. CH3CH2Cl
22.
The test tubes shown here contain equal amounts of the specified motor oils. Identical metal spheres were dropped at the same time into each of the tubes, and a brief moment later, the spheres had fallen to the heights indicated in the illustration.
Rank the motor oils in order of increasing viscosity, and explain your reasoning:
23.
Although steel is denser than water, a steel needle or paper clip placed carefully lengthwise on the surface of still water can be made to float. Explain at a molecular level how this is possible.
24.
The surface tension and viscosity values for diethyl ether, acetone, ethanol, and ethylene glycol are shown here.
1. Explain their differences in viscosity in terms of the size and shape of their molecules and their IMFs.
2. Explain their differences in surface tension in terms of the size and shape of their molecules and their IMFs:
25.
You may have heard someone use the figure of speech “slower than molasses in winter” to describe a process that occurs slowly. Explain why this is an apt idiom, using concepts of molecular size and shape, molecular interactions, and the effect of changing temperature.
26.
It is often recommended that you let your car engine run idle to warm up before driving, especially on cold winter days. While the benefit of prolonged idling is dubious, it is certainly true that a warm engine is more fuel efficient than a cold one. Explain the reason for this.
27.
The surface tension and viscosity of water at several different temperatures are given in this table.
Water Surface Tension (mN/m) Viscosity (mPa s)
0 °C 75.6 1.79
20 °C 72.8 1.00
60 °C 66.2 0.47
100 °C 58.9 0.28
1. As temperature increases, what happens to the surface tension of water? Explain why this occurs, in terms of molecular interactions and the effect of changing temperature.
2. As temperature increases, what happens to the viscosity of water? Explain why this occurs, in terms of molecular interactions and the effect of changing temperature.
28.
At 25 °C, how high will water rise in a glass capillary tube with an inner diameter of 0.63 mm? Refer to Example 10.4 for the required information.
29.
Water rises in a glass capillary tube to a height of 17 cm. What is the diameter of the capillary tube?
30.
Heat is added to boiling water. Explain why the temperature of the boiling water does not change. What does change?
31.
Heat is added to ice at 0 °C. Explain why the temperature of the ice does not change. What does change?
32.
What feature characterizes the dynamic equilibrium between a liquid and its vapor in a closed container?
33.
Identify two common observations indicating some liquids have sufficient vapor pressures to noticeably evaporate?
34.
Identify two common observations indicating some solids, such as dry ice and mothballs, have vapor pressures sufficient to sublime?
35.
What is the relationship between the intermolecular forces in a liquid and its vapor pressure?
36.
What is the relationship between the intermolecular forces in a solid and its melting temperature?
37.
Why does spilled gasoline evaporate more rapidly on a hot day than on a cold day?
38.
Carbon tetrachloride, CCl4, was once used as a dry cleaning solvent, but is no longer used because it is carcinogenic. At 57.8 °C, the vapor pressure of CCl4 is 54.0 kPa, and its enthalpy of vaporization is 33.05 kJ/mol. Use this information to estimate the normal boiling point for CCl4.
39.
When is the boiling point of a liquid equal to its normal boiling point?
40.
How does the boiling of a liquid differ from its evaporation?
41.
Use the information in Figure 10.24 to estimate the boiling point of water in Denver when the atmospheric pressure is 83.3 kPa.
42.
A syringe at a temperature of 20 °C is filled with liquid ether in such a way that there is no space for any vapor. If the temperature is kept constant and the plunger is withdrawn to create a volume that can be occupied by vapor, what would be the approximate pressure of the vapor produced?
43.
Explain the following observations:
1. It takes longer to cook an egg in Ft. Davis, Texas (altitude, 5000 feet above sea level) than it does in Boston (at sea level).
2. Perspiring is a mechanism for cooling the body.
44.
The enthalpy of vaporization of water is larger than its enthalpy of fusion. Explain why.
45.
Explain why the molar enthalpies of vaporization of the following substances increase in the order CH4 < C2H6 < C3H8, even though the type of IMF (dispersion) is the same.
46.
Explain why the enthalpies of vaporization of the following substances increase in the order CH4 < NH3 < H2O, even though all three substances have approximately the same molar mass.
47.
The enthalpy of vaporization of CO2(l) is 9.8 kJ/mol. Would you expect the enthalpy of vaporization of CS2(l) to be 28 kJ/mol, 9.8 kJ/mol, or −8.4 kJ/mol? Discuss the plausibility of each of these answers.
48.
The hydrogen fluoride molecule, HF, is more polar than a water molecule, H2O (for example, has a greater dipole moment), yet the molar enthalpy of vaporization for liquid hydrogen fluoride is lesser than that for water. Explain.
49.
Ethyl chloride (boiling point, 13 °C) is used as a local anesthetic. When the liquid is sprayed on the skin, it cools the skin enough to freeze and numb it. Explain the cooling effect of liquid ethyl chloride.
50.
Which contains the compounds listed correctly in order of increasing boiling points?
1. N2 < CS2 < H2O < KCl
2. H2O < N2 < CS2 < KCl
3. N2 < KCl < CS2 < H2O
4. CS2 < N2 < KCl < H2O
5. KCl < H2O < CS2 < N2
51.
How much heat is required to convert 422 g of liquid H2O at 23.5 °C into steam at 150 °C?
52.
Evaporation of sweat requires energy and thus take excess heat away from the body. Some of the water that you drink may eventually be converted into sweat and evaporate. If you drink a 20-ounce bottle of water that had been in the refrigerator at 3.8 °C, how much heat is needed to convert all of that water into sweat and then to vapor? (Note: Your body temperature is 36.6 °C. For the purpose of solving this problem, assume that the thermal properties of sweat are the same as for water.)
53.
Titanium tetrachloride, TiCl4, has a melting point of −23.2 °C and has a ΔH fusion = 9.37 kJ/mol.
1. How much energy is required to melt 263.1 g TiCl4?
2. For TiCl4, which will likely have the larger magnitude: ΔH fusion or ΔH vaporization? Explain your reasoning.
54.
From the phase diagram for water (Figure 10.31), determine the state of water at:
1. (f) 60 °C and 50 kPa
55.
What phase changes will take place when water is subjected to varying pressure at a constant temperature of 0.005 °C? At 40 °C? At −40 °C?
56.
Pressure cookers allow food to cook faster because the higher pressure inside the pressure cooker increases the boiling temperature of water. A particular pressure cooker has a safety valve that is set to vent steam if the pressure exceeds 3.4 atm. What is the approximate maximum temperature that can be reached inside this pressure cooker? Explain your reasoning.
57.
From the phase diagram for carbon dioxide in Figure 10.34, determine the state of CO2 at:
1. (f) −80 °C and 10 kPa
58.
Determine the phase changes that carbon dioxide undergoes as pressure is increased at a constant temperature of (a) −50 °C and (b) 50 °C. If the temperature is held at −40 °C? At 20 °C? (See the phase diagram in Figure 10.34.)
59.
Consider a cylinder containing a mixture of liquid carbon dioxide in equilibrium with gaseous carbon dioxide at an initial pressure of 65 atm and a temperature of 20 °C. Sketch a plot depicting the change in the cylinder pressure with time as gaseous carbon dioxide is released at constant temperature.
60.
Dry ice, CO2(s), does not melt at atmospheric pressure. It sublimes at a temperature of −78 °C. What is the lowest pressure at which CO2(s) will melt to give CO2(l)? At approximately what temperature will this occur? (See Figure 10.34 for the phase diagram.)
61.
If a severe storm results in the loss of electricity, it may be necessary to use a clothesline to dry laundry. In many parts of the country in the dead of winter, the clothes will quickly freeze when they are hung on the line. If it does not snow, will they dry anyway? Explain your answer.
62.
Is it possible to liquefy nitrogen at room temperature (about 25 °C)? Is it possible to liquefy sulfur dioxide at room temperature? Explain your answers.
63.
Elemental carbon has one gas phase, one liquid phase, and two different solid phases, as shown in the phase diagram:
1. (f) If the temperature of a sample of carbon increases from 3000 K to 5000 K at a constant pressure of 106 Pa, which phase transition occurs, if any?
64.
What types of liquids typically form amorphous solids?
65.
At very low temperatures oxygen, O2, freezes and forms a crystalline solid. Which best describes these crystals?
1. ionic
2. covalent network
3. metallic
4. amorphous
5. molecular crystals
66.
As it cools, olive oil slowly solidifies and forms a solid over a range of temperatures. Which best describes the solid?
1. ionic
2. covalent network
3. metallic
4. amorphous
5. molecular crystals
67.
Explain why ice, which is a crystalline solid, has a melting temperature of 0 °C, whereas butter, which is an amorphous solid, softens over a range of temperatures.
68.
Identify the type of crystalline solid (metallic, network covalent, ionic, or molecular) formed by each of the following substances:
1. (i) C2H5OH
69.
Identify the type of crystalline solid (metallic, network covalent, ionic, or molecular) formed by each of the following substances:
1. (i) K3PO4
70.
Classify each substance in the table as either a metallic, ionic, molecular, or covalent network solid:
Substance Appearance Melting Point Electrical Conductivity Solubility in Water
X lustrous, malleable 1500 °C high insoluble
Y soft, yellow 113 °C none insoluble
Z hard, white 800 °C only if melted/dissolved soluble
71.
Classify each substance in the table as either a metallic, ionic, molecular, or covalent network solid:
Substance Appearance Melting Point Electrical Conductivity Solubility in Water
X brittle, white 800 °C only if melted/dissolved soluble
Y shiny, malleable 1100 °C high insoluble
Z hard, colorless 3550 °C none insoluble
72.
Identify the following substances as ionic, metallic, covalent network, or molecular solids:
Substance A is malleable, ductile, conducts electricity well, and has a melting point of 1135 °C. Substance B is brittle, does not conduct electricity as a solid but does when molten, and has a melting point of 2072 °C. Substance C is very hard, does not conduct electricity, and has a melting point of 3440 °C. Substance D is soft, does not conduct electricity, and has a melting point of 185 °C.
73.
Substance A is shiny, conducts electricity well, and melts at 975 °C. Substance A is likely a(n):
1. ionic solid
2. metallic solid
3. molecular solid
4. covalent network solid
74.
Substance B is hard, does not conduct electricity, and melts at 1200 °C. Substance B is likely a(n):
1. ionic solid
2. metallic solid
3. molecular solid
4. covalent network solid
75.
Describe the crystal structure of iron, which crystallizes with two equivalent metal atoms in a cubic unit cell.
76.
Describe the crystal structure of Pt, which crystallizes with four equivalent metal atoms in a cubic unit cell.
77.
What is the coordination number of a chromium atom in the body-centered cubic structure of chromium?
78.
What is the coordination number of an aluminum atom in the face-centered cubic structure of aluminum?
79.
Cobalt metal crystallizes in a hexagonal closest packed structure. What is the coordination number of a cobalt atom?
80.
Nickel metal crystallizes in a cubic closest packed structure. What is the coordination number of a nickel atom?
81.
Tungsten crystallizes in a body-centered cubic unit cell with an edge length of 3.165 Å.
1. What is the atomic radius of tungsten in this structure?
2. Calculate the density of tungsten.
82.
Platinum (atomic radius = 1.38 Å) crystallizes in a cubic closely packed structure. Calculate the edge length of the face-centered cubic unit cell and the density of platinum.
83.
Barium crystallizes in a body-centered cubic unit cell with an edge length of 5.025 Å
1. What is the atomic radius of barium in this structure?
2. Calculate the density of barium.
84.
Aluminum (atomic radius = 1.43 Å) crystallizes in a cubic closely packed structure. Calculate the edge length of the face-centered cubic unit cell and the density of aluminum.
85.
The density of aluminum is 2.7 g/cm3; that of silicon is 2.3 g/cm3. Explain why Si has the lower density even though it has heavier atoms.
86.
The free space in a metal may be found by subtracting the volume of the atoms in a unit cell from the volume of the cell. Calculate the percentage of free space in each of the three cubic lattices if all atoms in each are of equal size and touch their nearest neighbors. Which of these structures represents the most efficient packing? That is, which packs with the least amount of unused space?
87.
Cadmium sulfide, sometimes used as a yellow pigment by artists, crystallizes with cadmium, occupying one-half of the tetrahedral holes in a closest packed array of sulfide ions. What is the formula of cadmium sulfide? Explain your answer.
88.
A compound of cadmium, tin, and phosphorus is used in the fabrication of some semiconductors. It crystallizes with cadmium occupying one-fourth of the tetrahedral holes and tin occupying one-fourth of the tetrahedral holes in a closest packed array of phosphide ions. What is the formula of the compound? Explain your answer.
89.
What is the formula of the magnetic oxide of cobalt, used in recording tapes, that crystallizes with cobalt atoms occupying one-eighth of the tetrahedral holes and one-half of the octahedral holes in a closely packed array of oxide ions?
90.
A compound containing zinc, aluminum, and sulfur crystallizes with a closest-packed array of sulfide ions. Zinc ions are found in one-eighth of the tetrahedral holes and aluminum ions in one-half of the octahedral holes. What is the empirical formula of the compound?
91.
A compound of thallium and iodine crystallizes in a simple cubic array of iodide ions with thallium ions in all of the cubic holes. What is the formula of this iodide? Explain your answer.
92.
Which of the following elements reacts with sulfur to form a solid in which the sulfur atoms form a closest-packed array with all of the octahedral holes occupied: Li, Na, Be, Ca, or Al?
93.
What is the percent by mass of titanium in rutile, a mineral that contains titanium and oxygen, if structure can be described as a closest packed array of oxide ions with titanium ions in one-half of the octahedral holes? What is the oxidation number of titanium?
94.
Explain why the chemically similar alkali metal chlorides NaCl and CsCl have different structures, whereas the chemically different NaCl and MnS have the same structure.
95.
As minerals were formed from the molten magma, different ions occupied the same cites in the crystals. Lithium often occurs along with magnesium in minerals despite the difference in the charge on their ions. Suggest an explanation.
96.
Rubidium iodide crystallizes with a cubic unit cell that contains iodide ions at the corners and a rubidium ion in the center. What is the formula of the compound?
97.
One of the various manganese oxides crystallizes with a cubic unit cell that contains manganese ions at the corners and in the center. Oxide ions are located at the center of each edge of the unit cell. What is the formula of the compound?
98.
NaH crystallizes with the same crystal structure as NaCl. The edge length of the cubic unit cell of NaH is 4.880 Å.
1. Calculate the ionic radius of H. (The ionic radius of Li+ is 0.0.95 Å.)
2. Calculate the density of NaH.
99.
Thallium(I) iodide crystallizes with the same structure as CsCl. The edge length of the unit cell of TlI is 4.20 Å. Calculate the ionic radius of TI+. (The ionic radius of I is 2.16 Å.)
100.
A cubic unit cell contains manganese ions at the corners and fluoride ions at the center of each edge.
1. What is the empirical formula of this compound? Explain your answer.
2. What is the coordination number of the Mn3+ ion?
3. Calculate the edge length of the unit cell if the radius of a Mn3+ ion is 0.65 A.
4. Calculate the density of the compound.
101.
What is the spacing between crystal planes that diffract X-rays with a wavelength of 1.541 nm at an angle θ of 15.55° (first order reflection)?
102.
A diffractometer using X-rays with a wavelength of 0.2287 nm produced first order diffraction peak for a crystal angle θ = 16.21°. Determine the spacing between the diffracting planes in this crystal.
103.
A metal with spacing between planes equal to 0.4164 nm diffracts X-rays with a wavelength of 0.2879 nm. What is the diffraction angle for the first order diffraction peak?
104.
Gold crystallizes in a face-centered cubic unit cell. The second-order reflection (n = 2) of X-rays for the planes that make up the tops and bottoms of the unit cells is at θ = 22.20°. The wavelength of the X-rays is 1.54 Å. What is the density of metallic gold?
105.
When an electron in an excited molybdenum atom falls from the L to the K shell, an X-ray is emitted. These X-rays are diffracted at an angle of 7.75° by planes with a separation of 2.64 Å. What is the difference in energy between the K shell and the L shell in molybdenum assuming a first order diffraction? | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/10%3A_Liquids_and_Solids/10.10%3A_Exercises.txt |
In this chapter, we will consider the nature of solutions, and examine factors that determine whether a solution will form and what properties it may have. In addition, we will discuss colloids—systems that resemble solutions but consist of dispersions of particles somewhat larger than ordinary molecules or ions.
• 11.0: Introduction
Solutions are crucial to the processes that sustain life and to many other processes involving chemical reactions. In this chapter, we will consider the nature of solutions, and examine factors that determine whether a solution will form and what properties it may have. In addition, we will discuss colloids—systems that resemble solutions but consist of dispersions of particles somewhat larger than ordinary molecules or ions.
• 11.1: The Dissolution Process
A solution forms when two or more substances combine physically to yield a mixture that is homogeneous at the molecular level. The solvent is the most concentrated component and determines the physical state of the solution. The solutes are the other components typically present at concentrations less than that of the solvent. Solutions may form endothermically or exothermically, depending upon the relative magnitudes of solute and solvent intermolecular attractive forces.
• 11.2: Electrolytes
Substances that dissolve in water to yield ions are called electrolytes. Electrolytes may be covalent compounds that chemically react with water to produce ions (for example, acids and bases), or they may be ionic compounds that dissociate to yield their constituent cations and anions, when dissolved. Dissolution of an ionic compound is facilitated by ion-dipole attractions between the ions of the compound and the polar water molecules.
• 11.3: Solubility
The extent to which one substance will dissolve in another is determined by several factors, including the types and relative strengths of intermolecular attractive forces that may exist between the substances’ atoms, ions, or molecules. This tendency to dissolve is quantified as substance’s solubility, its maximum concentration in a solution at equilibrium under specified conditions. A saturated solution contains solute at a concentration equal to its solubility.
• 11.4: Colligative Properties
Properties of a solution that depend only on the concentration of solute particles are called colligative properties. They include changes in the vapor pressure, boiling point, and freezing point of the solvent in the solution. The magnitudes of these properties depend only on the total concentration of solute particles in solution, not on the type of particles. The total concentration of solute particles in a solution also determines its osmotic pressure.
• 11.5: Colloids
Colloids are mixtures in which one or more substances are dispersed as relatively large solid particles or liquid droplets throughout a solid, liquid, or gaseous medium. The particles of a colloid remain dispersed and do not settle due to gravity, and they are often electrically charged. Colloids are widespread in nature and are involved in many technological applications.
• 11.6: Key Terms
• 11.7: Key Equations
• 11.8: Summary
• 11.9: Exercises
These are homework exercises to accompany the Textmap created for "Chemistry" by OpenStax. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
Thumbnail: Nile red solution. (CC BY-SA 3.0; Armin Kübelbeck).
11: Solutions and Colloids
Coral reefs are home to about 25% of all marine species. They are being threatened by climate change, oceanic acidification, and water pollution, all of which change the composition of the solution known as seawater. Dissolved oxygen in seawater is critical for sea creatures, but as the oceans warm, oxygen becomes less soluble. As the concentration of carbon dioxide in the atmosphere increases, the concentration of carbon dioxide in the oceans increases, contributing to oceanic acidification. Coral reefs are particularly sensitive to the acidification of the ocean, since the exoskeletons of the coral polyps are soluble in acidic solutions. Humans contribute to the changing of seawater composition by allowing agricultural runoff and other forms of pollution to affect our oceans.
Solutions are crucial to the processes that sustain life and to many other processes involving chemical reactions. This chapter considers the nature of solutions and examines factors that determine whether a solution will form and what properties it may have. The properties of colloids—mixtures containing dispersed particles larger than the molecules and ions of typical solutions—are also discussed. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/11%3A_Solutions_and_Colloids/11.00%3A_Introduction.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the basic properties of solutions and how they form
• Predict whether a given mixture will yield a solution based on molecular properties of its components
• Explain why some solutions either produce or absorb heat when they form
An earlier chapter of this text introduced solutions, defined as homogeneous mixtures of two or more substances. Often, one component of a solution is present at a significantly greater concentration, in which case it is called the solvent. The other components of the solution present in relatively lesser concentrations are called solutes. Sugar is a covalent solid composed of sucrose molecules, $\ce{C12H22O11}$. When this compound dissolves in water, its molecules become uniformly distributed among the molecules of water:
$\ce{C_{12}H_{22}O_{11}(s) \longrightarrow C_{12}H_{22}O_{11}(aq)} \nonumber$
The subscript “aq” in the equation signifies that the sucrose molecules are solutes and are therefore individually dispersed throughout the aqueous solution (water is the solvent). Although sucrose molecules are heavier than water molecules, they remain dispersed throughout the solution; gravity does not cause them to “settle out” over time.
Potassium dichromate, $\ce{K2Cr2O7}$, is an ionic compound composed of colorless potassium ions, $\ce{K^{+}}$, and orange dichromate ions, $\ce{Cr2O7^{2−}}$. When a small amount of solid potassium dichromate is added to water, the compound dissolves and dissociates to yield potassium ions and dichromate ions uniformly distributed throughout the mixture (Figure $1$), as indicated in this equation:
$\ce{K_2Cr_2O_7(s) -> 2 K^{+}(aq) + Cr2O7^{2-}(aq)} \nonumber$
As with the mixture of sugar and water, this mixture is also an aqueous solution. Its solutes, potassium and dichromate ions, remain individually dispersed among the solvent (water) molecules.
Link to Learning
Visit this virtual lab to view simulations of the dissolution of common covalent and ionic substances (sugar and salt) in water.
Water is used so often as a solvent that the word solution has come to imply an aqueous solution to many people. However, almost any gas, liquid, or solid can act as a solvent. Many alloys are solid solutions of one metal dissolved in another; for example, US five-cent coins contain nickel dissolved in copper. Air is a gaseous solution, a homogeneous mixture of nitrogen, oxygen, and several other gases. Oxygen (a gas), alcohol (a liquid), and sugar (a solid) all dissolve in water (a liquid) to form liquid solutions. Table $1$ gives examples of several different solutions and the phases of the solutes and solvents.
Table $1$: Different Types of Solutions
Solution Solute Solvent
air O2(g) N2(g)
soft drinks1 CO2(g) H2O(l)
hydrogen in palladium H2(g) Pd(s)
rubbing alcohol H2O(l) C3H8O(l) (2-propanol)
saltwater NaCl(s) H2O(l)
brass Zn(s) Cu(s)
Solutions exhibit these defining traits:
• They are homogeneous; after a solution is mixed, it has the same composition at all points throughout (its composition is uniform).
• The physical state of a solution—solid, liquid, or gas—is typically the same as that of the solvent, as demonstrated by the examples in Table $1$.
• The components of a solution are dispersed on a molecular scale; they consist of a mixture of separated solute particles (molecules, atoms, and/or ions) each closely surrounded by solvent species.
• The dissolved solute in a solution will not settle out or separate from the solvent.
• The composition of a solution, or the concentrations of its components, can be varied continuously (within limits determined by the solubility of the components, discussed in detail later in this chapter).
The Formation of Solutions
The formation of a solution is an example of a spontaneous process, a process that occurs under specified conditions without the requirement of energy from some external source. Sometimes a mixture is stirred to speed up the dissolution process, but this is not necessary; a homogeneous solution will form eventually. The topic of spontaneity is critically important to the study of chemical thermodynamics and is treated more thoroughly in a later chapter of this text. For purposes of this chapter’s discussion, it will suffice to consider two criteria that favor, but do not guarantee, the spontaneous formation of a solution:
1. a decrease in the internal energy of the system (an exothermic change, as discussed in the previous chapter on thermochemistry)
2. an increased dispersal of matter in the system (which indicates an increase in the entropy of the system, as you will learn about in the later chapter on thermodynamics)
In the process of dissolution, an internal energy change often, but not always, occurs as heat is absorbed or evolved. An increase in matter dispersal always results when a solution forms from the uniform distribution of solute molecules throughout a solvent.
When the strengths of the intermolecular forces of attraction between solute and solvent species in a solution are no different than those present in the separated components, the solution is formed with no accompanying energy change. Such a solution is called an ideal solution. A mixture of ideal gases (or gases such as helium and argon, which closely approach ideal behavior) is an example of an ideal solution, since the entities comprising these gases experience no significant intermolecular attractions.
When containers of helium and argon are connected, the gases spontaneously mix due to diffusion and form a solution (Figure $2$). The formation of this solution clearly involves an increase in matter dispersal, since the helium and argon atoms occupy a volume twice as large as that which each occupied before mixing.
Ideal solutions may also form when structurally similar liquids are mixed. For example, mixtures of the alcohols methanol (CH3OH) and ethanol (C2H5OH) form ideal solutions, as do mixtures of the hydrocarbons pentane, C5H12, and hexane, C6H14. Placing methanol and ethanol, or pentane and hexane, in the bulbs shown in Figure $2$: will result in the same diffusion and subsequent mixing of these liquids as is observed for the He and Ar gases (although at a much slower rate), yielding solutions with no significant change in energy. Unlike a mixture of gases, however, the components of these liquid-liquid solutions do, indeed, experience intermolecular attractive forces. But since the molecules of the two substances being mixed are structurally very similar, the intermolecular attractive forces between like and unlike molecules are essentially the same, and the dissolution process, therefore, does not entail any appreciable increase or decrease in energy. These examples illustrate how increased matter dispersal alone can provide the driving force required to cause the spontaneous formation of a solution. In some cases, however, the relative magnitudes of intermolecular forces of attraction between solute and solvent species may prevent dissolution.
Three types of intermolecular attractive forces are relevant to the dissolution process: solute-solute, solvent-solvent, and solute-solvent. As illustrated in Figure $3$: , the formation of a solution may be viewed as a stepwise process in which energy is consumed to overcome solute-solute and solvent-solvent attractions (endothermic processes) and released when solute-solvent attractions are established (an exothermic process referred to as solvation). The relative magnitudes of the energy changes associated with these stepwise processes determine whether the dissolution process overall will release or absorb energy. In some cases, solutions do not form because the energy required to separate solute and solvent species is so much greater than the energy released by solvation.
Consider the example of an ionic compound dissolving in water. Formation of the solution requires the electrostatic forces between the cations and anions of the compound (solute–solute) be overcome completely as attractive forces are established between these ions and water molecules (solute–solvent). Hydrogen bonding between a relatively small fraction of the water molecules must also be overcome to accommodate any dissolved solute. If the solute’s electrostatic forces are significantly greater than the solvation forces, the dissolution process is significantly endothermic and the compound may not dissolve to an appreciable extent. Calcium carbonate, the major component of coral reefs, is one example of such an “insoluble” ionic compound. On the other hand, if the solvation forces are much stronger than the compound’s electrostatic forces, the dissolution is significantly exothermic and the compound may be highly soluble. common example of this type of ionic compound is sodium hydroxide, commonly known as lye.
As noted at the beginning of this module, spontaneous solution formation is favored, but not guaranteed, by exothermic dissolution processes. While many soluble compounds do, indeed, dissolve with the release of heat, some dissolve endothermically. Ammonium nitrate (NH4NO3) is one such example and is used to make instant cold packs, like the one pictured in Figure $4$, which are used for treating injuries. A thin-walled plastic bag of water is sealed inside a larger bag with solid NH4NO3. When the smaller bag is broken, a solution of NH4NO3 forms, absorbing heat from the surroundings (the injured area to which the pack is applied) and providing a cold compress that decreases swelling. Endothermic dissolutions such as this one require a greater energy input to separate the solute species than is recovered when the solutes are solvated, but they are spontaneous nonetheless due to the increase in disorder that accompanies formation of the solution.
Link to Learning
Watch this brief video illustrating endothermic and exothermic dissolution processes.
Footnotes
• 1If bubbles of gas are observed within the liquid, the mixture is not homogeneous and, thus, not a solution. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/11%3A_Solutions_and_Colloids/11.01%3A_The_Dissolution_Process.txt |
Learning Objectives
By the end of this section, you will be able to:
• Define and give examples of electrolytes
• Distinguish between the physical and chemical changes that accompany dissolution of ionic and covalent electrolytes
• Relate electrolyte strength to solute-solvent attractive forces
When some substances are dissolved in water, they undergo either a physical or a chemical change that yields ions in solution. These substances constitute an important class of compounds called electrolytes. Substances that do not yield ions when dissolved are called nonelectrolytes. If the physical or chemical process that generates the ions is essentially 100% efficient (all of the dissolved compound yields ions), then the substance is known as a strong electrolyte. If only a relatively small fraction of the dissolved substance undergoes the ion-producing process, it is called a weak electrolyte.
Substances may be identified as strong, weak, or nonelectrolytes by measuring the electrical conductance of an aqueous solution containing the substance. To conduct electricity, a substance must contain freely mobile, charged species. Most familiar is the conduction of electricity through metallic wires, in which case the mobile, charged entities are electrons. Solutions may also conduct electricity if they contain dissolved ions, with conductivity increasing as ion concentration increases. Applying a voltage to electrodes immersed in a solution permits assessment of the relative concentration of dissolved ions, either quantitatively, by measuring the electrical current flow, or qualitatively, by observing the brightness of a light bulb included in the circuit (Figure $1$).
Ionic Electrolytes
Water and other polar molecules are attracted to ions, as shown in Figure $2$. The electrostatic attraction between an ion and a molecule with a dipole is called an ion-dipole attraction. These attractions play an important role in the dissolution of ionic compounds in water.
When ionic compounds dissolve in water, the ions in the solid separate and disperse uniformly throughout the solution because water molecules surround and solvate the ions, reducing the strong electrostatic forces between them. This process represents a physical change known as dissociation. Under most conditions, ionic compounds will dissociate nearly completely when dissolved, and so they are classified as strong electrolytes. Even sparingly, soluble ionic compounds are strong electrolytes, since the small amount that does dissolve will dissociate completely.
Consider what happens at the microscopic level when solid KCl is added to water. Ion-dipole forces attract the positive (hydrogen) end of the polar water molecules to the negative chloride ions at the surface of the solid, and they attract the negative (oxygen) ends to the positive potassium ions. The water molecules surround individual K+ and Cl ions, reducing the strong interionic forces that bind the ions together and letting them move off into solution as solvated ions, as Figure $2$ shows. Overcoming the electrostatic attraction permits the independent motion of each hydrated ion in a dilute solution as the ions transition from fixed positions in the undissolved compound to widely dispersed, solvated ions in solution.
Covalent Electrolytes
Pure water is an extremely poor conductor of electricity because it is only very slightly ionized—only about two out of every 1 billion molecules ionize at 25 °C. Water ionizes when one molecule of water gives up a proton (H+ ion) to another molecule of water, yielding hydronium and hydroxide ions.
$\ce{H2O(l) + H2O(l) \rightleftharpoons H3O^{+}(aq) + OH^{-}(aq)} \nonumber$
In some cases, solutions prepared from covalent compounds conduct electricity because the solute molecules react chemically with the solvent to produce ions. For example, pure hydrogen chloride is a gas consisting of covalent HCl molecules. This gas contains no ions. However, an aqueous solution of HCl is a very good conductor, indicating that an appreciable concentration of ions exists within the solution.
Because HCl is an acid, its molecules react with water, transferring H+ ions to form hydronium ions (H3O+) and chloride ions (Cl):
This reaction is essentially 100% complete for HCl (i.e., it is a strong acid and, consequently, a strong electrolyte). Likewise, weak acids and bases that only react partially generate relatively low concentrations of ions when dissolved in water and are classified as weak electrolytes. The reader may wish to review the discussion of strong and weak acids provided in the earlier chapter of this text on reaction classes and stoichiometry. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/11%3A_Solutions_and_Colloids/11.02%3A_Electrolytes.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the effects of temperature and pressure on solubility
• State Henry’s law and use it in calculations involving the solubility of a gas in a liquid
• Explain the degrees of solubility possible for liquid-liquid solutions
Imagine adding a small amount of sugar to a glass of water, stirring until all the sugar has dissolved, and then adding a bit more. You can repeat this process until the sugar concentration of the solution reaches its natural limit, a limit determined primarily by the relative strengths of the solute-solute, solute-solvent, and solvent-solvent attractive forces discussed in the previous two modules of this chapter. You can be certain that you have reached this limit because, no matter how long you stir the solution, undissolved sugar remains. The concentration of sugar in the solution at this point is known as its solubility.
The solubility of a solute in a particular solvent is the maximum concentration that may be achieved under given conditions when the dissolution process is at equilibrium.
When a solute’s concentration is equal to its solubility, the solution is said to be saturated with that solute. If the solute’s concentration is less than its solubility, the solution is said to be unsaturated. A solution that contains a relatively low concentration of solute is called dilute, and one with a relatively high concentration is called concentrated.
Link to Learning
Use this interactive simulation to prepare various saturated solutions.
Solutions may be prepared in which a solute concentration exceeds its solubility. Such solutions are said to be supersaturated, and they are interesting examples of nonequilibrium states (a detailed treatment of this important concept is provided in the text chapters on equilibrium). For example, the carbonated beverage in an open container that has not yet “gone flat” is supersaturated with carbon dioxide gas; given time, the CO2 concentration will decrease until it reaches its solubility.
Link to Learning
Watch this impressive video showing the precipitation of sodium acetate from a supersaturated solution.
Solutions of Gases in Liquids
As for any solution, the solubility of a gas in a liquid is affected by the intermolecular attractive forces between solute and solvent species. Unlike solid and liquid solutes, however, there is no solute-solute intermolecular attraction to overcome when a gaseous solute dissolves in a liquid solvent (see Figure 11.4) since the atoms or molecules comprising a gas are far separated and experience negligible interactions. Consequently, solute-solvent interactions are the sole energetic factor affecting solubility. For example, the water solubility of oxygen is approximately three times greater than that of helium (there are greater dispersion forces between water and the larger oxygen molecules) but 100 times less than the solubility of chloromethane, CHCl3 (polar chloromethane molecules experience dipole–dipole attraction to polar water molecules). Likewise note the solubility of oxygen in hexane, C6H14, is approximately 20 times greater than it is in water because greater dispersion forces exist between oxygen and the larger hexane molecules.
Temperature is another factor affecting solubility, with gas solubility typically decreasing as temperature increases (Figure $1$). This inverse relation between temperature and dissolved gas concentration is responsible for one of the major impacts of thermal pollution in natural waters.
When the temperature of a river, lake, or stream is raised, 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, $C_g$, and partial pressure, $P_g$, is a proportional one:
$C_{ g }=k P_{ g } \nonumber$
where k is a proportionality constant that depends on the identity of the gaseous solute, the identity of the solvent, and 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 is 1.38 \times 10^{−3} mol L−1. Use Henry’s law to determine the solubility of oxygen when its partial pressure is 20.7 kPa, 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 \times 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 in this case). Because both Cg and Pg are known, this relation can be rearranged and used to solve for k.
\begin{align*} C_{ g } & =k P_{ g } \[4pt] k & =\frac{C_{ g }}{P_{ g }} \[4pt] & =\frac{1.38 \times 10^{-3} mol L^{-1}}{101.3 kPa } \[4pt] & =1.36 \times 10^{-5} mol L^{-1} kPa^{-1} \end{align*}
Now, use $k$ to find the solubility at the lower pressure.
\begin{align*} C_{ g }= & k P_{ g } \[4pt] & 1.36 \times 10^{-5} mol L^{-1} kPa^{-1} \times 20.7 kPa \[4pt] = & 2.82 \times 10^{-4} mol L^{-1} \end{align*}
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 152 torr resulted in the dissolution of 1.45 \times 10^{−3} g of the solute. Use Henry’s law to determine the solubility of this gaseous solute when its pressure is 760 torr.
Answer
7.25 \times 10^{−3} in 100.0 mL or 0.0725 g/L
Example $2$: Thermal Pollution and Oxygen Solubility
A certain species of freshwater trout requires a dissolved oxygen concentration of 7.5 mg/L. Could these fish thrive in a thermally polluted mountain stream (water temperature is 30.0 °C, partial pressure of atmospheric oxygen is 0.17 atm)? Use the data in Figure $1$ to estimate a value for the Henry's law constant at this temperature.
Solution
First, estimate the Henry’s law constant for oxygen in water at the specified temperature of 30.0 °C (Figure $1$ indicates the solubility at this temperature is approximately ~1.2 mol/L).
$k=\frac{C_{ g }}{P_{ g }}=1.2 \times 10^{-3} mol / L / 1.00 atm =1.2 \times 10^{-3} mol / L atm] Then, use this $k$ value to compute the oxygen solubility at the specified oxygen partial pressure, 0.17 atm. \[C_g=k P_g=\left(1.2 \times 10^{-3} mol / L atm \right)(0.17 atm )=2.0 \times 10^{-4} mol / L \nonumber$
Finally, convert this dissolved oxygen concentration from mol/L to mg/L.
$\left(2.0 \times 10^{-4} mol / L \right)(32.0 g / 1 mol )(1000 mg / g )=6.4 mg / L \nonumber$
This concentration is lesser than the required minimum value of 7.5 mg/L, and so these trout would likely not thrive in the polluted stream.
Exercise $2$
What dissolved oxygen concentration is expected for the stream above when it returns to a normal summer time temperature of 15 °C?
Answer
8.2 mg/L
Chemistry in Everyday Life: Decompression Sickness or “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 $4$). Researchers are also investigating related body reactions and defenses in order to develop better testing and treatment for decompression sickness. For example, Ingrid Eftedal, a barophysiologist specializing in bodily reactions to diving, has shown that white blood cells undergo chemical and genetic changes as a result of the condition; these can potentially be used to create biomarker tests and other methods to manage decompression sickness.
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 increases more rapidly with increasing pressure than 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.
Solutions of Liquids in Liquids
Some liquids may be mixed in any proportions to yield solutions; in other words, they have infinite mutual solubility and are said to be miscible. Ethanol, sulfuric acid, and ethylene glycol (popular for use as antifreeze, pictured in Figure $6$) are examples of liquids that are completely miscible with water. Two-cycle motor oil is miscible with gasoline, mixtures of which are used as lubricating fuels for various types of outdoor power equipment (chainsaws, leaf blowers, and so on).
Miscible liquids are typically those with very similar polarities. Consider, for example, liquids that are polar or capable of hydrogen bonding. For such liquids, the dipole-dipole attractions (or hydrogen bonding) of the solute molecules with the solvent molecules are at least as strong as those between molecules in the pure solute or in the pure solvent. Hence, the two kinds of molecules mix easily. Likewise, nonpolar liquids are miscible with each other because there is no appreciable difference in the strengths of solute-solute, solvent-solvent, and solute-solvent intermolecular attractions. The solubility of polar molecules in polar solvents and of nonpolar molecules in nonpolar solvents is, again, an illustration of the chemical axiom “like dissolves like.”
Two liquids that do not mix to an appreciable extent are called immiscible. Separate layers are formed when immiscible liquids are poured into the same container. Gasoline, oil (Figure $7$), benzene, carbon tetrachloride, some paints, and many other nonpolar liquids are immiscible with water. Relatively weak attractive forces between the polar water molecules and the nonpolar liquid molecules are not adequate to overcome much stronger hydrogen bonding between water molecules. The distinction between immiscibility and miscibility is really one of extent, so that miscible liquids are of infinite mutual solubility, while liquids said to be immiscible are of very low (though not zero) mutual solubility.
Two liquids, such as bromine and water, that are of moderate mutual solubility are said to be partially miscible. Two partially miscible liquids usually form two layers when mixed. In the case of the bromine and water mixture, the upper layer is water, saturated with bromine, and the lower layer is bromine saturated with water. Since bromine is nonpolar, and, thus, not very soluble in water, the water layer is only slightly discolored by the bright orange bromine dissolved in it. Since the solubility of water in bromine is very low, there is no noticeable effect on the dark color of the bromine layer (Figure $8$).
Solutions of Solids in Liquids
The dependence of solubility on temperature for a number of solids in water is shown by the solubility curves in Figure $9$. Reviewing these data indicates a general trend of increasing solubility with temperature, although there are exceptions, as illustrated by the ionic compound cerium sulfate.
The temperature dependence of solubility can be exploited to prepare supersaturated solutions of certain compounds. A solution may be saturated with the compound at an elevated temperature (where the solute is more soluble) and subsequently cooled to a lower temperature without precipitating the solute. The resultant solution contains solute at a concentration greater than its equilibrium solubility at the lower temperature (i.e., it is supersaturated) and is relatively stable. Precipitation of the excess solute can be initiated by adding a seed crystal (see the video in the Link to Learning earlier in this module) or by mechanically agitating the solution. Some hand warmers, such as the one pictured in Figure $10$, take advantage of this behavior.
Link to Learning
This video shows the crystallization process occurring in a hand warmer. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/11%3A_Solutions_and_Colloids/11.03%3A_Solubility.txt |
Learning Objectives
By the end of this section, you will be able to:
• Express concentrations of solution components using mole fraction and molality
• Describe the effect of solute concentration on various solution properties (vapor pressure, boiling point, freezing point, and osmotic pressure)
• Perform calculations using the mathematical equations that describe these various colligative effects
• Describe the process of distillation and its practical applications
• Explain the process of osmosis and describe how it is applied industrially and in nature
The properties of a solution are different from those of either the pure solute(s) or solvent. Many solution properties are dependent upon the chemical identity of the solute. Compared to pure water, a solution of hydrogen chloride is more acidic, a solution of ammonia is more basic, a solution of sodium chloride is more dense, and a solution of sucrose is more viscous. There are a few solution properties, however, that depend only upon the total concentration of solute species, regardless of their identities. These colligative properties include vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure. This small set of properties is of central importance to many natural phenomena and technological applications, as will be described in this module.
Mole Fraction and Molality
Several units commonly used to express the concentrations of solution components were introduced in an earlier chapter of this text, each providing certain benefits for use in different applications. For example, molarity (M) is a convenient unit for use in stoichiometric calculations, since it is defined in terms of the molar amounts of solute species:
$M=\frac{\text { mol solute }}{\text { L solution }} \nonumber$
Because solution volumes vary with temperature, molar concentrations will likewise vary. When expressed as molarity, the concentration of a solution with identical numbers of solute and solvent species will be different at different temperatures, due to the contraction/expansion of the solution. More appropriate for calculations involving many colligative properties are mole-based concentration units whose values are not dependent on temperature. Two such units are mole fraction (introduced in the previous chapter on gases) and molality.
The mole fraction, $X$, of a component is the ratio of its molar amount to the total number of moles of all solution components:
$X_{ A }=\frac{ mol A }{\text { total mol of all components }} \nonumber$
By this definition, the sum of mole fractions for all solution components (the solvent and all solutes) is equal to one.
Molality is a concentration unit defined as the ratio of the numbers of moles of solute to the mass of the solvent in kilograms:
$m=\frac{\text { mol solute }}{ kg \text { solvent }} \nonumber$
Since these units are computed using only masses and molar amounts, they do not vary with temperature and, thus, are better suited for applications requiring temperature-independent concentrations, including several colligative properties, as will be described in this chapter module.
Example $1$: Calculating Mole Fraction and Molality
The antifreeze in most automobile radiators is a mixture of equal volumes of ethylene glycol and water, with minor amounts of other additives that prevent corrosion. What are the (a) mole fraction and (b) molality of ethylene glycol, C2H4(OH)2, in a solution prepared from 2.22 \times 10^{3} g of ethylene glycol and 2.00 \times 10^{3} g of water (approximately 2 L of glycol and 2 L of water)?
Solution
(a) The mole fraction of ethylene glycol may be computed by first deriving molar amounts of both solution components and then substituting these amounts into the definition of mole fraction.
\begin{align*} mol C_2 H_4( OH )_2=2.22 \times 10^3 g \times \frac{1 mol C_2 H_4( OH )_2}{62.07 g C_2 H_4( OH )_2}=35.8 mol C_2 H_4( OH )_2 \[4pt] mol H_2 O =2.00 \times 10^3 g \times \frac{1 mol H_2 O }{18.02 g H_2 O }=111 mol H_2 O \[4pt] X_{\text {ethylene glycol }}=\frac{35.8 mol C_2 H_4( OH )_2}{(35.8+111) \text { mol total }}=0.244 \end{align*}
Notice that mole fraction is a dimensionless property, being the ratio of properties with identical units (moles).
(b) Derive moles of solute and mass of solvent (in kg).
First, use the given mass of ethylene glycol and its molar mass to find the moles of solute:
$2.22 \times 10^3 g C_2 H_4( OH )_2\left(\frac{ mol C_2 H_2( OH )_2}{62.07 g }\right)=35.8 mol C_2 H_4( OH )_2 \nonumber$
Then, convert the mass of the water from grams to kilograms:
$2.00 \times 10^3 g H_2 O \left(\frac{1 kg }{1000 g }\right)=2.00 kg H_2 O \nonumber$
Finally, calculate molality per its definition:
\begin{align*} \text { molality }&=\frac{\text { mol solute }}{\text { kg solvent }} \[4pt] & =\frac{35.8 mol C_2 H_4( OH )_2}{2 kg H_2 O } \[4pt] &=17.9 m \end{align*}
Exercise $1$
What are the mole fraction and molality of a solution that contains 0.850 g of ammonia, NH3, dissolved in 125 g of water?
Answer
7.14 \times 10^{−3}; 0.399 m
Example $2$: Converting Mole Fraction and Molal Concentrations
Calculate the mole fraction of solute and solvent in a 3.0 m solution of sodium chloride.
Solution
Converting from one concentration unit to another is accomplished by first comparing the two unit definitions. In this case, both units have the same numerator (moles of solute) but different denominators. The provided molal concentration may be written as:
$\frac{3.0 mol NaCl }{1 kg H { }_2 O } \nonumber$
The numerator for this solution’s mole fraction is, therefore, 3.0 mol NaCl. The denominator may be computed by deriving the molar amount of water corresponding to 1.0 kg
$1.0 kg H_2 O \left(\frac{1000 g }{1 kg }\right)\left(\frac{ mol H_2 O }{18.02 g }\right)=55 mol H_2 O \nonumber$
and then substituting these molar amounts into the definition for mole fraction.
\begin{align*} X_{ H_2 O } & =\frac{ mol H_2 O }{ mol NaCl + mol H_2 O } \[4pt] & =\frac{55 mol H_2 O }{3.0 mol NaCl +55 mol H_2 O } \[4pt] & =0.95 \[4pt] & =\frac{ mol NaCl }{ mol NaCl + mol H H_2 O } \[4pt] & =\frac{3.0 mol NaCl }{3.0 mol NaCl +55 mol H_2 O } \[4pt] & =0.052 \end{align*}
Exercise $2$
The mole fraction of iodine, I2, dissolved in dichloromethane, CH2Cl2, is 0.115. What is the molal concentration, m, of iodine in this solution?
Answer
1.50 m
Example $3$: Molality and Molarity Conversions
Intravenous infusion of a 0.556 M aqueous solution of glucose (density of 1.04 g/mL) is part of some post-operative recovery therapies. What is the molal concentration of glucose in this solution?
Solution
The provided molal concentration may be explicitly written as:
$M=0.556 mol \text { glucose } / 1 L \text { solution } \nonumber$
Consider the definition of molality:
$m=\text { mol solute } / kg \text { solvent } \nonumber$
The amount of glucose in 1-L of this solution is 0.556 mol, so the mass of water in this volume of solution is needed.
First, compute the mass of 1.00 L of the solution:
$(1.0 L \text { soln })(1.04 g / mL )(1000 mL / 1 L )(1 kg / 1000 g )=1.04 kg \text { soln } \nonumber$
This is the mass of both the water and its solute, glucose, and so the mass of glucose must be subtracted. Compute the mass of glucose from its molar amount:
$(0.556 mol \text { glucose })(180.2 g / 1 mol )=100.2 g \text { or } 0.1002 kg \nonumber$
Subtracting the mass of glucose yields the mass of water in the solution:
$1.04 kg \text { solution }-0.1002 kg \text { glucose }=0.94 kg \text { water } \nonumber$
Finally, the molality of glucose in this solution is computed as:
$m=\dfrac{0.556 ~\text{mol glucose } }{0.94~ \text{kg water }}=0.59~ m \nonumber$
Exercise $3$
Nitric acid, $\ce{HNO3(aq)}$, is commercially available as a 33.7 m aqueous solution (density = 1.35 g/mL). What is the molarity of this solution?
Answer
14.6 M
Vapor Pressure Lowering
As described in the chapter on liquids and solids, the equilibrium vapor pressure of a liquid is the pressure exerted by its gaseous phase when vaporization and condensation are occurring at equal rates:
$\text { liquid } \rightleftharpoons \text { gas } \nonumber$
Dissolving a nonvolatile substance in a volatile liquid results in a lowering of the liquid’s vapor pressure. This phenomenon can be rationalized by considering the effect of added solute molecules on the liquid's vaporization and condensation processes. To vaporize, solvent molecules must be present at the surface of the solution. The presence of solute decreases the surface area available to solvent molecules and thereby reduces the rate of solvent vaporization. Since the rate of condensation is unaffected by the presence of solute, the net result is that the vaporization-condensation equilibrium is achieved with fewer solvent molecules in the vapor phase (i.e., at a lower vapor pressure) (Figure $1$). While this interpretation is useful, it does not account for several important aspects of the colligative nature of vapor pressure lowering. A more rigorous explanation involves the property of entropy, a topic of discussion in a later text chapter on thermodynamics. For purposes of understanding the lowering of a liquid's vapor pressure, it is adequate to note that the more dispersed nature of matter in a solution, compared to separate solvent and solute phases, serves to effectively stabilize the solvent molecules and hinder their vaporization. A lower vapor pressure results, and a correspondingly higher boiling point as described in the next section of this module.
The relationship between the vapor pressures of solution components and the concentrations of those components is described by Raoult’s law: The partial pressure exerted by any component of an ideal solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution.
$P_{ A }=X_{ A } P_{ A }^* \label{Raoult}$
where $P_A$ is the partial pressure exerted by component A in the solution, $P_A^{*}$ is the vapor pressure of pure $A$, and $X_A$ is the mole fraction of A in the solution.
Recalling that the total pressure of a gaseous mixture is equal to the sum of partial pressures for all its components (Dalton’s law of partial pressures), the total vapor pressure exerted by a solution containing i components is
$P_{\text {solution }}=\sum_i P_i=\sum_i X_i P_i^{*} \nonumber$
A nonvolatile substance is one whose vapor pressure is negligible ($P^{*} ≈ 0$), and so the vapor pressure above a solution containing only nonvolatile solutes is due only to the solvent:
$P_{\text {solution }}=X_{\text {solvent }} P_{\text {solvent }}^{*} \nonumber$
Example $4$: Calculation of a Vapor Pressure
Compute the vapor pressure of an ideal solution containing 92.1 g of glycerin, C3H5(OH)3, and 184.4 g of ethanol, C2H5OH, at 40 °C. The vapor pressure of pure ethanol is 0.178 atm at 40 °C. Glycerin is essentially nonvolatile at this temperature.
Solution
Since the solvent is the only volatile component of this solution, its vapor pressure may be computed per Raoult’s law (Equation \ref{Raoult}) as:
$P_{\text {solution }}=X_{\text {solvent }} P_{\text {solvent }}^{*} \nonumber$
First, calculate the molar amounts of each solution component using the provided mass data.
\begin{align*} & 92.1 g_3 H_5( OH )_3 \times \frac{1 mol C_3 H_5( OH )_3}{92.094 g_5 C_3 H_5( OH )_3}=1.00 mol C_3 H_5( OH )_3 \[4pt] & 184.4 g_2 C_2 H_5 OH \times \frac{1 mol C_2 H_5 OH }{46.069 g_2 CH_5 OH }=4.000 mol C_2 H_5 OH \end{align*}
Next, calculate the mole fraction of the solvent (ethanol) and use Raoult’s law to compute the solution’s vapor pressure.
\begin{align*} & X_{ C_2 H_5 OH }=\frac{4.000 mol }{(1.00 mol +4.000 mol )}=0.800 \[4pt] & P_{\text {solv }}=X_{\text {solv }} P_{\text {solv }}^*=0.800 \times 0.178 atm =0.142 atm \end{align*} \nonumber
Exercise $4$
A solution contains 5.00 g of urea, $\ce{CO(NH2)2}$ (a nonvolatile solute) and 0.100 kg of water. If the vapor pressure of pure water at 25 °C is 23.7 torr, what is the vapor pressure of the solution assuming ideal behavior?
Answer
23.4 torr
Distillation of Solutions
Solutions whose components have significantly different vapor pressures may be separated by a selective vaporization process known as distillation. Consider the simple case of a mixture of two volatile liquids, A and B, with A being the more volatile liquid. Raoult’s law can be used to show that the vapor above the solution is enriched in component A, that is, the mole fraction of A in the vapor is greater than the mole fraction of A in the liquid (see end-of-chapter Exercise 65). By appropriately heating the mixture, component A may be vaporized, condensed, and collected—effectively separating it from component B.
Distillation is widely applied in both laboratory and industrial settings, being used to refine petroleum, to isolate fermentation products, and to purify water. A typical apparatus for laboratory-scale distillations is shown in Figure $2$.
Oil refineries use large-scale fractional distillation to separate the components of crude oil. The crude oil is heated to high temperatures at the base of a tall fractionating column, vaporizing many of the components that rise within the column. As vaporized components reach adequately cool zones during their ascent, they condense and are collected. The collected liquids are simpler mixtures of hydrocarbons and other petroleum compounds that are of appropriate composition for various applications (e.g., diesel fuel, kerosene, gasoline), as depicted in Figure $3$.
Boiling Point Elevation
As described in the chapter on liquids and solids, the boiling point of a liquid is the temperature at which its vapor pressure is equal to ambient atmospheric pressure. Since the vapor pressure of a solution is lowered due to the presence of nonvolatile solutes, it stands to reason that the solution’s boiling point will subsequently be increased. Vapor pressure increases with temperature, and so a solution will require a higher temperature than will pure solvent to achieve any given vapor pressure, including one equivalent to that of the surrounding atmosphere. The increase in boiling point observed when nonvolatile solute is dissolved in a solvent, $ΔT_b$, is called boiling point elevation and is directly proportional to the molal concentration of solute species:
$\Delta T_{ b }=K_{ b } m \nonumber$
where Kb is the boiling point elevation constant, or the ebullioscopic constant and m is the molal concentration (molality) of all solute species.
Boiling point elevation constants are characteristic properties that depend on the identity of the solvent. Values of Kb for several solvents are listed in Table $1$.
Table $1$: Boiling Point Elevation and Freezing Point Depression Constants for Several Solvents
Solvent Boiling Point (°C at 1 atm) Kb (ºCm−1) Freezing Point (°C at 1 atm) Kf (ºCm−1)
water 100.0 0.512 0.0 1.86
hydrogen acetate 118.1 3.07 16.6 3.9
benzene 80.1 2.53 5.5 5.12
chloroform 61.26 3.63 −63.5 4.68
nitrobenzene 210.9 5.24 5.67 8.1
The extent to which the vapor pressure of a solvent is lowered and the boiling point is elevated depends on the total number of solute particles present in a given amount of solvent, not on the mass or size or chemical identities of the particles. A 1 m aqueous solution of sucrose (342 g/mol) and a 1 m aqueous solution of ethylene glycol (62 g/mol) will exhibit the same boiling point because each solution has one mole of solute particles (molecules) per kilogram of solvent.
Example $5$: Calculating the Boiling Point of a Solution
Assuming ideal solution behavior, what is the boiling point of a 0.33 m solution of a nonvolatile solute in benzene?
Solution
Use the equation relating boiling point elevation to solute molality to solve this problem in two steps.
Step 1: Calculate the change in boiling point.
$\Delta T_{ b }=K_{ b } m=2.53{ }^{\circ} C m^{-1} \times 0.33 m=0.83{ }^{\circ} C \nonumber$
Step 2: Add the boiling point elevation to the pure solvent’s boiling point.
$\text { Boiling temperature }=80.1{ }^{\circ} C +0.83^{\circ} C =80.9^{\circ} C \nonumber$
Exercise $1$
Assuming ideal solution behavior, what is the boiling point of the antifreeze described in Example $1$?
Answer
109.2 °C
Example $6$: The Boiling Point of an Iodine Solution
Find the boiling point of a solution of 92.1 g of iodine, I2, in 800.0 g of chloroform, CHCl3, assuming that the iodine is nonvolatile and that the solution is ideal.
Solution
A four-step approach to solving this problem is outlined below.
Step 1. Convert from grams to moles of I2 using the molar mass of I2 in the unit conversion factor.
Result: 0.363 mol
Step 2. Determine the molality of the solution from the number of moles of solute and the mass of solvent, in kilograms.
Result: 0.454 m
Step 3. Use the direct proportionality between the change in boiling point and molal concentration to determine how much the boiling point changes.
Result: 1.65 °C
Step 4. Determine the new boiling point from the boiling point of the pure solvent and the change.
Result: 62.91 °C
Check each result as a self-assessment.
Exercise $5$
What is the boiling point of a solution of 1.0 g of glycerin, C3H5(OH)3, in 47.8 g of water? Assume an ideal solution.
Answer
100.12 °C
Freezing Point Depression
Solutions freeze at lower temperatures than pure liquids. This phenomenon is exploited in “de-icing” schemes that use salt (Figure $4$), 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).
The decrease in freezing point of a dilute solution compared to that of the pure solvent, $ΔT_f$, is called the freezing point depression and is directly proportional to the molal concentration of the solute
$\Delta T_{ f }=K_{ f } m \nonumber$
where m is the molal concentration of the solute and Kf is called the freezing point depression constant (or cryoscopic constant). Just as for boiling point elevation constants, these are characteristic properties whose values depend on the chemical identity of the solvent. Values of Kf for several solvents are listed in Table $1$.
Example $7$: Calculation of the Freezing Point of a Solution
Assuming ideal solution behavior, what is the freezing point of the 0.33 m solution of a nonvolatile nonelectrolyte solute in benzene described in Example $2$?
Solution
Use the equation relating freezing point depression to solute molality to solve this problem in two steps.
Step 1: Calculate the change in freezing point.
$\Delta T_{ f }=K_{ f } m=5.12{ }^{\circ} C m^{-1} \times 0.33 m=1.7{ }^{\circ} C \nonumber$
Step 2: Subtract the freezing point change observed from the pure solvent’s freezing point.
$\text { Freezing Temperature }=5.5^{\circ} C -1.7^{\circ} C =3.8^{\circ} C \nonumber$
Exercise $1$
Assuming ideal solution behavior, what is the freezing point of a 1.85 m solution of a nonvolatile nonelectrolyte solute in nitrobenzene?
Answer
−9.3 °C
Chemistry in Everyday Life: Colligative Properties and De-Icing
Sodium chloride and its group 2 analogs calcium and magnesium chloride are often used to de-ice roadways and sidewalks, due to the fact that a solution of any one of these salts will have a freezing point lower than 0 °C, the freezing point of pure water. The group 2 metal salts are frequently mixed with the cheaper and more readily available sodium chloride (“rock salt”) for use on roads, since they tend to be somewhat less corrosive than the NaCl, and they provide a larger depression of the freezing point, since they dissociate to yield three particles per formula unit, rather than two particles like the sodium chloride.
Because these ionic compounds tend to hasten the corrosion of metal, they would not be a wise choice to use in antifreeze for the radiator in your car or to de-ice a plane prior to takeoff. For these applications, covalent compounds, such as ethylene or propylene glycol, are often used. The glycols used in radiator fluid not only lower the freezing point of the liquid, but they elevate the boiling point, making the fluid useful in both winter and summer. 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 (Figure $5$).
Phase Diagram for a Solution
The colligative effects on vapor pressure, boiling point, and freezing point described in the previous section are conveniently summarized by comparing the phase diagrams for a pure liquid and a solution derived from that liquid (Figure $6$).
The liquid-vapor curve for the solution is located beneath the corresponding curve for the solvent, depicting the vapor pressure lowering, ΔP, that results from the dissolution of nonvolatile solute. Consequently, at any given pressure, the solution’s boiling point is observed at a higher temperature than that for the pure solvent, reflecting the boiling point elevation, ΔTb, associated with the presence of nonvolatile solute. The solid-liquid curve for the solution is displaced left of that for the pure solvent, representing the freezing point depression, ΔTf, that accompanies solution formation. Finally, notice that the solid-gas curves for the solvent and its solution are identical. This is the case for many solutions comprising liquid solvents and nonvolatile solutes. Just as for vaporization, when a solution of this sort is frozen, it is actually just the solvent molecules that undergo the liquid-to-solid transition, forming pure solid solvent that excludes solute species. The solid and gaseous phases, therefore, are composed of solvent only, and so transitions between these phases are not subject to colligative effects.
Osmosis and Osmotic Pressure of Solutions
A number of natural and synthetic materials exhibit selective permeation, meaning that only molecules or ions of a certain size, shape, polarity, charge, and so forth, are capable of passing through (permeating) the material. Biological cell membranes provide elegant examples of selective permeation in nature, while dialysis tubing used to remove metabolic wastes from blood is a more simplistic technological example. Regardless of how they may be fabricated, these materials are generally referred to as semipermeable membranes.
Consider the apparatus illustrated in Figure $7$, in which samples of pure solvent and a solution are separated by a membrane that only solvent molecules may permeate. Solvent molecules will diffuse across the membrane in both directions. Since the concentration of solvent is greater in the pure solvent than the solution, these molecules will diffuse from the solvent side of the membrane to the solution side at a faster rate than they will in the reverse direction. The result is a net transfer of solvent molecules from the pure solvent to the solution. Diffusion-driven transfer of solvent molecules through a semipermeable membrane is a process known as osmosis.
When osmosis is carried out in an apparatus like that shown in Figure $7$, the volume of the solution increases as it becomes diluted by accumulation of solvent. This causes the level of the solution to rise, increasing its hydrostatic pressure (due to the weight of the column of solution in the tube) and resulting in a faster transfer of solvent molecules back to the pure solvent side. When the pressure reaches a value that yields a reverse solvent transfer rate equal to the osmosis rate, bulk transfer of solvent ceases. This pressure is called the osmotic pressure ($Π$) of the solution. The osmotic pressure of a dilute solution is related to its solute molarity, M, and absolute temperature, T, according to the equation
$\Pi=M R T \label{osmotic}$
where $R$ is the universal gas constant.
Example $8$: Calculation of Osmotic Pressure
Assuming ideal solution behavior, what is the osmotic pressure (atm) of a 0.30 M solution of glucose in water that is used for intravenous infusion at body temperature, 37 °C?
Solution
Find the osmotic pressure, $Π$, using Equation \ref{osmotic} ($\ce{Π = MRT$), where $T$ is on the Kelvin scale (310 K) and the value of $R$ is expressed in appropriate units (0.08206 L atm/mol K).
\begin{align*} \Pi & =M R T \[4pt] & =0.30 mol / L \times 0.08206 L atm / mol K \times 310 K \[4pt] & =7.6 atm \end{align*}
Exercise $8$
Assuming ideal solution behavior, what is the osmotic pressure (atm) a solution with a volume of 0.750 L that contains 5.0 g of methanol, CH3OH, in water at 37 °C?
Answer
5.3 atm
If a solution is placed in an apparatus like the one shown in Figure $8$, applying pressure greater than the osmotic pressure of the solution reverses the osmosis and pushes solvent molecules from the solution into the pure solvent. This technique of reverse osmosis is used for large-scale desalination of seawater and on smaller scales to produce high-purity tap water for drinking.
Chemistry in Everyday Life: Reverse Osmosis Water Purification
In the process of osmosis, diffusion serves to move water through a semipermeable membrane from a less concentrated solution to a more concentrated solution. Osmotic pressure is the amount of pressure that must be applied to the more concentrated solution to cause osmosis to stop. If greater pressure is applied, the water will go from the more concentrated solution to a less concentrated (more pure) solution. This is called reverse osmosis. Reverse osmosis (RO) is used to purify water in many applications, from desalination plants in coastal cities, to water-purifying machines in grocery stores (Figure $9$), and smaller reverse-osmosis household units. With a hand-operated pump, small RO units can be used in third-world countries, disaster areas, and in lifeboats. Our military forces have a variety of generator-operated RO units that can be transported in vehicles to remote locations.
Examples of osmosis are evident in many biological systems because cells are surrounded by semipermeable membranes. Carrots and celery that have become limp because they have lost water can be made crisp again by placing them in water. Water moves into the carrot or celery cells by osmosis. A cucumber placed in a concentrated salt solution loses water by osmosis and absorbs some salt to become a pickle. Osmosis can also affect animal cells. Solute concentrations are particularly important when solutions are injected into the body. Solutes in body cell fluids and blood serum give these solutions an osmotic pressure of approximately 7.7 atm. Solutions injected into the body must have the same osmotic pressure as blood serum; that is, they should be isotonic with blood serum. If a less concentrated solution, a hypotonic solution, is injected in sufficient quantity to dilute the blood serum, water from the diluted serum passes into the blood cells by osmosis, causing the cells to expand and rupture. This process is called hemolysis. When a more concentrated solution, a hypertonic solution, is injected, the cells lose water to the more concentrated solution, shrivel, and possibly die in a process called crenation. These effects are illustrated in Figure $9$.
Determination of Molar Masses
Osmotic pressure and changes in freezing point, boiling point, and vapor pressure are directly proportional to the number of solute species present in a given amount of solution. Consequently, measuring one of these properties for a solution prepared using a known mass of solute permits determination of the solute’s molar mass.
Example $9$: Determination of a Molar Mass from a 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. Assuming ideal solution behavior, what is the molar mass of this compound?
Solution
Solve this problem using the following steps.
Step 1. Determine the change in freezing point from the observed freezing point and the freezing point of pure benzene (Table $1$.
$\Delta T_{ f }=5.5{ }^{\circ} C -2.32{ }^{\circ} C =3.2{ }^{\circ} C \nonumber$
Step 2. Determine the molal concentration from Kf, the freezing point depression constant for benzene (Table $1$), and $ΔT_f$.
\begin{align*} \Delta T_{ f }=K_{ f } m \[4pt] m=\frac{\Delta T_{ f }}{K_{ f }}=\frac{3.2{ }^{\circ} C }{5.12{ }^{\circ} C m^{-1}}=0.63 m \end{align*}
Step 3. Determine the number of moles of compound in the solution from the molal concentration and the mass of solvent used to make the solution.
$\text { Moles of solute }=\frac{0.63 ~\text{mol solute }}{1.00 ~ \cancel{\text{kg solvent}}} \times 0.0550~ \cancel{\text{kg solvent }}=0.035 mol \nonumber$
Step 4. Determine the molar mass from the mass of the solute and the number of moles in that mass.
$\text { Molar mass }=\frac{4.00 ~\text{g} }{0.035~\text{mol} }=1.1 \times 10^2 ~\text{g / mol} \nonumber$
Exercise $9$
A solution of 35.7 g of a nonelectrolyte in 220.0 g of chloroform has a boiling point of 64.5 °C. Assuming ideal solution behavior, what is the molar mass of this compound?
Answer
1.8 \times 10^{2} g/mol
Example $10$: 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. Assuming ideal solution behavior, what is the molar mass of hemoglobin?
Solution
Here is one set of steps that can be used to solve the problem:
Step 1. Convert the osmotic pressure to atmospheres, then determine the molar concentration from the osmotic pressure.
\begin{align*} \Pi=\frac{5.9 \text { torr } \times 1 atm }{760 \text { torr }}=7.8 \times 10^{-3} atm \[4pt] \Pi=M R T \end{align*}
$M=\frac{\Pi}{R T}=\frac{7.8 \times 10^{-3} atm }{(0.08206 L atm / mol K )(295 K )}=3.2 \times 10^{-4} M \nonumber$
Step 2. Determine the number of moles of hemoglobin in the solution from the concentration and the volume of the solution.
$\text { moles of hemoglobin }=\frac{3.2 \times 10^{-4} mol }{1 ~ \cancel{ \text{L solution }}} \times 0.500 ~ \cancel{\text{L solution }} =1.6 \times 10^{-4} ~\text{mol} \nonumber$
Step 3. Determine the molar mass from the mass of hemoglobin and the number of moles in that mass.
$\text { molar mass }=\frac{10.0 ~\text{g}}{1.6 \times 10^{-4}~\text{mol}}=6.2 \times 10^4 ~\text{g / mol} \nonumber$
Exercise $10$
Assuming ideal solution behavior, 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
3 \times 10^{4} g/mol
Colligative Properties of Electrolytes
As noted previously in this module, the colligative properties of a solution depend only on the number, not on the identity, of solute species dissolved. The concentration terms in the equations for various colligative properties (freezing point depression, boiling point elevation, osmotic pressure) pertain to all solute species present in the solution. For the solutions considered thus far in this chapter, the solutes have been nonelectrolytes that dissolve physically without dissociation or any other accompanying process. Each molecule that dissolves yields one dissolved solute molecule. The dissolution of an electroyte, however, is not this simple, as illustrated by the two common examples below:
$\begin{array}{ll} \text { dissociation } & NaCl ( s ) \longrightarrow Na^{+}(aq) + Cl^{-}(aq) \[4pt] \text { ionization } & HCl (aq) + H_2 O ( l ) \longrightarrow Cl^{-}(aq) + H3O^{+}(aq) \end{array} \nonumber$
Considering the first of these examples, and assuming complete dissociation, a 1.0 m aqueous solution of NaCl contains 2.0 mole of ions (1.0 mol Na+ and 1.0 mol Cl) per each kilogram of water, and its freezing point depression is expected to be
$\Delta T_{ f }=2.0 mol \text { ions } / kg \text { water } \times 1.86^{\circ} C kg \text { water } / mol \text { ion }=3.7^{\circ} C \text {. } \nonumber$
When this solution is actually prepared and its freezing point depression measured, however, a value of 3.4 °C is obtained. Similar discrepancies are observed for other ionic compounds, and the differences between the measured and expected colligative property values typically become more significant as solute concentrations increase. These observations suggest that the ions of sodium chloride (and other strong electrolytes) are not completely dissociated in solution.
To account for this and avoid the errors accompanying the assumption of total dissociation, an experimentally measured parameter named in honor of Nobel Prize-winning German chemist Jacobus Henricus van ’t Hoff is used. The van ’t Hoff factor ($i$) is defined as the ratio of solute particles in solution to the number of formula units dissolved:
$i=\dfrac{\text { moles of particles in solution }}{\text { moles of formula units dissolved }} \nonumber$
Values for measured van ’t Hoff factors for several solutes, along with predicted values assuming complete dissociation, are shown in Table $2$.
Table $2$: Predicated and Measured van’t Hoff Factors for Several 0.050 m Aqueous Solutions
Formula unit Classification Dissolution products i (predicted) i (measured)
C12H22O11 (glucose) Nonelectrolyte C12H22O11 1 1.0
NaCl Strong electrolyte Na+, Cl 2 1.9
HCl Strong electrolyte (acid) H3O+, Cl 2 1.9
MgSO4 Strong electrolyte Mg2+, SO42, 2 1.3
MgCl2 Strong electrolyte Mg2+, 2Cl 3 2.7
FeCl3 Strong electrolyte Fe3+, 3Cl 4 3.4
In 1923, the chemists Peter Debye and Erich Hückel proposed a theory to explain the apparent incomplete ionization of strong electrolytes. They suggested that although interionic attraction in an aqueous solution is very greatly reduced by solvation of the ions and the insulating action of the polar solvent, it is not completely nullified. The residual attractions prevent the ions from behaving as totally independent particles (Figure $11$). In some cases, a positive and negative ion may actually touch, giving a solvated unit called an ion pair. Thus, the activity, or the effective concentration, of any particular kind of ion is less than that indicated by the actual concentration. Ions become more and more widely separated the more dilute the solution, and the residual interionic attractions become less and less. Thus, in extremely dilute solutions, the effective concentrations of the ions (their activities) are essentially equal to the actual concentrations. Note that the van’t Hoff factors for the electrolytes in Table $2$ are for 0.05 m solutions, at which concentration the value of i for NaCl is 1.9, as opposed to an ideal value of 2.
Example $11$: The Freezing Point of a Solution of an Electrolyte
The concentration of ions in seawater is approximately the same as that in a solution containing 4.2 g of NaCl dissolved in 125 g of water. Use this information and a predicted value for the van’t Hoff factor (Table $2$) to determine the freezing temperature the solution (assume ideal solution behavior).
Solution
Solve this problem using the following series of steps.
1. Step 1. Convert from grams to moles of NaCl using the molar mass of NaCl in the unit conversion factor.
Result: 0.072 mol NaCl
2. Step 2. Determine the number of moles of ions present in the solution using the number of moles of ions in 1 mole of NaCl as the conversion factor (2 mol ions/1 mol NaCl).
Result: 0.14 mol ions
3. Step 3. Determine the molality of the ions in the solution from the number of moles of ions and the mass of solvent, in kilograms.
Result: 1.2 m
4. Step 4. Use the direct proportionality between the change in freezing point and molal concentration to determine how much the freezing point changes.
Result: 2.1 °C
5. Step 5. Determine the new freezing point from the freezing point of the pure solvent and the change.
Result: −2.1 °C
Check each result as a self-assessment, taking care to avoid rounding errors by retaining guard digits in each step’s result for computing the next step’s result.
Exercise $11$
Assuming complete dissociation and ideal solution behavior, calculate the freezing point of a solution of 0.724 g of CaCl2 in 175 g of water.
Answer
−0.208 °C | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/11%3A_Solutions_and_Colloids/11.04%3A_Colligative_Properties.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the composition and properties of colloidal dispersions
• List and explain several technological applications of colloids
As a child, you may have made suspensions such as mixtures of mud and water, flour and water, or a suspension of solid pigments in water, known as tempera paint. These suspensions are heterogeneous mixtures composed of relatively large particles that are visible (or that can be seen with a magnifying glass). They are cloudy, and the suspended particles settle out after mixing. On the other hand, a solution is a homogeneous mixture in which no settling occurs and in which the dissolved species are molecules or ions. Solutions exhibit completely different behavior from suspensions. A solution may be colored, but it is transparent, the molecules or ions are invisible, and they do not settle out on standing. Another class of mixtures called colloids (or colloidal dispersions) exhibit properties intermediate between those of suspensions and solutions (Figure $1$). The particles in a colloid are larger than most simple molecules; however, colloidal particles are small enough that they do not settle out upon standing.
The particles in a colloid are large enough to scatter light, a phenomenon called the Tyndall effect. This can make colloidal mixtures appear cloudy or opaque, such as the searchlight beams shown in Figure $2$. Clouds are colloidal mixtures. They are composed of water droplets that are much larger than molecules, but that are small enough that they do not settle out.
The term “colloid”—from the Greek words kolla, meaning “glue,” and eidos, meaning “like”—was first used in 1861 by Thomas Graham to classify mixtures such as starch in water and gelatin. Many colloidal particles are aggregates of hundreds or thousands of molecules, but others (such as proteins and polymer molecules) consist of a single extremely large molecule. The protein and synthetic polymer molecules that form colloids may have molecular masses ranging from a few thousand to many million atomic mass units.
Analogous to the identification of solution components as “solute” and “solvent,” the components of a colloid are likewise classified according to their relative amounts. The particulate component typically present in a relatively minor amount is called the dispersed phase and the substance or solution throughout which the particulate is dispersed is called the dispersion medium. Colloids may involve virtually any combination of physical states (gas in liquid, liquid in solid, solid in gas, etc.), as illustrated by the examples of colloidal systems given in Table $1$.
Table $1$: Examples of Colloidal Systems
Dispersed Phase Dispersion Medium Common Examples Name
solid gas smoke, dust
solid liquid starch in water, some inks, paints, milk of magnesia sol
solid solid some colored gems, some alloys
liquid gas clouds, fogs, mists, sprays aerosol
liquid liquid milk, mayonnaise, butter emulsion
liquid solid jellies, gels, pearl, opal (H2O in SiO2) gel
gas liquid foams, whipped cream, beaten egg whites foam
gas solid pumice, floating soaps
Preparation of Colloidal Systems
Colloids are prepared by producing particles of colloidal dimensions and distributing these particles throughout a dispersion medium. Particles of colloidal size are formed by two methods:
1. Dispersion methods: breaking down larger particles. For example, paint pigments are produced by dispersing large particles by grinding in special mills.
2. Condensation methods: growth from smaller units, such as molecules or ions. For example, clouds form when water molecules condense and form very small droplets.
A few solid substances, when brought into contact with water, disperse spontaneously and form colloidal systems. Gelatin, glue, starch, and dehydrated milk powder behave in this manner. The particles are already of colloidal size; the water simply disperses them. Powdered milk particles of colloidal size are produced by dehydrating milk spray. Some atomizers produce colloidal dispersions of a liquid in air.
An emulsion may be prepared by shaking together or blending two immiscible liquids. This breaks one liquid into droplets of colloidal size, which then disperse throughout the other liquid. Oil spills in the ocean may be difficult to clean up, partly because wave action can cause the oil and water to form an emulsion. In many emulsions, however, the dispersed phase tends to coalesce, form large drops, and separate. Therefore, emulsions are usually stabilized by an emulsifying agent, a substance that inhibits the coalescence of the dispersed liquid. For example, a little soap will stabilize an emulsion of kerosene in water. Milk is an emulsion of butterfat in water, with the protein casein serving as the emulsifying agent. Mayonnaise is an emulsion of oil in vinegar, with egg yolk components as the emulsifying agents.
Condensation methods form colloidal particles by aggregation of molecules or ions. If the particles grow beyond the colloidal size range, drops or precipitates form, and no colloidal system results. Clouds form when water molecules aggregate and form colloid-sized particles. If these water particles coalesce to form adequately large water drops of liquid water or crystals of solid water, they settle from the sky as rain, sleet, or snow. Many condensation methods involve chemical reactions. A red colloidal suspension of iron(III) hydroxide may be prepared by mixing a concentrated solution of iron(III) chloride with hot water:
$\ce{Fe^{3+}(aq) + 3 Cl^{-}(aq) + 6 H2O(l) \longrightarrow Fe ( OH )_3(s) + H3O^{+}(aq) + 3 Cl^{-}(aq)} \nonumber$
A colloidal gold sol results from the reduction of a very dilute solution of gold(III) chloride by a reducing agent such as formaldehyde, tin(II) chloride, or iron(II) sulfate:
$\ce{Au^{3+} + 3 e^{-} -> Au} \nonumber$
Some gold sols prepared in 1857 are still intact (the particles have not coalesced and settled), illustrating the long-term stability of many colloids.
Soaps and Detergents
Pioneers made soap by boiling fats with a strongly basic solution made by leaching potassium carbonate, K2CO3, from wood ashes with hot water. Animal fats contain polyesters of fatty acids (long-chain carboxylic acids). When animal fats are treated with a base like potassium carbonate or sodium hydroxide, glycerol and salts of fatty acids such as palmitic, oleic, and stearic acid are formed. The salts of fatty acids are called soaps. The sodium salt of stearic acid, sodium stearate, has the formula C17H35CO2Na and contains an uncharged nonpolar hydrocarbon chain, the C17H35— unit, and an ionic carboxylate group, the —
Detergents (soap substitutes) also contain nonpolar hydrocarbon chains, such as C12H25—, and an ionic group, such as a sulfate—
The cleaning action of soaps and detergents can be explained in terms of the structures of the molecules involved. The hydrocarbon (nonpolar) end of a soap or detergent molecule dissolves in, or is attracted to, nonpolar substances such as oil, grease, or dirt particles. The ionic end is attracted by water (polar), illustrated in Figure $5$. As a result, the soap or detergent molecules become oriented at the interface between the dirt particles and the water so they act as a kind of bridge between two different kinds of matter, nonpolar and polar. Molecules such as this are termed amphiphilic since they have both a hydrophobic (“water-fearing”) part and a hydrophilic (“water-loving”) part. As a consequence, dirt particles become suspended as colloidal particles and are readily washed away.
Chemistry in Everyday Life: Deepwater Horizon Oil Spill
The blowout of the Deepwater Horizon oil drilling rig on April 20, 2010, in the Gulf of Mexico near Mississippi began the largest marine oil spill in the history of the petroleum industry. In the 87 days following the blowout, an estimated 4.9 million barrels (210 million gallons) of oil flowed from the ruptured well 5000 feet below the water’s surface. The well was finally declared sealed on September 19, 2010.
Crude oil is immiscible with and less dense than water, so the spilled oil rose to the surface of the water. Floating booms, skimmer ships, and controlled burns were used to remove oil from the water’s surface in an attempt to protect beaches and wetlands along the Gulf coast. In addition to removal of the oil, attempts were also made to lessen its environmental impact by rendering it “soluble” (in the loose sense of the term) and thus allowing it to be diluted to hopefully less harmful levels by the vast volume of ocean water. This approach used 1.84 million gallons of the oil dispersant Corexit 9527, most of which was injected underwater at the site of the leak, with small amounts being sprayed on top of the spill. Corexit 9527 contains 2-butoxyethanol (C6H14O2), an amphiphilic molecule whose polar and nonpolar ends are useful for emulsifying oil into small droplets, increasing the surface area of the oil and making it more available to marine bacteria for digestion (Figure $6$). While this approach avoids many of the immediate hazards that bulk oil poses to marine and coastal ecosystems, it introduces the possibility of long-term effects resulting from the introduction of the complex and potential toxic components of petroleum into the ocean’s food chain. A number of organizations are involved in monitoring the extended impact of this oil spill, including the National Oceanic and Atmospheric Administration (visit this website for additional details).
Electrical Properties of Colloidal Particles
Dispersed colloidal particles are often electrically charged. A colloidal particle of iron(III) hydroxide, for example, does not contain enough hydroxide ions to compensate exactly for the positive charges on the iron(III) ions. Thus, each individual colloidal particle bears a positive charge, and the colloidal dispersion consists of charged colloidal particles and some free hydroxide ions, which keep the dispersion electrically neutral. Most metal hydroxide colloids have positive charges, whereas most metals and metal sulfides form negatively charged dispersions. All colloidal particles in any one system have charges of the same sign. This helps keep them dispersed because particles containing like charges repel each other.
The charged nature of some colloidal particles may be exploited to remove them from a variety of mixtures. For example, the particles comprising smoke are often colloidally dispersed and electrically charged. Frederick Cottrell, an American chemist, developed a process to remove these particles. The charged particles are attracted to highly charged electrodes, where they are neutralized and deposited as dust (Figure $8$). This is one of the important methods used to clean up the smoke from a variety of industrial processes. The process is also important in the recovery of valuable products from the smoke and flue dust of smelters, furnaces, and kilns. There are also similar electrostatic air filters designed for home use to improve indoor air quality.
Portrait of a Chemist: Frederick Gardner Cottrell
Born in Oakland, CA, in 1877, Frederick Cottrell devoured textbooks as if they were novels and graduated from high school at the age of 16. He then entered the University of California (UC), Berkeley, completing a Bachelor’s degree in three years. He saved money from his \$1200 annual salary as a chemistry teacher at Oakland High School to fund his studies in chemistry in Berlin with Nobel prize winner Jacobus Henricus van’t Hoff, and in Leipzig with Wilhelm Ostwald, another Nobel awardee. After earning his PhD in physical chemistry, he returned to the United States to teach at UC Berkeley. He also consulted for the DuPont Company, where he developed the electrostatic precipitator, a device designed to curb air pollution by removing colloidal particles from air. Cottrell used the proceeds from his invention to fund a nonprofit research corporation to finance scientific research.
Gels
Gelatin desserts, such as Jell-O, are a type of colloid (Figure $9$). Gelatin sets on cooling because the hot aqueous mixture of gelatin coagulates as it cools, yielding an extremely viscous body known as a gel. A gel is a colloidal dispersion of a liquid phase throughout a solid phase. It appears that the fibers of the dispersing medium form a complex three-dimensional network, the interstices being filled with the liquid medium or a dilute solution of the dispersing medium.
Pectin, a carbohydrate from fruit juices, is a gel-forming substance important in jelly making. Silica gel, a colloidal dispersion of hydrated silicon dioxide, is formed when dilute hydrochloric acid is added to a dilute solution of sodium silicate. Canned Heat is a flammable gel made by mixing alcohol and a saturated aqueous solution of calcium acetate. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/11%3A_Solutions_and_Colloids/11.05%3A_Colloids.txt |
Example and Directions
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(Eg. "Genetic, Hereditary, DNA ...")(Eg. "Relating to genes or heredity")The infamous double helix https://bio.libretexts.org/CC-BY-SA; Delmar Larsen
Glossary Entries
Word(s)DefinitionImageCaptionLinkSource
alloysolid mixture of a metallic element and one or more additional elements
amphiphilicmolecules possessing both hydrophobic (nonpolar) and a hydrophilic (polar) parts
boiling point elevationelevation of the boiling point of a liquid by addition of a solute
boiling point elevation constantthe proportionality constant in the equation relating boiling point elevation to solute molality; also known as the ebullioscopic constant
colligative propertyproperty of a solution that depends only on the concentration of a solute species
colloid(also, colloidal dispersion) mixture in which relatively large solid or liquid particles are dispersed uniformly throughout a gas, liquid, or solid
crenationprocess whereby biological cells become shriveled due to loss of water by osmosis
dispersed phasesubstance present as relatively large solid or liquid particles in a colloid
dispersion mediumsolid, liquid, or gas in which colloidal particles are dispersed
dissociationphysical process accompanying the dissolution of an ionic compound in which the compound’s constituent ions are solvated and dispersed throughout the solution
electrolytesubstance that produces ions when dissolved in water
emulsifying agentamphiphilic substance used to stabilize the particles of some emulsions
emulsioncolloid formed from immiscible liquids
freezing point depressionlowering of the freezing point of a liquid by addition of a solute
freezing point depression constant(also, cryoscopic constant) proportionality constant in the equation relating freezing point depression to solute molality
gelcolloidal dispersion of a liquid in a solid
hemolysisrupture of red blood cells due to the accumulation of excess water by osmosis
Henry’s lawthe proportional relationship between the concentration of dissolved gas in a solution and the partial pressure of the gas in contact with the solution
hypertonicof greater osmotic pressure
hypotonicof less osmotic pressure
ideal solutionsolution that forms with no accompanying energy change
immiscibleof negligible mutual solubility; typically refers to liquid substances
ion pairsolvated anion/cation pair held together by moderate electrostatic attraction
ion-dipole attractionelectrostatic attraction between an ion and a polar molecule
isotonicof equal osmotic pressure
misciblemutually soluble in all proportions; typically refers to liquid substances
molality (m)a concentration unit defined as the ratio of the numbers of moles of solute to the mass of the solvent in kilograms
nonelectrolytesubstance that does not produce ions when dissolved in water
osmosisdiffusion of solvent molecules through a semipermeable membrane
osmotic pressure (Π)opposing pressure required to prevent bulk transfer of solvent molecules through a semipermeable membrane
partially miscibleof moderate mutual solubility; typically refers to liquid substances
Raoult’s lawthe relationship between a solution’s vapor pressure and the vapor pressures and concentrations of its components
saturatedof concentration equal to solubility; containing the maximum concentration of solute possible for a given temperature and pressure
semipermeable membranea membrane that selectively permits passage of certain ions or molecules
solubilityextent to which a solute may be dissolved in water, or any solvent
solvationexothermic process in which intermolecular attractive forces between the solute and solvent in a solution are established
spontaneous processphysical or chemical change that occurs without the addition of energy from an external source
strong electrolytesubstance that dissociates or ionizes completely when dissolved in water
supersaturatedof concentration that exceeds solubility; a nonequilibrium state
suspensionheterogeneous mixture in which relatively large component particles are temporarily dispersed but settle out over time
Tyndall effectscattering of visible light by a colloidal dispersion
unsaturatedof concentration less than solubility
van’t Hoff factor (i)the ratio of the number of moles of particles in a solution to the number of moles of formula units dissolved in the solution
weak electrolytesubstance that ionizes only partially when dissolved in water | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/11%3A_Solutions_and_Colloids/11.06%3A_Key_Terms.txt |
ΔTb = Kbm
ΔTf = Kfm
Π = MRT
11.08: Summary
A solution forms when two or more substances combine physically to yield a mixture that is homogeneous at the molecular level. The solvent is the most concentrated component and determines the physical state of the solution. The solutes are the other components typically present at concentrations less than that of the solvent. Solutions may form endothermically or exothermically, depending upon the relative magnitudes of solute and solvent intermolecular attractive forces. Ideal solutions form with no appreciable change in energy.
Substances that dissolve in water to yield ions are called electrolytes. Electrolytes may be covalent compounds that chemically react with water to produce ions (for example, acids and bases), or they may be ionic compounds that dissociate to yield their constituent cations and anions, when dissolved. Dissolution of an ionic compound is facilitated by ion-dipole attractions between the ions of the compound and the polar water molecules. Soluble ionic substances and strong acids ionize completely and are strong electrolytes, while weak acids and bases ionize to only a small extent and are weak electrolytes. Nonelectrolytes are substances that do not produce ions when dissolved in water.
The extent to which one substance will dissolve in another is determined by several factors, including the types and relative strengths of intermolecular attractive forces that may exist between the substances’ atoms, ions, or molecules. This tendency to dissolve is quantified as a substance’s solubility, its maximum concentration in a solution at equilibrium under specified conditions. A saturated solution contains solute at a concentration equal to its solubility. A supersaturated solution is one in which a solute’s concentration exceeds its solubility—a nonequilibrium (unstable) condition that will result in solute precipitation when the solution is appropriately perturbed. Miscible liquids are soluble in all proportions, and immiscible liquids exhibit very low mutual solubility. Solubilities for gaseous solutes decrease with increasing temperature, while those for most, but not all, solid solutes increase with temperature. The concentration of a gaseous solute in a solution is proportional to the partial pressure of the gas to which the solution is exposed, a relation known as Henry’s law.
Properties of a solution that depend only on the concentration of solute particles are called colligative properties. They include changes in the vapor pressure, boiling point, and freezing point of the solvent in the solution. The magnitudes of these properties depend only on the total concentration of solute particles in solution, not on the type of particles. The total concentration of solute particles in a solution also determines its osmotic pressure. This is the pressure that must be applied to the solution to prevent diffusion of molecules of pure solvent through a semipermeable membrane into the solution. Ionic compounds may not completely dissociate in solution due to activity effects, in which case observed colligative effects may be less than predicted.
Colloids are mixtures in which one or more substances are dispersed as relatively large solid particles or liquid droplets throughout a solid, liquid, or gaseous medium. The particles of a colloid remain dispersed and do not settle due to gravity, and they are often electrically charged. Colloids are widespread in nature and are involved in many technological applications. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/11%3A_Solutions_and_Colloids/11.07%3A_Key_Equations.txt |
1.
How do solutions differ from compounds? From other mixtures?
2.
Which of the principal characteristics of solutions are evident in the solutions of K2Cr2O7 shown in Figure 11.2?
3.
When KNO3 is dissolved in water, the resulting solution is significantly colder than the water was originally.
1. Is the dissolution of KNO3 an endothermic or an exothermic process?
2. What conclusions can you draw about the intermolecular attractions involved in the process?
3. Is the resulting solution an ideal solution?
4.
Give an example of each of the following types of solutions:
1. a gas in a liquid
2. a gas in a gas
3. a solid in a solid
5.
Indicate the most important types of intermolecular attractions in each of the following solutions:
1. The solution in Figure 11.2.
2. NO(l) in CO(l)
3. Cl2(g) in Br2(l)
4. HCl(g) in benzene C6H6(l)
5. Methanol CH3OH(l) in H2O(l)
6.
Predict whether each of the following substances would be more soluble in water (polar solvent) or in a hydrocarbon such as heptane (C7H16, nonpolar solvent):
1. vegetable oil (nonpolar)
2. isopropyl alcohol (polar)
3. potassium bromide (ionic)
7.
Heat is released when some solutions form; heat is absorbed when other solutions form. Provide a molecular explanation for the difference between these two types of spontaneous processes.
8.
Solutions of hydrogen in palladium may be formed by exposing Pd metal to H2 gas. The concentration of hydrogen in the palladium depends on the pressure of H2 gas applied, but in a more complex fashion than can be described by Henry’s law. Under certain conditions, 0.94 g of hydrogen gas is dissolved in 215 g of palladium metal (solution density = 10.8 g cm3).
1. Determine the molarity of this solution.
2. Determine the molality of this solution.
3. Determine the percent by mass of hydrogen atoms in this solution.
9.
Explain why the ions Na+ and Cl are strongly solvated in water but not in hexane, a solvent composed of nonpolar molecules.
10.
Explain why solutions of HBr in benzene (a nonpolar solvent) are nonconductive, while solutions in water (a polar solvent) are conductive.
11.
Consider the solutions presented:
(a) Which of the following sketches best represents the ions in a solution of Fe(NO3)3(aq)?
(b) Write a balanced chemical equation showing the products of the dissolution of Fe(NO3)3.
12.
Compare the processes that occur when methanol (CH3OH), hydrogen chloride (HCl), and sodium hydroxide (NaOH) dissolve in water. Write equations and prepare sketches showing the form in which each of these compounds is present in its respective solution.
13.
What is the expected electrical conductivity of the following solutions?
1. NaOH(aq)
2. HCl(aq)
3. C6H12O6(aq) (glucose)
4. NH3(aq)
14.
Why are most solid ionic compounds electrically nonconductive, whereas aqueous solutions of ionic compounds are good conductors? Would you expect a liquid (molten) ionic compound to be electrically conductive or nonconductive? Explain.
15.
Indicate the most important type of intermolecular attraction responsible for solvation in each of the following solutions:
1. the solutions in Figure 11.7
2. methanol, CH3OH, dissolved in ethanol, C2H5OH
3. methane, CH4, dissolved in benzene, C6H6
4. the polar halocarbon CF2Cl2 dissolved in the polar halocarbon CF2ClCFCl2
5. O2(l) in N2(l)
16.
Suppose you are presented with a clear solution of sodium thiosulfate, Na2S2O3. How could you determine whether the solution is unsaturated, saturated, or supersaturated?
17.
Supersaturated solutions of most solids in water are prepared by cooling saturated solutions. Supersaturated solutions of most gases in water are prepared by heating saturated solutions. Explain the reasons for the difference in the two procedures.
18.
Suggest an explanation for the observations that ethanol, C2H5OH, is completely miscible with water and that ethanethiol, C2H5SH, is soluble only to the extent of 1.5 g per 100 mL of water.
19.
Calculate the percent by mass of KBr in a saturated solution of KBr in water at 10 °C. See Figure 11.16 for useful data, and report the computed percentage to one significant digit.
20.
Which of the following gases is expected to be most soluble in water? Explain your reasoning.
1. CH4
2. CCl4
3. CHCl3
21.
At 0 °C and 1.00 atm, as much as 0.70 g of O2 can dissolve in 1 L of water. At 0 °C and 4.00 atm, how many grams of O2 dissolve in 1 L of water?
22.
Refer to Figure 11.10.
1. How did the concentration of dissolved CO2 in the beverage change when the bottle was opened?
2. What caused this change?
3. Is the beverage unsaturated, saturated, or supersaturated with CO2?
23.
The Henry’s law constant for CO2 is 3.4 \times 10^{−2} M/atm at 25 °C. Assuming ideal solution behavior, what pressure of carbon dioxide is needed to maintain a CO2 concentration of 0.10 M in a can of lemon-lime soda?
24.
The Henry’s law constant for O2 is 1.3 \times 10^{−3} M/atm at 25 °C. Assuming ideal solution behavior, what mass of oxygen would be dissolved in a 40-L aquarium at 25 °C, assuming an atmospheric pressure of 1.00 atm, and that the partial pressure of O2 is 0.21 atm?
25.
Assuming ideal solution behavior, how many liters of HCl gas, measured at 30.0 °C and 745 torr, are required to prepare 1.25 L of a 3.20-M solution of hydrochloric acid?
26.
Which is/are part of the macroscopic domain of solutions and which is/are part of the microscopic domain: boiling point elevation, Henry’s law, hydrogen bond, ion-dipole attraction, molarity, nonelectrolyte, nonstoichiometric compound, osmosis, solvated ion?
27.
What is the microscopic explanation for the macroscopic behavior illustrated in Figure 11.14?
28.
Sketch a qualitative graph of the pressure versus time for water vapor above a sample of pure water and a sugar solution, as the liquids evaporate to half their original volume.
29.
A solution of potassium nitrate, an electrolyte, and a solution of glycerin (C3H5(OH)3), a nonelectrolyte, both boil at 100.3 °C. What other physical properties of the two solutions are identical?
30.
What are the mole fractions of H3PO4 and water in a solution of 14.5 g of H3PO4 in 125 g of water?
1. Outline the steps necessary to answer the question.
2. Answer the question.
31.
What are the mole fractions of HNO3 and water in a concentrated solution of nitric acid (68.0% HNO3 by mass)?
1. Outline the steps necessary to answer the question.
2. Answer the question.
32.
Calculate the mole fraction of each solute and solvent:
1. 583 g of H2SO4 in 1.50 kg of water—the acid solution used in an automobile battery
2. 0.86 g of NaCl in 1.00 \times 10^{2} g of water—a solution of sodium chloride for intravenous injection
3. 46.85 g of codeine, C18H21NO3, in 125.5 g of ethanol, C2H5OH
4. 25 g of I2 in 125 g of ethanol, C2H5OH
33.
Calculate the mole fraction of each solute and solvent:
1. 0.710 kg of sodium carbonate (washing soda), Na2CO3, in 10.0 kg of water—a saturated solution at 0 °C
2. 125 g of NH4NO3 in 275 g of water—a mixture used to make an instant ice pack
3. 25 g of Cl2 in 125 g of dichloromethane, CH2Cl2
4. 0.372 g of tetrahydropyridine, C5H9N, in 125 g of chloroform, CHCl3
34.
Calculate the mole fractions of methanol, CH3OH; ethanol, C2H5OH; and water in a solution that is 40% methanol, 40% ethanol, and 20% water by mass. (Assume the data are good to two significant figures.)
35.
What is the difference between a 1 M solution and a 1 m solution?
36.
What is the molality of phosphoric acid, H3PO4, in a solution of 14.5 g of H3PO4 in 125 g of water?
1. Outline the steps necessary to answer the question.
2. Answer the question.
37.
What is the molality of nitric acid in a concentrated solution of nitric acid (68.0% HNO3 by mass)?
1. Outline the steps necessary to answer the question.
2. Answer the question.
38.
Calculate the molality of each of the following solutions:
1. 583 g of H2SO4 in 1.50 kg of water—the acid solution used in an automobile battery
2. 0.86 g of NaCl in 1.00 \times 10^{2} g of water—a solution of sodium chloride for intravenous injection
3. 46.85 g of codeine, C18H21NO3, in 125.5 g of ethanol, C2H5OH
4. 25 g of I2 in 125 g of ethanol, C2H5OH
39.
Calculate the molality of each of the following solutions:
1. 0.710 kg of sodium carbonate (washing soda), Na2CO3, in 10.0 kg of water—a saturated solution at 0°C
2. 125 g of NH4NO3 in 275 g of water—a mixture used to make an instant ice pack
3. 25 g of Cl2 in 125 g of dichloromethane, CH2Cl2
4. 0.372 g of tetrahydropyridine, C5H9N, in 125 g of chloroform, CHCl3
40.
The concentration of glucose, C6H12O6, in normal spinal fluid is What is the molality of the solution?
41.
A 13.0% solution of K2CO3 by mass has a density of 1.09 g/cm3. Calculate the molality of the solution.
42.
Why does 1 mol of sodium chloride depress the freezing point of 1 kg of water almost twice as much as 1 mol of glycerin?
43.
Assuming ideal solution behavior, what is the boiling point of a solution of 115.0 g of nonvolatile sucrose, C12H22O11, in 350.0 g of water?
1. Outline the steps necessary to answer the question
2. Answer the question
44.
Assuming ideal solution behavior, what is the boiling point of a solution of 9.04 g of I2 in 75.5 g of benzene, assuming the I2 is nonvolatile?
1. Outline the steps necessary to answer the question.
2. Answer the question.
45.
Assuming ideal solution behavior, what is the freezing temperature of a solution of 115.0 g of sucrose, C12H22O11, in 350.0 g of water?
1. Outline the steps necessary to answer the question.
2. Answer the question.
46.
Assuming ideal solution behavior, what is the freezing point of a solution of 9.04 g of I2 in 75.5 g of benzene?
1. Outline the steps necessary to answer the following question.
2. Answer the question.
47.
Assuming ideal solution behavior, what is the osmotic pressure of an aqueous solution of 1.64 g of Ca(NO3)2 in water at 25 °C? The volume of the solution is 275 mL.
1. Outline the steps necessary to answer the question.
2. Answer the question.
48.
Assuming ideal solution behavior, what is osmotic pressure of a solution of bovine insulin (molar mass, 5700 g mol−1) at 18 °C if 100.0 mL of the solution contains 0.103 g of the insulin?
1. Outline the steps necessary to answer the question.
2. Answer the question.
49.
Assuming ideal solution behavior, what is the molar mass of a solution of 5.00 g of a compound in 25.00 g of carbon tetrachloride (bp 76.8 °C; Kb = 5.02 °C/m) that boils at 81.5 °C at 1 atm?
1. Outline the steps necessary to answer the question.
2. Solve the problem.
50.
A sample of an organic compound (a nonelectrolyte) weighing 1.35 g lowered the freezing point of 10.0 g of benzene by 3.66 °C. Assuming ideal solution behavior, calculate the molar mass of the compound.
51.
A 1.0 m solution of HCl in benzene has a freezing point of 0.4 °C. Is HCl an electrolyte in benzene? Explain.
52.
A solution contains 5.00 g of urea, CO(NH2)2, a nonvolatile compound, dissolved in 0.100 kg of water. If the vapor pressure of pure water at 25 °C is 23.7 torr, what is the vapor pressure of the solution (assuming ideal solution behavior)?
53.
A 12.0-g sample of a nonelectrolyte is dissolved in 80.0 g of water. The solution freezes at −1.94 °C. Assuming ideal solution behavior, calculate the molar mass of the substance.
54.
Arrange the following solutions in order by their decreasing freezing points: 0.1 m Na3PO4, 0.1 m C2H5OH, 0.01 m CO2, 0.15 m NaCl, and 0.2 m CaCl2.
55.
Calculate the boiling point elevation of 0.100 kg of water containing 0.010 mol of NaCl, 0.020 mol of Na2SO4, and 0.030 mol of MgCl2, assuming complete dissociation of these electrolytes and ideal solution behavior.
56.
How could you prepare a 3.08 m aqueous solution of glycerin, C3H8O3? Assuming ideal solution behavior, what is the freezing point of this solution?
57.
A sample of sulfur weighing 0.210 g was dissolved in 17.8 g of carbon disulfide, CS2 (Kb = 2.34 °C/m). If the boiling point elevation was 0.107 °C, what is the formula of a sulfur molecule in carbon disulfide (assuming ideal solution behavior)?
58.
In a significant experiment performed many years ago, 5.6977 g of cadmium iodide in 44.69 g of water raised the boiling point 0.181 °C. What does this suggest about the nature of a solution of CdI2?
59.
Lysozyme is an enzyme that cleaves cell walls. A 0.100-L sample of a solution of lysozyme that contains 0.0750 g of the enzyme exhibits an osmotic pressure of 1.32 \times 10^{−3} atm at 25 °C. Assuming ideal solution behavior, what is the molar mass of lysozyme?
60.
The osmotic pressure of a solution containing 7.0 g of insulin per liter is 23 torr at 25 °C. Assuming ideal solution behavior, what is the molar mass of insulin?
61.
The osmotic pressure of human blood is 7.6 atm at 37 °C. What mass of glucose, C6H12O6, is required to make 1.00 L of aqueous solution for intravenous feeding if the solution must have the same osmotic pressure as blood at body temperature, 37 °C (assuming ideal solution behavior)?
62.
Assuming ideal solution behavior, what is the freezing point of a solution of dibromobenzene, C6H4Br2, in 0.250 kg of benzene, if the solution boils at 83.5 °C?
63.
Assuming ideal solution behavior, what is the boiling point of a solution of NaCl in water if the solution freezes at −0.93 °C?
64.
The sugar fructose contains 40.0% C, 6.7% H, and 53.3% O by mass. A solution of 11.7 g of fructose in 325 g of ethanol has a boiling point of 78.59 °C. The boiling point of ethanol is 78.35 °C, and Kb for ethanol is 1.20 °C/m. Assuming ideal solution behavior, what is the molecular formula of fructose?
65.
The vapor pressure of methanol, CH3OH, is 94 torr at 20 °C. The vapor pressure of ethanol, C2H5OH, is 44 torr at the same temperature.
1. Calculate the mole fraction of methanol and of ethanol in a solution of 50.0 g of methanol and 50.0 g of ethanol.
2. Ethanol and methanol form a solution that behaves like an ideal solution. Calculate the vapor pressure of methanol and of ethanol above the solution at 20 °C.
3. Calculate the mole fraction of methanol and of ethanol in the vapor above the solution.
66.
The triple point of air-free water is defined as 273.16 K. Why is it important that the water be free of air?
67.
Meat can be classified as fresh (not frozen) even though it is stored at −1 °C. Why wouldn’t meat freeze at this temperature?
68.
An organic compound has a composition of 93.46% C and 6.54% H by mass. A solution of 0.090 g of this compound in 1.10 g of camphor melts at 158.4 °C. The melting point of pure camphor is 178.4 °C. Kf for camphor is 37.7 °C/m. Assuming ideal solution behavior, what is the molecular formula of the solute? Show your calculations.
69.
A sample of HgCl2 weighing 9.41 g is dissolved in 32.75 g of ethanol, C2H5OH (Kb = 1.20 °C/m). The boiling point elevation of the solution is 1.27 °C. Is HgCl2 an electrolyte in ethanol? Show your calculations.
70.
A salt is known to be an alkali metal fluoride. A quick approximate determination of freezing point indicates that 4 g of the salt dissolved in 100 g of water produces a solution that freezes at about −1.4 °C. Assuming ideal solution behavior, what is the formula of the salt? Show your calculations.
71.
Identify the dispersed phase and the dispersion medium in each of the following colloidal systems: starch dispersion, smoke, fog, pearl, whipped cream, floating soap, jelly, milk, and ruby.
72.
Distinguish between dispersion methods and condensation methods for preparing colloidal systems.
73.
How do colloids differ from solutions with regard to dispersed particle size and homogeneity?
74.
Explain the cleansing action of soap.
75.
How can it be demonstrated that colloidal particles are electrically charged? | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/11%3A_Solutions_and_Colloids/11.09%3A_Exercises.txt |
Chemical kinetics is the study of rates of chemical processes and includes investigations of how different experimental conditions can influence the speed of a chemical reaction and yield information about the reaction's mechanism and transition states, as well as the construction of mathematical models that can describe the characteristics of a chemical reaction.
• 12.0: Introduction
The study of chemical kinetics concerns the rate at which a reaction yields products and the molecular-scale means by which a reaction occurs. In this chapter, we will examine the factors that influence the rates of chemical reactions, the mechanisms by which reactions proceed, and the quantitative techniques used to determine and describe the rate at which reactions occur.
• 12.1: Chemical Reaction Rates
The rate of a reaction can be expressed either in terms of the decrease in the amount of a reactant or the increase in the amount of a product per unit time. Relations between different rate expressions for a given reaction are derived directly from the stoichiometric coefficients of the equation representing the reaction.
• 12.2: Factors Affecting Reaction Rates
The rate of a chemical reaction is affected by several parameters. Reactions involving two phases proceed more rapidly when there is greater surface area contact. If temperature or reactant concentration is increased, the rate of a given reaction generally increases as well. A catalyst can increase the rate of a reaction by providing an alternative pathway that causes the activation energy of the reaction to decrease.
• 12.3: Rate Laws
Rate laws provide a mathematical description of how changes in the amount of a substance affect the rate of a chemical reaction. Rate laws are determined experimentally and cannot be predicted by reaction stoichiometry. The order of reaction describes how much a change in the amount of each substance affects the overall rate, and the overall order of a reaction is the sum of the orders for each substance present in the reaction.
• 12.4: Integrated Rate Laws
Differential rate laws can be determined by the method of initial rates or other methods. We measure values for the initial rates of a reaction at different concentrations of the reactants. From these measurements, we determine the order of the reaction in each reactant. Integrated rate laws are determined by integration of the corresponding differential rate laws. Rate constants for those rate laws are determined from measurements of concentration at various times during a reaction.
• 12.5: Collision Theory
Chemical reactions require collisions between reactant species. These reactant collisions must be of proper orientation and sufficient energy in order to result in product formation. Collision theory provides a simple but effective explanation for the effect of many experimental parameters on reaction rates. The Arrhenius equation describes the relation between a reaction’s rate constant and its activation energy, temperature, and dependence on collision orientation.
• 12.6: Reaction Mechanisms
The sequence of individual steps, or elementary reactions, by which reactants are converted into products during the course of a reaction is called the reaction mechanism. The overall rate of a reaction is determined by the rate of the slowest step, called the rate-determining step. Unimolecular elementary reactions have first-order rate laws, while bimolecular elementary reactions have second-order rate laws.
• 12.7: Catalysis
Catalysts affect the rate of a chemical reaction by altering its mechanism to provide a lower activation energy. Catalysts can be homogenous (in the same phase as the reactants) or heterogeneous (a different phase than the reactants).
• 12.8: Key Terms
• 12.9: Key Equations
• 12.10: Summary
• 12.11: Exercises
These are homework exercises to accompany the Textmap created for "Chemistry" by OpenStax. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
Thumbnail: Molecular collisions frequency. (Public Domain; Sadi Carnot via Wikipedia)
12: Kinetics
The lizard in the photograph is not simply enjoying the sunshine or working on its tan. The heat from the sun’s rays is critical to the lizard’s survival. A warm lizard can move faster than a cold one because the chemical reactions that allow its muscles to move occur more rapidly at higher temperatures. A cold lizard is a slower lizard and an easier meal for predators.
From baking a cake to determining the useful lifespan of a bridge, rates of chemical reactions play important roles in our understanding of processes that involve chemical changes. Two questions are typically posed when planning to carry out a chemical reaction. The first is: “Will the reaction produce the desired products in useful quantities?” The second question is: “How rapidly will the reaction occur?” A third question is often asked when investigating reactions in greater detail: “What specific molecular-level processes take place as the reaction occurs?” Knowing the answer to this question is of practical importance when the yield or rate of a reaction needs to be controlled.
The study of chemical kinetics concerns the second and third questions—that is, the rate at which a reaction yields products and the molecular-scale means by which a reaction occurs. This chapter examines the factors that influence the rates of chemical reactions, the mechanisms by which reactions proceed, and the quantitative techniques used to describe the rates at which reactions occur. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/12%3A_Kinetics/12.00%3A_Introduction.txt |
Learning Objectives
By the end of this section, you will be able to:
• Define chemical reaction rate
• Derive rate expressions from the balanced equation for a given chemical reaction
• Calculate reaction rates from experimental data
A rate is a measure of how some property varies with time. Speed is a familiar rate that expresses the distance traveled by an object in a given amount of time. Wage is a rate that represents the amount of money earned by a person working for a given amount of time. Likewise, the rate of a chemical reaction is a measure of how much reactant is consumed, or how much product is produced, by the reaction in a given amount of time.
The rate of reaction is the change in the amount of a reactant or product per unit time. Reaction rates are therefore determined by measuring the time dependence of some property that can be related to reactant or product amounts. Rates of reactions that consume or produce gaseous substances, for example, are conveniently determined by measuring changes in volume or pressure. For reactions involving one or more colored substances, rates may be monitored via measurements of light absorption. For reactions involving aqueous electrolytes, rates may be measured via changes in a solution’s conductivity.
For reactants and products in solution, their relative amounts (concentrations) are conveniently used for purposes of expressing reaction rates. For example, the concentration of hydrogen peroxide, H2O2, in an aqueous solution changes slowly over time as it decomposes according to the equation:
$2 H_2 O_2(aq) \longrightarrow 2 H_2 O (l) + O_2(g) \nonumber$
The rate at which the hydrogen peroxide decomposes can be expressed in terms of the rate of change of its concentration, as shown here:
\text { rate of decomposition of } \begin{align*} H_2O_2 & =-\dfrac{\text { change in concentration of reactant }}{\text { time interval }} \[4pt] & =-\dfrac{\left[ H_2 O_2\right]_{t_2}-\left[ H_2 O_2\right]_{t_1}}{t_2-t_1} \[4pt] & =-\dfrac{\Delta\left[ H_2 O_2\right]}{\Delta t} \end{align*}
This mathematical representation of the change in species concentration over time is the rate expression for the reaction. The brackets indicate molar concentrations, and the symbol delta ($Δ$) indicates “change in.” Thus, $[\ce{H2O2}]_{t_1}}$ represents the molar concentration of hydrogen peroxide at some time $t_1$; likewise, $[\ce{H2O2}]_{t_2}}$ represents the molar concentration of hydrogen peroxide at a later time t2; and Δ[H2O2] represents the change in molar concentration of hydrogen peroxide during the time interval $Δt$ (that is, $t_2 − t_1$). Since the reactant concentration decreases as the reaction proceeds, $Δ[\ce{H2O2}]$ is a negative quantity. Reaction rates are, by convention, positive quantities, and so this negative change in concentration is multiplied by −1. Figure $1$ provides an example of data collected during the decomposition of $\ce{H2O2}$.
To obtain the tabulated results for this decomposition, the concentration of hydrogen peroxide was measured every 6 hours over the course of a day at a constant temperature of 40 °C. Reaction rates were computed for each time interval by dividing the change in concentration by the corresponding time increment, as shown here for the first 6-hour period:
$\frac{-\Delta\left[ H_2 O_2\right]}{\Delta t}=\frac{-(0.500 mol / L -1.000 mol / L )}{(6.00 h -0.00 h )}=0.0833 mol L^{-1} h^{-1} \nonumber$
Notice that the reaction rates vary with time, decreasing as the reaction proceeds. Results for the last 6-hour period yield a reaction rate of:
$\frac{-\Delta\left[ H_2 O_2\right]}{\Delta t}=\frac{-(0.0625 mol / L -0.125 mol / L )}{(24.00 h -18.00 h )}=0.010 mol L^{-1} h^{-1} \nonumber$
This behavior indicates the reaction continually slows with time. Using the concentrations at the beginning and end of a time period over which the reaction rate is changing results in the calculation of an average rate for the reaction over this time interval. At any specific time, the rate at which a reaction is proceeding is known as its instantaneous rate. The instantaneous rate of a reaction at “time zero,” when the reaction commences, is its initial rate. Consider the analogy of a car slowing down as it approaches a stop sign. The vehicle’s initial rate—analogous to the beginning of a chemical reaction—would be the speedometer reading at the moment the driver begins pressing the brakes (t0). A few moments later, the instantaneous rate at a specific moment—call it t1—would be somewhat slower, as indicated by the speedometer reading at that point in time. As time passes, the instantaneous rate will continue to fall until it reaches zero, when the car (or reaction) stops. Unlike instantaneous speed, the car’s average speed is not indicated by the speedometer; but it can be calculated as the ratio of the distance traveled to the time required to bring the vehicle to a complete stop (Δt). Like the decelerating car, the average rate of a chemical reaction will fall somewhere between its initial and final rates.
The instantaneous rate of a reaction may be determined one of two ways. If experimental conditions permit the measurement of concentration changes over very short time intervals, then average rates computed as described earlier provide reasonably good approximations of instantaneous rates. Alternatively, a graphical procedure may be used that, in effect, yields the results that would be obtained if short time interval measurements were possible. In a plot of the concentration of hydrogen peroxide against time, the instantaneous rate of decomposition of $\ce{H2O2}$ at any time $t$ is given by the slope of a straight line that is tangent to the curve at that time (Figure $2$). These tangent line slopes may be evaluated using calculus, but the procedure for doing so is beyond the scope of this chapter.
Chemistry in Everyday Life: Reaction Rates in Analysis: Test Strips for Urinalysis
Physicians often use disposable test strips to measure the amounts of various substances in a patient’s urine (Figure $3$). These test strips contain various chemical reagents, embedded in small pads at various locations along the strip, which undergo changes in color upon exposure to sufficient concentrations of specific substances. The usage instructions for test strips often stress that proper read time is critical for optimal results. This emphasis on read time suggests that kinetic aspects of the chemical reactions occurring on the test strip are important considerations.
The test for urinary glucose relies on a two-step process represented by the chemical equations shown here:
$\begin{gathered} C_6 H_{12} O_6+ O_2 \stackrel{\text { catalyst }}{\longrightarrow} C_6 H_{10} O_6+ H_2 O_2 \[4pt] 2 H_2 O_2+2 I^{-} \stackrel{\text { catalyst }}{\longrightarrow} I_2+2 H_2 O + O_2 \end{gathered} \nonumber$
The first equation depicts the oxidation of glucose in the urine to yield glucolactone and hydrogen peroxide. The hydrogen peroxide produced subsequently oxidizes colorless iodide ion to yield brown iodine, which may be visually detected. Some strips include an additional substance that reacts with iodine to produce a more distinct color change.
The two test reactions shown above are inherently very slow, but their rates are increased by special enzymes embedded in the test strip pad. This is an example of catalysis, a topic discussed later in this chapter. A typical glucose test strip for use with urine requires approximately 30 seconds for completion of the color-forming reactions. Reading the result too soon might lead one to conclude that the glucose concentration of the urine sample is lower than it actually is (a false-negative result). Waiting too long to assess the color change can lead to a false positive due to the slower (not catalyzed) oxidation of iodide ion by other substances found in urine.
Relative Rates of Reaction
The rate of a reaction may be expressed as the change in concentration of any reactant or product. For any given reaction, these rate expressions are all related simply to one another according to the reaction stoichiometry. The rate of the general reaction
$aA \longrightarrow bB] can be expressed in terms of the decrease in the concentration of A or the increase in the concentration of B. These two rate expressions are related by the stoichiometry of the reaction: \[\text { rate }=-\left(\frac{1}{ a }\right)\left(\frac{\Delta A }{\Delta t}\right)=\left(\frac{1}{ b }\right)\left(\frac{\Delta B }{\Delta t}\right) \nonumber$
Consider the reaction represented by the following equation:
$2 NH_3(g) \longrightarrow N_2(g) + 3 H_2(g) \nonumber$
The relation between the reaction rates expressed in terms of nitrogen production and ammonia consumption, for example, is:
$-\frac{\Delta mol NH_3}{\Delta t} \times \frac{1 mol N_2}{2 mol NH_3}=\frac{\Delta mol N_2}{\Delta t} \nonumber$
This may be represented in an abbreviated format by omitting the units of the stoichiometric factor:
$-\frac{1}{2} \frac{\Delta mol NH_3}{\Delta t}=\frac{\Delta mol N_2}{\Delta t} \nonumber$
Note that a negative sign has been included as a factor to account for the opposite signs of the two amount changes (the reactant amount is decreasing while the product amount is increasing). For homogeneous reactions, both the reactants and products are present in the same solution and thus occupy the same volume, so the molar amounts may be replaced with molar concentrations:
$-\frac{1}{2} \frac{\Delta\left[ NH_3\right]}{\Delta t}=\frac{\Delta\left[ N_2\right]}{\Delta t} \nonumber$
Similarly, the rate of formation of H2 is three times the rate of formation of N2 because three moles of H2 are produced for each mole of N2 produced.
$\frac{1}{3} \frac{\Delta\left[ H_2\right]}{\Delta t}=\frac{\Delta\left[ N_2\right]}{\Delta t} \nonumber$
Figure $4$: illustrates the change in concentrations over time for the decomposition of ammonia into nitrogen and hydrogen at 1100 °C. Slopes of the tangent lines at t = 500 s show that the instantaneous rates derived from all three species involved in the reaction are related by their stoichiometric factors. The rate of hydrogen production, for example, is observed to be three times greater than that for nitrogen production:
$\frac{2.91 \times 10^{-6} M / s }{9.70 \times 10^{-7} M / s } \approx 3 \nonumber$
Example $1$: Expressions for Relative Reaction Rates
The first step in the production of nitric acid is the combustion of ammonia:
Exercise $1$
The rate of formation of Br2 is 6.0 10−6 mol/L/s in a reaction described by the following net ionic equation:
$5 Br^{-}+ BrO_3^{-}+6 H^{+} \longrightarrow 3 Br_2+3 H_2 O \nonumber$
Write the equations that relate the rates of consumption of the reactants and the rates of formation of the products.
Answer
$-\frac{1}{5} \frac{\Delta\left[ Br^{-}\right]}{\Delta t}=-\frac{\Delta\left[ BrO_3^{-}\right]}{\Delta t}=-\frac{1}{6} \frac{\Delta\left[ H^{+}\right]}{\Delta t}=\frac{1}{3} \frac{\Delta\left[ Br_2\right]}{\Delta t}=\frac{1}{3} \frac{\Delta\left[ H_2 O \right]}{\Delta t} \nonumber$
Example $2$: Reaction Rate Expressions for Decomposition of H2O2
The graph in Figure $2$: shows the rate of the decomposition of H2O2 over time:
$2 H_2 O_2 \longrightarrow 2 H_2 O + O_2 \nonumber$
Based on these data, the instantaneous rate of decomposition of H2O2 at t = 11.1 h is determined to be
3.20 10−2 mol/L/h, that is:
$-\frac{\Delta\left[ H_2 O_2\right]}{\Delta t}=3.20 \times 10^{-2} mol L^{-1} h^{-1} \nonumber$
What is the instantaneous rate of production of $\ce{H2O}$ and $\ce{O2}$?
Solution
The reaction stoichiometry shows that
$-\frac{1}{2} \frac{\Delta\left[ H_2 O_2\right]}{\Delta t}=\frac{1}{2} \frac{\Delta\left[ H_2 O \right]}{\Delta t}=\frac{\Delta\left[ O_2\right]}{\Delta t}] Therefore: \[\frac{1}{2} \times 3.20 \times 10^{-2} mol L^{-1} h^{-1}=\frac{\Delta\left[ O_2\right]}{\Delta t} \nonumber$
and
$\frac{\Delta\left[ O_2\right]}{\Delta t}=1.60 \times 10^{-2} mol L^{-1} h^{-1} \nonumber$
Exercise $2$
If the rate of decomposition of ammonia, NH3, at 1150 K is 2.10 10−6 mol/L/s, what is the rate of production of nitrogen and hydrogen?
Answer
1.05 10−6 mol/L/s, N2 and 3.15 10−6 mol/L/s, H2. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/12%3A_Kinetics/12.01%3A_Chemical_Reaction_Rates.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the effects of chemical nature, physical state, temperature, concentration, and catalysis on reaction rates
The rates at which reactants are consumed and products are formed during chemical reactions vary greatly. Five factors typically affecting the rates of chemical reactions will be explored in this section: the chemical nature of the reacting substances, the state of subdivision (one large lump versus many small particles) of the reactants, the temperature of the reactants, the concentration of the reactants, and the presence of a catalyst.
The Chemical Nature of the Reacting Substances
The rate of a reaction depends on the nature of the participating substances. Reactions that appear similar may have different rates under the same conditions, depending on the identity of the reactants. For example, when small pieces of the metals iron and sodium are exposed to air, the sodium reacts completely with air overnight, whereas the iron is barely affected. The active metals calcium and sodium both react with water to form hydrogen gas and a base. Yet calcium reacts at a moderate rate, whereas sodium reacts so rapidly that the reaction is almost explosive.
The Physical States of the Reactants
A chemical reaction between two or more substances requires intimate contact between the reactants. When reactants are in different physical states, or phases (solid, liquid, gaseous, dissolved), the reaction takes place only at the interface between the phases. Consider the heterogeneous reaction between a solid phase and either a liquid or gaseous phase. Compared with the reaction rate for large solid particles, the rate for smaller particles will be greater because the surface area in contact with the other reactant phase is greater. For example, large pieces of iron react more slowly with acids than they do with finely divided iron powder (Figure $1$). Large pieces of wood smolder, smaller pieces burn rapidly, and saw dust burns explosively.
Link to Learning
Watch this video to see the reaction of cesium with water in slow motion and a discussion of how the state of reactants and particle size affect reaction rates.
Temperature of the Reactants
Chemical reactions typically occur faster at higher temperatures. Food can spoil quickly when left on the kitchen counter. However, the lower temperature inside of a refrigerator slows that process so that the same food remains fresh for days. Gas burners, hot plates, and ovens are often used in the laboratory to increase the speed of reactions that proceed slowly at ordinary temperatures. For many chemical processes, reaction rates are approximately doubled when the temperature is raised by 10 °C.
Concentrations of the Reactants
The rates of many reactions depend on the concentrations of the reactants. Rates usually increase when the concentration of one or more of the reactants increases. For example, calcium carbonate (CaCO3) deteriorates as a result of its reaction with the pollutant sulfur dioxide. The rate of this reaction depends on the amount of sulfur dioxide in the air (Figure $2$). An acidic oxide, sulfur dioxide combines with water vapor in the air to produce sulfurous acid in the following reaction:
$\ce{SO_2(g) + H_2O(g) \longrightarrow H_2SO_3(aq)} \nonumber$
Calcium carbonate reacts with sulfurous acid as follows:
$\ce{CaCO_3(s) + H_2SO_3(aq) \longrightarrow CaSO_3(aq) + CO_2(g) + H_2O(l)} \nonumber$
In a polluted atmosphere where the concentration of sulfur dioxide is high, calcium carbonate deteriorates more rapidly than in less polluted air. Similarly, phosphorus burns much more rapidly in an atmosphere of pure oxygen than in air, which is only about 20% oxygen.
Link to Learning
Phosphorus burns rapidly in air, but it will burn even more rapidly if the concentration of oxygen is higher. Watch this video to see an example.
The Presence of a Catalyst
Relatively dilute aqueous solutions of hydrogen peroxide, $\ce{H2O2}$, are commonly used as topical antiseptics. Hydrogen peroxide decomposes to yield water and oxygen gas according to the equation:
$\ce{2 H_2O_2(aq) \longrightarrow 2 H_2O(l) + O_2(g)} \nonumber$
Under typical conditions, this decomposition occurs very slowly. When dilute $\ce{H2O2(aq)}$ is poured onto an open wound, however, the reaction occurs rapidly and the solution foams because of the vigorous production of oxygen gas. This dramatic difference is caused by the presence of substances within the wound’s exposed tissues that accelerate the decomposition process. Substances that function to increase the rate of a reaction are called catalysts, a topic treated in greater detail later in this chapter.
Link to Learning
Chemical reactions occur when molecules collide with each other and undergo a chemical transformation. Before physically performing a reaction in a laboratory, scientists can use molecular modeling simulations to predict how the parameters discussed earlier will influence the rate of a reaction. Use the PhET Reactions & Rates interactive to explore how temperature, concentration, and the nature of the reactants affect reaction rates. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/12%3A_Kinetics/12.02%3A_Factors_Affecting_Reaction_Rates.txt |
Learning Objectives
By the end of this section, you will be able to:
• Explain the form and function of a rate law
• Use rate laws to calculate reaction rates
• Use rate and concentration data to identify reaction orders and derive rate laws
As described in the previous module, the rate of a reaction is often affected by the concentrations of reactants. Rate laws (sometimes called differential rate laws) or rate equations are mathematical expressions that describe the relationship between the rate of a chemical reaction and the concentration of its reactants. As an example, consider the reaction described by the chemical equation
$a A+b B \longrightarrow \text { products } \nonumber$
where $a$ and $b$ are stoichiometric coefficients. The rate law for this reaction is written as:
$\text { rate }=k[A]^m[B]^n \nonumber$
in which [A] and [B] represent the molar concentrations of reactants, and k is the rate constant, which is specific for a particular reaction at a particular temperature. The exponents m and n are the reaction orders and are typically positive integers, though they can be fractions, negative, or zero. The rate constant k and the reaction orders m and n must be determined experimentally by observing how the rate of a reaction changes as the concentrations of the reactants are changed. The rate constant k is independent of the reactant concentrations, but it does vary with temperature.
The reaction orders in a rate law describe the mathematical dependence of the rate on reactant concentrations. Referring to the generic rate law above, the reaction is m order with respect to A and n order with respect to B. For example, if m = 1 and n = 2, the reaction is first order in A and second order in B. The overall reaction order is simply the sum of orders for each reactant. For the example rate law here, the reaction is third order overall (1 + 2 = 3). A few specific examples are shown below to further illustrate this concept.
The rate law:
$\text { rate }=k\left[ H_2 O_2\right] \nonumber$
describes a reaction that is first order in hydrogen peroxide and first order overall. The rate law:
$\text { rate }=k\left[ \ce{C4H6} \right]^2 \nonumber$
describes a reaction that is second order in $\ce{C4H6}$ and second order overall. The rate law:
$\text { rate }=k\left[ \ce{H^{+}} \right]\left[ \ce{OH^{-}} \right] \nonumber$
describes a reaction that is first order in $\ce{H^{+}}$, first order in $\ce{OH^{−}}$, and second order overall.
Example $1$: Writing Rate Laws from Reaction Orders
An experiment shows that the reaction of nitrogen dioxide with carbon monoxide:
$\ce{NO2(g) + CO(g) -> NO(g) + CO2(g)} \nonumber$
is second order in $\ce{NO2}$ and zero order in $\ce{CO}$ at 100 °C. What is the rate law for the reaction?
Solution
The reaction will have the form:
$\text { rate }=k\left[ \ce{NO_2} \right]^m[ \ce{CO} ]^n \nonumber$
The reaction is second order in $\ce{NO2}$; thus $m = 2$. The reaction is zero order in $\ce{CO}$; thus $n = 0$. The rate law is:
$\text { rate }=k\left[ \ce{NO_2} \right]^2[ \ce{CO} ]^0=k\left[ \ce{NO2} \right]^2 \nonumber$
Remember that a number raised to the zero power is equal to 1, thus [CO]0 = 1, which is why the $\ce{CO}$ concentration term may be omitted from the rate law: the rate of reaction is solely dependent on the concentration of $\ce{NO2}$. A later chapter section on reaction mechanisms will explain how a reactant’s concentration can have no effect on a reaction rate despite being involved in the reaction.
Exercise $\PageIndex{1A}$
The rate law for the reaction:
$\ce{H2(g) + 2 NO(g) -> N2O(g) + H2O(g)} \nonumber$
has been determined to be rate = k[NO]2[H2]. What are the orders with respect to each reactant, and what is the overall order of the reaction?
Answer
order in NO = 2; order in H2 = 1; overall order = 3
Exercise $\PageIndex{1B}$
In a transesterification reaction, a triglyceride reacts with an alcohol to form an ester and glycerol. Many students learn about the reaction between methanol ($\ce{CH3OH}$) and ethyl acetate ($\ce{CH3CH2OCOCH3}$) as a sample reaction before studying the chemical reactions that produce biodiesel:
$\ce{CH3OH + CH3CH2OCOCH3 -> CH3OCOCH3 + CH3CH2OH} \nonumber$
The rate law for the reaction between methanol and ethyl acetate is, under certain conditions, determined to be:
$\text { rate }=k\left[ \ce{CH3OH} \right] \nonumber$
What is the order of reaction with respect to methanol and ethyl acetate, and what is the overall order of reaction?
Answer
order in CH3OH = 1; order in CH3CH2OCOCH3 = 0; overall order = 1
A common experimental approach to the determination of rate laws is the method of initial rates. This method involves measuring reaction rates for multiple experimental trials carried out using different initial reactant concentrations. Comparing the measured rates for these trials permits determination of the reaction orders and, subsequently, the rate constant, which together are used to formulate a rate law. This approach is illustrated in the next two example exercises.
Example $2$: Determining a Rate Law from Initial Rates
Ozone in the upper atmosphere is depleted when it reacts with nitrogen oxides. The rates of the reactions of nitrogen oxides with ozone are important factors in deciding how significant these reactions are in the formation of the ozone hole over Antarctica (Figure $1$). One such reaction is the combination of nitric oxide, NO, with ozone, O3:
$\ce{NO(g) + O3(g) -> NO2(g) + O2(g)} \nonumber$
This reaction has been studied in the laboratory, and the following rate data were determined at 25 °C.
Trial [NO] (mol/L) [O3] (mol/L) $\dfrac{\Delta[ \ce{NO2}]}{\Delta t} (\text{mol}\text{L}^{-1} \text{s}^{-1})$
1 1.00 × 10−6 3.00 × 10−6 6.60 × 10−5
2 1.00 × 10−6 6.00 × 10−6 1.32 × 10−4
3 1.00 × 10−6 9.00 × 10−6 1.98 × 10−4
4 2.00 × 10−6 9.00 × 10−6 3.96 × 10−4
5 3.00 × 10−6 9.00 × 10−6 5.94 × 10−4
Determine the rate law and the rate constant for the reaction at 25 °C.
Solution
The rate law will have the form:
$\text { rate }=k[ NO ]^m\left[ O_3\right]^n \nonumber$
Determine the values of m, n, and k from the experimental data using the following three-part process:
Step 1:
Determine the value of m from the data in which [NO] varies and [O3] is constant. In the last three experiments, [NO] varies while [O3] remains constant. When [NO] doubles from trial 3 to 4, the rate doubles, and when [NO] triples from trial 3 to 5, the rate also triples. Thus, the rate is also directly proportional to [NO], and m in the rate law is equal to 1.
Step 2:
Determine the value of n from data in which [O3] varies and [NO] is constant. In the first three experiments, [NO] is constant and [O3] varies. The reaction rate changes in direct proportion to the change in [O3]. When [O3] doubles from trial 1 to 2, the rate doubles; when [O3] triples from trial 1 to 3, the rate increases also triples. Thus, the rate is directly proportional to [O3], and n is equal to 1.The rate law is thus:
$\text { rate }=k[ NO ]^1\left[ O_3\right]^1=k[ NO ]\left[ O_3\right] \nonumber$
Step 3:
Determine the value of k from one set of concentrations and the corresponding rate. The data from trial 1 are used below:
\begin{align*} k & =\dfrac{\text { rate }}{[ \ce{NO} ]\left[ \ce{O3} \right]} \[4pt] & =\dfrac{6.60 \times 10^{-5} ~\text{mol L}^{-1}\text{s}^{-1}}{\left(1.00 \times 10^{-6} ~\text{mol L}^{-1}\right)\left(3.00 \times 10^{-6}~\text{mol L}^{-1}\right)} \[4pt] & =2.20 \times 10^7~\text{L mol}^{-1}\text{s}^{-1} \end{align*}
Exercise $2$
Acetaldehyde decomposes when heated to yield methane and carbon monoxide according to the equation:
$\ce{CH3CHO(g) -> CH4(g) + CO(g)} \nonumber$
Determine the rate law and the rate constant for the reaction from the following experimental data:
Trial [CH3CHO] (mol/L) $-\dfrac{\Delta [\ce{CH3CHO}]}{\Delta t} (\text{mol}\,\text{L}^{-1}\text{s}^{-1})$
1 1.75 × 10−3 2.06 × 10−11
2 3.50 × 10−3 8.24 × 10−11
3 7.00 × 10−3 3.30 × 10−10
Answer
$\text { rate }=k\left[ \ce{CH3CHO} \right]^2$ with k = 6.7310−6 L/mol/s
Example $3$: Determining Rate Laws from Initial Rates
Using the initial rates method and the experimental data, determine the rate law and the value of the rate constant for this reaction:
$2 NO (g) + Cl_2(g) \longrightarrow 2 NOCl (g) \nonumber$
Trial [NO] (mol/L) [Cl2] (mol/L) $-\dfrac{\Delta [\ce{CO}]}{\Delta t} (\text{mol} ~L^{-1} \text{s}^{-1})$
1 0.10 0.10 0.00300
2 0.10 0.15 0.00450
3 0.15 0.10 0.00675
Solution
The rate law for this reaction will have the form:
$\text { rate }=k[ NO ]^m\left[ Cl_2\right]^n \nonumber$
As in Example $2$, approach this problem in a stepwise fashion, determining the values of m and n from the experimental data and then using these values to determine the value of k. In this example, however, an explicit algebraic approach (vs. the implicit approach of the previous example) will be used to determine the values of m and n:
Step 1
Determine the value of m from the data in which $[\ce{NO}]$ varies and $[\ce{Cl2}]$ is constant. Write the ratios with the subscripts x and y to indicate data from two different trials:
$\dfrac{\text { rate }_x}{\text { rate }_y}=\dfrac{k[ NO ]_x^m\left[ Cl_2\right]_x^n}{k[ NO ]_y^m\left[ Cl_2\right]_y^n} \nonumber$
Using the third trial and the first trial, in which $[\ce{Cl2}]$ does not vary, gives:
$\dfrac{\text { rate } 3}{\text { rate } 1}=\dfrac{0.00675}{0.00300}=\dfrac{k(0.15)^m(0.10)^n}{k(0.10)^m(0.10)^n} \nonumber$
Canceling equivalent terms in the numerator and denominator leaves:
$\dfrac{0.00675}{0.00300}=\dfrac{(0.15)^m}{(0.10)^m} \nonumber$
which simplifies to:
$2.25=(1.5)^m \nonumber$
Use logarithms to determine the value of the exponent m:
\begin{align*} \ln (2.25) & =m \ln (1.5) \[4pt] \dfrac{\ln (2.25)}{\ln (1.5)} & =m \[4pt] 2 & =m \end{align*}
Confirm the result
$1.5^2=2.25 \nonumber$
Step 2
Determine the value of $n$ from data in which $[\ce{Cl2}]$ varies and $[\ce{NO}]$ is constant.
$\dfrac{\text { rate } 2}{\text { rate } 1}=\dfrac{0.00450}{0.00300}=\dfrac{k0.10)^m(0.15)^n}{k(0.10)^m(0.10)^n} \nonumber$
Cancelation gives:
$\dfrac{0.0045}{0.0030}=\dfrac{(0.15)^n}{(0.10)^n} \nonumber$
which simplifies to:
$1.5=(1.5)^n \nonumber$
Thus $n$ must be 1, and the form of the rate law is:
$\text { rate }=k[ NO ]^m\left[ Cl_2\right]^n=k[ NO ]^2\left[ Cl_2\right] \nonumber$
Step 3.
Determine the numerical value of the constant with appropriate units. The units for the rate of a reaction are mol/L/s. The units for are whatever is needed so that substituting into the rate law expression affords the appropriate units for the rate. In this example, the concentration units are mol3/L3. The units for k should be mol L/s so that the rate is in terms of mol/L/s.
To determine the value of k once the rate law expression has been solved, simply plug in values from the first experimental trial and solve for k:
\begin{align*} 0.00300~\text{mol L}^{-1} \text{s}^{-1} & =k\left(0.10~\text{mol L}^{-1}\right)^2\left(0.10 ~\text{mol L}^{-1}\right)^1 \[4pt] k & =3.0 ~\text{mol}^{-2}\text{L}^2 \text{s}^{-1} \end{align*}
Exercise $3$
Use the provided initial rate data to derive the rate law for the reaction whose equation is:
$\ce{OCl^{-}(aq) + I^{-}(aq) -> OI^{-}(aq) + Cl^{-}(aq)} \nonumber$
Trial [OCl] (mol/L) [I] (mol/L) Initial Rate (mol/L/s)
1 0.0040 0.0020 0.00184
2 0.0020 0.0040 0.00092
3 0.0020 0.0020 0.00046
Determine the rate law expression and the value of the rate constant k with appropriate units for this reaction.
Answer
$\dfrac{\text { rate } 2}{\text { rate 3 }}=\dfrac{0.00092}{0.00046}=\dfrac{k(0.0020)^x(0.0040)^y}{k(0.0020)^x(0.0020)^y} \nonumber$
\begin{align*} 2.00 &=2.00^y \[4pt] y&=1 \end{align*}
$\dfrac{\text { rate } 1}{\text { rate } 2}=\dfrac{0.00184}{0.00092}=\dfrac{k(0.0040)^x(0.0020)^y}{k(0.0020)^x(0.0040)^y} \nonumber$
\begin{align*} 2.00 & =\dfrac{2^x}{2^y} \[4pt] 2.00 & =\dfrac{2^x}{2^1} \[4pt] 4.00 & =2^x \[4pt] x & =2 \end{align*}
Substituting the concentration data from trial 1 and solving for $k$ yields:
\begin{align*} \text { rate } & =k\left[ OCl^{-}\right]^2\left[ I^{-}\right]^1 \[4pt] 0.00184 & =k(0.0040)^2(0.0020)^1 \[4pt] k & =5.75 \times 10^4 mol^{-2} L^2 s^{-1} \end{align*}
Reaction Order and Rate Constant Units
In some of our examples, the reaction orders in the rate law happen to be the same as the coefficients in the chemical equation for the reaction. This is merely a coincidence and very often not the case.
Rate laws may exhibit fractional orders for some reactants, and negative reaction orders are sometimes observed when an increase in the concentration of one reactant causes a decrease in reaction rate. A few examples illustrating these points are provided:
\begin{align*} \ce{NO_2 + CO & -> NO + CO_2} && \text { rate }=k\left[ NO_2\right]^2 \[4pt] \ce{CH_3 CHO & -> CH_4 + CO} && \text { rate }=k\left[ CH_3 CHO \right]^2 \[4pt] \ce{2 N_2 O_5 & -> NO_2 + O_2} && \text { rate }=k\left[ N_2 O_5\right] \[4pt] \ce{2 NO_2 + F_2 & -> 2 NO_2 F} && \text { rate }=k\left[ NO_2\right]\left[ F_2\right] \[4pt] \ce{2 NO_2 Cl & -> 2 NO_2 + Cl_2} && \text { rate }=k\left[ NO_2 Cl \right] \end{align*} \nonumber
Note
It is important to note that rate laws are determined by experiment only and are not reliably predicted by reaction stoichiometry.
The units for a rate constant will vary as appropriate to accommodate the overall order of the reaction. The unit of the rate constant for the second-order reaction described in Example $2$ was determined to be $k$ was derived to be $\text{L}\,\text{mol}^{−1}\text{s}^{−1}$. For the third-order reaction, $k$ was derived to be $\text{L}^{2}\text{mol}^{−2}\text{s}^{−1}$. Dimensional analysis requires the rate constant unit for a reaction whose overall order is x to be $\text{L}^{x-1}\,\text{mol}^{1-x}\text{s}^{−1}$. Table $1$ summarizes the rate constant units for common reaction orders.
Table $1$: Rate Constant Units for Common Reaction Orders
Overall Reaction Order (x) Rate Constant Unit (Lx−1 mol1x s−1)
0 (zero) mol L−1 s−1
1 (first) s−1
2 (second) L mol−1 s−1
3 (third) L2 mol−2 s−1
Note that the units in this table were derived using specific units for concentration (mol/L) and time (s), though any valid units for these two properties may be used. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/12%3A_Kinetics/12.03%3A_Rate_Laws.txt |
Learning Objectives
By the end of this section, you will be able to:
• Explain the form and function of an integrated rate law
• Perform integrated rate law calculations for zero-, first-, and second-order reactions
• Define half-life and carry out related calculations
• Identify the order of a reaction from concentration/time data
The rate laws discussed thus far relate the rate and the concentrations of reactants. We can also determine a second form of each rate law that relates the concentrations of reactants and time. These are called integrated rate laws. We can use an integrated rate law to determine the amount of reactant or product present after a period of time or to estimate the time required for a reaction to proceed to a certain extent. For example, an integrated rate law is used to determine the length of time a radioactive material must be stored for its radioactivity to decay to a safe level.
Using calculus, the differential rate law for a chemical reaction can be integrated with respect to time to give an equation that relates the amount of reactant or product present in a reaction mixture to the elapsed time of the reaction. This process can either be very straightforward or very complex, depending on the complexity of the differential rate law. For purposes of discussion, we will focus on the resulting integrated rate laws for first-, second-, and zero-order reactions.
First-Order Reactions
Integration of the rate law for a simple first-order reaction (rate = k[A]) results in an equation describing how the reactant concentration varies with time:
$[A]_t=[A]_0 e^{-k t} \nonumber$
where [A]t is the concentration of A at any time t, [A]0 is the initial concentration of A, and k is the first-order rate constant.
For mathematical convenience, this equation may be rearranged to other formats, including direct and indirect proportionalities:
$\ln \left(\frac{[A]_t}{[A]_0}\right)=-k t \nonumber$
or
$\ln \left(\frac{[A]_0}{[A]_t}\right)=k t \nonumber$
and a format showing a linear dependence of concentration in time:
$\ln [A]_t=\ln [A]_0-k t \nonumber$
Example $1$: The Integrated Rate Law for a First-Order Reaction
The rate constant for the first-order decomposition of cyclobutane, $\ce{C4H8}$ at 500 °C is 9.2 10−3 s−1:
$\ce{C_4H_8 \longrightarrow 2 C_2H_4} \nonumber$
How long will it take for 80.0% of a sample of $\ce{C4H8}$ to decompose?
Solution
Since the relative change in reactant concentration is provided, a convenient format for the integrated rate law is:
$\ln \left(\frac{[A]_0}{[A]_t}\right)=k t \nonumber$
The initial concentration of $\ce{C4H8}$, [A]0, is not provided, but the provision that 80.0% of the sample has decomposed is enough information to solve this problem. Let x be the initial concentration, in which case the concentration after 80.0% decomposition is 20.0% of x or 0.200x. Rearranging the rate law to isolate t and substituting the provided quantities yields:
\begin{align*} t & =\ln \frac{[x]}{[0.200 x]} \times \frac{1}{k} \[4pt] & =\ln 5 \times \frac{1}{9.2 \times 10^{-3} ~\text{s}^{-1}} \[4pt] & =1.609 \times \frac{1}{9.2 \times 10^{-3} ~\text{s}^{-1}} \[4pt] & =1.7 \times 10^2 ~\text{s} \end{align*}
Exercise $1$
Iodine-131 is a radioactive isotope that is used to diagnose and treat some forms of thyroid cancer. Iodine-131 decays to xenon-131 according to the equation:
$\text{I-131} \longrightarrow \text{Xe-131} + \text{electron } \nonumber$
The decay is first-order with a rate constant of 0.138 d−1. How many days will it take for 90% of the iodine−131 in a 0.500 M solution of this substance to decay to Xe-131?
Answer
16.7 days
In the next example exercise, a linear format for the integrated rate law will be convenient:
\begin{align*} \ln [A]_t & =(-k)(t) + \ln [A]_0 \[4pt] y & =m x+b \end{align*}
A plot of ln[A]t versus t for a first-order reaction is a straight line with a slope of −k and a y-intercept of ln[A]0. If a set of rate data are plotted in this fashion but do not result in a straight line, the reaction is not first order in A.
Example $2$: Graphical Determination of Reaction Order and Rate Constant
Show that the data in Figure $1$ can be represented by a first-order rate law by graphing ln[H2O2] versus time. Determine the rate constant for the decomposition of H2O2 from these data.
Solution
The data from Figure $1$ are tabulated below, and a plot of ln[H2O2] is shown in Figure $1$.
Time (h) [H2O2] (M) ln[H2O2]
0.00 1.000 0.000
6.00 0.500 −0.693
12.00 0.250 −1.386
18.00 0.125 −2.079
24.00 0.0625 −2.772
The plot of ln[H2O2] versus time is linear, indicating that the reaction may be described by a first-order rate law.
According to the linear format of the first-order integrated rate law, the rate constant is given by the negative of this plot’s slope.
$\text { slope }=\frac{\text { change in } y}{\text { change in } x}=\frac{\Delta y}{\Delta x}=\frac{\Delta \ln \left[ H_2 O_2\right]}{\Delta t} \nonumber$
The slope of this line may be derived from two values of ln[H2O2] at different values of t (one near each end of the line is preferable). For example, the value of ln[H2O2] when t is 0.00 h is 0.000; the value when t = 24.00 h is −2.772
\begin{align*} \text { slope } & =\frac{-2.772-0.000}{24.00-0.00 h } \[4pt] & =\frac{-2.772}{24.00 h } \[4pt] & =-0.116 h^{-1} \[4pt] k & =- \text { slope }=-\left(-0.116 h^{-1}\right)=0.116 h^{-1} \end{align*}
Exercise $2$
Graph the following data to determine whether the reaction $A \longrightarrow B+C$ is first order.
Time (s) [A]
4.0 0.220
8.0 0.144
12.0 0.110
16.0 0.088
20.0 0.074
Answer
The plot of ln[A]t vs. t is not linear, indicating the reaction is not first order:
Second-Order Reactions
The equations that relate the concentrations of reactants and the rate constant of second-order reactions can be fairly complicated. To illustrate the point with minimal complexity, only the simplest second-order reactions will be described here, namely, those whose rates depend on the concentration of just one reactant. For these types of reactions, the differential rate law is written as:
$\text { rate }=k[A]^2 \nonumber$
For these second-order reactions, the integrated rate law is:
$\frac{1}{[A]_t}=k t+\frac{1}{[A]_0} \nonumber$
where the terms in the equation have their usual meanings as defined earlier.
Example $3$: The Integrated Rate Law for a Second-Order Reaction
The reaction of butadiene gas (C4H6) to yield C8H12 gas is described by the equation:
$2 C_4 H_6(g) \longrightarrow C_8 H_{12}(g) \nonumber$
This “dimerization” reaction is second order with a rate constant equal to 5.76 10−2 L mol−1 min−1 under certain conditions. If the initial concentration of butadiene is 0.200 M, what is the concentration after 10.0 min?
Solution
For a second-order reaction, the integrated rate law is written
$\frac{1}{[A]_t}=k t+\frac{1}{[A]_0} \nonumber$
We know three variables in this equation: [A]0 = 0.200 mol/L, k = 5.76 10−2 L/mol/min, and t = 10.0 min. Therefore, we can solve for [A], the fourth variable:
\begin{align*} \frac{1}{[A]_t} & =\left(5.76 \times 10^{-2} L mol^{-1} min^{-1}\right)(10 min ) + \dfrac{1}{0.200 mol^{-1}} \[4pt] & =\left(5.76 \times 10^{-1} L mol^{-1}\right) + 5.00 L mol^{-1} \[4pt] & =5.58 L mol^{-1} \[4pt] [A]_t & =1.79 \times 10^{-1} mol L^{-1} \end{align*}
Therefore 0.179 mol/L of butadiene remain at the end of 10.0 min, compared to the 0.200 mol/L that was originally present.
Exercise $3$
If the initial concentration of butadiene is 0.0200 M, what is the concentration remaining after 20.0 min?
Answer
0.0195 mol/L
The integrated rate law for second-order reactions has the form of the equation of a straight line:
\begin{align*} \frac{1}{[A]_t} & =k t+\frac{1}{[A]_0} \[4pt] y & =m x+b \end{align*} \nonumber
A plot of $\frac{1}{[A]_t}$ versus $t$ for a second-order reaction is a straight line with a slope of k and a y-intercept of $\frac{1}{[A]_0}$. If the plot is not a straight line, then the reaction is not second order.
Example $4$: Graphical Determination of Reaction Order and Rate Constant
The data below are for the same reaction described in Example $3$. Prepare and compare two appropriate data plots to identify the reaction as being either first or second order. After identifying the reaction order, estimate a value for the rate constant.
Solution
Time (s) [C4H6] (M)
0 1.00 × 10−2
1600 5.04 × 10−3
3200 3.37 × 10−3
4800 2.53 × 10−3
6200 2.08 × 10−3
In order to distinguish a first-order reaction from a second-order reaction, prepare a plot of ln[C4H6]t versus t and compare it to a plot of $\frac{1}{[\ce{C4H6}]_t}$ versus $t$. The values needed for these plots follow.
Time (s) $\frac{1}{[ \ce{C4H6}]} (\text{M}^{-1})$ $\ln[\ce{C4H6}]$
0 100 −4.605
1600 198 −5.289
3200 296 −5.692
4800 395 −5.978
6200 481 −6.175
The plots are shown in Figure $2$, which clearly shows the plot of $\ln[\ce{C4H6}]_t$ versus $t$ is not linear, therefore the reaction is not first order. The plot of $\frac{1}{[\ce{C4H6}]_t}$ versus $t$ is linear, indicating that the reaction is second order.
According to the second-order integrated rate law, the rate constant is equal to the slope of the versus t plot. Using the data for t = 0 s and t = 6200 s, the rate constant is estimated as follows:
$k=\text { slope }=\frac{\left(481 M^{-1}-100 M^{-1}\right)}{(6200 s -0 s )}=0.0614 M^{-1} s^{-1} \nonumber$
Exercise $4$
Do the following data fit a second-order rate law?
Time (s) [A] (M)
5 0.952
10 0.625
15 0.465
20 0.370
25 0.308
35 0.230
Answer
Yes. The plot of $\frac{1}{[A]_t}$ vs. $t$ is linear:
Zero-Order Reactions
For zero-order reactions, the differential rate law is:
$\text { rate }=k \nonumber$
A zero-order reaction thus exhibits a constant reaction rate, regardless of the concentration of its reactant(s). This may seem counterintuitive, since the reaction rate certainly can’t be finite when the reactant concentration is zero. For purposes of this introductory text, it will suffice to note that zero-order kinetics are observed for some reactions only under certain specific conditions. These same reactions exhibit different kinetic behaviors when the specific conditions aren’t met, and for this reason the more prudent term pseudo-zero-order is sometimes used.
The integrated rate law for a zero-order reaction is a linear function:
\begin{align*} {[A]_t } & =-k t+[A]_0 \[4pt] y & =m x+b \end{align*}
A plot of [A] versus t for a zero-order reaction is a straight line with a slope of −k and a y-intercept of [A]0. Figure $3$: shows a plot of [NH3] versus t for the thermal decomposition of ammonia at the surface of two different heated solids. The decomposition reaction exhibits first-order behavior at a quartz (SiO2) surface, as suggested by the exponentially decaying plot of concentration versus time. On a tungsten surface, however, the plot is linear, indicating zero-order kinetics.
Example $5$: Graphical Determination of Zero-Order Rate Constant
Use the data plot in Figure $3$ to graphically estimate the zero-order rate constant for ammonia decomposition at a tungsten surface.
Solution
The integrated rate law for zero-order kinetics describes a linear plot of reactant concentration, [A]t, versus time, t, with a slope equal to the negative of the rate constant, −k. Following the mathematical approach of previous examples, the slope of the linear data plot (for decomposition on W) is estimated from the graph. Using the ammonia concentrations at t = 0 and t = 1000 s:
$k=- \text { slope }=-\frac{\left(0.0015 mol L^{-1}-0.0028 mol L^{-1}\right)}{(1000 s -0 s )}=1.3 \times 10^{-6} mol L^{-1} s^{-1} \nonumber$
Exercise $5$
The zero-order plot in Figure $3$ shows an initial ammonia concentration of 0.0028 mol L−1 decreasing linearly with time for 1000 s. Assuming no change in this zero-order behavior, at what time (min) will the concentration reach 0.0001 mol L−1?
Answer
35 min
The Half-Life of a Reaction
The half-life of a reaction (t1/2) is the time required for one-half of a given amount of reactant to be consumed. In each succeeding half-life, half of the remaining concentration of the reactant is consumed. Using the decomposition of hydrogen peroxide (Figure $1$) as an example, we find that during the first half-life (from 0.00 hours to 6.00 hours), the concentration of H2O2 decreases from 1.000 M to 0.500 M. During the second half-life (from 6.00 hours to 12.00 hours), it decreases from 0.500 M to 0.250 M; during the third half-life, it decreases from 0.250 M to 0.125 M. The concentration of H2O2 decreases by half during each successive period of 6.00 hours. The decomposition of hydrogen peroxide is a first-order reaction, and, as can be shown, the half-life of a first-order reaction is independent of the concentration of the reactant. However, half-lives of reactions with other orders depend on the concentrations of the reactants.
First-Order Reactions
An equation relating the half-life of a first-order reaction to its rate constant may be derived from the integrated rate law as follows:
\begin{align*} \ln \frac{[A]_0}{[A]_t} & =k t \[4pt] t & =\ln \frac{[A]_0}{[A]_t} \times \frac{1}{k} \end{align*}
Invoking the definition of half-life, symbolized requires that the concentration of A at this point is one-half its initial concentration: $t=t_{1 / 2}$, $[A]_t=\frac{1}{2}[A]_0$
Substituting these terms into the rearranged integrated rate law and simplifying yields the equation for half-life:
\begin{align*} t_{1 / 2} & =\ln \frac{[A]_0}{\frac{1}{2}[A]_0} \times \frac{1}{k} \[4pt] & =\ln 2 \times \frac{1}{k}=0.693 \times \frac{1}{k} \[4pt] t_{1 / 2} & =\frac{0.693}{k} \end{align*}
This equation describes an expected inverse relation between the half-life of the reaction and its rate constant, k. Faster reactions exhibit larger rate constants and correspondingly shorter half-lives. Slower reactions exhibit smaller rate constants and longer half-lives.
Example $6$: Calculation of a First-order Rate Constant using Half-Life
Calculate the rate constant for the first-order decomposition of hydrogen peroxide in water at 40 °C, using the data given in Figure $4$.
Solution
Inspecting the concentration/time data in Figure $4$ shows the half-life for the decomposition of H2O2 is 2.16 × 104 s:
\begin{align*} t_{1 / 2} & =\frac{0.693}{k} \[4pt] k & =\frac{0.693}{t_{1 / 2}}=\frac{0.693}{2.16 \times 10^4 s }=3.21 \times 10^{-5} s^{-1} \end{align*}
Exercise $1$
The first-order radioactive decay of iodine-131 exhibits a rate constant of 0.138 d−1. What is the half-life for this decay?
Answer
5.02 d.
Second-Order Reactions
Following the same approach as used for first-order reactions, an equation relating the half-life of a second-order reaction to its rate constant and initial concentration may be derived from its integrated rate law:
$\frac{1}{[A]_t}=k t+\frac{1}{[A]_0} \nonumber$
or
$\frac{1}{[A]}-\frac{1}{[A]_0}=k t \nonumber$
Restrict t to t1/2
$t=t_{1 / 2} \nonumber$
define [A]t as one-half [A]0
$[A]_t=\frac{1}{2}[A]_0 \nonumber$
and then substitute into the integrated rate law and simplify:
\begin{align*} \frac{1}{\frac{1}{2}[A]_0}-\frac{1}{[A]_0} & =k t_{1 / 2} \[4pt] \frac{2}{[A]_0}-\frac{1}{[A]_0} & =k t_{1 / 2} \[4pt] \frac{1}{[A]_0} & =k t_{1 / 2} \[4pt] t_{1 / 2} & =\frac{1}{k[A]_0} \end{align*}
For a second-order reaction, $t_{1/2}$ is inversely proportional to the concentration of the reactant, and the half-life increases as the reaction proceeds because the concentration of reactant decreases. Unlike with first-order reactions, the rate constant of a second-order reaction cannot be calculated directly from the half-life unless the initial concentration is known.
Zero-Order Reactions
As for other reaction orders, an equation for zero-order half-life may be derived from the integrated rate law:
$[A]=-k t+[A]_0 \nonumber$
Restricting the time and concentrations to those defined by half-life: and Substituting these terms into the zero-order integrated rate law yields:
\begin{align*} \frac{[ A ]_0}{2} & =-k t_{1 / 2}+[ A ]_0 \[4pt] k t_{1 / 2} & =\frac{[ A ]_0}{2} \[4pt] t_{1 / 2} & =\frac{[A]_0}{2 k} \end{align*}
As for all reaction orders, the half-life for a zero-order reaction is inversely proportional to its rate constant. However, the half-life of a zero-order reaction increases as the initial concentration increases.
Equations for both differential and integrated rate laws and the corresponding half-lives for zero-, first-, and second-order reactions are summarized in Table $1$.
Table $1$: Summary of Rate Laws for Zero-, First-, and Second-Order Reactions
Zero-Order First-Order Second-Order
rate law rate = k rate = k[A] rate = k[A]2
units of rate constant M s−1 s−1 M−1 s−1
integrated rate law $[A]=-k t+[A]_0$ $\ln [A]=-k t+\ln [A]_0$ $\frac{1}{[A]}=k t+\left(\frac{1}{[A]_0}\right)$
plot needed for linear fit of rate data [A] vs. t ln[A] vs. t vs. t
relationship between slope of linear plot and rate constant k = −slope k = −slope k = slope
half-life $t_{1 / 2}=\frac{[A]_0}{2 k}$ $t_{1 / 2}=\frac{0.693}{k}$ $t_{1 / 2}=\frac{1}{[A]_0 k}$
Example $7$: Half-Life for Zero-Order and Second-Order Reactions
What is the half-life for the butadiene dimerization reaction described in Example $3$?
Solution
The reaction in question is second order, is initiated with a 0.200 mol L−1 reactant solution, and exhibits a rate constant of 0.0576 L mol−1 min−1. Substituting these quantities into the second-order half-life equation:
$t_{1 / 2}=\frac{1}{\left[\left(0.0576 L mol^{-1} min^{-1}\right)\left(0.200 mol L^{-1}\right)\right]}=86.8~\text{min} \nonumber$
Exercise $7$
What is the half-life (min) for the thermal decomposition of ammonia on tungsten (see Figure $3$)?
Answer
87 min | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/12%3A_Kinetics/12.04%3A_Integrated_Rate_Laws.txt |
Learning Objectives
By the end of this section, you will be able to:
• Use the postulates of collision theory to explain the effects of physical state, temperature, and concentration on reaction rates
• Define the concepts of activation energy and transition state
• Use the Arrhenius equation in calculations relating rate constants to temperature
We should not be surprised that atoms, molecules, or ions must collide before they can react with each other. Atoms must be close together to form chemical bonds. This simple premise is the basis for a very powerful theory that explains many observations regarding chemical kinetics, including factors affecting reaction rates.
Collision theory is based on the following postulates:
1. The rate of a reaction is proportional to the rate of reactant collisions: $\text { reaction rate } \propto \frac{\# \text { collisions }}{\text { time }} \nonumber$
2. The reacting species must collide in an orientation that allows contact between the atoms that will become bonded together in the product.
3. The collision must occur with adequate energy to permit mutual penetration of the reacting species’ valence shells so that the electrons can rearrange and form new bonds (and new chemical species).
We can see the importance of the two physical factors noted in postulates 2 and 3, the orientation and energy of collisions, when we consider the reaction of carbon monoxide with oxygen:
$\ce{2 CO (g) + O_2(g) \longrightarrow 2 CO_2(g)} \nonumber$
Carbon monoxide is a pollutant produced by the combustion of hydrocarbon fuels. To reduce this pollutant, automobiles have catalytic converters that use a catalyst to carry out this reaction. It is also a side reaction of the combustion of gunpowder that results in muzzle flash for many firearms. If carbon monoxide and oxygen are present in sufficient amounts, the reaction will occur at high temperature and pressure.
The first step in the gas-phase reaction between carbon monoxide and oxygen is a collision between the two molecules:
$\ce{CO (g) + O_2(g) \longrightarrow CO_2(g) + O (g)} \nonumber$
Although there are many different possible orientations the two molecules can have relative to each other, consider the two presented in Figure $1$. In the first case, the oxygen side of the carbon monoxide molecule collides with the oxygen molecule. In the second case, the carbon side of the carbon monoxide molecule collides with the oxygen molecule. The second case is clearly more likely to result in the formation of carbon dioxide, which has a central carbon atom bonded to two oxygen atoms (\ce{O=C=O}\). This is a rather simple example of how important the orientation of the collision is in terms of creating the desired product of the reaction.
If the collision does take place with the correct orientation, there is still no guarantee that the reaction will proceed to form carbon dioxide. In addition to a proper orientation, the collision must also occur with sufficient energy to result in product formation. When reactant species collide with both proper orientation and adequate energy, they combine to form an unstable species called an activated complex or a transition state. These species are very short lived and usually undetectable by most analytical instruments. In some cases, sophisticated spectral measurements have been used to observe transition states.
Collision theory explains why most reaction rates increase as concentrations increase. With an increase in the concentration of any reacting substance, the chances for collisions between molecules are increased because there are more molecules per unit of volume. More collisions mean a faster reaction rate, assuming the energy of the collisions is adequate.
Activation Energy and the Arrhenius Equation
The minimum energy necessary to form a product during a collision between reactants is called the activation energy (Ea). How this energy compares to the kinetic energy provided by colliding reactant molecules is a primary factor affecting the rate of a chemical reaction. If the activation energy is much larger than the average kinetic energy of the molecules, the reaction will occur slowly since only a few fast-moving molecules will have enough energy to react. If the activation energy is much smaller than the average kinetic energy of the molecules, a large fraction of molecules will be adequately energetic and the reaction will proceed rapidly.
Figure $2$ shows how the energy of a chemical system changes as it undergoes a reaction converting reactants to products according to the equation
$\ce{A+B \longrightarrow C+D} \nonumber$
These reaction diagrams are widely used in chemical kinetics to illustrate various properties of the reaction of interest. Viewing the diagram from left to right, the system initially comprises reactants only, $\ce{A + B}$. Reactant molecules with sufficient energy can collide to form a high-energy activated complex or transition state. The unstable transition state can then subsequently decay to yield stable products, C + D. The diagram depicts the reaction's activation energy, Ea, as the energy difference between the reactants and the transition state. Using a specific energy, the enthalpy (see chapter on thermochemistry), the enthalpy change of the reaction, ΔH, is estimated as the energy difference between the reactants and products. In this case, the reaction is exothermic (ΔH < 0) since it yields a decrease in system enthalpy.
The Arrhenius equation relates the activation energy and the rate constant, k, for many chemical reactions:
$k=A e^{-E_{ a } / R T} \nonumber$
In this equation, $R$ is the ideal gas constant, which has a value 8.314 J/mol/K, $T$ is temperature on the Kelvin scale, $E_a$ is the activation energy in joules per mole, e is the constant 2.7183, and A is a constant called the frequency factor, which is related to the frequency of collisions and the orientation of the reacting molecules.
Postulates of collision theory are nicely accommodated by the Arrhenius equation. The frequency factor, A, reflects how well the reaction conditions favor properly oriented collisions between reactant molecules. An increased probability of effectively oriented collisions results in larger values for A and faster reaction rates.
The exponential term, e−Ea/RT, describes the effect of activation energy on reaction rate. According to kinetic molecular theory (see chapter on gases), the temperature of matter is a measure of the average kinetic energy of its constituent atoms or molecules. The distribution of energies among the molecules composing a sample of matter at any given temperature is described by the plot shown in Figure $\PageIndex{3a}$. Two shaded areas under the curve represent the numbers of molecules possessing adequate energy (RT) to overcome the activation barriers (Ea). A lower activation energy results in a greater fraction of adequately energized molecules and a faster reaction.
The exponential term also describes the effect of temperature on reaction rate. A higher temperature represents a correspondingly greater fraction of molecules possessing sufficient energy (RT) to overcome the activation barrier (Ea), as shown in Figure $\PageIndex{3b}$. This yields a greater value for the rate constant and a correspondingly faster reaction rate.
A convenient approach for determining $E_a$ for a reaction involves the measurement of k at two or more different temperatures and using an alternate version of the Arrhenius equation that takes the form of a linear equation
\begin{align*} \ln k & =\left(\frac{-E_{ a }}{R}\right)\left(\frac{1}{T}\right) + \ln A \[4pt] y & =m x+b \end{align*}
A plot of $\ln k$ versus $\frac{1}{T}$ is linear with a slope equal to $\frac{-E_a}{R}$ and a y-intercept equal to $\ln A$.
Example $1$: Determination of Ea
The variation of the rate constant with temperature for the decomposition of $\ce{HI(g)}$ to $\ce{H2(g)}$ and $\ce{I2(g)}$ is given here. What is the activation energy for the reaction?
$\ce{2 HI (g) \longrightarrow H_2(g) + I_2(g)} \nonumber$
T (K) k (L/mol/s)
555 3.52 × 10−7
575 1.22 × 10−6
645 8.59 × 10−5
700 1.16 × 10−3
781 3.95 × 10−2
Solution
Use the provided data to derive values of $\frac{1}{T}$ and $\ln k$:
$\frac{1}{T}$ $\ln k$
1.80 × 10−3 −14.860
1.74 × 10−3 −13.617
1.55 × 10−3 −9.362
1.43 × 10−3 −6.759
1.28 × 10−3 −3.231
Figure $4$ is a graph of $\ln k$ versus $\frac{1}{T}$. In practice, the equation of the line (slope and y-intercept) that best fits these plotted data points would be derived using a statistical process called regression. This is helpful for most experimental data because a perfect fit of each data point with the line is rarely encountered. For the data here, the fit is nearly perfect and the slope may be estimated using any two of the provided data pairs. Using the first and last data points permits estimation of the slope.
\begin{align*} \text { Slope }= & \frac{\Delta(\ln k)}{\Delta\left(\frac{1}{T}\right)} \[4pt] = & \frac{(-14.860)-(-3.231)}{\left(1.80 \times 10^{-3} K^{-1}\right)-\left(1.28 \times 10^{-3} K^{-1}\right)} \[4pt] = & \frac{-11.629}{0.52 \times 10^{-3} K^{-1}}=-2.2 \times 10^4 K \[4pt] = & -\frac{E_{ a }}{R} \[4pt] E_{ a }= & - \text { slope } \times R=-\left(-2.2 \times 10^4 K \times 8.314 J mol^{-1} K^{-1}\right) \[4pt] & 1.8 \times 10^5 J mol^{-1} \text { or } 180 kJ mol^{-1} \end{align*}
Alternative approach: A more expedient approach involves deriving activation energy from measurements of the rate constant at just two temperatures. In this approach, the Arrhenius equation is rearranged to a convenient two-point form:
$\ln \frac{k_1}{k_2}=\frac{E_{ a }}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right) \nonumber$
Rearranging this equation to isolate activation energy yields:
$E_{ a }=-R\left(\frac{\ln k_2-\ln k_1}{\left(\frac{1}{T_2}\right)-\left(\frac{1}{T_1}\right)}\right) \nonumber$
Any two data pairs may be substituted into this equation—for example, the first and last entries from the above data table:
$E_{ a }=-8.314 J mol^{-1} K^{-1}\left(\frac{-3.231-(-14.860)}{1.28 \times 10^{-3} K^{-1}-1.80 \times 10^{-3} K^{-1}}\right) \nonumber$
and the result is Ea = 1.8 105 J mol−1 or 180 kJ mol−1
This approach yields the same result as the more rigorous graphical approach used above, as expected. In practice, the graphical approach typically provides more reliable results when working with actual experimental data.
Exercise $1$
The rate constant for the rate of decomposition of $\ce{N2O5}$ to $\ce{NO}$ and $\ce{O2}$ in the gas phase is 1.66 L/mol/s at 650 K and 7.39 L/mol/s at 700 K:
$\ce{2 N2O5(g) -> 4 NO(g) + 3 O2(g)} \nonumber$
Assuming the kinetics of this reaction are consistent with the Arrhenius equation, calculate the activation energy for this decomposition.
Answer
1.1 105 J mol−1 or 110 kJ mol−1 | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/12%3A_Kinetics/12.05%3A_Collision_Theory.txt |
Learning Objectives
By the end of this section, you will be able to:
• Distinguish net reactions from elementary reactions (steps)
• Identify the molecularity of elementary reactions
• Write a balanced chemical equation for a process given its reaction mechanism
• Derive the rate law consistent with a given reaction mechanism
Chemical reactions very often occur in a step-wise fashion, involving two or more distinct reactions taking place in sequence. A balanced equation indicates what is reacting and what is produced, but it reveals no details about how the reaction actually takes place. The reaction mechanism (or reaction path) provides details regarding the precise, step-by-step process by which a reaction occurs.
The decomposition of ozone, for example, appears to follow a mechanism with two steps:
\begin{align*} & O_3(g) \longrightarrow O_2(g) + O \[4pt] & O + O_3(g) \longrightarrow 2 O_2(g) \end{align*}
Each of the steps in a reaction mechanism is an elementary reaction. These elementary reactions occur precisely as represented in the step equations, and they must sum to yield the balanced chemical equation representing the overall reaction:
$2 O_3(g) \longrightarrow 3 O_2(g) \nonumber$
Notice that the oxygen atom produced in the first step of this mechanism is consumed in the second step and therefore does not appear as a product in the overall reaction. Species that are produced in one step and consumed in a subsequent step are called intermediates.
While the overall reaction equation for the decomposition of ozone indicates that two molecules of ozone react to give three molecules of oxygen, the mechanism of the reaction does not involve the direct collision and reaction of two ozone molecules. Instead, one O3 decomposes to yield O2 and an oxygen atom, and a second O3 molecule subsequently reacts with the oxygen atom to yield two additional O2 molecules.
Unlike balanced equations representing an overall reaction, the equations for elementary reactions are explicit representations of the chemical change taking place. The reactant(s) in an elementary reaction’s equation undergo only the bond-breaking and/or making events depicted to yield the product(s). For this reason, the rate law for an elementary reaction may be derived directly from the balanced chemical equation describing the reaction. This is not the case for typical chemical reactions, for which rate laws may be reliably determined only via experimentation.
Unimolecular Elementary Reactions
The molecularity of an elementary reaction is the number of reactant species (atoms, molecules, or ions). For example, a unimolecular reaction involves the reaction of a single reactant species to produce one or more molecules of product:
$A \longrightarrow \text { products } \nonumber$
The rate law for a unimolecular reaction is first order:
$\text { rate }=k[A] \nonumber$
A unimolecular reaction may be one of several elementary reactions in a complex mechanism. For example, the reaction:
$O_3 \longrightarrow O_2 + O \nonumber$
illustrates a unimolecular elementary reaction that occurs as one part of a two-step reaction mechanism as described above. However, some unimolecular reactions may be the only step of a single-step reaction mechanism. (In other words, an “overall” reaction may also be an elementary reaction in some cases.) For example, the gas-phase decomposition of cyclobutane, C4H8, to ethylene, C2H4, is represented by the following chemical equation:
This equation represents the overall reaction observed, and it might also represent a legitimate unimolecular elementary reaction. The rate law predicted from this equation, assuming it is an elementary reaction, turns out to be the same as the rate law derived experimentally for the overall reaction, namely, one showing first-order behavior:
$\text { rate }=-\frac{\Delta\left[ C_4 H_8\right]}{\Delta t}=k\left[ C_4 H_8\right] \nonumber$
This agreement between observed and predicted rate laws is interpreted to mean that the proposed unimolecular, single-step process is a reasonable mechanism for the butadiene reaction.
Bimolecular Elementary Reactions
A bimolecular reaction involves two reactant species, for example:
$A+B \longrightarrow \text { products } \nonumber$
and
$2 A \longrightarrow \text { products } \nonumber$
For the first type, in which the two reactant molecules are different, the rate law is first-order in A and first order in B (second-order overall):
$\text { rate }=k[A][B] \nonumber$
For the second type, in which two identical molecules collide and react, the rate law is second order in A:
$\text { rate }=k[A][A]=k[A]^2 \nonumber$
Some chemical reactions occur by mechanisms that consist of a single bimolecular elementary reaction. One example is the reaction of nitrogen dioxide with carbon monoxide:
$NO_2(g) + CO (g) \longrightarrow NO (g) + CO_2(g) \nonumber$
(see Figure $1$)
Bimolecular elementary reactions may also be involved as steps in a multistep reaction mechanism. The reaction of atomic oxygen with ozone is the second step of the two-step ozone decomposition mechanism discussed earlier in this section:
$O (g) + O_3(g) \longrightarrow 2 O_2(g) \nonumber$
Termolecular Elementary Reactions
An elementary termolecular reaction involves the simultaneous collision of three atoms, molecules, or ions. Termolecular elementary reactions are uncommon because the probability of three particles colliding simultaneously is less than one one-thousandth of the probability of two particles colliding. There are, however, a few established termolecular elementary reactions. The reaction of nitric oxide with oxygen appears to involve termolecular steps:
\begin{align*} & 2 NO + O_2 \longrightarrow 2 NO_2 \[4pt] & \text { rate }=k[ NO ]^2\left[ O_2\right] \end{align*}
Likewise, the reaction of nitric oxide with chlorine appears to involve termolecular steps:
\begin{align*} & 2 NO + Cl_2 \longrightarrow 2 NOCl \[4pt] & \text { rate }=k[ NO ]^2\left[ Cl_2\right] \end{align*}
Relating Reaction Mechanisms to Rate Laws
It’s often the case that one step in a multistep reaction mechanism is significantly slower than the others. Because a reaction cannot proceed faster than its slowest step, this step will limit the rate at which the overall reaction occurs. The slowest step is therefore called the rate-limiting step (or rate-determining step) of the reaction Figure $2$.
As described earlier, rate laws may be derived directly from the chemical equations for elementary reactions. This is not the case, however, for ordinary chemical reactions. The balanced equations most often encountered represent the overall change for some chemical system, and very often this is the result of some multistep reaction mechanisms. In every case, the rate law must be determined from experimental data and the reaction mechanism subsequently deduced from the rate law (and sometimes from other data). The reaction of NO2 and CO provides an illustrative example:
$NO_2(g) + CO (g) \longrightarrow CO_2(g) + NO (g) \nonumber$
For temperatures above 225 °C, the rate law has been found to be:
$\text { rate }=k\left[ NO_2\right][ CO ] \nonumber$
The reaction is first order with respect to NO2 and first-order with respect to CO. This is consistent with a single-step bimolecular mechanism and it is possible that this is the mechanism for this reaction at high temperatures.
At temperatures below 225 °C, the reaction is described by a rate law that is second order with respect to NO2:
$\text { rate }=k\left[ NO_2\right]^2 \nonumber$
This rate law is not consistent with the single-step mechanism, but is consistent with the following two-step mechanism:
\begin{align*} & NO_2(g) + NO_2(g) \longrightarrow NO_3(g) + NO (g) \text { (slow) } \[4pt] & NO_3(g) + CO (g) \longrightarrow NO_2(g) + CO_2(g) \text { (fast) } \end{align*}
The rate-determining (slower) step gives a rate law showing second-order dependence on the NO2 concentration, and the sum of the two equations gives the net overall reaction.
In general, when the rate-determining (slower) step is the first step in a mechanism, the rate law for the overall reaction is the same as the rate law for this step. However, when the rate-determining step is preceded by a step involving a rapidly reversible reaction the rate law for the overall reaction may be more difficult to derive.
As discussed in several chapters of this text, a reversible reaction is at equilibrium when the rates of the forward and reverse processes are equal. Consider the reversible elementary reaction in which NO dimerizes to yield an intermediate species N2O2.
$NO + NO \rightleftharpoons N_2 O_2 \nonumber$
When this reaction is at equilibrium:
\begin{align*} & \text { rate }_{\text {forward }}=\text { rate }_{\text {reverse }} \[4pt] & k_1[ NO ]^2=k_{-1}\left[ N_2 O_2\right] \end{align*}
This expression may be rearranged to express the concentration of the intermediate in terms of the reactant NO:
$\left(\frac{ k_1[ NO ]^2}{ k_{-1}}\right)=\left[ N_2 O_2\right] \nonumber$
Since intermediate species concentrations are not used in formulating rate laws for overall reactions, this approach is sometimes necessary, as illustrated in the following example exercise.
Example $1$: Deriving a Rate Law from a Reaction Mechanism
The two-step mechanism below has been proposed for a reaction between nitrogen monoxide and molecular chlorine:
\begin{align*} &\text{Step 1: } \quad \ce{NO(g) + Cl2(g) <=> NOCl2(g)} \tag{fast} \[4pt] &\text{Step 2: } \quad \ce{NOCl2(g) + NO(g) <=> 2 NOCl(g)} \tag{slow} \end{align*}
Use this mechanism to derive the equation and predicted rate law for the overall reaction.
Solution
The equation for the overall reaction is obtained by adding the two elementary reactions:
$2 NO (g) + Cl_2(g) \longrightarrow 2 NOCl (g) \nonumber$
To derive a rate law from this mechanism, first write rates laws for each of the two steps.
\begin{align*} & \text { rate }_1=k_1[ NO ]\left[ Cl_2\right] \text { for the forward reaction of step } 1 \[4pt] & \text { rate }_{-1}=k_{-1}\left[ NOCl_2\right] \text { for the reverse reaction of step } 1 \[4pt] & \text { rate }_2=k_2\left[ NOCl_2\right][ NO ] \text { for step } 2 \end{align*}
Step 2 is the rate-determining step, and so the rate law for the overall reaction should be the same as for this step. However, the step 2 rate law, as written, contains an intermediate species concentration, [NOCl2]. To remedy this, use the first step’s rate laws to derive an expression for the intermediate concentration in terms of the reactant concentrations.
Assuming step 1 is at equilibrium:
\begin{align*} \text { rate }_1 & =\text { rate }_{-1} \[4pt] k_1[ NO ]\left[ Cl_2\right] & =k_{-1}\left[ NOCl_2\right] \[4pt] {\left[ NOCl_2\right] } & =\left(\frac{k_1}{k_{-1}}\right)[ NO ]\left[ Cl_2\right] \end{align*}
Substituting this expression into the rate law for step 2 yields:
$\text { rate }_2=\text { rate }_{\text {overall }}=\left(\frac{k_2 k_1}{k_{-1}}\right)[ NO ]^2\left[ Cl_2\right] \nonumber$
Exercise $1$
The first step of a proposed multistep mechanism is:
$\ce{F2(g) <=> 2 F(g) \tag{fast}$
Derive the equation relating atomic fluorine concentration to molecular fluorine concentration.
Answer
$[\ce{F}]=\left(\frac{k_1\left[ F_2\right]}{k_{-1}}\right)^{1 / 2} \nonumber$ | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/12%3A_Kinetics/12.06%3A_Reaction_Mechanisms.txt |
Learning Objectives
By the end of this section, you will be able to:
• Explain the function of a catalyst in terms of reaction mechanisms and potential energy diagrams
• List examples of catalysis in natural and industrial processes
Among the factors affecting chemical reaction rates discussed earlier in this chapter was the presence of a catalyst, a substance that can increase the reaction rate without being consumed in the reaction. The concepts introduced in the previous section on reaction mechanisms provide the basis for understanding how catalysts are able to accomplish this very important function.
Figure $1$ shows reaction diagrams for a chemical process in the absence and presence of a catalyst. Inspection of the diagrams reveals several traits of these reactions. Consistent with the fact that the two diagrams represent the same overall reaction, both curves begin and end at the same energies (in this case, because products are more energetic than reactants, the reaction is endothermic). The reaction mechanisms, however, are clearly different. The uncatalyzed reaction proceeds via a one-step mechanism (one transition state observed), whereas the catalyzed reaction follows a two-step mechanism (two transition states observed) with a notably lesser activation energy. This difference illustrates the means by which a catalyst functions to accelerate reactions, namely, by providing an alternative reaction mechanism with a lower activation energy. Although the catalyzed reaction mechanism for a reaction needn’t necessarily involve a different number of steps than the uncatalyzed mechanism, it must provide a reaction path whose rate determining step is faster (lower Ea).
Example $1$: Reaction Diagrams for Catalyzed Reactions
The two reaction diagrams here represent the same reaction: one without a catalyst and one with a catalyst. Estimate the activation energy for each process, and identify which one involves a catalyst.
Solution
Activation energies are calculated by subtracting the reactant energy from the transition state energy.
\begin{align*} & \operatorname{diagram}( a ): E_{ a }=32 kJ -6 kJ =26 kJ \[4pt] & \operatorname{diagram}( b ): E_{ a }=20 kJ -6 kJ =14 kJ \end{align*}
The catalyzed reaction is the one with lesser activation energy, in this case represented by diagram b.
Exercise $1$
Reaction diagrams for a chemical process with and without a catalyst are shown below. Both reactions involve a two-step mechanism with a rate-determining first step. Compute activation energies for the first step of each mechanism, and identify which corresponds to the catalyzed reaction. How do the second steps of these two mechanisms compare?
Answer
For the first step, Ea = 80 kJ for (a) and 70 kJ for (b), so diagram (b) depicts the catalyzed reaction. Activation energies for the second steps of both mechanisms are the same, 20 kJ.
Homogeneous Catalysts
A homogeneous catalyst is present in the same phase as the reactants. It interacts with a reactant to form an intermediate substance, which then decomposes or reacts with another reactant in one or more steps to regenerate the original catalyst and form product.
As an important illustration of homogeneous catalysis, consider the earth’s ozone layer. Ozone in the upper atmosphere, which protects the earth from ultraviolet radiation, is formed when oxygen molecules absorb ultraviolet light and undergo the reaction:
$3 O_2(g) \stackrel{h v}{\longrightarrow} 2 O_3(g) \nonumber$
Ozone is a relatively unstable molecule that decomposes to yield diatomic oxygen by the reverse of this equation. This decomposition reaction is consistent with the following two-step mechanism:
\begin{align*} & O_3 \longrightarrow O_2+ O \[4pt] & O + O_3 \longrightarrow 2 O_2 \end{align*}
A number of substances can catalyze the decomposition of ozone. For example, the nitric oxide–catalyzed decomposition of ozone is believed to occur via the following three-step mechanism:
\begin{align*} & NO (g) + O_3(g) \longrightarrow NO_2(g) + O_2(g) \[4pt] & O_3(g) \longrightarrow O_2(g) + O (g) \[4pt] & NO_2(g) + O (g) \longrightarrow NO (g) + O_2(g) \end{align*}
As required, the overall reaction is the same for both the two-step uncatalyzed mechanism and the three-step NO-catalyzed mechanism:
$2 O_3(g) \longrightarrow 3 O_2(g) \nonumber$
Notice that NO is a reactant in the first step of the mechanism and a product in the last step. This is another characteristic trait of a catalyst: Though it participates in the chemical reaction, it is not consumed by the reaction.
Portrait of a Chemist: Mario J. Molina
The 1995 Nobel Prize in Chemistry was shared by Paul J. Crutzen, Mario J. Molina (Figure $2$), and F. Sherwood Rowland “for their work in atmospheric chemistry, particularly concerning the formation and decomposition of ozone.”1 Molina, a Mexican citizen, carried out the majority of his work at the Massachusetts Institute of Technology (MIT).
In 1974, Molina and Rowland published a paper in the journal Nature detailing the threat of chlorofluorocarbon gases to the stability of the ozone layer in earth’s upper atmosphere. The ozone layer protects earth from solar radiation by absorbing ultraviolet light. As chemical reactions deplete the amount of ozone in the upper atmosphere, a measurable “hole” forms above Antarctica, and an increase in the amount of solar ultraviolet radiation— strongly linked to the prevalence of skin cancers—reaches earth’s surface. The work of Molina and Rowland was instrumental in the adoption of the Montreal Protocol, an international treaty signed in 1987 that successfully began phasing out production of chemicals linked to ozone destruction.
Molina and Rowland demonstrated that chlorine atoms from human-made chemicals can catalyze ozone destruction in a process similar to that by which NO accelerates the depletion of ozone. Chlorine atoms are generated when chlorocarbons or chlorofluorocarbons—once widely used as refrigerants and propellants—are photochemically decomposed by ultraviolet light or react with hydroxyl radicals. A sample mechanism is shown here using methyl chloride:
CH_3 Cl + OH \longrightarrow Cl +\text { other products }] Chlorine radicals break down ozone and are regenerated by the following catalytic cycle: \[\begin{align*} & Cl + O_3 \longrightarrow ClO + O_2 \[4pt] & ClO + O \longrightarrow Cl + O_2 \end{align*} \nonumber
overall Reaction: $O_3+ O \longrightarrow 2 O_2$ \nonumber \]
A single monatomic chlorine can break down thousands of ozone molecules. Luckily, the majority of atmospheric chlorine exists as the catalytically inactive forms Cl2 and ClONO2.
Since receiving his portion of the Nobel Prize, Molina has continued his work in atmospheric chemistry at MIT.
Enzymes in the human body act as catalysts for important chemical reactions in cellular metabolism. As such, a deficiency of a particular enzyme can translate to a life-threatening disease. G6PD (glucose-6-phosphate dehydrogenase) deficiency, a genetic condition that results in a shortage of the enzyme glucose-6-phosphate dehydrogenase, is the most common enzyme deficiency in humans. This enzyme, shown in Figure 12.21, is the rate-limiting enzyme for the metabolic pathway that supplies NADPH to cells (Figure $4$).
Heterogeneous Catalysts
A heterogeneous catalyst is a catalyst that is present in a different phase (usually a solid) than the reactants. Such catalysts generally function by furnishing an active surface upon which a reaction can occur. Gas and liquid phase reactions catalyzed by heterogeneous catalysts occur on the surface of the catalyst rather than within the gas or liquid phase.
Heterogeneous catalysis typically involves the following processes:
1. Adsorption of the reactant(s) onto the surface of the catalyst
2. Activation of the adsorbed reactant(s)
3. Reaction of the adsorbed reactant(s)
Figure $5$: illustrates the steps of a mechanism for the reaction of compounds containing a carbon–carbon double bond with hydrogen on a nickel catalyst. Nickel is the catalyst used in the hydrogenation of polyunsaturated fats and oils (which contain several carbon–carbon double bonds) to produce saturated fats and oils (which contain only carbon–carbon single bonds).
Many important chemical products are prepared via industrial processes that use heterogeneous catalysts, including ammonia, nitric acid, sulfuric acid, and methanol. Heterogeneous catalysts are also used in the catalytic converters found on most gasoline-powered automobiles (Figure $6$).
Chemistry in Everyday Life: Automobile Catalytic Converters
Scientists developed catalytic converters to reduce the amount of toxic emissions produced by burning gasoline in internal combustion engines. By utilizing a carefully selected blend of catalytically active metals, it is possible to effect complete combustion of all carbon-containing compounds to carbon dioxide while also reducing the output of nitrogen oxides. This is particularly impressive when we consider that one step involves adding more oxygen to the molecule and the other involves removing the oxygen (Figure $6$).
Most modern, three-way catalytic converters possess a surface impregnated with a platinum-rhodium catalyst, which catalyzes the conversion of nitric oxide into dinitrogen and oxygen as well as the conversion of carbon monoxide and hydrocarbons such as octane into carbon dioxide and water vapor:
\begin{align*} & 2 NO_2(g) \longrightarrow N_2(g) + 2 O_2(g) \[4pt] & 2 CO (g) + O_2(g) \longrightarrow 2 CO_2(g) \[4pt] & 2 C_8 H_{18}(g) + 25 O_2(g) \longrightarrow 16 CO_2(g) + 18 H_2 O (g) \end{align*}
In order to be as efficient as possible, most catalytic converters are preheated by an electric heater. This ensures that the metals in the catalyst are fully active even before the automobile exhaust is hot enough to maintain appropriate reaction temperatures.
Link to Learning
The LibreTexts provides a thorough explanation of how catalytic converters work.
How Sciences Interconnect: Enzyme Structure and Function
The study of enzymes is an important interconnection between biology and chemistry. Enzymes are usually proteins (polypeptides) that help to control the rate of chemical reactions between biologically important compounds, particularly those that are involved in cellular metabolism. Different classes of enzymes perform a variety of functions, as shown in Table $1$.
Table $1$: Classes of Enzymes and Their Functions
Class Function
oxidoreductases redox reactions
transferases transfer of functional groups
hydrolases hydrolysis reactions
lyases group elimination to form double bonds
isomerases isomerization
ligases bond formation with ATP hydrolysis
Enzyme molecules possess an active site, a part of the molecule with a shape that allows it to bond to a specific substrate (a reactant molecule), forming an enzyme-substrate complex as a reaction intermediate. There are two models that attempt to explain how this active site works. The most simplistic model is referred to as the lock-and-key hypothesis, which suggests that the molecular shapes of the active site and substrate are complementary, fitting together like a key in a lock. The induced fit hypothesis, on the other hand, suggests that the enzyme molecule is flexible and changes shape to accommodate a bond with the substrate. This is not to suggest that an enzyme’s active site is completely malleable, however. Both the lock-and-key model and the induced fit model account for the fact that enzymes can only bind with specific substrates, since in general a particular enzyme only catalyzes a particular reaction (Figure $7$).
Link to Learning
The Royal Society of Chemistry provides an excellent introduction to enzymes for students and teachers. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/12%3A_Kinetics/12.07%3A_Catalysis.txt |
Example and Directions
Words (or words that have the same definition)The definition is case sensitive(Optional) Image to display with the definition [Not displayed in Glossary, only in pop-up on pages](Optional) Caption for Image(Optional) External or Internal Link(Optional) Source for Definition
(Eg. "Genetic, Hereditary, DNA ...")(Eg. "Relating to genes or heredity")The infamous double helix https://bio.libretexts.org/CC-BY-SA; Delmar Larsen
Glossary Entries
Word(s)DefinitionImageCaptionLinkSource
activated complex(also, transition state) unstable combination of reactant species formed during a chemical reaction
activation energy (Ea)minimum energy necessary in order for a reaction to take place
Arrhenius equationmathematical relationship between a reaction’s rate constant, activation energy, and temperature
average raterate of a chemical reaction computed as the ratio of a measured change in amount or concentration of substance to the time interval over which the change occurred
bimolecular reactionelementary reaction involving two reactant species
catalystsubstance that increases the rate of a reaction without itself being consumed by the reaction
collision theorymodel that emphasizes the energy and orientation of molecular collisions to explain and predict reaction kinetics
elementary reactionreaction that takes place in a single step, precisely as depicted in its chemical equation
frequency factor (A)proportionality constant in the Arrhenius equation, related to the relative number of collisions having an orientation capable of leading to product formation
half-life of a reaction (tl/2)time required for half of a given amount of reactant to be consumed
heterogeneous catalystcatalyst present in a different phase from the reactants, furnishing a surface at which a reaction can occur
homogeneous catalystcatalyst present in the same phase as the reactants
initial rateinstantaneous rate of a chemical reaction at t = 0 s (immediately after the reaction has begun)
instantaneous raterate of a chemical reaction at any instant in time, determined by the slope of the line tangential to a graph of concentration as a function of time
integrated rate lawequation that relates the concentration of a reactant to elapsed time of reaction
intermediatespecies produced in one step of a reaction mechanism and consumed in a subsequent step
method of initial ratescommon experimental approach to determining rate laws that involves measuring reaction rates at varying initial reactant concentrations
molecularitynumber of reactant species involved in an elementary reaction
overall reaction ordersum of the reaction orders for each substance represented in the rate law
rate constant (k)proportionality constant in a rate law
rate expressionmathematical representation defining reaction rate as change in amount, concentration, or pressure of reactant or product species per unit time
rate law(also, rate equation) (also, differential rate laws) mathematical equation showing the dependence of reaction rate on the rate constant and the concentration of one or more reactants
rate of reactionmeasure of the speed at which a chemical reaction takes place
rate-determining step(also, rate-limiting step) slowest elementary reaction in a reaction mechanism; determines the rate of the overall reaction
reaction diagramused in chemical kinetics to illustrate various properties of a reaction
reaction mechanismstepwise sequence of elementary reactions by which a chemical change takes place
reaction ordervalue of an exponent in a rate law (for example, zero order for 0, first order for 1, second order for 2, and so on)
termolecular reactionelementary reaction involving three reactant species
unimolecular reactionelementary reaction involving a single reactant species | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/12%3A_Kinetics/12.08%3A_Key_Terms.txt |
integrated rate law for zero-order reactions:
half-life for a zero-order reaction
integrated rate law for first-order reactions:
half-life for a first-order reaction
integrated rate law for second-order reactions:
half-life for a second-order reaction
12.10: Summary
The rate of a reaction can be expressed either in terms of the decrease in the amount of a reactant or the increase in the amount of a product per unit time. Relations between different rate expressions for a given reaction are derived directly from the stoichiometric coefficients of the equation representing the reaction.
The rate of a chemical reaction is affected by several parameters. Reactions involving two phases proceed more rapidly when there is greater surface area contact. If temperature or reactant concentration is increased, the rate of a given reaction generally increases as well. A catalyst can increase the rate of a reaction by providing an alternative pathway with a lower activation energy.
Rate laws (differential rate laws) provide a mathematical description of how changes in the concentration of a substance affect the rate of a chemical reaction. Rate laws are determined experimentally and cannot be predicted by reaction stoichiometry. The order of reaction describes how much a change in the concentration of each substance affects the overall rate, and the overall order of a reaction is the sum of the orders for each substance present in the reaction. Reaction orders are typically first order, second order, or zero order, but fractional and even negative orders are possible.
Integrated rate laws are mathematically derived from differential rate laws, and they describe the time dependence of reactant and product concentrations.
The half-life of a reaction is the time required to decrease the amount of a given reactant by one-half. A reaction’s half-life varies with rate constant and, for some reaction orders, reactant concentration. The half-life of a zero-order reaction decreases as the initial concentration of the reactant in the reaction decreases. The half-life of a first-order reaction is independent of concentration, and the half-life of a second-order reaction decreases as the concentration increases.
Chemical reactions typically require collisions between reactant species. These reactant collisions must be of proper orientation and sufficient energy in order to result in product formation. Collision theory provides a simple but effective explanation for the effect of many experimental parameters on reaction rates. The Arrhenius equation describes the relation between a reaction’s rate constant, activation energy, temperature, and dependence on collision orientation.
The sequence of individual steps, or elementary reactions, by which reactants are converted into products during the course of a reaction is called the reaction mechanism. The molecularity of an elementary reaction is the number of reactant species involved, typically one (unimolecular), two (bimolecular), or, less commonly, three (termolecular). The overall rate of a reaction is determined by the rate of the slowest in its mechanism, called the rate-determining step. Unimolecular elementary reactions have first-order rate laws, while bimolecular elementary reactions have second-order rate laws. By comparing the rate laws derived from a reaction mechanism to that determined experimentally, the mechanism may be deemed either incorrect or plausible.
Catalysts affect the rate of a chemical reaction by altering its mechanism to provide a lower activation energy. Catalysts can be homogenous (in the same phase as the reactants) or heterogeneous (a different phase than the reactants). | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/12%3A_Kinetics/12.09%3A_Key_Equations.txt |
1.
What is the difference between average rate, initial rate, and instantaneous rate?
2.
Ozone decomposes to oxygen according to the equation Write the equation that relates the rate expressions for this reaction in terms of the disappearance of O3 and the formation of oxygen.
3.
In the nuclear industry, chlorine trifluoride is used to prepare uranium hexafluoride, a volatile compound of uranium used in the separation of uranium isotopes. Chlorine trifluoride is prepared by the reaction Write the equation that relates the rate expressions for this reaction in terms of the disappearance of Cl2 and F2 and the formation of ClF3.
4.
A study of the rate of dimerization of C4H6 gave the data shown in the table:
Time (s) 0 1600 3200 4800 6200
[C4H6] (M) 1.00 10−2 5.04 10−3 3.37 10−3 2.53 10−3 2.08 10−3
1. Determine the average rate of dimerization between 0 s and 1600 s, and between 1600 s and 3200 s.
2. Estimate the instantaneous rate of dimerization at 3200 s from a graph of time versus [C4H6]. What are the units of this rate?
3. Determine the average rate of formation of C8H12 at 1600 s and the instantaneous rate of formation at 3200 s from the rates found in parts (a) and (b).
5.
A study of the rate of the reaction represented as gave the following data:
Time (s) 0.0 5.0 10.0 15.0 20.0 25.0 35.0
[A] (M) 1.00 0.775 0.625 0.465 0.360 0.285 0.230
1. Determine the average rate of disappearance of A between 0.0 s and 10.0 s, and between 10.0 s and 20.0 s.
2. Estimate the instantaneous rate of disappearance of A at 15.0 s from a graph of time versus [A]. What are the units of this rate?
3. Use the rates found in parts (a) and (b) to determine the average rate of formation of B between 0.00 s and 10.0 s, and the instantaneous rate of formation of B at 15.0 s.
6.
Consider the following reaction in aqueous solution:
If the rate of disappearance of Br(aq) at a particular moment during the reaction is 3.5 10−4 mol L−1 s−1, what is the rate of appearance of Br2(aq) at that moment?
7.
Describe the effect of each of the following on the rate of the reaction of magnesium metal with a solution of hydrochloric acid: the molarity of the hydrochloric acid, the temperature of the solution, and the size of the pieces of magnesium.
8.
Explain why an egg cooks more slowly in boiling water in Denver than in New York City. (Hint: Consider the effect of temperature on reaction rate and the effect of pressure on boiling point.)
9.
Go to the PhET Reactions & Rates interactive. Use the Single Collision tab to represent how the collision between monatomic oxygen (O) and carbon monoxide (CO) results in the breaking of one bond and the formation of another. Pull back on the red plunger to release the atom and observe the results. Then, click on “Reload Launcher” and change to “Angled shot” to see the difference.
1. What happens when the angle of the collision is changed?
2. Explain how this is relevant to rate of reaction.
10.
In the PhET Reactions & Rates interactive, use the “Many Collisions” tab to observe how multiple atoms and molecules interact under varying conditions. Select a molecule to pump into the chamber. Set the initial temperature and select the current amounts of each reactant. Select “Show bonds” under Options. How is the rate of the reaction affected by concentration and temperature?
11.
In the PhET Reactions & Rates interactive, on the Many Collisions tab, set up a simulation with 15 molecules of A and 10 molecules of BC. Select “Show Bonds” under Options.
1. Leave the Initial Temperature at the default setting. Observe the reaction. Is the rate of reaction fast or slow?
2. Click “Pause” and then “Reset All,” and then enter 15 molecules of A and 10 molecules of BC once again. Select “Show Bonds” under Options. This time, increase the initial temperature until, on the graph, the total average energy line is completely above the potential energy curve. Describe what happens to the reaction.
12.
How do the rate of a reaction and its rate constant differ?
13.
Doubling the concentration of a reactant increases the rate of a reaction four times. With this knowledge, answer the following questions:
1. What is the order of the reaction with respect to that reactant?
2. Tripling the concentration of a different reactant increases the rate of a reaction three times. What is the order of the reaction with respect to that reactant?
14.
Tripling the concentration of a reactant increases the rate of a reaction nine-fold. With this knowledge, answer the following questions:
1. What is the order of the reaction with respect to that reactant?
2. Increasing the concentration of a reactant by a factor of four increases the rate of a reaction four-fold. What is the order of the reaction with respect to that reactant?
15.
How will the rate of reaction change for the process: if the rate law for the reaction is
1. Decreasing the pressure of NO2 from 0.50 atm to 0.250 atm.
2. Increasing the concentration of CO from 0.01 M to 0.03 M.
16.
How will each of the following affect the rate of the reaction: if the rate law for the reaction is ?
1. Increasing the pressure of NO2 from 0.1 atm to 0.3 atm
2. Increasing the concentration of CO from 0.02 M to 0.06 M.
17.
Regular flights of supersonic aircraft in the stratosphere are of concern because such aircraft produce nitric oxide, NO, as a byproduct in the exhaust of their engines. Nitric oxide reacts with ozone, and it has been suggested that this could contribute to depletion of the ozone layer. The reaction is first order with respect to both NO and O3 with a rate constant of 2.20 107 L/mol/s. What is the instantaneous rate of disappearance of NO when [NO] = 3.3 10−6 M and [O3] = 5.9 10−7 M?
18.
Radioactive phosphorus is used in the study of biochemical reaction mechanisms because phosphorus atoms are components of many biochemical molecules. The location of the phosphorus (and the location of the molecule it is bound in) can be detected from the electrons (beta particles) it produces:
rate = 4.85 10−2
What is the instantaneous rate of production of electrons in a sample with a phosphorus concentration of 0.0033 M?
19.
The rate constant for the radioactive decay of 14C is 1.21 10−4 year−1. The products of the decay are nitrogen atoms and electrons (beta particles):
What is the instantaneous rate of production of N atoms in a sample with a carbon-14 content of 6.5 10−9 M?
20.
The decomposition of acetaldehyde is a second order reaction with a rate constant of 4.71 10−8 L mol−1 s−1. What is the instantaneous rate of decomposition of acetaldehyde in a solution with a concentration of 5.55 10−4 M?
21.
Alcohol is removed from the bloodstream by a series of metabolic reactions. The first reaction produces acetaldehyde; then other products are formed. The following data have been determined for the rate at which alcohol is removed from the blood of an average male, although individual rates can vary by 25–30%. Women metabolize alcohol a little more slowly than men:
[C2H5OH] (M) 4.4 10−2 3.3 10−2 2.2 10−2
Rate (mol L−1 h−1) 2.0 10−2 2.0 10−2 2.0 10−2
Determine the rate law, the rate constant, and the overall order for this reaction.
22.
Under certain conditions the decomposition of ammonia on a metal surface gives the following data:
[NH3] (M) 1.0 10−3 2.0 10−3 3.0 10−3
Rate (mol L−1 h−1) 1.5 10−6 1.5 10−6 1.5 10−6
Determine the rate law, the rate constant, and the overall order for this reaction.
23.
Nitrosyl chloride, NOCl, decomposes to NO and Cl2.
Determine the rate law, the rate constant, and the overall order for this reaction from the following data:
[NOCl] (M) 0.10 0.20 0.30
Rate (mol L−1 h−1) 8.0 10−10 3.2 10−9 7.2 10−9
24.
From the following data, determine the rate law, the rate constant, and the order with respect to A for the reaction
[A] (M) 1.33 10−2 2.66 10−2 3.99 10−2
Rate (mol L−1 h−1) 3.80 10−7 1.52 10−6 3.42 10−6
25.
Nitrogen monoxide reacts with chlorine according to the equation:
The following initial rates of reaction have been observed for certain reactant concentrations:
[NO] (mol/L) [Cl2] (mol/L) Rate (mol L−1 h−1)
0.50 0.50 1.14
1.00 0.50 4.56
1.00 1.00 9.12
What is the rate law that describes the rate’s dependence on the concentrations of NO and Cl2? What is the rate constant? What are the orders with respect to each reactant?
26.
Hydrogen reacts with nitrogen monoxide to form dinitrogen monoxide (laughing gas) according to the equation:
Determine the rate law, the rate constant, and the orders with respect to each reactant from the following data:
[NO] (M) 0.30 0.60 0.60
[H2] (M) 0.35 0.35 0.70
Rate (mol L−1 s−1) 2.835 10−3 1.134 10−2 2.268 10−2
27.
For the reaction the following data were obtained at 30 °C:
[A] (M) 0.230 0.356 0.557
Rate (mol L−1 s−1) 4.17 10−4 9.99 10−4 2.44 10−3
1. What is the order of the reaction with respect to [A], and what is the rate law?
2. What is the rate constant?
28.
For the reaction the following data were obtained at 30 °C:
[Q]initial (M) 0.170 0.212 0.357
Rate (mol L−1 s−1) 6.68 10−3 1.04 10−2 2.94 10−2
1. What is the order of the reaction with respect to [Q], and what is the rate law?
2. What is the rate constant?
29.
The rate constant for the first-order decomposition at 45 °C of dinitrogen pentoxide, N2O5, dissolved in chloroform, CHCl3, is 6.2 10−4 min−1.
What is the rate of the reaction when [N2O5] = 0.40 M?
30.
The annual production of HNO3 in 2013 was 60 million metric tons Most of that was prepared by the following sequence of reactions, each run in a separate reaction vessel.
1. The first reaction is run by burning ammonia in air over a platinum catalyst. This reaction is fast. The reaction in equation (c) is also fast. The second reaction limits the rate at which nitric acid can be prepared from ammonia. If equation (b) is second order in NO and first order in O2, what is the rate of formation of NO2 when the oxygen concentration is 0.50 M and the nitric oxide concentration is 0.75 M? The rate constant for the reaction is 5.8 10−6 L2 mol−2 s−1.
31.
The following data have been determined for the reaction:
1 2 3
(M) 0.10 0.20 0.30
(M) 0.050 0.050 0.010
Rate (mol L−1 s−1) 3.05 10−4 6.20 10−4 1.83 10−4
Determine the rate law and the rate constant for this reaction.
32.
Describe how graphical methods can be used to determine the order of a reaction and its rate constant from a series of data that includes the concentration of A at varying times.
33.
Use the data provided to graphically determine the order and rate constant of the following reaction:
Time (s) 0 5.00 103 1.00 104 1.50 104
[SO2Cl2] (M) 0.100 0.0896 0.0802 0.0719
Time (s) 2.50 104 3.00 104 4.00 104
[SO2Cl2] (M) 0.0577 0.0517 0.0415
34.
Pure ozone decomposes slowly to oxygen, Use the data provided in a graphical method and determine the order and rate constant of the reaction.
Time (h) 0 2.0 103 7.6 103 1.00 104
[O3] (M) 1.00 10−5 4.98 10−6 2.07 10−6 1.66 10−6
Time (h) 1.23 104 1.43 104 1.70 104
[O3] (M) 1.39 10−6 1.22 10−6 1.05 10−6
35.
From the given data, use a graphical method to determine the order and rate constant of the following reaction:
Time (s) 5.0 10.0 15.0 20.0 25.0 30.0 35.0 40.0
[X] (M) 0.0990 0.0497 0.0332 0.0249 0.0200 0.0166 0.0143 0.0125
36.
What is the half-life for the first-order decay of phosphorus-32? The rate constant for the decay is 4.85 10−2 day−1.
37.
What is the half-life for the first-order decay of carbon-14? The rate constant for the decay is 1.21 10−4 year−1.
38.
What is the half-life for the decomposition of NOCl when the concentration of NOCl is 0.15 M? The rate constant for this second-order reaction is 8.0 10−8 L mol−1 s−1.
39.
What is the half-life for the decomposition of O3 when the concentration of O3 is 2.35 10−6 M? The rate constant for this second-order reaction is 50.4 L mol−1 h−1.
40.
The reaction of compound A to give compounds C and D was found to be second-order in A. The rate constant for the reaction was determined to be 2.42 L mol−1 s−1. If the initial concentration is 0.500 mol/L, what is the value of t1/2?
41.
The half-life of a reaction of compound A to give compounds D and E is 8.50 min when the initial concentration of A is 0.150 M. How long will it take for the concentration to drop to 0.0300 M if the reaction is (a) first order with respect to A or (b) second order with respect to A?
42.
Some bacteria are resistant to the antibiotic penicillin because they produce penicillinase, an enzyme with a molecular weight of 3 104 g/mol that converts penicillin into inactive molecules. Although the kinetics of enzyme-catalyzed reactions can be complex, at low concentrations this reaction can be described by a rate law that is first order in the catalyst (penicillinase) and that also involves the concentration of penicillin. From the following data: 1.0 L of a solution containing 0.15 µg (0.15 10−6 g) of penicillinase, determine the order of the reaction with respect to penicillin and the value of the rate constant.
[Penicillin] (M) Rate (mol L−1 min−1)
2.0 10−6 1.0 10−10
3.0 10−6 1.5 10−10
4.0 10−6 2.0 10−10
43.
Both technetium-99 and thallium-201 are used to image heart muscle in patients with suspected heart problems. The half-lives are 6 h and 73 h, respectively. What percent of the radioactivity would remain for each of the isotopes after 2 days (48 h)?
44.
There are two molecules with the formula C3H6. Propene, is the monomer of the polymer polypropylene, which is used for indoor-outdoor carpets. Cyclopropane is used as an anesthetic:
When heated to 499 °C, cyclopropane rearranges (isomerizes) and forms propene with a rate constant of
5.95 10−4 s−1. What is the half-life of this reaction? What fraction of the cyclopropane remains after 0.75 h at 499 °C?
45.
Fluorine-18 is a radioactive isotope that decays by positron emission to form oxygen-18 with a half-life of 109.7 min. (A positron is a particle with the mass of an electron and a single unit of positive charge; the equation is Physicians use 18F to study the brain by injecting a quantity of fluoro-substituted glucose into the blood of a patient. The glucose accumulates in the regions where the brain is active and needs nourishment.
1. What is the rate constant for the decomposition of fluorine-18?
2. If a sample of glucose containing radioactive fluorine-18 is injected into the blood, what percent of the radioactivity will remain after 5.59 h?
3. How long does it take for 99.99% of the 18F to decay?
46.
Suppose that the half-life of steroids taken by an athlete is 42 days. Assuming that the steroids biodegrade by a first-order process, how long would it take for of the initial dose to remain in the athlete’s body?
47.
Recently, the skeleton of King Richard III was found under a parking lot in England. If tissue samples from the skeleton contain about 93.79% of the carbon-14 expected in living tissue, what year did King Richard III die? The half-life for carbon-14 is 5730 years.
48.
Nitroglycerine is an extremely sensitive explosive. In a series of carefully controlled experiments, samples of the explosive were heated to 160 °C and their first-order decomposition studied. Determine the average rate constants for each experiment using the following data:
Initial [C3H5N3O9] (M) 4.88 3.52 2.29 1.81 5.33 4.05 2.95 1.72
t (s) 300 300 300 300 180 180 180 180
% Decomposed 52.0 52.9 53.2 53.9 34.6 35.9 36.0 35.4
49.
For the past 10 years, the unsaturated hydrocarbon 1,3-butadiene has ranked 38th among the top 50 industrial chemicals. It is used primarily for the manufacture of synthetic rubber. An isomer exists also as cyclobutene:
The isomerization of cyclobutene to butadiene is first-order and the rate constant has been measured as 2.0 10−4 s−1 at 150 °C in a 0.53-L flask. Determine the partial pressure of cyclobutene and its concentration after 30.0 minutes if an isomerization reaction is carried out at 150 °C with an initial pressure of 55 torr.
50.
Chemical reactions occur when reactants collide. What are two factors that may prevent a collision from producing a chemical reaction?
51.
When every collision between reactants leads to a reaction, what determines the rate at which the reaction occurs?
52.
What is the activation energy of a reaction, and how is this energy related to the activated complex of the reaction?
53.
Account for the relationship between the rate of a reaction and its activation energy.
54.
Describe how graphical methods can be used to determine the activation energy of a reaction from a series of data that includes the rate of reaction at varying temperatures.
55.
How does an increase in temperature affect rate of reaction? Explain this effect in terms of the collision theory of the reaction rate.
56.
The rate of a certain reaction doubles for every 10 °C rise in temperature.
1. How much faster does the reaction proceed at 45 °C than at 25 °C?
2. How much faster does the reaction proceed at 95 °C than at 25 °C?
57.
In an experiment, a sample of NaClO3 was 90% decomposed in 48 min. Approximately how long would this decomposition have taken if the sample had been heated 20 °C higher? (Hint: Assume the rate doubles for each 10 °C rise in temperature.)
58.
The rate constant at 325 °C for the decomposition reaction is 6.1 10−8 s−1, and the activation energy is 261 kJ per mole of C4H8. Determine the frequency factor for the reaction.
59.
The rate constant for the decomposition of acetaldehyde, CH3CHO, to methane, CH4, and carbon monoxide, CO, in the gas phase is 1.1 10−2 L mol−1 s−1 at 703 K and 4.95 L mol−1 s−1 at 865 K. Determine the activation energy for this decomposition.
60.
An elevated level of the enzyme alkaline phosphatase (ALP) in human serum is an indication of possible liver or bone disorder. The level of serum ALP is so low that it is very difficult to measure directly. However, ALP catalyzes a number of reactions, and its relative concentration can be determined by measuring the rate of one of these reactions under controlled conditions. One such reaction is the conversion of p-nitrophenyl phosphate (PNPP) to p-nitrophenoxide ion (PNP) and phosphate ion. Control of temperature during the test is very important; the rate of the reaction increases 1.47 times if the temperature changes from 30 °C to 37 °C. What is the activation energy for the ALP–catalyzed conversion of PNPP to PNP and phosphate?
61.
In terms of collision theory, to which of the following is the rate of a chemical reaction proportional?
1. the change in free energy per second
2. the change in temperature per second
3. the number of collisions per second
4. the number of product molecules
62.
Hydrogen iodide, HI, decomposes in the gas phase to produce hydrogen, H2, and iodine, I2. The value of the rate constant, k, for the reaction was measured at several different temperatures and the data are shown here:
Temperature (K) k (L mol−1 s−1)
555 6.23 10−7
575 2.42 10−6
645 1.44 10−4
700 2.01 10−3
What is the value of the activation energy (in kJ/mol) for this reaction?
63.
The element Co exists in two oxidation states, Co(II) and Co(III), and the ions form many complexes. The rate at which one of the complexes of Co(III) was reduced by Fe(II) in water was measured. Determine the activation energy of the reaction from the following data:
T (K) k (s−1)
293 0.054
298 0.100
64.
The hydrolysis of the sugar sucrose to the sugars glucose and fructose,
follows a first-order rate law for the disappearance of sucrose: rate = k[C12H22O11] (The products of the reaction, glucose and fructose, have the same molecular formulas but differ in the arrangement of the atoms in their molecules.)
1. In neutral solution, k = 2.1 10−11 s−1 at 27 °C and 8.5 10−11 s−1 at 37 °C. Determine the activation energy, the frequency factor, and the rate constant for this equation at 47 °C (assuming the kinetics remain consistent with the Arrhenius equation at this temperature).
2. When a solution of sucrose with an initial concentration of 0.150 M reaches equilibrium, the concentration of sucrose is 1.65 10−7 M. How long will it take the solution to reach equilibrium at 27 °C in the absence of a catalyst? Because the concentration of sucrose at equilibrium is so low, assume that the reaction is irreversible.
3. Why does assuming that the reaction is irreversible simplify the calculation in part (b)?
65.
Use the PhET Reactions & Rates interactive simulation to simulate a system. On the “Single collision” tab of the simulation applet, enable the “Energy view” by clicking the “+” icon. Select the first reaction (A is yellow, B is purple, and C is navy blue). Using the “straight shot” default option, try launching the A atom with varying amounts of energy. What changes when the Total Energy line at launch is below the transition state of the Potential Energy line? Why? What happens when it is above the transition state? Why?
66.
Use the PhET Reactions & Rates interactive simulation to simulate a system. On the “Single collision” tab of the simulation applet, enable the “Energy view” by clicking the “+” icon. Select the first reaction (A is yellow, B is purple, and C is navy blue). Using the “angled shot” option, try launching the A atom with varying angles, but with more Total energy than the transition state. What happens when the A atom hits the BC molecule from different directions? Why?
67.
Why are elementary reactions involving three or more reactants very uncommon?
68.
In general, can we predict the effect of doubling the concentration of A on the rate of the overall reaction ? Can we predict the effect if the reaction is known to be an elementary reaction?
69.
Define these terms:
1. unimolecular reaction
2. bimolecular reaction
3. elementary reaction
4. overall reaction
70.
What is the rate law for the elementary termolecular reaction For
71.
Given the following reactions and the corresponding rate laws, in which of the reactions might the elementary reaction and the overall reaction be the same?
72.
Write the rate law for each of the following elementary reactions:
73.
Nitrogen monoxide, NO, reacts with hydrogen, H2, according to the following equation:
What would the rate law be if the mechanism for this reaction were:
74.
Experiments were conducted to study the rate of the reaction represented by this equation.2
Initial concentrations and rates of reaction are given here.
Experiment Initial Concentration [NO] (mol L−1) Initial Concentration, [H2] (mol L−1 min−1) Initial Rate of Formation of N2 (mol L−1 min−1)
1 0.0060 0.0010 1.8 10−4
2 0.0060 0.0020 3.6 10−4
3 0.0010 0.0060 0.30 10−4
4 0.0020 0.0060 1.2 10−4
Consider the following questions:
1. Step 3:
75.
The reaction of CO with Cl2 gives phosgene (COCl2), a nerve gas that was used in World War I. Use the mechanism shown here to complete the following exercises:
(fast, k1 represents the forward rate constant, k−1 the reverse rate constant)
(slow, k2 the rate constant)
(fast, k3 the rate constant)
1. Write the overall reaction.
2. Identify all intermediates.
3. Write the rate law for each elementary reaction.
4. Write the overall rate law expression.
76.
Account for the increase in reaction rate brought about by a catalyst.
77.
Compare the functions of homogeneous and heterogeneous catalysts.
78.
Consider this scenario and answer the following questions: Chlorine atoms resulting from decomposition of chlorofluoromethanes, such as CCl2F2, catalyze the decomposition of ozone in the atmosphere. One simplified mechanism for the decomposition is:
(a) Explain why chlorine atoms are catalysts in the gas-phase transformation:
(b) Nitric oxide is also involved in the decomposition of ozone by the mechanism:
Is NO a catalyst for the decomposition? Explain your answer.
79.
For each of the following pairs of reaction diagrams, identify which of the pair is catalyzed:
(a)
(b)
80.
For each of the following pairs of reaction diagrams, identify which of the pairs is catalyzed:
(a)
(b)
81.
For each of the following reaction diagrams, estimate the activation energy (Ea) of the reaction:
(a)
(b)
82.
For each of the following reaction diagrams, estimate the activation energy (Ea) of the reaction:
(a)
(b)
83.
Assuming the diagrams in Exercise 12.81 represent different mechanisms for the same reaction, which of the reactions has the faster rate?
84.
Consider the similarities and differences in the two reaction diagrams shown in Exercise 12.82. Do these diagrams represent two different overall reactions, or do they represent the same overall reaction taking place by two different mechanisms? Explain your answer.
Footnotes
• 2This question is taken from the Chemistry Advanced Placement Examination and is used with the permission of the Educational Testing Service. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/12%3A_Kinetics/12.11%3A_Exercises.txt |
In this chapter, you will learn how to predict the position of the balance and the yield of a product of a reaction under specific conditions, how to change a reaction's conditions to increase or reduce yield, and how to evaluate an equilibrium system's reaction to disturbances.
• 13.0: Introduction
Reactions can occur in both directions simultaneously (reactants to products and products to reactants) and eventually reach a state of balance. Reactions can occur in both directions simultaneously (reactants to products and products to reactants) and eventually reach a state of balance.
• 13.1: Chemical Equilibria
A reaction is at equilibrium when the amounts of reactants or products no longer change. Chemical equilibrium is a dynamic process, meaning the rate of formation of products by the forward reaction is equal to the rate at which the products re-form reactants by the reverse reaction.
• 13.2: Equilibrium Constants
For any reaction that is at equilibrium, the reaction quotient Q is equal to the equilibrium constant K for the reaction. If a reactant or product is a pure solid, a pure liquid, or the solvent in a dilute solution, the concentration of this component does not appear in the expression for the equilibrium constant. At equilibrium, the values of the concentrations of the reactants and products are constant and the reaction quotient will always equal K.
• 13.3: Shifting Equilibria - Le Chatelier’s Principle
Systems at equilibrium can be disturbed by changes to temperature, concentration, and, in some cases, volume and pressure; volume and pressure changes will disturb equilibrium if the number of moles of gas is different on the reactant and product sides of the reaction. The system's response to these disturbances is described by Le Châtelier's principle: The system will respond in a way that counteracts the disturbance. Not all changes to the system result in a disturbance of the equilibrium.
• 13.4: Equilibrium Calculations
The ratios of the rate of change in concentrations of a reaction are equal to the ratios of the coefficients in the balanced chemical equation. The sign of the coefficient of X is positive when the concentration increases and negative when it decreases. We learned to approach three basic types of equilibrium problems. When given the concentrations of the reactants and products at equilibrium, we can solve for the equilibrium constant.
• 13.5: Key Terms
• 13.6: Key Equations
• 13.7: Summary
• 13.8: Exercises
These are homework exercises to accompany the Textmap created for "Chemistry" by OpenStax. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
13: Fundamental Equilibrium Concepts
Imagine a beach populated with sunbathers and swimmers. As those basking in the sun get too hot, they enter the surf to swim and cool off. As the swimmers tire, they return to the beach to rest. If the rate at which sunbathers enter the surf were to equal the rate at which swimmers return to the sand, then the numbers (though not the identities) of sunbathers and swimmers would remain constant. This scenario illustrates a dynamic phenomenon known as equilibrium, in which opposing processes occur at equal rates. Chemical and physical processes are subject to this phenomenon; these processes are at equilibrium when the forward and reverse reaction rates are equal. Equilibrium systems are pervasive in nature; the various reactions involving carbon dioxide dissolved in blood are examples (see Figure \(1\)). This chapter provides a thorough introduction to the essential aspects of chemical equilibria. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/13%3A_Fundamental_Equilibrium_Concepts/13.00%3A_Introduction.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the nature of equilibrium systems
• Explain the dynamic nature of a chemical equilibrium
• Relate the magnitude of an equilibrium constant to properties of the chemical system
The convention for writing chemical equations involves placing reactant formulas on the left side of a reaction arrow and product formulas on the right side. By this convention, and the definitions of “reactant” and “product,” a chemical equation represents the reaction in question as proceeding from left to right. Reversible reactions, however, may proceed in both forward (left to right) and reverse (right to left) directions. When the rates of the forward and reverse reactions are equal, the concentrations of the reactant and product species remain constant over time and the system is at equilibrium. The relative concentrations of reactants and products in equilibrium systems vary greatly; some systems contain mostly products at equilibrium, some contain mostly reactants, and some contain appreciable amounts of both.
Figure $1$ illustrates fundamental equilibrium concepts using the reversible decomposition of colorless dinitrogen tetroxide to yield brown nitrogen dioxide, an elementary reaction described by the equation:
$N_2 O_4(g) \rightleftharpoons 2 NO_2(g) \nonumber$
Note that a special double arrow is used to emphasize the reversible nature of the reaction.
For this elementary process, rate laws for the forward and reverse reactions may be derived directly from the reaction stoichiometry:
\begin{aligned} &\text {rate}_f=k_f\left[ N_2 O_4\right]\[4pt] &\text { rate }_r=k_r\left[ NO_2\right]^2 \end{aligned} \nonumber
As the reaction begins (t = 0), the concentration of the N2O4 reactant is finite and that of the NO2 product is zero, so the forward reaction proceeds at a finite rate while the reverse reaction rate is zero. As time passes, N2O4 is consumed and its concentration falls, while NO2 is produced and its concentration increases (Figure $\PageIndex{1b}$). The decreasing concentration of the reactant slows the forward reaction rate, and the increasing product concentration speeds the reverse reaction rate (Figure $\PageIndex{1c}$). This process continues until the forward and reverse reaction rates become equal, at which time the reaction has reached equilibrium, as characterized by constant concentrations of its reactants and products (shaded areas of Figure $\PageIndex{1b}$ and Figure $\PageIndex{1c}$). It’s important to emphasize that chemical equilibria are dynamic; a reaction at equilibrium has not “stopped,” but is proceeding in the forward and reverse directions at the same rate. This dynamic nature is essential to understanding equilibrium behavior as discussed in this and subsequent chapters of the text.
Physical changes, such as phase transitions, are also reversible and may establish equilibria. This concept was introduced in another chapter of this text through discussion of the vapor pressure of a condensed phase (liquid or solid). As one example, consider the vaporization of bromine:
$\ce{Br2(l) <=> Br2(g)} \nonumber$
When liquid bromine is added to an otherwise empty container and the container is sealed, the forward process depicted above (vaporization) will commence and continue at a roughly constant rate as long as the exposed surface area of the liquid and its temperature remain constant. As increasing amounts of gaseous bromine are produced, the rate of the reverse process (condensation) will increase until it equals the rate of vaporization and equilibrium is established. A photograph showing this phase transition equilibrium is provided in Figure $3$. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/13%3A_Fundamental_Equilibrium_Concepts/13.01%3A_Chemical_Equilibria.txt |
Learning Objectives
By the end of this section, you will be able to:
• Derive reaction quotients from chemical equations representing homogeneous and heterogeneous reactions
• Calculate values of reaction quotients and equilibrium constants, using concentrations and pressures
• Relate the magnitude of an equilibrium constant to properties of the chemical system
The status of a reversible reaction is conveniently assessed by evaluating its reaction quotient (Q). For a reversible reaction described by
$m A +n B +\rightleftharpoons x C +y D \nonumber$
the reaction quotient is derived directly from the stoichiometry of the balanced equation as
$Q_c=\frac{[ C ]^x[ D ]^y}{[ A ]^m[ B ]^n} \nonumber$
where the subscript c denotes the use of molar concentrations in the expression. If the reactants and products are gaseous, a reaction quotient may be similarly derived using partial pressures:
$Q_p=\frac{P_{ C }^x P_{ D }^y}{P_{ A }{ }^m P_{ B }{ }^n} \nonumber$
Note that the reaction quotient equations above are a simplification of more rigorous expressions that use relative values for concentrations and pressures rather than absolute values. These relative concentration and pressure values are dimensionless (they have no units); consequently, so are the reaction quotients. For purposes of this introductory text, it will suffice to use the simplified equations and to disregard units when computing Q. In most cases, this will introduce only modest errors in calculations involving reaction quotients.
Example $1$: Writing Reaction Quotient Expressions
Write the concentration-based reaction quotient expression for each of the following reactions:
1. $\ce{3 O2(g) \rightleftharpoons 2 O3(g)}$
2. $\ce{N2(g) + 3 H2(g) \rightleftharpoons 2 NH3(g)}$
3. $\ce{4 NH3(g) + 7 O2(g) \rightleftharpoons 4 NO2(g) + 6 H2O(g)}$
Solution
1. $Q_c=\frac{\left[ \ce{O3} \right]^2}{\left[ \ce{O2} \right]^3} \nonumber$
2. $Q_c=\frac{\left[ \ce{NH3} \right]^2}{\left[ \ce{N2} \right]\left[ \ce{H2} \right]^3} \nonumber$
3. $Q_c=\frac{\left[ \ce{NO2} \right]^4\left[ \ce{H2O} \right]^6}{\left[ \ce{NH3} \right]^4\left[ \ce{O2} \right]^7} \nonumber$
Exercise $1$
Write the concentration-based reaction quotient expression for each of the following reactions:
1. $\ce{2SO2(g) + O2(g) \rightleftharpoons 2 SO3(g)}$
2. $\ce{C4H8(g) \rightleftharpoons 2 C2H4(g)}$
3. $\ce{2 C4H10(g) + 13 O2(g) \rightleftharpoons 8 CO2(g) + 10 H2O (g)}$
Answer
1. $Q_c=\frac{\left[ \ce{SO3} \right]^2}{\left[ \ce{SO2} \right]^2\left[ \ce{O2} \right]} \nonumber$
2. $Q_c=\frac{\left[ \ce{C2H4} \right]^2}{\left[ \ce{C4H8} \right]} \nonumber$
3. $Q_c=\frac{\left[ \ce{CO2} \right]^8\left[ \ce{H2O} \right]^{10}}{\left[ \ce{C4H10} \right]^2\left[ \ce{O2} \right]^{13}} \nonumber$
The numerical value of $Q$ varies as a reaction proceeds towards equilibrium; therefore, it can serve as a useful indicator of the reaction’s status. To illustrate this point, consider the oxidation of sulfur dioxide:
$\ce{2 SO2(g) + O2(g) \rightleftharpoons 2 SO3(g)} \nonumber$
Two different experimental scenarios are depicted in Figure $1$, one in which this reaction is initiated with a mixture of reactants only, SO2 and O2, and another that begins with only product, SO3. For the reaction that begins with a mixture of reactants only, Q is initially equal to zero:
$Q_c=\frac{\left[ \ce{SO3} \right]^2}{\left[ \ce{SO2} \right]^2\left[ \ce{O2} \right]}=\frac{0^2}{\left[ \ce{SO2} \right]^2\left[ \ce{O2} \right]}=0 \nonumber$
As the reaction proceeds toward equilibrium in the forward direction, reactant concentrations decrease (as does the denominator of Qc), product concentration increases (as does the numerator of Qc), and the reaction quotient consequently increases. When equilibrium is achieved, the concentrations of reactants and product remain constant, as does the value of Qc.
If the reaction begins with only product present, the value of Qc is initially undefined (immeasurably large, or infinite):
$Q_c=\frac{\left[ \ce{SO3} \right]^2}{\left[ \ce{SO2} \right]^2\left[ \ce{O2} \right]}=\frac{\left[ \ce{SO3} \right]^2}{0} \rightarrow \infty \nonumber$
In this case, the reaction proceeds toward equilibrium in the reverse direction. The product concentration and the numerator of Qc decrease with time, the reactant concentrations and the denominator of Qc increase, and the reaction quotient consequently decreases until it becomes constant at equilibrium.
The constant value of Q exhibited by a system at equilibrium is called the equilibrium constant, $K$:
$K \equiv Q \text { at equilibrium } \nonumber$
Comparison of the data plots in Figure $1$ shows that both experimental scenarios resulted in the same value for the equilibrium constant. This is a general observation for all equilibrium systems, known as the law of mass action.
Definition: Law of Mass Action
At a given temperature, the reaction quotient for a system at equilibrium is constant.
Example $2$: Evaluating a Reaction Quotient
Gaseous nitrogen dioxide forms dinitrogen tetroxide according to this equation:
$\ce{2 NO2(g) <=> N2O4(g)} \nonumber$
When 0.10 mol NO2 is added to a 1.0-L flask at 25 °C, the concentration changes so that at equilibrium, [NO2] = 0.016 M and [N2O4] = 0.042 M.
1. What is the value of the reaction quotient before any reaction occurs?
2. What is the value of the equilibrium constant for the reaction?
Solution
As for all equilibrium calculations in this text, use the simplified equations for $Q$ and $K$ and disregard any concentration or pressure units, as noted previously in this section.
(a) Before any product is formed
$\left[ \ce{NO2} \right]=\frac{0.10~\text{mol} }{1.0~\text{L} }=0.10~\text{M} \nonumber$
$[\ce{N2O4}] = 0~\text{M} \nonumber$
Thus
$Q_c=\frac{\left[ \ce{N2O4} \right]}{\left[ \ce{NO2} \right]^2}=\frac{0}{0.10^2}=0 \nonumber$
(b) At equilibrium,
$K_c=Q_c=\frac{\left[ \ce{N2O4} \right]}{\left[ \ce{NO2} \right]^2}=\frac{0.042}{0.016^2}=1.6 \times 10^2. \nonumber$
The equilibrium constant is $1.6 \times 10^{2}$.
Exercise $2$
For the reaction
$\ce{2 SO2(g) + O2(g) <=> 2 SO3(g)} \nonumber$
the equilibrium concentrations are [SO2] = 0.90 M, [O2] = 0.35 M, and [SO3] = 1.1 M. What is the value of the equilibrium constant, Kc?
Answer
Kc = 4.3
By its definition, the magnitude of an equilibrium constant explicitly reflects the composition of a reaction mixture at equilibrium, and it may be interpreted with regard to the extent of the forward reaction. A reaction exhibiting a large K will reach equilibrium when most of the reactant has been converted to product, whereas a small K indicates the reaction achieves equilibrium after very little reactant has been converted. It’s important to keep in mind that the magnitude of K does not indicate how rapidly or slowly equilibrium will be reached. Some equilibria are established so quickly as to be nearly instantaneous, and others so slowly that no perceptible change is observed over the course of days, years, or longer.
The equilibrium constant for a reaction can be used to predict the behavior of mixtures containing its reactants and/or products. As demonstrated by the sulfur dioxide oxidation process described above, a chemical reaction will proceed in whatever direction is necessary to achieve equilibrium. Comparing Q to K for an equilibrium system of interest allows prediction of what reaction (forward or reverse), if any, will occur.
To further illustrate this important point, consider the reversible reaction shown below:
$\ce{CO(g) + H2O(g) \rightleftharpoons CO2(g) + H2(g)} \quad K_c=0.640 \quad T =800{ }^{\circ} C \nonumber$
The bar charts in Figure $2$ represent changes in reactant and product concentrations for three different reaction mixtures. The reaction quotients for mixtures 1 and 3 are initially lesser than the reaction’s equilibrium constant, so each of these mixtures will experience a net forward reaction to achieve equilibrium. The reaction quotient for mixture 2 is initially greater than the equilibrium constant, so this mixture will proceed in the reverse direction until equilibrium is established.
Example $3$: Predicting the Direction of Reaction
Given here are the starting concentrations of reactants and products for three experiments involving this reaction:
$\ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)} \quad K_c=0.64 \nonumber$
Determine in which direction the reaction proceeds as it goes to equilibrium in each of the three experiments shown.
Reactants/Products Experiment 1 Experiment 2 Experiment 3
[CO]i 0.020 M 0.011 M 0.0094 M
[H2O]i 0.020 M 0.0011 M 0.0025 M
[CO2]i 0.0040 M 0.037 M 0.0015 M
[H2]i 0.0040 M 0.046 M 0.0076 M
Solution
Experiment 1:
$Q_c=\frac{\left[ \ce{CO2} \right]\left[ \ce{H2} \right]}{[ \ce{CO} ]\left[ \ce{H2O} \right]}=\frac{(0.0040)(0.0040)}{(0.020)(0.020)}=0.040 \nonumber$
Qc < Kc (0.040 < 0.64)
The reaction will proceed in the forward direction.
Experiment 2:
$Q_c=\frac{\left[ \ce{CO2} \right]\left[ \ce{H2} \right]}{[ \ce{CO} ]\left[ \ce{H2O} \right]}=\frac{(0.037)(0.046)}{(0.011)(0.0011)}=1.4 \times 10^2 \nonumber$
Qc > Kc (140 > 0.64)
The reaction will proceed in the reverse direction.
Experiment 3:
$Q_c=\frac{\left[ \ce{CO2} \right]\left[ \ce{H2} \right]}{[ \ce{CO} ]\left[ \ce{H2O} \right]}=\frac{(0.0015)(0.0076)}{(0.0094)(0.0025)}=0.48 \nonumber$
Qc < Kc (0.48 < 0.64)
The reaction will proceed in the forward direction.
Exercise $3$
Calculate the reaction quotient and determine the direction in which each of the following reactions will proceed to reach equilibrium.
1. A 1.00-L flask containing 0.0500 mol of NO(g), 0.0155 mol of Cl2(g), and 0.500 mol of NOCl: $\ce{2 NO(g) + Cl2(g) <=> 2 NOCl(g)} \quad K_c=4.6 \times 10^4 \nonumber$
2. A 5.0-L flask containing 17 g of NH3, 14 g of N2, and 12 g of H2: $\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)} \quad K_c=0.060 \nonumber$
3. A 2.00-L flask containing 230 g of SO3(g): $\ce{2 SO3(g) <=> 2 SO2(g) + O2(g)} \quad K_c=0.230 \nonumber$
Answer
(a) Qc = 6.45 \times 10^{3} DELMAR, forward. (b) Qc = 0.23, reverse. (c) Qc = 0, forward.
Homogeneous Equilibria
A homogeneous equilibrium is one in which all reactants and products (and any catalysts, if applicable) are present in the same phase. By this definition, homogeneous equilibria take place in solutions. These solutions are most commonly either liquid or gaseous phases, as shown by the examples below:
\begin{aligned} \ce{C2H2(aq) + 2 Br2(aq) & \rightleftharpoons C2H2Br4(aq)} & K_c & =\frac{\left[ C_2 H_2 Br_4\right]}{\left[ C_2 H_2\right]\left[ Br_2\right]^2} \[4pt] \ce{I2(aq) + I^{-}(aq) & \rightleftharpoons I_3^{-}(aq)} & K_c & =\frac{\left[ I_3-\right.}{\left[ I_2\right]\left[ I^{-}\right]} \[4pt] \ce{HF(aq) + H2O(l) & \rightleftharpoons H3O^{+}(aq) + F^{-}(aq)} & K_c & =\frac{\left[ H_3 O^{+}\right]\left[ F^{-}\right]}{[ HF ]} \[4pt] \ce{NH3(aq) + H2O(l) & \rightleftharpoons NH4^{+}(aq) + OH^{-}(aq)} & K_c & =\frac{\left[ NH_4^{+}\right]\left[ OH^{-}\right]}{\left[ NH_3\right]} \end{aligned} \nonumber
These examples all involve aqueous solutions, those in which water functions as the solvent. In the last two examples, water also functions as a reactant, but its concentration is not included in the reaction quotient. The reason for this omission is related to the more rigorous form of the Q (or K) expression mentioned previously in this chapter, in which relative concentrations for liquids and solids are equal to 1 and needn’t be included. Consequently, reaction quotients include concentration or pressure terms only for gaseous and solute species.
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.
The equilibria below all involve gas-phase solutions:
\begin{aligned} \ce{C2H6(g) & \rightleftharpoons C2H4(g) + H2(g)} & K_c & =\frac{\left[ C_2 H_4\right]\left[ H_2\right]}{\left[ C_2 H_6\right]} \[4pt] \ce{3 O2(g) & \rightleftharpoons 2 O3(g)} & K_c & =\frac{\left[ O_3\right]^2}{\left[ O_2\right]^3} \[4pt] \ce{N2(g) + 3 H2(g) & \rightleftharpoons 2 NH3(g)} & K_c & =\frac{\left[ NH_3\right]^2}{\left[ N_2\right]\left[ H_2\right]^3} \[4pt] \ce{C3H8(g) + 5 O2(g) & \rightleftharpoons 3 CO2(g) + 4 H2O(g)} & K_c & =\frac{\left[ CO_2\right]^3\left[ H_2 O \right]^4}{\left[ C_3 H_8\right]\left[ O_2\right]^5} \end{aligned} \nonumber
For gas-phase solutions, the equilibrium constant may be expressed in terms of either the molar concentrations (Kc) or partial pressures (Kp) of the reactants and products. A relation between these two K values may be simply derived from the ideal gas equation and the definition of molarity:
\begin{align*} P V &=n R T \[4pt] P & =\left(\frac{n}{V}\right) R T \[4pt] & =M R T \end{align*} \nonumber
where $P$ is partial pressure, $V$ is volume, $n$ is molar amount, $R$ is the gas constant, $T$ is temperature, and $M$ is molar concentration.
For the gas-phase reaction
$a A +b B \rightleftharpoons c C +d D \nonumber$
\begin{align*} K_P &=\frac{\left(P_C\right)^c\left(P_D\right)^d}{\left(P_A\right)^a\left(P_B\right)^b} \[4pt] &= \dfrac{([ C ] \times R T)^c([ D ] \times R T)^d}{([ A ] \times R T)^a([ B ] \times R T)^b} \[4pt] &= \dfrac{[ C ]^c[ D ]^d}{[ A ]^a[ B ]^b} \times \frac{(R T)^{c+d}}{(R T)^{a+b}} \[4pt] &= K_c(R T)^{(c+d)-(a+b)} \[4pt] &= K_c(R T)^{\Delta n} \end{align*} \nonumber
And so, the relationship between Kc and KP is
$K_P=K_c(R T)^{\Delta n} \nonumber$
where $Δn$ is the difference in the molar amounts of product and reactant gases, in this case:
$\Delta n=(c+d)-(a+b) \nonumber$
Example $4$: Calculation of KP
Write the equations relating Kc to KP for each of the following reactions:
1. $\ce{C2H6(g) <=> C2H4(g) + H2(g)}$
2. $\ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)}$
3. $\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)}$
4. Kc is equal to 0.28 for the following reaction at 900 °C: $\ce{CS2(g) + 4 H2(g) <=> CH4(g) + 2 H2S(g)} \nonumber$ What is KP at this temperature?
Solution
1. $Δn = (2) − (1) = 1$ $K_P = K_c (RT)^{Δn} = K_c (RT)^1 = K_c (RT) \nonumber$
2. $Δn = (2) − (2) = 0$ $K_P = K_c (RT)^{Δn} = K_c (RT)^0 = K_c \nonumber$
3. $Δn = (2) − (1 + 3) = −2$ $K_P = K_c (RT)^{Δn} = K_c (RT)^{−2} = \dfrac{K_c}{(R T)^2} \nonumber$
4. $K_P = K_c (RT)^{Δn} = (0.28)[(0.0821)(1173)]^{−2} = 3.0 \times 10^{−5} \nonumber$
Exercise $4$
Write the equations relating Kc to KP for each of the following reactions:
1. $\ce{2 SO2(g) + O2(g) <=> 2 SO3(g)}$
2. $\ce{N2O4(g) <=> 2 NO2(g)}$
3. $\ce{C3H8(g) + 5 O2(g) <=> 3 CO2(g) + 4 H2O (g)}$
4. At 227 °C, the following reaction has Kc = 0.0952: $\ce{CH3OH(g) <=> CO(g) + 2 H2(g)} \nonumber$ What would be the value of KP at this temperature?
Answer
(a) KP = Kc (RT)−1; (b) KP = Kc (RT); (c) KP = Kc (RT); (d) 160 or 1.6 \times 10^{2} DELMAR
Heterogeneous Equilibria
A heterogeneous equilibrium involves reactants and products in two or more different phases, as illustrated by the following examples:
\begin{align*} \ce{PbCl2(s) & \rightleftharpoons Pb^{2+}(aq) + 2 Cl^{-}(aq)} & K_c & =\left[ \ce{Pb^{2+}} \right]\left[ \ce{Cl^{-}} \right]^2 \[4pt] \ce{CaO(s) + CO2(g) & \rightleftharpoons CaCO3(s)} & K_c & =\frac{1}{\left[ \ce{CO_2}\right]} \[4pt] \ce{C(s) + 2S(g) & \rightleftharpoons CS2(g)} & K_c & =\frac{\left[ CS_2\right]}{\left[ \ce{S^2} \right.} \[4pt] \ce{Br2(l) & \rightleftharpoons Br2(g)} & K_c & =\left[ \ce{Br_2(g)} \right] \end{align*} \nonumber
Again, note that concentration terms are only included for gaseous and solute species, as discussed previously.
Two of the above examples include terms for gaseous species only in their equilibrium constants, and so Kp expressions may also be written:
\begin{align*} \ce{CaO(s) + CO2(g) & \rightleftharpoons CaCO3(s)} & K_P & =\frac{1}{P_{ \ce{CO2}}} \[4pt] \ce{C(s) + 2 S(g) & \rightleftharpoons CS2(g)} & K_P & =\frac{P_{ \ce{CS2}}}{\left(P_{ \ce{S} }\right)^2} \end{align*} \nonumber
Coupled Equilibria
The equilibrium systems discussed so far have all been relatively simple, involving just single reversible reactions. Many systems, however, involve two or more coupled equilibrium reactions, those which have in common one or more reactant or product species. Since the law of mass action allows for a straightforward derivation of equilibrium constant expressions from balanced chemical equations, the K value for a system involving coupled equilibria can be related to the K values of the individual reactions. Three basic manipulations are involved in this approach, as described below.
1. Changing the direction of a chemical equation essentially swaps the identities of “reactants” and “products,” and so the equilibrium constant for the reversed equation is simply the reciprocal of that for the forward equation.
$\begin{array}{ll} A \rightleftharpoons B & K_{ c }=\frac{[ B ]}{[ A ]} \[4pt] B \rightleftharpoons A & K_{ c^{\prime}}=\frac{[ A ]}{[ B ]} \end{array} \nonumber$
$K_{ c^{\prime}}=\frac{1}{ K_{ c }} \nonumber$
2. Changing the stoichiometric coefficients in an equation by some factor x results in an exponential change in the equilibrium constant by that same factor:
$\begin{array}{ll} A \rightleftharpoons B & K_{ c }=\frac{[ B ]}{[ A ]} \[4pt] xA \rightleftharpoons xB & K_{ c }=\frac{[ B ]^{ x }}{[ A ]^{ x }} \end{array} \nonumber$
$K_{ c^{\prime}}= K_{ c }{ }^{ x } \nonumber$
3. Adding two or more equilibrium equations together yields an overall equation whose equilibrium constant is the mathematical product of the individual reaction’s K values:
$\begin{array}{ll} A \rightleftharpoons B & K_{ c 1}=\frac{[ B ]}{[ A ]} \[4pt] B \rightleftharpoons C & K_{ c 2}=\frac{[ C ]}{[ B ]} \end{array} \nonumber$
The net reaction for these coupled equilibria is obtained by summing the two equilibrium equations and canceling any redundancies:
\begin{aligned} & A + B \rightleftharpoons B + C \[4pt] & A + B \rightleftharpoons B + C \[4pt] & A \rightleftharpoons C \end{aligned} \nonumber
$K_{ c^{\prime}}=\frac{[ C ]}{[ A ]} \nonumber$
Comparing the equilibrium constant for the net reaction to those for the two coupled equilibrium reactions reveals the following relationship:
$K_{ c 1} K_{ c 2}=\frac{[ B ]}{[ A ]} \times \frac{[ C ]}{[ B ]}=\frac{ \cancel{[ B ]}[ C ]}{[ A ] \cancel{[ B ]}}=\frac{[ C ]}{[ A ]}= K_{ c^{\prime}} \nonumber$
$K_{ c^{\prime}}= K_{ c 1} K_{ c 2} \nonumber$
Example $5$ demonstrates the use of this strategy in describing coupled equilibrium processes.
Example $5$: Equilibrium Constants for Coupled Reactions
A mixture containing nitrogen, hydrogen, and iodine established the following equilibrium at 400 °C:
$\ce{2 NH3(g) + 3 I2(g) \rightleftharpoons N2(g) + 6 HI(g)} \nonumber$
Use the information below to calculate Kc for this reaction.
\begin{align*} \ce{N2(g) + 3 H2(g) \rightleftharpoons 2 NH3(g)} & K_{ c 1}=0.50 \text { at } 400{ }^{\circ} C \[4pt] \ce{H2(g) + I2(g) \rightleftharpoons 2 HI(g)} & K_{ c 2}=50 \text { at } 400{ }^{\circ} C \end{align*} \nonumber
Solution
The equilibrium equation of interest and its K value may be derived from the equations for the two coupled reactions as follows.
Reverse the first coupled reaction equation:
$2 NH_3(g) \rightleftharpoons N_2(g) + 3 H_2(g) \quad K_{ c 1}{ }^{\prime}=\frac{1}{ K_{ c 1}}=\frac{1}{0.50}=2.0 \nonumber$
Multiply the second coupled reaction by 3:
$3 H_2(g) + 3 I_2(g) \rightleftharpoons 6 HI (g) \quad K_{ c 2}{ }^{\prime}= K_{ c 2}^3=50^3=1.2 \times 10^5 \nonumber$
Finally, add the two revised equations:
\begin{align*} \ce{2 NH3(g) + 3 H2(g) + 3 I2(g) &\rightleftharpoons N2(g) + 3 H2(g) + 6 HI(g)} \[4pt] \ce{2 NH3(g) + 3 I2(g) &\rightleftharpoons N2(g) + 6HI(g)} \end{align*} \nonumber
$K_{ c }= K_{ c 1}, K_{ c 2},=(2.0)\left(1.2 \times 10^5\right)=2.5 \times 10^5 \nonumber$
Exercise $5$
Use the provided information to calculate Kc for the following reaction at 550 °C:
\begin{align*} \ce{H2(g) + CO2(g) &\rightleftharpoons CO(g) + H2O(g)} && K_{ c }=? \[4pt] \ce{CoO(s) + CO(g) &\rightleftharpoons Co(s)+ CO2(g)} && K_{ c 1}=490 \[4pt] \ce{CoO(s) + H2(g) &\rightleftharpoons Co(s)+ H2O(g)} && K_{ c 1}=67 \end{align*} \nonumber
Answer
Kc = 0.14 | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/13%3A_Fundamental_Equilibrium_Concepts/13.02%3A_Equilibrium_Constants.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the ways in which an equilibrium system can be stressed
• Predict the response of a stressed equilibrium using Le Chatelier’s principle
A system at equilibrium is in a state of dynamic balance, with forward and reverse reactions taking place at equal rates. If an equilibrium system is subjected to a change in conditions that affects these reaction rates differently (a stress), then the rates are no longer equal and the system is not at equilibrium. The system will subsequently experience a net reaction in the direction of greater rate (a shift) that will re-establish the equilibrium. This phenomenon is summarized by Le Chatelier’s principle
Definition: Le Chatelier’s Principle
if an equilibrium system is stressed, the system will experience a shift in response to the stress that re-establishes equilibrium.
Reaction rates are affected primarily by concentrations, as described by the reaction’s rate law, and temperature, as described by the Arrhenius equation. Consequently, changes in concentration and temperature are the two stresses that can shift an equilibrium.
Effect of a Change in Concentration
If an equilibrium system is subjected to a change in the concentration of a reactant or product species, the rate of either the forward or the reverse reaction will change. As an example, consider the equilibrium reaction
$\ce{H2(g) + I2(g) <=> 2 HI (g)} \quad K_c=50.0 \text { at } 400{ }^{\circ} C \nonumber$
The rate laws for the forward and reverse reactions are
\begin{align*} &\text{forward} \quad \ce{H2(g) + I2(g) \rightarrow 2HI(g)} \quad &&\text{rate}_f=k_f[ \ce{H2}]^m [ \ce{I2} ]^n \[4pt] &\text{reverse} \quad \ce{2 HI(g) \rightarrow H2(g) + I2(g)} \quad &&\text{rate}_r=k_r[\ce{HI}]^x \end{align*} \nonumber
When this system is at equilibrium, the forward and reverse reaction rates are equal.
$\text { rate }_f=\text { rate }_r \nonumber$
If the system is stressed by adding reactant, either $\ce{H_2}$ or $\ce{I_2}$, the resulting increase in concentration causes the rate of the forward reaction to increase, exceeding that of the reverse reaction:
$\operatorname{rate}_f>\text { rate }_r \nonumber$
The system will experience a temporary net reaction in the forward direction to re-establish equilibrium (the equilibrium will shift right). This same shift will result if some product HI is removed from the system, which decreases the rate of the reverse reaction, again resulting in the same imbalance in rates.
The same logic can be used to explain the left shift that results from either removing reactant or adding product to an equilibrium system. These stresses both result in an increased rate for the reverse reaction
$\text { rate }_f<\text { rate }_r \nonumber$
and a temporary net reaction in the reverse direction to re-establish equilibrium.
As an alternative to this kinetic interpretation, the effect of changes in concentration on equilibria can be rationalized in terms of reaction quotients. When the system is at equilibrium,
$Q_c=\dfrac{[ \ce{HI} ]^2}{\left[ \ce{H2} \right]\left[ \ce{I2} \right]}=K_c \nonumber$
If reactant is added (increasing the denominator of the reaction quotient) or product is removed (decreasing the numerator), then Qc < Kc and the equilibrium will shift right. Note that the three different ways of inducing this stress result in three different changes in the composition of the equilibrium mixture. If H2 is added, the right shift will consume I2 and produce HI as equilibrium is re-established, yielding a mixture with a greater concentrations of H2 and HI and a lesser concentration of I2 than was present before. If I2 is added, the new equilibrium mixture will have greater concentrations of I2 and HI and a lesser concentration of H2. Finally, if HI is removed, the concentrations of all three species will be lower when equilibrium is reestablished. Despite these differences in composition, the value of the equilibrium constant will be the same after the stress as it was before (per the law of mass action). The same logic may be applied for stresses involving removing reactants or adding product, in which case Qc > Kc and the equilibrium will shift left.For gas-phase equilibria such as this one, some additional perspectives on changing the concentrations of reactants and products are worthy of mention. The partial pressure P of an ideal gas is proportional to its molar concentration M,
$M=\dfrac{n}{V}=\dfrac{P}{R T} \nonumber$
and so changes in the partial pressures of any reactant or product are essentially changes in concentrations and thus yield the same effects on equilibria. Aside from adding or removing reactant or product, the pressures (concentrations) of species in a gas-phase equilibrium can also be changed by changing the volume occupied by the system. Since
all species of a gas-phase equilibrium occupy the same volume, a given change in volume will cause the same change in concentration for both reactants and products. In order to discern what shift, if any, this type of stress will induce the stoichiometry of the reaction must be considered.
At equilibrium, the reaction
$\ce{H2(g) + I2(g) <=> 2HI(g)} \nonumber$
is described by the reaction quotient
$Q_P=\dfrac{(P_{\ce{HI}})^2}{P_{\ce{H_2}} P_{\ce{I2}}}=K_p \nonumber$
If the volume occupied by an equilibrium mixture of these species is decreased by a factor of 3, the partial pressures of all three species will be increased by a factor of 3:
\begin{align*} Q_p{ }^{\prime} &=\dfrac{\left(3 P_{ HI }\right)^2}{3 P_{ H_2} 3 P_{ I_2}}=\dfrac{9 P_{ H^2}}{9 P_{ H_2} P_{ I_2}}=\dfrac{P_{ HI^2}}{P_{ H_2} P_{ I_2}} \[4pt] &=Q_P=K_P \end{align*} \nonumber
And so, changing the volume of this gas-phase equilibrium mixture does not result in a shift of the equilibrium.
A similar treatment of a different system,
$\ce{2NO2(g) ⇌ 2 NO(g) + O2(g)} \nonumber$
however, yields a different result:
\begin{align*} Q_P &=\dfrac{(P_{NO})^2} {(P_{O_2})}{(P_{NO_2})^2} \[4pt] Q_P^{\prime} &= \dfrac{(3 P_{ \ce{NO}})^2 (3 P_{ \ce{O2}}) }{3 (P_{ \ce{NO2} })^2} = \dfrac{9 (P_{ \ce{NO}})^2 (3 P_{\ce{O_2}}) }{9 (P_{\ce{NO2}})^2} = \dfrac{27 (P_{\ce{NO}})^2 P_{ \ce{O_2}}}{9 (P_{ \ce{NO_2} })^2} \[4pt] &=3 Q_P>K_P \end{align*} \nonumber
In this case, the change in volume results in a reaction quotient greater than the equilibrium constant, and so the equilibrium will shift left.
These results illustrate the relationship between the stoichiometry of a gas-phase equilibrium and the effect of a volume-induced pressure (concentration) change. If the total molar amounts of reactants and products are equal, as in the first example, a change in volume does not shift the equilibrium. If the molar amounts of reactants and products are different, a change in volume will shift the equilibrium in a direction that better “accommodates” the volume change. In the second example, two moles of reactant (NO2) yield three moles of product (2NO + O2), and so decreasing the system volume causes the equilibrium to shift left since the reverse reaction produces less gas (2 mol) than the forward reaction (3 mol). Conversely, increasing the volume of this equilibrium system would result in a shift towards products.
Link to Learning
Check out this link to see a dramatic visual demonstration of how equilibrium changes with pressure changes.
Chemistry in Everyday Life: Equilibrium and Soft Drinks
The connection between chemistry and carbonated soft drinks goes back to 1767, when Joseph Priestley (1733–1804) developed a method of infusing water with carbon dioxide to make carbonated water. Priestley’s approach involved production of carbon dioxide by reacting oil of vitriol (sulfuric acid) with chalk (calcium carbonate).
The carbon dioxide was then dissolved in water, reacting to produce hydrogen carbonate, a weak acid that subsequently ionized to yield bicarbonate and hydrogen ions:
\begin{align*} &\text{dissolution} \quad &\ce{CO2(g) <=> CO2(aq)} \[4pt] &\text{hydrolysis} \quad &\ce{CO2(aq) + H2O(l) <=> H2CO3(aq)} \[4pt] &\text{ionization} \quad &\ce{H2CO3(aq) <=> HCO3^{-}(aq) + H^{+}(aq)} \end{align*} \nonumber
These same equilibrium reactions are the basis of today’s soft-drink carbonation process. Beverages are exposed to a high pressure of gaseous carbon dioxide during the process to shift the first equilibrium above to
the right, resulting in desirably high concentrations of dissolved carbon dioxide and, per similar shifts in the other two equilibria, its hydrolysis and ionization products. A bottle or can is then nearly filled with the carbonated beverage, leaving a relatively small volume of air in the container above the beverage surface (the headspace) before it is sealed. The pressure of carbon dioxide in the container headspace is very low immediately after sealing, but it rises as the dissolution equilibrium is re-established by shifting to the left. Since the volume of the beverage is significantly greater than the volume of the headspace, only a relatively small amount of dissolved carbon dioxide is lost to the headspace.
When a carbonated beverage container is opened, a hissing sound is heard as pressurized CO2 escapes from the headspace. This causes the dissolution equilibrium to shift left, resulting in a decrease in the concentration of dissolved CO2 and subsequent left-shifts of the hydrolysis and ionization equilibria. Fortunately for the consumer, the dissolution equilibrium is usually re-established slowly, and so the beverage may be enjoyed while its dissolved carbon dioxide concentration remains palatably high. Once the equilibria are re-established, the CO2(aq) concentration will be significantly lowered, and the beverage acquires a characteristic taste referred to as “flat.”
Effect of a Change in Temperature
Consistent with the law of mass action, an equilibrium stressed by a change in concentration will shift to re-establish equilibrium without any change in the value of the
equilibrium constant, K. When an equilibrium shifts in response to a temperature change, however, it is re-established with a different relative composition that exhibits a different value for the equilibrium constant.
To understand this phenomenon, consider the elementary reaction
$A \rightleftharpoons B \nonumber$
Since this is an elementary reaction, the rates laws for the forward and reverse may be derived directly from the balanced equation’s stoichiometry:
\begin{aligned} \operatorname{rate}_f & =k_f[ A ] \[4pt] \operatorname{rate}_r & =k_r[ B ] \end{aligned} \nonumber
When the system is at equilibrium,
$\text { rate }_r=\text { rate }_f \nonumber$
Substituting the rate laws into this equality and rearranging gives
\begin{aligned} & k_f[ A ]=k_r[ B ] \[4pt] & \dfrac{[B]}{[A]}=\dfrac{k_f}{k_r}=K_c \end{aligned} \nonumber
The equilibrium constant is seen to be a mathematical function of the rate constants for the forward and reverse reactions. Since the rate constants vary with temperature as
described by the Arrhenius equation, is stands to reason that the equilibrium constant will likewise vary with temperature (assuming the rate constants are affected to different extents by the temperature change). For more complex reactions involving multistep reaction mechanisms, a similar but more complex mathematical relation exists between the equilibrium constant and the rate constants of the steps in the mechanism. Regardless of how complex the reaction may be, the temperature-dependence of its equilibrium constant persists.
Predicting the shift an equilibrium will experience in response to a change in temperature is most conveniently accomplished by considering the enthalpy change of the reaction. For example, the decomposition of dinitrogen tetroxide is an endothermic (heat-consuming) process:
$\ce{N2O4(g) <=> 2 NO2(g)} \quad \Delta H=+57.20 ~\text{kJ} \nonumber$
For purposes of applying Le Chatelier’s principle, heat (q) may be viewed as a reactant:
$\ce{ heat + N2O4(g) <=> 2 NO2(g)} \nonumber$
Raising the temperature of the system is akin to increasing the amount of a reactant, and so the equilibrium will shift to the right. Lowering the system temperature will likewise cause the equilibrium to shift left. For exothermic processes, heat is viewed as a product of the reaction and so the opposite temperature dependence is observed.
Effect of a Catalyst
The kinetics chapter of this text identifies a catalyst as a substance that enables a reaction to proceed via a different mechanism with an accelerated rate. The catalyzed reaction mechanism involves a lower energy transition state than the uncatalyzed reaction, resulting in a lower activation energy, Ea, and a correspondingly greater rate constant.
To discern the effect of catalysis on an equilibrium system, consider the reaction diagram for a simple one-step (elementary) reaction shown in Figure $2$. The lowered transition state energy of the catalyzed reaction results in lowered activation energies for both the forward and the reverse reactions. Consequently, both forward and reverse reactions are accelerated, and equilibrium is achieved more quickly but without a change in the equilibrium constant.
An interesting case study highlighting these equilibrium concepts is the industrial production of ammonia, NH3. This substance is among the “top 10” industrial chemicals with regard to production, with roughly two billion pounds produced annually in the US. Ammonia is used as a chemical feedstock to synthesize a wide range of commercially useful compounds, including fertilizers, plastics, dyes, and explosives.
Most industrial production of ammonia uses the Haber-Bosch process based on the following equilibrium reaction:
$\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)} \quad \Delta H=-92.2 ~\text{kJ} \nonumber$
The traits of this reaction present challenges to its use in an efficient industrial process. The equilibrium constant is relatively small (Kp on the order of $10^{−5}$ at 25 °C), meaning very little ammonia is present in an equilibrium mixture. Also, the rate of this reaction is relatively slow at low temperatures. To raise the yield of ammonia, the industrial process is designed to operate under conditions favoring product formation:
• High pressures (concentrations) of reactants are used, ~150−250 atm, to shift the equilibrium right, favoring product formation.
• Ammonia is continually removed (collected) from the equilibrium mixture during the process, lowering its concentration and also shifting the equilibrium right.
• Although low temperatures favor product formation for this exothermic process, the reaction rate at low temperatures is inefficiently slow. A catalyst is used to accelerate the reaction to reasonable rates at relatively moderate temperatures (400−500 °C).
A diagram illustrating a typical industrial setup for production of ammonia via the Haber-Bosch process is shown in Figure $3$. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/13%3A_Fundamental_Equilibrium_Concepts/13.03%3A_Shifting_Equilibria_-_Le_Chateliers_Principle.txt |
Learning Objectives
By the end of this section, you will be able to:
• Identify the changes in concentration or pressure that occur for chemical species in equilibrium systems
• Calculate equilibrium concentrations or pressures and equilibrium constants, using various algebraic approaches
Having covered the essential concepts of chemical equilibria in the preceding sections of this chapter, this final section will demonstrate the more practical aspect of using these concepts and appropriate mathematical strategies to perform various equilibrium calculations. These types of computations are essential to many areas of science and technology—for example, in the formulation and dosing of pharmaceutical products. After a drug is ingested or injected, it is typically involved in several chemical equilibria that affect its ultimate concentration in the body system of interest. Knowledge of the quantitative aspects of these equilibria is required to compute a dosage amount that will solicit the desired therapeutic effect.
Many of the useful equilibrium calculations that will be demonstrated here require terms representing changes in reactant and product concentrations. These terms are derived from the stoichiometry of the reaction, as illustrated by decomposition of ammonia:
$\ce{2 NH3(g) <=> N2(g) + 3 H2(g)} \nonumber$
As shown earlier in this chapter, this equilibrium may be established within a sealed container that initially contains either NH3 only, or a mixture of any two of the three chemical species involved in the equilibrium. Regardless of its initial composition, a reaction mixture will show the same relationships between changes in the concentrations of the three species involved, as dictated by the reaction stoichiometry (see also the related content on expressing reaction rates in the chapter on kinetics). For example, if the nitrogen concentration increases by an amount x:
$\Delta\left[ N_2\right]=+x \nonumber$
the corresponding changes in the other species concentrations are
\begin{align*} \Delta\left[ \ce{H2} \right] &=\Delta\left[ \ce{N2} \right]\left(\frac{3 ~\text{mol} ~\ce{H2} }{1~\text{mol}~\ce{N2} }\right)=+3 x \[4pt] \Delta\left[ \ce{NH3} \right] &=-\Delta\left[ \ce{N2} \right]\left(\frac{2 ~\text{mol}~\ce{NH3}}{1~\text{mol} ~\ce{N2} }\right)=-2 x \end{align*}
where the negative sign indicates a decrease in concentration.
Example $1$: Determining Relative Changes in Concentration
Derive the missing terms representing concentration changes for each of the following reactions.
1. $\underset{x}{\ce{C2H2(g)}} + \underset{?}{\ce{2 Br2(g)}} \ce{<=>} \underset{?}{\ce{C2H2Br4(g)}} \nonumber$
2. $\underset{?}{\ce{I2(aq)}} + \underset{?}{\ce{I^{-}(aq)}} \ce{<=>} \underset{x}{\ce{I3^{-}(aq)}} \nonumber$
3. $\underset{x}{\ce{C3H8(g)}} + \underset{?}{\ce{5O2(g)}} \ce{<=>} \underset{?}{\ce{3CO2(g)}} + \underset{?}{\ce{4H2O(g)}} \nonumber$
Solution
1. $\underset{x}{\ce{C2H2(g)}} + \underset{2x}{\ce{2 Br2(g)}} \ce{<=>} \underset{-x}{\ce{C2H2Br4(g)}} \nonumber$
2. $\underset{-x}{\ce{I2(aq)}} + \underset{-x}{\ce{I^{-}(aq)}} \ce{<=>} \underset{x}{\ce{I3^{-}(aq)}} \nonumber$
3. $\underset{x}{\ce{C3H8(g)}} + \underset{5x}{\ce{5O2(g)}} \ce{<=>} \underset{-3x}{\ce{3CO2(g)}} + \underset{-4x}{\ce{4H2O(g)}} \nonumber$
Exercise $1$
Complete the changes in concentrations for each of the following reactions:
1. $\underset{?}{\ce{2SO2(g)}} + \underset{x}{\ce{O2(g)}} \ce{<=>} \underset{?}{\ce{2SO3(g)}} \nonumber$
2. $\underset{?}{\ce{C4H8(g)}} \ce{<=>} \underset{-2x}{\ce{2C2H4(g)}} \nonumber$
3. $\underset{?}{\ce{4NH3(g)}} + \underset{x}{\ce{7O2(g)}} \ce{<=>} \underset{?}{\ce{4NO2(g)}} + \underset{?}{\ce{6H2O(g)}} \nonumber$
Answer
(a) 2x, x, −2x; (b) x, −2x; (c) 4x, 7x, −4x, −6x or −4x, −7x, 4x, 6x
Calculation of an Equilibrium Constant
The equilibrium constant for a reaction is calculated from the equilibrium concentrations (or pressures) of its reactants and products. If these concentrations are known, the calculation simply involves their substitution into the K expression, as was illustrated by Example $1$. A slightly more challenging example is provided next, in which the reaction stoichiometry is used to derive equilibrium concentrations from the information provided. The basic strategy of this computation is helpful for many types of equilibrium computations and relies on the use of terms for the reactant and product concentrations initially present, for how they change as the reaction proceeds, and for what they are when the system reaches equilibrium. The acronym ICE is commonly used to refer to this mathematical approach, and the concentrations terms are usually gathered in a tabular format called an ICE table.
Example $2$: Calculation of an Equilibrium Constant
Iodine molecules react reversibly with iodide ions to produce triiodide ions.
$\ce{I2(aq) + I^{-}(aq) \rightleftharpoons I_3^{-}(aq)} \nonumber$
If a solution with the concentrations of $\ce{I2}$ and $\ce{I^{−}}$ both equal to $1.000 \times 10^{−3}~\text{M}$ before reaction gives an equilibrium concentration of $\ce{I2}$ of $6.61 \times 10^{−4} ~\text{M}$, what is the equilibrium constant for the reaction?
Solution
To calculate the equilibrium constants, equilibrium concentrations are needed for all the reactants and products:
$K_C=\frac{\left[ \ce{I3^{-}} \right]}{\left[ \ce{I2} \right]\left[ \ce{I^{-}} \right]} \nonumber$
Provided are the initial concentrations of the reactants and the equilibrium concentration of the product. Use this information to derive terms for the equilibrium concentrations of the reactants, presenting all the information in an ICE table.
At equilibrium the concentration of I2 is $6.61 \times 10^{−4}~\text{M}$ so that
\begin{align*} 1.000 \times 10^{-3}-x & =6.61 \times 10^{-4}\[4pt] x &=1.000 \times 10^{-3}-6.61 \times 10^{-4}\[4pt] &=3.39 \times 10^{-4} M \end{align*} \nonumber
The ICE table may now be updated with numerical values for all its concentrations:
Finally, substitute the equilibrium concentrations into the K expression and solve:
\begin{align*} K_c &=\frac{\left[ \ce{I3^{-}} \right]}{\left[ \ce{I2} \right]\left[ \ce{I^{-}} \right]} \[4pt] &=\frac{3.39 \times 10^{-4} M}{\left(6.61 \times 10^{-4} M\right)\left(6.61 \times 10^{-4} M\right)}=776 \end{align*} \nonumber
Exercise $2$
Ethanol and acetic acid react and form water and ethyl acetate, the solvent responsible for the odor of some nail polish removers.
$\ce{C2H5OH + CH3CO2H \rightleftharpoons CH3CO2C2H5 + H2O} \nonumber$
When 1 mol each of C2H5OH and CH3CO2H are allowed to react in 1 L of the solvent dioxane, equilibrium is established when $\frac{1}{3}$ mol of each of the reactants remains. Calculate the equilibrium constant for the reaction. (Note: Water is a solute in this reaction.)
Answer
Kc = 4
Calculation of a Missing Equilibrium Concentration
When the equilibrium constant and all but one equilibrium concentration are provided, the other equilibrium concentration(s) may be calculated. A computation of this sort is illustrated in the next example exercise.
Example $3$: Calculation of a Missing Equilibrium Concentration
Nitrogen oxides are air pollutants produced by the reaction of nitrogen and oxygen at high temperatures. At 2000 °C, the value of the Kc for the reaction,
$\ce{N2(g) + O2(g) \rightleftharpoons 2 NO(g)} \nonumber$
is $4.1 \times 10^{−4}$. Calculate the equilibrium concentration of NO(g) in air at 1 atm pressure and 2000 °C. The equilibrium concentrations of N2 and O2 at this pressure and temperature are 0.036 M and 0.0089 M, respectively.
Solution
Substitute the provided quantities into the equilibrium constant expression and solve for [NO]:
\begin{align*} K_c &=\frac{[ NO ]^2}{\left[ N_2\right]\left[ O_2\right]} \[4pt] [ NO ]^2&=K_c\left[ N_2\right]\left[ O_2\right] \[4pt] [ NO ]&=\sqrt{K_c\left[ N_2\right]\left[ O_2\right]} \[4pt] &=\sqrt{\left(4.1 \times 10^{-4}\right)(0.036)(0.0089)} \[4pt] &=\sqrt{1.31 \times 10^{-7}} \[4pt] &=3.6 \times 10^{-4} \end{align*} \nonumber
Thus [NO] is $3.6 \times 10^{−4} ~\text{mol/L}$ at equilibrium under these conditions.
To confirm this result, it may be used along with the provided equilibrium concentrations to calculate a value for K:
\begin{align*} K_c &= \frac{[ NO ]^2}{\left[ N_2\right]\left[ O_2\right]} \[4pt] &= \frac{\left(3.6 \times 10^{-4}\right)^2}{(0.036)(0.0089)} \[4pt] &= 4.0 \times 10^{-4} \end{align*} \nonumber
This result is consistent with the provided value for K within nominal uncertainty, differing by just 1 in the least significant digit’s place.
Exercise $3$
The equilibrium constant Kc for the reaction of nitrogen and hydrogen to produce ammonia at a certain temperature is $6.00 \times 10^{−2}$. Calculate the equilibrium concentration of ammonia if the equilibrium concentrations of nitrogen and hydrogen are 4.26 M and 2.09 M, respectively.
Answer
1.53 mol/L
Calculation of Equilibrium Concentrations from Initial Concentrations
Perhaps the most challenging type of equilibrium calculation can be one in which equilibrium concentrations are derived from initial concentrations and an equilibrium constant. For these calculations, a four-step approach is typically useful:
1. Identify the direction in which the reaction will proceed to reach equilibrium.
2. Develop an ICE table.
3. Calculate the concentration changes and, subsequently, the equilibrium concentrations.
4. Confirm the calculated equilibrium concentrations.
The last two example exercises of this chapter demonstrate the application of this strategy.
Example $4$: Calculation of Equilibrium Concentrations
Under certain conditions, the equilibrium constant Kc for the decomposition of $\ce{PCl5(g)}$ into $\ce{PCl3(g)}$ and $\ce{Cl2(g)}$ is 0.0211. What are the equilibrium concentrations of $\ce{PCl5}$, $\ce{PCl3}$, and $\ce{Cl2}$ in a mixture that initially contained only $\ce{PCl5}$ at a concentration of 1.00 M?
Solution
Use the stepwise process described earlier.
Step 1.
Determine the direction the reaction proceeds.
The balanced equation for the decomposition of $\ce{PCl5}$ is
$\ce{PCl5(g) \rightleftharpoons PCl3(g) + Cl2(g)} \nonumber$
Because only the reactant is present initially Qc = 0 and the reaction will proceed to the right.
Step 2:
Develop an ICE table.
Step 3.
Solve for the change and the equilibrium concentrations.
Substituting the equilibrium concentrations into the equilibrium constant equation gives
\begin{align*} K_c &=\frac{\left[ \ce{PCl3} \right]\left[ \ce{Cl2} \right]}{\left[ \ce{PCl5} \right]}=0.0211 \[4pt] &=\frac{(x)(x)}{(1.00-x)} \[4pt] 0.0211&=\frac{(x)(x)}{(1.00-x)} \[4pt] 0.0211(1.00-x)7 &=x^2 \[4pt] x^2+0.0211 x - 0.0211 &=0 \end{align*}
Appendix B shows an equation of the form $ax^2 + bx + c = 0$ can be rearranged to solve for $x$:
$x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \nonumber$
In this case, a = 1, b = 0.0211, and c = −0.0211. Substituting the appropriate values for a, b, and c yields:
\begin{align*} x&=\frac{-0.0211 \pm \sqrt{(0.0211)^2-4(1)(-0.0211)}}{2(1)} \[4pt] &=\frac{-0.0211 \pm \sqrt{\left(4.45 \times 10^{-4}\right)+\left(8.44 \times 10^{-2}\right)}}{2} \[4pt] &=\frac{-0.0211 \pm 0.291}{2} \end{align*}
The two roots of the quadratic are, therefore,
$x=\frac{-0.0211+0.291}{2}=0.135 \nonumber$
and
$x=\frac{-0.0211-0.291}{2}=-0.156 \nonumber$
For this scenario, only the positive root is physically meaningful (concentrations are either zero or positive), and so x = 0.135 M.
The equilibrium concentrations are
\begin{align*} [ \ce{PCl5}] &= 1.00-0.135 = 0.87 ~\text{M} \[4pt] [ \ce{PCl3} ] &= x=0.135 ~\text{M}= \[4pt] [ \ce{Cl2} ] &= x =0.135 ~\text{M} \end{align*} \nonumber
Step 4.
Confirm the calculated equilibrium concentrations.
Substitution into the expression for Kc (to check the calculation) gives
$K_c=\frac{\left[ \ce{PCl3} \right]\left[ \ce{Cl2}\right]}{\left[ \ce{PCl5} \right]}=\frac{(0.135)(0.135)}{0.87}=0.021 \nonumber$
The equilibrium constant calculated from the equilibrium concentrations is equal to the value of Kc given in the problem (when rounded to the proper number of significant figures).
Exercise $\PageIndex{4A}$
Acetic acid, CH3CO2H, reacts with ethanol, C2H5OH, to form water and ethyl acetate, CH3CO2C2H5.
$\ce{CH3CO2H + C2H5OH <=> CH3CO2C2H5 + H2O} \nonumber$
The equilibrium constant for this reaction with dioxane as a solvent is 4.0. What are the equilibrium concentrations for a mixture that is initially 0.15 M in CH3CO2H, 0.15 M in C2H5OH, 0.40 M in CH3CO2C2H5, and 0.40 M in H2O?
Answer
[CH3CO2H] = 0.18 M, [C2H5OH] = 0.18 M, [CH3CO2C2H5] = 0.37 M, [H2O] = 0.37 M
Exercise $\PageIndex{4B}$
A 1.00-L flask is filled with 1.00 mole of H2 and 2.00 moles of I2. The value of the equilibrium constant for the reaction of hydrogen and iodine reacting to form hydrogen iodide is 50.5 under the given conditions. What are the equilibrium concentrations of H2, I2, and HI in moles/L?
$\ce{H2(g) + I2(g) <=> 2 HI(g)} \nonumber$
Answer
[H2] = 0.06 M, [I2] = 1.06 M, [HI] = 1.88 M
Example $5$: Calculation of Equilibrium Concentrations Using an Algebra-Simplifying Assumption
What are the concentrations at equilibrium of a 0.15 M solution of HCN?
$\ce{HCN(aq) <=> H^{+}(aq) + CN^{-}(aq)} \quad \quad K_c=4.9 \times 10^{-10} \nonumber$
Solution
Using “x” to represent the concentration of each product at equilibrium gives this ICE table.
Substitute the equilibrium concentration terms into the Kc expression
$K_c=\frac{(x)(x)}{0.15-x} \nonumber$
rearrange to the quadratic form and solve for x
\begin{align*} &x^2+4.9 \times 10^{-10}-7.35 \times 10^{-11}=0 \[4pt] &x=8.56 \times 10^{-6} M(3 \text { sig. figs. })=8.6 \times 10^{-6} M(2 \text { sig. figs. }) \end{align*}
Thus $[\ce{H^{+}}] = [\ce{CN^{-}}] = x = 8.6 \times 10^{-6} ~\text{M}$ and $[\ce{HCN}] = 0.15 – x = 0.15 ~\text{M}$.
Note in this case that the change in concentration is significantly less than the initial concentration (a consequence of the small K), and so the initial concentration experiences a negligible change:
$\text { if } x \ll 0.15 M \text {, then }(0.15-x) \approx 0.15 \nonumber$
This approximation allows for a more expedient mathematical approach to the calculation that avoids the need to solve for the roots of a quadratic equation:
\begin{align*} K_c &=\frac{(x)(x)}{0.15-x} \approx \frac{x^2}{0.15}\[4pt] 4.9 \times 10^{-10}&=\frac{x^2}{0.15}\[4pt] x^2 &=(0.15)\left(4.9 \times 10^{-10}\right) \[4pt] &=7.4 \times 10^{-11} \[4pt] x&=\sqrt{7.4 \times 10^{-11}} \[4pt] &=8.6 \times 10^{-6} ~\text{M} \end{align*} \nonumber
The value of $x$ calculated is, indeed, much less than the initial concentration
$8.6 \times 10^{-6} \ll 0.15 \nonumber$
and so the approximation was justified. If this simplified approach were to yield a value for x that did not justify the approximation, the calculation would need to be repeated without making the approximation.
Exercise $5$
What are the equilibrium concentrations in a 0.25 M NH3 solution?
$\ce{NH3(aq) + H2O(l) <=> NH4^{+}(aq) + OH^{-}(aq)} \quad K_{ c }=1.8 \times 10^{-5} \nonumber$
Answer
$\left[ \ce{OH^{-}} \right]=\left[ \ce{NH_4^{+}}\right]=0.0021 ~\text{M}$
$[\ce{NH3}] = 0.25 ~\text{M}$
13.05: Key Terms
Example and Directions
Words (or words that have the same definition)The definition is case sensitive(Optional) Image to display with the definition [Not displayed in Glossary, only in pop-up on pages](Optional) Caption for Image(Optional) External or Internal Link(Optional) Source for Definition
(Eg. "Genetic, Hereditary, DNA ...")(Eg. "Relating to genes or heredity")The infamous double helix https://bio.libretexts.org/CC-BY-SA; Delmar Larsen
Glossary Entries
Word(s)DefinitionImageCaptionLinkSource
equilibriumstate of a reversible reaction in which the forward and reverse processes occur at equal rates
equilibrium constant (K)value of the reaction quotient for a system at equilibrium; may be expressed using concentrations (Kc) or partial pressures (Kp)
heterogeneous equilibriaequilibria in which reactants and products occupy two or more different phases
homogeneous equilibriaequilibria in which all reactants and products occupy the same phase
law of mass actionwhen a reversible reaction has attained equilibrium at a given temperature, the reaction quotient remains constant
Le Chatelier’s principlean equilibrium subjected to stress will shift in a way to counter the stress and re-establish equilibrium
reaction quotient (Q)mathematical function describing the relative amounts of reactants and products in a reaction mixture; may be expressed in terms of concentrations (Qc) or pressures (Qp)
reversible reactionchemical reaction that can proceed in both the forward and reverse directions under given conditions
13.06: Key Equations
P = MRT
Kc = Qc at equilibrium
Kp = Qp at equilibrium
KP = Kc (RT)Δn
13.07: Summary
A reversible reaction is at equilibrium when the forward and reverse processes occur at equal rates. Chemical equilibria are dynamic processes characterized by constant amounts of reactant and product species.
The composition of a reaction mixture may be represented by a mathematical function known as the reaction quotient, Q. For a reaction at equilibrium, the composition is constant, and Q is called the equilibrium constant, K.
A homogeneous equilibrium is an equilibrium in which all components are in the same phase. A heterogeneous equilibrium is an equilibrium in which components are in two or more phases.
Systems at equilibrium can be disturbed by changes to temperature, concentration, and, in some cases, volume and pressure. The system’s response to these disturbances is described by Le Chatelier’s principle: An equilibrium system subjected to a disturbance will shift in a way that counters the disturbance and re-establishes equilibrium. A catalyst will increase the rate of both the forward and reverse reactions of a reversible process, increasing the rate at which equilibrium is reached but not altering the equilibrium mixture’s composition (K does not change).
Calculating values for equilibrium constants and/or equilibrium concentrations is of practical benefit to many applications. A mathematical strategy that uses initial concentrations, changes in concentrations, and equilibrium concentrations (and goes by the acronym ICE) is useful for several types of equilibrium calculations. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/13%3A_Fundamental_Equilibrium_Concepts/13.04%3A_Equilibrium_Calculations.txt |
1.
What does it mean to describe a reaction as “reversible”?
2.
When writing an equation, how is a reversible reaction distinguished from a nonreversible reaction?
3.
If a reaction is reversible, when can it be said to have reached equilibrium?
4.
Is a system at equilibrium if the rate constants of the forward and reverse reactions are equal?
5.
If the concentrations of products and reactants are equal, is the system at equilibrium?
6.
Explain why there may be an infinite number of values for the reaction quotient of a reaction at a given temperature but there can be only one value for the equilibrium constant at that temperature.
7.
Explain why an equilibrium between Br2(l) and Br2(g) would not be established if the container were not a closed vessel shown in Figure 13.4.
8.
If you observe the following reaction at equilibrium, is it possible to tell whether the reaction started with pure NO2 or with pure N2O4?
9.
Among the solubility rules previously discussed is the statement: All chlorides are soluble except Hg2Cl2, AgCl, PbCl2, and CuCl.
1. Write the expression for the equilibrium constant for the reaction represented by the equation Is Kc > 1, < 1, or ≈ 1? Explain your answer.
2. Write the expression for the equilibrium constant for the reaction represented by the equation Is Kc > 1, < 1, or ≈ 1? Explain your answer.
10.
Among the solubility rules previously discussed is the statement: Carbonates, phosphates, borates, and arsenates—except those of the ammonium ion and the alkali metals—are insoluble.
1. Write the expression for the equilibrium constant for the reaction represented by the equation Is Kc > 1, < 1, or ≈ 1? Explain your answer.
2. Write the expression for the equilibrium constant for the reaction represented by the equation Is Kc > 1, < 1, or ≈ 1? Explain your answer.
11.
Benzene is one of the compounds used as octane enhancers in unleaded gasoline. It is manufactured by the catalytic conversion of acetylene to benzene: Which value of Kc would make this reaction most useful commercially? Kc ≈ 0.01, Kc ≈ 1, or Kc ≈ 10. Explain your answer.
12.
Show that the complete chemical equation, the total ionic equation, and the net ionic equation for the reaction represented by the equation give the same expression for the reaction quotient. KI3 is composed of the ions K+ and
13.
For a titration to be effective, the reaction must be rapid and the yield of the reaction must essentially be 100%. Is Kc > 1, < 1, or ≈ 1 for a titration reaction?
14.
For a precipitation reaction to be useful in a gravimetric analysis, the product of the reaction must be insoluble. Is Kc > 1, < 1, or ≈ 1 for a useful precipitation reaction?
15.
Write the mathematical expression for the reaction quotient, Qc, for each of the following reactions:
16.
Write the mathematical expression for the reaction quotient, Qc, for each of the following reactions:
17.
The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in which each system will proceed to reach equilibrium.
1. (f) NO = 10.0 atm, N2 = O2 = 5 atm
18.
The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in which each system will proceed to reach equilibrium.
1. (f) [N2] = 0.100 M, [O2] = 0.200 M, [NO] = 1.00 M
19.
The following reaction has KP = 4.50 \times 10^{−5} DELMAR at 720 K.
If a reaction vessel is filled with each gas to the partial pressures listed, in which direction will it shift to reach equilibrium? P(NH3) = 93 atm, P(N2) = 48 atm, and P(H2) = 52 atm
20.
Determine if the following system is at equilibrium. If not, in which direction will the system need to shift to reach equilibrium?
[SO2Cl2] = 0.12 M, [Cl2] = 0.16 M and [SO2] = 0.050 M. Kc for the reaction is 0.078.
21.
Which of the systems described in Exercise 13.15 are homogeneous equilibria? Which are heterogeneous equilibria?
22.
Which of the systems described in Exercise 13.16 are homogeneous equilibria? Which are heterogeneous equilibria?
23.
For which of the reactions in Exercise 13.15 does Kc (calculated using concentrations) equal KP (calculated using pressures)?
24.
For which of the reactions in Exercise 13.16 does Kc (calculated using concentrations) equal KP (calculated using pressures)?
25.
Convert the values of Kc to values of KP or the values of KP to values of Kc.
26.
Convert the values of Kc to values of KP or the values of KP to values of Kc.
27.
What is the value of the equilibrium constant expression for the change
28.
Write the expression of the reaction quotient for the ionization of HOCN in water.
29.
Write the reaction quotient expression for the ionization of NH3 in water.
30.
What is the approximate value of the equilibrium constant KP for the change at 25 °C. (The equilibrium vapor pressure for this substance is 570 torr at 25 °C.)
31.
The following equation represents a reversible decomposition:
Under what conditions will decomposition in a closed container proceed to completion so that no CaCO3 remains?
32.
Explain how to recognize the conditions under which changes in volume will affect gas-phase systems at equilibrium.
33.
What property of a reaction can we use to predict the effect of a change in temperature on the value of an equilibrium constant?
34.
The following reaction occurs when a burner on a gas stove is lit:
Is an equilibrium among CH4, O2, CO2, and H2O established under these conditions? Explain your answer.
35.
A necessary step in the manufacture of sulfuric acid is the formation of sulfur trioxide, SO3, from sulfur dioxide, SO2, and oxygen, O2, shown here. At high temperatures, the rate of formation of SO3 is higher, but the equilibrium amount (concentration or partial pressure) of SO3 is lower than it would be at lower temperatures.
1. Does the equilibrium constant for the reaction increase, decrease, or remain about the same as the temperature increases?
2. Is the reaction endothermic or exothermic?
36.
Suggest four ways in which the concentration of hydrazine, N2H4, could be increased in an equilibrium described by the following equation:
37.
Suggest four ways in which the concentration of PH3 could be increased in an equilibrium described by the following equation:
38.
How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each?
39.
How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each?
40.
Methanol can be prepared from carbon monoxide and hydrogen at high temperature and pressure in the presence of a suitable catalyst.
(a) Write the expression for the equilibrium constant (Kc) for the reversible reaction
(b) What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if more H2 is added?
(c) What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if CO is removed?
(d) What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if CH3OH is added?
(e) What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if the temperature of the system is increased?
(f) What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if more catalyst is added?
41.
Nitrogen and oxygen react at high temperatures.
(a) Write the expression for the equilibrium constant (Kc) for the reversible reaction
(b) What will happen to the concentrations of N2, O2, and NO at equilibrium if more O2 is added?
(c) What will happen to the concentrations of N2, O2, and NO at equilibrium if N2 is removed?
(d) What will happen to the concentrations of N2, O2, and NO at equilibrium if NO is added?
(e) What will happen to the concentrations of N2, O2, and NO at equilibrium if the volume of the reaction vessel is decreased?
(f) What will happen to the concentrations of N2, O2, and NO at equilibrium if the temperature of the system is increased?
(g) What will happen to the concentrations of N2, O2, and NO at equilibrium if a catalyst is added?
42.
Water gas, a mixture of H2 and CO, is an important industrial fuel produced by the reaction of steam with red hot coke, essentially pure carbon.
(a) Write the expression for the equilibrium constant for the reversible reaction
(b) What will happen to the concentration of each reactant and product at equilibrium if more C is added?
(c) What will happen to the concentration of each reactant and product at equilibrium if H2O is removed?
(d) What will happen to the concentration of each reactant and product at equilibrium if CO is added?
(e) What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased?
43.
Pure iron metal can be produced by the reduction of iron(III) oxide with hydrogen gas.
(a) Write the expression for the equilibrium constant (Kc) for the reversible reaction
(b) What will happen to the concentration of each reactant and product at equilibrium if more Fe is added?
(c) What will happen to the concentration of each reactant and product at equilibrium if H2O is removed?
(d) What will happen to the concentration of each reactant and product at equilibrium if H2 is added?
(e) What will happen to the concentration of each reactant and product at equilibrium if the volume of the reaction vessel is decreased?
(f) What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased?
44.
Ammonia is a weak base that reacts with water according to this equation:
Will any of the following increase the percent of ammonia that is converted to the ammonium ion in water?
1. Addition of NaOH
2. Addition of HCl
3. Addition of NH4Cl
45.
Acetic acid is a weak acid that reacts with water according to this equation:
Will any of the following increase the percent of acetic acid that reacts and produces ion?
1. Addition of HCl
2. Addition of NaOH
3. Addition of NaCH3CO2
46.
Suggest two ways in which the equilibrium concentration of Ag+ can be reduced in a solution of Na+, Cl, Ag+, and in contact with solid AgCl.
47.
How can the pressure of water vapor be increased in the following equilibrium?
48.
A solution is saturated with silver sulfate and contains excess solid silver sulfate:
A small amount of solid silver sulfate containing a radioactive isotope of silver is added to this solution. Within a few minutes, a portion of the solution phase is sampled and tests positive for radioactive Ag+ ions. Explain this observation.
49.
When equal molar amounts of HCl and HOCl are dissolved separately in equal amounts of water, the solution of HCl freezes at a lower temperature. Which compound has the larger equilibrium constant for acid ionization?
1. HCl
2. H+ + Cl
3. HOCl
4. H+ + OCl
50.
A reaction is represented by this equation:
1. Write the mathematical expression for the equilibrium constant.
2. Using concentrations ≤1 M, identify two sets of concentrations that describe a mixture of A, B, and C at equilibrium.
51.
A reaction is represented by this equation:
1. Write the mathematical expression for the equilibrium constant.
2. Using concentrations of ≤1 M, identify two sets of concentrations that describe a mixture of W, X, and Y at equilibrium.
52.
What is the value of the equilibrium constant at 500 °C for the formation of NH3 according to the following equation?
An equilibrium mixture of NH3(g), H2(g), and N2(g) at 500 °C was found to contain 1.35 M H2, 1.15 M N2, and 4.12 \times 10^{−1} DELMAR M NH3.
53.
Hydrogen is prepared commercially by the reaction of methane and water vapor at elevated temperatures.
What is the equilibrium constant for the reaction if a mixture at equilibrium contains gases with the following concentrations: CH4, 0.126 M; H2O, 0.242 M; CO, 0.126 M; H2 1.15 M, at a temperature of 760 °C?
54.
A 0.72-mol sample of PCl5 is put into a 1.00-L vessel and heated. At equilibrium, the vessel contains 0.40 mol of PCl3(g) and 0.40 mol of Cl2(g). Calculate the value of the equilibrium constant for the decomposition of PCl5 to PCl3 and Cl2 at this temperature.
55.
At 1 atm and 25 °C, NO2 with an initial concentration of 1.00 M is 0.0033% decomposed into NO and O2. Calculate the value of the equilibrium constant for the reaction.
56.
Calculate the value of the equilibrium constant KP for the reaction from these equilibrium pressures: NO, 0.050 atm; Cl2, 0.30 atm; NOCl, 1.2 atm.
57.
When heated, iodine vapor dissociates according to this equation:
At 1274 K, a sample exhibits a partial pressure of I2 of 0.1122 atm and a partial pressure due to I atoms of 0.1378 atm. Determine the value of the equilibrium constant, KP, for the decomposition at 1274 K.
58.
A sample of ammonium chloride was heated in a closed container.
At equilibrium, the pressure of NH3(g) was found to be 1.75 atm. What is the value of the equilibrium constant KP for the decomposition at this temperature?
59.
At a temperature of 60 °C, the vapor pressure of water is 0.196 atm. What is the value of the equilibrium constant KP for the vaporization equilibrium at 60 °C?
60.
Complete the following partial ICE tables.
(a)
(b)
(c)
(d)
(e)
(f)
61.
Complete the following partial ICE tables.
(a)
(b)
(c)
(d)
(e)
(f)
62.
Why are there no changes specified for Ni in Exercise 13.60, part (f)? What property of Ni does change?
63.
Why are there no changes specified for NH4HS in Exercise 13.61, part (e)? What property of NH4HS does change?
64.
Analysis of the gases in a sealed reaction vessel containing NH3, N2, and H2 at equilibrium at 400 °C established the concentration of N2 to be 1.2 M and the concentration of H2 to be 0.24 M.
Calculate the equilibrium molar concentration of NH3.
65.
Calculate the number of moles of HI that are at equilibrium with 1.25 mol of H2 and 1.25 mol of I2 in a 5.00−L flask at 448 °C.
66.
What is the pressure of BrCl in an equilibrium mixture of Cl2, Br2, and BrCl if the pressure of Cl2 in the mixture is 0.115 atm and the pressure of Br2 in the mixture is 0.450 atm?
67.
What is the pressure of CO2 in a mixture at equilibrium that contains 0.50 atm H2, 2.0 atm of H2O, and 1.0 atm of CO at 990 °C?
68.
Cobalt metal can be prepared by reducing cobalt(II) oxide with carbon monoxide.
What concentration of CO remains in an equilibrium mixture with [CO2] = 0.100 M?
69.
Carbon reacts with water vapor at elevated temperatures.
Assuming a reaction mixture initially contains only reactants, what is the concentration of CO in an equilibrium mixture with [H2O] = 0.500 M at 1000 °C?
70.
Sodium sulfate 10−hydrate, Na2SO4·10H2O, dehydrates according to the equation
What is the pressure of water vapor at equilibrium with a mixture of Na2SO4·10H2O and NaSO4?
71.
Calcium chloride 6−hydrate, CaCl2·6H2O, dehydrates according to the equation
What is the pressure of water vapor at equilibrium with a mixture of CaCl2·6H2O and CaCl2 at 25 °C?
72.
A student solved the following problem and found the equilibrium concentrations to be [SO2] = 0.590 M, [O2] = 0.0450 M, and [SO3] = 0.260 M. How could this student check the work without reworking the problem? The problem was: For the following reaction at 600 °C:
73.
A student solved the following problem and found [N2O4] = 0.16 M at equilibrium. How could this student recognize that the answer was wrong without reworking the problem? The problem was: What is the equilibrium concentration of N2O4 in a mixture formed from a sample of NO2 with a concentration of 0.10 M?
74.
Assume that the change in concentration of N2O4 is small enough to be neglected in the following problem.
(a) Calculate the equilibrium concentration of both species in 1.00 L of a solution prepared from 0.129 mol of N2O4 with chloroform as the solvent.
in chloroform
(b) Confirm that the change is small enough to be neglected.
75.
Assume that the change in concentration of COCl2 is small enough to be neglected in the following problem.
(a) Calculate the equilibrium concentration of all species in an equilibrium mixture that results from the decomposition of COCl2 with an initial concentration of 0.3166 M.
(b) Confirm that the change is small enough to be neglected.
76.
Assume that the change in pressure of H2S is small enough to be neglected in the following problem.
(a) Calculate the equilibrium pressures of all species in an equilibrium mixture that results from the decomposition of H2S with an initial pressure of 0.824 atm.
(b) Confirm that the change is small enough to be neglected.
77.
What are all concentrations after a mixture that contains [H2O] = 1.00 M and [Cl2O] = 1.00 M comes to equilibrium at 25 °C?
78.
What are the concentrations of PCl5, PCl3, and Cl2 in an equilibrium mixture produced by the decomposition of a sample of pure PCl5 with [PCl5] = 2.00 M?
79.
Calculate the number of grams of HI that are at equilibrium with 1.25 mol of H2 and 63.5 g of iodine at 448 °C.
80.
Butane exists as two isomers, n−butane and isobutane.
KP = 2.5 at 25 °C
What is the pressure of isobutane in a container of the two isomers at equilibrium with a total pressure of 1.22 atm?
81.
What is the minimum mass of CaCO3 required to establish equilibrium at a certain temperature in a 6.50-L container if the equilibrium constant (Kc) is 0.50 for the decomposition reaction of CaCO3 at that temperature?
82.
The equilibrium constant (Kc) for this reaction is 1.60 at 990 °C:
Calculate the number of moles of each component in the final equilibrium mixture obtained from adding 1.00 mol of H2, 2.00 mol of CO2, 0.750 mol of H2O, and 1.00 mol of CO to a 5.00-L container at 990 °C.
83.
In a 3.0-L vessel, the following equilibrium partial pressures are measured: N2, 190 torr; H2, 317 torr; NH3, 1.00 \times 10^{3} DELMAR torr.
1. How will the partial pressures of H2, N2, and NH3 change if H2 is removed from the system? Will they increase, decrease, or remain the same?
2. Hydrogen is removed from the vessel until the partial pressure of nitrogen, at equilibrium, is 250 torr. Calculate the partial pressures of the other substances under the new conditions.
84.
The equilibrium constant (Kc) for this reaction is 5.0 at a given temperature.
1. On analysis, an equilibrium mixture of the substances present at the given temperature was found to contain 0.20 mol of CO, 0.30 mol of water vapor, and 0.90 mol of H2 in a liter. How many moles of CO2 were there in the equilibrium mixture?
2. Maintaining the same temperature, additional H2 was added to the system, and some water vapor was removed by drying. A new equilibrium mixture was thereby established containing 0.40 mol of CO, 0.30 mol of water vapor, and 1.2 mol of H2 in a liter. How many moles of CO2 were in the new equilibrium mixture? Compare this with the quantity in part (a), and discuss whether the second value is reasonable. Explain how it is possible for the water vapor concentration to be the same in the two equilibrium solutions even though some vapor was removed before the second equilibrium was established.
85.
Antimony pentachloride decomposes according to this equation:
An equilibrium mixture in a 5.00-L flask at 448 °C contains 3.85 g of SbCl5, 9.14 g of SbCl3, and 2.84 g of Cl2. How many grams of each will be found if the mixture is transferred into a 2.00-L flask at the same temperature?
86.
Consider the equilibrium
1. What is the expression for the equilibrium constant (Kc) of the reaction?
2. How must the concentration of NH3 change to reach equilibrium if the reaction quotient is less than the equilibrium constant?
3. If the reaction were at equilibrium, how would an increase in the volume of the reaction vessel affect the pressure of NO2?
4. If the change in the pressure of NO2 is 28 torr as a mixture of the four gases reaches equilibrium, how much will the pressure of O2 change?
87.
The binding of oxygen by hemoglobin (Hb), giving oxyhemoglobin (HbO2), is partially regulated by the concentration of H3O+ and dissolved CO2 in the blood. Although the equilibrium is complicated, it can be summarized as
1. Write the equilibrium constant expression for this reaction.
2. Explain why the production of lactic acid and CO2 in a muscle during exertion stimulates release of O2 from the oxyhemoglobin in the blood passing through the muscle.
88.
Liquid N2O3 is dark blue at low temperatures, but the color fades and becomes greenish at higher temperatures as the compound decomposes to NO and NO2. At 25 °C, a value of KP = 1.91 has been established for this decomposition. If 0.236 moles of N2O3 are placed in a 1.52-L vessel at 25 °C, calculate the equilibrium partial pressures of N2O3(g), NO2(g), and NO(g).
89.
A 1.00-L vessel at 400 °C contains the following equilibrium concentrations: N2, 1.00 M; H2, 0.50 M; and NH3, 0.25 M. How many moles of hydrogen must be removed from the vessel to increase the concentration of nitrogen to 1.1 M? The equilibrium reaction is | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/13%3A_Fundamental_Equilibrium_Concepts/13.08%3A_Exercises.txt |
This chapter will illustrate the chemistry of acid-base reactions and equilibria, and provide you with tools for quantifying the concentrations of acids and bases in solutions.
• 14.0: Introduction
• 14.1: Brønsted-Lowry Acids and Bases
Compounds that 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. Amphiprotic species can act as both proton donors and proton acceptors. Water is the most important amphiprotic species.
• 14.2: pH and pOH
The concentration of hydronium ion in a solution of an acid in water is greater than 1.0×10⁻⁷M at 25 °C. The concentration of hydroxide ion in a solution of a base in water is greater than 1.0×10⁻⁷M M at 25 °C. The concentration of H₃O⁺ in a solution can be expressed as the pH of the solution; pH=−log H₃O⁺. The concentration of OH⁻ can be expressed as the pOH of the solution: pOH=−log[OH⁻].
• 14.3: Relative Strengths of Acids and Bases
The strengths of Brønsted-Lowry acids and bases in aqueous solutions can be determined by their acid or base ionization constants. Stronger acids form weaker conjugate bases, and weaker acids form stronger conjugate bases. Thus strong acids are completely ionized in aqueous solution because their conjugate bases are weaker bases than water. Weak acids are only partially ionized because their conjugate bases are compete successfully with water for possession of protons.
• 14.4: Hydrolysis of Salt Solutions
The characteristic properties of aqueous solutions of Brønsted-Lowry acids are due to the presence of hydronium ions; those of aqueous solutions of Brønsted-Lowry bases are due to the presence of hydroxide ions. The neutralization that occurs when aqueous solutions of acids and bases are combined results from the reaction of the hydronium and hydroxide ions to form water. Some salts formed in neutralization reactions may make the product solutions slightly acidic or slightly basic.
• 14.5: 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.
• 14.6: Buffers
A solution containing a mixture of an acid and its conjugate base, or of a base and its conjugate acid, is called a buffer solution. Unlike in the case of an acid, base, or salt solution, the hydronium ion concentration of a buffer solution does not change greatly when a small amount of acid or base is added to the buffer solution. The base (or acid) in the buffer reacts with the added acid (or base).
• 14.7: Acid-Base Titrations
A titration curve is a graph that relates the change in pH of an acidic or basic solution to the volume of added titrant. The characteristics of the titration curve are dependent on the specific solutions being titrated. The pH of the solution at the equivalence point may be greater than, equal to, or less than 7.00. The choice of an indicator for a given titration depends on the expected pH at the equivalence point of the titration, and the range of the color change of the indicator.
• 14.8: Key Terms
• 14.9: Key Equations
• 14.10: Summary
• 14.11: Exercises
These are homework exercises to accompany the Textmap created for "Chemistry" by OpenStax.
14: Acid-Base Equilibria
Liquid water is essential to life on our planet, and chemistry involving the characteristic ions of water, \(\ce{H^{+}}\) and \(\ce{OH^{–}}\), is widely encountered in nature and society. As introduced in another chapter of this text, acid-base chemistry involves the transfer of hydrogen ions from donors (acids) to acceptors (bases). These \(\ce{H^{+}}\) transfer reactions are reversible, and the equilibria established by acid-base systems are essential aspects of phenomena ranging from sinkhole formation (Figure \(1\)) to oxygen transport in the human body. This chapter will further explore acid-base chemistry with an emphasis on the equilibrium aspects of this important reaction class. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/14%3A_Acid-Base_Equilibria/14.00%3A_Introduction.txt |
Learning Objectives
By the end of this section, you will be able to:
• 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 acid-base reaction class has been studied for quite some time. In 1680, Robert Boyle reported traits of acid solutions that included their ability to dissolve many substances, to change the colors of certain natural dyes, and to lose these traits after coming in contact with alkali (base) solutions. 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 $\ce{CO2}$), 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.
Johannes Brønsted and Thomas Lowry proposed a more general description in 1923 in which acids and bases were defined in terms of the transfer of hydrogen ions, H+. (Note that these hydrogen ions are often referred to simply as protons, since that subatomic particle is the only component of cations derived from the most abundant hydrogen isotope, 1H.) 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, thus, the transfer of a proton from a donor (acid) to an acceptor (base).
The concept of conjugate pairs is useful in describing Brønsted-Lowry acid-base reactions (and other reversible reactions, as well). When an acid donates $\ce{H^{+}}$, the species that remains is called the conjugate base of the acid because it reacts as a proton acceptor in the reverse reaction. Likewise, when a base accepts $\ce{H^{+}}$, it is converted to its conjugate acid. The reaction between water and ammonia illustrates this idea. In the forward direction, water acts as an acid by donating a proton to ammonia and subsequently becoming a hydroxide ion, $\ce{OH^{−}}$, the conjugate base of water. The ammonia acts as a base in accepting this proton, becoming an ammonium ion, $\ce{NH4^{+}}$, the conjugate acid of ammonia. In the reverse direction, a hydroxide ion acts as a base in accepting a proton from ammonium ion, which acts as an acid.
The reaction between a Brønsted-Lowry acid and water is called acid ionization. For example, when hydrogen fluoride dissolves in water and ionizes, protons are transferred from hydrogen fluoride molecules to water molecules, yielding hydronium ions and fluoride ions:
Base ionization of a species occurs when it accepts protons from water molecules. In the example below, pyridine molecules, $\ce{C5NH5}$, undergo base ionization when dissolved in water, yielding hydroxide and pyridinium ions:
The preceding ionization reactions suggest that water may function as both a base (as in its reaction with hydrogen fluoride) and an acid (as in its reaction with ammonia). Species capable of either donating or accepting protons are called amphiprotic, or more generally, amphoteric, a term that may be used for acids and bases per definitions other than the Brønsted-Lowry one. The equations below show the two possible acid-base reactions for two amphiprotic species, bicarbonate ion and water:
$\ce{HCO3^{–}(aq) + H2O(l) <=> CO3^{2–}(aq) + H3O^{+} (aq)} \nonumber$
$\ce{HCO3^{-}(aq) + H2O(l) <=> H2CO3(aq) + OH^{-} (aq)} \nonumber$
The first equation represents the reaction of bicarbonate as an acid with water as a base, whereas the second represents reaction of bicarbonate as a base with water as an acid. When bicarbonate is added to water, both these equilibria are established simultaneously and the composition of the resulting solution may be determined through appropriate equilibrium calculations, as described later in this chapter.
In the liquid state, molecules of an amphiprotic substance can react with one another as illustrated for water in the equations below:
The process in which like molecules react to yield ions is called autoionization. Liquid water undergoes autoionization to a very slight extent; at 25 °C, approximately two out of every billion water molecules are ionized. The extent of the water autoionization process is reflected in the value of its equilibrium constant, the ion-product constant for water, $K_w$:
$\ce{2H2O(l) <=> H3O^{+}(aq) + OH^{-}(aq)} \label{autoionization}$
with
$K_w=[\ce{H3O^{+}}][\ce{OH^{-}}] \label{Kw}$
The slight ionization of pure water is reflected in the small value of the equilibrium constant; at 25 °C, $K_w$ has a value of $1.0 \times 10^{−14}$. The process is endothermic, and so the extent of ionization and the resulting concentrations of hydronium ion and hydroxide ion increase with temperature. For example, at 100 °C, the value for $K_w$ is about $5.6 \times 10^{−13}$, roughly 50 times larger than the value at 25 °C.
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, [H3O+] = [OH] = x. At 25 °C:
$K_{ w }=\left[ H_3 O^{+}\right]\left[ OH^{-}\right]=(x)(x)=x^2=1.0 \times 10^{-14} \nonumber$
So:
$x=\left[ H_3 O^{+}\right]=\left[ OH^{-}\right]=\sqrt{1.0 \times 10^{-14}}=1.0 \times 10^{-7} M \nonumber$
The hydronium ion concentration and the hydroxide ion concentration are the same, $1.0 \times 10^{−7}\, M$.
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
$[\ce{H3O^{+}}] = [\ce{OH^{-}}] = 4.9 \times 10^{−7}\, M \nonumber$
Example $2$: The Inverse Relation between [H3O+] and [OH−]
A solution of an acid 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
Use the value of the ion-product constant for water at 25 °C (Equations \ref{autoionization} and \ref{Kw}) to calculate the missing equilibrium concentration.
$\ce{2H2O(l) <=> H3O^{+}(aq) + OH^{-}(aq)} \nonumber$
with $K_w=[\ce{H3O^{+}}][\ce{OH^{-}}] \nonumber$.
Rearrangement of the Kw expression shows that [OH] is inversely proportional to [H3O+]:
$K_{ w }=\left[\ce{H3O^{+}}\right]\left[\ce{OH^{-}}\right]=\left(2.0 \times 10^{-6}\right)\left(5.0 \times 10^{-9}\right)=1.0 \times 10^{-14} \nonumber$
Compared with pure water, a solution of acid exhibits a higher concentration of hydronium ions (due to ionization of the acid) and a proportionally lower concentration of hydroxide ions. This may be explained via Le Chatelier’s principle as a left shift in the water autoionization equilibrium resulting from the stress of increased hydronium ion concentration.
Substituting the ion concentrations into the Kw expression confirms this calculation, resulting in the expected value:
$K_w=[\ce{H3O^{+}}][\ce{OH^{-}}]=(2.0 \times 10^{−6})(5.0 \times 10^{−9})=1.0 \times 10^{−14} \nonumber$
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
$\ce{[H3O+] = 1 \times 10^{−11}\, M} \nonumber$
Example $3$: Representing 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. $\ce{HSO3^{-}(aq) + HI(aq) <=> H2SO3(aq) + I^{-}(aq)}$
Exercise $3$
Write separate equations representing the reaction of $\ce{H2PO4^{-}}$
1. as a base with $\ce{HBr}$
2. as an acid with $\ce{OH^{-}}$
Answer
1. $\ce{H2PO4^{-}(aq) + HBr(aq) <=> H3PO4(aq) + Br^{-}(aq)}$
2. $\ce{H2PO4^{-}(aq) + OH^{-}(aq) <=> HPO4^{2-}(aq) + H2O(l)}$ | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/14%3A_Acid-Base_Equilibria/14.01%3A_Brnsted-Lowry_Acids_and_Bases.txt |
Learning Objectives
By the end of this section, you will be able to:
• Explain the characterization of aqueous solutions as acidic, basic, or neutral
• Express hydronium and hydroxide ion concentrations on the pH and pOH scales
• Perform calculations relating pH and pOH
As discussed earlier, hydronium and hydroxide ions are present both in pure water and in all aqueous solutions, and their concentrations are inversely proportional as determined by the ion product of water ($K_w$). The concentrations of these ions in a solution are often critical determinants of the solution’s properties and the chemical behaviors of its other solutes, and specific vocabulary has been developed to describe these concentrations in relative terms. A solution is neutral if it contains equal concentrations of hydronium and hydroxide ions; acidic if it contains a greater concentration of hydronium ions than hydroxide ions; and basic if it contains a lesser concentration of hydronium ions than hydroxide ions.
A common means of expressing quantities that may span many orders of magnitude is to use a logarithmic scale. One such scale that is very popular for chemical concentrations and equilibrium constants is based on the p-function, defined as shown where “$X$” is the quantity of interest and “log” is the base-10 logarithm:
$pX =-\log X \nonumber$
The pH of a solution is therefore defined as shown here, where [H3O+] is the molar concentration of hydronium ion in the solution:
$pH =-\log \left[ H_3 O^{+}\right] \nonumber$
Rearranging this equation to isolate the hydronium ion molarity yields the equivalent expression:
$\left[ H_3 O^{+}\right]=10^{- pH } \nonumber$
Likewise, the hydroxide ion molarity may be expressed as a p-function, or pOH:
$pOH =-\log \left[ OH^{-}\right] \nonumber$
or
$\left[ OH^{-}\right]=10^{- pOH } \nonumber$
Finally, the relation between these two ion concentration expressed as p-functions is easily derived from the $K_w$ expression:
\begin{align} K_{ w } &=\left[ \ce{H3O^{+}} \right]\left[\ce{OH^{-}}\right] \nonumber \[4pt] -\log K_{ w } &=-\log \left(\left[ \ce{H_3O^{+}}\right]\left[\ce{OH^{-}}\right]\right) \nonumber \[4pt] &=-\log \left[ \ce{H3O^{+}}\right] + -\log \left[\ce{OH^{-}}\right] \nonumber \[4pt] pK_{ w } &= pH + pOH \label{pHpOH} \end{align}
At 25 °C, the value of $K_w$ is $1.0 \times 10^{−14}$, and so:
$14.00= pH + pOH \nonumber$
As was demonstrated previously, the hydronium ion molarity in pure water (or any neutral solution) is $1.0 \times 10^{−7}~\text{M}$ at 25 °C. The pH and pOH of a neutral solution at this temperature are therefore:
$pH =-\log \left[ H_3 O^{+}\right]=-\log \left(1.0 \times 10^{-7}\right)=7.00 \nonumber$
$pOH =-\log \left[ OH^{-}\right]=-\log \left(1.0 \times 10^{-7}\right)=7.00 \nonumber$
And so, at this temperature, acidic solutions are those with hydronium ion molarities greater than $1.0 \times 10^{−7} ~\text{M}$ and hydroxide ion molarities less than $1.0 \times 10^{−7} ~\text{M}$ (corresponding to pH values less than 7.00 and pOH values greater than 7.00). Basic solutions are those with hydronium ion molarities less than $1.0 \times 10^{−7} ~\text{M}$ and hydroxide ion molarities greater than $1.0 \times 10^{−7} ~\text{M}$ (corresponding to pH values greater than 7.00 and pOH values less than 7.00).
Since the autoionization constant Kw is temperature dependent, these correlations between pH values and the acidic/neutral/basic adjectives will be different at temperatures other than 25 °C. For example, Exercise $1$ showed the hydronium molarity of pure water at 80 °C is $4.9 \times 10^{−7} ~\text{M}$, which corresponds to pH and pOH values of:
$pH =-\log \left[ H_3 O^{+}\right]=-\log \left(4.9 \times 10^{-7}\right)=6.31 \nonumber$
$pOH =-\log \left[ OH^{-}\right]=-\log \left(4.9 \times 10^{-7}\right)=6.31 \nonumber$
At this temperature, then, neutral solutions exhibit pH = pOH = 6.31, acidic solutions exhibit pH less than 6.31 and pOH greater than 6.31, whereas basic solutions exhibit pH greater than 6.31 and pOH less than 6.31. This distinction can be important when studying certain processes that occur at other temperatures, such as enzyme reactions in warm-blooded organisms at a temperature around 36–40 °C. Unless otherwise noted, references to pH values are presumed to be those at 25 °C (Table $1$).
Table $1$: Summary of Relations for Acidic, Basic and Neutral Solutions
Classification Relative Ion Concentrations pH at 25 °C
acidic [H3O+] > [OH] pH < 7
neutral [H3O+] = [OH] pH = 7
basic [H3O+] < [OH] pH > 7
Figure $1$: shows the relationships between [H3O+], [OH], pH, and pOH for solutions classified as acidic, basic, and neutral.
Example $1$: Calculation of pH from [H3O+]
What is the pH of stomach acid, a solution of $\ce{HCl}$ with a hydronium ion concentration of $1.2 \times 10^{−3}\, M$?
Solution
\begin{align*} pH &=-\log \left[\ce{H3O^{+}}\right] \[4pt] &=-\log \left(1.2 \times 10^{-3}\right) \[4pt] &=-(-2.92)=2.92 \end{align*} \nonumber
(The use of logarithms is explained in Appendix B. When taking the log of a value, keep as many decimal places in the result as there are significant figures in the value.)
Exercise $1$
Water exposed to air contains carbonic acid, $\ce{H2CO3}$, due to the reaction between carbon dioxide and water:
$\ce{CO2(aq) + H2O(l) <=> H2CO3(aq)} \nonumber$
Air-saturated water has a hydronium ion concentration caused by the dissolved $\ce{CO2}$ of $2.0 \times 10^{−6} ~\text{M}$, about 20-times larger than that of pure water. Calculate the pH of the solution at 25 °C.
Answer
5.70
Example $2$: Calculation of Hydronium Ion Concentration from pH
Calculate the hydronium ion concentration of blood, the pH of which is 7.3.
Solution
\begin{align*} pH =-\log \left[ H_3 O^{+}\right] &=7.3 \[4pt] \log \left[ H_3O^{+}\right] &=-7.3 \[4pt] \left[ H_3 O^{+}\right] &=10^{-7.3} \end{align*} \nonumber
or
\begin{align*} [ \ce{H3O^{+}} ] &= \text {antilog of } -7.3 \[4pt] &=5 \times 10^{-8} M \end{align*} \nonumber
(On a calculator take the antilog, or the “inverse” log, of −7.3, or calculate $10^{−7.3}$.)
Exercise $2$
Calculate the hydronium ion concentration of a solution with a pH of −1.07.
Answer
12 M
How Sciences Interconnect: Environmental Science
Normal rainwater has a pH between 5 and 6 due to the presence of dissolved $\ce{CO2}$ which forms carbonic acid:
\begin{align*} \ce{H2O(l) + CO2(g) & -> H2CO3(aq)} \[4pt] \ce{H2CO3(aq) &<=> H^{+}(aq) + HCO3^{-}(aq)}\end{align*} \nonumber
Acid rain is rainwater that has a pH of less than 5, due to a variety of nonmetal oxides, including $\ce{CO2}$, $\ce{SO2}$, $\ce{SO3}$, $\ce{NO}$, and $\ce{NO2}$ being dissolved in the water and reacting with it to form not only carbonic acid, but sulfuric acid and nitric acid. The formation and subsequent ionization of sulfuric acid are shown here:
\begin{align*} \ce{H2O(l) + SO3(g) &-> H2SO4(aq)} \[4pt] \ce{H2SO4(aq) &-> H^{+}(aq) + HSO_4^{-}(aq)} \end{align*} \nonumber
Carbon dioxide is naturally present in the atmosphere because most organisms produce it as a waste product of metabolism. Carbon dioxide is also formed when fires release carbon stored in vegetation or fossil fuels. Sulfur trioxide in the atmosphere is naturally produced by volcanic activity, but it also originates from burning fossil fuels, which have traces of sulfur, and from the process of “roasting” ores of metal sulfides in metal-refining processes. Oxides of nitrogen are formed in internal combustion engines where the high temperatures make it possible for the nitrogen and oxygen in air to chemically combine.
Acid rain is a particular problem in industrial areas where the products of combustion and smelting are released into the air without being stripped of sulfur and nitrogen oxides. In North America and Europe until the 1980s, it was responsible for the destruction of forests and freshwater lakes, when the acidity of the rain actually killed trees, damaged soil, and made lakes uninhabitable for all but the most acid-tolerant species. Acid rain also corrodes statuary and building facades that are made of marble and limestone (Figure $2$). Regulations limiting the amount of sulfur and nitrogen oxides that can be released into the atmosphere by industry and automobiles have reduced the severity of acid damage to both natural and manmade environments in North America and Europe. It is now a growing problem in industrial areas of China and India.
For further information on acid rain, visit this website hosted by the US Environmental Protection Agency.
Example $3$: Calculation of pOH
What are the $\text{pOH}$ and the $\text{pH} of a 0.0125-M solution of potassium hydroxide, \(\ce{KOH}$?
Solution
Potassium hydroxide is a highly soluble ionic compound and completely dissociates when dissolved in dilute solution, yielding [OH] = 0.0125 M:
\begin{align*} \text{pOH} & =-\log \left[ OH^{-}\right]=-\log 0.0125 \[4pt] & =-(-1.903)=1.903 \end{align*} \nonumber
The pH can be found from the pOH:
\begin{align*} \text{pH} + \text{pOH} &=14.00 \[4pt] pH &=14.00- pOH \[4pt] &=14.00-1.903=12.10 \end{align*} \nonumber
Exercise $3$
The hydronium ion concentration of vinegar is approximately $4 \times 10^{−3} ~\text{M}$. What are the corresponding values of $\text{pOH}$ and $\text{pH}$?
Answer
pOH = 11.6, pH = 2.4
The acidity of a solution is typically assessed experimentally by measurement of its pH. The pOH of a solution is not usually measured, as it is easily calculated from an experimentally determined pH value. The pH of a solution can be directly measured using a pH meter (Figure $3$).
The pH of a solution may also be visually estimated using colored indicators (Figure $4$). The acid-base equilibria that enable use of these indicator dyes for pH measurements are described in a later section of this chapter. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/14%3A_Acid-Base_Equilibria/14.02%3A_pH_and_pOH.txt |
Learning Objectives
By the end of this section, you will be able to:
• Assess the relative strengths of acids and bases according to their ionization constants
• Rationalize trends in acid–base strength in relation to molecular structure
• Carry out equilibrium calculations for weak acid–base systems
Acid and Base Ionization Constants
The relative strength of an acid or base is the extent to which it ionizes when dissolved in water. If the ionization reaction is essentially complete, the acid or base is termed strong; if relatively little ionization occurs, the acid or base is weak. As will be evident throughout the remainder of this chapter, there are many more weak acids and bases than strong ones. The most common strong acids and bases are listed in Figure $1$.
The relative strengths of acids may be quantified by measuring their equilibrium constants in aqueous solutions. In solutions of the same concentration, stronger acids ionize to a greater extent, and so yield higher concentrations of hydronium ions than do weaker acids. The equilibrium constant for an acid is called the acid-ionization constant, $K_a$. For the reaction of an acid $\ce{HA}$:
$\ce{HA(aq) + H2O(l) <=> H3O^{+}(aq) + A^{-}(aq)} \nonumber$
the acid ionization constant is written
$K_{ a }=\frac{[\ce{H3O^{+}}] [\ce{A^{-}}]}{[\ce{HA}]} \nonumber$
where the concentrations are those at equilibrium. Although water is a reactant in the reaction, it is the solvent as well, so we do not include $[\ce{H2O}]$ in the equation.
Note
Equilibrium constant expressions are actually ratios of activities, and the value of K is determined at the limit of infinite dilution of the solutes. In these very dilute solutions, the activity of the solvent has a value of unity (1) and the activity of each solute can be approximated by its molar concentration.
The larger the Ka of an acid, the larger the concentration of $\ce{H3O^{+}}$ and $\ce{A^{-}}$ relative to the concentration of the nonionized acid, $\ce{HA}$, in an equilibrium mixture, and the stronger the acid. An acid is classified as “strong” when it undergoes complete ionization, in which case the concentration of HA is zero and the acid ionization constant is immeasurably large ($K_a ≈ ∞$). Acids that are partially ionized are called “weak,” and their acid ionization constants may be experimentally measured. A table of ionization constants for weak acids is provided in Appendix H.
To illustrate this idea, three acid ionization equations and Ka values are shown below. The ionization constants increase from first to last of the listed equations, indicating the relative acid strength increases in the order:
$\ce{CH3CO2H < HNO2 < HSO4^{−}} \nonumber$
as demonstrated with the equations below:
$\ce{CH3CO2H(aq) + H2O(l) <=> H3O^{+}(aq) + CH3CO2^{-}(aq)} \quad \quad Ka=1.8 \times 10^{−5} \nonumber$
$\ce{HNO2(aq) + H2O(l) <=> H3O^{+}(aq) + NO2^{-}(aq)} \quad \quad K_a=4.6 \times 10^{-4} \nonumber$
$\ce{HSO4^{-}(aq) + H2O(aq) <=> H3O^{+}(aq) + SO4^{2-}(aq) } \quad\quad K_a=1.2 \times 10^{-2} \nonumber$
Another measure of the strength of an acid is its percent ionization. The percent ionization of a weak acid is defined in terms of the composition of an equilibrium mixture:
$\% \text { ionization }=\frac{\left[ \ce{H3O^{+}} \right]_{ eq }}{[ \ce{HA} ]_0} \times 100 \nonumber$
where the numerator is equivalent to the concentration of the acid's conjugate base (per stoichiometry, [A] = [H3O+]). Unlike the Ka value, the percent ionization of a weak acid varies with the initial concentration of acid, typically decreasing as concentration increases. Equilibrium calculations of the sort described later in this chapter can be used to confirm this behavior.
Example $1$: Calculation of Percent Ionization from pH
Calculate the percent ionization of a 0.125-M solution of nitrous acid (a weak acid), with a pH of 2.09.
Solution
The percent ionization for an acid is:
$\frac{\left[ \ce{H3O^{+}} \right]_{ eq }}{\left[ \ce{HNO2} \right]_0} \times 100 \nonumber$
Converting the provided pH to hydronium ion molarity yields
$\left[ \ce{H3O^{+}} \right]=10^{-2.09}=0.0081 M \nonumber$
Substituting this value and the provided initial acid concentration into the percent ionization equation gives
$\frac{8.1 \times 10^{-3}}{0.125} \times 100=6.5 \% \nonumber$
(Recall the provided pH value of 2.09 is logarithmic, and so it contains just two significant digits, limiting the certainty of the computed percent ionization.)
Exercise $1$
Calculate the percent ionization of a 0.10-M solution of acetic acid with a pH of 2.89.
Answer
1.3% ionized
Link to Learning
View the simulation of strong and weak acids and bases at the molecular level.
Just as for acids, the relative strength of a base is reflected in the magnitude of its base-ionization constant (Kb) in aqueous solutions. In solutions of the same concentration, stronger bases ionize to a greater extent, and so yield higher hydroxide ion concentrations than do weaker bases. A stronger base has a larger ionization constant than does a weaker base. For the reaction of a base, $\ce{B}$:
$\ce{B(aq) + H2O(l) <=> HB^{+}(aq) + OH^{-}(aq)}, \nonumber$
the ionization constant is written as
$K_{ b }=\frac{\left[ \ce{HB^{+}} \right]\left[ \ce{OH^{-}} \right]}{[ \ce{B} ]} \nonumber$
Inspection of the data for three weak bases presented below shows the base strength increases in the order: $\ce{NO2^{−} < CH2CO2^{−} < NH3}$
\begin{align*} \ce{NO_2^{-}(aq) + H2O(l) & <=> HNO_2(aq) + OH^{-}(aq)} & K_{ b }=2.17 \times 10^{-11} \[4pt] \ce{CH_3 CO_2^{-}(aq) + H2O(l) & <=> CH_3 CO_2 H(aq) + OH^{-}(aq)} & K_{ b }=5.6 \times 10^{-10} \[4pt] \ce{NH_3(aq) + H2O(l) & <=> NH_4^{+}(aq) + OH^{-}(aq)} & K_{ b }=1.8 \times 10^{-5} \end{align*} \nonumber
A table of ionization constants for weak bases appears in Appendix I. As for acids, the relative strength of a base is also reflected in its percent ionization, computed as
$\% \text { ionization }= \left( \dfrac{[ \ce{OH^{-}}]_{eq}}{[ \ce{B} ]_0} \right) \times 100 \% \nonumber$
but will vary depending on the base ionization constant and the initial concentration of the solution.
Relative Strengths of Conjugate Acid-Base Pairs
Brønsted-Lowry acid-base chemistry is the transfer of protons; thus, logic suggests a relation between the relative strengths of conjugate acid-base pairs. The strength of an acid or base is quantified in its ionization constant, $K_a$ or $K_b$, which represents the extent of the acid or base ionization reaction. For the conjugate acid-base pair $\ce{HA / A^{-}}$, ionization equilibrium equations and ionization constant expressions are
$\ce{HA(aq) + H2O(l) <=> H3O^{+}(aq) + A^{-}(aq)} \nonumber$
with
$K_a=\ce{\dfrac{[H3O^{+}][A^{−}]}{[HA]}} \nonumber$
and
$\ce{A^{−}(aq) + H2O(l) <=> OH^{-}(aq) + HA(aq)} \nonumber$
with
$K_b=\ce{\dfrac{[HA][OH–]}{[A−]}} \nonumber$
Adding these two chemical equations yields the equation for the autoionization for water:
\begin{align*} \ce{\cancel{HA(aq)} + H2O(l) + \cancel{A^{−}(aq)} + H2O(l) &<=> H3O^{+}(aq) + \cancel{A^{-}(aq)} + OH^{-}(aq) + \cancel{HA(aq)}} \[4pt] \ce{2H2O(l) &<=> H3O+(aq) + OH^{−}(aq)} \end{align*} \nonumber
As discussed in another chapter on equilibrium, the equilibrium constant for a summed reaction is equal to the mathematical product of the equilibrium constants for the added reactions, and so
$K_a \times K_b= \dfrac{[\ce{H3O^{+}}][\ce{A^{−}}]}{[\ce{HA}]} × \dfrac{[\ce{HA}][\ce{OH^{-}}]}{[\ce{A^{-}}]}=[\ce{H3O^{+}}][\ce{OH^{-}}]=K_w \nonumber$
This equation states the relation between ionization constants for any conjugate acid-base pair, namely, their mathematical product is equal to the ion product of water, $K_w$. By rearranging this equation, a reciprocal relation between the strengths of a conjugate acid-base pair becomes evident:
$K_{ a }= \dfrac{K_{ w }}{ K_{ b }} \nonumber$
or
$K_{ b }=\dfrac{K_{ w }}{K_{ a }} \nonumber$
The inverse proportional relation between $K_a$ and $K_b$ means the stronger the acid or base, the weaker its conjugate partner. Figure $2$: illustrates this relation for several conjugate acid-base pairs.
The listing of conjugate acid–base pairs shown in Figure $3$: is arranged to show the relative strength of each species as compared with water, whose entries are highlighted in each of the table’s columns. In the acid column, those species listed below water are weaker acids than water. These species do not undergo acid ionization in water; they are not Bronsted-Lowry acids. All the species listed above water are stronger acids, transferring protons to water to some extent when dissolved in an aqueous solution to generate hydronium ions. Species above water but below hydronium ion are weak acids, undergoing partial acid ionization, wheres those above hydronium ion are strong acids that are completely ionized in aqueous solution.
If all these strong acids are completely ionized in water, why does the column indicate they vary in strength, with nitric acid being the weakest and perchloric acid the strongest? Notice that the sole acid species present in an aqueous solution of any strong acid is H3O+(aq), meaning that hydronium ion is the strongest acid that may exist in water; any stronger acid will react completely with water to generate hydronium ions. This limit on the acid strength of solutes in a solution is called a leveling effect. To measure the differences in acid strength for “strong” acids, the acids must be dissolved in a solvent that is less basic than water. In such solvents, the acids will be “weak,” and so any differences in the extent of their ionization can be determined. For example, the binary hydrogen halides HCl, HBr, and HI are strong acids in water but weak acids in ethanol (strength increasing HCl < HBr < HI).
The right column of Figure $3$ lists a number of substances in order of increasing base strength from top to bottom. Following the same logic as for the left column, species listed above water are weaker bases and so they don’t undergo base ionization when dissolved in water. Species listed between water and its conjugate base, hydroxide ion, are weak bases that partially ionize. Species listed below hydroxide ion are strong bases that completely ionize in water to yield hydroxide ions (i.e., they are leveled to hydroxide). A comparison of the acid and base columns in this table supports the reciprocal relation between the strengths of conjugate acid-base pairs. For example, the conjugate bases of the strong acids (top of table) are all of negligible strength. A strong acid exhibits an immeasurably large Ka, and so its conjugate base will exhibit a Kb that is essentially zero:
• strong acid: $\quad K_{ a } \approx \infty$
• conjugate base: $K_{ b }=K_{ w } / K_{ a }=K_{ w } / \infty \approx 0$
A similar approach can be used to support the observation that conjugate acids of strong bases (Kb ≈ ∞) are of negligible strength (Ka ≈ 0).
Example $2$: Calculating Ionization Constants for Conjugate Acid-Base Pairs
Use the Kb for the nitrite ion, $\ce{NO2^{-}}$, to calculate the Ka for its conjugate acid.
Solution
$K_b$ for $\ce{NO2^{-}}$ is given in this section as $2.17 \times 10^{−11}$. The conjugate acid of $\ce{NO2^{-}}$ is $\ce{HNO2}$; $K_a$ for $\ce{HNO2}$ can be calculated using the relationship:
$K_{ a } \times K_{ b }=1.0 \times 10^{-14}=K_{ w } \nonumber$
Solving for Ka yields
$K_{ a }=\frac{K_{ w }}{K_{ b }}=\frac{1.0 \times 10^{-14}}{2.17 \times 10^{-11}}=4.6 \times 10^{-4} \nonumber$
This answer can be verified by finding the Ka for $\ce{HNO2}$ in Appendix H.
Exercise $2$
Determine the relative acid strengths of $\ce{ NH 4^{ +}}$ and $\ce{HCN}$ by comparing their ionization constants. The ionization constant of $\ce{HCN}$ is given in Appendix H as $4.9 \times 10^{−10}$. The ionization constant of $\ce{ NH 4^{ +}}$ is not listed, but the ionization constant of its conjugate base, $\ce{NH3}$, is listed as $1.8 \times 10^{−5}$.
Answer
$\ce{NH4^{+}}$ is the slightly stronger acid ($K_a$ for $\ce{NH4^{+}}$ is $5.6 \times 10^{−10}$).
Acid-Base Equilibrium Calculations
The chapter on chemical equilibria introduced several types of equilibrium calculations and the various mathematical strategies that are helpful in performing them. These strategies are generally useful for equilibrium systems regardless of chemical reaction class, and so they may be effectively applied to acid-base equilibrium problems. This section presents several example exercises involving equilibrium calculations for acid-base systems.
Example $3$: Determination of Ka from Equilibrium Concentrations
Acetic acid is the principal ingredient in vinegar (Figure $4$) that provides its sour taste. At equilibrium, a solution contains $\ce{[CH3CO2H]} = 0.0787\, M$: and $\ce{H3O^+}]=[\ce{CH3CO2^{-}}]=0.00118\,M$. What is the value of $K_a$ for acetic acid?
Solution
The relevant equilibrium equation and its equilibrium constant expression are shown below. Substitution of the provided equilibrium concentrations permits a straightforward calculation of the Ka for acetic acid.
$\ce{CH3CO2H(aq) + H2O(l) <=> H3O^{+}(aq) + CH3CO2^{-}(aq)} \nonumber$
$K_{ a }=\frac{\left[ \ce{H3O^{+}} \right]\left[ \ce{CH3CO2^{-}} \right]}{\left[ \ce{CH3CO2H} \right]}=\frac{(0.00118)(0.00118)}{0.0787}=1.77 \times 10^{-5} \nonumber$
Exercise $3$
The $\ce{HSO4^{−}}$ ion, weak acid used in some household cleansers:
$\ce{HSO4^{-}(aq) + H2O(l) <=> H3O^{+}(aq) + SO4^{2-}(aq)} \nonumber$
What is the acid ionization constant for this weak acid if an equilibrium mixture has the following composition: $[\ce{H3O^{+}}] = 0.027\, M$; $[\ce{HSO4^{-}}]=0.29\,M$; $[\ce{HSO4^{-}}] =0.29\,M$; and $[\ce{SO4^{2-}}]=0.13\,M$?
Answer
Ka for $\ce{HSO4^{-}} = 1.2 \times 10^{−2}$
Example $4$: Determination of Kb from Equilibrium Concentrations
Caffeine, $\ce{C8H10N4O2}$ is a weak base. What is the value of $K_b$ for caffeine if a solution at equilibrium has $[\ce{C8H10N4O2}] = 0.050~\text{M}$, $[\ce{C8H10N4O2H^{+}}] = 5.0 \times 10^{−3} ~\text{M}$, and $[\ce{OH^{-}}] = 2.5 \times 10^{−3} ~\text{M}$?
Solution
The relevant equilibrium equation and its equilibrium constant expression are shown below. Substitution of the provided equilibrium concentrations permits a straightforward calculation of the $K_b$ for caffeine.
$\ce{C8H10N4O2(aq) + H2O(l)⇌C8H10N4O2H^{+}(aq) + OH^{−}(aq)} \nonumber$
\begin{align*} K_b &=\ce{[C8H10N4O2H^{+}][OH^{-}][C8H10N4O2]} \[4pt] &=(5.0 \times 10^{-3})(2.5 \times 10^{-3}) (0.050) \[4pt]&=2.5 \times 10^{-4} \end{align*} \nonumber
Exercise $4$
What is the equilibrium constant for the ionization of the $\ce{HPO4^{2−}}$ ion, a weak base
$\ce{HPO4^{2-}(aq) + H2O(l) <=> H2PO4^{−}(aq) + OH^{-}(aq)} \nonumber$
if the composition of an equilibrium mixture is as follows: $\ce{[OH^{-}}] = 1.3 \times 10^{−6} M}$; $\ce{H2PO4^{-}}]=0.042\,M$; and $[\ce{HPO4^{2-}}]=0.341\,M$?
Answer
$K_b$ for $\ce{HPO4^{2-}}$ is $1.6 \times 10^{−7}$
Example $5$: Determination of Ka or Kb from pH
The pH of a 0.0516-M solution of nitrous acid, $\ce{HNO2}$, is 2.34. What is its $K_a$?
$\ce{HNO2(aq) + H2O(l) <=> H3O^{+}(aq) + NO2^{−}(aq)} \nonumber$
Solution
The nitrous acid concentration provided is a formal concentration, one that does not account for any chemical equilibria that may be established in solution. Such concentrations are treated as “initial” values for equilibrium calculations using the ICE table approach. Notice the initial value of hydronium ion is listed as approximately zero because a small concentration of H3O+ is present ($1 × \times 10^{−7}~ \text{M}$) due to the autoprotolysis of water. In many cases, such as all the ones presented in this chapter, this concentration is much less than that generated by ionization of the acid (or base) in question and may be neglected.
The pH provided is a logarithmic measure of the hydronium ion concentration resulting from the acid ionization of the nitrous acid, and so it represents an “equilibrium” value for the ICE table:
$\left[ \ce{H3O^{+}} \right]=10^{-2.34}=0.0046 ~\text{M} \nonumber$
The ICE table for this system is then
Finally, calculate the value of the equilibrium constant using the data in the table:
$K_{ a }=\frac{\left[ \ce{H3O^{+}} \right]\left[ \ce{NO2^{-}} \right]}{\left[ \ce{HNO2} \right]}=\frac{(0.0046)(0.0046)}{(0.0470)}=4.6 \times 10^{-4} \nonumber$
Exercise $5$
The pH of a solution of household ammonia, a 0.950-M solution of $\ce{NH3}$, is 11.612. What is $K_b$ for $\ce{NH3}$.
Answer
$K_b = 1.8 \times 10^{−5}$
Example $6$: Calculating Equilibrium Concentrations in a Weak Acid Solution
Formic acid, HCO2H, is one irritant that causes the body’s reaction to some ant bites and stings (Figure $5$).
What is the concentration of hydronium ion and the pH of a 0.534-M solution of formic acid?
$\ce{HCO2H(aq) + H2O(l) <=> H3O^{+}(aq) + HCO2^{-}(aq)} \quad K_{ a }=1.8 \times 10^{-4} \nonumber$
Solution
The ICE table for this system is
Substituting the equilibrium concentration terms into the Ka expression gives
\begin{align*} K_{ a }&=1.8 \times 10^{-4}=\frac{\left[ \ce{H3O^{+}} \right]\left[ \ce{HCO2^{-}} \right]}{\left[ \ce{HCO2H} \right]}\[4pt] &=\frac{(x)(x)}{0.534-x}=1.8 \times 10^{-4} \end{align*} \nonumber
The relatively large initial concentration and small equilibrium constant permits the simplifying assumption that x will be much lesser than 0.534, and so the equation becomes
$K_{ a }=1.8 \times 10^{-4}=\frac{x^2}{0.534} \nonumber$
Solving the equation for $x$ yields
\begin{align*} x^2 &=0.534 \times\left(1.8 \times 10^{-4}\right)=9.6 \times 10^{-5}\[4pt] x&=\sqrt{9.6 \times 10^{-5}}\[4pt] &=9.8 \times 10^{-3} M \end{align*} \nonumber
To check the assumption that x is small compared to 0.534, its relative magnitude can be estimated:
$\frac{x}{0.534}=\frac{9.8 \times 10^{-3}}{0.534}=1.8 \times 10^{-2}(1.8 \% \text { of } 0.534) \nonumber$
Because x is less than 5% of the initial concentration, the assumption is valid.
As defined in the ICE table, x is equal to the equilibrium concentration of hydronium ion:
$x=\left[ \ce{H3O^{+}} \right]=0.0098 ~\text{M} \nonumber$
Finally, the pH is calculated to be
$\text{pH} =-\log \left[ \ce{H3O^{+}} \right]=-\log (0.0098)=2.01 \nonumber$
Exercise $6$
Only a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of a 0.100-M solution of acetic acid, CH3CO2H?
$\ce{CH3CO2H(aq) + H2O(l) <=> H3O^{+}(aq) + CH3CO2^{-}(aq)} \quad K_{ a }=1.8 \times 10^{-5} \nonumber$
Answer
percent ionization = 1.3%
Example $7$: Calculating Equilibrium Concentrations in a Weak Base Solution
Find the concentration of hydroxide ion, the pOH, and the pH of a 0.25-M solution of trimethylamine, a weak base:
$\ce{(CH3)3N(aq) + H2O(l) <=> (CH3)3NH^{+}(aq) + OH^{-}(aq)} \quad K_{ b }=6.3 \times 10^{-5} \nonumber$
Solution
The ICE table for this system is
Substituting the equilibrium concentration terms into the Kb expression gives
$K_{ b }=\frac{\left[ \ce{(CH3)3NH^{+}} \right]\left[ \ce{OH^{-}} \right]}{\left[ \ce{(CH3)3N} \right]}=\frac{(x)(x)}{0.25-x}=6.3 \times 10^{-5} \nonumber$
Assuming $x \ll 0.25$ and solving for $x$ yields
$x=4.0 \times 10^{-3}~\text{M}\nonumber$
This value is less than 5% of the initial concentration (0.25), so the assumption is justified. As defined in the ICE table, $x$ is equal to the equilibrium concentration of hydroxide ion:
\begin{align*} \left[ OH^{-}\right] &=\sim 0+x \[4pt] &=x=4.0 \times 10^{-3} M\[4pt] \end{align*} \nonumber
The pOH is calculated to be
$\text{pOH} =-\log \left(4.0 \times 10^{-3}\right)=2.40 \nonumber$
Using the relation introduced in the previous section of this chapter:
$\text{pH} + \text{pOH} = p K_{ w }=14.00 \nonumber$
permits the computation of pH:
$\text{pH} = 14.00 - \text{pOH} =14.00-2.40=11.60 \nonumber$
Exercise $7$
Calculate the hydroxide ion concentration and the percent ionization of a 0.0325-M solution of ammonia, a weak base with a Kb of $1.76 \times 10^{−5}$.
Answer
$7.56 \times 10^{−4}~ \text{M}$, 2.33%
In some cases, the strength of the weak acid or base and its formal (initial) concentration result in an appreciable ionization. Though the ICE strategy remains effective for these systems, the algebra is a bit more involved because the simplifying assumption that x is negligible cannot be made. Calculations of this sort are demonstrated in Example $8$ below.
Example $8$: Calculating Equilibrium Concentrations without Simplifying Assumptions
Sodium bisulfate, NaHSO4, is used in some household cleansers as a source of the $\ce{HSO4^{−}}$ ion, a weak acid. What is the pH of a 0.50-M solution of $\ce{HSO4^{-}}$?
$\ce{HSO4^{-}(aq) + H2O(l) <=> H3O^{+}(aq) + SO4^{2-}(aq)} \quad K_{ a }=1.2 \times 10^{-2} \nonumber$
Solution
The ICE table for this system is
Substituting the equilibrium concentration terms into the Ka expression gives
$K_{ a }=1.2 \times 10^{-2}=\frac{\left[ \ce{H3O^{+}} \right]\left[ \ce{SO4^{2-}} \right]}{\left[ \ce{HSO4^{-}} \right]}=\frac{(x)(x)}{0.50-x} \nonumber$
If the assumption that $x \ll 0.5$ is made, simplifying and solving the above equation yields
$x=0.077~\text{M} \nonumber$
This value of $x$ is clearly not significantly less than 0.50 M; rather, it is approximately 15% of the initial concentration:
When we check the assumption, we calculate:
$\frac{x}{\left[ \ce{HSO4^{-}} \right]_{ i }} \nonumber$
$\frac{x}{0.50}=\frac{7.7 \times 10^{-2}}{0.50}=0.15(15 \%) \nonumber$
Because the simplifying assumption is not valid for this system, the equilibrium constant expression is solved as follows:
$K_{ a }=1.2 \times 10^{-2}=\frac{(x)(x)}{0.50-x} \nonumber$
Rearranging this equation yields
$6.0 \times 10^{-3}-1.2 \times 10^{-2} x=x^2 \nonumber$
Writing the equation in quadratic form gives
$x^2+1.2 \times 10^{-2} x-6.0 \times 10^{-3}=0 \nonumber$
Solving for the two roots of this quadratic equation results in a negative value that may be discarded as physically irrelevant and a positive value equal to x. As defined in the ICE table, $x$ is equal to the hydronium concentration.
$x=\left[ H3O^{+}\right]=0.072~\text{M} \nonumber$
$\text{pH} =-\log \left[ \ce{H3O^{+}} \right]=-\log (0.072)=1.14 \nonumber$
Exercise $8$
Calculate the pH in a 0.010-M solution of caffeine, a weak base:
$\ce{C8H{10}N4O2(aq) + H2O(l) <=> C8H{10}N4O2H^{+}(aq) + OH^{-}(aq)} \quad K_{ b }=2.5 \times 10^{-4} \nonumber$
Answer
pH = 11.16
Effect of Molecular Structure on Acid-Base Strength
Binary Acids and Bases
In the absence of any leveling effect, the acid strength of binary compounds of hydrogen with nonmetals (A) increases as the H-A bond strength decreases down a group in the periodic table. For group 17, the order of increasing acidity is
$\ce{HF < HCl < HBr < HI}. \nonumber$
Likewise, for group 16, the order of increasing acid strength is
$\ce{H2O < H2S < H2Se < H2Te.} \nonumber$
Across a row in the periodic table, the acid strength of binary hydrogen compounds increases with increasing electronegativity of the nonmetal atom because the polarity of the H-A bond increases. Thus, the order of increasing acidity (for removal of one proton) across the second row is $\ce{CH4 < NH3 < H2O < HF}$; across the third row, it is $\ce{SiH4 < PH3 < H2S < HCl}$ (see Figure $6$).
Ternary Acids and Bases
Ternary compounds composed of hydrogen, oxygen, and some third element (“E”) may be structured as depicted in the image below. In these compounds, the central E atom is bonded to one or more O atoms, and at least one of the O atoms is also bonded to an H atom, corresponding to the general molecular formula OmE(OH)n. These compounds may be acidic, basic, or amphoteric depending on the properties of the central E atom. Examples of such compounds include sulfuric acid, O2S(OH)2, sulfurous acid, OS(OH)2, nitric acid, O2NOH, perchloric acid, O3ClOH, aluminum hydroxide, Al(OH)3, calcium hydroxide, Ca(OH)2, and potassium hydroxide, KOH:
If the central atom, E, has a low electronegativity, its attraction for electrons is low. Little tendency exists for the central atom to form a strong covalent bond with the oxygen atom, and bond a between the element and oxygen is more readily broken than bond b between oxygen and hydrogen. Hence bond a is ionic, hydroxide ions are released to the solution, and the material behaves as a base—this is the case with Ca(OH)2 and KOH. Lower electronegativity is characteristic of the more metallic elements; hence, the metallic elements form ionic hydroxides that are by definition basic compounds.
If, on the other hand, the atom E has a relatively high electronegativity, it strongly attracts the electrons it shares with the oxygen atom, making bond a relatively strongly covalent. The oxygen-hydrogen bond, bond b, is thereby weakened because electrons are displaced toward E. Bond b is polar and readily releases hydrogen ions to the solution, so the material behaves as an acid. High electronegativities are characteristic of the more nonmetallic elements. Thus, nonmetallic elements form covalent compounds containing acidic −OH groups that are called oxyacids.
Increasing the oxidation number of the central atom E also increases the acidity of an oxyacid because this increases the attraction of E for the electrons it shares with oxygen and thereby weakens the O-H bond. Sulfuric acid, H2SO4, or O2S(OH)2 (with a sulfur oxidation number of +6), is more acidic than sulfurous acid, H2SO3, or OS(OH)2 (with a sulfur oxidation number of +4). Likewise nitric acid, HNO3, or O2NOH (N oxidation number = +5), is more acidic than nitrous acid, HNO2, or ONOH (N oxidation number = +3). In each of these pairs, the oxidation number of the central atom is larger for the stronger acid (Figure $7$).
Hydroxy compounds of elements with intermediate electronegativities and relatively high oxidation numbers (for example, elements near the diagonal line separating the metals from the nonmetals in the periodic table) are usually amphoteric. This means that the hydroxy compounds act as acids when they react with strong bases and as bases when they react with strong acids. The amphoterism of aluminum hydroxide, which commonly exists as the hydrate $\ce{Al(H2O)3(OH)3}$, is reflected in its solubility in both strong acids and strong bases. In strong bases, the relatively insoluble hydrated aluminum hydroxide, $\ce{Al(H2O)3(OH)3}$, is converted into the soluble ion, $\ce{[Al(H2O)2(OH)4]^{-}}$, by reaction with hydroxide ion:
$\ce{Al(H_2O)_3(OH)3(aq) + OH^{-}(aq) <=> H2O(l) + [Al(H2O)2(OH)4]^{-}(aq)} \nonumber$
In this reaction, a proton is transferred from one of the aluminum-bound H2O molecules to a hydroxide ion in solution. The $\ce{Al(H2O)3(OH)3}$ compound thus acts as an acid under these conditions. On the other hand, when dissolved in strong acids, it is converted to the soluble ion $\ce{[Al(H2O)6]^{3+}}$ by reaction with hydronium ion:
$\ce{3 H3O^{+}(aq) + Al(H2O)3(OH)3(aq) <=> Al(H2O)6^{3+}(aq) + 3 H2O(l)} \nonumber$
In this case, protons are transferred from hydronium ions in solution to $\ce{Al(H2O)3(OH)3}$, and the compound functions as a base. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/14%3A_Acid-Base_Equilibria/14.03%3A_Relative_Strengths_of_Acids_and_Bases.txt |
Learning Objectives
By the end of this section, you will be able to:
• Predict whether a salt solution will be acidic, basic, or neutral
• Calculate the concentrations of the various species in a salt solution
• Describe the acid ionization of hydrated metal ions
Salts with Acidic Ions
Salts are ionic compounds composed of cations and anions, either of which may be capable of undergoing an acid or base ionization reaction with water. Aqueous salt solutions, therefore, may be acidic, basic, or neutral, depending on the relative acid-base strengths of the salt's constituent ions. For example, dissolving ammonium chloride in water results in its dissociation, as described by the equation
$\ce{NH4Cl(s) <=> NH4^{+}(aq) + Cl^{-}(aq)} \nonumber$
The ammonium ion is the conjugate acid of the base ammonia, NH3; its acid ionization (or acid hydrolysis) reaction is represented by
$\ce{NH4^{+}(aq) + H2O(l) <=> H3O^{+}(aq) + NH3(aq)} \quad K_{ a }=K_{ w } / K_{ b } \nonumber$
Since ammonia is a weak base, Kb is measurable and Ka > 0 (ammonium ion is a weak acid).
The chloride ion is the conjugate base of hydrochloric acid, and so its base ionization (or base hydrolysis) reaction is represented by
$\ce{Cl^{-}(aq) + H2O(l) <=> HCl(aq) + OH^{-}(aq)} \quad K_{ b }=K_{ w } / K_{ a } \nonumber$
Since $\ce{HCl}$ is a strong acid, Ka is immeasurably large and Kb ≈ 0 (chloride ions don’t undergo appreciable hydrolysis).
Thus, dissolving ammonium chloride in water yields a solution of weak acid cations () and inert anions (Cl), resulting in an acidic solution.
Example $1$: Calculating the pH of an Acidic Salt Solution
Aniline is an amine that is used to manufacture dyes. It is isolated as anilinium chloride, a salt prepared by the reaction of the weak base aniline and hydrochloric acid. What is the pH of a 0.233 M solution of anilinium chloride
$\ce{C6H5NH3^{+}(aq) + H2O(l) <=> H3O^{+}(aq) + C6H5NH2(aq)} \nonumber$
Solution
The Ka for anilinium ion is derived from the Kb for its conjugate base, aniline (see Appendix H):
$K_{ a }=\frac{K_{ w }}{K_{ b }}=\frac{1.0 \times 10^{-14}}{4.3 \times 10^{-10}}=2.3 \times 10^{-5} \nonumber$
Using the provided information, an ICE table for this system is prepared:
Substituting these equilibrium concentration terms into the Ka expression gives
\begin{align*} K_a & =\dfrac{\left[ \ce{C6H5NH2} \right]\left[ \ce{H3O^{+}} \right] }{\left[ \ce{C6H5NH3^{+}}\right]} \[4pt] 2.3 \times 10^{-5} & = \dfrac{(x)(x)}{0.233-x} \end{align*} \nonumber
Assuming $x \ll 0.233$, the equation is simplified and solved for $x$:
\begin{align*} & 2.3 \times 10^{-5}= \dfrac{x^2}{0.233} \[4pt] & x=0.0023 ~\text{M} \end{align*} \nonumber
The ICE table defines $x$ as the hydronium ion molarity, and so the pH is computed as
$\text{pH} =-\log \left[ \ce{H3O^{+}} \right]=-\log (0.0023)=2.64 \nonumber$
Exercise $1$
What is the hydronium ion concentration in a 0.100-M solution of ammonium nitrate, $\ce{NH4NO3}$, a salt composed of the ions $NH_4^{+}$ and $NO_3^{-}$ and.
Which is the stronger acid $\ce{C6H5NH3+}$ or $\ce{NH4+}$?
Answer
$[\ce{H3O^{+}}] = 7.5 \times 10^{−6} ~\text{M}$; is the stronger acid.
Salts with Basic Ions
As another example, consider dissolving sodium acetate in water:
$\ce{NaCH_3CO_2(s) <=> Na^{+}(aq) + CH_3CO_2^{-}(aq)} \nonumber$
The sodium ion does not undergo appreciable acid or base ionization and has no effect on the solution pH. This may seem obvious from the ion's formula, which indicates no hydrogen or oxygen atoms, but some dissolved metal ions function as weak acids, as addressed later in this section.
The acetate ion, $\ce{CH3CO2^{-}(aq)}$, is the conjugate base of acetic acid, $\ce{CH3CO2H}$, and so its base ionization (or base hydrolysis) reaction is represented by
$\ce{CH3CO2^{-}(aq) + H2O(l) <=> CH_3 CO_2 H (aq) + OH -(aq)} \quad K_{ b }=K_{ w } / K_{ a } \nonumber$
Because acetic acid is a weak acid, its Ka is measurable and Kb > 0 (acetate ion is a weak base).
Dissolving sodium acetate in water yields a solution of inert cations (Na+) and weak base anions $( \ce{CH3CO2^{-}})$, resulting in a basic solution.
Example $2$: Equilibrium in a Solution of a Salt of a Weak Acid and a Strong Base
Determine the acetic acid concentration in a solution with and $[\ce{OH^{-}}] = 0.050~\text{M}$ and $[\ce{OH^{-}}] = 2.5 \times 10^{−6} M$ at equilibrium. The reaction is:
$\ce{CH_3CO_2^{-}(aq) + H2O(l) <=> CH_3CO_2H(aq) + OH^{-}(aq)} \nonumber$
Solution
The provided equilibrium concentrations and a value for the equilibrium constant will permit calculation of the missing equilibrium concentration. The process in question is the base ionization of acetate ion, for which
$\ce{CH_3CO_2^{-}(aq) + H2O(l) <=> CH_3CO_2H(aq) + OH^{-}(aq)} \nonumber$
Substituting the available values into the Kb expression gives
\begin{align*} K_{ b } &=\frac{\left[ CH_3 CO_2 H \right]\left[ \ce{OH^{-}} \right]}{\left[ \ce{CH3CO2^{-}} \right]}=5.6 \times 10^{-10} \[4pt] &=\frac{\left[ \ce{CH3CO2H} \right]\left(2.5 \times 10^{-6}\right)}{(0.050)}=5.6 \times 10^{-10} \end{align*} \nonumber
Solving the above equation for the acetic acid molarity yields $[\ce{CH3CO2H}] = 1.1 \times 10^{−5}~ \text{M}$.
Exercise $2$
What is the pH of a 0.083-M solution of $\ce{NaCN}$?
Answer
11.11
Salts with Acidic and Basic Ions
Some salts are composed of both acidic and basic ions, and so the pH of their solutions will depend on the relative strengths of these two species. Likewise, some salts contain a single ion that is amphiprotic, and so the relative strengths of this ion’s acid and base character will determine its effect on solution pH. For both types of salts, a comparison of the Ka and Kb values allows prediction of the solution’s acid-base status, as illustrated in the following example exercise.
Example $3$: Determining the Acidic or Basic Nature of Salts
Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
1. KBr
2. NaHCO3
3. Na2HPO4
4. NH4F
Solution
Consider each of the ions separately in terms of its effect on the pH of the solution, as shown here:
(a) The $\ce{K^{+}}$ cation is inert and will not affect pH. The bromide ion is the conjugate base of a strong acid, and so it is of negligible base strength (no appreciable base ionization). The solution is neutral.
(b) The $\ce{Na^{+}}$ cation is inert and will not affect the pH of the solution; while the $\ce{HCO3^{−}}$ anion is amphiprotic. The $K_a$ of $\ce{HCO3^{-}}$ is $4.7 \times 10^{−11}$, and its $K_b$ is:
$\frac{1.0 \times 10^{-14}}{4.3 \times 10^{-7}}=2.3 \times 10^{-8} \nonumber$
Since $K_b \gg K_a$, the solution is basic.
(c) The $\ce{Na^{+}}$ cation is inert and will not affect the pH of the solution, while the $\ce{HPO4^{2−}}$ anion is amphiprotic. The $K_a$ of $\ce{HPO4^{2−}}$ is $4.2 \times 10^{−13}$, and its $K_b$ is:
$\frac{1.0 \times 10^{-14}}{6.2 \times 10^{-8}}=1.6 \times 10^{-7} \nonumber$
Because $K_b \gg K_a$, the solution is basic.
(d) The $\ce{NH4^{+}}$ ion is acidic (see above discussion) and the $\ce{F^{−}}$ ion is basic (conjugate base of the weak acid $\ce{HF}$). Comparing the two ionization constants: $K_a$ of $\ce{NH4^{+}}$ is 5.6 × 10−10 and the $K_b$ of $\ce{F^{−}}$ is $1.6 \times 10^{−11}$, so the solution is acidic, since $K_a > K_b$.
Exercise $3$
Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
1. K2CO3
2. CaCl2
3. KH2PO4
4. (NH4)2CO3
Answer
(a) basic; (b) neutral; (c) acidic; (d) basic
The Ionization of Hydrated Metal Ions
Unlike the group 1 and 2 metal ions of the preceding examples (Na+, Ca2+, etc.), some metal ions function as acids in aqueous solutions. These ions are not just loosely solvated by water molecules when dissolved, instead they are covalently bonded to a fixed number of water molecules to yield a complex ion (see chapter on coordination chemistry). As an example, the dissolution of aluminum nitrate in water is typically represented as
$\ce{Al(NO3)3(s) <=> Al^{3+}(aq) +3 NO3^{-}(aq)} \nonumber$
However, the aluminum(III) ion actually reacts with six water molecules to form a stable complex ion, and so the more explicit representation of the dissolution process is
$\ce{Al(NO3)3(s) +6 H2O(l) <=> Al(H2O)6^{3+}(aq) +3 NO_3^{-}(aq)} \nonumber$
As shown in Figure $1$, the ions involve bonds between a central Al atom and the O atoms of the six water molecules. Consequently, the bonded water molecules' O–H bonds are more polar than in nonbonded water molecules, making the bonded molecules more prone to donation of a hydrogen ion:
$\ce{Al(H2O)6^{3+}(aq) + H2O(l) <=> H3O^{+}(aq) + Al(H2O)5(OH)^{2+}(aq)} \quad \quad K_{ a }=1.4 \times 10^{-5} \nonumber$
The conjugate base produced by this process contains five other bonded water molecules capable of acting as acids, and so the sequential or step-wise transfer of protons is possible as depicted in few equations below:
\begin{align*} \ce{Al(H2O)_6^{3+}(aq) + H2O(l) &<=> H3O^{+}(aq) + Al(H2O)_5(OH)^{2+}(aq)} \[4pt] \ce{Al(H2O)_5(OH)^{2+}(aq) + H2O(l) &<=> H3O^{+}(aq) + Al(H2O)_4(OH)_2^{+}(aq)} \[4pt] \ce{Al(H2O)_4(OH)_2^{+}(aq) + H2O(l) &<=> H3O^{+}(aq) + Al(H2O)_3(OH)_3(aq)} \end{align*} \nonumber
This is an example of a polyprotic acid, the topic of discussion in a later section of this chapter.
Aside from the alkali metals (group 1) and some alkaline earth metals (group 2), most other metal ions will undergo acid ionization to some extent when dissolved in water. The acid strength of these complex ions typically increases with increasing charge and decreasing size of the metal ions. The first-step acid ionization equations for a few other acidic metal ions are shown below:
\begin{align*} \ce{Fe(H2O)_6^{3+}(aq) + H2O(l) &<=> H3O^{+}(aq) + Fe(H2O)_5(OH)^{2+}(aq)} && p K_{ a }=2.74 \[4pt] \ce{Cu(H2O)_6^{2+}(aq) + H2O(l) &<=> H3O^{+}(aq) + Cu(H2O)_5(OH)^{+}(aq)} && p K_{ a }=\sim 6.3 \[4pt] \ce{Zn(H2O)_4^{2+}(aq) + H2O(l) &<=> H3O^{+}(aq) + Zn(H2O)_3(OH)^{+}(aq)} && p K_{ a }=9.6 \end{align*} \nonumber
Example $4$: Hydrolysis of [Al(H2O)6]3+
Calculate the pH of a 0.10-M solution of aluminum chloride, which dissolves completely to give the hydrated aluminum ion ($\ce{[Al(H2O)6]^{3+}}$) in solution.
Solution
The equation for the reaction and Ka are:
$\ce{Al(H2O)6^{3+}(aq) + H2O(l) <=> H3O^{+}(aq) + Al(H2O)5(OH)^{2+}(aq)} \quad K_{ a }=1.4 \times 10^{-5} \nonumber$
An ICE table with the provided information is
Substituting the expressions for the equilibrium concentrations into the equation for the ionization constant yields:
\begin{align*} K_{ a } &=\frac{\left[ \ce{H3O^{+}} \right]\left[ \ce{Al(H2O)5(OH)^{2+}} \right]}{\left[ \ce{Al(H2O)6^{3+}} \right]}\[4pt] &=\frac{(x)(x)}{0.10-x}=1.4 \times 10^{-5} \end{align*} \nonumber
Assuming $x \ll 0.10$ and solving the simplified equation gives:
$x=1.2 \times 10^{-3} ~\text{M} \nonumber$
The ICE table defined $x$ as equal to the hydronium ion concentration, and so the pH is calculated to be
$[\ce{H3O^{+}}]=0+x=1.2 \times 10^{-3}~\text{M} \nonumber$
$\text{pH} =-\log \left[ \ce{H3O^{+}} \right]=2.92 \nonumber$
Therefore this is an acidic solution.
Exercise $4$
What is $[Al(H2O)5(OH)^{2+}}]$ in a 0.15-M solution of $\ce{Al(NO3)3}$ that contains enough of the strong acid $\ce{HNO3}$ to bring [$\ce{H3O^{+}}$] to 0.10 M?
Answer
$2.1 \times 10^{−5} M$ | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/14%3A_Acid-Base_Equilibria/14.04%3A_Hydrolysis_of_Salt_Solutions.txt |
Learning Objectives
By the end of this section, you will be able to:
• Extend previously introduced equilibrium concepts to acids and bases that may donate or accept more than one proton
Acids are classified 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:
\begin{align*} \ce{HCl (aq) + H_2 O (l)} & \ce{\longrightarrow H3O^{+}(aq) + Cl^{-}(aq) } \[4pt] \ce{HNO_3(aq) + H_2 O (l)} & \ce{\longrightarrow H3O^{+}(aq) + NO_3^{-}(aq) } \[4pt] \ce{HCN (aq) + H_2 O (l)} & \ce{\rightleftharpoons H3O^{+}(aq) + CN^{-}(aq) } \end{align*} \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 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:
$\tag{ First ionization} \ce{H_2 SO_4 (aq) + H2O (l) \rightleftharpoons H_3O^{+}(aq) + HSO_4^{-}(aq)}$
with $K_{ a 1} > 10^2$.
$\tag{ Second ionization} \ce{HSO_4^{-}(aq) + H_2 O (l) \rightleftharpoons H3O^{+}(aq) + SO_4^{2-}(aq) }$
with $K_{ a 2}=1.2 \times 10^{-2}$.
This stepwise ionization process occurs for all polyprotic 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.
$\tag{First ionization} \ce{H_2 CO_3(aq) + H_2 O (l) \rightleftharpoons H_3O^{+}(aq) + HCO_3^{-}(aq) }$
with
$K_{ H_2 CO_3}=\frac{\left[\ce{H3O^{+}} \right]\left[ \ce{HCO_3^{-}} \right]}{\left[ \ce{H_2CO_3} \right]}=4.3 \times 10^{-7} \nonumber$
The bicarbonate ion can also act as an acid. It ionizes and forms hydronium ions and carbonate ions in even smaller quantities.
$\tag{Second ionization} \ce{ HCO_3^{-}(aq) + H_2 O (l) \rightleftharpoons H3O^{+}(aq) + CO_3^{2-}(aq) }$
with
$K_{\ce{HCO_3^{-}}}=\frac{\left[ \ce{H_3O^{+}}\right]\left[ \ce{CO_3^{2-}}\right]}{\left[ \ce{HCO_3^{-}}\right]}=4.7 \times 10^{-11} \nonumber$
$K_{\ce{H_2CO_3}}$ is larger than $K_{\ce{HCO_3^{-}}}$ by a factor of $10^{4}$, so $\ce{H2CO3}$ is the dominant producer of hydronium ion in the solution. This means that little of the formed by the ionization of $\ce{H2CO3}$ ionizes to give hydronium ions (and carbonate ions), and the concentrations of $\ce{H3O^{+}}$ and are practically equal in a pure aqueous solution of $\ce{H2CO3}$.
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 approach is demonstrated in the following example and exercise.
Example $1$: Ionization of a Diprotic Acid
“Carbonated water” contains a palatable amount of dissolved carbon dioxide. The solution is acidic because $\ce{CO2}$ reacts with water to form carbonic acid, $\ce{H2CO3}$. What are and in a saturated solution of $\ce{CO2}$ with an initial $\ce{[H2CO3] = 0.033 \,M}$?
$\ce{H_2 CO_3(aq) + H_2 O (l) \rightleftharpoons H3O^{+}(aq) + HCO_3^{-}(aq)} \quad K_{ a 1}=4.3 \times 10^{-7} \nonumber$
$\ce{HCO_3^{-}(aq) + H_2 O (l) \rightleftharpoons H3O^{+}(aq) + CO_3^{2-}(aq)} \quad K_{ a 2}=4.7 \times 10^{-11} \nonumber$
Solution
As indicated by the ionization constants, $\ce{H2CO3}$ is a much stronger acid than so the stepwise ionization reactions may be treated separately.
The first ionization reaction is
$\ce{H_2 CO_3(aq) + H_2 O (l) \rightleftharpoons H3O^{+}(aq) + HCO_3^{-}(aq)} \quad K_{ a 1}=4.3 \times 10^{-7} \nonumber$
Using provided information, an ICE table for this first step is prepared:
Substituting the equilibrium concentrations into the equilibrium equation gives
$K_{ H_2 CO_3}=\frac{\left[ H3O^{+}\right]\left[ HCO_3^{-}\right]}{\left[ H_2 CO_3\right]}=\frac{(x)(x)}{0.033-x}=4.3 \times 10^{-7} \nonumber$
Assuming $x \ll 0.033$ and solving the simplified equation yields
$x=1.2 \times 10^{-4} \nonumber$
The ICE table defined x as equal to the bicarbonate ion molarity and the hydronium ion molarity:
$\left[ H_2 CO_3\right]=0.033 M \nonumber$
$\left[ H3O^{+}\right]=\left[ HCO_3^{-}\right]=1.2 \times 10^{-4} M \nonumber$
Using the bicarbonate ion concentration computed above, the second ionization is subjected to a similar equilibrium calculation:
$\ce{HCO_3^{-}(aq) + H_2 O (l) \rightleftharpoons H3O^{+}(aq) + CO_3^{2-}(aq)} \nonumber$
$K_{ HCO_{3^{-}}}=\frac{\left[ H3O^{+}\right]\left[ CO_3^{2-}\right]}{\left[ HCO_3^{-}\right]}=\frac{\left(1.2 \times 10^{-4}\right)\left[ CO_3^{2-}\right]}{1.2 \times 10^{-4}} \nonumber$
$\left[ CO_3^{2-}\right]=\frac{\left(4.7 \times 10^{-11}\right)\left(1.2 \times 10^{-4}\right)}{1.2 \times 10^{-4}}=4.7 \times 10^{-11} M \nonumber$
To summarize: at equilibrium $\ce{[H2CO3]} = 0.033 M$; $\ce{[H3O^{+}]} = 1.2 \times 10^{−4}$; $\ce{[HCO3^{-}]}=1.2 \times 10^{−4}\,M$;
$\ce{[CO32^{−}]}=4.7 \times 10^{−11}\,M$.
Exercise $1$
The concentration of $\ce{H2S}$ 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:
$\ce{H_2 S (aq) + H_2 O (l) \rightleftharpoons H3O^{+}(aq) + HS^{-}(aq)} \quad K_{ a 1}=8.9 \times 10^{-8} \nonumber$
$\ce{HS^{-}(aq) + H_2 O (l) \rightleftharpoons H3O^{+}(aq) + S^{2-}(aq)} \quad K_{ a 2}=1.0 \times 10^{-19} \nonumber$
Answer
$[\ce{H2S}] = 0.1\, M$
$[\ce{H3O^{+}}] = [\ce{HS^{-}}] = 0.000094\, M$
$[\ce{S^{2-}}] = 1 \times 10^{−19}\, M$
A triprotic acid is an acid that has three ionizable H atoms. Phosphoric acid is one example:
$\tag{First ionization} \ce{H3PO_4(aq) + H2O (l) \rightleftharpoons H3O^{+}(aq) + H_2 PO_4^{-}(aq)} \quad K_{ a 1}=7.5 \times 10^{-3}$
$\tag{Second ionization} \ce{H2PO4^{-}(aq) + H2O (l) \rightleftharpoons H3O^{+}(aq) + HPO_4^{2-}(aq)} \quad K_{ a 2}=6.2 \times 10^{-8}$
$\tag{Third ionization} \ce{HPO4^{2-}(aq) + H2O (l) \rightleftharpoons H3O^{+}(aq) + PO_4^{3-}(aq)} \quad K_{ a 3}=4.2 \times 10^{-13}$
As for the diprotic acid examples, each successive ionization reaction is less extensive than the former, reflected in decreasing values for the stepwise acid ionization constants. This is a general characteristic of polyprotic acids and successive ionization constants often differ by a factor of about $10^{5}$ to $10^{6}$.
This set of three dissociation reactions may appear to make calculations of equilibrium concentrations in a solution of $\ce{H3PO4}$ complicated. However, because the successive ionization constants differ by a factor of $10^{5}$ to $10^{6}$, large differences exist in the small changes in concentration accompanying the ionization reactions. This allows the use of math-simplifying assumptions and processes, as demonstrated in the examples above.
Polyprotic bases are capable of accepting more than one hydrogen ion. The carbonate ion is an example of a diprotic base, because it can accept two protons, as shown below. Similar to the case for polyprotic acids, note the ionization constants decrease with ionization step. Likewise, equilibrium calculations involving polyprotic bases follow the same approaches as those for polyprotic acids.
$\ce{H_2 O (l) + CO_3^{2-}(aq) \rightleftharpoons HCO_3^{-}(aq) + OH^{-}(aq)} \quad K_{ b 1}=2.1 \times 10^{-4} \nonumber$
$\ce{H_2 O (l) + HCO_3^{-}(aq) \rightleftharpoons H_2 CO_3(aq) + OH^{-}(aq)} \quad K_{ b 2}=2.3 \times 10^{-8} \nonumber$ | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/14%3A_Acid-Base_Equilibria/14.05%3A_Polyprotic_Acids.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the composition and function of acid–base buffers
• Calculate the pH of a buffer before and after the addition of added acid or base
A solution containing appreciable amounts of a weak conjugate acid-base pair is called a buffer solution, or a buffer. Buffer solutions resist a change in pH when small amounts of a strong acid or a strong base are added (Figure $1$). A solution of acetic acid and sodium acetate (CH3COOH + CH3COONa) is an example of a buffer that consists of a weak acid and its salt. An example of a buffer that consists of a weak base and its salt is a solution of ammonia and ammonium chloride (NH3(aq) + NH4Cl(aq)).
How Buffers Work
To illustrate the function of a buffer solution, consider a mixture of roughly equal amounts of acetic acid and sodium acetate. The presence of a weak conjugate acid-base pair in the solution imparts the ability to neutralize modest amounts of added strong acid or base. For example, strong base added to this solution will neutralize hydronium ion, causing the acetic acid ionization equilibrium to shift to the right and generate additional amounts of the weak conjugate base (acetate ion):
$\ce{CH_3 CO_2 H (aq) + H2O(l) <=> H3O^{+}(aq) + CH_3 CO_2^{-}(aq)} \nonumber$
Likewise, strong acid added to this buffer solution will shift the above ionization equilibrium left, producing additional amounts of the weak conjugate acid (acetic acid). Figure $2$ provides a graphical illustration of the changes in conjugate-partner concentration that occur in this buffer solution when strong acid and base are added. The buffering action of the solution is essentially a result of the added strong acid and base being converted to the weak acid and base that make up the buffer's conjugate pair. The weaker acid and base undergo only slight ionization, as compared with the complete ionization of the strong acid and base, and the solution pH, therefore, changes much less drastically than it would in an unbuffered solution.
Example 14.20: pH Changes in Buffered and Unbuffered Solutions
Acetate buffers are used in biochemical studies of enzymes and other chemical components of cells to prevent pH changes that might affect the biochemical activity of these compounds.
1. Calculate the pH of an acetate buffer that is a mixture with 0.10 M acetic acid and 0.10 M sodium acetate.
2. Calculate the pH after 1.0 mL of 0.10 NaOH is added to 100 mL of this buffer.
3. For comparison, calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of a solution of an unbuffered solution with a pH of 4.74.
Solution
(a) Following the ICE approach to this equilibrium calculation yields the following:
Substituting the equilibrium concentration terms into the Ka expression, assuming $x \ll 0.10$, and solving the simplified equation for $x$ yields
$x=1.8 \times 10^{-5}~\text{M} \nonumber$
$\left[ \ce{H3O^{+}} \right]=0+x=1.8 \times 10^{-5}~\text{M} \nonumber$
\begin{align*} pH &=-\log \left[ \ce{H3O^{+}} \right] \[4pt] &=-\log \left(1.8 \times 10^{-5}\right) \[4pt] &= 4.74 \end{align*} \nonumber
(b) Calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of this buffer.
Adding strong base will neutralize some of the acetic acid, yielding the conjugate base acetate ion. Compute the new concentrations of these two buffer components, then repeat the equilibrium calculation of part (a) using these new concentrations.
$0.0010~ \cancel{\text{L}} \times \left(\dfrac{0.10 ~\text{mol} ~\ce{NaOH}}{1 ~ \text{L}}\right)=1.0 \times 10^{-4}~\text{mol} ~ \ce{NaOH} \nonumber$
The initial molar amount of acetic acid is
$0.100 ~ \cancel{\text{L}} \times\left(\frac{0.100~ \text{mol} ~ \ce{CH3CO2H} }{1 ~ \cancel{\text{L}} }\right)=1.00 \times 10^{-2} ~\text{mol} ~ \ce{CH3CO2H} \nonumber$
The amount of acetic acid remaining after some is neutralized by the added base is
$\left(1.0 \times 10^{-2}\right)-\left(0.01 \times 10^{-2}\right)=0.99 \times 10^{-2} ~\text{mol} ~ \ce{CH3CO2H} \nonumber$
The newly formed acetate ion, along with the initially present acetate, gives a final acetate concentration of
$\left(1.0 \times 10^{-2}\right) + \left(0.01 \times 10^{-2}\right)=1.01 \times 10^{-2}~ \text{mol} ~ \ce{NaCH3CO2} \nonumber$
Compute molar concentrations for the two buffer components:
\begin{align*} \left[ \ce{CH3CO2H} \right] & =\frac{9.9 \times 10^{-3} ~\text{mol} }{0.101~\text{L} }=0.098 ~\text{M} \[4pt] \left[ \ce{NaCH3CO2} \right] & =\frac{1.01 \times 10^{-2} ~\text{mol} }{0.101 ~\text{L}}=0.100~\text{M} \end{align*} \nonumber
Using these concentrations, the pH of the solution may be computed as in part (a) above, yielding pH = 4.75 (only slightly different from that prior to adding the strong base).
(c) For comparison, calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of a solution of an unbuffered solution with a pH of 4.74.
The amount of hydronium ion initially present in the solution is
\begin{align*} \left[ \ce{H3O^{+}} \right] &=10^{-4.74}=1.8 \times 10^{-5} ~ \text{M} \[4pt] \text{mol} ~\ce{H3O^{+}} & =(0.100~\text{L})\left(1.8 \times 10^{-5}~\text{M} \right)=1.8 \times 10^{-6}~ \text{mol}~\ce{H3O^{+}} \end{align*} \nonumber
The amount of hydroxide ion added to the solution is
$mol OH^{-}=(0.0010 L)(0.10 M)=1.0 \times 10^{-4} mol \ce{OH^{-}} \nonumber$
The added hydroxide will neutralize hydronium ion via the reaction
$\ce{H3O^{+}(aq) + OH^{-}(aq) \leftrightharpoons 2 H2O(l)} \nonumber$
The 1:1 stoichiometry of this reaction shows that an excess of hydroxide has been added (greater molar amount than the initially present hydronium ion).
The amount of hydroxide ion remaining is
$1.0 \times 10^{-4} mol -1.8 \times 10^{-6} mol =9.8 \times 10^{-5} mol OH^{-} \nonumber$
corresponding to a hydroxide molarity of
$\dfrac{9.8 \times 10^{-5}~\text{mol}~\ce{OH^{-}}}{0.101~\text{L}}=9.7 \times 10^{-4} M \nonumber$
The pH of the solution is then calculated to be
$pH =14.00- pOH =14.00 - -\log \left(9.7 \times 10^{-4}\right)=10.99 \nonumber$
In this unbuffered solution, addition of the base results in a significant rise in pH (from 4.74 to 10.99) compared with the very slight increase observed for the buffer solution in part (b) (from 4.74 to 4.75).
Exercise $1$
Show that adding 1.0 mL of $0.10~\text{M}~\ce{HCl}$ changes the pH of 100 mL of a $1.8 \times 10^{−5} ~\text{M}~\ce{HCl}$ solution from 4.74 to 3.00.
Answer
Initial pH of $1.8 \times 10^{−5} ~\text{M}~\ce{HCl}$;
$\text{pH} = −\log[\ce{H3O^{+}}] = −\log[1.8 \times 10^{−5}] = 4.74 \nonumber$
Moles of H3O+ in 100 mL $1.8 \times 10^{−5} ~\text{M}~\ce{HCl}$; $(1.8 \times 10^{−5}~\text{moles/L})\times (0.100 ~\text{L} = 1.8 \times 10^{−6}~\text{mol}\nonumber$
Moles of H3O+ added by addition of 1.0 mL of 0.10 M HCl: $(0.10~\text{moles/L}) \times (0.0010~ \text{L}) = 1.0 \times 10^{−4}~\text{mol} \nonumber$
Final pH after addition of 1.0 mL of 0.10 M HCl:
$pH =-\log \left[ \ce{H3O^{+}} \right]=-\log \left(\frac{\text { total moles } \ce{H3O^{+}}}{\text {total volume }}\right)=-\log \left(\frac{1.0 \times 10^{-4} mol +1.8 \times 10^{-6}~\text{mol}}{101~\text{mL} \left(\frac{1~\text{L} }{1000~\text{mL}}\right)}\right) = 3.00 \nonumber$
Buffer Capacity
Buffer solutions do not have an unlimited capacity to keep the pH relatively constant (Figure $3$). Instead, the ability of a buffer solution to resist changes in pH relies on the presence of appreciable amounts of its conjugate weak acid-base pair. When enough strong acid or base is added to substantially lower the concentration of either member of the buffer pair, the buffering action within the solution is compromised.
The buffer capacity is the amount of acid or base that can be added to a given volume of a buffer solution before the pH changes significantly, usually by one unit. Buffer capacity depends on the amounts of the weak acid and its conjugate base that are in a buffer mixture. For example, 1 L of a solution that is 1.0 M in acetic acid and 1.0 M in sodium acetate has a greater buffer capacity than 1 L of a solution that is 0.10 M in acetic acid and 0.10 M in sodium acetate even though both solutions have the same pH. The first solution has more buffer capacity because it contains more acetic acid and acetate ion.
Selection of Suitable Buffer Mixtures
There are two useful rules of thumb for selecting buffer mixtures:
1. A good buffer mixture should have about equal concentrations of both of its components. A buffer solution has generally lost its usefulness when one component of the buffer pair is less than about 10% of the other. Figure $4$ shows how pH changes for an acetic acid-acetate ion buffer as base is added. The initial pH is 4.74. A change of 1 pH unit occurs when the acetic acid concentration is reduced to 11% of the acetate ion concentration.
2. Weak acids and their salts are better as buffers for pHs less than 7; weak bases and their salts are better as buffers for pHs greater than 7.
Blood is an important example of a buffered solution, with the principal acid and ion responsible for the buffering action being carbonic acid, $\ce{H2CO3}$, and the bicarbonate ion, $\ce{HCO3^{-}}$. When a hydronium ion is introduced to the blood stream, it is removed primarily by the reaction:
$\ce{H3O^{+}(aq) + HCO_3^{-}(aq) \longrightarrow H_2 CO_3(aq) + H2O(l)} \nonumber$
An added hydroxide ion is removed by the reaction:
$\ce{ OH^{-}(aq) + H_2CO_3(aq) \longrightarrow HCO_3^{-}(aq) + H2O(l)} \nonumber$
The added strong acid or base is thus effectively converted to the much weaker acid or base of the buffer pair (H3O+ is converted to H2CO3 and OH- is converted to HCO3-). The pH of human blood thus remains very near the value determined by the buffer pairs pKa, in this case, 7.35. Normal variations in blood pH are usually less than 0.1, and pH changes of 0.4 or greater are likely to be fatal.
The Henderson-Hasselbalch Equation
The ionization-constant expression for a solution of a weak acid can be written as:
$K_{ a }=\frac{\left[ \ce{H3O^{+}} \right]\left[ \ce{A^{-}} \right]}{[ \ce{HA} ]} \nonumber$
Rearranging to solve for [H3O+] yields:
$\left[ \ce{H3O^{+}} \right]=K_{ a } \times \frac{[ \ce{HA} ]}{\left[ \ce{A^{-}} \right]} \nonumber$
Taking the negative logarithm of both sides of this equation gives
$-\log \left[ \ce{H3O^{+}} \right]=-\log K_{ a }-\log \frac{[ \ce{HA} ]}{\left[ \ce{A^{-}} \right]} \nonumber$
which can be written as
$pH = p K_{ a }+\log \frac{\left[ \ce{A^{-}} \right]}{[ \ce{HA} ]} \nonumber$
where pKa is the negative of the logarithm of the ionization constant of the weak acid (pKa = −log Ka). This equation relates the pH, the ionization constant of a weak acid, and the concentrations of the weak conjugate acid-base pair in a buffered solution. Scientists often use this expression, called the Henderson-Hasselbalch equation, to calculate the pH of buffer solutions. It is important to note that the “x is small” assumption must be valid to use this equation.
Portrait of a Chemist: Lawrence Joseph Henderson and Karl Albert Hasselbalch
Lawrence Joseph Henderson (1878–1942) was an American physician, biochemist and physiologist, to name only a few of his many pursuits. He obtained a medical degree from Harvard and then spent 2 years studying in Strasbourg, then a part of Germany, before returning to take a lecturer position at Harvard. He eventually became a professor at Harvard and worked there his entire life. He discovered that the acid-base balance in human blood is regulated by a buffer system formed by the dissolved carbon dioxide in blood. He wrote an equation in 1908 to describe the carbonic acid-carbonate buffer system in blood. Henderson was broadly knowledgeable; in addition to his important research on the physiology of blood, he also wrote on the adaptations of organisms and their fit with their environments, on sociology and on university education. He also founded the Fatigue Laboratory, at the Harvard Business School, which examined human physiology with specific focus on work in industry, exercise, and nutrition.
In 1916, Karl Albert Hasselbalch (1874–1962), a Danish physician and chemist, shared authorship in a paper with Christian Bohr in 1904 that described the Bohr effect, which showed that the ability of hemoglobin in the blood to bind with oxygen was inversely related to the acidity of the blood and the concentration of carbon dioxide. The pH scale was introduced in 1909 by another Dane, Sørensen, and in 1912, Hasselbalch published measurements of the pH of blood. In 1916, Hasselbalch expressed Henderson’s equation in logarithmic terms, consistent with the logarithmic scale of pH, and thus the Henderson-Hasselbalch equation was born.
How Sciences Interconnect: Medicine: The Buffer System in Blood
The normal pH of human blood is about 7.4. The carbonate buffer system in the blood uses the following equilibrium reaction:
$\ce{CO_2(g) + 2 H2O(l) <=> H2CO3(aq) <=> HCO_3^{-}(aq) + H3O^{+}(aq)} \nonumber$
The concentration of carbonic acid, $\ce{H2CO3}$ is approximately 0.0012 M, and the concentration of the hydrogen carbonate ion, is around 0.024 M. Using the Henderson-Hasselbalch equation and the pKa of carbonic acid at body temperature, we can calculate the pH of blood:
$pH = pK_{a} + \log \dfrac{[\text { base }]}{[\text { acid }]} = 6.4 + \log \frac{0.024}{0.0012} = 7.7 \nonumber$
The fact that the $\ce{H2CO3}$ concentration is significantly lower than that of the $\ce{HCO3^{-}}$ ion may seem unusual, but this imbalance is due to the fact that most of the by-products of our metabolism that enter our bloodstream are acidic. Therefore, there must be a larger proportion of base than acid, so that the capacity of the buffer will not be exceeded.
Lactic acid is produced in our muscles when we exercise. As the lactic acid enters the bloodstream, it is neutralized by the ion, producing H2CO3. An enzyme then accelerates the breakdown of the excess carbonic acid to carbon dioxide and water, which can be eliminated by breathing. In fact, in addition to the regulating effects of the carbonate buffering system on the pH of blood, the body uses breathing to regulate blood pH. If the pH of the blood decreases too far, an increase in breathing removes CO2 from the blood through the lungs driving the equilibrium reaction such that [H3O+] is lowered. If the blood is too alkaline, a lower breath rate increases CO2 concentration in the blood, driving the equilibrium reaction the other way, increasing [H+] and restoring an appropriate pH.
Link to Learning
View information on the buffer system encountered in natural waters. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/14%3A_Acid-Base_Equilibria/14.06%3A_Buffers.txt |
Learning Objectives
By the end of this section, you will be able to:
• Interpret titration curves for strong and weak acid-base systems
• Compute sample pH at important stages of a titration
• Explain the function of acid-base indicators
As seen in the chapter on the stoichiometry of chemical reactions, titrations can be used to quantitatively analyze solutions for their acid or base concentrations. In this section, we will explore the underlying chemical equilibria that make acid-base titrimetry a useful analytical technique.
Titration Curves
A titration curve is a plot of some solution property versus the amount of added titrant. For acid-base titrations, solution pH is a useful property to monitor because it varies predictably with the solution composition and, therefore, may be used to monitor the titration’s progress and detect its end point. The following example exercise demonstrates the computation of pH for a titration solution after additions of several specified titrant volumes. The first example involves a strong acid titration that requires only stoichiometric calculations to derive the solution pH. The second example addresses a weak acid titration requiring equilibrium calculations.
Example $1$: Calculating pH for Titration Solutions: Strong Acid/Strong Base
A titration is carried out for 25.00 mL of 0.100 M HCl (strong acid) with 0.100 M of a strong base NaOH (the titration curve is shown in Figure $1$). Calculate the pH at these volumes of added base solution:
1. 0.00 mL
2. 12.50 mL
3. 25.00 mL
4. 37.50 mL
Solution
(a) Titrant volume = 0 mL. The solution pH is due to the acid ionization of HCl. Because this is a strong acid, the ionization is complete and the hydronium ion molarity is 0.100 M. The pH of the solution is then
$\text{pH} =-\log (0.100)=1.000 \nonumber$
(b) Titrant volume = 12.50 mL. Since the acid sample and the base titrant are both monoprotic and equally concentrated, this titrant addition involves less than a stoichiometric amount of base, and so it is completely consumed by reaction with the excess acid in the sample. The concentration of acid remaining is computed by subtracting the consumed amount from the initial amount and then dividing by the solution volume:
$\left[ \ce{H3O^{+}} \right]=\frac{ n \left( \ce{H^{+}} \right)}{V}=\frac{0.002500~\text{mol} \times\left(\frac{1000 ~mL }{1 ~L }\right)-0.100 ~M \times 12.50 ~mL }{25.00~ mL +12.50~ mL }=0.0333~ M \nonumber$
(c) Titrant volume = 25.00 mL. This titrant addition involves a stoichiometric amount of base (the equivalence point), and so only products of the neutralization reaction are in solution (water and NaCl). Neither the cation nor the anion of this salt undergo acid-base ionization; the only process generating hydronium ions is the autoprotolysis of water. The solution is neutral, having a pH = 7.00.
(d) Titrant volume = 37.50 mL. This involves the addition of titrant in excess of the equivalence point. The solution pH is then calculated using the concentration of hydroxide ion:
$n \left( OH^{-}\right)_0> n \left( H^{+}\right)_0 \nonumber$
$\left[ \ce{OH^{-}} \right]=\frac{ n \left( \ce{OH^{-}} \right)}{\text{V}}=\dfrac{0.100~\text{M} \times 37.50~\text{mL} - 0.002500 ~\text{mol} \times\left(\frac{1000~\text{mL}}{1~\text{L}}\right)}{25.00~\text{mL} + 37.50~\text{mL}}=0.0200~\text{M} \nonumber$
$\text{pH} = 14 − \text{pOH} = 14 + \log([\ce{OH^{-}}]) = 14 + \log(0.0200) = 12.30 \nonumber$
Exercise $1$
Calculate the pH for the strong acid/strong base titration between 50.0 mL of 0.100 M HNO3(aq) and 0.200 M NaOH (titrant) at the listed volumes of added base: 0.00 mL, 15.0 mL, 25.0 mL, and 40.0 mL.
Answer
0.00: 1.000; 15.0: 1.5111; 25.0: 7; 40.0: 12.523
Example $2$: Titration of a Weak Acid with a Strong Base
Consider the titration of 25.00 mL of 0.100 M CH3CO2H with 0.100 M NaOH. The reaction can be represented as:
$\ce{CH3CO2H + OH^{-} -> CH3CO2^{-} + H2O} \nonumber$
Calculate the pH of the titration solution after the addition of the following volumes of NaOH titrant:
1. 0.00 mL
2. 25.00 mL
3. 12.50 mL
4. 37.50 mL
Solution
(a) The initial pH is computed for the acetic acid solution in the usual ICE approach:
$K_{ a }=\frac{\left[ H3O^{+}\right]\left[ CH_3 CO_2^{-}\right]}{\left[ CH_3 CO_2 H \right]} \approx \frac{\left[ H3O^{+}\right]^2}{\left[ CH_3 CO_2 H \right]_0} \nonumber$
and
$\left[ H3O^{+}\right]=\sqrt{K_a \times\left[ CH_3 CO_2 H \right]}=\sqrt{1.8 \times 10^{-5} \times 0.100}=1.3 \times 10^{-3} \nonumber$
$pH =-\log \left(1.3 \times 10^{-3}\right)=2.87 \nonumber$
(b) The acid and titrant are both monoprotic and the sample and titrant solutions are equally concentrated; thus, this volume of titrant represents the equivalence point. Unlike the strong-acid example above, however, the reaction mixture in this case contains a weak conjugate base (acetate ion). The solution pH is computed considering the base ionization of acetate, which is present at a concentration of
$\frac{0.00250~\text{mol} }{0.0500~\text{L}}=0.0500~\text{M}~\ce{CH3CO2^{-}} \nonumber$
Base ionization of acetate is represented by the equation
$\ce{CH3CO2^{-}(aq) + H2O(l) <=> CH3CO2H(aq) + OH^{-}(aq)} \nonumber$
$K_{ b }=\frac{\left[ H^{+}\right]\left[ OH^{-}\right]}{K_{ a }}=\frac{K_{ w }}{K_{ a }}=\frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}}=5.6 \times 10^{-10} \nonumber$
Assuming $x \ll 0.0500$, the pH may be calculated via the usual ICE approach:
$K_{ b }=\frac{x^2}{0.0500 M} \nonumber$
$x=\left[ OH^{-}\right]=5.3 \times 10^{-6} \nonumber$
\begin{align*} \text{pOH} &=-\log \left(5.3 \times 10^{-6}\right)=5.28 \ \text{pH} &=14.00-5.28=8.72 \end{align*}
Note that the pH at the equivalence point of this titration is significantly greater than 7, as expected when titrating a weak acid with a strong base.
(c) Titrant volume = 12.50 mL. This volume represents one-half of the stoichiometric amount of titrant, and so one-half of the acetic acid has been neutralized to yield an equivalent amount of acetate ion. The concentrations of these conjugate acid-base partners, therefore, are equal. A convenient approach to computing the pH is use of the Henderson-Hasselbalch equation:
$pH =p K_{ a }+\log \frac{[\text { Base }]}{[\text { Acid }]}=-\log \left(K_{ a }\right)+\log \frac{\left[ \ce{CH3CO2^{-}} \right]}{\left[ \ce{CH3CO2H} \right]}=-\log \left(1.8 \times 10^{-5}\right)+\log (1) \nonumber$
$pH =-\log \left(1.8 \times 10^{-5}\right)=4.74 \nonumber$
(pH = pKa at the half-equivalence point in a titration of a weak acid)
(d) Titrant volume = 37.50 mL. This volume represents a stoichiometric excess of titrant, and a reaction solution containing both the titration product, acetate ion, and the excess strong titrant. In such solutions, the solution pH is determined primarily by the amount of excess strong base:
$\left[ \ce{OH^{-}} \right]=\frac{(0.003750 mol -0.00250 mol )}{0.06250 L }=2.00 \times 10^{-2} M \nonumber$
\begin{align*} \text{pOH} &=-\log \left(2.00 \times 10^{-2}\right)=1.70 \[4pt] \text{pH} &=14.00-1.70=12.30 \end{align*}
Exercise $1$
Calculate the pH for the weak acid/strong base titration between 50.0 mL of 0.100 M HCOOH(aq) (formic acid) and 0.200 M NaOH (titrant) at the listed volumes of added base: 0.00 mL, 15.0 mL, 25.0 mL, and 30.0 mL.
Answer
0.00 mL: 2.37; 15.0 mL: 3.92; 25.00 mL: 8.29; 30.0 mL: 12.097
Performing additional calculations similar to those in the preceding example permits a more full assessment of titration curves. A summary of pH/volume data pairs for the strong and weak acid titrations is provided in Table $1$ and plotted as titration curves in Figure $1$. A comparison of these two curves illustrates several important concepts that are best addressed by identifying the four stages of a titration:
• initial state (added titrant volume = 0 mL): pH is determined by the acid being titrated; because the two acid samples are equally concentrated, the weak acid will exhibit a greater initial pH
• pre-equivalence point (0 mL < V < 25 mL): solution pH increases gradually and the acid is consumed by reaction with added titrant; composition includes unreacted acid and the reaction product, its conjugate base
• equivalence point (V = 25 mL): a drastic rise in pH is observed as the solution composition transitions from acidic to either neutral (for the strong acid sample) or basic (for the weak acid sample), with pH determined by ionization of the conjugate base of the acid
• postequivalence point (V > 25 mL): pH is determined by the amount of excess strong base titrant added; since both samples are titrated with the same titrant, both titration curves appear similar at this stage.
Table $1$: pH Values in the Titrations of a Strong Acid and of a Weak Acid
Volume of 0.100 M NaOH Added (mL) Moles of NaOH Added pH Values 0.100 M HCl1 pH Values 0.100 M CH3CO2H2
0.0 0.0 1.00 2.87
5.0 0.00050 1.18 4.14
10.0 0.00100 1.37 4.57
15.0 0.00150 1.60 4.92
20.0 0.00200 1.95 5.35
22.0 0.00220 2.20 5.61
24.0 0.00240 2.69 6.13
24.5 0.00245 3.00 6.44
24.9 0.00249 3.70 7.14
25.0 0.00250 7.00 8.72
25.1 0.00251 10.30 10.30
25.5 0.00255 11.00 11.00
26.0 0.00260 11.29 11.29
28.0 0.00280 11.75 11.75
30.0 0.00300 11.96 11.96
35.0 0.00350 12.22 12.22
40.0 0.00400 12.36 12.36
45.0 0.00450 12.46 12.46
50.0 0.00500 12.52 12.52
Acid-Base 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 $5.0 \times 10^{−9} ~\text{M}$ (pH < 8.3). In more basic solutions where the hydronium ion concentration is less than $5.0 \times 10^{−9} ~\text{M}$ (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:
$\underset{\text { red }}{ \ce{HIn(aq) + H2O(l)} \ce{ <=> } \underset{\text { yellow }}{ \ce{H3O^{+}(aq)} + In^{-}(aq)}} \nonumber$
$K_a=\dfrac{\left[ \ce{H3O^{+}}\right]\left[ \ce{In^{-}} \right]}{[ \ce{HIn} ]}=4.0 \times 10^{-4} \nonumber$
The anion of methyl orange, $\ce{In^{−}}$, is yellow, and the nonionized form, $\ce{HIn}$, is red. When we add acid to a solution of methyl orange, the increased hydronium ion concentration shifts the equilibrium toward the nonionized red form, in accordance with Le Chatelier’s principle. If we add base, we shift the equilibrium towards the yellow form. This behavior is completely analogous to the action of buffers.
The perceived color of an indicator solution is determined by the ratio of the concentrations of the two species $\ce{In^{−}}$ and $\ce{HIn}$. If most of the indicator (typically about 60−90% or more) is present as $\ce{In^{−}}$, the perceived color of the solution is yellow. If most is present as $\ce{HIn}$, then the solution color appears red. The Henderson-Hasselbalch equation is useful for understanding the relationship between the pH of an indicator solution and its composition (thus, perceived color):
$pH = p K a +\log \left(\frac{\left[ \ce{In^{-}} \right]}{[ \ce{HIn} ]}\right) \nonumber$
In solutions where pH > pKa, the logarithmic term must be positive, indicating an excess of the conjugate base form of the indicator (yellow solution). When pH < pKa, the log term must be negative, indicating an excess of the conjugate acid (red solution). When the solution pH is close to the indicator pKa, appreciable amounts of both conjugate partners are present, and the solution color is that of an additive combination of each (yellow and red, yielding orange). The color change interval (or pH interval) for an acid-base indicator is defined as the range of pH values over which a change in color is observed, and for most indicators this range is approximately pKa ± 1.
There are many different acid-base indicators that cover a wide range of pH values and can be used to determine the approximate pH of an unknown solution by a process of elimination. Universal indicators and pH paper contain a mixture of indicators and exhibit different colors at different pHs. Figure $2$ presents several indicators, their colors, and their color-change intervals.
The titration curves shown in Figure $3$ illustrate the choice of a suitable indicator for specific titrations. In the strong acid titration, use of any of the three indicators should yield reasonably sharp color changes and accurate end point determinations. For this titration, the solution pH reaches the lower limit of the methyl orange color change interval after addition of ~24 mL of titrant, at which point the initially red solution would begin to appear orange. When 25 mL of titrant has been added (the equivalence point), the pH is well above the upper limit and the solution will appear yellow. The titration's end point may then be estimated as the volume of titrant that yields a distinct orange-to-yellow color change. This color change would be challenging for most human eyes to precisely discern. More-accurate estimates of the titration end point are possible using either litmus or phenolphthalein, both of which exhibit color change intervals that are encompassed by the steep rise in pH that occurs around the 25.00 mL equivalence point.
The weak acid titration curve in Figure $3$ shows that only one of the three indicators is suitable for end point detection. If methyl orange is used in this titration, the solution will undergo a gradual red-to-orange-to-yellow color change over a relatively large volume interval (0–6 mL), completing the color change well before the equivalence point (25 mL) has been reached. Use of litmus would show a color change that begins after adding 7–8 mL of titrant and ends just before the equivalence point. Phenolphthalein, on the other hand, exhibits a color change interval that nicely brackets the abrupt change in pH occurring at the titration's equivalence point. A sharp color change from colorless to pink will be observed within a very small volume interval around the equivalence point.
Footnotes
• 1Titration of 25.00 mL of 0.100 M HCl (0.00250 mol of HCI) with 0.100 M NaOH.
• 2Titration of 25.00 mL of 0.100 M CH3CO2H (0.00250 mol of CH3CO2H) with 0.100 M NaOH. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/14%3A_Acid-Base_Equilibria/14.07%3A_Acid-Base_Titrations.txt |
Example and Directions
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Glossary Entries
Word(s)DefinitionImageCaptionLinkSource
acid ionizationreaction involving the transfer of a proton from an acid to water, yielding hydronium ions and the conjugate base of the acid
acid ionization constant (Ka)equilibrium constant for an acid ionization reaction
acid-base indicatorweak acid or base whose conjugate partner imparts a different solution color; used in visual assessments of solution pH
acidica solution in which [H3O+] > [OH]
amphiproticspecies that may either donate or accept a proton in a Bronsted-Lowry acid-base reaction
amphotericspecies that can act as either an acid or a base
autoionizationreaction between identical species yielding ionic products; for water, this reaction involves transfer of protons to yield hydronium and hydroxide ions
base ionizationreaction involving the transfer of a proton from water to a base, yielding hydroxide ions and the conjugate acid of the base
base ionization constant (Kb)equilibrium constant for a base ionization reaction
basica solution in which [H3O+] < [OH]
Brønsted-Lowry acidproton donor
Brønsted-Lowry baseproton acceptor
buffermixture of appreciable amounts of a weak acid-base pair the pH of a buffer resists change when small amounts of acid or base are added
buffer capacityamount of an acid or base that can be added to a volume of a buffer solution before its pH changes significantly (usually by one pH unit)
color-change intervalrange in pH over which the color change of an indicator is observed
conjugate acidsubstance formed when a base gains a proton
conjugate basesubstance formed when an acid loses a proton
diprotic acidacid containing two ionizable hydrogen atoms per molecule
diprotic basebase capable of accepting two protons
Henderson-Hasselbalch equationlogarithmic version of the acid ionization constant expression, conveniently formatted for calculating the pH of buffer solutions
ion-product constant for water (Kw)equilibrium constant for the autoionization of water
leveling effectobservation that acid-base strength of solutes in a given solvent is limited to that of the solvent’s characteristic acid and base species (in water, hydronium and hydroxide ions, respectively)
monoprotic acidacid containing one ionizable hydrogen atom per molecule
neutraldescribes a solution in which [H3O+] = [OH]
oxyacidternary compound with acidic properties, molecules of which contain a central nonmetallic atom bonded to one or more O atoms, at least one of which is bonded to an ionizable H atom
percent ionizationratio of the concentration of ionized acid to initial acid concentration expressed as a percentage
pHlogarithmic measure of the concentration of hydronium ions in a solution
pOHlogarithmic measure of the concentration of hydroxide ions in a solution
stepwise ionizationprocess in which a polyprotic acid is ionized by losing protons sequentially
titration curveplot of some sample property (such as pH) versus volume of added titrant
triprotic acidacid that contains three ionizable hydrogen atoms per molecule
14.09: Key Equations
Kw = [H3O+][OH] = 1.0 \times 10^{−14} (at 25 °C)
pOH = −log[OH]
[H3O+] = \times 10^{−pH}
[OH] = \times 10^{−pOH}
pH + pOH = pKw = 14.00 at 25 °C
Ka Kb = 1.0 \times 10^{−14} = Kw
pKa = −log Ka
pKb = −log Kb | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/14%3A_Acid-Base_Equilibria/14.08%3A_Key_Terms.txt |
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, OH when it undergoes autoionization:
The ion product of water, Kw is the equilibrium constant for the autoionization reaction:
Concentrations of hydronium and hydroxide ions in aqueous media are often represented as logarithmic pH and pOH values, respectively. At 25 °C, the autoprotolysis equilibrium for water requires the sum of pH and pOH to equal 14 for any aqueous solution. The relative concentrations of hydronium and hydroxide ion in a solution define its status as acidic ([H3O+] > [OH]), basic ([H3O+] < [OH]), or neutral ([H3O+] = [OH]). At 25 °C, a pH < 7 indicates an acidic solution, a pH > 7 a basic solution, and a pH = 7 a neutral solution.
The relative strengths of acids and bases are reflected in the magnitudes of their ionization constants; the stronger the acid or base, the larger its ionization constant. A reciprocal relation exists between the strengths of a conjugate acid-base pair: the stronger the acid, the weaker its conjugate base. Water exerts a leveling effect on dissolved acids or bases, reacting completely to generate its characteristic hydronium and hydroxide ions (the strongest acid and base that may exist in water). The strengths of the binary acids increase from left to right across a period of the periodic table (CH4 < NH3 < H2O < HF), and they increase down a group (HF < HCl < HBr < HI). The strengths of oxyacids that contain the same central element increase as the oxidation number of the element increases (H2SO3 < H2SO4). The strengths of oxyacids also increase as the electronegativity of the central element increases [H2SeO4 < H2SO4].
The ions composing salts may possess acidic or basic character, ionizing when dissolved in water to yield acidic or basic solutions. Acidic cations are typically the conjugate partners of weak bases, and basic anions are the conjugate partners of weak acids. Many metal ions bond to water molecules when dissolved to yield complex ions that may function as acids.
An acid that contains more than one ionizable proton is a polyprotic acid. These acids undergo stepwise ionization reactions involving the transfer of single protons. The ionization constants for polyprotic acids decrease with each subsequent step; these decreases typically are large enough to permit simple equilibrium calculations that treat each step separately.
Solutions that contain appreciable amounts of a weak conjugate acid-base pair are called buffers. A buffered solution will experience only slight changes in pH when small amounts of acid or base are added. Addition of large amounts of acid or base can exceed the buffer capacity, consuming most of one conjugate partner and preventing further buffering action.
The titration curve for an acid-base titration is typically a plot of pH versus volume of added titrant. These curves are useful in selecting appropriate acid-base indicators that will permit accurate determinations of titration end points. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/14%3A_Acid-Base_Equilibria/14.10%3A_Summary.txt |
1.
Write equations that show NH3 as both a conjugate acid and a conjugate base.
2.
Write equations that show acting both as an acid and as a base.
3.
Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry acid:
1. (f)
4.
Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry acid:
1. (f) HS
5.
Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry base:
1. (f)
6.
Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry base:
1. (f)
7.
What is the conjugate acid of each of the following? What is the conjugate base of each?
1. OH
2. H2O
3. NH3
4. H2O2
5. HS
8.
What is the conjugate acid of each of the following? What is the conjugate base of each?
1. H2S
2. PH3
3. HS
4. H4N2
5. CH3OH
9.
Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations:
1. (f)
10.
Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations:
1. (f)
11.
What are amphiprotic species? Illustrate with suitable equations.
12.
State which of the following species are amphiprotic and write chemical equations illustrating the amphiprotic character of these species:
1. H2O
2. S2−
13.
State which of the following species are amphiprotic and write chemical equations illustrating the amphiprotic character of these species.
1. NH3
2. Br
14.
Is the self-ionization of water endothermic or exothermic? The ionization constant for water (Kw) is 2.9 \times 10^{−14} at 40 °C and 9.3 \times 10^{−14} at 60 °C.
15.
Explain why a sample of pure water at 40 °C is neutral even though [H3O+] = 1.7 \times 10^{−7} M. Kw is 2.9 \times 10^{−14} at 40 °C.
16.
The ionization constant for water (Kw) is 2.9 \times 10^{−14} at 40 °C. Calculate [H3O+], [OH], pH, and pOH for pure water at 40 °C.
17.
The ionization constant for water (Kw) is 9.311 \times 10^{−14} at 60 °C. Calculate [H3O+], [OH], pH, and pOH for pure water at 60 °C.
18.
Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely:
1. 0.200 M HCl
2. 0.0143 M NaOH
3. 3.0 M HNO3
4. 0.0031 M Ca(OH)2
19.
Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely:
1. 0.000259 M HClO4
2. 0.21 M NaOH
3. 0.000071 M Ba(OH)2
4. 2.5 M KOH
20.
What are the pH and pOH of a solution of 2.0 M HCl, which ionizes completely?
21.
What are the hydronium and hydroxide ion concentrations in a solution whose pH is 6.52?
22.
Calculate the hydrogen ion concentration and the hydroxide ion concentration in wine from its pH. See Figure 14.2 for useful information.
23.
Calculate the hydronium ion concentration and the hydroxide ion concentration in lime juice from its pH. See Figure 14.2 for useful information.
24.
The hydronium ion concentration in a sample of rainwater is found to be 1.7 \times 10^{−6} M at 25 °C. What is the concentration of hydroxide ions in the rainwater?
25.
The hydroxide ion concentration in household ammonia is 3.2 \times 10^{−3} M at 25 °C. What is the concentration of hydronium ions in the solution?
26.
Explain why the neutralization reaction of a strong acid and a weak base gives a weakly acidic solution.
27.
Explain why the neutralization reaction of a weak acid and a strong base gives a weakly basic solution.
28.
Use this list of important industrial compounds (and Figure 14.8) to answer the following questions regarding: Ca(OH)2, CH3CO2H, HCl, H2CO3, HF, HNO2, HNO3, H3PO4, H2SO4, NH3, NaOH, Na2CO3.
1. Identify the strong Brønsted-Lowry acids and strong Brønsted-Lowry bases.
2. Identify the compounds that can behave as Brønsted-Lowry acids with strengths lying between those of H3O+ and H2O.
3. Identify the compounds that can behave as Brønsted-Lowry bases with strengths lying between those of H2O and OH.
29.
The odor of vinegar is due to the presence of acetic acid, CH3CO2H, a weak acid. List, in order of descending concentration, all of the ionic and molecular species present in a 1-M aqueous solution of this acid.
30.
Household ammonia is a solution of the weak base NH3 in water. List, in order of descending concentration, all of the ionic and molecular species present in a 1-M aqueous solution of this base.
31.
Explain why the ionization constant, Ka, for H2SO4 is larger than the ionization constant for H2SO3.
32.
Explain why the ionization constant, Ka, for HI is larger than the ionization constant for HF.
33.
Gastric juice, the digestive fluid produced in the stomach, contains hydrochloric acid, HCl. Milk of Magnesia, a suspension of solid Mg(OH)2 in an aqueous medium, is sometimes used to neutralize excess stomach acid. Write a complete balanced equation for the neutralization reaction, and identify the conjugate acid-base pairs.
34.
Nitric acid reacts with insoluble copper(II) oxide to form soluble copper(II) nitrate, Cu(NO3)2, a compound that has been used to prevent the growth of algae in swimming pools. Write the balanced chemical equation for the reaction of an aqueous solution of HNO3 with CuO.
35.
What is the ionization constant at 25 °C for the weak acid the conjugate acid of the weak base CH3NH2, Kb = 4.4 \times 10^{−4}.
36.
What is the ionization constant at 25 °C for the weak acid the conjugate acid of the weak base (CH3)2NH, Kb = 5.9 \times 10^{−4}?
37.
Which base, CH3NH2 or (CH3)2NH, is the stronger base? Which conjugate acid, or , is the stronger acid?
38.
Which is the stronger acid, or HBrO?
39.
Which is the stronger base, (CH3)3N or
40.
Predict which acid in each of the following pairs is the stronger and explain your reasoning for each.
1. H2O or HF
2. B(OH)3 or Al(OH)3
3. or
4. NH3 or H2S
5. H2O or H2Te
41.
Predict which compound in each of the following pairs of compounds is more acidic and explain your reasoning for each.
1. or
2. NH3 or H2O
3. PH3 or HI
4. NH3 or PH3
5. H2S or HBr
42.
Rank the compounds in each of the following groups in order of increasing acidity or basicity, as indicated, and explain the order you assign.
1. acidity: HCl, HBr, HI
2. basicity: H2O, OH, H, Cl
3. basicity: Mg(OH)2, Si(OH)4, ClO3(OH) (Hint: Formula could also be written as HClO4.)
4. acidity: HF, H2O, NH3, CH4
43.
Rank the compounds in each of the following groups in order of increasing acidity or basicity, as indicated, and explain the order you assign.
1. (f) basicity: BrO,
44.
Both HF and HCN ionize in water to a limited extent. Which of the conjugate bases, F or CN, is the stronger base?
45.
The active ingredient formed by aspirin in the body is salicylic acid, C6H4OH(CO2H). The carboxyl group (−CO2H) acts as a weak acid. The phenol group (an OH group bonded to an aromatic ring) also acts as an acid but a much weaker acid. List, in order of descending concentration, all of the ionic and molecular species present in a 0.001-M aqueous solution of C6H4OH(CO2H).
46.
Are the concentrations of hydronium ion and hydroxide ion in a solution of an acid or a base in water directly proportional or inversely proportional? Explain your answer.
47.
What two common assumptions can simplify calculation of equilibrium concentrations in a solution of a weak acid or base?
48.
Which of the following will increase the percent of NH3 that is converted to the ammonium ion in water?
1. addition of NaOH
2. addition of HCl
3. addition of NH4Cl
49.
Which of the following will increase the percentage of HF that is converted to the fluoride ion in water?
1. addition of NaOH
2. addition of HCl
3. addition of NaF
50.
What is the effect on the concentrations of HNO2, and OH when the following are added to a solution of KNO2 in water:
1. HCl
2. HNO2
3. NaOH
4. NaCl
5. KNO
51.
What is the effect on the concentration of hydrofluoric acid, hydronium ion, and fluoride ion when the following are added to separate solutions of hydrofluoric acid?
1. HCl
2. KF
3. NaCl
4. KOH
5. HF
52.
Why is the hydronium ion concentration in a solution that is 0.10 M in HCl and 0.10 M in HCOOH determined by the concentration of HCl?
53.
From the equilibrium concentrations given, calculate Ka for each of the weak acids and Kb for each of the weak bases.
(a) CH3CO2H: = 1.34 \times 10^{−3} M;
= 1.34 \times 10^{−3} M;
[CH3CO2H] = 9.866 \times 10^{−2} M;
(b) ClO: [OH] = 4.0 \times 10^{−4} M;
[HClO] = 2.38 \times 10^{−4} M;
[ClO] = 0.273 M;
(c) HCO2H: [HCO2H] = 0.524 M;
= 9.8 \times 10^{−3} M;
= 9.8 \times 10^{−3} M;
(d) = 0.233 M;
[C6H5NH2] = 2.3 \times 10^{−3} M;
= 2.3 \times 10^{−3} M
54.
From the equilibrium concentrations given, calculate Ka for each of the weak acids and Kb for each of the weak bases.
(a) NH3: [OH] = 3.1 \times 10^{−3} M;
= 3.1 \times 10^{−3} M;
[NH3] = 0.533 M;
(b) HNO2: = 0.011 M;
= 0.0438 M;
[HNO2] = 1.07 M;
(c) (CH3)3N: [(CH3)3N] = 0.25 M;
[(CH3)3NH+] = 4.3 \times 10^{−3} M;
[OH] = 3.7 \times 10^{−3} M;
(d) = 0.100 M;
[NH3] = 7.5 \times 10^{−6} M;
[H3O+] = 7.5 \times 10^{−6} M
55.
Determine Kb for the nitrite ion, In a 0.10-M solution this base is 0.0015% ionized.
56.
Determine Ka for hydrogen sulfate ion, In a 0.10-M solution the acid is 29% ionized.
57.
Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:
1. (f) (as a base)
58.
Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:
1. (f) (as a base)
59.
Using the Ka value of 1.4
60.
Calculate the concentration of all solute species in each of the following solutions of acids or bases. Assume that the ionization of water can be neglected, and show that the change in the initial concentrations can be neglected.
1. 0.0092 M HClO, a weak acid
2. 0.0784 M C6H5NH2, a weak base
3. 0.0810 M HCN, a weak acid
4. 0.11 M (CH3)3N, a weak base
5. 0.120 M a weak acid, Ka = 1.6 \times 10^{−7}
61.
Propionic acid, C2H5CO2H (Ka = 1.34 \times 10^{−5}), is used in the manufacture of calcium propionate, a food preservative. What is the pH of a 0.698-M solution of C2H5CO2H?
62.
White vinegar is a 5.0% by mass solution of acetic acid in water. If the density of white vinegar is 1.007 g/cm3, what is the pH?
63.
The ionization constant of lactic acid, CH3CH(OH)CO2H, an acid found in the blood after strenuous exercise, is 1.36 \times 10^{−4}. If 20.0 g of lactic acid is used to make a solution with a volume of 1.00 L, what is the concentration of hydronium ion in the solution?
64.
Nicotine, C10H14N2, is a base that will accept two protons (Kb1 = 7 \times 10^{−7}, Kb2 = 1.4 \times 10^{−11}). What is the concentration of each species present in a 0.050-M solution of nicotine?
65.
The pH of a 0.23-M solution of HF is 1.92. Determine Ka for HF from these data.
66.
The pH of a 0.15-M solution of is 1.43. Determine Ka for from these data.
67.
The pH of a 0.10-M solution of caffeine is 11.70. Determine Kb for caffeine from these data:
68.
The pH of a solution of household ammonia, a 0.950 M solution of NH3, is 11.612. Determine Kb for NH3 from these data.
69.
Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
1. Al(NO3)3
2. RbI
3. KHCO2
4. CH3NH3Br
70.
Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
1. FeCl3
2. K2CO3
3. NH4Br
4. KClO4
71.
Novocaine, C13H21O2N2Cl, is the salt of the base procaine and hydrochloric acid. The ionization constant for procaine is 7 \times 10^{−6}. Is a solution of novocaine acidic or basic? What are [H3O+], [OH], and pH of a 2.0% solution by mass of novocaine, assuming that the density of the solution is 1.0 g/mL.
72.
Which of the following concentrations would be practically equal in a calculation of the equilibrium concentrations in a 0.134-M solution of H2CO3, a diprotic acid: [OH], [H2CO3], No calculations are needed to answer this question.
73.
Calculate the concentration of each species present in a 0.050-M solution of H2S.
74.
Calculate the concentration of each species present in a 0.010-M solution of phthalic acid, C6H4(CO2H)2.
75.
Salicylic acid, HOC6H4CO2H, and its derivatives have been used as pain relievers for a long time. Salicylic acid occurs in small amounts in the leaves, bark, and roots of some vegetation (most notably historically in the bark of the willow tree). Extracts of these plants have been used as medications for centuries. The acid was first isolated in the laboratory in 1838.
1. Both functional groups of salicylic acid ionize in water, with Ka = 1.0 \times 10^{−3} for the—CO2H group and 4.2 \times 10^{−13} for the −OH group. What is the pH of a saturated solution of the acid (solubility = 1.8 g/L).
2. Aspirin was discovered as a result of efforts to produce a derivative of salicylic acid that would not be irritating to the stomach lining. Aspirin is acetylsalicylic acid, CH3CO2C6H4CO2H. The −CO2H functional group is still present, but its acidity is reduced, Ka = 3.0 \times 10^{−4}. What is the pH of a solution of aspirin with the same concentration as a saturated solution of salicylic acid (See Part a).
76.
The ion HTe is an amphiprotic species; it can act as either an acid or a base.
1. What is Ka for the acid reaction of HTe with H2O?
2. What is Kb for the reaction in which HTe functions as a base in water?
3. Demonstrate whether or not the second ionization of H2Te can be neglected in the calculation of [HTe] in a 0.10 M solution of H2Te.
77.
Explain why a buffer can be prepared from a mixture of NH4Cl and NaOH but not from NH3 and NaOH.
78.
Explain why the pH does not change significantly when a small amount of an acid or a base is added to a solution that contains equal amounts of the acid H3PO4 and a salt of its conjugate base NaH2PO4.
79.
Explain why the pH does not change significantly when a small amount of an acid or a base is added to a solution that contains equal amounts of the base NH3 and a salt of its conjugate acid NH4Cl.
80.
What is [H3O+] in a solution of 0.25 M CH3CO2H and 0.030 M NaCH3CO2?
81.
What is [H3O+] in a solution of 0.075 M HNO2 and 0.030 M NaNO2?
82.
What is [OH] in a solution of 0.125 M CH3NH2 and 0.130 M CH3NH3Cl?
83.
What is [OH] in a solution of 1.25 M NH3 and 0.78 M NH4NO3?
84.
What is the effect on the concentration of acetic acid, hydronium ion, and acetate ion when the following are added to an acidic buffer solution of equal concentrations of acetic acid and sodium acetate:
1. HCl
2. KCH3CO2
3. NaCl
4. KOH
5. CH3CO2H
85.
What is the effect on the concentration of ammonia, hydroxide ion, and ammonium ion when the following are added to a basic buffer solution of equal concentrations of ammonia and ammonium nitrate:
1. KI
2. NH3
3. HI
4. NaOH
5. NH4Cl
86.
What will be the pH of a buffer solution prepared from 0.20 mol NH3, 0.40 mol NH4NO3, and just enough water to give 1.00 L of solution?
87.
Calculate the pH of a buffer solution prepared from 0.155 mol of phosphoric acid, 0.250 mole of KH2PO4, and enough water to make 0.500 L of solution.
88.
How much solid NaCH3CO2•3H2O must be added to 0.300 L of a 0.50-M acetic acid solution to give a buffer with a pH of 5.00? (Hint: Assume a negligible change in volume as the solid is added.)
89.
What mass of NH4Cl must be added to 0.750 L of a 0.100-M solution of NH3 to give a buffer solution with a pH of 9.26? (Hint: Assume a negligible change in volume as the solid is added.)
90.
A buffer solution is prepared from equal volumes of 0.200 M acetic acid and 0.600 M sodium acetate. Use 1.80 \times 10^{−5} as Ka for acetic acid.
1. What is the pH of the solution?
2. Is the solution acidic or basic?
3. What is the pH of a solution that results when 3.00 mL of 0.034 M HCl is added to 0.200 L of the original buffer?
91.
A 5.36–g sample of NH4Cl was added to 25.0 mL of 1.00 M NaOH and the resulting solution
diluted to 0.100 L.
1. What is the pH of this buffer solution?
2. Is the solution acidic or basic?
3. What is the pH of a solution that results when 3.00 mL of 0.034 M HCl is added to the solution?
92.
Explain how to choose the appropriate acid-base indicator for the titration of a weak base with a strong acid.
93.
Explain why an acid-base indicator changes color over a range of pH values rather than at a specific pH.
94.
Calculate the pH at the following points in a titration of 40 mL (0.040 L) of 0.100 M barbituric acid (Ka = 9.8 \times 10^{−5}) with 0.100 M KOH.
1. no KOH added
2. 20 mL of KOH solution added
3. 39 mL of KOH solution added
4. 40 mL of KOH solution added
5. 41 mL of KOH solution added
95.
The indicator dinitrophenol is an acid with a Ka of 1.1 \times 10^{−4}. In a 1.0 \times 10^{−4}-M solution, it is colorless in acid and yellow in base. Calculate the pH range over which it goes from 10% ionized (colorless) to 90% ionized (yellow). | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/14%3A_Acid-Base_Equilibria/14.11%3A_Exercises.txt |
We previously learned about aqueous solutions and their importance, as well as about solubility rules. While this gives us a picture of solubility, that picture is not complete if we look at the rules alone. Solubility equilibrium, which we will explore in this chapter, is a more complex topic that allows us to determine the extent to which a slightly soluble ionic solid will dissolve, and the conditions under which precipitation.
• 15.0: Introduction
• 15.1: Precipitation and Dissolution
The equilibrium constant for an equilibrium involving the precipitation or dissolution of a slightly soluble ionic solid is called the solubility product, Ksp, of the solid. The solubility product of a slightly soluble electrolyte can be calculated from its solubility; conversely, its solubility can be calculated from its Ksp, provided the only significant reaction that occurs when the solid dissolves is the formation of its ions.
• 15.2: Lewis Acids and Bases
A Lewis acid is a species that can accept an electron pair, whereas a Lewis base has an electron pair available for donation to a Lewis acid. Complex ions are examples of Lewis acid-base adducts. In a complex ion, we have a central atom, often consisting of a transition metal cation, which acts as a Lewis acid, and several neutral molecules or ions surrounding them called ligands that act as Lewis bases. Complex ions form by sharing electron pairs to form coordinate covalent bonds.
• 15.3: Coupled Equilibria
Several systems we encounter consist of multiple equilibria, systems where two or more equilibria processes are occurring simultaneously. Some common examples include acid rain, fluoridation, and dissolution of carbon dioxide in sea water. When looking at these systems, we need to consider each equilibrium separately and then combine the individual equilibrium constants into one solubility product or reaction quotient expression using the tools from the first equilibrium chapter.
• 15.4: Key Terms
• 15.5: Key Equations
• 15.6: Summary
• 15.7: Exercises
These are homework exercises to accompany the Textmap created for "Chemistry" by OpenStax. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
Thumbnail: Lead (II) iodide precipitates when potassium iodide is mixed with lead (II) nitrate. (CC BY-SA 3.0 Unported; PRHaney via Wikipedia)
15: Equilibria of Other Reaction Classes
The mineral fluorite, CaF2 Figure \(1\): , is commonly used as a semiprecious stone in many types of jewelry because of its striking appearance. Deposits of fluorite are formed through a process called hydrothermal precipitation in which calcium and fluoride ions dissolved in groundwater combine to produce insoluble CaF2 in response to some change in solution conditions. For example, a decrease in temperature may trigger fluorite precipitation if its solubility is exceeded at the lower temperature. Because fluoride ion is a weak base, its solubility is also affected by solution pH, and so geologic or other processes that change groundwater pH will also affect the precipitation of fluorite. This chapter extends the equilibrium discussion of other chapters by addressing some additional reaction classes (including precipitation) and systems involving coupled equilibrium reactions. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/15%3A_Equilibria_of_Other_Reaction_Classes/15.00%3A_Introduction.txt |
Learning Objectives
By the end of this section, you will be able to:
• Write chemical equations and equilibrium expressions representing solubility equilibria
• Carry out equilibrium computations involving solubility, equilibrium expressions, and solute concentrations
Solubility equilibria are established when the dissolution and precipitation of a solute species occur at equal rates. These equilibria underlie many natural and technological processes, ranging from tooth decay to water purification. An understanding of the factors affecting compound solubility is, therefore, essential to the effective management of these processes. This section applies previously introduced equilibrium concepts and tools to systems involving dissolution and precipitation.
The Solubility Product
Recall from the chapter on solutions that the solubility of a substance can vary from essentially zero (insoluble or sparingly soluble) to infinity (miscible). A solute with finite solubility can yield a saturated solution when it is added to a solvent in an amount exceeding its solubility, resulting in a heterogeneous mixture of the saturated solution and the excess, undissolved solute. For example, a saturated solution of silver chloride is one in which the equilibrium shown below has been established.
$\ce{AgCl(s) <=>[\text{dissolution}][\text{precipitation}] Ag^{+}(aq) + Cl^{-}(aq)} \nonumber$
In this solution, an excess of solid AgCl dissolves and dissociates to produce aqueous $\ce{Ag^{+}}$ and $\ce{Cl^{-}}$ ions at the same rate that these aqueous ions combine and precipitate to form solid $\ce{AgCl}$ (Figure $1$). Because silver chloride is a sparingly soluble salt, the equilibrium concentration of its dissolved ions in the solution is relatively low.
The equilibrium constant for solubility equilibria such as this one is called the solubility product constant, Ksp, in this case
$\ce{AgCl(s) <=> Ag^{+}(aq) + Cl^{-}(aq)} \quad \quad K_{ sp }=\left[\ce{Ag^{+}(aq)}\right]\left[\ce{Cl^{-}(aq)}\right] \nonumber$
Recall that only gases and solutes are represented in equilibrium constant expressions, so the Ksp does not include a term for the undissolved AgCl. A listing of solubility product constants for several sparingly soluble compounds is provided in Appendix J.
Example $1$: Writing Equations and Solubility Products
Write the dissolution equation and the solubility product expression for each of the following slightly soluble ionic compounds:
1. AgI, silver iodide, a solid with antiseptic properties
2. CaCO3, calcium carbonate, the active ingredient in many over-the-counter chewable antacids
3. Mg(OH)2, magnesium hydroxide, the active ingredient in Milk of Magnesia
4. Mg(NH4)PO4, magnesium ammonium phosphate, an essentially insoluble substance used in tests for magnesium
5. Ca5(PO4)3OH, the mineral apatite, a source of phosphate for fertilizers
Solution
1. $\ce{AgI(s) <=> Ag^{+}(aq) + I^{-}(aq)}$ $K_{ sp }=\left[\ce{Ag^{+}}\right]\left[\ce{I^{-}}\right] \nonumber$
2. $\ce{CaCO_3(s) <=> Ca^{2+}(aq) + CO3^{2-}(aq)}$ $K_{\text {sp}}=\left[\ce{Ca^{2+}} \right]\left[\ce{CO3^{2-}}\right] \nonumber$
3. $\ce{Mg(OH)_2(s) <=> Mg^{2+}(aq) + 2 OH^{-}(aq)}$ $K_{\text {sp}}=\left[\ce{Mg^{2+}}\right]\left[ \ce{OH^{-}}\right]^2 \nonumber$
4. $\ce{Mg(NH4)PO4(s) <=> Mg^{2+}(aq) + NH4^{+}(aq) + PO4^{3-}(aq)}$ $K_{\text{sp}}=\left[\ce{Mg^{2+}}\right]\left[\ce{NH4^{+}}\right]\left[\ce{ PO4^{3-}}\right] \nonumber$
5. $\ce{Ca5(PO4) 3OH(s) <=> 5 Ca^{2+}(aq) + 3 PO4^{3-}(aq) + OH^{-}(aq)}$ $K_{ sp }=\left[\ce{Ca^{2+}}\right]^5\left[\ce{PO4^{3-}}\right]^3[\ce{OH^{-}}] \nonumber$
Exercise $1$
Write the dissolution equation and the solubility product for each of the following slightly soluble compounds:
1. BaSO4
2. Ag2SO4
3. Al(OH)3
4. Pb(OH)Cl
Answer
1. $\ce{BaSO4(s) <=> Ba^{2+}(aq) + SO4^{2-}(aq)}$ $K_{ sp }=\left[ \ce{Ba^{2+}}\right]\left[ \ce{SO4^{2-}}\right] \nonumber$
2. $\ce{Ag2SO4(s) <=> 2 Ag^{+}(aq) + SO4^{2-}(aq)}$ $K_{ sp }=\left[ \ce{Ag^{+}}\right]^2\left[ \ce{SO4^{2-}}\right] \nonumber$
3. $\ce{Al(OH)3(s) <=> Al^{3+}(aq) + 3 OH^{-}(aq)}$ $K_{ sp }=\left[ \ce{Al^{3+}}\right]\left[ \ce{OH^{-}}\right]^3 \nonumber$
4. $\ce{Pb(OH)Cl(s) <=> Pb^{2+}(aq) + OH^{-}(aq) + Cl^{-}(aq)}$ $K_{ sp }=\left[ \ce{Pb^{2+}}\right]\left[ \ce{OH^{-}} \right]\left[ \ce{Cl^{-}}\right] \nonumber$
Ksp and Solubility
The Ksp of a slightly soluble ionic compound may be simply related to its measured solubility provided the dissolution process involves only dissociation and solvation, for example:
$\ce{M_{p}X_{q}(s) <=> p~M^{m+}(aq) + q~X^{n-}(aq)} \nonumber$
For cases such as these, one may derive Ksp values from provided solubilities, or vice-versa. Calculations of this sort are most conveniently performed using a compound’s molar solubility, measured as moles of dissolved solute per liter of saturated solution.
Example $2$: Calculation of Ksp from Equilibrium Concentrations
Fluorite, $\ce{CaF2}$, is a slightly soluble solid that dissolves according to the equation:
$\ce{CaF_2(s) <=> Ca^{2+}(aq) + 2 F^{-}(aq)} \nonumber$
The concentration of $\ce{Ca^{2+}}$ in a saturated solution of $\ce{CaF2}$ is $2.15 \times 10^{–4} ~\text{M}$. What is the solubility product of fluorite?
Solution
According to the stoichiometry of the dissolution equation, the fluoride ion molarity of a $\ce{CaF2}$ solution is equal to twice its calcium ion molarity:
$\left[ F^{-}\right]=\left(2~\text{mol}~\ce{F^{-}} / 1~\text{mol}~\ce{Ca^{2+}}\right)=(2)\left(2.15 \times 10^{-4} M\right)=4.30 \times 10^{-4} M \nonumber$
Substituting the ion concentrations into the Ksp expression gives
$K_{ sp }=\left[ \ce{Ca^{2+}}\right]\left[ \ce{F^{-}}\right]^2=\left(2.15 \times 10^{-4}\right)\left(4.30 \times 10^{-4}\right)^2=3.98 \times 10^{-11} \nonumber$
Exercise $2$
In a saturated solution of Mg(OH)2, the concentration of Mg2+ is $1.31 \times 10^{–4} ~\text{M}$. What is the solubility product for Mg(OH)2?
$\ce{Mg(OH)2(s) <=> Mg^{2+}(aq) + 2 OH^{-}(aq)} \nonumber$
Answer
$8.99 \times 10^{–12}$
Example $3$: Determination of Molar Solubility from Ksp
The Ksp of copper(I) bromide, CuBr, is $6.3 \times 10^{–9}$. Calculate the molar solubility of copper bromide.
Solution
The dissolution equation and solubility product expression are
$\ce{CuBr(s) <=> Cu^{+}(aq)+ Br^{-}(aq)} \nonumber$
with
$K_{ sp }=\left[ \ce{Cu^{+}}\right]\left[ \ce{Br^{-}}\right] \nonumber$
Following the ICE approach to this calculation yields the table
Substituting the equilibrium concentration terms into the solubility product expression and solving for x yields
\begin{align*} K_{ sp }&=\left[ \ce{Cu^{+}}\right]\left[ \ce{Br^{-}}\right] \[4pt] 6.3 \times 10^{-9} &=(x)(x)=x^2 \[4pt] x&=\sqrt{6.3 \times 10^{-9}}\[4pt] &=7.9 \times 10^{-5}\,\text{M} \end{align*} \nonumber
Since the dissolution stoichiometry shows one mole of copper(I) ion and one mole of bromide ion are produced for each moles of $\ce{Br}$ dissolved, the molar solubility of $\ce{CuBr}$ is $7.9 \times 10^{–5} ~\text{M}$.
Exercise $3$
The Ksp of AgI is $1.5 \times 10^{–16}$. Calculate the molar solubility of silver iodide.
Answer
$1.2 \times 10^{–8}~ \text{M}$
Example $4$: Determination of Molar Solubility from Ksp
The Ksp of calcium hydroxide, Ca(OH)2, is $1.3 \times 10^{–6}$. Calculate the molar solubility of calcium hydroxide.
Solution
The dissolution equation and solubility product expression are
$\ce{Ca(OH)2(s) <=> Ca^{2+}(aq) +2 OH^{-}(aq)} \nonumber$
with
$K_{ sp }=\left[ \ce{Ca^{2+}}\right]\left[ \ce{OH^{-}}\right]^2 \nonumber$
The ICE table for this system is
Substituting terms for the equilibrium concentrations into the solubility product expression and solving for $x$ gives
\begin{aligned} K_{ sp } &=\left[ \ce{Ca^{2+}}\right]\left[ \ce{OH^{-}}\right]^2\[4pt] 1.3 \times 10^{-6}&=(x)(2 x)^2=(x)\left(4 x^2\right)=4 x^3\[4pt] x&=\sqrt[3]{\frac{1.3 \times 10^{-6}}{4}}\[4pt] &=6.9 \times 10^{-3}\,\text{M} \end{aligned} \nonumber
As defined in the ICE table, $x$ is the molarity of calcium ion in the saturated solution. The dissolution stoichiometry shows a 1:1 relation between moles of calcium ion in solution and moles of compound dissolved, and so, the molar solubility of Ca(OH)2 is $6.9 \times 10^{–3}~\text{M}$.
Exercise $4$
The Ksp of PbI2 is $1.4 \times 10^{–8}$. Calculate the molar solubility of lead(II) iodide.
Answer
$1.5 \times 10^{–3}~ \text{M}$
Example $5$: Determination of Ksp from Gram Solubility
Many of the pigments used by artists in oil-based paints (Figure $2$) are sparingly soluble in water. For example, the solubility of the artist’s pigment chrome yellow, PbCrO4, is $4.6 \times 10^{–6}~\text{g/L}$. Determine the solubility product for PbCrO4.
Solution
Before calculating the solubility product, the provided solubility must be converted to molarity:
\begin{align*} [\ce{PbCrO4}] &= \dfrac{4.6 \times 10^{-6}~\text{g}~\text{PbCrO4}}{1~\text{L} } \times \dfrac{1~ \text{mol}~ \ce{PbCrO4}}{323.2~ \text{g} ~ \ce{PbCrO4}} \[4pt] &= \frac{1.4 \times 10^{-8}~\text{mol} ~ \ce{PbCrO4}}{1\,\text{L} } \[4pt] &= 1.4 \times 10^{-8}\,\text{M} \end{align*} \nonumber
The dissolution equation for this compound is
$\ce{PbCrO4(s) <=> Pb^{2+}(aq) + CrO4^{2-}(aq)} \nonumber$
The dissolution stoichiometry shows a 1:1 relation between the molar amounts of compound and its two ions, and so both $[\ce{Pb^{2+}}]$ and $[\ce{CrO4^{2-}}]$ are equal to the molar solubility of $\ce{PbCrO4}$:
$[\ce{Pb^{2+}}]= [\ce{CrO4^{2-}}] =1.4 \times 10^{-8}~ \text{M} \nonumber$
Therefore
$K_{sp} = [\ce{Pb^{2+}}][\ce{CrO4^{2-}}] = (1.4 \times 10^{–8})(1.4 \times 10^{–8}) = 2.0 \times 10^{–16} \nonumber$
Exercise $5$
The solubility of TlCl [thallium(I) chloride], an intermediate formed when thallium is being isolated from ores, is 3.12 grams per liter at 20 °C. What is its solubility product?
Answer
$1.69 \times 10^{–4}$
Example $6$: Calculating the Solubility of Hg2Cl2
Calomel, Hg2Cl2, is a compound composed of the diatomic ion of mercury(I), and chloride ions, Cl. Although most mercury compounds are now known to be poisonous, eighteenth-century physicians used calomel as a medication. Their patients rarely suffered any mercury poisoning from the treatments because calomel has a very low solubility, as suggested by its very small Ksp:
$\ce{Hg2Cl2(s) <=> Hg2^{2+}(aq) + 2 Cl^{-}(aq)} \quad K_{ sp }=1.1 \times 10^{-18} \nonumber$
Calculate the molar solubility of $\ce{Hg2Cl2}$.
Solution
The dissolution stoichiometry shows a 1:1 relation between the amount of compound dissolved and the amount of mercury(I) ions, and so the molar solubility of Hg2Cl2 is equal to the concentration of ions
Following the ICE approach results in
Substituting the equilibrium concentration terms into the solubility product expression and solving for x gives
\begin{align*}
K_{ sp } &=\left[ \ce{Hg2^{2+}} \right]\left[ \ce{Cl^{-}} \right]^2 \4pt] 1.1 \times 10^{-18} &=(x)(2 x)^2 \[4pt] 4 x^3& =1.1 \times 10^{-18} \[4pt] x&=\sqrt[3]{\left(\frac{1.1 \times 10^{-18}}{4}\right)} \[4pt] &=6.5 \times 10^{-7}~ \text{M} \end{align*} Therefore \begin{align*} \left[ \ce{Hg2^{2+}}\right] &=6.5 \times 10^{-7}~\text{M} \[4pt]&=6.5 \times 10^{-7}~ \text{M} \[4pt] \left[ \ce{Cl^{-}}\right] &=2 x=2\left(6.5 \times 10^{-7} ~\text{M}\right) \[4pt] &=1.3 \times 10^{-6} M \end{align*} The dissolution stoichiometry shows the molar solubility of $\ce{Hg2Cl2}$ is equal to $[ \ce{Hg2^{2+}}]$ or $6.5 \times 10^{–7}~ \text{M}$. Exercise $6$ Determine the molar solubility of MgF2 from its solubility product: $K_{sp} = 6.4 \times 10^{–9}$. Answer $1.2 \times 10^{–3} ~\text{M}$ How Sciences Interconnect: Using Barium Sulfate for Medical Imaging Various types of medical imaging techniques are used to aid diagnoses of illnesses in a noninvasive manner. One such technique utilizes the ingestion of a barium compound before taking an X-ray image. A suspension of barium sulfate, a chalky powder, is ingested by the patient. Since the $K_{sp}$ of barium sulfate is $2.3 \times 10^{–8}$, very little of it dissolves as it coats the lining of the patient’s intestinal tract. Barium-coated areas of the digestive tract then appear on an X-ray as white, allowing for greater visual detail than a traditional X-ray (Figure $3$). Medical imaging using barium sulfate can be used to diagnose acid reflux disease, Crohn’s disease, and ulcers in addition to other conditions. Visit this website for more information on how barium is used in medical diagnoses and which conditions it is used to diagnose. Predicting Precipitation The equation that describes the equilibrium between solid calcium carbonate and its solvated ions is: \[\ce{CaCO3(s) <=> Ca^{2+}(aq) + CO3^{2-}(aq)} \nonumber
with
$K_{s p}=\left[ \ce{Ca^{2+}} \right]\left[ \ce{CO3^{2-}} \right]=8.7 \times 10^{-9} \nonumber$
It is important to realize that this equilibrium is established in any aqueous solution containing Ca2+ and CO32 ions, not just in a solution formed by saturating water with calcium carbonate. Consider, for example, mixing aqueous solutions of the soluble compounds sodium carbonate and calcium nitrate. If the concentrations of calcium and carbonate ions in the mixture do not yield a reaction quotient, Qsp, that exceeds the solubility product, Ksp, then no precipitation will occur. If the ion concentrations yield a reaction quotient greater than the solubility product, then precipitation will occur, lowering those concentrations until equilibrium is established (Qsp = Ksp). The comparison of Qsp to Ksp to predict precipitation is an example of the general approach to predicting the direction of a reaction first introduced in the chapter on equilibrium. For the specific case of solubility equilibria:
• Qsp < Ksp: the reaction proceeds in the forward direction (solution is not saturated; no precipitation observed)
• Qsp > Ksp: the reaction proceeds in the reverse direction (solution is supersaturated; precipitation will occur)
This predictive strategy and related calculations are demonstrated in the next few example exercises.
Example $7$: Precipitation of Mg(OH)2
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:
$\ce{Mg(OH)2(s) <=> Mg^{2+}(aq) + 2 OH^{-}(aq)} \quad K_{ sp }=8.9 \times 10^{-12} \nonumber$
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
Calculation of the reaction quotient under these conditions is shown here:
$Q=\left[ \ce{Mg^{2+}}\right]\left[ \ce{OH^{-}}\right]^2=(0.0537)(0.0010)^2=5.4 \times 10^{-8} \nonumber$
Because Q is greater than Ksp ($Q = 5.4 \times 10^{–8}$ is larger than $K_{sp} = 8.9 \times 10^{–12}$), the reverse reaction will proceed, precipitating magnesium hydroxide until the dissolved ion concentrations have been sufficiently lowered, so that $Q_{sp} = K_{sp}$.
Exercise $7$
Predict whether $\ce{CaHPO4}$ will precipitate from a solution with [Ca2+] = 0.0001 M and = 0.001 M.
Answer
No precipitation of $\ce{CaHPO4}$; $Q = 1 \times 10^{–7}$, which is less than Ksp ($7 \times 10^{–7}$)
Example $8$: Precipitation of AgCl
Does silver chloride precipitate when equal volumes of a $2.0 \times 10^{–4}$-M solution of AgNO3 and a $2.0 \times 10^{–4}$-M solution of NaCl are mixed?
Solution
The equation for the equilibrium between solid silver chloride, silver ion, and chloride ion is:
$\ce{AgCl (s) <=> Ag^{+}(aq) + Cl^{-}(aq)} \nonumber$
The solubility product is $1.6 \times 10^{-10}$ (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. Because the volume doubles when equal volumes of AgNO3 and NaCl solutions are mixed, each concentration is reduced to half its initial value
$\frac{1}{2}\left(2.0 \times 10^{-4}\right) M=1.0 \times 10^{-4}~\text{M} \nonumber$
The reaction quotient, Q, is greater than Ksp for AgCl, so a supersaturated solution is formed:
$Q=\left[ \ce{Ag^{+}} \right]\left[ \ce{Cl^{-}} \right]=\left(1.0 \times 10^{-4}\right)\left(1.0 \times 10^{-4}\right)=1.0 \times 10^{-8}>K_{ sp } \nonumber$
AgCl will precipitate from the mixture until the dissolution equilibrium is established, with Q equal to Ksp.
Exercise $8$
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 $\ce{ClO4^{-}}$? (Hint: Use the dilution equation to calculate the concentrations of potassium and perchlorate ions in the mixture.)
Answer
No, $Q = 4.0 \times 10^{–3}$, which is less than $K_{sp} = 1.05 \times 10^{–2}$
Example $9$: 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, $\ce{CaC2O4 \cdot H2O}$ (calcium oxalate monohydrate). The concentration of $\ce{Ca^{2+}}$ in a sample of blood serum is $2.2 \times 10^{–3} M$. What concentration of $\ce{C2O4^{2−}}$ ion must be established before $\ce{CaC2O4 \cdot H2O}$ begins to precipitate?
Solution
The equilibrium expression is:
$\ce{CaC2O4(s) <=> Ca^{2+}(aq) + C2O4^{2-}(aq)} \nonumber$
For this reaction:
$K_{\text {sp }}=\left[ \ce{Ca^{2+}} \right]\left[ \ce{C2O4^{2-}} \right]=1.96 \times 10^{-9} \nonumber$
(see Appendix J)
Substitute the provided calcium ion concentration into the solubility product expression and solve for oxalate concentration:
\begin{aligned}
Q=K_{ sp }=\left[ \ce{Ca^{2+}}\right]\left[ \ce{C2O4^{2-}}\right]&=1.96 \times 10^{-9}\4pt] \left(2.2 \times 10^{-3}\right)\left[ C_2 O_4{ }^{2-}\right] &=1.96 \times 10^{-9}\[4pt] \left[ \ce{C2O4^{2-}}\right] &=\frac{1.96 \times 10^{-9}}{2.2 \times 10^{-3}} \[4pt]&=8.9 \times 10^{-7}~\text{M} \end{aligned} A concentration of $[\ce{C2O4^{2-}}] = 8.9 \times 10^{-7}~ \text{M}$ is necessary to initiate the precipitation of $\ce{CaC2O4}$ under these conditions. Exercise $9$ 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 $6.7 \time 10^{–5}~\text{M}$ Example $10$: Concentrations Following Precipitation Clothing washed in water that has a manganese $[\ce{Mn^{2+}(aq)}]$ concentration exceeding 0.1 mg/L ($1.8 \times 10^{–6} ~\text{M}$) may be stained by the manganese upon oxidation, but the amount of Mn2+ in the water can be decreased by adding a base to precipitate Mn(OH)2. What pH is required to keep [Mn2+] equal to $1.8 \times 10^{–6}~ \text{M}$? Solution The dissolution of Mn(OH)2 is described by the equation: \[\ce{Mn(OH)2(s) <=> Mn^{2+}(aq) + 2 OH^{-}(aq)} \quad \quad K_{ sp }=2 \times 10^{-13} \nonumber
At equilibrium:
$K_{ sp }=\left[ \ce{Mn^{2+}} \right]\left[ \ce{OH^{-}} \right]^2 \nonumber$
or
$\left(1.8 \times 10^{-6}\right)\left[ \ce{OH^{-}} \right]^2=2 \times 10^{-13} \nonumber$
so
$\left[ \ce{OH^{-}} \right]=3.3 \times 10^{-4} M \nonumber$
Calculate the pH from the pOH:
\begin{align*} \text{pOH} &=-\log \left[ \ce{OH^{-}}\right] \[4pt] &=-\log (3.3 \times 10^{-4})=3.48 \[4pt] \text{pH} &=14.00 - \text{pOH} \[4pt] &=14.00-3.48 \[4pt] &=10.52\end{align*} \nonumber
(final result rounded to one significant digit, limited by the certainty of the Ksp)
Exercise $10$
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 $5.37 \times 10^{–2}~\text{M}$. Calculate the pH at which [Mg2+] is decreased to $1.0 \times 10^{–5}~\text{M}$
Answer
10.97
In solutions containing two or more ions that may form insoluble compounds with the same counter ion, an experimental strategy called selective precipitation may be used to remove individual ions from solution. By increasing the counter ion concentration in a controlled manner, ions in solution may be precipitated individually, assuming their compound solubilities are adequately different. In solutions with equal concentrations of target ions, the ion forming the least soluble compound will precipitate first (at the lowest concentration of counter ion), with the other ions subsequently precipitating as their compound’s solubilities are reached. As an illustration of this technique, the next example exercise describes separation of a two halide ions via precipitation of one as a silver salt.
Chemistry in Everyday Life: The Role of Precipitation in Wastewater Treatment
Solubility equilibria are useful tools in the treatment of wastewater carried out in facilities that may treat the municipal water in your city or town (Figure $5$). Specifically, selective precipitation is used to remove contaminants from wastewater before it is released back into natural bodies of water. For example, phosphate ions are often present in the water discharged from manufacturing facilities. An abundance of phosphate causes excess algae to grow, which impacts the amount of oxygen available for marine life as well as making water unsuitable for human consumption.
One common way to remove phosphates from water is by the addition of calcium hydroxide, or lime, Ca(OH)2. As the water is made more basic, the calcium ions react with phosphate ions to produce hydroxylapatite, Ca5(PO4)3OH, which then precipitates out of the solution:
$\ce{5 Ca^{2+} + 3 PO4^{3-} + OH^{-} <=> Ca5 (PO4)3 \cdot OH(s)} \nonumber$
Because the amount of calcium ion added does not result in exceeding the solubility products for other calcium salts, the anions of those salts remain behind in the wastewater. The precipitate is then removed by filtration and the water is brought back to a neutral pH by the addition of CO2 in a recarbonation process. Other chemicals can also be used for the removal of phosphates by precipitation, including iron(III) chloride and aluminum sulfate.
View this site for more information on how phosphorus is removed from wastewater.
Example $11$: Precipitation of Silver Halides
A solution contains 0.00010 mol of KBr and 0.10 mol of KCl per liter. AgNO3 is gradually added to this solution. Which forms first, solid AgBr or solid AgCl?
Solution
The two equilibria involved are:
\begin{align*} \ce{AgCl(s) & <=> Ag^{+}(aq) + Cl^{-}(aq)} & K_{ sp } & =1.6 \times 10^{-10} \[4pt] \ce{AgBr(s) & <=> Ag^{+}(aq) + Br^{-}(aq)} & K_{ sp } & =5.0 \times 10^{-13} \end{align*} \nonumber
If the solution contained about equal concentrations of Cl and Br, then the silver salt with the smaller Ksp (AgBr) would precipitate first. The concentrations are not equal, however, so the [Ag+] at which AgCl begins to precipitate and the [Ag+] at which AgBr begins to precipitate must be calculated. The salt that forms at the lower [Ag+] precipitates first.
AgBr precipitates when Q equals Ksp for AgBr
\begin{align*} Q_{ sp }=K_{ sp }=\left[ \ce{Ag^{+}} \right]\left[ \ce{Br^{-}}\right]=\left[ \ce{Ag^{+}}\right](0.00010)&=5.0 \times 10^{-13} \[4pt] \left[ \ce{Ag^{+}}\right] &=\frac{5.0 \times 10^{-13}}{0.00010} \[4pt] &=5.0 \times 10^{-9}~\text{M} \end{align*} \nonumber
AgBr begins to precipitate when [Ag+] is $5.0 \times 10^{–9} ~\text{M}$.
For AgCl: AgCl precipitates when Q equals Ksp for AgCl ($1.6 \times 10^{–10}$). When [Cl] = 0.10 M:
\begin{align*} Q_{ sp }=K_{ sp }=\left[ \ce{Ag^{+}}\right]\left[ \ce{Cl^{-}}\right]=\left[ \ce{Ag^{+}}\right](0.10)&=1.6 \times 10^{-10} \[4pt] \left[ \ce{Ag^{+}}\right] &=\frac{1.6 \times 10^{-10}}{0.10} \[4pt] &=1.6 \times 10^{-9}~\text{M} \end{align*} \nonumber
AgCl begins to precipitate when [Ag+] is $1.6 \times 10^{–9} ~\text{M}$.
AgCl begins to precipitate at a lower [Ag+] than AgBr, so AgCl begins to precipitate first. Note the chloride ion concentration of the initial mixture was significantly greater than the bromide ion concentration, and so silver chloride precipitated first despite having a Ksp greater than that of silver bromide.
Exercise $11$
If silver nitrate solution is added to a solution which is 0.050 M in both Cl and Br ions, at what [Ag+] would precipitation begin, and what would be the formula of the precipitate?
Answer
$[\ce{Ag^{+}}] = 1.0 \times 10^{-11}~ \text{M}$; $\ce{AgBr}$ precipitates first
Common Ion Effect
Compared with pure water, the solubility of an ionic compound is less in aqueous solutions containing a common ion (one also produced by dissolution of the ionic compound). This is an example of a phenomenon known as the common ion effect, which is a consequence of the law of mass action that may be explained using Le Chatelier’s principle. Consider the dissolution of silver iodide:
$\ce{AgI(s) <=> Ag^{+}(aq) + I^{-}(aq)} \nonumber$
This solubility equilibrium may be shifted left by the addition of either silver(I) or iodide ions, resulting in the precipitation of AgI and lowered concentrations of dissolved Ag+ and I. In solutions that already contain either of these ions, less AgI may be dissolved than in solutions without these ions.
This effect may also be explained in terms of mass action as represented in the solubility product expression:
$K_{ sp }=\left[ \ce{Ag^{+}} \right]\left[ \ce{I^{-}}\right] \nonumber$
The mathematical product of silver(I) and iodide ion molarities is constant in an equilibrium mixture regardless of the source of the ions, and so an increase in one ion’s concentration must be balanced by a proportional decrease in the other.
Link to Learning
View this simulation to explore various aspects of the common ion effect.
Example $12$: Common Ion Effect on Solubility
What is the effect on the amount of solid Mg(OH)2 and the concentrations of Mg2+ and OH when each of the following are added to a saturated solution of Mg(OH)2?
1. MgCl2
2. KOH
3. NaNO3
4. Mg(OH)2
Solution
The solubility equilibrium is
$\ce{Mg(OH)2(s) <=> Mg^{2+}(aq) + 2 OH^{-}(aq)} \nonumber$
1. Adding a common ion, $\ce{Mg^{2+}}$, will increase the concentration of this ion and shift the solubility equilibrium to the left, decreasing the concentration of hydroxide ion and increasing the amount of undissolved magnesium hydroxide.
2. Adding a common ion, $\ce{OH^{-}}$, will increase the concentration of this ion and shift the solubility equilibrium to the left, decreasing the concentration of magnesium ion and increasing the amount of undissolved magnesium hydroxide.
3. The added compound does not contain a common ion, and no effect on the magnesium hydroxide solubility equilibrium is expected.
4. Adding more solid magnesium hydroxide will increase the amount of undissolved compound in the mixture. The solution is already saturated, though, so the concentrations of dissolved magnesium and hydroxide ions will remain the same. $Q=\left[ \ce{Mg^{2+}} \right]\left[ \ce{OH^{-}}\right]^2 \nonumber$
Thus, changing the amount of solid magnesium hydroxide in the mixture has no effect on the value of $Q$, and no shift is required to restore $Q$ to the value of the equilibrium constant.
Exercise $12$
What is the effect on the amount of solid $\ce{NiCO3}$ and the concentrations of $\ce{Ni^{2+}}$ and $\ce{CO3^{2-}}$ when each of the following are added to a saturated solution of $\ce{NiCO3}$
1. Ni(NO3)2
2. KClO4
3. NiCO3
4. K2CO3
Answer
(a) mass of NiCO3(s) increases, [Ni2+] increases, $\ce{CO3^{2-}}$ decreases; (b) no appreciable effect; (c) no effect except to increase the amount of solid NiCO3; (d) mass of NiCO3(s) increases, [Ni2+] decreases, $\ce{CO3^{2-}}$ increases;
Example $13$: Common Ion Effect
Calculate the molar solubility of cadmium sulfide ($\ce{CdS}$) in a 0.010-M solution of cadmium bromide ($\ce{CdBr2}$). The Ksp of $\ce{CdS}$ is $1.0 \times 10^{–28}$.
Solution
This calculation can be performed using the ICE approach:
$\ce{CdS(s) <=> Cd^{2+}(aq) + S^{2-}(aq)} \nonumber$
\begin{align*} K_{\text {sp }}=\left[ \ce{Cd^{2+}} \right]\left[ \ce{S^{2-}}\right] &=1.0 \times 10^{-28} \[4pt] (0.010+x)(x) &=1.0 \times 10^{-28} \end{align*} \nonumber
Because Ksp is very small, assume $x \ll 0.010$ and solve the simplified equation for $x$:
\begin{align*} (0.010)(x) &=1.0 \times 10^{-28}\[4pt] x&=1.0 \times 10^{-26}~ \text{M} \end{align*} \nonumber
The molar solubility of $\ce{CdS}$ in this solution is $1.0 \times 10^{–26} ~\text{M}$.
Exercise $13$
Calculate the molar solubility of aluminum hydroxide, Al(OH)3, in a 0.015-M solution of aluminum nitrate, Al(NO3)3. The Ksp of Al(OH)3 is $2 \times 10^{–32}$.
Answer
$4 \times 10^{–11} ~\text{M}$ | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/15%3A_Equilibria_of_Other_Reaction_Classes/15.01%3A_Precipitation_and_Dissolution.txt |
Learning Objectives
By the end of this section, you will be able to:
• Explain the Lewis model of acid-base chemistry
• Write equations for the formation of adducts and complex ions
In 1923, G. N. Lewis proposed a generalized definition of acid-base behavior in which acids and bases are identified by their ability to accept or to donate a pair of electrons and form a coordinate covalent bond.
A coordinate covalent bond (or dative bond) occurs when one of the atoms in the bond provides both bonding electrons. For example, a coordinate covalent bond occurs when a water molecule combines with a hydrogen ion to form a hydronium ion. A coordinate covalent bond also results when an ammonia molecule combines with a hydrogen ion to form an ammonium ion. Both of these equations are shown here.
Reactions involving the formation of coordinate covalent bonds are classified as Lewis acid-base chemistry. The species donating the electron pair that compose the bond is a Lewis base, the species accepting the electron pair is a Lewis acid, and the product of the reaction is a Lewis acid-base adduct. As the two examples above illustrate, Brønsted-Lowry acid-base reactions represent a subcategory of Lewis acid reactions, specifically, those in which the acid species is H+. A few examples involving other Lewis acids and bases are described below.
The boron atom in boron trifluoride, BF3, has only six electrons in its valence shell. Being short of the preferred octet, BF3 is a very good Lewis acid and reacts with many Lewis bases; a fluoride ion is the Lewis base in this reaction, donating one of its lone pairs:
In the following reaction, each of two ammonia molecules, Lewis bases, donates a pair of electrons to a silver ion, the Lewis acid:
Nonmetal oxides act as Lewis acids and react with oxide ions, Lewis bases, to form oxyanions:
Many Lewis acid-base reactions are displacement reactions in which one Lewis base displaces another Lewis base from an acid-base adduct, or in which one Lewis acid displaces another Lewis acid:
Another type of Lewis acid-base chemistry involves the formation of a complex ion (or a coordination complex) comprising a central atom, typically a transition metal cation, surrounded by ions or molecules called ligands. These ligands can be neutral molecules like H2O or NH3, or ions such as CN or OH. Often, the ligands act as Lewis bases, donating a pair of electrons to the central atom. These types of Lewis acid-base reactions are examples of a broad subdiscipline called coordination chemistry—the topic of another chapter in this text.
The equilibrium constant for the reaction of a metal ion with one or more ligands to form a coordination complex is called a formation constant (Kf) (sometimes called a stability constant). For example, the complex ion
is produced by the reaction
$\ce{Cu^{+}(aq) + 2 CN^{-}(aq) <=> Cu(CN)2^{-}(aq)} \nonumber$
The formation constant for this reaction is
$K_{ f }=\frac{\left[ \ce{Cu(CN)2^{-}} \right]}{\left[ \ce{Cu^{+}} \right]\left[ \ce{CN^{-}}\right]^2} \nonumber$
Alternatively, the reverse reaction (decomposition of the complex ion) can be considered, in which case the equilibrium constant is a dissociation constant (Kd). Per the relation between equilibrium constants for reciprocal reactions described, the dissociation constant is the mathematical inverse of the formation constant, Kd = Kf–1. A tabulation of formation constants is provided in Appendix K.
As an example of dissolution by complex ion formation, let us consider what happens when we add aqueous ammonia to a mixture of silver chloride and water. Silver chloride dissolves slightly in water, giving a small concentration of Ag+ ($[\ce{Ag^{+}}] = 1.3 \times 10^{–5} ~\text{M}$):
$\ce{AgCl(s) <=> Ag^{+}(aq) + Cl^{-}(aq)} \nonumber$
However, if NH3 is present in the water, the complex ion, $\ce{Ag(NH3)2^{+}}$ can form according to the equation:
$\ce{Ag^{+}(aq) + 2 NH3(aq) <=> Ag(NH3)2^{+}(aq)} \nonumber$
with
$K_{ f }=\dfrac{[ \ce{Ag(NH3)2^{+}} ]}{[ \ce{Ag^{+}} ][\ce{NH3}]^2}=1.7 \times 10^7 \nonumber$
The large size of this formation constant indicates that most of the free silver ions produced by the dissolution of AgCl combine with NH 3 to form $\ce{Ag(NH3)2^{+}}$. As a consequence, the concentration of silver ions, $\ce{[Ag^{+}]}$, is reduced, and the reaction quotient for the dissolution of silver chloride, [Ag + ][Cl ], falls below the solubility product of AgCl:
$Q=\left[ \ce{Ag^{+}} \right]\left[ \ce{Cl^{-}} \right]<K_{ sp } \nonumber$
More silver chloride then dissolves. If the concentration of ammonia is great enough, all of the silver chloride dissolves.
Example $1$: Dissociation of a Complex Ion
Calculate the concentration of the silver ion in a solution that initially is 0.10 M with respect to $\ce{Ag(NH3)2^{+}}$).
Solution
Applying the standard ICE approach to this reaction yields the following:
Substituting these equilibrium concentration terms into the Kf expression gives
\begin{align*} K_{ f } &=\dfrac{[ \ce{Ag(NH3)2^{+}} ]}{[ \ce{Ag^{+}} ][ \ce{NH3} ]^2} \[4pt] 1.7 \times 10^7 &=\frac{0.10-x}{(x)(2 x)^2} \end{align*} \nonumber
The very large equilibrium constant means the amount of the complex ion that will dissociate, x, will be very small. Assuming $x \ll 0.1$ permits simplifying the above equation:
\begin{align*} 1.7 \times 10^7 &=\frac{0.10}{(x)(2 x)^2} \[4pt] x^3&=\frac{0.10}{4\left(1.7 \times 10^7\right)}=1.5 \times 10^{-9} \[4pt] x&=\sqrt[3]{1.5 \times 10^{-9}}=1.1 \times 10^{-3} \end{align*} \nonumber
Because only 1.1% of the $\ce{Ag(NH3)2^{+}}$) dissociates into $\ce{Ag^{+}}$ and $\ce{NH3}$, the assumption that x is small is justified.
Using this value of x and the relations in the above ICE table allows calculation of all species’ equilibrium concentrations:
\begin{align*} [ \ce{Ag^{+}} ] &=0+x=1.1 \times 10^{-3}~\text{M} \[4pt] [ \ce{NH3}] &=0+2 x=2.2 \times 10^{-3} ~\text{M} \[4pt] [ \ce{Ag(NH3)2^{+}}] &=0.10-x=0.10-0.0011 \[4pt] &=0.099 ~\text{M} \end{align*} \nonumber
The concentration of free silver ion in the solution is 0.0011 M.
Exercise $1$
Calculate the silver ion concentration, [Ag+], of a solution prepared by dissolving 1.00 g of $\ce{AgNO3}$ and 10.0 g of $\ce{KCN}$ in sufficient water to make 1.00 L of solution. (Hint: Because Kf is very large, assume the reaction goes to completion then calculate the [Ag+] produced by dissociation of the complex.)
Answer
$2.9 \times 10^{–22} ~\text{M}$ | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/15%3A_Equilibria_of_Other_Reaction_Classes/15.02%3A_Lewis_Acids_and_Bases.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe examples of systems involving two (or more) coupled chemical equilibria
• Calculate reactant and product concentrations for coupled equilibrium systems
As discussed in preceding chapters on equilibrium, coupled equilibria involve two or more separate chemical reactions that share one or more reactants or products. This section of this chapter will address solubility equilibria coupled with acid-base and complex-formation reactions.
An environmentally relevant example illustrating the coupling of solubility and acid-base equilibria is the impact of ocean acidification on the health of the ocean’s coral reefs. These reefs are built upon skeletons of sparingly soluble calcium carbonate excreted by colonies of corals (small marine invertebrates). The relevant dissolution equilibrium is
$\ce{CaCO_3(s) <=> Ca^{2+}(aq) + CO_3^{-2}(aq)} \quad K_{\text {sp }}=8.7 \times 10^{-9} \nonumber$
Rising concentrations of atmospheric carbon dioxide contribute to an increased acidity of ocean waters due to the dissolution, hydrolysis, and acid ionization of carbon dioxide:
\begin{align*} \ce{CO_2(g) &<=> CO_2(aq)} \[4pt] \ce{CO_2(aq) + H2O(l) &<=> H_2 CO_3(aq)} \[4pt] \ce{H_2 CO_3(aq) + H2O(l) &<=> HCO_3^{-}(aq) + H3O^{+}(aq)} \quad K_{ a 1}=4.3 \times 10^{-7} \[4pt] \ce{HCO_3^{-}(aq) + H2O(l) &<=> CO_3^{2-}(aq) + H3O^{+}(aq)} \quad K_{ a 2}=4.7 \times 10^{-11} \end{align*}
Inspection of these equilibria shows the carbonate ion is involved in the calcium carbonate dissolution and the acid hydrolysis of bicarbonate ion. Combining the dissolution equation with the reverse of the acid hydrolysis equation yields
$\ce{CaCO_3(s) + H3O^{+}(aq) <=> Ca^{2+}(aq) + HCO_3^{-}(aq) + H2O(l)} \quad \quad K=K_{\text {sp }} / K_{ a 2}=180 \nonumber$
The equilibrium constant for this net reaction is much greater than the Ksp for calcium carbonate, indicating its solubility is markedly increased in acidic solutions. As rising carbon dioxide levels in the atmosphere increase the acidity of ocean waters, the calcium carbonate skeletons of coral reefs become more prone to dissolution and subsequently less healthy (Figure $1$).
Link to Learning
Learn more about ocean acidification and how it affects other marine creatures.
This site has detailed information about how ocean acidification specifically affects coral reefs.
The dramatic increase in solubility with increasing acidity described above for calcium carbonate is typical of salts containing basic anions (e.g., carbonate, fluoride, hydroxide, sulfide). Another familiar example is the formation of dental cavities in tooth enamel. The major mineral component of enamel is calcium hydroxyapatite (Figure $2$), a sparingly soluble ionic compound whose dissolution equilibrium is
$\ce{Ca5(PO4)3OH(s) <=> 5 Ca^{2+}(aq) +3 PO4^{3-}(aq) + OH^{-}(aq)} \nonumber$
This compound dissolved to yield two different basic ions: triprotic phosphate ions
\begin{align*} \ce{PO_4^{3-}(aq) + H3O^{+}(aq) &-> H_2 PO_4^{2-}(aq) + H2O(l)} \[4pt] \ce{H_2 PO_4^{2-}(aq) + H3O^{+}(aq) &-> H_2 PO_4^{-}(aq) + H2O(l)} \[4pt] \ce{H_2 PO_4^{-}(aq) + H3O^{+}(aq) &-> H_3 PO_4(aq) + H2O(l)} \end{align*}
and monoprotic hydroxide ions:
$\ce{OH^{-}(aq) + H3O^{+} -> 2 H2O} \nonumber$
Of the two basic productions, the hydroxide is, of course, by far the stronger base (it’s the strongest base that can exist in aqueous solution), and so it is the dominant factor providing the compound an acid-dependent solubility. Dental cavities form when the acid waste of bacteria growing on the surface of teeth hastens the dissolution of tooth enamel by reacting completely with the strong base hydroxide, shifting the hydroxyapatite solubility equilibrium to the right. Some toothpastes and mouth rinses contain added NaF or SnF2 that make enamel more acid resistant by replacing the strong base hydroxide with the weak base fluoride:
$\ce{NaF + Ca5(PO4)3OH <=> Ca5(PO4)3F + Na^{+} + OH^{-}} \nonumber$
The weak base fluoride ion reacts only partially with the bacterial acid waste, resulting in a less extensive shift in the solubility equilibrium and an increased resistance to acid dissolution. See the Chemistry in Everyday Life feature on the role of fluoride in preventing tooth decay for more information.
Chemistry in Everyday Life: Role of Fluoride in Preventing Tooth Decay
As we saw previously, fluoride ions help protect our teeth by reacting with hydroxylapatite to form fluorapatite, Ca5(PO4)3F. Since it lacks a hydroxide ion, fluorapatite is more resistant to attacks by acids in our mouths and is thus less soluble, protecting our teeth. Scientists discovered that naturally fluorinated water could be beneficial to your teeth, and so it became common practice to add fluoride to drinking water. Toothpastes and mouthwashes also contain amounts of fluoride (Figure $3$).
Unfortunately, excess fluoride can negate its advantages. Natural sources of drinking water in various parts of the world have varying concentrations of fluoride, and places where that concentration is high are prone to certain health risks when there is no other source of drinking water. The most serious side effect of excess fluoride is the bone disease, skeletal fluorosis. When excess fluoride is in the body, it can cause the joints to stiffen and the bones to thicken. It can severely impact mobility and can negatively affect the thyroid gland. Skeletal fluorosis is a condition that over 2.7 million people suffer from across the world. So while fluoride can protect our teeth from decay, the US Environmental Protection Agency sets a maximum level of 4 ppm (4 mg/L) of fluoride in drinking water in the US. Fluoride levels in water are not regulated in all countries, so fluorosis is a problem in areas with high levels of fluoride in the groundwater.
The solubility of ionic compounds may also be increased when dissolution is coupled to the formation of a complex ion. For example, aluminum hydroxide dissolves in a solution of sodium hydroxide or another strong base because of the formation of the complex ion
The equations for the dissolution of aluminum hydroxide, the formation of the complex ion, and the combined (net) equation are shown below. As indicated by the relatively large value of K for the net reaction, coupling complex formation with dissolution drastically increases the solubility of Al(OH)3.
\begin{align*} \ce{Al ( OH )_3(s) &<=> Al^{3+}(aq) +3 OH^{-}(aq)} & K_{ sp }=2 \times 10^{-32} \[4pt] \ce{Al^{3+}(aq) +4 OH^{-}(aq) &<=> Al ( OH )_4^{-}(aq) }& K_{ f }=1.1 \times 10^{33} \[4pt] \text { Net: } \ce{Al ( OH )_3(s) + OH^{-}(aq) &<=> Al ( OH )_4^{-}(aq)} & K=K_{ sp } K_{ f }=22 \end{align*} \nonumber
Example $1$: Increased Solubility in Acidic Solutions
Compute and compare the molar solubilities for aluminum hydroxide, Al(OH)3, dissolved in (a) pure water and (b) a buffer containing 0.100 M acetic acid and 0.100 M sodium acetate.
Solution
(a) The molar solubility of aluminum hydroxide in water is computed considering the dissolution equilibrium only as demonstrated in several previous examples:
$\ce{Al ( OH )_3(s) <=> Al^{3+}(aq) +3 OH^{-}(aq)} \quad \quad K_{ sp }=2 \times 10^{-32} \nonumber] \[\text{molar solubility in water} =\left[ \ce{Al^{3+}} \right]= \left(\dfrac{2 \times 10^{-32}}{27}\right)^{1/4} =5 \times 10^{-9} ~\text{M} \nonumber$
(b) The concentration of hydroxide ion of the buffered solution is conveniently calculated by the Henderson-Hasselbalch equation:
\begin{align*} \text{pH} &= pK_{ a }+\log \dfrac{\left[ \ce{CH3COO^{-}} \right] }{ \left[\ce{CH3COOH} \right]} \[4pt] &=4.74+\log \left(\dfrac{0.100}{0.100}\right)=4.74 \end{align*} \nonumber
At this pH, the concentration of hydroxide ion is
\begin{align*} & \text{pOH} =14.00-4.74=9.26 \[4pt] & {\left[ \ce{OH^{-}} \right]=10^{-9.26}=5.5 \times 10^{-10}} \end{align*} \nonumber
The solubility of $\ce{Al(OH)3}$ in this buffer is then calculated from its solubility product expression:
$K_{ sp }=\left[ \ce{Al^{3+}}\right]\left[ \ce{OH^{-}}\right]^3 \nonumber$
Via
\begin{align*} \text { molar solubility in buffer }&=\left[ \ce{Al^{3+}} \right] \[4pt] &= \dfrac{K_{ sp } }{\left[ \ce{OH^{-}} \right]^3} \[4pt] &=\dfrac{2 \times 10^{-32}}{(5.5 \times 10^{-10})^3} \[4pt] &= 1.2 \times 10^{-4}~\text{M} \end{align*}
Compared to pure water, the solubility of aluminum hydroxide in this mildly acidic buffer is approximately ten million times greater (though still relatively low).
Exercise $1$
What is the solubility of aluminum hydroxide in a buffer comprised of 0.100 M formic acid and 0.100 M sodium formate?
Answer
0.1 M
Example $2$: Multiple Equilibria
Unexposed silver halides are removed from photographic film when they react with sodium thiosulfate (Na2S2O3, called hypo) to form the complex ion $\ce{Ag(S2O3)2^{3-}}$ with $K_f = 4.7 \times 10^{13}$.
What mass of Na2S2O3 is required to prepare 1.00 L of a solution that will dissolve 1.00 g of AgBr by the formation of $\ce{Ag(S2O3)2^{3-}}$
Solution
Two equilibria are involved when silver bromide dissolves in an aqueous thiosulfate solution containing the ion:
• dissolution: $\ce{AgBr(s) <=> Ag^{+}(aq) + Br^{-}(aq)} \quad \quad K_{ sp }=5.0 \times 10^{-13} \nonumber$
• complexation: $\nonumber \ce{Ag^{+}(aq) +2 S2O3^{2-}(aq) <=> Ag ( S2O3)_2^{3-}(aq)}\quad \quad K_{ f }=4.7 \times 10^{13} \nonumber$
Combining these two equilibrium equations yields
$\ce{AgBr(s) +2S2O3^{2-}(aq) <=> Ag(S2O3)2^{3-}(aq) + Br^{-}(aq)} \nonumber$
with
$K=\frac{\left[ \ce{Ag(S2O3)2^3} \right][ \ce{Br^{-}} ]}{\left[ \ce{S2O3^{-2}} \right]^2}=K_{ sp } K_{ f }=24 \nonumber$
The concentration of bromide resulting from dissolution of 1.00 g of AgBr in 1.00 L of solution is
$\left[ Br^{-}\right]=\frac{1.00 ~\text{g} ~\ce{AgBr} \times \dfrac{1 ~ \text{mol} ~\ce{AgBr} }{187.77\, \text{g / mol} } \times \dfrac{1 ~ \text{mol} ~\ce{Br^{-}}}{1~\text{mol} ~\ce{AgBr^{-}}}}{1.00~\text{L} }=0.00532~\text{M} \nonumber$
The stoichiometry of the dissolution equilibrium indicates the same concentration of aqueous silver ion will result, 0.00532 M, and the very large value of ensures that essentially all the dissolved silver ion will be complexed by thiosulfate ion:
$\left[ \ce{Ag(S2O3)2^{3-}}\right]=0.00532 ~\text{M} \nonumber$
Rearranging the K expression for the combined equilibrium equations and solving for the concentration of thiosulfate ion yields
$\left[ \ce{S2O3^{2-}} \right]=\frac{\left[ \ce{Ag(S2O3)2^{3-}} \right]\left[ \ce{Br^{-}} \right]}{K}=\frac{(0.00532 M)(0.00532~\text{M})}{24}=0.0011 ~ \text{M} \nonumber$
Finally, the total mass of $\ce{Na2S2O3}$ required to provide enough thiosulfate to yield the concentrations cited above can be calculated.
Mass of $\ce{Na2S2O3}$ required to yield 0.00532 M $\ce{Ag(2S2O3)2^{3-}}$
$0.00532 \dfrac{ \text{mol} ~\ce{Ag(S2O3)2^{3-}}}{1.00~\text{L}} \times \dfrac{2~\text{mol}~\ce{S2O3^{2-}}}{1~\text{mol} ~\ce{Ag(S2O3)2^{3-}}} \times \dfrac{1~\text{mol}~\ce{Na2S2O3}}{1 ~\text{mol}~\ce{S2O3^{2-}}} \times \dfrac{158.1 ~ \text{g}~\ce{Na2S2O3}}{1 ~\text{mol}~\ce{Na2S2O3}}=1.68~\text{g} \nonumber$
Mass of $\ce{Na2S2O3}$ required to yield 0.00110 M $\ce{S2O3^{2-}}$
$0.0011 \frac{ \text{mol} ~\ce{S2O3^{2-}}}{1.00~\text{L} } \times \frac{1~\text{mol} ~ \ce{S2O3^{2-}}}{1 ~ \text{mol} ~\ce{Na2S2O3}} \times \frac{158.1 ~\text{g}~\ce{Na2S2O3}}{1~\text{mol} ~\ce{Na2S2O3}}=0.17~\text{g} \nonumber$
The mass of required to dissolve 1.00 g of AgBr in 1.00 L of water is thus 1.68 g + 0.17 g = 1.85 g
Exercise $2$
AgCl(s), silver chloride, has a very low solubility:
$\ce{AgCl (s) <=> Ag^{+}(aq) + Cl^{-}(aq)} \quad\quad K_{ so }=1.6 \times 10^{-10} \nonumber$
Adding ammonia significantly increases the solubility of AgCl because a complex ion is formed:
$\ce{Ag^{+}(aq) +2 NH_3(aq) <=> Ag(NH3)2^{+}(aq)} \quad\quad K_{ f }=1.7 \times 10^7 \nonumber$
What mass of $\ce{NH3}$ is required to prepare 1.00 L of solution that will dissolve 2.00 g of AgCl by formation of $\ce{Ag(NH3)^{2+}}$?
Answer
1.00 L of a solution prepared with 4.81 g NH3 dissolves 2.0 g of $\ce{AgCl}$. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/15%3A_Equilibria_of_Other_Reaction_Classes/15.03%3A_Coupled_Equilibria.txt |
Example and Directions
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Glossary Entries
Word(s)DefinitionImageCaptionLinkSource
common ion effecteffect 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
complex ionion consisting of a central atom surrounding molecules or ions called ligands via coordinate covalent bonds
coordinate covalent bond(also, dative bond) covalent bond in which both electrons originated from the same atom
coupled equilibriasystem characterized the simultaneous establishment of two or more equilibrium reactions sharing one or more reactant or product
dissociation constant(Kd) equilibrium constant for the decomposition of a complex ion into its components
formation constant(Kf) (also, stability constant) equilibrium constant for the formation of a complex ion from its components
Lewis acidany species that can accept a pair of electrons and form a coordinate covalent bond
Lewis acid-base adductcompound or ion that contains a coordinate covalent bond between a Lewis acid and a Lewis base
Lewis acid-base chemistryreactions involving the formation of coordinate covalent bonds
Lewis baseany species that can donate a pair of electrons and form a coordinate covalent bond
ligandmolecule or ion acting as a Lewis base in complex ion formation; bonds to the central atom of the complex
molar solubilitysolubility of a compound expressed in units of moles per liter (mol/L)
selective precipitationprocess in which ions are separated using differences in their solubility with a given precipitating reagent
solubility product constant (Ksp)equilibrium constant for the dissolution of an ionic compound
15.06: Summary
The equilibrium constant for an equilibrium involving the precipitation or dissolution of a slightly soluble ionic solid is called the solubility product, Ksp, of the solid. For a heterogeneous equilibrium involving the slightly soluble solid MpXq and its ions Mm+ and Xn–:
the solubility product expression is:
The solubility product of a slightly soluble electrolyte can be calculated from its solubility; conversely, its solubility can be calculated from its Ksp, provided the only significant reaction that occurs when the solid dissolves is the formation of its ions.
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 Lewis acid is a species that can accept an electron pair, whereas a Lewis base has an electron pair available for donation to a Lewis acid. Complex ions are examples of Lewis acid-base adducts and comprise central metal atoms or ions acting as Lewis acids bonded to molecules or ions called ligands that act as Lewis bases. The equilibrium constant for the reaction between a metal ion and ligands produces a complex ion called a formation constant; for the reverse reaction, it is called a dissociation constant.
Systems involving two or more chemical equilibria that share one or more reactant or product are called coupled equilibria. Common examples of coupled equilibria include the increased solubility of some compounds in acidic solutions (coupled dissolution and neutralization equilibria) and in solutions containing ligands (coupled dissolution and complex formation). The equilibrium tools from other chapters may be applied to describe and perform calculations on these systems. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/15%3A_Equilibria_of_Other_Reaction_Classes/15.04%3A_Key_Terms.txt |
1.
Complete the changes in concentrations for each of the following reactions:
(a)
(b)
(c)
(d)
(e)
2.
Complete the changes in concentrations for each of the following reactions:
(a)
(b
(c)
(d)
(e)
3.
How do the concentrations of Ag+ and in a saturated solution above 1.0 g of solid Ag2CrO4 change when 100 g of solid Ag2CrO4 is added to the system? Explain.
4.
How do the concentrations of Pb2+ and S2– change when K2S is added to a saturated solution of PbS?
5.
What additional information do we need to answer the following question: How is the equilibrium of solid silver bromide with a saturated solution of its ions affected when the temperature is raised?
6.
Which of the following slightly soluble compounds has a solubility greater than that calculated from its solubility product because of hydrolysis of the anion present: CoSO3, CuI, PbCO3, PbCl2, Tl2S, KClO4?
7.
Which of the following slightly soluble compounds has a solubility greater than that calculated from its solubility product because of hydrolysis of the anion present: AgCl, BaSO4, CaF2, Hg2I2, MnCO3, and ZnS?
8.
Write the ionic equation for dissolution and the solubility product (Ksp) expression for each of the following slightly soluble ionic compounds:
1. PbCl2
2. Ag2S
3. Sr3(PO4)2
4. SrSO4
9.
Write the ionic equation for the dissolution and the Ksp expression for each of the following slightly soluble ionic compounds:
1. LaF3
2. CaCO3
3. Ag2SO4
4. Pb(OH)2
10.
The Handbook of Chemistry and Physics gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each.
1. BaSiF6, 0.026 g/100 mL (contains ions)
2. Ce(IO3)4, 1.5 10–2 g/100 mL
3. Gd2(SO4)3, 3.98 g/100 mL
4. (NH4)2PtBr6, 0.59 g/100 mL (contains ions)
11.
The Handbook of Chemistry and Physics gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each.
1. BaSeO4, 0.0118 g/100 mL
2. Ba(BrO3)2·H2O, 0.30 g/100 mL
3. NH4MgAsO4·6H2O, 0.038 g/100 mL
4. La2(MoO4)3, 0.00179 g/100 mL
12.
Use solubility products and predict which of the following salts is the most soluble, in terms of moles per liter, in pure water: CaF2, Hg2Cl2, PbI2, or Sn(OH)2.
13.
Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product:
1. KHC4H4O6
2. PbI2
3. Ag4[Fe(CN)6], a salt containing the ion
4. Hg2I2
14.
Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product:
1. Ag2SO4
2. PbBr2
3. AgI
4. CaC2O4·H2O
15.
Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected.
1. AgCl(s) in 0.025 M NaCl
2. CaF2(s) in 0.00133 M KF
3. Ag2SO4(s) in 0.500 L of a solution containing 19.50 g of K2SO4
4. Zn(OH)2(s) in a solution buffered at a pH of 11.45
16.
Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected.
1. TlCl(s) in 1.250 M HCl
2. PbI2(s) in 0.0355 M CaI2
3. Ag2CrO4(s) in 0.225 L of a solution containing 0.856 g of K2CrO4
4. Cd(OH)2(s) in a solution buffered at a pH of 10.995
17.
Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that it is not appropriate to neglect the changes in the initial concentrations of the common ions.
1. TlCl(s) in 0.025 M TlNO3
2. BaF2(s) in 0.0313 M KF
3. MgC2O4 in 2.250 L of a solution containing 8.156 g of Mg(NO3)2
4. Ca(OH)2(s) in an unbuffered solution initially with a pH of 12.700
18.
Explain why the changes in concentrations of the common ions in Exercise 15.17 can be neglected.
19.
Explain why the changes in concentrations of the common ions in Exercise 15.18 cannot be neglected.
20.
Calculate the solubility of aluminum hydroxide, Al(OH)3, in a solution buffered at pH 11.00.
21.
Refer to Appendix J for solubility products for calcium salts. Determine which of the calcium salts listed is most soluble in moles per liter and which is most soluble in grams per liter.
22.
Most barium compounds are very poisonous; however, barium sulfate is often administered internally as an aid in the X-ray examination of the lower intestinal tract (Figure 15.4). This use of BaSO4 is possible because of its low solubility. Calculate the molar solubility of BaSO4 and the mass of barium present in 1.00 L of water saturated with BaSO4.
23.
Public Health Service standards for drinking water set a maximum of 250 mg/L (2.60 10–3 M) of because of its cathartic action (it is a laxative). Does natural water that is saturated with CaSO4 (“gyp” water) as a result or passing through soil containing gypsum, CaSO4·2H2O, meet these standards? What is the concentration of in such water?
24.
Perform the following calculations:
1. Calculate [Ag+] in a saturated aqueous solution of AgBr.
2. What will [Ag+] be when enough KBr has been added to make [Br] = 0.050 M?
3. What will [Br] be when enough AgNO3 has been added to make [Ag+] = 0.020 M?
25.
The solubility product of CaSO4·2H2O is 2.4 10–5. What mass of this salt will dissolve in 1.0 L of 0.010 M
26.
Assuming that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the following (see Appendix J for solubility products).
1. TlCl
2. BaF2
3. Ag2CrO4
4. CaC2O4·H2O
5. the mineral anglesite, PbSO4
27.
Assuming that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the following (see Appendix J for solubility products):
1. AgI
2. Ag2SO4
3. Mn(OH)2
4. Sr(OH)2·8H2O
5. the mineral brucite, Mg(OH)2
28.
The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate Ksp for each of the slightly soluble solids indicated:
1. AgBr: [Ag+] = 5.7 10–7 M, [Br] = 5.7 10–7 M
2. CaCO3: [Ca2+] = 5.3 10–3 M, = 9.0 10–7 M
3. PbF2: [Pb2+] = 2.1 10–3 M, [F] = 4.2 10–3 M
4. Ag2CrO4: [Ag+] = 5.3 10–5 M, 3.2 10–3 M
5. InF3: [In3+] = 2.3 10–3 M, [F] = 7.0 10–3 M
29.
The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate Ksp for each of the slightly soluble solids indicated:
1. TlCl: [Tl+] = 1.21 10–2 M, [Cl] = 1.2 10–2 M
2. Ce(IO3)4: [Ce4+] = 1.8 10–4 M, = 2.6 10–13 M
3. Gd2(SO4)3: [Gd3+] = 0.132 M, = 0.198 M
4. Ag2SO4: [Ag+] = 2.40 10–2 M, = 2.05 10–2 M
5. BaSO4: [Ba2+] = 0.500 M, = 4.6 10−8 M
30.
Which of the following compounds precipitates from a solution that has the concentrations indicated? (See Appendix J for Ksp values.)
1. KClO4: [K+] = 0.01 M, = 0.01 M
2. K2PtCl6: [K+] = 0.01 M, = 0.01 M
3. PbI2: [Pb2+] = 0.003 M, [I] = 1.3 10–3 M
4. Ag2S: [Ag+] = 1 10–10 M, [S2–] = 1 10–13 M
31.
Which of the following compounds precipitates from a solution that has the concentrations indicated? (See Appendix J for Ksp values.)
1. CaCO3: [Ca2+] = 0.003 M, = 0.003 M
2. Co(OH)2: [Co2+] = 0.01 M, [OH] = 1 10–7 M
3. CaHPO4: [Ca2+] = 0.01 M, = 2 10–6 M
4. Pb3(PO4)2: [Pb2+] = 0.01 M, = 1 10–13 M
32.
Calculate the concentration of Tl+ when TlCl just begins to precipitate from a solution that is 0.0250 M in Cl.
33.
Calculate the concentration of sulfate ion when BaSO4 just begins to precipitate from a solution that is 0.0758 M in Ba2+.
34.
Calculate the concentration of Sr2+ when SrCrO4 starts to precipitate from a solution that is 0.0025 M in CrO42–.
35.
Calculate the concentration of when Ag3PO4 starts to precipitate from a solution that is 0.0125 M in Ag+.
36.
Calculate the concentration of F required to begin precipitation of CaF2 in a solution that is 0.010 M in Ca2+.
37.
Calculate the concentration of Ag+ required to begin precipitation of Ag2CO3 in a solution that is 2.50 10–6 M in
38.
What [Ag+] is required to reduce to 8.2 10–4 M by precipitation of Ag2CO3?
39.
What [F] is required to reduce [Ca2+] to 1.0 10–4 M by precipitation of CaF2?
40.
A volume of 0.800 L of a 2 10–4-M Ba(NO3)2 solution is added to 0.200 L of 5 10–4 M Li2SO4. Does BaSO4 precipitate? Explain your answer.
41.
Perform these calculations for nickel(II) carbonate. (a) With what volume of water must a precipitate containing NiCO3 be washed to dissolve 0.100 g of this compound? Assume that the wash water becomes saturated with NiCO3 (Ksp = 1.36 10–7).
(b) If the NiCO3 were a contaminant in a sample of CoCO3 (Ksp = 1.0 10–12), what mass of CoCO3 would have been lost? Keep in mind that both NiCO3 and CoCO3 dissolve in the same solution.
42.
Iron concentrations greater than 5.4 10–6 M in water used for laundry purposes can cause staining. What [OH] is required to reduce [Fe2+] to this level by precipitation of Fe(OH)2?
43.
A solution is 0.010 M in both Cu2+ and Cd2+. What percentage of Cd2+ remains in the solution when 99.9% of the Cu2+ has been precipitated as CuS by adding sulfide?
44.
A solution is 0.15 M in both Pb2+ and Ag+. If Cl is added to this solution, what is [Ag+] when PbCl2 begins to precipitate?
45.
What reagent might be used to separate the ions in each of the following mixtures, which are 0.1 M with respect to each ion? In some cases it may be necessary to control the pH. (Hint: Consider the Ksp values given in Appendix J.)
1. (f) and OH
46.
A solution contains 1.0 10–5 mol of KBr and 0.10 mol of KCl per liter. AgNO3 is gradually added to this solution. Which forms first, solid AgBr or solid AgCl?
47.
A solution contains 1.0 10–2 mol of KI and 0.10 mol of KCl per liter. AgNO3 is gradually added to this solution. Which forms first, solid AgI or solid AgCl?
48.
The calcium ions in human blood serum are necessary for coagulation (Figure 15.5). Potassium oxalate, K2C2O4, is used as an anticoagulant when a blood sample is drawn for laboratory tests because it removes the calcium as a precipitate of CaC2O4·H2O. It is necessary to remove all but 1.0% of the Ca2+ in serum in order to prevent coagulation. If normal blood serum with a buffered pH of 7.40 contains 9.5 mg of Ca2+ per 100 mL of serum, what mass of K2C2O4 is required to prevent the coagulation of a 10 mL blood sample that is 55% serum by volume? (All volumes are accurate to two significant figures. Note that the volume of serum in a 10-mL blood sample is 5.5 mL. Assume that the Ksp value for CaC2O4 in serum is the same as in water.)
49.
About 50% of urinary calculi (kidney stones) consist of calcium phosphate, Ca3(PO4)2. The normal mid range calcium content excreted in the urine is 0.10 g of Ca2+ per day. The normal mid range amount of urine passed may be taken as 1.4 L per day. What is the maximum concentration of phosphate ion that urine can contain before a calculus begins to form?
50.
The pH of normal urine is 6.30, and the total phosphate concentration
51.
Magnesium metal (a component of alloys used in aircraft and a reducing agent used in the production of uranium, titanium, and other active metals) is isolated from sea water by the following sequence of reactions:
Sea water has a density of 1.026 g/cm3 and contains 1272 parts per million of magnesium as Mg2+(aq) by mass. What mass, in kilograms, of Ca(OH)2 is required to precipitate 99.9% of the magnesium in 1.00 103 L of sea water?
52.
Hydrogen sulfide is bubbled into a solution that is 0.10 M in both Pb2+ and Fe2+ and 0.30 M in HCl. After the solution has come to equilibrium it is saturated with H2S ([H2S] = 0.10 M). What concentrations of Pb2+ and Fe2+ remain in the solution? For a saturated solution of H2S we can use the equilibrium:
(Hint: The changes as metal sulfides precipitate.)
53.
Perform the following calculations involving concentrations of iodate ions:
1. The iodate ion concentration of a saturated solution of La(IO3)3 was found to be 3.1 10–3 mol/L. Find the Ksp.
2. Find the concentration of iodate ions in a saturated solution of Cu(IO3)2 (Ksp = 7.4 10–8).
54.
Calculate the molar solubility of AgBr in 0.035 M NaBr (Ksp = 5 10–13).
55.
How many grams of Pb(OH)2 will dissolve in 500 mL of a 0.050-M PbCl2 solution (Ksp = 1.2 10–15)?
56.
Use the simulation from the earlier Link to Learning to complete the following exercise. Using 0.01 g CaF2, give the Ksp values found in a 0.2-M solution of each of the salts. Discuss why the values change as you change soluble salts.
57.
How many grams of Milk of Magnesia, Mg(OH)2 (s) (58.3 g/mol), would be soluble in 200 mL of water. Ksp = 7.1 10–12. Include the ionic reaction and the expression for Ksp in your answer. (Kw = 1 10–14 = [H3O+][OH])
58.
Two hypothetical salts, LM2 and LQ, have the same molar solubility in H2O. If Ksp for LM2 is 3.20 10–5, what is the Ksp value for LQ?
59.
The carbonate ion concentration is gradually increased in a solution containing equal concentrations of the divalent cations of magnesium, calcium, strontium, barium, and manganese. Which of the following carbonates will precipitate first? Which will precipitate last? Explain.
60.
How many grams of Zn(CN)2(s) (117.44 g/mol) would be soluble in 100 mL of H2O? Include the balanced reaction and the expression for Ksp in your answer. The Ksp value for Zn(CN)2(s) is 3.0 10–16.
61.
Even though Ca(OH)2 is an inexpensive base, its limited solubility restricts its use. What is the pH of a saturated solution of Ca(OH)2?
62.
Under what circumstances, if any, does a sample of solid AgCl completely dissolve in pure water?
63.
Explain why the addition of NH3 or HNO3 to a saturated solution of Ag2CO3 in contact with solid Ag2CO3 increases the solubility of the solid.
64.
Calculate the cadmium ion concentration, [Cd2+], in a solution prepared by mixing 0.100 L of 0.0100 M Cd(NO3)2 with 0.150 L of 0.100 NH3(aq).
65.
Explain why addition of NH3 or HNO3 to a saturated solution of Cu(OH)2 in contact with solid Cu(OH)2 increases the solubility of the solid.
66.
Sometimes equilibria for complex ions are described in terms of dissociation constants, Kd. For the complex ion the dissociation reaction is:
and
Calculate the value of the formation constant, Kf, for
67.
Using the value of the formation constant for the complex ion calculate the dissociation constant.
68.
Using the dissociation constant, Kd = 7.8 10–18, calculate the equilibrium concentrations of Cd2+ and CN in a 0.250-M solution of
69.
Using the dissociation constant, Kd = 3.4 10–15, calculate the equilibrium concentrations of Zn2+ and OH in a 0.0465-M solution of
70.
Using the dissociation constant, Kd = 2.2 10–34, calculate the equilibrium concentrations of Co3+ and NH3 in a 0.500-M solution of
71.
Using the dissociation constant, Kd = 1 10–44, calculate the equilibrium concentrations of Fe3+ and CN in a 0.333 M solution of
72.
Calculate the mass of potassium cyanide ion that must be added to 100 mL of solution to dissolve 2.0 10–2 mol of silver cyanide, AgCN.
73.
Calculate the minimum concentration of ammonia needed in 1.0 L of solution to dissolve 3.0 10–3 mol of silver bromide.
74.
A roll of 35-mm black and white photographic film contains about 0.27 g of unexposed AgBr before developing. What mass of Na2S2O3·5H2O (sodium thiosulfate pentahydrate or hypo) in 1.0 L of developer is required to dissolve the AgBr as (Kf = 4.7 1013)?
75.
We have seen an introductory definition of an acid: An acid is a compound that reacts with water and increases the amount of hydronium ion present. In the chapter on acids and bases, we saw two more definitions of acids: a compound that donates a proton (a hydrogen ion, H+) to another compound is called a Brønsted-Lowry acid, and a Lewis acid is any species that can accept a pair of electrons. Explain why the introductory definition is a macroscopic definition, while the Brønsted-Lowry definition and the Lewis definition are microscopic definitions.
76.
Write the Lewis structures of the reactants and product of each of the following equations, and identify the Lewis acid and the Lewis base in each:
1. (use Al-Cl single bonds)
77.
Write the Lewis structures of the reactants and product of each of the following equations, and identify the Lewis acid and the Lewis base in each:
78.
Using Lewis structures, write balanced equations for the following reactions:
79.
Calculate in a solution prepared by adding 0.0200 mol of NaCl to 0.250 L of a 0.100-M HgCl2 solution.
80.
In a titration of cyanide ion, 28.72 mL of 0.0100 M AgNO3 is added before precipitation begins. [The reaction of Ag+ with CN goes to completion, producing the complex.] Precipitation of solid AgCN takes place when excess Ag+ is added to the solution, above the amount needed to complete the formation of How many grams of NaCN were in the original sample?
81.
What are the concentrations of Ag+, CN, and in a saturated solution of AgCN?
82.
In dilute aqueous solution HF acts as a weak acid. However, pure liquid HF (boiling point = 19.5 °C) is a strong acid. In liquid HF, HNO3 acts like a base and accepts protons. The acidity of liquid HF can be increased by adding one of several inorganic fluorides that are Lewis acids and accept F ion (for example, BF3 or SbF5). Write balanced chemical equations for the reaction of pure HNO3 with pure HF and of pure HF with BF3.
83.
The simplest amino acid is glycine, H2NCH2CO2H. The common feature of amino acids is that they contain the functional groups: an amine group, –NH2, and a carboxylic acid group, –CO2H. An amino acid can function as either an acid or a base. For glycine, the acid strength of the carboxyl group is about the same as that of acetic acid, CH3CO2H, and the base strength of the amino group is slightly greater than that of ammonia, NH3.
1. Write the Lewis structures of the ions that form when glycine is dissolved in 1 M HCl and in 1 M KOH.
2. Write the Lewis structure of glycine when this amino acid is dissolved in water. (Hint: Consider the relative base strengths of the –NH2 and groups.)
84.
Boric acid, H3BO3, is not a Brønsted-Lowry acid but a Lewis acid.
1. Write an equation for its reaction with water.
2. Predict the shape of the anion thus formed.
3. What is the hybridization on the boron consistent with the shape you have predicted?
85.
A saturated solution of a slightly soluble electrolyte in contact with some of the solid electrolyte is said to be a system in equilibrium. Explain. Why is such a system called a heterogeneous equilibrium?
86.
Calculate the equilibrium concentration of Ni2+ in a 1.0-M solution [Ni(NH3)6](NO3)2.
87.
Calculate the equilibrium concentration of Zn2+ in a 0.30-M solution of
88.
Calculate the equilibrium concentration of Cu2+ in a solution initially with 0.050 M Cu2+ and 1.00 M NH3.
89.
Calculate the equilibrium concentration of Zn2+ in a solution initially with 0.150 M Zn2+ and 2.50 M CN.
90.
Calculate the Fe3+ equilibrium concentration when 0.0888 mole of K3[Fe(CN)6] is added to a solution with 0.0.00010 M CN.
91.
Calculate the Co2+ equilibrium concentration when 0.010 mole of [Co(NH3)6](NO3)2 is added to a solution with 0.25 M NH3. Assume the volume is 1.00 L.
92.
Calculate the molar solubility of Sn(OH)2 in a buffer solution containing equal concentrations of NH3 and
93.
Calculate the molar solubility of Al(OH)3 in a buffer solution with 0.100 M NH3 and 0.400 M
94.
What is the molar solubility of CaF2 in a 0.100-M solution of HF? Ka for HF = 6.4 10–4.
95.
What is the molar solubility of BaSO4 in a 0.250-M solution of NaHSO4? Ka for = 1.2 10–2.
96.
What is the molar solubility of Tl(OH)3 in a 0.10-M solution of NH3?
97.
What is the molar solubility of Pb(OH)2 in a 0.138-M solution of CH3NH2?
98.
A solution of 0.075 M CoBr2 is saturated with H2S ([H2S] = 0.10 M). What is the minimum pH at which CoS begins to precipitate?
99.
A 0.125-M solution of Mn(NO3)2 is saturated with H2S ([H2S] = 0.10 M). At what pH does MnS begin to precipitate?
100.
Both AgCl and AgI dissolve in NH3.
1. What mass of AgI dissolves in 1.0 L of 1.0 M NH3?
2. What mass of AgCl dissolves in 1.0 L of 1.0 M NH3?
101.
The following question is taken from a Chemistry Advanced Placement Examination and is used with the permission of the Educational Testing Service.
Solve the following problem:
In a saturated solution of MgF2 at 18 °C, the concentration of Mg2+ is 1.21 10–3 M. The equilibrium is represented by the preceding equation.
1. Write the expression for the solubility-product constant, Ksp, and calculate its value at 18 °C.
2. Calculate the equilibrium concentration of Mg2+ in 1.000 L of saturated MgF2 solution at 18 °C to which 0.100 mol of solid KF has been added. The KF dissolves completely. Assume the volume change is negligible.
3. Predict whether a precipitate of MgF2 will form when 100.0 mL of a 3.00 10–3-M solution of Mg(NO3)2 is mixed with 200.0 mL of a 2.00 10–3-M solution of NaF at 18 °C. Show the calculations to support your prediction.
4. At 27 °C the concentration of Mg2+ in a saturated solution of MgF2 is 1.17 10–3 M. Is the dissolving of MgF2 in water an endothermic or an exothermic process? Give an explanation to support your conclusion.
102.
Which of the following compounds, when dissolved in a 0.01-M solution of HClO4, has a solubility greater than in pure water: CuCl, CaCO3, MnS, PbBr2, CaF2? Explain your answer.
103.
Which of the following compounds, when dissolved in a 0.01-M solution of HClO4, has a solubility greater than in pure water: AgBr, BaF2, Ca3(PO4)2, ZnS, PbI2? Explain your answer.
104.
What is the effect on the amount of solid Mg(OH)2 that dissolves and the concentrations of Mg2+ and OH when each of the following are added to a mixture of solid Mg(OH)2 and water at equilibrium?
1. MgCl2
2. KOH
3. HClO4
4. NaNO3
5. Mg(OH)2
105.
What is the effect on the amount of CaHPO4 that dissolves and the concentrations of Ca2+ and when each of the following are added to a mixture of solid CaHPO4 and water at equilibrium?
1. CaCl2
2. HCl
3. KClO4
4. NaOH
5. CaHPO4
106.
Identify all chemical species present in an aqueous solution of Ca3(PO4)2 and list these species in decreasing order of their concentrations. (Hint: Remember that the ion is a weak base.) | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/15%3A_Equilibria_of_Other_Reaction_Classes/15.07%3A_Exercises.txt |
Among the many capabilities of chemistry is its ability to predict if a process will occur under specified conditions. Thermodynamics, the study of relationships between the energy and work associated with chemical and physical processes, provides this predictive ability. Previous chapters in this text have described various applications of thermochemistry, an important aspect of thermodynamics concerned with the heat flow accompanying chemical reactions and phase transitions. This chapter will introduce additional thermodynamic concepts, including those that enable the prediction of any chemical or physical changes under a given set of conditions.
• 16.0: Introduction
Previous chapters in this text have described various applications of thermochemistry, an important aspect of thermodynamics concerned with the heat flow accompanying chemical reactions and phase transitions. This chapter will introduce additional thermodynamic concepts, including those that enable the prediction of any chemical or physical changes under a given set of conditions.
• 16.1: Spontaneity
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.
• 16.2: Entropy
Entropy (S) is a state function that can be related to the number of microstates for a system (the number of ways the system can be arranged) and to the ratio of reversible heat to Kelvin temperature. It 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<SgasSsolid<Sliquid<SgasS_{solid} < S_{liquid} < S_{gas} in a given physical state at
• 16.3: The Second and Third Laws of Thermodynamics
The second law of thermodynamics states spontaneous processes increases the entropy of the universe. If a process would decrease the entropy of the universe, then the process is nonspontaneous, and if no change occurs, the system is at equilibrium. The third law of thermodynamics establishes the zero for entropy at 0 J/Kelvin for a perfect, pure crystalline solid at 0 K with only one possible microstate.
• 16.4: Free Energy
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 that the reaction will proceed in the forward direction to reach equilibrium; a positive ΔG indicates that the reaction will proceed in the reverse direction to reach equilibrium; 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.
• 16.5: Key Terms
• 16.6: Key Equations
• 16.7: Summary
• 16.8: Exercises
These are homework exercises to accompany the Textmap created for "Chemistry" by OpenStax.
16: Thermodynamics
Among the many capabilities of chemistry is its ability to predict if a process will occur under specified conditions. Thermodynamics, the study of relationships between the energy and work associated with chemical and physical processes, provides this predictive ability. Previous chapters in this text have described various applications of thermochemistry, an important aspect of thermodynamics concerned with the heat flow accompanying chemical reactions and phase transitions. This chapter will introduce additional thermodynamic concepts, including those that enable the prediction of any chemical or physical changes under a given set of conditions. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/16%3A_Thermodynamics/16.00%3A_Introduction.txt |
Learning Objectives
By the end of this section, you will be able to:
• Distinguish between spontaneous and nonspontaneous processes
• Describe the dispersal of matter and energy that accompanies certain spontaneous 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. Iron exposed to the earth’s atmosphere will corrode, but rust is not converted to iron without intentional chemical treatment. 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 $2$).
$C (s \text {, diamond }) \longrightarrow C (s \text {, graphite }) \nonumber$
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; 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
Extending the 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 in one flask and the other flask is empty ($P = 0$) (Figure $3$). When the valve is opened, the gas spontaneously expands to fill both flasks equally. 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.
$w=-P \Delta V=0 \quad \quad (P=0 \text { in a vacuum })|] 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. \[\Delta U=q+w=0+0=0 \nonumber$
The spontaneity of this process is therefore not a consequence of any change in energy that accompanies the process. Instead, the driving force 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 expansion 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.
$q_{ X }<0 \quad \text { and } \quad q_{ Y }=-q_{ X }>0 \nonumber$
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 is 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.
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.
3. The process in question is diffusion. This process yields a more uniform dispersal of matter, since the initial state of the system involves two regions of different dye concentrations (high in the drop of dye, zero in the water), and the final state of the system contains a single dye concentration throughout.
Exercise $1$
Describe how energy is redistributed when a spoon at room temperature is placed in a cup of hot coffee.
Answer
Heat will spontaneously flow from the hotter object (coffee) to the colder object (spoon), resulting in a more uniform distribution of thermal energy as the spoon warms and the coffee cools. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/16%3A_Thermodynamics/16.01%3A_Spontaneity.txt |
Learning Objectives
By the end of this section, you will be able to:
• Define entropy
• Explain the relationship between entropy and the number of microstates
• Predict the sign of the entropy change for chemical and physical processes
In 1824, at the age of 28, Nicolas Léonard Sadi Carnot (Figure $1$) published the results of an extensive study regarding the efficiency of steam heat engines. A later review of Carnot’s findings by 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). In thermodynamics, a reversible process is one 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 in 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, 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:
$\Delta S=\frac{q_{ rev }}{T} \nonumber$
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.
Entropy and Microstates
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 (W) possible for the system. A microstate is a specific configuration of all the locations and energies of the atoms or molecules that make up a system. The relation between a system’s entropy and the number of possible microstates is
$S=k \ln W \nonumber$
where k is the Boltzmann constant, 1.38 10−23 J/K.
As for other state functions, the change in entropy for a process is the difference between its final ($S_f$) and initial ($S_i$) values:
$\Delta S=S_{ f }-S_{ i }=k \ln W_{ f }-k \ln W_{ i }=k \ln \frac{W_{ f }}{W_{ i }} \nonumber$
For processes involving an increase in the number of microstates, Wf > Wi, the entropy of the system increases and 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 $2$. 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 $\frac{6}{16}$ or $\frac{3}{8}$ The least probable configuration of the system is one in which all four particles are in one box, corresponding to distributions (a) and (e), each with a probability of The probability of finding all particles in only one box (either the left box or right box) is then or
$\left(\frac{1}{16}+\frac{1}{16}\right)=\frac{2}{16} \text { or } \frac{1}{8} \nonumber$
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.
This matter dispersal model of entropy is often described qualitatively in terms of the disorder of the system. By this description, microstates in which all the particles are in a single box are the most ordered, thus possessing the least entropy. Microstates in which the particles are more evenly distributed among the boxes are more disordered, possessing greater entropy.
The previous description of an ideal gas expanding into a vacuum (Figure 16.4) 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. Initially, the gas molecules are confined to just one of the two flasks. Opening the valve between the flasks increases the volume available to the gas molecules and, correspondingly, the number of microstates possible for the system. Since Wf > Wi, the expansion process involves an increase in entropy (ΔS > 0) and is spontaneous.
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 thermal energy (represented as “*”) in Figure $3$. 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. Thus, 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 $\frac{3}{10}$. More likely is the flow of heat to yield one of the other two distribution, the combined probability being $\frac{7}{10}$. 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 $\frac{4}{10}$. 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.
Example $1$: Determination of ΔS
Calculate the change in entropy for the process depicted below.
Solution
The initial number of microstates is one, the final six:
$\Delta S=k \ln \frac{W_{ c }}{W_{ a }}=1.38 \times 10^{-23} J / K \times \ln \frac{6}{1}=2.47 \times 10^{-23} J / K \nonumber$
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
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 $4$. 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 gaseous phase, in which a given number of atoms or molecules occupy a much greater volume than in the liquid phase. Each atom or molecule can be found in many more locations, corresponding to a much greater number of microstates. 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 $5$).
Link to Learning
Try this simulator with interactive visualization of the dependence of particle location and freedom of motion on physical state and temperature.
The entropy of a substance is influenced by the 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 (a topic beyond the scope of this text). For molecules, greater numbers of atoms increase the number of ways in which the molecules can vibrate and thus the number of possible microstates and the entropy of the system.
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 $2$.
Example $2$: Predicting the Sign of ∆S
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. $Ag^{+}(aq)+ Cl^{-}(aq) \longrightarrow AgCl (s)$
3. $C_6 H_6(l)+\frac{15}{2} O_2(g) \longrightarrow 6 CO_2(g)+3 H_2O(l)) 4. \(NH_3(s) \longrightarrow NH_3(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 $1$
Predict the sign of the entropy change for the following processes. Give a reason for your prediction.
1. $NaNO_3(s) \longrightarrow Na^{+}(aq)+ NO_3{ }^{-}(aq)$
2. the freezing of liquid water
3. $CO_2(s) \longrightarrow CO_2(g)$
4. $CaCO_3(s) \longrightarrow CaO (s)+ CO_2(g)$
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 increase in the amount of gaseous species. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/16%3A_Thermodynamics/16.02%3A_Entropy.txt |
Learning Objectives
By the end of this section, you will be able to:
• State and explain the second and third laws of thermodynamics
• Calculate entropy changes for phase transitions and chemical reactions under standard conditions
The Second Law of Thermodynamics
In the quest to identify a property that may reliably predict the spontaneity of a process, a promising candidate has been identified: 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:
$\Delta S_{\text {univ }}=\Delta S_{\text {sys }}+\Delta S_{\text {surr }} \nonumber$
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:
$\Delta S_{ sys }=\frac{-q_{ rev }}{T_{ sys }} \quad \text { and } \quad \Delta S_{ surr }=\frac{q_{ rev }}{T_{ surr }} \nonumber$
The magnitudes of −qrev and qrev are equal, their opposite arithmetic signs denoting loss of heat by the system and gain of heat by the surroundings. Since Tsys > Tsurr in this scenario, the entropy decrease of the system will be less than the entropy increase of the surroundings, and so the entropy of the universe will increase:
\begin{aligned} & \left|\Delta S_{\text {sys }}\right|<\left|\Delta S_{\text {surr }}\right| \[4pt] & \Delta S_{\text {univ }}=\Delta S_{\text {sys }}+\Delta S_{\text {surr }}>0 \end{aligned} \nonumber
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:
$\Delta S_{ sys }=\frac{q_{ rev }}{T_{ sys }} \quad \text { and } \quad \Delta S_{ surr }=\frac{-q_{ rev }}{T_{ surr }} \nonumber$
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 (that is, the direction of the heat flow) will yield a negative value for ΔSuniv. This process involves a decrease in the entropy of the universe.
3. The objects are at essentially the same temperature, TsysTsurr, and so the magnitudes of the entropy changes are essentially the same for both the system and the surroundings. In this case, the entropy change of the universe is zero, and the system is at equilibrium.
\begin{aligned} & \left|\Delta S_{\text {sys }}\right| \approx\left|\Delta S_{\text {surr }}\right| \[4pt] & \Delta S_{\text {univ }}=\Delta S_{\text {sys }}+\Delta S_{\text {surr }}=0 \end{aligned} \nonumber
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 at equilibrium
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:
$\Delta S_{ univ }=\Delta S_{ sys }+\Delta S_{ surr }=\Delta S_{ sys }+\frac{q_{ surr }}{T} \nonumber$
We may use this equation to predict the spontaneity of a process as illustrated in Example $1$.
Example $1$: Will Ice Spontaneously Melt?
The entropy change for the process
$H_2O(s) \longrightarrow H_2O(l) \nonumber$
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{aligned} \Delta S_{\text {univ }} & =\Delta S_{\text {sys }}+\Delta S_{\text {surr }}=\Delta S_{\text {sys }}+\frac{q_{\text {surr }}}{T} \[4pt] & =22.1 J / K +\frac{-6.00 \times 10^3 J }{263.15 K }=-0.7 J / K \end{aligned} \nonumber
$S_{univ} < 0$, so melting is nonspontaneous (not spontaneous) at −10.0 °C.
At 10.00 °C (283.15 K), the following is true:
\begin{align*} \Delta S_{\text {univ }} &=\Delta S_{\text {sys }}+\frac{q_{\text {surr }}}{T} \[4pt] &=22.1 J / K +\frac{-6.00 \times 10^3 J }{283.15 K }=+0.9 J / K \end{align*} \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, so ΔSfreezing = −ΔSmelting = −22.1 J/K and qsurr = +6.00 kJ. At −10.00 °C spontaneous, +0.7 J/K; at +10.00 °C nonspontaneous, −0.9 J/K.
The Third Law of Thermodynamics
The previous section described the various contributions of matter and energy dispersal that contribute to the entropy of a system. With these contributions in mind, consider the entropy of a pure, perfectly crystalline solid possessing no kinetic energy (that is, at a temperature of absolute zero, 0 K). This system may be described by a single microstate, as its purity, perfect crystallinity and complete lack of motion means there is but one possible location for each identical atom or molecule comprising the crystal (W = 1). According to the Boltzmann equation, the entropy of this system is zero.
$S=k \ln W=k \ln (1)=0 \nonumber$
This limiting condition for a system’s entropy represents the third law of thermodynamics: the entropy of a pure, perfect crystalline substance at 0 K is zero.
Careful calorimetric measurements can be made to determine the temperature dependence of a substance’s entropy and to derive absolute entropy values under specific conditions. Standard entropies (S°) are for one mole of substance under standard conditions (a pressure of 1 bar and a temperature of 298.15 K; see details regarding standard conditions in the thermochemistry chapter of this text). The standard entropy changeS°) for a reaction may be computed using standard entropies as shown below:
$\Delta S^{\circ}=\sum \nu S^{\circ}(\text { products })-\sum \nu S^{\circ}(\text { reactants }) \nonumber$
where ν represents stoichiometric coefficients in the balanced equation representing the process. For example, ΔS° for the following reaction at room temperature
$m A +n B \longrightarrow x C +y D \nonumber$
is computed as:
$=\left[x S^{\circ}( C )+y S^{\circ}( D )\right]-\left[m S^{\circ}( A )+n S^{\circ}( B )\right] \nonumber$
A partial listing of standard entropies is provided in Table $2$, and additional values are provided in Appendix G. The example exercises that follow demonstrate the use of S° values in calculating standard entropy changes for physical and chemical processes.
Table $2$: Standard entropies for selected substances measured at 1 atm and 298.15 K. (Values are approximately equal to those measured at 1 bar, the currently accepted standard state pressure.)
Substance (J mol−1 K−1)
carbon
C(s, graphite) 5.740
C(s, diamond) 2.38
CO(g) 197.7
CO2(g) 213.8
CH4(g) 186.3
C2H4(g) 219.5
C2H6(g) 229.5
CH3OH(l) 126.8
C2H5OH(l) 160.7
hydrogen
H2(g) 130.57
H(g) 114.6
H2O(g) 188.71
H2O(l) 69.91
HCI(g) 186.8
H2S(g) 205.7
oxygen
O2(g) 205.03
Example $2$: Determination of ΔS°
Calculate the standard entropy change for the following process:
$H_2O(g) \longrightarrow H_2O(l) \nonumber$
Solution
Calculate the entropy change using standard entropies as shown above:
$\Delta S^{\circ}=(1 mol )\left(70.0 J mol^{-1} K^{-1}\right)-(1 mol )\left(188.8 J mol^{-1} K^{-1}\right)=-118.8 J / K \nonumber$
The value for ΔS° is negative, as expected for this phase transition (condensation), which the previous section discussed.
Exercise $2$
Calculate the standard entropy change for the following process:
$H_2(g)+ C_2 H_4(g) \longrightarrow C_2 H_6(g) \nonumber$
Answer
−120.6 J K–1 mol–1
Example $3$: Determination of ΔS°
Calculate the standard entropy change for the combustion of methanol, CH3OH:
$2 CH_3 OH (l)+3 O_2(g) \longrightarrow 2 CO_2(g)+4 H_2O(l) \nonumber$
Solution
Calculate the entropy change using standard entropies as shown above:
$\begin{gathered} \Delta S^{\circ}=\sum \nu S^{\circ}(\text { products })-\sum \nu S^{\circ}(\text { reactants }) \[4pt] {\left[2 mol \times S^{\circ}\left( CO_2(g)\right)+4 mol \times S^{\circ}\left( H_2O(l)\right)\right]-\left[2 mol \times S^{\circ}\left( CH_3 OH (l)\right)+3 mol \times S^{\circ}\left( O_2(g)]\right)\right.} \[4pt] =\{[2(213.8)+4 \times 70.0]-[2(126.8)+3(205.2)]\}=-161.6 J / K \end{gathered} \nonumber$
Exercise $3$
Calculate the standard entropy change for the following reaction:
$Ca ( OH )_2( s ) \longrightarrow CaO (s)+ H_2O(l) \nonumber$
Answer
24.7 J/K | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/16%3A_Thermodynamics/16.03%3A_The_Second_and_Third_Laws_of_Thermodynamics.txt |
Learning Objectives
By the end of this section, you will be able to:
• 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
• Explain how temperature affects the spontaneity of some processes
• Relate standard free energy changes to equilibrium constants
One of the challenges of using the second law of thermodynamics to determine if a process is spontaneous is that it requires measurements of 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 ($G$) (or simply the free energy), and it is defined in terms of a system’s enthalpy and entropy as the following:
$G=H-T S \nonumber$
Free energy is a state function, and at constant temperature and pressure, the free energy change ($ΔG$) may be expressed as the following:
$\Delta G=\Delta H-T \Delta S \nonumber$
(For simplicity’s sake, the subscript “sys” will be omitted henceforth.)
The relationship between this system property and the spontaneity of a process may be understood by recalling the previously derived second law expression:
$\Delta S_{\text {univ }}=\Delta S+\frac{q_{\text {surr }}}{T} \nonumber$
The first law requires that $q_{surr} = −q_{sys}$, and at constant pressure $q_{sys} = ΔH$, so this expression may be rewritten as:
$\Delta S_{\text {univ }}=\Delta S-\frac{\Delta H}{T} \nonumber$
Multiplying both sides of this equation by $−T$, and rearranging yields the following:
$-T \Delta S_{\text {univ }}=\Delta H-T \Delta S \nonumber$
Comparing this equation to the previous one for free energy change shows the following relation:
$\Delta G=-T \Delta S_{\text {univ }} \nonumber$
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 $1$ summarizes the relation between the spontaneity of a process and the arithmetic signs of these indicators.
Table $1$: Relation between Process Spontaneity and Signs of Thermodynamic Properties
$ΔS_{univ} > 0$ ΔG < 0 moves spontaneously in the forward direction, as written, to reach equilibrium
$ΔS_{univ} < 0$ ΔG > 0
nonspontaneous in the forward direction, as written, but moves spontaneously in the reverse direction, as written, to reach equilibrium
$ΔS_{univ} = 0$ ΔG = 0 reversible (at equilibrium)
What’s “Free” about ΔG?
In addition to indicating spontaneity, the free energy change also provides information regarding the amount of useful work ($w$) that may be accomplished by a spontaneous process. Although a rigorous treatment of this subject is beyond the scope of an introductory chemistry text, a brief discussion is helpful for gaining a better perspective on this important thermodynamic property.
For this purpose, consider a spontaneous, exothermic process that involves a decrease in entropy. The free energy, as defined by
$\Delta G=\Delta H-T \Delta S \nonumber$
may be interpreted as representing the difference between the energy produced by the process, ΔH, and the energy lost to the surroundings, TΔS. The difference between the energy produced and the energy lost is the energy available (or “free”) to do useful work by the process, ΔG. If the process somehow could be made to take place under conditions of thermodynamic reversibility, the amount of work that could be done would be maximal:
$\Delta G=w_{max } \nonumber$
where $w_{max}$ refers to all types of work except expansion (pressure-volume) work.
However, as noted previously in this chapter, such conditions are not realistic. In addition, the technologies used to extract work from a spontaneous process (e.g., batteries) are never 100% efficient, and so the work done by these processes is always less than the theoretical maximum. Similar reasoning may be applied to a nonspontaneous process, for which the free energy change represents the minimum amount of work that must be done on the system to carry out the process.
Calculating Free Energy Change
Free energy is a state function, so its value depends only on the conditions of the initial and final states of the system. 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, $ΔG^o$, according to the following relation.
$\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ} \nonumber$
It is important to understand that for phase changes, $\Delta G^º$ tells you if the phase change is spontaneous or not; will it happen, or not happen. For chemical reactions, $\Delta G^º$ tells you the extent of a reaction. In other words, $\Delta G^º$ for a reaction tells you how much product will be present at equilibrium. A reaction with $\Delta G^º$ < 0 is considered product-favored at equilibrium; there will be more products than reactants when the reaction reaches equilibrium. A reaction with $\Delta G^º$ > 0 is considered reactant-favored at equilibrium; there will be more reactants than products when the reaction reaches equilibrium.
Example $1$: Using Standard Enthalpy and Entropy Changes to Calculate ΔG°
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 $ΔG^o$ say about the spontaneity of this process?
Solution
The process of interest is the following:
$\ce{H_2O(l) \longrightarrow H_2O(g)} \nonumber$
The standard change in free energy may be calculated using the following equation:
$\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ} \nonumber$
From Appendix G:
Substance $\Delta H_{ f }^{\circ} (\text{kJ/mol})$ $S^{\circ} (\text{kJ/K mol})$
H2O(l) −285.83 70.0
H2O(g) −241.82 188.8
Using the appendix data to calculate the standard enthalpy and entropy changes yields:
\begin{align*} \Delta H^{\circ} &=\Delta H_{ f }^{\circ}\left( \ce{H2O(g)} \right)-\Delta H_{ f }^{\circ}\left( \ce{H2O(l)} \right) \[4pt] &=[-241.82 ~\text{kJ/mol} -(-285.83)] ~\text{kJ/mol} =44.01 ~\text{kJ/mol} \[4pt] \Delta S^{\circ} &=1 ~\text{mol} \times S^{\circ}\left( \ce{H2O(g)} \right)-1 ~\text{mol} \times S^{\circ}\left( \ce{H2O(l)} \right) \[4pt] &=(1 ~\text{mol}) ~188.8 J / mol \cdot K -(1 ~\text{mol}) ~ 70.0 ~ \text{J} / \text{mol K} =118.8 ~ \text{J / K} \[4pt] \Delta G^{\circ} &=\Delta H^{\circ}-T \Delta S^{\circ} \end{align*} \nonumber
Substitution into the standard free energy equation yields:
\begin{align*} \Delta G^{\circ} &=\Delta H^{\circ}-T \Delta S^{\circ} \[4pt] &=44.01~\text{kJ} -(298 K \times 118.8~\text{J/K}) \times \frac{1 ~\text{kJ}}{1000~\text{J} } \[4pt] &=44.01~\text{kJ} -35.4~\text{kJ} \[4pt] &=8.6~\text{kJ} \end{align*} \nonumber
At 298 K (25 °C) so boiling is nonspontaneous (not spontaneous).
Exercise $1$
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?
$\ce{C2H6(g) \longrightarrow H2(g) + C2H4(g)} \nonumber$
Answer
the reaction is nonspontaneous (not spontaneous) at 25 °C.
The standard free energy change for a reaction may also be calculated from standard free energy of formation $ΔG_f^o$ values 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 enthalpy of formation, $ΔG_f^o$ is by definition zero for elemental substances in their standard states. The approach used to calculate $ΔG^o$ for a reaction from $ΔG_f^o$ values is the same as that demonstrated previously for enthalpy and entropy changes. For the reaction
$m A +n B \longrightarrow x C +y D \nonumber$
the standard free energy change at room temperature may be calculated as
\begin{align*} \Delta G^{\circ} &= \sum \nu \Delta G^{\circ}(\text { products })-\sum \nu \Delta G^{\circ}(\text { reactants }) \[4pt] &= \left[x \Delta G_{ f }^{\circ}( C )+y \Delta G_{ f }^{\circ}( D )\right]-\left[m \Delta G_{ f }^{\circ}( A )+n \Delta G_{ f }^{\circ}( B )\right] . \end{align*} \nonumber
Example $2$: Using Standard Free Energies of Formation to Calculate ΔG°
Consider the decomposition of yellow mercury(II) oxide.
$\ce{HgO (s, \text { yellow }) -> Hg (l) + 1/2 O2(g)} \nonumber$
Calculate the standard free energy change at room temperature, 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 $\Delta G_{ f }^{\circ}( \text{kJ / mol} )$ $\Delta H_{ f }^{\circ}( \text{kJ / mol} )$ $S^{\circ}( \text{kJ / K mol} )$
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:
\begin{align*} \Delta G^{\circ} & =\sum \nu G_{ f }^{\circ}(\text { products })-\sum \nu \Delta G_{ f }^{\circ}(\text { reactants }) \[4pt] &=\left[1 \Delta G_{ f }^{\circ}~ \ce{Hg(l)} + \dfrac{1}{2} \Delta G_{f}^{\circ} ~\ce{O2(g)} \right]-1 \Delta G_{ f }^{\circ} ~\ce{HgO(s, yellow)} \[4pt] &=\left[1 ~\text{mol} (0 ~\text{kJ / mol} )+\frac{1}{2} ~\text{mol} (0 ~\text{kJ / mol} )\right] - 1 ~\text{mol} (-58.43 ~\text{kJ / mol} )=58.43 ~\text{kJ / mol} \end{align*} \nonumber
(b) Using enthalpies and entropies of formation:
\begin{align*} \Delta H^{\circ}&=\sum \nu \Delta H_{ f }^{\circ}(\text { products })-\sum \nu \Delta H_{ f }^{\circ}(\text { reactants }) \[4pt] &=\left[1 \Delta H_{ f }^{\circ}~\ce{Hg(l)} +\frac{1}{2} \Delta H_{ f }^{\circ}~\ce{O2(g)} \right]-1 \Delta H_{ f }^{\circ} ~\ce{HgO (s, yellow )} \[4pt] &=\left[1 ~\text{mol} (0~ \text{kJ / mol} ) + \frac{1}{2} ~\text{mol} (0 ~\text{kJ / mol} )\right]-1 ~\text{mol} (-90.46~\text{kJ / mol} ) = 90.46 ~\text{kJ / mol} \[8pt] \Delta S^{\circ} &=\sum \nu \Delta S^{\circ}(\text { products })-\sum \nu \Delta S^{\circ}(\text { reactants }) \[4pt] &=\left[1 ~\Delta S^{\circ} ~\ce{Hg (l)} + \frac{1}{2} \Delta S^{\circ} ~\ce{O2(g)} \right]-1 \Delta S^{\circ} ~\ce{HgO (s, yellow )} \[4pt] & =\left[1 ~\text{mol} (75.9~\text{J / mol K} ) + \frac{1}{2} ~\text{mol} (205.2~\text{J / mol K} )\right] -1 \text{mol} (71.13 ~\text{J / mol K} )=107.4 ~\text{ J / mol K} \[8pt] \Delta G^{\circ} &=\Delta H^{\circ}-T \Delta S^{\circ}=90.46 ~\text{kJ} - 298.15 ~\text{K} \times 107.4 ~\text{J / K} \cdot \text{mol} \times \frac{1 ~\text{ kJ} }{1000 ~\text{J} } \[4pt] & =(90.46 - 32.01) ~\text{kJ / mol} =58.45 ~\text{kJ / mol} \end{align*} \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 $1$
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?
$\ce{C2H4(g) \longrightarrow H2(g) + C2H2(g)} \nonumber$
Answer
1. 140.8 kJ/mol, nonspontaneous
2. 141.5 kJ/mol, nonspontaneous
Free Energy Changes for Coupled Reactions
The use of free energies of formation to compute free energy changes for reactions as described above is possible because $ΔG$ is a state function, and the approach is analogous to the use of Hess’ Law in computing enthalpy changes (see the chapter on thermochemistry). Consider the vaporization of water as an example:
$\ce{H_2O(l) -> H_2O(g)} \nonumber$
An equation representing this process may be derived by adding the formation reactions for the two phases of water (necessarily reversing the reaction for the liquid phase). The free energy change for the sum reaction is the sum of free energy changes for the two added reactions:
\begin{align*} \ce{H2(g) + 1/2 O2(g) &-> H2O(g)} && \Delta G_{ f }^{\circ} \text { gas } \[4pt] \ce{H2O(l) &-> H2(g) + 1/2 O2(g)} && - \Delta G_{ f }^{ o } \text { liquid } \[4pt] \hline \ce{H2O(l) &-> H2O(g)} && \Delta G^{\circ}=\Delta G_{ f }^{\circ} ~\text{gas} - \Delta G_{ f }^{\circ} ~\text{liquid} \end{align*}
This approach may also be used in cases where a nonspontaneous reaction is enabled by coupling it to a spontaneous reaction. For example, the production of elemental zinc from zinc sulfide is thermodynamically unfavorable, as indicated by a positive value for $ΔG^o$:
$\ce{ZnS (s) \rightarrow Zn(s) + S(s)} \quad \Delta G_1^{\circ}=201.3 kJ \nonumber$
The industrial process for production of zinc from sulfidic ores involves coupling this decomposition reaction to the thermodynamically favorable oxidation of sulfur:
$\ce{S(s) + O2(g) \rightarrow SO2(g)} \quad \Delta G_2^{\circ}=-300.1 kJ \nonumber$
The coupled reaction exhibits a negative free energy change and is spontaneous:
$\ce{ZnS(s) + O2(g) \rightarrow Zn(s) + SO2(g)} \nonumber$
$\Delta G^{\circ}=201.3 ~\text{kJ} + -300.1 ~\text{kJ} =-98.8 ~\text{kJ} \nonumber$
This process is typically carried out at elevated temperatures, so this result obtained using standard free energy values is just an estimate. The gist of the calculation, however, holds true.
Example $3$: Calculating Free Energy Change for a Coupled Reaction
Is a reaction coupling the decomposition of $\ce{ZnS}$ to the formation of $\ce{H2S}$ expected to be spontaneous under standard conditions?
Solution
Following the approach outlined above and using free energy values from Appendix G:
\begin{align*} &\text{Decomposition of zinc sulfide:} & \ce{ZnS (s) \rightarrow Zn(s) + S(s)} && \Delta G_1^{\circ}=201.3 ~\text{kJ}\[4pt] &\text{Formation of hydrogen sulfide:} & \ce{S(s) + H2(g) \rightarrow H2S(g)} && \Delta G_1^{\circ}=-33.4 ~\text{kJ} \[4pt] \hline &\text{Coupled reaction:} & \ce{ZnS(s) + H2(g) \rightarrow Zn(s) + H2S(g)} && \Delta G^{\circ}=201.3 ~\text{kJ} +-33.4 ~\text{kJ} =167.9 ~\text{kJ} \end{align*}
The coupled reaction exhibits a positive free energy change and is thus nonspontaneous.
Exercise $3$
What is the standard free energy change for the reaction below? Is the reaction expected to be spontaneous under standard conditions?
$\ce{FeS(s) \, + O2(g) \rightarrow Fe(s) \, + SO2(g)} \nonumber$
Answer
−199.7 kJ; spontaneous
Temperature Dependence of Spontaneity and Extent of Reaction
As was previously demonstrated in this chapter’s section on entropy, 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. In a similar, but not identical fashion, some chemical reactions can switch from being product-favored at equilibrium, to being reactant-favored at equilibrium, depending on the temperature.
Note
The numerical value of $\Delta G^º$ is always dependent on the temperature. In this section we are determining whether or not the sign of $\Delta G^º$ is dependent on the temperature.
To illustrate this concept, the equation relating free energy change to the enthalpy and entropy changes for the process is considered:
$ΔG^º=ΔH^º−TΔS^º \nonumber$
The extent of a process, as reflected in the arithmetic sign of its standard 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 ΔHº and ΔSº are positive. This condition describes an endothermic process that involves an increase in system entropy. In this case, ΔGº will be negative if the magnitude of the TΔSº term is greater than ΔHº. If the TΔSº term is less than ΔHº, the free energy change will be positive. Such a process is product-favored at equilibrium at high temperatures and reactant-favored at equilibrium at low temperatures.
2. Both ΔHº and ΔSº are negative. This condition describes an exothermic process that involves a decrease in system entropy. In this case, ΔGº will be negative if the magnitude of the TΔSº term is less than ΔHº. If the TΔSº term’s magnitude is greater than ΔHº, the free energy change will be positive. Such a process is product-favored at equilibrium at low temperatures and reactant-favored at equilibrium at high temperatures.
3. ΔHº is positive and ΔSº is negative. This condition describes an endothermic process that involves a decrease in system entropy. In this case, ΔGº will be positive regardless of the temperature. Such a process is reactant-favored at equilibrium at all temperatures.
4. ΔHº is negative and ΔSº is positive. This condition describes an exothermic process that involves an increase in system entropy. In this case, ΔGº will be negative regardless of the temperature. Such a process is product-favored at equilibrium at all temperatures.
These four scenarios are summarized in Table $1$
Table $1$
Sign of $\Delta H^o$
Sign of $\Delta S^o$
Sign of $\Delta G^o$
Temperature Dependence of $\Delta G^o$
- + - The sign of $\Delta G^o$ does not depend on the temperature.The reaction is product-favored at equilibrium at all temperatures.
+ - + The sign of $\Delta G^o$ does not depend on the temperature.The reaction is reactant-favored at equilibrium at all temperatures.
- - - or + The sign of $\Delta G^o$ does depend on the temperature. The reaction will be product-favored at equilibrium at lower temperatures.
+ + - or +
The sign of $\Delta G^o$ does depend on the temperature. The reaction will be product-favored at equilibrium at lower temperatures.
Example $3$: Predicting the Temperature Dependence of Spontaneity
The incomplete combustion of carbon is described by the following equation:
$\ce{2C}(s)+\ce{O2}(g)⟶\ce{2CO}(g) \nonumber$
Does the sign of $\Delta G^º$ of this process depend upon temperature?
Solution
Combustion processes are exothermic ($ΔH^º < 0$). 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, $ΔS^º > 0$). The reaction is therefore product-favored at equilibrium ($ΔG^º < 0$) at all temperatures.
Exercise $3$
Popular chemical hand warmers generate heat by the air-oxidation of iron:
$\ce{4Fe}(s)+\ce{3O2}(g)⟶\ce{2Fe2O3}(s) \nonumber$
Does the sign of $\Delta G^o$ of this process depend upon temperature?
Answer
ΔHº and ΔSº are both negative; the reaction is product-favored at equilibrium at low temperatures.
When considering the conclusions drawn regarding the temperature dependence of the sign of ΔGº, 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 reactant-favored at equilibrium at one temperature but product-favored at equilibrium at another temperature will necessarily undergo a change in “extent” (as reflected by its ΔGº) as temperature varies. This is clearly illustrated by a graphical presentation of the free energy change equation, in which ΔGº is plotted on the y axis versus T on the x axis:
$ΔG^º=ΔH^º−TΔS^º \nonumber$
$y=b+mx \nonumber$
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-dependence for the sign of ΔGº as depicted by the two yellow lines in the plot. Each line crosses from one spontaneity domain (positive or negative ΔGº) 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 ΔGº is zero:
$ΔG^º=0=ΔH^º−TΔS^º \nonumber$
$T=\dfrac{ΔH^º}{ΔS^º} \nonumber$
Thus, saying a process is product-favored at equilibrium at “high” or “low” temperatures means the temperature is above or below, respectively, that temperature at which ΔGº for the process is zero.
Note
In this discussion, we have used two different descriptions for the meaning of the sign of ΔGº. You should be aware of the meaning of each description.
a) Extent of Reaction: This description is used to predict the ratio of the product and reactant concentrations at equilibrium. In this description, we use the thermodynamic term ΔGº to tell us the same information as the equilibrium constant, K. When ΔGº < 0, K > 1, and the reaction will be product-favored at equilibrium. When ΔGº > 0, K< 1, and the reaction is reactant-favored at equilibrium. When ΔGº = 0, K =1, and the reaction will have roughly equal amounts of products and reactants at equilibrium. In all cases, the reaction will form a mixture of products and reactants at equilibrium. We use the sign and magnitude of ΔGº to tell us how much product will be made if the reaction is allowed to reach equilibrium.
b) Spontaneity: This description is much more complicated because it involves two different interpretations of how a reaction at standard state occurs. One interpretation involves the hypothetical process in which the reaction proceeds from a starting point of pure reactants to a finishing point of pure products, with all substances isolated in their own containers under standard state conditions. In the second, more realistic interpretation, the reaction starts with all reactants and all products in their standard state in one container. We then allow this specific mixture to react an infinitesimally small amount so that we can obtain a rate of change in free energy with respect to the extent of reaction when all reactants and products are mixed and (essentially) in their standard states. Although each interpretation describes a different reaction scenario, the value of the difference in free energy and the value of the rate of change in free energy are the same number. If ΔGº < 0, we say that the reaction is spontaneous, meaning that the reaction would proceed in the forward direction, as written, to form pure products in their standard state. If ΔGº > 0, we say that the reaction is nonspontaneous, meaning that the reaction would proceed in the reverse direction, as written, to form pure reactants in their standard state. If ΔGº = 0, we say that the neither the reactants nor the products are favored to be formed.
A detailed treatment of the meaning of ΔGº can be found in the paper, "Free Energy versus Extent of Reaction" by Richard S. Treptow, Journal of Chemical Education, 1996, Volume 73 (1), 51-54.
Example $4$: 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 Tables T1 or T2 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 Tables T1 or T2,
\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*} \nonumber
\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*} \nonumber
$T=\dfrac{ΔH°}{ΔS°}=\mathrm{\dfrac{44.01×10^3\:J/mol}{118.8\:J/K⋅mol}=370.5\:K=97.3\:°C} \nonumber$
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 (Tables T1 or T2.). 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 Tables T1 or T2 to estimate the boiling point of CS2.
Answer
313 K (accepted value 319 K).
Free Energy and Equilibrium
The free energy change for a process may be viewed as a measure of its driving force. A negative value for ΔG represents a driving force for the process in the forward direction, while a positive value represents a driving force for the process in the reverse direction. When ΔG is zero, the forward and reverse driving forces are equal, and the process occurs in both directions at the same rate (the system is at equilibrium).
In the chapter on equilibrium the reaction quotient, Q, was introduced as a convenient measure of the status of an equilibrium system. Recall that Q is the numerical value of the mass action expression for the system, and that you may use its value to identify the direction in which a reaction will proceed in order to achieve equilibrium. When Q is lesser than the equilibrium constant, K, the reaction will proceed in the forward direction until equilibrium is reached and Q = K. Conversely, if Q > K, the process will proceed in the reverse direction until equilibrium is achieved.
The free energy change for a process taking place with reactants and products present under nonstandard conditions (pressures other than 1 bar; concentrations other than 1 M) is related to the standard free energy change according to this equation:
$\Delta G=\Delta G^{\circ}+R T \ln Q \nonumber$
R is the gas constant (8.314 J/K mol), T is the kelvin or absolute temperature, and Q is the reaction quotient. For gas phase equilibria, the pressure-based reaction quotient, QP, is used. The concentration-based reaction quotient, QC, is used for condensed phase equilibria. This equation may be used to predict the spontaneity for a process under any given set of conditions as illustrated in Example $6$.
Example $6$: Calculating ΔG under Nonstandard Conditions
What is the free energy change for the process shown here
$2 NH_3(g) \longrightarrow 3 H_2(g)+ N_2(g) \nonumber$
under the specified conditions?
• $T=25^{\circ} C ,$
• $P_{ N_2}=0.870 ~\text{atm}$
• $P_{ H_2}=0.250 ~\text{atm}$
• $P_{ NH_3}=12.9 ~\text{atm}$
$\Delta G^{\circ}=33.0 kJ / mol \nonumber$
Solution
The equation relating free energy change to standard free energy change and reaction quotient may be used directly:
\begin{align*} \Delta G &=\Delta G^{\circ}+R T \ln Q \[4pt] &=33.0 \frac{ \text{kJ} }{ \text{mol} }+\left(8.314 \frac{ \text{J} }{ \text{mol K} } \times 298 K \times \ln \frac{\left(0.250^3\right) \times 0.870}{12.9^2}\right) \[4pt] &=9680 \frac{ \text{J} }{ \text{mol} } \text { or } 9.68 ~\text{kJ/mol} \end{align*} \nonumber
Since the computed value for ΔG is positive, the reaction is nonspontaneous under these conditions.
Exercise $6$
Calculate the free energy change for this same reaction at 875 °C in a 5.00 L mixture containing 0.100 mol of each gas. Is the reaction spontaneous under these conditions?
Answer
ΔG = –123.5 kJ/mol; yes
For a system at equilibrium, Q = K and ΔG = 0, and the previous equation may be written as
$0=\Delta G^{\circ}+R T \ln K \quad \text { (at equilibrium) } \nonumber$
$\Delta G^{\circ}=-R T \ln K \nonumber$
or
$K=e^{-\frac{\Delta G}{R T}} \nonumber$
This form of the equation provides a useful link between these two essential thermodynamic properties, and it can be used to derive equilibrium constants from standard free energy changes and vice versa. The relations between standard free energy changes and equilibrium constants are summarized in Table $2$.
Table $2$: Relations between Standard Free Energy Changes and Equilibrium Constants
K ΔG° Composition of an Equilibrium Mixture
> 1 < 0 Products are more abundant
< 1 > 0 Reactants are more abundant
= 1 = 0 Reactants and products are comparably abundant
Example $7$: Calculating an Equilibrium Constant using Standard Free Energy Change
Given that the standard free energies of formation of Ag+(aq), Cl(aq), and AgCl(s) are 77.1 kJ/mol, −131.2 kJ/mol, and −109.8 kJ/mol, respectively, calculate the solubility product, Ksp, for AgCl.
Solution
The reaction of interest is the following:
$AgCl (s) \rightleftharpoons Ag^{+}(aq)+ Cl^{-}(aq) \quad K_{ sp }=\left[ Ag^{+}\right]\left[ Cl^{-}\right] \nonumber$
The standard free energy change for this reaction is first computed using standard free energies of formation for its reactants and products:
\begin{align*} \Delta G^{\circ} &=\left[\Delta G_{ f }^{\circ}\left( \ce{Ag^{+}(aq)} \right)+\Delta G_{ f }^{\circ}\left( \ce{Cl^{-}(aq)} \right)\right]-\left[\Delta G_{ f }^{\circ}( \ce{AgCl(s)} )\right] \[4pt] & =[77.1 ~\text{kJ/mol} -131.2 ~\text{kJ/mol} ]-[-109.8 ~\text{kJ/mol} ]=55.7 ~\text{kJ/mol} \end{align*} \nonumber
The equilibrium constant for the reaction may then be derived from its standard free energy change:
$K_{ sp }=e^{-\frac{\Delta G^{\circ}}{R T}}=\exp \left(-\frac{\Delta G^{\circ}}{R T}\right)=\exp \left(-\frac{55.7 \times 10^3 J / mol }{8.314 J / mol \cdot K \times 298.15 K }\right) =\exp (-22.470)=e^{-22.470}=1.74 \times 10^{-10} \nonumber$
This result is in reasonable agreement with the value provided in Appendix J.
Exercise $7$
Use the thermodynamic data provided in Appendix G to calculate the equilibrium constant for the dissociation of dinitrogen tetroxide at 25 °C.
$2 NO_2(g) \rightleftharpoons N_2 O_4(g) \nonumber$
Answer
K = 3.1
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 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 $3$). If a system consists of reactants and products in nonequilibrium amounts (QK), the reaction will proceed spontaneously in the direction necessary to establish equilibrium. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/16%3A_Thermodynamics/16.04%3A_Free_Energy.txt |
Example and Directions
Words (or words that have the same definition)The definition is case sensitive(Optional) Image to display with the definition [Not displayed in Glossary, only in pop-up on pages](Optional) Caption for Image(Optional) External or Internal Link(Optional) Source for Definition
(Eg. "Genetic, Hereditary, DNA ...")(Eg. "Relating to genes or heredity")The infamous double helix https://bio.libretexts.org/CC-BY-SA; Delmar Larsen
Glossary Entries
Word(s)DefinitionImageCaptionLinkSource
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
Gibbs free energy change (G)thermodynamic property defined in terms of system enthalpy and entropy; all spontaneous processes involve a decrease in G
microstatepossible configuration or arrangement of matter and energy within a system
nonspontaneous processprocess that requires continual input of energy from an external source
reversible processprocess 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
second law of thermodynamicsall spontaneous processes involve an increase in the entropy of the universe
spontaneous changeprocess that takes place without a continuous input of energy from an external source
standard entropy (S°)entropy for one mole of a substance at 1 bar pressure; tabulated values are usually determined at 298.15 K
standard entropy change (ΔS°)change in entropy for a reaction calculated using the standard entropies
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 change in free energy accompanying the formation of one mole of substance from its elements in their standard states
third law of thermodynamicsentropy of a perfect crystal at absolute zero (0 K) is zero
16.06: Key Equations
S = k ln W
ΔSuniv = ΔSsys + ΔSsurr
ΔG = ΔHTΔS
16.07: 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.
Entropy (S) is a state function that can be related to the number of microstates for a system (the number of ways the system can be arranged) and to the ratio of reversible heat to kelvin temperature. It 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, entropy depends on phase with Ssolid < Sliquid < Sgas. For different substances in the same 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 and physical changes may be reliably predicted.
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. The third law of thermodynamics establishes the zero for entropy as that of a perfect, pure crystalline solid at 0 K. With only one possible microstate, the entropy is zero. We may compute the standard entropy change for a process by using standard entropy values for the reactants and products involved in the process.
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. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/16%3A_Thermodynamics/16.05%3A_Key_Terms.txt |
1.
What is a spontaneous reaction?
2.
What is a nonspontaneous reaction?
3.
Indicate whether the following processes are spontaneous or nonspontaneous.
1. (f) Iron rusting in a moist atmosphere
4.
A helium-filled balloon spontaneously deflates overnight as He atoms diffuse through the wall of the balloon. Describe the redistribution of matter and/or energy that accompanies this process.
5.
Many plastic materials are organic polymers that contain carbon and hydrogen. The oxidation of these plastics in air to form carbon dioxide and water is a spontaneous process; however, plastic materials tend to persist in the environment. Explain.
6.
In Figure 16.8 all possible distributions and microstates are shown for four different particles shared between two boxes. Determine the entropy change, ΔS, if the particles are initially evenly distributed between the two boxes, but upon redistribution all end up in Box (b).
7.
In Figure 16.8 all of the possible distributions and microstates are shown for four different particles shared between two boxes. Determine the entropy change, ΔS, for the system when it is converted from distribution (b) to distribution (d).
8.
How does the process described in the previous item relate to the system shown in Figure 16.4?
9.
Consider a system similar to the one in Figure 16.8, except that it contains six particles instead of four. What is the probability of having all the particles in only one of the two boxes in the case? Compare this with the similar probability for the system of four particles that we have derived to be equal to What does this comparison tell us about even larger systems?
10.
Consider the system shown in Figure 16.9. What is the change in entropy for the process where the energy is initially associated only with particle A, but in the final state the energy is distributed between two different particles?
11.
Consider the system shown in Figure 16.9. What is the change in entropy for the process where the energy is initially associated with particles A and B, and the energy is distributed between two particles in different boxes (one in A-B, the other in C-D)?
12.
Arrange the following sets of systems in order of increasing entropy. Assume one mole of each substance and the same temperature for each member of a set.
1. H2(g), HBrO4(g), HBr(g)
2. H2O(l), H2O(g), H2O(s)
3. He(g), Cl2(g), P4(g)
13.
At room temperature, the entropy of the halogens increases from I2 to Br2 to Cl2. Explain.
14.
Consider two processes: sublimation of I2(s) and melting of I2(s) (Note: the latter process can occur at the same temperature but somewhat higher pressure).
Is ΔS positive or negative in these processes? In which of the processes will the magnitude of the entropy change be greater?
15.
Indicate which substance in the given pairs has the higher entropy value. Explain your choices.
1. C2H5OH(l) or C3H7OH(l)
2. C2H5OH(l) or C2H5OH(g)
3. 2H(g) or H(g)
16.
Predict the sign of the entropy change for the following processes.
1. An ice cube is warmed to near its melting point.
2. Exhaled breath forms fog on a cold morning.
3. Snow melts.
17.
Predict the sign of the entropy change for the following processes. Give a reason for your prediction.
18.
Write the balanced chemical equation for the combustion of methane, CH4(g), to give carbon dioxide and water vapor. Explain why it is difficult to predict whether ΔS is positive or negative for this chemical reaction.
19.
Write the balanced chemical equation for the combustion of benzene, C6H6(l), to give carbon dioxide and water vapor. Would you expect ΔS to be positive or negative in this process?
20.
What is the difference between ΔS and ΔS° for a chemical change?
21.
Calculate for the following changes.
1. (f)
22.
Determine the entropy change for the combustion of liquid ethanol, C2H5OH, under the standard conditions to give gaseous carbon dioxide and liquid water.
23.
Determine the entropy change for the combustion of gaseous propane, C3H8, under the standard conditions to give gaseous carbon dioxide and water.
24.
“Thermite” reactions have been used for welding metal parts such as railway rails and in metal refining. One such thermite reaction is Is the reaction spontaneous at room temperature under standard conditions? During the reaction, the surroundings absorb 851.8 kJ/mol of heat.
25.
Using the relevant values listed in Appendix G, calculate for the following changes:
26.
From the following information, determine for the following:
27.
By calculating ΔSuniv at each temperature, determine if the melting of 1 mole of NaCl(s) is spontaneous at 500 °C and at 700 °C.
What assumptions are made about the thermodynamic information (entropy and enthalpy values) used to solve this problem?
28.
Use the standard entropy data in Appendix G to determine the change in entropy for each of the following reactions. All the processes occur at the standard conditions and 25 °C.
1. (f)
29.
Use the standard entropy data in Appendix G to determine the change in entropy for each of the following reactions. All the processes occur at the standard conditions and 25 °C.
1. (f)
30.
What is the difference between ΔG and ΔG° for a chemical change?
31.
A reaction has = 100 kJ/mol and Is the reaction spontaneous at room temperature? If not, under what temperature conditions will it become spontaneous?
32.
Explain what happens as a reaction starts with ΔG < 0 (negative) and reaches the point where ΔG = 0.
33.
Use the standard free energy of formation data in Appendix G to determine the free energy change for each of the following reactions, which are run under standard state conditions and 25 °C. Identify each as either spontaneous or nonspontaneous at these conditions.
1. (f)
34.
Use the standard free energy data in Appendix G to determine the free energy change for each of the following reactions, which are run under standard state conditions and 25 °C. Identify each as either spontaneous or nonspontaneous at these conditions.
1. (f)
35.
Given:
1. Determine the standard free energy of formation, for phosphoric acid.
2. How does your calculated result compare to the value in Appendix G? Explain.
36.
Is the formation of ozone (O3(g)) from oxygen (O2(g)) spontaneous at room temperature under standard state conditions?
37.
Consider the decomposition of red mercury(II) oxide under standard state conditions.
1. Is the decomposition spontaneous under standard state conditions?
2. Above what temperature does the reaction become spontaneous?
38.
Among other things, an ideal fuel for the control thrusters of a space vehicle should decompose in a spontaneous exothermic reaction when exposed to the appropriate catalyst. Evaluate the following substances under standard state conditions as suitable candidates for fuels.
1. Ammonia:
2. Diborane:
3. Hydrazine:
4. Hydrogen peroxide:
39.
Calculate ΔG° for each of the following reactions from the equilibrium constant at the temperature given.
1. (f)
40.
Calculate ΔG° for each of the following reactions from the equilibrium constant at the temperature given.
1. (f)
41.
Calculate the equilibrium constant at 25 °C for each of the following reactions from the value of ΔG° given.
42.
Calculate the equilibrium constant at 25 °C for each of the following reactions from the value of ΔG° given.
43.
Calculate the equilibrium constant at the temperature given.
44.
Calculate the equilibrium constant at the temperature given.
45.
Consider the following reaction at 298 K:
What is the standard free energy change at this temperature? Describe what happens to the initial system, where the reactants and products are in standard states, as it approaches equilibrium.
46.
Determine the normal boiling point (in kelvin) of dichloromethane, CH2Cl2. Find the actual boiling point using the Internet or some other source, and calculate the percent error in the temperature. Explain the differences, if any, between the two values.
47.
Under what conditions is spontaneous?
48.
At room temperature, the equilibrium constant (Kw) for the self-ionization of water is 1.00 10−14. Using this information, calculate the standard free energy change for the aqueous reaction of hydrogen ion with hydroxide ion to produce water. (Hint: The reaction is the reverse of the self-ionization reaction.)
49.
Hydrogen sulfide is a pollutant found in natural gas. Following its removal, it is converted to sulfur by the reaction What is the equilibrium constant for this reaction? Is the reaction endothermic or exothermic?
50.
Consider the decomposition of CaCO3(s) into CaO(s) and CO2(g). What is the equilibrium partial pressure of CO2 at room temperature?
51.
In the laboratory, hydrogen chloride (HCl(g)) and ammonia (NH3(g)) often escape from bottles of their solutions and react to form the ammonium chloride (NH4Cl(s)), the white glaze often seen on glassware. Assuming that the number of moles of each gas that escapes into the room is the same, what is the maximum partial pressure of HCl and NH3 in the laboratory at room temperature? (Hint: The partial pressures will be equal and are at their maximum value when at equilibrium.)
52.
Benzene can be prepared from acetylene. Determine the equilibrium constant at 25 °C and at 850 °C. Is the reaction spontaneous at either of these temperatures? Why is all acetylene not found as benzene?
53.
Carbon dioxide decomposes into CO and O2 at elevated temperatures. What is the equilibrium partial pressure of oxygen in a sample at 1000 °C for which the initial pressure of CO2 was 1.15 atm?
54.
Carbon tetrachloride, an important industrial solvent, is prepared by the chlorination of methane at 850 K.
What is the equilibrium constant for the reaction at 850 K? Would the reaction vessel need to be heated or cooled to keep the temperature of the reaction constant?
55.
Acetic acid, CH3CO2H, can form a dimer, (CH3CO2H)2, in the gas phase.
The dimer is held together by two hydrogen bonds with a total strength of 66.5 kJ per mole of dimer.
At 25 °C, the equilibrium constant for the dimerization is 1.3 103 (pressure in atm). What is ΔS° for the reaction?
56.
Determine ΔGº for the following reactions.
(a) Antimony pentachloride decomposes at 448 °C. The reaction is:
An equilibrium mixture in a 5.00 L flask at 448 °C contains 3.85 g of SbCl5, 9.14 g of SbCl3, and 2.84 g of Cl2.
(b) Chlorine molecules dissociate according to this reaction:
1.00% of Cl2 molecules dissociate at 975 K and a pressure of 1.00 atm.
57.
Given that the for Pb2+(aq) and Cl(aq) is −24.3 kJ/mole and −131.2 kJ/mole respectively, determine the solubility product, Ksp, for PbCl2(s).
58.
Determine the standard free energy change, for the formation of S2−(aq) given that the for Ag+(aq) and Ag2S(s) are 77.1 kJ/mole and −39.5 kJ/mole respectively, and the solubility product for Ag2S(s) is 8 10−51.
59.
Determine the standard enthalpy change, entropy change, and free energy change for the conversion of diamond to graphite. Discuss the spontaneity of the conversion with respect to the enthalpy and entropy changes. Explain why diamond spontaneously changing into graphite is not observed.
60.
The evaporation of one mole of water at 298 K has a standard free energy change of 8.58 kJ.
1. Is the evaporation of water under standard thermodynamic conditions spontaneous?
2. Determine the equilibrium constant, KP, for this physical process.
3. By calculating ∆G, determine if the evaporation of water at 298 K is spontaneous when the partial pressure of water, is 0.011 atm.
4. If the evaporation of water were always nonspontaneous at room temperature, wet laundry would never dry when placed outside. In order for laundry to dry, what must be the value of in the air?
61.
In glycolysis, the reaction of glucose (Glu) to form glucose-6-phosphate (G6P) requires ATP to be present as described by the following equation:
In this process, ATP becomes ADP summarized by the following equation:
Determine the standard free energy change for the following reaction, and explain why ATP is necessary to drive this process:
62.
One of the important reactions in the biochemical pathway glycolysis is the reaction of glucose-6-phosphate (G6P) to form fructose-6-phosphate (F6P):
1. Is the reaction spontaneous or nonspontaneous under standard thermodynamic conditions?
2. Standard thermodynamic conditions imply the concentrations of G6P and F6P to be 1 M, however, in a typical cell, they are not even close to these values. Calculate ΔG when the concentrations of G6P and F6P are 120 μM and 28 μM respectively, and discuss the spontaneity of the forward reaction under these conditions. Assume the temperature is 37 °C.
63.
Without doing a numerical calculation, determine which of the following will reduce the free energy change for the reaction, that is, make it less positive or more negative, when the temperature is increased. Explain.
64.
When ammonium chloride is added to water and stirred, it dissolves spontaneously and the resulting solution feels cold. Without doing any calculations, deduce the signs of ΔG, ΔH, and ΔS for this process, and justify your choices.
65.
An important source of copper is from the copper ore, chalcocite, a form of copper(I) sulfide. When heated, the Cu2S decomposes to form copper and sulfur described by the following equation:
1. Determine for the decomposition of Cu2S(s).
2. The reaction of sulfur with oxygen yields sulfur dioxide as the only product. Write an equation that describes this reaction, and determine for the process.
3. The production of copper from chalcocite is performed by roasting the Cu2S in air to produce the Cu. By combining the equations from Parts (a) and (b), write the equation that describes the roasting of the chalcocite, and explain why coupling these reactions together makes for a more efficient process for the production of the copper.
66.
What happens to (becomes more negative or more positive) for the following chemical reactions when the partial pressure of oxygen is increased? | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/16%3A_Thermodynamics/16.08%3A_Exercises.txt |
Electrochemistry deals with chemical reactions that produce electricity and the changes associated with the passage of electrical current through matter. The reactions involve electron transfer, and so they are oxidation-reduction (or redox) reactions. Many metals may be purified or electroplated using electrochemical methods. Devices such as automobiles, smartphones, electronic tablets, watches, pacemakers, and many others use batteries for power. Batteries use chemical reactions that produce electricity spontaneously and that can be converted into useful work. All electrochemical systems involve the transfer of electrons in a reacting system. In many systems, the reactions occur in a region known as the cell, where the transfer of electrons occurs at electrodes.
• 17.0: Introduction
• 17.1: Review of Redox Chemistry
An electric current consists of moving charge. The charge may be in the form of electrons or ions. Current flows through an unbroken or closed circular path called a circuit. The current flows through a conducting medium as a result of a difference in electrical potential between two points in a circuit. Electrical potential has the units of energy per charge. In SI units, charge is measured in coulombs (C), current in amperes, and electrical potential in volts.
• 17.2: Galvanic Cells
Electrochemical cells typically consist of two half-cells. The half-cells separate the oxidation half-reaction from the reduction half-reaction and make it possible for current to flow through an external wire. One half-cell contains the anode. Oxidation occurs at the anode. The anode is connected to the cathode in the other half-cell. Reduction occurs at the cathode. Adding a salt bridge completes the circuit allowing current to flow.
• 17.3: Electrode and Cell Potentials
Assigning the potential of the standard hydrogen electrode (SHE) as zero volts allows the determination of standard reduction potentials, E°, for half-reactions in electrochemical cells. As the name implies, standard reduction potentials use standard states (1 bar or 1 atm for gases; 1 M for solutes, often at 298.15 K) and are written as reductions (where electrons appear on the left side of the equation).
• 17.4: Potential, Free Energy, and Equilibrium
Electrical work is the negative of the product of the total charge (Q) and the cell potential (Ecell). The total charge can be calculated as the number of moles of electrons (n) times the Faraday constant (F = 96,485 C/mol e−). Electrical work is the maximum work that the system can produce and so is equal to the change in free energy. Thus, anything that can be done with or to a free energy change can also be done to or with a cell potential.
• 17.5: Batteries and Fuel Cells
Batteries are galvanic cells, or a series of cells, that produce an electric current. When cells are combined into batteries, the potential of the battery is an integer multiple of the potential of a single cell. There are two basic types of batteries: primary and secondary. Primary batteries are “single use” and cannot be recharged. Dry cells and (most) alkaline batteries are examples of primary batteries. The second type is rechargeable and is called a secondary battery.
• 17.6: Corrosion
Corrosion is the degradation of a metal caused by an electrochemical process. Large sums of money are spent each year repairing the effects of, or preventing, corrosion. Some metals, such as aluminum and copper, produce a protective layer when they corrode in air. The thin layer that forms on the surface of the metal prevents oxygen from coming into contact with more of the metal atoms and thus “protects” the remaining metal from further corrosion. Iron corrodes (forms rust) when exposed to wate
• 17.7: Electrolysis
Using electricity to force a nonspontaneous process to occur is electrolysis. Electrolytic cells are electrochemical cells with negative cell potentials (meaning a positive Gibbs free energy), and so are nonspontaneous. Electrolysis can occur in electrolytic cells by introducing a power supply, which supplies the energy to force the electrons to flow in the nonspontaneous direction. Electrolysis is done in solutions, which contain enough ions so current can flow.
• 17.8: Key Terms
• 17.9: Key Equations
• 17.10: Summary
• 17.11: Exercises
These are homework exercises to accompany the Textmap created for "Chemistry" by OpenStax. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
17: Electrochemistry
Another chapter in this text introduced the chemistry of reduction-oxidation (redox) reactions. This important reaction class is defined by changes in oxidation states for one or more reactant elements, and it includes a subset of reactions involving the transfer of electrons between reactant species. Around the turn of the nineteenth century, chemists began exploring ways these electrons could be transferred indirectly via an external circuit rather than directly via intimate contact of redox reactants. In the two centuries since, the field of electrochemistry has evolved to yield significant insights on the fundamental aspects of redox chemistry as well as a wealth of technologies ranging from industrial-scale metallurgical processes to robust, rechargeable batteries for electric vehicles (Figure \(1\)). In this chapter, the essential concepts of electrochemistry will be addressed. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/17%3A_Electrochemistry/17.00%3A_Introduction.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe defining traits of redox chemistry
• Identify the oxidant and reductant of a redox reaction
• Balance chemical equations for redox reactions using the half-reaction method
Since reactions involving electron transfer are essential to the topic of electrochemistry, a brief review of redox chemistry is provided here that summarizes and extends the content of an earlier text chapter (see chapter on reaction stoichiometry). Readers wishing additional review are referred to the text chapter on reaction stoichiometry.
Oxidation Numbers
By definition, a redox reaction is one that entails changes in oxidation number (or oxidation state) for one or more of the elements involved. The oxidation number of an element in a compound is essentially an assessment of how the electronic environment of its atoms is different in comparison to atoms of the pure element. By this description, the oxidation number of an atom in an element is equal to zero. For an atom in a compound, the oxidation number is equal to the charge the atom would have in the compound if the compound were ionic. Consequential to these rules, the sum of oxidation numbers for all atoms in a molecule is equal to the charge on the molecule. To illustrate this formalism, examples from the two compound classes, ionic and covalent, will be considered.
Simple ionic compounds present the simplest examples to illustrate this formalism, since by definition the elements’ oxidation numbers are numerically equivalent to ionic charges. Sodium chloride, NaCl, is comprised of Na+ cations and Cl anions, and so oxidation numbers for sodium and chlorine are, +1 and −1, respectively. Calcium fluoride, CaF2, is comprised of Ca2+ cations and F anions, and so oxidation numbers for calcium and fluorine are, +2 and −1, respectively.
Covalent compounds require a more challenging use of the formalism. Water is a covalent compound whose molecules consist of two H atoms bonded separately to a central O atom via polar covalent O−H bonds. The shared electrons comprising an O−H bond are more strongly attracted to the more electronegative O atom, and so it acquires a partial negative charge in the water molecule (relative to an O atom in elemental oxygen). Consequently, H atoms in a water molecule exhibit partial positive charges compared to H atoms in elemental hydrogen. The sum of the partial negative and partial positive charges for each water molecule is zero, and the water molecule is neutral.
Imagine that the polarization of shared electrons within the O−H bonds of water were 100% complete—the result would be transfer of electrons from H to O, and water would be an ionic compound comprised of O2− anions and H+ cations. And so, the oxidations numbers for oxygen and hydrogen in water are −2 and +1, respectively. Applying this same logic to carbon tetrachloride, CCl4, yields oxidation numbers of +4 for carbon and −1 for chlorine. In the nitrate ion, $\ce{NO3^{-}}$, the oxidation number for nitrogen is +5 and that for oxygen is −2, summing to equal the 1− charge on the molecule:
$(1 N \text { atom })\left(\frac{+5}{ N \text { atom }}\right)+(3 O \text { atoms })\left(\frac{-2}{ O \text { atom }}\right)=+5+-6=-1 \nonumber$
Balancing Redox Equations
The unbalanced equation below describes the decomposition of molten sodium chloride:
$\ce{NaCl(l) -> Na(l) + Cl2(g)} \tag{unbalanced}$
This reaction satisfies the criterion for redox classification, since the oxidation number for Na is decreased from +1 to 0 (it undergoes reduction) and that for Cl is increased from −1 to 0 (it undergoes oxidation). The equation in this case is easily balanced by inspection, requiring stoichiometric coefficients of 2 for the NaCl and Na:
$\ce{2 NaCl(l) -> 2 Na(l) + Cl2(g)} \tag{balanced}$
Redox reactions that take place in aqueous solutions are commonly encountered in electrochemistry, and many involve water or its characteristic ions, H+(aq) and OH(aq), as reactants or products. In these cases, equations representing the redox reaction can be very challenging to balance by inspection, and the use of a systematic approach called the half-reaction method is helpful.
Balancing REDOX reactions via the half-reaction Method
This approach involves the following steps:
1. Write skeletal equations for the oxidation and reduction half-reactions.
2. Balance each half-reaction for all elements except H and O.
3. Balance each half-reaction for O by adding H2O.
4. Balance each half-reaction for H by adding H+.
5. Balance each half-reaction for charge by adding electrons.
6. If necessary, multiply one or both half-reactions so that the number of electrons consumed in one is equal to the number produced in the other.
7. Add the two half-reactions and simplify.
8. If the reaction takes place in a basic medium, add OH ions the equation obtained in step 7 to neutralize the H+ ions (add in equal numbers to both sides of the equation) and simplify.
The examples below demonstrate the application of this method to balancing equations for aqueous redox reactions.
Example $1$: Balancing Equations for Redox Reactions in Acidic Solutions
Write the balanced equation representing reaction between solid copper and nitric acid to yield aqueous copper(II) ions and nitrogen monoxide gas.
Solution
Following the steps of the half-reaction method:
Step 1: Write skeletal equations for the oxidation and reduction half-reactions.
\begin{align*} &\text { oxidation: } \quad \ce{Cu (s) -> Cu^{2+}(aq)} \[4pt] &\text { reduction: } \quad \ce{HNO3(aq) -> NO(g)} \end{align*} \nonumber
Step 2: Balance each half-reaction for all elements except $\ce{H}$ and $\ce{O}$.
\begin{align*} &\text { oxidation: } \quad\ce{Cu(s) -> Cu^{2+}(aq)} \[4pt] &\text { reduction: } \quad\ce{HNO3(aq) -> NO(g)} \end{align*} \nonumber
Step 3: Balance each half-reaction for $\ce{O}$ by adding $\ce{H2O}$.
\begin{align*} &\text { oxidation: } \quad \ce{Cu (s) -> Cu^{2+}(aq)} \[4pt] &\text { reduction: } \quad \ce{HNO3(aq) -> NO(g) + \mathbf{2 H2O(l)}} \end{align*} \nonumber
Step 4: Balance each half-reaction for $\ce{H}$ by adding $\ce{H^{+}}$.
\begin{align*} &\text { oxidation: } \quad \ce{Cu(s) -> Cu^{2+}(aq)} \[4pt] &\text { reduction: } \quad \ce{\mathbf{3 H^{+}(aq)} + HNO3(aq) -> NO(g) + 2 H2O(l)} \end{align*} \nonumber
Step 5: Balance each half-reaction for charge by adding electrons.
\begin{align*} &\text { oxidation: } \quad \ce{Cu(s) -> Cu^{2+}(aq) + \mathbf{ 2 e^{-}}} \[4pt] &\text {reduction: } \quad \ce{\mathbf{3 e^{-}} + 3 H^{+}(aq) + HNO3(aq) -> NO(g) + 2 H2O(l)} \end{align*} \nonumber
Step 6: If necessary, multiply one or both half-reactions so that the number of electrons consumed in one is equal to the number produced in the other.
\begin{align*} &\text { oxidation }(\times 3): \quad \ce{\mathbf{3} Cu(s) -> \mathbf{3} Cu^{2+}(aq) + \mathbf{6} \cancel{2} e^{-}} \[4pt] &\text {reduction }(\times 2): \quad \ce{\mathbf{6} \cancel{3} e^{-} + \mathbf{6} \cancel{3} H^{+}(aq) + \mathbf{2} HNO3(aq) -> \mathbf{2} NO(g) + \mathbf{4} \cancel{2} H2O (l)} \end{align*} \nonumber
Step 7: Add the two half-reactions and simplify.
\begin{align*} \ce{3 Cu(s) + \cancel{6 e^{-}} + 6 H^{+}(aq) +2 HNO3(aq) &-> 3 Cu^{2+}(aq) + \cancel{6 e^{-}} + 2 NO(g) + 4 H2O (l)} \[4pt] \ce{3 Cu(s) + 6 H^{+}(aq) + 2 HNO3(aq) &-> 3 Cu^{2+}(aq) + 2 NO(g) + 4 H2O (l)} \end{align*} \nonumber
Step 8: If the reaction takes place in a basic medium, add $\ce{OH^{−}}$ ions the equation obtained in step 7 to neutralize the $\ce{H^{+}}$ ions (add in equal numbers to both sides of the equation) and simplify.
This step not necessary since the solution is stipulated to be acidic.
The balanced equation for the reaction in an acidic solution is then
$\ce{3 Cu(s) + 6 H^{+}(aq) + 2 HNO3(aq) -> 3 Cu^{2+}(aq) + 2 NO(g) + 4 H2O(l)} \nonumber$
Exercise $1$
The reaction above results when using relatively diluted nitric acid. If concentrated nitric acid is used, nitrogen dioxide is produced instead of nitrogen monoxide. Write a balanced equation for this reaction.
Answer
$\ce{Cu(s) + 2 H^{+}(aq) + 2 HNO3(aq) -> Cu^{2+}(aq) + 2 NO2(g) + 2 H2O(l)} \nonumber$
Example $2$: Balancing Equations for Redox Reactions in Basic Solutions
Write the balanced equation representing reaction between aqueous permanganate ion, , and solid chromium(III) hydroxide, Cr(OH)3, to yield solid manganese(IV) oxide, MnO2, and aqueous chromate ion, The reaction takes place in a basic solution.
Solution
Step 1: Write skeletal equations for the oxidation and reduction half-reactions.
\begin{align*} &\text { oxidation: } \quad \ce{Cr(OH)3(s) -> CrO4^{2-}(aq)} \[4pt] &\text { reduction: } \quad \ce{MnO4^{-}(aq) -> MnO2(s)} \end{align*} \nonumber
Step 2: Balance each half-reaction for all elements except $\ce{H}$ and $\ce{O}$.
\begin{align*} &\text { oxidation: } \quad \ce{Cr(OH)3(s) -> CrO4^{2-}(aq)} \[4pt] &\text { reduction: } \quad \ce{MnO4^{-}(aq) -> MnO2(s)} \end{align*} \nonumber
Step 3: Balance each half-reaction for $\ce{O}$ by adding $\ce{H2O}$.
\begin{align*} &\text { oxidation: } \quad \ce{\mathbf{H2O(l)} + Cr(OH)3(s) -> CrO4^{2-}(aq)} \[4pt] &\text { reduction: } \quad \ce{MnO4^{-}(aq) -> MnO2(s) + \mathbf{2 H2O(l)}} \end{align*} \nonumber
Step 4: Balance each half-reaction for $\ce{H}$ by adding $\ce{H^{+}}$.
\begin{align*} &\text { oxidation: } \quad \ce{H2O(l) + Cr(OH)3(s) -> CrO4^{2-}(aq) + \mathbf{5H^{+}(aq)} } \[4pt] &\text { reduction: } \quad \ce{\mathbf{4H^{+}(aq)} + MnO4^{-}(aq) -> MnO2(s) + 2 H2O(l)} \end{align*} \nonumber
Step 5: Balance each half-reaction for charge by adding electrons.
\begin{align*} &\text { oxidation: } \quad \ce{H2O(l) + Cr(OH)3(s) -> CrO4^{2-}(aq) + 5H^{+}(aq) + \mathbf{3 e^{-}}} \[4pt] &\text { reduction: } \quad \ce{\mathbf{3 e^{-}} + 4H^{+}(aq) + MnO4^{-}(aq) -> MnO2(s) + 2 H2O(l)} \end{align*} \nonumber
Step 6: If necessary, multiply one or both half-reactions so that the number of electrons consumed in one is equal to the number produced in the other.
This step is not necessary since the number of electrons is already in balance.
Step 7: Add the two half-reactions and simplify.
\begin{align*} \ce{\cancel{H2O(l)} + Cr(OH)3(s) &-> CrO4^{2-}(aq) + \cancelto{1}{5}H^{+}(aq) + \bcancel{3 e^{-}}} \[4pt] + \quad \quad \ce{\bcancel{3 e^{-}} + \cancel{4H^{+}(aq)} + MnO4^{-}(aq) &-> MnO2(s) + \cancelto{1}{2} H2O(l)} \[4pt] \hline \ce{Cr(OH)3(s) + MnO4^{-}(aq) &-> CrO4^{2-}(aq) + MnO2(s) + H^{+}(aq) + H2O(l)} \end{align*} \nonumber
Step 8: If the reaction takes place in a basic medium, add $\ce{OH^{−}}$ ions the equation obtained in step 7 to neutralize the $\ce{H^{+}}$ ions (add in equal numbers to both sides of the equation) and simplify.
$\ce{OH^{-}(aq) + Cr(OH)3(s) + MnO4^{-}(aq) -> CrO4^{2-}(aq) + MnO2(s) + \cancel{H^{+}(aq)} + H2O(l) + \cancel{OH^{-}(aq)}} \nonumber$
after neutralizing the $\ce{H^{+}}$, we get the balanced equation for the reaction in a basic solution
$\ce{OH^{-}(aq) + Cr(OH)3(s) + MnO4^{-}(aq) -> CrO4^{2-}(aq) + MnO2(s) + 2H2O(l)} \nonumber$
Exercise $2$
Aqueous permanganate ion may also be reduced using aqueous bromide ion, $\ce{Br^{-}}$, the products of this reaction being solid manganese(IV) oxide and aqueous bromate ion, $\ce{BrO3^{-}}$. Write the balanced equation for this reaction occurring in a basic medium.
Answer
$\ce{H2O(l) + 2 MnO4^{-}(aq) + Br^{-}(aq) -> 2 MnO2(s) + BrO3^{-}(aq) + 2 OH^{-}(aq)} \nonumber$ | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/17%3A_Electrochemistry/17.01%3A_Review_of_Redox_Chemistry.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the function of a galvanic cell and its components
• Use cell notation to symbolize the composition and construction of galvanic cells
As demonstration of spontaneous chemical change, Figure \(1\): shows the result of immersing a coiled wire of copper into an aqueous solution of silver nitrate. A gradual but visually impressive change spontaneously occurs as the initially colorless solution becomes increasingly blue, and the initially smooth copper wire becomes covered with a porous gray solid.
These observations are consistent with (i) the oxidation of elemental copper to yield copper(II) ions, Cu2+(aq), which impart a blue color to the solution, and (ii) the reduction of silver(I) ions to yield elemental silver, which deposits as a fluffy solid on the copper wire surface. And so, the direct transfer of electrons from the copper wire to the aqueous silver ions is spontaneous under the employed conditions. A summary of this redox system is provided by these equations:
Consider the construction of a device that contains all the reactants and products of a redox system like the one here, but prevents physical contact between the reactants. Direct transfer of electrons is, therefore, prevented; transfer, instead, takes place indirectly through an external circuit that contacts the separated reactants. Devices of this sort are generally referred to as electrochemical cells, and those in which a spontaneous redox reaction takes place are called galvanic cells (or voltaic cells).
A galvanic cell based on the spontaneous reaction between copper and silver(I) is depicted in Figure \(2\). The cell is comprised of two half-cells, each containing the redox conjugate pair (“couple”) of a single reactant. The half-cell shown at the left contains the Cu(0)/Cu(II) couple in the form of a solid copper foil and an aqueous solution of copper nitrate. The right half-cell contains the Ag(I)/Ag(0) couple as solid silver foil and an aqueous silver nitrate solution. An external circuit is connected to each half-cell at its solid foil, meaning the Cu and Ag foil each function as an electrode. By definition, the anode of an electrochemical cell is the electrode at which oxidation occurs (in this case, the Cu foil) and the cathode is the electrode where reduction occurs (the Ag foil). The redox reactions in a galvanic cell occur only at the interface between each half-cell’s reaction mixture and its electrode. To keep the reactants separate while maintaining charge-balance, the two half-cell solutions are connected by a tube filled with inert electrolyte solution called a salt bridge. The spontaneous reaction in this cell produces Cu2+ cations in the anode half-cell and consumes Ag+ ions in the cathode half-cell, resulting in a compensatory flow of inert ions from the salt bridge that maintains charge balance. Increasing concentrations of Cu2+ in the anode half-cell are balanced by an influx of NO3 from the salt bridge, while a flow of Na+ into the cathode half-cell compensates for the decreasing Ag+ concentration.
Cell Notation
Abbreviated symbolism is commonly used to represent a galvanic cell by providing essential information on its composition and structure. These symbolic representations are called cell notations or cell schematics, and they are written following a few guidelines:
• The relevant components of each half-cell are represented by their chemical formulas or element symbols
• All interfaces between component phases are represented by vertical parallel lines; if two or more components are present in the same phase, their formulas are separated by commas
• By convention, the schematic begins with the anode and proceeds left-to-right identifying phases and interfaces encountered within the cell, ending with the cathode
A verbal description of the cell as viewed from anode-to-cathode is often a useful first-step in writing its schematic. For example, the galvanic cell shown in Figure \(2\): consists of a solid copper anode immersed in an aqueous solution of copper(II) nitrate that is connected via a salt bridge to an aqueous silver(I) nitrate solution, immersed in which is a solid silver cathode. Converting this statement to symbolism following the above guidelines results in the cell schematic:
Consider a different galvanic cell (see Figure \(3\)) based on the spontaneous reaction between solid magnesium and aqueous iron(III) ions:
In this cell, a solid magnesium anode is immersed in an aqueous solution of magnesium chloride that is connected via a salt bridge to an aqueous solution containing a mixture of iron(III) chloride and iron(II) chloride, immersed in which is a platinum cathode. The cell schematic is then written as
Notice the cathode half-cell is different from the others considered thus far in that its electrode is comprised of a substance (Pt) that is neither a reactant nor a product of the cell reaction. This is required when neither member of the half-cell’s redox couple can reasonably function as an electrode, which must be electrically conductive and in a phase separate from the half-cell solution. In this case, both members of the redox couple are solute species, and so Pt is used as an inert electrode that can simply provide or accept electrons to redox species in solution. Electrodes constructed from a member of the redox couple, such as the Mg anode in this cell, are called active electrodes.
Example \(1\): Writing Galvanic Cell Schematics
A galvanic cell is fabricated by connecting two half-cells with a salt bridge, one in which a chromium wire is immersed in a 1 M CrCl3 solution and another in which a copper wire is immersed in 1 M CuCl2. Assuming the chromium wire functions as an anode, write the schematic for this cell along with equations for the anode half-reaction, the cathode half-reaction, and the overall cell reaction.
Solution
Since the chromium wire is stipulated to be the anode, the schematic begins with it and proceeds left-to-right, symbolizing the other cell components until ending with the copper wire cathode:
The half-reactions for this cell are
Multiplying to make the number of electrons lost by Cr and gained by Cu2+ equal yields
Adding the half-reaction equations and simplifying yields an equation for the cell reaction:
Exercise \(1\)
Omitting solute concentrations and spectator ion identities, write the schematic for a galvanic cell whose net cell reaction is shown below.
Answer | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/17%3A_Electrochemistry/17.02%3A_Galvanic_Cells.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe and relate the definitions of electrode and cell potentials
• Interpret electrode potentials in terms of relative oxidant and reductant strengths
• Calculate cell potentials and predict redox spontaneity using standard electrode potentials
Unlike the spontaneous oxidation of copper by aqueous silver(I) ions described previously, immersing a copper wire in an aqueous solution of lead(II) ions yields no reaction. The two species, Ag+(aq) and Pb2+(aq), thus show a distinct difference in their redox activity towards copper: the silver ion spontaneously oxidized copper, but the lead ion did not. Electrochemical cells permit this relative redox activity to be quantified by an easily measured property, potential. This property is more commonly called voltage when referenced in regard to electrical applications, and it is a measure of energy accompanying the transfer of charge. Potentials are measured in the volt unit, defined as one joule of energy per one coulomb of charge, V = J/C.
When measured for purposes of electrochemistry, a potential reflects the driving force for a specific type of charge transfer process, namely, the transfer of electrons between redox reactants. Considering the nature of potential in this context, it is clear that the potential of a single half-cell or a single electrode can’t be measured; “transfer” of electrons requires both a donor and recipient, in this case a reductant and an oxidant, respectively. Instead, a half-cell potential may only be assessed relative to that of another half-cell. It is only the difference in potential between two half-cells that may be measured, and these measured potentials are called cell potentials, Ecell, defined as
$E_{\text {cell }}= E_{\text {cathode }}- E_{\text {anode }} \nonumber$
where $E_{cathode}$ and $E_{anode}$ are the potentials of two different half-cells functioning as specified in the subscripts. As for other thermodynamic quantities, the standard cell potential, E°cell, is a cell potential measured when both half-cells are under standard-state conditions (1 M concentrations, 1 bar pressures, 298 K):
$E_{\text {cell }}^{\circ}= E_{\text {cathode }}^{\circ}- E_{\text {anode }}^{\circ} \nonumber$
To simplify the collection and sharing of potential data for half-reactions, the scientific community has designated one particular half-cell to serve as a universal reference for cell potential measurements, assigning it a potential of exactly 0 V. This half-cell is the standard hydrogen electrode (SHE) and it is based on half-reaction below:
$\ce{2 H^{+}(aq) + 2 e^{-} -> H2(g)} \nonumber$
A typical SHE contains an inert platinum electrode immersed in precisely 1 M aqueous H+ and a stream of bubbling H2 gas at 1 bar pressure, all maintained at a temperature of 298 K (see Figure $1$).
The assigned potential of the SHE permits the definition of a conveniently measured potential for a single half-cell. The electrode potential (EX) for a half-cell X is defined as the potential measured for a cell comprised of X acting as cathode and the SHE acting as anode:
\begin{aligned} & E_{\text {cell }}= E_{ X }- E_{ SHE } \[4pt] & E_{ SHE }=0 V \text { (defined) } \[4pt] & E_{\text {cell }}= E_{ X } \end{aligned} \nonumber
When the half-cell X is under standard-state conditions, its potential is the standard electrode potential, E°X. Since the definition of cell potential requires the half-cells function as cathodes, these potentials are sometimes called standard reduction potentials.
This approach to measuring electrode potentials is illustrated in Figure $2$: , which depicts a cell comprised of an SHE connected to a copper(II)/copper(0) half-cell under standard-state conditions. A voltmeter in the external circuit allows measurement of the potential difference between the two half-cells. Since the Cu half-cell is designated as the cathode in the definition of cell potential, it is connected to the red (positive) input of the voltmeter, while the designated SHE anode is connected to the black (negative) input. These connections insure that the sign of the measured potential will be consistent with the sign conventions of electrochemistry per the various definitions discussed above. A cell potential of +0.337 V is measured, and so
$E_{\text {cell }}^{\circ}= E_{\ce{Cu}}^{\circ}=+0.337 V \nonumber$
Tabulations of E° values for other half-cells measured in a similar fashion are available as reference literature to permit calculations of cell potentials and the prediction of the spontaneity of redox processes.
Table $1$ provides a listing of standard electrode potentials for a selection of half-reactions in numerical order, and a more extensive alphabetical listing is given in Appendix L.
Table $1$: Selected Standard Reduction Potentials at 25 °C
Half-Reaction E° (V)
$\ce{F2(g) + 2e^{-}-> 2F^{-}(aq)}$ +2.866
$\ce{PbO2(s) + SO4^{2-}(aq) + 4H^{+}(aq) + 2e^{-}-> PbSO4(s) + 2H2O(l)}$ +1.69
$\ce{MnO4^{-}(aq) + 8H^{+}(aq) + 5e^{-}-> Mn^{2+}(aq) + 4H2O(l)}$ +1.507
$\ce{Au^{3+}(aq) + 3e^{-}-> Au(s)}$ +1.498
$\ce{Cl2(g) + 2e^{-}-> 2Cl^{-}(aq)}$ +1.35827
$\ce{O2(g) + 4H^{+}(aq) + 4e^{-}-> 2H2O(l)}$ +1.229
$\ce{Pt^{2+}(aq) + 2e^{-}-> Pt(s)}$ +1.20
$\ce{Br2(aq) + 2e^{-}-> 2Br^{-}(aq)}$ +1.0873
$\ce{Ag^{+}(aq) + e^{-}-> Ag(s)}$ +0.7996
$\ce{Hg2^{2+}(aq) + 2e^{-}-> 2Hg(l)}$ +0.7973
$\ce{Fe^{3+}(aq) + e^{-} -> Fe^{2+}(aq)}$ +0.771
$\ce{MnO4^{-}(aq) + 2H2O(l) + 3e^{-}-> MnO2(s) + 4OH^{-}(aq)}$ +0.558
$\ce{I2(s) + 2e^{-}-> 2I^{-}(aq)}$ +0.5355
$\ce{NiO2(s) + 2H2O(l) + 2e^{-}-> Ni(OH)2(s) + 2OH^{-}(aq)}$ +0.49
$\ce{Cu^{2+}(aq) + 2e^{-}-> Cu(s)}$ +0.34
$\ce{Hg2Cl2(s) + 2e^{-}-> 2Hg(l) + 2Cl^{-}(aq)}$ +0.26808
$\ce{AgCl(s) + e^{-}-> Ag(s) + Cl^{-}(aq)}$ +0.22233
$\ce{Sn^{4+}(aq) + 2e^{-}-> Sn^{2+}(aq)}$ +0.151
$\ce{2H^{+}(aq) + 2e^{-}-> H2(g)}$ 0.00
$\ce{Pb^{2+}(aq) + 2e^{-}-> Pb(s)}$ -0.1262
$\ce{Sn^{2+}(aq) + 2e^{-}-> Sn(s)}$ -0.1375
$\ce{Ni^{2+}(aq) + 2e^{-}-> Ni(s)}$ -0.257
$\ce{Co^{2+}(aq) + 2e^{-}-> Co(s)}$ -0.28
$\ce{PbSO4(s) + 2e^{-}-> Pb(s) + SO4^{2-}(aq)}$ -0.3505
$\ce{Cd^{2+}(aq) + 2e^{-}-> Cd(s)}$ -0.4030
$\ce{Fe^{2+}(aq) + 2e^{-}-> Fe(s)}$ -0.447
$\ce{Cr^{3+}(aq) + 3e^{-}-> Cr(s)}$ -0.744
$\ce{Mn^{2+}(aq) + 2e^{-}-> Mn(s)}$ -1.185
$\ce{Zn(OH)2(s) + 2e^{-}-> Zn(s) + 2OH^{-}(aq)}$ -1.245
$\ce{Zn^{2+}(aq) + 2e^{-}-> Zn(s)}$ -0.7618
$\ce{Al^{3+}(aq) + 3e^{-}-> Al(s)}$ -1.662
$\ce{Mg2(aq) + 2e^{-}-> Mg(s)}$ -2.372
$\ce{Na^{+}(aq) + e^{-}-> Na(s)}$ -2.71
$\ce{Ca^{2+}(aq) + 2e^{-}-> Ca(s)}$ -2.868
$\ce{Ba^{2+}(aq) + 2e^{-}-> Ba(s)}$ -2.912
$\ce{K^{+}(aq) + e^{-}-> K(s)}$ -2.931
$\ce{K^{+}(aq) + e^{-}-> K(s)}$ -3.04
Example $1$: Calculating Standard Cell Potentials
What is the standard potential of the galvanic cell shown in Figure 17.3?
Solution
The cell in Figure 17.3 is galvanic, the spontaneous cell reaction involving oxidation of its copper anode and reduction of silver(I) ions at its silver cathode:
\begin{align*} & \text {anode: } \quad && \ce{Cu(s) -> Cu^{2+} + 2 e^{-}} \[4pt] & \text {cathode: } \quad && \ce{2Ag^{+}(aq) + 2 e^{-} -> 2Ag(s)}\[4pt] \hline &\text { cell: } \quad && \ce{Cu(s) +2Ag^{+}(aq) -> Cu^{2+} + 2Ag(s) } && E^o_{\text {cell }} = 0\,\text{V} \end{align*} \nonumber
The standard cell potential computed as
\begin{aligned} E_{\text {cell }}^{\circ} & = E_{\text {cathode }}^{\circ}- E_{\text {anode }}^{\circ} \ & = E_{ Ag }^{\circ}- E_{ Cu }^{\circ} \ & =0.7996 V -0.34 V \ & =+0.46 V \end{aligned} \nonumber
Exercise $1$
What is the standard cell potential expected if the silver cathode half-cell in Figure 17.3 is replaced with a lead half-cell:
$\ce{Pb^{2+}(aq) + 2 e^{-} -> Pb(s)} \nonumber$
Answer
−0. 47 V
Intrepreting Electrode and Cell Potentials
Thinking carefully about the definitions of cell and electrode potentials and the observations of spontaneous redox change presented thus far, a significant relation is noted. The previous section described the spontaneous oxidation of copper by aqueous silver(I) ions, but no observed reaction with aqueous lead(II) ions. Results of the calculations in Example $1$ have just shown the spontaneous process is described by a positive cell potential while the nonspontaneous process exhibits a negative cell potential. And so, with regard to the relative effectiveness (“strength”) with which aqueous Ag+ and Pb2+ ions oxidize Cu under standard conditions, the stronger oxidant is the one exhibiting the greater standard electrode potential, E°. Since by convention electrode potentials are for reduction processes, an increased value of $E^o$ corresponds to an increased driving force behind the reduction of the species (hence increased effectiveness of its action as an oxidizing agent on some other species). Negative values for electrode potentials are simply a consequence of assigning a value of 0 V to the SHE, indicating the reactant of the half-reaction is a weaker oxidant than aqueous hydrogen ions.
Applying this logic to the numerically ordered listing of standard electrode potentials in Table $1$ shows this listing to be likewise in order of the oxidizing strength of the half-reaction’s reactant species, decreasing from strongest oxidant (most positive E°) to weakest oxidant (most negative E°). Predictions regarding the spontaneity of redox reactions under standard state conditions can then be easily made by simply comparing the relative positions of their table entries. By definition, E°cell is positive when $E^o_{\text{cathode}} > E^o_{\text{anode}}$, and so any redox reaction in which the oxidant’s entry is above the reductant’s entry is predicted to be spontaneous.
Reconsideration of the two redox reactions in Example $1$ provides support for this fact. The entry for the silver(I)/silver(0) half-reaction is above that for the copper(II)/copper(0) half-reaction, and so the oxidation of Cu by Ag+ is predicted to be spontaneous (E°cathode > E°anode and so E°cell > 0). Conversely, the entry for the lead(II)/lead(0) half-cell is beneath that for copper(II)/copper(0), and the oxidation of Cu by Pb2+ is nonspontaneous (E°cathode < E°anode and so E°cell < 0).
Recalling the chapter on thermodynamics, the spontaneities of the forward and reverse reactions of a reversible process show a reciprocal relationship: if a process is spontaneous in one direction, it is non-spontaneous in the opposite direction. As an indicator of spontaneity for redox reactions, the potential of a cell reaction shows a consequential relationship in its arithmetic sign. The spontaneous oxidation of copper by lead(II) ions is not observed,
$\ce{Cu(s) + Pb^2+(aq) -> Cu^2+(aq) + Pb(s)} \hspace{20px} E^\circ_\mathrm{forward}=\mathrm{−0.47\: V (negative, non-spontaneous)} \nonumber$
and so the reverse reaction, the oxidation of lead by copper(II) ions, is predicted to occur spontaneously:
$\ce{Pb(s) + Cu^2+(aq) -> Pb^2+(aq) + Cu(s)} \hspace{20px} E^\circ_\mathrm{forward}=\mathrm{+0.47\: V (positive, spontaneous)} \nonumber$
Note that reversing the direction of a redox reaction effectively interchanges the identities of the cathode and anode half-reactions, and so the cell potential is calculated from electrode potentials in the reverse subtraction order than that for the forward reaction. In practice, a voltmeter would report a potential of −0.47 V with its red and black inputs connected to the Pb and Cu electrodes, respectively. If the inputs were swapped, the reported voltage would be +0.47 V.
Example $2$: Predicting Redox Spontaneity
Are aqueous iron(II) ions predicted to spontaneously oxidize elemental chromium under standard state conditions? Assume the half-reactions to be those available in Table $1$.
Solution
Referring to the tabulated half-reactions, the redox reaction in question can be represented by the equations below:
$\ce{Cr(s) + Fe^{2+}(aq) -> Cr^{3+}(aq) + Fe(s)} \nonumber$
The entry for the putative oxidant, Fe2+, appears above the entry for the reductant, Cr, and so a spontaneous reaction is predicted per the quick approach described above. Supporting this predication by calculating the standard cell potential for this reaction gives
\begin{aligned} E_{\text {cell }}^{\circ} & = E_{\text {cathode }}^{\circ}- E_{\text {anode }}^{\circ} \ & = E_{\text {Fe(II) }}^{\circ}- E_{\text {Cr }}^{\circ} \ & =-0.447 V --0.744 V =+0.297 V \end{aligned} \nonumber
The positive value for the standard cell potential indicates the process is spontaneous under standard state conditions.
Exercise $2$
Use the data in Table $1$ to predict the spontaneity of the oxidation of bromide ion by molecular iodine under standard state conditions, supporting the prediction by calculating the standard cell potential for the reaction. Repeat for the oxidation of iodide ion by molecular bromine.
Answer
\begin{align*} \ce{I2(s) + 2 Br^{-}(aq) &-> 2 I^{-}(aq) + Br2(l)} && E_{\text {cell }}=-0.5518\,\text{V} \text { (nonspontaneous) } \ \ce{Br2(s) + 2 I^{-}(aq) &-> 2 Br^{-}(aq) + I2(l)} && E_{\text {cell }}^{\circ}=+0.5518\,\text{V} \text { (spontaneous) } \end{align*} \nonumber | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/17%3A_Electrochemistry/17.03%3A_Electrode_and_Cell_Potentials.txt |
Learning Objectives
By the end of this section, you will be able to:
• Explain the relations between potential, free energy change, and equilibrium constants
• Perform calculations involving the relations between cell potentials, free energy changes, and equilibrium
• Use the Nernst equation to determine cell potentials under nonstandard conditions
So far in this chapter, the relationship between the cell potential and reaction spontaneity has been described, suggesting a link to the free energy change for the reaction (see chapter on thermodynamics). The interpretation of potentials as measures of oxidant strength was presented, bringing to mind similar measures of acid-base strength as reflected in equilibrium constants (see the chapter on acid-base equilibria). This section provides a summary of the relationships between potential and the related thermodynamic properties ΔG and K.
E° and ΔG°
The standard free energy change of a process, $ΔG^°$, was defined in a previous chapter as the maximum work that could be performed by a system, $w_{max}$. In the case of a redox reaction taking place within a galvanic cell under standard state conditions, essentially all the work is associated with transferring the electrons from reductant-to-oxidant, $w_{elec}$:
$\Delta G^{\circ}=w_{\max }=w_{\text {elec }} \nonumber$
The work associated with transferring electrons is determined by the total amount of charge (coulombs) transferred and the cell potential:
$\Delta G^{\circ}=w_{\text {elec }}=-n F E_{\text {cell }}^{\circ} \nonumber$
where $n$ is the number of moles of electrons transferred, $F$ is Faraday’s constant, and $E^°_{cell}$ is the standard cell potential. The relation between free energy change and standard cell potential confirms the sign conventions and spontaneity criteria previously discussed for both of these properties: spontaneous redox reactions exhibit positive potentials and negative free energy changes.
E° and K
Combining a previously derived relation between ΔG° and K (see the chapter on thermodynamics) and the equation above relating ΔG° and E°cell yields the following:
$\Delta G^{\circ}=-R T \ln K=-n F E_{\text {cell }}^{\circ} \label{eq1}$
with
$E_{\text {coll }}^{\circ}=\left(\frac{R T}{n F}\right) \ln K \nonumber$
This equation indicates redox reactions with large (positive) standard cell potentials will proceed far towards completion, reaching equilibrium when the majority of reactant has been converted to product. A summary of the relations between E°, ΔG° and K is depicted in Figure $1$, and a table correlating reaction spontaneity to values of these properties is provided in Table $1$.
Table $1$
K ΔG° E°cell Comment
> 1 < 0 > 0
Reaction is spontaneous under standard conditions
Products more abundant at equilibrium
< 1 > 0 < 0
Reaction is non-spontaneous under standard conditions
Reactants more abundant at equilibrium
= 1 = 0 = 0
Reaction is at equilibrium under standard conditions
Reactants and products equally abundant
Example $1$: Equilibrium Constants, Standard Cell Potentials, and Standard Free Energy Changes
Use data from Appendix L to calculate the standard cell potential, standard free energy change, and equilibrium constant for the following reaction at 25 °C. Comment on the spontaneity of the forward reaction and the composition of an equilibrium mixture of reactants and products.
$\ce{2 Ag^{+}(aq) + Fe(s) <=> 2 Ag(s) + Fe^{2+}(aq)} \nonumber$
Solution
The reaction involves an oxidation-reduction reaction, so the standard cell potential can be calculated using the data in Appendix L.
\begin{align*} & \text { anode: } \quad && \ce{Fe(s) -> Fe^{2+}(aq) + 2 e^{-}} &&E^o_{\ce{Fe^{2+}/Fe}} =-0.447\,\text{V}\[4pt] & \text {cathode: } \quad && \ce{2 \times (Ag^{+}(aq) + e^{-} -> Ag(s))} &&E^o_{\ce{Ag^{+}/Ag}} = +0.7996\,\text{V}\[4pt] \hline &\text { cell: } \quad && \ce{Fe(s) + 2Ag^{+}(aq) -> Fe^{2+}(aq) + 2Ag(s)} && E^o_{\text {cell }} = +1.2247\,\text{V} \end{align*} \nonumber
with
$E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}= E^o_{\ce{Fe^{2+}/Fe}} - E^o_{\ce{Ag^{+}/Ag}} = +1.2247\,\text{V} \nonumber$
With $n = 2$, the equilibrium constant is then
\begin{aligned} E_{\text {cell }}^{\circ} &=\frac{0.0592 V }{n} \log K \[4pt] K&=10^{n \times E_{\text {cal }}^o / 0.0592 V } \[4pt] &=10^{2 \times 1.247 V / 0.0592 V } \[4pt] &=10^{42.128} \[4pt] &=1.3 \times 10^{42} \end{aligned} \nonumber
The standard free energy is then Equation \ref{eq1}
\begin{aligned} & \Delta G^{\circ}=-n F E_{\text {cell }}^{\circ} \ & =-2 \times 96,485 \frac{ C }{ mol } \times 1.247 \frac{ J }{ C }=-240.6 \frac{ kJ }{ mol } \end{aligned} \nonumber
The reaction is spontaneous, as indicated by a negative free energy change and a positive cell potential. The K value is very large, indicating the reaction proceeds to near completion to yield an equilibrium mixture containing mostly products.
Exercise $1$
What is the standard free energy change and the equilibrium constant for the following reaction at room temperature? Is the reaction spontaneous?
$\ce{Sn(s) + 2 Cu^{2+}(aq) <=> Sn^{2+}(aq) + 2 Cu^{+}(aq)} \nonumber$
Answer
Spontaneous
$n = 2$
$E_{\text {cell }}^{\circ}=+0.291 V$
$[\Delta G^{\circ}=-56.2 \frac{ kJ }{ mol }$
K = 6.8 109.
Potentials at Nonstandard Conditions: The Nernst Equation
Most of the redox processes that interest science and society do not occur under standard state conditions, and so the potentials of these systems under nonstandard conditions are a property worthy of attention. Having established the relationship between potential and free energy change in this section, the previously discussed relation between free energy change and reaction mixture composition can be used for this purpose.
$\Delta G=\Delta G^{\circ}+R T \ln Q \nonumber$
Notice the reaction quotient, Q, appears in this equation, making the free energy change dependent upon the composition of the reaction mixture. Substituting the equation relating free energy change to cell potential yields the Nernst equation:
\begin{align*} -n F E_{\text {cell }}&=-n F E_{\text {cell }}^{\circ}+R T \ln Q \[4pt] E_{\text {cell }}&=E_{\text {cell }}^{\circ}-\frac{R T}{n F} \ln Q \end{align*} \nonumber
This equation describes how the potential of a redox system (such as a galvanic cell) varies from its standard state value, specifically, showing it to be a function of the number of electrons transferred, n, the temperature, T, and the reaction mixture composition as reflected in Q. A convenient form of the Nernst equation for most work is one in which values for the fundamental constants (R and F) and standard temperature (298) K), along with a factor converting from natural to base-10 logarithms, have been included:
$E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.0592 V }{n} \log Q \label{Nernst at room temperature}$
Example $2$: Predicting Redox Spontaneity Under Nonstandard Conditions
Use the Nernst equation to predict the spontaneity of the redox reaction shown below at room temperature.
$\ce{Co(s)} + \ce{Fe^{2+}} \text{(aq, 1.94 M)} \longrightarrow \ce{Co^{2+}}\text{(aq, 0.15 M)} + \ce{Fe (s)} \nonumber$
Solution
Collecting information from Appendix L and the problem,
\begin{align*} & \text { anode: } \quad && \ce{Co(s) -> Co^{2+}(aq) + 2 e^{-}} &&E^o_{\ce{Co^{2+}/Co}} =-0.28\,\text{V}\[4pt] & \text {cathode: } \quad && \ce{Fe^{2+}(aq) + 2 e^{-} -> Fe(s)} &&E^o_{\ce{Fe^{+}/Fe}} = -0.447\,\text{V}\[4pt] \hline &\text { cell: } \quad && \ce{Co(s) + Fe^{2+}(Aq) -> Fe(s) + Co^{2+}(aq)} && E^o_{\text {cell }} = -0.17\,\text{V} \end{align*} \nonumber
with
$E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}=-0.447 V -(-0.28 V )=-0.17 V \nonumber$
Notice the negative value of the standard cell potential indicates the process is not spontaneous under standard conditions. Substitution of the Nernst equation terms for the nonstandard conditions requires calculating the reaction quotient $Q$
$Q =\frac{\left[\ce{Co^{2+}}\right]}{\left[\ce{Fe^{2+}}\right]}=\dfrac{0.15 M}{1.94 M}=0.077 \nonumber$
them $Q$ and $n$ are substituted into Equation \red{Nernst at room temperature}
\begin{aligned} E_{\text {cell }} & =E_{\text {cell }}^{\circ}-\frac{0.0592 \,\text{V} }{n} \log Q \[4pt] & =-0.17 \,\text{V} -\frac{0.0592 \,\text{V} }{2} \log 0.077 \[4pt] & =-0.17 \,\text{V} +0.033 \,\text{V} \[4pt] &=-0.14\,\text{V} \end{aligned} \nonumber
The cell potential remains negative (slightly) under the specified conditions, and so the reaction remains nonspontaneous.
Exercise $2$
For the cell schematic below at room temperature, identify values for n and Q, and calculate the cell potential, Ecell.
$\ce{Al(s)} | \ce{Al^{3+}}\text{(aq, 0.15 M)} | \ce{Cu^{2+}}\text{(aq, 0.025 M)} | \ce{Cu(s)} \nonumber$
Answer
n = 6; Q = 1440; Ecell = +1.97 V, spontaneous.
A concentration cell is constructed by connecting two nearly identical half-cells, each based on the same half-reaction and using the same electrode, varying only in the concentration of one redox species. The potential of a concentration cell, therefore, is determined only by the difference in concentration of the chosen redox species. The example problem below illustrates the use of the Nernst equation in calculations involving concentration cells.
Example $3$: Concentration Cells
What is the cell potential of the concentration cell operating at room temperature described by
$\ce{Zn(s)} | \ce{Zn^{2+}}\text{(aq, 0.10 M)} | \ce{Zn^{2+}} \text{(aq, 0.50 M)} | \ce{Zn(s)} \nonumber$
Solution
From the information given:
\begin{align*} & \text { anode: } \quad && \ce{Zn(s) -> Zn^{2+}}\text{(aq, 0.10 M)} + \ce{2 e^{-}} &&E^o_{\ce{Co^{2+}/Co}} =-0.7618\,\text{V}\[4pt] & \text {cathode: } \quad && \ce{Zn^{2+}}\text{(aq, 0.50 M)} + \ce{2 e^{-} -> Zn(s)} &&E^o_{\ce{Fe^{+}/Fe}} = -0,7618\,\text{V}\[4pt] \hline &\text { cell: } \quad && \ce{\cancel{Zn(s)}} + \ce{Zn^{2+}} \text{(aq, 0.50 M)} \ce{-> Zn^{2+}}\text{(aq, 0.10 M)} + \cancel{\ce{Zn(s)}} && E^o_{\text {cell }} = 0\,\text{V} \end{align*} \nonumber
Substituting into the Nernst equation (Equation \ref{Nernst at room temperature}),
$E_{\text {cell }}=0.000\,\text{V} -\frac{0.0592\,\text{V} }{2} \log \dfrac{0.10}{0.50} = +0.021\, \text{V} \nonumber$
The positive value for cell potential indicates the overall cell reaction (see above) is spontaneous. This spontaneous reaction is one in which the zinc ion concentration in the cathode falls (it is reduced to elemental zinc) while that in the anode rises (it is produced by oxidation of the zinc anode). A greater driving force for zinc reduction is present in the cathode, where the zinc(II) ion concentration is greater (Ecathode > Eanode).
Exercise $1$
The concentration cell above was allowed to operate until the cell reaction reached equilibrium at room temperature. What are the cell potential and the concentrations of zinc(II) in each half-cell for the cell now?
Answer
Ecell = 0.000 V; [Zn2+]cathode = [Zn2+]anode = 0.30 M | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/17%3A_Electrochemistry/17.04%3A_Potential_Free_Energy_and_Equilibrium.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the electrochemistry associated with several common batteries
• Distinguish the operation of a fuel cell from that of a battery
There are many technological products associated with the past two centuries of electrochemistry research, none more immediately obvious than the battery. A battery is a galvanic cell that has been specially designed and constructed in a way that best suits its intended use a source of electrical power for specific applications. Among the first successful batteries was the Daniell cell, which relied on the spontaneous oxidation of zinc by copper(II) ions (Figure $1$):
$\ce{Zn(s) + Cu^{2+}(aq) -> Zn^{2+}(aq) + Cu(s)} \nonumber$
Modern batteries exist in a multitude of forms to accommodate various applications, from tiny button batteries that provide the modest power needs of a wristwatch to the very large batteries used to supply backup energy to municipal power grids. Some batteries are designed for single-use applications and cannot be recharged (primary cells), while others are based on conveniently reversible cell reactions that allow recharging by an external power source (secondary cells). This section will provide a summary of the basic electrochemical aspects of several batteries familiar to most consumers, and will introduce a related electrochemical device called a fuel cell that can offer improved performance in certain applications.
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Single-Use Batteries
A common primary battery is the dry cell, which uses a zinc can as both container and anode (“–” terminal) and a graphite rod as the cathode (“+” terminal). The Zn can is filled with an electrolyte paste containing manganese(IV) oxide, zinc(II) chloride, ammonium chloride, and water. A graphite rod is immersed in the electrolyte paste to complete the cell. The spontaneous cell reaction involves the oxidation of zinc:
$\ce{Zn(s) -> Zn^{2+}(aq) +2 e^{-}} \tag{anode}$
and the reduction of manganese(IV)
$\ce{2 MnO2(s) + 2 NH4Cl(aq) + 2 e^{-} -> Mn2O3(s) + 2 NH3(aq) + H2O(l) + 2 Cl^{-}} \tag{cathode}$
which together yield the cell reaction:
$\ce{2 MnO2(s) + 2NH4Cl(aq) + Zn(s) -> Zn^{2+}(aq) + Mn2O3(s) + 2 NH3(aq) + H2O(l) + 2 Cl^{-}} \tag{cell}$
The voltage (cell potential) of a dry cell is approximately 1.5 V. Dry cells are available in various sizes (e.g., D, C, AA, AAA). All sizes of dry cells comprise the same components, and so they exhibit the same voltage, but larger cells contain greater amounts of the redox reactants and therefore are capable of transferring correspondingly greater amounts of charge. Like other galvanic cells, dry cells may be connected in series to yield batteries with greater voltage outputs, if needed.
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Alkaline batteries (Figure $3$) were developed in the 1950s to improve on the performance of the dry cell, and they were designed around the same redox couples. As their name suggests, these types of batteries use alkaline electrolytes, often potassium hydroxide. The reactions are
\begin{align*}
& \text { anode: } \quad && \ce{Zn(s) + 2OH^{-}(aq) -> ZnO(s) + H2O(l) + 2e^{-}} \4pt] & \text {cathode: } \quad && \ce{2 MnO2(s) + H2O(l) + 2 e^{-} -> Mn2O3(s) + 2 OH^{-}(aq)} \[4pt] \hline &\text { cell: } \quad && \ce{Zn(s) + 2 MnO2(s) -> ZnO(s) + Mn2O3(s)} \quad \quad \quad E_{\text {cell }}=+1.43\,\text{V} \end{align*} An alkaline battery can deliver about three to five times the energy of a zinc-carbon dry cell of similar size. Alkaline batteries are prone to leaking potassium hydroxide, so they should be removed from devices for long-term storage. While some alkaline batteries are rechargeable, most are not. Attempts to recharge an alkaline battery that is not rechargeable often leads to rupture of the battery and leakage of the potassium hydroxide electrolyte. Link to Learning Visit this site to learn more about alkaline batteries. Rechargeable (Secondary) Batteries Nickel-cadmium, or NiCd, batteries (Figure $4$) consist of a nickel-plated cathode, cadmium-plated anode, and a potassium hydroxide electrode. The positive and negative plates, which are prevented from shorting by the separator, are rolled together and put into the case. This is a “jelly-roll” design and allows the NiCd cell to deliver much more current than a similar-sized alkaline battery. The reactions are \[\begin{align*} & \text { anode: } \quad && \ce{Cd(s) + 2OH^{-}(aq) -> Cd(OH)2(s) + 2 e^{-}} \[4pt] & \text {cathode: } \quad && \ce{NiO2(s) + 2 H2O(l) + 2 e^{-} -> Ni(OH)2(s) + 2OH^{-}(aq)} \[4pt] \hline &\text { cell: } \quad && \ce{Cd(s) + NiO2(s) + 2 H2O(l) -> Cd(OH)2(s) + Ni(OH)2(s)} \quad \quad \quad E_{\text {cell }}\approx+1.2 \,\text{V} \end{align*} \nonumber
When properly treated, a NiCd battery can be recharged about 1000 times. Cadmium is a toxic heavy metal so NiCd batteries should never be ruptured or incinerated, and they should be disposed of in accordance with relevant toxic waste guidelines.
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Lithium ion batteries (Figure $5$) are among the most popular rechargeable batteries and are used in many portable electronic devices. The reactions are
\begin{align*} & \text { anode: } \quad && \ce{LiCoO2 -> Li_{1-x}CoO2 + x~ Li^{+} + x~ e^{-}} \[4pt] & \text {cathode: } \quad && \ce{x~ Li^{+} + x~e^{-} + x~ C6 -> x~ LiC6} \[4pt] \hline &\text { cell: } \quad && \ce{LiCoO2 + x~ C6 -> Li_{1-x}CoO2 + x~ LiC6} \quad \quad \quad E_{\text {cell }}\approx+3.7\,\text{V} \end{align*} \nonumber
The variable stoichiometry of the cell reaction leads to variation in cell voltages, but for typical conditions, x is usually no more than 0.5 and the cell voltage is approximately 3.7 V. Lithium batteries are popular because they can provide a large amount current, are lighter than comparable batteries of other types, produce a nearly constant voltage as they discharge, and only slowly lose their charge when stored.
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The lead acid battery (Figure $6$) is the type of secondary battery commonly used in automobiles. It is inexpensive and capable of producing the high current required by automobile starter motors. The reactions for a lead acid battery are
\begin{align*} & \text { anode: } \quad && \ce{Pb(s) + HSO4^{-}(aq) -> PbSO4(s) + H^{+}(aq) + 2 e^{-}} \[4pt] & \text {cathode: } \quad && \ce{PbO2(s) + HSO4^{-}(aq) + 3 H^{+}(aq) + 2 e^{-} -> PbSO4(s) + 2 H2O(l)} \[4pt] \hline &\text { cell: } \quad && \ce{Pb(s) + PbO2(s) + 2 H2SO4(aq) -> 2 PbSO4(s) + 2 H2O (l)} \quad \quad \quad E_{\text {cell }}\approx +2\,\text{V} \end{align*} \nonumber
Each cell produces 2 V, so six cells are connected in series to produce a 12-V car battery. Lead acid batteries are heavy and contain a caustic liquid electrolyte, H2SO4(aq), but are often still the battery of choice because of their high current density. Since these batteries contain a significant amount of lead, they must always be disposed of properly.
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Fuel Cells
A fuel cell is a galvanic cell that uses traditional combustive fuels, most often hydrogen or methane, that are continuously fed into the cell along with an oxidant. (An alternative, but not very popular, name for a fuel cell is a flow battery.) Within the cell, fuel and oxidant undergo the same redox chemistry as when they are combusted, but via a catalyzed electrochemical that is significantly more efficient. For example, a typical hydrogen fuel cell uses graphite electrodes embedded with platinum-based catalysts to accelerate the two half-cell reactions:
\begin{align*} & \text { anode: } \quad && \ce{2 H2(g) -> 4 H^{+}(aq) + 4 e^{-}} \[4pt] & \text {cathode: } \quad && \ce{O2(g) + 4 H^{+}(aq) + 4 e^{-} -> 2 H2O (g)} \[4pt] \hline &\text { cell: } \quad && \ce{2 H2(g) + O2(g) -> 2 H2O (g)} \quad \quad \quad E_{\text {cell }}\approx+1.2\,\text{V} \end{align*} \nonumber
These types of fuel cells generally produce voltages of approximately 1.2 V. Compared to an internal combustion engine, the energy efficiency of a fuel cell using the same redox reaction is typically more than double (~20%–25% for an engine versus ~50%–75% for a fuel cell). Hydrogen fuel cells are commonly used on extended space missions, and prototypes for personal vehicles have been developed, though the technology remains relatively immature.
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Learning Objectives
By the end of this section, you will be able to:
• Define corrosion
• List some of the methods used to prevent or slow corrosion
Corrosion is usually defined as the degradation of metals by a naturally occurring electrochemical process. The formation of rust on iron, tarnish on silver, and the blue-green patina that develops on copper are all examples of corrosion. The total cost of corrosion remediation in the United States is significant, with estimates in excess of half a trillion dollars a year.
Chemistry in Everyday Life: Statue of Liberty: Changing Colors
The Statue of Liberty is a landmark every American recognizes. The Statue of Liberty is easily identified by its height, stance, and unique blue-green color (Figure $1$). When this statue was first delivered from France, its appearance was not green. It was brown, the color of its copper “skin.” So how did the Statue of Liberty change colors? The change in appearance was a direct result of corrosion. The copper that is the primary component of the statue slowly underwent oxidation from the air. The oxidation-reduction reactions of copper metal in the environment occur in several steps. Copper metal is oxidized to copper(I) oxide (Cu2O), which is red, and then to copper(II) oxide, which is black
\begin{align} \ce{2 Cu(s) + 1/2 O2(g) -> Cu_2 O (s)} \tag{red} \[4pt] \ce{2 Cu2O(s) + 1/2 O2(g) -> 2CuO (s)} \tag{black} \end{align}
Coal, which was often high in sulfur, was burned extensively in the early part of the last century. As a result, atmospheric sulfur trioxide, carbon dioxide, and water all reacted with the $\ce{CuO}$:
\begin{align*} \ce{2 CuO(s) + CO2(g) + H2O (l) &-> Cu2CO3(OH)2(s)} \tag{green}\[4pt] \ce{3 CuO(s) +2 CO2(g) + H2O (l) &-> Cu2(CO3)2(OH)2(s)} \tag{blue} \[4pt] \ce{4 CuO(s) + SO3(g) + 3 H2O (l) &-> Cu4SO4(OH)6(s)} \tag{green} \end{align*} \nonumber
These three compounds are responsible for the characteristic blue-green patina seen on the Statue of Liberty (and other outdoor copper structures). Fortunately, formation of patina creates a protective layer on the copper surface, preventing further corrosion of the underlying copper. The formation of the protective layer is called passivation, a phenomenon discussed further in another chapter of this text.
Perhaps the most familiar example of corrosion is the formation of rust on iron. Iron will rust when it is exposed to oxygen and water. Rust formation involves the creation of a galvanic cell at an iron surface, as illustrated in Figure $1$. The relevant redox reactions are described by the following equations:
\begin{align*} &\text { anode: } & \ce{Fe(s) \longrightarrow Fe^{2+}(aq) + 2 e^{-}} && E_{\ce{Fe^{2+} / Fe} }^{\circ}=-0.44\,\text{V} \[4pt] &\text { cathode: } & \ce{O2(g) + 4 H^{+}(aq) + 4 e^{-} -> 2 H2O (l)} && E_{\ce{O2 /O^{2-}}}^{\circ}=+1.23\,\text{V} \[4pt] \hline &\text { overall: } & \ce{2 Fe(s) + O2(g) + 4 H^{+}(aq) -> 2 Fe^{2+}(aq) + 2 H2O (l)} && E_{\text {cell }}^{\circ}=+1.67\,\text{V} \end{align*} \nonumber
Further reaction of the iron(II) product in humid air results in the production of an iron(III) oxide hydrate known as rust:
$\ce{4 Fe^{2+}(aq) + O2(g) + (4+2 x) ~H2O(l) -> 2 Fe2O3 \cdot x~ H2O(s) + 8 H^{+}(aq)} \nonumber$
The stoichiometry of the hydrate varies, as indicated by the use of x in the compound formula. Unlike the patina on copper, the formation of rust does not create a protective layer and so corrosion of the iron continues as the rust flakes off and exposes fresh iron to the atmosphere.
One way to keep iron from corroding is to keep it painted. The layer of paint prevents the water and oxygen necessary for rust formation from coming into contact with the iron. As long as the paint remains intact, the iron is protected from corrosion.
Other strategies include alloying the iron with other metals. For example, stainless steel is an alloy of iron containing a small amount of chromium. The chromium tends to collect near the surface, where it corrodes and forms a passivating an oxide layer that protects the iron.
Iron and other metals may also be protected from corrosion by galvanization, a process in which the metal to be protected is coated with a layer of a more readily oxidized metal, usually zinc. When the zinc layer is intact, it prevents air from contacting the underlying iron and thus prevents corrosion. If the zinc layer is breached by either corrosion or mechanical abrasion, the iron may still be protected from corrosion by a cathodic protection process, which is described in the next paragraph.
Another important way to protect metal is to make it the cathode in a galvanic cell. This is cathodic protection and can be used for metals other than just iron. For example, the rusting of underground iron storage tanks and pipes can be prevented or greatly reduced by connecting them to a more active metal such as zinc or magnesium (Figure $3$). This is also used to protect the metal parts in water heaters. The more active metals (lower reduction potential) are called sacrificial anodes because as they get used up as they corrode (oxidize) at the anode. The metal being protected serves as the cathode for the reduction of oxygen in air, and so it simply serves to conduct (not react with) the electrons being transferred. When the anodes are properly monitored and periodically replaced, the useful lifetime of the iron storage tank can be greatly extended. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/17%3A_Electrochemistry/17.06%3A_Corrosion.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the process of electrolysis
• Compare the operation of electrolytic cells with that of galvanic cells
• Perform stoichiometric calculations for electrolytic processes
Electrochemical cells in which spontaneous redox reactions take place (galvanic cells) have been the topic of discussion so far in this chapter. In these cells, electrical work is done by a redox system on its surroundings as electrons produced by the redox reaction are transferred through an external circuit. This final section of the chapter will address an alternative scenario in which an external circuit does work on a redox system by imposing a voltage sufficient to drive an otherwise nonspontaneous reaction, a process known as electrolysis. A familiar example of electrolysis is recharging a battery, which involves use of an external power source to drive the spontaneous (discharge) cell reaction in the reverse direction, restoring to some extent the composition of the half-cells and the voltage of the battery. Perhaps less familiar is the use of electrolysis in the refinement of metallic ores, the manufacture of commodity chemicals, and the electroplating of metallic coatings on various products (e.g., jewelry, utensils, auto parts). To illustrate the essential concepts of electrolysis, a few specific processes will be considered.
The Electrolysis of Molten Sodium Chloride
Metallic sodium, Na, and chlorine gas, Cl2, are used in numerous applications, and their industrial production relies on the large-scale electrolysis of molten sodium chloride, NaCl(l). The industrial process typically uses a Downs cell similar to the simplified illustration shown in Figure $1$. The reactions associated with this process are:
\begin{aligned} &\text { oxidation: } &&\ce{2Cl^{-}(l) &&-> Cl2(g} + 2e^{-}} \[4pt] + \quad &\text { reduction: } && \ce{Na^{+}(l) + e^{-} &&-> Na(l)} \[4pt] \hline &\text{cell:} &&\ce{2Cl^{-}(l) + 2Na^{+}(l) &&-> 2Na(l) + Cl2(g} + 2e^{-}} \end{aligned} \nonumber
The cell potential for the above process is negative, indicating the reaction as written (decomposition of liquid NaCl) is not spontaneous. To force this reaction, a positive potential of magnitude greater than the negative cell potential must be applied to the cell.
The Electrolysis of Water
Water may be electrolytically decomposed in a cell similar to the one illustrated in Figure $2$. To improve electrical conductivity without introducing a different redox species, the hydrogen ion concentration of the water is typically increased by addition of a strong acid. The redox processes associated with this cell are
\begin{aligned} &\text { oxidation: } &&\ce{2H2(l) &&-> O2(g) + 4H^{+} + 4e^{-}} &&E^o_{anode} = +1.299\,\text{V} \[4pt] + \quad &\text { reduction: } && \ce{2H^{+}(aq) + 2e^{-} &&-> H2(g)} &&E^o_{anode} = 0\,\text{V} \[4pt] \hline &\text{cell:} &&\ce{2H2O(l) &&-> 2H2(g) + O2(g}} &&E^o_{cell} = -1.299\,\text{V} \end{aligned} \nonumber
Again, the cell potential as written is negative, indicating a nonspontaneous cell reaction that must be driven by imposing a cell voltage greater than +1.229 V. Keep in mind that standard electrode potentials are used to inform thermodynamic predictions here, though the cell is not operating under standard state conditions. Therefore, at best, calculated cell potentials should be considered ballpark estimates.
The Electrolysis of Aqueous Sodium Chloride
When aqueous solutions of ionic compounds are electrolyzed, the anode and cathode half-reactions may involve the electrolysis of either water species (H2O, H+, OH-) or solute species (the cations and anions of the compound). As an example, the electrolysis of aqueous sodium chloride could involve either of these two anode reactions:
$\ce{2 Cl^{-}(aq) -> Cl2(g) + 2 e^{-}} \quad E_{\text {anode}}^{\circ}=+1.35827\,\text{V} \tag{i}$
or
$\ce{2 H2O(l) -> O2(g) + 4 H^{+}(aq) + 4 e^{-}} \quad E_{\text {anode }}^{\circ}=+1.229\,\text{V} \tag{ii}$
The standard electrode (reduction) potentials of these two half-reactions indicate water may be oxidized at a less negative/more positive potential (–1.229 V) than chloride ion (–1.358 V). Thermodynamics thus predicts that water would be more readily oxidized, though in practice it is observed that both water and chloride ion are oxidized under typical conditions, producing a mixture of oxygen and chlorine gas.
Turning attention to the cathode, the possibilities for reduction are:
$\ce{2 H^{+}(aq) +2 e^{-} -> H2(g)} \quad E_{\text {anode}}^{\circ}=0\,\text{V} \tag{iii}$
or
$\ce{2 H2O(l) +2 e^{-} -> H2(g) + 4 OH^{-}(aq)} \quad E_{\text {anode }}^{\circ}=-0.8277\,\text{V} \tag{iv}$
or
$\ce{Na^{+}(aq) + e^{-} -> Na(s)} \quad E_{\text {anode }}^{\circ}=-2.71\,\text{V} \tag{v}$
Comparison of these standard half-reaction potentials suggests the reduction of hydrogen ion is thermodynamically favored. However, in a neutral aqueous sodium chloride solution, the concentration of hydrogen ion is far below the standard state value of 1 M (approximately 10-7 M), and so the observed cathode reaction is actually reduction of water. The net cell reaction in this case is then
$\ce{ 2 H2O(l) + 2 Cl^{-}(aq) -> H2(g) + Cl2(g) +2 OH^{-}(aq)} \quad E_{\text {cell }}^{\circ}=-2.186\,\text{V} \nonumber$
This electrolysis reaction is part of the chlor-alkali process used by industry to produce chlorine and sodium hydroxide (lye).
Chemistry in Everyday Life: Electroplating
An important use for electrolytic cells is in electroplating. Electroplating results in a thin coating of one metal on top of a conducting surface. Reasons for electroplating include making the object more corrosion resistant, strengthening the surface, producing a more attractive finish, or for purifying metal. The metals commonly used in electroplating include cadmium, chromium, copper, gold, nickel, silver, and tin. Common consumer products include silver-plated or gold-plated tableware, chrome-plated automobile parts, and jewelry. The silver plating of eating utensils is used here to illustrate the process. (Figure $3$).
In the figure, the anode consists of a silver electrode, shown on the left. The cathode is located on the right and is the spoon, which is made from inexpensive metal. Both electrodes are immersed in a solution of silver nitrate. Applying a sufficient potential results in the oxidation of the silver anode
$\ce{Ag(s) -> Ag^{+}(aq) + e^{-}} \tag{anode}$
and reduction of silver ion at the (spoon) cathode:
$\ce{Ag^{+}(aq) + e^{-} -> Ag(s)} \tag{cathode}$
The net result is the transfer of silver metal from the anode to the cathode. Several experimental factors must be carefully controlled to obtain high-quality silver coatings, including the exact composition of the electrolyte solution, the cell voltage applied, and the rate of the electrolysis reaction (electrical current).
Quantitative Aspects of Electrolysis
Electrical current is defined as the rate of flow for any charged species. Most relevant to this discussion is the flow of electrons. Current is measured in a composite unit called an ampere, defined as one coulomb per second (A = 1 C/s). The charge transferred, Q, by passage of a constant current, I, over a specified time interval, t, is then given by the simple mathematical product
$Q=I t \nonumber$
When electrons are transferred during a redox process, the stoichiometry of the reaction may be used to derive the total amount of (electronic) charge involved. For example, the generic reduction process
$\ce{M^{n+}(aq) + n\,e^{-} -> M(s)} \nonumber$
involves the transfer of $n$ mole of electrons. The total charge transferred is, therefore,
$Q=n F \nonumber$
where $F$ is Faraday’s constant, the charge in coulombs for one mole of electrons. If the reaction takes place in an electrochemical cell, the current flow is conveniently measured, and it may be used to assist in stoichiometric calculations related to the cell reaction.
Example $1$: Converting Current to Moles of Electrons
In one process used for electroplating silver, a current of 10.23 A was passed through an electrolytic cell for exactly 1 hour. How many moles of electrons passed through the cell? What mass of silver was deposited at the cathode from the silver nitrate solution?
Solution
Faraday’s constant can be used to convert the charge (Q) into moles of electrons (n). The charge is the current (I) multiplied by the time
$n=\frac{Q}{F}=\frac{\frac{10.23 C }{ s } \times 1 hr \times \frac{60 min }{ hr } \times \frac{60 s }{\min }}{96,485 C / mol e^{-}}=\frac{36,830 C }{96,485 C / mol e^{-}}=0.3817 mol \textrm {e }^ { - } \nonumber$
From the problem, the solution contains AgNO3, so the reaction at the cathode involves 1 mole of electrons for each mole of silver
$\ce{Ag^{+}(aq) + e^{-} -> Ag(s)} \tag{cathode}$
The atomic mass of silver is 107.9 g/mol, so
$\text{mass}\, Ag =0.3817\,\text{mol e}^{-} \times \dfrac{1\, \text{mol}\,\ce{Ag}}{1\,\text{mol}\,\ce{e}^{-}} \times \frac{107.9\,\text{g}\,\ce{Ag}}{1\,\text{mol}\,\ce{Ag}}=41.19\,\text{g}\,\ce{Ag} \nonumber$
Exercise $1$
Aluminum metal can be made from aluminum(III) ions by electrolysis. What is the half-reaction at the cathode? What mass of aluminum metal would be recovered if a current of 25.0 A passed through the solution for 15.0 minutes?
Answer
$Al^{3+}(aq) +3 e^{-} \longrightarrow Al (s) \nonumber$
0.0777 mol Al = 2.10 g Al.
Example $2$: Time Required for Deposition
In one application, a 0.010-mm layer of chromium must be deposited on a part with a total surface area of 3.3 m2 from a solution of containing chromium(III) ions. How long would it take to deposit the layer of chromium if the current was 33.46 A? The density of chromium (metal) is 7.19 g/cm3.
Solution
First, compute the volume of chromium that must be produced (equal to the product of surface area and thickness):
$\text { volume }=\left(0.010 mm \times \frac{1 cm }{10 mm }\right) \times\left(3.3 m^2 \times\left(\frac{10,000 cm^2}{1 m^2}\right)\right)=33 cm^3 \nonumber$
Use the computed volume and the provided density to calculate the molar amount of chromium required:
$\text { mass }=\text { volume } \times \text { density }=33 cm^3 \times \frac{7.19 g }{ cm^3}=237 g Cr \nonumber$
$mol Cr =237 g Cr \times \frac{1 mol Cr }{52.00 g Cr }=4.56 mol Cr \nonumber$
The stoichiometry of the chromium(III) reduction process requires three moles of electrons for each mole of chromium(0) produced, and so the total charge required is:
$Q=4.56 mol Cr \times \frac{3 mol e^{-}}{1 mol Cr } \times \frac{96485 C }{ mol e^{-}}=1.32 \times 10^6 C \nonumber$
Finally, if this charge is passed at a rate of 33.46 C/s, the required time is:
$t=\frac{Q}{I}=\frac{1.32 \times 10^6 C }{33.46 C / s }=3.95 \times 10^4 s =11.0 hr \nonumber$
Exercise $2$
What mass of zinc is required to galvanize the top of a 3.00 m 5.50 m sheet of iron to a thickness of 0.100 mm of zinc? If the zinc comes from a solution of Zn(NO3)2 and the current is 25.5 A, how long will it take to galvanize the top of the iron? The density of zinc is 7.140 g/cm3.
Answer
11.8 kg Zn requires 382 hours. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/17%3A_Electrochemistry/17.07%3A_Electrolysis.txt |
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